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0. ds' ds' rp(r2, -12) Hence we must have 0
< dstfdsip < 1.
{6.17)
If these two curves have a second point of intersection Q, then at Q we should have ds1 > ds ds' - ds'
0 > ' which implies that
~:
L
2::: 1.
{6.18)
But this is impossible because we can see from {6.16) that as 8 increases ft increases whereas f2 decreases, and so we would have dstfdslq < dstfdslp, which, together with {6.17), indicates that (6.18) cannot hold. This proves the uniqueness of the fixed point; that is, system {6.5) has a unique limit cycle. From (6.17) we can see (according to Theorem 2.1 in §2) that it should be a stable cycle. Case 2. 0 < h1 :::::; h2 < 1. In this case if the two curves s = f(s') and s1 = g(s') intersect, then at the first point of intersection we would have dstfds = e2h2r2-"''t1't) :::::; 1, but this is impossible since f2 > 1r >ft. and 12 2::: 11 when h2 2::: h1. We can prove that when h1 2::: 1 > h2 > 0, system (6.5) also has a unique stable limit cycle, while (0, 0) is a stable nodal point for the trajectory in region (II). Hence {6.6) cannot be used again. The proof is similar to the first case, hence it is omitted here. For general nonlinear equations, since there is no way to give a clear analytic expression for the point transformation, we cannot prove the uniqueness of the limit cycle in the same way as we have done before. However, according
THEORY OF LIMIT CYCLES
124
• FIGURE 6.8 to the idea of the point transformation, we can use the following method to prove the uniqueness. (a) We first use the qualitative method to prove that any trajectory starting from any point A on a half-ray l passing through the singular point 0 must return and meet l again at B (Figure 6.8). (b) Next we prove that IDAI: lOBI (or the difference fOAI-1 0 B I) increases (decreases) monotonically as IDAI increases. Then IOAI : lOBI can equal 1 at most once (I'OAI-IOBI can at most equal 0 once); this shows that the limit cycle, if it exists, must be unique. The authors of [63] and [64] used this method to study the uniqueness of a limit cycle, and obtained better results. II. The Poincare method. This method was given in Poincare's classical paper Integral curves defined by differential equations [1]. We first prove LEMMA. Let the differential equation in polar coordinates
dp/ dw = cp(p, w)
(6.19)
be given, where the origin 0 is a singular point. Let 1/J(p) be any single-valued continuous function of p. If equation (6.19) has two closed trajectories r1 and r2. then there exists a point between rl and r2 such that one of the following five relations holds: d'ljJ
-=oo, dp
cp(p,w) = oo, ocp -=oo,
ap
~"" dp2 = oo.
(6.20)
PROOF. If one of r 1 and r 2 does not contain the other, or if they do and 0 lies in the exterior of these two trajectories or between them, or if they do and 0 is in the interior of the inner trajectory but there exists a half-ray through 0 which has mor.e than one point of intersection with one of them, then for
§6. UNIQUENESS
125
all these cases it is easy to prove between rl and r2 there must exist a point such that cp(p, w) = oo-for this we should use the well-known formula dp
tanT = p/ dw' where T represents the angle between the radius vector passing through a point and the trajectory of (6.19) passing through this point. Now suppose that 0 c rl c r2 and any half-ray w = Wo starting from 0 intersects r i at only one point (Wo' p~i)) (i = 1' 2). Study the function
~(wo) = 1/J(p~2)) - 1/J(p~t>). It is easy to see that this is a continuous periodic function of wa; hence there exist an w1 at which ~ takes its maximum value. At w = Wt we have d~(w)/dwlw=w 1 = 0;
that is,
dp) - (d"' dp) ( df/J dp . dw (wi,P~2)(wi)) dp • dw (wl,p~l)(wi))
(6.21)
We now fix w11 and consider (df/l/dp)(dp/dw) as a function of p; the interval of variation of pis [p&1>(wt),p&2 >(wt)]. From (6.21) and Rolle's theorem we know there exists p*, p&1>(wt) < p* < p~2 >(wt), such that at the point (wt,p*) at least one of the following three equalities holds: df/J dp dp dw
=.00
'
!_
ap
(d"'
dp) dp dw
=
0,
!_ ap
(d"'
dp) dp dw
= oo.
Substituting dp/dw = cp(p,w) in the above formulas, we get (6.20). The lemma is completely proved.THEOREM 6.1. If for equation (6.19) we can find a function 1/J(p) such that in some simply connected region G none of the five relations in (6.20) hold, then there exists at most one closed traiectory in G, and it must be a limit cycle.
This is obvious from the lemma. COROLLARY. If 1/J(p,w) is a quotient of two continuous functions, and the denominator in the region G is not equal to zero, and along any half-ray w = wo, cp(p,w) is a monotonically increasing (decreasing) function of p, then there exists at most one closed traiectory in G. PROOF. By hypothesis there is no point in G such that !P(p,w) = oo. Hence from the proof of the lemma we can see that if there exist two closed trajectories r 1 and r 2t then they must contain the origin in their interior, and
THEORY OF LIMIT CYCLES
126
any half-ray w = wo can only meet ri at one point. Since cp(p,w) is monotone with respect to p, we get immediately
0 = rf.. cp(p1(6), 0) dO > ftl (<)
J.
fr~
cp(p2(6), 0) dO= 0.
This is a contradiction. Hence there cannot ~xist two closed trajectories. EXAMPLE 2. Consider the system of equations
dxjdt
= -y + xF(x, y),
dyjdt
=x+
yF(x, y),
(6.22)
where
F(x, y) = a[(x- xo) 2
+ (y- Yo) 2 -
a f 0, k > 0.
k],
Then when k > x~ + y~, system (6.22) has a unique limit cycle, which is stable (unstable) when a< 0 (> 0); if k ~ x~ + y~, then (6.22) does not have a closed trajectory [105].( 5 ) PROOF. When k > x~ + y~, the circle F(x, y) = 0 contains the origin in its interior. Let V(x, y) = !(x2 + y 2 ). Then we can calculate
dV dx +ydt dy =ax ( 2 +y2)[( x-xo )2 + (y-yo )2 -k.] dt=xdt Take r > 0 so small that the circle Cr : x 2 + y 2 = r 2 lies entirely in the interior of the circle F(x, y) = 0; thus dvjdt and a have different signs on Cr. Similarly, takeR > 0 so large that the circle CR: x 2 + y2 = R 2 contains the circle F(x, y) = 0 in its interior; thus dvjdt and a have the same sign on CR. Hence there must exist a limit cycle between Cr and CR, which is stable (unstable) when a < 0 (> 0). To prove its uniqueness, we can transform (6.22) into polar coordinates, and obtain (6.23) dpjdw = pF(pcosw,psinw) = cp(p,w), where F(pcosw,psinw) = a[p2 Now take 1/J(p)
= 1/p.
d.,Pjdp-:; oo,
-
2p(xocosw+yosinw) +
pg- k],
pg = x& +Y&·
Then it is easy to see that between Cr and CR we have
0,
1, the equation does not have a periodic solution [79]. 0, the function P2(x) is obviously always negative. In fact, let x + 1 = t; we get P2(t- 1), whose discriminant is
§7. Existence of Several Limit Cycles H a system of equations has more than one limit cycle, then they naturally
can be distributed in many different ways. For example, three limit cycles can be distributed in any of the four ways shown in Figure 7.1. In this section we discuss the first kind of distribution, since the equations we are going to discuss have only a unique singular point and the interior of each limit cycle must contain the singular point. Other kinds of distribution will be discussed with respect to the quadratic differential systems in the second half of this book.
0
0
@
(a)
00 (c)
(b)
(d)
FIGURE 7.1
Back in 1958, M. I. Vollokov [127] pointed out that if we added suitable requirements to the geometrical figure of the function y = F(x), we could guarantee that the system of equations dxjdt = y- F(x),
dyjdt = -x
(7.1)
would have exactly n limit cycles, where n is any given positive integer. However, since the system involves too many parameters, it is rather difficult in that paper to determine that a given system (7.1) has exactly n limit cycles. In 1966, G. S. Rychkov [128] gave a condition to guarantee that the system of equations dxjdt = y- F(x), dyjdt = -g(x) (7.2) 155
156
THEORY OF LIMIT CYCLES
has at least n limit cycles. As far as the problem of limit cycles is concerned, it is much more useful to be able to answer the following question: when the system of equations is given, how can we show that there exists a limit cycle? If limit cycles exist, how many are there? In 1975, Rychkov [129] proved that when F(x) = a2x 5 + a1x 3 + aox in (7.1), the system has at most two limit cycles, and also pointed out that if F(x) = e-(x 5 -~-tx 3 + x), where E > 0 and IJ > 2.5, then (7.1) satisfies the condition of his previous paper [128], and so this system has exactly two limit cycles. This result can be said to be the earliest one on the existence of two and only two limit cycles. The idea of the proof is to study dh/da1 (the rate of change of h(L) with respect to the ordinate of a11 where h(L) is the integral value of the divergence along a spiral L leaving from, for example, a point a1 on the positive y-axis, going around the singular point (0, 0) and returning to another point a2 on the positive y-axis) and to prove that dhjda1 has a fixed sign, e.g. always positive. Since two adjacent sides of two adjacent cycles have different stability, we know at once that the system of equations has at most two cycles, the integral value of the divergence along the inner cycle is less than zero (or :50), and the integral value along the outer cycle is at least 0 (or > 0). In the proof the technique of proving step (8) in the proof of Theorem 6.11 in §6 is used. Since 1979 the Chinese mathematicians Huang Ke-cheng [130], Zhang Zhifen and Ke Qi-min [131], [132], Chen Xiu-dong [133], Huang Qi-chang and Yang Si-ren [134], and Ding Sun-hong [135] have given sufficient conditions for system (7.2) to have at most or at least n limit cycles; Zhang's result [131] solved a previous conjecture. In the following discussion we assume that F(x) E Cl, F(O) = 0, g(x) E C, and xg(x) > 0 when x '# 0. It is easy to see that (0, 0) is a unique singular g(e) de. We first point of (7.2). As before let F'(x) = f(x) and G(x) = present the result of [130].
J;
LEMMA 7 .1. Suppose there exist constants a, a', b, and b' (b' < a' < 0 < a
§7. EXISTENCE OF SEVERAL CYCLES
FIGURE
'II= F(x) at P1 and
PI·
157
7.2
Suppose yp1 =-Nand YP: = M. Let
.>t2(x, y) = !(Y- M) 2 + G(x), and let .>t,(x, y) = .>t,(A) if (x, y) are the coordinates of A. We compute ~arately the total derivatives of .>tt(x,y) and ..X2(x,y) with respect tot (along the trajectory of (7.2)):
d.>tl/dt = -g(x)[N + F(x)], d.>t2/dt = g(x)[M- F(x)].
(7.3) {7.4)
From condition 2) we know that F(x) ::; M when xp: ::; x::; 0, and F(x) ~ M when b' ::; X ::; X p:. We inspect the parts of r 1 and r 2 on the left half-plane because
{7.5)
Similarly, we can see that
158
THEORY OF LIMIT CYCLES
and (7.6) From (7.5) and (7.6) we at once get ..\2(A2) - ..\2(D2) ;::: ..\2(A1) - ..\2(D1); that is,
(YA~- M) 2 - (YD~- M) 2 ;::: (YAt- M) 2 - (YDt- M) 2.
(7.7)
Similarly, ..\1(D2)- ..\1(A2);::: ..\1(D1)- ..\1(Al); that is,
(YD~
+ N) 2 -
(YA~
+ N) 2 ~ (YDt + N) 2 -
(YAt
+ N) 2.
(7.8)
From condition 3) we know that equality cannot hold simultaneously in (7.7) and (7.8). Adding these two inequalities and simplifying, we get
But M + N ;::: 0, YA 1 - YA~ < 0, and YD 1 - YD~ > 0, which is a contradiction. The lemma is proved. Corresponding to Lemma 7.1, we obviously have LEMMA 7. 2. Suppose there exist constants b' < a' < 0 < a < b such that the following conditions hold: 1) F(x) ~ F(a) when 0 ~ x ~a, and F(x) is monotonically nondecreasing on [a, b]. 2) F(x) ;::: F(a') when a'~ x ~ 0, and F(x) is monotonically nondecreasing on [b', a']. 3) F(x) ¢. 0 when a' ~ x ~a. Then in the strip b' ~ x ~ b, system (7.2) has at most one limit cycle which can intersect both the lines x = a and x = a'.
Next we prove another lemma. LEMMA 7. 3.
Suppose there exist constants N ;::: 0, a > 0, and b' < 0 such
that 1) F(x) ;::: -N when 0 ~ x ~ a, and
2) F(b') ~ -N- y'2G[cij. Then the limit cycle of (7.2) which intersects the line x the line x =a.
= b'
must intersect
PROOF. Let the trajectory starting from A( a, -N) as t increases intersect the negative y-axis at A', and let the trajectory starting from B(b', F(b')) when
§7. EXISTENCE OF SEVERAL CYCLES
159
FIGURE 7.3
t' decreases intersect the negative y-axis at B' (Figure 7.3). From condition 1) and (7.3) we know that YA' ~ -N- J2G(a) and YB' < YB = F(b'); hence
YB' < YA', i.e., B' is below A'. From this we can see at once that the lemma holds. Similarly, we can establish the following three lemmas. LEMMA 7. 4. Suppose there exist constants M ~ 0, a
> 0, and b' < 0
such that 1) F(x) ~ M when 0 ~ x ~a, and 2) F(b') ~ M + ~J2G(a). Then the limit cycle of system (7.2) which intersects the line x intersect the line x = a.
= b'
must
LEMMA 7. 5. Suppose there exist constants N ~ 0, a' < 0, and b > 0 such that 1) F(x) ~ -N when a'~ x ~ 0, and 2) F(b) ~ -NThen the limit cycle of system (7.2) which intersects the line x = b must intersect the line x = a'.
JWM.
LEMMA 7. 6. Suppose there exist constants M ~ 0, a' < 0, and b > 0 such that 1) F(x) ~ M when a' ~ x ~ 0, and 2) F(b) ~ M + y'2G(a'). Then the limit cycle of system (7.2) which intersects the line x = b must intersect the line x = a'.
THEORY OF LIMIT CYCLES
160
From the above lemm88 it is not difficult to give sufficient conditions for system (7.2) to have at most n limit cycles. For example, THEOREM 7 .1. Suppose in system (7.2) F(x) and g(x) satisfy the follow-
ing conditions: 1) F( -x) = -F(x) and g( -x) = -g(x). 2) In the interval (0, b), f(x) has only n zeros 0 < 01 < 02 <···
(k
= 1, ... , n- 1),
where .Bk+l E (nk+t. ak+2) and F(.Bk+t) = F(ak)· Then in the strip lxl ~ b, system (7.2) has at most n cycles.
PROOF. Since F(x) and g(x) are odd functions, the closed trajectory of system (7.2) is symmetric with respect to the origin. From the conditions of the theorem we know that (-1)k+1F(x) is monotonically incre88ing when x E [.Bk,Ok+t],
(-1)kF(x) ~ (-1)kF(.8k) when x E [O,,Bk],
(7.9)
where k = 1, ... , n, .BE (nt.n2), and F(.Bt) = 0. It is e88y to see for lxl ~ .B {7.2) does not have limit cycles. From (7.9), condition 3) and Lemm88 7.1-7.4, we know that (7.2) h88 at most one limit cycle in the strip lxl ~ 02. Any limit cycle intersecting the line x = Ok must intersect the line x = -.Bk (k = 2, ... , n), and in the strip lxl ~ Ok+l there is at most one limit cycle which can intersect x = Ok. Thus we can see that in the strip lxl ~ b there exist at most n limit cycles of system (7.2). In §5 we indicated that the systems of equations
dxfdt = y + x2 sinx,
dyfdt
= -x
and
dx/dt = y + x2 sin x, dyfdt = -2x/(1 + x2) 2 have at le88t n limit cycles in the strip lxl ~ mr+ ~'11'. Now from Theorem 7.1
we also know that every system h88 at most n limit cycles in lxl ~ n'll' + !11'· Hence, every system in this strip h88 exactly n limit cycles. From Lemm88 7.1-7.6 we know that even if F(x) and g(x) are not odd functions, it is not difficult to obtain sufficient conditions for system (7.2) to have at most n limit cycles. In the following we introduce again the work of Rychkov [129] and Zhang [254]. We first prove some lemm88.
§7. EXISTENCE OF SEVERAL CYCLES
161
LEMMA 7. 7. Suppose there exist constants 0 ~ a < e < {3 such that 1) F(a) = F({3); and 2) f(x) > 0 (/(x) < 0) when X E (a, e), and f(x) < 0 (/(x) > 0) and X E (e, {3). Then along any arc 8 of the trajectory of the system
dxjdt in the strip a
~
x
~
= v,
{3, v = v(x)
dvjdt = -x- f(x)v ~ 0,
l-f(x)dt > 0
(7.10)
x E [a, ,8], we have (l-f(x)dt
PROOF. We only consider the case outside the parentheses. Let z = F(x), x E [a,,B]. Suppose F(a) = F(,8) =a and F(e) =b.. From condition 2) we
know there exists an inverse function
X= Xt(z) E [a, eJ, and Xt(a) =a, Xt(b) = e; X2(a) = {3, X2(b) =e. Thus we obtain
re r
{ f(x) dt = [ 13 f(x) dx = f(x) dx + [ 13 f(x) dx la la v(x) la v(x) Je v(x) b dz dz = a v(xt(z)) + Jb v(x2(z))
1
=
1
1
b[
a
v(xt(z))
-
1
v(x2(z))
] dz
(7.11)
< 0,
since
v(x2(z))- v(xt(z))
=
1
z2(z)
zl(z)
dv -d dx X
=
1z2(z) Zl(z)
X
--( ) dx < 0 V X
(7.12)
when z E [a, b]. The lemma is completely proved. Similarly we can prove LEMMA 7. 8. Suppose there exist constants a < e < {3 ~ 0 such that 1) F(a) = F(,B); and 2) f(x) > 0 (/(x) < 0) when x E (a, e), and f(x) < 0 (/(x) > 0) when X E (e,,B). Then along any arc 8 of any trajectory of system (7.10) in the strip a ~ t ~ ,8, v = v(x), x E [a, ,8], we have
-1
f(x)dt < 0
( -1/(x) dt >
0) .
THEORY OF LIMIT CYCLES
162
Also we prove
7.9. If the conditions of Lemma 7.7 hold, then along two arcs and s2: v = v2(x), v2(x) > v1(x) ~ 0 when x E [o:,,B], of any two trajectories of system (7.10) in the strip o: ~ x ~ ,8, we ha'!e LEMMA
= v1(x)
s1: v
f - f(x) dt > f - f(x) dt
181
182
( f - f(x) dt < f - f(x) dt).
181
(7.13)
182
PROOF. We consider the case outside the parentheses. From (7.11) we get
f - f(x) dt- f - f(x) dt
1s1
1s2
= =
lb [v1(x~(z))- v1(x~(z))] lb [v2 (x~(z))- v2 (x~(z))] lb [(v1(x~(z))- v2(x~(z)))- (v1(x:(z))- v2(x:(z)))] dz-
dz
dz,
and
d(v2(x)- v1(x)) = __ x_ + _x_ = x v2(x)- v1(x) > O, dx v2(x) v1(x) v1(x)v2(x) when x E ( o:, ,B). Again from (7.12) we get (7.13) at once. The proof is complete. Similarly, we have
-
LEMMA 7 .10. If the conditions of Lemma 7.8 hold, then along two arcs = v1(x) and s2: v = v2(x), v2(x) > v1(x) ~ 0 when x E [o:,,B], of any
s1: v
two trajectories of system (7.10) in the strip a
1 1
-f(x)dt <
1 2
-f(x)dt
(1
1
~
x
~
-f(x)dt
,8, we have
>
1 2
-f(x)dt).
We still need two more lemmas. LEMMA 7 .11. Suppose f(x) ~ 0 (f(x) ~ 0), and f(x) ¢ 0 when x [o:, ,8]. Then along two arcs s1: v = v1(x) and s2: v = v2(x), v2(x) > v1(x) 0 when x E [o:, ,8], of any two trajectories of system (7.10) in the strip o:
x
~
.B
we have
1 1
-f(x)dt >
1 2
-f(x)dt
(1
1
-f(x)dt <
1 2
-f(x)dt).
PROOF. Since
f - f(x) dt- f - f(x) dt = ff3 - f((x)) dx- [(3 - f((x)) dx
1s
1
1s
2
1a
=-
we know at once that the lemma holds.
V1 X
1:
f(x)
1a
V2 X
[v1 ~x)- v2~x)] dx,
E ~ ~
§7. EXISTENCE OF SEVERAL CYCLES
163
FIGURE 7.4
LEMMA 7. 12. Suppose when x E (a, /3] (a
> 0)
1) f(x) > 0 (f(x) < 0), and 2) f(x) is monotonically nondecreasing (nonincreasing). Then along any two arcs of traiectories 1i (i = 1, 2) of system (7.10) passing through the points (.B;, 0) (i = 1, 2) and intersecting the line x = a twice, we have l
-f(x)dt "'11
where a
> l-f(x)dt "'12
(l
-f(x)dt
"'11
-f(x)dt), "'12
< f3i < /32 ::; f3.
PROOF. We consider the case outside the parentheses. As shown in Figure
7.4, 11 get
= A~
and 12
= AB = A~ U .4;B'; U B;B.
r_ -
-f(x)dt
From condition 1) we
< 0.
jAA2UB2B
Let x = p cos (} and v = p sin (}. System (7.10) is transformed to
= -psin2 Of(pcosO), dOjdt = -1- cosOsinOf(pcosO). dp/dt
(7.14)
THEORY OF LIMIT CYCLES
164
From condition 2) we know that ..(Bi can be expressed asp= Pi(9) (i and p 1 (9) < P2(9) when 61 ~ 9 ~ 9'1. (Figure 7.4). Thus
f
-f(x)dt-
1A2B2
f
= 1, 2},
-f(x)dt
1A1B1
{61
= 19
2
-f(P2(9) cos9) d9 -1- cos9sin0/(P2(9)cos9)
{61
-f(p 1 ( 9) cos 6) d9
(7.15)
- 19 -1- cos6 sin6/(p1 (9) cos6) = [92 f(Pt (9) cos 9) -f(P2(0) cos 6) dB < 191 { ·[-1-cos9ein9f(P1(9}coe9}] [-l-coe9sin9f(P2(9)coe9}]} - 0. · 2
From (7.14) and (7.15) we at once get
1
-f(x)dt>j-l(x)dt.
"'tl
"12
Now we present Rychkov's theorem, but the proof we give is that of Zhang Zhi-fen; moreover, the conditions are somewhat weakened. THEOREM 7. 2. Suppose that when x E ( -d, d) the following conditions hold:
1} /(-x)=/(x). 2) /(x) has only positive zeros a1 and a'J., 0 < a1 < a2
PROOF. We consider system (7.10} in the strip lxl < d. Since f(x) is an even function, the closed trajectory of (7.10} is symmetric with respect to the origin. Let {3 be the smallest positive zero of F(x) (at < {3 < a2). It is easy to see that system (7.10) has no limit cycles in the strip lxl ~ {3. Moreover, f(x) < 0 when x E ({3, o:2); hence, according to Lemmas 7.7 and 7.8, if (7.10) has a limit cycle in the strip lxl ~ 0:2, then it must be a unique unstable cycle. On the other hand, from Lemmas 7.9-7.12 we know that if system (7.10) has two limit cycles Lt C £2, both intersecting the line x = a'J., then
,/.. - f(x) dt > rf- - f(x) dt.
~1
~2
(7.16)
From this we know that system (7.10) does not have a compound cycle and a periodic cycle. In the following we divide our discussion into two cases.
§7. EXISTENCE OF SEVERAL CYCLES
165
(I) In the strip lxl ~ a~, system (7.10) has a unique unstable limit cycle
Lt. H there exists a limit cycle of (7.10) intersecting the line x = a~, we assume the cycle closest to L1 is L~ (:::) L 1 ); then L~ must be internally stable. In the following we prove that L~ must be externally stable; otherwise L~ should be an internally stable but externally unstable semistable cycle. We examine the system of equations
dxfdt = v,
dvfdt = -x- [/(x) + a-y(x)]v,
(7.17)
where a> 0,-y(x) = 0 when lxl < a2, and -y(x) = (lxl-a2)~ when a~ :Sixl < d. From the theory of generalized rotated vector fields we know that, when a is suitably small, system (7.17) has two cycles L~1 ) C L~~> in the strip lxl < d which intersect the line x = a 2 such that
~ < > -It (x) dt?:. 0, ~
h2
2
(ll
h-2
-/t(x) dt :50,
where ft(x) = f(x) +a-y(x), which still satisfies the conditions of the theorem. This contradicts (7.16). Hence L~ must be a stable cycle. From (7.16) again, we know there is no cycle outside L~. (II) System (7.10) does not have limit cycles in the strip lxl :5 a~. Since the singular point 0 is stable, if (7.10) has cycles, then the cycle L 1 closest to 0 must be internally unstable. H L 1 is also externally unstable, then, similarly to (I), we can prove that the exterior of L1 has at most one more cycle. H Lt is externally stable, then~ 1 -f(x) dt = 0. Hthere is also a limit cycle L2 outside Lt, then L~ must be internally unstable, which would contradict (7.16). Hence system (7.10) has only a unique semistable cycle L1. Considering the above discussion, we see that (7.10) has at most two limit cycles. COROLLARY.
If F(x) = a2 x5 + a1x3 + aox, then system (7.1) has at most
two limit cycles. PROOF. We may as well assume a~ > 0. When F(x) has at most one positive zero (a multiple zero is counted as one), system (7.1) has at most only one limit cycle; if F(x) has two different positive zeros, then from Theorem 7.2 we know that system (7.1) has at most two limit cycles. By Theorem 5.2 we can prove that if F(x) = x 5 - 50x3 /3+45x, then (7.1) has exactly two limit cycles. Many persons (see [136]-[139]) have studied the special case of system (7.1)
dxfdt = y + psinx,
dyfdt = -x
(7.18)
THEORY OF LIMIT CYCLES
166
and conjectured that, for allp. ¥= 0, system (7.18) has exactly n limit cycles in the strip lxl ~ (n + 1)11". Zhang Zhi-fen [131] completely solved this problem. Her result is as follows: THEOREM
7. 3. The system of equations
dxfdt = v,
dvfdt
= -x + (p.cosx)v
(7.19)
(equivalent to (7.18)) has exactly n limit cycles in the strip lxl
~
(n + 1)1r.
It is clear that we only have to discuss the case p. > 0. For convenience of presentation, we still let
f(x) = -p.cosx,
F(x)
= foz f(s)ds = -p.sinx.
First we prove some lemmas. LEMMA 7 .13. In the half-plane v ~ 0, along the same trajectory v = v(x) of system (7.19), we have v(x) > v( -1r-x) when x ~ 0, and v( -x) > v(1r+x) when x ~ 0. PROOF. From
(7.19) we get
v(1r+x) -v(-x)
11"+z e
= 1-z
v(e) de
1.,.+z --ed e = l -zo + 1.,. + o .,. v(e) = e[v(7r +e)- v( -e)I de
r lo
v(-e)v(1r+e)
r v(e) e de- lor
- lo
11" v(7r +e) de,
X~
0.
Differentiating both sides yields
d dx[v(1r+x) -v(-x)]
=
x[v(1r+x)-v(-x)] 1r v(-x)v(1r+x) - -,v(:-1r-+-x"7)
<
x[v(1r + x)- v( -x)] v( -x)v(1r + x) '
X~
0,
thus
d:
[(v(7r+x)-v(-x))exp(-
foz v(-e~~!+e))] <0 whenx~O.
But v(1r) - v(O) < 0, and so v(1r + x) - v( -x) < 0 when x ~ 0. Similarly we can prove the other inequality. Now we construct a positive definite state function >.(x, v) = x 2 /2 + (v+F(x)) 2 /2. Differentiating along the direction field defined by (7.19) yields
§7. EXISTENCE OF SEVERAL CYCLES
167
d>..jdt = -xF(x). For simplicity, let >..(x) = >..(x,v(x)), where v = v(x) is the equation of the trajectory of (7.19). In the following we study the variation of the state function along the trajectory of (7.19). LEMMA 7.14. If the trajectory v = v(x) of system (7.19) intersects the lines x = ±m1r in the half-plane v ~ 0, then along this section of the trajectory we have m~l.
(7.20)
= A.((k + 1)1r)- A.(k1r) and Kk = >..( -k1l")- >..( -(k + 1)1r) 0 is an integer). First we prove the inequalities
PROOF. Let .dk
(k
~
.6o > 0,
Since d>..jdt
Ko > 0, 1 (-1)k+ (.6k + Kk+1) > o, k ~
(7.21)
( -1)k+ 1 (Kk
(7.23)
o, + .6k+l) > o, k ~ o.
(7.22)
> 0 when 0 < lxl < 1r, inequality (7.21) holds. Moreover,
.6k + Kk+1 =
=(-1)k
l
(k+l),.. -xF(x)
k71"
( )
V X
r[ t1r+x)_ k1r +X
}0
V
V
dx +
~-(k+ 1 ),..
-xF(x)
-(k+2)11"
V X
()
dx
(k1r+1r+x )]IF(x)idx, -k1r- 1l"- X
k~O.
From Lemma 7.13 we know that the function under the integral sign is less than zero; hence (7.22) is proved. Similarly we can prove (7.23). Now let ,6_1 = K-1 = 0. Then from (7.21)-(7.23) we have m-1 ( -1)m+l[>..(m7r)- >..( -m1r)] = ( -1)m+ 1
L (.6k + .6k)
k=O [(m-1)/2)
= {-1)m+l
L
(.6m-2k-2
+ 'Km-2k-d
k=O [(m-1)/2)
+ (-1)m+l
L
('Km-2k-2
+ .6m-2k-d
k=O
>0. Lemma 7.14 is completely proved. Now let f[a,/3) f(x(t)) dt denote the integral of f(x) along a section of the trajectory of (7.19), where [a, .8] is an interval of variation of x. Let x(tt) =a and x(t 2 ) == ,8. Then the interval of variation oft is [t 1 , t 2 ].
168
THEORY OF LIMIT CYCLES LEMMA
7.15. In the half-plane v
(-1)m- 1
~
0
r
f(x(t)) dt > 0,
(7.24)
1[-m'lr,m'lr]
where m is a positive integer. PROOF. Let
d1c =
i
f(x(t)) dt =
(k1r,(lc+1)1r]
= (-1)/c
1
(k+1)1r
v( x)
k1r
r v(k1rf(x)+ x) dx
f(x)
-
dx
lo
= (-1)/c r/2 v(krr + x)- v(krr + 11"- x) if(x)l dx, )0 v(krr + x)v(krr + 1r- x)
dk
=r
f(x(t)) dt
= ~-k'lr
1(-(lc+1)1r,-lc'lr]
= ( -1)/c
r v(-krrf(x) dx x)
-(k+l)'lr
k
~ 0,
(7.25)
f(x) dx v(x)
)0
= (-1)/c r/2 v(-krr- x)- v(-krr -'II"+ x) if(x)i dx, k ~ 0. (7.26) lo v( -krr- x)v( -k1r- 1r + x) First we prove the inequalities
+ d~c) > 0, (-1)/c+l((llc+l + d~c) > 0,
(7.27) (7.28)
(-1)/c+l(dlc+1
It is easy to see that
d~c 1 +die= (-1)/c+l +
r 12 v(k1r + + x) -v(k1r + 2 x) if(x)l dx v(krr + + x)v(krr + 2rr- x) 11"-
11"
lo
1r
- ( -1)/c+l r/2 v( -k'll"- x) - v( -krr- 11" + x) lf(x)l dx.
lo
v( -krr- x)v( -krr- 1r + x)
(7.29)
From Lemma 7.13 we have
v(krr + 1r + x) < v( -krr- x), 0 ~ x ~ rr/2, k ~ 0, v(krr + 2rr- x) < v(-krr -'II"+ x), 0 ~ x ~ rr/2, k ~ 0.
(7.30)
§7. EXISTENCE OF SEVERAL CYCLES
169
From (7.19) and Lemma 7.13 we have
v(k1r + 1r + x)- v(k1r + 27r- x) = fk1r+21r-x ( (x) + f(x)) dx
=
Jk1r+1r+x
1 1
k1r+1r-X 7r + X
V X
dx k1r+x v(1r + x) k1r+11'-X X > --dx k1r+x v( -x)
= v(-k1r- x)- v(-k1r -1r + x) > 0, 0 ~X~ 1rj2,
k;::: 0.
(7.31)
From (7.29)-(7.31) we obtain (7.27). Similarly we can prove (7.28). From Lemmas 7.7 and 7.8 we can get
do = f
f(x(t)) dt > 0,
(7.32)
][0,7r)
do = Let d-1
= iL1 = 0.
f(x(t)) dt > 0.
f
(7.33)
j[-1r,O)
From (7.27)-(7.33) we get m-1
f(x(t)) dt = ( -1)m-l
(-1)m-l {
I: (dk + dk) k=O
J[-m1r,m7r)
[(m-1)/2)
I:
= ( -1)m-l
(dm-2k-l + dm-2k-2)
k=O [(m-1}/2) + (-1)m-l
I:
(dm-2k-l +dm-2k-2)
k=O
> 0,
m;::: 1.
Lemma 7.15 is completely proved. Since the closed trajectory of system (7.19) is symmetric with respect to the origin and (-1)m-l f(x);::: 0 when m1r ~ x ~(2m+ 1)7r/2, from Lemma 7.15 we have LEMMA 7 .16. If the closed trajectory Lm of system (7.19) intersects the interval [m1r, (2m+ 1)7r/2] in the positive x-axis, then
( -1)m-l J.
h ...
f(x(t)) dt > 0,
m;:::
1.
THEORY OF LIMIT CYCLES
170
Next we prove LEMMA 7.17. If system (7.19) has two closed trajectories Lt C L2, and
= ±(2m+ 1}rr/2 in the half-plane v ~ 0,
they both intersect the lines x
f
(-1)m
(f(x2(t))- /(xt(t))] dt > 0,
m
then
> 0.
1(-(2m+1)11'/2,(2m+l)11'/2]
PROOF. First, from Lemmas 7.9 and 7.10 we get
(-1)kDk = (-1)k
f
[f(x2(t))- f(xt(t))]dt
1((2k+ 1)11' /2,(2k+5)11' /2]
> 0,
k ~ 0,
(-1)k Dk = ( -1)k
(7.34)
f
[f(x2(t))- /(xt(t))] dt
1(-(2k+5)11'/2,-(2k+l)11'/2]
> 0,
k ~ 0,
f
(7.35)
[f(x2(t))- f(xt(t))] dt
< 0.
(7.36)
1(-11',11']
> 0 when
When m is odd, by (7.34)-(7.36) we have f(x) and by Lemma 7.11 we have
f
1r
~ lxl
< 3rr/2,
[j(x2(t))- /(xt(t))] dt
1[-(2m+1)11'/2,(2m+1)11'/2]
f
=
+f
1[-311'/2,-x] (m-1)/2
+
E
+f
1(-11',11']
(D2k-1
[f(x 2 (t))- f(x 1(t))] dt
1[11',311'/2]
+ D2k-d < 0.
k=1
When m is even, by (7.34) and (7.35) we have f(x) and by Lemma 7.11 we have
f
< 0 when 0 ~ lxl < rr/2,
[j(x2(t))- f(x1(t))] dt
1(-(2m+1)11'/2,(2m+1)11'/2] ·
=f
m/2-1
[f(x2(t))- f(xt(t))] dt +
1(-11'/2,11'/2]
E
(D2k
+ D2k) > 0.
k=O
Lemma 7.17 is completely proved. From Lemmas 7.12 and 7.17 we get
*
LEMMA 7.18. If system (7.19) has two closed trajectories L1 C L2 and both intersect the interval [(2m+ 1)rr /2, (m + 1)rr] in the positive x-axis, then
{-l)m
[.fr:
3
j(x2(t))dt-
1
f(x1(t))dt]
> 0,
m;:::
1.
§7. EXISTENCE
OF SEVERAL CYCLES
171
PROOF OF THEOREM 7.3. (I) We show that system (7.19) has at least n limit cycles in the strip lxl ~ (n + 1)11'. Suppose the trajectory starting from a point Pm(-m1r,O) of the negative x-axis, after passing through the half-plane v ~ 0, intersects the positive xaxis at Qm. From Lemma 7.14 we know that when m > 0 is odd we must have XQm > m1r; when m > 0 is even, XQm < m1r. Since the directional field is symmetric, we know that when the trajectory from the point Pm(m1r,O), after passing through the half-plane v ~ 0, intersects the negative x-axis at a point 'lJm, we must have XQm = -XQm· Let PmQ:n and PmlJ,;: represent the segments of the trajectories passing through the points Pm,Qm and Pm,Qm respectively; let QmPm and QmPm represent the line segments joining Qm, Pm and 7Jm, P m respectively. Set
r m = -p;;:Q;, U QmPm U PmQm U QmPm. Then in r 1, r 2 , ••• , r m, ... , r n+ 1 every pair of adjacent simple closed curves form the inner and outer boundary curves of a Poincare annular region, and there exists at least one closed trajectory between them. Hence in the strip lxl ~ (n + 1)11', there exist at least n limit cycles. (II) We show that there exist exactly n limit cycles of system (7.19) in the strip lxl ~ (n + 1)11'. We divide this into two cases. 1. There exists a closed trajectory Lm which intersects the interval [m1r, (2m+ 1)11'/2] in the x-axis. By Lemma 7.16 we know that when m is odd (even), Lm is stable (unstable) and there are no more trajectories intersecting (m1r, (2m+1)11'/2]. Now we prove there is no closed trajectory which intersects the interval ((2m+1)11'/2, (m+1)11']. Suppose the contrary. Then the trajectory L~ closest to Lm and containing it must be internally unstable (stable). Next we prove L~ cannot be semistable. In fact, suppose L~ is semistable. Consider the system of equations dxfdt
where fa(x) =
-JLCOSX
= v,
dvfdt = -x- !a(x)v,
(7.19*)
+ O')'m(x), a> 0, and when when
lxl ~(2m+ 1)11'/2, Ixi > (2m+ 1)11'/2.
System (7.19)* forms a family of generalized rotated vector fields with respect to a, and when a is very small, there exist closed trajectories L!.!> C L~) of (7.19)* which intersect ((2m+ 1)11'/2, (m + 1)11'] in the x-axis, and, moreover, (-1)mrf. h~l
fa(x)dt~O.
172
THEORY OF LIMIT CYCLES
This contradicts Lemmas 7.12 and 7.17. Hence L~ must be an unstable (stable) cycle. From the proof of (I), at the same time there exists at least one one-sided, or even both-sided stable (unstable) closed trajectory outside L~ which intersects the interval ((2m+ 1}n/2, (m + 1)1r]. This contradicts Lemma 7.18. Hence there cannot be another closed trajectory which intersects
((2m+ 1)11"/2, (m + 1)1r]. 2. There is no closed trajectory which intersects [m1r, (2m+1)?r/2] on the xaxis. From the proof of (I), we know there exists at least one closed trajectory which intersects ((2m+ 1)11"/2, (m + 1)1r]. Suppose Lm is the one closest to the origin; when m is odd (even), it must be internally stable (unstable). Lm cannot be semistable, for otherwise by the proof of (I) there would exist at least one one-sided or even both-sided stable (unstable) closed trajectory outside Lm which intersects ((2m+ 1)11"/2, (m+ 1)1r]; this contradicts Lemma 7.18. Hence, Lm must be a stable (unstable) closed trajectory. Then, similarly to case 1 we can prove there is no other closed trajectory except Lm which intersects the interval ((2m+ 1)11"/2, (m + 1)11"]. From the above discussion, we know that when n is odd (even), system (7.19) has a unique stable (unstable) closed trajectory which intersects [m1r, (m + 1)1r] in the positive x-axis. Also, in the strip lxl ~ 1r, system (7.19) does not have a closed trajectory; hence in the strip lxl ~ (n + 1)11" there are exactly n limit cycles, and stable and unstable limit cycles are arranged alternately. Theorem 7.3 is completely proved. After proving this beautiful theorem, it is fitting for us to introduce the history of this problem. We know that the van der Pol equation
x + J.tf(x)
+ x = o,
(7.37)
where f(x) = -x+x3 /3,(1) has a unique limit cycle with respect to all J.t =f:. 0. Similarly, we can prove that if f(x) is a polynomial of degree 2n + 1 which contains only terms of odd degrees, then for sufficiently small IJ.tl equation {7.37) has at least n limit cycles. Eckweiler [136] first noticed that if we take f(x) = sinx, then in order to obtain a solution from the family of solutions z:l + z2 = A 2 for p = 0 that will produce a periodic solution for J.t =/:- 0, the amplitude A must satisfy the equation( 2 )
Jor1f sin~sin(Asin ~) d~ = 21rJ1 (A) = 0, =
(7.38)
( 1 )Differentiating (7.37) with respect to t, and letting :i y, we obtain the well-known form. Equation (7.37) is sometimes called the Rayleigh equation; it first appeared in Rayleigh's book Theory of aound (1894), and was much earlier than van der Pol. (2 )See (4.18) in §4.
§7. EXISTENCE OF SEVERAL CYCLES
173
where J1(A) is a first-order Bessel function. It is easy to calculate the derivative of the above formula, and get
{211:
lo
sin2 tp cos(A sin tp) dtp
= 21r[J0 (A) -
J2 (A)].
(7.39)
By the properties of Bessel functions,( 3 ) Jt(A) has an infinite number of positive zeros, and the large zeros obey the asymptotic law
An= n1r + 11'/4 + 0(1/n).
(7.40)
Under this condition, the right side of (7.39) becomes
4( -1Y'.../211'/An + 0(1/n).
(7.41)
Hence we see that this value is different from zero provided that n is sufficiently large. Thus [136] affirmed that the equation
x +~sin±+ x = 0
(7.42)
has an infinite number of limit cycles when 1~1 « 1. Later, Hochstadt and Stephan [138) pointed out that this affirmation is not rigorous because it may not be possible to find an interval for the same ~ such that the value of~ in this interval and the infinite number of limit cycles corresponding to the zeros of J1(A) can coexist. The authors of [138) applied more asymptotic properties of J 1 (z) and the method of constructing the Poincare annular region, first proving that the asymptotic equation
d1 / d8 = J.t sin 8 sin("Y sin 8) of the equation (7.42) in polar coordinates d""f _ ~sinOsin(""(sinO) dO- 1 + (~h)cos8sin("Ysin8)
(7.43)
has an infinite number of limit cycles when 1~1 > 0 is sufficiently small, and then showing that this conclusion also holds for (7.43) provided 1~1 is taken slightly smaller. Later, R. N. D'heedene [139) improved the method of constructing the annular region in [138] to obtain the above conclusion when 0 < IJ.tl < 2, but for IJ.tl ~ 2 he only proved that outside a sufficiently large neighborhood of the origin there exist an infinite number of limit cycles, and he conjectured that Theorem 7.3 should also hold. Zhang Zhi-fen in [131) completely proved his conjecture. REMARK 1. For the van der Pol equation
dxjdt = y, (3)Refer to [140].
dyjdt = -x + JJ(Y- y 3 /3),
(7.44)
174
THEORY OF LIMIT CYCLES
FIGURE 7.5
if we let x = -J.Lv, then we get
dy dv
= J.L
2y
- y 3 /3 + v
,;;.____;;__"""y--'
(7.45)
and we can prove( 4 ) that as J.L ---+ +oo the limit position of this periodic solution of (7.45) is a discontinuous periodic solution, as shown by the bold line in Figure 7.5. Applying to (7.42) the transformation x = J.LV, or
dx/dt = y,
dy/dt
= -x- J.Lsiny,
we get
dy/dv = -J.L2 (v+siny)jy.
(7.46)
We conjecture that as J.L---+ +oo we can obtain some results on (7.46) similar to those for (7.45). REMARK 2. Exercise 5 of §6 of the first edition of this book (1965) (Exercise 3 in this edition) has shown that when F(x) in (7.1) is a cubic polynomial of x, the system has at most one limit cycle. This result was also obtained independently by A. Lins, W. de Melo and C. C. Pugh [141]. They also conjectured that if F(x) is a polynomial of degree 2n + 1 or 2n + 2, then (7.1) has at most n limit cycles. This conjecture has not yet been confirmed or disproved. We think it is correct.
( 4 )See
Appendix IV in (137).
§8. Structural Stability of Differential Systems Briefly speaking, a property possessed by a differential system is called stable if this property remains after a slight change of the system itself. For example, in §4 we have already seen that if an ordinary differential system (P(no),Q(no)) possesses a single limit cycle r 0 , then in the family ofrotated vector fields with a as parameter, for all a in a sufficiently small neighborhood of n 0 , the system (P(a), Q(n)) also has a unique single cycle in the vicinity of fo; hence the property of "possessing a single limit cycle" with respect to the rotation of vector fields is stable. However, if (P(nt), Q(nt)) possesses a semistable cycle f 1 , then for a close to n 1 on one side, (P(a), Q(n)) does not have a limit cycle in the vicinity of r 1 , but for a close to nt on the other side, (P(a), Q(n)) has at least two limit cycles in the vicinity of ft: this illustrates that the property of ''possessing a semistable limit cycle" is unstable with respect to the rotation of vector fields. In dynamical systems in the plane, the study of the existence and the number of limit cycles is an important part of the global qualitative theory; hence it must involve the problem of whether the global qualitative structure is stable. This is the structural stability to be discussed in this section. The importance of structural stability in practical applications is very obvious, since mathematical models abstracted from practical problems (which are in general represented by algebraic, ordinary or partial differential equations) have gone through processes of approximations and simplifications. To make sure the properties obtained from the study of mathematical systems can correctly reflect the state of the practical problems, We must demand that these models be structurally stable. In this section we first introduce a rigorous definition of structural stability of plane ordinary differential systems, and give topological characteristics of phase-portraits of structurally stable systems; then we discuss the problem of structural stability of polynomial systems. This section is preparatory for a discussion on the problems of limit cycles and global structure for quadratic differential systems in the following sections. 175
THEORY OF LIMIT CYCLES
176
Let the plane system
dxjdt = P(x, y),
dyjdt
= Q(x, y)
(I)
be defined in a region B of the (x, y)-plane, let P, Q E C 1 (i.e., continuously differentiable), and let the boundary of B be a simple closed curve having no points of contact with respect to {I). For definiteness, we may as well assume the trajectory of (I) crossing the boundary of B does so from outside to inside. All the systems whose P and Q satisfy the above conditions form a set X. Now in X we introduce a metric p as follows: Let another system in X be
dx dt = P(x, y) + p(x, y),
dy dt = Q(x, y)
+ q(x, y).
(II)
Then we define the distance between (I) and (II) in X to be
p(I,II)
+
= m;x (IPI lql +~~~~+I ~:I+ ~~!I+ ~~:1).
We can prove that, under such a definition of the metric 1 X becomes a Banach space. DEFINITION 8.1. If there exists a 8 > 0 such that there is a topological map T from B into itself which can carry trajectories of {I) into trajectories of (II) provided that p(I, II) < 8, then we call system (I) a structurally stable system or coarse system in B, and system (II) a permissible perturbed system of (I); and p and q are called perturbations. The concept of coarse systems was first introduced in 1937 by A. A. Andronov and L. S. Pontryagin [142]. They assumed P and Q were analytic functions, and for the map T, besides the requirements of Definition 8.1, they added the following restriction: for any c > 0, we can make d(M, T(M)) < c(l) for any point M in B provided 8 > 0 is sufficiently small; here d( , ) represents the distance in R 2 • They have pointed out that the necessary and sufficient conditions for system (I) to be structurally stable are 1. It has only a finite number of elementary singular points, and the characteristic roots of its corresponding linear approximate system do not have zero real parts. 2. It has only a finite number of closed trajectories, which are all single limit cycles. 3. It does not have a trajectory from a saddle point to another saddle point. (1 )We shall call any map satisfying this conditione-homeomorphic.
(III)
§8. STRUCTURAL STABILITY
177
The details of their proof, however, have never been published. In 1952, H. F. De Baggis (143) weakened the requirement of analyticity of P and Q to P, Q E C 1 , and gave a detailed proof for the necessary and sufficient conditions. In 1959, M. M. Peixoto [144] further proved that Definition 8.1 and Andronov's original definition are equivalent, and generalized to n-dimensional syste1ns. In 1962, Peixoto [145) gave and proved necessary and sufficient conditions for structural stability of differential dynamical syste1ns on compact two-dimensional manifolds. His results will be mentioned at the end of this section. Since Peixoto has proved the equivalence of the two definitions of structural stability, we may as well add the requirement d(M, T(M)) < e in Definition 8.1 in order to simplify our proofs. Because we only have to consider perturbations p and q whose absolute values are very small, in order to prove the necessity of (ill) we can assume that the singular point of (I) under the map T becomes the singular point of (II) in its vicinity; the same hypothesis is also applied to the limit cycle. These hypotheses greatly shorten the proof of the following theorem. THEOREM 8. 1. The conditions (III) are necessary conditions for system
(I) to be structurally stable. The proof will be divided into the following four lemmas, in all of which we assume that (I) is structurally stable. LEMMA 8. 1. System (I) can only have a finite number of singular points and a finite numbe1> of closed trajectories. PROOF. From the Weierstrass approximation theorem, in any neighborhood of (I) there always exists a polynomial system (II) (that is, the right sides of the system are polynomials) such that the two polynomials of (II) do not have a common factor; hence the number of singular points of (II) can only be finite. Also, from Dulac's well-known theorem [146], [147], we know that (II) can only have a finite number of closed trajectories; hence (I) can only have a finite number of singular points and a finite number of closed trajectories. LEMMA 8. 2. System (I) can only have elementary singular points, and the characteristic roots of its corresponding linear approximate system do not have zero real parts. PROOF. Take any singular point, which we may assume to be (0, 0). If it is not an elementary singular point, then a(P,Q)
I -o
l a(x, y) (0,0) -
,
178
THEORY OF LIMIT CYCLES
which shows that the two curves P = 0 and Q = 0 have a point of intersection at (0, 0) which is not simple.(2) Thus we can make a slight perturbation of one of them such that its singular point changes from one to more than one. This contradicts the hypothesis that (I) is structurally stable. We now prove that the characteristic roots of its linear approximate system at (0, 0) do not have zero real parts. Expand (I) into dx dt =ax+ by+ P1(x,y),
dy dt =ex+ dy
+ Q 1(x,y),
Jx
2 + y 2 ). Suppose its characteristic root has zero where P 1 and Q1 are o( real part, i.e., a+d = 0; thus system (I) takes the origin as its center or focus. Now we add a perturbed term o:x, 0 < lo:l « 1, on the right side of dxfdt; then the perturbed sytem still takes the origin as its singular point and the real part of its characteristic root is -o:. If (0, 0) is a center of (I) but is a· focus with respect to the perturbed system, it is clear that the center and the neighborhood of the focus cannot be €-homeomorphic, which contradicts the hypothesis that (I) is structurally stable. If (0, 0) is a stable (unstable) focus of (I), then we use Theorem 3.7 in §3 and take o: < 0 (> 0), which can make the origin become an unstable (stable) focus of the perturbed system and can generate new limit cycles in its vicinity; hence these two systems are not €-homeomorphic. In short, the real part of the characteristic root of the linear approximate system of (0, 0) cannot be zero. This lemma is completely proved.
LEMMA 8. 3.
There is no trajectory connecting two saddle points of sys-
tem (I). PROOF. To get a contradiction, assume there is a trajectory of (I) which
connects two saddle points A and B (which may coincide). Construct a family of rotated vector fields dxfdt = P- o:Q,
dyfdt = Q
+ o:P;
(IV)
when o: < 0, the vector field rotates in the clockwise direction and the arrows in Figure 8.1 indicate the directions of the trajectories of (IV). It is easy to see that (IV) still has A and B as its saddle points, but at the same time the separatrix entering B does not come from A, and the separatrix starting from A does not enter B; that is, (I) and (IV) are not €-homeomorphic, which contradicts the hypothesis that (I) is structurally stable. This completes the proof. (2)This implies that these two curves touch tangentially at (0,0), or at least one curve has (0, 0) as a multiple point.
§8. STRUCTURAL STABILITY
FIGURE
179
8.1
LEMMA 8. 4. Closed traJectories of system (I) are single limit cycles. PROOF.
Let "ft, ... ,
"'N be all the limit cycles of (I), and set
h("tt) =
i( 'll
aP ax
+ aQ) ay
dt = 0.
Now we shall construct a function cp 1(x, y) E C 1 such that 'Pt(x, y) = 0 on i= 0 outside an q-neighborhood N.,., of "ft. where 17 > 0 is small enough that N.,., does not meet with any "fi (i :j: 1). For this, we only have to take "fl, acptfax ~ 0 on "ft. and 'Pl
• 2 1rn(x, y) + ( '1f"2 sm n x, y ) exp [217
'Pt(x,y) =
t
y)] an 2 1rn(x, 211
when (x, y) EN.,.,, 1f"
when (x, y) ¢.
2
N.,.,,
"'l·
where n(x,y) represents the distance of a point (x,y) of N.,., from From 1 n(x, y) E C it is easy to prove that 'Pt(x, y) indeed satisfies the above conditions. Similarly, for any "fi we can define 'Pi(x, y). Now we study the system of equations dy ~ acpl (V) - = P(x y) + c'P1'P2 · · · 'PNdt = Q(x,y). dt ' ax' It is easy to see that (V) and (I) both take "ft. ... , as closed trajectories, and when c > 0 is sufficiently small, they are permissible perturbed systems. Using the fact that h("tt) = 0 and the properties of 'Pl (x, y) we can compute that, for system (V),
"'N
hy("tt) =
i
ccp2 · · · 'PN (
"'ll
a;:l)
2
dt.
"'ll
Since acptfax ~ 0 and is continuous on and 'Pi(x, y) > 0 on the right side of the above formula is not equal to 0, and has the same sign as c. Now assume that 11 is an externally stable cycle of system (I). From the theory of rotated vector fields in §3, we know that in the outer neighborhood of there exists a curve r such that all the trajectories of (I) intersecting
"'t
THEORY OF LIMIT CYCLES
180
r
move in the exterior-to-interior direction. Take E > 0 small enough so that r is still a curve without contact with respect to (V), and (V) does not have singular points between "Y1 and r. On the other hand, since E > 0, "11 has become an unstable cycle of (V); hence in the outer neighborhood of "11 there appear closed trajectories of (V) different from 11 ,, .. , "fN, which is impossible. The lemma is completely proved. Combining Lemmas 8.1-8.4, we know immediately that conditions (III) are necessary for system (I) to be structurally stable. REMARK. Lemma 8.4 also tells us that systems possessing a multiple cycle can, after a slight perturbation, make the multiple cycle split into at least two cycles. This important fact has not been mentioned in §§2 and 4. Using similar methods, Ding Tong-ren (148] proved that if a multiple cycle has even (odd) multiplicity, then the system can, after any arbitrarily small perturbation, split the multiple cycle into an even (odd) number of single cycles (not less than 2). THEOREM 8. 2. Conditions (III) are sufficient for system (I) to be structurally stable.
For this we first introduce the concept of singular trajectory and regular trajectory. DEFINITION 8.2. If the positive direction and the negative direction of a trajectory are orbitally stable, or the positive direction of this trajectory is orbitally stable and the negative direction runs out of the region B, then the trajectory is called a regular traj'ectory. If a trajectory has at least one direction in B which is orbitally unstable, then it is called a singular traj'ectory. It is easy to see that a center or central focus is a regular trajectory, but any other singular point is a singular trajectory since it always has one direction which is orbitally unstable. Periodic cycles and compound limit cycles are regular trajectories, but other closed trajectories are singular trajectories. A separatrix entering or leaving a saddle point must be a singular trajectory. If both ends of a trajectory run to a saddle point or a focus or approach a limit cycle, or if its positive direction does so but its negative direction runs out of the region B, then this trajectory must be regular, since all the trajectories in its vicinity possess the same properties. Thus, as far as a structurally stable system is concerned, all its singular points, limit cycles and separatrices are singular trajectories, and there are no other singular trajectories. We now turn to the proof of Theorem 8.2. Since (I) satisfies conditions (III), there exist only a finite number of singular trajectories (including singular points) in the region B, which divide B into a finite number of subregions, called normal regions; and the interior of each normal region is completely
§8. STRUCTURAL STABILITY
181
filled with regular trajectories. There are two kinds of normal regions: one kind takes the boundary of B (or part of it) as part of its boundary, and another kind does not contain any boundary point of B on its boundary. DEFINITION 8.3. Stable nodal points, foci, and limit cycles are called sinks, and unstable nodal points and foci and the boundary of B are all called sources. It is easy to see the boundary of every normal region has at least one source and one sink; for otherwise the regular trajectory in this normal region would have nowhere to go as t-+ ±oo.( 3 ) LEMMA 8. 5. The boundary of every normal region has one and only one source and one sink.
Suppose that on the boundary of the normal region G there exist two sinks M1 and M2. We shall obtain a contradiction. Use a smooth arc C lying entirely in G to connect Mt and M2. Since Mt is stable, the regular trajectories passing through points on C close to M1 all enter M1 (as t -+ +oo); the totality of all the points on C entering M 1 form a corresponding open set 81. Similarly, the totality of all the points on Centering M2 form a corresponding open set 82. The sum of 82 and S 1 cannot be equal to C, since there must exist a point N on C such that the trajectory passing through N as t -+ +oo enters a third sink M3. We may as well assume N is a boundary point of the open set 8 1 ; but the totality of all the points on C entering M3 also form an open set, and so the trajectory passing through the points in the vicinity of N in 81 will enter M3. This contradicts the definition of N. The lemma is completely proved. From this lemma it is easy to see that normal regions of structurally stable systems can only have the following three types: 1. There are one source and one sink on the boundary (Figure 8.2(a)-(d)). 2. There are one source, one sink, two saddle points, and four separatrices (Figure 8.3). 3. There are one source, one sink, and three separatrices (Figure 8.4(a), (b)).(4) (In Figures 8.3 and 8.4, S=source and H=sink.) PROOF.
LEMMA 8. 6. If system (I) satisfies conditions (III), then for sufficiently
small 8 a system (II) which satisfies p(I, II) < 8 also satisfies conditions (III) and has the same number and same type of singular trajectories as (I). ( 3 )Since we have assumed that conditions (III) hold, the regular trajectory in the normal region cannot be a closed trajectory. ( 4 )The source in Figures 8.3 and 8.4 can be a limit cycle or boundary of B; the sink lnay also be a limit cycle. For simplicity, here they are each represented by one point.
