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Xn 0 for any 6> 0.
From the second comparison theorem (cf Theorem 3 2, Chapter 1) it follows that (x, p) has not less than n +1 zeros in the interval (0, Therefore for sufficiently large b the function (x p) has not less n +1 zeros in the interval (0, b), which contradicts the oscillation theorem The interlacing of the zeros of the functions (n = 0, 1,2, )
follows from (2 11) and the second comparison theorem This the theorem. §3. Investigation of the spectrum for the case q(x) E
(0,
cx,)
I. The following result will be needed in the sequel. LRInIi 3 1 that the functions h1(x), h2(x), gj(x) and g2(x) on the interval 0 x X, and suppose that h1(x) and we h,(x) we continuow,, and g1(x) and g2(x) integrable, on this interval
If (3.1)
(t)g1
(t)+ h2(t)g2(t)]dt,
where C is a constant, then one has the inequalities
0
x
o)
§3. CASE q(x) E (3.2)
h1 (x), h2 (x)
205
(t) + g2 (t)]} dt, 0
C exp
x
X.
PROOF. We put y (z)
(3.3)
(3.3')
=
J [h1(s) g1 (s) + h2 (s) g2 (s)J ds,
_—h1 (x) g1 (x) + h2 (x) g2 (z).
Multiplying the first of the inequalities (3.1) by g1(x), the second by g2(x), and adding, we obtain by virtue of (3.3) and (3.3') C[g1 (z) +g2(z)1 + g(x)[g1 (x)+ g2(x)].
This inequality can be rewritten in the form x (a,)
(s) + g2 (s)) ds]}
exp
C [g1 (x) +
g2 (x)}
exp
[g1 (s) + g2 (s)) ds}.
Integrating this inequality from 0 to x, we find that y (x) exp
[g1 (s)
c — c exp
+
[g1 (s) + g2 (s)] ds}
Hence
(34)
y (x)
c exp
[g1 (s) + g2 (s)1 ds} — c {
By virtue of (3 3), the inequalities (3 2) follow at once from (3 1) and
(3 4), which proves the lemma (1 1) satisfymg the conditions 2 Let (x, X) be the solution of = —cosa Rewriting equation (1 1) m the =slna,
form y" +Ay =q(x)y, applymg the method of variation of constants, and putting X =s2, we obtain p (ar, X)
cos SX Sin
sin ax
—
cos
r35)
+
sin (s (x —
(t) p.(t, X) dt.
206
IV. INVESTIGATION OF SPECI!RUM
Let s
0. We introduce the notation
Then it follows from (3.5) that sin
(z,
sin {s (z —t)) q(t) Pi (t, A) dt. Since cos sx equality
e
X
sm sx
e
the last inequality implies the
X
Jp1(x,
Applying Lemma 3.1 (putting h2(x) =g2(x) (3.6)
(t ±
(x, A) I
I
exp
Smce by hypothesis q(t) E equality that the function
0), we obtain I
q (t) I
dt).
it follows from the last ml is bounded for 0 x < for
(0, cx'), A)
Let us first consider real positive values of s Then for $ function (x, A) is bounded. Therefore from formula (3.5) we (for x—+co) p. (x,
A) =
cos
sz. sin a —
+ (3.7)
sinsx
cos
sin {s (x — t)) q (t) p (t, A) dt 0
X)dt
=p.(A)cossx+v(A)sinix+o(1), I
p.(A)
sinst .q(t)p(t, A)dt, 0
(3.9)
_
q(t)p(t, A)dt 0
Sincetheintegralsm (38) and (39) converge uniformly for it follows that p(A) and ,(A) are continuous functions of S
p
I
th4j
o)
207
Similarly if 9(x, A) is the solution of equation (1.1) satisfying the conditions O(0,X) =cosa, O'(O,X) = —sina, then for (3.7')
O(x,A)
where
(A)=
(3.8')
cos
sin
q(t) O(t, X)dt,
.q(t)O(t, X)dt.
v1(X)=
Further, differentiating (3.5) with respect to x, we find that (x, X) = —s sin sx (3.10)
sin — cos sx cos cos {s(x—t)}q(t)p(t, X)dt
Estimating this function in the same way as was done above, we obtain X—*
=
(310') imilarly,
= —s,ii(A)sinsx-Fsvi(A)cossx+o(1).
Let us consider the Wronskian determinant of
(x, A)
and 0 (x, A).
equation (1.1) does not contain the first derivative, it follows from a well-known formula of Liouville that the Wronskian determinant is ual to a constant. Therefore
W{p, O)=W{p,
=W
(A)
cos sx + v (A) sin sx,
(A)
cos sx + v1 (A)
sin
sx)
rom the equality =1, ing
11)
into account that the left side does not depend upon x we obtain
=1/V5
From this last formula it follows, in particular, that the functions (A) d v(X) cannot both vanish for one value of A 3. Let us now consider complex values of s. For a fixed positive r it flows from (3.5) (for x—* cx) that
208
IV. INVESTIGATION OF SPECTRUM —'Ix
p (t,
X) =
Since
o
X) dt + 0
j
q (t) p (t, A)
dt).
we have (putting, for example, ô = x/2) q (t) p (t, A) I dt))
=0
dt)
+o
=o On the other hand,
i:
I q (t) I
(t) p (t, A) dt = 0
I
q(t) j dt)
I q (t)
j
I
dt)
q (t)j
dt) = o
(e.
= o (eu).
Therefore
+0(1) where
M(A)==4- sin
(3.13)
+
(t)p(t, X)dt.
—
If we use formula (3.10), we obtain
= —ise'IM(X) +0(1)
(3.14)
Similarly
(3.12')
O(x,X)
+0(1)
where A)dt.
now have at our disposal all the auxiliary tools for the investi of the spectrum of a Sturm-Liouville operator for the case q(i Let b > 0. We will consider the boundary-value probler the finite interval [0, b]. igating the negative part of the spectrum It follow and (3 13) that for every fixed A0 <0 the numbç =0 which he in the interval —
209
cc)
§3.
bounded, uniformly with respect to b. Therefore the number of points of increase of the function p6(X) (for the definition of pb(X) see Chapter 2, §1) in the interval — cc <x is bounded, uniformly with respect to b, and so the limit function p (X) has the same property. For the case in which q (x) 0 as x —b cc, this fact easily follows from Sturmian theory (see the remark following the proof of Lemma 1.1). If the number of zeros of the function (x, 0) on the halfline [0, cx) is finite, then the number of negative eigenvalues is also finite. In fact, it follows in this case from Sturm's oscillation theorem that for any b> 0 the problem (2.1) —(2.3) has a bounded number of negative eigenvalues. We have thus proved: is
(0, cc), then the negative part of the spectrum THEOREM 3.1. If q(x) of the problem (1.1) —(1.2) is discrete and bounded from below.
The following lemma gives a simple sufficient criterion for the function 0) to have a finite number of zeros. (0, cc), then the function LEMMA 3.2. If (1 +x2)q(x) E a finite number of zeros in the interval [0, 03) and consequently of negative eigenvalues is finite.
if (1 +x2)q(x) E
Indeed, as is
(0, cc), then
the
has number
there exists a
fundamental system of solutions y1 and Y2 of the equation y" — q (x) y =0
such that limy1(x) =1,
lim[y2(x) —x]=O.
(x, 0) = C1 Yi (x) + C2 Y2 (x), it follows that
= and
limç,(x,0) =1 C2
0) has a finite number of zeros on the half.
= 0. In either case
ne [0, cc).
Let us now consider the positive part of the spectrum. Since by virtue the functions ,u (A) and v (A) cannot both equal zero for one value r x, we have i2(X) +v2(A) > 0. Therefore, putting r (3.11)
—
+v2(A)]"
-
sin
(A),
(X)
,
+V2(X)]/2
= cos
(A),
Cf. V. V. Stepanov, Course in differential equations, Moscow, 1945, p. 246.
210
IV.
we can rewrite (&7)
INVESTIGATION OF SPECTRUM
in the form (for x —*
a')
= [M2(X) +v2(X)111"2sin[sx+5(X)}+o(1).
(3.15) If we assume
(3 5) with respect to s,
we obtain after some sims
sin [sx + (A)i) + o(1) + v2 .f.. Let us denote by b a large positive number and assume, for simplicit that = 0 in the boundary condition (2.3) (the case of an arbitrary j3 can be handled similarly) Then the positive eigenvalues of the proble (2.1) —(2.3) are determined from the equation (3 17)
p (x, A)
{[p.2 (A)
=o(1).
sin[sb
(3.18)
Let s1 be a positive root of the equation (3.18). We note that o(X) = 00 Then (3.19)
s1b
=m,r +0(1)
(Si
Let s2 be the next root, after s1, of (3.18). Then either
s2b+ô(X2) =mir+o(1)
(3.20)
or
=(m+1)ir+o(1).
(3.21)
We will show that the case (3.20) is impossible. In fact, if (3.20) hel then it would follow from Rolle's theorem that the equation (b, A) = has a root s3 lying between s1 and s2, and consequently satisfying the conc
tion s3b + ô (A3) = mir + o(1), which is impossible by virtue of the asyin
totic formula (3.17). Subtracting (3.19) from (3.21) and taking into account that 5(X) is continuous function, we obtain for two successive eigenvalues A1 aJ A2 of the problem (2.1) —(2.3) the asymptotic formula (3 22)
—
= j- + 0(4.)
(s, = SIx,,
= 1,
2)
rid ofthS condition (3.16). will subsequently The boundedness of (x X) for large x follows from Lemma 3 1
E
j3.
3.2. If q(x) E
(0,
co)
211
a'), A >0 and
0, then
(A)=.
p
.e.thespectrumoftheproblem (1.1) —(1.2) is continuous in the interval (0, a').
Let us denote by A1
PROOF
the eigenvalues of the problem
b
(x, X2,,j,... the corresponding eigenUnctions. Let us first assume that q(x) satisfies the condition (3.16). Then the definition of the function (cf. Chapter 2, §1) and the asymp-
2.1) —(2.3), and by
(x, A1,,,),
ptlc formula (3 22) we have (sn,, = p,,
Pb (X) =
(A +
b
dx
=
b
1,
dx
b — 8n, b) $ p2 (x,
_f
Afl÷1
p2(x,
,,)
,,) dx+'O (1)
It follows from the asymptotic formula (3.7) that
fl p2(x,
+
+ o(1).
herefore
P,, (A + — P,, b
,,)+ v2 (As,
± o (t)} X
to the limit b
b
—
1'n, ii)•
a', we obtain the assertion of the theorem for
e case in which q(x) satisfies the condition (3.16). Suppose now that the condition (3.16) is not satisfied. We put
x>n.
212
IV. INVESTICATIONOF SPECrRUM
The function q,(x) obviously satisfies the condition (3.16). denoting by we obtain
the functions
M,(X) and
+— Passing to the limit n —
(A)+
S we
obtain (3.23).
§4. Transformation of the basic equation
In certain cases equation (1.1) with an unboundedly increasing, q(x be transformed, by means of an appropriate change of variables, another form which makes it possiile to deduce asymptotic formulas. Let us first assume that X is a real number, and that q' (x) and q" (x are continuous functions. We introduce the notation
can
(x)
(A
(x)
—q
= {). — q (x)
dx 1
q
(x)i"I
Therefore
q"(x)
5
q'2(x)
]
I
(
equation (1.1) can be transformed to the form
(41) This equation is of the same form as (1.1), but in (4.1) the coefficiei of 'p in the last term turns out to be small for large A. It follows from (4.1) that each of its solutions = satisfies
intagralequation
=p (0) cos + p' (0) sin (42)
—
sin —
[A—q(t)]3
p
where r =E(t). This equation can be used to obtain asymptotic formub
213
§4. TRANSFORMATION OF BASIC EQUATION
For nonreal X or for q(x) > A the function E assumes nonreal values, and therefore formula (4.2) contains an integral along a complex path. There is, however, no necessity to introduce such integrals, since one can obtain a similar equation with a real independent variable, which in fact turns out to be x. To obtain this equation we assume that q (0) =0 (this can always be achieved by an appropriate choice of the origin in the Aplane). We then put
P (x) =
(4.3)
(x)
—q
q
(4.3')
q
d2y
(x)1"4--_4.
q"(x)
51
q'2(x)
L
1
,
dx2
By virtue of (1.1) and (4.3)
I=
sin
=
(x) — (t)}
sin
—q(t))"
{tx ._
(x) — (t)}
+ sin
dt dt
+
(x) — (t)}. {X —
Integrating by parts twice, we obtain
= {sin
(x) — E (t)} [X — q
+ cos (0)
—
sin
4f}
(x) — (t)]
sin (x) + {cos
dt (x) — (t)]
(x) — (t)J [A — q (0)
(t))
(t) dt
sin (x) + (z) — (0) cos (x) — 12
Therefore
= Le
satisfies the following integral equation
+ I,
214
sin (4.4k
sin
4-
(#))
(x)
R (t) (t) dt,
where
fl
(A
+16
_q(t)}312
(A
In ftrmulas (4.4) and (4.5) we assume, for ImA > 0, that 0 <argx
suppose that
x>
q'(x)
(4.6) where c
is some fixed number satisfying the condition 0
(47) converges uniformly with respect to A in every region in which [x
—q
(x)
a• By the estimate (4.6)
PROOF.
q'2(x)
$
{q (x))
2
dx==O $
[q(x)] '
xo
xo
Moreover, we have c
(q(x)) /2
dx=
[q(z)J/2
2
ç
q'2(x) {q(.r)} /'
dx=O(1).
The lemma follows from these two estimates by virtue of (4.5). Thecaseq(x)—*— q(x)—*--c-o q'(x) <0 for 5.1. Suppose that q(x) i'mdq'(x) is of fixed sign for large x. Moreover, suppose that the condition (4.8) 1. sailafied. If the integral
(5.1)
j
q (x)
J_1/2
dx
215
CASE q(x) —p —
+ ). PROOF. We remark that the conditions of the theorem imply the convergence of the integral (4.7). Therefore it follows from equation diverges, then the spectrum is continuous and fills up the entire axis (—
(4.4) that for positive A (x) = (0) cos (x)
+
sin (x)
(0)
Hence, putting in §3, we obtain
o(1).
is the same as
=[x —q(x)]1"4co(x,A), where
+v(A)sinE(x) } +o(1),
={A —q(x)
(5.2)
(t)dt +
(x)— (t))
sin
where (X) =
(53)
=
(5.4)
sin a —
sin (t)R(t) (X — q
q' (0)sin a
+
(t, A) dt,
cos (t)R (t) {A —
q
(t, A) dt.
Similarly, if O(x, A) is that solution of equation (1.1) for which 0(0, A) =cosa, O'(O,A) =sina, then sin (x) + + o(1)), where iij(A) and v1 (A) are obtained from (A) and ii (A) by replacing sin a, — cos a and (t, A) by cos a, sm a and 0 (t, A) respectively From these arguments it also follows that the integrals in (5 3) and (5 4) converge uniformly with respect to A, and so and v(A) are continuous functions of A (5.5)
0
(x, A)
—
q(x)}—'14
Further, differentiating (4 4), we find that (x)
(A — q (x))'1' { X—'/a
(0) cos (x) —
(0)
sin (x)
x
cos
40
Therefore, proceeding as above, we obtam ([A — q
(x, A)) {X — q (x))"[v (A) cos
(x) —
(A)
from which, by virtue of (5 2) and (4 6), follows
sin (x) + o (I)],
216
IV. INVESTIGATION OF
A) =
(5.6)
A
—q(x) }1"4{v(A)cosE(x) —
+0(1) 1.
Similarly, (5.7)
8'(x,X) = $A —q(x)
+0(1) }.
Thus
lim Wk,O}
=1.
—Mi(X)v(X) =
Consequently and v(X) cannot both equal zero for the same positive value of A.
The above arguments require some modification for A <0. However we can choose X so that for x X inequality A — q(x) > 0 holds, and repeat all of the preceding argument for the interval (X, co). As a consequence we arrive at the very same conclusions concerning the behaviour of the functions (A) and v (A) for A 0. Let us first calculate the spectrum for the case A 0. Let b> 0 be arbitrary. We consider the boundary condition
=0.
(5.8)
If sm
0, then as b —* co the condition (5 8) can be rewritten, usmg
(5.2) and (5.6), in the form =o(1).
(5.9)
For sin $ = 0 we obtain o(1)
(5 9')
The analysis of these last two equations is identical Let us consider Putting (A)+v2(X)}
we
/2
==sin w(X)
v(A)
'
(X)+v2 (A))
,
/2
=cosw(X)
can rewrite it in the form
(5,10)
+w(X)} =o(1).
successive positive roots of the equation (b =0 (or what is the same, of the equation (5 10)) Using the asymptotic formula (5 2) and arguing in the same way in the analysis of the equation (3 18), one can show that Let A1 and A2 be two
A,J—E(b,
(x,
{A_q(t))t/sdt)
§5. CASE q(x)
—.>
217
—
Therefore WA2 — q
(t) —
— q (t)} dt
(5.11)
Further b
dt 2VX1—q(X)
0
dt
j 0
b
dt VX1—q(t) [VX2—q(t) + VX1—q (t)}2 0
it follows from (5.11) that 1
+
b
—q (t)
$0•2VA1—q Ve will now show that for b b
A))
5.13)
VX—q(t)
$
dt=O(1).
Indeed, let us consider the integral with the sine (the integral with he cosine can be similarly estimated). Since (t, A) =
— q (u) dii,
ye have b
b
sin
ç 3
0
%1X—q(t)
3
0
A—q(t)
t'
Sil)
—
A—-q(t)
drL
0
14) A))
i
increases, then t increases, and consequently [x — q (t)
decreases.
218
iv. INVESTIGATION OF SPECTRUM
Therefore (k+1)7t/2
(k+i)z/2
q(t)
k
dTj
I
—
q (k
$
-r-
sin
increases, the last expression decreases monotonicaiiy,.
and the estimate (5 13) follows from the expansion (5 14), which is a sum with alternating signs formula (5 2) and the From the estimates (5 13), the divergence of the integral (5.1) it follows that dt (A)
(t, X) dt =
(X)
+ v2 (X)
Hence we conclude from (5.12) that for X > 0,
>0
b
5 p2 (t,
Afl+1,b—An,b
=
ban, ?<Xn,
b) dt
5 p2 (t,
dt I
{
0 the spectrum is continuous If A <0, then choosing X accordmg to the condition that A —
1 e for A
>0 for t> X, and replacmg
[x — q(t) ]1"2dt by
fx1 [x — q(t)
Il/Il
everywhere in the asymptotic formulas, we arrive at the previo4
I
zesult This proves the theorem THEOREM 5 2 If all the hypotheses of Theorem 5 1 are satisfied, bl (5 1) conuerges, then the spectrum is discrete and has a
bmd point at infinity PROOF It follows from the convergence of the integral (5 1) the ds6nition of the function (x, A) that as x —*
219
§6. DIRAC SYSTEM WITH SUMMABLE COEFFICIENTS
— {—q (t))'1'] dt
(x, A) — (x, 0) = [(A — q Xdt
ç
3 (A —q
Adt
ç
(A — q (t))'12
+ (—q (t))112
+ (—q (j))V2
<
the imaginary part of the function A) is uniformly bounded in every bounded portion of the A-plane, in particular in every Therefore
finite interval of the real axis. It follows from (5.2) that the integral ?)dt
(5 15)
is uniformly bounded in every finite interval of variation of the real variable A.
Let us assume that in some finite interval (A, A + the spectrum consists of an infinite set of points. Obviously in this case there must exist an increasmg unbounded sequence
such that the number of eigenvalues
}
of positive numbers
in the interval
Aft
(A, A +
grows unboundedly with k Therefore since the integral (5 15) is bounded, the sum f p2 (t,
X<X,,,
dt
which contradicts Lemma 1 1 of
would grow unboundedly as k 2
§6. Investigation of the spectrum of a Dirac system for the case of summable coefficients
We will consider the problem —{A+qj(x)}y2=0,
6.1)
=0,
6.2)
y'2(x)
6.3)
y1(0)cosa +y2(0)sina =0
i
the
halfline [0, ix).
belong to
We will assume that the coefficients q1 (x) and
(0, co)
We will denote by
=
A)
220
IV. INVESTIGATION OF SPECTRUM
the solution of the system (6 1) —(6 2) satisfying the initial conditionsi
=cosa. = —sina, Let us rewrite the system (6.1) —(6.2) in the following form: (6.4)
(6.1')
y'j(x) —Xy2=qj(x)y2, y'2(x) +Ay1
(6.2')
—q2(x)y1.
the right sides of (6 1') and (6 2') as known and the method of variation of constants, we find that Regarding
p1(z,
X)q1(s)cosX(x—s)ds 0
(6.5)
X)q2(s)
p2(x,
X)=
sin X(x—s)ds
5p2(s, 0
(6.6)
(s,
—5
A) q2 (s) cos
A(x — s)
(r>0) and put
Let
A) = h1(x)e",
(6.7)
A) =
Then it follows from (6.5) and (6.6) that h1
=
sin (Ax —
(6 5') ——5
sin A (x — s) ds,
(s) q2 (s)
h1
cosX(x—s) ds
+ 5h2(s)q1
(s) q2 (s)
cos A (x — s) ds.
5
Since Isin(Ax —a)I
Icos(Ax —a)!
the absolute values of both sides of (6 5') and (6 6') we obti
§6. DIRAC SYSTEM WITH SUMMABLE COEFFICIENTS
221
I
I+
{fh1(s)J fq2(s)j + !h2(s)
1q1(s)I} ds.
Therefore Lemma 3.1 is applicable, and we obtain Jh1(x)I,
But since by hypothesis q1 (x), q2 (x) E
(0, cx), it follows from the
'ast bounds that the vector-function h(x) is
/ h1(x)
A bounded for 0 x < p > 0 and r of (6.7) we have for large x
0. Consequently, by
j
(x, X) =0 (eu),
[6.8)
X) = 0
Let us first consider real values of A. Then by (6.8) the functions p1(x, A) and p2(x, A) are bounded. Therefore we obtain from (6.5) and (6.6) that for x—* ?1(x,
A)q1(s)cosk(x—s) A)g2(s)sinX(x—s)}ds
6.9)
sin Ax+o(I), X)q1(s)sinA(x—s)
+p1(s, X)q9(s)cosX(x—s))ds 6.10)
= v (A) cos Ax — (A)
Ax ± o(1),
222
IV. INVESTIGATION OF SPEcFRUM
where (6 11)
(6.12
—sin
(A)
(A)
+
(p2 (s, A) q1 (s) cos 1+ p1 (s, A) q2 (s) sin As) ds,
= cos +
(s, A) q1 (s) sin As —
p1
(s, A)
cos As) ds.
mtegrals in (6 11) and (6 12) converge uniformly with the functions (A) and v (A) are continuous functions of A. Similarly, if
Since the
to
A,
= /01(x, A) \02(X,X)
is the solution of the system (6.1) —(6.2) satisfying the conditions (6.4')
01(0,A) = —cosa,
82(0,A) = —sina,
then (6.13)
01(x,A) 02(x, A)
(6.14)
= ,j(A)cosAx —
+0(1),
where (6.13')E
(A)=
(6.14')
(A)
A)q2(s)sinAs}
(02(8, A)q1 (s) cos
= —sin +
(s, A) q2 (s) cos As) d4
(s, A) q1 (s) sin
Then from (6.9), (6.10), (6.13) and (6.14) we have W{p, 0)=p1(x, A)02(x, A)—p2(x, A)01(x,
A)
(6.15)
Since by the initial conditions (6.4) and (6.4')
= 1, follows from (6
15) that for real A
=1 It
follows from (6 16) that
and v(A) cannot
both
vanish for t
§6. DIRAC SYSTEM WITH SUMMABLE COEFFICIENTS
Let us now consider complex values of A. For fixed positive r obtain from (6.5), for
we
.
p1(x,
(s, A) q1 (s) — ip1 (s,
(6.17)
223
q2 (s)) ds
5
Jp2(s, A)q1(s)—ip1(s,
Further, since by virtue of (6.8) we have, for x
o{
A) q1 (s) —
I
(s, X)q2(s) ds
}
=0
I
I + q2 (s)
1
ds
{
+0 {P2 (s, A) q] (s) — 1p1
5
(s, A)
(s)}
=o(ex) we have from (6.17) (for x—4 cz)
+0(1)1, M k6.19)
2 1
11 (s, A)q1 (s) — iq2(s)
(s, A))
L
from (6 6) and also from the analogous formulas for the
iv. INVESTIGATION OF SPECTRUM
224
solution O(x, A) we obtain (6.20)
4,2(x,A)
+o(l)},
(6.21)
01(x,X)
+0(1)
(6.22)
82(x,A)
+o(1)},
where Af
r\
Sifla
(6.23) {ip2 (s, X)q1 (s) + p1
(s,
X) q2 (s)) ds,
(6.24) (02
(s,
A)q1 (s) —
(s,
X)q2(s)) ds,
(A\_SIflaCOsa (6.25)
2
2i
{iO2(s, X)q1 (s) +
(s,
X)q2 (s)) ds.
THEOREM 6 1 If the coefficients q1(x) and q2(x) belong to 9 (0, cx), the spectrum of the problem (6 1) + (6 2) + (6 3) is continuous and o). fills up the entire axis (— then
Let A) = 0 (x, A) + m (A) (x, A) be that solution system (6.1) —(6.2) which belongs to 92(0, Q). Then using (6.18), (6.20) —(6.22), we have PROOF.
of th
4i1(x,A) =01(x,A) +m(A)4,1(x,A)
+m(A)M1(A) +0(1)
!l/2(x,A) =02(x,A)
+m(A)M2(A) +o(1)}.
for every nonreal A there exists only one solution of the systen (61) —(6 2) which belongs to 92(0, a), it is obvious that 4,(x, A) l(x,A) do not belong to 92(0, co) But smce, by hypothesis, A 2(0, we must have E Since
m'A'— 'M1(A)M2(X) '
when A tends to a real limit (i e r — 0), it follow
—(625) and (611), (612), (613') and (614') tha
§7. TRANSFORMATION OF BASIC SYSTEM
4.{v(A)
—ip.(X)),
225
N2(X)—+
Consequently
limm'X'— Therefore for real A Im
Since the functions (A) and v (A) do not both vanish for one value of A, it follows that 1mm (A) is a continuous function of A on the entire Chapter 3, we have axis (— a, a). Then by virtue of
which proves the theorem. §7. Transformation of the basic system
In certain cases the system (7.1) (7.2)
coefficients q1(x) and q2(x) which increase unboundedly, can by means of an appropriate change of variables be transformed to a form which enables us to derive asymptotic formulas. Let us assume that A is a real number, and that the coefficients q1 (x) and q2(x) are twice continuously differentiable. We put with
(z) = {R + q1 (x)] [A + q2
(73)
dx,
u(x)=F(x, X)y2(x),
(74)
v(x)_—G(x,
A)y1(x),
where
(7 5)
Then I
F (x,
A)
={
G (x, ))
226
IV. INVESTIGATION OFSPECTRUM
du
du
I
F
I?
—
j Now inserting this expression for
Y2
into the equation (7.2), we obtain
orby virtue of (7.3)—(7.5)
(7) A)y1(x)—_t'(x).
We further have
= whence
+ Gy +
—
F1y}
by virtue of (7.2) we find that
=—
+
Fcc'
+ q2 (x)J —
{GF [A
}
from this equality in equation (7.1) and taking account of the first of the equalities (7 4), we obtain Inserting the value of dv
'
da
A+qj(x) a'F'
Since by (7 3) and (7 5) the coefficient of u in the first term on the right side of (7.7) equals — 1, and the coefficient of Yi is zero, equatioiiT (7.7) assumes the form (7.8,
dv
X)
Thus the system (7.1) —(7.2) is transformed to the form (cf. (7.6) and
du/da = (710)
dv/da—u+R(x
(7.11)
R(x,
or,
(73) and (74)
G'(x,
A)u, X)
X)
§7. TRANSFORMATION OF BASIC SYSTEM
227
q'(x)
I?'
I
7
[A+qj(x)]IX±q9(x)1 LEMMA 7.1. If the vector-function
= is a solution of the system (7.9)—(7.10), i.e. if (7.9')
'12,
di2/da =
(7.10')
then it satisfies the system of integral equations
X)
X)
(s,
(s) —a(x)]ds,
X)
where
P(x,
(7.14)
X).a'
PROOF. In fact, by equations (7.10') and
(7.14)
we have
x
X)sin[cz(s)—c'.(x)}ds
{
(s)
+
(s, X)
(s)} sin
(s) — (x)J ds
x
X)x'(s)sin[a(s)—o(x)Jds.
Now integrating the first integral in the right side of (7.15) by parts, then using equation (7.9'), and then integrating by parts once more,
228
IV. INVEGATION OF SPECTRUM
we find that sin [a(s) —a(x)]da(s)
=
sin a (x) —
cos [a (s)
— a (z)] da
(s)
(x, A) sin [a (s) — a (x)] dcc (s).
The validity of (7.12) follows from (7.15) by virtue of the preceding equality. The proof of (7.13) is similar. 7.2. Suppose that and q2(x) are monotonic functions, the following conditions is satisfied:
and that one of
q1(x)—3a,q2(x)-—*—co;
1. 2.
4. Suppose also that
q1(x),
>0, <0,
3.
the following
conditions are
satisfied:
5.
qç(x) =O{
(7.16) where
=Ot
0
6. qç (x) and
(x) preserve their
signs for large
(in absolute value)
values of x. Then the integral (7.17)
A
in
every
region IA +q1(x)
for
By virtue of (7.11'), (7.14) and (7.3)
PROOF.
P(x, A)= 4 P. + q1
(x) 8
(x)]" [A
±
[q2 (.c))4
7
+W
[X+qj
I,
IA +q2(x)
I
229
§7. TRANSFORMATION OF BASIC SYSTEM
Obviously to prove the lemma it is sufficient to show that each of the terms is integrable on the semiaxis [0, co). Suppose that the conditions 1, 5 and 6 are satisfied (the proof for the remaining cases can be carried out in a completely analogous way). Then by (7.16) we have x
x
{q(x))2
ç
{q2 (x))" Since
dx—O —
c
{q2 (X))'/2
{q1
.1
dx
by hypothesis q2(x) is a monotomc function, applying the theorem
of the mean to the last integral we obtain (X0 <E <X)
I {qi
(7.19)
{q2 (x))t/2
I
dx=O[(
= 0 [(q2
dq1
(E)}'1' ç
(x)
{q1
(q1 (x)}°"]L = 0 (1).
Similarly we obtain the estimate (720)
{q(x))Z
c
dx=0(1).
(x)) /2 {q2 (x))
Further, again using the theorem of the mean and then integrating by
parts, we find that (X0 (x))
{q2 (x)}
<X)
,
{q1 (x)) 12
dx.
qj(x)
— —{
c
d
Therefore from the estimates (7.16) and (7.19) we conclude that
(721)
{qi (x))
(q2 (x)}'1'
dx=0(1).
We can similarly establish that q(x)
(7.22)
1{q1 (x))' {q2 (x)) 12
dx—O(1).
Finally, using the estimate (7.16) and the theorem of the mean, we obtain
I
(7.23)
Ix {q1 (x))'1' {q2
c
dX
—
{q1 (X)}'1'2 (q2
230
IV. INVESTIGATION OF SPECTRUM
=0
= 0 ('1).
{q1
The convergence of the mtegral (7 17) follows from the estimates (7.19) —(7.23) by virtue of (7.18). This proves the lemma. §8. The case of a pure point spectrum
Our goal in this section is to establish various sufficient conditions for the discreteness of the spectrum of the problem yc—{x+q1(x)}y2=o,
(8.1) (8.2)
y1(0)cosa +y2(O)sina
(8.3)
0.
THEOREM 8 1 If the coefficients q1 (x) and q2 (x) satisfy the conditions 1 +5 +6 or 2 +5 +6 of Lemma 7 2, then the problem (8 1) + (8 2) + (8 3) has a pure point spectrum PROOF
For Im A > 0 we will assume that 0 <arg A
thosecasesmwhich q1(x) orq2(x)
—
and arg[x +q2(x)] vary from argx to
+
arg
arg X
(x)}'IB
we will
and for
assume that arg[x +qi(x)]
Therefore arg {X +
q2
(x))'/2
j.
It follows that the imaginary part of the function (x)
{[X + q1 (s)] [A + q2 (s)}}V2 ds 5
is positive. Further, if condition 1 or 2 of Lemma 7.2 is satisfied, A is bounded and x a, then i
(x)
(x)}1/2 {q2 (s)}t18 ds,
as x—* that from (7.12) and (7.13) that
Let
so
E1 (x, A) =
1h
—
A)=
cos +
(0) (s,
A)
ii
e'
A) =
Then it followS
sin (x) (s, A) sin
(s)
5
cos
sin A) cos
— (x)) ds,
231
§8. PURE POINT SPECTRUM
Hence, since 1,
1,
it follows that X)!ds,
k1(0)l +
1E2(x, X)I
(s, X)J
IP(s, X) tds,
and so by Lemma 3.1, which is applicable by virtue of Lemma 7.2, X)f,
Consequently
ki(x,
X)l,
(8.4)
(0)1) exp
(0)1 + I
(lip (s, X) ds}
=0
0 {eImu(x)).
Further, for fixed A, ImA > 0, and for X—* co it follows from (7.12) that
+0
(x, X) = (8.5)
(s, X)
P(s,
(s, X) I
X)
(x)-u (8)Jds}
(s, X)
(8.6)
I
=0 = 0 {e1mE )_21fl
I
P (s,
I
P (s, X) I ds}
+0
I
P (s, X) ds}.
IV. INVESTIGATION OF SPECTRUM
232
Since
{[X+q1
it follows that —Im
(x) — 2a (x —
=
{[X +q1 (s)J
Therefore exp{ —Im[a(x) —2a(x —o)]}
(x—*
we obtain from (8.6)
Consequently, using Lemma 7.2 again, for x
the estimate (8.6')
X)
)_a(8))ds}
P(s,
=o
On the other hand, by Lemma 7.2 we have for x •
(s, X) P (s, X)
(8
(8
the equality (8 5) can
rewritten in the form (8.8)
i1(x,X) _—e
+0(1)1,
where
A)P(s,
we obtain from (7.13) ,2(x,X)
+0(1) A)
P (a, A)
233
§8. PURE POINT SPECTRUM
—* then and consequently as x Now if Im a (x) —* — proceeding in the same way as above, we can give (7.12) —(7.13) the form
(8.12)
+0(1)},
,11(x,A)
+o(1)},
(8.13)
where (8.14)
M (A) =
(8.15)
N
(s, A) P (s, A)
A)P(s,
Let us denote by /c01(x, A)
=
ç2(x, A)
the solution of the system (8.1) —(8.2) satisfying the conditions
= —sina,
(8.16)
ç2(O,A) =cosa,
and by O(x,A)
=
/01(x, A)
02(x, A)
the solution satisfying the conditions (8.17)
01(O,A) = —cosa,
02(O,A)
= —sina.
Further, since it follows from (7.4) that =F(O)y2(O),
(8.18)
=G(O)y2(O) —F'(O)y1(O),
then by (8.16) we have
(8.18')
,,2(O)
=F(O)cosa, F'(O)sina +G(O)cosa.
On the other hand, thanks to (7.4)
A) = G(x, A),71(x, A) — F(x, (8.19)
=F'(x,A),71(x,A).
A),
234
IV. INVEST1GA'FION OF SPECTRUM
Therefore, using (88)ahd (8.10), we obtain for the sóiutión
ip
(x, A)
as
(820)
F'(x,
4,t(x, A) =
(8.21) where
+0(1)1,
,,t(x,X) —e
+0(1)
by virtue of (8 18') and (8 19), using (89) and (8 11),
± [iF (0) — G (0)] cos
(0) sin
M (A)
X)P(s, X)F(s, (F_i (0)
(A) =
— [iF (0) — G (O)rcos
(8.23)
A)P(s, X)F(s,
We can similarly obtain the following asymptotic formula for the solutionO(x,X) (8.24)
OjF(x, A)
=
—
=e
(8.25)
F(x,
+0(1) },
+0(1)
and Nt(A) are obtained respectively from and by cosa, —sina and Let #(x,A) =O(x,A) +m(A)p(x,A) be that solution of the system 2(0 x). Then using the asymptotic (8.1) —(8.2) which belongs to
where
by replacing sina, COsa and
formulas (8.20) —(8.21) and (8.24) —(8.25), we obtain A) = 01 (x,
=
+
(A)
{G (x, A)
(x, A) (A) — F (x, A)
(A) — F (x, A)
+ m"' (A) [G (x, A) (x, A)
=
(x, A) + {F—1 (x, A)
A) E
A'—
+
2(0, co),
(A)
(A)
(X)] + 0(1)),
(x, A)
(A) +
(A) F—1 (x,
(A) + o
it follows from the last two formulas that G(x
A)N4(X)
that we are considering the case in which Ima(x) —' +
235
49. OTHER CASES
Similar tormulas. also hold for the case in which Im a (x) —+ — with and in formulas (8.26) replaced, respectively, by m — (A), and where the last two functions are obtained from (8.14) and (8.15) in the same way that and were obtained from (8.9) and (8.11).
Let A be a fixed real number. According to condition 1 of Lemma 7.2, q1 (x) tends monotonically to + and q2 (x) tends monotonically Let us first assume that for all x we have A-f-q1(x) >0 and to — A +q2(x) <0. Then the function a(x) is pure imaginary, so that a(x) = a (x) I. Let us assume the same thing if condition 2 is satisfied. Then, as is easily seen, F(x, A) =
and G(x, A) =
F(x, A)
G(x, A)
Consequently it follows from (8.22) that the function real. One similarly proves that the functions M,, (A), Mr (A), or by
is
(A) and NI (A) become real when they are multiplied
We have thus established that the function
is
in fact
real and continuous on the entire real axis. Since, moreover, M7(x) is regular in the upper A-halfplane, it follows from the symmetry principle that is an entire function. We can similarly conclude that is an entire function. Consequently, by formula (8.26) is meromorphic. The eigenvalues coincide with its poles, and consequently the spectrum is a pure point spectrum. The remaining cases can be worked out in similar fashion. The preceding arguments can all be carried through from the same point of view for the case in which, when q1(x) + q2(x) — or
when
q1(x) —p
—
q2(x)
—p
+
the inequalities
A +q1(x) > 0,
A-j--q2(x) <0 or A+q1(x) <0, A+q2(x) >0 do not hold for all x. It is obvious that one can then choose X such that for all x> X and for fixed real A we will have A +qi(x) > 0, A +q2(x) <0 (for the case of condition 1) and A ±q1(x) <0, A +q2(x) > 0 (for the case of condition 2), following which all the preceding deductions can be repeated for the interval (X, ). The proof of the theorem is complete. §9. Other cases for the spectrum
In this section we will continue the study of the spectrum of the problem (8.1) +(8.2) +(8.3). Let the functions \ and O(x, A) = / 01(x,A) 4,(x, A) = ; A) / \02(X, A)
236
IV. INVESTIGATION OF SPECTRUM
A) and A) the solutions of the be as before. We denote by system (7.W) —(7.10') which are obtained from (7.4) by replacing y(x) = by 4,(x A) and O(x A), respectively Then by virtue Y2(X)
of (8.16) and (8.17) it follows from (8.18) that (9.1)
= F(0)cosa, = G(0)cosa +F'(O)sina,
(92)
= —F(0)sma, = — G(0)sina +F1(0)cosa.
Further, from (8.19) we obtain the following expressions for and O(0,A): A) = 4'2(0,A) 01(0,A)
=G(0),110(0)
•
By virtue of the conditions (8 16) and (8 17) we have
=1. and (9.4), we obtain
Therefore, using the expressions W (9.5)
(p,
0)
— [G
[G (0) (0)
(0)
—F
(0) — F
(0)
(0)] F1 (0)
(0) '120 (0)) F1 (0)
(0)
= '11,(0)'120(0)'12,(0)'11e(0) 5
(0)
1.
THEOREM 9.1. If the coefficients q1(i) and q2(x) satisfy conditions 3, and 6 or 4, 5 and 6 of Lemma 7.2, then the spectrum of the problem (8.1)
+ (8.2) + (8.3) is continuous and fills up the entire axis (— PROOF.
d
co).
Let A be a fixed real number. Since, by conditions 3 and 7.2, as x—'
4
the functions q1(x) and q2(x) simultaneously
or to — we will assume first that for all x we have either A >0, x +q2(x) >0 or A +q1(x) <0, ) +q,(x) <0. In both cases we obtain that both a (x) and F(x, A) (cf. fonnulas (7.3) and (7.5)) are real. Thus Im a (x) = 0. Therefore it follows are bounded. On from the estimate (8.4) that ,71(x, A) and n2(x, A)
taideither to +
237
§9. OTHER CASES
the other hand, the hypotheses of the theorem guarantee the convergence of the integral (7.18). Then formulas (7.12) and (7.13) can be rewritten in the following form for x =.u(A)sina(x) +v(A)cosa(x) +0(1),
(9.6)
+0(1),
= —v(X)sina(x)
(9.7)
where (9.8)
(0)
(X)
+
(s, A) P (s, A) cog a (s) ds,
A)P(s, X)sincL(s)ds.
(9.9)
Since the integrals in (9.8) and (9.9) converge uniformly with respect to A, it follows that (A) and v (A) are continuous and bounded functions of A. Further, by virtue of the asymptotic formulas (9.6) and (9.7), it follows from (9.5) that for x W{p, (A) sin a
(A)
sin a
(A)
cos a (x)J (A) siu a (c) + vax) cos a (x)] + 0(1) where
and p8(A), ve(A) are
(9.9) by replacing
obtained from formulas (9.8) and
,2(x) respectively by F(x,
A)
A)
A) —
F'(x, X)01(x, A).
can
vanish for one
Consequently, for x—* 1,
and
therefore neither
v9(A)
vç(A) nor
value of A.
By (9.1) and (9.2) we have from (8.22) X)eia(8)ds]
(0)
or —
238
IV. INvESrIGATIóN OF SPECTRUM
We can similarly obtam the formulas (A) = — 4.
(A)
—
(A)], (A)],
(A)
= 4. [p.s (A) — jv8 (A)].
Similar expressions hold for the functions M (A) and N (A). Then from formula (8.26) we obtain for the expression
—________ Consequently
—_________ Similarly
imm T
1
In both cases Imm(A) is a continuous and bounded function of A. There-
fore the assertion of the theorem, i.e. the continuity of the spectrum, follows from (3.19), Chapter 3, since by this formula
For the case in which the inequalities A +q1(x) > 0, A +q2(x) > 0 or A +qj(x) <0, A -+-q2(x) <0 do not hold for all x, the proof can be carried
out m exactly the same way as that of Theorem 8 1 This proves the theorem Bibhographacal references
§1 Lemma 1 1 is due to Levitan [1], and Lemma 1 2 is due to the authors and is published here for the first time §2 Theorem 2 1 under the condition q(x) + was obtamed by [ii, and under the condition (1 5) of A M. the theorem wes proved by the authors and is published here for the first time results of this section are essentially due to Weyl [1], cf also §3. [I] The niethod discussed here for calculating the function
p(A) s due to Levitan [ii
REFERENCES
239
§4. The results of this section are due. to Titchmarsh [1]. §5. Theorems 5.1 and 5.2 are due to Titchmarsh [ii. The proofs of these theorems given here are due to Levitan [1]. §6. The results of this section are due to Conte and Sangren [2]. This section was written with the help of their work. §*7, 8, 9. These sections were written with the help of the paper by Roos and Sangren [1]. The discussion of their results presented here is due to the authors.
________________
CHAPTER 5 EXAMPLES
§1. The classical Fourier integral
In this chapter we will consider a number of examples which illustrate the theory discussed above. For the Sturm-Liouvifie problem the simplest example is the case of the classical Fourier integral, corresponding to q(x) 0. If the interval
under consideration is the real semiaxis [0, ), and the boundary conditions at zero are given by (2 3), Chapter 2, then in this case
+
0 (x, A) = cos a cos
a sin
p(x, A)=srna The function co)
does
only
not belong to
Consequently
m(A)=
If
for ImA>0 can
A) =O(x,X)
by a constant factor, as
differ from
sin a — i cos a +
cos a sin a
0, then for A <0 the denominator of this function cannot vanish Therefore m this case ctg a
A
—Imm(X) =
> 0'
(ctga<0),
X<0
0,
and since a negative spectrum is absent the inversion formulas hav€ the form (1.1)
(sin a cos
F(X)
f (x)
= --
(sill
F
-—
—I, '
cos a
—
COS
cos
a'. sin
dx,
\//X ,—
Stfl yAXJ
.
cos2a-1—Xsin2a I
If ctga h> 0, then for x = — h2 the denominator of the function m(A) vsinishes. Consequently in this case there is one negative eigen240
_________________
________
______
§2. FOURIER-BESSEL SERIES
value A =
241
and the corresponding normalized eigenfunction is to the right side of the expansion (1.1) one has
— h2,
Therefore
to add the term 1(t)
and the expansion assumes the form
f(t) (1.2) 1
. F(X) (Sm
,—
.
605 \IXX — A
COS
•
Sill \ AX) cos2a + A Sjfl22
The expansions (1.1) and (1.2) can, of course, be justified under the usual assumptions for Fourier integrals by a direct consideration of the integrand. We note that formula (1.2) can also be obtained from formula (1.12), Chapter 2, for q(x) 0. In fact, in this case jz(A)
=sina,
z'(A) =
and consequently —
/
p (X)
A
*2. Expansion in a FoUrier-Bessel series
1. We consider the Bessel equation (2.1)
where A, x andp are real numbers. By means of the substitution y
equation (2.1) reduces to the form /
p-4
(2.2) dx2
a:2
J
This equation has the following linearly independent solutions
= s):
y2(x,A)
y1(x,A)
where J7,(t) is the Bessel function of the first kind and -
Y it'
.1 •P
(t) cos
— .f.. P (t)
is the Bessel function of the second kind. From well-known powerseries expansions of Bessel functions it follows that for x —i 0
V.EXAMPLES
242
y2=O(x
It follows from these estimates that for p
).
1 only Yi is square integrable
in a neighborhood ofzero. On the other hand, if 0 p < 1, then both solutions aze square integrable. arbitrary real numbers. We adjoin to (2.1) Let 0
(2.4) Since
the equation (2.2) does not have singularities in the interval
with the boundary conditions (2 3) and (2 4) it determmes a regular Sturm-Liouville problem, to which the classical theory disin Chapter 1 is applicable. . Let us denote by X1, X2,... the eigenvalues, and by q,i(x), corresponding eigenfunctions of this problem. In order to obtain a we will find the cendental equation whose roots are the numbers solution of equation (2.2) which satisfies the boundary condition It is easily seen that this solution is the function [a, b],
together
y(x,X)
=Cy1(x,X) +y2(x,X).
where
c—— Therefore
y2(b,
A) A)
the eigenvalues X,, are
the
roots of the equation
=0,
y(a,X)cosa
which can be written in the form (26) defined by (2.5). Since y2(x,X)—* as x—'O, and —, 0, for every fixed A the number of zeros of the function y(x, A) in interval [a, bJ remains bounded as a 0 It follows (cf Chapter 4) tha C
is
equation (2 2) and the boundary condition (2 4) determine in (0, b] a discrete spectrum of eigenvalues with a unique hmi
,r
The correspondmg eigenfunctions must have mtegrahj p 1 this condition umquely replaces the
emce in this case only yi(x, A) determine
is squar
the eigenvalues in the interval
243
§2. FOURIER-BESSEL SERIES
we have to insert for y(x, X), in the boundary condition (2.4), the funcand we obtain the following transcendental equation: tion (2.7)
+
1,, (Sb)
sin
+
(zb) sin
= 0.
0, then we obtain the simpler transcenciental equation
If sin 13
(2.8)
=0.
Let us denote the eigenvalues in the latter case (i.e. the roots of equation (2.8)) by
By a well-known
(sx) dx Putting
here s =
(sb)]2 +
(t — P2)11
we obtain
(sax) dx = 4j- [1,
Therefore the normalized eigenfunctions have the form (sub),
and a Fourier-Bessel expansion can be written in the form (2 9)
/(x)
(set)
1(t)
dt
2 For p < 1 all the solutions of equation (2 2) are square mtegrable Therefore m this case the requirement of square integrabthty does not replace the boundary condition (2 3) We will show that, suitably changmg a m (2 3) together with a (which we will require to tend to zero), one can obtam the missmg boundary condition Smce the equation (2 2) does not contam a first derivative, the Wronskian determmant of y1(x, X) and y2(x, A) is constant, i e
W{y1, y2}=y1(x,
=
(sx)
= sx
X)—g2(x, (sx))'
—
A)
(sx)
(sx) Y, (sx) — V7,
(sx) 1, (sx)) = const
To complete the calculation of W{ Y1,Y2 } we will use the well known See for example Watson El]
§5 1
244
V. EXAMPLES
asymptotic formulas for Bessel functions of a large argument (cf., for
example, Watson [1], §7.21):
(t —-ç-----j-) (210)
(t) =
[sin (t
(i)],
[sin (t (210') Insertmg these expressions in the formula for W{ Yi, Y2 }, and then letting x —*
we obtain W{ Yi, Y2 =
for definiteness, that fi it!2, i e y' (b) =0 We can also assume without loss of generality that y(b) = 1. It is easily shown thati Suppose,
y(x, X)=
W{y1, Y2}
(y1(x,
X)y2(x, X))
(2.11)
=4.8
Y; (sb) —
VP
(sx) 1, (sb)).
The eigenvalues X. are now determined from the equation y(a,X)ctga +y'(a,X)
0.
Suppose first that 0
11')
y(x,
(sx)
=—
(sx) J;(os)}
(bs) —
For fixed s and /,, (sx)
=
+0
2Pr(t±
Therefore
+0
.7; (sx) — -
X)
+ {
sPJ'
(bs)
2-?r(t—p)
+
F p——p
I
\
+P
§2. FOURIER-BESSEL SERIES
245
Let us denote by c an arbitrary constant, and let a satisfy the equation 1
a22'ctg a
p)a 2—Pr(i_p)
(2.13)
—
a4+P ctg a + (1 + p1a
2Pr(1+p)
Solving for ctga, we obtain ctga —
2'r
p)a
2
(1
o(!).
c2Pr(f
p)
—
—p c2—Pr(f_p)a2
—a
1
—2Pr(1+p)a
and consequently we can neglect the 0-terms in (2.12) if 3/2 —p > p — i.e. p <1. From (2.12) and (2.13) there follows a transcendental equation
which replaces the boundary condition (2.3): =0, or (2.14)
=0.
The numbers We denote the roots of the equation (2.14) by equation (2.2) with the boundary conditions are the eigenvalues of the (2.4) (with f3 =ir/2) and (2.14). We obtain the corresponding eigenby replacing s in (2.11') by functions 2
2
sinp-It
{J p(sax) 1
i,
(bs,,) —
(2.15) —
T
—
If c = a, then we obtain the usual Fourier-Bessel series expansion of order p, and if c =0, of order —p.
Now let p =0. In this case
2J
1'0 (sx) = —' 2
where -y
is the Euler constant. Therefore
248
v(a,
A)ctga±y'(a, A)
>(
(a +O(a2
lnaj).
Putting + In 4-) (a4
ctg CL
+4- a
4)
+
ctg
CL
+4- a
carrying out computations similar to the preceding, we obtain the following equation for the determination of the eigenvalues: and
(bs) —
For c =
we
(bs) in s} = 0.
(bs) —
obtain the usual Fourier-Bessel series expansion.
One can similarly study the expansion for $ =0. We obtain the following equation for the determination of the eigenvalues: 0), (bs) — (bs) = 0 cj0 (bs) —
(bs) —
L
(bs) In s} = 0.
3. Using ordinary methods, one can show that eigenfunctions corre-
spending to distmct eigenvalues are orthogonal Further, for every function 1(x) which is square mtegrable over the mterval [0, b] the Parseval equality holds (cf Theorem 5 1, Chapter 1) From the Parseval equality it easily follows (cf Theorem 5 2, Chapter 1) that if the Fourier Bessel sense of a continuous function f(x) converges uniformly, then
x>Olta sum of the
(2.9)
the senes (29) equal zero Iff(O) zero at x = 0 it cannot equal f(0)
then since the
§2 FOURIER BESSEL SERIES
247
We will give a simple criterion for the uniform (and absolute) con vergence of a Fourier Bessel series For definiteness we will consider the ordmary expansion m a Fourier Bessel series If the function f(x) satisfies the following conditions 1. f(x) has a continuous second derivative in the interval [0, b], THEOREM 2 1
2 f(b)=0, 3 f(0) =f'(O) =0, then the Fourier-Bessel series of f(x) converges absolutely and uniformly. PROOF Let (2 16)
/ (x) =
(s,,x) -4
ViI,, (set) 1(t) dt,
where
By the asymptotic formula (2 10)
Further, mtegrating by parts twice we obtain
(2 17)
= — 4—
—
p2; 1(t) dt
where the mtegrated terms vanish by conditions 2 and 3 It is also easy to see that condition 3 implies the boundedness of the function f(t) It2 It therefore follows from (2 17) that (218) From the asymptotic formula (2 10) it easily follows that n—*
= 0(n2)
248
V. EXAMPLES
Therefore it follows from (2.18) that the Fourier-Bessel series (2.16) converges absolutely and uniformly, and the theorem is proved. Using the asymptotic formulas (2.10), one can show, just as fOr the classical Sturm-Liouville problem (cf. Chapter 1, §9) that the FourierBessel series converges under the same conditions as does the ordinary Fourier cosine series, for a Fourier-Hankel integral
1. Let us now consider an expansion with respect to the eigenfunctions As was shown in §1, of equation (2.2) in the infinite interval (0, Chapter 2, an eigenfunction expansion in the interval (0, cn) can be obtamed from an expansion m a finite mterval [0, b] as b cu We will first consider the case p> 1, and suppose for definiteness that $ = ir/2 in the boundary condition (2.4).
Using the asymptotic formulas for the Bessel functions
and
(x),
one can show that It
— S,, = + 0 s,, is the nth root of the equation =0. Arguing further as in §1, Chapter 2 (see also subsection 2 of the
where
present section), one can obtain the formulas
/ (x) = sF(s) (3.1)
(sx) ds,
0
and the Parseval formula (3.2)
(x)
dx = sF2 (s) ds.
Formulas (3.1) are called the Fourier-Hankel inversion formulas. 2. Suppose now that 0
249
§3 FOURIER HANKEL INTEGRAL
from the equation
=0,
(33)
where c is an arbitrary real number. Let us denote by A1 and A2 successive positive roots of the equation (3.3). From the asymptotic formulas (2.10) and (2.10') it easily follows that for b—p \ "2
(34)
V
Ai
°
—T
/1
By (2 15) and the fact that the boundary value problem is homo
geneous, the functions
=
—
also are eigenfunctions To determme the spectrum of the boundary value problem under study for the interval [0, cc), we need to calculate the limit (x) dx
hin
It follows from the asymptotic formulas (2 10) that for fixed A > 0 P*and large x
(x) 2
cos (sax
—
—
=
s;*
cos
(sax —
—
+ pit) +0
—
— / cos pit) cos
erefore
lim
(x)
{(c —
cos
pit)2 +
sin2 pit)
b-÷cx
= tting, as in §1, Chapter 2,
(c2
—
COS pit +
V. EXAMPLES.
250
(x) dx
we obtain, by virtue of (3.4), A
V
(
8n+1 — b
{* + —
(x) dx
o
—s- + o(1) }
I
sds
-p.'
ç
Consequently,
—*
—
2c1-,, cos
'1
s4P
for X> 0 the spectrum is continuous. If there is no
negative spectrum, then the expansion formula has the form
/ (x)= S
s
ds.\ \/t (ts)} /
—
(ts) cit.
Let us now investigate, for which c there will be a negative spectrum. = icr (cr> 0). From equation (3.3) it follows that Let A <0, (3.6)
As is known, —
(tz) = e2
pi:i
I,, (z),
(3.7)
where
I,, (z)=
P+2k
Z
I I'(p ±k +
t)&)
Differentiating (3.7), we obtain
J;(tz)=e2 from equation (3.6) follows (ba) — ce2
-
or
= 0, i.e.
i (ba) = 0
_____
§3 FOURIER EANKEL INTEGRAL
251
=
(38) From
the asymptotic formula for a Bessel function with a pure
imaginary argument it follows (cf, for example, Watson [1], §7 21) that the left side of (3 8) tends to unity Therefore if c <0, then ss b for sufficiently large b equation (3.8) has no solutions, and consequently
in this case a negative spectrum is absent, and the expansion formula can be written in the form (3 5) For c > 0 we obtain one negative eigenvalue u = Let us determine the corresponding eigenfunction Since = it follows from (2 11) that 6(x,
__! sin pr
Vx
factor — (2c/ir)smpir can be discarded, and we obtain the eigen
To normalize this function we use a well ifimction known formula from the theory of Bessel functions (t) dt =
dx 5
5 Thus, for tcie case c> 0 we have to add to the right side of (3 5)
iie term 2cP sin
(t) dt.
Similar calculations can be carried out for the case p =0. We give mly the final rv;{CJO (xs) —
.f(x)=j
Jo (xs) ins}
Y0 (xs)
1
2
12
sds
+1
0
Y0
x5
+2 Titchmarsh [31,
§8.19.
jo (ts)Ins}/ (t) dt dt.
V. EXAMPLES
252
Thus, in this caae the spectrum contains one negative value for any + For c = + the additional term to the integral drops out, and we obtain the ordinary expansion in a Fourier-Hankel integral with respect to Bessel functions of zero order. 3. From (3.2) there easily follows the generalized Parseval equality c
<
1(x) g (x)dx= F (s) G (s) s ds,
(3.9)
where G(s) is defined by g(x) in the same way that F(s) was defined by f(x). Just as in §1, Chapter 2 (cf. Theorem 1.2) the following theorem can be obtained from the generalized Parseval equality (3.9). THEOREM 3.1. If
f(x) is continuous for all x> 0
(3.10)
sF(s)
I,
and
the
(sx) ds
converges uniformly in every finite interval, then
1(x) = sF(s)
(3.11)
(sx) ds.
Similarly to the way in which Theorem 2.1 was proven, one can prove the following theorem. THEOREM 3.2. If
1. f(x)
has
for all x
a continuous
0
second derivative,
2. f(0) =f'(O) =0, 3. f(+ cx,) =0, and (3.10) converges expansion
1(x),
ce), then
f'(x), f"(x) E
absolutely and uniformly
for all x
0,
and
the
integral
therefore the
(3.11) is valid.
4. As is known, many differential equations can be reduced to the Bessel equation by an appropriate change of variables Let us consider, in particular, the equation d2y
(3.12)
—o
Tlua equation was already considered by Euler If we make the sub 2y, we obtain the equation stitution 2
253
§3. FOURIER-HANKEL INTEGRAL
d2z
(
jz=O.
As we have seen, the solutions of this equation are the functions and Therefore equation (3.12) has the solutions ",
/
/
".
Let us put (
r(,)+1)
1k )
k—O
satisfies the initial condi-
It is easily seen that the function
j(0) =0.
tions
Let us write down the formula for expansion in a Fourier-Hankel To this end we will first write down the ordinary Fourier-Hankel inversion formulas, i.e. the integral with fespect to the functions formulas (3.1),
Let f(x) = (s) =
Then / (x)
(xs) dx
(s)
+
(sx) dx
p4
= 22Pr2 (p ± I) S
(s)
(sx)
______
EXAMPLES
254
Thus the expansion in terms of the functions t' (s) =
(sx)
has the form
g (x) /,, (sx)
(sx)
(s)
g (x)=22pr2 (p + 1)
14. Expansion in terms of the orthogonal
functions
1. To begin with, we construct the so-called polynomials. To this end we consider the function
=
(x, t) =
(x, t) in powers of t, we obtain
Expanding
=
(x, t) = from which it follows at once that H (x) = ô"4' (x, t)
(4.1)
—
i
dx"
are called the polynomials. We will derive a differential equation for these polynomials. From the t)/ôx = t) follows relation The functions
(n
(4.2)
and from the relation
1),
(x, t) / ôt + 2 (t — x) (x,
t) =0 follows
=0 (n Combining the last two formulas, we obtain the
1).
following linear
homogeneous second-order differential equation with respect to
H(x)
=0
The first few
(n
0).
Hermite polynomials have the form H0(i) =1,
113(x)
H1(x) =2x,
8x3 — 12,
H2(x) =4x2—2,
H4(x) = 16x4
A general expression for the (2xy'2-J—
—
48x2
+ 12.
Hermite polynomial n (n— 1) (n.—2) (n
(2x)"4 +
is
§4. ORTHOGONAL CEBYSEV-HERMITE FUNCTIONS
255
The last term is equal to (—1)
(n/2)I
forevenn,
2x foroddn. We will show that the over the entire line (—
cz)
polynomials are orthogonal with respect to the weight p (x) = exp( — x2).
In fact, by formula (4.1) we have 5
=
(x)
(x)
(x)
5
d d
dx.
Integrating the last integral by parts for n> m and taking account of (4.2) and the find that
that all -derivatives of exp ( — x2) vanish as x
(x) H1, (x) 5
= (— tr' 2m 5
dx =
(x)
_x2
= (—ly'-tm 2mmt
(x) 5
dx:-rn
=0
For n = m one can compute m the same way that
r
=
5
Therefore the functions
JJ(x)e
—
(n=O,
1, 2,
...)
ccnstitute an orthonormal system on the real line (— called the orthonormal functions.
2. Let q(x) = x2, —
<x <
), and are
In this case we are led to solve the
óquation
y" +(x —x2)y=O. Since q (x) = x2 + as x —* ± we know from the results of bhapter 4 that the spectrum of the operator 1(y) — y" ± x2y on the (4.5)
256
V. EXAMPLES
entire line is discrete. Let us denote by Xi, X2,... the spectrum of the equation (4.5) on the real line. We will show that the orthonormalized eigenfunctions, i.e. the solutions (i.e. the square integrable solutions of (4.5)) of (4.5) in coincide with the functions defined by (4.4). In fact, let y = exp ( — x2/2) z. Then equation (4.5) assumes the form
z" —2xz' +(X —1)z =0.
(4.6)
If (47) is a
solution of equation (4.6), we obtain from (4.6) the following recursion
relation for the nonvanishing coefficients: 2k+J—X ak÷2 (4.8)
The
(k4-1)(k-j-2)
relation (4.8) shows that equation (4.6) has two solutions; an
even one zj(x) and an odd one z2(x). So in the power series (4.7) repre-
senting these solutions there appear either solely even or solely odd powers of x. From (4.8) it follows, first of all, that the series (4.7) either
terminates (namely, if A =2k +1 is an odd nonnegative integer), and in this case equation (4.6) coincides with equation (4.3), with the solution thus being a polynomial, or has an infinite number of
nonvanishing coefficients and converges for all values of x. In the second case all ak from some k on have the same sign. Suppose they are positive; then they satisfy the inequality ak+2 > (k + 2) and, as is easily seen, for sufficiently large x C=/=O,
and consequently y =zexp(— x2/2) is not square mtegrable over the real line This establishes that the orthogonal Hermite functions 2( — x) of our problem are the umque solutions in The completeness of the orthogonal functions in 2( —
follows
from the general theory discussed in Chapter 2
Expansion in terms of the Legendre polynomials and the associated Legendre functions
1 The Legendre polynomials are defined by the following formula P0 (z)
1,
=
I
d'(x2—I)" dx'
(n
=
1,
2,
.
.
.).
257
§5. LEGENDRE POLYNOMIALS
Obviously the function
is a polynomial of degree n. We will prove that the polynomials form an orthogonal system = (x2 — on the interval [— 1,11. In fact, putting we have
for any nonnegative integer m
=
ut,") (x) xmdx =0,
which is easily verified by repeated integration by parts, bearing in mind that all the derivatives of of order up to n — 1 vanish at the endpoints of the interval of integration. It follows that
i.e. two distinct polynomials are indeed orthogonal. To compute the normalization constants we obtain by repeated integration by parts (x)
=
dx
(x)
(x) dx
(x)
dx
— (2n)!(2n
it follows that
dx=
22n (n
1'
1)2
Therefore the normalized polynomials have the form (5.1)
=v
i/2n+1 2
v
I dxl*
(n=O, 1, 2, . the property that for any a .
The Legendre polynomials
have
258
V. EXAMPLEs
(5.2)
=1,
which is easily seen by forming the nth derivative of the expression (x — +1)" by Leibnitz' rule and then putting x =1. 2. The Legendre polynomials play an important role in potential theory, where they appear as the coefficients in an expansion of the socalled generating function. In fact, the generating function in potential theory is defined as the reciprocal of the distance between two points, one of which is at unit distance from the origin, the other at a distance ii <1 from the origin, and whose radius vectors form an angle arc cos x, i.e.
Now, inserting in the expansion
+
+
(2n—1)U
t = — 2xu + u2 and grouping like powers of u, we obtain an expansion of the generating function in the form
u)= Vi_2xu+u2
(5.3)
Let us show that the functions are identical with the Legendre polynomials defined earlier. To do this we will first prove that the form an orthogonal system over the interval [— 1,11. It follows from (5.3) that I Vt — 2xu
I
+
u2
Vt — 2xv
±
v2
=
nfl, Q (x)Q (x)uu
Integrating the left side of (5.4) with respect to x from obtain dx
J whence,
by virtue of the expansion In
—1
to 1, we
259
§5. LEGENDRE POLYNOMIALS 1
'v"
dx
(55)
J v'(l
— 2xu + U2) (1 — 2xv
2
2n
+
+I
UV
Now, integrating the right side of (5.4) with respect to x from — to 1 and comparing the result with the expansion (5.5), we obtain 1 0, for mq=n, Q,, (x) Q,, (x)
dx
2
2n-+1'
—1
for
1
— rn—n.
i.e. the functions form an orthogonal system on the interval [— 1,11. It follows from the construction of the that they are polynomials. the On the other hand, just like the Legendre polynomials polynomials can be regarded as a system obtained from the system 1, x, x2,... by orthogonalizing the latter. Since the process of orthogonali
zation shows that, aside from constant factors, there exists only one system of orthogonal polynomials which contains polynomials of all to show that for every n it suffices to note that, like the poiynomials the polynomials Q,, (x) satisfy the condition (5.2), i.e. = 1. In fact, putting x = 1 in (5.3), we obtain U2 +.
.. +
+ ... =
which proves our assertion.
3. We will obtain a recursive formula and a differential equation for the Legendre polynomials. Differentiating (5.3) with respect to u and then replacing by we obtain x—u
u) ÔU
Vt —2xu
(1
i.e.
—2xu + from which, equating the coefficients of we obtain the following formula for three successive Legendre polynomials:
=0. See, for example, Courant and Hilbert [1], Chapter II, §1.2.
260
V. EXAMPLES
The first few Legendre polynomials have the form
P0(x)=1,
P1(x)=x, —-ix,
The Legendre polynomials I
dx"
satisfy a homogeneous second-order linear differential equation: (5.6)
(x2—l)y"+2xy' —n(n-j-1)y=rO
or
(5.6')
[(x2—1)y'}' —n(n+1)y=0.
This can be proved by differentiating the obvious equality (x2 — 1) u' = 2nxu, where u = (x2 — 1)", n + 1 times, and then replacing by 2"n!y. It follows from equation (5.6) that all roots of the polynomial is a solution of a second-order equation. It are distinct, since on the basis further follows from the definition of the polynomial of Rolle's theorem, that all of its roots are real and lie in the interval [—1,11.
4. Let us consider a Sturm-Liouville problem in standard form:
o. — c° as Since here q(t) = — ± wI2, we have the + singular case. Let us find the eigenvalues and eigenfunctions of the problem (5.7). By means of the change of variables
(5.8)
x=sint,
the problem (5.7) transforms into the equation (5.9)
(1 —x2)y" —2xy' +Xy =0,
the interval [— ir/2, ir/2] transforms into the interval — 1, 1], and the boundary conditions become the condition that y(x) remain finite at both singular points of equation (5.9), i.e. at the points ± 1. We already know that the numbers A = n(n + 1) are eigenvalues, and the eigenfunctions, of this problem. Legendre {
261
§5. LEGENDRE POLYNOMIALS
We wifi show that the Legendre polynomials are the only solutions of this eigenvalue problem which are bounded on the interval [— 1,1]. Note that if u = f(x) satisfies the differential equation (5.9), so does the function f( — x), and consequently so do the functions f(x) +f( — x) and f(x) — — x), the first of which is even, and the second, odd, and at least one of these does not vanish identically, since by hypothesis the function u = f(x) is not identically zero. Thus it is sufficient to show that every even and every odd solution u of the equation (5.9) which is continuous for —1 <x < 1 is a Legendre polynomial and that A
must be a number of the form n(n + 1). Representing u as a power series u we obtain from (5.9) the following recursion formula for the coefficients:
(n=O, 1, 2, . (n±1)(n±2) If u (x) is an even function, then all the with odd n equal zero; if u (x) is an odd function, then the coefficients with even n are zero. For the case n — 2h > Owe have from (5.10) (5.10)
.
ir (5.11)
A
A
(n—3)(n--4)
(n—1)(n—.2)JL1
x
1
...[t_(fl2h±l)(fl2h)Jkak,
where k = n — 2h. The series for u (x) terminates if and only if A is a number of the form n (n + 1). Obviously in this case u (x) is the nth Legendre polynomial. For all other values of A we obtain an infinite series, which from elementary tests is seen to converge for I x <1. Let us give k a fixed value so large that all the factors in the product (5.11) are positive (ak can be assumed positive). Since as n increases the product of the square brackets in (5.11), according to well-known theorems, tends c/n, where c is to a positive limit, it follows that for n > k surely cannot be bounded in the a positive constant. Therefore neighborhood of the point x =1. It follows that a number A n (n +1) cannot be an eigenvalue.
From (5.1) and (5.8) we conclude that the eigenvalues of the problem (5.7) are the numbers = n (n + 1), and the normalized eigenfunctions are (5.12)
(t) =
(sin t)
j/
2n
±1
•
2( The completeness in — 1, 1) of the system (5.1) follows from the Weierstrass approximation theorem.
262
Y.FXAMPI,8
5. From the differential equation (5.9) for the Legendre polynomials it is easy to derive other orthogonal systems Differentiating (5 9)
with respect to x, we obtain a differential equation for the function u', only for A = n (n +1) does there and just as before, it turns out exist a solution which is regular at both endpoints of [—1,11, namely Ph(x). The equation thus obtained for (x) is no longer self adjoint, but. it can be transformed to a selfadjoint one by introducing the func-
as the unknown. Then the new equation
tion assumes the form
{(1—x2)z')'—1
Its eigenvalues are the numbers X,n(n = + 1) (n = 1,2,...), and its = (x). The functions are called the associated Legendre = functions of order one. The functions will in this connection be called the Legendre functions of zero order. The Legendre functions satisfy the orthogonality condition eigenfunctions,
for
In the same way, differentiating equation (5.9) h times, we obtain for the functions
_x2)h.
P,Zh(x)
the differential equation
(5.12')
((1 —x2)
—
+Az
0
having the eigenvalues = n(n +1) (n = h, h + 1, . . •) and the corresponding eigenfunctions Pfl,h(x), which are also pairwise orthogonal over the interval [— 1, 1] and are called the associated Legendre functions
of order h. Their normalization can be carried out by means of the following easily proved equality:
,(x)dx= That ill the eigenvalues and eigenfunctions of the differential equation obtained in this way can be proved in the same way (5.12') he,e as for the functions of zero order.