182
THEORY OF LIMIT CYCLES
(a)
(b)
(c)
(d) FIGURE 8.2 H
s FIGURE 8.3 PROOF. For a singular point Oi we can construct a very small simple closed curve ci which contains oi but does not contain any other singular points; the index of Ci is the index of Oi. When 6 > 0 is sufficiently small, the vector fields determined by (II) and (I) on Ci do not take opposite directions. Hence the indices of Ci are the same in these two systems. Moreover, since we
§8. STRUCTURAL STABILITY
(a)
183
(b) FIGURE 8.4
o
only have to take the radius of Ci and very small, for any singular point of (II) in the interior of Ci, the real part of the characteristic roots of its linear approximate system is not zero, and has the same sign as the characteristic root of (I) at Oi; hence (II) has one and only one elementary singular point in the interior of Ci, whose index is the same as that of Oi. However, we should note that when Oi is a (critical) nodal point of (I), there may appear in the interior of Ci a focus of (II). After we remove the interiors of Ci from the region B, IP(x, y)l + IQ(x, y)l has a positive lower bound in the remaining region; hence, when is sufficiently small, the quantity
o
IP(x, y) + p(x, Y)l + IQ(x, y) + q(x, y)l also has a positive lower bound in that region. Thus we have proved that (II) and (I) have the same number of elementary singular points possessing the same index in B. For every limit cycle ri of (I), we can construct a small neighborhood ~ of r i such that its inner and outer boundary curves are curves without contact. When o > 0 is sufficiently small, the boundary curves of R; with respect to (II) are still curves without contact possessing the same properties. Moreover, in R; there are no singular points of (II); hence (II) has at least one limit cycler; in every R;. Consider that H(s) ds does not change sign (for fJ > 0 sufficiently small), and the adjacent limit cycles possess different stability. We know (II) has only one limit cycle r; in every R;. Moreover, when fJ > 0 is sufficiently small, in the vicinity of an elementary singular point of (I) there are no limit cycles of (II), and in the other parts of B also no limit cycles of (II) appear; hence (I) and (II) have the same number and the same type of limit cycles. Finally, since a solution is continuous with respect to its initial conditions and its parameter, we can easily see that if (I) has a separatrix "Yl from a
J
THEORY OF LIMIT CYCLES
184
B
D*0
A FIGURE 8.5
s
FIGURE 8.6
source A 1 to a saddle point B and separatrices "Y2 and "Y3 from the saddle point B to the sinks A2 and Aa respectively, and the singular points of (II) in the neighborhood of Ai and B are Ai and B* respectively, then (II) also has a separatrix "Yi from Ai to the saddle point B*, and separatrices "Y2 and "Y3 from the saddle point B* to A2 and Aj respectively. Except for these separatrices, which correspond one-to-one to the separatrices of (I), (II) cannot have any other separatrices. The lemma is completely proved. According to Lemma 8.6, there is a one-to-one correspondence between the finite number of normal regions of (I) and (II), and corresponding normal regions have the same type. In the following we construct a homeomorphic transformation which maps a normal region G1 of (I) to the corresponding normal region G 2 of (II) in such a way that their trajectories correspond to each other, and then combine the point transformations of different normal regions to form a homeomorphic transformation of the region B into itself. Now we use the above-mentioned normal region of type 2 to illustrate the construction of this transformation; the construction for regions of types 1 and 3 is about the same.
-------
. B on the boundary~ Df the Suppose first that the source A and the smk normal region G 1 of type 2 are not limit cycles, and let the saddle points in G 1 be Do and D1 (Figure 8.5). Let b~} (0:::;; .X:::;; 17) represent the totality of trajectories in G 1 , where "Yo and "Y., represent the separatrices from A to Do and D 1 respectively, which have extensions "Yb and "Y~ (also separatrices) to B respectively. From a point M 0 on "YO to a point M., on "Y., we construct an arc without contact Co. in this normal region, and let the point of intersection of
§8. STRUCTURAL STABILITY
185
"t>. and Co. be M>., and denote by A the arc length from Mo toM>. along Ca. We now define a homeomorphic transformation ~ from the normal region G 1 (excluding the points A and B) to a curvilinear rectangle 61 in the (s, A)plane such that the image of a trajectory 1>.• in G1 is a line segment A= A* in 6 1 . Suppose we measure an arc length s• from M>.· on 1>.• to a point M~: (negative towards A, and positive towards B), and set the image of M~: to be (s*, A*). No matter whether A and Bare foci or nodal points, the arc length along 1>. towards A orB always approaches a finite limit; hence the left and right ends of 61 are bounded. Suppose that under this transformation ~ the image of Do is a point D0 on A = 0, and the image of D 1 is a point Di on the line A = TJ. AB before, we can define a homeomorphic transformation "Ill on the normal region G2 (except the two points, source A and sink B) of type 2 of (II) to a curvilinear rectangle 6 2 in the (s, X)-plane. Suppose that under this transformation the images of the saddle points Do and D1 in G2 are the point D~ on the line X= 0 and the point D~ on the line X= Tj respectively. Now it is not difficult to establish a homeomorphic transformation 8 between 61 and 62 such that A = 0, D0, A = TJ, and Di correspond to X= 0, D~, X= Tj, and D~, and the line segment parallel to the s-axis in 61 corresponds to the line segment parallel to the s-axis in 6 2 • Combining the three homeomorphic transformations ~. 8, and w- 1 , and setting the images of A and B to be A and B respectively, we get a homeomorphic mapping between the normal regions G1 and G2. When A is a limit cycle, if we follow the above method of construction, then the left sides of 61 and 62 will become unbounded, and so it is not obvious how to define the transformation between points of A and the points of the corresponding limit cycle A in order for the map between G1 and G2 to be homeomorphic. Now suppose R is a very small open neighborhood of A, whose boundary r (two simple closed curves) is a curve without contact with respect to (I) and (II) (the directions of trajectories crossing it are the same), and the interior of R has a unique limit cycle A of (II). We would like to establish a homeomorphic transformation from R into itself such that the image of A is A, and the trajectories of (I) become the trajectories of (II). On A we measure from a fixed point to a point whose arc length is a, and construct a normal line T/ai the points of intersection of trajectories of (I) with T/a are arranged in order. Let i = ±1, ±2, ... represent the points of intersection of TJa with every trajectory 1b (exterior of A) or 1b' (interior of A) in this order. Here the first point of intersection is on r, the positive integer i represents the point of intersection of 1b and TJa, and the negative
THEORY OF LIMIT CYCLES
186
"'a·
integer -i represents the point of intersection of 1b' and Thus in R U r all the points except A can be represented by (a,b,i) and (a,b',-i), where i is a positive integer. Suppose we have defined a homeomorphic transformation T 1 of B \ R( 5 ) into itself, which changes trajectories of (I) into trajectories of (II), and T1 on the boundary r of R determines the transformation
(a, b, 1) -+ (
(a,b,i)-+ (n,,P(b),l), (a,b',-i)-+ (n,w(b'),-z), where n =
!{1 + 1/i)
1. represents the ith point of intersection of the line segment of TlA outside A
-z
with the trajectory 1,p(b) of (II), and represents the ith point of intersection of the line segment of inside A with the trajectory 1w(b') of (II). First we can see T2 is a one-to-one transformation. Otherwise, suppose a2 > a1 are such that the points (at. b, i) and (a2, b, i) correspond to the same point (n, ,P(b), 1.); then (1 + 1/i)[
"'a
which contradicts the fact that both
\ R represents the remainder set of R in B.
§8. STRUCTURAL STABILITY
187
The above theorems were extended by M. M. Peixoto [145] in 1962 to the two-dimensional compact differential manifold M 2 • He proved THEOREM 8. 3. The necessary and sufficient conditions for differentiable systems defined on M 2 to be structurally stable are conditions (III) and the follwing condition: 4. The a- and w-limit set of every trajectory can only be a singular point or a closed trajectory.
A more important contribution of Peixoto, however, is another theorem in the same paper.
8. 4. In the Banach space B of differentiable systems on M 2 , the set E of all structurally stable systems is open and dense in B.( 6 ) THEOREM
People who are interested in the theory of limit cycles naturally will consider the following problem: if we limit the functions P and Q on the right sides of the system (I) to a narrower class of functions (e.g. polynomials), will conditions (III) still be necessary and sufficient for structural stability? It is easy to see that if we study polynomial systems(1) on the closed Poincare hemispherical surface 0 (that is, the hemispherical surface with equator), then the equator E often becomes a trajectory connecting two saddle points, and so number 3 in the conditions (III) cannot hold. Hence, the characteristics of structurally stable polynomial systems in 0 are as follows: 1. There are only a finite number of singular points in the interior of 0 and on E, and they are all hyperbolic singular points (that is, elementary singular points of the linear approximate system whose characteristic roots do not possess zero real parts). 2. In the interior of 0 there are only a finite number of closed trajectories; E can also be a closed trajectory when it does not contain a singular point. These closed trajectories are single limit cycles. 3. There is no trajectory connecting two saddle points, except, perhaps, E.
(VI)
Next, in the Banach space of polynomial systems (assuming the degrees of P and Q do not exceed some fixed natural number n), is the set Xs of all the ( 6 )In 1978, Carlos Gutierrez [149] pointed out that M should be assumed to be orientable. (T)The reader will see later that it is essential to proceed in this way. Conversely, in 0 we can only study polynomial systems, since only to such a system can we apply the spherical projection and homogeneous coordinates to map it into n.
THEORY OF LIMIT CYCLES
188
structurally stable systems possessing properties (VI) open and dense? For this problem, we can prove THEOREM 8. 5. In the above notation, Xs is an open and dense subset of X. PROOF. The fact that Xs is open can be obtained at once from the structural stability. We now prove that it is dense. Thus, we introduce some intermediate sets between Xs and X: Xoo = {(I) E X: If (I) has a singular point on E, it can only be a hyperbolic singular point; if E is a closed trajectory of (I), then it must be a single limit cycle}. XH ={(I) E X 00 : (I) only has hyperbolic singular points in 0\ E}. Xc ={(I) E XH: (I) does not have a nonequatorial trajectory connecting two saddle points}. From the definitions, it is easy to see that Xs C Xc C XH C Xoo C X. Hence we have to prove that Xoo is dense in X (see [151]), XH is dense in X 00 , and Xc is dense in XH (see [152]). These are rather easy to prove, and we leave them as exercises. In the following we shall only prove that Xs is dense in Xc. On this point, the proof in [152] is questionable. Take any point in Xc, i.e., a polynomial system, which possesses the properties of (VI) except that it may have a multiple limit cycle r in the finite plane. Now we make a slight perturbation of this system (but this perturbed system should still be a polynomial system of degree ~ n) and eliminate this property; then we obtain a point in Xs. Take the perturbed system to be (P- aQ, Q + aP), where 0 < lal «: 1. When r is a periodic cycle or a multiple cycle of even multiplicity, from §3 we know that when lal > 0 and a takes suitable sign, (P- aQ, Q + aP) will not have a limit cycle in the vicinity off; then our goal has been achieved. However, if r is a multiple cycle of odd multiplicity, then the proof of Lemma 8.4 is of no use, since its perturbed system (V) may fail to be a polynomial system. Hence we have to use other methods to prove this lemma. Let the parametric equations of r be X = cp(t) andy = '1/;(t), 0 ~ t ~ T. In a neighborhood of r we can construct a successor function which can be expanded as a power series in no and a:
,y,( ) k ~ no, a = akon 0 =
+ ao1a + ak+t,on 0k+l + auano + ao2a 2
+ terms of higher degrees (ako + ot(1))n~ + (ao1 + o2(1))o,
§8. STRUCTURAL STABILITY
189
where k is an odd number greater than 1, 01(1) and o2 (1) represent the terms which go to zero as no, a - 0, and ako =F 0 since we have assumed r is a k-multiple cycle. From §4 it is easy to prove that
....
~ ,..(0)! ""'(0) exp ( f (P. + Q,) dt) loT exp (- fo (Px + Qy) dt) !! ds. X
8
I'
Since the family of vector fields (P- aQ, Q + aP) forms a family of rotated vector fields with respect to a, aoI aa ~ 0; therefore aOl "' 0. Hence, if we only take lal and lnol suitably small, the number of roots of 1/J(no, a) = 0 can be completely determined from the equation akon~ + ao1 a = 0. It is easy to see that this equation has a unique simple root no, which corresponds to the unique single cycle of (P - aQ, Q + aP). This shows that in the vicinity of the vector field (P, Q) there always exists a vector field (P- aQ, Q + aP), which has a single cycle in the vicinity of r. Hence Xs is dense in Xc. The theorem is completely proved. For the study of the global properties of differential dynamical systems on differentiable manifolds, after the above-mentioned paper of Peixoto [145], the schools led by Stephen Smale in the United States, D. V. Anosov and V. I. Amol'd in the Soviet Union, Liao Shan-tao in China have done a lot of important work, which has become an important part of the new branch of mathematics known as "global analysis". In recent years, people have used some of these more abstract results to study bifurcation theory and the qualitative theory of concrete ordinary differential equations, which deserve serious consideration. Interested readers can refer to [153]-[158].
§9. Work of M. Frommer and N. N. Bautin Starting now we proceed to study problems of limit cycles and the global structure of trajectories for polynomial systems
dxjdt
= Pn(X, y),
dyjdt
= Qn(X, y).
(9.1)
However, when n is a general positive integer, only fragmentary work has been done, except for a long paper of Henri Dulac in 1923 [4]. Dulac's paper was translated into Russian and published in book form in 1980 [147], and a Chinese translation directly from the French paper by S. W. Ye will appear soon; hence its material will not be presented here, although we shall mention a little of this work either in the text or in the exercises of subsequent sections. Hence from this section on we shall mainly discuss the problems of limit cycles and the global structure of trajectories of quadratic polynomial differential systems (which will hereafter be called quadratic systems): •
dxfdt =
2
L
2
dyfdt =
aikxiyk,
i+k=O
L
bikXiyk
(9.2)
i+k=O
We can say that the study of limit cycles of system (9.2) is the most interesting and inspiring part in the theory of limit cycles. The reason is very simple: the system of first degree (i.e., the linear system) always has a general integral; hence it does not have a limit cycle, and detailed discussions of the qualitative state of its trajectories have been presented in textbooks on ordinary differential equations. Hence (9.2) is the simplest nonlinear system. Generally speaking, except for some special cases, there does not exist a general solution of (9.2); even a first integral cannot be obtained. However, the horizontal and vertical isoclines of (9.2) are all quadratic curves; hence their limit cycles and trajectories can possess some rather special properties, and it is possible to study these curves more deeply; moreover, from this we can derive many general problems worthy of consideration, and solving these problems will undoubtedly promote the development of qualitative theory in 191
192
THEORY OF LIMIT CYCLES
the plane. On the other hand, from the viewpoint of practical applications, there are a large number of problems in engineering, technology, and the natural sciences whose mathematical models can be transformed into quadratic and cubic systems. The readers have already seen some of these in §§5 and 6. Hence the qualitative study of (9.2) has important practical significance. For the quadratic system, the earliest research was to find necessary and sufficient conditions for the coefficients of the system to satisfy in order for it to have a center. Both Dulac [5] and M. Frommer [6] studied this problem; in [6] not only were necessary and sufficient conditions obtained, but also the figures of the corresponding trajectories were drawn. Finally, in [159], the figures of the trajectories in the projective plane for all the cases involving the appearance of a center were drawn. In this section, we shall narrate this problem mainly following [6], but we also adopt the method in [5] of obtaining the first integral. Dividing the two equations in (9.2) and eliminating dt, we rewrite (9.2) as dy x + ax 2 + (2b + a)xy + cy 2 (9.3) dx = - y + bx2 + (2c + {J)xy + dy2 ' which is permissible for the problem of studying its center. Since (9.2) takes the origin (0, 0) as its singular point, if its right sides do not have linear terms, or even if they have linear terms but the characteristic roots of the linear parts are not pure imaginary, then (0, 0) cannot be a center; when the characteristic roots of the linear parts of (9.2) are a pair of pure imaginary roots, then we can make (9.3) possess the above form through a linear transformation of x andy. I. If a = {J = 0, then (9.3) can be integrated to obtain a family of cubic algebraic curves:
~(x 2 + y2 ) + ix3 + bx2y + cxy2 + ~y 3 = const.
(9.4)
For the figures of the family of curves (9.4), we can classify them according to the number of their asymptotes and the number of singular points of (9.3). For example, in Figures 9.1-9.4 the families of curves all possess only a real asymptote; Figure 9.1 has one real singular point other than the origin but Figures 9.2 and 9.3 have three different real singular points other than the origin, and equation (9.3) at the same time has two centers. By continuous deformation we can obtain Figure 9.4 from Figure 9.2, where Figure 9.4 has a double singular point and two single singular points. In Figure 9.5 there is a triple singular point other than the origin, which can be seen from the equation dy dx
=
x+xy
(9.5)
§9. WORK OF FROMMER AND BAUTIN
FIGURE 9.1
FIGURE 9.3
FIGURE 9.2
FIGURE 9.4
where (0, -1) is a triple point of intersection of y + ~x 2 + y 2 = 0 and x(1 + y) = 0. Figures 9.6-9.8 all have three real asymptotes, and the boundaries of the families of closed curves surrounding the center have three saddle points, two Saddle points and one saddle point respectively, but there are three saddle Points in each figure. Let the saddle point on the lower-left corner of Figure 9.8 go to infinity; then two asymptotic directions coincide, and the line at infinity is a double asymptotic direction, on which two elementary singular Points of index +1 and the saddle point which goes to infinity coincide to become a singular point of higher order and index +1; then we obtain Figure 9.9.
193
194
THEORY OF LIMIT CYCLES
FIGURE 9.5
FIGURE 9.6
FIGURE 9.7
FIGURE 9.8
FIGURE 9.9
FIGURE 9.10
§9. WORK OF FROMMER AND BAUTIN
195
Figure 9.10 is the case when three asymptotic directions coincide, and this case can be realized from dyjdx
= -xj(y- y 2).
(9.6)
II. Ia: I+I.BI =/:- 0. According to the well-known method of computing the first focal quantity of the singular point (0, 0) of (9.3) (see [67], Part II, Chapter 2), we get (9.7) D1 = [a:( a+ c)- ,B(b +d)].
i
Hence, in order to have a center, we must have a(a+ c)= ,B(b +d).
(9.8)
In the following, we discuss three cases. Il1. a+ c = b + d = 0. It is easy to prove that this property is invariant under the rotation of axes. We rotate the coordinate axes by an angle
a= acos3
(9.9)
hence we can always select
=
x +a'xy Y + b'x2 + c'xy- b'y2.
(9.10)
From this we can see that this equation has another singular point (0, 1/b') besides (0,0), and the integral line y = -1/a'. H there are other singular points, they should lie on this line. Now we apply the transformation ~ x=--
71
1 + b'77'
y = 1 + b'77'
(9.11)
in order to move the singular point (0, -1/b') to infinity. Thus (9.10) becomes d77
de
e[1 +(a'+ 2b')77 + b'(a' + b')77 2 ] =77[1 + b'(a' + b')~ 2 ]
c'e-
(9.12)
We can obtain a general integral of (9.12) by separation of variables, and, generally speaking, it has four integral lines, one of which is 1+b'77 = 0. Hence (9.10), generally speaking, has three integral lines, but when c'2
+ 4b'(a' + b') < 0
there is only one. Of course, the line at infinity can be considered as the fourth integral line of (9.10).
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THEORY OF LIMIT CYCLES
FIGURE 9.12
FIGURE 9.11
Integrating (9.12), we obtain a general integral in the form of a power series
Returning to the (x, y) coordinates, we get !{x2 + y2 ) - ![c'x3
-
3b'x2 y +(a'- b')y3 ] + · · · = const.
(9.13)
This power series has a nonzero radius of convergence; hence (0, 0) is a center because the locus of (9.13) in the vicinity of (0, 0) is a closed curve. Of course, the general integral of (9.12), and then (9.10), can be represented by a finite combination of power functions, inverse trigonometric functions and exponential functions. Hence it is possible to determine the figure for the global structure of the integral curves. The family of integral curves can be classified according to the number of integral lines, and whether there are two or three integral lines which coincide. Figure 9.6 (which can appear as the figure of both case I and case ITt) represents the case when the three integral lines are distinct;(!) Figure 9.11 represents the case when there exist two integral lines which coincide;(2} Figure 9.12 represents the case when three integral lines mutually coincide; and Figure 9.13 represents the case when all the integral lines coincide at infinity (i.e., a'= b' = c' = 0) and the family of integral curves is a family of concentric circles. H there is only one real integral line at infinity (that is, the previously mentioned case c'2 + 4b'(a' + b') < 0}, then we have Figure 9.14. (1 )All the figures in this section can be considered as projective figures, in which every line can be treated as a line at infinity. (2)The original figure of Frommer is wrong; we follow A. N. Berlinski1 [160] for its correction.
§9. WORK OF FROMMER AND BAUTIN
197
FIGURE 9.14
FIGURE 9.13
lh. a+ c = 0, b + d # 0. Here we must have {3 = 0. Now if a= c = 0, then (9.3) can be rewritten as x + o:xy = - y + f3x2 + "YY2 .
dy dx
(9.14)
The plane vector field determined from this equation is symmetric with respect to the y-axis. This case can also be further divided into several different cases, but these will be discussed in detail in the following case III. Now suppose a= -c # 0. Then we can use a similarity transformation to make a = -c = 1. Thus (9.3) becomes
dy dx =
x + x 2 + (2b + o:)xy- y 2 y + bx2 - 2xy + dy2
(9 .15)
The second focal quantity of the focus of (0, 0) can be computed as
(9.16) since b + d # 0. To make D2 = 0, we must have o: = 0 oro:= -5(b + d)1r. The case o: = 0 (we already know that {3 = 0) has previously been discussed. Hence we assume o: = -5(b +d). Then (9.15) becomes
dy dx =
x
+ x2 -
(3b + 5d)xy- y 2 y + bx2 - 2xy + dy2
(9 .17)
Next we compute the third focal quantity of the focus as
D3
= (b + d) 2(bd + 2d2 + 1).
(9.18)
Since b + d # 0, the necessary condition for existence of its center is
bd + 2~ + 1 = 0,
or
b = - 2d2d+ 1 .
(9.19)
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198
Here we can assume d :f:. 0, since when d = 0, to make Da = 0 we must have b = 0, but we have already assumed b + d :f:. 0. In the following we use Dulac's method [5] to prove that when D 1 = D 2 = D 3 = 0, (0, 0) is a center. Rewrite (9.17) as [x + x 2
-
(3b + 5d)xy- y2 ] dx + [y + bx 2
2xy + d1J2J dy = 0,
-
(9.20)
and then apply the transformation 1( I I) X= 2i X - y '
y
1
= 2(xl + yl),
(9.21)
to obtain the equation
b+ d b + d 12] d [12y + (b +3d 1) X12 + -4-X y X I
1 1
- 4 - - 2i
+ [!2 X
I _
b + d l2 b + d 1 1 2 X + 4 Xy +
I
-2-y
(
b + 3d 4
.!..) y12] dy - 0•
+ 2i
1 _
Again we apply the transformation I 2 X =-b+dX1,
to obtain
2
2i [Y1 - b +b3d ++ d X1 - X1Y1
+ [x1 + 2x 2 1 -
X1Y1-
+ 2y2] 1
dx1
2]
b + 3d - 2i b + d Y1 dy1 = 0,
or simply [Yl where
+ 2y~ -
X1Y1 +ax~] dx1
+ [x1 + 2x~ -
b+ 3d+ 2i a=- b+d
X1Y1 +by~] dy1 = 0,
b- b+3d- 2i -
-(b+ d) .
(9.22) (9.23)
Making use of condition (9.19), we can easily prove that a and b satisfy the condition ab = 1. (9.24) Moreover, it is easy to prove that equation (9.22) under condition (9.24) has an integrating factor [/(xl, yt)]- 512 = (1 + 2xl
+ 2yl +ax~+ 2X1Y1 + byn- 5/ 2 ,
and the general integral (9.25)
§9. WORK OF FROMMER AND BAUTIN
199
where
F(x1, yt) = 1 + 3(xl
+ yt)
+~[(a+ 1)x~ +(a+ b + 2)x1y1
+ (b + 1)y~] +! [a(a + 1)x~ + 3(a + 1)x~y1 + 3(b + 1)x1y~ + b(b + 1)yf]. Finally, using the transformation X1
= -b +2 -d( -zx. -
y ),
b+d(. Yl = - zx-y ) 2
to return to the original variables x andy, we see at once that (9.17) has a general integral
[tf + 2d(d2 + 1)y +(if+ 1)(x- dy) 2 ] 3 = C[d2 + 3d(d2 + 1)y + 3d2 (d 2 + 1)y2 - 3d(d2 + 1)xy + (d 2 + 1)(dy- x) 3 ] 2 •
(9.26)
Hence the origin is a center, and the family of integral curves is a family of algebraic c1,1rves.( 3 ) Ila. a+ c # 0, b + d = 0. This case is the same as Ih. The condition for getting a center is
b + d =a= .B + 5(a +c)= ac + 2a2 + d2 = 0. III. The case of a symmetric vector field (9.14). In this case the equation has the following special integral curves: 1) the line at infinity; 2) the line y = -1/a; and 3) the quadratic curve x 2 = Ay2 + 2By + C, where
"Y-a-.8 , a+ .B (a+ .B)( a+ 2.8) C-"Y+a+.B - .B(a +.B)( a+ 2.8)"
"Y A=---,
B=
The singular points are (0, 0) and P(O, -1h) on the axis of symmetry (yaxis), and possible points of intersection of the integral lines and the quadratic integral curve. The figures of the family of integral curves are determined mainly from the following three quantities: i) {a- "Y)h: when this is greater {less) than 0, Pis a center {saddle point); when it is equal to 0 or oo, P is a singular point of higher order; ( 3 )The
global phase-portrait of this case can be found in [159].
THEORY OF LIMIT CYCLES
200
FIGURE 9.15
FIGURE 9.16
FIGURE 9.17
FIGURE 9.18
FIGURE 9.19
ii) (o:- "t)/{3: when this is greater (less) than 0, y = -1/o: has (does not have) real singular points; when it is zero, two real singular points coincide, i.e., the integral line and the quadratic integral curve touch each other tangentially; iii) "1/(o: + [3), which decides the type of the quadratic curve: when it is greater (less) than zero, it is an ellipse (hyperbola); if it equals oo, it is a double integral line y = -1/ o:. The figures of the family of integral curves are Figures 9.12, 9.7, 9.3, and 9.15-9.19.
§9.
WORK OF FROMMER AND BAUTIN
201
IV. a+ c # 0, b + d # 0, but (9.8) still holds. In this case we can rotate an angle tp so that a+c 1 tantp = - - - = - b+d k' and require that k satisfies a' = ak 3
-
(3b + a)k 2 + (3c + {J)k- d = 0.
Thus it is easy to see that the equation after rotation of axes has a' = c' = {31 = 0; that is, it takes the origin as its center. Summarizing the above results, we get THEOREM 9 .1. Equation (9.3) has a center if and only if one of the following conditions holds: 1) a= (3 = 0. 2) a+ c = b + d = 0. 3) a=c=f3=0 (orb=d=a=O). (9.27) 2 4 4) a+ c = {3 =a+ 5(b +d) = bd + 2~ + a = 0, but b + d # 0( ) (orb+ d =a= (3 + 5(a +c)= ac + 2a2 + ~ = 0, but a+ c # 0). 5) a/(3 = (b +d)/( a+ c)= k, ak 3 - (3b + a)k 2 + (3c + (3)k- d = 0. REMARK. By Theorem 9.1 and the above several figures we know that a quadratic system can have at most two centers; when it has one center, although it may have a focus (Figure 9.14), it cannot have a limit cycle in the vicinity of tl:re focus.( 5 ) V. T. Borukhov [161] gave a concrete example to show that for a quadric polynomial system a center and a limit cyc'le can coexist. His equations are dyjdt =ax- x 3 ,
(9.28)
where a = 2A + A 2 and 0 < A < 5 · w- 5 • He proved that (9.28) takes 0(0, 0) as its center, and there exists limit cycles in the vicinity of each of the two foci (±ya, a 2 j(a-A)). V. M. Dolov [162] obtained the result that for a cubic polynomial system, a center and a limit cycle can also coexist. His equations are dxjdt = y- y3 , (9.29) where IJ. < 0. He proved that this system takes (0, 0) as its center, but the foci (y'=JL,±l) change their stability when IJ. varies and passes through -1; hence their neighborhoods can also have limit cycles (see Theorem 3.7 in §3). ( 4 )Note ( 5 )The
that before we derived condition (9.19) we changed a to 1. rigorous proof is given in [159].
THEORY OF LIMIT CYCLES
202
The conditions for existence of a center for quadratic systems, besides (9.27), can also be given in the form introduced by N. N. Bautin. For this we write the quadratic system in a form different from (9.3):
dx/dt = A1X- y- A3X 2 + (2>.2 + As)xy + A6y 2, dy/dt = x + >.1y + A2X 2 + (2A3 + A4)xy- >.2y 2.
(9.30)
Here 0(0, 0) is a coarse focus when A1 =I 0, but in the second equation the coefficient of x2 and the coefficient of y2 only differ by a sign. From this it appears that (9.30) is more special than (9.3). In fact, (9.2), after a suitable coordinate transformation, can always be changed to (9.30) provided that (9.31) that is, 0(0, 0) is a focus or a center of (9.2). For this we can first use the transformation
1 x = - -bb [(a10- a1)17 + b1~], 1 10
1 y = --TJ bt
(9.32)
to change it to
d17jdt1 = bt~ + a111 + B2oe + Bu~TJ + Bo217 2, d~jdt1 = a1~- b111 + A2oe + Au~TJ + Ao211 2,
(9.33)
where a1 ± ib1 (b1 =I 0) are characteristic roots of the linear approximate system of (9.2). Then, after a suitable rotation of axes which leaves a1 and b1 unchanged and makes the sum of the new B2o and Bo2 equal to zero, on dividing both sides of (9.33) by b1 and making t = b1t1. we can rewrite (9.33) as (9.30). For (9.30), we can easily deduce from Theorem 9.1 THEOREM 9. 2. System (9.30) takes the origin as its center if and only if at least one of the following four conditions holds: 1) A1 = A4 =As = 0. 2) A1 = A3 - A6 = 0. 3) A1 = A2 = As = 0. 4) At = As = >.4 + 5A3 - 5A6 = A3A5 - 2A~ - A~ = 0. An important contribution of Bautin [21] is THEOREM 9. 3. When 0(0, 0) is a center of (9.30), after a slight variation of its coefficients there exist at most three limit cycles in the vicinity of 0; for a center of type 4) it is possible to generate three limit cycles.
§9. W
203
For the proof, we introduce the polar coordinates x = p cos cp, y and get from (9.30) dpjdcp = p{A1 + p[->.a cos3 cp + (3>.2 + >.s) cos 2 cpsin cp
= p sin cp,
+ (2>.a + >.4 + >.6) cos cp sin 2 cp- >.2 sin3 cp]} · {1 + p[>.2 cos3 cp + (3>.3 + >.4) cos2 cp sin cp - (3>.2 + >.s) cos cp sin 2 cp- >. 6 sin3 cp]} -l = p[>.t + pA(cp)]/[1 + pB(cp)]
(9.34)
Rt = >.1, R2 = A(cp)- >.tB(cp), Ra = -A(cp)B(cp) + >. 1 B 2 (cp), ... , Rk = (-1)kA(cp)B(cp)k-2 + (-1)k-1>.tB(cp)k-1,
(9.35)
where
} ....
For all the >.i in the vicinity of a fixed point>.; in the >.-space (for example, all the >.i satisfying ll>.i- >.;II < e) and for all cp, there always exists a "f(e, >.i) such that when IPI < "((e, >.i), the series on the right side of (9.34) converges. It is well known that the solution p(cp) of (9.34) satisfying the initial condition p(O) = Po can be expanded as
p = PoVt (cp, >.i) + p~v2( cp, >.i) + pgva( cp, >.i) + · · ·, (9.36). where vk(cp, >.k) satisfies the conditions v 1(0, >.i) = 1 and vk(O, >.i) = 0 when k 2:: 2. Substituting (9.36) into (9.34) and comparing the coefficients of like powers of p, we can determine the equations of all the Vk: dvtfdcp = VtRt. dv2jdcp = v2R1 + vf R2, } dvafdcp = vaRt + 2v1v2R2 + vrRa, ....
(9.37)
For all sufficiently small p and all >.i satisfying ll>.i- >.ill < e, some segment of the line cp = 0 is a segment without contact. Letting cp = 211" in (9.36), we can obtain a successor function P(Po, 2rr) = povt(211", >.i) + p~v2(21r, >.i) (9.38) in a sufficiently small line segment 0 < p < p1 , cp = 0, within the radius of convergence of the power series. LEMMA 9.1. vk(211", >.i) is an integral function of all >.i and is a homogeneous polynomial of degree k- 1 for >.2, ... , >.6 when >.1 = 0. PROOF. From (9.37) and the theorem for dependence of a solution on the parameters, we know that vk(21r, >.i) is an integral function of all >.i. In
204
THEORY OF LIMIT CYCLES
particular, if >.1 = 0, then Rt = 0, and Rk is a homogeneous polynomial of degree k- 1 of >. 2, ... , >.6. Since at the same time every equation of (9.37) is separable, vk(21r, >.i) is also a homogeneous polynomial ?f degree k- 1 of ).2,····).6· LEMMA
9.2.
Vt(2rr, Ai) = e 2 ,..~ 1 , v2(2rr, Ai) = >.10~ 1 ) 1 v3(2rr1 Ai) = v3 + >.10~ 1 ), v4(2rr1 Ai) = v39i3>+ >.tOi1 >, vs(2rr, >.i) =tis+ v3(}~3 > + >.10~ 1 ) 1 v5(2rr1 >.i) = tls9~5 ) + v30~3 ) + >. 1 9~ 1 ), \ ) - +-V5 (J(S) +-V3 0(3) + .1\1 \ 0(1) V7 (2 1r 1 Ai = V7 7 7 7 1 vk(2rr1>.i) = v19r> + vsOks) + v30k3>+ >.t0k1}
(9.39)
(k
> 7),
where
= -i>.s(>.3- >.6)1 } tis = ~ >.2>.4(>.3 - >.6)(>.4 + 5>.3 - 5>.6), 117 = -~~rr>.2>.4(>.3- >.6) 2(>.3>.6- 2>.~- >.~),
V3
and
oii>
(9.40)
is an integral function of >.i.
Vt(2rr,>.i) can be obtained at once from (9.33). According to Lemma 9.1 1 we can write PROOF. The formula for
vk(2rr, >.i) = vi0 > + >.t0k1)
(k > 1),
where vi0 ) is a homogeneous polynomial of degree k - 1 of all the Ai (i == 2, ... , 6), without containing >.1. From the fourth condition for existence of a center in Theorem 9.2, we know that vi0 ) = 0 when >.s = A4 + 5>.3 - 5>.6 == >.3>.6- 2>.~ - >.~ = 0. Hence
vk(2rr, Ai) = (>.3>.6- 2>.~- >.~)()~'
(9.41)
+ (>.4 + 5>.3 - 5>.6)9~ + .As9A: + .At9k1),
but when >. 1 = .As = >.2 = 0 and .At = >. 5 = A4 = 0, a center also appears; that is, the vk(2rr, Ai) (k -#1) should all be equal to zero, and so
We then use the second condition for the existence of a center, >.1 0, and get -
-(7)
Ok = (>.3 - .A5)0k ,
-.
(5)
Ok = (>.3 - .Aa)Ok ,
= A3 -
9A: = (>.3 - .Aa)Ok(3) .
>. 6 ==
§9. WORK OF FROMMER AND BAUTIN
205
Substituting all this in (9.41) yields vk(27r, .xi) = (..\3..\s- 2.x~ - .x~)..\2..\4(..\3 - .Xs)ofl
+ (..\3- ..\s)(..\4 + 5,\3- 5..\s)..\2,\40k5 ) + ..\s(..\3 - ..\s)Ok3l + ..\10k1l (k :/: 1). Now suppose k = 2. Then from Lemma 9.1 we know that ofl 0~3 ) = 0 on the right side of (9.42); hence
(9.42)
= 0~5 ) =
V2(27r, Ai) = ,\10~ 1 ). When k compute
= 3,
as before, we have ofl ::::: 0~5 )
:::::
0, and from (9.37) we can
hence
+ ..\ 1 (}~1 ) = v3 + ,\ 1 0~ 1 ). When k = 4, as before, we know that Oi7 l :::::: oi5 l = 0, and hence v4(21r, Ai) = ..\s(,\3- ..\s)Oi3l + >.10i1l :::: v30i3l + >. 10i1l, v3(21r, Ai) = ( -71"/4)..\s(..\3- ..\s)
where the coefficient of 1!3 can be denoted by 0i3). We then compute vs(27r, Ai)· From Lemma IU we have note that
ofl = 0.
Next we
= R1vs + R2(2v2v3 + 2v1~1 ) + R3(3v1v~ + 3v~v3) + 1R 4 v~v 2 + R 5 v~. then R1 = 0 and v1 0; thus
dvsfdtp If we let ..\1 = 0,
=
dv 5 jdtp = 2R2v2v3 + 2R2v4 + 3R3v~ + 3~~'!13 + 4R4v2 + R5 .
(9.43)
Note that here we have dv2jdtp
= R2,
dv3jdtp
=
dv4jdtp = R2(v~ + 2v3) + 3~~p3 + R 4; hence the right side of (9.43) can be written a&
R3 v3
+ 2R4v2 + Rs + 4v~ ~v2 + 2v3 dv3 '1-tp
dtp
2d(v2v4) _ dtp
3d(v3v~) dtp
.
THEORY OF LIMIT CYCLES
206
Integrate both sides of (9.43) from 0 to 271", and take .As :;::; 0. Since under the condition .A1 =.As = 0 we have
it follows that vs(27r, .Ai) =
1 2
1r
(R3v3
+ 2R4v2 + Rs) dcp.
Using this formula to compute vs(27r, .Ai), we get (when .At = .As = 0) vs(27r, .Ai) = (7r/24).A2.A4(>.3- >.6)(>.4
+ 5.A3- 5.A6)·
Hence for >.1 :f. 0 and .As :f. 0 we have vs(27r, .Ai) = (7r/24)(.A2.A4(.A3- .A5)(.A4
+ 5>.3- 5.A6)
+ .As(>.3 - >.6)8~3 ) + .A18P)
-- -Vs +-V3 (}(3) \ ()(1) s + "1 S · For k = 6, note that by Lyapunov's classical theory the first nonzero vk(21r, Ai) (k > 1) must have odd subscript, and when .At = .As = >.4 + 5).3 - 5>.6 = 0 we have V2 = ... = Vs = 0; hence ofl = 0 by (9.42). Thus v6(21r, .Ai) = vs8~s)
+ v3 8~3 ) + >. 1 0~ 1 ).
Fork= 7, from (9.42) we have V7(21r, Ai) = .A2>.4(A3 - >.6)(>.3>.6 - 2>.~+-V3 ()(3) + 1\1 \ ()(1) + -Vs (J(S) 7 7 7 • As before, we let >.1 = >.s = >.4 we can determine that
+ 5>. 3 -
.A~)()fl
5.A6 = 0 to compute v7(21r, .Ai)· Thus
(}7(7) = -(25/32)11" ( >.3- >.6)· Hence '\ ) =V7 V7 (2 11",1\i
+-Vs (J(S) +-V3 ()(3) + V1 ()(1) 7 7 7 •
Finally, we prove the representation formula for vk(27r, .Ai) for k > 7. By (9.42) it suffices to show that ofl contains a factor >. 3 - >. 6 . We know that efl is a homogeneous form of degree k- 6 of >. 2, ... , >. 6, and, making use of the indeterminate nature of the forms of efl, ois), and (Ji3), we can always move the terms involving >.4 and >.s of efl to oi5) and oi3), so that ei7 ) does not contain >.4 and >.5. In this way we can prove that, under the conditions
§9. WORK OF FROMMER AND BAUTIN
or)
At =As = A4 + 5A3 - 5A6 = 0, contains a factor J.t = A3 same time, since A4 = -5(A3 - A6), we have
207 -
A5.( 6) At the
and
(9.44)
=
Hence this problem turns into the problem of proving that 1/J(po, 0) 0 on some line segment of rp = 0. The proof of this fact is very difficult, and the method used by Bautin deserves our careful attention. Consider the system of equations
dx/dt
= -H~(x,y) + J.tp(x,y),
dyjdt
= H~(x, y) + JJ-Q(X, y),
{9.45)
where
When h > 0 is very small, H(x, y) = h represents a family of closed curves near the origin which is first integral of
dx / dt = -y - A5X 2 + 2A2XY + A6Y 2, dyjdt =X+ A2X 2 + 2A6XY- A2y 2
(9.47)
0
(9.47) can be obtained from system (9.30) by letting At = As = A3 - A6 = ..\4 = 0. p(x, y) and q(x, y) are polynomials in x and y. If we take p = -x 2 and q = -3xy, we immediately obtain (9.30) with At = As = A4 + 5A3 - 5A6 = 0. Now in the region of convergence of the series (9.36), we can take a curve Cho from the family H(x, y) = h, and move the coordinate system to the curve Cho and use the curvilinear coordinates (8, h). Here his a parameter corresponding to the curve of the family H(x, y) = h, and 8 represents the periodic coordinate on ch such that d8/dt = 1 on cho; but on ch we have dsjdt = Th 0 /Th, where Th is a period of Ch, Tho is a period of Cho and 8 is measured from the x-axis. From this we can see that any ch in the vicinity of Cho has a fixed period Tho, and the transformation of coordinates
x=ft(s,h), is analytic. When h
(9.48)
= ho, we get the equations of Cho, denoted as x = /t(s, ho) = rp(s),
-(7)
y = h(s,h)
Y = h(s,ho)
= 1/J(s).
(9.49)
(6)1£ (}k contains a factor A4, then when A4 + 5A3 - 5A6 = 0 we can produce a factor >. 3 - A6 from A4, although A3 - A6 originally is not a factor of iifl .
THEORY OF LIMIT CYCLES
208
Using the relation formulas H~fL~
H(ft(s, h), h(s, h))= h, H~f~h
+ H~f~a =
0,
+ H~f~h = 1,
we obtain the equation (9.50) in the new coordinate system. In this equation we let h = ho
+ 6, and obtain
d6 /ds = R(6, s, J.L);
(9.51)
the right side of this equation is analytic in 6 and J.L when 161 and IJ.LI are very small, and is also an analytic periodic function of s. Expand the solution 6 = ,5(s) of (9.51) in a power series in J.L and its initial value 60 = 6(0), and denote it as
6 = Cw(s)8o + Co1(s)J.L + C2o(s)85
(9.52)
+ Cu(s)8oJ.L + Co2(s)J.L2 + ···.
In this formula we lets= T (= Th 0 ), and note that Cw(r) = 1 and Cko(r) = 0 for k 2:: 2, and we obtain at once the successor function on the x-axis:
8(r) = 6o + Co1(r)J.L + Cu(r)8oJ.L + C 0 2(r)j£2 + · · ·. LEMMA 9. 3. 1/J(po, 0)
(9.53)
=0 is equivalent to Co2 (r) 2=0, which in turn shows
that no matter how Cko is chosen, the coefficient of J.L in the expansion (9.53) is always equal to zero.
PROOF. Let p0 denote the abscissa of the point of intersection of Cho and
cp = 0. Expanding the right side of (9.44) in powers of Po- p0, we obtain
p- Po= J.L 2[1/J(p0,o) + .,p~o (p0,O)(po- p0) + J.L'I/J~(p0 , o) + .. ·].
=
(9.54)
Since p0 and p0 are arbitrary, .,P(po, 0) 0 is clearly equivalent to .,P(p0,0) Now we prove that the latter identity is equivalent to Co2(r) 0. From the equation H(p, 0) = ho + 8 we get !P2 + -!A2P3 = ho
=
=0.
+ 8.
(9.55)
hO·
(9 · 56)
When 8 = 0 and p = p0, we have 1 •2 2Po
+ gi\2Po 1\ •3 -
Subtracting (9.56) from (9.55), we get {j
= .\32 (p- Po)3 + ~(1 + 2.\2p'Q)(p- p'Q)2 + (p- p'Q)(p'Q + A2Po2).
(9.57)
§9. WORK OF FROMMER AND BAUTIN
209
Let p = po, and get
t5o = >.32 (Po- p(i) 3 +
~ (1 + 2>.2p())(Po- p(i) 2 +(Po- p(;)(p(; + >.2Po 2).
(9.58)
Now subtracting (9.58) from (9.57), we get
t5-t5o= ~2 (p-po) 3
+ [~(1+2>.2Po)+>.2(Po-Po)] (P-Po) 2 (9.59)
+ [(p(j + >.2p() 2) + (1 + 2>.2p())(Po- Po) + >.2(Po - Po) 2](p- Po).
Substituting (9.54) in the right side of (9.59), we get a power series expansion of t5 - t5o in terms of po - p0 and J.t. But this expansion can also be obtained by substituting (9.58) into the right side of (9.53). Comparing the coefficient of J.t2 in these two power series expansions, we get
(p0+ >.2Po2)1/J(p(),O) = Co2(r). From this we at once get the proof of the lemma. In the following we shall prove that Co2 (T) 0. Expanding the right side of (9.51) in power series of t5 and J.t, and noting that
=
R(O,s,O) = R6(0, s,O) = RZ~(O,s,O) = · · · = 0, we obtain
~! =R~(O,s,O)J.t+2 [nz~'(O,s,O)t5J.t+~R~2(0,s,O)JL 2 ]
+ .. ·.
Substituting (9.52) into both sides of the above formula, and comparing the coefficients of like powers of t50 and J.t, we obtain
Cw = 0,
Cot = R~ (0, s, 0),
Cu = 2CwRZ~' (0, s, 0),
C2o = 0,
Co2 = 2CotRZ~' (0, s, 0) + R~2 (0, s, 0),
... '
(9.60)
where Cii represents the derivative of Cii with respect to t. It is clear that has the initial conditions
cij
Cw(O) = 1,
Cij(O) = 0 when (i,j) =P (1,0).
(9.61)
From (9.50) it is easy to compute R~(O, s, 0)
= ~P- tj:;q,
R~2 (0, s, 0) = 2(~p- tj:;q)(f~hP- f~hq) lo=O'
RZ/'(0, s, 0)
=
! (!~
. . P-
f~ . . q) IJ'=O=O
= [J;hsP- f~1hsq
+ f~..,(p~J~h + P~f~h) - f~ ... (q~f~h + q~f~h)lo=O·
(9.62)
THEORY OF LIMIT CYCLES
210
From (9.60)-(9.62) we get
C10(s)
= 1,
Co1(r) =loT (-Jp- cpq) ds,
Cu(r) = loT 2RZil ds =
2[J~hP- f~hq](i + 2 loT[- J~h(P~f~s + P~f~s) + f~h(q~f~s + q~f~s) + f~s(p~f~h + P~f~h)- f~s(q~f~h + q~f~h)]8=0 ds
= 2 loT
U~sf~h- f~hf~s)8=0(P~ + q~) ds, (9.63)
But
' !'lh (!2s
-
!'2h !'ls )8=0 =
(!'lh dy , dx) dt - f2h dt Jl=c5=0
= (H~f~h + H~f~h)8=o and so
Cu(r) = 2 loT (p~
=1,
+ q~) ds.
(9.64)
Finally,
Co2(r) = 2 loT (los (-Jp- cpq) ds)
[f~1hsP- f~'hsq + f~s(P~f~h + P~f~h) - f~s(q~f~h
+ q~f~h)]8=0 ds
+ 2 loT (-Jp- cpq)(f~hP- f~hq)8=o ds = 2 loT
(foB (-Jp- cpq) dS) · [:s
U~hP- f~hq) + U~sf~h- !~hf~sHP~ +. q~)L=o ds
+ 2 loT (-Jp- cpq)(/~hp- ~~hQ)8=0 ds = 2 loT [los (p-J - cpq) ds]
(p~ + q~) ds
+ (f~h(r)p- f~h(r)q)Col(r).
(9.65)
Now we pick p and q so that system {9.45) still takes the origin as its center; then all the coefficients of the successor function should be zero. From this we can obtain some identities, and using these we can prove Co2 (r) 0. 1. Take p 0 and q = cp ·1/J = x · y. Then system (9.45) corresponds to (9.30) under the condition -" 1 = A3 - -"6 = 0; hence the origin is a center.
=
=
§9. WORK OF FROMMER AND BAUTIN
211
Then from (9.63)-(9.65) we get
C01 (r)
=-loT xyxds = 0,
C 11 (r) = 2
C02 (r)
=-loT (x loa xyxdt)
ds = 0.
loT xds = 0, (9.66)
=
2. Take p = xy = rp · '1/J and q 0. System (9.45) still corresponds to (9.30) under the condition At = A3- A6 = 0. As before,
Cu(r)
= 2 loT yds = 0.
(9.67)
3. Take p = -x2 = -rp2 and q = 2xy = 2rp'I/J. Then system (9.45) corresponds to (9.30) when At =As = A4 = 0. Hence the origin is a center. From this we obtain Cot(r)
=-loT (x iJ + 2xyx) ds = 0. 2
Using the first formula of (9.66), we get
loT x
2
y ds =
0.
(9.68)
4. Take p = -x 2 and q = -3xy. We get the case A1 = As = A4 +5A3 -5A6 = 0. But now the origin is not necessarily a center; hence it is not certain whether Co2 (T) 0. Now we use formulas (9.66)-(9.68) which we have already obtained, and the obvious identities
=
loT iJ ds = loT x ds = loT x x ds = loT xx ds = loT yiJ ds =loT (xy + xiJ) ds = 0 2
(9.69)
=
to prove C 0 2(r} 0. First we note that by (9.66) and (9.68) we know the right side of (9.65}:
Cot(r)
=loT (-x iJ + 3xyx) ds = 0. 2
lienee from (9.65} we get
Co2(r) = -5
loT [x lo 8(3xyx- x 2y)dt] ds.
Dsing the third formula of (9.61), we see that
~Co2(r) =loT (x los x2 ydt)
ds.
(9.70)
THEORY OF LIMIT CYCLES
212
But
loa x2iJ dt = [x y] 0- 2 loa xyx dt, 2
and so the right side of (9. 70) becomes
loT x 3 yds-x 2(0)y(O) loT xds-2 loT (x los xyxdt)
ds.
By (9.66) the second and third terms of the above expression are equal to zero; hence it suffices to prove that
loT x 3 y ds = 0. We use (9.47) to compute
(9.71)
J; x 2yds and J; x 2x ds, and obtain
loT x 2(x + >.2x2 + 2>.6xy- >.2y 2) ds = 0,
(9.72)
loT x 2 ( -y- >.6x2 + 2>.2xy + A6Y 2 ) ds = 0.
(9.73)
Multiplying (9.72) by >.6, (9.73) by >.2 and adding, we get
2(>.~ + >.~) loT x 3 y ds + >.6 loT x3 ds -
>. 2
loT x 2y ds = 0.
(9.74)
Moreover, using (9.66) and (9.67), we also have
loT xds = ->.6 loT x 2 ds + 2>.2 loT xy ds + >.6 loT y2 ds = 0,
(9.75)
loT iJ ds = >.2 loT x2 ds + 2>.6 loT xy ds- >.2 loT y2 ds = 0.
(9.76)
Multiplying (9.75) by >.2, (9.76) by >.6 and adding, we get
2(>.~+>.~) loT xyds=O. We may as well assume >.~ + >.~ Co2(r) = 0 clearly holds. Hence
=f. 0,
for otherwise the origin is a center and
loT xyds = 0, and, substituting in the right side of (9.75), we get
->.6
loT (x 2 -
y 2) ds = 0.
(9.77)
213
§9. WORK OF FROMMER AND BAUTIN
Moreover, we also have
loT xx ds = - ).61T x 3 ds + 2).2 loT x 2y ds + ).6loT xy 2ds = 0,
r yyds = ~r x .yds + ~r xy ds- lor 2
).2
~
2).6
2
).2
y 3 ds = 0,
faT (xy + xy) ds = -).61T x 2y ds + 2). 21T xy 2ds + ).61T y 3 ds + ..X21T x3 ds + 2).6 loT x 2y ds Eliminating
J; xy 2(
2
ds and
).2
loT xy 2 ds = 0.
J; y 3 ds (using (9.77)), we get
).~ + ,\~) [,\61ar x3 ds -
.\2
loT x 2y ds] = 0.
Using [ ] = 0 in the above formula, we can obtain (9.71) immediately from (9.74). Lemma (9.2) is completely proved. Now we prove Theorem 9.3 stated earlier in this section. By Lemma 9.2 we can write p- Po= Po[211'.\1(1 + A1
+ Pot/Jl(po, Ai))
+ V"g(l + Pot/Jg(po, .\i))P~ + Us(l + PotPs(po, ).i))P~
+ V"7(l + PotP7(Po, Ai))pg), or simply {9.78) where all the 1/Jj are power series of po,(1) and their coefficients are integral functions of all the ).i· These power series converge when I -Xi -..Xi II < e and Po < -y(e, ..x:). Now suppose system (9.30) corresponding to -Xi takes the origin as its center. We prove we can always choose eo and t5o such that when I -Xi- . Xi II < co the equation p - Po = 0 cannot have more than three positive roots in a <5o-neighborhood of the origin; this also shows that the system (9.30) cannot have more than three limit cycles in a !50 -neighborhood of the origin. (7)By virtue of this formula, if Al f. 0, then (0, 0) is a coarse focus, which is equivalent to the divergence being nonzero at (0, 0). If Al = 0 but li3 f. 0, then (0, 0) is a fine focus of first order; if Al = li3 = 0 but tis "! 0, then (0, 0) is a fine focus of second order; if >.1 ""'ii3 = ti 0 = 0 but i/7 f. 0, then (0, 0) is a fine focus of third order. Hence the focus of a QUadratic differential system is at most of third order. This definition will often be used in the sections of §11.
THEORY OF LIMIT CYCLES
214
0
First, we can find c:1 < c: and a1 such that when I!Ai - >.iII < c:1 and < p0 ~ a1 we have 1/Jj 2:: ~ (j = 1, 3, 5, 7). We rewrite (9. 78) as (9.79)
when Po is sufficiently small, every 1/Jj /1/Ji can be expanded into a power series in Po and >.t. and
= 1 + >.1 cp~i) + Pol/J~i) = 1/Jj* (j = 3, 5, 7). Then we take c:2 and a2 such that for I >.i- >.iII < c:2 and 0 < Po ~ 62 we have 1/Jj /1/Ji
1/Jj* 2:: ~, and it is easy to see that when 0 < Po ~ 62, the positive roots of the equation p - Po = 0 are positive roots of 1/Jo = 21r >.1 + 1i31/J3* P~ + 1is1/J5* P~ + 'ih1/J1* pg = 0. Let
1fi =~~~/Po = 2V3(1 + -\1cp~3 ) + Pol/J~~)) + 41is(1 + >.1cp~5 ) + pol/JW)pg + 61:17(1 + -\1 cp~7) + Pol/JW)p~.
(9·80 )
We know that for 0 < Po ~ 62 the number of positive zeros of the function 1/Jo can be at most one greater than the number of positive zeros of tfi. Now we note that the right side of (9.80) and the terms in the square brackets on the right side of (9. 79) have the same form, and their coefficients also possess similar properties; the degree of Po in the latter (the functions in the parentheses are taken as coefficients) is lower than that of the former by two. Hence for 1fi we apply the previous method two more times, and deduce that there must exist a 63 > 0 such that when 0 < po ~ 03 the number of positive zeros of the function tij can be at most two greater than the number of positive zeros of the function =;;J
=
48v7(1 + >.1cp&7) + Pol/J~il).
1fi it is clear that there exist C:4 and 64 such that for all Ai satisfying I!Ai --'iII < C:4 and 0 < Po ~ a4, =;;J does not have zeros; thus for all these Ai
But for
and Po the function 1/Jo has at most three positive zeros. Similarly we can prove that if system (9.30) corresponding to -'i does not take the origin as its center (that is, not all the functions vk(21r, >.:) are equal to zero), then when the first term not equal to zero in the parenthesis on the right side of (9. 78) is >.i, U3, Us or V7 respectively, we can find c:0 and 6o such that for all >.i satisfying ll>.i- >.;II <eo and 0
215
§9. WORK OF FROMMER AND BAUTIN
equation (9.20) does not have any, or has at most one, two, or three, limit cycles in the Do-neighborhood of the origin respectively. Next we prove the second part of the theorem. Suppose the system corresponding to .At takes the origin as its center, and .Ai
= .A5 = ).: + 5).j -
5).6
= ).j.A6 -
2.A6 2
-
.A2 2
= 0,
but .A4 ::/:- 0, .A2 ::/:- 0 and .Aj - .A{i ::/:- 0; that is, the origin is a center of the fourth type in Lemma 9.1. Now we vary .A2 slightly to make ).2 = 0, and make .Aj.A6 - 2.A6 2
-
).~
::/:-
so that here v7 ::/:- 0 (we may as well assume v7
0,
> 0), and
vs = va = 1It = 0. We know that for any e and 8, 0 < e <eo, 0 < 8 < 8o, when 0 < p < 8 and II.Ai- .At II < e, the condition 1/Jj 2:: always holds. Now we vary .A4 to become ). 4 in such a way that va becomes negative, but we still have 1J7 > 0; and we also require lvsl to be so small that for all PoE {8d1),8d2 )) C (8/2,8), the inequality p - Po > 0 still holds. On the other hand, when IPo I is very small, the right side of {9. 78) and v5 have the same sign; hence in (0, 8) the equation p - Po = 0 has at least one positive root. Suppose p = p' is the smallest root among them. Now we vary .A5 to become ). 5 in such a way that va > 0 but Ivai is so small that for all Po E (8p>, 8i 2 )) C (8d 1 ), Od 2 )) the inequality p- Po > 0 still holds, and for all po E (8~ 1 ), 8~ 2 )) C (0, p') the inequality p- Po < 0 holds. Since, for IPol sufficiently small, p- Po will have the same positive sign as va, we know that p- Po has at least two roots in (0, 8); suppose p = p" is the smallest root among them. Finally, vary .Ai to become ). 1 < 0, so that 2 >), the inequality p- Po > 0 holds, 1 ), 1. for all PoE (8~ 1 ), 8~ 2 )) C 2 1 2. for all PoE (oJ•>,oJ >) c (o~ >,oJ 2 >), the inequality p-p0 < 0 holds, and 3. for all PoE (8~ 1 >,8~ 2 )) c (O,p''), the inequality p-Po > 0 holds. Moreover, we also know that when IPol is sufficiently small, p- Po has the same negative sign as .At; hence now in (0, 8) the equation p - Po = 0 has at least three roots. For the other types of centers, it is not certain whether we can prove they can generate three limit cycles; the method of discussion is the same as before, and is therefore omitted. Similarly, we can also prove that the focus of (9. 78) which makes the first nonzero coefficient on its right side be v7 can generate three limit cycles, the focus with the first nonzero coefficient v5 can generate two limit cycles, and the focus with the first nonzero coefficient v3 can generate one limit cycle. The proof is omitted.
!
(oi oi
THEORY OF LIMIT CYCLES
216
Summarizing the above, we can see that the difficulty in the proof of Lemma
9.2 lies in proving the last formula of (9.39); that is, proving that 1/J(po, 0) in (9.44) is identically zero in some line segment of tp = 0.(8 ) Bautin's method can also be divided into the following three steps: 1. Seek a family of closed curves H(x, y) = h, and with this introduce curvilinear coordinates (c5,s) to replace the original polar coordinates (p,tp), and write the successor function (9.53) of system (9.45) in this new coordinate system. 2. Expand c5 (T) - c5o in a power series in I" and po - Po by two different methods. Hence we can obtain the proof of equivalence between 1/J(po, 0) = 0 and Co2(r) 0. 3. Select some special p(x, y) and q(x, y) in (9.45) such that the origin is still its center, and, at the same time, for the corresponding (9.53) c5(r)-c5o = 0 should hold. From this we can obtain the identities (9.66)-(9.68) depending only on the family of curves H(x, y) = h, and then use these identities and some obvious identities (9.69) to prove that the successor function of system (9.45) corresponding to system (9.30) with At = As = A4 + 5A3 - 5A6 = 0 under the new coordinate system has Co2 (T) 0 in its expansion. Necessary and sufficient conditions for the existence of a center for quadratic systems, other than those given in Theorems 9.1 and 9.2 in this section, will later be provided in other forms for convenience of application. Similar problems for cubic systems which do not contain terms of second degree have been solved by N. A. Sakharnikov [163].