263
CEBYSEV-LAGUERRE POLYNOMIALS
§6. Expansion in ternis of
polynomials
polynomial of degree n is defined by
1. The
It is easily seen that (x) is in fact a polynomial of degree n. Indeed, by Leibnitz' formula we have d'1
(n) n(n—1) ... (+) n (ii —
.1)
l)xk (n — k +
t)
[n(n— 1) ..n —k + j)2fl_2 —
+
—
+ (—1)nn!).
polynomials have the form
The first few
(x) = —x3 + 9x2 — 18x + 6, 2'4 (x) = — 16x3 + 72x2 — 96x +
24.
In order to obtain recursion formulas and a differential equation for (x), we consider the function the functions t)
We will prove that for any n = In fact, by the definition of the function S4 (x) we have (_1)k n=O k=O
(l)kxk
(f)kxk
(6.1) k=O
k!
n=lc tic) \ /
=
k! k=O
xt e
(x, t),
264
V. EXAMPLES
which proves our assertion by virtue of the uniqueness of the expansion of
t) in powers of t. From the obvious relation (1
there follows, by virtue of (6.1), the identity (1
Equating coefficients of like powers of t, we arrive at the recursive formula —
= 0 (n 1).
(2n + 1 — x)
This formula, and the formula (n
1) ,
=
which follows in similar fashion from the relation (1 —t)
=
(n 1) and to
leads to the formula
the homogeneous second-order linear differential equation
xy"+(l—x)y'+ny=O
(6.2)
for the The
polynomials.
polynomials form an orthogonal system on
the halfline (0, co) with weight e_x, i.e.
(mKn). This follows from the equality (x) dx =
xk
dx = —k
dx
dx
k
The nonnabzmg coefficients can be computed in the following way
= (n!)2.
§6. CEBYSEV-LAGUERRE POLYNOMIALS
265
Therefore the functions x e
1, 2,
...)
form an orthonormal system on the halfline (0, cx'), and are called the
orthonormal
functions.
2. Let us consider the equation (6.3)
d'+{X
0
(0
By means of the substitutions x = t2/4, y(x) = u (t) t
< 1"2exp (t2/8),
equa-
tion (6.3) reduces to the form (6.4)
xy"+(l_x)y'+Ay=O
Since, in equation (6.3), q(t) = t2/16 — 1/4t2 — 1/2 tends inonotonically the spectrum of (6.3) is discrete. On the other hand, to + cx' as t —* equation (6.4) coincides with equation (6.2) • for which the eigenvalues polynomials. are X. = n and the eigenfunctions are the That the equation (6.4) does not have any eigenfunctions in addition to those just indicated can be proved in the same way as for the Hermite = n, and polynomials. Consequently the eigenvalues of (6.3) are the eigenfunctions are
; (t)
Vie T2 (4W).
3. As for the Legendre functions in §5, the process of differentiation and multiplication by a suitable factor leads here to the higher-order Laguerre functions (generalized Laguerre functions), which satisfy similar differential equations. First of all, differentiating equation shows that the functions (6.4) m
satisfy the differential equation (6.5)
xu" + (m + 1 — x) u' + (x
— m)
u = 0,
which can be written in the following selfadjoint form: (xm+le_xuF)/
—m)u
=0.
The corresponding orthogonal functions = xm/2e satisfy the following equation of Sturm-Liouville type:
266
V. EXAMPLES
/1—rn
x
rn2\
2
In Order to show that these equations do not have any other eigenvalues and eigenfunctions, we msert the power series u = E a x in equation (65) and obtam, by means of recursion relations, the following expression for the coefficients: a0 (m—X)...(rn—A+v—1) (rn—j—I) ...
W note that for an arbitrary given value of A the coefficients a, have fixed sign, starting with some value of v, and the series converges for all values of x; consequently it does actually represent a solution of For A = n a positive integer (n > m) (6.5) which is regular for 0 x < the series terminates and is therefore a polynomial. For every other value of A it is easy to obtain a bound of the form
>c/v!V, where c is a suitable constant, and r is a positive integer. But from this our solution tends ii follows that as x infinity not slower than We have thus proved that it cannot be an eigenfunction of the I
problem being considered. §7. The "hydrogen atom"
1. We will consider the equation (7.1)
(O
where 1 is a positive integer or zero. In quantum mechanics the study of the energy levels of the hydrogen This substitution R = y/r reduces equation atom leads to this (7.1) to the form (7.2)
Just as m the case of Bessel's equation (cf §2), one can show that finite mterval [0, b] the spectrum is discrete consider the infinite interval (0, co). We will show that for .E the spectrum remauis discrete Let E <0 Obviously there a number F—rE such that
a
Bloluncev El] §50
267
§7. "HYDROGEN ATOM"
outside the interval (0, TE). Therefore every solution of (7.2) has at most one zero in the interval (TE, co). The number of zeros in the interval (0, rE) is finite. From the finiteness, for E <0, of the number of zeros of any solution follows the discreteness of the spectrum. To find the negative eigenvalues we will seek a solution of (7.2) in the form y(r) E
tr
'
dr
dr2
r2
J
We now represent the solution v(r) of (7.2') in the form of a power series (7.3)
and, inserting it in equation (7.2'), we obtain for the coefficients the recursion relation —
(7.4)
a
'v—O — ' I
It is not difficult to see that the series (7.3) will converge for all values of r, but that for r—*
the solution of (7.1), R(r)
(7.5)
tends to infinity. In fact, putting in the form
= 1/X, s = 21 + 1, we rewrite (7.4)
( s
+ 1)
Hence it is evident that a,+i/a,—*2X/(v +1) as v—* can choose ii — v' such that
Further, we
s+1 2
f
<1. Starting with this value of ii, the coefficients a, grow faster than the coefficients of a series defined by the recursion where
0,
relation b
—
A(t+E)
b
The series with these coefficients is v1 (r) =
268
v. EXAMPLES
Therefore v(r) grows faster than v1 (r), and consequently the function (7.5) tends to infinity as r —+ Thus, by virtue of (7.5) the solution of equation (7.1) can be square integrable over the haifline (0, co) only if the series (7.3) terminates. Then vfr) will be a polynomial. Obviously the series (7.3) can terminate at aome point only for particular values of the parameter A (cf. (7.4)). Suppoee that a given coefficient is not zero. In order that the next 1/ (n + 1 + 1). It is coefficient be zero, it is necessary that A clear that when this is the case not only but all subsequent co-
efficients will equal zero, since all of them are proportional to Putting k = n + 1 +1 and taking into account that E = — A2, we conclude that the negative eigenvalues are the numbers Ek = — 1/k2 (k 1,2,...). Let us consider the form of the eigenfunctions corresponding to the = 1/k, we obtain from (7.4) eigenvalues Ek = — 1/k2. Since Ak the following relation for the coefficients a,,: 2
k.—(v+l+f)
Computing one coefficient from the other (ao is arbitrary) and inserting
their values in (7.3), we obtain k—i—f u(.r)_—a0r
2r
'
2!(21+2)(21+3)
/
(k—i.—1)(k—i—-2),..3.2.1 f2r'\fl
X (k—I)! (21+2)(21H-3)...
Putting E = 2r/k, by virtue of (7.5) and the preceding result we obtain
the following definitive expression for the function R(r): RkJ (r) where
C,,,
is some constant, and
(i), is
function of order 21 +1 (cf. §6.3). 3. We now consider positive values of E; let
a generalized Laguerre s.
As is known,7' for a solution of (7.2) which is bounded at zero one baa the following asymptotic formula for s co: (7.6)
whee
.
fr(i+i +7)f a'=argr(l+l+i/s). Fok til.
p. 155.
269
§7. "HYDROGEN ATOM"
It easily follows from (7.6) that as b
the distance between us denote two of them by successive zeros of the function y(b, s) —let s1 and s2 —tends to zero. We remark that, as is not hard to show, y(b, s) does not have multiple roots. Using these facts, one can obtain a more precise estimate for s2 —
(s2— si)b+_L
(7.7)
cx
+ o(1),
(7.8)
Inserting (7.8) in (7.7), we obtain (s2 —
b
+
) In
b = (s2 —
b+
In b + o
=
The final estimate for s2 — s1 is (7.9)
We also need the following estimate for b
cos(2sr +lIn2sr+ 2a)dr=O(1).
(7.10)
Let a> 0 be arbitrary. We have cos (2sr+!ln
=
2sr + 2a)dr cos
2sr.
cos
(7ln 2sr
— sin 2sr. sin (-_ lu 2sr + 2a) dr. Both integrals on the right side can be estimated in the same way. Let us consider, for example, the first integral. Integrating by parts twice, we obtain cos 2sr.
cos
/2 ln 2sr +
dr
cos (4
ln2sr +a)}dr=O(1).
270
From the estimates (7 9) and (7 10) follows
dr
S y2 (r,
(7.11)
x
+o(1)J
6
(1
+ a]dr
I
Using
the well-known properties of the r-fnnction =2r(z), r(z)r(1 —z) =lr/sin7rz,
we obtain
V(1+
.
.
i(!)r(i ii
(i
Therefore we finally obtain from (7.11) I iXp
(s) —
=J
a)
i.e. the spectrum is continuous. §8. A Dirac system
We will consider the
(
q
+ (x)
(x)
(y1
)
(Ocx< cc).
(x))
(z)
271
§8. A DIRAC SYSTEM
Let p (x) = a, q(x) =0, r(x) = — a, where a is an arbitrary positive number. Then equation (8.1) assumes the form yç+(a+A)y2=O,
(8.2)
Let us denote by
(:)
= the solution of the system (8.2) satisfying the initial conditions
=cosa, where a is an arbitrary real number. It follows from the system (8.2) = —sina,
(8.3)
that the functions 4,1(x, A) and 4,2(x, A) satisfy the same second-order equation, namely y" =(a2—A2)y.
(8.4)
Further, from the system (8.2) and the conditions (8.3) follows
=(a—A)sina.
=—(a+X)cosa,
(8.5)
Now, solving equation (8.4) for the initial conditions (8.3) and (8.5), we obtain the following explicit expressions for (x, A) and (x, A): p1(x,
X—a
(8.6)
A)==cosa- cos
cp2(x,
2
2
Now let O(x,X) =
f01(x,
A)
\02(X, A)
be the solution of the system (8.2) satisfying the initial conditions 01(O,A) =COsa,
02(0,X) =sina.
Proceeding as above, we obtain the following explicit expressions for 01(x,X) and 02(x,X): fj,
sing • sin
y,
(8.7) 02(x,
sina
.
(SOS
.
Sill
2
272
V. EXAMPLES
All the square roots in (8.6) and (8.7) real and greater than a.
are
real and positive if A is
From the general theory (cf. Chapter 3, §2) it follows that the function for ImA >0. Consebelongs to =O(x,X) 2(0 quently, the function O1(x,A) +m(X)cci(x,A) also belongs to for ImA > 0, and so it is a multiple of —a2 }, i.e. we have
—
{sin
+m
cos
= C {cos (x
2
sin
— a2}
+ i sin {x
a2)
— a2)].
Hence we obtain
or /
sin a±
rn(A)==—
ens a
If A > a, it follows from (8.8) that %/X2_a2
If — a a), then +a= — —a2
If cosa and sina have the same sign, then m(A) has no poles on the real axis between — a and a, and therefore in this case the expansion -
formula has the form
/ (x) = $
(x, -—p.) F (—p.) dp. ç
s/A2_a2
X±acos2a
p(x, A)F(A)dX,
the Fourier transform of 1(x) with respect to the func-
If pomt
ai$ ana have opposite signs, then m (A) has a pole at the with residue — 2a sm 2a Since
REFERENCES
—acos2a) = —sina .
273
—acos2a) =cosa
one has to add to the right of the expansion (8.9) the term —2a sin
(x, — a
(t)
sin
+ 12(t) cos
Bibliographical references
§1. The results of this section are classical.
§2. Cf. Watson [1], Chapter 18, Titchmarsh [1], Chapter 4. The method discussed here is due to Levitan. §3. Cf. Weyl [3], Watson [1], Chapter 14, Titchmarsh [1], Chapter 4, Levitan [5]
§4. Szego [1], Titchrnarsh [1], Chapter 4, Levitan [1]. §5. Szego [1], Titchmarsh [1], Chapter 4, Courant and Hilbert [1]. §6. Courant and Hilbert [1]. §7. Schrodinger [1], Titchmarsh [1], Chapter 4. The method discussed here is due to Levitan [1]. §8. This example is taken from a paper by Titchmarsh {5].
CHAPTER 6
SOLUTION OF THE CAUCHY PROBLEM FOR THE ONE-DIMENSIONAL WAVE EQUATION §1. Application of the method of successive approximations
Let q(x) be a real function on the line (— over every finite interval, and let f(x) be a also defined on the line (— Under
which is summable
these assumptions on the functions q(x) and f(x), we will
consider the following Cauchy problem:
= ö2u/ôx2 — q(x) u,
(1.1)
=f(x), =0. Let us construct an integral equation for the solution of this Cauchy
(1.2)
problem. To this end we rewrite equation (1.1) in the form (1.1')
If q(x)
92u/9t2 —02u/0x2 = —q(x)u.
0, then (1.1') assumes the form ö2u/ät2 — .32u/0x2
(1.3)
= 0.
As is known, the solution of the problem (1.3) + (1.2) has the form
+f(x—t)]. Let us now consider the inhomogeneous equation uo(x,t)
(1.4)
(1.5)
—
ô2u/äx2
=g(x, t)
with known right side g(x, t). We denote by ü(x, t) the solution of equation (1.5) which satisfies the initial conditions (1.6)
It
üIt=e=0, is well known that
(1.7) where
il(x,
t)dyd'c,
is the triangle in the (y r) plane whose vertices are the points
(x—t,O), (x,t) and (x+t,O). 274
275
§1. SUCCESSIVE APPROXIMATIONS
Using formulas (1.4) and (1.7), it is not hard to construct an integral equation for the solution of the problem (1.1') + (1.2). In fact, considering the right side of (1.1') as a known function and applying formulas (1.4) and (1.7), we obtain the following integral equation, which is equivalent to the problem (1 1') + (1 2) I
u(x t)=4-{f(x+t)+f(x_t))__45
q(y)u(y, ¶)dg
5
x—(t—t)
0
(1 8)
=u0(x,
5
q(y)u(g, t)dy.
d't
.0
The integral equation (1.8) is an equation of Volterra type and can therefore be solved by the method of successive approximations. Let the function uo(x, t) be defined by (1.4). For k 1 we put I
Uk (z, t)
q (y)
d'c
4-
5
dy
(y,
(k = 1,
2,
.
z—(t—r)
0
Then from the uniform convergence of the successive approximations,
proof we omit, assuming it to be well known, it follows that the solution u(x, t) of the integral equation (1.8) is given by
I whose
u(x,t) =u0(x,t) —u1(x,t) +u2(x,t) —
(1.9)
.
We will show that each of the functions Uk(X, t) (k = 1, 2,...) be represented in the form
can
x+t (1 10)
Ufr (x, t)
= 4— 5i (s)
t, s) ds
(a,,
For k=1 we have I
(x,
t) =4-5
q 5
=
(z) 4-
[f (z + + / (z —
x—(I—r)
0
I
I
5 dt 0 I
=4- 5
q 5
(z) / (z + dz +
q 5
x—(t-—t)
x±I
0
(z) f (z —
dz
x—(I-----)
I 5
5
q(s±
nterchanging the order of integration in both integrals, we obtain
VI. CAUCHYPROBLEM FOR WAVE EQUATION
276
a1 (x, t)
q (s — t)
= 4- j f(s) ds + 4-
5
f(s) ds
q 5
(s + 'c) dc
x+i
=4.
4.
f(s) 4-
5
q(t)d'c 5
x—t
(X+8—t)
Thus we obtain formula (1.10), if we put 4-
w1(x, t, s)=4
(1.11)
5
4-
Let us now assume that for k = 1,2,
. •, n — 1 formula (1.10) has been proved. We will prove it for k = n. We have already
t
(1.12)
=
I
x+(t—c)
4- 5dv
5
q(z)
x—(t—r)
0
1 5
Change the order of integration so that the outer integral is with respect
s x + t. Therefore the limits of integration with respect to s are x—tand x+t. We thus obtain x+t (x, t)
= 4-
5
f(s) 4-
55
(z, 't, s)q (z)
dz ds,
is some region in the (r, z) plane (depending on x and t), whose exact form will not be needed in what follows. Thus, formula (1 10) has been proved for k = n, where
where
tLi3)
(x,
t, s)=4. 55
t,
From (1 8) and (1 9) follows the formula x+t
(114)
u(z, t)=4.{f
t, s)f(s)ds,
277
§1. SUCCESSIVE APPROXIMATIONS
where we have put (1.15)
+
w(x,t,s) = —w1(x,t,s) +w2(x,t,s) —
We now derive various estimates for the function w (x t s), valid for small t. LEMMA 1.1. Suppose that q(x) is summable over every finite interval. Then for t> 0 we have the estimate x±t
x±t
Iw(z, t,
(1.16)
PROOF. We will show that under the stated assumptions one has the inequalities (1.17)
Iwk(x, t,
(k=1,2,...).
(k—i)!
2,...). Let us first consider the case k =1. We find from formula (1.11) that
t t (cf the formula preceding (1 11)), the pre ceding formula implies the estimate I.
t,
Ite
(1 17)
Let us now assume that (1.17) has been proved for all k = 1,2, . . ., n
We will prove it for k = n From (1 13) and the estimate (1 17) follows the estimate t, (
i
n—I
I
(n2)!
d'cdz.
Since (cf (112)) x + t. Therefore the last estimate implies the inequality
-_ 1.
278
VI. CAUCHY PROBLEM FOR WAVE EQUATION
(1.18)
t,
n—i
Further, since 0 <
r t and x — t + r z x + t — r, we have x — t lies in the rectangle 0 x + t, and consequently the region t; x — t z < x + t. Therefore from the inequality (1.18) follows the estimate
z
x+t
n—i
t, n
in—i
I
(n—I)! i.e. (1.17). The inequality (1.16) follows at once from (1.17).. In fact, by virtue of formula (1.15) w(x, t,
s)I
(k-i)! x+t
=
as
I q (t) I
t
I
was to be proved. By making various assumptions concerning the behaviour of the
function q(x), we can obtain various estimates for the function w(x, t, s) for small t. THEOREM 1.1. If the function q(x) is summable over every finite interval,
then for every fixed finite interval (x0, Xi) and any fixed number to> 0 a constant C = C(x0, x1; t0) such that for x E (x0, x1) and O
The proof follows at once from (1.16).
THEOREM 1.2. Suppose that in some interval the function q(x) satisfies the following condition there exist positive constants C = C(x0, x1) and a a(xo,x1) such that for t t0 is a fixed number) and x E (x0, x1) Mw the inequality
(119)
q ('c)l d.t
279
§2 REDUCTION TO GOURSAT PROBLEM
Then there exists a constant C which depends upon the constants in the inequality (1 19) but not upon t such that for x E (x0, x1) and t t0 we have Iw(x t,s)l <
(120)
The proof follows at once from (1 16) REMARK For the bound (1 19) it is sufficient that there exist constants C(xo,x1, = C and p =p(x0, x1, t0)> i. such that for x E (x0 — t0, x1 + t0) one has the inequality fX+fo
In fact, by Holder's inequality (1/p + lip' =
1)
(XThI
we have for t
d'r)
x—t
v—i
From the differentiability of the functions wi(x t s) with respect to x and from the uniform convergence of the series obtained by termwise differentiation of the series (1 15), which results from the uniform convergence of the senes obtained by termwise (multiple) differentiation
of (1 15) with respect to x, follows LEMMA 1 2 If the function q(x) has a derivative of order k — 1, then the function w(x, t, s) is k times differentiable with respect to x and s
§2 Reduct,on to the Goursat problem
In this section we will show that the function w(x t s) is the solution
of the Goursat problem We will carry out the argument under the assumption that formula (1 14) has already been established, and will denve the Goursat problem for the function w(x t s) By a modification of the argument we could at the same time obtam a new justification of the formula (1 14), but we will not do this Thus, we will consider the Cauchy problem (2 1)
a2u/at2 =ô2u/&2 —q(x)u,
=f(x), (23)
Suppose that the solution of this problem is given by (1 14), i e x+i
(2 4)
u (x, t) = 4. {f (x + t) + / (z — t))
w (x, t,
+ 4. x—t
s) f(s) ds
280
VI. CAUCHY PROBLEM FOR WAVE EQUATION
We wish to clarify the conditions that the function w(x, t,s) must satisfy in order that the function u (x, t), defined by (2.4), be a sOlution
of the problem (2.1) +(2.2) +(2.3). It follows from
(2.4)
that the
condition (2.2) is automatically fulfilled. Further, differentiating (2.4)
with respect to t and then putting t =0 and using the condition (2.3), we obtain
+w(x,t,x —t)f(x
0=
—t)
=w(x,0,x)f(x).
Therefore
w(x,0,x) =0.
(2.5)
The condition (2.5) will be used further on. We will now derive a differential equation and supplementary conditions
for the function w (x, t, s) (we will see that the G oursat problem is obtained for the function w (x, t, s)). Let us denote by u(x, t; f) the solution of the problem (2.1) + (2.2) +(2.3), and suppose we already know that u(x, t; 1) is represented the operator the operator d2/dt2, and by by (2.4). Denote by d2/dx2 — q(x). We will show that
In fact, the left and right sides of this equality satisfy the same Cauchy problem
=0. Therefore the required equality follows from the uniqueness of the solution of the Cauchy problem. Thus equation (2.1) can be written in the form (2.6)
Using (2.4), we obtain
{f' (x + t) + I" (x — t)) (2.7)
+/(x—t +4 {f(x + t)
t, x+t)+W(x, x+t)f'(x+t). t, x—t)—w(x, t, x_t)f(x_t)} s)
(x — t) x+i
Ow (x, t, s)
ô2w(x,t,
s)
ds.
§3. MIXED PROBLEM ON HALFLINE
281
lurther, agam using (2 4) and integrating by parts twice, we obtain
u(x, t; Mxf)—_T{f'(x+t) x+t
—
q (s) w (x,
— w (x,
t, s) / (s) ds +
{w
(x, t, x + t) /' (x + t)
t, x_t)f(x_t)}__4 {f(x + t)
(x,t, s)
x+i }
the function 1(x) is arbitrary, the coefficients in (2.7) and (2.8) + t) and f(x — t) and the integrands must coincide (the coefficients derivatives of f(x) cancel each other). Equating the coefficients of :+t) in (2.7) and (2.8), we obtain jw(z,t,s) dw(x, t, x±t) dt
at
(X
s=x+t —
dw(x, t, x+t)
+t)
ö
I
dt
egrating with respect to t from 0 to t and using the condition we obtain
w(x, t, 'i the same way, equating the coefficients of f(x — t) leads to
w(x, t, x—t)= Finally, equating the integrands in (2.7) and (2.8), we obtain for
1 function w(x, t, s), as a function of the variables t and s, the following equation: 82w/8t2 = ô2w/äs2
—
q(s) w.
equation (2.11), together with the conditions (2.9) and (2.10), .mes the Goursat problem for the function w(x, t, s). §3. The solution of a mixed problem on the haifline
q(x) be a real function, defined on the haifline [0, c') and summable
every finite interval, and let f(x) be a twice differentiable function
282
VI. CAUCHY PROBLEM FOR WAVE EQUATION
on the same haifline. We extend the function q(x) to the negative haifline so that it is still sununable over every finite interval, but otherwise arbitrary, and we assume f(x) to be extended to the negative haifline in an as yet unspecified way (further on we will make precise
the way in which f(x) is to be extended). Wth these assumptions concerning the functions q(x) and 1(x), we the following problem: (3.1) (3.2)
ô2u/8t2 =02u/ôx2 —q(x)u,
ulj..o—f(x),
(3.3) (t) is some real function defined for all positive t. The conditions imposed upon will be clarified later.
where
We will denote the solution of the problem (3.1) + (3.2) + (3.3) by u(x,t). As we saw in §1, under the stated assumptions regarding the functions
q(x) and 1(x) the solution of the Cauchy problem (3.1) ± (3.2) can be represented in the form x+t
(3.4)
u(z, t)=4-{t(x+t)+f(x—t))
t, s)f(s)ds.
Using the condition (3.3), we obtain from (3.4) (3.5)
f(t)+f(—-t)+ 5 w(O, t, s)f(s)ds=2p(t).
are given for t 0, then equation If we assume that f(t) and (3.5) can be taken to define f(t) for t <0. In fact, let us rewrite (3.5) in the following form:
/ (—t) +
w (0,
t, s) f(s) ds
(3.6)
w(0, t, We obviously know g(t) for t
0. Further,
t, s)/(s)ds=5K(t, s)f(—s)ds,
__
283
§3. MIXED PROBLEM ON HALFLINE
there K(t, s) = w(O, t, — s) is a known function. Thus equation (3.6) the form
K(t, s)/(—s)ds= g(t)
(37)
Equation (3.7) is an integral equation of Volterra type. Solving it
f( — t), we find that
(38)
s)g(s)ds
Suppose that f(t) and g(t) have continuous second derivatives on the haifline 0 t < cx• We extend f(t) to the negative haifline in accordance with (3.8). Let us clarify under what conditions the extended function also has a continuous second derivative on the entire real line. follows from (3.8) that for t <0 the function f(t) has a continuous second derivative. Thus we only have to study the behaviour of the extended function and its first two derivatives at the point t =0. Obviously, in order that the problem (3.1) —(3.3) have a solution which is continuous at zero, it is necessary that the condition (3.9)
be fulfilled. Therefore if this condition is fulfilled, then passing to the hmit
+0 in equation (3.6), we obtain f(—0) =f(+0),
extended function is continuous at zero. Further, differentiating (3 6) with respect to t, —f' (—t)
+ w (0,
we
i.e. the
obtam
i(s) ds
t, —t) / (--t) +
(3.10)
=2p' (t) —/'(t) —f(t) w (0, t, [setting (3.10')
(Ot
/(s)ds
+0 in (3 10), we obtain —f' (—0) + / (—0) w (0, 0, 0) — 2p' (+0)
—1' (+0) — / (+0) w (0, 0, 0).
Smce w(0,0,0) =0 (cf (25)), from (3 10') we obtain (311)
f'(+O) —f'(—O)
Further, it follows from the conditions (3 2) and (3 3) that
284
VI. CAUCHY PROBLEM FOR WAVE EQUATION
=0.
(3.12)
Therefore from (3.11) followsf'(+O) =f'(—O), and so the first derivative is continuous. Finally, let us consider the second derivative. To this end we differentiate (3.10) with respect to t, obtaining I" (—t) + / (—t)
—t)
dw
ow (0, t, a)
+ /(—t)
(—i) w (0, t, —t) 02w
Lt + ç
(0,t, a)
f(s) ds = 2p" (t)
t)_/1(t)w(o, t, t,s)
—,
('
Ot
t)
02w(O, t, a)
1s'ds / /
0t2
3
0
Letting t + 0 and taking the preceding results concerning f(0) and f' (0) into account, we obtain (3.13) 2p" (+0) — /" (+0) — f' (—0) (0, t, t) dt
'L
(0, t,
(0, t, _t) .LOw (0, t, s) dt
Ot
'
a)
Ot
Let us compute the term in the brackets on the right side of (3.13). To this end we differentiate (3.4) twice with respect to t, obtaining
+f"(z —t)}+4- {f'(x+ t) w (x, t, x + dw(x,t,x+t) —f'(x—t)w(x, t, z—t) t)
+t(x+t)
t,x—t) t, a)
t)
(x,t,
+
a)
f(s) ds.
Putting t =0, by virtue of (2.5)we obtain W
0t2 t=o
—
'
"
/—2 /" / t, Ot
8)
+
t,
t, z—t)
dt
dt
Ow(z, t, a) Ot
,=x--t
J t—O
Further, using the first of conditions (3.2), we obtain —q(x)f(x).
285
§3. MIXED PROBLEM ON HALFLINE
lence 82u/öt2It.o —f"(x) = —q(x)f(x).
k15)
From (3.14) and (3.15) follows (x,
dt
3.16)
t3 17)
t, x + t)
+
dw (z, t, x — t) dt Ow (x, t, s) 1 ot
ow (x, t, s) r3t
= —2q (x) 1(x).
x =0 in (3 16) and msertmg the result m (3 13), we obtain = — 2q(0)f(0) —f"(+O)
bxi the other hand, differentiating the condition (3.3) twice with respect
to t, insertmg the result in (3 1) and then putting t = + 0, we obtain —q(0)f(0).
(3.18)
From (3.17) and (3.18) follows f"(+0) =f"(—O), i.e. the second tierivative of the extended function is continuous. We have thus proved the following: belong to the THEOREM 3.1. Suppose that the functions f(x) and C2(0, cx)). In order that the extension of the function f(x) to the negative given by
1(x) + / (—x)
—
w
(0, x, s) f(s) ds = 2p (x)
cx), it is necessary and sufficient that the elong to the class C2( — onditions (3.9) and (3.12) hold, i.e. =0. =f(+0),
We will now consider a more general problem, namely: Ô2u/0t2 = 02u/0x2
3.1)
—
=f(x),
3.2)
q(x)u,
=0,
33') Let us assume that the function q (x) satisfies the previous requirenents and extend it to the entire real line, preserving the condition hat it be locally summable. We will seek a solution of the problem 3.1) +(3.2) +(3.3') in the form x+t
3.4)
u (x, t)
(f (x +. t) + / (x — t)) +
w (x,
t, s)f (s) ds.
286
VI. CAUCHY PROBLEM FOR WAVE EQUATION
It follows from the condiand Let f(x) E C2(0, E C2(O, tions (3.2) and (3.3') that in order for the problem (3.1) — (3.3') to have a solution which is continuous at zero, it 'is necessary that 4,(+0)f'(+O)—hf(+O),
cc"(+O)O. and u(0,t) from (3.4) and insert their If we determine values in the boundary condition (3.3'), we obtain f(t)+/'(—t) +f(t)w(0, t, t) —/(—t)w(0, t, —t) —h{f(t) +I(—t)) (3.19)
(3.20)
+
àw (x,t, s)
•
i
f(s) ds— h
w (0,
t, s) f(s) ds = 2p (t).
Let us rewrite equation (3.20) in the form
t, —t) + h} / (—t).
1' (—t) — [w (0,
+ [ ow
ha, (0, t, s)] f(s) ds
—t
(3:20')
=2p(t)— f(t)—[w(0, t, t) —hJ/(t) Ow (x,t, s)
±
+ hit
(0,
t, s)]/ (s) ds.
The right side of (3.20') is a known function; denote it by g(t). By a change of variables in the integral on the left, we can put (3.20') in the form
/'(—t) —[w(0, t, —t) +h]/(—t)
(3.21)
K(t, s)f(—s)ds =g (t),
where K (t, s)
— Ow (z,t, —s)
+hw (0, t, —s).
Equation (3.21) is an integro-differential equation in the unknown function f( — t). If we integrate it once, we obtain an integral equation ófVolterra type for f( — t). Integrating (3.21) from 0 to t with respect to t, we obtain
---/(—t)±/(---O)—
{w(0, t, —t)+h) /(—t)dt
287
*3. MIXED PROBLEM ON HALFLINE
fter some simple transformations, this relation can be rewritten in le form L22)
/ (—t)—f (—0) + K1 (t, s)/ (—s) ds = g1 (t).
Ve put f( —0) = C and write equation (3.22) in the form
K1(t, s)/(—s)ds=g1(t) + C.
3.23)
(3.23) has a solution for any C, and so it will have a solution
or C =f(+0). On the other hand, for any C the solution of equation 3.23) is also a solution of the integro-differential equation (3.21). rherefore, putting C = f( + 0) in (3.23) and then t =0, we obtain 3.24) f( —0) = C f(+ 0), the extended function is continuous at x =0. Further, putting t = 0 in (3.20') and taking into account (2.5), i.e. 0, x) 0 for any x, we obtain —f'(+O) +hf(+0).
f'(—O) —hf(—0)
lence by virtue of (3.24) and the first of conditions (3.19) there follows
f'(—O) =f'(+O), e. the first derivative of the extended function is also continuous at
3.25)
r=0. Finally, let us consider the second derivative of the extension. To end we differentiate (3.20') with respect to t, obtaining —f' (—t) — / (—t)
t)
dw
s)
+/(—t){
—h(0, t,
+ f(s)![
— hw (0, t,
(3.26)
—
ow
8)
—hw (0,
[
/ (s)
s)]ds
,t, t) +f'(t)[—w (0, t, t) + hj
—/"(t) _f (t) — 1(t)
(—t) [w (0, t, —t) + hJ
ô [OW (x,t,
s)
t, s)]
— hw (0,
t, s)]
288
VI. CAUCHY PROBLEM FOR WAVE EQUATION
Putting t = 0 in (3.26) and taking into account (2.5), (3.24), (3.25) and the second of the equalities (3.19), we obtain dw (O,t—t) dw (O,t, t) (—0)
—P(+0) =f(O) {
1"
(3.27)
(x, t,
ow (x, t, s)
s)
OX
1
OX
t=t )
Further, since (cf. formulas (2.9) and (2.10)) dw (0, t. t)
dw (0,
I
dt
—t)
I
dt
we have
[dw(0
t
dw(0 t ._t)]
t)
_4-{q(O) —q(O)} =
0
Therefore (3.27) assumes the form (3.27')
1:
s)
Ow
f'(—O) —/,'(-+-O) =f(O)
s)
Ow
+
I} ..•—t
a—.t
{
Let us compute the term in brackets on the right side of (3.27'). To this end, differentiating (3.4) once with respect to x and then once with respect to t, we obtain
+4- {f'(x+t)w(x, t, x+t)+/'(x—t)w(x,
tx+t) +4- {t (x — t)
Ow
(xt
s)
+ / (x — t) I
Ow
t,
v—t))
tx_t)}
(xt
s)
x+t r O2w(x, t, s)
f(s) ds
Puttmg t = 0 and taking account of the condition (3 2) and (2 5), (2.9) and (2.10), we obtain '3 •28'/
50w (x, t, s)
Ow (x, t, s)
Ox
Ox
1
s_.x—ifi=o
— —
which holds for all x. Therefore from (3.27) follows f"(—O) =f"(+O),:
which gives the continuity of the second derivative of the extended function at x=0. We have thus proved the following
§4. MIXED PROBLEM ON FINITE INTERVAL
289
THEOREM 3 2 Let f(t) E C2(0, cx') In order that the extension of f(t) the negative halfline, defined by (3.20'), belong to C2( — cx>, cx>), it is necessary and sufficient that condition (3 19) hold ie ( + 0) = f' (+ 0)
Lhf(+0),
=0
§4 Solution of a mixed problem on a fimte interval We now consider a mixed problem for the one-dimensional wave equation on a finite interval of the real line which, without loss of generality, we may assume to be the interval [0, lr]:
ô2u/i9t2 =82u/8x2 —q(x)u,
(4.1)
44.3)
(4.4)
We wifi show that the problem (4.1) —(4.4) can be solved by the method which we already used in the preceding section, without
resorting to an eigenfunction expansion. However, in contrast with the considerations of the preceding section for the case of the haifline, for the case of a finite interval the extension has to be carried out both the left of the point x =0 and to the nght of the point x = Let the function q(x) be contmuous on [0, ir] We extend q(x) to the entire real line, preservmg its contmuity but otherwise arbitranly, and we seek a solution of the problejn (4.1) — (4 5)
u (x, t) =
{/ (x + t) + / (x — t))
+
w (x, t, s) f(s) ds,
where the function f(t) has to be extended outside the interval [0, ir}, Using the boundary conditions (4 3) and (4 4) We will first derive certam necessary conditions which the functions and have to satisfy. It follows from the condition (4 3) that (46)
From condition (4 3) and the second of conditions (4 2) it follows that
=0. Similarly, from condition (4 4) and the second of conditions (4 2) It follows that
290
VI. CAUCHY PROBLEM FOR WAVE EQUATION
(4.8)
—0)
+Hf(,r —0),
(4.9)
=0.
Using
the boundary condition (4.3), we extend the function 1(t) from
the interval (0,7) to the interval (—7,0). Suppose that 1(t), E C2(0, ir). We saw in the preceding section that the conditions (4.6)
and (4.7) are necessary and sufficient for the extended function f(t) to belong to C2( — ir, 7). To extend the function f(t) into the interval (,r, 27) we wifi use the boundary condition (4.4), from which, as we saw above, follow the conditions (4.8) and (4.9). We will show that for the function f(t), after extension to (ir, 27), to belong to C2(0, 27), conditions
(4.8) and (4.9) are not only necessary, but also sufficient. In fact, in the boundary condition (4.4), we obtain
inserting
(4.10)
+w(n,
t, Tv+t)/(n+t)—w(ic, t,
ôw(x,t,
it—t)/(it--—t)
s)
t,
We have obtained an integro-differential equation for the determina4. If this equation is integrated once, then we obtain an integral equation of Volterra type for 1(7 + t). Let us first rewrite (4.10) in the form tion of 1(7 + t) (0 t
t,
(4.11)
(z,t, s)
+
{
(t)
— I'
— t)
ôw(x,t,
+ Hw (it,
— {H + w
t,
t,
s)}
— t))
f(s) ds /
— t)
+Hw(n, t,
or after a change of variables in the integral on the left side
t, it+t)}/(ic+t) (4.11')
ow (x,t, s)
+ 0
Puttmg
{
+ Hw
t,
+
+ a)da=g1
291
§4. MIXED PROBLEM ON FINITE INTERVAL
K1 (t, c) = Ow (x,t, s)
+ Hw (iv, t,
iv
+
integrating equation (4.11') from 0 to t with respect to t, we obtain
/(iv+t)_f(it+O)
t,
after simple transformations 4.12)
f(iv+t)-.—f(ic+0)
Let us consider the integral equation 4.13)
/ (iv + t) + K (t, s) /
+ s) ds = g (t) +
C,
rhere C is an arbitrary constant. It is obvious that for any C a solution (the integral equatiion (4.13) is also a solution of the integro-differential quation (4.11). Therefore for C—f(ir —0) equation (4.13) also has solution, which is in turn a solution of equation (4.11). Putting —0) and t =0 in (4.13), we obtain
f(7r+0) =f(ir—0), the extended function f(t) is continuous at the point t = ir. putting t = 0 in (4.11) and taking into account (2.5), (4.8) fld (4.14), we find that 4.14)
415)
f'(7r+0) rf'(ir—O),
e. the first derivative ot the function f(t) extended according to (4.11)
i continuous at the point t = ir. let us examine the second derivative of the extended function U). To
this end, we differentiate equation (4.11) with respect to t,
staining
f'(iv+t)+f(iv+t){H+w(iv, t,
__
292
VI. CAUCHY PROBLEM FOR WAVE EQUATION
+ / (ic
(x,t, s)
+ t)
+ 11w (it,
t,
+ t)}
{
1c+t
'
t, s)
s)lI(\d
t,
_LH
ôxot
3
J' '
/
t, n—t))
(4.16) t,
t1
dt a)
(:,t,
a)
+H
ow (z,t, a)
f(s) ds.
Putting t =0 and taking account of (2.5), (4.9), (4.14) and (4.15), we obtain it—t)
(4.17) f'(it+O)—/"(it—O)= J(
dw(it, t,
Further, by virtue of (2.9) and (2.10) we have t, dt
dw(it, t, it—t)
— 1.
/
Therefore from (4.17) it finally follows that f"(7 —0)
f"(ir +0), i.e.
the second derivative of the function 1(t), extended according to (4.11),
is continuous at the point t = Thus we have extended the original function f(t) from the interval (0,7) to the intervals (— ir,0) and (ii-, 27). This extension is of class C2( —7,27). The integral equations (3.23) and (4.13) make it possible to extend f(t) to the entire real line. Let us show, for example, how to obtain an extension of f(t) to the interval (— 2ir, —
To do this we have
to assume that t varies in equation (3.23) in the interval (0,27). Since the solution of a Volterra equation is differentiable as many times as is the free term (assuming the kernel is sufficiently differentiable, which is the case here), our extension is of class C2(— 27,0) Since equation (3.23) is a Volterra equation, the original function, which is undefined m (—7,0), does not appear Therefore the extension which we obtam is of class C2(—21r,21r).
If we now assume that t varies from 0 to 37 in equation (4.13), then we obtam an extension of the original function f(t) to the interval
REFERENCES
4ir). This extension is of class C2( — 2ir,
293
Thus step by step we
r±end the original function to the entire real axis. This extension is of oo). If the function thus obtained is inserted in (4.5), lass C2(— hen by virtue of the very process used to extend it we obtain a solution f the problem (4.1) —(4.4). Bibliographical references
§1 Formula (1 14) can be obtained by Riemann's method Indeed is derived in this way in a paper by Povzner [1] The method which we have presented here is also, in essence, close to that of Riemann As concerns the estimate (1 16) and what follows, see the paper by Levitan [14] See also the joint paper by the authors [3] §2 A similar method, but m another connection, was applied in the paper by Gel'fand and Levitan [1] **3, 4 So far as we know, the extension method has been previously apphed only for the case of mixed problems for the simple wave equation
= uxx
CHAPTER 7
EIGENFUNCTION EXPANSION OF A STURM-LIOUVILLE OPERATOR §1. Derivation of auxiliary formulas
We will study the equation y" +[A —q(x)]y =0,
(1.1)
defined
on the entire real line.
Concerning
the function q(x)
we
will
that it is real and summable over every finite interval. Denote A) the solution of (1.1) satisfying the initial conditions =1, (1.2) =0, and by (x, A) the solution of (1.1) satisfying the conditions assume by
(1.2')
4/(0,A) =1.
=0,
Together
with equation (1.1) we will consider the partial differential
equation
= 02u/ôx2 — q(x) u.
(1.3)
Suppose that the function f(x) has a continuous second derivative. We denote by u (x, t; f) the solution of equation (1.3) satisfying the initial conditions (1.3') As was shown in the preceding represented
ou/otIt.o=0. chapter, the function u(x, t; f) can
be
in the form x+t
(1.4)
u(x,
Putting"
t; f) f(x)
t, s)/(s)ds.
{f in
(1.4) equal successively to p(x, A) and
A),
we
obtain the equalities
X) precisely, putting f(x) in the conditions (1.3') equal successively to then solving the problem (1.3)—(1.3') by Fourier's method, from (1.4) we obtain, by the uniqueness of the solution of this problem, (1.5) and (1.5').
294
296
§1. AUXILIARY FORMULAS
(1.5) 5
=
(z, X) cos
iv(x, t, s)p(s, X)ds,
(x + t, X) + (x — t,
X)}
(1.5') X)ds.
t,
Let us denote by g, (t) a function satisfying the following conditions: 1. is even, i.e. g,(t) vanishes outside the interval 3. g, (t)
has a piecewise continuous and piecewise monotonic first
j 4erivative
Further, let us denote by i.e. we put
the Fourier cosine transform of the
g,(t)cos p.tdt
(16)
It follows from condition 3 that for large
one has the estimate
1(17)
and integrating with respect
Multiplying both sides of (1.5) by to t from 0 to we obtain (18)
(x
s)
X)
(x,
s)
dt
the same way, it follows from (1 5') that (1 8')
{g8 (x— s)
s) is defined by (1.9). 2(_ cx) We put Let f(x),g(x)
+
(x, s))
(s, ?) ds,
296
VII. EXPANSION OF A STURM-LIOUVILLE OPERATOR A)dx,
A)dx, (1.10)
g(x)p(x, A)dx, A)dx.
The functions F1(X), F2(X), G1(A) and G2(X) are called the Fourier transforms of the functions 1(x) and g(x). If f(x) and g(x) vanish- outside
some finite interval, then the integrals in formulas (1.10) exist in the ordinary sense. In the contrary case (1.10) is to be understood in a generalized sense (cf. §6, Chapter 2). Further, as is known, there exist monotone functions and which are bounded in every finite interval, and a function 77(X) which is of bounded variation over every finite interval, such that if 1(x), g(x) E 2( — a>, cx>), then one has the Parseval equality
/ (x) g (x) dx = (1.11)
F1 (A)
(A)
(A)
{F1 (A) G2 (A) + F2 (A) G1 (A))
(A)
(A) G2 (A)
(A).
Formulas (1.8) and (1.8') show that for fixed x the functions and are the Fourier transforms of the function which is equal to [g, (x — s) + (x, s)] in the interval (x —
and equal to zero outside this interval. Therefore, replacing x by y in (1.8') and applying the Parseval equality (1.11), we obtain x
A)p(y,
(112)
=1
A)
s))
s))ds,
297
§1. AUXILIARY FORMULAS
is the intersection of the intervals (x — x + and (y —
where
y+
(1.12) assumes a simpler form if we introduce the spectral function of equation (1.1): Formula
o
(x, y, A)
A)p(y, X)dE(A)+[p(x, (y, A)
—
(y,
(p (x, A) p
A) dE (A)
+
A)
A)
(A))
A)
(y,
(A
A)
(A
Using the spectral function O(x, y,
(x, y,
A),
we can write (1.12) in the form
A)
(1.12')
(x—s)
=4. In particular, if x = y,
+
+ x8 (y, s))
s)}
then
(114)
s)}2ds
x,
The function O(x y, A) yields the resolution of the identity of the
operator (1 1) and therefore has the following properties (cf, for example, Chapter 13) and put Let denote the interval (A, A + 1 POSITIVITY —O(x,y,A).
For
0
2 ORTHOGONALITY
For
any two intervals
one has (1 15)
where
0 (x,
t,
0 (t, y,
dt = 0 (x, y,
denotes the mtersection of
and
= (A, A +
and
VII. EXPANSION OF A STURM-LIOUVILLE OPERATOR
298
is', then from (1.15) and the Cauchy-Bunjakovskii If we put inequality there follows the bound
\
1/2
0
(x, g,
dt)
92
(L16)
=
[0
(x, x,
(i
dt)
02 (t, y,
(ii,
110
1/
(x, x,
+ 0 (y, y,
§2. Preliminary bound on the spectral function. The case of the entire line
As we already mentioned, the function O(x, y, A) is called the spectral
function of equation (1.1). In this section, making use of (1.14), we will obtain several important estimates for the spectral function. Subsequently, using these preliminary estimates, we will obtain the asymptotic
behaviour of the spectral function of equation (1.1) for A LEMMA 2.1.
be an
+
arbitrary positive number and (x0, x1) an arbitrary
finite interval on the real line. There exists a constant C = C(x0, x1, such that if x and y belong to the interval (x0, x1), then one has the inequality (x, y, X)
In particular, for any finite
x
C.
and y y;
PROOF. We take for the function g, (t) the function
g1(t)=
for ti
lo
Then as is easily seen,
=
=
(2.1)
g,
(t)
cos
srn
\/X t dt
0
the value of
Ar ( (2.2)
I
sh-
j
Vffl
from (2 1) into (1 14), we obtain .
<
Put — sh
§2. PRELIMINARY BOUND ON SPECTRAL FUNCTION
299
= a. It follows from the mean value theorem that sh u = 0 = u ch Ou> a. We therefore deduce from (2.2) that
sh u
(1
r (2.3)
Let = +1 It is easily seen that for
and ()
(x — s)
(x, x, A) < 4.
,j
f
A
+ L (x,
s))2 ds.
> 1 we have the inequality
therefore the inequality (2.3) implies the bound
._!. 2
(2.4)
x,A)
{g8 (x
s) + x. (x, s))2 ds +
(x, x,
The bound (2.4) proves the lemma for x = y. The case x y reduces to the case x = y by means of the mequality (1 16) This completely proves the lemma. I Let us denote by a an arbitrary real number and put X cos at. Since the function g. (t, a) satisfies all the requirements placed on the function g (t), replacing g (t) by g (t, a) in (1 14) we obtain (25)
s, a))2ds,
x,
where
(26)
s, a)_—
w(x, t,
(t, a) dt,
(27)
(28)
H(x, y, a,
y, A)
If the function g (t) is chosçn as before, then from
VII. EXPANSION OF A STURM-LIOUVILLE OPERATOR
300
a) =
-
(s
— t I
cos
at• cos
t dt
there follows for A <0 the bound therefore y, x)I.
IH(x, y, a,
From this bound and the preceding lemma there at once follows:
LEMMA 2.2. If the coefficient q(x) in equation (1.1) is summable over every finite interval and if (x0, x1) is an arbitrary finite interval of the real is an arbitrary positive number, then there exists a constant line and C = C(x0, x1, to), which does not depend upon a, such that for x, y E (x0, x1) and we have the inequality
IH(x,y,a,f)I
(2.9)
y,
(2.10)
=4;
{g1(x—s,
s, a))ds
s, a))
— H (x, y, a, s),
is defined by (2.6). Let us put A =,L2, U(x,y,A) =8(x,y,L2) —O1(x,y,,i) for A >0 and for fixed x and y extend the function 01(x, y, i) to the negative semiaxis as an odd function. Then (2.10) assumes the form where
-
a)
(2.10')
(x, y,
a)+X1(x, s, —2H(x, y, a, e).
In particular, for x = y we obtain
a)+Xe(y, s, a))ds.
§2. PRELIMINARY BOUND ON SPECTRAL FUNCTION
a)dp.01(x, X,
301
p.)
s, a)2 ds—2H (x, x, a,
2.3. If the coefficient q(x) in equation (1.1) is summable over ,y finite interval and (x0, x1) is an arbitrary finite interval of the real then there exists a constant C = C(x0, x1) such that for x, y E (x0, x1) has the inequality
LEMMA
y,}L))
Put =2 in (2.11). Then we obtain (by virtue of (2.1)) sin2 (p_a)lsd
ç
)
1
(p — a)2
+ a)2
f
p.
Q
X,
1
)
s, a)2ds—211(x, x, a, 1).
By Theorem 1.1 of Chapter 6 and Lemma 2.2 there exists a constant C(x0,x1) such that if xE (x0,x1), then the right side of (2.13) does exceed C. Therefore from (2.13) follows the bound
I
dp.Oi (x, x, p.) <
C,
X,
a)
dp.61(x, x,
a)
I so a fortiori
in2 one has the mequality
i is known, for 0
x,
x,
x, a)).
302
VII. EXPANSION OF A STURM-LIOUVILLE OPERATOR
The inequalities (2.14) and (2.14') prove the lemma for x =y. With the help of (1.16) the case x y reduces to the case x = y. 2(_ co), and let Suppose that f(s) E +X.(x,s) for s E (x — x + and zero otherwise. Then by the Parseval equality (1.11) it follows from (1.8), (1.8') and the first two of equalities (1.10) that (F1 (A) p (x, A)
+ [F1 (A)
(2.15)
(A)
(x, A) + F, (A) p (x, A)]
(A)
+
(A)
(x,
A)
(A))
Using the spectral function O(x,y,A) (cf. (1.13)) and the definition of the functions F1(X) and F,(A) (ci. (1.10)), we can write (2.15) in the form f(s)O(x, s, A)ds (2 16)
=4f(s) is arbitrary, it follows from (2 16) that ( 21 '7)
s),
1
0,
The convergence of the mtegral on the left follows from Lemmas 2 1 and 2 2, and also from the bound (1 7) The formula (2 16) can be written m the form (x, fL)
(2.18)
/ (s) {g1 (x — s) +
where we have put
(x, s)} ds —
(x, A),
§2 PRELIMINARY BOUND ON SPECTRAL FUNCTION
303
s, i)ds,
S(x,
(FL>O),
Similarly, (2.17) can
be
<0).
= —S1 (x, —p.)
S1 (x,
written in the form s,
g1 (x — a)
+
(z, a)
(2.19)
—2
x—a
0 in (1.1), then w(x,t,s) 0, and consequently Therefore (2.18) and (2.19) assume the form
If q(x)
x+s
(2.20)
(ii)
(x,
=
f(s)
(z — a) ds,
x-e
(2.21)
a
0,
where we have put (x, s
=!
(x,
=
/(s)
(x s, u) da
Subtracting (2.20) from (2.18), we obtain I)
(s)
where we have put R(x from (2 19), we obtam
(x, s) ds—2
f) =S1(x,,i)
d,S (x, X),
Subtracting (2 21)
304
VII. EXPANSION OF A STURM-LIOUVILLE OPERATOR
s,
(2.23)
(x, s, X),
=
s, A),
Formulas (2.22)
where 4'(x,s,j) =01(x,s,4u)
and (2.23)
play a fundamental role in what follows. §3. Asymptotic behaviour of the spectral function. Case of the entire line
In this section we will study the asymptotic behaviour of the spectral function O(x,y,X) of equation (1.1). We will here place upon the function g, (t) more rigid restrictions than were imposed earlier. We will, namely, assume that g, (t) satisfies the following conditions:
=g(—t). =1 if ti 3. g,(t) =0 if iti 4. g, (t) is bounded as a function of t and c has a continuous fourth derivative. 5. 1.
2.
LEMMA 3.1. Let us assume that the coefficient q(x) in equation (1.1)
is summable over every finite interval. Let (xo, x1) denote an arbitrary finite interval. There exists a constant C = C(xo, x1) such that if x, s E (x0, x1), then (3.1)
(x, s, A) 1< C.
PROOF. Since
I
cos
0
O(e&ITJ+1)s),
the bound (3.1) follows at once from Lemma 2.1.
Let us now proceed to the investigation of the asymptotic behaviour We will first transform the right side
of the spectral function 0 (x, s,
Let us first consider the term
§3. ASYMPTOTICS OF SPECTRAL FUNCTION. ENTIRE LINE
(x,
x8
s)=
305
w(x, t, s)g, (t) dt. I
I
dt.
w (x, t, s) cos
(x, s, l.x—sI
the Fourier inversion formula, we have
follows from the Parseval equality for the ordinary Fourier that s,
w(x, t, IX—81
0.
Using the evenness of the function a(x, S. p), this last formula can transformed to the form s),
s,
tO, the term corresponding to the negative spectrum. We have hanging the order of integration) cos
(x, s, X)
h (x, s, t) =
cos
(x, s,
=
h
(t)
h (x, s, t) dt,
(x, s, X).
(x, s, t) cos
dt.
306
EXPANSION OF A
OPERATOR
It follows from the Parseval equality for the ordinary Fourier integral that (cf. the derivation of (3.2))
s, t)dt.
s,
(3.3)
(3.2) and (3.3) follows the important equality
From
(ii)
(3.4)
(x, s, t.L) = 0,
where
v)dv+
s,
(35)
In
s;
+
By virtue of the estimates
IL
IL
s, v)dv
V IL
co)
o
o
and Lemma 2.3, the following estimate (for function
=0(1),
s,
V
+
co)
holds for
s, ii):
(3.6)
where the constant C depends only upon the interval of variation x and s. Therefore by virtue of (3.4) a Tauberian theorem of Levitai (cf. Chapter 14, Theorem 4.1) is applicable, from which it follows tha as —* we have, uniformly for x and s in any finite interval, = 0(1).
(3.7)
From (3.7) it is not hard to obtain the asymptotic behaviour of t$ spectral function. 3 1
If q(x) satisfies
the conditions
(1 19), then for
/ixedxands I
(3.8)
sinp.(x—s)
0(1).
§3. ASYMPTOTICS OF SPECTRAL FUNCTION. ENTIRE LINE
307
estimate (3.8) holds uniformly as x and s vary over any finite interval.
ROOF. By virtue of the asymptotic formula (3.7) and formula (3.5), •ie (3.8) it suffices to show that as + s,
v)dv=—O(x,s, iformiy in every finite region. will first prove (3.9). We have c
w(x, t,
cz(x, s, v)dv =
dt.
x—sJ
denote
an arbitrary positive number not exceeding
w(x, t,
+
dt =i{
t,
and unity,
dt=11 + '2
y virtue of the bound (1.20) I
Cir.
C
erefore for any positive number we will have
one can choose
i,
so small that
r all
IIiI
ther, it follows from the Riemann-Lebesgue lemma that for fixed can choose so large that for > we have 1121
..,i the bounds (3.11) and (3.12) it follows that for ii> r I
w(x, t,
s)
sirtfLt
dt
—
I
proves (3.9) since the number t us prove (3.10). We have
is arbitrary.
VII. EXPANS!ON OF A STURM-LIOUVILLE OPERATOR
308
2r
(x, s, v) dv
= 2ç h (x, s, t) —i--- dt. 0
0
Since s,
s, t) is an entire analytic function of t, and so, in particular, it is differentiable in a neighborhood of t =0. Therefore by a well-known theorem of the theory of Fourier series it follows that
the function h (x,
IL S
Ii.
2
2c h(x, s,
s,
dt
t)
=h(x, s, 0)+o(l)=—O(z, s, —oD)+o(1), as was to be proved. The uniformity of (3.10) follows from the bounded ness of the functions h(x, s, t) and äh(x, S. t)/at. REMARK. From a Tauberian theorem of (cf. Chapter 14 Theorem 4.2, special case) there follows a result which is more than (3.8), namely, (3.13)
I
I urn tOl(x, s;
s; —co).
}L
For x = s the equality (3.13) follows directly from the identity (2.23
and the indicated theorem of the monotonic function P(x,s;M) and
apply
For x
s one has to conside
=O1(x,x;IL) +201(x,s;IL) +O1(s,s;IL)
theorem to it (agam using the identity (2 23)j
§4 Asymptotic behaviour of the spectral function The case of the haifline
Let the equation
y"+[x—q(x)]y=O be defined on the haifline [0, co). We will assume the function summable over every finite interval. We adjoin to equation (4.1) boundary condition at x = 0: (4.1)
(4.2)
y'(O) —hy(0) =0,
309
§4. ASYMPTOTICS OF SPECTRAL FUNCTION. HALFLINE
is an arbitrary real number. The case h = will not be conhere. It is rather completely treated in the paper [15] by Levitan.
h
denote the solution of equation
A)
(4.1)
satisfying the
conditions =h.
=1,
is known, for a given h there
exists at least one function p (A)
is monotonic, bounded in every finite interval, continuous from left, and which generates an isometric mapping of the space
the space
(—
F(X)=lim 2(Q
cxx, cx>)
according
(x)ph(x, X)dx,
2(0,
x>)
to the formula
f(x)=lim
the integrals converge in the metrics of (— ccx) respectively, and one has the Parseval equality
cxx)
and
Fa (A) dp(A). 0
will call the function Oh(x, y,
spectral function of equation (4.1) (for the initial conditions (4.3)). Further, we by (x, y, A) the spectral function of the same oblem for q(x) 0. Let us first assume that h> 0. In this case the y" + Xy = 0
not have a negative spectrum. For x > 0 put x =
Then
Oh(X,y,A) =Oh(X,y,,z2) =Oh,1(x,y,,4,
the arguments 'inptotic
formula
of the preceding section,2> we obtain the
>x>)
Oh,1(X,y,,L) —Uhl(X,Y,/.L) =Oh(X,y, — x>) +o(1).
the Remark followmg the proof of Theorem 3 1
310
VII. EXPANSION OF A
OPERATOR
Let us find the principal term of
0
1(x y,
As is known (cf.
(CoSvx+! Sin vx)(cnc COSVX
COS
p.
2k ç sin (x + y)
_L 2h2 ç sin
v2+h2 d VT
0 p.
(4.6)
d
0
r
2
vx• sin
3
p.
ç cos vx • cos vy v2 + h2
cos vx • cos vy
o
o
sin vz. sin vy
2h2
2
cos vx 5
2k
sin (x +
cos vy dv +
+ h2
2h2ccosV(x_Y)d
2h25cogv(x_y)
v
d
y'
dv+o(1)
vdv
(x+y>O),
0
where
y,
=
5 cos vx
cos vy dv
is the spectral function of the problem (4.1) + (4.2) for q(x)
From the asymptotic formula (4.5) and the relation
follows at once:
0 and h (4.6) th
THEOREM 4.1. Let the function q(x) satisfy the condition (1.19) Chapter 6 For every fixed x 0 and y 0 lim {0A 1(x, y,
(47) )
y'
y
311
§5. REISZ SUMMABILITY OF SPECTRAL FUNCTION
relation (4.7) holds uniformly as x and y vary in any bounded region
>0). x=y=0 we obtain from (4.5) >
.1
—h.
Eiiii IL
us consider the case h <0. In this case equation (4.4) has one eigenvalue Ao = — h2. Indeed, the eigenfunctions corresponding 'e negative eigenvalues are square integrable. For A <0 the solutions
have the form *
—
(x, X) = cos h \11 X x + h
+
=
cz), then I =0, i.e. A =A0 = —h2. E Thus, corresponding normalized eigenfunction has the form .a comparing the spectral function Oh(X, y, A) with the spectral funcA)
an
additional term —
corresponding
to the
ative eigenvalue of the problem (4.4) —(4.2), appears.
instead of formula (4.7) we obtain (x ±y> 0) Jim {Ohl(x, y,
it
3
± h2
3
0
0
us now consider the case x +y = 0. Instead of (4.5) we now obtain Oh,1(X,Y,,U)
—
co) —2h+o(1),
which for x =y =0 we again obtain (4.8). §5. Riesz summability of the spectral function
results obtained in §3 enable us to study the asymptotic ber of the Riesz means of spectral functions. We will consider the of the entire line. The case of the haifline and also the case of a interval can be studied analogously. virtue of (3.7) and (3.9), there follows from the Tauberian Theorem Chapter 14) the bound (s
0 is an arbitrary number)
VIL
312
OF A STURM-LIOUVILLE OPERATOR
(5.1) where
the constant C depends only upon the domain of variation of
the point (x, y).
Using the bound (5.1), it is not hard to obtain the asymptotic beof the integral
haviour as
y, v).
r(s+1)
In fact, by (3.8) and from the definition of the functions F*(x,y,v) we have
and jL
r
1)
p.
(i
(x, y, v) —r
r(s±1)
(i —
±1)
(
(x, y, v)
0
(5.2)
y, v)dv
g,
By a well-known formula (cf. Titchmarsh [3], §7.1) i (st)
p.
(5.3)
2 0 ( ii.
Therefore
t)2
'17cr(s±j) 1
1
[p.(x—y)]
2
2
2
(i — V2)8 (x, y, v) dv
(5;.4)
t)
w (x, t,
cos vtdv] dt 0 1
2
2
.1
ç 3
2
w(x, t, y)
1
8+— (p.t)
di.
313
§5. REISZ SUMMABILITY OF SPECTRAL FUNCTION
prom the last formula, by Theorem 1.2, Chapter 6, i.e. by virtue of bound (1.20) of Chapter 6, there follows the bound (t)
=
s), c
8,
a = S. we have (p.1)
—_______
dt
t)
(x,
—
2
6;)
I
(3/
1
/
+
2
2
(p.t)
—
0
(x, y, —a)
cJ
dt
(p.t)
—
—
(t)
1
2
2
1
dt + o(1).
2
0
rrom formulas (5.2), (5.4), (5.5), (5.6) and the bound (5.1) follows: 5.1. Let the function q(x) satisfy the condition (1.19) of 6. Then for every fixed x and y, and p.
y,
r(s±1) r)
[p.(x—y)]
0
'C Y,
v = mm
(a, s), a
2
S.
,a=s
s relation holds uniformly as x and y vary over any bounded region.
314
VII. EXPANSIOWOFA STURM-LIOUVILLE OPERATOR
§6. Proof of a theorem on equiconvergence
Making use of formula (2.22) and the Tauberian Theorem 4.1 Chapter 14, we will in this section give a complete derivation of a theorem concerning the equiconvergence of an expansion in a generalized Fourier integral and an expansion in an ordinary Fourier integral under
minimal requirements on the coefficient q(x) —its summability over every finite interval. We will consider the case of the entire line. The case of the halfline and the case of a finite interval can be studied analogously
Let us turn first to the transformation of the nght side of (2 22) Changing the order of integration we obtain t,
s)g€(t)dt}ds
t, s)/(s)
ds} dt.
Put h (x,
t) =
w (x,
t, s) f(s) ds,
t)cosp.tdt.
cc(x,
From the Parseval equality for the ordinary Fourier integral
t)dt (6.1)
=
f(s) x—a
w
(x, t, s)
(t) dt} ds.
{ 1x—,I
Further, we have (as a consequence of changing the order of
t)dt,
s,
where
k(x, t)=
s, X)f(s)ds.
315
§6. EQUICONVERGENCE
rrom the Parseval equality for the ordinary Fourier integral follows
(2.22), (6.1) and (6.2) we finally obtain
R*(x,
v)dv.
ii., 0
U
ZMMA 6.1. For every fixed x
urn V {fr(x, relation holds uniformly with respect to x over every finite interval. F. It is sufficient to prove a relation analogous to (6.5) for term in the right side of (6.4) separately. We have
S1(x,g). Let ia,bJ and consider the problem
will prove
denote
an arbitrary finite
y"+{X—q(x)}g=O (b) sin = 0, are arbitrary real numbers. We denote by A,, =
y (b) cos P +
,hich a and ye eigenvalues of this problem, arranged in increasing order, by ,t',,(x) corresponding normalized eigenfunctions. Further, put 01,
a
(x, s, p) =
(x)
s,
Si,a,ö(X, a
(s),
VII. EXPANSION OF A STURM-LIOUVILLE OPERATOR
316
It is known (cf., for example, Chapter 2) that as (a, b) —* 01,O,b(X, s, ,.i) —p
(x, s, iz),
(—
co)
—* S1 (x,
S1,a,b(X,
Further, 1
V Ii.
(65') x,
+ 1)0i,a, b(X,
urn
(a,
b)-*(--co, o)
=
E2 (t)
(t) + F2 (t) dc(t),
(t) + 2E (t) F (t)
(6.5) follows from the bound (6.5') by passing to the limit, first witt respect to (a, b), and then with respect to For the function p) the relation (6.5) can be obtained in similai fashion or, even simpler, starting from the explicit form of this
by an application of the Cauchy-Bunjakovskii inequality. Further, we have IL+1
V
(x,
t) dt =
(x,
dt =0(1)
o
and similarly for the integral with
i'). Thus
the lemma is
proved. From this lemma and from the Tauberian Theorem 4d Chapter 14 for Fourier integrals there follows at once LEMMA 6.2. For every fixed x 0.
(6.6)
This relation holds uniformly with respect to x over every finite interval. For the case in which the spectral function is nonunique, one has to pass to limit along some sequence
{ (ak, bk) }.
317
§6. EQUICONVERGENCE
F prove the theorem on equiconvergence, it remains to study the
of the functions
as
and
6.3. Suppose that the coefficient q(x) is summable over every Then for every fixed x we have
v)dv=O.
urn 0
i relation holds uniformly over every finite interval. PROOF. We have
h(x, t)
cit.
irther, from the bound !w(x,t,s)I
x+t
Jw(x, t,
I h(x,
.
x—t
x—t 1
1
denote an arbitrary positive number and put 1
t)
dt=11+19.
virtue of the bound (6.8) 1
,-,i dt —=L/ll 0
o for any number wifi have
0 we can choose
so small that for all
IIiI :g
we
then take
so large that for all
>
we will
From the proof of the Riemann-Lebesgue lemma it follows that the bound (6.10) uniformly over every finite interval.
318
VII.
OF.A STURM-LIOUVILLE OPERATOR
(6.10)
1121
which is possible by the Riemann-Lebesgue lemma. The relation follows from the bounds (6.9) and (6.10) and the arbitrariness of number LEMMA
6.5. For every fixed x 2
urn —
r
(x, —co).
o
This relation holds uniformly over every finite interval. PROOF. The function k (x, t) (for whose definition see the beginnin of this section) is analytic (and, in particular, differentiable) with respec
to t. Therefore it follows from a well-known theorem of the theory c Fourier series that k (x, t)
(x, v) dv = Tim
urn
Sifl
dt = k (x, 0) = —s (x, —o
o
o
as was to be From Lemmas 6.2, 6.3 and 6.4 follows: THEOREM 6.1 (ON EQUICONVERGENCE). Suppose that the coefficient
is summable over every finite interval and that f(x) E Then we have, uniformly in every finite interval, urn {
f(s) 01 (x, s;
ds
It
/ (s)
=
SIfl X
S
ds }
/(s)0(x, s; —co)ds,
i.e. the difference between an expansion with respect to the eigenfunctid,
of a Sturm-Liouville operator and the ordinary Fourier integral teru to zero uniformly in every finite interval. That (6.11) holds uniformly in every finite interval follows from the boundedne of the function ôk(x t)/ot
319
§7. CONVERGENCE AND SUMMABILITY
§7. Convergence and summabiity of an eigenfunction expansion
am the theorem on equiconvergence, proved in §6, there immediately
ws a theorem on the convergence of an expansion in the eigenof a Sturm-Liouville operator. 7.1. If the coefficient q(x) is summable over every finite interval a, x,), then at every point at which f(x) satisfies local (x) E
ions for its expandibility in an ordinary Fourier integral, we have urn A-.,.
5
s, —X)]f(s)ds=f(x).
[O(x, s,
—a)
this section we will study the Riesz summability of the generalized
er integral.
virtue of (6.3) and the bound (6.5), it follows from the Tauberian
4.1 of Chapter 14 that (O<s
1),
the constant C depends upon the interval in which x varies but and the function R*(X,v) is defined by (6.4). to study the question of the Riesz summability of an exi.e. to estimate FL
r(s±1) rave
5
(i
to estimate the integrals F'-
'd)dv,
us start with the first integral. We have
ii
J
'1
\ 0
h(x, t)
8+
2
dt.
320
VU. EXPANSION OF A STURM-LIOUVILLE OPERATOR
Suppose that the bound (1.20) of Chapter 6 holds. Then
jf(s) ds <
h (x, t)
•
C
/
+
at
1
dt}<
',"
+*.
Further, we have k(x,t) =k(x,0) +0(t) = —S(x, 1
1.,
=
2
k (x, t)
p dt
8+1
I
'-1
cx) +0(t). Therefori
i
(st)
I
—
—
I
+ 0 (t) J
8+1
dt
1(t)
(7.2) 2
2
S(x,
j
8+1
I
r(s+1) S(x, From the estimates (7.1) and (7.2) follows:
THEOREM 7.2. Suppose that the coefficient q(x) is summable over ever finite interval, and suppose that condition (1.19) of Chapter 6 is ), then we have, uniformly in every finite interva If f(x) E 2( —
V)=p( where
S(x,
min(s, a + D.
§8. Asymptotic behaviour of the spectral function of an equation of order 2n
We will consider a formally symmetric differential expression defined on the real line or the halfline, of the form
§8. EQUATION OF ORDER 2n
1(y) =
+ ...
(—
321
+po(x)y.
assume the coefficients Pk (x) to be real. L be one of its selfadjoint extensions, which exist since the co-
cients are real and therefore the deficiency indices are equal. The family EA of the operator L is in this case a family of integral rs with kernel O(x, E; A), which is called the spectral function. ;her, equivalent definition of the spectral function can be given. A) be the solutions of the equation l(y) =Ay ich satisfy the conditions
(j,s=1,2,...,2n).
=6js
for the case of the entire line the spectral function O(x, 0 (x,
A2)
— 0 (x,
A)
is
A1) = 0 (x,
j,k=1 re cjk(A)
is the
matrix introduced in §16, Chapter 1, i.e. the spectral
of the operator L. the operator L is defined on the nalfline [0, ce), then
u,(x,
O(x,
A)
da,k (A),
j, k=1
A) are certain solutions of equation (8.1) which satisfy the tl conditions =
(j,s = 1,2, . . ., n).
proof of the equivalence of these two definitions of the spectral O(x, E; A) can be found in the monograph by Naimark [1]. from the representations (8.3) and (8.3') that tYJk(A)
=D
= 0j—1,k—1 (0,0; I
A)
is the question of the asymptotic behaviour of the spectral matrix )
can be reduced to the question of finding the asymptotic
of the spectral function 8(x, E; A) and its derivatives. now formulate a result concerning the asymptotic behaviour of *ctral function of the operator L on the entire line and the haifline.