=
=
Exercises 1. Prove that (9.25) is a general integral of (9.22). 2. Move [ ] on the right side of (9.26) to the left and then expand it near the origin to prove that the origin is a center. 3. Prove in detail that vs(27r, >.i) = (11"/24)>.2>.4(>.3- >.6)(>.4 + 5>.3- 5>.6)· 4. Prove the quadratic curve x2 = Ay 2 + 2By + C is indeed an integral curve of equation (9.14). 5. Without using the transformation (9.11), prove that (9.10) has an integral factor 1 1+a'y
--[(a'+ b')b'x2 - (1 + b'y) 2 + c'x(1 + b'y)]- 1 • 6. Prove that (9.14) has an integrating factor (1 + o.y) 2f3/o.-l. ( 8 )Note that in order to prove Theorem 9.3, practically we only need the representation formula for vk(27r, .>.;) given just before (9.44).
§9. WORK OF FROMMER AND BAUTIN
217
7. Prove in detail that there exist limit cycles in the exterior of each singular point (±JC.i, a 2 f(a- A)) (use the Annular Region Theorem). 8. Prove in detail that system (9.29) has limit cycles in every neighborhood of (J=j:t,±l) for some J.L < 0. 9. Prove that the system of equations dxjdt = y,
takes (0, 0) as its center, and for some value of o: there exists a limit cycle in every neighborhood of (±2, 0) (see [162]).
§10. Global Structural Analysis of Some Quadratic Systems without Limit Cycles
For a given plane polynomial system, if we know the number of its singular points on the finite plane and the equator, the topological properties of each singular point, the existence or nonexistence, number, and relative positions of closed trajectories, and the directions of separatrices passing through the singular points, then the global structure of this system can be determined. Since the fifties, people have studied the global analysis of quadratic systems without llm1t cycles, and have drawn their global phase-portraits. In this section we present some of the work in this area. In §13, the reader will see that if we add one term to a system without limit cycles so that limit cycles can be generated, then the number of limit cycles and their relative positions are closely related to the global structure of the original system without limit cycles. I. Global topological classification of homogeneous quadratic systems. The system 2 2 { dxjdt = aux + a12XY + a22Y , dyjdt = bux 2 + b12XY + b22Y 2
{10.1)
is called a homogeneous quadratic system. Clearly it does not have a limit cycle. When the right sides of (10.1) do not have a common factor, {0, 0) is its unique singular point. L. S. Lyagina [164] studied the structure of {10.1) near the origin, showed that there were sixteen possible cases, and gave methods for classification. Lawrence Markus [165] studied the global topological classification of {10.1), and, according to whether it had a straight line filled with singular points and the number of straight line solutions, obtained a standard form under the linear transformation. It is worth mentioning that this paper uses the methods of nonassociative algebra. Recently, N. I. Vulpe 219
220
THEORY OF LIMIT CYCLES
and K. S. Sibirskil [166] studied the global topological classification and geometrical classification of systems (10.1) whose right sides have and do not have a common factor, and pointed out some criteria on their coefficients for classification. In the present section we introduce all possible structures for systems (10.1) whose right sides do not have a common factor. For the case with a common factor the problem is much simpler, and is left to the reader as exercises.(!) We first do some preparatory work, which is always given in general textbooks on qualitative theory. The exceptional direction 9 = 9o of (10.1) is determined from the equation
G(9) = - a22 sin3 9 + (b22- a12) sin2 9cos9 + (b12- au) sin9cos2 9 + bu cos3 9 = 0.
(10.2)
It is easy to see that equation (10.2) has at least a pair of real roots 9 = 90 and 9 = 9o + rr in [0, 2rr). We may as well assume 9o = 0; thus bu = 0. Replacing {10.1), we only have to discuss the system
dx I dt = aux 2 + a12xy + a22Y 2, dy I dt = bl2XY + b22Y 2. Here we must have au common factor. Now (10.2) becomes
(10.3)
=I 0; otherwise the right sides of (10.3) would have a
G(9) = sin9[-a22 sin2 9+(b22-a12) sin9cos9+(b12-au) cos 2 9] = 0. (10.4) The discriminant inside the brackets in the above formula is
(10.5) If we introduce polar coordinates x = rcos9, y = rsin9, then {10.3) b&comes
d9 G(9) r dr = H(9)'
where
H(9) =au cos3 9 + a12 cos2 9sin9
+ (b12 + a22) cos 9 sin2 9 + b22 sin3 9. Integrating (10.6), we get r = r1 exp
---1
(i: ~~:~
d9)
(10.6)
(10.7)
(10.8)
( )Recently, Nikola Samardzija (see [299]) pointed out that (10.1) also has characteristic values and characteristic vectors which are used to study the stability of the singular point
(0,0).
§10. QUADRATIC SYSTEMS WITHOUT LIMIT CYCLES
221
In order to study the infinite singular point, we change (10.3) into homogeneous coordinates, again letting x = 1 and dtjdr = z; we get
+ a12Y + a22y 2), -a22Y3 + (b22- a12)y 2 + (b12- au)y.
dz/dr = -z(au dyjdr =
(10.9)
The Yi of the infinite singular point (1, Yi, 0) is a root of the equation (10.10) whose discriminant is the same as (10.5). The product of the two roots A1 and A2 of the characteristic equation at the infinite singular point (1, Yi, 0) is A1A2
=-
(au+ a12Yi + a22Yl)
x [-3a22Yl
+ 2(b22 -
a12)Yi
+ b12 -au].
(10.11)
Now we begin our discussion of (10.3). 1. It has a pair of exceptional directions. Here again we can divide into two cases. (i) () = 0 and () = 1r are the unique pair of single exceptional directions; that is, suppose
a < 0,
and hence a22(bt2- au)
< 0.
(10.12)
Now y = 0 is a unique integral line. In order to study the state of the integral curves in the sector neighborhood of the exceptional direction () = 0, let the point (rt, Ot) in (10.8) lie in a sector neighborhood of()= 0, and expand G(O) and H(O) into power series of 0; then we get
H(O) G(O)
=
au [1 + .. ·]. (b12 - au)O
From (10.8) we know that if au(bt2 - au) > 0, then r -+ 0 as () -+ 0. In this case () = 0 is a so-called ray of nodal type, as in Figure 10.1. If a 11 (b 12 - au) < 0, then r-+ 0 as()-+ 0, and()= 0 is a ray of isolated type, as in Figure 10.2. On the other hand, from (10.10) we know that in this case there is a unique singular point (1, 0, 0) at infinity. Again from (10.11) we know that if au(bt2- au)> 0, then (1,0,0) is a saddle point; if au(bt2- au)< 0, then (1, 0, 0) is a nodal point. The conclusion for () = 1r is the same. Although the above discussion is local, we note that (10.3) is a homogeneous system, does not have a limit cycle, has only the above-mentioned singular
222 THEORY OF LIMIT CYCLES
~e-o 0 FIGURE 10.2
FIGURE 10.1
point, and y = 0 is the unique integral line. Hence we get THEoREM 10.1. For system (10.3), suppose the right sides of the two equations do not have a common factor, and suppose 6 < 0. Then when au(b 12 - au)> 0 its global phase-portrait is shown in Figure 10.5(a), below, and for a 11 (b12- au) < 0 its global portrait is shown in Figure 10.5(b).
(ii) (} == 0 and (} = 71" is a pair of triple exceptional directions; that is, suppose (10.13) Thus we must have a22 =f. 0, for otherwise the right sides of the two equations of (10.3) would have a common factor. Now y = 0 is still a unique integral line, and H(O) au · G(O) = - a2203 {1 + .. ·]. From (10.8) we know that if aua22 > 0, then r ~ oo as(}~ 0, and(}= 0 is a ray of isolated type as in Figure 10.2; if au a22 < 0, then r ~ 0 as (} ~ 0, and (} = 0 is a ray of nodal type as in Figure 10.1. Under the above assumptions, system (10.9) becomes dz/dr = -z(au
+ a12Y + a22Y 2),
dy/dr = -a22Y 3,
whose unique infinite singular point (1, 0, 0) is of higher order. From the theory of higher order singular points, we know that if aua22 > 0, then (1, 0, 0) is a nodal point, and if aua22 < 0, then (1, 0, 0) is a saddle point. The conclusion for (} = 71" is still the same. Thus we have THEOREM 10. 2. For the system ( 10.3) whose right sides do not have a common factor, suppose b22- a12 = b12- au = 0. Then when aua22 > 0 its global phase-portrait is as shown in Figure 10.5(b), and when aua22 < 0, its global phase-portrait is as shown in Figure 10.5(a). 2. If (10.3) has only two pairs of exceptional directions, then among them at least one pair is double. We may as well assume (} = 0 and (} = 71" are double exceptional directions, while (} = 71" /2 and (} = 371" /2 are single exceptional
§10.
QUADRATIC SYSTEMS WITHOUT LIMIT CYCLES
223
J--/0 0 FIGURE 10.3
FIGURE 10.4
directions; that is, suppose a22
= 0,
= 0, x = 0
b12- au
hence b22 ~ a12·
(10.14)
Thus there are two integral lines, and y = 0. In the following we shall investigate the state of integral lines in the sector neighborhood of the exceptional directions 0 = 0 (0 = 1r) and 0 = 71'/2 (0 = 371'/2). First we examine the exceptional direction 0 = 0, in which case we have
H(O) G(O)
au [ J a12)0 2 1 + ....
= (~2 -
From (10.8) we know that if au(b22- a12) > 0, then r--+ 0 as 0 --+ o+ and r --+ oo as 0 --+ o-; then 0 = 0 is the so-called altered ray of the second kind as in Figure 10.3. If au(~2- a12) < 0, then r--+ oo as 0--+ o+ and r--+ 0 as 0 --+ o-; then 0 = 0 is the so-called altered ray of the first kind as in Figure 10.4. Under the above assumptions, system (10.9) becomes
dzjdr
= -z(au + a12Y),
From the theory of higher order singular points we know that (1, 0, 0) is a semisaddle nodal point; when au(~2 - a12) > 0 the half of the positive yaxis of (1, 0, 0) is a hyperbolic region, and when au (~2- a12) < 0 the half of the positive y-axis of (1, 0, 0) is a parabolic region. For 0 = 1r and the singular point ( -1, 0, 0) we have a similar conclusion. For the exceptional direction 0 = 1rj2 (0 = 371'/2), our discussion is similar to 1(i) provided we interchange the coefficients of a and band interchange the subscripts 1 and 2, but let a12 = a21 and b12 = ~1· Hence we conclude that if ~2(a12- ~2) > 0, then 0 = 1rj2 (0 = 371'/2) is a ray of nodal type and the infinite singular point (0, 1, 0) is a saddle point; if ~2 (a12 - ~2) < 0, then 8 = 1rj2 (8 = 371'/2) is a ray of isolated type and the infinite singular point (0, 1, 0) is a nodal point. Combining the two cases au (~2 - a12) ~ 0 and the two cases ~2(a12 - b22) ~ 0, we can get four cases. Through analysis, we find if we disregard different directions of the x-axis, then the two global phase-portraits
THEORY OF LIMIT CYCLES
224
of b22 (a 12 - b22) > 0 are in fact the same, and the two global phase-portraits for b22(a12- b22) < 0 are also the same. Thus we have THEOREM 10.3. For system (10.3) whose right sides do not have a common factor, suppose we can make a22 = 0 and b12 = au. Then for b22 (a12 - b22) > 0 the global phase-portrait is as shown in Figure 10.5(c), below, and for b22(a12 - b22) < 0 the global phase-portrait is as shown in Figure 10.5(d). 3. Now suppose (10.3) has three pairs of single exceptional directions. We may as well assume these three pairs of exceptional directions are 0 = 0 (0 = 1r), 0 = 1rj2 (0 = 311"/2), and 0 = Oo (0 = Oo + 1r), 0 < Oo < 1rj2. That is to say, we may assume tanOo =
b12- au b a12- 22
> 0.
(10.15)
Now we have three integral lines x = 0, y = 0, and y - x tan Oo = 0. From 1(i) we know that when au(b12- au) > 0, 0 = 0 (0 = 1r) is a ray of nodal type and the infinite singular point (1, 0, 0) is a saddle point; when au(b12- au)< 0, 0 = 0 (0 = 1r) is a ray of isolated type, and (1,0,0) is a nodal point. Also from 2 we know that when ~2(a12 - b22) > 0, 0 = 11"/2 (0 = 311" /2) is a ray of nodal type and the infinite singular point (0, 1, 0) is a saddle point; when b22(au- b22) < 0, 0 = 7r/2 (0 = 37r/2) is a ray of isolated type and (0, 1, 0) is a nodal point. For 0 = Oo, we expand G(O) and H(O) into power series of 0- Oo, and get
H(O) H(Oo) G(O) = 01(0- 00 ) [ 1 + .. ·], where 2 2 2 01 = G'(Oo) = - sin0ocos Oo[(a12- b22) + (b12 - a 11 ) ] a12- b22 3 H(Oo) = cos Oo[au + a12 tan Oo + b12 tan 2 Oo + b22 tan3 Oo] =
3 Oo ) (aub22- a12b12) [(a12- b22 )2 ( cos b 3
a12-
22
# O,
+ (b12- au )2] ..J.r 0.
Hence OtH(Oo) and aub22 -a 12b12 have the same sign. From (10.8) we know that if aub22 - b12a12 > 0, then 0 = 00 is a ray of nodal type; then we compute the product of two characteristic roots at the infinite singular point (1,tan0o,O) and get
§10. QUADRATIC SYSTEMS WITHOUT LIMIT CYCLES
(a)
(b)
225
tc)
(d)
(e)
(f)
(g) FIGURE 10.5
Hence (1, tan Oo, 0) is a saddle point. If aub22 - b12a12 < 0, then(}= Oo is a ray of isolated type, and AlA2 > 0; then (1, tan Oo, 0) is a nodal point. Thus we have THEOREM 10.4. For system (10.3) whose right sides do not have a com-
mon factor, suppose a22 = 0 and (b12 - au)(a12 - b22) > 0. Then when au(b12- au), B b22(a12- b22), and C aub22- a12b12 are all negative, its global phase-portrait is as shown in Figure 10.5(e); when two of .4, B and C are negative and the other is positive, its global phase-portrait is as shown in Figure 10.5(f); and when two of them are positive and the other is negative, its global phase-portrait is as shown in Figure 10.5(g).
.4
=
=
=
226
THEORY OF LIMIT CYCLES
REMARK. A, B, and C cannot be all positive, because, when A, B > 0, since tan Oo > 0 we know that au and b22 have the same sign. At the same time, if au > 0, then b 12 > au > 0 and a12 > b22 > 01 and so C < 0; if au < 0, then we also have C < 0. Summarizing the above four theorems, we get THEOREM 10.5. The global phase-portraits for homogeneous quadratic systems whose right sides do not have a common factor have altogether seven different topological structures, given in Figure 10.5. The arrows in each figure show only one of the two possible cases; for the other case, the arrows should be reversed. II. Global structure of quadratic systems possessing a star nodal point. A. N. Berlinskil [167] studied a quadratic system having a star nodal point, and through his analysis proved that this system does not have a limit cycle, and there are altogether seventeen global phase-portraits of its topological structures. He constructed the phase-portraits, and gave some methods to distinguish them. We now present his work, but simplify his proof. We first point out (from the theory of Jordan systems) that the following two lemmas clearly hold. LEMMA 10. 1. . The form of the system
+ box 2 + b1xy + b2y 2 , dyfdt = Y + aox 2 + a1XY + a2y 2
dxfdt = x
(10.16)
remains unchanged after any nonsingular real linear transformation.
LEMMA 1 0. 2. For a quadratic system, a necessary and sufficient condition for the origin to be a star nodal point is that the system possess the form (10.16). In the sequel we carry out our discussion separately according to the number of finite singular points. 1. The case of four finite singular points. Suppose a general quadratic system has four finite singular points. The following lemma shows that the question of whether the quadrilateral with the four singular points as vertices is convex or concave has a close relationship with the properties of the singular points [169]. The following simple proof is taken from [168]. LEMMA 10.3. Suppose a quadratic system has four singular points. If the quadrilateral with these points as vertices is convex, then two opposite singular points are saddle points, and the other two opposite singular points are nonsaddle points (that is, nodal points, foci or centers); if the quadrilateral is
§10. QUADRATIC SYSTEMS WITHOUT LIMIT CYCLES
227
concave, then the three outside singular points are saddle points (non-saddle points), and the inside singular point is a non-saddle point (saddle point). PROOF. After a linear transformation, we can let these singular points lie at the origin 0(0,0) and the points A1(1,0), A2(a:,,B), and A3(0,1), where a > 0, ,8 > 0, and a + ,8 =1 1. Thus the coefficients of the quadratic system dxjdt = a1x + a2y + aux 2 + a12XY + a22y 2, (10.17) dyjdt = b1x + b2y + bux 2 + b12XY + b22Y 2
should satisfy the following relations:
a1 + au
= 0,
= 0,
a2 + a22 = 0, ~ + ~2 2 a1a: + a2.B + aua: + a12a:.B + a22.8 2 = 0, b1a + ~,8 + bua:2 + b12a:.B + ~2.82 = 0.
b1 + bu
= 0,
Thus (10.17) can be written as
= PtX(x- 1) + P2Y(Y- 1) + P3XY, dyjdt = QtX(x- 1) + Q2Y(Y- 1) + Q3Xy, dxjdt
(10.18)
where
a-1
= - - -,8p 1 -
,8-1
(10.19) a ' Since this system has exactly four singular points, each one is elementary. Computing the constant term of the characteristic equation of the linear part of the point 0, we iihould get Do= PtQ2- P2Ql =I 0. It is easy to prove that for the other three singular points At. A2, and A3, similarly,
P3
Dt
=
--P2
a+!- 1 Do,
D2
=(a+ ,8 -1)Do, D3 = a+;- 1 Do.
Whether the singular point Ai is a saddle point or not can be decided by whether Di < 0 or > 0. If a + ,8 > 1, then the quadrilateral is convex, D 0 and D 2 have the same sign, and D1 and D3 have the same sign, but have sign opposite to Do. If a:+,B < 1, then the quadrilateral is concave, O,At, and A3 are three outside singular points, their Do, Dt. and D3 have the same sign, but have sign opposite to D2. The proof is complete. REMARK. Although this proof is succinct, it does not show why the property of the dynamical system of these four singular points has such a delicate relationship with the geometrical property of the quadrilateral formed by them. To understand this point, please refer to Lemma 11.3 and the remark after it in §11. In the following we return to the quadratic system possessing a star nodal point. As before, we may assume its four singular points are 0(0, 0), At(1, 0),
THEORY OF LIMIT CYCLES
228
A2(o:, /3), and Aa(O, 1), where 0 is a star nodal point, o: '# 0, {3 '# 0, and o: + {3 '# 1. From Lemma 10.2, we know that now P2 = Qt = 0 and Pt = Q2 = -1 in system (10.18), thus the system can be written as dx dt
= X (1 -
Again let (o: - 1) I {3 written as
dxldt
X
+
0: -
{3
1 y) I
dy dt
= y (1 +
Q:
= b and (/3 - 1)I o: = a;
= x(l- x +by),
dyldt
{3 - 1 X
-
y) '
thus the above system can be
= y(l +ax- y),
(10.20)
where a'# -1, b '# -1, and ab '# 1. The four singular points of (10.20) are 0 (star nodal point), At(l, 0), A2((1+b)l(l-ab), (l+a)l(l-ab)), and Aa(O, 1); and the three integral lines are
OAt: y = 0;
OA2: (1 + b)y- (1 + a)x = 0;
OAa: x = 0.
From this we can see that for every singular point there is an integral line passing through it, and so (10.20) does not have a limit cycle;(2) thus whether At, A2, and A3 are saddle points or nodal points depends on whether the quantities
Dt =-(a+ 1),
D _ (1 + a)(1 +b) 2 l-ab'
Da
= -(b+ 1)
are less than zero or greater than zero. Next, if we change (10.20) into homogeneous coordinates, we can easily prove that it has three infinite singular points:
Bt(l,O,O),
B2 (1,
~::.o),
Ba(O,l,O),
and the quantities corresponding to the above-mentioned D, are now
(1 + a)(l- ab) Ja = 1 +b. l+b Since D, = -h we know that if A, is a nodal point (saddle point), then B, is a saddle point (nodal point). The three fixed lines x = 0, y = 0, and x + y = 1 passing through the three singular points O,At, and Aa respectively divide the (x,y)-plane into seven regions. From Lemma 10.3 we know that, given which region A2 is in, we can tell whether A, (hence Bi), i = 1, 2, 3, is a nodal point or a saddle point. Moreover, since other integral curves cannot cross the three lines OA 11 OA2, and OAa outside 0, A1. A2, and Aa, the global phase-portraits can be ( 2 )This
point can also be known from Exercise 9 in §1.
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QUADRATIC SYSTEMS WITHOUT LIMIT CYCLES
229
b
Cl
b--1
FIGURE 10.6 completely determined, and easily drawn. The position of A2 in the (x, y)plane can then be determined from the values of the parameters a and b. On the (a,b)-parametric plane, three bifurcation curves ab = 1, a= -1, and b = -1 divide the parametric plane into seven regions (Figure 10.6). The global topological structure of system (10.20) corresponding to the point in every region is the same. But it is easy to see that when the point (a, b) passes through the line a= -1 orb= -1, the figure of the global topological stucture of (10.20) does not change, only a certain pair of finite singular points and the corresponding pair of infinite singular points interchange their relative positions, and at the same time the labels of the saddle points and nodal points are also interchanged; hence the global topological structure is not affected. Thus there are in fact only three different global topological phas&portraits as shown in Figure 10.7(a), (b), and (c) respectively. They correspond respectively to: 1) ab > 1, a< 0, 2) ab < 1, and 3) ab > 1, a> 0, respectively. Thus we get THEOREM 10.6. If a quadratic system possessing a star nodal point has four singular points, then: (i) When these four singular points form the vertices of a concave quadrilateral, and the star nodal point is an outside vertex, the other three singular points are nodal point, nodal point, and saddle point (Figure 10.7(a)). (ii) When these four singular points form the vertices of a concave quadrilateral, and the star nodal point is an inside vertex, the other three singular points are saddle points (Figure 10.7(c)).
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THEORY OF LIMIT CYCLES
(iii) When the four singular points form the vertices of a convex quadrilateral, the other singular points are nodal point, saddle point, and saddle point (Figure 10.7(b)).
If we instead use the coefficients of the system to indicate our results, then this theorem can be rewritten as THEOREM 10. 6'. For a quadratic system ( 10.20) possessing a star nodal point where (a+ 1)(b + 1)(ab -1) -:f. 0, then corresponding to Theorem 10.6(i), (ii), and (iii), we have (i)' ab > 1, a< 0, (ii)' ab < 1, and (iii)' ab > 1, a> 0, respectively. 2. The case of only three finite singular points. We may as well assume that except for its star nodal point (the origin), there are two finite singular points A1(1,0) and A3(0, 1). Following the method of deriving {10.20), we know that the system to be discussed still possesses the form {10.20), but its parameters a and b correspond to the bifurcation values mentioned in 1; that is, a = -1, or b = -1, or ab = 1 but not a = b = -1; or else they correspond to A2 = A1. or A2 = A3, or A2 = B2 in 1, respectively. It is easy to see that in our present case this system still has no limit cycles. In the following we study these cases separately. (i) Suppose a= -1 (b -:f. -1); that is, A2 = A1. and hence B2 = B1. The two higher order singular points obtained from this are semi-saddle nodal points. It is easy to see that the equator is a middle separatrix of two hyperbolic regions of the singular point B1 (1, 0, 0) and the type of the other singular point remains unchanged. When a = -1 and b < -1, the corresponding global phase-portrait is as shown in Figure 10.7(d); when a = -1 and b > -1, the corresponding global phase-portrait is as shown in Figure 10.7(e).
(ii) Suppose b = -1 (a -:f. -1). Similarly to (i), when b = -1 and a< -1, the corresponding global phase-portrait is as shown in Figure 10.7(d); when b = -1 and a > -1, the corresponding global phase-portrait is as shown in Figure 10.7(e). (iii) Suppose ab = 1, but a = b = -1 does not hold. This corresponds to A2 = B2; B2 becomes a semisaddle nodal point, but the equator separates the hyperbolic region of B2 from its parabolic region. The other singular point keeps the original type. When ab = 1 and a < 0, a :f. -1, the corresponding global phase-portrait is as shown in Figure 10.7(f); when ab = 1 and a> 0, the corresponding global phase-portrait is as shown in Figure 10.7(g).
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QUADRATIC SYSTEMS WITHOUT LIMIT CYCLES
231
The above results are written as follows: THEOREM 10.7. For a quadratic system (10.20) possessing a star nodal point, if there are only three finite singular points, then t.he other two finite singular points can only be: (i) a nodal point and a semisaddl~ nodal point; here a= -1 and b < -1, orb= -1 and a< -1 (Figure 10.7(d)); or (ii) a saddle point ant/. a semisaddle nodal point; here a= ~1 and b > -1, or a> -1 and b = -1 (Figure 10.7(e)); or (iii) a saddle point and a nodal point; here ab = 1 and a< 0, a-=/= -1 (Figure 10.7(f)); or (iv) a saddle point and a saddle point; here ab = 1 and a > 0 (Figure 10.7(g)). 3. The case of having only two finite singular points. We may as well assume the origin is a star nodal point and the other singular point is At ( 1, 0). Then the system (10.16) can be written as dxfdt = x- x 2 + btXY + b2y 2,
dyjdt = y(1 + atx + a2y).
(10.21)
In the following we divide the discussion into three cases. (i) Suppose At is a triple singular point. Then the following two subcases can arise: (a) The line Lt : y = 0 and the curve £2 : x - x 2 + bt xy + ~y 2 = 0 intersect at At. and the line La: 1 + atX + a2y = 0 and L2 touch tangentially at At. Then we can deduce that at = -1 and a2 = bt; thus (10.21) can be written as
.
dxfdt
= x(1- x + bty) + ~y 2 ,
dyjdt = y(1- X+ bty).
(10.22)
It is clear that b2 -=/= 0 in (10.22). In order to investigate the characteristics of At, we apply the transformation X= 1 - X+ bty,
y = y,
and still denote X andY as x andy; thus (10.22) is changed to dxfdt
=
-x + x 2 - b2y 2,
dyjdt
= xy.
(10.23)
From the theory of higher order singular points(3) we know that At is a nodal point for b2 > 0, At is a saddle point for ~ < 0, and the system has no limit cycles. Transform the system again in homogeneous coordinates. It is easy to know there is a unique singular point Bt (1, 0, 0) at infinity, and Bt is a saddle point when ~ > 0; Bt is a nodal point when ~ < 0. Moreover, since y = 0 is an integral line passing through 0 and At, it is easy to construct the global (3)For example, see Theorem 65 in [170], §21.
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THEORY OF LIMIT CYCLES
phase-portrait of system {10.23). When~> 0 it is as shown in Figure 10.7(h); < 0, it is as shown in Figure 10.7(i). (b) L 3 and L2 intersect at At, and Lt and L2 touch tangentially at At. This is impossible. (ii) Suppose At is a double singular point. Following the same discussion as in (i), we know that {10.21) should have at = -1, b2 = a2(a2- bt), and a2- bt 'I 0. Thus {10.21) can be changed to when~
dxjdt = x(1- x + ay),
dyjdt = y(1- x),
(10.24)
where a = bt - a2 'I 0. It is easy to see that At (1, 0) and the infinite singular points (1, 0, 0) and (0, 1, 0) are all semisaddle nodal points, and do not have limit cycles. Its global phase-portrait is as shown in Figure 10.7U). (iii) Suppose At is a simple singular point. Then (10.21) when a2 'I 0 can be changed to
dyfdt
= y(1 +ax- y),
(10.25)
where at= a, b = -bt/a2, and c = b2/a~. Since At is a simple singular point, we can deduce that these coefficients satisfy either o) (b+ 1) 2 + 4(a + 1) < 0, or {3) a2c + ab - 1 = 2ac + b + 1 = 0, c 'I 0. If a2 = 0, then (10.21) becomes
dxjdt = x- x2 + bxy + cy2 ,
dyjdt
= y(1 +ax),
(10.26)
where a = at, b = bt. and c = b2. Since At is a simple singular point, we can deduce that these coefficients either satisfy o) b2 + 4(a + 1)c < 0, or {J) b = c = 0, a -:J -1, or 7) a= 0, b2 + 4c ~ 0. Study case o) of (10.25). It is easy to see at this time that we must have a< -1, the point At(1, 0) is a nodal point, the unique infinite singular point (1, 0, 0) is a saddle point, and the system does not have a limit cycle. Its global phase-portrait is shown in Figure 10.7(h). For the case {3) of (10.25), we must have a 'I 0; the system can be written as dx a+2 a+1 2 dy 2 - = x- x + - - x y - - - y dt = y(1 +ax- y). (10.27) dt a a2 ' It is easy to see that this system has three integral lines, y = 0, y = ax, and y = a(x- 1). Since for each singular point there is an integral line passing through it, the system does not have a limit cycle. When a < -1 (> -1), the singular point At is a nodal point (saddle point), the infinite singular point (1, 0, 0) is a saddle point, and the other infinite singular point (1, a, 0) is always a semisaddle nodal point; its global phase-portrait is as shown in Figure 10.7(k) (Figure 10.7(1)).
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233
For system (10.26) we can carry out a similar study, and the results are as follows: In case a), when a< -1 (> -1), its global phase-portrait is as shown in Figure 10.7(h) (10.7(i)). In case {3), there are three integral lines; two of them pass through the origin, and the third does not. When a< -1 (> -1), we get Figure 10.7(k) (10.7(1)). In case-y), there are also three integral lines, the singular point At(1,0) is a saddle point and the system does not have a limit cycle. When b2 + 4c = 0, two of the three integral lines pass through the origin. Two infinite singular points are (star) nodal point and semisaddle point respectively. Its global phase-portrait is shown in Figure 10.7(1). When b2 + 4c > 0, three integral lines always pass through the origin; among the three infinite singular points, one is a nodal point and the other two are semisaddle nodal points. Its global phase-portrait is shown in Figure 10.7(m). Summarizing the above discussion, we get THEOREM 10. 8. If a quadratic system possessing a star nodal point has only two finite singular points and if there is only one integral line passing through the origin, then another singular point is either a nodal point or a saddle point; the global structures of this system are as shown in Figures 10. 7(h), (i). If this system has only two integral lines passing through the origin, then the singular point is a semisaddle nodal point, nodal point, or saddle point; the global structures of the system are as shown in Figures10.7(j), (k), (1). If the system has three integral lines passing through the origin, then the other singular point must be a saddle point, and its global structure is as shown in Figure 10.7(m). In all these cases, there is no limit cycle.
4. The case of having only one finite singular point. In this case the system clearly does not have a limit cycle. Through a suitable rotation of axes, we can make ~ = 0 in (10.16). From the uniqueness of the singular point, we know that a2 = 0. Thus we have
dxfdt = x(1 +box+ bty),
dyfdt = y + aox 2
+ a1 xy.
(10.28)
For the same reason, the coefficients of (10.28) should satisfy one of the following conditions: a) (at - bo) 2 + 4aobt < 0; {3) bt = 0, at = bo =I 0, ao =f:. 0; 1) bt = bo = o, ao =I O; 6) bt = bo = ao = 0.
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234
(a)
,
(b)
(c)
"
, (d)
(e)
(f)
(g)
(h)
(i)
FIGURE 10.7 When a) holds, there is a unique singular point (0, 1, 0) at infinity, which is a semisaddle point. The equator separates the parabolic region from the hyperbolic region. There is only one integral line x = 0, and its global phaseportrait is as shown in Figure 10.7(n). When (3) holds, (10.28) has an integral line x = -1/b besides they-axis. It is easy to see that the infinite singular point (0, 1, 0) is a semisaddle nodal point. Two hyperbolic regions are all on the same side of the equator, but on this side there is a part belonging to the parabolic region; its global phaseportrait is as shown in Figure 10.7(o).
§10. QUADRATIC SYSTEMS WITHOUT LIMIT CYCLES
235
Bli
(k)
(I)
(o)
(n)
(m)
(q)
(p) FIGURE 10.7
When 1) holds, if a 1 =/; 0, then there are two infinite singular points, one of which is (1, -a0 fat.O), a semisaddle nodal point; the equator separates the parabolic region from the hyperbolic region. The other is (0, 1, 0), which is also a semisaddle nodal point. There are two integral lines which pass through the origin. The global phase-portrait is in Figure 10.7(p). If a1 = 0, then there is a unique infinite singular point (0, 1, 0) which is a semisaddle nodal point; the equator separates the parabolic region from the hyperbolic
236
THEORY OF LIMIT CYCLES
region. There is only one integral line passing through the origin, and its global ph~portrait is as shown in Figure 10.7{n). When c5) holds, {10.28) becomes dxjdt
= x,
dyjdt
= y(1 + a1x).
When at i= 0, there are two infinite singular points {1, 0, 0) and {0, 1, 0) which are semisaddle nodal points. Its global phase-portrait is shown in Figure 10.7{p). When at = 0, we have the phase-portrait of Figure 10.7{q), and {10.28) degenerates to a linear system. Summarizing the above disCussion, we get THEOREM 10. 9. If a quadratic system possessing a star nodal point has a unique finite singular point, then there are four and only four cases: {i) it has only one integral line, which passes through the origin; (ii) it has only two integral lines, one of which passes through the origin; {iii) it has only two integral lines, both of which pass through the origin; or (iv) all the integral lines are lines passing through the origin. This system does not have a limit cycle. Phase-portraits are shown in Figures 10.7(n)-(q). Combining the above Theorems 10.6--10.9, we have THEOREM 10. 10. The finite singular points of a quadratic system possessing a star nodal point whose right sides do not have a common factor are only saddle points, nodal points, and semisaddle nodal points. The system does not have a limit cycle. The global phase-portraits have seventeen and only seventeen different topological structures, as shown in Figures 10.7{a)-(q). REMARK. In all the parts of Figure 10.7, s =saddle point, n =nodal point, en =star nodal point, sn =semisaddle nodal point, and db =double singular point. III. Topological classification of structurally stable quadratic systems without limit cycles. Let X denote the Banach space of polynomial differentiable systems of degrees not greater than n, let En be the set of structurally stable systems in X, and let Xs be the set of all systems satisfying the conditions {VI) of §8. We have already seen that {1) Xs is open and dense in X; and (2) a necessary and sufficient condition for a system a to be in Xs is that a is in En.
§10.
QUADRATIC SYSTEMS WITHOUT LIMIT CYCLES
237
In this subsection we shall study the problem of topological classification of E2 without limit cycles. The problem was first mentioned and discussed by G. Tavares dos Santos in [171]; and, in the end, he obtained 25 different topological structures. More recently, Cai Sui-lin [172] pointed out that the classification of [171] was incomplete, and added eight more examples; hence, altogether, there should be at least 33 different topological structures. He also studied the direction of a separatrix passing a saddle point and from this determined that under certain conditions there could not be other different topological structures. Independently of [172], Shi Song-ling [173] concluded by studying the logical possibilities that the structurally stable systems E2 can only have at most 65 different topological structures, but he did not prove whether these 65 kinds could be realized in quadratic systems. In fact, we shall see later that among the 65 kinds in [173], there are several which definitely cannot be realized in quadratic systems. Hence the problem of determining how many topological structures of structurally stable quadratic systems without limit cycles there are is up to now unsolved. Now we shall briefly present the above work. In order to-study the topological classification of this system, we first have to study all the possible combinations of its singular points in the interior of the Poincare closed hemispherical surface 0 (equivalent to the Euclidean plane) and on the equator E. In the following we shall adopt the following notation: s: saddle point in the interior of 0; p: c non-saddle point in the interior of (elementary singular point of index +1); 8: saddle point onE; F: source onE; P: (deep) sink on E; o:,/3,"'(,0, and c: elements of E2.
n
THEOREM 10 .11. If o: is in E2, then the combinations of singular points of o: on 0 must be one of the following twelve cases: type a: (1) F; (2) psF; (3) P1P2s1s2F; type b: (1) P1P28; (2) P1P2P3s8; type c: (1) Sls2FlPF2; (2) PSlS2S3FlPF2; typed: (1) SPF; (2) psSPF; (3) P1P2s1s2SPF; type e: (1) P1P28182F; (2) P1P2P3S8182F. PROOF. The proof is divided into four steps: (i) Since Xs = E 2 when we know 0: only has elementary singular points in the interior of n and on E. (ii) The coordinates of a singular point onE are roots of a cubic
n.
= 2,
238
THEORY OF LIMIT CYCLES
p
p
FIGURE 10.8
FIGURE 10.9
equation, and so there exist only one or three singular points on E. (iii) Since Xs is open and dense in X, the zero isocline and infinite isocline can be considered as nondegenerate elliptic or hyperbolic curves; thus it is not possible between them to have one or three points of intersection, i.e. the number of finite singular points is 0, 2, or 4. (iv) Since the sum of indices of the singular points on 0 is equal to 1, we can again use Lemma 10.3 to prove the conclusion of this theorem. The proof is omitted and left to the reader as an exercise. In the following, for convenience of presentation, we adopt the notation 8(Pt.P2,P3,P) to represent four separatrices (Lt,Lt,L1,L2) of a saddle point 8 connecting the singular points Pt. P2, p3, and P respectively, in which the two separatrices starting from the first two points Pt and P2 enters s, and the other two separatrices starting from 8 enter P3 and P respectively (Figure 10.8); and we use the notation S(F,p) for a saddle point onE using separatrices [+ and [- not along the equator to connect the singular points F and p respectively (Figure 10.9). The other notation has similar meaning. Since the system under discussion does not have a limit cycle, and it is not possible to use a separatrix to connect two saddle points, hence, from connecting the separatrices of saddle points and the singular points whose indices are +1, we can determine the topological structure of the corresponding phase-portrait. We now carry out our discussion according to the five types and twelve forms in Theorem 10.11. 1. Type a(1). Form F. It is easy to see that there is only one topological structure as shown in Figure 10.10. Its realizable example is 01:
dx/dt = -2xy, dyjdt = x 2
+ y2 + c
(c > 0).
2. Type a(2). Form p8F. It is easy to see that this case also has only one topological structure, whose characteristic is 8(F, F, p, F') or 8(F', F', p, F);
§10. QUADRATIC SYSTEMS WITHOUT LIMIT CYCLES
239
F'
F'
F
F
FIGURE 10.10
FIGURE 10.11
the former is shown in Figure 10.11. Its realizable example is o:2: dx/dt = P- OQ, dyfdt = Q
which is obtained by rotating an angle(} (0
< (} « 1)
=
dxfdt = P -y(v + 2x), 0:' { 3 . dy I dt = Q = (X + ~) + y 2
+ fJP,
2
-
T
from (~-t
< 0 < v).
Now we set o: 2 = o: 9. In fact, the singular points of the system o: belong to form psF, but the characteristic root of p has zero real part. Since there exists a first integral 'P(x, y) = x(x 3 /3 + y 2) + ~-tx 2 /2 + vy 2 /2, pis a center. After rotating through an angle (}, the center becomes a focus, and a family of closed trajectories of a becomes an arc without contact of o: 2; hence 0:2 does not have a limit cycle. 02 still belongs to the form psF, and its phase-portrait is shown in Figure 10.11. 3. Type a(3). Form PtP2S1s2F. For this kind we have the following theorem: THEOREM 10. 12. Structurally stable quadratic systems without limit cycles of form P1P2s1s2F have five and only five different kinds of topological structures, whose characteristics are
s1(F,F,Pt.P2),s2(F,F,p2,F'); s1(F,F,p1,F'),s2(F,P2,F',F'); o:s: s1 (F, F, Pt. F'), s2(F, F, p2, F'); o:6: s1 (F,p2, Pt. F'), s2(F, P2• Pt. F'); 0:7: sl(F,p2,Pt.Pi), s2(F, F,p2, F'). o:3: o:4:
PROOF. We first prove there are at most five different kinds of topological structures as mentioned above.
240
THEORY OF LIMIT CYCLES 8'
8'
8
FIGURE 10.12
FIGURE 10.13
FIGURE 10.14
(1) Suppose Pt and P2 are sinks (sources). Then the directions of separatrices of a saddle point 8 1 can only have the following three possibilities (if it is a source, then we just interchange the first two elements and the last two elements in the following):
8t(F, F,Pt.P2),
8t(F, F,pt, F'),
8t(F, F,p2, F').
It is the same for 82. After combinations of 8t and 82, it is clear that we can only have two different kinds of topological structures, a 3 and a 5 • (2) Suppose one of Pt and P2 is a sink and the other is a source. Then the directions of separatrices of 8t can only be of the following five kinds:
8t(F,F,pt,F'),
8t(F,p2,Pt.Pt),
St (F, P2. F'' F'),
8t(F,P2,Pt.F'),
St (P2. P2. Pt. F').
There are the same five kinds for 82. After the combinations of 8t and 82, it is clear that only three different kinds of topological structures a4, a5, and a7 are possible. For details see [172]. On the other hand, the above five different topological structures can all be realized. Examples can be found in [171] and [172]. 4. Type b(1). Form P1P2S. It is easy to see that this case has only one topological structure, whose characteristic is S(Pt. P2), as shown in Figure 10.12. A realizable example is
f3t
= {3:
(0
dxjdt = -2xy, < (} « 1), where {3E:: { d y jdt =y 2 - X 2 + c
(c > 0).
In fact, the singular points of {3E: belong to the form P1P2S, where Pt and P2 are centers and x = 0 is an integral line. After a rotation through an angle 0, its center becomes a focus, and it does not have a limit cycle nor any curve connecting two saddle points. Its phase-portrait is shown in Figure 10.12.
§10.
QUADRATIC SYSTEMS WITHOUT LIMIT CYCLES
241
p
p
P'
P'
FIGURE 10.15
FIGURE 10.16
5. Type b(2). Form PlP2P3SS. It is easy to see that this case has only one topological structure, whose characteristics are s(PI.P2.P3,Pa) and S(pt,Pa), as shown in Figure 10.13. A realizable example is /h = tYJ.. (0 < () ~ 1), where
/3>.. { dxfdt = -.Xx + 2xy, ·
dy / dt
= .Xy + y 2 -
x2
(.X> 0).
In fact, the singular points of the system /3>.' belong to the form P1P2P3sS, the value of whose divergence is 4y; but x = 0 is a line without contact (except the origin), and so /3>. does not have a limit cycle. After a rotation through an angle() (0 < (} ~ 1), the separatrix connecting Sand S' (which originally had two singular points s and pa) splits. It still has no limit cycles, as shown in Figure 10.13. 6. Type c(1). Form s 1s2F1PF2. It is easy to see that this case has only one topological structure, whose characteristics are 81 (F1, F2, P, F{) and s2(F1. P', F~, Fi), and its phase-portrait is as shown in Figure 10.14. A realizable example is 11: dxfdt
= x 2 + 2xy,
dyfdt
= -2xy- y2
-
c
(c > 0).
The concrete discussion is omitted and left as an exercise. 7. Type c(2). Form ps1s2saFtPF2. For this kind, we have the following: THEOREM 10.13. Structurally stable quadratic systems without limit cycles of the form ps1s2saF1PF2 have four and only four different topological structures, whose characteristics are 12: Bt(Ft,p,P,F~), s2(P',p,Ff,F~), sa(F2,p,P,Fi); "/3: St(Ft,p, P, P), s2(Ft. P', P, F~), s 3 (F2 , P', P, Fi); 14: s1(F1, F2, P, F~), s2(P',p, Ff, F~), s 3 (F2, p, F{, F~); 1s: s1(Ft.F2,P,F~), s2(F2,P',F{,F~), s 3 (F2 ,p,P,P).
THEORY OF LIMIT CYCLES
242
We briefly indicate the method of proof. First we note that since the system under discussion is quadratic, any line cannot have more than two points of contact with the trajectory of this system (including the singular pointW) unless this line is also a trajectory. Hence the following two cases cannot be realized in the quadratic system: (1) Bt(Ft, p, F~', F~), s2(Ft, P', P, F~), s3(F2, P', P, F{), as shown in Figure 10.15, and (2) St(P',p,P,P), s2(Ft,P',P,F~), s3(F2,P',P,Ff), as shown in Figure 10.16. Next, from a discussion similar to Theorem 10.12, we know that in the present case only four different topological structures 12, "Y3, "Y4, and "'(5 are possible. We can prove that they can be realized by quadratic systems; the examples can be found in [171], and so are omitted here. 8. Type d(1). Form SPF. It is easy to see that in this case there is only one topological structure S(F, F'), and it can be realized. The example is in [171]. 9. Type d(2). Form psSPF. This case can only have three different topological structures, whose characteristics are 62: s(F,P',p,P), S(P',p); 63: s(F,F,p,P), S(P',P); 64: s(F, F,p, P), S(F, F').
They can all be realized; examples are in [171]. 10. Type d(3). Form PtP2B1s2SPF. For this there can only be at most 30 different topological structures,( 5 ) in which there are 8 kinds which have been realized (see [171] and [172]). The remaining 22 kinds are still to be realized or eliminated. Since there are too many kinds, we do not list them here. 11. Type e(1). Form PtP2S1S2F. It is easy to see that this case has only three different topological structures, whose characteristics are ct: St(Pt,P2), S2(F,F'); C2: St(Pt,F'), S2(P2,F'); C3: St (Pt. P2), S2(Pb P2)· They all have realizable examples. See [171]. 12. Type e(2). Form PtP2P3BStS2F. For this there are at most 9 different kinds of topological structures.(6 ) There are four kinds which have been realized, and the remaining five have still to be realized or eliminated. ( 4 )The proof is given ( 5 )See (173], but No.
in §11, Lemma 11.1. 7 and No.8 of that paper and C10 and C13 have the same topt; logical structures. Hence there are at most 30 different kinds of topological structures. ( 6 )See (173]; but we can prove that Nos. 8, 9, and 10 in Table 2 of that paper cannot be realized (see Exetcise 9); hence this form can have at most 9 kinds.
§10. QUADRATIC SYSTEMS WITHOUT LIMIT CYCLES
243
Summarizing the above analysis, we get THEOREM 10.14. Structurally stable quadratic systems without limit cy-
cles can have at most 60 different kinds of topological structures, of which at least 33 can be realized. (7 )
Exercises 1. We have seen that the system dxfdt
= x(A1x + B1y + Cl),
dyjdt = y(Ax +By+ C)
does not have a limit cycle. For the case when the system possesses four finite singular points, use the inequalities among the coefficients to discuss its global topological classification [175]. 2. Suppose a cubic system has integral curves xy = 0 and x 2 + y 2 = 1. Discuss its global topological phase-portrait. 3. We have seen that the quadratic system possessing a fine focus of third order tlzfdt = -y + lx 2 + xy, dyjdt = x + ~x 2 + 3lxy does not have a limit cycle. Discuss its global topological phase-portrait [174]. 4. Use the inequalities among the coefficients of the system (10.28) to present different topological structures of this system. 5. Give a detailed proof of Theorem 10.11. 6. Give a detailed proof of Theorem 10.13. 8'
FIGURE 10.17
7. Suppose the singular points on the equator E of a quadratic system belong to form S P F and S (F, F'). If the system in the region SF' P' S (the region on the side of the figure) and S' F P S' do not have any singular points, then this system does not have a singular point in the interior of 11. Prove it. ( 7 )Recently, D. M. Zhu has shown that the number of possible different topological structures is less than 60, but at least 36 could be realized.
244
THEORY OF LIMIT CYCLES
8. For singular points belonging to type e(2), form PtP2PasS1S2F, prove that the following three topological structures cannot be realized: (1) s(p~, F, F', F'), St (pa, P2), S2(F, F'); (2) s(p~,F,F',F'), St(Pa,F'), S2(p2,F'); (3) s(F,F,pbF'), St(P3,F'), S2(P2,F'). 9. Study the global structure of
dxjdt
= y -lx2 + y 2 ,
=
dyjdt
= -x(1 +ax+ by),
where a < 0, l > 0, D -4(b- l) 3 - 27a3 > 0, -a < 2v'l, a+ bv'l 2 1 < b < 1 + a /4l. For example, take a= -2, b = 1.1, and l = 5. 10. Let
~~ = x( -6x + 4y + 2), ~~ = y(7x + y- 2) + sx, Study the global structures of o8 and o; 6 (0 < (} « 1). Oe:
> 0,
O<s«l.
11. Study all possible cases of global phase-portraits of system (10.1) whose right sides have common linear factors, and draw the figures.
§11. General Properties and Relative Positions of Limit Cycles in Quadratic Differential Systems
For a quadratic differential system
dyjdt = Q2(x, y),
dxjdt = P2(x, y),
(11.1)
except for the questions of the existence of a center and the nonexistence of closed trajectories discussed in the previous two sections, the most important question is the existence of a limit cycle. Its importance is beyond doubt no matter whetlier it is considered from a theoretical or a practical viewpoint. Hence, starting from this section, we formally turn to the study of limit cycles. First we study the general properties and relative positions of limit cycles of (11.1). Some of the so-called general properties discussed here may hold for periodic cycles, and some hold for nth-order differential systems; but, generally speaking, most of the properties only hold for limit cycles of quadratic differential systems. The research covered here first appeared in [11], [12], [14], and [17]; then [19] and [20] obtained a new breakthrough, and, recently, [176] and [177] did some summarization and extension. When we present these results, we shall give the methods of proof that are the simplest and easiest to understand.
LEMMA 11.1. Any line can have at most two points of contact with the trajectory of(11.1) (which can include the singular points of(11.1)) unless the line itself is a trajectory of (11.1). PROOF. Suppose the given line is ax+ by+ c = 0. Then the coordinates of the point of contact should satisfy
-a b
ax+by+c = 0,
Q2(x,y) P2(X, y).
The system in general has two solutions (xi, Yi) (i = 1, 2) unless the first equation is part of the second equation, in which case ax + by + c = 0 is a 245
246
THEORY OF LIMIT CYCLES
trajectory. In particular, if Q2(xi, Yi)
= P2(xi, Yi) = 0,
axi
+ byi + c = 0,
then (xi, Yi) is a singular point of the line ax+ by+ c = 0.' REMARK. This lemma can be generalized to nth order differential equations, to show the number of points of contact is at most n; and if we change the line to an algebraic curve of degree m, then the number of points of contact is at most m(m- 1 + n). LEMMA 11. 2. Suppose 0 and 0' are two adiacent elementary singular
points along one branch of P2(x, y) = 0 (on both its sides, P2(x, y) has different signs) or a branch of Q(x, y) = 0. Then the index of one is +1, and the index of the other is -1, provided that between 0 and 0' there does not exist a point of intersection of this branch of P2(x, y) = 0 (or Q2(x, y) = 0) with its other branch; conversely, if between 0 and 0' there still exists a point of intersection of this branch of P 2(x, y) = 0 (or Q2(x, y) = 0) and its other branch, then the indices of 0 and 0' are either both +1 or both -1. PROOF. Suppose 0 is (xo, Yo). Then the characteristic equation of the linear approximate system of ( 11.1) at 0 is
8P2 _A ax
8P2 8y
8Q2 ax
8Q2 _A By
=0. (xo,yo)
Since we assume 0 is an elementary singular point, ( aP2 BQ2 _ aP2 BQ2) ax By By ax (xo,yo)
#0 •
This shows that P2(x, y) = 0 and Q2(x, y) = 0 have different slopes at 0; hence they intersect, but do not touch each other tangentially at this point. Similarly, it is the same at 0'. Now we study two adjacent elementary singular points 0 and 0' along one branch of P2 (x, y) = 0. Suppose between 0 and 0' there does not exist a point of intersection of this branch of P2(x,y) = 0 with the other branch (Figure 11.1). Construct a smooth simple closed curve C containing 0 and 0' such that the other branches of P2(x, y) = 0 and Q2(x, y) = 0, which do not pass 0 and 0', do not intersect with C, and the interior of C does not contain any singular points other than 0 and 0'. Suppose above this branch of P2 = 0 (or on its left if this branch is a vertical trajectory) P2 > 0, and below it {or on its right) P2 < 0. Moreover, between two branches of Q2 = 0 (they may join in the exterior of C) we have Q2 > 0, and outside these two branches we have
§11.
LIMIT CYCLES IN QUADRATIC DIFFERENTIAL SYSTEMS
FIGURE 11.1
247
FIGURE 11.2
Q2 < 0. Thus we can draw several points Ai, Bi, Di, Ei, Fi, and Gi (i = 1, 2) on C from the directions of the vector fields of system (11.1). From this we can see at once that when a point runs from A1 along A~ towards D1, the direction of the vector field rotates through an angle 1r /2; when the point runs from D 1 along ~ towards F 1 , its direction rotates through an angle 0; when the point runs from F1 along F~ towards A2, its direction rotates through an angle -11"/2. Hence from A1 to A2 along A1D1F1A2 the net rotation of the vector field is 0. Similarly, along A2D2F2A1 from A2 to A 1 , the net rotation of the vector field is 0. In other words, the index of the vector field determined from the closed curve C with respect to (11.1) is zero; hence the indices of 0 and 0' are +1 and -1 respectively because they are elementary singular points. If between 0 and 0' there is a point of intersection of this branch of P 2 = 0 with the other branch (in this case P2 = 0 must be two intersecting lines and OCJi is one of the lines) (Figure 11.2), then, as before, we can prove that the index of the vector field determined from C with respect to (11.1) is +2 or -2; hence the indices of 0 and 0' are both +1 or both -1. REMARK 1. This property can be generalized not only to an nth order differential system, but also to plane autonomous systems. REMARK 2. If the polygonal line OMo" in Figure 11.2 is taken as one branch of P 2 = 0, and the other polygonal line NMO' containing 0' is considered as another branch of P2 = 0, then we can prove as before that the
248
THEORY OF LIMIT CYCLES
indices of the elementary singular points 0 and 0" have opposite signs. Because of this, later on, when the hyperbola degenerates into a pair of lines, we can always take two adjacent half-lines starting from the point of intersection as one branch of the hyperbola, and the remaining two adjacent half-lines are considered as the other branch. LEMMA 11.3. System (11.1) has at most three elementary singular points whose indices are +1 (or -1). PROOF. We may as well assume that P2(x,y) = 0 and Q2(x,y) = 0 do not have a common factor, for otherwise there exists a line or a curve which is filled with higher order singular points, and the other elementary singular point is at most one. It is easy to see that system (11.1) can have at most four elementary singular points. If they are all on the same branch of P2 = 0 or Q2 = 0, then by Lemma 11.2 the indices of two of them are +1, and the others are -1. Hence if P 2 = 0 or Q 2 = 0 is an ellipse or a parabola, the lemma is clearly established. Now suppose both P2 = 0 and Q2 = 0 are hyperbolas, and the four singular points lie neither on the same branch of P 2 = 0, nor on the same branch of Q2 = 0. Then either some two singular points lie on the same branch of P2 = 0 (or Q2 = 0), and the other two on the other branch, in which case there can only exist two singular points of index +1 ( -1); or some three singular points lie on one branch, and the other lies on the other branch, in which case it is possible that there are three singular points with index equal to +1 (or -1), but the fourth singular point must be -1 (or +1). REMARK. Following the method of proof of Lemma 11.3, it is not difficult to give a new geometrical proof to Lemma 10.3 in §10, and from this it is easy to see why there exists such a delicate relationship between the properties of the singular points of a dynamical system and the geometrical properties of the quadrilateral formed by them [176]. LEMMA 11.4. If the line passing through two singular points 81 and 82 of (11.1) is not an integral line, then it must be formed by three line segments without contact ooS1, 8182, S2oo; here the trajectories cross ooS1 and S2oo in one direction, and cross 8182 in the opposite direction. PROOF. Since any singular point can be considered as a point of contact, the first half of this lemma can be deduced at once from Lemma 11.1. In order to prove the second half, we first rotate the axes to make the line 8182 become the x-axis, and suppose that on ooS1 the trajectories of (11.1) all cross it from above to below (Figure 11.3). This shows that ooS1 lies in the region Q 2(x,y) < 0. Now Q2(x,y) = 0 must pass the two points 8 1 and 82, but it cannot have the x-axis as part of it; from the properties of quadratic
§11. LIMIT CYCLES IN QUADRATIC DIFFERENTIAL SYSTEMS
I
I \
249
I I
FIGURE 11.3
curves we know that Q2(x, y) = 0, and the x-axis must cross 81 and 82. Thus ~ must lie in the region Q 2 (x, y) > 0, and S2oo must lie in the region
Q2 (x, y) < 0. The lemma is completely proved. We can further prove [177]
The line connecting a finite singular point and its infinite singular point of system (11.1) is either a trajectory or a line without contact (except that finite singular point). LEMMA 11.5.
PROOF. Suppose this finite singular point is the origin (0, 0). Thus (11.1)
can be
wri~ten
as dxjdt
= awx + ao1Y + a2ox 2 + auxy + ao2y 2,
dyjdt = bwx + bo1Y + b2ox 2 + buxy + bo2y 2 • Its characteristic equation at (0, 0) is
(11.2)
(aw + bol)..\ + (awbo1 - a01bw) = 0. If (0, 0) is not a focus or a center, we have ..\ 2 -
(11.3) (aw + bo1) 2 - 4(awbol- a01bw) = (aw- bo1) 2 + 4aolbw ~ 0. Suppose the infinite singular point of (11.2) is (1, TJ, 0). Then it is easy to see that 11 should satisfy
L
(aij'f/- bij)TJi
= 0,
(11.4)
i+j=2
which is a cubic equation in TJ, and has at least one real root. Thus the line mentioned in the lemma can be written as L = y - TJX = 0. Finding the rate of change of L along the trajectory of (11.2), it is easy to see (since 1J satisfies (11.4)) that
I
ddL = x(bw t L=O
+ (bo1 -
aw)TJ - ao111 2 ).
(11.5)
From this we can see that if the value inside the parentheses on the right side of (11.5) is zero, then y -TJX = 0 is a trajectory; otherwise, it is a line without contact, but on both sides of the singular point the directions of trajectories crossing them are different.
250
THEORY OF LIMIT CYCLES
LEMMA 11.6. The interior of the closed trajectory r of system (11.1) cannot contain a nodal point, a saddle point, or a higher order singular point [178], [179]. PROOF. Suppose the interior of r contains a singular point 0. We may as well assume 0 is the origin of the coordinate system, and it is a nodal point, a saddle point, or a higher order singular point. Thus (11.1) can be written in the form (11.2). Since now (11.3) holds, we can pick T/ to make the value of the parentheses on the right of (11.5) equal to zero. Then take the line L = y - qx = 0 passing through 0, and find the rate of change of L along the trajectory of (11.2). It is easy to see that now we have
I
dL dt L=O
= x2
[
I: (bi;- ai;T/ )qi] .
(11.6)
i+j=2
From this we can see that if the value in the parentheses on the right of (11.6) is zero, then y- TJX = 0 is a trajectory, for otherwise the trajectories intersecting the line would always cross it from the same direction. In either case, the vicinity of 0 cannot contain a closed trajectory. This contradicts the assumption. COROLLARY. The interior of a limit cycle r of system (11.1) can only have a unique singular point, which must be a focus. REMARK. Lemma 11.6 and its corollary cannot hold even for a cubic differential system. Li Ji-bin [180] gave an example of a cubic system which has a limit cycle whose interior contains three singular points. LEMMA 11.7. A closed trajectory of system (11.1) or a singular closed trajectory which contains only one singular point cannot contain a line segment. PROOF. Suppose the closed trajectory r of system (11.1) contains a line segment. Then the line l defined by this line segment has an infinite number of points of contact with r. _By Lemma 11.1 we know that l itself is also a trajectory of (11.1). But the points on r leaving l (at least two points) will violate the uniqueness of solution of the differential equations; hence it is not possible. When r contains only one singular point, it is also impossible. (I) REMARK. From the proof of Lemma 11.7, we know that if r contains two singular points, then the conclusion does not necessarily hold. In fact this is easy to prove [14]. (1 )If a singular closed trajectory r can have the equator as its part, then one finite singular point and also a half-line. An example is given in [179].
r
can contain
§11. LIMIT CYCLES IN QUADRATIC DIFFERENTIAL SYSTEMS
D
251
I
FIGURE 11.4
FIGURE 11.5
LEMMA 11. 8. The line segment connecting two saddle points on a singular closed trajectory r must belong to r' but r does not have any other singular point. PROOF. If this line segment does not belong to r, then by Lemma 11.4 we know it is a line segment without contact, and it will form a region with some section of the arc of r whose interior does not contain a singular point and which intersects the line segment. As t increases (or decreases), all the trajectories entering this region have nowhere to go. A concrete example is dx 2 dt = -y- y '
dy = x- x2 dt .
The figure of its global trajectory can be drawn by the reader. THEOREM 11.1. A closed trajectory of system (11.1) or its singular closed trajectory containing only one saddle point and lying in a bounded region must be a strictly convex closed curve which intersects one branch each of P2(x, y) = 0 and Q2(x, y) = 0 at only two points [11]. PROOF. If r is not a strictly convex closed curve, then, since r does not contain a line segment, we can always find a line l which has at least four points of intersection with r, as in Figure 11.4. According to the directions of crossing of r and l, on the line segments AB, BC, and CD, each has at least a point of contact with the trajectory of (11.1) (which may be a singular point). But by Lemma 11.1, we know this is impossible.
THEORY OF LIMIT CYCLES
252
Next, since r does not contain a line segment, r can have one highest point H, one lowest point G, one extreme right point Rand one extreme left point L (Figure 11.5). According to whether the section of arc of is rising or falling, it is clear H and G are the only two points of intersection of the same branch of Q2(x, y) = 0 with r, orR and L are the only two points of intersection of the same branch of P2(x,y) = 0 with r. If two adjacent points out of these four, such as H and R, coincide then H = R is a singular point; thus r becomes singular closed trajectory. At the same time, by the hypothesis of this theorem, L and G cannot coincide with each other, and they cannot coincide with H. REMARK. Similarly we can prove that any singular closed trajectory containing two or three saddle points can also be a convex closed curve, but it must contain a line segment. Moreover, Theorem 11.1 does not necessarily hold for cubic differential systems; for example, the limit cycle of the van der Pol equation is not convex when 11. is rather large.
:r
THEOREM 11.2. For system (11.1), the following two kinds of relative positions cannot exist [11]:
(a)
(b)
FIGURE 11.6 PROOF. Since in Figure 11.6(a) the interior of r1 should contain more than one singular point, this is impossible. In Figure 11.6(b), the line segments 0102,0203, and 0301 connecting the singular points should be line segments without contact. Now suppose the positive direction on r 1 (i.e. the direction of increasing t) is counterclockwise; thus the positive direction on r2 should be clockwise, and so the positive direction on r3 should be counterclockwise. Then D;Ol cannot be a line segment without contact. Similarly, we can prove COROLLARY. The total number of centers and foci of system (11.1) is at most 2.
§11.
LIMIT CYCLES IN QUADRATIC DIFFERENTIAL SYSTEMS
253
THEOREM 11.3. The relative positions of limit cycles of system (11.1) can only have the following two cases [11]: 1) There exist one or more limit cycles in the vicinity of only one focus. 2) Limit cycles appear in the vicinity of each of the two different foci. Based on Theorem 11.3, the further problems are specific realizations of all possible distributions. In this area, [11], [14], [17], [19], and [20] have done some work. Since in [19] we have given examples to illustrate that system (11.1) can possibly have four limit cycles, we should at least discuss whether the distributions of (1, 0), (2, 0), (3, 0), (4, 0), (1, 1), (1, 2), (1, 3), and (2, 2) can indeed be realized. Here (m, n) represents the distribution of m limit cycles in the vicinity of the first focus and of n limit cycles in the vicinity of the second focus (if it exists).( 2 ) For this problem, past research can be divided into two groups: 1) Give an example or prove from theory that the quadratic differential systems can have distributions (1, 0), (2, 0), and (3, 0) [21], (1,1) [11], (1, 2) [14], and (1, 3) [19], [20]; but there is no way to prove whether the limit cycles of the system under discussion have the exact number or have more than that number. 2) Prove rigorously that the limit cycles of some quadratic system have (1,0) distribution (see Example 3) and the limit cycles of another quadratic differential system must have (1, 1) distribution [180). For distributions (4,0) and (2, 2), it is not clear right now whether they can be realized. We conjecture that (4,x), x ~ 0 and (y,z), y ~ 2, z ~ 2 cannot be realized. EXAMPLE 1 [181]. The system
~; = 6x- y- ~x 2 + xy, when, 0
~~ = x ( 1- ~)
(11.7)
< 6 < ! has a unique limit cycle in the vicinity of (0, 0) and of
(2, 2- 26). When 6 ~ 0, there is no limit cycle; when 6 =!,these two cycles expand and become two separatrix cycles formed by the equator and the line X= 1. For the proof we use the Annular Region Theorem, the theory of rotated vector fields, and the uniqueness theorem of [111). The rest of the proof is left to the reader. EXAMPLE 2 [19). For the system dx dt = -y- 62x- 3x 3 + (1 - 61 ) xy + y 2 , when 0 < ~
«
61
«
1, the distribution (1,3) appears.
(2)For quadratic differential systems we pointed out in §9 that the center and the limit cycle cannot coexist, no matter whether they link together or not.
THEORY OF LIMIT CYCLES
254
FIGURE 11.7 PROOF. First consider the system of equations 2 dx dt = -y - 3x
dy ( 2 ) dt = x 1 + gx - 3y .
+ xy + y 2 ,
{11.9)
It is easy to see that {11.9) has only two finite singular points, 0(0, 0) and N(O, 1); the latter is an unstable coarse focus. After a coordinate transformation, we change {11.9) into a standard form of Bautin (9.30). We get
At = 0,
\
1\4
A 2 -
82 = - v. rnn
o~,
77
A -
1152
3 - 41v'82' 41v'82' \ \ _ 3644 0
1\5
= ,
1\6 -
369 J82.
From this, according to the formulas of §9, we can compute that 1h = ii3 = 0 and iis < 0; hence 0(0, 0) is a stable fine focus of second order of {11.9). The infinite singular point {1, .,, 0) of {11.9) can be determined from the equation 'T/ 3 + ., 2 - 2/9 = 0, which has only one real (positive) root. It is easy to prove that the corresponding singular point is a saddle point, and the direction of the separatrix passing through the saddle point is shown in Figure
11.7. Also on the line 1- 3y = 0, we have dy / dt = 2x 2 /9 ~ 0. This indicates that the trajectory of {11.9) always crosses the line from below to above; hence by the Annular Region Theorem we know that system {11.9) has limit cycles in the vicinity of both 0 and N; that is, there exists a (1, 1) distribution {but it is not known whether system (11.9) has exactly two limit cycles). It is easy to see that the limit cycle r 1 closest to N must be an internally stable cycle, and the limit cycle r 2 closest to 0 must be an internally unstable cycle. Then we apply the method of Bautin in §9. We change the right sides of {11.9), to get
dx dt = -y - 3x 2
+ (1 -
£) xy + y 2 ,
v1
dy 2 x - 3y) , dt = x ( 1 + 9
(11.10)
§11. LIMIT CYCLES IN QUADRATIC DIFFERENTIAL SYSTEMS
255
where 0 < 81 « 1. For (11.10), since ii3 = 2St > 0, 0(0, 0) has become an unstable first-order fine focus, and from Theorem 3. 7 in §3 we know that in its vicinity there will appear again a stable limit cycle r3 (c r2). Since when 81 is sufficiently small both r 1 and r2 do not disappear, the (1, 2) distribution exists for system (11.10). Finally, we add one term -82x to the right side of the first equation of (11.10) to change it into (11.8), where 0 < ~ « 81 < 1. Since 0 changes again from an unstable fine focus to become a stable coarse focus, in a smaller neighborhood, there appears again one unstable limit cycle r 4, and at the same time r1.r2, and r3 still exist. Hence system (11.8) has at least three limit cycles in the vicinity of 0, and there exists at least one limit cycle in the vicinity of N. The proof is completed. A similar example can be obtained in [20]. But the starting system of equations dx dt
= -y -
10x2
dy 2 = x+x -25xy dt
+ 5xy + y 2 ,
-
(11.11)
has no cycles in the vicinity of the point 0 and takes 0 as a third order fine focus. Although this example is not as good as Example 2, it yet corrects a mistake of a symbol of ii7 in Bautin's formula (9.40) for the convenience of our readers. More recently, [182] and [183] extended the range of quadratic systems with distribution (1, 3) so that the coefficients of the starting equations are not all fixed numbers. For example, [183] obtained the following result: Suppose the system of equations dx (2l + b)a 2 2 = -y+lx + xy+ny dt l+n '
~~ = x + ax 2 + bxy
-
(11.12)
satisfies the following conditions: 1) a '=I 0; 2) 3n(l + 2n) :$ n(n +b) < 0; and 3) the infinite singular point is unique. Then we can prove that in the vicinity of (0, 1/n) there exists at least one limit cycle and 0(0,0) is a third-order unstable fine focus, or a second-order stable fine focus. Thus if 0 < -..X « -11 « -€ « 1, then the system (2l + b)a
dx dt
= ..Xx - y + lx 2 + (
dy dt
= x + ax2 + (b + e(l + n) + "') xy
l +n
+ e)
2
xy + ny , (11.13)
a
has at least three limit cycles in the vicinity of 0 and at least one in the vicinity of M.
THEORY OF LIMIT CYCLES
256
Similarly, the system of equations in Example 1 has been extended to become the problem of studying under what conditions the system
dxfdt
= -y +ox+ lx 2 + mxy,
dyfdt
= x(1 +ax)
has two (and only two) limit cycles which do not contain each other.( 3 ) Finally, we give a most obvious example of the distribution (1,0) [17]. EXAMPLE 3. If system (11.1) takes a circle or an ellipse r as a limit cycle, then r is the unique limit cycle of system ( 11.1), and is a single cycle. PROOF. First, after an affine transformation, we can changer to the unit circle x2 + y 2 = 1. It is easy to see that (11.1) should be changed to
= -y(ax +by+ c)- k2(x 2 + y 2 - 1), dyfdt = x(ax +by+ c)+ kt(x 2 + y 2 - 1),
dxfdt
(11.14)
where k~ + k~ i= 0. This system has two singular points, which both lie on the line k2x = k 1 y. After a rotation of axes, we make this line the vertical axis; then (11.14) becomes (we still use x and y to denote the rectangular coordinates)
~~ = -y(a'x + b'y + c')- k(x 2 + y 2 -
~~ = x(a'x + b'y + c'),
1),
where k i= 0. The above system can be changed to
:~ = -y(ax +,By+')')- (x 2 + y2 -
1),
dy dr
= x(ax +,By+')').
(11.15)
According to the hypothesis, the unit circle x 2 + y 2 = 1 should be a limit cycle, and it is not possible to have a singular point on it; that is, we should have (11.16) Moreover, the singular point inside the circle cannot be a center; hence (11.17)
Now we analyze the system with center
~~ = -y(,By +I')- (x2 + y2 -
1),
(11.18)
This system and (11.15) have the same closed trajectory x2 + y 2 = 1 and the same singular points. It is easy to draw the complete graph of its trajectory as in Figures 11.8 (.B :::= -1), 11.9 (.8 < -1), 11.10 ( -1 < .8 < 0), 11.11 (.B > 0), and 11.12 (.8 = 0). (3)Simila.r results were also obtained in [185].
§11. LIMIT CYCLES IN QUADRATIC DIFFERENTIAL SYSTEMS
257
.8<-1 FIGURE 11.8
FIGURE 11.9
a:
,8-0 ,B>D
-1<,8<0 FIGURE 11.10
FIGURE 11.11
FIGURE 11.12
Now we consider o in the system (11.15) as a parameter. Then, as ovaries, (11.15) forms a family of equations, and for o = 0 we get (11.18). Computing (11.19) we know that (11.15) forms families of generalized rotated vector fields with opposite directions of rotation outside and inside the circle x 2 + y 2 = 1 respectively, and that x 2 + y 2 = 1 is a closed trajectory of every equation in the family. Since (11.18) has a family of closed trajectories surrounding one or two centers of index +1, from Theorem 3.2 in §3 we know that foro =F 0 the system (11.15) does not have a closed trajectory outside and inside the unit
258
THEORY OF LIMIT CYCLES
circle. Moreover, from (11.19) we can also see that if the trajectory of (11.15) near the outside of the unit circle gets out of (into) the closed trajectory of (11.18) as 1' increases, then the trajectory of (11.15) near the inside of the unit circle should get into (out of) the closed trajectory of (11.1'8) as 1' increases; that is, x 2 + y 2 = 1 must be a stable or unstable cycle of system (11.15), and cannot be a semistable cycle. Finally, in order to prove that x 2 + y 2 = 1 is a single cycle of (11.15), we can assume it has the parametric equations
x=cost(T),
y=sint(T),
where t(T) is easily seen to satisfy the equation
dtjdT =a. cost+ {3sin t + "Y (> 0 when "Y > 0). Thus, along the unit circle, x and y are periodic functions of t with period 211'. We then calculate the value of the integral of the divergence with respect to 1' around the unit circle once, and it is easy to prove it is not equal to zero; hence the unit circle is a single limit cycle. A necessary and sufficient condition for a cubic system to have a quadratic algebraic limit cycle can be seen in [18] and [187]. Xu Shi-long [188] showed that a necessary and sufficient condition for the algebraic curve yn + xm = 1 to be a limit cycle of a quadratic differential system is n = m = 2; a necessary and sufficient condition for the curve to be a limit cycle of a cubic differential system is either n = m = 2 or m = 2n = 4 or n =2m= 4. Moreover, in [189] and [190] the conditions for existence of a quadratic algebraic curve solution for a quadratic differential system were studied. Since in this section we introduce the general properties of limit cycles of quadratic differential systems, we list all the important properties of quadratic systems which have been proved and will be proved later, for the reader's convenience. 1. A quadratic differential system cannot have a center and a limit cycle (§9). 2. A quadratic differential system whose linear part takes the origin (a singular point) as a star nodal point does not have a limit cycle (§10). 3. A quadratic differential system with two integral lines does not have a limit cycle. 4. A quadratic differential system with one integral line and one fine focus does not have a limit cycle (§15). 5. A quadratic differential system with one integral line can have at most one limit cycle (§15). 6. The locus of the points whose divergence apI ax + aQ I ay = 0 is a line, and any closed trajectory, if it exists, must meet the line.
§11. LIMIT CYCLES IN QUADRATIC DIFFERENTIAL SYSTEMS
259
7. If the divergences of two singular points of a quadratic differential system are both equal to zero, then the system does not have a limit cycle (§15). 8. If the number of singular points of a quadratic differential system is more than two, and one of them is a fine focus, then its limit cycle can only appear in the vicinity of one focus [19]. 9. If a quadratic differential system has two fine foci, they can only be fine foci of first order [191]. 10. A quadratic differential system whose vector field has a center of symmetry can have at most two limit cycles, which should have distribution (1, 1)
(§15). 11. A quadratic differential system possessing a third-order fine focus does not have a line solution [176]. 12. If a singular closed trajectory has three saddle points on it, then it must be a triangle with the saddle points at its vertices [14]. 13. If r l and r2 both are singular closed trajectories with one saddle point on each of them, then they cannot have this saddle point as their common singular point [14]. Exercises 1. Prove that when P2(x, y) = 0 or Q2(x, y) = 0 of system (11.1) is an ellipse, its section of arc contained in the interior of a closed trajectory of (11.1) cannot be greater than half the length of the ellipse. 2. Starting from the system of equations
d d; = -338x + 32y + 169x2
-
16y2 ,
d
d~
= -288x + 18y + 144x2 - 9y2
and constructing a family of uniform rotated vector fields, prove that when the angle of rotation passes through 1r/4 the neighborhood of each of the singular points (0,0) and (1, 1) can generate a limit cycle [11]. 3. First analyze the global structure of the trajectory of the system
dx = xy dy = _!(x- 1)(x + 2) + !y2 + !xy- !y dt , dt 3 2 3 3 and then construct a family of uniform rotated vector fields starting from this system. Prove that when 0 < -0 ~ 1, the (1, 2) distribution of limit cycles can appear [14]. 4. Prove the limit cycles of the system of equations dx dy 2 dt =-y+6x+lx 2 +mxy+ny, dt =x(1+ax+by) when 1) b + n =I= 0 and 2) b + n = 6 = m +a= 0, even if they exist, cannot be monotonically close to each other.
THEORY OF LIMIT CYCLES
260
5. Prove the conclusion of Example 1. 6. Prove that a quadratic differential system with two integral lines does not have a limit cycle. 7. Prove that the singular closed trajectory of a quadr~tic differential system with three saddle points on it should be a triangle with these saddle points as its vertices. 8. Prove the second half of Lemma 11.2 and the conclusion of Remark 2 after that lemma. 9. Draw the graph of the trajectory of the system of equations following Lemma 11.8. 10. Prove the conclusion of the system (11.13) in this book. 11. Prove that r in Example 3 is a single cycle. 12. Prove that if in the system of equations dxjdt
= -y +ox+ lx 2 + mxy,
dyjdt
= x(1 +ax)
we have 2a + m = O,a < O,l < O,l 2 - 8a2 < 0, and 0 < 6 < l/2a, then there are two and only two limit cycles which do not contain each other in the whole plane.
§12. Classification of Quadratic Differential Systems. Limit Cycles of Equations of Class I
For a quadratic differential system dxjdt = P2(x, y),
dyjdt = Q2(x, y),
(12.1)
aside from the relative positions of its limit cycles, more important questions are: For a given quadratic differential system, how can we determine whether it has a limit cycle? If it has one, does it have more, and how many? In studying these problems, either in the form (12.1), where P2 and Q2 are general quadratic polynomials or the form of Bautin (9.30), which we have seen in §9, we always run into some inconvenience. This is because we have to find the coordinates of the singular point; sometimes it is very troublesome to solve for x or y from a quadratic equation. Hence we now first introduce a method of classification; that is, we apply some simple transformations to system (12.1), which may have a limit cycle, to bring it into one of three standard forms, and then we proceed to study them one by one [12]. We may assume P2(x, y) and Q2(x, y) do not have a common factor, for otherwise (12.1) can be simplified to a linear system, which obviously does not have a limit cycle. From the theory of quadratic curves, we know there exists at least one real ~ making the equation ~P2(x,y)
(12.2)
+Q2(x,y) = 0
into a degenerate quadratic curve (if Q2(x, y) = 0 is degenerate, then we take ~ = 0; if P2(x, y) = 0 is degenerate, then we can take~= oo; see footnote 1). When this degenerate quadratic curve represents a point or does not have a real locus, the system obtained from (12.1) by the transformation y' = ~x+y, x'=xis dy' dt
= AP2
+ Q2 =
Q2' (X ','Y) ,
dx' dt
nl('') = r2 X ,y ,
( ) 12.3
where Q~(x',y') = 0 represents a point or does not have a real focus. According to the theory of §11, this system cannot have a closed trajectory. Hence, 261
THEORY OF LIMIT CYCLES
262
we may as well assume that >.P2 + Q2 = R1R2, where~ (i ~ 1, 2) is a real polynomial with degrees of x andy not higher than one, and at least one of them is not a constant. It is easy to prove (do so as an exercise) that if for i = 1, 2 the determinant of the transformation y'
= Y + >.x,
x' =~
(12.4)
is always zero, then the system has one or two integral lines, and also does not have a closed trajectory. Hence we only have to discuss the case when the determinant of transformation (12.4) is not zero for i = 1 or 2. In this case the system (12.1) under this transformation(!) becomes dx' fdt
= P~(x', y'),
dy' fdt = x'(ax' +by'+ c).
(12.5)
We still write x', y' as x, y, and, depending on the values of a, b, and c, we can divide the systems (12.5) into three classes: I. a = b = 0, c # 0, dx dt
= k + Dx + ey + lx 2 + mxy + ny
2
,
dy dt =ex.
(12.6)
II. a # 0, b = 0, c # 0,
~~ = k + lJ x + ey + lx2 + mxy + ny2 ,
dy dt = x(ax +c).
(12.7)
dy dt = x(ax +by+ c).
(12.8)
Ill. b # 0,
~~
= k+8x+ey+lx 2 +mxy+ny2,
There is no need to discuss the case when b = c = 0 but a # 0, since in this case Q2(x, y) ~ 0 holds in the whole plane, and it is not possible to have a closed trajectory. According to the theory of §11, if system (12.6), (12.7), or (12.8) has a closed trajectory, then its interior must contain a unique focus or center with index +1. We translate the origin to the singular point and then apply a suitable transformation x = J.LX 1 , y = vy', t = >.t' to change (12.6), (12.7), and (12.8) into I. dy (I) dt = x,
II.
2 { dxfdt = -y +ox+ lx + mxy + ny2, dyfdt = x(1 +ax), a# 0,
=
(II)
( 1 )If.>.. oo, we first interchange x and y in (12.1) and then apply the transformation (12.4) with .>.. = 0.
§12 .. QUADRATIC SYSTEMS OF CLASS I
263
dxjdt = -y + 8x + lx 2 + mxy + ny2 , dyjdt = x(1 +ax+ by), b ~ 0,
(III)
III. {
respectively. Later we shall study systems of classes I, II, and III in detail one by one. Aside from the method of classification, devised by Chinese scholars, there is another method, used in the Soviet Union. Cherkas [192] divided the systems (12.1) into two classes:(2) {
dxjdt dyjdt
=boo+ xy, = aoo + awx + ao1Y + a2ox 2 + auxy + ao2Y2 = Q2(x, y), dxfdt = ~ox 2
+ y,
dyjdt
(A)
= Q2(x,y).
(B)
For systems of classes II and III, under certain conditions, the nonexistence or uniqueness of limit cycles and the interesting property 7 listed at the end of §11 can be proved by first changing the system into (A) or (B), and then changing it again into an equation of Lienard type. Since there are two systems of classification, one naturally asks whether one can once and for all obtain the relationship between the coefficients of these two classes so that we do not have to carry through the transformation from one class to another every time the problem comes up. Part of this work was done in [193], where Liang Zh~jun obtained the formula for the coefficients of (A) or (B) in terms of the coefficients of (III). Specific examples are as follows: (1) n ~ 0. Let k denote a nonzero root( 3 ) of the equation a+ (b- l)k - mk 2 - nk3 = 0,
where o: = k 3 - 8k 2 + k and (3 = kl- a- nk3. When (3 can be changed to (B), where
= -ko: + k2 -
aoo
= 0,
a2o
= - nk3 -
8k 2 - k [ 3 nk3 2a + bk - nk
a01
= 8k 2 ,
a 11 =
aw ao:
n~ 3
[2a
{12.9)
= 0 and o: ~ 0,
(III)
8k 3 = -k4 ,
1 2 +a (a + bk) (8 k -
+ bk +~(a+ bk)(8k 2 -
k
)]
'
k)],
-1
ao2 = k (a+ bk). n 3 o:
- (2)We --shall later give a detailed introduction to the method of changing (12.1) into (A) or (B). ( 3 )The case when (12.9) has only a (real) zero root was not discussed in [193]; this case is left to the reader as an exercise.
THEORY OF LIMIT CYCLES
264
When {3
i
boo=
Q
0, (III) can be changed to (A), where
[~(a- kl) + k- b'k 2] '
1 a00 = ""fi [ko{3 + ao 2 - b10(k{3 + o(2a + bk)) + b~ 0 (a + bk) + boo(a- kl)], 1 a10 = ""fi[-(k{3 + 2ao)- b2o(k{3 + o(2a + bk))
+ b10(2a + bk) + 2b10b2o(a + bk)], a2o
=~[a+ b2o(2a + bk) +(a+ bk)b~ 0 ],
1 aot = ""fi[k{3 + o(2a + bk)- 2b10(a + bk)], -1
an= lf[a + kl + bk + 2(a + bk)b2o],
1 ao2 = ""fi(a. + bk),
in which blO = b'k2 - k- 2o(a- kl)/{3 and b2o =(a...., kl)/{3. Here we do not consider the case o = {3 = 0, since for this case (III) cannot have a limit cycle. (2) n = 0. When m = 0, we leave this case to the reader as an exercise.( 4 ) Now suppose m i 0. Then we can prove that (III) can be changed to (A), where
boo = l + mb', aoo = m 2 + ma- b(2l + mb') + l(l + mb'), a10 = m 2 + 2am- b(3l + mb'), aot = b, a2o = ma- bl,
au = b + l,
ao2 = 0.
Of course, conversely, we can use the coefficients of (A) or (B) to express the coefficients of (III); and work can also be carried out along this line. Next, how to use b', l, m, n, a, and b under our method of classification to express the quantities tit, v3, tis, and 'V7 from §9 is also an important problem.( 5 ) Here we should first find the formulas for At, ... , As from §9. This can be done following the method mentioned in the paragraph before Theorem 9.2 in §9; that is, we first apply the transformation
_.,,
x=~-e,
-2,
y=~,
(12.10)
( 4 )It was incorrectly observed in [193] that the case n = m = 0 did not have a limit cycle, and the formula was not given there. ( 5 }The proof of the last theorem in §16 illustrating Bautin's formulas for 1i3, tis, and 1i7, and the system (12.1} in Bautin's form are sometimes quite convenient for studying certain properties of quadratic differential systems.
§12. QUADRATIC SYSTEMS OF CLASS I
265
and then rotate the axes, taking tan tp =
-(2a+bt5)~ M2 + 26(a- m) - 4{1 + n)
-:-=-':-:-:---'-.,....-~,.---......,..
(12.11)
and e=xcostp-ysintp,
77=Xsintp+ycostp,
r=!v'4-6 2t.
(12.12)
We obtain dxjdt = AtX- y- A3x 2 + (2A2 + As)xy + A6y 2, dyjdt = x + AtY + A2x 2 + (2A3 + A4)xy- A2y 2,
(12.13)
where At= 6/~. Moreover, when 6 ::F 0 the formulas for A2, ... , A6 are too complicated, and there is not much use for them; hence they will not be written out in detail here. Now it is important to know the values of all the coefficients of (12.13) when At = 0; that is, when 6 = 0 and (0, 0) is a fine focus. In this case a tan tp = -1 - , +n
sintp=
cos tp =
Ja2 +~l+n)2'
l+n Ja 2 +(l+n)2'
-;:=;r=::::::;:;=~
At =0,
t
A3 = [a2 + (l + n) 2 312 [l(l + n) 3 + a(m + a)(l + n) 2 + a 2(l + n)(b + n)], A6 = [a2 + (l + n) 2]-312[ma(l + n) 2 + a 2(b -l)(l + n) - a 4 - n(l + n) 3], 2A2 +As= [a2 + (l + n) 2t 312[(2a + m)(l + n)a 2 + ba 3 + a(2l- 2n- b)(l + n) 2 - m(l + n) 3 ], A2 = [a2 + (l + n) 2t 312[na 3 + ma 2(l + n) + a(l- b)(l + n) 2 - a(l + n) 3], 2A3 + A4 = [a2 + (l + n) 2t 312[a2(l + n)(b- 2l + 2n) + a(2a + m)(l + n) 2 - ma3
-
b(l + n) 3 ]. (12.14)
From this we can also obtain A -ma- (l + n)(b + 2l) 4= Ja2+(l+n)2 '
A _ a(b + 2l)- m(l + n) sJa2+(l+n)2 ' ..\a- ..\6 = Ja 2
(12.15)
+ (l + n)2.
Thus, the formulas for At, ... , A6 in terms of 6, l, m, n, a, and b can be completely obtained. From this it is not difficult to further obtain the formulas for ila, Us, and v7.
THEORY OF LIMIT CYCLES
266
Comparing the conditions for the origin being a center in Theorem 9.2 in §9, we get THEOREM 12. 1. The four groups of conditions for system (III) to have the origin as its center are: 1) = 0, m(l + n)- a(b + 2l) = 0, and ma + (l + n)(b + 2l) = 0 (equivalent to >.1 = >.4 = >.s = 0); 2) = 0 and a= l + n = 0 (equivalent to >.1 = >.3- >.s = 0); 3) = 0, m(l + n) = a(b + 2l), and a[a 2(n + b + 21) - (b + n)(l + n) 2] = 0 (equivalent to >.1 = >.2 = >.s = 0); and 4) = 0, m = 5a, b = 3l + 5n, and ln + 2n2 + 2a = 0 (equivalent to >.1 = >.s = >.4 + 5>.3- 5>.s = >.3>.s- 2>.~- >.~ = 0).
o
o o
o
The proof is omitted, and can be used as an exercise. REMARK 1. When we transform the equations, cp can differ by 1r and the corresponding >.2, ... , >.s all change signs, but >.1 and Vi (i = 3, 5, 7) remain unchanged. REMARK 2. Although in theory we can obtain formulas for 'Vs and v7 to distinguish the stability of the origin and the order of its being a fine focus, these formulas are very complicated, and it is not easy to simplify them (however, this is not difficult for v3). Li Cheng-zi [183] first studied necessary and sufficient conditions for the quadratic system of the form dxjdt
= -y + a2ox2 + auxy + ao2Y 2,
dyjdt = x + b2ox 2 + buxy + bo2Y 2
(12.16)
with (0, 0) as its kth-order fine focus (k = 1, 2, 3) and then derived necessary and sufficient conditions for (0, 0) to be a center. These two groups of results are rather simple and easy to apply. In the following we shall introduce several theorems of [183], but their proof is omitted. THEOREM 12.2. For system (12.16) introduce the quantities
W1 = Aa- B/3, W2 = [f3(5A- /3}
+ a(5B- a)b,
(12.17)
W3 = (A/3 + Ba)lo, where
A= a2o + ao2, B = b2o + bo2, a= au+ 2bo2, f3 = bu + 2a2o, 1 = b2oA3 - (a2o - bu)A2B + (bo2 - au)AB 2 - ao2B 3,
o=
a~ 2 + b~o + ao2A + b2oB.
(12.18)
§12. QUADRATIC SYSTEMS OF CLASS I
Then the following assertions are true : 1) (0, 0) is a kth-order fine focus (k kth group of conditions holds: 1°) Wt =1=
o;
2°) Wt =
o, w2 =1= o;
267
= 1, 2, 3)
if and only if the following
3°) Wt =
w2 = o,
w3
=1=
o.
{12.19)
2) The stability of the kth-order fine focus is decided by the sign ofWk: it is stable for wk < 0 and unstable for wk > 0. 3) The origin is a center if and only if Wt = W2 = W3 = 0. In order to study the stability of (0, 0), when only Wt = 0 we have to examine W2, and when only Wt = W2 = 0 we have to examine W3. Hence if we use the quantities Wt =Aa-B/3, when A=/: 0, when B =/: 0, when A= B =0; when A=/: 0, when B =/: 0, when A= B =0
(5A- f3)/3"'f,
W~ = { (5B - a)a"'f,
0, A/316, W~ = { Ba"'fO, 0,
(12.20)
to replace W., W2, and W3 in (12.17), then the conclusion of Theorem 12.2 remains unchanged. COROLLARY. System (12.16) takes the origin as its center if and only if at least one of the following four groups of conditions holds: 1) A= B = 0 (equivalent to At= A3- As= 0). 2) a= /3 = 0 (equivalent to At= A4 =As= 0). 3) Aa- B/3 = "Y = 0 (equivalent to At= A2 = -Xs = 0). 4) 5A- /3 = 5B- a= 6 = 0 (equivalent to At =As= A4 + 5A3- 5-Xs = >.3-Xs - 2-X~- A~ = 0). THEOREM 12.3. For system (III), introduce the quantities
W t = m(l + n)- a(b + 21), W 2 = ma(5a- m)[(l + n) 2 (n +b)- a 2 (b + 2l + n)],
(12.21)
W 3 = ma 2 [2a 2 + n(l + 2n)][(l + n) 2 (n +b)- a 2 (b + 21 + n)]. Then system (12.2) (with Wi changed to Wi) holds for system (III). System
(III) takes the origin as its center if and only if at least one of the following groups of conditions holds: 1) a= l + n = 0. 2) m(l + n) = a(b + 2l) and a[(l + n) 2 (n +b)- a2 (b + 2l + n)] = 0, a =f. 0.
THEORY OF LIMIT CYCLES
268
3) m = b + 21 = 0. = 5a, b = 3l + 5n, and 2a2 + n(l + 2n)
4) m
= 0.
In the following we study nonexistence, existence and uniqueness of limit cycles of equations of class I. First we prove a nonexistence theorem. THEOREM 12. 4. The system of equations
dx/dt = -y + lx 2 + mxy + ny 2 ,
dyjdt = x
(I)o=O
takes the origin as its center when m(l + n) = 0, and does not have a closed trajectory or a singular closed trajectory when m(l + n) -=f. 0. PROOF. We use the method of Dulac functions to prove this theorem. First, for the case of n = 0, we take the Dulac function as
B(x, y) = exp(mx- 2ly- !m2 y 2 ). Then we have
a
ax (BP2)
a
+ ay (BQ2) =
(12.22)
mlx 2 exp(mx- 2ly- !m2y 2).
When ml = 0, the right side of the above formula is equal to zero, and {12.22) becomes an integrating factor of system (I)o=Oi it is clear that the origin is a center. When ml ::I 0, the right side of the above formula keeps a constant sign in the whole plane; hence (I)o=O does not have a closed trajectory or a singular closed trajectory. If n ::I 0 in system (I)o=O• then we take
B(x,y)
= e(o:mn- 2I)Y(x- any+ a)"m,
(12.23)
where a = (m + v4n2 + m 2)/2n2 is a positive root of n 2a 2 - ma- 1 We can compute
a
-(BP2)
= 0.
a + -(BQ2) = am(l + n)x2 (x- nay+ a)mo:-le(o:mn- 2l)y.
ax ay Hence, when m(l +n) = 0, the origin is a center; when m(l +n) -=f. 0, the right side of the above formula keeps a constant sign in the half-plane x-nay+a > 0. Note that the origin lies in this half-plane, and the line x- nay+ a= 0 is crossed by the trajectories of system (I)o=O all in the same direction; hence (I)o=O does not have a closed trajectory. Moreover, since the saddle point (0, 1/n) lies on the line x- nay+ a= 0, if a singular closed trajectory exists, then it must pass this saddle point, and part of it consists of a certain pair of separatrices passing this saddle point. From the previous discussion, we know this is impossible. The theorem is completely proved. The figures showing the global trajectories of (I)o=O which were clearly described in [194] are shown in Figures 12.1-12.7, in which we have assumed
§12. QUADRATIC SYSTEMS OF CLASS I
FIGURE 12.1
FIGURE 12.2
FIGURE 12.3
FIGURE 12.4
m = 1. The case of a center (m = 0 or l and the figures will not be given here.
+ n = 0)
269
can be deduced from §9,
THEOREM 12. 5. System (I) does not have a closed or singular closed trajectory when m(l + n) = 0 but 6 =I 0, nor for 6m(l + n) > 0; however, for Om(l + n) < 0 and 161 sufficiently small, system (I) has a unique limit cycle. PROOF. When 6 varies, system (I) forms a complete family of generalized rotated vector fields with 6 as parameter. Since (I) has a family of closed trajectories when m(l + n) = 6 = 0, it does not have a closed or singular closed trajectory when m(l + n) = 0 but 6 =I 0. Next, if 6m(l + n) < 0, then when 161 increases from 0, the stability of the origin happens to change. lienee from Theorem 3. 7 in §3 we know that limit cycles will appear near the
270
THEORY OF LIMIT CYCLES
FIGURE 12.5
FIGURE 12.6
n>-1>0 FIGURE 12.7
ongm. Moreover from §6, VII we know that when 181 is sufficiently small, there exists a unique limit cycle in the vicinity of the origin. (This conclusion is also true even for system (III) provided that 8[m(l +n)- o:(b+ 2l)] < 0; but for (III), 8 varies and does not form a family of rotated vector fields.) Since this limit cycle monotonically expands and covers some neighborhood as 161 increases from zero (if there is more than one limit cycle, some limit cycle may monotonically contract; but we shall see later that no matter how large 161 is, the limit cycle, if it exists, is unique), from the nonintersecting theorem of §3 (Theorem 3.2) we know that, when 8m(l + n) > 0, (I) does not have a limit cycle. The theorem is completely proved.(6 ) (6)When 6m(l+n) > 0, the fact that (I) does not have a limit cycle can be seen from the comparison theorem for differential equations. For example, suppose m(l + n) > 0; then we have shown that the origin is an unstable focus of (I)o=O• and the system does not have a limit cycle. Hence the origin of (I)o=O is also an unstable focus, and, when t increases, the
§12. QUADRATIC SYSTEMS OF CLASS I
271
The following work is to remove the hypothesis "lhl is sufficiently small" in Theorem 12.5; that is, we have to prove that if system (I) has a limit cycle, it is unique. In China this problem had been completely solved by 1967 (see [16], [195], [196], and [197], although the last two were only published in 1975 and 1978 respectively). In the Soviet Union there were [118] and [198]. Our method of proof shows that in order to solve the problem of uniqueness of a limit cycle for equations of class I, the uniqueness theorem of [111] is all we need. Since m f. 0 is a necessary condition for existence of a limit cycle of equations of class I, without loss of generality, we shall assume that(1) m
= 1,
l + n > 0,
h < 0.
(12.24)
LEMMA 12.1. When 1) n > 0 and h + 1/2n ~ 0; or 2) n ~ 0 and h + l ~ 0; or 3) l ~ 0 and h + n ~ 0, (I) does not have a closed or singular closed trajectory. PROOF. 1) We adopt the method similar to Theorem 1.13 of §1. Take
1- 4nl ( x )] M(x,y) = exp [ 2n y + 2n '
B(x,y) = 2nM(x,y),
P=Q2B-P2M = exp [ 1
~:nl (y + 2:)] {2nx- ( -y + hx + lx 2 + xy + ny 2)},
Q = -P2B
= -2n( -y +ox+ lx 2 + xy + ny 2) exp [ 1 ~:nl
(y + 2:)].
If (I) has a closed or singular closed trajectory r in the neighborhood of the origin, it must be positively oriented; thus from Green's formula we have
But it is easy to compute the value on the left side of the above formula:
1i(BQ2- P2M)P2dt- BP2 · Q2dt =-
i
MPfdt < 0,
trajectories of (1)6>0 starting from some regular point lie on the outside of the trajectories of {1)6=0i hence (1)6>0 also does not have a limit cycle. (1)In order to study any quadratic system, we can in general apply a suitable similarity transformation of :z:, y, and t to three fixed nonzero coefficients of system {III) for some fixed values.
THEORY OF LIMIT CYCLES
272
and the value of the right side:
- j j (1 + 2n6)M(x, y) dx dy > 0. intr
This contradiction shows that r does not exist.( 8 ) 2) When n :::; 0, l > 0 must hold. Taking a Dulac function B(x, y) (1- x)- 1 for system (I) we get !_(BP.)
ax
2
!_(BQ ) = 6 + l + ny2 -l(x- 1)2 2 (1- x) 2 •
+ ay
=
(12.25)
When 6 + l :::; 0, the right side of the above formula keeps a constant sign :::; 0, and the line 1- x = 0 is a line without contact of (I). Hence, the conclusion of the lemma holds. In fact we can relax the condition on 2) to ben:::; 0 and 6(1 + n6) + l :::; 0. But in this case the proof is very complicated, and so we omit it here (it can be found in the first edition of this book). 3) When n > 0 and l :::; 0 but 6 = 0, the trajectories of (I)o=O are shown in Figures 12.5 and 12.7. From this we can see that the separatrix passing through the saddle point (0, 1/n) from the right lies below the line through the saddle point and touches the line tangentially at that point. Now if 6 + n = 0, then the tangent line to the separatrix at (0, 1/n) is ny+x = 1. Let V = ny+x. We compute the rate of change of V along the trajectory of (I), and get
I
dVd = (6 + n)x + lx 2 = lx 2 :::; 0. t V=l From this we see that as 6 decreases from 0 to -n, in Figures 12.5 and 12.7 the separatrix entering (0, 1/n) from its right not only turns to the outside of the!separatrix leaving (0, 1/n) from its left, but also has turned to the upper part of the line ny + x = 1. Thus in this case the limit cycle, of course, does not exist. When 8 < -n, the same is true. The lemma is completely proved.
COROLLARY. When n > 0 and l :::; 0 but n have a limit cycle as long as 6 :::; -1/-/2.
+ l > 0,
system (I) does not
PROOF. This is so because there is· a number not less than 1/-/2 between nand 1/2n. LEMMA 12.2. Suppose 6 system of equations
< 0, l ~ 0, 1- al > 0, and a -l > 0. Then the
dxjdt = -y + 8x + lx 2 does not have
+ xy, dyjdt = x + ay a closed trajectory when 8 + l = 0.
(12.26)
( 8 )This result was first obtained in [194], but the proof was rather complicated, and also there was an additional condition I > 0. Here we adopt the method of proof in [27],
273
§12. QUADRATIC SYSTEMS OF CLASS I PROOF. Apply the transformation x = 1- e-x, y =
y. Then (12.26)
becomes dyjdt = 1- e-x+ ay. Change the system again into a second order differential equation and get
i +(le-x- a)i + (1- al)(1- e-x) = 0. Finally, changing it to the Lienard plane yields
-
~
~
dt = -z + (ax+ le-x - l),
-
dt = (1 - al)(1- e-x).
(12.27)
Now suppose (12.27) has a closed trajectory r. We may as well assume r is the one closest to the origin. Since a -l < 0, it is easy to see that (0, 0) is a stable focus or nodal point of (12.27). However, on the other hand, computing the integral of the divergence along r once, we get
i
(a -le-x) dt
= =
that is,
r
i i
(a -l) dt +
i
l(1- e-x) dt
(a - l) dt < 0;
is also a stable cycle, which is impossible. The lemma is proved.
LEMMA 12.3.
When l > 0, n > 0, and o ~ -1/l, the system of equations dxjdt = -y- F(x),
dyjdt = g(x)
(12.28)
does not have a limit cycle in the half-plane x < 1/n, where -1
F(x) = p[(lx + ol + 1)ePROOF. When
o=
l X-
ol- 1],
-1/l, (12.28) becomes
dxjdt = -y + (xjl)e-lx, Let
dyjdt = (x- nx 2 )e- 21 x.
(12.29)
r
1 H(x, y) = 2y2 + Jo (s- ns2)e-2ls ds.
Then H(x,y) = Cis a family of closed curves containing (0,0) which is a family of closed trajectories of the system of equations dxfdt = -y,
(12.30)
Computing dH/dt along the trajectory of (12.29), we get dH _ aH dx dt - ax dt
+
8H dy _ x 2 (1 ) -3lx O ay dt - l nx e > ,
1 when x < -. n
THEORY OF LIMIT CYCLES
274
Hence (12.29) does not have a limit cycle. Since (12.28) can be obtained from (I) by a change of variables,( 9 ) and (I) forms a family of generalized rotated vector fields with respect to the parameter 6, (12.28) does not have a limit cycle when 6 < -1/l. The lemma is completely proved.(1°} THEOREM 12.6. For arbitrary 6,l, and n, system (I) has at most one limit cycle. PROOF. By the previous discussion, we can assume that (12.24) holds; that is, l + n > 0 and 6 < 0. In the following, we consider several cases. (1) The case when n = 0 or l = 0. When n = 0, (I) becomes
= -y + 6x + lx 2 + xy, dyjdt = x. y = -y', and t = -r. Then (12.31) becomes
dxjdt
Let x
= 1- e-x',
dx' I I dr = -y - [(6 + l)ex - (6 dy' dr
I
+ 2l) +le-x] =
(12.31)
~y'- F(x'),
I
= 1- e-x = g(x'),
where g(x') = 1- e-x' is continuous, x'g(x')
G(±oo)
= fo±oo g(x) dx = +oo.
Moreover, since f(x') = F'(x') continuous, f(O) = c5 < 0, and
=
(c5
_:!:___ [f(x')] = e-x'[-6
dx'
g(x')
> 0 when x' # 0, and
+ l)ex'
-le-x' we see that f(x') is
+ (6 + l)(ex'- 1) 2 ] O (1-e-x')2 > '
since by Lemma 12.1 we know that when a limit cycle exists, c5 + l > 0. Thus, f(x')/g(x') is a nondecreasing function in both (-oo,O) and (O,+oo), and by Theorem 6.4 in §6 we can prove the uniqueness of a limit cycle.( 11 ) When l = 0, (I) becomes dx/dt
= -y + 6x + xy + ny2 ,
Let x' = y, y' = x - 6y dx' /dr
h 2 , and r = = -y'- F(x'),
dyjdt = x.
(12.32)
-t. Then (12.32) becomes dy' /dr
= g(x'),
(12.33)
(9)See system (13.34). (IO)See footnote 6. (11 )In fact what we have used is only a special case of it; that is, the uniqueness theorem in [111], whose conditions can be seen in the remarks after Theorem 6.4. Whenever we mention Theorem 6.4 in this book, we mean the uniqueness theorem in [111] unless stated otherwise.
§12. QUADRATIC SYSTEMS OF CLASS I
275
where F( x') = 8x' + x'2 /2 and g( x') = x' - nx'2. It is easy to see that if a limit cycle exists for (12.33), it must lie in the half-plane x < 1/n. We note that 1 + n8 > 0, and hence, as in the case n = 0, it is easy to verify that f(O) = F'(O) = 8 < 0 and f Jg is a nondecreasing function in ( -oo, 0) and (0, 1/n). This proves that (12.33), and hence (12.32), has at most one limit cycle. (2) The case when l > 0 and n > 0. Apply the transformation X=
Uelx',
dtjdr = e-lx'
y=x',
to (I). We get
au = dr
( -x'
+ nx'2 )e- 21 x' + (8 + x')ue-lx''
dx' dr
=u.
x', we get x' = (-x' + nx'2)e- 21 x' + (8 + x')e-lx' x'.
Changing it again into a second order equation in
Finally transforming it into the Lienard plane, and changing r to-T, we get dx' dr
d
= -y' + lx' + 81 + 1 e-lx' _ 81 + 1 [2
I
= -y' _ F(x')
[2
'
(12.34)
I
_]f__ = (x' - nx'2)e-2lx = g(x'). dr If a limit cycle exists for (12.34), it must lie in the half-plane x Then we have f(O) = F'(O) = 8
< 0,
x'g(x') = x' 2 (1- nx')e- 21 x'
< 1/n.
> 0 when x' =/:: 0.
To prove that (I) has at most one limit cycle, we only have to show that in (-oo, 0) and (0, 1/n), the inequality
_.!!.._ [f(x')] = W(l, x') elx' > 0 dx' g(x') (x' - nx' 2 ) 2
(12.35)
holds, where
+ (l + n- ln8)x' 2 + (2n8 + l8)x'- 8 = ( -8- x')(lx'- 1)(nx'- 1) + (1 + n8)x' = W1(x') + W2(x'), = (-8- x')(lx'- 1)(nx'- 1} and W2(x') = (1 + n8)x'.
W(l, x') = -lnx'3
(12.36)
with W 1(x') In order to prove (12.35), except for the case when x < 1/n, l > 0, and n > 0, owing to Lemma 12.1 and Lemma 12.3, it is sufficient to prove that W(l,x') > 0 under the conditions -1/n < 8 < 0 and -1/l < 8 < 0. i) When x' ~ 0, from the first line of (12.36) we see at once that W(l,x') > 0.
THEORY OF LIMIT CYCLES
276
ii) When 0 < x1 ~ -6, since W1 (x1) > 0 and W2(x 1) > 0, W(l, x1) > 0. iii) When -6 < x1 < 1/n, we note that W(l, -6)
a
= W(O, -6) = -6(1 + n6) > 0, a
ax1W(l, -6) = -l6(1 + n6) > ax1W(O, -6) = 0;
hence there is an
E
> 0 such that when -6 < x1 < -6 + E we have W(l, x1) > W(O, x1).
(12.37)
But W(l, x1) - W(O, x1) = lx1( -6 - x1)(nx1 - 1) has only the three zeros x1 = 0, x1 = -6, and X1 = 1/n. Hence (12.37) holds in the whole interval -6 < x1 < 1/n. It is clear that W(O, x1) = nx12 + 2n6x1- 6 = n(x1+ 6) 2 - 6(1 + n6) > 0;
hence when -6 < x1 < 1/n we have W(l, x1 ) > 0. Combining i), ii), and iii), we get (12.35) at once; (3) The case n < 0. Apply the transformation x = x1+ >..y1,
I
y = y'
(12.38)
= -1+v'1-4nl >0 2l
(12.39)
where the number >.
is a positive real root of
(12.40) Then (I) changes to dx1/dt = (6- >..)x1+ (6>..- >.. 2 - 1)y1 + lx 12 + (2l>.. + 1)x'y1, dy1/dt = x1+ >.y1.
(12.41)
Again let Xt=
2).l+1 I >..2-6>..+1x,
2).[+1
I
Yt= v'>..2-o>..+1y,
7'=VA2-6).+1t; (12.42)
then (12.41) changes to dxl dr = = ~~
lv'>..2- o>.. + 1 2 x1 + 2 ).[ + 1
6- >..
V..\ 2 _ 6..\ + 1 Xt - Yl + -yl + 6'x1 + l'x~ + XtYll
dr = Xt + v..\2
A
_
o).
XtYl
(12.43) I
+ 1 Yl = Xt +a Yt,
§12. QUADRATIC SYSTEMS OF CLASS I
277
where
6'
=
a'=
6 -,\
v'-\ 2 -6,\+1 ,\
y',\2-6,\+1
Note that
<0
'
(12.44)
> 0.
6, + l'
= (,\l +
1)6 + (l + n) (2,\l + 1)y,\2- 6,\ + 1'
(12.45)
and when 6 = 0 we have 6' + l'
a , ,
86 (6
+ l) =
> 0. Moreover, A3 l- A2 l6 + 3,\l + ,\ 2 2(2,\l + 1)(,\2- 6,\
6,\ + 2
+ 1)3/2
> o.
(12.46)
We can see that 6' + l' is a monotonically increasing function of 6; as 6 goes from 0 to negative values, 6' + l' monotonically decreases. At the same time
a' _ l'
=
,\l6 - (l + n) (2,\l + 1)y',\2- 6,\ + 1
< O.
From Lemma 12.2 and (12.45) we know that when 6' = -(l + n)f(,\l + 1) the system (I) does not have a closed trajectory. But (I) forms a family of generalized rotated vector fields with respect to the parameter 6; hence (I) does not have a closed trajectory when 6 ~ -(l + n)f(,\l + 1).(1 2 ) Thus, from (12.46) we know that (12.43) does not have a closed trajectory when {/ + l' ~ 0. Hence, later on, in order to study the uniqueness of a limit cycle of (12.43), we may as well assume that 6' + l' > 0. In order to apply Theorem 6.4 in §6 to prove the uniqueness of a limit cycle of system (I), we can first let Yl = -jj and x1 = x in (12.43), and change the sign of r so that the origin becomes unstable. Thus we get dx/dr = -ii- 6'i -l'x 2
+ xjj,
djj/dr
= i - a'jj.
(12.47)
Then, similarly to Lemma 12.2 for system (12.26), we can apply the same change of variables to (12.47), change it again to a second order differential equation, and finally transform it into the Lienard plane to get
~;
= -y-
((6' + l')eJ: + l'e-x + a'x- 6'- 2l'J = -y- F(x),
~~ = (1- a'l')(l- e-J:) + a'(6' + l')(ex -1) = g(x)
(12.48)
(we still use the independent variable t and the dependent variables x and y). ( 12 )See
footnote 6.
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THEORY OF LIMIT CYCLES
Now we only have to examine whether the conditions of Theorem 6.4 are satisfied. We have
W + l')ex -l'e-x +a',
f(x) =
f(O) = a' + o' < Ol
(12.49)
where f(x) = F'(x). Moreover,
~ [f(x)] dx g(x)
{
(1- 2a'l')(o' + l')(ex/ 2
-
e-x/ 2 ) 2
+ (a'l'- 1)(a' + o')e-x- a'(O' + l')(ai + o')ex [(1- e-x)(l- a'l') + a'(O' + l')(ex- 1)]2
-
}
>0 (12.50)
since
o' +l' > o,
1 1 - 2a'l' = 2Al + 1
-Al-l
> 0' a'l' - 1 = 2Al + 1 < 0' a' + vr:l < 0.
In addition, from the equality g(x) = (1-e-x)(1-a'l' +a'(O' +l')ex) we know that the locus of g(x) = 0 is the vertical line x = 0; hence xg(x) > 0 when x =f. 0. Combining the above results, we know at once that all the conditions of the uniqueness theorem of [111] are satisfied. (4) The case l < 0. Here we still take A ( < 0) as indicated in (12.39). Then (12.41)-(12.49) can be similarly established. For example, the transformation (12.42) still has meaning, because A2 -OA+ 1 is positive when 161 is very small, and when A2 - OA + 1 = 0 system (12.41) has an integral line x' = 0 passing through the origin, and it clearly does not have a closed trajectory. Hence in order to study the uniqueness of a limit cycle, we may as well assume that A2
-
OA + 1 > 0 or
o
> (A 2 + 1)/A.
(12.51)
Using (12.51), it is easy to see A3 l- A2 lo + 3Al + A2
-
OA + 2 =(A- 6)(A 2 l +A)+ 2(1 + 2Al)- Al
= -n(A- 6) + 2(1 + 2Al)- Al = -A(l + n) + 2(1 + 2Al) +no
> -A(l+n)+ = 1 + 2Al
nA 2 + n n -lA 2 A +2(1+2Al) = A +2(1+2Al)
> 0.
Hence (12.46) also holds. Finally, since clearly a' - l' < 0, Lemma 12.2 also holds. By Lemma 12.1, we can assume f/ > -1/2n, so that the factor on the right of (12.50) has a positive minimum; therefore (12.50) also holds. The theorem is completely proved. With this theorem and the theory of rotated vector fields, we know at once that when om(l + n) < 0, as 161 increases from zero the unique limit cycle which is generated because of change of stability of the origin will expand monotonically; finally, when 6 takes the value 6* = f(l, m, n), it will meet
§12.
279
QUADRATIC SYSTEMS OF CLASS I
4 l
FIGURE 12.8 a finite or an infinite saddle point to become a separatrix cycle and then disappear. From the several cases in Figures 12.1-12.7 it is easy to see that for n ~ 0 there are two infinite saddle points and a section of the equator on the separatrix cycle, but for n > 0 there is only one finite saddle point (0, 1/n) on the separatrix cycle. = f(l, m, n) is algebraic or transcendental, and Whether the function whether the 'Separatrix cycle is an algebraic curve or a transcendental curve, are two problems which have not yet been solved. In this area, I. G. Rozet [199] did some calculations. He transformed (I)m=l to
o•
~~
= -y +ox+ px 2
+ qxy- y 2 ,
dy dt
-=X
and then, using the results of many calculations, sketched the branch curve in Figure 12.8 (the case n = 0) and the branch surface in Figure 12.9. However, even in the case n = 0 he did not obtain an approximate representation formula for the curve fJ = f (l). There are a few results on the range of variation of fJ to guarantee existence of a limit cycle of (I) by qualitative methods, some of which can be found in the exercises for this section. REMARK 1. Equations of class I can be directly transformed to a secondorder nonlinear equation; hence their practical value is very high. REMARK 2. By Theorem 12.4 we know that a focus of equations of class I can at most be a first-order fine focus, but for equations of class II it can be a third-order fine focus. (See Exercise 4.) REMARK 3. In [200] and [201], Russian mathematicians gave a new proof of Theorem 12.4. This method of proving nonexistence of a closed trajectory was generalized and applied in [202] and [203]. In [202] a result equivalent to part 1 of Lemma 12.1 was obtained. When l > 0 it is better than 1), but when l < 0 it is not as good as 1).
280
THEORY OF LIMIT CYCLES
FIGURE 12.9
Exercises 1. Prove the inference before formula (12.4). 2. Discuss the relationship between the two methods of classification when (12.9) has zero as its only real root and when m = n = 0. 3. Prove Theorem 12.1. 4. Use Bautin's method of §9 to prove that equations of class II can have three limit cycles near the origin. 5. Prove that (12.17) of Theorem 12.2 can be replaced by (12.20). 6. Prove the validity of Figures 12.1-12.7 and sketch the global figure of (I)h"=O when m(l + n) = 0. 7. Complete the proof of condition 2) of Lemma 12.1. 8. Suppose that in (I), l = 0, m = 1, 8 < 0, and 0 < n ~ 1/3, ..j-8/n ~ n/2; then show that there does not exist an unstable limit cycle near the origin. 9. Suppose 4ln = 1 in (I)m=l· First prove that when 8 = -2l = -1/2n, (I)m=l does not have a limit cycle, and then show that if a limit cycle exists, it must be unique [195]. 10. Prove that when a: = f3 = 0 system (III) does not have a limit cycle, and when m = n = 0 limit cycles may exist. 11. Use a method similar to the proof of Lemma 12.2 to prove that system (I) does not have a limit cycle when n > 0 and 8 ~ -1/n.
§13. Global Structure of Trajectories of Equations of Class II without Limit Cycles
In the previous section we have seen that there are two main problems for equations of class I: one is to determine the range of variation of the coefficient of x (i.e., o) in P2 (x, y) which guarantee the existence of a limit cycle, and the other is the problem of uniqueness of a limit cycle. For the equations of class II, in addition to the above two problems, there are three other important questions: 1) H we already know that the system does not have a closed trajectory, how can we determine the global structure of its trajectory? 2) When the system has two singular points of index +1, how do the generation and disappearance of limit cycles affect each other? 3) When there may be more than one limit cycle in the neighborhood of a singular point, how can we solve the problem of having at most two or at most three limit cycles? In this section we start by studying the first problem. It is easy to establish the following fact. THEOREM.
The system
dx/dt
= -y + mxy + ny2 ,
dyjdt = x(1 + ux)
(13.1)
has one or two centers when mn = 0, and does not have a closed trajectory or a singular closed trajectory when mn # 0.
In the proof we take the Dulac function B(x, y) = 1/(1- mx).
(13.2)
The details are omitted (consider it as an exercise). When mn = 0, the global structure of the trajectory of (13.1) is easily determined (form= 0, see Figure 13.5; for n = 0, consider it as an exercise). Now suppose mn # 0. Then after a suitable affine transformation of x, y, and t, we can assume n = -1 and a < 0. Thus we obtain the system of equations dxjdt = -y(l + y- mx), 281
dyfdt = x(l +ax).
{13.3)
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THEORY OF LIMIT CYCLES
This system has four singular points: 0(0, 0) is a focus and is stable when m > 0, unstable when m < 0; M(O, -1) is a saddle point, and the others are N( -1/a, 0) and R( -1/a, -(a+ m)/2). It is easy to compute that the characteristic roots of the linear approximate system of (13.3) at N are ±J(a + m)/a; hence when m <-a, N is a saddle point, and so R is a focus or a nodal point, lying below N; when m = -a, R = N becomes a higherorder singular point; and when m > -a, N is a focus and R is a saddle point lying above N. In the following we first study the global structure of the trajectory of (13.3) for the case m < -a. First, we study the direction of crossing of the trajectory on the line MN : x + yja = -1/a. Let U = x + yja. Then on MN we have y -dU = ( x +dt a =
1)
+ -a
(m- ~+a)
(x- ay) +
(m- -1+a) a
xy
xy.
From this we can see that MN is a trajectory for m = 1/a- a, and for # 1/a - a this line is divided into three segments by the two singular points M and N, and on each segment all the trajectories have the same direction of crossing. Next we study the infinite singular points. Transforming (13.3) into homogeneous coordinates and letting x = 1 and dt / dr = z, we get
m
~; = z(yz + y2 -
my),
~~ = z +a+ y2 z + y3 -
my 2.