OPERATOR
VII. EXPANSIONOF.A
322
<x
THEOREM 8.1. Let the operator L be defined on the entire line — and be semibounded, and suppose that its coefficients pk(x)
that is a piecewise are locally 8ummable
smooth function, while
such
the remaining coefficients
plane, and as A —* Then in every bounded region of the (x, we have for the spectral function O(x, A) of any semibounded selfadjoint extension i e for any O(x, A) generated by an arbitrary spectral matrix = 0, A
(8.4)
+0(1),
where Oo(x,
A)
is
on
the spectral function of the operator (—
the
entire line, and is given by sin
If the operator L is defined on the halfline [0, cx) with the boundary conditions (8.5)
B1
(0)
=
(j = 1, 2,
.
.
., n),
then a theorem analogous to that just stated holds, if the conditions satisfy certain specific conditions. Namely, let be the roots of the equation upper halfplane and let
Let us denote by Bf (r, A) =
b(3
— A =0
lying in th
the remainder resulting
We require that the division of by 1. 0 (condition of solvability). 2 The elements c3, of the mverse of satisfy the inequa1ity where rj is the highest order of differentiation Al the jth boundary condition. We then have the following result: II
THEOREM 8.2. For the spectral function O(x, E; A) of any selfadjoint extension of the operator L on the halfline [0, ), if the conditioni
I a,id
2
are satisfied, we have the following asymptotic formula as A O(x,E;A)
uniformly with respect to x and
=Oo(x,E;A) +0(1), in any bounded region Here Oo(x
function of the operator (— with the same boundary conditions (8.5). is the
on the halfline
[0,
§8. EQUATION OF ORDER 2n
323
the case n = 1 Theorems 8.1 and 8.2 imply Theorems 3.1 and - ectively, of the present chapter. r the case n> 1 the scheme for deriving the asymptotic formulas ) and (8.6), based on a consideration of the Cauchy problem for the equation, cannot be carried through here. This is connected with fact that the equation ô2u/9t2 = Lu is hyperbolic only for the case and this hyperbolicity is the determming factor in the consideration operator L with arbitrary (increasing) coefficients. It is also not to show that for n> 1 there does not exist a polynomial P(r, A) in two variables r and A such that P(i9/at, L) u = 0 is a hyperbolic ion.
wifi now indicate the scheme for proving Theorem 8.1.
first studies the spectral function ê coefficients have compact support,
of an operator L
(x,
A)
i.e.
vanish outside some
interval. The asymptotic behaviour of
A) can be established
a direct way. Namely, let us consider the equation for the kernel A) of the resolvent operator of L:
K(x, A) is the kernel of the resolvent of the operator (— entire line, and P(x, d/dx) is the operator containing all lower s of the operator L (its coefficients have compact support), i.e.
K0(x,
e
= i.— (—
not hard to show that for large Aj equation (8.7) can be solved method of successive approximations and that as A —* uniformly with respect to (x, E),
-
one
=Ko(x,E;A) +o(1)/IAI.
egrating the left and right sides of (8.8) along a closed contour 1',, complex plane A = + jr which passes through the point on axis, we obtain as Oo(x,E;n) —Oo(x,E;n) +o(1), .....e
00(x,
is
the spectral function of the operator (—
suppose now that we have to establish (8.4) in the region S (a
x b + h with a—h x 0 and consider the "reduced" operator L, whose coefficients coincide the coefficients pi(x) of the operator L in the region u, and vanish b1)
>
We choose some region u ={ x
STURM-LIOUVILLE OPERATOR
VII.
324
outside Let G0(x,
—h—c
t) denote the Green's function of the Cauchy problem for the parabolic equation öu/öt = — Lu. t) is the kernel of the integral operator i.e. thel Green's function of the Cauchy problem for the operator equation au/at— —Lu, then as can be seen by means of Green's formula, wel
the region S G (x,
(8.10)
t) —
(x,
t) = 0(1)
where c is a constant which does not depend upon On the other hand, as is not difficult to see, G(x,
or t.
h
t)=5
Taking account of this representation, (8.10) can be rewritten in the form (8.11)
(x, is
X)
the spectral function of the "reduced" operator L.
After multiplying the left and right sides of (8.11) by with respect to t from 0 to we obtain
X)ç
çdO (x,
1
3
I
I
and integrating
1 + 0 ()
is a constant which depends only upon c, and h* = Using this last relation, we can establish the following useful where
(x, E; A) =
(8.12)
(x,
A) + G
where G(s) is analytic with respect to s = + jr in the strip
I
and G(0) =0. Putting p(X) =O(x,x; we obtain
o(A)
(A)
(8 13)
(8,9)
on
(A) + G (s)
find that
A2n)=!X+o(1).
the basis of (8.12
§8. EQUATION OF ORDER 2n
325
on the basis of
Tauberian theorem Chapter 14), ,hich makes possible a sharper estimate of the remainder in Ikehara's theorem, we obtain from (8.13)
—
(A+O)—
(X)
P
2
an absolute constant Since the number h can be chosen arbitrarily large, and h, along pith h, is arbitrarily large, taking into account (8.7) and the fact that =0, we obtain the assertion of Theorem 8.1 for the case = x. this, the proof for the case x is carried out by considerakm of the nondecreasing function there A
is
=O(x,x;x) +26(x,E;X) 1'he proof of Theorem 8.2 is technically rather complicated, but can e carried out according to the same scheme [I] on estimates for the We remark that applying results of pectral function of general effiptic operators on the negative spectrum, can also establish Theorems 8 1 and 82 for operators which are Ot semibounded. At least for operators with sufficiently smooth coftIcients, this can be done rather easily. Namely, one has to consider square of the non-semibounded operator L which is now a positive perator. One further has to use Theorem 8.1 or 8.2.
Theorem 8.1 follows: THEOREM 8.3
(ON EQUIcONVERGENcE). Every function 1(x)
with
mpact support which can be expanded in a Fourier integral can also panded as an integral with respect to eigenfunctions, and conversely,
be
We are concerned here with the pointwise convergence of the expansions.
Fhis theorem follows imthediately from Theorem 8.1, since the partial
ins of an eigenfunction expansion and a Fourier integral expansion equal respectively to b
S(x, f;
0(x,
X)
d- consequently
S(x,
(
/; X)__j a
. S'11
(x,
X) /
326
VII. EXPANSION GFASTURM-LIOUVILLE OPERATOR
By this
one. can. also obtain the corresponding asymptotic (x, A), which enabled D/ 0 (x X) = one to write down the asymptotic behaviour of the spectral matrix: behaviour of
derivatives
Bibliographical references
§1. The basic results concerning eigenfunction expansions are pre-j sented in the monographs by Titchmarsh [1], Levitan [1] and The identities (1.12) and (1.14) were obtained in a paper by Levitan [14])
*2. Lemmas 2.1 and 2.2 for the case of the halfline [0, x') are part due to [1,2]. The proof of Lemma 2.1 here is due to Levitan [14]. §3. The asymptotic formula (3.8) was obtained in papers by [14, 15]. Cf. also
[3].
§4. Formula (4.5) was derived in a paper by Levitan [15]; a paper by [3]. §5. The Riesz summabiity of the spectral function was in papers by Levitan [14,15]. 7. The theorem on equiconvergence was first proved by
cf.
[14]; cf. also his paper [15] and [1,2]. The Riesz of an eigenfunction expansion was considered by Levitan [14, 15].
§8 The results of this section are due to
[1,2]
CHAPTER 8
DIFFERENTIATION OF AN EIGENFUNCTION EXPANSION §1. Preliminary estimates of the derivatives of the spectral function
In this chapter we will study the question of the differentiation s expansion with respect to the eigenfunctions of the selfadjoint sd-order differential operator
y" +[x —q(x)]y=0, med on the halfline [0, with the initial
conditions
y'(O) =0.
y(0) =1,
that q (x) is a real function defined summable over every finite interval.
assume
on the halfline
[0,
hroughout the chapter we will assume that the spectrum of the (1.1) —(1.2) is nonnegative or, if a negative spectrum exists, it is bounded from below. The latter case can easily be reduced former. Thus, let A the solution (A> 0). We denote by problem (1.1) —(1.2). s is known, there exists a monotonic function p (,z) which is bounded
finite interval, such that for every function f(x) E has the Parseval equality
every
co)
(FL),
FL)ds.
;t us denote by O(x, s,
the spectral function of equation (1.1) (for
initial data (1.2)), i.e. we put v)p(s, v)dp(v),
O(x, s, p.)= 0,
327
328
VIII. DIFFERENTIATION OF AN EIGENFUNCTION EXPANSION
In this section we will obtain various preliminary estimates for the derivatives of the spectral function O(x, s, As we have seen in the preceding chapter, for arbitrary real t we have I
(1.5)
x+t
t, s)cp(s,
Differentiating (1.5) with respect to x, we obtain
t, (x,
t, x — I
Suppose that the function
x+i
(x
ow(x, t,
t, u.)) s)
p (s,
ds.
satisfies the conditions indicated on and then integrate with
p. 296. If we multiply both sides of (1.6) by respect to t from 0 to we obtain
p.')}g8(t)dt
+
(w (x,
t, x + t) p (x + t,
(1.7)
— w (x,
t, x — t) p (x — t,
(t) dt
Let us transform the right side of (1.7). Consider the first term. Integrating by parts, we find that
=—p(x,
fl. PRELIMINARY E&FIMATES
329
p(x—t, using the evenness of g, (t) with respect to t, we obtain
Ii = — p (x + t,
(t) dt + p (x — t, x)
(s,
—
gç (t) dt
x)
(s,
ds
x
ag (s—x)
ds
We now transform the second term of (1.7). By virtue of (2.9) and 2.10) of Chapter 6 we obtain
t, x+t)p(x+t,
x+B
—4
p (s,
g1 (s— x) {
= —4 —5
w(x,
{g1
q (x+
(s — x)
q (jt)
p (s,
t, x—t)p(x—t,
=45p(x_t,
=4
5
p(s,
5
dt} p(s,
ds
330
VIII. DIFFERENTIATION OF AN EIGENFUNC'rION EXPANSION
Consequently, (1.9)
p(s,
Finally, changing the order of integration in the last integral in (1.7), obtain
we
=
x+s p (s,
X,
(x, s) ds,
where
(x,i, s) dt.
(x, s)
(1.11)
By virtue of (1.8), (1.9) and (1.10), the relation (1.7) assumes the form (x,
(1 12)
= —4 (s — x)
q
p (s,
ds
p.)
p (s,
p.) ds + 4
i.(x, s) p (s
s)—
ts
p)ds
It follows from (1.12) that for fixed x the function be regarded as the generalized Fourier transform (with respect to eigenfunctions (x, of the function which is equal to (113)
s)—
—
s E (x and is equal .to zero outside this interval. Therefore by the Parseval equality we obtain from (1.12)
(1.13')
(p.)
(x,
p.)
(p.) dp (p.)
=
/ (s)
(x, s) ds.
Hence, by the arbitrariness of f(s) and by (1.13) we obtain •(p.)p(s,
p.)dp(p.)=
331
§1. PRELIMINARY ESTIMATES
(x, s) —
(s— x)
x)
(s
J 1
J Obtain
q (t) 6,
X—
Ix—s(>s.
0,
the identity (1.14) with respect to s 1) and using (1.11),
for x> s
ow (z, t, s) x
+
O2g, (s— x)
s)
g, (t)
=1 x —sJ
IX—St>E.
0,
the definition of the spectral function O(x, S.
Le. (1.4), the identity
assumes the form
s)1
+ f
0,
(t)
02:,t, r
—
(s— x)
I
q(s),
Jx—sI>s.
assume here that the function q(x) is absolutely continuous.
332
VIII. DIFFERENTIATION OF AN EIGENFUNCTION EXPANSION
Extending the function O(x, s, in the variable odd function, for a = x we obtarn from (1 15) d
ö20 (x
S
p.)
— g2 t s)
(i., t
( )
so as to obtain an
s)
(x)
J
2 We will now prove some simple lemmas concerning the behaviour of the derivatives of the eigenfunctions LEMMA 1 1
If the function q(x) is absolutely continuous in every finite
interval, then as a —* {a20
(1 17)
have the following estimate
we P.)}
=
(x,
dp (it)
0(a2)
5
This estimate holds uniformly as x varies in any finite interval PROOF From well-known asymptotic formulas for the eigenfunctions it follows that for x in a finite interval and —*
(118)
Hence (1 17) follows from Lemma 2 3 of Chapter 7 One can prove the following lemma in a similar way LEMMA 1 2 If the function q(x) has a derivative of order 2k which is summable over every finite interval then for every fixed x we have as a
a',
a-i-i
s,
(1.19)
a
=5
t.
a
(k=1,2,...). This estimate holds uniformly as x varies over any finite interval.
From Lemma 2.3 of Chapter 7, Lemma 1.1 and the CauchyBunjakovskii inequality follows: LEMMA 1.3. If q(x) has a first derivative which is summable over every finite interval then for every fixed x and s we have as a—* co,
(120)
P1=
a-4--i
§2. ASYMPTOTICS
333
estimate holds uniformly as x and s vary over any finite intervals.
s
!rom Lemma 1.2 and the
Cauchy-Bunjakovskii inequality follows:
LEMMA 1.4. Under the conditions of Lemma 1.2 we have the estimate ct+1
y
A)
{a2k6 (x, S,
(z, " i
(s,
0
I
(1+1=2k). s estimate holds uniformly as x and s mzzy over any finite intervals.
t
denote the spectral function of the problem (1.1) —(1.2)
q(x)
0, i.e. we put s,
the definition of O*(x, 1.5. As
we have the estimate
(k+f=n=O, 1,
àxôsi
a
s
there immediately follows:
.
estimate holds uniformly over the entire halfline (0, ce.), §2. Asymptotic behaviour of the derivatives of the spectral function
ormula (1.14) enables us to study the asymptotic behaviour of the of the spectral functions 8(x,s,,z). In order to study the •;otic behaviour of the derivatives of the spectral function of the oble a (1.1) + (1.2), we wifi compare them with the derivatives of function of a simpler problem, namely the problem
y"+Xy=O y(0) =1,
this end we will transform the
=
X8
0,
—
(A>0), y'(O)
=0.
right side of (1.14). We have
x)
334
VIII. DIFFERENTIATION OF ANEIGENFUNCTION EXPANSION
we can give (2.3) the form
Using the spectral function
(2.3') —x)
Jx — x
0,
i.e. for the spectral function1 Writing down for the function of the problem (2.1) + (2.2), a formula analogous to (2.3'), subtracting1 it from (2.3') and taking into account that for the problem (2.1) + one has w(x, t, s) 0, we obtain {ôO
ôO*(x,s,
(x,s,
} (2.4)
= Ix—sI>e,
0, where
(2.5)
(x,
s) =
(t)
5
ow (x,t, s) dt.
Put Ow(x,t,
(2.6)
5 tx—al
Then by the Parseval equality for the ordinary Fourier integral it
from (1.6), Chapter 7, and (2.6) that (2.7)
5
0
(x, s,
=
Ow
(x,t, s)
gg (t) dt.
5
tx—il
Further, by virtue of the inversion formula we find from (1.6), ChaptE
7,that (2.8)
' 'd
335
§2. ASYMPTOTICS
since the functions and a(x,s,,2) are even with respect it follows from (2.7) and (2.8) that (2.4) can be rewritten in the wing form: s,
}=O. x
2.1. If q(x) is absolutely continuous in every finite interval, for every fixed x and s we have (p cx) s, ax
S.
J
0
asymptotic relation holds uniformly as x and s vary over any finite ervals.
IL
V
s,
v) dv = (i')'
V IL
0
a by (1.20) and (2.9) the Tauberian Theorem 4.1 of Chapter 14 is ?, from which it follows that for every fixed x and s the Riesz hans of order 1 of the function ôO*(x,s,
dO(x,s,
s, v)dv
cosv(s—x)dv i.e. we have
nain bounded as IL
j\
j.L2
d,
{ao
(x,:,
ôÜ*
s
.v) _!
s, a)da
0
cos a (s —x) da}
= 0(1).
the theorem it remains to compute the integrals (1
s, v)dv,
336
EIGENFUNCTION EXPANSION
VIII. 8
'2=
q
(i
(t) d—
x
cos
—
0
(s
— x)
Let us compute the first one. By the definition of the function.i a(X,S,v), i.e. (2.6), we have
= ni\/ I ——- is
ow (x, t, s) cos i ———— Ox r
ç
Changing
J
dt
the order of integration, we obtain
(2.12)
=
s)
Ow
(i {0
IX—81
cos vt dv}
—
dt.
Further, by a well-known theorem (cf. Titchmarsh [3], §7.1) (2.13)
cos vt dv
—
=
172
From the last two formulas we finally obtain
'2 14'
Ow (x, t, s)
I— —
dt
Ox
3
For the second integral we obtain, by (2.13), j
8 18
(s —
(2.15) the theorem follows from (2.11). Similar asymptotic formulas can be obtained for the higher-ordei derivatives of the spectral function O(x,s,ii). However we will not dc this here, restricting ourselves to stating a theorem for the second
derivatives.
If q(x) has a first derivative which is absolutely in every finite interval, then for every fixed x and s we have the asymptotic formula THEOREM 2 2
v)
(216)
O2Q*(x: v)]O(1)
Ths formula holds uniformly as x and s vary over any finite intervals
§3. EQUISUMMABILITY
337
§3. Equisummability of differentiated eigenfunction expansions
this section, using the preliminary estimates obtained in §1 for derivatives of the spectral function O(x, s, hz), we will prove a theorem equisummability of differentiated expansions in the eigenfunctions
Sturm-Liouville operator and in an ordinary Fourier cosine integral Linctions in
will first prove some lemmas. formula (1.13') follows (p.)
(x, p.) dp (p.) =
F (p.)
f(s)
(x, s) ds,
and F (p.)
1
.
i. m f(s)
(s, p.) ds,
A-+co
ow (x,t, s)
(x, s)
g1
(t) di, +
0g. (s— x)
Ix—sf a
—4g1(s—x)
introduce the notation S(x, s, p.)ds.
the definition of the function S(x,,z), (3.1) assumes the form (p.) d
OS
=
s)
(s) {
(t) dt
Ix—'I
+
g. (s — x)
q(t) dt} ds.
function S (x, is a segment of the expansion in the eigenfunctions a Sturm-Liouville operator, and S*(x, is a segment of the expansion ordinary Fourier cosine integral, of the function f(x) E 2(0, cx>). (x, riting for a formula analogous to (3.6), taking into account for this case q(x) 0, w(x,t,s) 0, and subtracting it from (3.6), obtain -
338
VIII. DiFFERENTIATION OF AN EIGENFUNCTION EXPANSION
fôS(x,
r
j
ax
üx
37)
=
f(s)
g5 (t) dt —
g1 (s — x)
q
ds
2(0 If q(x) is absolutely continuous ir LZMMA 3.1. Let f(x) E the every finite interval, then for each fixed x we have, as p.+i
(3.8) This
estimate holds uniformly as x varies in any finite interval.
i.e. by (3.4) we have
PROOF. By the definition of II.
p.)F(p.)dp(p.)
Therefore from the Cauchy Bunjakovsku inequality it follows {OS (x,
}=
I
(x, p.) F (p.) I dp (p.)
( . )
'/,
p.+i
F2 (p.) dp
(x, p.)]2 dp (p))
(p))
By Lemma 1.1 we have (x, p.)]2 dp (p.) = 0
(3 10)
Further, since J'2.
dp (ii) <
(p.2)
as u —÷
we
have
F2(p.)dp(p.)=o(1).
(3.11)
Therefore the lemma follows from the inequality (3.9) by virtue the estimates (3 10) and (3 11) One can
prove:
2(0 aD) If q(x) has a kth derivative LEMMA 3 2 Let f(x) E is sumrnable over every finite interval, then for every fixed x we have
the estimate
J
-!
339
§3. EQUISUMMABILITY
(k=1, 2, ...). estimate holds uniformly as x varies over every finite interval. the definition of S*(x, IL) there immediately follows:
3.3. Let f(s) E IL+1
V
Then as IL—*
akS* 1x
(k=t, 2,
IC
.
..).
estimate holds uniformly over the halfline [0, cx).
3.1. Let f(x) E 2(0, co). If q(x) is absolutely continuous wery finite interval, then we have, uniformly in every finite interval, rztion
urn the
J\(i
—
(z,
(x,
d
j
difference between the Riesz means of order one of the first derivatives
a Sturmexpansion of f(x) E 2(0, iwzlle operator and of the expansion of f(x) in an ordinary Fourier integral tends to zero uniformly in every finite interval. RooF. We put x+t s)
h(x,t) = a(x,IL)
/(s)ds,
h(x,
t)
Parseval equality for ordinary Fourier integrals we obtain from
—(3.16) and (3.17) —(3.18) respectively
340
VIII. DIFFERENTIATION OF AN EIGENFUNCTION EXPANSION
t)dt 0
(x,t, s)
(3.19)
I
{I
/ (s) ds} dt
àw(x,t, s)
t)dt t)
(x
Therefore by virtue of (3.19) and written in the form
(t) dt} dt
the identity (3.7) can be
(3.20)
—L
v)dv 0
(3.21)
For
the validity of the following estimates is obvious:
(3.22)
V
p.+1
IL
(x, v) dv
o (1), V IL
o
P (x, v) dv
= o (1).
0
by (3.8), (3.13), (3.22) and (3.21) the Tauberian Theon applicable, on the basis of which the 4.1 of Chapter 14 Riesz mean of the function Therefore
ÔS (x
__ÔS*(x_t.L)_L
{a(x, v) — (x, v)) civ [3
in every finite interval, i e
341
§3. EQUISUMMABILITY
ç (i
aS
d,
—
aS*(:,
v)
—
(x,
v)
— (x,
v)1
dv}
{
IL
urn c
(i
—d faS'(x, v)
ÔS*(x, v)
ax
U
IL
v2\
I =-Iim
v)} dv.
p.)
, prove the theorem it suffices to show that 'v)—fI(x, v)}dv==O
urn
uniformly in every finite interval. By the definition of the funca a(x,L), i.e. (3.15), we have
)cosvtdt]dv.
v)dv=S(i Lnging the order of integration, we obtain
ç(i nce by hypothesis q(x) is absolutely continuous in every finite val, it is bounded in every finite interval. It then follows from ma 1.2, Chapter 7, that the function aw(x,t,s)/ax is bounded in finite interval, so that from (3.14) we find that x+t
h(x,
which by the Cauchy-Bunjakovskii inequality we obtain 1/,
x+t Ih
(x, t)
/2(s) ds)
x-I-t
'I,
ds)
f(s) is square-integrable. Then from (3.25), (3.26) and (2.14)
342
VIII. DIFFERENTIATION OF AN EIGENFUNCTION EXPANSION
there follows for p
the estimate
<
'v)
(3.27)
=
13/(U)
ç
du
=0
Further, arguing analogously, we obtain from (3.15), (3.16) and
ç(i
v)dv
=
t (x, t)
(3.28)
{T q
=0
estimates (3 27) and (3 28) prove (3 24), and thus the theorem, Starting from Lemmas 3.2 and 3.3, we can prove in a completely
The
analogous way: ccx). If q(x) has a derivative of THEOREM 3.2. Let f(x) E 1 which is absolutely continuous in every finite interval, then we have
k—
uniformly in every finite interval, urn i.e.
ç(i
1
I
the difference between the Riesz means of order k of the kth
of the expansion of f(x) with respect to the eigenfunctions of a Sturm -Liouvil
operator and as an ordinary Fourier cosine integral tends to zero in every finite interval. §4. Summation of differentiated ordinary and generalized Fourier integral expansions
From the theorem on the equisummabthty of a differentiated e pansion in the eigenfunctions of a Sturm Liouville operator and Øpansion as an ordinary Fourier cosine mtegral, which was proved the preceding section, it is easy to obtain theorems concerning ti expansions in the eigenfunctions of a Liouville cperator to the values of the corresponding derivatives of ti fiincbmi Which is being expanded, if one has such theorems for t]
343
§4. FOURIER INTEGRAL EXPANSIONS
of the derivatives of the ordinary Fourier integral. Concerning
summation of the first derivative of an ordinary Fourier series, e is the well-known theorem of Fejer-Lebesgue (cf., for example, [1]). In this section, using our method, we will prove a analogous to the Fejér-Lebesgue theorem. REOREM 4.1. Let F{ f(x) denote the ordinary Fourier series of the f(x). The series F' f(x) is summable at the point x0 by the firstr Riesz method to the value f' (x0) if the following conditions are s/led at the point x0. f(x) is absolutely continuous in some neighborhood of x0. f' (x) is continuous at x0. PROOF. Suppose that f(x) is absolutely continuous in the interval —&
xo + o) (o >
0). Further, let
be a sequence of smooth
ctions, defined on the interval (0, 2ir), which converges to f(x) in 1(0 2ir). By virtue of the smoothness of metric of we have
(n=t, 2, .e
and are the Fourier coefficients of converges uniformly.
and the series
t be an arbitrary real number satisfying ItI <&/2. From (4.1) -ows
coskt
(n=l,2 iplying both sides of (4.2) by g, (t) and integrating with respect to 0 to by (3.18) we obtain sin kx}
(n=1, 2, us
introduce the notation sin/cr) (4.3)
assumes the form
(n==1, 2,
.
.
.
. .).
344
VIII.
OF AN EIGENFUNCTION EXPANSION
0
(n=1,2,
(44)
to f(x) allows us to pass The convergence of the sequence { the limit under the integral sign in the right side of (4.4), and the bourn
(1.7) allows us to pass to the limit in the left side of (4.4). As a consequence we obtain
=ç{f(x+ t) +/(x—t))g1(t)dt,
(4.5)
where
S(x,
are the Fourier coefficients of f(x). ak and Differentiating (4.5) with respect to x for x E (x0 —
x0 + 5/2)
we find that (4.6)
Put
=
(x,
(4.7)
(x
+ t) +
(x — t)} cos
dt.
By the Parseval equality for the ordinary Fourier integral, it follo%i from (4.6) and (4.7) that (4.8)
Since
(x
+ t) + 4
are
(x — t)}
(t) dt = 4 a (x,
even with respect to
it follows from
and (4.8) that (49)
Jurther fqllows from the definition of S(x, 4 we have the estimates
and a (x
4
that
345
§4. FOURIER INTEGRAL EXPANSIONS
m (4.9), (4.10) and the Tauberian Theorem 4.1, Chapter 14, it ws that the Riesz mean of order one of the function 1
t
ox 0
to zero uniformly in the interval (x0 — o/2, x0 + o/2), i.e.
t follows from (4.11) that to prove the theorem it suffices to prove validity of the following
0
the definition of the function
we have
the order of integration, we obtain
{ç(i from (2.14) and (4.13) follows dt.
transform the integral in the right side of (4.14) as follows:
I=
+ t) +
+
t) — 2/' (xe)]
(tLt)
2
346
VIII. DIFFERENTIATION OF AN EIGENFUNCTION EXPANSION
By a change of variables in the integrals in the right side of (4.15) can transform it to the form 1
=
[ix
(4,16)
(; +
2/' (xe)]
+ ix (;
dz
—
+
2
dz.
1' (xe)
now wish to estimate the first integral in the right side of (4 Let ii denote an arbitrary positive number and put
(; + (417)
— 2/' (xo)]
+/
/,
(z) 3/
+
=
dz
— 2/' (xo)]
dz ='l+ '2 f' (x) is continuous at x0, for any ô> 0 and fixed so large that for all > we will have
Since
we can cho
(4 18)
I'll
On the other hand, choosing
sufficiently large, we obtain the inequali
(419)
1121
From (4.18), (4.19) and the arbitrariness of a
we
obtain, by vir
of (4.17), (420)
ix
(; +
+ ix (; — — 2/'
Ja(Z) (xe)
dz
=0
passmg to the limit in (4 16), we obtain by (4 20) (4.21)
Jim 1=2
1' (xe)
dz
by a well-known formula (cf Watson [1], §13 24) 7J3,(z) 3
0
—
i
347
§5. CONVERGENCE
together with (4.21) proves (4.12) and thus the theorem. By iuivalence of the first-order Cesàro and Riesz summation methods, of Fejér-Lebesgue follows from the theorem just proved. not difficult to see from the proof that the preceding theorem is true Fourier integrals under the same assumptions, i.e. we have: 4.2. Suppose that the following conditions are satisfied: x). f(x) E 1(x) is absolutely continuous in some neighborhood of x0.
f'(x) is continuous at x0. the first derivative of the expansion of f(x) as an ordinary Fourier is summable at the point x0 by the Riesz method to the e f'(xo).
a Theorems 3.1 and 4.2 follows: 4.3. Let q(x) be absolutely continuous in every finite interval, I suppose that the following conditions are satisfied:
.f(x) E
CX>).
1(x) is absolutely continuous in some neighborhood of
f'(x) is continuous at x0 Ilin
'
dS(x,v)
0'
first derivative of the expansion of f(x) with respect to the eigenions of the Sturm-Liouville operator is summable at the point x0 by rst-order Riesz method to the value f'(x0).
ar theorem holds for the higher-order derivatives. §5. Convergence of a differentiated eigenfunction expansion
5.1. Suppose that q(x) is summable over every finite interval, wppose that f(x) satisfies the following conditions for x 0:
f(x) and
belong to
..J)cosa+f'(O)sina=O. W{ (x, A), f(x) } = 0 for all A from the spectrum.
the eigenf unction expansion of f(x) converges absolutely.
OF. From the hypotheses of the theorem and from Green's follows
=
1
1(x)
p (x,
X) dx =
VIII DIFFERENVATION OF AN EIGENFUNCTION EXPANSION
348
i .m
—
{p (x, A)) dx
1(x)
=—
0
(A) =1.i• m 2 {/ (x)) p (x,
A) dx.
0
From Lemma 2.1, Chapter 7 and the Cauchy-Bunjakovskii inequali
it follows that
Let us consider the integral JF(A)fJp(x, A)Jdp(A).
If we replace A by F(A) this
and put p(A)
F1(u),
integral assumes the form
It further follows from (5.1), Lemma 2.3
of
Chapter 7
and
the
Bunjakovskir inequality that a+l F1
I p1
(x,
dp1
(FL)
J
a+1
where
(z, x, a +1) —
dp1 (p))
( the constant C
does
01
(x,
x, a))
4,
not depend upon a Therefore from 1
p4jdp1(p.)< o
o
a—
349
§6. JUSTIFICATION OF FOURIER'S METHOD
together with (5.2) proves the theorem. purpose of this section is the proof of the following theorem. i 5.2. Suppose that f(x) satisfies the conditions of the previous em and, moreover, that the function g(x) =f"(x) — q(x)f(x) can be nded with respect to the eigenfunctions ce'(x, A) at the point x0
f'(x0) = urn
0.2) Then
(x0, A) dp (A).
F (A)
By hypothesis tim
(A) p (x0, A) dp (A)
F (A)
—urn
AF (A) p (x0, A) dp (X)
(x0, A) dp (A)_q (x0) tim
F (A) p (x0, A) dp (A).
—m e
g(x0) ==f"(x0) —q(x0)f(x0) and urn
F (A) p (x0, A) dp (A)
/(x0),
(5.4) follows from (5.5). §6. Justification of Founer's method for the one-dimensional wave equation
section we will show that the results of §4 of the preceding r, in which we investigated the solution of a mixed problem for Due-dimensional wave equation on a finite interval, together with rem 5.2, enable us to justify Fourier's method for the one-dimensional
equation under weaker conditions on the initial function than ones.
will consider the following mixed problem on an interval: ô2u/ôt2 =ô2u/ôx2 —q(x)u,
ulto =f(x), a follows from Theorem 7.1, Chapter 7, for this it is sufficient that the function satisfy any local conditions which suffice for it to be expandable in a trigonometric er series. Since, by hypothesis, f(x) has a second derivative for x 0, it is for
le sufficient that f" (x) and q(x) be of bounded variation in a neighborhood of hit Xe•
350
VIII. DIFFERENTIATION OF AN EIGENFUNCTION EXPANSION
=0,
(6.3) (6.4)
[ôu/Ox
(6.5)
[Ou/äx
=0, = 0.
(cf. §4, Chapter 6), if f(x) has a continuous derivative and satisfies the conditions f'(+O) —hf(+0) =0, f'(7r —0) +Hf(ir —0) =0, (6.7) then the mixed problem (6.1) —(6.5) has a solution for all t> 0. Let ui denote this solution by u0(x, t). For fixed t this solution satisfies boundary conditions and has a continuous second derivative. by a classical theorem u0(x, t) can for any fixed t be expanded in a1 absolutely and uniformly convergent series with respect to the functions of the Sturin-Liouville operator (6.8) (6.9)
y'(lr) +Hy(7r)
y'(O) —hy(0) =0,
0,
i.e. (6.10)
(x,
c,,
"=1
(t)
p,, (x),
where
(6.11)
(t)
=
u0
(x, t)
(x) dx,
are the eigenfunctions of the problem (6.8) + (6.9). and we will first determme To this end we use the initial conditions (6.2) —(6.3). Putting t =0! (6.11), by virtue of (6.2) we obtain
and the
determine the coefficients
(6.12)
(0) = 1(x)
(x) dx
=
Differentiating (6.11) with respect to t and then putting t = 0, by
(6 11) twice with respect to t, by (6 1) we obtal
351
§6. JUSTIFICATION OF FOURIER'S METHOD
c (t)
ô2u0(x,
t)
(x) dx =
—
q
(x) u0 (x,
(x) dx.
utegrating the last integral by parts twice and using the boundary nditions (6.4) —(6.5) and (6.9), we obtain
(t) =
(x, t)
(x)} dx,
(x) — q (x)
which by (6.8) follows (t)
=
u0 (x, t)
(x) dx
=
(t).
by virtue of (6.12), (6.13) and (6.14) we obtain the following auchy problem for the determination of
=0,
=0.
c,,(0)
solution of this problem has the form Inserting this value of
in (6.10), we obtain
u0(x,
the function f" (x) satisfies a Lipschitz condition with exponent on the the interval then it follows from (4.5), Chapter 6, Theorem that the series (6.10') can be differentiated twice respect to x, i.e. ô2u
0=
cos
tp (x)
the series (6.16) converges uniformly. will prove that the series obtained from (6.10') by differentiating e with respect to t converges absolutely. In fact, put ieorem 5.2 considers the halfline [0, ). However it is easily seen that this theorem .e proved by the same methods for the case of a regular Sturm-Liouville problem. as is evident from the proof, for the regular case Theorem 5.2 follows from a theorem on the equiconvergence of the expansion of a function in a Sturmuville series and the expansion of this function in a Fourier cosine series (cf. §9, —
Ler 1).
352
VIII.
OF AN EIGENFUNCTION EXPANSION
Integrating the last integral by parts twice, by the conditions (6.6)
—
(6.'3
and (69) we obtain
0
=
(x) dx =
/ (x)
from which (6.17)
Differentiating the series (6 10') with respect to t, we obtain ——
(6 18)
(x)
sin
Further, by virtue of (6.17) and the Cauchy-Bunjakovskii we obtain (619)
n=i
n=1
which follows from the fact that f"(x) —q(x)f(x) E
It follows from the bound (6.19) that the series (6.18)
is
convergent.
of the second derivative Let us now clarify the respect to t. By virtue of the equation (6.8) we have —
t
cos
=
(r) cos m=1
(x)
— q(x) n=1
(x) p,, (x))
(x) cos
the convergence of the series (6 10) and (6 16) follows
of the series
(x=x0)
353
REFERENCES
now consider the case
=f(x). u(x,0) =0, u1(x, t) denote the solution of the problem (6.1) + (6.20) + (6.4) Then
u1(x, t)=çu0(x, now put u1(x, t) =
c1(t)==
t)
dx,
n=O
we obtain for
c(t) +
(t) =
sin
t.
rest of this investigation can be carried out analogously to the investigation. Bibliographical references
2,3,4. The results of these sections are due to Sargsjan [1—4]. Theorem 5.2 is due to the authors and was published in their papers [1—2]. Another proof is given in the joint paper [3] by
authors.
A justification of Fourier's method was given in the paper by itan [17]. Cf. also the joint paper [3] of the authors.