(13.4)
From this we can see that the y-coordinate of the infinite singular point Ai(1, Yi, 0) satisfies the equation y3 - my 2 + a = -0,
(13.5)
when a(27a- 4m3) < 0 or 27a > 4m3, (13.5) has three different real roots, and the corresponding infinite singular points are Ai(1, Yi, 0) (i = 1, 2, 3). It is easy to see that we should have Yl < Y2 < 0 < Y3. If 27a = 4m3, then A1 = A2 becomes an infinite higher-order singular point; if 27a <4m3, then A1 and A2 disappear, and only A3 remains. It is easy to compute the two characteristic roots of the linear approximate system at the singular point Ai :
Using (13.5), we see that
>.1 = -afyi < 0 if i
= 1, 2;
> 0 if i
= 3.
(13.6)
§13. CLASS II EQUATIONS WITHOUT LIMIT CYCLES
283
Moreover, A2 represents the slope of the curve u = yg- my2 +a (a< 0, m < -a) in the (y,u)-plane at its point of intersection (Yi,O) with the y-axis. Hence, for m < 0,
A2
>0
for i = 1, 3;
< 0 fori= 2.
(13.7)
When m ~ 0, there is only one Ag(1,yg,O), for which we have A2 > 0. From (13.6) and (13.7) we know that At is a saddle point, A2 is a stable nodal point, and Ag is an unstable nodal point. Finally, we study the direction of crossing of the trajectory through the line M'Ai whose equation is YiX- y- 1 = 0. Let V = YiX- y. Then on '1YAi we have dV
dt
= (YiX- y -1) (-~ + YiY) + (myi- ~- Yl) xyYi Yi
(1 +~)X
Yi+a =---x. Yi
Yi
{13.8)
From this we can see that M~ is a trajectory of {13.3) if and only if Yi = -a; that is, yg = -a > 0. At the same time, -a should satisfy {13.5), and so m = 1/a- a; that is, MAg and MN coincide at m = 1/a- a to become a trajectory. In order to determine the sign of the right side of (13.8), we note that equation {13.5), satisfied by }i, can be written as (Yi
+ a)g -
(3a + m)(yi + a) 2 + (3a 2 + 2am)(yi +a) - a 2 ( m-
~ +a) = 0.
Under the conditioll m < -a, we have 3a + m < 0 and 3a 2 + 2am > 0. From this we can see that 1. Yi +a< 0 fori= 1, 2, 3 when m < 1/a- a, and 2. Yt +a< 0, Y2 +a< 0, and yg +a> 0 when m > 1/a- a. Thus the direction of crossing of the trajectory through Mfii can be determined from (13.8). Moreover, when m '# 1/a- a, we can also determine the relative positions of N and R with respect to the line 'MAi as follows: 1. N is between MA2 and MAg when m > 1/a - a; 2. N is above MAg when m < 1/a- a; and 3. R is always below all M'fli. The picture in Figure 13.1 is the distribution of trajectories for the case m < 1/a- a, 27a >4m3 • Using the isoclines P2 (x,y) = 0 and Q2 (x,y) = 0 and the directions of trajectories crossing the lines MN and 'MAi, we can determine completely all the directions of all the separatrices passing the saddle points M, N, A 1 , and A 1 with the exception of two. For example, the separatrix starting from M and entering the right half-plane must lie
284
THEORY OF LIMIT CYCLES
1 m
FIGURE 13.1
.!.-cs<m
FIGURE 13.2
between MA3 and MA2, because if it lies above MA 3 (or below MA2), then the adjacent trajectory below it (above it) will also be above MA3 (or below MA2) which is impossible. Thus the separatrix will finally enter A2· Moreover, the separatrix entering N from the upper right side must come from A3, the separatrix entering M from the lower right side and entering A1 must come from R, the separatrix leaving N to the lower right side must enter A2, the separatrix leaving M to the lower left side and the separatrix coming from A1 must all enter A:3 • These are quite obvious. Now we show that the separatrix entering N from the lower left side must come from 0 · If it comes from A3, then the separatrix starting from N to the upper left side
§13. CLASS II EQUATIONS WITHOUT LIMIT CYCLES
285
can only surround 0; but 0 is an unstable singular point when m < 0, and so a stable limit cycle will appear in the vicinity of 0, which is impossible. Next, the two separatrices on the left of N cannot coincide to form a separatrix cycle surrounding the point 0, because as we mentioned in Remark 2 after Theorem 12.1, system (13.3) does not have a singular closed trajectory.(!) Denying the above two possibilities, and noting the direction of the trajectory crossing MN, we see that the separatrix entering N from the lower left side must come from 0. In order to determine the global structure of the trajectory of system {13.3), we still have to know the relative positions of the separatrix from N going to the upper left side and the separatrix entering M from the upper left side. Making use of the method mentioned above, under the conditions m < 1Ja- a and 27a > 4m3 there is no way for us to be sure of their relative positions. Here there exist two possibilities; that is, 1. Under the conditions m < 1/a-a and 27a >4m3 , the relative positions of these two separatrices are in fact completely determined, but we cannot find any method to prove it. 2. Under the conditions m < 1/a- a and 27a > 4m3 , there may appear three different cases for the relative positions of these two separatrices. Which of these is correct? This is the first problem we have to solve in this section. Before we solve this problem, we have to see the global structure of the trajectories of (13.3) when m and a satisfy other similar conditions.(2 ) Using the previous method, we can draw the figure of the global structure (Figure 13.2) under the conditions 1/a-a < m < 0 and 27a >4m 3 , and the problem of Figure 13.1 does not appear. The main reason is that the direction of (1) We explain more clearly the proof of nonexistence of singular closed trajectories in this specific case. Comparing the absolute values of the slopes of the trajectories of system (13.3) at the points Ft(:z:,y) and F2(x,-y), where y > 0 and F1 and F2 all lie above the line 1 - m:z: + y = 0, we get
jdxdyj
F1
l:z:(1 + a:z:)l = y(l + y- m:z:)
l:z:(l + a:z:)l
< y(l- y- m:z:)
Idyl = dx F 2
•
Hence, by the comparison theorem, we see that for the closed trajectory or the singular closed trajectory above the line 1 - m:z: + y = 0 the curve symmetric to the part of r above the :z:-axis with respect to the :z:-axis will completely contain the part of r between points and the :z:-axis and below the :z:-axis. Suppose G is an inner region of r. Then it is easy to see that
II(~~+ aa~2) G
dxdy=
II
mydxdy>O,
G
which contradicts Bendixson's theorem. Hence r does not exist. (2)For the coexistence of these conditions, refer to Figure 13.11.
286
THEORY OF LIMIT CYCLES
m .... .l-a
"
m
"
(a)
27a<4m3
(b)
FIGURE 13.3 the trajectory crossing the line segment M N is different from Figure 13.1. Similarly, we can draw the global figure of the trajectory when m = 1/ a-a < 0 and 27a >4m3 ; this can be done by changing MN in Figure 13.2 to become a trajectory, and the directions of other separatrices are the same as in Figure 13.2. Hence we do not draw this figure. If in the above three cases we keep the remaining inequalities, but change 27a > 4m3 to 27a = 4m3 , then the figure of the corresponding global structure can be obtained from the original figure by letting At= A2 ; here At= A2 is a semisaddle nodal point. If we again change 27a = 4m3 to 27a < 4m3 (but still assume m < 0), then At = A2 disappears, and the problem of how to determine the relative positions of the separatrix coming from N and the separatrix going to M in the fourth quadrant is created. However, since now R is a stable focus or nodal point, and we already know that system (13.3) does not have a closed trajectory, the relative positions of these two separatrices can still be determined when m < 1/a- a, as indicated in Figures 13.3(a) and (b). Only in the case when 1/a- a < m < 0 and 27a < 4m3 does the problem ofrelative positions of the above two separatrices arise. For convenience, let lt, lt, l1, and 12 denote the four separatrices passing through M, and let Lt, Lt, L1, and L2 denote the four separatrices passing through N (when m <-a) orR (when m >-a), as indicated in Figure 13.4. Now we change the condition m < 0 to m = 0. Then system {13.3) has two centers and one infinite singular point A3. The conditions m:;; 1/a- a become a :;; -1. Figures 13.5(a), (b), and (c) show the global structure of system (13.3) in three cases: m = 0 and a < -1, a = -1, and a > -1 respectively.
§13. CLASS II EQUATIONS WITHOUT LIMIT CYCLES
11
Lt
Li
0
:li
L:
Li
li
287
B
z:
FIGURE 13.4
Next, suppose 0 < m <-a. Then 27a <4m3 , but there are three possible ways of ordering between a- 1/a and m. Since the stability of 0 and R has been changed, now the figures corresponding to 13.1 and 13.2 are 13.6 and 13. 7. In Figure 13.6, only the relative positions of Lt and cannot be determined; in Figure 13. 7, only the relative positions of L! and l! cannot be determined. But if m = 1/a- a, then the global structure of the trajectories can be completely determined as indicated in Figure 13.8. H we suppose m = -a, then R = N is a higher-order singular point; then m > 1/a- a and 27a <4m3 must hold. Hence it is only on Figure 13.9 that the relative positions of L! and l! cannot be determined. Finally, suppose m > -a. Then m > 1/a- a and 27a <4m3 still hold. Here we still denote the singular point above the line 1 + ax = 0 by N, which is a saddle point; the singular point below the line is R, which is an unstable focus, the global structure is as shown in Figure 13.10. In Figure 13.10, except for the relative positions of Lt and L! and l! cannot be determined, and we may still have the problem of determining the relative positions of Lt and l! because Lt may possibly cross through the left side of MN. In order to show clearly and completely whether the indeterminate nature of the relative positions of some separatrices in the above figures is due to the nonexistence of this case or the deficiency of the method used, we now introduce an (a, m)-parametric plane, and a bifurcation curve in this plane. The so-called bifurcation curve is a curve in the (a, m )-plane such that for any point (a*, m*) in this curve, the graph of the trajectory of its corresponding system (13.3) is structurally unstable. We should note that by a structurally unstable system in this section we mean that kind of system (13.3) for which the topological structure of the trajectories on the projective plane can change
zt
zt,
THEORY OF LIMIT CYCLES
288
m-o, a<-1
m-0, a--1
(a)
(b)
m ... o, a>-1
(c) FIGURE 13.5
when the coefficients a and m on its right sides vary slightly. Hence the definition is slightly different from the one given in §8. Since we already know that (13.3) cannot have a limit cycle, the unstable structure can only appear in the following cases: 1. System (13.3) has a center. 2. System (13.3) has a higher-order singular point (finite or infinite). 3. System (13.3) has a separatrix connecting two saddle points.( 3 ) We know that case 1 can only appear when m = 0; hence the horizontal axis is a bifurcation curve in the (a, m)-plane. Since we have assumed a< 0, (3)Strictly speaking, under the definition of structural stability in this section, whether case 3 is a sufficient condition for structural stability has not yet been definitely proved.
§13. CLASS II EQUATIONS WITHOUT LIMIT CYCLES
1
O<m< -a, m
O<m-..!.-a<-a CJ
FIGURE 13.8
289
1 O<m<-a, m>--a a
FIGURE 13.7
1 m- -a>-a, 27a<4m' a
FIGURE 13.9
we may as well just confine ourselves to the left half-plane (Figure 13.11). The global figure of system (13.3) corresponding tom= 0 has three different topological structures as seen in Figures 13.5(a), (b), and (c). Case 2 can only appear when m =-a (with R = N) or 27a =4m3 (with At = A2 ). Hence the line m = -a and the curve 27a = 4m3 are bifurcation curves. If we also include system (13.3) for a= 0 in our considerations, then
290
THEORY OF LIMIT CYCLES
m> -a>.!.-a, 27a<4m*
a
FIGURE 13.10
FIGURE 13.11
the line a= 0 is also a bifurcation curve, and N = A1 = A2, R = Ag (when m > 0), or N = A2 = Ag, R = A1 (when m < 0). Case 3 is the most complicated one. According to Figures 13.1-13.10, we can see that Lt always comes from Ag, 12 always runs to Ag, and these two separatrices cannot coincide with other separatrices. Also from Figures 13.1 and 13.2 we can see that when A1 and A1 exist, the separatrix entering A1 must come from R. R is not a saddle point, and the separatrix leaving A1 must enter A3 ; hence these two separatrices -eannot coincide with other separatrices. There are six separatrices Lt, L1, Li, zt, zt, and 11 remaining; among them, any two separatrices with different stability (this means leaving or entering the saddle point as t increases) may coincide. Hence we can divide case 3 into two subcases: 3a. A separatrix starting from one saddle point and returning to the same saddle point. 3b. A separatrix starting from one saddle point and returning to another saddle point. The cases Li = Li I Lt = Li I zt = l1, and zt = zt belong to case 3a (for simplicity, from now, = means "coincides with"), but from Remark 2 after Theorem 12.1 we know that all these cases can only occur when m = 0, and so there is no new bifurcation curve to be added.
§13. CLASS II EQUATIONS WITHOUT LIMIT CYCLES
291
There are five different possible cases belonging to 3b:
lf = Lz, L! = l!, Lt = lt, Lt = l!, L! = lt. From Figure 13.5, we know that when m = 0 and a = -1 the equalities L! = l!, Lt = zt, Lz = lf appear simultaneously; but when m = 0 and a < -1 these three equalities cannot hold, nor can they form= 0 and a> -1. Moreover, when a< -1, the relative positions of these three pairs of separatrices are just opposite to the case when a> -1. From now on let Ct.C2, and Ca denote the bifurcation curves in the (a, m )-plane such that L! = l!, Lt = zt, and Lz = lf respectively. Then at least one branch of C1, C2, and C3 must pass through the point P( -1, 0) so as to separate the line segment (-oo, -1) on the aaxis from the segment (-1,0) in which two ends of C1 and Ca should go to infinity; and the upper half-branch of c2 must go to infinity;(4 ) for otherwise we can use a continuous curve 8 in the second or the third quadrant to connect a point Q* in the segment (-co, -1) on the negative a-axis to a point Q' on the segment ( -1,0) without intersecting C1 , C2 , or Ca. Thus when a point moves from Q* to Q' along 8, the relative positions of three pairs of separatrices of (13.3) remain unchanged in the corresponding (x, y)-plane, which is impossible. But it is sufficient for the lower branch of C2 to lie above the curve 27a = 4m3 , since for any point (a, m) below the curve 27a = 4m3 its corresponding system (13.3) has three infinite singular points. From Figures 13.1 and 13.2, we see that then Lt enters A2, but zt comes from R. It is clear we can take Lt to surround zt from its outside. From the discussion at the beginning of this section, we see at once that the bifurcation curve Ca which makes Lz = lf has only a unique branch m = 1/a - a (the part of a > 0 is not considered); that is to say, if L2 = lf, then they must coincide on the line segment liN, for otherwise, as in Figure 13.12, a contradiction will arise. Next it is easy to prove that in the unbounded angular region surrounded by the upper half-branch of Ca and the line segment (-co, -1) on the negative a-axis there is no locus of C1 , for otherwise, as indicated in Figure 13.13, L! = l! will appear, 0 will be stable, and all the trajectories meeting the line segment liN will always cross it from left to right, which is impossible. Similarly, we can prove that the unbounded angular region in the third quadrant with the same vertical angle as the above angular region does not contain any locus of C 1 • Also, below the line m = -a, except for the above two angular ( 4 )They
cannot reach the positive m-axis, for the reasons given in Theorem 13.1.
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THEORY OF LIMIT CYCLES y
N
M
M
FIGURE 13.13
FIGURE 13.12
regions, the remaining two angular regions with the same vertical angle do not contain the locus of c2. Not only this, we can prove still more: the angular region bounded by the lines m =-a and m = -2a does not have C2 in it, and the angular region bounded by the lines m = -2a and the positive m-axis does not have cl in it. To see this, we only have to note that when -a< m:::; -2a, the ordinate of R, -(a+m)ja, is in the interval (0, 1]. Using the method of comparing the slopes of the trajectories (Figure 13.14), it is easy to prove that the curve of symmetry (shown by the dashed curve) of the section of the arc of Lt inside the region {y :::; 0, x ~ -1/ a} with respect to the x-axis lies entirely below the upper half-branch of Lt. From this we can see that the absolute value of the ordinate of the other point of intersection Q of Lt and 1 + ax = 0 is less than 1. However, on the other hand, zt has a negative slope on the left side of 1 + ax = 0, and the ordinate of M is -1; hence the point of intersection P of lt and 1 + ax = 0 must lie below Q; that is, lt and Lt cannot coincide. Similarly, we can prove that the angular region bounded by the positive m-axis and m -2a does not have C1 in it (Figure 13.15).(5 ) Summarizing the above discussion, we get
=
THEOREM 13 .1. The bifurcation curve Ca for L2 = lt has a unique branch m = 1/a- a. The bifurcation curve C1 for L! = l! can only lie in the region {m > 1/a- a,O < m < -2a} and the region {m < 1/a- a< 0}. The bifurcation curve C2 for Lt = zt can only lie in the regions {0 < m < 1/a- a}, {1/a- a< m < 0, 27a <4m3 }, and {0 < -2a < m}. ( 6 )From this we can see that in Figure 13.10, Li = 11 and Lt = It cannot appear simultaneously; that is, the singular closed trajectories formed by these four separatrices cannot exist.
§13. CLASS II EQUATIONS WITHOUT LIMIT CYCLES
FIGURE 13.14
293
FIGURE 13.15
For the branches of C1 and C2 paBSing through P(-1,0), we believe they both exist and are unique, as shown in the graph of Figure 13.11, and both cl and c2 should be smooth curves and not contain interior points when they are considered as point sets in two-dimensional Euclidean space; yet we have no way to prove this. From now on, we assume we only have unique branches ofC1 and C2 passing through P(-1,0). In Figure 13.9 we see that when m = -a and 27a < 4m3 , the relative positions of L1 and 11 cannot be determined. The following shows that this is indeed the situation. THEOREM 13. 2. The upper half-branch of C not only lies above m = 1fa- a, but also crosses through the line m =-a to its upper half. PROOF. In (13.3), let x
d
I
= -x fa, y = -y' fa,
...J!.... = x1(1- x 1 ) dt '
1
1
and m
dX = -yI (1 Tt
= -a.
1 1- x1) . ay
Then (13.9)
Now we prove that when lal is sufficiently large, for system (13.9), L1 should run to the right of 11; this kind of relative positions is the same as the relative positions of L1 and 11 corresponding to the point on (-co, -1) on the negative a-axis. Hence this shows that the point on m = -a must lie below
Cl-(6) ( 6 )Does the following possibility exist: C1 has an even number of branches passing through P(-1,0), but all lie below m = -a and moreover there are an odd number of branches lying in the angular region -a< m < -2a? According to the previously mentioned theory that cl has to separate the two line segments (-oo,-1) and (-1,0) on the
THEORY OF LIMIT CYCLES
294
y'
FIGURE
13.16
First, in the region {y' > 0, 0 < x' < 1} we compare the slopes of the trajectory of (13.9) and the trajectory of the system
dx' jdt = -y',
dy' fdt
= x'
(13.10)
(Figure 13.16), and we have
-x'(1- x') y'(1- y' fa-x')
x' -x'y' = > 0. y' ay'(1- y' fa-x')
---:--"""-.,..--'---:- + -
Hence the trajectory of system (13.10) passing through N'(1, 0) (a circle with center at the origin) is above the trajectory of (13.9) passing through the same point. Suppose they meet the positive y 1-axis at A and B respectively. Next, in the region {x' < 0, a < y' < 0}, we compare the slopes of the trajectory of (13.9) and the trajectory of the system
dx' fdt = -y1 (1- y' fa),
dy' fdt
= x'(1- x')
(13.11)
and get
-x1(1- x') y'(1- y' fa-x')
x'(1- x')
-x'2 (1- x')
+ y'(1- y' fa) = y'(1- y' fa- x')(1- y' fa) > O.
negative a-axis from each other, we know that the above possibility does not exist. In other words, there must exist an odd number of branches of C passing through P( -1, 0) to cross through the line m = -a and then running to the upper half of this line.
§13. CLASS II EQUATIONS WITHOUT LIMIT CYCLES
295
Hence the trajectory of (13.11) passing through M'(O, a) lies on the right of the separatrix of (13.9) passing through the same point. Suppose they intersect the negative x'-axis at C and D respectively. It is easy to compute the equation of the trajectory M0 of (13.11):
x'2
x'3
y'2
2-3+2-
y'3 3a
a2
=6.
When iai > 1, the equation x'2 f2- x'3 f3 = a2f6 has a unique (negative) real root; that is, the abscissa of C, X6. From this we can see that if ial is very large, then (13.12) X~ :::: - {la272. But xi:, < x0; hence as long as ial is sufficiently large, lxiJI can be greater than any given positive number. Finally, in the second quadrant, we compare the slopes of the trajectory of (13.9) and that of the system
dx'fdt=a-y',
dy' fdt
= x'
(13.13)
and get
-x'(1- x') x' y'(1- y' fa-x') --a--y'
=
-=-x'[a2 (1- x')- y'2 ] ay1 (1- y' fa- x')(a- y') > O.
The term inside the square brackets on the right of the above formula can be assumed to be positive, since a 2 (1- x')- y'2 = 0 is a parabola with its vertex at N'(1, 0) and passing through the points (0, ±a). We have mentioned previously (see Figure 13.15) that the separatrix M' DH passing through M' of system (13.9) must lie in the region a 2 (1- x')- y'2 > 0, and now we have to compare the trajectories CF and CE of (13.9) and (13.13) passing through C respectively; they must also lie in the region a 2 (1- x')- y'2 > 0, and CF should be above CE. System (13.13) has a first integral
x'2 +(a- y')2 = k2; hence we know the equation of 6E is
x'2 +(a- y')2 =
Xc + a2,
and so the ordinate of E is
Yk = Ja'J. + x'6 +a. From (13.12) we know that, when Ia! is very large,
Ys::::
Ja2 + V4 a{0,4 +a:::: _1_1all/3 2~ ·
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296
When a varies, the ordinate of A remains unchanged, but y~ can become greater than 1 provided that lal is sufficiently large. Hence E is above A, and therefore His above B, which is what we had to prove. We can prove that as lal- oo, Ct must enter the infinite singular point G on m = - 2a. For the lower branch of C 1 we can also prove that along any line a= ao < -1, when m < 0 and lml is sufficiently large, the point (ao, m) must lie below C1 [204]. Moreover, we can also prove, going to infinity along the lower half-branch of Ct. that mfa- oo [205]; the proof is omitted. From this we can see that Ct cannot be symmetric with respect to the a-axis, because the upper half-branch of Ct must remain below the line m = -2a, and along this branch, as a- -oo, lm/al cannot approach oo. AB for the bifurcation curve C2 , not only must its upper half-branch remain below the hyperbola m = 1/a- a going to infinity, but also we can show that mfa- 0 along C2 as a- -oo (the proof is omitted). On the other hand, the lower half-branch of C2 must lie between the curve 27a = 4m3 and the negative a-axis and must approach the origin, because for the system (13.3) corresponding to the point (a,m) which makes 27a ~4m 3 , Lt all enter A2, but zt all come from R. From this we see that when (a, m) and 27a = 4m3 are very close but satisfy 27a <4m3 , Lt must run below zt; conversely, for the system (13.3) corresponding to the point on the segment (-1,0) of the negative a-axis, its Lt should run above lt. Hence there must exist C2 which separates the negative a-axis from the curve 27a = 4m3 • In Theorem 13.1 we proved that the angular region bounded by the line m = -2a and the positive m-axis cannot have C1 in it, but it can have C2. Now we shall prove that in this angular region there exists a branch of C2 which connects with the C2 passing through P( -1, 0) at the origin a = m = 0. For convenience, from now on we denote this branch by 02. THEOREM 13.3. There exists a C2 in the angular region -m/2
= -y,
dyjdt
=X
in the region x ~ -1/a, 0 :5 y :5 -(a+ m)/a, and get
-x(l+ax) +~-x[y-(a+m)x] + y- mx) y - y(l + y _ mx) > 0.
Y(l
(13.14)
§13. CLASS II EQUATIONS WITHOUT LIMIT CYCLES
297
FIGURE 13.17
From this we can see that the circular arc iB with its center at the origin and passing through R is below the separatrix Lt of (13.3). It is easy to compute XB
= _!J1 +(a+ m)2. a
Let the point of intersection of Lt and the x-axis be D. Then x D > x B. Next, we compare the slopes of the trajectory of system (13.3) and that of the system
dxjdt =my, in the region x
~
-1/a, -1
~
y
~
dyjdt = 1 +ax
(13.15)
0, and get
-x(1+ax) _1+ax y(1+y-mx) my
= -(1+y)(1+ax)
>O·
my(1+y-mx)
'
hence the trajectory of (13.15) passing through B lies on the left side of the trajectory of (13.3) through B. It is easy to see that the former trajectory has the equation
and the abscissa x E of the point of intersection E of the trajectory and y = -1 satisfies the equation (13.16)
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298
Again in the region y :5 -1, 0 :5 x :5 -1/a we compare the slopes of the trajectory of (13.3) and that of the system
dxjdt
= -y(1 + y),
-x(1 +ax)
x(1 +ax)
dyjdt = x(1 +ax) ,
(13.17)
and get -mx 2(1 +ax)
0
= y(1+y-mx)(1+y) > ·
y(1+y-mx) + y(1+y)
From this we can see that the separatrix of (13.17) entering M(O, -1) from It is easy to see that the former has the lower right part must be below the equation !2 x2 + ~3 x3 + !2 y2 + !3 y3 1 and the ordinate YA of its point of intersection A with 1 +ax = 0 satisfies the equation(T)
zt.
=s
12 -YA 2
13 = + -yA 3
-1(1 - -1) 6 a2
•
(13.18)
Finally, we compare the slopes of the trajectory of (13.3) and that of the system (13.19) dyjdt = -ay~(1 +ax) in the region x
~
-1/a, YA :5 y :5 -1, and get
-x(1+ax) +ay~(1+ax) y(1+y-mx) m2y = (1 + ax)[-m(m + ay~)x + ay~(1 + y)] > O, m2y(1 + y- mx) provided m is sufficiently large. Thus, the arc AS of the trajectory of (13.3) must be on the left side of the arc AQ of the trajectory of (13.19). It is easy to see that the equation of AQ is y~(1
thus the abscissa
XQ
+ ax)2 + m2y2 = m2y~;
of Q should satisfy
(1 + axQ) 2 = m 2 (1- 1/y~).
(13.20)
Now if we can prove that XE > XQ for sufficiently large m, then the separatrix Lt of (13.3) will run below lt; this kind of relative positions of Lt and is just opposite to their relative positions when -a < m < - 2a. From this
zt
(7)From this we can see that when a is fixed, the value of YA is also fixed, independently ofm.
§13. CLASS II EQUATIONS WITHOUT LIMIT CYCLES
299
a-o FIGURE 13.18
we can deduce that C2 exists in the angular region m > -2a > 0. In order to get XE > XQ, we know by (13.16) and (13.20) that we only need show that [1-
Ji +(a+ m)2] 2 +am> m2 ( 1- ~);
that is, 2 + (a+m)2 +am- 2J1 + (a+m)2
m
2
1
> 1 -2· YA
This formula clearly holds when a is fixed and m is sufficiently large, since its right side is a fixed number less than 1, whereas its left side approaches 1 as m--+ oo. Finally, we have to prove that C2 passes through the origin. For this, we note that when a= 0, system (13.3) becomes dxfdt
= -y(1 + y- mx),
dyfdt
= x.
The global structure of its trajectory is as shown in Figure 13.18; here N = A1 = A2 is an infinite higher-order singular point (1, 0, 0) and R = A3 is a semisaddle nodal point (1, m, 0). From Figure 13.18, we see that Lt will run below zt; their relative positions are just opposite to the relative positions of Lt and zt which correspond to the points to the right of the point (-1, 0) on the negative a-axis, and so C2 must separate the positive m-axis from the negative a-axis; that is, it should pass through the origin. The theorem is completely proved. We can further prove m/a--+ -2 along C2 as a--+ -oo; that is, C2 must pass the infinite point of the line m = -2a. The proof is omitted (see [204]). In the following we discuss again the possibility of establishing the equalities Lt = l'l and Li = lt.
THEORY OF LIMIT CYCLES
300
'Y
FIGURE 13.19
zt
FIGURE 13.20
THEOREM 13. 4. There does not exist a point (a, m) which makes L1 = hold.
PROOF. If there exists a point (a*, m*) such that its corresponding system (13.3) has L! = then, as indicated in Figure 13.19, 0 should be an unstable singular point and R should be a stable singular pojnt; this is not possible under any circumstances when a < 0.
zt,
THEOREM 13. 5. The bifurcation curve C4 for Lt = l1 is a part of the curve which lies above the line m = -a, and connects the origin and the point of intersection S of Ct and m = -a. PROOF. As in Figure 13.20, when Lt = l1, first we know from the direction of the trajectory crossing MN that R must be unstable and 0 must be stable. Hence C4 must be above the line m = -a.( 8 ) We note again in Figure 13.20 the relative positions of Lt and and the relative positions of L1 and 11; we know that C4 must lie below 02 and on the right side of Ct. Now from Figure 13.21 we see that corresponding to the points of 02, we have Lt = lt. Since 0 is stable, if we consider to be a continuation of lt, then Lt must run to the right side of l!. In fact, according to the continuity of a solution with respect to its initial value, for the points below and close to 02 this is actually the case. Again from Figure 13.22, we see that, corresponding to the points on the line segment m = -a on the right side of C1, R = N is a higher-order singular point, Lt = L2 = 0, and L1 will
zt
zt
( 8 )When 0 < m <-a and m > 1/a- a, Figure 13.7 shows that R is a stable singular point with no closed trajectory near it; hence Lt must enter R and cannot cross through the left side of MN.
§13. CLASS II EQUATIONS WITHOUT LIMIT CYCLES
301
N
------~~----~~;r--~
a;
M
FIGURE 13.21
FIGURE 13.22
FIGURE 13.23
run to the left side of 11. Hence from the continuity of a solution with respect to its initial value, for the points close to and above this finite line segment, Lt of the corresponding system (13.3) will run to the left of 11. Thus, there must be a c4 separating from m = -a; that is, c4 must pass through the origin. Also, for the points of C 1 above m =-a we have L1 = l1 in the figure of the trajectory of the corresponding system (13.3), and Lt crosses through M N and enters 0 (Figure 13.23). The relative positions of Lt and 11 are just opposite to the relative positions corresponding to the points close to and above the line m = -a and on the right side of C1 ; hence C4 must terminate at a point of intersectionS of C 1 and m =-a. In fact, the system
ct
THEORY OF LIMIT CYCLES
302
(13.3) corresponding to S has R=N, Considering Lt as a point and Li as a continuation of Lt: we then have The theorem is completely proved. We believe the bifurcation curves on the parametric (a, m)-plane have all been found (Figure 13.11), and they are all differentiable curves. For C~, C2 , and C4 , we still do not know whether they are algebraic or transcendental curves, nor by what formulas they are represented. Going back from Figure 13.11 to Figure 13.1, we find that under the conditions 27a >4m3 m < 1/a-a,
Lt = l1.
the relative positions of L1 and li have in fact three possibilities; which one appears depends on whether the point (a, m) is on the left side of C~, on the right side of C1 or above C1. For Figures 13.6, 13.7, 13.9, and 13.10, the situations are the same. Up to here, the initial steps in the problem of the global structure of the trajectories of system (13.3) have been taken. Now we have some results on whether there exists a unique branch for each C1 , C2 or C4 as shown in Figure 13.11. In [206], after transformation, system (13.3) becomes dx/dt
= -y(1 +y- mx) + mx,
dyjdt
= x(1 +ax).
Then, using the theory of rotated vector fields, we prove that there is only one C2 which does not have an interior point. Moreover, if in (13.3) we let m = ka (k > 0 is fixed), and let the parameter vary, then we can use the theory of rotated vector fields again to prove that in Figure 13.11 there is only one C1 below the a-axis, and that it does not have an interior point. The problem of uniqueness of C~, C2, and C4 can be solved by using the formula for differentiating the solution with respect to its parameter and the implicit function theorem. Cao Yu-lin has given a careful proof of this assertion. Also, for the problem of bifurcation curves of equations of class II with two fine foci (hence without a limit cycle) dx m(m+ 2a) 2 dt = -y + 4 x
2
+ mxy + Y
'
~~ =
x(1 +ax)
(13.21)
for most cases, quite satisfactory results have been obtained. Note that Figure 13.10 also belongs to the case of two fine foci, but does not belong to (13.21). Also the bifurcation curve corresponding to this figure is just the curvilinear triangle formed by c~. C2, and 04 above the line m = -a in Figure 13.11.
§13. CLASS II EQUATIONS WITHOUT LIMIT CYCLES
303
There is also the following work on the study of global structures of trajectories and bifurcation curves of quadratic systems (not necessarily equations of class II): For the system
dxfdt
= -y + lx 2 + ny2 ,
dyjdt
= x(1 +ax+ by)
Luo Ding-jun [208] and Sun Kai-jun [209] studied the global structure and bifurcation surface when -y+lx2 +ny2 = 0 is an ellipse, parabola or hyperbola with a(b + 2l) ::/: 0 (hence the system does not have a closed trajectory, and the proof can be seen at the beginning of §15), but their results are not as complete as in this section. Cao Zhen-zhong [206] studied the global structure and bifurcation surface of the system of equations
dx/dt = -y +ox+ mxy- y 2 ,
dyjdt
= x(1 +ax).
Cao Xian-tong [210] studied the same problem for the system of equations
dxjdt = -y + lx 2 + mxy- y 2 ,
dyjdt = x(1 +ax).
Ren Yong-iai and Suo Guang-jian [211] studied the global structure of a quadratic system with three straight line solutions. Liang Zh~jun [174] studied the global structure and phase-portrait of the system of equations
dxjdt
= -y + lx 2 + 5axy,
dyfdt
= x + ax 2 + 3lxy
without cycles, but with a third-order fine focus. Exercises 1. Prove Theorem 13.1. 2. Construct the figures of all the trajectories of system (13.1) when n = 0. 3. Prove the uniqueness of the lower half-branch of C1 and C2 at the end
of this section. 4. Prove if there is no higher-order singular point except a fine focus (0, 0) for system (13.3), then it must have a fine saddle point (that is, the value of its divergence at this saddle point is zero) or a fine focus. 5. Use the transformation
= -a'y', y = a'x' to change (13.3) into equations of class m, and use the method of analysis on a'= 1/a,
b'
= m/b,
x
global structure to prove that along the lower half-branch of the bifurcation curve cl going to oo, m/a-+ +oo. 6. Prove that along the upper half-branch of the bifurcation curve C2, mja--+ 0 as a--+ -oo.
THEORY OF LIMIT CYCLES
304
7. Prove that in the angular region 0 < -a < m in Figure 13.11 we can construct a topological transformation T, which does not keep a constant orientation, to change every half-ray starting from the origin to another halfray, to change an open arc §G of C 1 into C2, and to change C4 into itself; and prove T is an identity transformation on the line m = - 2a. 8. For the system of equations
dxjdt
= -y + lx 2 + mx + ny2 ,
dyjdt
= x(1 +ax)
find necessary and sufficient conditions for a line connecting two saddle points to be an integral line. 9. Prove that when 8 satisfies
a8 3
-
2am8 2
+ (am2 + 3a- m)8 + 1- am- a2 = 0
the system of equations
dx/dt
= -y + 8x + mxy- y 2 ,
dyjdt = x(1 +ax)
takes a certain M~ as an integral line; hereM is a saddle point (0, -1) and ~ is an infinite singular point. Moreover, the condition for this system to have three real roots is 27a- 4m3 > 0 (suppose a< 0). 10. Find a parabola solution of system (13.3) passing through M and N with its principal axis through A2, and prove this solution exists only when a= and m = -5/v'B (hence the lower half-branch of Ct passes the point ( -5/.;6)).
-J312 -J312,
§14. Relative Positions of Limit Cycles and Conditions for Having at Most One and Two Limit Cycles in Equations of Class II
In this section we shall study the second and third problems mentioned at the beginning of §13. We first look at the simpler case. Suppose that in equations of class II,
~; =
-y + 8x + lx 2 + mxy + ny 2 ,
dy dt = x(1 +ax),
(14.1)
two of the coefficients of the terms of second degree on the right side of the first equation are zero. Then we have
dxjdt = -y + 8x + ny 2 , dxjdt = -y + 8x + mxy, dxjdt = -y + 8x + lx 2 ,
dyjdt = x(1 +ax); dyjdt = x(1 +ax); dyjdt
= x(1 +ax).
(14.2) (14.3) (14.4)
It is easy to see that (14.2) and (14.3) can be integrated when 8 = 0, and they take (0, 0) as their center; and when 8 :f. 0 they have no limit cycles because the divergence of (14.2) is a constant 8, and (14.3) can be proved to have no limit cycles by the Dulac function (1 - mx)- 1 • For (14.4), the situation is not the sante. We may as well assume l > 0 and a > 0; and it is not difficult to use the well-known method to prove that when 8 ~ 0 or 6 2:: lja, (14.4) does not have a limit cycle, but when 8al > 0 and 8 lies in some interval (0, 8*), (14.4) has a unique limit cycle. In the following we study mainly the case where the coefficients of the quadratic terms on the right side of the first equation of (14.1) have only one zero. (I) l = 0. In this case we have the system (we may as well assume n = -1)
dxjdt = -y + 8x + mxy- y 2 ,
dyjdt = x(1 +ax).
First we prove a useful theorem for nonexistence singular closed trajectory. 305
(14.5)
of a closed trajectory and a
THEORY OF LIMIT CYCLES
306
THEOREM 14 .1. System (14.5) (we may as well assume a < 0) cannot have a closed trajectory or a singular closed trajectory passing a saddle point in either of the following cases: 1) m8 $ 0, lml + 181 -::/= 0; 2) 8(m- 8) $ 0, lml + 181 i= 0.( 1 ) PROOF. When the first group of conditions holds, it is easy to see that any closed trajectory or any singular closed trajectory passing a saddle point of (14.5) cannot intersect the line 1- mx = 0. Now we take the Dulac function to be B = 1/(1- mx); then we have
a a 8 -my 2 ax (BP) + ay (BQ) = (1- mx) 2 • The right side of this formula always keeps a constant sign on any side of the line 1 - mx = 0; hence the theorem is proved. Now we suppose the second group of conditions holds. We translate the x-axis to the line y = -8/m (the case of m = 0 has been seen in (14.2); hence we may as well assume m i= 0) and keep they-axis unchanged. Then (14.5) becomes dx = !._ ( 1 + ( 1 + 28) y+mxy-y2 , dt m m m (14.6) dy dt = x(1 +ax).
!._)
The system of equations whose vector field is symmetric to that of (14.6) with respect to the new x-axis is dx dt
= !._ ( 1 m
!._) _(1 + 28) y _ mxy _ y2, m m
(14. 7)
dy = -x(1 +ax). dt The locus of points of contact of the trajectories of these two systems is easily seen to be x = 0,
1 + ax = 0,
and
8 (1 - m8) -
m
y 2 = 0.
(14.8)
Since the divergence of (14.6) is only zero on the x-axis, any closed or singular closed trajectory r must intersect the x-axis. Also, from the theorems of §11, we know that if r appears in the vicinity of (0, 0), then it cannot meet 1 +ax= 0. Moreover, for 8(m- 8) < 0 the last equation of (14.8) does not have a real locus, and for 8(m - 8) = 0 its locus is the x-axis. From this (1 )This theorem was first obtained in [212]. But here we use the method of proof in [124], which was first seen in [16].
307
§14. CLASS II EQUATIONS WITH 1 OR 2 CYCLES
we can see that the curve symmetric to r+ (the part of r above the x-ax:is) with respect to the x-ax:is, and r- (the part of r below the x-ax:is) do not have a common point except on the x-ax:is; that is, the curve symmetric to r+ lies entirely above (or below) r-. Thus, for any closed or singular closed trajectory r of (14.6), we must have
11 (~= + ~~)
11
dxdy =
intr
mydxdy #
o.
int r
This contradiction shows that r does not exist. Similarly, we know that the vicinity of another singular point with index +1 on 1 + ax = 0 cannot contain a closed or singular closed trajectory. The theorem is completely proved. The following transformation of coordinates is useful for our later discussion. Note the two singular points on the line 1 + ax = 0 are
R( N
-~,4 [-1-:- J(u ~)'- ¥]),
of
+ J(u :)'- ~]),
of
H· ~ [-1-:
index index -1. +1;
Moving the origin toR, we get
~~ = dy = dt
(8
+my2)x- ( 1+: +2y2) y +mxy- y3,
~x+ax 2
(14.9)
,
where Y2 represents the ordinate of R. Now we apply the transformation x=- [
m 2 48] 3 / 4 (1+a) -~ x,
m 2 48] l/ 2 (1+a) -~ y, 48] - t m t=- [ (1+a) -~ 1
y=-
[
1
1/ 4
2
(14.10)
1
to (14.9) and get dx1/ dt 1 = -y' + 81x' + m 1x' y' - y12 , dy 1jdt' = x'(1 + a 1x 1),
(14.11)
where
,
8 = [(1
-(8 + my2) + mja)2- 48/ajl/4' ml = m [ ( 1 +
1
[(
a +a
:f- ~]
m)2 1 +-; 1/4
48] 3/ 4
- ~
'
(14.12)
THEORY OF LIMIT CYCLES
308
According to Theorem 14.1, it is only possible for system (14.1) to have limit cycles near the two singular points of index +1 when
m6 > 0 but 161 < lml.
(14.13)
In the following we want to explain whether limit cycles can coexist near both singular points of index +1. This problem is closely related to the order between m and a. 1. m > -a > 0. As shown in Figure 13.10, for 6 = 0, 0 is stable, R is unstable, and M and N are saddle points. Let 6 increase from zero. Then 0 becomes unstable. By the proof of Theorem 3. 7 of §3 and Theorem 12.5 of §12, there exists a unique stable cycle near 0. On the other hand, the two singular points on the line 1 + ax = 0 move far apart (we denote them by R' and N'). Now we prove that the stability of R' is different from R, since near R' there also appears a unique unstable cycle.(2) Hence we note that for m > -a> 0 and 161 sufficiently small,
H m112
~ 6-
; [ 1+ :
= 6- m [1
2
+
J(
1+ : )' -
+ m- (1 + m) (1a
a
~
l
26 )] a(1+m/a)2
+ 0(62)
a8 + 0(6 2 ) < 0, a+m where Y2 is the ordinate of R'. Hence, from (14.11) and (14.12) we know that R' is a stable focus of (14.9).(3) Hence when m > -a > 0 and 0 < 6 « 1, limit cycles coexist near the singular points 0 and R', but their stability is different. When 6 continuously increases, it is not certain whether these limit cycles disappear simultaneously. Now suppose near 0 the limit cycle, it exists, is always unique, and at 6 = 6 it expands and becomes a separatrix cycle passing through M (note that when 6 varies, (14.5) forms a family of generalized rotated vector fields on-each side of the line 1 + ax = 0; hence, when 6 increases, the two separatrices lt an~ li in Figure 13.10 can certainly come close together and coincide). Thus 6 should clearly be a function 6 = f(m,a) of m and a. HE>nce from (14.11) we know that the value 6' of 6' which makes the limit cycle near R' expand and becorne a separatrix cycle through N' must be the same function of rn' = --
--;--
lc51. ( )The Uniqueness of the limit cycle indicated here is only limited to sufficiently small
l'<Jt)~~ fact can also be seen from the figure of isoclines, since now the upper branch of th x, yh roug
A{ 0 Passes through 0
and N', and is above c5 +my = 0; its lower branch passes and R', and is below c5 + my = 0.
§14. CLASS II EQUATIONS WITH 1 OR 2 CYCLES
309
and a'; that is, 'E' = f(m', a'). Whether this function f can be determined is a problem worthy of our consideration. Of course, the uniqueness of a limit cycle in the vicinity of every singular point has to be proved. In [215] the uniqueness of a limit cycle is proved only when m > - 2a and 1 + 4am ~ 0. Aside from this, there is no other result. 2. 0 < m ~ -a. First we examine the case when 0 < m < -a.(4 ) When 6 = 0, N( -1/a, 0) is a saddle point and R( -1/a, -(a+ m)fa) below N is a focus. If 6 increases from 0, then N' and R' move far apart. Since for 6 = 0
6 + my2
= -m ( 1 + :) < 0,
initially R' remains unstable. Now we ask: Does the stability of R' change as 6 continuously increases? When does it happen? Clearly we can see that a necessary condition for R' to change its stability is 6' = 0; that is, li + ffl!h
= li -
;
[1+ :
+
J(
1+ : )' -
~] = 0.
Solving this equation, we get 6 = m, and 86' 88
I
= _ [a- 38 + 3m/2 + m2 /2a + (m/2)J(1 + m/a) 2 a[{1 + mfa)2- 48jaj31 2
.5=m
48/al 6=m
-1
= {1- m/a)3/2 < 0. {14.14) Hence when 6 increases from m, 8' decreases from 0; that is, R' as a singular point of (14.11) changes from stable to unstable, and so, as the singular point of the original system {14.5), it must change from stable to unstable. However, according to Theorem 14.1, we know that for 8 ~ m there is no closed trajectory in the vicinity of R'. Hence from Theorem 3. 7 in §3, we know that as 8 increases from less than m to m, there is an unstable limit cycle which shrinks (not necessarily monotonically) and approaches R'. How is this unstable cycle generated? There are several possibilities:( 5 ) (i) It is generated from a separatrix cycle through N' and surrounding R'. (ii) It is generated from a separatrix cycle passing through M and N' and surrounding R'. (iii) It is generated by splitting a semistable cycle which suddenly appears in the vicinity of R'. ( 4 )The global figure of the trajectory at 6 = 0 has three possible cases as in Figures 13.6, 13.7, and 13.8. ( 5 )This unstable cycle cannot be generated from a separatrix cycle passing through M and surrounding R'. The reason is seen in formula (14.16), below.
THEORY OF LIMIT CYCLES
310
FIGURE 14.1
FIGURE 14.2
FIGURE 14.3
Which case it belongs to depends on the order relation between m and
1/a- a and the order relation between m and 1 +a. In fact, when 6 = m = 1 +a, the coordinates of R' are ( -1/a, -1), and at this time the coefficients in (14.11) are
m'
= mv'-1/a = (1 + a)v-1/a,
a'= a(-1/a) 3 12 = -~.
From this we can see that m' = 1/a'- a'; that is, in this case MN' has become an integral line, and when o= m, from m' < 1/a' -a' we can deduce that m > 1 +a, and from m' > 1/a'- a' we can deduce that m < 1 +a. Thus similarly, as before, we can draw three figures (Figures 14.1-14.3) for 0 < m < -a, which correspond to m = o < 1 + a, m = o = 1 + a, and m = 8 > 1 + a, respectively.
§14. CLASS II EQUATIONS WITH 1 OR 2 CYCLES
311
Moreover, it is easy to see that when 0 < m $ 1/a- a we must have a< -1, whence m > 1 +a; and when m > 1/a- a, and m > 0, (but 1/a- a is not necessarily positive), then the three relations m > 1 +a, m = 1 +a, and m < 1 + a can all possibly exist. Now when Figures 13.6 and 13.8 for the case 0 < m $ 1/a-a, 6 = 0 change to Figure 14.3 for the case m = 6 > 1+a, we observe the change of limit cycles and separatrices in the neighborhoods of the two singular points. Since when 6 varies, (14.5) on any side of 1 +ax = 0 forms a family of generalized rotated vector fields, hence when 6 increases from 0, on the one hand, the point 0 changes from stable to unstable, and a stable limit cycle is generated near it; on the other hand, the two separatrices passing through N' and surrounding . 0 then approach each other and coincide. Notice that for
aQ2)
( aP2 ax + ay
N' =
6 + my1
> 0,
(14.15)
where Y1 > 0 is the ordinate of N', we can see that when two separatrices coincide and form a singular closed trajectory, this trajectory should be internally unstable. Hence it cannot be formed by the expansion of a stable cycle in the vicinity of 0 after arriving at N'. On the contrary, the separatrix passing through N' is first formed; then the two separatrices interchange their positions and an unstable cycle is generated from the separatrix cycle, which shrinks inwards as 6 increases, and finally coincides with the limit cycle, which lies inside and expands outwards to become a semistable cycle, and then disappears. Of course, here we assume that during the process of increase of 6 there do not suddenly appear one or more semistable cycles in the neighborhood of 0 which then split, some expanding and some shrinking (we conjecture this case does not happen). Hence in the neighborhood of 0, there must exist two values of 6, 0 < 61 < ~ < m, in the interval (0, m) such that when 6 = 61 a separatrix cycle passing through N' is formed, but when 6 = ~ a semistable cycle is formed. Hence for 6 in (0, 61), the vicinity of 0 has a unique stable cycle; for 6 in (61, 62) this vicinity has two and only two limit cycles, the outside one unstable, and the inside one stable; for 6 > 62 , there is no limit cycle in the neighborhood of 0. Next we observe how the separatrices in the vicinity of R' vary. Since the direction of the trajectory crossing M N' moves from left to right, it is easy to see that during the process of change of 6 from 0 to m there exists at least one value 6i < m such that the two separatrices passing through M and surrounding R' coincide and become a separatrix cycle. On the other hand, we can compute = 6 -m
ax
ay
(0,-1)
312
THEORY OF LIMIT CYCLES
Hence this separatrix cycle must be internally stable. From this we can see that as 6 j m, the unstable cycle shrinking towards R' cannot be generated by the above separatrix cycle. On the contrary, there must exist a 62 (< 6i) for which the vicinity of R' suddenly generates a semistable cycle, which splits into at least two limit cycles when 6 > 62, the outer one being a stable cycle which expands gradually and becomes a separatrix cycle passing through M at 6 = 6i, and then disappears; the inner one is an unstable cycle which shrinks into one point R' (-1 I a, -1) and changes the stability of R'. · However, the above analysis cannot determine whether limit cycles can coexist in the vicinity of 0 or R'; it is useless to compare the values of ~ and 62 since even though system (14.11) has the same form as (14.5), the values of m', a' and m, a, and the values of m' I a' and ml a are not the same. In order to solve this problem, note that when 2 6 = m +2am 4a
(14.17)
we have J(1 + mla)2- 46la = 1; hence a'= a, m' = m, and 6' = 6; that is, at this time (14.11) and (14.5) are identical. Hence the structure of the trajectory in the vicinity of 0 is the same as in the vicinity of R. But this fact can only occur in the following four cases: a) The vicinity of 0 does not have a limit cycle, and the periphery of R has not yet generated a limit cycle. b) The vicinity of 0 and the vicinity of R have a unique single cycle. c) The vicinity of 0 and the vicinity of R have a semistable cycle. d) There are two cycles each in the vicinity of 0 and R, and their forms are identical. If case a) occurs, then it is clear that the limit cycles cannot coexist in the vicinities of 0 and R for any 6; if case c) or d) occurs, then we have an example of (2.2) distribution since even in case c) a semistable cycle can be considered as a combination of two single cycles just as the multiple root of an algebraic equation. This problem has recently been solved in [216], and the answer is that a) holds. The author first transformed (14.5) into equations of the class (A) of Cherkas [192] introduced in §12
dxldt = 1 + xy,
(14.18)
and then according to the method of [192] applied the transformations Y = y 3 jx + 1/x2 , = 1/x and y = lell-ao 2 z, = to change (14.18) into an
e
e e
§14. CLASS II EQUATIONS WITH 1 OR 2 CYCLES
313
equation of Lienard type:
dz/dt
= P4(~)1~12ao2-3sgn ~ + zP2(~)1~1ao2-2sgn ~.
dUdt = z,
(14.19)
where P2(~) and P4(~) are quadratic and quadric polynomials respectively. Finally, [216] made use of the condition (14.17) to show that in this ca.se the divergence of (14.19) ha.s constant sign and hence (14.19) does not have a limit cycle. This explains that for any 8, limit cycles cannot coexist in the vicinities of both 0 and R. Next we see, for 0 < 1/a- a< m < -a (here we must have a< -1 and hence must have m > 1 +a), a.s 8 increa.ses from 0 tom, how the figure of the trajectory changes from Figure 13.7 to Figure 14.3. At this time a stable cycle is generated from 0, and continuously expands. On the other hand, the two separatrices through M and surrounding 0 come close together and coincide. Since the divergence of (14.5) always takes negative value at the point M (when 0 :5: 8 < m), there must be an odd number of limit cycles in the vicinity of 0. We conjecture that when a limit cycle exists, it should be unique; that is, the stable cycle generated from 0 continuously expands and becomes a separatrix cycle through M, and then disappears. Afterwards, the two separatrices lf and l1 through M interchange their positions, and in the vicinity of 0 there is no closed or singular closed trajectory. But the uniqueness ha.s not yet been proved. The situation is more complicated in the neighborhood of R'. When 8 increa.ses from 0, the two separatrices through N' and surrounding R' rotate in different directions; they may coincide to form a separatrix cycle through N', or it may be possible that the separatrix going directly to the left side coincides with another separatrix starting from M going to the right side to become an integral line M N', the separatrix from the right not yet having arrived to coincide with it. In the former ca.se, from (14.5) we know that the separatrix cycle will generate an unstable cycle, and when 8 j m, it shrinks towards R'; this is case (i) mentioned previously. In the latter ca.se, there can be two different situations: One is: when M N' becomes an integral line, the separatrix from N' going to the lower right side ha.s not yet interchanged its position with the separatrix entering M from the lower right side; then the two separatrices Lt and lf all enter R', but tt is still on the outside of Lt, a.s shown in Figure 14.4. As 8continues to increase, Lt and tt coincide, and then interchange their positions; hence lt and lf together surround R'. According to (14.16), as discussed previously, we know that now in the vicinity of R' a semistable cycle, externally stable but internally unstable, will suddenly appear, and this is case (iii) as previously mentioned. The other situation is
THEORY OF LIMIT CYCLES
314
Cl'
0
FIGURE
FIGURE
14.4
14.5
zt
that when M N' becomes an integral line, Lt and also just coincide. Then as 8 continues to increase, the separatrix cycle through the two saddle points generates an unstable cycle, and this is case (ii) mentioned previously. We can give examples showing that the above three different cases can appear [21], and it is clear that case (ii) is the transitional case from case (i) to case (iii). It is obvious that Figure 13.7 with m > 1/a- a, m > 0, and 8 = 0 can change into Figures 14.1 and 14.2 form= 8 ~ 1 +a in case (i) as previously mentioned. In particular, when 1/a- a > 0 (i.e., a < -1}, Liang Zhao-jun [217] used the previous method to prove the limit cycles cannot coexist in the vicinity of both 0 and R'. But whether this conclusion holds when a ~ -1 is still unknown.( 6 ) The above demonstration is also suitable for m = -a, because when 6 = 0, even though R = N is a higher order singular point, as soon as 8 becomes positive, N' immediately separates from R', and R' of index + 1 lies below and it is a stable singular point; moreover, when 8 = m, it becomes unstable again. The above method of discussion by dividing into the three cases m > 1 +a, m = 1 + a, and m < 1 + a can also be applied to the case m > -a, because for 8 = m the figures of its global structures are exactly the same as Figures 14.1-14.3, save that the starting figure of the global structure is not Figure 13.6, 13.7 or 13.8, but Figure 13.10. Readers can analyze for themselves when ( 6 }If we limit ourselves to the case 0 < 6 < m < -a, then, using the method of [27], we can prove that when m/2 ~ 6 < m or 0 < 6 < !m + !m2 /a, the limit cycles of (14.5) are centrally distributed; here there is a gap between these two intervals.
315
§14. CLASS II EQUATIONS WITH 1 OR 2 CYCLES
this change takes place. We should note that when 6 = 0 and 6 = m, we have = 0, and also
6'
how 6' changes with 6 is shown in Figure 14.5. 3. a< m < 0. When 6 = 0, 0 and R(-1la, -(a+m)la) are unstable foci. If we let 6 decrease from 0, then the point 0 becomes stable and a unique unstable cycle appears near it. As before, the stability of R' begins to change at 6 = m, and R' and N' following the decrease of 6 come close together; at 6 = m, R' arrives at ( -1la, -1), and there is a stable cycle which shrinks and approaches R'. In the following we prove that the limit cycles cannot coexist in the vicinities of 0 and R'. LEMMA 14 .1. When 6I m ~ ~, the vicinity of the point 0 does not con-
tain a closed or singular closed trajectory. The method of proof is almost the same as the second part in the proof of Theorem 14.1, and is omitted. Now we note that by (14.12) we have ~-!+ (1+mla)l2-6lm m'- 2 J(1+mla)2 -46la'
(14.20)
From this we can see that 6' I m' > ~ for 6I m < ~. Hence applying Lemma 14.1 to system (14.11), we see that here the vicinity of R' does not contain a closed or singular closed trajectory. In other words, if there may exist a limit cycle near the point 0, the vicinity of R' must have a limit cycle. The case of an even or odd number of limit cycles in the vicinities of these two singular points can be discussed as before. In short, for system (14.5), we conjecture that if there exist an odd (even) number of limit cycles in the vicinity of a singular point, there must be at most one (at most two). 4. m ~ a. Similarly to 3, if we let 6 decrease from 0, then an unstable cycle appears near the point 0. Moreover, R' and N' come close together as 6 decreases. If m = a, then at 6 = m, R' and N' arrive simultaneously at ( -1la, -1) and become a higher-order singular point; when 6 decreases again, R' and N' disappear. If m < a, then at 6 = m the saddle point N' arrives at (-11a, -1), which corresponds to 6' = 0, but R' corresponds to o' = m(l- mla) > 0, and so R' does not change its stability. As ocontinues to decrease, R' and N' come into coincidence, and then disappear. From (14.20), since now mla ~ 1, it is easy to see that we have o' lm' ~ for all 6 in [m, 0]; that is, a limit cycle never appears near R'.
!
THEORY OF LIMIT CYCLES
316
Synthesizing the above analysis, we get THEOREM 14. 2. For system (II)t=O (a < 0, m f. 0), we have: (i) If m > -a > 0, limit cycles can coexist in the vicinities of two singular
points of index +1. (ii) IfO < m < 1/a-a, then, for 8 in some interval (81,82), in the vicinity of the point 0 there are at least two limit cycles; for 8 in another interval (62,61), the vicinity of R' has at least two limit cycles. However, limit cycles cannot coexist in the vicinity of 0 or R' if 0 < m ~ -a and -a > 1. (iii) If a < m < 0, then limit cycles cannot coexist in the vicinity of 0 orR'. (iv) If m ~ a, then there does not exist a limit cycle in the vicinity of R'.
(II) m = 0. Here we have the system dx dt
= -y +ox+ lx 2 + ny2 ,
dy dt
= x(1 +ax),
a f. 0.
(14.21)
We may as well assume a< 0 and taken= 1; and from now on we only have to study( 7 ) dy dt = x(1 +ax),
a< 0.
(14.22)
LEMMA 14.2. When lo ~ 0 but Ill+ 181 f. 0, system (14.22) does not have a closed trajectory or a singular closed trajectory. PROOF. We can take the Dulac function as e- 21 Y, and the proof is omitted.
It is easy to compute that when 8 = 0, we have v3 = -!7l'al for system > 0 (< 0) the origin is an unstable (stable) focus, and when 8 < 0 (> 0) and 181 is sufficiently small there exists a unique unstable (stable) limit cycle near the origin. (14.22); that is, when l
LEMMA 14.3. When
8 ~ l/a (l > 0)
or
8 ~ 1/a (l < 0)
the limit cycle in the vicinity of the origin disappears, and the case for the neighborhood of another singular point of index +1 is similar but the inequality is reversed. PROOF. When 8 = lja, from (14.22) we get
dx -y+ y 2 dy = x(1 +ax)
l
+ ~·
(14.23)
(1)At this point we mainly present several theorems from (179]. For system (14.22), L. I. Zhilevich (218] did similar research, but the results were not the same.