CHAPTER 9 SOLUTION OF THE CAUCHY PROBLEM
FOR A ONE-DIMENSIONAL DIRAC SYSTEM §1. Derivation of formulas for the solution of the Cauchy problem
Let p (x) and r(x) be real functions defined on the entire line summable over every finite and let the vector-function f(x) be
=
real and continuously differentiable. Let u1(x,t) u(x,t)=(I\ u2(x, t)
We will consider the Cauchy problem (1.1)
01
p(x)
(0
(1.2)
0
0
ufr, 0)=/(x).
Let us first consider the problem (1.1) +(1.2) for p(x) =r(x) the problem (1.3)
0,
u°(x, 0)=/(x),
where we have put
f10\
B
/01
0
The problem (1.3) is equivalent to two independent problems i t) of the vector-function u°(x, t) and
volving the components
Further on, depending upon the circumstance, other smoothness conditions be impoeed upon the functions p(x) and r(x). 354
355
§1. FORMULAS FOR THE SOLUTION àx2 —
ot2 1.4)
(x, 0) =
àx2 —
àt2
1.5)
(x, 0\
= ., (xj,
(x),
= —i/
= /2(x),
(z).
J
the problems (1.4) and (1.5) by D 'Alembert's method, we obtain —if2(x—t)},
i4(x,t)
t6)
—t) +f2(x —t) —ifi(x+t) +f2(x+t)I the solution of the problem (1.3) is given by
}.
u0(x t)
Liere
Hr" is the transpose of H. Let us now consider an inhomogeneous problem with homogeneous itial data àü
7)
iZ(x, 0)=0. were
g(x,t)
= differentiable vector-function. It is not hard to see that this problem o breaks up into two independent problems involving the components t) of the vector-function ü(x, t), namely: (x, t) and —
ôt2
(x,
.
àx2 —
Ot
02a2
fl)=0,
ox
= —1g1 (x, 0),
0) = 0, Ox2 —
+
.
0g2
iIg1
Ot
Ox
0),
IX. CAUCHY PROBLEM FORA DIRAC SYSTEM
356
As is known,2> the solution of the problem 32v/ôt2 —ô2v/ôx2 =h(x,t),
v(x,0) =0, is given by
t)=4.
v(x,
x-f-(t—t)
z+t
q('r)d'r
h(s,
d'r 5
5
x—(i—-r)
0
applying this formula to the problems (1.8) and (1.9), find that the solutions of these problems are given respectively by Therefore,
_ig1(x-__t—j—t, 'c)—g2(x—t+r, t)}dt, (x,
t) =
{—g1 (x + t — 't, t) — tg2 (x + t —
t)
+g(x—t+;
t)}d.
Thus the solution of (1.7) is given by (1.10)
II (x, t)
+
= —4 5 {Hg (x + t —
(x — t
+
We now turn to the problem (1.1) + (1.2). We will rewrite it in the forni (1.1') (1.2)
where
t),
u(x, O)=/(x), fp(x)
0
o
r(x)
Let us assume that p (x) and r(x) have continuous derivatives. Putting now g(x, t) = — Q(x)u(x, t), assuming g(x, t) to be and then taking into account that the solution of the problem (1.1' s the sum of the solutions of the problems (1 3) and (1 7), vi4ueof (1 1) and 10) we obtam
*çL
kilibert [1] Chapter IV §2 3 be weakened
357
§1. FORMULAS FOR THE SOLUTION
u(x, t)=4{Hf(x+t)+Hnf(x_t)}
+4
{HQ(x+t—c)u(x+t—'c,
r, changing the variable of integration, u(x,
+4
111)
HQ(s)u(s, x+t—s)ds
s—x+t)ds. Thus, we have constructed for the problem (1.1) + (1.2) an equivalent
rstem of integral equations (1.11). These equations are of Volterra rpe and can therefore be solved by the method of successive approxiations. Let u0(x, t) be defined by u0(x,t)
put x+t (x, .12)
t)
=
i 4-
(s, x + t — s)
BQ (s)
+4-i
ds
s—x+t)ds.
from the uniform convergence of the successive approximations, proof we omit, it follows that the solution of equation (1.11) is yen by .13)
+
u(x,t) —u0(x,t) +ui(x,t) +
We wifi show that each of the vector-functions
t), p =
n be represented in the form u,, (x, .14)
t) =
(x, t) / (x + t) + I,,, (x, t) / (x — t) x4-/ 1 5
t, s)/(s)ds,
1,
2, 3,...,
_____
358
IX. CAUCHY PROBLEM FOR A DIRAC SYSTEM
t) and
t, s) are 2 X 2 matrices. Indeed, for p = 1 it follows from (1.12), by virtue of (1.6), that
where
t),
x+t
(x, t)
=
4- i
+4-is
HQ (s) (HI (x + t) + HT/ (2s — x — t)) ds HTQ
(s) (H/(2s— x + t) + HT! (x—t)} ds,
or, changing variables. t)
=(4-t
x+t
HQ(s)Hds)f(x+t)
+ (4-i
(1.15)
z+t
+4
(x --
HTQ (s)
I
from which (1.14) follows if we put K1 (x,
t) = 4-i
(s) H ds,
L1
(x, t) =4- i
11TQ (s) HTds.
(1.16) H1 (x, t,
Let us now assume that (1.14) has been proved for p = 1,2, . ., q — will prove it for p = q. Putting p = q —1 in (1.14) and inserting t expression for Uq_i(x, t) in (1.12), we obtain
We
Uq (x,
t) =4 i
HQ (s) {Kq_i (s, x + t —s) / (x + t)
.x+t—s)/(2s—--x—t)}ds.
s—x+t)/(2s—x+t) s—x+t)/(x—t)}ds
359
§1. FORMULAS FOR THE SOLUTION
x+i
+ 41
HQ (s) x
1
(s,
4 28—x—i
z
x + t — s,
'c)
ds
f (jc)
28—x+t
(s){4
ds
t,
=11+12+13+14. will transform each of the integrals 'k, k = 1,2,3,4. Making a change we obtain f variables in the second part of the integral 1.17')
z+t
+4-i
x+t_s)ds)/(x+t'
HQ(
bnilarly, we obtain 12
= (4 i5 HTQ (s)
s — x + t) ds) f (x — t) +
x+t
+4-i inally, changing the order of integration in the integrals 13 and e obtain
x+t—s, 'c)ds
+ we now put
(s)
(s, s — x
+ t,
't) ds
/
d-:
360
IX. CAUCBY PROBLEM FOR A DIRAC SYSTEM
x+t
x
(1.18)
(1.19)
Lq (x, t)
"
= 4- i
—'
t
(s)
(s, s — x
(x+t+s\L fx+t+s 2
t — s) ds,
2
1
+ t) ds,
'
2
(z4-t+a)
+4-i
x+t—'c, s)ds+
+ 4i
4 HTQ (ic) Hq_j ('c,
—x+
t,
s) d'c,
then (1.17) can be rewritten in the form (.; t) =
Kg (x,
t) / (x + t) + Lq (x, t) / (x — t) x+1
t, s)/(s)ds, which is (1.14) for p =q. Thus (1.14) is proved for any p 1. , m (1 13), Now, msertmg the expression for (x, t), p =0,1,2, find that the solution of the problem (1.1) + (1.2) is given by U
(x,
t) = K (x, t) f (x + t) + L (x,
t) / (x — t)
(1.20)
+4
H(x, t, s)f(s)ds,
where we have put
t)=
K(x, (x,
t),
(x, t, s).
The proof of the uniform convergence of these series is analogous
d
361
§1. FORMULAS FOR THE SOLUTION
he proof of the uniform convergence of the successive approximations, nd will be omitted. t) t) and Let us point out certain properties of the matrices thich will subsequently enable us to explicitly determine them. From he definition of the matrices Q(s) and H it follows that
p(s)+r(s), lip (s)—f-r(s)j
HQ(s)H= (
p(s)+r(s),
121)
=1P(s)+r(s)J( ['herefore by the definition of the matrix K1(x,t), i.e. (1.16), we have K1(x,t) =k1(x,OH,
1.22)
rhere
z+f k1 (x,
t)
[p (s) + r (s) I ds.
rurther,
p(s)+r(s), —i[p(s)+r(s)} p(s)+r(s) \z[p(s)+r(s)J,
/ HQ(s)H ==I. 1.23
onsequently, by virtue of (1.16) we obtain for the matrix L1(x, t) the L24)
L1(x, t)
t)HT,
'here
re
will
show that each of the matrices in the form t)
L25) r__lp(X,t)HT
t)
and
t)
can be
t)H, (p =
1,2,..
for p = 1 these formulas follow from (1.22) and (1.24). Let us that the formulas (1.25) have been proved for p = 1,2, . . ., q — 1. wifi prove them for p = q. Putting p = q —1 in (1.25) and inserting ie expression for Kqi(X, t) in (1.18), we obtain
362
IX. CAUCHY PROBLEM FOR A DIRAC SYSTEM
x+i K9 (x,
t) = 4- 1
HQ (s)
k9_1
(s, x + t — s) H ds
.x+t
Inserting the expression for HQ(s) H from (1.21), we find that x+t
K9(x,
t)=4-
k9_1(s, x+t—s)[p(s)+r(s)]ds. H,
which proves (1.25) for the matrix Kq(X, t), if we put x±t kg(X,t)
=-j-
k9_1(s, z+t—s)[p(s)+r(s)Jds.
The formula (1.25) for Lq(x, t) can be proved analogously by mean of (1.19) and (1.23). Now, inserting the expressions for t) and t) from (1.25) in (1.20) and putting k (x,
t) =
(x, t),
1 (x, t)
(x, t),
the formula (1.20) can be rewritten in the form ii (x,
(1.26)
t) = k (x, t) Hf (x + t) +
+
1
(x, t) HTI (x — t)
H (x, t, s) f(s) ds.
Formula (1.26) yields the solution of the problem (1.1) + (1.2) will play an important role in what follows.
ani
§2. Reduction to the Goursat problem
We again consider the problem (1.1) + (1.2). We will clarify that the functions k (x, t), 1 (x, t) and the matrix H(x, t, i mint satisfy m order that the vector-function u(x t), defined by (1 shall be a solution of the problem (1 1) + (1 2) In this section we matrix H(x, t, s) can be obtained from a Goursat we are able to explicitly determine the functions k (x, t)
of (1
'42,8), (2 10) below) the solution of (1 1) + (1 2) is given by (1 26) By ko(x,t) = 4 = t), and from (1 22), (1 24) and (1 25)
363
§2. REDUCTION TO GOURSAT PROBLEM
blows from the expressions for
t)
and
t)
that
0) =
0)
=0 (p =1,2,...). Therefore k(x,0) =l(x,0)
2.1)
It follows from (1.26) and (2.1) that the initial condition (1.2)
is
mtomatically satisfied.
Let u(x, t; f) denote the solution of (1.1) + (1.2), and suppose we know that the vector-function u(x, t; f) can be represented by (1.26). denote the operator denote the operator T1 = — iIO/ôt, and
In fact, it is not hard (x, t; f) u (x, t; D see that the left and right sides of this relation satisfy the same
We will show that problem
v(x,0)
the required equality follows from the uniqueness of the )lution of a Cauchy problem.
equation (1.1) can be written in the form
t,f) =u(x
12)
By the definition of the operator (x,
1; f)
= —II
[u (x,
and
formula (1 26) we have
t; j)J
t)Hf(x±t)—tk(x, t)H/'(x+t) —
i1(x,
t)
t) t,
t)
H (x,
s)f(s)ds.
the other hand, by the definition of the operator 26) we have
and
t)HB/'(x+t)
u(x, t;
+k
t) / (x
t, x
(x,
t) HQ (x +
t) / (x + t) + 1 (x, t) HTB/' (x — t)
+
formula
IX. CAUCHY PROBLEM FOR A DIRAC SYSTEM
364
± l(x, t)HTQ (x—t)f (x—t)
(x, t,
s) B/'(s)ds
t, s)Q(s)/(s)ds. the first integral in the right side of this relation by — iH, HTB = iHT, this relatioij can be rewritten in the form
and taking into account that HB = a (x, t; Baf)
+
1
(2.4)
-4
(x,
=
(x, t)
Hf' (x
1)
+/c(x, t)HQ(x±t)f(x-{-t) ffT/! (x t) HTQ (x — t) / (x — t) + ii (x, t) — t) + {IJ (x, t, x + t) Bf (x ± t) — H (x, t, X — t) Bf (x -— t, s)B—H(x, t, s)Q(s)}/(s)ds.
.
Since
the vector-function f(x) is arbitrary, it follows by virtue 4
(2.2) that the coefficients of f(x + t) and f(x — t), as well as the in the expressions (2.3) and (2.4), must coincide (the coefficients of the
derivatives I' (x + t) and f' (x — t) cancel each other). Thus, the coefficients of f(x + t) and f(x — t) respectively, we obtain (2.5)
—ik (x, t) H —4 H (x, t, x + t)
=k(x, t)HQ(x+t)+411(x, t, (2.6)
t, x—t) =l(x, t)HTQ(x_t)_4H(.x, t, x—t)B.
We will rewrite (2.5) in the form
kf(x t)H —zk(x,t)HQ(x+t) =
+t) [iB —
I]
The left and right sides of (2 5') are 2 x 2 matrices, and their the equality of their corresponding elements Let
x+t)=
(H11 (z, t, a; ± t)
t,
H12 (a;, t,
a; + t)'\
t, x±t))
365
§2. REDUCTION TO GOURSAT PROBLEM
the values of the matrices H, Q(x +t), B and H(x,t,x +t) (2.5'), we obtain Ikt (x t) + k (x t) r (x + t)
(x t) — ik (x t) p (s ± t)
(x 1)— ik (x t) r (x + t)
(x t) — k (x t) p (x + t) (x,
t, x +
t) — H11 (x, t, x -f- t), t, x -f- t) "12 t) — (a,, t, x --f- t),
t, x ±
iH11 (a,,
—f (x, t, x
±
t, x
4—
t) —
(a,, t, x
t)
t)
—i--
ince, in the matrix on the right, the first element of the first row, then multiplied by i, is equal to the negative of the second element I the first row, the matrix on the left must have the same property. ore
+k(x,t)r(x+tfl.
=
k(x,
r(x+t)}k(x, t).
olving this equation with due regard for the initial condition (2.1), we obtain L8)
k (x, t)
+ r (x +
[p (x +
exp
will now transform equation (2.6). We first write it in the form 6')
t) HTQ(x — t) = — 4-H(x, t x — t)
l( (x, t) Hr —
[iB + I]
r, in greater detail, (l't(x. t) 1 (x,
t)
(x,
t) —
t) +
ill21 (a,,
t)
t)
t,
t)
—iH22 (x, t, x —
(x
t,
t)
H21
t, x — t),
-4-—
t, x — 1)
H12 (x, t, a, —
—4--
(a,'
t, x — t)
since the elements of the first column of the matrix on the right, when by i, are equal to the corresponding elements of the second Olumn, the matrix on the left must have the same property Therefore i{4'(x,t) —ip(x —t)l(x,t) = —il[(x,t) —r(x —t)l(x,t),
366
(29)
IX. CAUCHY PROBLEM FOR A DIRAC SYSTEM
lax,
± r (x— t)) l(x, t).
Integrating this equation wijh due regard for the initial condition (2.1), we obtain the following explicit expression for the function 1 (x, t): (2.10)
1(x,
Further, by virtue of equation (2.7) we obtain from (2.5')
H(x, t, x+t)[iB—IJ =ik(x, t){p(x±t)±r(x±t)JH—2HQ(x±t)}. Since
iB_I=_HT, [p(x±t)+r(x+t)IH_2HQ(x±t)
= I [p (x + t) — r (x + t)]E,
where
it follows from (2.11) that (2.12)
H(x,t,x+t)HT=k(x,t) [p(x+t) —r(x+t)]E.
By (2.9) we obtain from (2.6') (2 13)
H (x, t, x — t) [lB
+ Il
—_—il(x, t){[p(x—t)+r(x—-t)]IJT—2HTQ(x—-t)).
But since by the definition of B, HT and Q(x — t)
IB+I—=H, [p(x_t)+r(x_t)}HT—2HTQ(x—t)
= I [r (x — t) — p (x — t)] E',
f—i 1\ (2.13) assumes the form H(x,t,x—t)H=l(x,t)[r(x---t) —p(x—t)]E'. (2.14)
Finally, equating the mtegrands in (2 3) and (2 4), we obtain for matrii li(x, t,s) the equation (2.15)
t, s) — H8' (x, t, s) B + H(x, t, s) Q(s) = 0.
367
§3. OPERATOR-MATRIX TRANSFORMATIONS
equation, together with (2.12) and (2.14), defines the Goursat roblem for the matrix H(x,t,s). From (2.5") and (2.6") we obtain another relation, which we will further on. Equating, in each of these equalities, the elements of first row and column, taking (2.7) and (2.9) into consideration, then putting t=O, by virtue of (2.1) we obtain
H11(x, 0, x)+iH12(x, 0, x)=±4[p(x)—r(x)], H11(x, 0, x)—tH12(x, 0, x)=4[p(x)—r(x)}, H11(x, 0, x)_—4[p(x)—r(x)],
H12(x, 0, x)=O.
equating the elements of the second row and first column, H21(x, 0,
0,
H21(x, 0,
0, x)==—4[p(x)—r(x)],
H21(x,
0, x)=0, H02(x, 0, x)=—4[p(x)—r(x)]. §3. Operator-matrix transformations
Let A1 and A2 be two linear differential operators, and L1 and L2, linear function spaces. )EFINITION. A continuous Linear operator X, which maps L1 into is called an operator transformation if it satisfies the following two 1. A1X = XA2. 2. The inverse operator X1 exists.
(p1(x)
(
0
d
o)
(p1(x) U
0
368
IX. CAUCHY PROBLEM FOR A DIRAC SYSTEM
/ —
(33)
( P2
421
(x) d
\— r2(x)j
0 1'\ d
1
(p2(x)
0
where p1(x), r1 (x), p2(x) and r2(x) are real functions which are summable
over every finite interval (0 < x b < cx>). We take for L1 the set of all continuously differentiable vector-functions
f(x) = \f2(x)
defined on the interval [0, b) and satisfying the boundary (34)
12(0) —h1f1(0) =0,
where h1 is an arbitrary finite real number. We take for L2 the set of all continuously differentiable vector-function1
g(x) = g2(x)
which are defined on the same interval [0, b) and satisfy the boundar condition g2(0) —h2g1(0) =0,
(35)
where h2 is an arbirary finite real number
We will seek the operator matrix X in the form (f(x) E L1) (3.6)
X [/
(z)J
R (x) / (x)
+ K (x,
s) / (s)
where R(x) and K(x,s) are second-order square matrices. By virtue of (3.2) and (3.6) we have
A1X[/(x)}=BR'(x)/(x)+ BR(x)/'(x)
+ BK (x, x) 1(x) +
(x,
+ Q1 (x) K (x, s)} f(s) d
by virtue of (3 3) and (3 6) II (x) Bf' (x) + R (x) Q2 (x)f (x) K (x, s) {Bf'
+ Q2 (s) /
369
§3. OPERATOR-MATRIX TRANSFORMATIONS
by parts, this relation can be rewritten in the form XA2 [1(x)]
R (x) Bf' (x) + R (x) Q2 (x) 1(x).
+K(x, x)B/(x)—K(x, O)Bf(O) s) B) / (s) ds.
(x,
ince f(x) is an arbitrary vector-function from L1, by virtue of (3.1) coefficients of f(x) and f' (x), and the integrands in the expressions
and (3.8), must be equal. Therefore, equating the coefficients we obtain BR(x)
if we put
\i(x) from (3.9) 5(x) =a(x) and has the form
/
= —fl(x),
i.e.
the matrix
13(x)
ct(x),
us determine the functions a (x) and (x). To do this, by equating coefficients of f(x) in (3.7) and (3.8), we obtain the following equation
the determination of the matrix R(x) 1)
BR'(x) +Q1(x)R(x) —R(x)Q2(x) =K(x,x)B —BK(x,x).
K(x,
(K11(x, s), K12(x, s)' s)
niing the explicit forms of the matrices Q1(x), Q2(x), B and R(x) J.1O)), we can write (3.11) in the following form: (x), (x) + ( —[3' (x) + [p1 (x) — p2 —cu (x) + [p2 (x) — (x)] [3(x), —[3'(x)
— (—[K12
(x, x) + K21(x,
x)
[3(x)
(x) —
T2
(x, x) x)
the matrix on the right side of (3.12) the elements on the principal 1 are equal except for having opposite signs, while the two offonal elements are equal. The same is therefore true of the matrix
DIRAC SYSTEM
370
on the left side of (3. 12), and so it follows that 4-[p1(z)
r2 (x)]
(x) =
(x)
+
(x) —
—P'(x)+
(x)1
— r2(x)] B(x),
i.e.
2a'(x)
(3.13)
2fl'(x) —q(x)a(x)
+q(x)(3(x) =0,
0,
q(x) =p1(x) —p2(x) +ri(x) —r2(x).
the system
From
ta2(x) +fl2(x)
= 0,
(3 13) follows 2a(x)a'(x) +2fl(x)fl'(x) = 0, ie from which we obtain
(315)
+132(x)
Suppose
2(0)
that the vector-function f(x) =
is
continuously differentiable and satisfies the conditions
=1,
(3.16)
12(0) =h1.
satisfies the boundary condition (3.4), and then L1. Further, let
Then f(x) obviously
fore f(x)
X[f(x)] =g(x),
(3.17) where
the vector-function g(x)
=
an element of the space L2 and so satisfies the boundary condit (3.5). Then for x =0 there follows from (3.17) by the definition of I is
operator-matrix X, i.e. by virtue of (3.6) and (3.10),
g1(0) =a(0)fj(0) +fl(0)f2(0),
g2(0) =
+a(0)f2(0).
the first of these equations by h2, subtracting it and takmg into account the boundary condition (3 5) condition (3.16), we find that P (0)
Put
§3. OPERATOR-MATRIX TRANSFORMATIONS
371
a(0) =1. =(h1—h2)/(1+h1h2). (1±hi)2(i_1_h2)2 _.x2 solving the system (3.13) and taking account of the expressions ), (3.18) —(3.20), we obtain the following explicit expressions for and $(x):
cL(x)=x sin
+ arcsin
4 q (t)
P (x) = x cos
+ arcsin 4-
the function q(x) is defined by (3.14), and the number ic is defined (3.20).
equating the integrands in (3.7) and (3.8), we obtain the owing equation for the matrix-kernel K(x, a): K,' (x,s)
B
(x, s) =
K(x,s)Qa(a) — Q1(x)K(x,s).
the term in (3.8) which contains f(O) must be zero in view of absence of a similar term in (3.7). Thus K(x, 0) Bf(0) =0, i.e. (—K12 (x, 0), K11 (x, 0)'\ (x, 0), K21 (x, 0))
/ 0 =0 is equivalent to the system of equations K12 (x, 0)
(0)
K11 (x,
0)/2(0),
(0) = K21 (x, 0)12 K22 (x, by virtue of the boundary condition (3.4) we finally obtain 0)/i
K12(x, 0)
= h1K11(x, 0),
K22(x, 0)
= h1K21(x, 0).
is further put K11(x,0)
K21(x,0)
are for the present arbitrary continuously differand functions. The conditions (3.24) and (3.25) for the matrix-
372
ix. CAUCHY PROBLEM FOR A DIRAC SYSTEM
kernel K(x,s) combine to give the condition p(x), h1p(x)\ (3.26) K(x,
and this condition together with equation (3.23) defines a Cau problem for K(x, s). This problem is Expressing the solution of the problem (3 23) + (3 26) by means ( (1.26) and inserting it in (3.11), we obtain an integral equation for
determination of the vector-function
(2x)
/
The quantities
.
and (0), which also appear in this equation, can be determined I the equation (3.12) for x = 0. Therefore, carrying out the calculations
the opposite order, we can prove that the operator-matrix X, &. by (3.6), in which the matrix-kernel K(x, s) is the solution of the
problem (3.23) + (3.26), satisfies the relation (3.1). The existence the inverse X 1 follows from the form of the operator X. §4. Solution of a mixed problem on the halfline
Let p(x), r(x) and the vector-function f(x) =
satisfy
previous requirements. We extend the functions p (x) and r(x) to negative halfline in such a way that they are still suminable over e finite interval, but otherwise arbitrary, and extend fi(x) and f2 a manner which for the present we leave unspecified (the ms which they are to be extended will be made precise further on). Under these assumptions we will consider the mixed problem (4.1) (4.2)
u(x, O)=/(x),
(4.3)
u2 (0, t) — ha1 (0, t)
where,
=0,
as before, I is the unit matrix,
/ Ot\
/p(x)
o)'
arbitrary finite
0
number.
0
\
§4. MIXED PROBLEM ON HALFLINE
373
Let
u(x,t) = / u1(x,t)
\U2(X,t)
the solution of the problem (4.1) —(4.3). According to the iula (1.26) the solution of the problem (4.1) + (4.2) can be represented the form u (x, t) k (x, t) Hf (x + t) + 1 (x, t) HTf (x — t)
t. s)f(s)ds, by the results of §2 k(x,
I
r(x+
l(x,
Let us now consider the application of operator transformations in to extend the solution of (4.1) to the negative haifline. We will w how to express the solution u(x, t) at the point — x as the result I a linear operator acting on u (a, t) (0 s x). Let the matrices A1 have the form 4 A2==B$—+Q(x)
will assume that the functions p(x) and r(x) have been extended negative halfline in such a way that p (—0) = p ( + 0), r( —0) i.e.
Q(—0) =Q(+0). ier words, we wifi assume that the extension of the matrix Q(x) ;inuous at zero. Further, let us denote by X an operator-matrix isformation which maps the space Lh of continuously differentiable f(x) satisfying the boundary condition (4.3), i.e. f2(0) =0, onto the space Lh. jose that the vector-function U
+1
I A
I
+(
_ \ ui4(x, t)
374
IX CAUCHY PROBLEM FOR A DIRAC SYSTEM
is the solution of the mixed problem (4 1) —(4 3) on the halfime [0, ccx) By virtue of (4 3), for every fixed t the vector-function u+(x, t) belongs
to the space Lh; therefore the operator X can be applied to The extension of the solution t) to the negative halfline be defined by the formula (4.7)
u(x, t) = X
u( — x, t)
t)J,
i.e. by the definition of X (cf. (3.6)) (4.8)
(x, t)] = u
t) = B (x)
(x,
(x, t) + K (x, s)
(s, t) ds.
We will show that the vector-function u — (x, t) satisfies the equatü (4.1) on the negative halfline. By virtue of the definition of the i
A1 we have to show that —
(49)
A1 [u (x, t)]
—ii àu .
Since for every fixed t the solution t) belongs to the space I; by the definition of the operator-matrix transformation X and (3.1 we have (4.10)
t)] =
A1X
t)].
According to (4.7) the left side of (4.10) coincides with the left of (4.9). Let us calculate the right side of (4.10). By virtue of (4 which can be rewritten in the form A2[u+(x,
t)] = X [—ii àu+] = —iIX
It follows in obvious fashion from (4.8) that the operator-matrix and the operator of differentiation ä/8t commute; therefore by the above relation implies that [n+ (x, t)] i.e.
= —il-i- X
(x, t)] =
the right sides of (4.9) and (4.10) coincide, which proves our asseri
We will now show that the extension of the solution u+(x, t) to negative is continuous together with its first derivative respect to x.
375
§4. MIXED PROBLEM ON HALFLINE
fact, putting x =0 in (4.8), we obtain
u(—0,t) definition (cf. (3.10)) (
(0))'
a(0) = 1, and can be computed from (3.19). Since in the .t instance the spaces L1 and L2 which were considered in the iious section coincide, L1 = L2 = Lk, it follows that h1 = h2 = h, and = 0. Hence R(0) = I, and the continuity at zero of the efore follows from (4.11).
of
ther, using the explicit form of the operators A1, A2 and X, by of (4.7) the relation (4.10) can be rewritten in the form B
+ Q (—x) u = R (x) [B
+ Q (x) u÷]
t)]ds.
utting x =0, we obtain B
Ou (—x, t)
+ Q (—0) u (—0, t) t)
àzi
+ Q(+0)u(+0,
R(0) = I, it follows by (4.6) and (4.11) that on (—x, t)
On (x, t)
Ox
Ox
first derivative is continuous at zero. ormula (4.8) can be used to extend the original vector-function e
•
In fact, putting t = 0 in (4.8) and taking account of the initial i (4.2), we obtain
f(—x)=R(x)/(x)+
s)f(s)ds.
continuity at zero of the extension of f(x) and its derivative follows (4.11) and (4.12), if we put t = +0 in them. t us now return to the problem (4.1) —(4.3).
x> t > 0, then the solution of this problem coincides with the
376
IX. CAUCHY PROBLEM FOR A DIRAC SYSTEM
solution of the problem (4.1) + (4.2), and consequently is given (4.4). Now if 0 <x
0), x — t <0, we replace the variable of integration s by — (4.13) for the value of the function f( — s). We similarly chang
and use
the value of the function f(x — t) in (4.4). Then following simple tranS
formations we obtain the following formula for the solution of problem (4 1) —(4 3) for 0 <x
t) = Ic (x, t) Hf (x + t) +
1
(x, t) HTR (t — x) / (t — x)
x+t
(4.14)
+
1
H (x, t, s) f(s) ds +
H1 (x, t, s) f(s) ds.
where H1 (x, t,
s) = 21 (x, t) HTK (t — x, s)+
+ H (x,
t, —s) R (s) ±
H (x, t,
K (t,
s)
§5. Solution of the problem (1.1) + (1.2) fort <0
We will again consider the problem (1.1) + (1.2), i.e. the problem .7ôu
w.
,,
u(x, O)=f(x),
(5.2)
assuming assumes
now however, that t <0 Put t =
=B
(5.1')
Let
—r
Then equation (5 1
the form
+ Q(x)u(x,
—jt).
denote the matrix
/0
1t\1 = 11u(x,
—
r). Since
1
0
= I,
u(x —r) of u(x, — r) in equation (5 1') we obtain
§5. SOLUTION FOR t <0 ày
+
377
Q(x)11v(x, t).
Itiplying both sides of this equation from the left by
I,
—
that Ov
Q1
(x)
= —11Q (z) Ii = (t—r(x)
0
(x)
virtue of (5.3), the initial condition (5.2) becomes v(x,O) =11f(x),
together with equation (5.4) defines a Cauchy problem for
4.
Since
the problem (5.4) + (5.6) is of the same form as the problem
applying (4.4) we obtain v(x,
; according to (4.4) k1(x
11(x,
by the definition of the matrix Q1(x), i.e. (5.5), tr Q1(x) = — tr Q(x),
1y follows from the preceding expressions that
ki(x,r) =1(x, —4,
11(x,r)
=k(x, —4,
k(x, —4 and 1(x, —4 are given by (4.4'). Therefore (5.7) can
written in the form
v(x, ¶)1(Z, —t)11111(x+'t)+k(x, _t)HTI1/(x_t)
378
IX. CAUCHY PROBLEM FOR A DIRAC SYSTEM
Hence, multiplying both sides from the left by
by virtue of (5.
we obtain
ufr,
.—t)Hf(x—c)+1(x, —-c)HTJ(x+-.)
+
(x, ; s) 11f (s) ds, 5
since 11H11 = HT, 11HT11 = H. Putting now — r = t, we will have
u(x, t)_—k(x, t)IJf(x+t)+i(x, t)HTI(x—t) 5
where
R(x, t, s)/(s)ds,
x+t
t, s) = 11H1(x,
(5.10)
—
t, s)
Let us now take up the derivation of an equation for the H(x,t,s). According to (2.15) the matrix-kernel H1(x,r,s) in satisfies the equation
+Hi(x,r,s)Qi(s) 0. Replacmg r by — t and takmg account of (5 5), we obtain —t,s)B —H1(x, —t,s)I Q(s)11 =0,
—t,s)
from which, multiplying both sides from the left and right by find that —
t,s)11
+
—
t,s)B11 + 11H1(x, —t,s) 11Q(s) =0.
It follows from the definition of the matrices and B that B11 = — therefore by (5.10) the preceding equation can be rewritten in the foi (5.11)
t, s) — IL' (x,
t, s)B +
t, s) Q(s) = 0.
We now compute H(x 0, x) Let
H1 (x, ; s) =
Ha))
Then by the definition of the matrix Q1(x), i.e. (5.5), and accoil to (2.16) and (2 17) we have
flu)
0, x) = —Ha) (x, 0, x) = 4 [p(x) — r(x)J, (x, 0, x) =
(x,
0, x) =0
__
379
§5. SOLUTION FOR t <0
other hand, by (5.10) we have H (x, t, s)
1'
11H1 (x,
=
s)
=
fore, using (5.12) we obtain 1111(x, 0,
x) = U .y, by virtue of (5.10), (5.5) and (5.8) it is not hard to obtain from and (2.14) the following conditions for the matrix H(x, t, s) on x)
(x,
characteristics:
1i(x, t, x+t)H2'==k(x, t)[r(x±t)—p(x+t)]E, R(x, t, x—t)H=l(x, t)jp(x—t)—r(x—t)]E', together with equation (5.11) define the Goursat problem for matrix kernel H(x t s) [ARK. The formulas obtained in this section for the solution of Cauchy problem for t <0 can also be obtamed on the basis of the s in §1.
ummarizing all the results of this chapter, we arrive at the following orem, which is important for the next two chapters. IIEOREM 5.1. Let p(x) and r(x) be real functions, defined on the haifline ')and continuously differentiable,0 and let f(x) be a real vector-function and satisfies the is defined and continuously differentiable on [0, _ion f2(0) — hf1(0) = 0, where h is an arbitrary real number. Then solution of the problem
u(x, O)=/(x),
I
a0(O, t)— hu1(0, t)==U.
is the 2 X 2 unit matrix, and
01'\ B
o)' Q(x)=
(p(X)
by the following formulas:
For 0
o
0
i(x)
ix. CAUCHY PROBLEM FOR A DIRAC SYSTEM
380
u(x, t)—_k(x, t)Hf(x+t)+t(x, t)HT/(x_t) x+i
(5.13)
1
H(x, t, s)f(s)ds; x—t
2.fort
k (x,
t) Hf (x + t) + (x, t) HTI (x — t) 1
t, s)/(s)ds. idsere
k(x,
l(x,
H=(. and
the 2 x
2
matrices H(x, t, s) and
t, s) are the solutions of
same matrix equation —
W(x,t,s)B+W(x,t,s)Q(s) =0,
satisfying the following conditions along the characteristics:
H(x, t, x+t)HT__k(x, t)(p(x+t)—r(x+t))E, H(x, t, x—t)H=l(x, t){r(x—t)—p(x—t)}E', 17(x, t, x+t)HT=k(x, t){r(x+t)—p(x+t))E, .!!(x, t, x—t)H—_l(x, t){p(x—t)—r(x--—t))E', where
I—i 1\ i .), 3. for 0 <x
/ (—x) = R (x) 1(x) +
K (x, s) f(s) 4s,
381
REFERENCES
(
tr [Q
(x) = cos
—Q
the 2 X 2 matrix K(x,s) is the solution of the problem
+Q(x)K(x,s) —K(x,s)Q(—s) =0, K(x, x)B — BK(x, x) = BR'(x) ±Q(x)R(x) — R(x)Q(— x),
K(x,0)Bf(0) =0. the solution is given by a (x,
t) = k (x, t) Hf (x + t) + I
H(x, t, s)/(s)ds+
t)HTR (1 — x) / (t — x)
H1(x,t,s)f(s)ds,
H1(x, t, s)=21(x, t)HTK(t_x, s)+
+ II (x,
t, —s) R (s) +
H (x, t,
K
s)
Bibliographical references
results of this chapter are due to Sargsjan [8, ii]. concerns §3, cf. the paper by Prats and Toll [1]; and for §4, the paper [4] by the authors.
CHAPTER 10
ASYMPTOTIC BEHAVIOUR OF THE SPECTRAL KERNEL AND ITS DERIVATIVES FOR THE CASE OF A DIRAC SYSTEM: §1. Derivation of the basic identities
We will study the problem (1.1)
B4J.-+Q(x)y—_X1y,
(1.2)
y2(0)—hy1(0)_—0,
where
I is the 2 X 2 unit matrix,
B=(101
Q(x)=(
p(x)
0\
0
the functions p (x) and r(x) satisfy the conditions of the prececi chapter, i.e. are continuously differentiable real functions defined the halfline [0, ), and h is an arbitrary real number. Together with the problem (1.1) —(1.2) we will consider the problem (1.3) (1.4)
•
ôu
ôu
u(x, 0)==/(x),
where f(x)
Ifi(x) =
is a real and continuously differentiable vector-function define 10, co) which satisfies the condition 12(0) — hf1 (0) =0 extend the functions p (x) and r(x) to the negative halfline, their differentiability but otherwise arbitrarily, and fix) according to (5 15) of the precedmg chapter, i e s)f(s)ds. 382
§1. BASIC IDENTITIES
= the
383
7 ço1(x, A) A)
solution of equation (Li) which satisfies the initial conditions 4,1(O,A)
A) satisfies
us now put 1(x) =
=h.
1,
the boundary condition A)
(1.2).
and consider the problem (1.3) + (1.4).
the one hand, the solution of this problem is given, for 0
erefore by the uniqueness of the solution of the problem (1.3) + (1.4),
g the right sides of (5.13), (5.14) and (5.16) of the preceding pter with the right side of (1.6), respectively, we arrive at the wing identities:
for 0
= k (x,
t) Hp (x
+ t,
A) + I (x,
A) HT,
(x
— t,
A)
x+t
+4-
H(x, t, s)p(s,
A)ds;
for t<0, 0< —t<x p
(x, A)
k (x, t) Hp (x
+ t, C
for
p (x,
==k(x, t)Hp(x+ t,
+
B (t
(x, t) Hr
H(x, t, s)p(s, —
(x — t,
A)
A)ds;
.v+t
0 <x
1
+ 1 (x, t)
A)
— x) p (t — x,A)+ 4+4-
H (x, t, s) p (s, A) '-Is
111(x, t, s)p(s, X)ds.
X. ASYMPTOTICS OF THE SPECTRAL KERNEL
384
Let denote an arbitrary positive number and let g, (t) be a functia satisfying the following conditions: 1. g,(t) vanishes outside the interval (— 2. has a piecewise continuous first derivative. Fuxthér, let 4', (A) denote the Fourier transform of g, (t), i.e. we (X)
(1,10)
=
(t)
We denote by x an arbitrary point of the halfline [0, Let us assume that x
Multiplying both sides of the identit (1 7) by g (t) and rntegratmg with respect to t from 0 to we have p (x, X)
(t)
=H + Hr
(1.11)
A)
(t) dt
1 (x, t) p (x — t, A)
(t) dt
k (x, t) p (x + t,
x+t
+ fl
t, s) p (s,
A)
ds}
(t) dt.
Changing variables in the first two integrals on the right and changi the order of integrating in the third integral, we can write the identi (1.11) in the form A)ds
A)
+HT
(1.11')
+ T{ X—S
x—s)g,(x—s)p(s, H (x, t, s)
X)
ds
(t) dt} p (s, A) ds.
JX—8J
Multiplying both sides of the identity (1.8) by g,(t) and integrati with respect to t from — e to 0, and then proceeding as we did in
to obtain the identity (1 11'), we obtain (t)
=H
the mudy of the case x
k (x, s — x) g8 (s — x) p (s, A) ds
is analogous, we will not consider it.
§2. PRELIMINARY ESTIMATION
385
X)ds
+fl{
R(x,
dt} p(s,
s)
1,
jnally, adding the identities (1.11') and (1.12) and taking account of 1.10), we obtain the identity L13)
p (x, A)
(A)
=
W (z,
s; a) p (s,
A)
here
H(x, t,
L14)
R(x,
t,
(t)dt,
tx—si
hich wifi be important in what follows.
Further, since by condition 1 on the function it follows from that W(x, x + e) = W(x, x — a; a) = 0, differentiating the identity with respect to x, we obtain a second identity which will be in what follows: p
A) ds.
§2. Preliminary estimation of the spectral kernel and its derivatives Let
f(x) = (f1(x)
a), and let as an arbitrary vector-function from fore, be the solution of the problem (1.1) + (1.5). we have already proved in Chapter 3, for a given h there exists a which is bounded in p (A), — ndecreasing
386
OF THE SPECTRAL KERNEL
X.
isometric mapping of the space 9 2(0, co) of vector-functions ont co, co) according to
the space (2.1)
X)dx,
(2.2)
X)dp(X),
Where the integrals (2.1) and (2.2) converge in the metrics of the co, co) and 92(0, aD) respectively, and one has the Parsevi equality (2.3)
Further, if f(x),g(x) E 92(0 co), and F(X), GOt) are their transforms, then as was proved in Chapter 3, one has the followifl generalized Parseval equality: (24) 0
introduce the following terminology: 1. The function p (X) will be called the spectral function of the (1.1) +(1.2). 2. The 2 X 2 matrix We
(2.5)
will be called the spectral kernel of the problem (1.1) + (1.2). In this section we will obtain various estimates for the spectral
and its derivatives which we will need further on. the preceding section we obtained the identity (1.13), s,
A)
A)ds,
2 matrix =
W(x, s; e)= II
(x, s; €)
s; s)
387
§2. PRELIMINARY ESTIMATION
The identity (2.6) says that for any x
0 the functions
=1,2) (the components of the vector-function
are
te Fourier transforms of the vector-functions which are equal to (W1(x,s; I x — sJ
and equal to zero for x — sI >
Therefore if
I
(x)\ cx>),
\J2tXi
by the generalized Parseval equality (2.4) we have s;
+ W12(x, s; e)12(s)} ds
(i = 1,
2),
in matrix form 5
-w
0
REMARK The identity (27'), which was derived under the assumption
p(x) and r(x) are continuously differentiable functions, is valid summable p(x) and r(x). This is easily obtained by a passage to limit.
ILEMMA 2.1. If the elements p(x) and r(x) of the matrix Q(x) of the (1 1) are summable over every finite interval and if (x0, x1) is an
finite interval of the real tine, then there exists a constant C 1:C(xo, x1) such that for any x and s from the interval (x0, x1) and for a we have the bounds +1
a+1
(x, s, ')} =
1)11
51
(s, ')l di (1) <
Let the function g (t) be defined by
g,(t)=
!(sItI), 0,
C
(ik
1,2)
388
X. ASYMPTOTICS OF THE SPECTRAL KERNEL
=
(t)
/ Sill
I (
-E
Leta be an arbitrary positive number, and put (2.9)
(X, a)
= Therefore
j
(t, a)
=
/ sin
2
a)
(
)
from (2.6) and (2.9) we have, by the Parseval equality, p (x, A) 2 dp (A)
(2.10)
II W (x, s; €, a)
J2
ds,
a) 112 denotes the sum of the squares of the element
where H W(x, s;
of the matrix W(x, s; a), where the latter is defined by (1.14), in v' But since g, (t) L the function g, (t) is replaced by g, (t, a) = g, (t) compact support, so does the matrix W(x, s; a), and since the elemeni
p(x) and r(x) of the matrix Q(x) are summable, so is the fund a) 2 Therefore the integral in the right side of (2.10) 2 there follows Then from (2.10) for II W(x, s;
{sin(X—_ a)}4
I p (x, A) J2 dp (A)
and so a fortiori a+1
(2.11)
Further, since for 0
sin4(X—a)
x
X)I2dp(A)
1 one has sinx/x> 2/ir, the bound
implies the bounds a+1
(2.12)
which prove the lemma for the functions The bounds (2 8) for these functions for s
(1=1,2), s; A) (i = 1, 2) for s x follow from (2 12)
the Cauchy-Bunjakovskii inequality. The completion of the for the functions (x, s; A) for i j is similar. This proves the len
389
§2. PRELIMINARY ESTIMATION
LEMMA 2.2. Under the conditions of Lemma 2.1 we have, for all a,
8')
p(a + 1) — p(a)
The proof of this lemma follows from the bounds (2.12) if we put = 1, x =0, and take into account that 9,1(0,X) = 1. We now take up the estimation of the derivatives of the spectral O(x,s; A). To this end let us consider the formula (1.15); 2.13)
(x, A) (j) (A)
oW
(s, A)
ds.
L follows from this relation that for fixed x the functions (x, A) (A) = 1,2) can be regarded as the generalized Fourier transformations of he vector-functions which equal (x, s; s),
2)
I
s; e)
+ and equal zero outside this interval. Therefore the Parseval equality we have s
2.14)
(x —
x
(A) J2
(x.s; t)
(x,
Let g,(t) be defined in the same way as in the proof of Lemma 2.1. in (2.14) g, (t) by g, (t, a) = and taking account f (2.9), we can rewrite (2.14) in the form s(X—a) {srn p' (x, A) J2 dp (A) =
OW
(xis;
(X—a)
£)
j2ds,
}
is as in the proof of Lemma 2.1. We now the integral in the right side of (2.14) for large a. Since the W(x,s; is defined by (1.14) with g,(t) replaced by
here the matrix W(x,s;
s; s, a) = Hk(x, s — x — s)
x) (x — —k—si
H(x, t,
+
I?(x, t,
X. ASYMPTOTICSOF THE SPECTRAL KERNEL
390
has compact support, so do the matrices W(x, s; a) ÔW(x,s; E,a)/Ox. On the other hand, by direct differentiation of matrix W(x, s; a) with respect to x it is easily seen that for large its elements are of order 0(a). But if the elements p (x) and r(x) of t matrix Q(x) have summable first derivatives, then for a
Since
X+8
(2.15)
s; €, a)D2ds__O(a2).
2.3. If the elements p(x) and r(x) of the matrix Q(x) of system (1.1) are absolutely continuous in every finite interval, then we have
(s, s; X)}
V
(2.16)
a-fl
(p (x, X)
J
(s, X) I dp (X) =
(1, k= 1, 2), where
s; A)
(i, k = 1,2) are
the
elements of
the
matrix O(x,
(cf. (2.5)). The
estimates (2.16) hold uniformly in every finite interval.
PROOF. By virtue of the estimate (2.15) we can obtain from similarly to what was done in the proof of Lemma 2.1,
jp(z, X)j2dp(X)=O(a2) (i=1, 2),
(2.17)
which proves the lemma for i = k. For the other cases the lemma foll from (2.17) by the Cauchy-Bunjakovskir inequality. From Lemmas 2.1 and 2.3 and the Cauchy-Bunjakovskll inequal follows: LEMMA 2.4.
When the conditions of Lemma 2.3 are satisfied, wet
for a —* ti+1
p(x, A)jIpk(s, a
(i,
k= 1,2).
estimates hold urnformly in every finite interval One can sumlarly prove the following lemma
§2. PRELIMINARY ESTIMATION
391
LEMMA 2.5. If the elements p(x) and r(x) of the matrix Q(x) have 1 — 1 in every finite interval,
bsolutely continuous derivatives of order for a—*
y
2 19)
ômp1(x 1)11 ã'9k(S
dp (A) = 0
(a'),
j4sere m +n 1, i, k = 1,2. These estimates hold uniformly as x and s over any finite intervals.
Let us now consider the problem (1 1) + (1 2) under the conditions = r(x) 0, i.e. the problem
.21)
(0)=O.
Let
the solution of equation (2.20) which satisfies the initial conditions
=1,
.22)
#2(0,x) =h.
is easily seen, the problem (2.20) + (2.22) breaks up into two indeproblems for the functions A), whose solutions A) and e given by =cosAx—hsinAx, now
4'2(x,A) =sinAx+hcosXx.
determine the spectral function and spectral kernel of the
blem (2.20) +(2.21). us adjoin to the boundary condition (2.21) the condition y2(b) —Hyi(b) =0,
24) b
is
an arbitrary positive number, and H is an arbitrary real
mber, and determine the eigenvalues of the boundary-value problem satisfies the boundary condition +(2.21) +(2.24). Since 21), to determine the eigenvalues we have to insert the expressions the functions and from (2.23) in the boundary (2.24). We have sinAb +hcosAb —Hcosxb +hHsinAb =0,
X. ASYMPTOTICS OF THE SPECTRAL KERNEL
392
i.e. (2.25)
(1 ±hH) sin xb + (h — H) cos Ab = 0.
If we put cOsw=
I
+ hH
S'(l +h2) (I + /12)
then equation (2.25) can be rewritten in the form sin(Xb +w) =0, froxi Consequently, we obtait +w = nir (n = 0, ±1, ± 2, ) fbr the eagenvalues the expression which (2.26)
(n0,±1,±2,
Xnb=fl
)
Further, since
we have by the definition of the spectral function pb(X)
'V
*
(cf.
Chapter 3, §1
I
Xn, 1
*
(1±h2)b' b>X
us transform p * (A) for A 0 (for A <0 the transformation analogous). Using (2.26), we obtain Let
*
1
'VI 2
I
An,
2
1t(1+h2)
For A <0 the result is the same. the spectral function p* (A) of the boundary-value has the form p*(A) =A/R-(1
spectral kernel has the form
P(x, s, A)
+h2),
393
§3. ASYMPTOTICS OF SPECTRAL FUNCTION
From the definition of the matrix O*(x, s; A) there immediately follows: LEMMA 2.6. For a —*
we a+1
\/
(2.28) a+1.(
have, uniformly for all positive values of x and s,
(x, s; X))
I
iS
(2.29) a
)}=_—O(a')
are the elements §3.
(l=m+n;i,k==1,2),
of the matrix O*(x,s;A).
Asymptotic behaviour of the spectral kernel and spectral function
Let /f1(x) f(x)
2((), an arbitrary vector-function from which vanishes outside some finite interval. Then its Fourier transform be
X)ds
[3.1)
in the ordinary sense. Now, inserting the expression for from (3.1) in (2.7), and then the order of integration in the left side, which is permissible
virtue of the bound (2.8') and Fubini's theorem, we obtain
1{ip(z,.x)pT(s,
s; s)/(s)ds.
X)
Due to the arbitrariness of the vector-function f(x), the last relation mplies the identity cp(x,
there
is
A)pT(s,
s; s),
given by (1.14).
Now, putting —jx—g'
W*(x, s;
H (x, t, s)g, (t)dt +
.J7 (x, t,
s)
(t)
dt
Ix—si
taking account of (1.14) and the definition of the spectral kernel x,s; A), the identity (3.2) can be rewritten in the form
rid
394
x. ASYMPTOTICS OF THE SPECTRAL KERNEL
X)=
s;
(3.2')
=
+W*(x, s;
s),
0,
We will now write down an identity analogous to (3 2') for the spectral matrix ,*(x,s, X) For p(x) =r(x) 0 it follows from the equalities (4,4') of the precedmg chapter that k(x, t) = l(x t) On the other band, as easily follows from the problems (2 15) + (2 12) + (2 14) and (511)+(5 11') of the preceding chapter, for p(x) =r(x) 0 the matrices H(x, t, s) and H(x, t, s) are the zero matrix, and consequently is also the zero matrix Therefore for the case in this case W*(x s
being considered the identity (3 2') assumes the form (X) dxO* (x, s; 5
(3.4)
x—s
1
—2-Hg1(s—
Ix—sI>e. Now subtracting (3.4) from (3.2'), we obtain (X) dx {0 (x, s; X) —
(3.5)
0*
(x, s; X))
H{k(x, s —
(x,
s; E), 0,
now take up the transformation of the right side of the identiti (3.5). To this end we put We
(x,
11 5
(x,
t, s) et?1 dt,
Ix—s!
(3.6)
(x, s;
A)
=
(x, s; A) =
On the other hand (cf (1 10)),
5
H (x, t, s) (x, s; A) +
(x, s;
A).
§3. ASYMPTOTIC S OF SPECTRAL FUNCTION
395
= g8 (t)
(3.7)
Then by the Parseval equality for the classical Fourier transform as well as (3.3), it follows from the last two equalities that 1) (A)
(3.8)
(x, s; A) dA = W* (x, s; s).
inverting (3.7), we obtain g (t)
(3.9)
On the basis of (3.8) and (3.9) the identity (3.5) assumes the final form (3.10)
(A)
(x, s;
A)
= 0,
vhere we have put
s; X)=O(x, s; A)_0(x, I
3.11)
I
s—z)—T} I
'he
s; v)dv
I
x__s)—ry
-—
I
i(z—s)
identity (3.10) and Taubenan theorem for Fourier (cf. Chapter 14) enable us to study the behaviour of the kernel for A We will first prove the following lemma. I
LEMMA 3.1. If the coefficients p(x) and r(x) of the system (1.1) are immable over every finite interval, then for any fixed x, s we have for all a a+1
3.12)
\J
(x,
s; A)) < C
(t, k= 1,'2),
A) are the elements of the matrix 4(x,s; A) defined by (3.11). bounds (3.12) hold uniformly as x and s vary over any finite intervals.
PROOF. To prove the lemma it is sufficient to obtain the bound 3.12) for every element of the matrix on the right side of (3.11). For Lie elements of the matrices O(x,s; A) and O*(x,s; A) the bound (3.12) Dincides with the bounds (2.8) and (2.28). Further, from the definition r the matrix a(x,s; A), i.e. from (3.6), it follows from the summability
_______________
396
X. ASYMPTOTICS OF THE SPECTRAL KERNEL
of the elements of the matrices H(x, t, s) and I(x, t, s) and the wells known Riemann-Lebesgue lemma that (i,k=1,2), =0 uniformly for x and s in any bounded regions, where a1k(x, s; A) are the elements of the matrix a (x, s; A). Therefore for a afi
Vanç
(x, s, v)dv
=
= 0 (1)
(x, s, C.
Finally, for the elements of the matrices H r {1(x,
— 1-
1
2itz.(s—x)
1
2ici(x—j
the bounds (3.12) can be obtained at once from the expressions these matrices, if we write them in the form —I
I
I
s_x)_T}
i(s—x) I
2
(3.13) I
-
'1.
i(x—s) sinX
x — s)}
—Hi' (1 (x,
2
(s — x)
*
This proves the lemma.
LEMMA 3.2. Suppose thatp(x) and r(x) satisfy a Dini condition uniforrnj
in every finite interval. Then for any fixed x and s we have (3.14)
urn
(x, s; v) dv
0.
This relation holds uniformly as x and s vary over any finite interva
By the definition of the matrix a(x
s, v),
fl(x, t, s)e'
H(x, t, —X
—2
i e (3 6), we
{ tx—al
—1
(H (x, t, s) + .:
—t, s)}
Sifl?
dt
397
§3. ASYMPTOTIC S OF SPECTRAL FUNCTION
the form of the matrices H(x, t, s) and H(x, t, s) it follows that Ft(x,0,s) +H(x,0,s) =0. Therefore (3.15) can be written in the form {F(x, t, s)—F(x, 0,
s; v)dv==
5
5
t, s) t, s)]. From the conditions of s) be lemma and the form of the matrices H(x, t, s) and H(x, t, s) it follows
hat F(x, t, s), as a function of t, satisfies a Dini condition at the point =0. Therefore urn
{F (x,
5 I
t, s)— F(x, 0, s)} SmtAt dt,
I
the proof of the lemma follows by virtue of (3.16). We will now prove the basic theorem of this section.
vhence
THEOREM 3.1. If p(x) and r(x) satisfy a Dini condition uniformly in have the asymptotic very finite interval, then for any fixed x and s ;rmula
O(x, s; X)—0(x, s; 3.17)
s;
s; —A)
+[Hk(x, s—x)+HTL(x, x—s)— I)
=o(1). over any finite intervals.
This formula holds uniformly as x and s
PROOF. Let (f,g) denote the scalar product of the vectors f and g,
(f,g) =fT.g• Let f and g be arbitrary constant vectors. It Ilows from (3.5) that 5
L18)
(A)
(x, s, X; f, g)
Ja(x, s; /,
(x, s, X; /, g)) s, s;
I, g),
=10, we have put o1(x,s,x;f,g) =(O(x,s;X)I,g), ff2(x,s,X;f,g) = (o*(x,s;x)f,g), I
a(x, s; f,g) =([Hk (x, s — x)
—
i, g) + ([Hnl (x, x — s) —
I, g),
398
X. ASYMPTOTIC S OF THE SPECTRAL KERNEL
For s = x and g = I it is obvious that a1 and are monotomc of X, and that a(x, x;f, f) 0. Therefore it follows from the Tauberian theorem (cf. Chapter 14, (3.18), by virtue of Theorem 42, special case) as well as (3.14), that {a1(x, x, A; I, /)—a1(x, x, —A; f, f)
(3,18'); $
/, /)—a2(x, x, —A, /, /)]}=O x and g = I let us consider the monotonic functions of A —[a2(x,
A) = a1(x,
A; f,f) +2o-i(x,
A; f,f) +ai(s,s, A; f,f
i-2(x,s;X) =a2(x,x,A;.f,f) +2u2(x,s,A;f,f) +a2(s,s,,A;f,f)
and again apply the aforementioned Tauberian theorem of As a consequence we obtain, by (3.18'), urn {a1(x, s, A; /, /) —a1(x, s, —X; /, /) —[a2(x, s, A; /, /)—a2(x, s, —A;
/)]
Now, transformmg the quadratic form to a bthnear form, we from the preceding formula Jim
(x, s, A; /, g) — — [a2 (x,
(x, s, —A;
s, A; L g) — —a
Since
a2 (x,
g)
s, —A; I, g)]
(x, s; b g)
sin A (x — s) r (x -— s)
the vectors f and g are arbitrary, (3.17) follows from this equality
This proves the theorem. From the asymptotic formula (3.17) there follows: THEOREM 3.2. Under the conditions of Theorem 3.1 we have
I
Jim
In fact, from the definition of the spectral kernel 0* (x, s, A), i e
(2g*fl, and from the explicit form of the vector function = we
(3.20)
find that O*(O, 0; A)
h2)
(1 ±h2)
frol
399
§3. ASYMPTOTICS OF SPECTRAL FUNCTION
p Further, by virtue of (2.1), Chapter 9, k(x,s =1(x,x
=f.
taking account of this and of the definition of the matrix ? (cf. (1.6'), Chapter 9), we obtain [Hk(x,s —x) +HTI(x,x —s) — 3.21) =0. it follows from the definition of the spectral kernel 0 (x, s; A) cf. (2.5)) and the condition (1.5) that 3.22)
0 (0, 0; A) — 0(0,
0; —A)=
(p
'ow, putting s =x =0 in (3.18) and t*king nd (3.22), we obtain
/1 h \
—p (—A) —
(A)
h2)
(1
—p
account of (3.20), (3.21)
+M) x}
This proves the theorem. Formula (3.19) gives the asymptotic behaviour of the odd component f the spectral function p(A). To determine the asymptotic behaviour f p(A) itself, we obviously have to find the asymptotic behaviour of even component. We have the following: thich proves the asymptotic formula
THEOREM 3.3. Suppose that p(x) and r(x) satisfy a Lipschitz condition niformly in euery finite intervaL Then for A —' zve have p(A) +p(—A) =o(1).
3.23)
PROOF. By virtue of (3.3), it follows from (3.5) for x —s =0 that 3.24)
0; A) —
(A)
rhere f
(0, 0; A)) =
G (t)
(t) dt,
H11(0, t, 0),
R11(0, t, 0), 0
y
OOA
A
)'
0*OOX "(
Tauberian theorem (cf. Chapter 14, Theorem 4.2') we
Ltain from (3.24) L25)
A
(A) + P (—A) =
A
G(+O)+G(—O) 2
400
X.- ASYMPTOTICS OF THE SPECTRAL KERNEL
On the other hand, it follows from (2.16) and (5.12'), Chapter 9, tho Then (3.23) follows from (3.25),which G(+0) +G(—O) the theorem. From this theorem and Theorem 3.2 follows: TaRokaM 3.4. Suppose that the hypotheses
of Theorem 3.3
are
satisfieG
Thenforjxj—*o we have (1).
*4. Asymptotic behaviour of the derivatives of the spectral kernel
Iii §1 we derived the formula (cf. (1.15)) (4.1)
il (a;,
X) p (A)
=
OW
(x,s;
£)
(s A) ds.
It follows from this formula that for fixed x the components of vector-function (x, A) (A), i.e. the functions and
A)
4,2(x, A),
can be regarded as the Fourier transforms of the vector-functions are equal respectively to i0 W11 'a W21 (x, s; e) s; E)
j to W12
for x —
\
(x,
s;
and zero for
/
1'
J
x
—
W22 (x, s;
e)
sj > E. Therefore if
E = and equals zero outside some finite interval, then by the generali2 Parseval equality (2.4) it follows from (3.1) and (4.1) that f(x)
=
ds.
Hence, since the vector-function f(x) is arbitrary, we obtain the (4.2)
s,
r),
____________
401
§4. DERIVATIVES OF SPECTRAL KERNEL
where
W(x, s; €)=Hk(x,
+ Hnl (x,
4.3)
(x —
x — s)
H (x, t,
+4
s)
Ii (x,
+4
t, s)
(t) dt
s)
(t) dt.
rurther, taking account of the definition of the spectral kernel 0 (x, s; A),
identity (4.2) can be rewritten in the form
he
X){
4.2')
s;
jx—sj
0,
)ifferentiating (3.4) with respect to x and subtracting the result from 4.2'), we obtain d
44)
(x,
(x, s;
s
—
W (x, s, e)
=
—4
(s — x)
(3
wifi now transform the right side of (4.4). From the definition of matrix W(x, s;
follows
tV (x, s;
(x, x — s) —
s—
_stgn(x_s)4{H(z, k—si, s)g8(jx—sj) R(x, e
+4
(x, t, s)
(t) dt +
(aS,
t, s)
(t) dt
X—8J
11+ 12 + 13 + 14.
402
X. ASYMPTOTICS OF THE SPECTRAL KERNEL
From the definition of the functions k(x, t) and l(x, t) (cf. (2.8) (2.10), Chapter 9) it follows that
+
—x) =
(4.6)
+
= j-1p(x)+r(x)1
(4.6')
On the other hand, from (3.9) we have [± (s — x)]
(4.7)
(X) 5
from (4.5) can be represented in the form
Therefore the term
=
=
(x, s — x)
(X) 5
±
(4.8)
(x, x — (X)
—
5
s) 5
{tIIkv (x, s — x)
r,
j
x—:s)
Since —
=2e
—1
=2e
•
•
e
s—x can
2
X —p---
-
s—x
be rewritten as 5
s—x)e
(4.8')
x—s)e
ax1 2 j
2
§4. DERIVATIVES OF SPECTRAL KERNEL
403
q(x, s; ,
inserting the expressions for s — x) and and (4.6') in (4.8') and using (4.9), we obtain
= p (x) +r (x)
eui(x, a;
a;
(X) d)
(x, x — s)
from
sin X)]
irther, taking into account the definition of the matrix
ii
a; X) —
2i
rjT
—
sinq(x, s; \—COS q(x, s;
( ...g (4.9), we
a; X)
X)
eosq(x, s; ?.)\
A)
sin q (x, s; A) J
s;
find that
A(x,s; A) = A (z, s) cos X
— A (x,
s) T sin X
A(x,s)=A(x,s;O); erting the expression for A (x, s; A) from (4.10') in (4.8") and putting s;
p(x)±r(z)
L
btain for
S—X
A(z, s)
—r 1—co5X(s—s)1 S—X
J'
the final form 8
A).
now transform the term '2 from the expression (4.5). us assume that the function g, (t) has a piecewise continuous and monotone second derivative. Then it easily follows from (3.7)
by parts and taking account of the conditions imposed
405
§4. DERIVATIVES OF SPECTRAL KERNEL
there
=
(x,
jX—81
— sign (x— s)
{H (x,
—s
4.18)
I' s)
.
Slfl
+I1(x, —k—si,
s)e
A
2
k—si
x—sI
rinally, we transform 14. To do this we put (s,
s;
A)
= 4-
(x, t, s) —fx--aj
(x, s; A)
(x,
4-
t, s) eiXtdt.
'hen by the classical Parseval equality there follows from (3.7) and 4.19) respectively
1)
(x, t, s)
4-
(t) dt
= 4—
(x s, A) di
(A)
Ix—81
—Ix—aI
(x, t, s) g1 (t) dt =
4-
(x, s; A)
p, (A)
4—
we obtain for 14 the final form 420)
s,
there
421)
(x, s, A) =
(x
s, v)
+
(x, s, v)J dv
luis, taking account of (4.13), (4.16), (4.17) and (4.20), the identity 4.4) can be rewritten in the form (A)
s;
A) =0,
there we have put R(x,s,A)
s, ?)_O*(x, s, A))—
s, i)
7he identity (4 22) and a Tauberian theorem of Levitan make it
_______________
406
X. ASYMPTOTICS OF THE SPECTRAL KERNEL
possible, similariy to what was done in obtaining (3.18), to study asymptotic behaviour of the derivative with respect to x of the spectra We first prove the following lemma. kernel O(x, a; A) as A —* LEMMA 4.1. If the functions p(x) and r(x) have derivatives which we have summabie over every finite interval, then for a a+1
\J {RIk (x, s; X)) = 0 (a)
(4.24)
(i, k = 1, 2),
where Rth(x,s; A) are the elementsof the matrix R(x,s; A) defined by The estimates (4.24) hold uniformly as x and s vary over any finite intervai
PROOF. To prove the lemma it is sufficient to obtain the (4.24) for every element of every term in the expression (4.23).
the elements of the matrices öO(x,s; A)/ôx and aO*(x,s; A)/ôx 1 =1 estimate (4.24) follows from (2.18) and (2.29) upon them. For the elements of the matrices ai(x, s; A) (k = 1,2,3) estimate (4.24) follows immediately from the explicit form of ti.. matrices. As for the matrix a4(x,s; A), by virtue of the continuity (x, t; s) and (x, t; s) (cf. (4.19)) and the Riemani t of the matrices Lebesgue lemma it follows that Jim
(x, s; A) = Jim
(x, s; A)
0
(1,
j
= 1, 2),
uniformly for x and s in any finite intervals, where 4(x,s; A) A) are the elements of the matrices a a-f-I
V
a-f-i
(x, s;
A)) = \J
(x, s; v)
+
—___________
s;
a+i s;
s;
which proves the lemma. THEOREM 4.1. If the functions p(x) and r(x) have derivatives over every finite interval, then for every fixed x and s we '. (4q35)
0 (x, s,
s,
lbs (I$)*pWSW formula holds uniformly as x and s vary over any
407
§4. DERIVATIVES OF SPECTRAL KERNEL
PROOF. By virtue of the estimates (4.24) and the identity (4.22), Tauberian theorem is applicable, according to which the gymptotic formula —
s,
v)=O(1)
olds uniformly as x and s vary over any finite intervals, or by virtue f the expression (4.23) for R(x,s; r), $.26)
et
s;
v)}=O(I).
us now calculate each of the integrals Ak =
1, 2, 3. 4).
s; v)
—
rom the definition of aj(x, s; ,),
from (4.12), there follows
o
p(x)±r(x) p
(x) ± r (x)
A (x, s)
(i —
[cos v (s — x) — T sin v (s — x)] dv.
ürther, v(s—x) — T sin v(s—x)}d'i
—
a well-known formula (cf Titchmarsh [3], §7 1) 127)
—
berefore we obtain for A1 the final expression 128)
A1 = p (x) +r(x) )C1/2A (z s) 2V2it
"12
[A (s — x)]
(S_X)'z
408
x. ASYMPTOTICS OF THE SPECTRAL KERNEL
From (4.15) we obtain for A2 the expression [A
A9
2
s) + TI
[e
r I
s—x
— 2
V
1 (s — x) 1
8—x
2
8—x
2
e
I
2
sin
dv,
it follows that (x, s) + T].
(x, X) + T]
(4.29)
We now determine A3. Using (4.18), we have 2
2
2
]
from which after simple computations we obtain for A3 the (H (x, x — s
(4.30)
,
s)
+ fi (x,
— x—s
j,
s)}.
Finally, we must determine A4. By (4.19) and (4.21) we have
(i —A
(x, t,
dt + —Ix—81
the order of integration and proceeding as above, we
409
REFERENCES —IX—81
+
(x,
(i —
t, s)
cos vt
dv] dt,
om which by virtue of (4 27) we find that 1
A4
=
(x, 2
t, s)
1, (At)
dt
L31)
+
2
(x, t, s)
Içt) dt. -,,
irom (2.12), (2.14) and (5.11'), Chapter 9, it easily follows that — A2, and from the asymptotic behaviour of the Bessel functions e obtain A2 + A4 = 0(1). Therefore the asymptotic formula (4.25), consequently the theorem, follows from (4,26) by virtue of (4.28) 14.31). This proves the theorem,
Bibliographical references
The results of this chapter are due to Sargsjan [5,9,12].
CHAPTER 11
EXPANSION, AND DIFFERENTIATION OF AN EXPANSION, WITH RESPECT TO THE EIGENFUNCTIONS OF A DIRAC
§1. Some preliminary estimates We will consider the eigenvalue problem
°1d
(1.1)
p(x\
\
0
o
(1.2)
=0.
I is the 2 X 2 unit matrix, and h is an arbitrary real number T functions p (x) and r(x) are real functions, defined on the halfline and summable over every finite interval. In the first section of the preceding chapter we obtained the identi (ci. formula (1.13), Chapter 10)
where
(1.3)
p(x,
s,
where the vector-function
\ \4,2(x, A) i
before, denotes the solution of equation (1.1) which satisfies the hill conditions as
(1.4)
çpi(0,A)
=1,
4'2(0,A) =h,
and the matrix W(x,s,E) is defined by (cf. (1.14), Chapter 10) W (x, s,
€)
A (x, s)
(x — s)
(1.5)
+ 4. Let
(H (x, t,
s) + II (x, t,
1x81
f(x)
(fi(x) \f2(x) 410
s))
g (t) dt.
411
§1. PRELIMINARY ESTIMATES
cx>). We will denote by F(X) an arbitrary vector-function from generalized Fourier transform of f(x), i.e. we put
X)ds.
and follows from the identity (1.3) that the functions are, for each fixed x, the generalized Fourier transforms I the vector-functions which are equal respectively to (x,
s, s)\ s,
e))
(x, s,
and
e)
s x + and equal to zero outside this interval. Therefore the generalized Parseval equality (2.4) of Chapter 10, it follows from
rx—
) and (1.6) that
W(x, s, e)/(s)ds. t us introduce the vector-function
= F (X) p (x, X)
= ::n (1.7) assumes the form ;?.
X)=
W(x, s, s)f(s)ds.
wifi now write down the formula analogous to (1.9) for the problem Li) + (1.2) with p(x) = r(x) 0. To this end we wifi find a new repre- )fl for the vector-function S(x, X). Inserting the value of F(X) ii (1.6) into the expression (1.8) and then changing the order of intelion, we obtain
S(x,
'(s) p (s,
(x,
X)dp
=
(s)
(s, A) +12(s)
p1(x, X) (s, A))
A))
— (/1(s)
(s,
X)+12(s)p2(s, A)p1(x, A)
f, (s)
(s,
A) +12(s) P2 (s,
A)
XI. EIGENFUNCTIONS OF A DIRAC SYSTEM
412
A) p1(x. A)
p2 (s, A) p1 (x,
X)'\ (li(s)
(s, A) P2 (x, X)
P2 (s, A) P2 (x,
A))
(s,
X)\
Therefore
p2(s, A)) t\f(s)) =(p (x, A) cpT(s, A))j
A),
\p2(x, A))
(s)
by the definition of the spectral kernel O(x, s; A) (cf.
Chapter 10) (1.10) assumes the form
S(x, X)=50(x,
(1.11)
s;
X)/(s)ds.
Consequently the vector-function analogous to S(x, A) for the probi..
0 can be defined by
(1.1) +(1.2) with p(x) =r(x)
=
(1.12)
(x, A)
= 50* (x,
s; A) I (s) ds,
where the matrix O*(x, s; A) is given by (2.27), Chapter 10. Further, for p (x) = r(x) 0 it follows from the ddlnition of the ii tionsk (x, t) and 1 (x, t) that k (x, t) =1 (x, t) = f, and the matrices H(x, and H(x, t, s), as solutions of the Goursat problem which equal zero
the characteristics, are identically equal to the zero matrix Thus I the case being considered, (1.9) has the form
X)=4-H5g,(s—x)f(s)ds (1.13) X 4-I
5
(x —-s)f(s)ds.