§14. CLASS II EQUATIONS WITH 1 OR 2 CYCLES
317
but the family of integral curves of the equation dx -y+y 2 dy = x{1 +ax)
in the vicinity of each of the two singular points of index +1 is a family of closed curves {see Figure 13.5 of §13). From {14.23) we can see that they are arcs without contact for the trajectory of system {14.22). Hence in this case {14.22) cannot have a limit cycle. Then from the theory of rotated vector fields we know at once that when 8 ~ lfa (l > 0) or 8 ~ lfa (l < 0) the vicinity of the origin does not have a limit cycle. In order to consider the other singular point of index +1, we first apply a transformation similar to {14.10) and then proceed with our discussion, and, similarly to the case of 0 < m < 1/a- a in {I), the following theorem will hold: THEOREM 14.3. Limit cycles cannot coexist in the vicinities of two singular points of index +1 of system {14.22).
The detailed proof is left to the reader. System {14.22) has at most four singular points, and they are denoted by 0{0, 0), N(O, 1), R( -1/a, yt), and A( -1/a, Y2), where Y1,2
=
a=FJa2-4(l-ao) · 2a
It is easy to see tkat N and A are saddle points, and 0 and R are nonsaddle points of index +1. In the following we use Theorem 6.4 (still its special form, i.e. the uniqueness theorem of [111]) to discuss the uniqueness of a limit cycle of system {14.22). From Lemma 14.3 we know that we only have to consider the interval (lja, 0) or {0, lfa) of o. If we rewrite {14.22) as dxfdt = -y + y 2 +ox+ lx 2 = -cp(y)- F(x), dyfdt = x + ax 2 = g(x),
where cp(y) = y- y 2 and cp'(y) = 1- 2y, then we can see that cp(y) monotonically increases in the interval -oo < y ~ !· Let G denote the plane region {-oo < x < -1/a, -oo < y ~ 1/2}, and F(x) =-ox -lx 2, f(x) = F'(x) = -8- 2lx. Then we have f(x))' _ 1 2 ( g(x) - (x + ax 2 ) 2 [2lax + 2a8x + 8].
THEORY OF LIMIT CYCLES
318
Hence when l > 0 and 6 < 0 we have (f(x)fg(x))' ~ 0 provided that 6 ~ 2l/a, but when l < 0 and 6 > 0 we have (f(x)fg(x))' ~ 0 provided that 6 ~ 2lfa. Hence, provided there exist parameters (a, l, 6) that make the limit cycles of the corresponding system (14.22) lie entirely in the region G, we are allowed to use Theorem 6.4 to prove the uniqueness of a limit cycle. In the following we shall give some conditions to be satisfied by the parameters (a, l, 6) in order to guarantee that the limit cycles of system (14.22) surrounding the origin must lie entirely in the region G.(8 ) THEOREM 14.4. When l > 0 and a 2 (a 2 - 4l) -16 ~ 0 (i.e., a 2 ~ 2l + 2y'i2TI), system (14.22) has at most one limit cycle (unstable cycle). PROOF. Let the two lines without contact passing through the saddle point A( -1/a, Y2) be Lt ,2
= y- Kt ,2
K1,2
=
(x + !) a -
Y2
= 0,
2~ ( -{3± Jf32 -4a),
where a= 2y2 -1 and f3 = 6- 21/a; K 1 and K 2 are two roots of the equation aK2 + {3K + 1 = 0. Since a< 0 and f3 > 0 (by Lemma 14.3, when l > 0, we may as well assume 6 > l/a; hence f3 > -l/a > 0), we have Kt < 0 and K2 > 0. Also it is easy to see when 6 E (l/a, 0), we have Y2 ~ 0. Suppose the line Lt without contact and they-axis intersect at Ytoi clearly Yto > 0. Since the highest point of the limit cycle in the vicinity of the origin must lie on the y-axis, and it cannot intersect Lt, hence if we require Yto ~ we can assure that the limit cycles surrounding the origin must lie entirely in the region G. For Yto ~ ! , it is easy to see that we only have to require
!,
H( 6) = 63
4l) - 16 < O. (14.24) 16a Let H'(8) = 0. We get its two roots to be 81 = (2l- a 2)/3a and 82 = -a/2; and, under the conditions of the theorem, 81 ,82 > 0, and since H(O) ~ 0, for all8 < 0 inequality (14.24) holds. The theorem is completely proved. For the case of a 2 (a2 -4l) -16 < 0, since at this time H(O) > 0, inequality (14.24) cannot hold for all 8 > 0. Suppose 83 is a unique negative root of H(6) = 0. Then when 8 < 83 we have H(6) < 0. Using a suitable estimate on 83, we can get the following theorem:
+
5a2 - 4l 62 4a
+
a2
-
2
21 6
+
a 2 (a 2
-
THEOREM 14.5. When l > 0 and 6 -l/a ~ -a(1 + (l- 4)/(a2 system (14.22) has at most one limit cycle (unstable). The proof is omitted; see the original paper of the author [179]. (B)The following five theorems are given in [179] and [218].
+ l))/8,
§14.
CLASS II EQUATIONS WITH 1 OR 2 CYCLES
319
FIGURE 14.7
FIGURE 14.6 THEOREM 14.6. When l one limit cycle (stable cycle).
< 0 and a2 +4l ~ 4,
system (14.22) has at most
PROOF. Let the line L1 without contact and they-axis intersect at 'fhi it is easy to see that 'fh > 0. As before, we only have to prove that 'fit ~ ~. By the proof of Theorem 14.4 we only have to establish inequality (14.24), and 6 ~ min(61. ~). Under the conditions of this theorem, it is clear that a 2 (a 2 - 4l)- 16 > 0; hence H(O) < 0. Hence H(6) has at least one positive root; we suppose its smallest positive root is 64. From the conditions of this theorem, we know that H (l /a) < 0, and
l
;i <
21- a 2 3a <
a
-2,
and it is easy to compute H( -a/2) > 0; thus -a/2 is a minimum point of H, and (21- a 2 )/3a is a maximum point of H; hence 64 should be between l/a and (21- a 2 )/3a (see Figure 14.6). Hence when 6 < 1/a, inequality (14.24) must hold, and 6 ~ min(61. ~). Again by Lemma 14.3, the conclusion of this theorem holds. Estimating suitably the approximate value of 64, we can also get THEOREM 14. 7. When l < 0 and 6 + a/8 ~ (1 (14.22) has at most one (stable) limit cycle. The proof is omitted. ·
+ Jl2"+4)j4a,
system
THEORY OF LIMIT CYCLES
320
The uniqueness Theorems 14.4 and 14.6 obtained above can only solve part of the problem on the number of limit cycles for system (14.22). Considered from the parametric (a, l)-plane, the above two theorems only prove that when (a,l) is in region I (Figure 14.7), the limit cycle of (14.22) is unique. Nothing is known outside this region. But we can definitely say that the points in (a, i)plane not in the region I, the uniqueness of a limit cycle does not necessarily hold. THEOREM 14. 8. When l ~ 2a2 , if there exists a limit cycle surrounding the origin, then for some 6 there are at least two [179]. PROOF. Suppose r is a limit cycle of (14.22) surrounding the origin. Calculating the integral of divergence along r' we get
J(f)
= =
i (~: i
[6
+ ~~)
dt
=
i
+ l ~a!6(y2- y)]
=~
1
(6
+ 2lx) dt =
i
dt
[y2-y+6(l-a6)] dt
!r 2al = ~ 1 [( _ !) 2 26(l- a6)- all l - a6 !r Y 2 + 4al l- a6
(6- 2alx 2) dt
d t.
Hence when 6 (we may as well assume 6 > lja) satisfies 26(l- a6)- al ~ 0, we have J(f) < 0. From this we can see that in this case r in fact does not exist, since. if r exists and is not unique, then we can always find one r such that J(f) < 0 does not hold; conversely, if r exists and is unique, the stability of r and the stability of the origin are the same, which is not possible. On the other hand, in order to make 6 satisfy 26(l- a6)- al ~ 0, we only require that
6
~ 6' = 2~ ( 1 -
J
7
1- 2
2
)
0
Also, only when 6 ~ "G = (4l- a 2 )/4a do the singular points Rand A exist. When l ~ 2a2 , it is clear that 6' > 8. Hence in the range of variation of the parameter from generation of a limit cycle to its disappearance, R and A do not exist. Now suppose the limit cycle r exists and is unique. Then it must be an unstable cycle, since the origin is stable. r continues to expand as 6 decreases, until it passes through the saddle point N and becomes an internally unstable separatrix cycle. But we compute that the divergence of N equals 6 < 0, and this requires that the separatrix cycle should be internally stable. The
321
§14. CLASS II EQUATIONS WITH 1 OR 2 CYCLES
above contradiction explains why for some 6, the number of limit cycles in the vicinity of 0 is at least two. From this we can see that if the number of limit cycles in the vicinity of 0 can be at most two, then this case is the same as the case of 0 < m < 1/a- a in (I), when the stable cycle in the vicinity of the origin has not yet expanded to arrive at the point N, but the two separatrices passing through the point N have already come close together and coincide to form a separatrix cycle. Afterwards, the separatrix cycle disappears, and generates a stable cycle which shrinks inwards; here the vicinity of the origin has exactly two limit cycles, and they finally coincide to become a semistable cycle and disappear. Theorems 14.4, 14.6, and 14.8 show that in the (a, i)-parametric half-plane, the problem of the number of limit cycles has been solved only in regions I and n, but the problem in region n has not yet been completely solved since the property of having at most two limit cycles has not yet been proved. Up to now, those theorems in §7 have not been used for system (14.22), since its function ~(y) is a monotonic function only when y < For anywhere outside regions I and IT, how many limit cycles does the cor~ responding system (14.22) have? This is still an unsolved problem. However, we still believe that if system (14.22) has limit cycles, it should not have more than two. (Ill) n = 0. Here we have the system
!·
dx dt
= -y+6x+lx 2 +mxy,
dy
dt = x(1 +ax).
(14.25)
Without loss of g~nerality, we assume that a = 1 and l > 0. Thus we have
dx
dt = -y + 6x + lx
2
+ mxy,
dy
dt = x(1 + x),
l > 0.
(14.26)
It is easy to see that this system has two singular points 0(0, 0) and R( -1, (l- 6)/(m + 1)). 0 is a nonsaddle point; R is a nonsaddle point when m + 1 :::; 0, and is a saddle point when m + 1 > 0. · LEMMA 14.4. When 6 = 0 and 0:::; a limit cycle in the whole plane [219].
m:::; 2, system (14.26) does not have
PROOF. We note that 1- mx = 0 is a line without contact. Construct a Dulac function
B(x,y)
= (1- mx)2/m
2
+2/m-tex 2 -my 2 +2(1/m+t)x-2l 11 •
Then when 6 = 0, we have
!_(BP2 ) + 88 (BQ2)
ax
y
=
-l(2mx2 + 2- m)x 2 (1- mx)- 1B(x,y).
THEORY OF LIMIT CYCLES
322
It is obvious that the right side of the above formula is negative when 0 $ m $ 2. The lemma is proved.(9 )
LEMMA 14. 5. When 6 = 0 and m $ 0, system (14.26) does not have a limit cycle surrounding the origin. PROOF. Construct a Dulac function B(x,y)
= (1/(mx -1))e- 21Y.
Then when 6 = 0, we have :x (BP2)
+ :y (BQ2)
= -l(2mx- 2 + m)x2(mx- 1) 2 e- 21 Y.
Clearly the right side of the above equation has a constant sign when x > 1/m. Since 1 - mx = 0 is a line without contact, any closed trajectory surrounding the origin must lie in the region x > 1/m. The lemma is proved.
LEMMA 14.6. When l(l + m6) $ 0, system (14.26) does not have a limit cycle. PROOF. Construct a Dulac function B(x,y) = (1- mx)- 1 . Then
:. (BP2 ) +
~ (BQ,) = ! [-1m
2 ( •-
! )' + + l
l
rM (1 -
=)-'.
Under the condition of the lemma, the right side of the above formula has a constant sign. The lemma is proved.
LEMMA 14. 7. When 6/l ~ 1, system (14.26) does not have a limit cycle surrounding the origin.(1°) PROOF. Consider the system dxjdt
= -y + mxy,
(14.27)
which takes the origin as its center. Comparing the slopes of the trajectories of (14.27) and (14.26), we get -y + mxy + lx 2 + 6x -y + mxy lx + 6 X + x2 X+ x 2 = 1 +X • From this we can see that the locus of points of contact of the trajectories of these two systems consists of 1 + x = 0 and lx + 6 = 0. Any closed trajectory r of system (14.26) cannot intersect 1 + x = 0; if 6/l ~ 1, then the line lx + 6 = 0 is on the left side of x + 1 = 0, and cannot intersect r. But this means that r cannot exist. ( 9 )[210] gave another method of proving this lemma. (1°)Since :Vs and l(m- 2) have the same sign, we only have to prove Lemma 14.7 for m < 2; when m ;::: 2, the conclusion is obvious. Moreover, this lemma also holds when l < o.
§14.
CLASS II EQUATIONS WITH 1 OR 2 CYCLES
323
LEMMA 14.8. When 6 ~ l, there does not exist a limit cycle of system (14.26) surrounding the singular point R. PROOF. Move the origin of the coordinate system to R. The system becomes dx ( m(l- 6) ) 2 dt = -(1 + m)y + 6 + m + 1 - 2l x + lx + mxy, (14.28) dy dt = x(x -1). We may as well assume 1 + m < 0, for otherwise R is a saddle point or does not exist. Apply the transformation y , t = t/ v' -1 - m; v'-1- m then (14.28) becomes (we still write x, y, and t as x, y, and t) x
= -x, y =
dxfdt = -y + 6'x + l'x 2 + m'xy,
dyfdt = x(1 + x),
(14.29)
where
6- (2 + m)l l' _ -l 6, _ - (m+1)../-1-m' - v-1-m' (14.30) m' = -m(1+m)- 1 • Then we obtain the proof of this lemma from Lemma 14. 7, since from 6' /l' ~ 1 we can deduce 6 ~ l. We have seen from the previous discussion that when 6 = 0 the focal quantity of the origin has the same sign as l(m- 2). Also from {12.21) it is easy to see that when 6 = 0 and m = 2, the focal quantity of the origin has the same sign as -l; hence from the theory of Bautin we can get THEOREM 14.9. (i) When 6 = 0, m > 2, and m- 2 « 1, there exists at least one limit cycle in the vicinity of the origin. (ii) When m > 2, 6 < 0, and 0 < 161 « m - 2 « 1, there exist at least two limit cycles in the vicinity of the origin. (iii) When m < 2, and 0 < 6 « 1, there exists at least one limit cycle in the vicinity of the origin. (iv) When 6 = 0, m < -2, but lm + 21 « 1, there exists at least one limit cycle in the vicinity of R. (v) When m = -2 and 0 < 6 « 1, there exists at least one limit cycle in the vicinity of R. COROLLARY 14.1. By (iii) and (v) of the theorem, when m = -2 and 0 < 6 « 1, limit cycles can coexist in the vicinities of 0 and R. In the following we shall study the uniqueness problem of a limit cycle in several cases. Since we already know that the line 1 - mx = 0 is without
THEORY OF LIMIT CYCLES
324
contact, we can introduce the change of variables dtfdr = 1/(1- mx). Hence system (14.26) becomes (we still denote r as t):
dx lx 2 +ox dt = -y + 1- mx '
dy x'+ x 2 dt = 1- mx ·
( 14·31 )
In order to study the uniqueness of a limit cycle, we apply the transformation _y
=-
y + lx 2 + ox 1-mx'
x=x
to system (14.31). Then the system becomes (we still denote andy) dxfdt = y, dyfdt = -g(x)- f(x)y, where
x + x2 ( ) gx =1-mx'
f(x)
x andy
as x
(14.32)
= lmx 2 -
2lx - o (1~mx) 2
System (14.32) has two singular points: 0(0, 0) is a focus when R(-1,0) is a saddle point when 1 + m > 0.
161 < 2, and
THEOREM 14.10. When m < 0, system (14.26) has at most one limit cycle surrounding the origin [220]. PROOF. We shall use Theorem 6.4 of [112}, §6, to prove this theorem. Clearly we only have to verify that when m < 0, f(x)fg(x) is a nondecreasing function in the intervals (1/m, x 1 ) and (0, +oo ), where Xt
= ( 1 - J1 + 6mjl) jm
is a zero of f(x) (from Lemma 14.6, we know that when 1 + 6m/l ~ 0 the system does not have a limit cycle; hence we may as well assume 1+6mfl > 0). Here we can consider two cases: when m ~ -1, we consider the intervals (1/m,xt) and (O,+oo), and when 0 > m > -1, we consider the intervals (-1,x 1 ) and {O,+oo). It is easy to compute
d (l(x)) dx g(x) where
=
P4(x) (1- mx)2(x + x2) 2 '
+ (21- ml- 3mo)x2 + (26 - 2m6)x + o. Since we have assumed l > 0, we may as well assume o > 0 when m < 0, for P4 (x) = m 2lx 4
-
4mlx 3
otherwise the vicinity of 0 cannot have a limit cycle. Obviously P4(x) > 0 when x > 0. In the following we only have to prove that in the interval (1/m,xt) (when m ~ -1) or in (-1,x 1 ) (when 0 > m > -1), P4(x) > 0.
§14. CLASS II EQUATIONS WITH 1 OR 2 CYCLES
325
First we consider x $ -6 jl, i.e. x E (1/m, -6 fl) or x E ( -1, -6 /l). Rewrite P4(x) as P4(x) = (x + 6/l) lx( -m + 2- 4mx + m 2 x 2 ) - 6m(x- 1/m)(mx2 + 1); then it is easy to see that no matter what the value of x E (1/m, -6/l) (when m $ -1), or x E ( -1, -6/l) (when 0 > m > -1) is, we have P4(x) > 0. In order to determine the sign of P4(x) in (-6/l,xt), we can rewrite it as
2lx- 6)x(mx- 1)- mlx 3 + (-ml- 2m6)x 2 + (6- 2m6)x + 6.
P4(x) = (mlx 2
-
Suppose
cp(x)
= -mlx 3 + (-ml- 2m6)x 2 + (6- 2m6)x + 6.
Then
= (1 -
+ m6 /l) > 0, cp'(x) = -3mlx 2 + 2( -ml- 2m6)x + 6- 2m6.
6 fl)8(1
(14.33) (14.34)
Form< 0, if cp'(x) = 0 has a real root, they must be two negative roots, the larger of which is
xo = 1 + 26/l + J(1- 6~~
+ 36(6/l + 1/m)/l
<
-~.
(14 _35)
From (14.34) and (14.35) we can see that cp(x) is an increasing function in (-6/l,xt), and from (14.33) we know that cp(x) > 0 in this interval. Hence when m < 0 and x.E (-6/l,xt), P4(x) > 0. The theorem is proved. From Corollary 14.1 and Theorem 14.10, we again consider system (14.29), and get COROLLARY 14.2. When m = -2 and 0 < 6 « 1, system (14.26) has one unique limit cycle each in the vicinity of 0 and R. Cherkas and Zhilevich [221] modified the uniqueness theorem of [118] to obtain the following Lemma 14.9, which is used to study the uniqueness of a limit cycle of system (14.26) when m > 0. LEMMA 14.9. Suppose in the strip Xt < 0 < x2 we are given an equation dy g(x) (14.36) dx = F(x)- y"
J; g(x) dx,
Introduce Filippov's transformation z = into dyfdz = 1/(Ft(z)- y), dyfdz = 1/(F2 (z)- y),
to change the equation
when x > 0, when x < 0.
(14.37) (14.38)
THEORY OF LIMIT CYCLES
326
Equations (14.37) and (14.38) are defined in the intervals (0, zot) and (0, zo2) respectively, where Zoi = J;• g(x) dx. Suppose the following conditions are satisfied: 1) xg(x) > 0 when x '=I 0, and g'(O) '=I 0. 2) Fi(z) < 0 when 0 < z < zot· 3) There exist two numbers z 0 and z* (0 ~ zo < z* < Zoi) such that (a) FHz)(z- zo) < 0 when z '=I zo, 0 < z < zo2; (b) Ft(z") = F2(z*); (c) F~'(z) < 0 when z" < z < Zo2i (d) either F~'(z) < 0 when z0 < z < z"; or when zo = 0 and limz--+0 o:i(z) = o:0 > 0 (where O:i ( z) = Fi ( z) - 1/ FI( z)), the following inequalities hold: F~'(z)(z-
F~'(z)
<0
z) < 0 when z '=I z < z*, 0 < z < z*, when 0 < z
< zoti
(e) (djdy)[F2(y)- Ft(Y)] > 0 when (3 < y < Ft(z"), where (3
=Max [ lim Fi(z)] . 1=1,2 Z--+Zo\
Here Fi(Y) is the inverse function of Fi(z). Then equation (14.36) has at most one limit cycle; if this cycle exists, it must be a single cycle.
The proof is similar to Theorem 6.11, and is therefore omitted. THEOREM 14.11. Let one of the following three conditions hold:
1) o~ = 0·,
2) 0 < m
-1 + v'4m - 7 <_ 2·, 3) m > 2' 1~ > ---::---2m ·
Then system (14.26) has at most one limit cycle; if it exists, it must be a stable cycle [221].
PROOF. From Lemma 14.7 we know that when~~ l, system (14.26) does not have a limit cycle; also from Lemma 14.4 and the theory of rotated vector fields we know that when ~ ~ 0 and 0 ~ m ~ 2, the system does not have a limit cycle. In the following we may as well just consider the system of equations dx lx2 +~x dy x+ x 2 (14.31) dt = -y + 1 - mx ' dt = 1-mx·
§14. CLASS II EQUATIONS WITH 1 OR 2 CYCLES
327
Comparing with (14.36), we know now
F( ) _ lx 2 + 6x x - 1-mx' z x+x 2 z= 1 -mx dx, 0 = F~ x' = (lx~ + 6xi)' 1-mxi az z 1- mx·I :l:i Xi +x~
( ) _ x + x2 gx -1-mx'
1
F~u
_ -mlx~(z) + 2lxi(z) + 6 - (1- mxi(z))(xi(z) + x~(z)) ·
'
Here Xt(z) (x2(z)) is the inverse function of z(x) for x > 0 (x < 0). From the above formula we get that zo = 0 when 6 = 0, and in the region in which a limit cycle may exist we have
F'(z) r=
l(-mx1 + 2) (1- mxt)(1 + ~1?·
When 5 > 0, the sign of F~ (x2 (z)) at the point X2(ZO) =
Xo
=
1- J1-m6/l
----!....----:.....
m
changes from positive to negative, and Fi(z) < 0 for 0 < z < ZOl· In the following we have to prove that the curves Ft(z) and F2(z) have a common point in the interval 0 < z < min(zot, zo2); for this we only need to show that the system of equations
lx~ + 6x2 _ lx~ + 8x1 1 - mx2 - 1 - mx1 ' :~:2 x + x2 x + x2 dx = dx 10 -:--1-mx 1-mx 0
1zt
(14 .39 ) (14.40)
has a unique nontrivial solution when x2 < 0 and Xt > 0. We note that if Ft(z) and F2(z) do not have a common point, or (14.39) and (14.40) do not have a real solution for x 2 < 0 and Xt > 0, then system (14.31) does not have a limit cycle. From (14.39) we can solve Xt
-6 -lx2 mx2)'
= cp(x2) = 1{1 -
and again in (14.40) we consider Xt as a function of x 2. From this we get
X~
= '1/J(Xt, x2) = X2(X2 + 1)(1- mxt). Xt(Xt + 1)(1- mx2)
THEORY OF LIMIT CYCLES
328
Study the difference of the two derivatives:
where
P4(x2) = m 2 l 2 x~
+ (-2ml 2 + m2 l 2 )x~ + (2l 2 + (2lo + ml8)x 2 + 62 - 8l.
ml 2 )x~
In order to prove that the system of equations (14.39) and (14.40) has a unique nontrivial solution, we only have to prove that P4(x2) has only one zero for -1 < x2 < 0. Now suppose not, i.e. P 4 (x2) has at least two zeros in -1 < x2 < 0. Clearly this is not possible at o = 0. Hence we can assume o > 0. Since P4( -1) > 0 and P4(0) < 0, P4(x2) has at least three zeros in (-1,0). Thus P4(x2) has at least two zeros in (-1,0). But from P4(-1) < 0 and P~(O) > 0, we can deduce that P4(x 2) has at least three zeros in (-1,0). Thus P~'(x 2 ) must have at least two zeros in (-1,0); but this contradicts the direct calculations. Thus the system of equations (14.39) and (14.40) has a unique nontrivial solution when x2 < 0 and Xt > 0. In the following we verify (c) and (d) of condition 3) in Lemma 14.9. We can directly compute
where
Q4(xi) = m 2lx1 - 4mlx~ + (2l- ml- 3mo)x~
+ 2(1- m)8xi + 0. It is easy to see that when 8 = 0 for all z E (0, zot) we have FJ.' < o. and as for F~'(z), there exists a unique z < z• such that when z = z, F~'(z) changes from positive to negative. When 8 > 0, we only have to study F~' (x2 (z)) for x2 E ( -1, xo). First we consider x2 E (-8/l,xo), and rewrite Q4(x2) as
Q4 (x2) = (mlx~ - 2lx2 - h')x2(mx2 -
1)
+ (x 2 + 1)( -mlx~- 2mox2 + h'). It is easy to see that Q4 (x 2 ) > 0 when F~'(x2(z))
< 0.
X2 E
(-ojl, xo), and consequently
§14. CLASS II EQUATIONS WITH 1 OR 2 CYCLES
329
Next consider -1 < x2 ~ -6/l, and rewrite Q4(x2) as Q4(x2) = ( x2
+ ~) [m 2 lx~- (4ml + m 2 6)x~ +
m2 62) mt52 m2t53] ( mt5 + 21 - ml + - 1- X2 - mO - - 1- - l l
mt52 +o + -~-
mt53
m2t54
+ l2 + ---za· It is clear that when mo + 2l - ml + m 2t5 2/l (-1 + v'4m -7)/2m) we have Q4(x2) > 0.
~ 0 (that is, 6 /l
>
In order to verify condition 3)(e) of Lemma 14.9, it suffices to prove that the equation
has only one zero when condition (14.39) holds. It is easy to compute that
Ll _ 1 -
where -y(u)
( -l- m0)(x1 - X2)'y(x1 + X2) ( -mlx~ + 2lx2 + o)( -mlx~ + 2lxt + o)'
= u 2 + (1- 2/m)u- 2ojml.
Since (14.39) holds,
-8 -lmx~
Xt
+ X2 = l(1- mx2) < 0,
and so Llt has only one zero in the region under study. The conditions of Lemma 14.9 have all been verified. THEOREM 14. 12 [222]. When t5 ~ 0 there exists at most one limit cycle of system (14.26) surrounding the origin. PROOF. By Lemma 14.7 we only have to study the case when m l
> 2 and
> o > 0. Changing the signs of y and t in (14.31), we obtain dy dt
=
x + x2 1 - mx'
dx dt
lx 2 + ox = -y - 1 - mx ·
We have to use Theorem 6.11 to prove this theorem. By the proof of Theorem 4.11, we now only have to verify the condition
2F{'(z)z + F{(z) < 0, since
z=
1 x
0
-1
x+x 2 dx 1-mx
1( m1) x -m1- ( m1) lnll-mxl,
2 -=-X
2m
when 0 < z < zot.
m
1--
2
1+-
THEORY OF LIMIT CYCLES
330
F'( ) -mlxHz) + 2lxt(z) + 6 1 z = (1- mxt(z))(xt(z) + xHz))' where Xt(z)
> 0 is the inverse function of z
=
1 z
0
x+x2 1 -mx dx,
and
where
P4(xt) = m 2 lxf - 4mlx~ + (2l- ml- 3m6)x~
+ 26(1- m)xt + 6. Hence we only have to prove that when 1/m > x
Q(x)
>0
= 2P4(x)z(x) + (mlx 2 - 2lx- 6)(x + x2 ) 2 < 0;
here z(x) can be expanded into a power series
z=
1
m+1
m(m+1) 4 x 4 mn- 3(m+ 1) xn+···, n
2 3 2x + 3 -x +
+···+
and so the above formula can be written as
Q(x)
= x3 { Q1(x) + Qo(x)- 2m6- 2l + i(m + 1)6 + x(ao(26- 2m6) + at6) + x2[at{26- 2m6) + a26] + x3[a2{26- 2m6) + aa6] + · · ·},
where
Q1(x) = [m2 lx 4 - 4mlx 3 + (2l- ml- 3ml)x2 ] · [ 32 (m + 1) + -m+1 4 -2mx + ... + 2(mn+ 1) mn-3xn-3 + .. ·] < 0, Q0 (x)
= x(-2l- 6- 3mo) + x 2 ( -2ml- 2l) + ml(m + 1)x2 < 0,
an= !(m + 1)mn-axn. n Since 0 < x < 1/m, an < an-t· From this it is easy to prove that Q(x) < 0, and the theorem is thus proved.
§14. CLASS IT EQUATIONS WITH 1 OR 2 CYCLES
331
Starting from the above three theorems, many cases of the problem of uniqueness of a limit cycle of system (14.26) have been basically solved, but no results on the problem of having at most two solutions have been obtained. The problem oflimit cycles of system (14.26) is also studied in [223] and [224]. The complete study of equations (14.1) of class II has been given in [209], as mentioned in §13; that is, the global structure and bifurcation curves of the system possessing two fine foci of third order (hence without cycles) was studied. Also, the author of [223] used the transformation x = x, y = y - (JX, but {3 there was a coefficient to be determined. After the above transformation, system (II) becomes (we still denote x and y by x and y)
dxjdt
=-
dyjdt =
y + (t5 + f3)x + (l- mf3 + nf3 2 )x2
+ (m- 2nf3)xy + ny2 = {3P(x, y) + x(1 +ax).
P(x, y),
(14.41)
= x(1 +ax),
(14.42)
Comparing this with the system
dxjdt
= P(x,y),
dyjdt
we can derive from the qualitative properties of (14.42) some qualitative properties of (14.41). For example, we can prove THEOREM 14.13. If (14.42) does not have a limit cycle in the vicinity of 0(0, 0), and 0 is a stable (unstable) singular point of (14.42), then when {3 > 0 ( < 0), (14.41.) does not have a limit cycle in the vicinity of 0. Suppose the vicinity of another singular point M of index +1 of (14.42) does not have a limit cycle, and M is an unstable (stable) singular point of (14.42). Then when {3 < 0 (> 0), (14.41) does not have a limit cycle in the vicinity of M. Also if (14.42) does not have a limit cycle in the whole plane, and 0 and M have the same stability, then limit cycles of (14.41) cannot coexist in the vicinity of 0 or M.
[223] also applied the above transformation to equations of class I, and chose a suitable {3 to prove THEOREM 14.14. 1) Suppose l
< 0, l + n > 0, 0 < 3n ~ y1- 4nl, and
~ =!4 ( v'1-n 4nl ) 3 +
1 2n (1- v'1- 4nl) > 0.
Then (I)m=l does not have an unstable limit cycle for 0 < -D ~ ~. 2) Suppose one of the following conditions holds: (i) n > 0, t5 + 1/2n ~ 0.
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332
(ii) 0 < 4n(l + n) < 1 and 8 + [1- y'1- 4n(l + n)]/2n ~ 0. (iii) n < 0 and 8 + 21/(1 + v'1 - 4nl) ~ 0. Then (I)m=l does not have a limit cycle. Also if in (14.41) we choose (3 = m/2n and choose a Dulac function e-2(1-m 2 / 4 n)y for (14.42), and apply Theorem 14.13, then we can prove that if a
(8 + ;:) (1-7:) ~ o
then limit cycles of (II), if they exist, must be centrally distributed. Moreover, Chen Lan-sun [224] also studied (14.1) and obtained theorems on nonexistence, existence, and number of limit cycles, for examples, THEOREM 14.15. I/8
m
'# 0,
= m(1+n)-2a = 0, n '# ~1, n(4-m2)-2m2 '# 0,
then system (14.1) has at most one limit cycle in the whole plane.
THEOREM 14.16. When 5n + 3 > 0, m(1 + n)- 2a > 0, 8 < 0, and l5n+ 31, lm(1 +n)- 2al, and lSI are suitably small, then system (14.1)1=1,6#0 has at least three limit cycles near the origin. THEOREM 14. 1 7. I/ 8 = l + n = 0 and a '# 0, then the vicinity of 0 does not contain a limit cycle. If 0 < -a- (m 2 + 4)/2m ¢:: 1, then the vicinity of R has at least one limit cycle. If a < 0 or -a- (m 2 + 4)/2m < 0, then the vicinity of R does not have a limit cycle.
Exercises 1. Prove the conclusion on the uniqueness of a limit cycle of system (14.1) at the beginning of this section. 2. Prove that when m < 0, n > 0, and a > 0, system (14.1)6=0 does not have a limit cycle in the vicinity of 0 [224]. (Hint. Introduce the time transformation dt/dr = e- 2 '11 /(mx- 1).) 3. Prove Lemma 14.1. 4. Prove Lemma 14.2. 5. Prove that the conclusion of Lemma 14.3 also holds for the singular point of index + 1 other than 0. 6. Prove Theorem 14.3. 7. Prove that Y10 ~ in Theorem 14.4 is equivalent to the inequality in (14.24). 8. Prove the equality in (14.30). 9. Prove that the system
!
dxjdt = -y + x 2 + mxy- y 2 ,
dyjdt = x(1 +ax)
§14. CLASS II EQUATIONS WITH 1 OR 2 CYCLES
333
does not have a closed or singular closed trajectory near the origin, and that near the other singular point R of index +1 when a ! -(4 + m 2 )/2m the system has an unstable cycle which shrinks and approaches R, and when a~ -(4 + m 2 )/2m there does not exist a closed or singular closed trajectory near R. 10. Prove that if in the system of equations
dxjdt
= -y +ox+ lx 2 + mxy + y 2 ,
we have a < 0, l > 0 and 82 + m8 + l in the vicinity of the origin.
dyjdt = x(l +ax)
< 0, then there exists at most one cycle
§15. Some Local and Global Properties of Equations of Class III
In this section we study some local and global properties of the system
dxjdt
= -y + 6x + lx2 + mxy + ny2 ,
dyjdt = x(1 +ax+ by),
(15.1)
which may have a limit cycle. H we do not require that b '::/: 0, then (15.1) represents the most general quadratic system which may have a limit cycle, and the equations of classes I and IT discussed in §§12-14 are in fact special examples of (15.1). AB we mentioned in §12, Soviet mathematicians have another way of classifying quadratic systems. No matter whether they are equations of Class A or B, after transformation into our classification, generally speaking, they all possess the characteristic of b '::/: 0. Hence in this section we shall also introduce some results on quadratic systems obtained by Soviet mathematicians in the seventies and shall see that their method of classification has its own good points. First we introduce a simple but important theorem on the nonexistence of a limit cycle. THEOREM 15 .1. The system
dxjdt
= -y + lx2 + ny 2 ,
dyjdt = x(1 +ax+ by)
(15.2)
has one or two centers when a(b + 21) = 0, and does not have a closed or singular closed trajectory when a(b + 21) '::/: 0. PROOF. For system (15.2) we can take(l)
B(x, y) = ( 1 + by)-21/b-1., (1 )Note that the right side of (15.3) can take a complex value when 1 +by < 0. 335
(15.3)
THEORY OF LIMIT CYCLES
336
Then we can compute :x (BP2)
+
:y
'
(BQ2) = -a(b + 2l){1 + by)- 21 /b- 2 .
{15.4)
When a(b + 2l) = 0, the right side of (15.4) is always zero, and system (15.2) has a first integral; hence it has one or two centers. When a(b + 2l) ::/: 0, the right side of (1.54) keeps a constant sign in the half-plane 1 +by > 0, and 1 +by = 0 is a line without contact for the system (15.2); hence (15.2) does not have a closed or singular closed trajectory in the vicinity of the origin. In order to prove that (15.2) is the same in the half-plane 1 +by < 0, we only have to move the origin to the singular point of index +1 in the half-plane 1 + by < 0 (it is easy to see that if this singular point exists, it must be (0, 1/n)); then, using the new function B(x, y) of the form (15.3), we can get the proof as before. When a( b + 2l) ::/: 0, results on the global phase-portrait and bifurcation surface of system (15.2) are given in [208] and [209], as mentioned at the end of §13. Later, Ju Nai-dan attempted in 1965 to prove for special equations of class 111(2)
~~
= -y + 8x
+ lx 2 + mxy + ny2 ,
~~ = x(1 +by)
(15.5)
some results completely parallel to Theorems 12.4 and 12.6 for equations of class I, but his conjecture was only partially proved. He obtained the following two theorems. THEOREM 15. 2. The system of equations
dx dt
= -y + lx 2 + mxy + ny2 ,
~~ = x(1 +by)
(15.6)
has a center when m(l + n) = 0, and if the algebraic equation n(n + b)fP - mO- 1 = 0
(15.7)
has a real root for 9, then (15.6) does not have a limit cycle when m(l+n) ::/:0. THEOREM 15.3. When one of the conditions 1) n = 0, 2) l = 0, 3) b = -n, or 4) b = l holds, the limit cycle of system (15.5), if it exists, must be unique. (2)His results given here were not published by him, but were given in §14 of the first edition of this book.
337
§15. EQUATIONS OF CLASS III
The first theorem can still be proved by the method of the Dulac function, and will be discussed in the next section. The second theorem can be proved by the uniqueness theorem of [111]. During the ten-year period after 1966, our work was stopped completely. But during the same period, Soviet mathematicians made remarkable progress in studying quadratic systems. At the end of §12 we mentioned the new method of Kukles and Shakhova [200] for proving Theorem 12.4. In 1970, Cherkas [203) used the method of [200) to prove Theorem 15.2, and did not need to add more conditions to system (15.6). After that Cherkas, Zhilevich, and Rychkov (see [225], [221] and [292]) also used [112], [128] and a uniqueness theorem similar to that in [118] to prove that if system (15.5) has a limit cycle, then it must be unique. Hence Ju Nai-dan's conjectures have been completely proved. Now we introduce new equations with center constructed in [203] to compare a given quadratic system, and use it to prove the nonexistence of a closed trajectory of system (15.6) (Theorem 15.2), in which we do not demand that (15.7) has a real root for 0. For convenience, we adopt the method of the original paper to prove that the equation dy -x+ax 2 +bxy+cy2 (15.8) = dx y(cx + 1) (which is the system (15.6)) does not have a limit cycle. We may as well assume c ¥- 0, and then assume c = 1. Also, we may as well assume a ~ 0 or -1 < a < 0. If a ~ -1, then we can use the transformation
1+xt
1 = -1 - ,
y Yt = 1 +x
+x
to change (15.8) to dy1 dx1
in which we have -(a+ 1) dy dt
+ bx1Y1 + (1- c)y~ Yt(Xt + 1)
-x1 - (a+ 1)x~
=--~~--~~--~~~--~~
~
·
0. Rewrite (15.8) as a system of equations
= -x + ax 2 + bxy + cy2
'
dx dt = (x
+ 1)y.
(15.9)
The stability of (0, 0) can be determined from the sign of b(a +c). When b(a +c)> 0 (< 0), (0, 0) is an unstable (stable) focus, and when b(a +c)= 0, (0, 0) is a center. It is clear that a limit cycle can only appear in the half-plane :t > -1. First we prove that when ac ~ 0, a 2 + c2 > 0, and b ¥:- 0, system
338
THEORY OF LIMIT CYCLES
(15.9) does not have a limit cycle. This because (here T is a period of the cycle)
h=
~loT [bx + {2c + 1)y] dt
=~loT [b (ax 2 +bxy+cy2- ~~) + (2c+ 1)y] dt = !!._
{T (ax 2 + cy 2 ) dt, T lo and the right side of the above equation has the same sign as b( a + c); hence there is no limit cycle. Hence from now on we assume one of the following two conditions holds in order to prove our conclusion: a
> 0, c < 0, b '!- 0;
-1 < a < 0,
c > 0,
(15.10)
b '!- 0.
{15.11)
Consider the equation
(x + 1)yy1 = -x + ax2 + [bx + 1/J(x)]y + cy2 •
(15.12)
We choose a suitable 1/J(x) to make {0, 0) become a center of (15.12). In {15.12) we apply the transformation y = cp(u), where cp(x) is also a function to be fixed. We get
(x + 1)cp2uu' = -x + ax2 + [bx + 1/J(x)]cpu + [ccp- (x + 1)cp']cpu2. Now choose
+ 1)cp']cp = a( -x + ax2), (bx + 1/J)cp = [3( -x + ax 2),
[ccp - (x
{15.13) (15.14)
where a and {3 are constants. From (15.13) we see that cp(x) satisfies an equation of the form
(x + 1)yy' = cy2 + ax(1- ax) whose general integral is 2
2c [
=
(x + 1Y [ M
y = (x + 1)
M
(15.15)
r (x +
+ Jo
2ax(1- ax) ] 1)2c+1 dx .
(15.16)
Now we take
cp(x)
p. ( )
+ 2a (x; ~) 2 c
]
- 2aP2(0)
1/2
,
§15. EQUATIONS OF CLASS III
339
where P2 (x) is the numerator of the rational function obtained from the integration of {15.16), which equals 1 +a
1 + 2a
a
~ + 1- 2c (x + 1) - 2 - 2c (x + 1)
2
1 +a+ {1 + 2a)(x + 1) ln(x + 1)- a(x + 1) 2 1+a - 2 - - (1 + 2a)(x + 1) - a(x + 1) 2 In(x + 1)
(c = 1).
M is a constant to be determined later. Then from {15.14) we get
'1/J(x)
X
= cp(x) [.8(-1 + o:x)- b
Note that o:, ,8, and M have not yet been fixed. H we can make '1/J(x) have a constant sign in some interval [m, n], except for a finite number of points, then comparing the equation (15.12) with center and the given equation {15.9), we see that {15.9) does not have a limit cycle in the strip m ~ x ~ n, -oo < y < +oo. In order to make the function '1/J(x) have the same sign in a neighborhood of x = 0, we require that
.8 = -b
(a+ c)(1 +a- c) (1- 2c)2c(1- c) Now we take an arbitrary o:
0
< ·
< 0 and take M
= 2o:P2(0); then the graph of
cp(x) is a branch of a hyperbola. Thus '1/J(x) has a constant sign in (-1, 1/a). At this time, the qualitative figure of the integral curve of equation (15.15) between the branches of the hyperbola is as shown in Figure 15.1. Note that the hyperbola is a separatrix passing through a saddle point on the line X= -1. (ii) When a+ c < 0 we still take o: < 0, and we choose
P2(1ja)
M = 2o:P2(0)- 2o: ( 1/a + 1)2c,
so that y = cp(x) is a separatrix of equation (15.15) passing through (1/a,O). Thus the line y = VM (1 - ax) intersects the separatrix at two points, whose
THEORY OF LIMIT CYCLES
340
y, cp(z)
FIGURE
15.1
y.
FIGURE
15.2
abscissas are 0 and 1/a, and there are no other points of intersection because the number of contacts of any line with the vector field of (15.15) is not more than two. The qualitative figure of the integral curve of (15.15) is as shown in Figure 15.2. From this we can see that ,P(x) keeps a constant sign in ( -1, 1/a). Now we look at the case (15.11). Take a> 0; then under condition (15.11), equation (15.15) has two saddle points (0, 0) and (1/a, 0) on the x-axis, which are on opposite sides of x = -1. It is easy to prove that the infinite point on the y-axis is a saddle point. At the same time there are still two finite nodal points, lying on the line x = -1. One separatrix starting from (0, 0) enters the nodal point Pt(-1, y'o(a + 1)/c), and another separatrix runs to the infinite nodal point. From this we can see all the trajectories above these two separatrices all connect the nodal point ( -1, y' o( a + 1)/c) and the infinite nodal point; that is, the graph of the single-valued function y = cp(x) obtained previously when M > 0. Now take M = afc(a + 1). Then the line y = v'M(1- ax) will pass through the saddle point (1/a, 0) and the nodal point (-1,y'o(a+l)/c), and this line is not tangent to the vector field of (15.15) when x > -1. If we take the function cp(x) determined by M (see Figure 15.3) and substitute it in the formula for ,P(x), we can see that ,P(x) keeps a constant sign when x > -1. Thus, we have already proved (15.8) does not have a limit cycle. In the following, we again prove the uniqueness of a limit cycle of equation
(15.5). THEOREM
15.4. If equation (15.5) has a limit cycle, it has at most one.
§15. EQUATIONS OF CLASS lli
341
FIGURE 15.3 PROOF. For the convenience of those readers who wish to refer to the original papers, we shall use their notation; that is, we rewrite {15.5) as
du dx
=
-x + ax 2 + (bx + c)u + du 2 u(x + 1)
{15.17)
Using the transformation u = (x + 1)dYl {when d = 0, it is an identity transformation), equation {15.17) can be changed to
dy1 -x + ax 2 + (bx + c)(x + 1)dY1 dx = (x + 1)2d+ly1 ' or to the equivalent system of equations
dy
dt
= (x- ax 2 )(x + 1)-'Jd-l = g(x),
~: = -y +
Lz (bx + c)(x +
1)-d- 1 dx
= -y
-foz f(x) dx.
(15.18)
First we study the uniqueness of a limit cycle in the vicinity of the origin. From Theorem 15.2 and the theory of rotated vector fields, we know that for cb(a +d)> 0 there does not exist a limit cycle surrounding the origin; hence in the following we may as well assume cb( a + d) < 0. Also, without loss of generality, we assume that b < 0, and (as in the proof of Theorem 15.2) that a > -1. Hence, from now on, we always have
b < 0,
c( a + d) > 0,
a > -1.
In the following, we carry out our discussion in six different cases: 1) a ~ 0, c > 0, d ~ 0; 2) a > 0, c > 0, d < 0, a + d > 0; 3) a > 0, c < 0, d < 0, a + d < 0;
THEORY OF LIMIT CYCLES
342
4) -1 0, a+ d < 0; 6) -1 < a ~ 0, c > 0, d > 0, a+ d > 0. I. First we study cases 1) and 4), which are characterized by ad~ 0. From the position of the singular points and integral lines of {15.17), we see that limit cycles can only lie in the strip -1 < x < 1/a (or x > -1, for case 4)) or, when a= 0, in the half-plane -1 < x. Next, on computing the divergence for (15.18), we get 8Pf8x + 8Qj8y = (bx
+ c)(x + 1)-d-l,
and any limit cycle cannot intersect with the integral line x + 1 = 0; hence it must intersect the line bx + c = 0. Thus only for b + ac < 0 does there exist a limit cycle. From this it is not difficult to prove that for system (15.18), in case 1), when x E (-1,0) or (-cjb, 1/a), we have( 3 ) d (') (x dx =
g
+ 1)(-abx2 - 2acx +c)+ dx(bx + c)(ax- 1) (ax2- x)2(x + 1)1-d
(15.19)
>0. Use x2 to denote the abscissa of the extreme right point of a limit cycle L1 near the origin. Then we must have X2 > -c/b. Construct a function !1(x) = f(x)- f((x 2)) g(x). g
X2
In the region surrounded by L1 we can find a point (x 1 < 0, 0) such that ft(xt) = 0, and it is clear that ft(x2) = 0. Since ft(x)fg(x) is monotone in the intervals (-1,0) and (-c/b, 1/a), we know that ft(x) < 0 when x E [x1,x2]. Now for system {15.18) we apply Theorem 6.4 of §6 to obtain the uniqueness of a limit cycle. In case 4) it is not hard to see that a limit cycle surrounding the origin can exist only when c > b,( 4 ) and as before it must contain the point ( -cjb, 0) of the x-axis. Here we again use (15.19) to prove that in the intervals (-1, -c/b) and (0, +oo) we have d(f fg)fdx < 0; the rest of the proof is similar to case 1). II. Next we study cases 2) and 3), the ones in which a> 0 but ad< 0. As before, we still assume b < 0. Hence we have to slightly modify the uniqueness theorem (Theorem 6.11) of Rychkov before we can use it (see [221]). that here -aln: 2 - 2acx +cis positive definite. first use the transformation dt/dT = 1/(1 + x), and prove that the integral of the divergence along any closed trajectory is always negative when c = b. Thus the origin is also stable, and there does not exist a closed tr~ectory. ( 3 )Note
( 4 )We
343
§15. EQUATIONS OF CLASS III
LEMMA 1. Suppose that for d1 < x < d2, dtd2 equation dy g(x) dx = F(x) -y'
<
0, we are given an
which can be transformed to dy dz
1 F1 (z ) - y
when x
> 0,
dy dz
1
= F2(z)- y
when x
< 0,
J;
by means of the Filippov transformation z = g(x) dx. The above two equations are defined for 0 < z < Zot and 0 < z < zo2 respectively, where Zoi = J;• g(x) dx. Suppose the following conditions are satisfied: 1) xg(x) > 0 when x =/:- 0, and g'(O) =/:- 0. 2) FHz) < 0 when 0 < z < zot· 3) There exist unique numbers zo and z• (0 ~ zo < z• < ZOi) such that a) F~(z)(z- zo) < 0 and z =/:- zo, 0 < z < zo2i b) Ft(z•) = F2(z•); c) F~'(z) ~ 0 when z• < z < Z
!~ (F,(z)- Ff~z)) =no> 0 the inequality F~'(z)(z- z) < 0 holds when z =/:also, Ff'(z) < 0 when 0 < z < zot; e) ddY [F2(Y)- Ft(Y)] > 0 for Ft(z*) > y > {3
z < z•, 0 < z < z•; and
=~ax [ lim t=1,2
Fi(z)] ,
Z-+Zoi
where Fi(Y) is the inverse function of Fi(z). Then the equation has at most one limit cycle; if it exists, it must be a single cycle.
The proof of this lemma, except a few individual cases, differs very little from Theorem 6.11; hence it is omitted. In the following we use this lemma to prove for a> 0 (i.e. cases 2) and 3)) the uniqueness of a limit cycle of system (15.18) surrounding the origin. We only prove case 2); the proof of 3) is similar. In the (y, z)-plane, we examine the curves y = Fi(z) (i = 1, 2) and use the following parametric equations to represent them: (i=1,2).
THEORY OF LIMIT CYCLES
344
Here
F(x)
=-
foz f(x) dx
c - b] b c- b = - (x + 1) -d [ d-b 1 (x + 1) + -d+ d- 1 + -d-, G(x)
= foz g(x) dx = (x + 1)
_ 2d [
1
1]
-1
< X2 < 0 < Xl
a 2a + a+ 2 2d - 2 (X + 1) + 1 - 2i X + 1) + 2d
a 2a + 1 a+ 1 - 2d- 2- 1- 2d- 2d'
1
The point of intersection of these two curves can be determined from the equations F(xt) = F(x 2) and G(xt) = G(x2). We can prove existence and uniqueness of the solution of this system as in [118], i.e., the curves y = F1 (z) and y = F2 (z) have only one point of intersection (z*, F 1 ( z*)). We omit the proof. From the method of differentiation of compound functions, we know that F{'(z) < 0 (when 0 < z < zot) in the lemma is equivalent to
.!!._ dx
(L) > o g
when 0 < x < 1/a. Now we rewrite the numerator of the fraction of (15.19) P(x) = (x + 1)( -abx 2 - 2acx +c)+ d(bx +c)( -x + ax 2) as
P(x)
= -a+d - ( x + 1)(-abx2 a
2acx +c)+ d(b + ac)(a + 1)x
3
de
--(ax -1) 2 [(a + 1)x + 1]
a
when a+ d > 0, d < 0, b + ac < 0, a> 0 and c > 0. From 0 < x < 1/a, we can deduce that P(x) > 0. Hence we obtain the proof that Fi'(z) < 0 when 0 < z < ZOl· In order to verify the condition
:y[Ft-l(y)- F2-t(y)] < 0 in the corresponding region, we only have to prove that the two curves z = [Fi- 1 (y)]~ (i = 1, 2) have at most one point of intersection for y > 0. The point of intersection can be determined from the system of equations G'(xt) G'(x2) F'(xt) = F'(x2)' F(xt) = F(x2),
§15. EQUATIONS OF CLASS III
345
which is equivalent to
A ( X1 + ~) ( X2 + ~)
= ~ + 0,
(15.20)
where
A= b(-ba d-1
+
B=ac(d~1 +c~b),
c-b) d
+
b(b+ac) d-1 '
C=c(d~1 +c~b)·
If we consider condition 2) and the necessary conditions for existence of a limit cycle (b + ac < 0, b- c < 0), we see that
B 2 + AC = c(b + ac)[ac(d- 1) + b(a + d)][c(d- 1) + bJ, d- 2 (d- 1)- 2 < 0, i.e., the first equation of (15.20) determines the increasing function x2 = cp(xl). Next, starting from the second equation of (15.20), we compute dx2/dx1 at z = z•, and get
dx2 dx1
F'(xl)
Fi(z*)dzl/dx
= F'(x2) = F~(z•)dz2/dx'
where
.
dzl/dx = g(xl) > 0, and Fi(z•) < 0 by condition 2), while FHz*) < 0 by condition 3a). Hence dx2/dx1 < 0, and from F(xt) = F(x2) we determine the decreasing function x2 = .,P(xt). Thus it is clear that (15.20) has a unique solution. The proof for case 3) is similar. III. Finally, there remain cases 5) and 6), in which ad < 0 but d > 0. In order to prove the uniqueness of a limit cycle of system (15.81) for this case, G. S. Rychkov [292] improved Lemma 1 and obtained the criterion for uniqueness:
zo, z• and y• (0 < zo < < zo1. /3 < y• < 0), such that the following conditions hold: (5 ) 1) F~(z) < 0, 0 < z < zo2, and F1(0) = F2(0) = 0. 2) (zo- z)Fi(z) > 0 when z '# zo,O < z < zo1i and F1(z*) = F2(z*). 3) 2F~'(z)z + FHz) < 0, 0 < z < Zo2· 4) (y•- y)d(F1(y)- F2(y))jdy > 0 when y # y•, /3 < y < 0, LEMMA 2. Suppose that there exist unique points
z•
( 5 )The
meaning of {3, zoi. and Fi(Y) is the same ss before.
THEORY OF LIMIT CYCLES
346
Then the equation dy g(x) dx = F(x)-y
cannot have more than one limit cycle; if it exists, it must be unique. There is a great difference between the proof of this lemma and the proof of Lemma 1 or Theorem 6.11. In order to verify that system (15.18) in cases 5) and 6) satisfies condition 3) of the lemma, the calculations are very complicated; hence the details are not given here. Interested readers can refer to Rychkov's paper [292]. The above analysis only proves the uniqueness of a limit cycle of system (15.5) in the vicinity of the origin. Note that if (0, 0) and another singular point (0, 1/n) are on the same side of 1 +by = 0, then (0, 1/n) is a saddle point; but if 1 +by = 0 separates (0, 1/n) from (0, 0), then (0, 1/n) is also a singular point of index +1. The uniqueness of a limit cycle in its vicinity can be solved by the method of translation of the origin, because at this time the form of system (15.5) does not change. Finally, take B(x, y) = (1 + by)- 21 /b-l; then we can compute
a
ax (BP)
a + ay (BQ) =(my+ o)B(x, y).
But y = -8/m is a line parallel to 1 +by= 0 which does not coincide with 1+by = 0; therefore it must be on one side of 1+by = 0, and any limit cycle of (15.5) cannot intersect the line 1 +by= 0. Hence limit cycles cannot coexist in the vicinities of (0,0) and (0, 1/n). Theorem 15.4 is completely proved. REMARK. The reader has already seen (Theorem 12.6 in §12) that in order to prove the uniqueness of a limit cycle of class (I), we have to consider four different cases, using different transformations of variables to change them into equations of Lienard type, and then apply the uniqueness theorem of [111] to solve the problem. For Theorem 15.4 of this section, the situation is similar. However, the form of equations, no matter whether it belongs to class (I) or (Ill)a=O (i.e. system (15.5)) is much simpler than the form of equations after transformation; moreover, the difference between these two systems is only one factor (1 + by) on the right side of the second equation, and the conclusions for the nonexistence and uniqueness of a limit cycle are exactly the same. Hence we have reasons to mention the following questions, which deserve the reader's consideration for further deep investigation. PROBLEM I. Can we prove the uniqueness of a limit cycle of class (I) directly without transformation of variables? PROBLEM II. Can we deduce from the uniqueness of a limit cycle of class (I) the uniqueness of a limit cycle of class (Ill)a=O?
347
§15. EQUATIONS OF CLASS III
Now we present two important corollaries of Theorem 15.4: COROLLARY 15 .1. If a quadratic system has an integral line and a fine focus, then it does not have a limit cycle.
This holds because we can change the system into (15.6) by transformation of coordinates. COROLLARY 15. 2. If a quadratic differential system has one integral line, then it has at most one limit cycle.
This holds because we can change the system into (15.5) by transformation of coordinates. The above two corollaries were mentioned at the end of §11. For the uniqueness of a limit cycle of equations of class m, we also have the following results. Chen Lan-sun [226] proved that the system
dxjdt = -y + 6x + lx 2 ,
dyjdt = x(1 +ax+ by)
has one and only one limit cycle when 6a(b + 21) > 0 and 6 varies in some interval (0, 6*) or (6*, 0). Zou Ying [227] proved the uniqueness of a limit cycle of the system
dxjdt
= -y + 6x + bx 2 + Mxy- by2 ,
dyjdt
= x(1 +ax+ by).
Liu Jun [228] extended these results, but assumed the first equation does not contain the y 2 term. Ren Yong-tai [229] studied the existence, uniqueness and stability of a limit cycle of the system
dxfdt = ly + mx -ly2
+ nxy- mx2 + a(x + bny- x2 ),
dyjdt = x(l +by- x). For the problem of centralized distribution, except the equations of class
(ill) discussed in §14, we still have the method of proof first used in [25]: when [2am- b(b + 2l)][m2 - 2n(b + 21)] ~ 0
(15.21)
limit cycles of system {15.1) must be centrally distributed (i.e. it is not possible that there exist limit cycles in the vicinities of two singular points of index +1). Yang Zong-pei [27] extended the method of [25] and also improved the result, showing that when (15.21) does not necessarily hold, the limit cycles of system (15.1) must be centrally distributed provided that [2am- b(b + 2l)][m2
-
2n(b + 2l)]
+ m 2 b2
~ 0.
(15.22)
348
THEORY OF LIMIT CYCLES
Li Xiao-gui [230] applied the transformation x = z, fj = ax/b+y to system (15.1) to obtain the system of equations (we still denote x and ii by x andy)
= -y + 6'x + l'x::~ + m'xy + ny2 = P(x, y), dyjdt = (a/b)P(x, y) + x(1 +by) = Q(z, y),
dxjdt
(15.23)
where
d'
= d + ~b'
l'
= l- ~b (m- ~n) b
(15.24)
I
Then comparing (15.23) with the system
~~ = -y + 6' z + l' x::~ + m' xy + ny2 ,
dy dt = x(1 +by)
(i5.25)
he obtained many results about the existence and centralized distribution of limit cycles. For example, he proved
15.5. If system (15.25) has a limit cycle or a separatrix cycle r surrounding a singular point 0, and if, when a6/b > 0, system (15.23) has a limit cycle in and only in the interior of r (this cycle must be stable (unstable) if a/b > 0 (< 0)), and (15.25) does not have a cycle in the vicinity of 0, then when (ajb)(6 +a/b) < 0 (15.23) does not have a cycle in the vicinity of 0. If (15.25) has a limit cycle or a separatrix cycle f' in the vicinity of another singular point M of index +1, then when (a/b)(6 + m/n) < 0 system (15.23) has a limit cycle in and only in the interior of r' (this cycle must be stable (unstable) ifa/b < 0 (> 0)), and if(15.25) does not have a cycle in the vicinity of M, then when THEOREM
i (6+ ~ -i) >0
(15.23) does not have a cycle in the vicinity of M. In addition, in the same paper Li also obtained some results on the centralized distribution of limit cycles. There is a very good result on the centralized distribution of limit cycles of class III obtained by Chen Lan-sun and Wang Ming-shu [19]. They proved THEOREM 15.6. If, in addition to two singular points of focal type of index +1, (15.1) has a third finite singular point, then its limit cycles must be centrally distributed; conversely, if (15.1) has only two finite singular points of focal type, then its limit cycles cannot be centrally distributed.
The proof is by the method of the Dulac function, whose details are in §16. In addition, in [231] Liu Nan-gen discussed the problem of centralized distribution of equations of class II.
§15. EQUATIONS OF CLASS III
349
The range of variation of 6 (the coefficient of x on the right side of the first equation of (15.1)) to guarantee existence of a limit cycle is an important problem of theoretical significance and practical value. We have already given a simple example in §11 (181]. Also in §12 we pointed out that for equations of class I, if the range ofvariationoU is (0,6*) or (6*,0), then 6* = f(l,tn,n), but this function and the equations of its separatrix cycles are unknown to us. For system (15.5), Yu Shu-xiang [179] obtained under given conditions four results on the precise interval of 6 to guarantee existence of a limit cycle, as follows: THEOREM 15.7. 1) /fn+l > 0, m > 0, l < 0, 0 < n S 1 andmll4n(l + 1) > 0 in system (15.5), then there exiats a limit cycle surrounding the origin if and only if
0 > 6 > 2:(::1) (-m + .../m2- 4n(l + 1)). 2) Ifl + n < 0, m > 0, l > -1, 0 < n < 1 and mil + 4l(1 - n) > 0 in (15.5), then there exists a limit cycle surrounding the origin if and only if
0 < 6 < 2:(::1) ( -m + .../m2 - 4n(l + 1)). 3) lfn = 0, m > 0, -1:::; l < 0 and mil+ l > 0 in (15.5) then there exists a limit cycle if and only if 0 < 6 < -1/m. 4) lfn = 0, m > 0, -1:::; l < 0 and mil+ l:::; 0 in (15.5), then there exists a limit cycle if and only if 0 < 6 < -m + 2A. Similar results were also obtained in [232] and (233]. For example, I. G. Rozet (233] proved THEOREM 15.8. When l = 0, b > -n, 6 = mjb, and mil+ 4nb < 0, system (15.5) has two separatrix cycles formed by the line 1 +by = 0 and the semi-equator, which surround two finite singular points of index +1, respectively. Similarly to [119], Kukles and Rozet (232]showed that the separatrix cycle corresponding to the endpoint of the interval of existence of 6 is also formed by a section of the arc on the equator and two half-ray segments. Wang Huifeng (234] studied the range of variation of 6 for the existence of a limit cycle and obtained the following succinct result: THEOREM 15.9. Suppose that b = -1 in (15.5}, and suppose that m > 0, 2l + 1 > 0, and n = 1 + m 2 /(2(21 + 1)). Then the system has a limit cycle if and only if (1 - n)/m < 6 < 0.
THEORY OF LIMIT CYCLES
350
Here when 6 = (1-n)/m, the separatrix cycle obtained is no longer formed by a line segment or a section of arc on the equator, but is an elliptic trajectory through the saddle point (1/n, 0). Moreover, Suo Guang-jian and Du Xing-fu [235] also studied the conditions for system (15.5) to take a curve formed by one branch of a hyperbola and an arc of the equator as its singular closed trajectory (they also obtained an admissible interval of variation for some varying parameter to guarantee existence of a limit cycle) and the conditions which take a curve formed by a parabola and one line segment to be its singular closed trajectory. Suo [236] studied similar problems for a quadratic system with symmetric center dy dt =
b
+
""' W
bijX
i
.
y3.
(15.26)
i+j=2
For example, he obtained THEOREM 15.10. If in the system of equations
dx 2 dt = 1- x +xy,
~~ = l-lx2 + 2xy + ny2 + a(1- x2 + xy)
(15.27)
we have l - 2 > n ~ 0, ln - l + 2 < 0, and ln - l + n + 3 > 0, then (15.27) has two limit cycles which do not contain each other if and only if l-ln- 2 . n+ 1 In addition, there are several papers which study the qualitative properties of a quadratic system with a third-order fine focus (0, 0):
O
~~ = -y + lx 2 + 5axy + ny2 ,
~~ = x + ax 2 + (3l + 5n)xy.
(15.28)
Cai Sui-lin [237] and Wang Ming-shu and Lin Ying-ju [238] proved independently that when n = 0, (15.28) does not have a limit cycle in the whole plane.( 6 ) Chin Yuan-shun et al. [182] further proved THEOREM 15 .11. If (15.28) has only one infinite elementary singular
point, then this system has only two finite singular points, one of which must be a coarse focus. There are an odd number of limit cycles surrounding the coarse focus, and an even number surrounding the origin.
In the following we present some nice results obtained by Soviet mathematicians on the study of quadratic systems according to their method of classification. ( 6 )Recently Li Cheng-zhi proved that when n =f. 0, (15.28) does not have a cycle in the whole plane (the paper will be published in Chinese Ann. Math.)
351
§15. EQUATIONS OF CLASS III
Consider the equation ' - Q2(x, y) - Lo
=
= y -TJX,
Yl
(15.29)
= y; then the coefficient
(~, ~) -17P2 (~. ~)] ele=o
~[(a~ob~oTJ) + (a~l- b~117)17 + (a~2- b~217)TJ 2 ]. 1]
We take 17 to be the zero of the quadratic expression of 17 in [ ] on the right side of the formula, and we still denote x1 and Yl by x andy. Thus we obtain
dy Q2(X, y) dx = boo+ b10x + ~ox 2 + buxy + bo1Y ·
(15.30)
1. If bu =/: 0, then we can further use a linear transformation through x to make bo1 ;:= 0, and assuming y' = b10 +~ox+ buy and x' = x, we get (we still denote the numerator by Q 2, and x' andy' by x andy) dy dx
Q2(x,y)
= boo+xy·
(15.31)
Here if boo = 0, then the equation and (III)a=O are essentially the same; if boo=/: 0, then (15.31) can be changed to
dy dx
a~ + a10x + a01y + a2ox 2 + auxy + ao2Y 2
=------~--~~--------~~--~
1+xy
If bu = 0, then (15.30) can have a limit cycle only when bo1 case (15.30), after a linear transformation of x, becomes
dy Q2(x,y) dx = ~ox 2 +y"
(15.32)
=/: 0. In this (15.33)
Here if ~o = 0, then the equation is of class I, and there is no need for further consideration. If ~o =/: 0, then (15.33) becomes
dy Q2(x,y) dx= y+x2 •
{15.34)
(15.32) and (15.34) are the two main classes in the Soviet method of classification mentioned in §11. Now we derive some useful results for (15.32) In (15.32) we make the transformation y 1 Y=-+-2
x
x
'
(15.35)
THEORY OF LIMIT CYCLES
352
Then we get (15.36) where P4({) = -a2o- a10{ +(au- aoo)e P2({) = (1 + 2ao2){2 - aot{- au.
+ ao1e- ao2e\
(15.37)
The system of equations corresponding to (15.36) is d{ = cy: dt ...
{15.38)
From (15.32) we can see that x = 0 is a line without contact of the system; hence it cannot intersect any limit cycle. Thus, when it is changed to (15.36), the number of limit cycles and their positions do not change. Since { = 0 is an integral line of (15.38), it cannot intersect any limit cycle; thus we can make a transformation {15.39) and (15.38) becomes Y ~~
= (P4(x)l{l 2ao 2- 3 + P2({)1elao 2- 2 y)sgn {,
{15.40)
or the corresponding system of equations d{/dt
where
= y,
dyfdt = -g({) -f({)y,
= -P4({)1{12ao2-3sgn {, /({) = -P2({)1elao 2sgn {.
g({)
(15.41)
{15.42)
2-
The divergence of system {15.41) is - f({). From this we can see that if P 2({) keeps a constant sign in the region { > 0 or { < 0, then system (15.41), and hence (15.32), does not have a limit cycle in that half-plane. In fact, this conclusion is also valid in the strip {15.43) For this we first prove THEOREM 15.12. If P4({) = Q2({)P2({) in (15.37), i.e. P2({) is a quadratic factor of P4({), then equation (15.32) does not have a limit cycle. PROOF. We first assume P2({) has two nonzero roots, i.e. au "I 0 and 1 + 2ao2 -I 0. Then Q 2({) = -a02 (1 + 2ao2)- 1 + ao1(1 + ao2)(1 + 2ao2)- 2{ + a2oail· (15.44)
e
§15. EQUATIONS OF CLASS III
353
Now we transform (15.41) into the Lienard plane, and obtain drJ/dt
= -g(~),
(15.45)
where F(~) =If.: J(€) de, and (eo, 0) is a singular point offocal type of (15.45) or (15.41). In order to prove (15.45) does not have a limit cycle in the vicinity ~f ( ~o, 0), we may as well confine our discussion to the strip < here eo_(€o) is greater than (less than) ~0 and is a root of g(e) = 0 closest to eo; if (eo) does not exist, we just take -oo (+oo) instead. Using Filippov's transformation z =If.: g(€) de, (15.45) can be transformed to two equations
eo e<eo;
eo
1 dTJ dz - F1(z)- TJ
(corresponding to
~0 ~ ~ < eo),
{15.45)1
1 drJ dz - F2(z) - TJ
(corresponding to
eo < ~ ~ ~o).
(15.45)2
= F1(z)
In the following we shall prove that the curves TJ has one common point (0, 0) in the region
O$z:$olln
[t
and TJ
= F2(z)
g(€)d€,t g(€)d€].
From this, by Theorem 5.4 of §5, we know at once that (15.45) does not have a limit cycle in the vicinity of (~o, 0); to see this we only have to prove that the quantity (15.46) (where Fi(TJ) is the inverse function of Fi(z)) does not change sign in the corresponding region, or( 7 ) that the two simultaneous equations
1
(1
Q2(6)~fo2-1 = Q2(~2)~~o2-1, P2(~)eao2-2 de=
f.o
1(2
(15.47)
P2(~)eao2-2 d~
(15.48)
f.o
eo·
to
have a unique system of solutions 6 = 6 = ~0 when > 6 ~ ~0 ~ 6 > It is easy to see that when ao2 ::F -1, 0, 1, condition (15.48) can be changed to 1 + 2a02 (€002+1 _ ~Go2+1) _ ao1 ( ~002 _ ~002) ao2 + 1 1 2 ao2 1 2 (15.49) _ au (CGD2-1 _ ~ao2-1) = 0 ao2
( 7)I.e., g(e<J)/ /({<J)
el = e'l =eo.
-1 "' 1
2
'
= g(6)//(6) and F(el) = F(e2) has only one system of solutions:
THEORY OF LIMIT CYCLES
354
and conditions (15.47) can be written as
- ao2(1 + 2ao2)(e~o 2 +1
e;o2+1) + ao1(1 + ao2)(e~oa - e;02 )
-
+ a2o (1 + 2ao2)2(e~o2-1- e;o2-1) = 0.
(15.50)
au Substituting the formula for eio 2+1- e;o 2+1 of (15.49) into (15.50), we get (eio2-1 - e;o2-1 )6.
= 0,
(15.51)
where
6. = -ao2au (1 + ao2) + a2o (1 + 2a 02 )2. ao2 -1 au Since except when 6 = 6 = eo we always have 6 < et. equation (15.51) has a unique solution 6 = 6 = when 6. ::/:- 0, i.e., in the vicinity of (eo, 0) there is no limit cycle. H 6. = 0, then (eo, 0) is a center and hence does not have a limit cycle. For the case ao2 = -1, 0 or 1, we can make a similar transformation and get an equation equivalent to (15.51) as (a2o/au)[e1 2 - G 2] = O,
eo
(a2o/au)[e1 1 - e2 1] = 0 or - 2auln{6/6) hence we can get the same conclusion as before. For the case au = 0, Q2 (e) has the formula
= 0;
Q2(e) = -ao2(1 + 2ao2)-le + a01(1 + ao2)(1 + 2ao2)- 2e + alO j aOl
the remaining calculations are similar to the case au ::/:- 0. Moreover, if P2(e) has a pair of zero roots, then from the fixed sign of the divergence we know at once that it does not have a limit cycle. The theorem is completely proved. From this theorem we can immediately deduce an important property of a quadratic system mentioned at the end of §11: THEOREM 15.13. If the curve of system (15.38) whose divergence i3 zero passes through two elementary singular points,( 8 ) then it does not have a limit cycle. PROOF. By our previous discussion, the elementary singular points should have the forms of singular points (6,0) and (6,0). Also from (15.38), we can see that the characteristic equation of these points is
>.2 ( 8 )In
-
P2(ei)>.- ei1'4(ei) =
o.
this case these two singular points are either fine foci or fine saddle points, or one is a fine focus and the other is a fine saddle point; and their common characterizing feature is: the linear part of the characteristic root has the same absolute value.
§15. EQUATIONS OF CLASS III
355
By the hypothesis of the theorem we must have P2 (6) = P2 (6) = 0. But on the other hand, 6 and 6 must be the roots of P4(€) = 0; hence P2(€) is a factor of P 4 (€). Application of Theorem 15.12 will complete the proof of this theorem. Now we return to (15.43). Suppose P2(€) has a constant sign in this strip region. Then it is clear that system (15.41) does not have a limit cycle in this strip. Hence if P2(€t) = P2(6) = 0, can the system have a limit cycle surrounding (6,0) or (€2,0)? From Theorem 15.13 we know immediately that the answer is negative. Finally, we note that when Bol = a10 = 0, the phase-portrait of system (15.38) has symmetric properties with respect to = 0; in this case in order to study the existence and number of limit cycles, it is sufficient to confine ourselves to the half-plane € ~ 0. Applying the transformation €2 = X, (15.36) can then be changed to a simpler form: 2XY(dY/dX) = - a2o +(au- aoo)X- ao2X2 (15.52) + [(1 + 2ao2)X- au]Y + (1- ao2)Y 2.
e
Hence in this case the system has changed into (15.5) of class (TII)a=O.· Thus, in the half-plane > 0 (or < 0), there exists at most one limit cycle, and if the system has a fine focus, it does not have a limit cycle.(9 ) Similar results were obtained later independently in [239]. REMARK. The results of [237] and [238] mentioned before for system (15.28) are proved by the method of proof of Theorem 15.12. In addition, in §14 we mentioned that the nonexistence of a limit cycle of system (14.5) under condition (14.17) is also proved by the same method [216]. Theorem 15.13 is only proved for equation (15.32). But Cherkas and Zhilevich in [222] carried out a similar study for (15.34). For this, they first used a transformation of variables
e
(15.53) to change (15.34) into
dfi dx
=
- f(x)jj- g(x) y
(15.54)
where
f(x)
= -P2(x)e-ao2x = -(ao1 + (2 + au)x- 2ao2x2)e-ao 2 x,
g(x)
= -P4(x)e-ao2x = -(aoo +
a10X
+ (a20- aol)x 2 - aux 3 + Bo2X4je- 2a 02 x.
(15.55)
--:----(9)Note that from (15.32) we can see that the vector field determined by this system indeed has a center of symmetry (0, 0).
THEORY OF LIMIT CYCLES
356
Then they used a similar method to show that Theorem 15.13 also holds for (15.54), and hence for (15.34). Finally we note that (15.31}, when boo 0, belongs to the class (Ill)o=o, and so Theorem 15.4 also holds for (15.31). Hence we obtain
=
THEOREM 15 .14. If the quadratic system (15.29) has two fine foci or two fine saddle points or one fine focus and one fine saddle point, then it does not
have a limit cycle. REMARK. According to the above discussion, all quadratic systems can be transformed into systems of Lienard type by an appropriate transformation of variables. But this method of transformation is not unique. Liu Jun [228] gave a transformation of variables suitable for systems more general than quadratic systems. THEOREM 15 .15. Equations of the form
dxjdt = fo(x) -!l(x)y, dyjdt
= go(x) + g1(x)y + g2(x)y2
(!l(x) #: 0)
(15.56)
can be transformed into equations of Lienard type by a change of variables. PROOF. First let x
dxjdt
= e,
= x and e= fo(x) -!l(x)y. de/dt
where
t/Jo(x)
Then
= -1/lo(x) -1/ll(x)e -11'2(x)e,
= !l(x)go(x) + fo(x)gl(x) + ~:~=~f~(x), 1/12(x) = g2(x) _ fi(x) !l(x) !l(x)'
t/J1(x) Again let
= ~:~:~fo(x) -fMx)- g1(x)- 2~:~=~fo(x).
e= uexp(- J; t/J2(z) dz). Then (15.57) becomes ~~ = uexp (- foz 11'2(z) dz), ~~ exp (- foz 11'2(z) dz) - u 2t/J2 exp ( -2 foz 11'2(z) dz) = -t/Jo(x) -1/11 (x )u exp (- foz tP2(z) dz) - 1P2(x)u2exp ( -2 foz t/J2(z) dz) .
(15.57)
357
§15. EQUATIONS OF CLASS III
The above formulas are the same as
~: = uexp (-los ~2 dz) ,
~; = -~ exp (los~ dz) - 1P1 u.
Next let
u loz 1P1 exp (lo' ~ dz) ds,
V= + Then we get
~: =V- los ~~(z)exp(Lz ~(s)ds)
dz, {15.58)
~ = -~(z)exp (2 los ~(z)dz),
which are the required forms of the equations. The theorem is completely proved. It is clear that (ill)n=O has already possessed the form of (15.56). In order to apply thi15 transformation to (ill)n~o, we first assume n 1, and apply the transformation z = i/(1 + y), y = fi/(1 + fi),
=
to the system
dz/dt = -y(1- y) + 6z + lz 2 + mzy,
dy/dt = z(1 + az +by)
which maps the line y = 1 containing the singular point (0, 1) to infinity (correspondingly, the equator becomes fi = -1), and thus we obtain a system of equations of the form (15.56). Finally, for singular closed tr9Jectories of quadratic systems, in addition to [14], we also mention the Soviet papers [192], [240], and [241]. In [192] the classification of singular closed tr9Jectories was tabulated, and examples were given to illustrate the possibility of being realized. In [240], L. H. Cherkas concentrated on the existence and stability of singular cycles of equation (15.31) when boo = 0, and in [241] he proved that when the coefficients of (15.31) satisfy the conditions
boo= 0, ag 1 - 4aooao2 < 0, (-1 + ao2)/ao2 < 0,
a~ 1 - 4a2o(ao2 -1)
< 0,
aooa2o < 0,
the equation has two critical singular cycles formed by the upper (lower) halfequator and the z-axis, but whose stability cannot be determined from the characteristic values of two saddle points in the singular cycle, and he obtained the sufficient conditions to distinguish its stability. But Zhou Kong-rong (242]
THEORY OF LIMIT CYCLES
358
pointed out mistakes in the above calculations, and corrected them, and also gave a simpler sufficient condition to distinguish its stability. Exercises
1. Prove Theorem 15.3. 2. Prove formula (15.16) and the formula for P2(x) after it. 3. Verify Figure 15.1, and construct under condition (15.11) the global phase-portrait of equation (15.15). 4. Use the method of computing the integral of the divergence around a closed trajectory once to prove that the equation
dyjdx = ( -x + ax 2 + bxy + cy 2 )Jy does not have a limit cycle provided one of the following three groups of conditions holds: (i) a < 0, c = 0; (ii) a = 0; (iii) a > 0, c = 0. 5. Prove (15.36) and (15.37). 6. Prove (15.40) and (15.44). 7. When ao2 = -1, 0 or 1, prove Theorem 15.12. 8. Prove (15.54) and (15.55). 9. Give details on the derivation of the formulas in the proof of Theorem 15.15. 10. Prove that the system of equations dx/dt = -y +ox+ lx 2 + mxy, dyjdt can be transformed into the Lienard equation dx dr
= -vdv dr
= x(1 +ax+ by)
[b b ox+ lx 2 ] m x + m 2 1n( 1 - mx)- 1- mx '
-=
x + ax 2 1-mx
bx(ox + lx (1-mx)2 '
2) +~'-----~
by the transformation
v
= y- !:x- _!_2 ln(1 m
m
mx) '
dt 1 = -:---dr 1-mx·
11. For the system of equations
dyjdt = x(1 +ax- y), prove that when o ~ m > 0, a(2l + 1) ~ 0 or o~ m < 0, a(2l + 1) ~ 0 it does not have a limit cycle, and if 0 ~ a - o ~ m and l + ma ~ 0, then it has one and only one limit cycle. 12. Study under the transformation X= 1/x, Y = (1 + xy)/x2 the corresponding relation between the regions in the projective plane (x, y, z) and (X, Y, Z), and point out in what regions they do not have a one-to-one correspondence. dxjdt = -y +ox+ lx 2 + mxy,
§16. The Method of Dulac Functions in the Qualitative Study of Quadratic Systems
In previous sections, in studying the nonexistence of a limit cycle for equations of classes I, II, and ill, we have already seen that the method of Dulac functions seems to have special advantages. For example in the proof of nonexistence of a limit cycle of (1) 6 = 0 (Theorem 12.4), we simply use two different Dulac functions (12.22) and (12.23), and then compute their divergences and immediately solve the problem. But [201] and [202] use three or four pages to prove the satne result. Similarly, for nonexistence of cycles of class (ill)a=6=0 in the previous section we presented the proof of [203] and also :used several pages. In the following we shall see the proof of Chen Lan-sun [243] and Ju Nai-dan: THEOREM 16 .1. The system of equations
~: =-y+lx2 +mxy+ny2, ~~ =x(1+by) center when m(l + n) = 0, and does not have a limit
has a m(l + n)
(16.1) cycle when
-1 0.
PROOF. 1) The case m 2 + 4n(n +b) ~ 0. Here n(n + b)02 - mO -1 = 0 has two real roots. Take any root of this equation, and let
Bl(x, y)
= (1 + by)(mn9-b-2l)fb(n0y- X- O)m9.
(16.2)
We compute 8(B1P)
ax
+ 8(B1Q) 8y
= -m(l + n)Ox2(1
+ by)(mnB-b-21)/b(nOy- X- o)m9-1;
(16.3)
also, we note that nOy- x- 0 = 0 is an integral line passing to the saddle point or nodal point (0, 1/n) and touching tangentially to one exceptional direction at this point, or all the trajectories intersecting it all cross through it in the 359
THEORY OF LIMIT CYCLES
360
same direction. Hence a limit cycle, if it exists, can only lie on one side of this line, and so from (16.3) we obtain the proof of this theorem. 2) The case m 2 + 4n(n + b) < 0. Here (16.2) cannot be used, sj.nce n(n + b)6 2 - m6- 1 = 0 has only a pair of conjugate complex roots. Now we follow [243) and take B 2(x,y)=(1+by)- 21 /b- 1 exp{ 2:tan- 1 [ where u = J-m 2
-
~in_+:;)- ;]}•
(16.4)
4n(n +b). Then we compute
:x (B2P) + :y (B2Q) = m(l +
n)~2(1 ;- ny) (1 +
x exp (2m tan_ 1 u
[-
by)-21/b-1
2n(n + b)x _ m]) u(1- ny) u '
(16.5)
where L = (1- ny) 2 - n(n + b)x 2 - mx( 1 - ny) is a positive definite quadratic form, and now 1 +by = 0 separates (0, 0) from (0, 1/n); hence any limit cycle in the vicinity of the origin cannot intersect 1- ny = 0. Thus from (16.5) we obtain the proof of this theorem. REMARK 1. Let b- 0 in (16.2). Then the limit function B(x,y) is the Dulac function in (12.23). REMARK 2. L can be factored as [(ny- 1)61 - x][(ny- 1)62 - x), where 61 and 62 are a pair of conjugate complex roots of n(n + b)6 2 - m6- 1 = 0. REMARK 3. H we use the well-known formula for the inverse trigonometric function in (16.4): tan- 1 z =
~ ln ~ = ln G~ :) 112i,
and rewrite -iu = Jm 2 + 4n(n +b)=
VIS., then we get
B' (x y) = (-1)mf.../X(1 + by)-21/b-1 2
'
m- ..{K
(16.6)
x [ x+2n(n+b)(1-ny)
[x + 2n(n m + VfS. (1- ny)]m/.../K +b) ] -m/.../K.
(16.7) If we drop the factor (-1)m/.../K. in front, then we can prove that this B~(x,y) can also be used in the case m 2 + 4n(n +b) ~ 0, replacing Bt(z,y) as the Dulac function there. We look at another example. In [201), the proof of nonexistence of cycles starts from the study of the properties of the system
dyfdt = z,
dxfdt = -y + lx 2 + mxy -ly2 + f(y).
(16.8)
§16.
THE METHOD OF DULAC FUNCTIONS
361
The author (G. S. Makar-Limanov) proved that if J(y) does not change sign, it is not identically zero in any neighborhood of y = 0, and the uniqueness of the initial value problem of (16.8) can be guaranteed, then (16.8) does not have a limit cycle. After choosing a suitable J(y), he also proved that (I).s=o does not have a limit cycle. Now we use the method of Dulac functions to prove the following theorem, which contains more information than the nonexistence of cycles for (I)o=O· THEOREM 16. 2. Suppose in the system dy (16.9) - =x=Q, dt where(} is a constant, J(y), in addition to satisfying the above conditions, also satisfies (l + n)6J(y) 2: 0. Then (16.9) does not have a limit cycle.
: = -y + lx 2 + mxy + ny2 + 6J(y) = P,
PROOF. Take a Dulac function (still the function in (12.23)), (16.10)
where a = (m + v'4n2 We can compute
a
a positive root of n 2 a2- ma- 1
= 0.
a
+ oy (B3Q) am[(l + n)x 2 + 6f(y)](x- any+ a)am-le(amn- 21 )11.
ox (B3P)
=
+ m2)/2n 2 is
(16.11)
Again we note that under the conditions of the theorem, the line L: x-any+ a = 0 is a line without contact, since
~~ IL=O = (l + n)x
2
+ 6/(y).
From this we immediately obtain the proof of this theorem. In particular, if J(y) 0, then this theorem reduces to Theorem 12.4; if l + n = 0, it reduces to the theorem in [201]. After reading about these Dulac functions, more complicated and more useful than those in the preceding sections, the reader will definitely ask some questions; How are these Dulac functions obtained? Does there exist a general method for constructing them? Indeed, at the beginning of the sixties, when we wanted to prove nonexistence of a limit cycle for some simple quadratic system, we depended completely on luck. At that time, the functions used were mainly exponential functions whose exponents were polynomials of lower degrees, or rational functions such as the function (12.22) of §12, (1-mx )- 1 in the proof of Theorem 14.1, and the function (15.3) of §15, etc. The function B3(x, y) was also obtained by luck. But once we obtain B3(x, y), we have obtained the inspiration: x - any + a = 0 is a line tangent to the separatrix
=
362
THEORY OF LIMIT CYCLES
through a saddle point at the saddle point, and is also a line without contact of the system. Moreover, when l + n = 0, on the one hand, x - o.ny + o. = 0 is an integral line; and on the other hand, B3 (x, y) is an integrating factor of (I).s=O· Hence, we realize that in Dulac functions we should extend fractional functions to the power functions as (ax+ by+ c)k where ax+ by+ c = 0 is either a line without contact or an integral line or a line passing a singular point in an exceptional direction-of course, this singular point can be a nodal point since a nodal point also has an exceptional direction. At the same time, to find a Dulac function, we can first find an integrating factor with center. Moreover, the passage from B 1 (x,y) to B2(x,y) shows that if a Dulac function can be suitable in even one case, then from the relation between the elementary functions of complex variables we can obtain another Dulac function suitable to other cases. (I) Finally we note the divergence in the above examples is either of constant sign or identically zero in the region where a limit cycle may exist. However, if the divergence has a linear function Ax + By + C of variable sign in addition to the part of constant sign, then the line Ax + By + C = 0 can have at most two points of contact with the trajectory of the quadratic system. H we have known this line passes through a singular point of nonfocal type of the quadratic system, then it has at most one point of contact on it; from this we can prove that limit cycles of this sytem, if they exist, must be centrally distributed. The above discussion is the main line of thought in the proof of Theorem 15.6 in §15. The Dulac function to be used will be given later. In the following we attempt to use the method of an integrating factor in the case of finding a center to obtain the Dulac function B 1 (x, y) and B2(x, y). Consider (III)a=cS=O:
dxfdt = -y + lx2 + mxy + ny2 ,
dyfdt
= x(1 +by),
(16.12)
which, when l + n = 0, has an integrating factor
JLt(x,y) =
v1- 1 v2- 1 ,
(16.13)
where Vt = 1- 2ny- mx + n 2 y 2 + mnxy- n(n + b)x2 = L (see (16.5)) and v2 = 1 +by. In order to make the formula for the divergence as simple as possible when l + n =I 0, we take the Dulac function of (16.12) as 71
= _1 _
2(l + n). b
(16.14)
( 1 )It is worthwhile to study the meaning of this fact in the qualitative theory in the complex region.
§16. THE METHOD OF DULAC FUNCTIONS
363
Thus we compute the divergence
a
a
ax (B4P) + By (B4Q)
= m(l + n)x2 [1- ny + 2n(:+ b) x] V1- 2 V2",
{16.15)
where V1
= L = -n(n + b)(ny- 191 -
x)(ny- 102 - x),
and 81 and 02 are the two roots ofn(n+b)92 -m0-1 then 81 and 02 are real, and the two lines
01(ny- 1)- x
=0
= 0.
Hm 2 +4(n+b) ~ 0,
and 02 (ny- 1) - x
=0
{16.16)
all pass through the singular point (0, 1/n), which is either a saddle point or a nodal point. The points of intersection of {16.16) with the x-axis are (-81,0) and (-82,0) respectively. Moreover 1- ny + 2n(n + b)x/m = 0 also passes through (0, 1/n), and its point of intersection with the x-axis ( -m/2n(n +b), 0) then lies between ( -Ot. 0) and (-02 , 0). From this we can see that if (/t8 2 < 0, i.e. n(n +b) > 0 (here (0, 1/n) is a saddle point), then 1- ny + 2n(n + b)x/m = 0 may intersect a limit cycle and does not lead to a contradiction. Hence (16.15) cannot be used to prove that (16.12) does not have a limit cycle when m( l + n) # 0. But if 8182 > 0, i.e. n(n + b) < 0 (here (0, 1/n) is a nodal point (or a focus)), then the two lines (16.16) intersect the x-a.xis on the same side of the origin (or they are both imaginary lines). When (0, 1/n) is a nodal point, the point of intersection of
1 -ny+ 2n(n +b) x= 0 m
with the x-a.xis lies between (Ot.O) and (82,0), and, since (16.16) is a line without contact, we know it is not possible to have a limit cycle. When (0, 1/n) is a focus but (16.16) and the x-axis do not have a real point of intersection, (16.15) still cannot be used to prove that (16.12) does not have a limit cycle when m(l + n) '::/:- 0. In the following we shall attempt to use
= ln [v,1/2n(n+b)(1 + by)1fb(n+b)] + 1 =C,
m
2u1n(n +b)
ln L+ L_
(16.17)
364
where
THEORY OF LIMIT CYCLES
= vrn2 + 4n(n +b) and L_ = -2n(n + b)x- (m- u1)(1- ny),
0'1
L+ = -2n(n + b)x- (m + ut)(1- ny),
(16.18)
v;
L+L1 =- 4n(n+b)"
Formula(16.17) isonlysuitableform2+4n(n+b) ~ 0. Ifm 2 +4n(n+b) < 0, then complex coefficients will appear in L+ and L_. Still using (16.6), we can get another general integral of (16.12) from (16.17): U:.~(x, y)
= ln[V11/2n(n+b){1 + by)l/b(n+b)]
m
+
u2n(n +b)
tan-l [-2n{n + b)x- m(1- ny)] u:~(1- ny)
( 16.19)
=0, where 0'2 = v-m2- 4n(n +b). Now when m 2 + 4n(n +b) ~ 0, we instead take the Dulac function as
B5 (x, y)
b)t
= exp ( 2n(n + u1- rn =
L~a1(1
Ut(x, y)) B"(x, y)
+ by)(mMl-21-b)/b 1
where a 1 = (m + ut)f2n(n +b) is a root of n(n + b)a 2 - ma- 1 = 0. Then clearly B 5 (x,y) is the B 1 (x,y) in (16.2); hence it can be used. Similarly, for m 2 + 4n(n +b) < 0, we instead take the Dulac function as B6(x, y) = e 2n(n+b)U2 (:c,y) • B4(x, y)
= {1 + by)-21/b-1 exp (2m tan-• [- 2n(n + b)x - .!!!.] ) i u2(1- ny)
0'2
{16.20)
0'2
then it is just B2{x, y) in (16.4) and therefore it can be used. From this we can see that the Dulac functions B 1 (x,y) and B2(x,y) can be constructed in a definite way, and are not completely obtained by chance. Now we give an example. Consider the system of equations [219]:
dxfdt 11-
= -y + lx 2 + mxy,
dyfdt = x(1 +ax).
It is easy to see that when l = 0 the system = 1/(1- mx) and a general integral U3
2 + _!_ (1 + .!_) X- !y2 + = ~X 2m m m 2
(1 +
(16.21)
has an integrating factor
~) ...!__ ln(1 m m2
mz).
Let (16.22)
§16. THE METHOD OF DULAC FUNCTIONS
365
It is obvious that
u1J.I. = (1- mx)2a/m +2/m-lea:z: -m11 +2(1+a/m)x 2
2
2
is still an integrating factor of (16.21), but this formula does not contain l. Similar to the previous example, in order to obtain the Dulac function at l ~ 0, we should take B(x,y)
= u1 J.I.e-21y = (1 -
mx)2a/m 2 +2/m-leax 2 -m11 2 +2(a/m+l)x-21 11 • (16.23) Here the new factor is just the integrating factor of {16.21) when a= m = 0. Then for system (16.21) we can compute
a
ax (BP)
a
+ ay (BQ) = -l[2a 2 mx:~ + 2a- m]x2B(1- mx)
_1 •
(16.24)
From (16.24) we can see that when l = 0 or a = m = 0, (16.21) has a center. When l ~ 0 and m ~ 2a, the origin is a focus. If m = 2a, then the origin is a fine focus, ii3 = 0, and the sign of iis can be determined from the sign of l (see (12.21) in §12); hence it does not have a limit cycle. Sometimes, for some equations, we can obtain at the same time several Dulac functions of different types. For example, the system (I)c=n=O dxfdt = -y + lx:l
has a center when l
+ mxy,
dyfdt = x
{16.25)
= 0, and has integrating factors
J.l.l;::: emz-m 2 y2 / 2 and J.l.2 = 1/(1- mx) and a general integral U = (1- mx)emx-m,y,/2.
Moreover, when m e- 21 11. The functions Bl
= 0, the system has a center and an integrating factor
= emx-my,/:l-2ly
and B2
= (1- mx)-le-21y
can also be used as Dulac functions of {16.25):
! :x
(BlP)
+ :y (Bl Q) = mlx2emz-21y-m,y, /2,
(B2 P) + :Y (B2Q)
= mlx 2(1- mx)-le- 2111 .
The above formulas show that for ml ~ 0, (16.25) does not have a limit cycle. For the above equations of class III, in fact we discuss only one case which may satisfy conditions of a center, i.e. {16.1). For quadratic systems which
THEORY OF LIMIT CYCLES
366
may satisfy other conditions of center, we say nothing here, but readers can refer to [244]. In the following we use the method of the Dulac function to prove Theorem 15.6 in §15. 1. The case m 2 + 4n(n +b) 2:: 0. Take L(al +m)b/2nat-1L(at-m)b/2nat-1 _ B 1_- eb(n+b)U1 /1-1+ , where U1 and 11- are given by {16.17) and {16.13) respectively, and L+ and L_ are given by (16.18). For the system of equations
dx dt
dy dt =x(1+ax+by)
2 = -y+lx2 +mxy+ny,
{16.26)
we compute its divergence, and obtain
a-
a 115
ax(B1P) + ay\BtQ = [Cx + A(1-
)
{16.27)
ny)]x2 L~t+m)b/2nat-2 L~t-m)b/2na.1 -2
where
A= m(l + n)- a(2l +b),
C = am(2l + b + n)- (2l + b)(n + b)(n + l). (16.28) The line l: Cx + A(1- ny) = 0 passes through (0, 1/n), which is in this case a saddle point or a nodal point. Hence there exists at most one more point of contact with the trajectory of (16.26). This shows that if (16.26) has a limit cycle (it must intersect with l, and hence its interior contains at least one point of contact on l), the limit cycles must be centrally distributed in the vicinity of one singular point of focal type of index +1. It may be (0,0), but it may be a singular point on 1 + ax + by = 0. 2. The case m 2 + 4n(n +b)< 0. Take
(2m(l+n)+mb t _ 1 [-2n(n+b)x-m(1-ny)]) B 2_TT(b+21)/2n - v1 ·exp · an ( ) • tw2 u2 1 - ny and compute its divergence. We get
a-
a-
ax (B2P) + ay (B2Q) = [Cx
+ (1- ny)AJVP'+b)/2n-t
·exp (
2m(l+n)+mb no-2
tan
_ 1 [-2n(n+b)x-m(1-ny)]) . u2(1- ny)
Note now that (0, 1/n) is a focus. Hence, if in the vicinity of (0, 1/n) there is no limit cycle, then as before we can prove the limit cycle, then as before we can prove that limit cycles of (16.26) must be centrally distributed. But if the
367
§16. THE METHOD OF DULAC FUNCTIONS
vicinity of (0, 1/n) does contain a limit cycle, then the vicinity of (0, 0) may still have a limit cycle and does not contradict the fact that on Cx+(1-ny)A = 0 there exist at most two points of contact with the trajectory of (16.26). In this case limit cycles need not be centrally distributed. An example was seen in system (11.10) of §11. But if there is some finite singula'r point (x11 yt) in addition to the two foci, then it must have an exceptional direction. We can then use this singular point to replace (0, 1/n) and use [19]:(2)
(i = 1,2) to replace L+ and L_ respectively, where fh and fh are two roots of (-1 + mx1
+ 2yt)0 2 + [(21- b)x1 + my1]0- ax1 = 0;
in fact, they are the slopes of two exceptional directions of (x1, Y1)· Moreover, we use B3 = [y- Y1- 01(x- xl)]k 1 [y- Y1- 02(x- xl)]k 2 to replace the above B1 and B2, where
k1 _ (01x1- YtHm- (b + 2l)02] 02-01 I k2
(02x1- YtHm- (b + 2l)01J. 01-02
=
Thus we can compute
:x
(B3P) + :y (B3Q)
= KL~'- 1 L~2 - 1 (ytz- z 1 y) 2 (A'x + B'y + C'),
where
A' = al(b + 2l) x 1 +am+ am(b + 2l), Y1
,
my1
B = -
2
+ mb + a(b + 2l) -
x1 C' = m(l + 1) - a(b + 2l).
m 2
x1
m2
+ -, x1
(16.29)
It is easy to verify that the line A' x + B' y + C' = 0 must pass through the singular point (xt. yt); thus, as before, we can show that in this case limit cycles cannot coexist in the vicinity of (0, 0) or (0, 1). It is worthwhile to consider: Can one use the method of Dulac functions to prove Theorem 15.12 of §15? We would guess that it is possible. (2)In the following calulations we assume n = 1; for all n :f: 0 we can always make this assumption.
THEORY OF LIMIT CYCLES
368
Finally, we would like to prove an important property of fineness of a focus for the quadratic system mentioned at the end of §11. THEOREM 16.3. If a quadratic system has two fine foci, then they can only be first-order fine foci. This theorem was first proved in [191] by the method of Dulac functions. But Yu Ren-chang [245] found a more simple and convenient method of proof was given. Hence we adopt the method of proof of [245]. As for the proof by means of the Dulac function, it is similar to the above approach and is left as an exercise. PROOF. We adopt the standard form of
= .X1x- y- .Xax 2 + (2.X2 + .Xs)xy + .Xay2, dyjdt = x + AlY + .X2x2 + (2-Xa + .X4)xy- .X2y2.
dxfdt
(16.30)
Suppose 0(0, 0) is at least a second-order fine focus. We would like to prove that if there exists a singular point N of focal type, then either N is a coarse focus, or both 0 and N are centers. From the formulas for tia and iis in (9.40) of §9, we know that if 0(0, 0) is at least a second-order fine focus, then we must have .X 1 = .X 5 = 0. At this time (16.30) becomes
dxjdt
= -y- .Xax2 + 2.X2xy + .Xay2 = P,
dyjdt
= x + .X2x2 + (2-Xa + .X4)xy- .X2y2 = Q,
(16.31)
and the divergence of the system is
D=
aP
aQ
+8y - =A4X. ax
H A4 = 0, then (16.31} is an exact equation, whose elementary singular point can only be a center or a saddle point. Hence we may as well assume .X 4 =# 0. At the same time, if another singular point N of focal type does not lie on the y-axis, then the value of D at N is not zero, and N must be a coarse focus. Now we suppose N is on they-axis, and we would like to obtain the coordinates of the singular point on the y-axis. Let x = 0 be substituted in P = 0, and obtain y = 0 and y = 1/ .Xa. Substituting Q = 0, we obtain .X2y2 = 0. Hence the possible coordinates of N are (0, 0} or (0, 1/.Xa}. In the first case, N = 0 and does not have a second fine focus. In the second case we must have .X2 = 0. But when .X 2 = 0, (16.31} can be integrated, and N can still be a center.
369
§16. THE METHOD OF DULAC FUNCTIONS
Of course, it ig possible for a quadratic system to have two first-order fine foci, such as the system (13.3) of §13, when m > -a > 1/a-a and 27a < 4m3 ; its global structure was shown in Figure 13.10; here 0(0, 0) and R( -1/a, 0) are both first-order fine foci. The theorem is completely proved. In [245], it was further proved that when a quadratic system has two firstorder fine foci, they must possess different stability. The proof is omitted, and left as an exercise for the reader.
Exercises 1. 2. 3. 4.
Prove (16.3) and (16.5). Prove Remark 1 after Theorem 16.1 and (16.7). Verify (16.27) and (16.29). Suppose that in the system
dx/dt =ax+ by+ x(x 2 + y 2 ),
dy/dt =ex+ dy- y(x 2
+ y2 )
we have a+ d > 0 and (a- d) 2 + 4bc < 0. Use the Dulac function [by 2 + (a- d)xy- cx 2 ]- 1 to prove that the system has a unique limit cycle and a stable cycle [170]. 5. Use the Dulac function 1/x(x + y) to prove that the system
dx/dt
= 3xy,
dy/dt
= y2 -
2x 2
-
4xy + 2x
does not have a limit cycle. 6. Prove that in the system of equations
dx/dt
= -y(1 + y- mx) + 6x,
dyfdt = x(1 +ax)
ifO < -1/a < m, then for 6 > m(1-y'1 + 1/ma)/2 the system does not have a limit cycle [246]. (Hint. Use the Dulac function [m 2 x- my+ d- m]- 1 .) 7. Prove that the system dx/dt = x(y- k), dy/dt = ""fX + (3y + gy 2 does not have a limit cycle [170]. 8. Prove that the system
dx/dt = y 2
-
(x + 1)[(x- 1) 2
+ ..\],
dy/dt
= -xy
does not have a closed trajectory when .,\ ;::: 1 [170]. 9. Prove that the system dx/dt = 2xy, dy/dt = 1 + y- x 2 have a closed trajectory [170]. 10. Prove that the system
dx/dt = 2x(1 + x2
-
2y 2 ),
does not have a closed trajectory [170}.
dyjdt = -y(1- 4x 2
+ y2
+ 3y2 )
does not
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THEORY OF LIMIT CYCLES
11. Take the Dulac function B = dx _ 1 - f3 dt - y- 1 + {3x'
e2e:z:,
and prove that the system
dy 2 2) dt = -E(x + Y
(where E > 0 and 0 < {3 < 1/3) does not have a closed trajectory in the half-plane 1 + {3x > 0 [170]. 12. Prove that the system dx
~
= xy
'
dy
&
= _!(x -1)(x + 2) + !y2 + !xy + !y 3
2
3
3
does not have a closed trajectory [170]. 13. Use the method of Dulac functions to prove Theorem 16.3.
§17. Limit Cycles in Bounded Quadratic Systems
The s~called bounded quadratic systems are quadratic systems whose trajectories remain bounded for t ~ 0. The research work in this area began with Dickson and Perko (247], who studied autonomous quadratic bounded differential systems in n-dimensional space, and then turned to a deep and detailed study of plane quadratic systems; the main part of the latter work was published in (248]. The authors made use of the results of [165] mentioned in §10 to give a detailed classification of bounded quadratic systems, and drew their global phase-portraits. However, there was no discussion at all on the existence and the number of limit cycles. Independently of (248], Levakov and Shpigel'man (249] obtained many sufficient conditions for a quadratic system to be bounded, but they too did not consider the problem of limit cycles. In this section we present the work of [250] and (253], which were the first papers to study existence and uniqueness of a limit cycle for a bounded quadratic system, and in which m;my results were obtained although the problems were not yet completely solved. In this area we also mention the recent papers (300] and [301], which among other things corrected a mistake in [250]; moreover, [300] added a new phase-portrait and the corresponding conditions to [248]. It is easy to see that any bounded quadratic system has at least one singular point; changing the origin to the singular point, we can rewrite the system as (17.1) where Pto,Pol!Qto, and Qol are constants, and X2(x,y) and Y2(x,y) are homogeneous quadratic polynomials. We study the boundedness of the quadratic system (17.1), to which the properties of the corresponding homogeneous quadratic system dyfdt = Y2(x, y)
dx/dt = X2(x, y), 371
(17.2)
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372
have a definite relationship. Changing (17.2) into polar coordinates, we get
~~
= r 2[X2(cos0, sinO) cosO+ Y2(cosO, sinO) sinO]
= r 2 F(O),
~: = r[Y2 (cos 0, sin 9) cos 0 = rG(O).
{17.3) X 2 (cos 0, sin 0) sin 0]
Let 9 = 00 be a root of G(O) = 0, and F(9o) # 0. Since G(IJ0 + 1r) =0, we may as well assume F(Oo) > 0. Then the solution of (17.3) which satisfies the initial conditions r(to) = ro and O(to) = 9o is 1
r(t) = F(Oo)
1
to+ (1/roF(Bo))- t'
9 = 90 '
(17.4)
It is a ray in the (z, y)-plane, and r(t) -+ +oo as t-+ to+ 1/roF(9o). DEFINITION 17.1. Suppose r = r(t) is a solution of (17.1) such that t approaches a finite value as r-+ +oo. Then r(t) is a solution possessing finite escape time. For (17.3), if there exists a Oo such that G(90 ) = 0 and F(90 ) # 0, and (17.4) is a solution possessing finite escape time, then {17.4) is called a ray solution. THEOREM 17 .1. If (17.2) has a ray solution, then (17.1) has an unbounded solution possessing finite escape time. PROOF. Apply the transformation z = rcos9, y (17.1) becomes
= rsinO, dr/dt = r. Then
~; = PlO cos2 9 +(Pot+ QtO) cos0sin9 + Qol sin2 0 + rF(O) = Pt(O) + rF(9), dO = ![Qto cos2 0 + (flot- PlO) cos8sin9- Pot sin2 9] + G(9) d r r
(17.5)
= !Qt(O) +G(O). r
From the hypothesis, we know there exists a 9 = 90 such that G(Oo) = 0 and F(Oo) # 0. We may assume F(Oo) = >. > 0. Let u = r- 1 • Then (17.5) becomes iJ = uGt(O) + G(O), which has a singular point (u = 0, 0 = 90 ) and the first equation has a negative characteristic root->.. Hence from the classical theorem ofLya.punov we know that the system of equations has a trajectory approaching to (u = 0, 9 = 00 ); correspondingly, system (17.5) has a trajectory (r(r),O(x))-+ (+oo,Oo).
§17.
LIMIT CYCLES IN BOUNDED QUADRATIC SYSTEMS
373
Applying the formula of variation of constants to the first equation of (17.5), we see that as r( r) -+ +oo, we must have r -+ +oo. Finally, from
r
ds t=Jo r(s) we can see that as r -+ +oo we have t < oo .. The proof is complete. From the necessary and sufficient conditions for a quadratic system to be bounded in [248], and the supplement from [300], we obtain the following: THEOREM 17.2. All the trajectories of quadratic system (17.1) are bounded when t ~ 0 if and only if there exists a linear transformation which changes (17.1) into one of the following types: (A) • 2 • 2 x = aux + a12Y + y , y = a21x + ~2Y- xy + cy ,
where lei < 2 and the other coefficients satisfy one of the following groups of conditions: (A1) au < 0,
(B)
(A2)
au = a21 = 0,
(A3)
au
= 0, a21 = -a12 =F 0, ca21 + a22 ~ 0;
x = aux,
where au < 0 and a22 ~ 0; or (C) . . x = aux + a12Y + y2 , Y• = a22Y, where au ~ 0, a22 ~ 0 and au + a22 < 0, or au < 0, a22 < 0, and 2a22 - au > 0. The proof depends on the results of [165] and Theorem 17.1, which includes many pages, and is therefore omitted. In the following, we plan to prove a rather special case of Theorem 17.3, and draw the global structural figures corresponding to several possible cases, in addition to the existence and number of limit cycles. REMARK. It is clear that under conditions (A2), (B), and (C) we can write the equation of the family of trajectories of (A), and using this we can prove it is a bounded system and does not have a limit cycle. Hence, we are only interested in the case when (A) satisfies (A1) or (A3). Now we change (A) and the corresponding conditions (A1) or (A3) into our familiar form of equations of class III. Without loss of generality, we can assume (0, 0) is an elementary singular point of (A) of index +1; hence aua22 - a21a12 = .!l > 0.
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374
Now in (A) we apply the change of variables
v = aux- a2 1y,
u = x or y = (auu- v)/a21,
x = u.
Then we get
u= -v +(au+ a22)u- u(auu- v)/a21 + cu 2, v=
u[(aua22- a21a12)- au(auu- v)/a21
(17.6)
+ (cau- a21)u].
Again let
Then the above system of equations is changed to
dxfdr = -jj +a -l/ 2(au + a22)x + xya - 112/a21 + (ca21- au)x2a - 1/a21. djjfdr = x + [a21(cau- a21)- a~ 1 ]a - 312x 2/a2 1 + auxya - 1/a21· 1t already possesses the form
~!
= -jj + ox + lx 2 + mxjj + ny 2,
djj dr
= x(1 +ax+ by).
(17.7)
Here we have
o=
(au + a22)a - 1/ 2, l = (ca2 1 - au)a - 1/a211 m=a- 112/a21, n=O,
a= [a2t(cau- a2t)- a~1]a- 3 1 2 /a21.
b = aua- 1/a21·
Note now that 4ma + (b -l) 2 = (c 2 - 4)a - 2,
mb = aua - 3 / 2 fa~1·
Hence condition (A1) of Theorem 17.2 and the inequality rewritten as
n = 0, and condition (A3) and n = 0,
(b- l) 2 + 4ma < 0,
lei <
2 can be
mb < 0,
lei < 2 together can be rewritten as
(b- l) 2 + 4ma < 0,
b = m +a = 0,
m(l + mo) ~ 0.
In the above two conditions we haven= 0. Hence we may rewrite (17.7)
as dx 2 dt = -y + ox + lx
+ mxy,
~~ = x(1 +ax+ by).
(17.8)
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375
Thus we have THEOREM 17.3. A necessary and sufficient condition for system (17.8) to be bounded is that one of the following conditions holds:
+ 4ma < 0 and mb < 0.
(1)
(b -l) 2
(2)
(b- l) 2 + 4ma < 0, b = m +a = 0, m(l + mc5) :5 0, and m
# 0.
(17.9)
PROOF. Sufficiency. Examine the state of the trajectories of system (17.8) near the infinite singular point. Apply the Poincare transformation
x
= ujz,
y = 1/z,
= z.
dtjdr
Then (17.8) becomes
du/dr = mu- z+ (l- b)u2
+ c5uz- au3 -
u2 z,
dz/dr = -uz(b +au+ z).
(17.10)
From (17.9) we know that (17.8) can have a unique infinite singular point
(0, ±1,0). Since m # 0, we may as well assume m > 0 (otherwise we change the sign of x and y to achieve this). Let
x = -z,
jj
= mu- z,
dr/dr'
= 1/m.
Then (17.10) becomes
dx = 2x 1 _(.:x- y_) [b m
-d1 T
djj
1(
dr' = jj + m =
x- jj
a -] -1 (a+ m )x_ + -y m
m
= .rn 3 ( x,_ y_) ,
) [ mc5 + l_
l - b _ m + a__ a _2] ----,:n-x- m y - ---:;ri2xy- m2 y
ii + Q3(x, y).
Suppose from jj + Q 3 (x, jj) = 0 we solve for jj = cp(x) which satisfies cp(O) = = 0. Substituting in P3(f, jj), we get
cp'(O) 1/J(x)
= P3(x, cp(x)) =
-;x +m m 2
-4lb(mc5 + l)- m(a + m)]x 3
+ mc5 ~ l [b(mc5 + 2l- b) m
m(m + 2a)]x4
(17.11)
+ o(x4 ).
From (17.11) we know that if b # 0, orb= m +a= 0 and mc5 + l # 0, then (0, ±1, 0) is a semisaddle nodal point, and when b < 0, or b = m +a = 0, mc5 + l < 0, the state of the trajectory of (17.8) near the infinite singular
THEORY OF LIMIT CYCLES
376
o>O:
a-o
m>o, b
b-m+a-~+l-0
FIGURE 17.1
FIGURE 17.2(a)
b=m+a-~+l-0
o>0,-2
b-m+a-~+l-0
o>O, 0<
~
< 2
FIGURE 17.2(c)
FIGURE 17.2(b)
point (0,±1,0) is as shown in Figure 17.1; hence all the solutions of (17.8) are bounded when t ~ 0.( 1 ) Also, if b = a+ m = m6 + l = 0, then from (17.9) we must have lol < 2; then (17.8) can be written as
dxjdt
= (1 + ax)(ox- y),
dyjdt
= x(1 +ax).
The line 1 +ax= 0 is filled with singular points. The global phase-portrait of the system is as shown in Figure 17.2; hence all the solutions of (17.8) are bounded. do not assume m > 0, then when m < 0, b > 0 or m < 0, b = m +a= 0, 0, the figure of the trajectory near the infinite singular point can be obtained from Figure 17.1 by symmetry with respect to (0,0). Here the arrow does not reverse its direction; hence all solutions are still bounded when t ~ 0. ( 1 )If we
mo + I >
§17. LIMIT CYCLES IN BOUNDED QUADRATIC SYSTEMS
377
Necessity. We change the corresponding homogeneous quadratic system of (17.8) by the transformation used in the proof of Theorem 17.1 into polar coordinates:
= r[l cos 2 9 + (m +a) cos9sin9 + bsin 2 9] cos9 = rF(9), d9fdr = [acos2 9 + (b -l) cos9sin9- msin 2 9] cos9 = G(9). drfdr
If (b-l) 2 +4ma ~ 0, then G(9) = 0 has one or two real roots in [0, 1r). Since the eliminant of the two trigonometric polynomials G(9) = 0 and F(9) = 0 is
R = (bl- ma)[(a + m) 2 + (b -1) 2 ], to have a common root, we need R = 0. When bl- am = 0, we first assume b/m = afl = k ;f: oo. Thus G(9) and F(9) become cos 9(kcos9- sin9)(l cos 9 + msin9) and
cos9(cos9 + ksin9)(lcos9 + msin9), respectively. Take 90 = tan- 1 k. Then G(9o) = 0 and F(9o) '1: 0, i.e. (17.8) still has an unbounded solution. If k = oo, then m = l = 0; the proof of existence of unbounded solutions will be seen in the following Theorem 17.4. When (a+ m) = (b -l) = 0, (b -l) 2 + 4ma ~ 0 becomes -4m2 ~ 0; hence we also have m = 0. From the following Theorem 17.4, we know (17.8) still has an unbounded solution. Hence (b -l) 2 + 4ma < 0 is a necessary condition for (17.8) to be a bounded system. At the same time, we must have m ;f: 0. If mb > 0, or b = m + a = 0 and m(m6 + l) > 0, then from (17.11) we know that (0, ±1,0) is still a semisaddle nodal point, but the state of the trajectory of (17.8) near the singular point is as shown in Figure 17.3(a); hence (17.8) has an unbounded solution. If m + a = 0 and b '1: 0, then when b # l, to make R = 0, we need bl = ma. As before, if m # 0, then it has an unbounded solution. If m = 0, then, by the following Theorem 17.4, we know that (17.8) still has an unbounded solution. Also, if b = 0 and m +a ;f: 0, then from (17.11) we know that (0, ±1, 0) is a nodal point or a saddle point; hence as t -+ +oo, the trajectory must enter the singular point from the finite plane, i.e. (17.8) has an unbounded solution. The theorem is completely proved. THEOREM 1 7. 4.
The quadratic system
x = -y + ox + lx2 , has at least an unbounded solution.
1i = x(1 +ax+ by)
(17.12)
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378
m>o,li>O or m>O, tna+Z>O FIGURE 17.3(a) u
0
U=O
FIGURE 17.3(b) PROOF. When l =f. 0 and b =f. l, we can take 80 to satisfy acosOo+ (b - l) sin Oo = 0. Thus G(Oo) = 0 and F(Oo) :j.: 0. Hence (17.12) has an
unbounded solution. H b = l =f. 0, we may as well assume l > 0. Apply the change of variables v = -y- !bx2 , x = x; then (17.12) becomes
:i: = v + 6x + 3lx2 /2, Let V
= v + 3lx2 f4.
Then along the trajectory of (17.13) we have
(17.13)
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379
Now we study the region D in the fourth quadrant in Figure 17.3(b). It is bounded by the parabola U = 0 and the horizontal line L. The point of intersection of L and U = 0 is P, whose abscissa is xo. It is easy to see, as long as xo is sufficiently large, that for x ~ 0 we have
dUfdtlu=o > 0,
viL <0,
i.e. the trajectory of (17.13), when entering D, must be unbounded. When l = 0 and b '# 0, we may as well assume b > 0. The proof is similar as before. The choice of v is unchanged, but we take U = v + kbx 2 , the sign of k depends on the sign of a+M, and the magnitude of lkl is chosen in order to make sure the coefficient of x 3 in the representation formula dU fdtlu=o has the same sign as a + bt5. The details are omitted. When l = b = 0, it is easy to see that there exists a region similar to D in the (x, y)-plane. The theorem is completely proved. In the following we begin to discuss the limit cycles of bounded quadratic systems. When b = m +a = m6 + l = 0, m '# 0, (17.8) does not have a limit cycle, even if 161 < 2, because in this case the right sides of the system have a common factor 1 - mx, and every point on the line 1 - mx = 0 is a singular point. Hence we only have to discuss the case when (17.8) has only a finite number of finite singular points. Clearly (17.8) has at most three finite singular points. In the following we carry out our discussion separately according to the number of singular points. I. The case of three finite singular points. We solve 1 +ax+ by= 0 for y, and substitute in the first equation of (17.8); we get
Dx 2 + Bx + 1 = 0,
(17.14)
where D = lb - ma and B = a + bt5 - m. A necessary and sufficient condition for (17.8) to have three finite singular points is B 2 > 4D. When this condition is satisfied the three singular points are (0,0), (x1,yl), and (x2,y2) where Xl
=
-B+JB2-4D 2D
a 1 Yl = -bx1- b'
- B - v,..B.,....2----:-4D=
a 1 Y2 = -bX2- b' 2D Note that when b = 0, by Theorem 17.3, we know that the condition B 2 > 4D does not hold; hence in this case we have only a finite singular point (0, 0). X2
=
LEMMA 17.1. When (b -l) 2 + 4ma < 0, mb < 0, and l + m6 '# 0, the structure of separatrices is homeomorphic to Figure 17.4 or 17.5; when (b-l) 2 +4ma < 0, mb < 0, and 1+m6 = 0, then the structure of separatrices
THEORY OF LIMIT CYCLES
380
(a)
(b)
(c)
(d) FIGURE 17.4
of (17.8) is homeomorphic to Figure 17.4(a) or 17.6. The symbols- 0 and 0 - in the figures represent the direction of the trajectory in the vicinity of a singular point of index +1, and whether there exists any limit cycle can only be determined after further analysis. PROOF. We may as well assume b > 0, for otherwise we can change the sign of x andy. Since when m =I 0 we have m < 0 and a> 0, from (17.9) we can get (b + l) 2 < 4D; hence x1x2 = 1/ D > 0. When B > 0, we have x2 < x1 < 0. Move the origin to (x1, Yl) and still use x, y to represent the new coordinates of (17.8). We have
dx 1 2 dt = b[B- a+ (lb+ D)x1]x + (mx1 -1)y + lx + mxy,
~~ = ax1x + bx1y + ax2 + bxy.
(17.15)
It is easy to calculate the product of the two characteristic roots of its linear approximate system to be
§17. LIMIT CYCLES IN BOUNDED QUADRATIC SYSTEMS
381
hence, the new origin (original singular point (Z1, Yl)) is a saddle point and the indices of the other two singular points 01 (-z1, -yl) and 02 (z2 - z1, Y2 - Yl) are + 1. Similarly, when B < 0 and z1 > z2 > 0, as before we can prove that (z2, Y2) is a saddle point, and the other two singular points are singular points of index +1. We first consider the case when (b-l) 2 +4ma < 0, mb < 0, and l+m6 '=f. 0. If B > 0, let v2 = y + (ajb)x, (17.16) and differentiate
vl and v2 along the trajectory of (17.15) respectively; then
we-
.
where
Note that
+ 6z1 + lx21 + mZ1Y1 = 0,
1 + az1 + by1
~
-yl
= 0,
and it is not difficult to simplify A11 A2, B1, and B2 as
A1
= __!_(myl + (l- b)x1Y1- axn Z1
= --A(myl + (l- b)x1Y1- ax~], zl B1 = (aDjb 2)(x1- z2), B2 = aD/b 2. A2
Hence (17.17) (17.18) and
dxl dt
= m z=O
(xl- ..!.) y, m
dyl dt
= ax(x + xt). 11=0
(17.19)
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382
(b)
(a)
(c)
(d) FIGURE 17.5
Note that (l- b) 2 +4ma < 0, m < 0; my~+ (l- b)x 1 y 1 - ax~< 0. 02 should fall into the region V1 > 0; hence we can assume y1 < 0. When l + mS #- 0, we have x1 - 1/m #- 0. Hence when x1 - 1/m > 0, the structure of separatrices of (17.15), and hence (17.8}, is homeomorphic to Figure 17.4; when x 1 -1/m < 0 the structure of separatrices of (17.8) is homeomorphic to Figure 17.5. H B < 0, we translate the origin to (x2, Y2) and still use x andy to represent the dependent variables of (17.8) after the translation; we then get dx 1 dt = b [B- a+ (lb + D)x2]x + (mx2- 1)y + lx 2 + mxy,
dy dt
= ax2x + bx2y + ax2 + bxy.
Apply the transformation
Y2
u1 =y- -x, X2
(17.20)
§17.
LIMIT CYCLES IN BOUNDED QUADRATIC SYSTEMS
383
FIGURE 17.6 and differentiate along the trajectory of (17.20). We get ddu 1 1
t u1=0
= - 12 [my~+ (l- b)x2Y2 X2
-du21 dt
U2=0
ay~]x(x + x2),
aD = -x[x-x1 +x2].
b2
The singular point (x1 - X2, Yt - Y2) always falls in the region Ut !xl t
=(mx2-1)y, z=O
> 0. Also,
!~I v=O =ax(x+x2)i
since mx2 -1 < 0, the structure of (17.20), and then (17.8), is homeomorphic to Figure 17.4 or 17.5. (The case of B < 0 can be included in the case of B > 0 by changing the signs of x andy.) Similarly, when B < 0, the structure of the separatrices of (17.8) is homeomorphic to Figure 17.4. Finally, consider the case when (b-l) 2+4ma < 0, mb < 0, and l+m.5 = 0. Under the hypothesis b > 0, we must have m < 0 and D > 0. Note that D = -m(a+bO); hence a+M > 0 and B = a+M -m > 0. H a+ M + m > 0, then Xt = m/ D, x2 = 1/m, and x = 1/m- m/ D are integral lines of (17.15), but the structure of the separatrices of (17.15), and hence (17.8), is still homeomorphic to Figure 17.4. H a+ M + m < 0, then Xt = 1/m, x2 = m/D, and x = 0 is an integral line of {17.15); hence the structure of separatrices of {17.8) is homeomorphic to Figure 17.6. The lemma is completely proved. Now we discuss the problem of limit cycles of bounded quadratic systems with three finite singular points. When l + mb = 0, from Lemma 17.1 we know the structure of the separatrices of (17.8) is homeomorphic to Figure 17.6 or 17.4. It has an integral line x = 1/m; hence, by the corollary to Theorem 15.4, (17.8) has at least one
THEORY OF LIMIT CYCLES
384
limit cycle. Also, the origin (0,0) is definitely of index +1, but the indices of the other two singular points
01 (;,
a~;D)
and 02
(!, -a~m)
have not yet been determined. We already know that the characteristic equation of (17.8) at (0, 0) is >. 2 - 6>. + 1 = 0; hence: 1) When 161 ~ 2, 01 is a nodal point and there is no limit cycle in its vicinity. 2) When -2 < 6 < 0, 01 is a stable focus, and if there are limit cycles in its vicinity, the number of limit cycles must be even; hence there are none. 3) When 6 = 0, 01 is a fine focus, and hence (17.8) does not have a limit cycle (Theorem 15.4). 4) When 0 < 6 « 1, 01 is an unstable focus, there is a unique limit cycle in its vicinity, and there is none in the vicinity of 0 2 • But the precise range of variation of 6 to guarantee existence of a limit cycle in the vicinity of 0 1 may be smaller than (0, 2), even though all the conditions which guarantee the system to be bounded and l + m6 = 0 remain the same. (See Example 1, below.) In order to study the properties of the other two singular points when l + m6 = 0, we can move the origin to (xi, Yi), and then we get the system of equations ~
dt
1 2 = b[Ba+ (lb + D)xi]x + (mxi- 1)y + lx + mxy,
~~ =
(17.21)
axix + bxiy +
ax 2
+ bxy.
First replacing (xi, Yi) by 02(1/m, (-a- m)/mb) we get ~
_ m(a+M)(a+M+m)
dtdy _ dt -
l
2
bD x+ x +mxy, a( a + M) _ b( a + M) 2 b x D y + ax + xy. D
From this we see that the two roots of the characteristic equation are (note = lb- rna= -m(a + M)):
D
>.1 =-(a+ lx5 + m)/b, Then replacing (xi,Yi) by 01(mjD, -(am+ D)/bD), we get
dx _ c5(a+M+m) _ m+a+M l 2+ dta+M x a+M y+ x mxy, dy a b 2 dt =- a+Mx- a+My+ax +bxy.
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385
\
\
1+~+2y-O
FIGURE 17.7
From this we can get the characteristic equation ,\ 2 _
c(a+M+m)-b,\- (a+M)(a+M+m) _ 0 a+M (a+b6)2 - ·
(17.22)
According to the previous hypotheses, b > 0. Hence, by condition 1) of Theorem 17.3, m < 0 and a > 0; but the sign of a + M + m cannot be determined yet. H a+ M + m > 0, then from (17.21) we know that the two characteristic roots at 02(1/m, (-a- m)/mb) are negative, i.e., it is a stable nodal point. On the other hand, from (17.22) we can see that the two characteristic roots at 01(mjD, -(am+ D)/D) have opposite signs; hence 01 is a saddle point. Conversely, if a+M +m < 0, then 0 2 is a saddle point. Also from (b-l)2+4ma < 0, we can deduce that (b+l) 2 < 4(bl-ma) = -4m(a+b6); hence a+ M > 0, and so by (17.22) 01 has index +1. In short, as far as the abscissa is concerned, the saddle point must lie between two singular points of index +1, but under the above conditions, 0 1 may be a focus or a nodal point. When 0 1 is a focus, we can discuss as before the existence of a limit cycle. H it exists, then it must be unique, and cannot coexist in the vicinity ofO. EXAMPLE 1. Study the system of equations £ £ 2 -dx = -y + vX + vX dt
-
xy,
dy
dt
= x(l + 6x + 2y),
(17.23)
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THEORY OF LIMIT CYCLES
FIGURE 17.8 where 0 ~ 6 ~ 2, l We have
FIGURE 17.9
FIGURE 17.10
= 6, m = -1, a= 6, b = 2, B = 7 + 26, and D = 26 + 6.
l + m6 = 0,
(b -1) 2
+ 4ma < 0,
mb < 0,
B 2 > 4D.
System (17.23) always has three finite singular points. When 6 = 0, they are (0, 0), (-1/6, 0), and (-1, 5/2); when 6 = 2, they are (0, 0), (-1/10, -1/5), and ( -1, 5/2). As 6 increases from 0, 0 changes from stable to unstable, and there exists a unique stable cycle in its vicinity. When 6 varies to 61 ~ 2, there exists a separatrix cycle in the vicinity of 0. It is formed by a separatrix h starting from the saddle point ( -1/10, -1/5) advancing to its upper left side, and finally entering ( -1, 5/2), and a separatrix h starting from (-1/10, -1/5), advancing to the lower right side, and finally entering (-1,5/2), as shown in Figure 17.7. When 8 > 2, 0 becomes an unstable nodal point, and there is no limit cycle in its vicinity. Note that in this example x = -1 is always an integral line, (0,0) and (-1,5/2) remain fixed, but the saddle point moves as 6 varies. Also, it is not known whether 61 equals 2. (In the original manuscript there were some errors in this example, which were corrected according to the suggestion of Han Mao-an.) When l + m6 :/:0, we can start from (17.15) and study similar problems of limit cycles as before. But since now the system does not necessarily have an integral line, even though under certain conditions we can prove there exists a limit cycle, we are not sure of its uniqueness. The details are omitted. II. The case of two finite singular points. From (17.14) we know that the necessary and sufficient condition for system (17.8) to have two finite singular points is B 2 = 4D. At the same time, 0(0, 0) is an elementary singular point of index +1, and O*(xo, Yo) is a higher-order singular point, where Yo=
2a-B bB .
§17. LIMIT CYCLES IN BOUNDED QUADRATIC SYSTEMS
(a)
(b)
(c)
(d)
387
FIGURE 17.11
FIGURE 17.12
Let
x = x- xo andy= y- Yo·
Then (17.8) becomes
~: =-b~(a+M+m)x- !
2a _
2b _
_2
dt = - B x - By + ax
b- _
+ xy.
Denote u =
-(1/bB)[a(a + M + m) + 2b2 ].
.(17.24)
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THEORY OF LIMIT CYCLES
(a)
(b)
(c)
(d)
FIGURE 17.13 We have the following lemma. LEMMA 17 .2. If
B 2 = 4D, and (b -1) 2 + 4ma < 0, mb < 0, then the
structure of the separatrices of (17.8) is homeomorphic to 1) Figure 17.12 when a( a+ M + m) + 2b2 = 0; 2) Figure 17.8 when B[a(a + M + m) + 2b2 ] < 0; 3) Figure 17.9 when B[a(a + M + m) + 2b2 ] > 0 and a( a+ bt5 + m) 4) Figure 17.11 when a( a+ M + m) > 0; and 5) Figure 17.10 or 17.13 when a+ M + m = 0. PROOF. We first consider the case u
= 0.
Then a(a + bt5
and system (17.24) can be written as dx 2b_ 2b2 _ 1_2 __ dt = Bx+ Bay+ x +mxy, • djj 2a _ 2b _ _2 b- _ dt = - B x - By + ax + xy. Let
x=u,
_ v lu2 + 2bu/B y = 2b2jBa+mu- 2b2jBa + mu·
< 0;
+ m) = -2b2 ,
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389
We get
du dt =v, dv = 2D u2 [1- B u+o(u)] dt B 2
+
~+~+D b
2 [ +· maB 2b2 v 1 -
&mD ] uv 1- 2b(D+b2+lb)u+o(u) [
(17.25)
maB ] 2b2 u + o( u) .
Therefore 0* is a higher-order degenerate singular point; hence the structure of the separatrices of (17.25) (and so also (17.8)) is homeomorphic to Figure 17.12. Next, suppose u :f:. 0. Applying the transformation X
*
b_ =X+ ay,
y
*
a ( ~.~ ) = 2a_ B X- bB a+ + m 'f} 00
to (17.24), we get dx* a dt = ux*- 2bBu2 (a+ bt5 + m)[2u(b + l) + B]x* 2
bD 1 + """'22y* 2 + 2 ~._ 2 [a(a+ bt5 + m)B- 2bu(b 2 + ma)]x*y*,
aq auu dy* a3 dt = B 3b2u2 (a+ bt5 + m)[(a + bt5 + m)(Bu + 2l) + 4mb]x* 2
(17.26)
2bD 2 a ( 2 + Boo2 y* + B 2bu2 [a+ bt5 + m)(bBa + 2D + 2bl) +4mb ]x*y*. Again applying the transformation x* = v, y* = u, dtfdr = 1/u to (17.26), we get du 2bD 2 a 2 dr = Boo3 u + bB 2 u3 [(a+ bt5 + m)(bBu + 2D + 2bl) +4mb ]uv a3 + B 3b2q3 (a+ bt5 + m)[(a + bt5 + m)(uB + 2l) + 4mb]v 2 = P2(u,v), dv bD 1 dr = v + a2u3 u2 + 2abu3 [a(a+ bt5 + m)B- 2bu(b2 + ma)]uv
- 2b;u3 (a+ bt5 + m)[2u(b + l) + B]v 2 = v + Q2(u,v). Since v + Q2 (u, v) = 0, we can solve
v = cp(u) ==- bD u 2 + o(u 2 ), a2q3
(17.27)
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and substituting it in the first equation of (17.27), we get
.,P(u)
=P2(u, cp(u)) = Auu + o(u 2
2 ),
(17.28)
where Au= 2bD/Bau 3 ; hence 0* is a semisaddle nodal point. We may as well assume b > 0; if not we just change the sign of x and y in (17.8). Since B 2 = 4D, we get 1 mx1 -1 =- B(a + b8 + m). (17.29) Note that under the condition B 2 > 4D, when B > 0 we have x2 < x 1 < 0; and when B < 0 we have 0 < x2 < x1. Hence when B 2 > 4D changes to B 2 = 4D, if B > 0 (< 0), then the higher-order singular point corresponds to the coalescing of the saddle point and the elementary singular point on its left (right) of index + 1. Now we consider the case B[a(a+bt5+m)+2~] < 0. If a(a+bt5+m)+2b2 < 0, then B > 0 and a( a + b8 + m) < 0; hence, we must have m < 0. Thus a> 0 and a+ b8 + m < 0. From (17.29) we know that 0* compounded of the saddle point in Figure 17.5 and the elementary singular point of index +1 on its left. Also by (17.28) the hyperbolic region near 0* contains the negative u-axis (in this case Au > 0). Returning to the xO*y plane, we see that 0* is compounded of the saddle point in Figure 17.5(a) and the elementary singular point of index + 1 on its left. Similarly, we can prove that if a( a+ bC + m) + 2b2 > 0, then 0* is compounded of the saddle point in Figure 17.4(a) and the elementary singular point of index +1 on its right. In these two cases, the structure of separatrices of (17.8) is homeomorphic to Figure 17 .8. Next, we consider the case when B[a(a + b8 + m) + 2b2 ] > 0 and a( a+ M + m) < 0. If a( a+ b8 + m) + 2b2 > 0, then B > 0, and so m < 0. 0* is compounded of the saddle point in Figure 17.5 and the elementary singular point of index + 1 on its left. But since Au < 0, the hyperbolic region near 0* contains the positive u-axis. Thus in the xO*y plane, 0* is compounded of the saddle point in Figure 17.5(d) and the elementary singular point of index + 1 on its left. Similarly, if a( a+ bC + m) + 2b2 < 0, then 0* is compounded of the saddle point in Figure 17.4(d) and the elementary singular point of index +1 on its right. In these two cases, the structure of the separatrices of (17.8) is homeomorphic to Figure 17.9. Third, consider the case a( a + b8 + m) > 0. Then a( a + b8 + m) + 2b2 > 0, m < 0, B =a+ bC + m- 2m > 0, and from (17.29) we know that 0* is compounded of the saddle point in Figure 17.4 and the elementary singular point of index +1 on its left. Hence the structure of the separatrices of {17.8) is homeomorphic to Figure 17.11.
§17. LIMIT CYCLES IN BOUNDED QUADRATIC SYSTEMS
391
FIGURE 17.14
Finally, if a + b8 + m = 0, then m < 0, for otherwise we would have B = 0, which is impossible. Since B > 0, x = 1/m is an integral line. H a+ b8 + m changes from negative to 0, then 0* is compounded of the saddle point in Figure 17.6 and the elementary singular point of index +1 on its left; hence the structure of separatrices of (17.8) is homeomorphic to Figure 17.10. If a + b8 + m changes from positive to 0, then 0* is compounded of the saddle point in Figure 17.4 and the elementary singular point of index +1 on its left; hence the structure of the separatrices of (17.8) is homeomorphic to Figure 17.13. The lemma is completely proved. We now turn to the problem of limit cycles. First of all, when 161 ~ 2, the unique elementary singular point of index +1 of (17.8) is a nodal point; hence (17.8) does not have a limit cycle. Thus we can assume 161 < 2. Under this condition, if a( a + b8 + m) + 2b2 ::/: 0, then B > 0; otherwise, let
- YOV =y=-x, xo -2/ Band Yo= (2a- B)/bB. Differentiating V and xo along the
where xo = trajectory of (17.24), we get
~I
t V=O
= 8~ [(4- 62 )b2 +(a+ m) 2 ]x(x + x0 ),
=~-x=O =-!(a+b8+m)y,
:1_
y=O
=ax(x+xo),
when B < 0 and xo > 0. Thus the structure of part of the separatrix of (17.24) is as shown in Figure 17.24 (the figure corresponds to the case of a > 0); but this is impossible. Since the shaded sector in the figure does not have a singular point, the trajectory starting from the semisaddle nodal point (0, 0) cannot get out of the sector as t - +oo.
THEORY OF LIMIT CYCLES
392
Thus, under the condition J8J < 2, conditions 2) and 3) of Lemma 17.2 can in fact be written as 2') a( a+ b8 + m) + 2b2 < 0, B > 0; 3') a( a+ b8 + m) + 2b2 > 0, B > 0, a( a+ b8 + m) < 0, respectively. Now it is not hard to see that under conditions 2') and 3') plus conditions 1) and 5) of Lemma 17.2, we have m '=/: 0. Hence to discuss existence and number of limit cycles, we can assume m '=/: 0 in the above cases. In fact, for system (17.8), if m = 0, then the uniqueness of a limit cycle was proved earlier in [226], as we mentioned in §15. Since we can assume b > 0, we must have m < 0; for convenience, we take m = -1. Thus (17.8) can be written as dx 2 dt = -y + 8x + lx - xy,
~~ = x(1 +ax+ by).
(17.30)
The boundedness condition 1) of Theorem 17.2 and the condition B 2 = 4D for (17.8) to have two finite singular points can be written respectively as (b - l) 2
< 4a,
b > 0,
(17.31) (17.32)
(a+ b8 + 1) 2 = 4(lb +a),
and conditions 1), 2), 3), and 5) of Lemma 17.2 can be written respectively as a( a+ b8 -1)
+ 2b2 =
0, a(a + b8 -1) + < 0, a+ b8 + 1 > 0, 2 a( a+ b8- 1) + 2b > 0, a( a+ b8 -1) < 0, B > 0, a+b8 -1 = 0.
2b2
(17.33) (17.34) (17.35) (17.36)
THEOREM 17.5. Suppose (17.30) satisfies (17.31) and (17.32), and when 0 < 8 < 2 it satisfies one of the conditions (17.33), (17.34) or (17.36). Then · the system has exactly one limit cycle, and this cycle is stable.
PROOF. Let x
= x and e = lx2 + 8x- (1 + x)y.
= e, de/dt = -
Then (17.30) becomes
dx/dt
x(1 + ax)(1 + x)- bx(lx 2 + 8x)
+ [o + (21 + b)x- x(~x++x6)] e + 1 ~ x'
§17. LIMIT CYCLES IN BOUNDED QUADRATIC SYSTEMS
Again applying the transformation x
393
= x, u = U(1 + x), we get
dx dt = u(1 + x), du dt
= - x(1 +ax)- bx2(lx + 6) + [6 + (2l + b)x-
Finally, let
x = x,
(17.37)
1+x
x(~x++x6 )] u.
r 6 + (2l +(1b)s+ s)2+ (l + b)s ds, 2
y = -u + lo
dt 1 dr = 1+x" Then we obtain an equation of Lienard type:
dyfdr = g(x),
dxfdr = -y- F(x),
(17.38)
· where
g(x)
= (1 +x x)2
F( ) - -
x-
(
1+
r 6 + (2l +(1+s)2 b)s + (l + b)s
lo
It is easy to see that xg(x) > 0 when x then /(0) = -6 < 0. At the same time,
~ [!(x)] dx g(x)
a+M+1 ) 2 x
# 0,
2
' 2
d
s.
and if we assume f(x)
= F'(x),
= _1_1 +(a+ b6 + 1)x/2W(x) g2(x)
(1 + x) 4
'
where
W(x)
= !O + b)(a + b6 + 1)x3 +
[l(a + b6 + 1) + (l + b)(a + M)]x 2
(
17.39)
+ ~6(a + b6 + 1)x + 6.
From (17.31) and (17.32) we can get a> 0 and (a+ b6- 1) 2 = 4b(l- 6); hence (17.40) l > 6 > 0. Since 6 > 0, (17.30) has at least one limit cycle. It is easy to prove that the line x = 1 is a line without contact; hence the limit cycle must lie in the region x > -1. Now suppose (17.33) holds. Then W(x) in (17.39) takes the form W(x) = ( 1 + a+M+1) x Z(x), 2
THEORY OF LIMIT CYCLES
394
where
Z(:o:)
= (l + b)x 2 + 2[(2l +b)~+ b:)- b] x
a+ + 2 [(2l + b)(a + M)- b] + (a+ M + 1)2 It is easy to compute the minimum point of Z = Z(x) to be xo=-
(2l+b)(a+M)-b a+M+1 '
and its minimum value is
Z(x ) = [(2l + b)(a + M)- b][-(2l + b)(a + M- 1) + 2b] 0 (l + b)(a + M + 1)2 Since a + M - 1 = - 2b2 /a
< 0, we have
-(2l + b)(a + M- 1) + 2b > 0. Also because
4(l +b) - 2(2l +b) 6= 0 (a+M+1)2 a+M+1 + ' or
2(2l + b)(a + M + 1}- 4(l +b)= 6(a + M + 1} 2 , we deduce that
(2l + b)(a + M) - b = !6(a + M + 1) 2 > 0, and then Z(x) ~ Z(xo)
.!!__ [f(x)] dx g(x)
> 0 when x > -1; thus
= _1_ (1 +(a+ M + 1)x/2} 2 Z(x) > 0 (x + 1} 4
g2(x)
'
when x > -1. According to Theorem 6.4 of §6, system (17.30} has exactly one limit cycle, and it is a stable cycle. Next, if (17.34} holds, then a+ M- 1 < 0, and then
-
a+
2 M
+ 1 < -1.
Since
[6 + (2l + b)x + (l + b}x2 ]x=-2/(a+bc5+1)
= -b\::~ ~ ~) [a(a+
M-1} + 2b < 0, e) 2]
it follows that
4(l +b)- 2(2l + b)(a + M + 1) + 6(a + M + 1) 2 < 0, 2This inequality does not hold under condition (17.35).
§17. LIMIT CYCLES IN BOUNDED QUADRATIC SYSTEMS
395
or
4(1 +b)- 2(21 + b)(a + M + 1) + 48(1b +a) < 0. Thus
b(a + M- 1) > -2a(l- 8).
(17.41)
From (17 139) we have
W'(x) = ~(1 + b)(a + M + 1)x2 + 2[l(a + M + 1) + (l + b)(a + M)]x
+ ~8(a + M + 1), and then the roots of equation W'(x) = 0 are =
-2[l(a + M + 1) + (l + b)(a + M)]- .fK 3(1 + b)(a + M + 1) '
x2 =
-2[l(a + M + 1) + (1 + b)(a + M)] + .fK 3(1 + b)(a + M + 1) '
Xt
where
~
= 4[1(a + M + 1) + (l + b)(a + M)] 2 - 98(1 + b)(a + M + 1) 2.
First we suppose~> 0. Hence Xt < x2 < 0 and W(x) takes its maximum value W(xt) and minimum value W(x2) at Xt and x2 respectively; moreover, W(x2) < W(xt). We have
W(x2)
=! [1(a + M + 1) + (l + b)(a + M)]x~ +8(a+M + 1)x2 + 8.
Consider
1/J(x)
= l[1(a + M + 1) + (l + b)(a + M)]x2 + o(a+ M + 1)x +8,
1/J'(x)
= i[l(a + M + 1) + (l + b)(a + M)]x +8(a+M+l).
The root of the equation
1/J' (x) = 0 is 38(a+M + 1}
xo =- 2[l(a +
M + 1) + (1 + b)(a + M))'
and 1/J(x) takes its minimum value at x = xo:
1/J(xo)
=
1
2o(a + M + 1)x0 + 8 + 1} + (l + b)(a + M) - 38(lb +a)] l(a + b8 + 1) + (l + b)(a + M)
_ 8[l(a + M
-
THEORY OF LIMIT CYCLES
396
Since a(l- o)
> 0, we have -o(lb +a) > -l(a + bO),
and then
l(a+ bO + 1) + (l + b)(a+ bO)- 3o(lb+ a)
> l +(a+ b0)(2l + b- 31) = b(a + M) -l(a + bO -1) >0; hence W(x2) = ,P(x2)
.,P(x2) > 0. Moreover, W(-1) = ~(l- o)(a + M) + !(l- o) +~(a+ bO -1). ~
According to (17.41),
W(-1)
> J(l- o)(a+ bO) + !(l- o)- a(l- o)
= !(l- o)(a +3M+ 1) > 0. Now, if~ < 0, then W(x) is a monotone increasing function of x (when x > -1), and W(-1) > 0. Hence, in either case, W(x) > 0 when x > -1. Thus
[!(x)] > 0
d dx g(x)
when x
> -1.
According to Theorem 6.4 in §6, system (17.30) has exactly one limit cycle, and it is a stable cycle. Finally, if (17.36) holds, then l = o > 0, a+ bO = 1, and a+ bO + 1 = 2. Then xg(x) > 0 when x :1: 0; /(0) = -o < 0, and (17.39) becomes
W(x)
= (b + 6)x3 + (36 + b)x2 + 3ox + o = (x + 1)[(b + o)x2 + 2ox + 6].
Since~= 46 2 -4o(b+o) = -4b0 < 0, it follows that W(x) > 0 when x > -1. Thus as before we can deduce the uniqueness of a limit cycle, and this cycle is stable. The theorem is completely proved. REMARK. This theorem requires that condition (17.22) hold. Hence for fixed a, b, and l, there exists only one value of 6. Hence the conclusion of this theorem cannot guarantee that system (17.30) has a limit cycle for all in (0,2). As mentioned before, when 161 ~ 2, (17.8) does not have a limit cycle; hence we only have to discuss the nonexistence of a limit cycle under the condition . 161 < 2.
o
THEOREM 17.6. Suppose that system (17.30) satisfies the conditions (17.31) and (17.32) when ~ 0, and one of (17.33), (17.34) or (17.36) holds.
o
Then it does not have a limit cycle. PROOF. Here we only prove the case when (17.34) holds, since the remaining two cases are similar. First we suppose o = 0; then
1Ja
= ~ [m(l + n) -
a(b + 21)]
= -~ [1 + a(b + 21)] < 0;
§17. LIMIT CYCLES IN BOUNDED QUADRATIC SYSTEMS
397
that is, the elementary singular point of (17.30) is a stable first order fine focus. Hence if limit cycles exist, the number of limit cycles must be even, and these cycles must lie in the region x > -1. But now W(x) in (17.39) becomes
2[l(a+1)+a(l+b)]] W( ) -!(l b)( 1) 2 [ X - 2 + a+ X X+ (l + b)(a + 1) ' and (17.41) becomes b(a- 1) > -2al; thus 1 2[l(a + 1) + a(l +b)] _ 1 = [l(a + 1) + 2al + b(a- 1)] (l + b)(a + 1) (l + b)(a + 1) l(a + 1) l > (l + b)(a + 1) = l + b > O. Hence
W(x) > !(l + b)(a + 1)x2 (x + 1) > 0, x > -1; that is, system (17.30) has at most one limit cycle. This contradiction shows that (17.30) does not have a limit cycle. Next we consider the case when -2 < 6 < 0. Here the elementary singular point (0, 0) o£{17.30) is a stable coarse focus. As before, if limit cycles exist, the number of limit cycles must be even and they must lie in the region x > -1. In the following we discuss the two cases l + b ~ 0 and l + b > 0. (1) The case l + b ~ 0. 1 Denote -y- F(x) P(x, y) and g(x) Q(x, y), where the meaning of F(x) and g(x) is given in (17.38). Now suppose (17.30), and thus (17.38), has limit cycles, and suppose r is any one of them; then
=
D=
rf,. (8P + 8Q)
1;
dr = rh_ (l
ax ay V(x) = ']{. (1 + x)2 dr,
'}{.
=
+ b)x2 + (2l + b)x + 6 dr (1 + x)2
rf...
where V(x) = (l + b)x2 + (2l + b)x + 6. H l + b = 0, then from (17.40) we know that V(x) = lx + 6 = l(x + 1)(l- 6) < 0 when x > -1. H l + b < 0, then it is easy to compute a maximum point of V(x): 2l +b b xo = 2(l+b) = - 1 + 2(l+b) < - 1. Since a + b6 - 1 < 0 and a + b6 + 1 > 0, we see that
-2
a+ lx5 + 1 < -1. Moreover, V(-1) = -(l- 6) < 0. This shows that V(x) < 0 when x < -1, and thus D < 0 when x > -1. That is, system (17.30) has at most one limit cycle, contradicting the statement that the number of limit cycles is even.
~
THEORY OF LIMIT CYCLES
398
{2) The case l + b > 0. First suppose {21 + b)(a + b8 + 1)- (1 +b) ~ 0, and suppose r is any limit cycle of {17.30) and f* is a limit cycle of the system corresponding to r. Then
D= 1 (aP + aQ) lr· ax ay aP + -a aQ) = -a
i( r·
1
= Tr·
x Y u(x) (x + 1)2 dr,
dr dr -
i
l+b b8 g( x) dr a+ + 1 r•
where
u(x)
l+b 3 = -4-(a + M + 1)x +
( 21 + b- a+1+b M + 1 ) x + 8.
In this case,
and
l-8 u{-1) =a+ M + 1 [-{a+ b8 + 1) + b(1 +b)] 1 a(a + M -1) + lb]
= - 41 (1- 8)(a- b8 + 3) < 0. Hence D < 0 when x > -1. Thus system {17.30) has at least one limit cycle, and so {17.30) in fact does not have a limit cycle. Next, if {21 + b)(a + b8 + 1) - (l +b) > 0, then 21 + b > 0, and
D= 1 (aP + aQ) lr· ax ay =
£. (~= + ~~) 1
=
dr dr- (21 +b)£. g(x)dr
K(x)
Tr• (x + 1)2 dr,
where
K(x) = -(21 + b)(lb + a)x 3 + [(1 +b)- (21 + b)(a + M + 1)]:t:1 + 8. Then Xl
=0
and
X2
=
2[(21 + b)(a + M + 1) - (1 +b)] -3(21 + b)(1b +a)
§17. LIMIT CYCLES IN BOUNDED QUADRATIC SYSTEMS
are maximum and minimum points of K K(x 1 ) = 6 < 0. It is easy to compute K( -1)
= K(x)
399
respectively, and K(x2)
<
= (l- 6)[b(2l +b) -1].
Since
0 < ((1- b6)/2)(a + b6 + 1) - b6
= 1- 2b6- !(a+ b6- 1)(b6- 1)
= 1- 2b6- !(a+ b6- 1) 2 + ia(a + b6- 1) = 1- 2b6- 2b(l- 6) +!a(a+ b6- 1) < 1- 2b6- 2bl + 2b6- b2 = 1- b(b + 21), we conclude that K( -1) < 0, and thus D < 0 when x > -1. AB before, (17.30) does not have a limit cycle. The theorem is completely proved. III. The case of one finite singular point. When b = 0, the necessary and sufficient conditions for (17.8) to have one finite singular point are m +a = 0 and l +m6 i- 0. Ubi- 0, then by (17.14) a necessary and sufficient condition for (17.8) to have a finite singular point is that B 2 < 4D or B = D = 0. Consider the boundedness condition of Theorem 17.3. We get LEMMA
1 7. 3. Suppose system ( 17.8) satisfies one of the following
conditions: (1) b = m +a= 0, 12 + 4ma < 0, and m(l + m6) < 0; (2) (b -1) 2 + 4ma < 0, mb < 0, and B 2 < 4D. Then it is a bounded system, and the structure of its separatrices possesses an infinite saddle poinl and a finite focus or nodal point. (The meaning of B and D is given in (17.14).) For case (1) of Lemma 17.3, we have the following result. THEOREM 17.5. Under condition (1) of Lemma 17.3,(3 ) system {17.8)
has exactly one limit cycle when 0 cycle when 6 ::; 0 or 6 ~ -lfm.
< 6 < -l/m, and does not have a limit
The proof is not difficult, and is left as an exercise. For case (2) of Lemma 17.3, we can only get partial results, i.e. when 0 < 161 ¢:: 1 with appropriate sign, there exists a unique limit cycle near (0, 0). Chen Guang-qing [252] and Yang Xin-an [253] used the original equations in [248] to study the problem of limit cycles for bounded quadratic systems. Chen [252] obtained some sufficient conditions to guarantee that system {A) has at most one limit cycle, but when system (A) is changed to the form ( 3 )We
may as well assume m
> 0 and l < 0, and thus tJs
= 3ml < 0 when 6 = 0.
THEORY OF LIMIT CYCLES
400
of (17.8), its results duplicated part of the results of [250] mentioned at the beginning of this section. [253] contains work of Yang Xin-an done before [250], in which (with the notation of [248]) he proved that a bounded quadratic system possessing two finite singular points (one of which is a degenerate higher-order singular point, which satisfies lei < 2, au < 0, a21 < 0, aua 22 = a12a211 and au + a22 = 0) has one and only one limit cycle when cau + a21 > 0, and does not have a limit cycle when cau + a21 ~ 0.
Exercises 1. Prove the remaining two cases of Theorem 17.4. 2. Give a detailed proof of Theorem 17.5. 3. Prove that the system dxjdt = -x - y- xy, dyjdt unbounded solution [249]. 4. Prove that the system
dxjdt
= -y + xy,
dy / dt
=x -
= x + xy has an
y + x2 + 2xy
has an unbounded solution [249]. 5. Prove that in the system
dxjdt
= aux + a12Y,
except for the case when a22 < 0 and a21 # 0, if along any trajectory t can be continued to +oo, then this trajectory must be bounded [249]. 6. Use (17.32) to deduce that (a+ lx5- 1) 2 = 4b(l- 6), and use these two equalities to prove the equality on the fourth line after (17.40):
[6 + (2l + b)x + (l + b)x2 ]x=-2/(a+b6+1) -(a+ lx5- 1) 2 = b(a + lx5 + 1) [a( a+ lx5 + 1) + 2b ]. 7. Change the several cases of the bounded quadratic system in this section to the several cases of [248]. 8. Use Bautin's method to prove that for the bounded quadratic system (17.8) there may exist two limit cycles in the vicinity of the origin, but (17.8) cannot have (0, 0) as a third-order fine focus.
§18. Appendix
The goal of this section is to give a brief presentation of some new results on the theory of limit cycles from recently received journals or preprints, and also to indicate the direction of research on limit cycles tried in some other papers, not necessarily recent ones, whose contents did not fit into the previous seventeen sections.
I. Plane polynomial systems. In [255], Cai Sui-lin and Wang Zhong-wei studied the sys.tem of equations dxjdt
= -y + lx 2 + mxy,
dyjdt
= x(1 +ax+ by),
(1)
assuming that a 'I 0, ml - a(b + 2l) = 0, m
(mlja 2 )(ma 2
-
'I 5a, and ml 2 + 2al 2 ) of 0.
Then (0, 0) is not a first- or third-order fine focus and is not a center, but is a genuine second-order fine focus, and when m varies the system can change its stability and generate a limit cycle; when ajm < 0 or ajm > 1/5, system (1) does not have a limit cycle in the whole plane; when 0 < 1/5- afm « 1 and 1-15(l/m) 2 of 0 there is at least one limit cycle in the vicinity of 0(0, 0) but there is no limit cycle surrounding any other singular point. But whether the precise range of variation of ajm which guarantees the existence of a limit cycle is 0 < afm < 1/5 or not is still unknown. Han Mao-an [304] proved that if there exists a limit cycle of (1) around its second-order fine focus (0, 0), then it must be unique. Moreover, Chen Wei-feng [305] also obtained many results for the existence and nonexistence of limit cycles around the secondorder fine focus (0, 0) of (1), when the right-hand side of the first equation in (1) contains an additional term ny2 • Ye Bai-ying [256] pointed out that [231] did not completely solve the pro~ lem of centralized distribution of limit cycles of equations of class II, and also gave some other sufficient conditions, and thus extended the range of variation 401
THEORY OF LIMIT CYCLES
402
of b to guarantee the centralized distribution of limit cycles of system {14.5) in §14,(1) but the problem has not yet been completely solved. [257], [293], and [294] all studied the problem of whether quadratic sy~ terns with parabolic solutions can have a limit cycle. The answer is affirmative. Thus the situation is different from the case of quadratic systems with hyperbolic solutions [122]. Wang Dong-da [259] proved that when the parameter A varies, the equation dy dx
-(2Ax + 1)(ax +by+ c)+ o:(Ax2 + y 2 + x + 1) 2y(ax+by+c)+,8(Ax 2 +y2+x+1)
(2)
can have several forms of orbits which are quadratic algebraic curves such as hyperbola, parabola, real ellipse, point ellipse, and imaginary ellipse; but if (2) has a limit cycle, it must be unique. Thus the results in {17] were extended. P. Holmes and D. Rand [260] studied the phase-portraits and bifurcation curves of the nonlinear oscillation equation
(3) In particular, they applied the theory of bifurcation to point out this equation can have @ distribution of limit cycles in the phase-plane. Li Ji-bin [261] continued to study, after [180] and [260], the general properties of a closed trajectory of the cubic system dxjdt = y,
dyjdt = Q3(x, y)
(4)
and pointed out that there are many properties of (4) which are not possessed by quadratic systems. He rigorously proved that system (4) has @ as its distribution of limit cycles. He also studied at the same time the. cubic system possessing nine singular points dyjdt
= Q3(x, y),
(5)
0
and proved that the system has ~ as its distribution of limit cycles, and gave concrete numerical examples. Also he gave sufficient conditions for a general system of Lienard equations dxjdt = y- F(x),
dyjdt = -g(x)
(6)
for the existence of a limit cycle whose interior contains three singular points. L. A. Cherkas [262] then got a sufficient condition for the equation yy' = -g(x)- f(x)y
{7)
(where f and g are rational functions) to have a family of closed trajectories containing three singular points, and proved that in this case the absolute (1 )See footnote 6 in §14.
403
§18. APPENDIX
values of the two characteristic roots of a saddle point in the interior of the closed trajectory must be equal. Li Ji-bin [263] first studied the perturbed system of a Hamiltonian system containing two parameters J.L and A, fJH dy dx fJH (8) dt = - ox - J.LY(q(x, y)- A), dt = fJy - J.LXp(x, y), and the question of generation of limit cycles when 0 < IIJ.LII « 1, and then discussed the question of limit cycles following the change of A and J.L in the cubic system
~: = dy dt
y(ax 2 + by 2 +c),
(9)
= -x (x 2
m 2 - A) • + ay - d - J.LY lx + '3y 2
)
(
2
His results are far better than those of [260] and [264], and the equations he considered are more general. We should point out that for the branching problem of cubic systems, there is work earlier than [260] and [264] on the interesting system x = y- (ax 3 + bx2 +ex), iJ = ( 1 - AX- y (10) in which the existence of S type distribution of limit cycles was pointed out and the branching curves in the (A, u)-parametric plane were sketched (see [265]). Huang Ke-cheng (266] studied the system of equations 2m
2m+l
i+j=O
i=l
dy
dx dt = y,
.. ~ . dt = -y L..... aiix'y3 - L..... aix', ~
(11)
and gave a very succinct sufficient condition to guarantee existence of limit cycles, and he also pointed out some errors in the proof of Theorem 2 in [267]. Note that system (11) is more general than the system
x=y, studied in [268]. Ye Yan-qian [269] extended several general properties of limit cycles of quadratic systems to the quadratic system
dxfdt
= P(x, y),
dyfdt
= Q(x, y)
(12)
with real coefficients, but with complex unknowns x and y, and the equivalent quadratic system in four-dimensional space
dxddt = Pr(x,y),
dx2/dt = Pj(x, y),
= Qr(x,y),
dy2jdt = Qj(X, y);
dytfdt
(13)
THEORY OF LIMIT CYCLES
here x = Xt +ix2 andy= Yl +iy2, while Pr, Pj and Qr, Qj represent the real and imaginary parts of P(x, y) and Q(x, y). Tsutomu Date [270] used the theory of nonvariables to classify homogeneous differential systems, and corrected some wrong figures in [166], and pointed out that their method of classification was not altogether independent of earlier methods. Shi Song-ling [271] obtained sufficient conditions for the existence of k limit cycles containing (0, 0) for the nth degree polynomial system dx dy dt = >.x- y + P(x, y), dt = x + >.y + Q(x, y) (14) and extended the results of [19] and [20]. Jorge Sotomayor [272] proved that if every singular point (including the singular point on the equator) of a plane polynomial system is an elementary singular point and the skeleton figure on the Poincare hemispherical surface is a simple figure, then the polynomial system can only have a finite number of closed trajectories. (A skeleton figure is formed by saddle points and separatrices, in which the w-limit set and the a-limit set of every separatrix consists of saddle points, and in the figure for every saddle point there exist at least one separatrix entering it and one separatrix leaving it; suppose the two characteristic roots at the saddle point Pi are JLi < 0 < Ai· If for all the saddle points P1, ... , Pk in the figure we have k
p
= II IJLil ~ 1,
i=l >.i then this figure is a simple figure.) Based on [272], Sotomayor and R. R. Paterlini [273] further proved that in a vector field x2 formed by all the vector fields of quadratic systems, there exists a nonobvious function R: X2 - R such that if X E X2 satisfies R(X) ~ 0, then X has only a finite number of closed trajectories. We should point out there are several lemmas in this paper which had been obtained earlier in [14]. Rodrigo Bam6n [274] proved that in a vector space X2 formed by the totality of vector fields of quadratic systems, there exists an analytic submanifold 8 of codimension 1 such that every vector field in 8 has one separatrix cycle surrounding one limit cycle. However, Bam6n's claim that this structural figure had not previously been discovered is wrong, since we saw the possibility of its appearance many times in §14. Liang Zhao-jun [275] studied several possibilities of phase-portraits, top~ logical structures, and branching curves of the systems of class II whose first equation contains two quadratic terms
dxjdt = -y + lx 2
+ mxy,
dyjdt = x- x 2 •
(15)
§18. APPENDIX
405
Note that, unlike system {13.3) of §13, {15) may have a limit cycle or a separatrix cycle. Cai Sui-lin and Zhong Ping-guang [302] proved that the limit cycles of quadratic systems possessing a fine saddle point, if they exist, must be centrally distributed in the vicinity of a focus. Also, if the quadratic system has a fine saddle point and a line solution not passing through this saddle point, then this system does not have a limit cycle. There are some other new results, different from the above results, which can be found in [303]. Moreover, we mention the papers [126], [214], and [295]-[297] on the study of quadratic systems.
II. Limit cycles and the typical integral and Darboux integral of differential equations. In the mid-fifties, N. F. Otrokov, K. S. Sibirski'i, and M. V. Dolov in the Soviet Union considered the system of equations
dxjdt = P(x, y),
dyjdt
= Q(x, y)
(16)
(P and Q are analytic functions) and its corresponding first-order partial differential equation
pal +Qaf =O.
(17) ax 8y They obtained a solution /o(x, y) of (17) which satisfies the initial condition [f(x, y)]., = c
{18)
in a neighborhood of a closed trajectory of {16) (where 8 is a segment of the line without contact x = xo +ac, y = Yo+bc, lei :::; co of lata point (xo, Yo) on it). Using the analytic function f(x, y) (generally speaking, it is multi-valued), obtained through all analytic continuations of the solution /o(x, y), as a tool, they studied the decomposition of multiple limit cycles and the existence of Darboux integral, and obtained some significant results, as follows. Suppose h (x, y) is a function obtained from fo(x, y) after continuation for one complete ~ound along a fixed direction in a neighborhood of l. Since on the segment 8 of the line without contact we have 00
[!l(x,y)].,
=w(c) =I:
"'fkCk
(19)
k=l
{here w(c) is a Poincare successor function), from the uniqueness of the solution we get It (x, y) = w(fo(x, y)), (20)
THEORY OF LIMIT CYCLES
406
and it is well known that 11 = exp h = exp
/,(P~ + Q~) dt = exp loT (P~ + Q~) dt.e)
(~1)
We call the analytic solution f(x, y) of (17) satisfying condition (18) a normal integral of system ( 16) along the closed trajectory l. It is well known that there exists a neighborhood S(e, l) of lin which either the equation
w(Jo)- fo
=0
(22)
has a unique solution (then lis a limit cycle of (16)), or the left side of (22) is identically equal to zero (then the neighborhood of l is completely filled with closed trajectories). Otrokov [277], [276] proved that l is an n-tuple limit cycle( 3 ) of (16) i( and only if there exists a first integral f(x, y) = c defined as above, where f(x,y) is single-valued along land has order n. Here "f(x,y) has order n on l" means that f and all its partial derivatives up to order n - 1 have period T with respect to t, and at least one nth order partial derivative is not periodic with respect to t. Hence it is easy to see that in this case any integrating factor J.L(x, y) of (16) on l is a single-valued analytic function of order n - 1. Moreover, Otrokov used the normal integral as a tool to solve, under given conditions, the problem of decomposition of multiple cycle of the original system (16) into single cycles with the aid of the perturbed system
dx dt
= P(x, y) + p(x, y),
dy
dt = Q(x, y)
+ q(x, y).
(23)
For example, in [276] he proved that if l is an n-tuple limit cycle of {16), and we take n-1
p(x, y)
=L
n-1
.\knk(x, y),
q(x, y)
=L
k=O
).kf3k(x, y),
k=O
where nk(x, y) and !3k(x, y) are analytic functions, and the ).k are small independent parameters, then no matter how p and q are chosen, l can at most be decomposed into n single cycles of (23). In [277) he further proved that we can take p and q to be rational functions such that the above conclusion remains the same. Dolov [278] first proved that equation (17) has a real analytic integral, whose first and second branches satisfy the simple relation:
F1(x,y) = Fo(x,y)exph.
(24)
(2) Here we suppose that the equations of l: x = cp(t), y = tl>(t) have period T with respect to t. ( 3 )I.e., in (19} we have '"'11 = l,-y2 = '"'13 = ·· · = '"'fn-1 = 0, and "'n ::/; 0.
§18. APPENDIX
407
This kind of integral is called a typical integral. He then proved: For a single limit cycle l of system (16), there exists a neighborhood S(l, c) such that the equations (25) aQ - (3P = P~ + Q~ and a~
+ (3~ = 0
(26) possess unique single-valued analytic solutions a( x, y) and (3( x, y). But if for every trajectory l we have h = 0, then these a and (3 exist if and only if all the trajectories of system (16) belonging to S(l, c) are closed. From this we can see that the existence of a and (3 is characteristic for a limit cycle to be single. Also, from (26) and the equality h=
i
a dy - (3 dx
we know that a and (3 in the region enclosed by l must have a singular point. For a and (3 which satisfy (25) but not (26), Dolov [278] proved that if in a simply connected region G there exist single-valued continuously differentiable solutions a art
THEORY OF LIMIT CYCLES
408
Dolov in [280] discussed the problem of whether there exists a typical integral in a neighborhood of a separatrix cycle h passing through a saddle point, and obtained several similar results as in [279]. Here we use the integral ht = /_:
[P~ (rp1 (t), 1/J1 (t)) + Q~(rpt (t), 1/Jt (t))] dt
to replace h in (21), where x = rp1(t) andy= 1/Jt(t) are the equations of l 1. In [281] Dolov studied the polynomial system
dxjdt
= y + p(x, y),
dyjdt = -x + q(x, y),
(27)
with (0, 0) as its center. Suppose it has a Darboux integral k
II 41j; = c
(28)
j=l
(where the 41; are polynomials of complex coefficients, the /3; are complex numbers, and 41; and 41, have no common factors when j '#l). He proved that if in this case (27) still has a limit cycle l, then l must be a closed branch of a real algebraic curve, this cycle must be structurally stable, and the irreducible polynomials which determine l must be contained in (28). If l is a single cycle, then in a neighborhood of l there exists a typical integral F(x, y, l), which is analytic in a neighborhood of the origin, and F(O, 0, l) ::1 0. Later, in [282] Dolov studied the formula for the typical integral corresponding to l in the case when a single limit cycle l of system (16) contains a unique coarse focus in its interior. He also proved that if in this case the system has a Darboux integral (28), and also has a limit cycle, then there are two algebraic curve solutions passing through the coarse focus, and they are complex conjugates. In [283] Dolov proved that if system (16) has a limit cycle l whose interior contains a unique singular point which is a nodal point 0, and the ratio of its two characteristic roots is not an integer, then (16) does not have a Darboux integral. Moreover, he gave some examples to show that Sibirskil's conjecture in [286], "there exists a Darboux integral (28) in which the set of the totality of polynomial systems in which 41;(xt, Yt) ::1 0 is dense in the space of all polynomial systems possessing elementary singular points of center-type" is incorrect. In [284] Dolov proved that for a quadratic system (16) with a fine focus, the Darboux integral cannot exist and the system cannot have more than two different irreducible algebraic curve solutions passing through this fine focus which are complex.
§18. APPENDIX
409
Finally, in [285] Dolov studied the polynomial system (16) whose Darboux integral has the form k
G=
m
II ~:; exp II w:· = c
j=l
(29)
s=l
(where the /3; and n, are in general complex-valued, the ~j and W, are in general polynomials with complex coefficients, and there is no functional relatiOn between u = n~l w:• and r = n~=l ~:; 1 there are nO COmmOn factors among the ~;, and there are no common factors among the W,); and the problems he solved are similar to the previous several papers. P. S. Atamanov and V. P. Zakharov in 1976 proved that the single-valued integrating factor M(x, y) of system (16) is unbounded near a limit cycle l. Dolov [287] further studied the relation between the order of increase of M(x, y) and the multiplicity of l. Since recently the study of polynomial systems, especially in the theory of limit cycles of quadratic systems, has become much deeper, we feel that the above work should be especially mentioned.
m. Equations of limit cycles.
For some artificially constructed systems
such as
(30) we can write down not only the equation of the limit cycle x2 + y 2 also its first integral
= 1, but (31)
and the formula for its solution. Many similar examples can be constructed. However, for the quadratic system studied in §11,
~: = -y(ax + {3y + 1)- (x 2 + y 2 -1), dy dt
(32)
= x(ax + /3y +"f),
we can only write down the equation of its limit cycle x 2 + y 2 = 1 (when > a 2 + {32 ) and the first integral of (32), but not the formula for its solution. As for other equations of class I studied in §12, although we have used qualitative methods to prove that its limit cycle, if it exists, must be unique, we cannot even write down the equation of the limit cycle. 12
410
THEORY OF LIMIT CYCLES
When the equation of the limit cycle of a dynamical system can be written down, not only can we determine the precise or approximate position of its limit cycle, but also we can study, when the limit cycle disappears owing to the variation of the parameter in the differential equations, whether it runs to the two-dimensional complex space. [269] gave examples to show that this situation is just like the case when a pair of real roots in an algebraic equation with real coefficients coincide and disappear (because of the variation of the coefficients): we can then obtain a pair of conjugate complex roots. When the limit cycle in the real plane shrinks towards the singular point in its interior, or coincides with another limit cycle and then disappears, we can often get it back in the two-dimensional plane or two-dimensional integral manifold of the fourdimensional vector space corresponding to two-dimensional complex space. Hence the study of equations of limit cycles is an important and interesting problem. For this problem, even for the limit cycle of the well-known van der Pol equation, in the past mathematicians have confined themselves to obtaining the first one or two terms of the approximate representation formula by approximation methods when the parameter is very small. Sun Shun-hua [55] first noted that if in the system
dxjdt
= y,
dyjdt
= -x + J.t(l- x 2)y
(J.t ::/: 0)
(33)
we write J.t as J.t 2 , and apply the transformation x' = J.tX, y' = J.tY, and get (we still denote x', y' as x, y) (34)
dxjdt = y,
then the system of equations forms a family of rotated vector fields with respect to JJ. 2 • As far as system (34) is concerned, when JJ. 2 changes from 0 to positive, the limit cycle is generated because of change of stability of the origin, not because of the circle of radius 2 as in (33). In the complete family of rotated vector fields, a closed trajectory monotonically expands or contracts according to the variation of its parameter and trajectories of the system
dx/dt
= P(x, y, JJ.),
dyjdt
= Q(x, y, J.t)
(35)
corresponding to different parameters do not intersect. Hence in the region G covered by the changing trajectories whose union and outer boundaries all contain singular points, we can define a function JJ.(x, y), whose value at the point (x, y) is the value of J1. of the system (35) which has a limit cycle passing through this point, and J.t(x, y) = JJ.(xo, Yo) = JJ.o is obviously the
§18. APPENDIX
411
equation of all the closed trajectories of {35) 110 , one of which passes through the point (xo, Yo). Sun [55] proved that when P and Q in {35) are continuously differentiable with respect to x, y, and J.t, this function is also corrfiiiuously differentiable in G, and satisfies the first order quasilinear partial differential equation a~-'
Q(x, y, J.t(x, y)) ay
a~-' + P(x, y, J.t(X, y)) ax = 0.
{36)
Finally, seeking the equation of the limit cycle of system (34) in the form 00
1-'2
= ~ Ok(cp)p2k k=l
in polar coordinates (where nk(cp) is a periodic function of period 211"), and returning to system (33), we obtain the equation of the unique limit cycle of {33) in power series form: 1-
~p2
+:,.
+ 16~4
(2sin2cp
+
sin4cp)p4
( 3 cos 6cp + 5 cos 4cp + 5 cos 2cp -
2!~4) + ... = 0
(37)
p6
(which converges when 11-'1 « 1). L. A. Cherkas [288] obtained system (36) again under the condition that P and Q are real analytic with respect to x, y, and J.t, and proved that in this case J.t(x, y) is also real analytic in G. Moreover, he sought the limit cycle 1-'2 = F(x,y) = !(x2 + y2) + F3
+ F4 + · · ·
in the form of a power series solution for the partial differential equation corresponding to system (34), and got F _ 1( 2 2) 1 3 1 a 1 2 4 - 4 x + y - Bx y- 6144 y - 2048 x y
+
+
143 x4 2 - -...!:.!._xs 1 3 5 385 5 3 2048 y 6144 + 4608 X y - 1152 X y terms of higher degrees.
(38)
Returning to (33), he got the equation of its unique limit cycle as 1( 2 2) 1 3 1-'2 0- 1- 4 x + Y + Bp,x Y + 6144(y6 + 3x2y4- 429x4y2 1-'3
+ 6144
(6160 5 3 4 3 5 ) -3-x y - 3x Y -+- .•.•
+ 17x6) (39)
412
THEORY OF LIMIT CYCLES
Ye Wei-lin [289], independently of [288], applied the transformation x'
=
J.I.X, y' = J.I.Y to the Rayleigh equation equivalent to (33)
dyfdt
= x,
dxfdt = -y + J.1. 2 (x- x3 /3)
(40)
and obtained (we still denote x' and y' by x and y) dyfdt
= x,
(41)
and then assumed the equation of the limit cycle of (41) to be J.1. 2 = F(x, y). From [55], we know that F(x, y) satisfies the partial differential equation
aF ( -y+xF-x ) -=0. aF x-+ By 3 ax 3
(42)
Suppose F(x, y) can be expanded into a power series F(x, y)
= a1x + b1y + a2x 2 + ~xy + c2y2 + · · ·,
and substitute it into (42). Then Ye [289] proved that all the coefficients in the expansion formula can be uniquely determined. All the coefficients up to the terms of 8th degree in x and y were computed, and the equation x2+y2 1 J.l. 2 = { + - - (29x6 4 18432 +
X
-
9x4 y2 - 153x2y4
3 { 1 ( -46 2 2 34 y 24 + 18432 X y - 18432 y
-
51y6 ) + · · ·
4) + .. •}
} (43)
= S1(x) + x 3 y82(x) was obtained. Since there was no way to obtain the formula for the general term of the coefficient, he used a computer to compute the approximate values of all the terms up to the homogeneous expression of 20th degree, and observed that S1(x) and S2(x) seem to have majorant series sinh(x2 +y 2) and cosh(x2 + y2). Hence (43) seems to be convergent in the whole plane. Ye Wei-lin [289] also discovered that the calculation of the fourth term on the left side of (37) was wrong: the correct answer should have been J.l.2 (7 1024 2 + cos 2
s
s]
413
§18. APPENDIX
From this we can see that the terms of 8th degree in (39) also have calculation errors. Finally, it is worth pointing out that the method of G. V. Kamenkov used in Chapter 14 of [290] to seek the equation of the limit cycle is not as simple and convenient as the method of undetermined coefficients introduced before in the case of the family of plane rotated vector fields. Moreover, from the theorems concerning equations of class I in §12 of this book, we know that the conclusion of §5 in Chapter 14 of [290] concerning the existence of a limit cycle for the system of equations dx dt = y,
at b = 0 is wrong.
dy [ 2 dt = -x + 1-' ax+ 2by + a2ox
+ a11 xy + ao2y2]
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416
THEORY OF LIMIT CYCLES
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=
I:
aikXiyk, dyfdt
=
I:
bikxiyk,
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qoo + q10x + Qo1Y + q2ox 2 + quxy + Qo2Y2
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=
y + aox2 + a1xy + a2y2 x+box 2 +b1xy+~y 2 '
(1)
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qoo + qtoX + qotY + q2ox 2 + quxy + qo2Y 2 = Poo + PtoX + PotY + P2oX2 + Puxy + Po2Y2 .
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