Now, subtracting (1.13) from (1.9), we obtain
X)_S*(x, A)) (1.14)
=
5
{W(x, s; e)—
LEMMA 1.1. Ifthecoefficientsp(x) andr(x) in equation (1.1) are corn then for any fixed x we have, as a —*
§1. PRELIMINARY ESTIMATES
413
a-fl
a+l
V
5
F(X)Hpk(x, X)Idp(X)=o(I)
(k=1, 2). estimates hold uniformly in every finite interval.
By the definition of the vector-function S (x X), i e (1 8),
PROOF
have a+l
V
=
(x,
a+l
F (A)
J
(x,
A) I
(k
dp (A)
=1, 2).
rierefore according to the Cauchy-Bunjakovskii inequality it follows that (a-fl
V
\'12 (a+1
IF(X)
A))
12dp
(k=1, on
(x, A)j2dp(A))
(A)) 5
2).
the one hand, by virtue of Lemma 2.1, Chapter 10, a+1
(k=1, 2), 5
4 on the other hand, since by the Parseval equality
F (A) 2 dp (A) = If (x) 2 dx s
<
obvious that as a —i a1
-
F
(A)
dp (A) =o (1).
5
estimates (1.15) follow from the estimates (1.17) and (1.18), which
oves the lemma. prom the definition of the vector-function S*(x, X), i.e. (1.12), and the explicit form of the matrix O*(x,s; X) (cf. (2.27) and (2.23), 10) there immediately follows: LEMMA 1.2. 14k have, uniformly over the entire haifline [0, w) as a—* cx, a+l
A))=o(1)
(k=1, 2).
414
XI. EIGENFUNCTIONS OF A DIRAC SYSTEM
We will now take up the question of estimating the variation of derivatives of the vector-functions S(x, A) and S*(x, A) with x. To this end we wifi consider the identity (1.15), Chapter 10, i.e. i identity (x, A)
(1.20)
ÔW(x s•
(A)
ds.
p (s,
It follows from this identity that for fixed x the functions (x, (k =1,2) can be regarded as the generalized Fourier transforms of t vector-functions which are equal respectively to
(/.Wkl(x, s; I
(k=1,
I
2)
s;
x + and equal to zero outside this interval. Therei by the generalized Parseval equality (2.4), Chapter 10, it follows f the identity (1.20) and the equality (1.6) that for x — < s
(x, A)F (A)
(x,s; s)
(A) dp (A)
f(s) ds,
which, using the definition of the vector-function S(x, A), assumes the i (1.21)
(A)
For the case p(x) = r(x) (1.22)
(x, s; s)
= = lB
og, (s— x)
where (1.23) We
B—
have taken into account here that —x)
ax
— —
f(s) ds.
0 the analogue of (1.21) is, by virtue of (1 OS*(:, A)
(A)
oW
A)
—s)
and
subtracting (1.22) from (1.21), we obtain
f(s) ds,.
1
415
§1. PRELIMINARY ESTIMATES
d
ÔS* (x,
(x, A)
A)
LEMMA 1.3. If the coefficients p(x) and r(x) in equation (1.1) have which are absolutely continuous in every finite interval, then
a,
Dr every fixed x we have, as a÷1
X)jdp(X)=o(a),
IF(X)I .
{
L.25)
(k=1, 2). estimates hold uniformly in every finite interval.
By the Cauchy-Bunjakovskii inequality it follows from the efinition of the vector-function S(x, X) that a+1
\ 1/
JFQ)12dp(A))
[26)
(k=1,2). hen the estimates (1.25) follow from the inequalities (1.26) by virtue (the estimates (1.18) and (2.17), Chapter 10. This proves the lemma. Proceeding in a similar way and making use of the estimates (2.19), rhapter 10, and (1.18), it is easy to prove the following:
1.3. If the coefficients p(x) and r(x) in equation (1.1) have which are absolutely continuous in every finite interval, then
r
every
fixed x we have, as a—4 a+1
IF(X) H
(k=1,
X)I
2).
estimates hold uniformly over every finite interval.
From the definition of the vector-function Liows:
(x, A) there immediately
416 LEMMA 1.5.
(1.28)
XI. EIGENFUNCrI0NS OF A DIRAC SYSTEM
have, uniformly over the entire haifline [0, co) a+1
,,
v
*
)}
_—o(a")
as a—* a
(k=1, 2).
§2. Proof of a theorem on equiconvergence
Basing our argument on formula (1.14) and on Levitan's Tauberii theorem (cf §4, Chapter 14), we will m this section give a proof of a theore on the equiconvergence of the expansion with respect to the eigenfunctio
of the problem (1.1) + (1.2) and of the simpler problem (2.1)
for an arbitrary vector-function which is square integrable on [0, We remark that if h = 0 in (2.1), then from (2.23), Chapter 10, we =sinXx,
=cosAx,
and therefore in this case the expansion of a vector-function from ..V2(0,
with respect to the eigenfunctions of the Dirac system is the same as expansion as an ordinary Fourier integral. We now take up the problem of transforming the right side of (LI Inserting the expression for the matrix W(x, s; in (1.14), we obtain {S (x, X) — S* (x, X)}
(x)
s_x)_—4]g,(s._-x) (2.2)
+HT[l(x, x—
+4
ds H (x, t, s) g, (t) dt
X—!
{ Jx—8I
X—8
+
0 (x,
Changing variables, we have
t, s) g8 (t) dt f(s) ds = + '2
417
§2. EQUICONVERGENCE
11==H
we now put
A)=
{l(x, l)_4}f(x
(x,
(x, X) ri
(x,
—4--
(x, )),
by the Parseval equality for the classical Fourier transform we find, (2.4) and the equality (cf. (1.10), Chapter 10)
('i) =
(t)
hanging the order of integration in the second integral in the right side (2.2), we bring it to the form £
H(x, t, s)f(s)ds x—t
-0 x—t
17(x, t, s)f(s)ds —s
t us further put
H(x,t,s)f(s)ds, (b1(x, t)
(x, t))
=
x-t
R(x, t, s)f(s)ds,
418
XI. EIGENFUNCTIONS OF A DIRAC SYSTEM
and b(x,
(2.8')
Applying the Parseval equality for the classical Fourier transform, obtain from (2.8') and (2.5) (2.9)
B
(x,
Therefore, taking into account (2.3), (2.6), (2.7) and (2.9), formi (2.2) can be rewritten in the final form (2.10)
where v)
(2.11)
+
v)} dv.
We will now prove several lemmas.
LEMMA 2.1. If the coefficients p(x) and r(x) are summable over finite interval, then for every fixed x we have, for a —* ai-1
(2.12)
V{Rk(x, 'A))=o(I)
(k=1, 2),
'R1 (x,
R(x,
These estimates hold uniformly in every finite interval.
PROOF. Since the total variation of a sum does not exceed the the total variations of the summands, to prove the estimates (2.12) sufficient to show that these estimates hold for each term of the (2.11). For the elements of the vector-functions S(x, A) and S*(x, A estimates (2.12) coincide with the estimates (1.15) and (1.19). Eu since the vector-functions
H[k(x, t) — ]f(x +t) and HT[l(x, t) — ]f(x
— t)
are summable, the Riemann-Lebesgue lemma is applicable, accu to which urn X÷co
X)=0
(k=1, 2),
/cL (x,
X)= (\
X)
'U,
419
§2. EQUICONVERGENCE
Therefore
(k=l, 2).
(2.13)
Since the vector-function b(x, t) is even differentiable, the estimate (2.12) for the functions
v)dv
(x,
(k=1, 2),
32(x,
be obtained from the Riemann-Lebesgue lemma in the same way t the estimates (2.13) were obtained. This proves the lemma. Let p(x) and r(x) be measurable and locally bounded, and Then for every f(x) be an arbitrary vector-function from
LEMMA 2.2.
<x
we have
Ibk(x,t)I
t)
is
(k =1,2).
We wifi prove (2.14) for the function b,(x,t);
the
proof for
analogous. By definition b, (x,
{H11(x, t, s)f,(s) + H12(x, t, s)f, (s)}ds.
Let x be an arbitrary point of the halfline [0, cx). Since the integral x we obtain h respect to t in (2.8') has the limits choosing 0, x — t 0. Therefore, applying the Cauchy-Bunjakovskii into the integral in the right side of (2.15), we find that
-
x+t
jb,(x,
(J
ds)
+
/2(s) ds)
= Ct12,
we have put
max IH,k(x,t,sH
x—tSsSx+t
(k=1,2).
EMMA 2.3. Under the hypotheses of the preceding lemma we have, for ry fixed x,
xi.
420
OF A DIRAC SYSTEM
(k=1, 2).
v)dv=O
urn
(216)
These relations hold uniformly in every finite interval.
PROOF. As in the proof of Lemma 2.2, we wifi prove (2.16) only (x, v). From its definition, i.e. (2.8), follows
the function
v) dv
b1
(x, t)
dv
(2.17)
dt.
Further, since it follows from (2.15) that b1(x, 0) = 0, by virtue of (2.1
we have b1(x, t)
— b1(x,
> 0). Therefore the assertion
0) =
follows from (2.17), thanks to (3.17), Cha
the lemma regarding
10. This proves the lemma.
2.4. If the coefficients p(x) and r(x) are summable over finite interval, then for any fixed x we have (2.18)
(k=1, 2).
ak(x,. v)dv=0
urn —x
These relations holds uniformly in every finite interval.
PROOF. Once again we will prove the lemma only for a1 (x, v). the definition of ai(x, v), i.e. (2.4) —(2.4'), it follows that (X, .v)
(x, v) .4..
t)
(x, v).
i
—
(x,
v) dv:
=
t)
—
(x —
421
§2. EQUICONVERGENCE
=2 5
{[k (x, t)
+ 25{[l (x,
(x
—
+ t) + i [k (x, t) —
dt
12 (x + t)}
(x— t) — i [i (x,
t)
dt.
Changing variables in the second integral in the right side, we obtain
=2 5
(x,
1(2.19)
+2
L Since k (x, t) + 1 (x, — t)
[k
5
(x, t) +1 (x, —t) — ] /1 (x + t)
[k(x, t) — lxx, —t)]/2(x+ t)
= cos q (x, t),
k (x, t) — 1 (x, — t)
SthXt
dt
dt.
= i sin q (x, t),
where we have put q(x,
can be rewritten in the form (x, v) dv = —4 5
sin2 5
{4-
+2i 5
q
(x, t)} /1 (x + t)
SlflXt
sin (q(x, t)}12(x+ t)
dt SlflAt
dt,
rom which (2.18) for the function ai(X, v) follows on the basis of the lemma. This proves the lemma. Lemmas 2.1 —2.4 easily imply the following, which is basic for this section. cx). If the THEOREM 2.1 (oN EQUICONVERGENCE). Let f(x) E aefficientsp (x) and r(x) in equation (1.1) are measurable and locally bounded,
we have, uniformly in every finite interval, L20)
[Sk(x,x) —Sk(x, —X)]
—x)fl =0,
difference between the expansions of the vector-function f(x) as a Fourier integral with respect to the eigenfunctions of the problems Li) —(1.2) and (2.1) tends to zero uniformly in every finite interval.
422
XI. EIGENFUNCPIONS OF A DIRAC
PROOF. By virtue of the estimates (2.12) and the equality (2.lt Levitan's Tauberian theorem (cf. Theorem 4.1, Chapter 14) is on the basis of which we have for every fixed x Jim {
A)
—
A) } = 0
(k
1,2),
i.e. according to (2.11)
lim {[Sk (x, A) — Sk (x, —A)] —
(x, A) —
(x, —A)])
v)}dv.
Then (2.20) follows by virtue of Lemmas 2.3 and 2.4. This proves theorem. Theorem 2.1 gives a definitive solution to the question of the conv
of the expansion of a vector-function, which is square integrable c [0, ix), with respect to the eigen-vector-functions of a one-dimensioi Dirac system. Indeed, if we put h =0 in the problem (2.1), then an pansion as a generalized Fourier integral with respect to the eigen-vecl
functions of this problem is just an ordmary Founer integral since, as was noted at the beginning of this section, for h = 0 the I vector-functions of this problem are #(x,A) =
/ cos Ax \ sm Ax
0 Theorem 2.1 can be formulated as aD) and that the hypo THEOREM 2.2. Suppose that f(x) E
Consequently, for h
of Theorem 2.1 are satisfied. Then the difference between the expansi f(x) with respect to the eigen-vector-functions of a one-dimensional
system and its expansion as an ordinary Fourier integral tends k uniformly in every finite interval.
In particular, there follows from Theorem 2.2: THEOREM 2.3. Suppose that the hypotheses of Theorem 2.1 are Then at every point x0 at which local conditions for the expandibility c asan ordinary Fourier integral are satisfied, we have lirn{S(xo,A) —S(x0, —A) } 'f(xo),
z e the expansion of f(x) as a generalized Fourier integral with resj*
423
§3. EQUISUMMABILITY
the eigen-vector-functions of a one-dimensional Dirac system tends at the point x0 to the value f(x0). §3. Proof of a theorem on the equisummabiity of differentiated expansions
In this section, basing our argument on (1.25) and the aforementioned Tauberian theorem of Levitan, we will prove a theorem on the Riesz equisummability of the differentiated expansions with respect to the eigen-vector-functions of the problem (1.1) + (1.2) and of the simpler problem (2.1). We will first take up the problem of transforming the right side of (1.24). inserting the expression for the matrix W(x, s; from (1.5)" in (1.24),
ñrst differentiating with respect to x, we obtain 'X' d 31's\)
(x, X) X1,
Ox
+
OS* (x
X)
—
[A(x, s) —
ii (x, —t, s))
i
f(s) g; (s — x) ds
I
+
(x, X—I
t, s) +
s)J
(t) dt f(s) ds
= '1 + '2 + 13 +
where
(3.1')
(x,
JX—8l
(cosq(x, s), —sinq(x, s)
A(x, s)= \ sin q (x, s),
cos q (x, s)
and
(311 1)
q(x,
For our purposes we can restrict ourselves in the sequel to the case of an even function For this case the expression for W(x,s; simplifies somewhat.
424
XI. EIGENFUNCTIONS OF A DIRAC SYSTEM
Suppose that the vector-function f(x) is differentiable in an E-neighborho of the point x. Then integrating by parts in 12, we can bring it to the L.. X+B
ÔA
'2
f(s) g8 (s — x) ds
[A (x, s) — I] /'(s) g, (s — x) a
+
Therefore
+ =
{[OA (x, s)
s)
+
+[A(x, s)_J]f'(s)}g,(s-_x)ds, from which we obtain, by the change of variables s — x = t,
+ =
{[ÔÁ
s)
(x, s)
+
/ (x
+ t)
/ (x
+ t)
(3.2)
+[A(x, If we now put (3.3)
(x,
{[ÔA
s)
=
8)]
+
+[A(x, then it follows from (2.5) by the Parseval equality for the Fourier t. form (3.4)
Ii+12=
X)dX=
Further, by a change of variables the integral 13 can be represer in the form {ff(x, JtJ,
x+t)+fl(x,
— Jtf,
and therefore, putting 2)
Note that for even g,(t) the relation (2.5) assumes the form
=
(t) cos At dt.
425
§3. EQUISUMMABILITY
X)=—
{H(x, ltI,x+t)
±fl(x, —jtj, x+t))/(x+t)cosXtdt and again applying the Parseval equality, we obtain from (3.5) and (2.5) X)dX
(3.6)
Finally, we transform the integral 14. Changing the order of integration, we find that 14
=
I
x+t
[H (x, t, s) + .R (x, —t, s)J f(s) ds g, (t) dt. x—t
Now, putting I
(37) y(x, X)=
./—[H(x, t, s) +R(x, —t, s)J/(s) ds} cos Atdt —s
and
applying the Parseval equality, we obtain from (3.7) and (2.5)
i(38)
14
(A)T(x, X)dXz
Consequently, by virtue of (3 4), (3 6) and (3 8) we can rewrite (3 1) the form X)==O,
(3.9) where X)
— ÔS*(x
:(3.1O) (x, v) dv
*
y (x, v) dv
We will precede the proof of the basic theorem of this section with veral lemmas.
LEMMA 3.1. If the coefficients p(x) and r(x) and the vector-function x) have locally bounded derivatives in every finite interval, then for every
d x we have, as
XI. EIGENFUNCTIONS OF A DIRAC SYSTEM
426
0+1
V
= o (a)
(x,
(3.11)
(k
= 1, 2),
(x, X)
X)=
(I)(x A) These estimates hold uniformly in every finite interval.
It is obvious that to prove the estimates (3.11) it is
PROOF.
to prove them for eaëh term appearing in (3.10). For the component of the vector-functions ÔS(x, A)/öx and aS*(x,A)/ax the estimatC (3.11) follow from the estimates (1.25) and (1.28). Further, since vector-function ÔA(x, S)]
s) L
(3.12)
ox
Os
f(z+t) t)
+[A(x,
is summable (in a neighborhood of the point t = 0), the Riemanr Lebesgue lemma is applicable, according to which hmak (x, A) =
(k = 1, 2),
0
(x, A) =
Therefore a-f.
v)
V
=
a-f-i
Jah(x, is)Id'.=o(1) (k=1, 2).
Further, proceeding similarly, we conclude from the definition of ti vector-functionsui(x, A) and 7(x, A) and the Riemann-Lebesgue lemma th
X)dv)=o(I)
(lc=l, 2). This proves the lemma. LEMMA 3.2. Suppose that the vector-function f(x) has a p derivative in an i-neighborhood of the point x, and that the equation (1 1) have locally bounded first derivatives Then we wzdr(x) in
(3.13)
(i
tim ft
PROOF
—
(x,
=0
(k = 1,
2).
From the definition of the matrix A (x, s), i e from
and (3.1"), there follows
427
§3. EQUISUMMABILITY
_!h(x)(
(3.14)oA(x, 8)
(3.15)
—
àÁ (x, s) us
8X+t
T
sinq(x, x+t) cosq(x: q (x, x + t)J'
x + t)
h (x+ h ) (
sin q (x, x + t) cos q (x, x +
'\COS q (x, x + t) sin q (x, x + t))'
where (3.16)
h(x+c)d'c, h(x)=p(x)+r(x).
q(x,
Further, by virtue again of
(3.17)
we have
A (x, x + t) — I / 2 sin .211 q (x, x +
t)
tT
q (x, x + t)
q
(x, x + t)}
We wifi now prove (3.13) for the function a1(x, v). The proof for a2(X, v)
is similar since, as is not hard to see, these functions contain terms of same type. From the definition of ai(x,v) (cf. (3.3)) it follows that it is the Fourier cosine transform of the first component of the vectoriunction (3 12), which by (3 14), (3 15-) and (3 17) has the form
x+t)
B1(x,
x+t) x+
t)
t)
= + b2 + b3 + b4 herefore aj(x, X) can be represented in the form t)cosXtdi
onsequently, to prove the lemma we have to prove the relation 319) e
Iirn±
will prove (3.19) for each term separately. We have
428
XI. EIGENFUNCTIONS OF A DIRAC SYSTEM
t)
from (3.18) and taking account
(x
+ t) sin q (z, x + t)
(4.27), Chapter 10, we find that
)<
I,
(Xt)
1's,
dt.
Since by hypothesis the functions p(x) and r(x) have locally derivatives, it follows from the definition of the function h(x) that ast—40
h(x+t) —h(x) = 0(t).
(3.20)
Therefore, using also the boundedness of the function fi (x + t), we obtair
from which it follows that as A (3.21)
I1—+0.
For the term in (3.19) which contains b2(x, t), the proof is compleU analogous. The proofs for the last two terms are similar. We will car
out the proof for the last term. By definition we have
14(1 =2
b4(x,
t)J/
Is..
dt.
Inserting the expression for b4(x, t) from (3.18), we obtain
14=2
f (x + t) sin q (x, x ± t)
dt.
429
§3. EQUISUMMABILITY
Since by (3.16) we have sin q(x, x + t) = 0(t) for small t, it follows that (x
I,1(Xt)
+ t) I It I
dt
4(xt)
If' (x + t) 2 dt)
/
2
(u)
c
I
u
3
Hence
14
as A
0
dt) \t/, d
<
w, which completes the proof of the lemma.
cx). If the coefficients p(x) and r(x) in LEMMA 3.3. Let f(x) E equation (1.1) have locally bounded derivatives, then for every fixed x we have (3.22)
v)dv=O
urn
(k=1, 2).
These relations hold uniformly in every finite interval.
PRooF By the differentiabthty of the matrix
H(x, ItI,x+t) —ItI,x+t) with respect to t, as well as (2 16), (2 17) and (5 12), Chapter 9, we have (for small t)
—ItI,x+t) =Ct Therefore the proof of (3.22) can be carried out in exactly the same way as was the proof of (3.21). cx). If the coefficients p(x) and r(x) in LEMMA 3.4. Let f(x) E equation (1.1) have locally bounded derivatives, then for every fixed x we have
v)dv=O
urn
(k=1, 2).
These relations hold uniformly in every finite interval.
PROOF. Proceeding as in the proof of Lemma 2.2, we establish hat as t—40 Ici(x,t)I =2Ct
(k=1,2),
where
C (x, t)=
=
t, s) + 17 (x, t, s)j f(s) ds,
430
XL EIGENFUNCTIONS OF k DIRAC SYSTEM
{H (x, t, s) + ii (a,, t, s))
max
C
Then using (3.24), the proof can be completed in the same way as
of (3.21).
Erom Lemmas 3.1 —3.4 and (3.9) it easily follows that: 3.1 (oN EQuIsuMMABILITY) Suppose that the vector-functi a) and satisfies the following conditions: 1(x) belongs to 1. the vector-function 1' (x) is square -summable in some neighborhood
of the point x0, 2. the coefficients p(x) and r(x) in equation (1.1) have locally bouna first derivatives. Then we ha ye (3.25)
3
k. —
\d
J ôSk (x, A) •?l dx
—
(x, ax
A) J
(k==1, 2), i.e. the difference between the Riesz means of order one of the derivatiL of the expansions of f(x) with respect to the eigen-vector-functions of problems (1.1) —(1.2) and (2.1) tends to zero at the point x0.
PROOF. By Lemma 3.1 and (3.9), Levitan's Tauberian applicable, according to which (3.26)
Urn Co —
(1
—
(x,
v)
=0
(Ic =
1,
2).
The relation (3.25) and thus the theorem follows from (3.26), by
of Lemmas 3.2, 3.3 and 3.4.
If we put h =0 in the problem (2.1), then, as was noted in expansion with respect to the eigen-vector-functions of this p coincides with an ordinary Fourier integral expansion. Therefore Theorem 3.1 and Theorem 4.2, Chapter 8, follows: THEOREM 32 (oN SUMMABILITY) Suppose that the coefficients afld:r(x) in equation (1.1) have locally bounded first derivatives, and the following conditions are satisfied: cx>). 1. 1(x) E The trector-function f' (x) is continuous at some point x0.
2.
1
431
REFERENCES
Then
/ —
ox
=I
(x0)'
i.e. the first derivative at the point x0 of the expansion of f(x) with respect to the eigen-vector-functions of a one-dimensional Dirac system is summable by the Riesz method of order one to the value f' (x0). BibliographicaJ references
The results of this chapter are due to Sargsjan [6,10,12].
CHAPTER 12
ASYMPTOTIC BEHAVIOUR OF THE NUMBER OF EIGENVALUES OF A STURM-LIOUVILLE OPERATOR An integral equation for the Green's function
1 From the results of Chapter 4 we know that if the function q(4 m the Sturm-Liouville operator
Ly=—y"+q(x)y
(11)
(a<x
bounded from below and tends to + as x —* a or as x b (or both), then the spectrum of L is discrete 1) In this chapter we will give another proof of this fact for the cases: Besides the condition and 2) a =0, b = + 1) a — b=+ is
+
(for x—-, ± o) on the function q(x) we will impose conditions which will enable us not only to prove the discreteness of q(x)
spectrum of the operator L, but also to obtam an asymptotic
for the number of
for
Smce by hypothesis the iunction q(x) is bounded from below, WI may assume without loss of generality that q(x) > 1 for all x E (a b)4 We will consider m detail the case of the entire ime, 1 e the case a — b = + The changes required for the case of the halfinie not comphcated, and will be mdicated briefly We put for fixed x and p>O (1.2)
ii
and consider the mtegral equation
(13)G(x,
It will be shown below that for sufficiently large
equation (IJ
can be solved by an iteration method and that the solution is the Green Under the assumption that at least one of the boundary points is singular In connection if one of the boundary points is regular then we have to specify a
condition at this point. 432
§1. INTEGRAL EQUATION FOR THE GREEN'S FUNCTION
433
function of the operator (1.1). (The uniqueness of the Green's function follows from Theorem 2.3, Chapter 2.) 2. Let us denote by X the Banach of numerically valued functions A (x, (— <x, n < + cx) with the norm I!A(x,
and define an operator N in X by (14)
NA (x,
where the function g(x, is given by (1.2). Further on an important role will be played by the following: LEMMA 1.1. Suppose
that
q(x)
satisfies the
conditions:
1. For (1.5)
K> 0 and 0 1
B is
a constant.
Then for sufficiently large
the
operator N is a contraction
PROOF. Put
g(x,
=
+
g (x,
[q
(x)] A
m
>
en .7)
M(x, M2(x,
urther, 2)
For the definition of a Banach space cf. Chapter 13.
p)).
(x,
(i.e.
434
XII. ASYMPTOTIC NUMBER OF EIGENVALUES
a2(x,
(1.8)
g (x,
Ii
—q
(x)
IA
I
{
By virtue of the condition (1.5) we have for x —
1
—xl.
Therefore g(x, (
I
qa(x)g(x,
(K2 (
/
Inserting the value of the function g(x,
from (1.2), we obtain
(x)
Since ic = [q(x) +
]1/2,
we have as M —, qa(x)K_2+e
(1.10) where
0
+
=
since by virtue of 0< a <3/2 one can find <1 such
=1 + — a wifi be a positive number.
Then it follows from the estimates (1.8), (1.9) and (1.10) that a2(x,
Ix
(1.11)
Integrating the inequality (1.11) with respect to x from —
to
we obtain, following an obvious change of variables and
of the limits of integration (using also the
Cauchy-Bunjakovski
equality), a2 (x, 5
dx
5
Iu
du
5
I u'
dii'
435
§1. INTEGRAL EQUATION FOR THE GREEN'S FUNCTION
dii'
u'
dii
A (x,
12
dx}
{
Integrating now with respect to
is a constant. Let us now estimate $(x,
to
from —
+
we
obtain
where C
lix as
iz —p
By the definition of
+
have (x,
p.) =
g (x, I
— q (x)] A
p.) [q
>1
=
g (x, — q(x)
p.) q
g (x,
A
= +
A
prom (1 6) and (1 2) follows
x
exp
I
—
i) IA
I
x Iu'i>l ow,
just as in the derivation of the estimate (1.12), integrating the we obtain to + inequality with respect to x from —
integrating this inequality with respect to obtain the needed estimate for n; lix
from
by
as
—
—p +
436
XII. ASYMPTOTIC NUMBER OF EIGENVALUES
where C is a constant, and r is an arbitrarily large positive number. A similar estimate holds for Then by virtue of the estimates (1.12) and (1.13), we obtain (1.7) the following estimate for NA (x, ,) as + a': INA(x, which proves the lemma.
REMARK. Let us introduce the Banach spaces 1; r is an arbitrary real number), whose elements are ii valued functions A (x, <x, < cx), with the norms define (— respectively by: (p
(1.14)
A (x,
(1.15)
IA (x,
(1.16)
fJP(p) = sup
=
51 A (x,
5i A(x,
5
= sup
A (x,
dx,
12
{
51 A (x,
12
q2't
dii.
it is easily seen that a lemma analogous to Lemma 1.1 holds f: of these Banach spaces. 3. We will again consider the integral equation (1.3), which ca written, using the definition of the operator N (cf. (1.4)), in the I (1.17)
G(x,i7;1i)
Since for sufficiently large i the operator N is a contraction i of the Banach spaces considered above, it follows that if g(x, belongs to one of these spaces, then equation (1.17) can be solved b
method of iteration and its solution belongs to that space. In the following lemmas we wifi indicate sufficient conditions i function g(x, to belong, respectively, to the spaces X, and (r> 0). LEMMA 1.2. If q(x) >0, then the functions (p > 0), i.e.
and
belong to the space (1.18) (1.18')
sup sup
5
{g.(x,
5{fg(x,
< cx,
437
§1. INTEGRAL EQUATION FOR THE GREEN'S FUNCTION
i.e. from (1.2),
PROOF. From the definition of the function g(x, follows
=
2
whence (1.18) follows at once. The proof of (1.18') is analogous. LEMMA 1.3. If the condition g(x, belongs to the space X, i.e.
{ q(x)
}
312dx
<
is satisfied, then
(1.19)
PROOF. In fact, for
> 0 we have
CX)
CX)
p.))2 dij =
{g (x,
exp (-.--x j x —
2x3
(1.19) follows from the hypothesis of the lemma. LEMMA 1.4. If for x —
1 the function q(x) satisfies
(1 20)
the
condition
C,
Where C is a constant, then g(x,
p.21)
belongs to the space
ie
sup
PRooF. By the definition of g(x,
=
4 I
+
we
exp (—21 x —
have
J {q
(x) + p.]t/2}
I
4 [q
(—21 x —
From the condition (1.20) follows
I [q (x) + p.]V2)
=a(x)+b(x).
438
XII.ASYM?rOTIC NUMBER OF EIGENVALUES
(1.22)
exp(—2xJuf)du
=sup
Further, by virtue of (1.6) we have for x — I > I
1
Therefore, proceeding as we did to obtain the bound (1.22), we which together with (1.22) proves (1.21) and thus <
t
lemma. LEMMA 1.5. Suppose that for Ix —,,I <1 we
(1.20')
Then the function 82g(x,
belongs to the space
(x
z)
sup
(1.23) PROOF.
If
then
= q(x)
where
(1.23) can be obtained in the same i
Therefore
as was (1.21). LEMMA 1.6. Suppose
that q(x) satisfies the condition (1.20') and
the condition
a).
(1.24)
belongs to the space
Then
i.e.
a bounded n instead of the condition (1.20) one assumes that from above, but also from below by a positive number, then we obtain consequence.
By deñnition 2
5
{
g'I'2 (ii) } dii = lim
2
à2g 5
(ii) }
102g
+ Inn 51
-
439
§2. FIRST DERIVATIVE OF THE GREEN'S FUNCTION
PROOF. Put {g(x,
5
{g(x,
5
Ix—1
+
{g (x,
I
1
=a
p.))2
5
From the definition of g(x, 1 it follows for > 0 that
± (x).
and the condition (1.20') for
{g(x,n;M)
x—
—21x
Therefore (x)
C {q(x))
from which on the basis of (1.24) follows a(x) E Further, for Ix—nI >1
(—
2
cx,).
Cexp(—21x Therefore C 5 exp (—2ux)
b (x)
du =
C1e2z
cx'). Consequently a(x) +b(x) so b(x) also belongs to cx'), which proves the lemma. (—
§2. The first derivative of the function G(x,
Differentiating the equation (1.3) formally with respect to ij, we obtain
= ft —
:2.1)
5 g (21,
p.) [g
by K(x,
teplacing OG(x,
— q (x)]
in equation
(2.1), we obtain
he integral equation
K(x,
p.)
2
—
g (x,
p.)
[g
— q (x)} K
p.)
Therefore from Lemma 1 1 (cf Lemma 1 2, Og(x, !2)/8,) E he Remark followmg this lemma, and §1 3) it follows that K(x, x the function K(x,i,, It is also not hard to verify that for E contmuous with respect to
XII. ASYMPTOTIC NUMBER OF EIGENVALUES
440
Integrating (2.2) with respect to
from
—
to
we
obtain
K(x, 2
2'
= x the function ôg(x, /ô,i, and consequently K(x, has a discontinuity of the first kind, which does not prevent the cation of the Newton-Leibnitz formula. The existence of the i For
in the right side of equation (2.2'), and also the validity of interchangir
the order of integration, follow from the fact that K(x, i.e. from the fact that
E
sup
Equation (2.2') coincides with the integral equation (1.3). TI it follows from the uniqueness of the solution of (1.3) that
K(x, —0:)
Differentiating this relation with respect to
K(x,
we
obtain for
= aG(x,
Therefore the integral equation (2.2) can be written in the form —-a- = —
g (x,
[q
— q (x)]
g
(2.3) —
g(x,
loG
ogl
Applying estimates similar to those which we used in the proof the basic Lemma 1.1, one can show that the function l(x, . To do this it is sufficient to the improper integral which defines the function l(x, uniformly with respect to i, which follows easily from (1 6)
is contifluous with respect to
con
§3. SECOND DERIVATIVE OF THE GREEN'S FUNCTION
441
Now, solving the equation (2.3) by iteration, it is not hard to show the continuity with respect to of the function ,2)/a?, — ag(x,,,;
Therefore the function OG(x, has the same discontinuities as the function Og(x, The latter has a unique discontinuity for = x. In fact, if
=
x=
if
Therefore g (x,
(2.4)
Thus the function ÔG(x,
where except for the pomt first kind, and (2.5)
= —L
g (x,
—
is continuous with respect to every= x, at which it has a discontmuity of the
-—G(x,
'ri;
*3 The second derivative of the function G(x, i,,
In the preceding section we proved that the first derivative of G(x, i,; as an element of the space for any p> 1 (for the defimtion cf the Remark followmg the proof of Lemma 1 1 m §1), is of x, and satisfies the integral equation continuous with respect to for exists
g (x,
[q
— q (x)J
Fhis equation can be reduced to the form
there a
2?(x,
a
3.2)
l(x,
IJ.):
\
g(X,
OG
442
XIj. ASYMPTOTIC NUMBER OF EIGENVALUES
Differentiating the equation (3.1) formally with respect to ii, we obtain (3
(3
i)
(3.3)
Further, differentiating (3.2), again formally for the moment, respect to and using (2.4) (cf. also the footnote on p. 000), we
Let
us now consider the integral equation M(x,
(3.4)
\
g(x,
belongs to the space We wifi show that It thei follows from Lemma 1.1 that the solution of (3.4) also belongs to Banach space. We have 11(x,
g(x, —
g (x,
q (x)]
[q
—
From the conditions (1.5) and (1.6) it is not hard to deduce and hence to belongs to the space
Let us now consider '2.
Assuming
that the condition (1.20')
satisfied, we obtam
g(x,
exp { — — g(x,
q(x)t
I
+
+ _IL]"41
§3. SECOND DERIVATIVE OF THE GREEN'S FUNCTION
443
is a constant, depending only upon where Finally, using the condition (1.6), we obtain the following bound for 13: 13
g(x, >1
Therefore
('2 +13)q112(n)
The integral on the right can be bounded as in the proof of Lemma 1.1.
As a consequence it is easily proved that 11(x, space
sup We
belongs
(x,
(ii)
<
wifi now show that M(x, i.e. the solution of the integral is (3 4) (which, as was proved, belongs to the space for x to
Integrating the equation (3.4) with respect to
from
number), we obtain
Further, from equation (3.1) follows v.6) g (x,
if we show that 1.7)
to the
i.e.
E;
[q
q
(E,
—
to
is
a
444
XII. ASYMPTOTIC NUMBER OF EIGENVALUES
then from the uniqueness of the solution of equations (3.5) and it would follow that M(x, i.e.
M(x, (3.8)
p.)—%g(x,
Now since 82g(x,
/ô172
exists for
a2G(x,
x, it follows from (3.8) is an element of the Banach
exists for x and
Thus it remains for us to prove (3.7). It is easy to show from the condition (1.6) that the function
=
(3.9)
—
g(x,
can be differentiated under the integral sign. In fact, we choose N large that the point lies in the interval (— N, N), and represent t function l(x, in the form l(x,
p.).
and a3(x, ,1; can be differentiated u the integral sign, since following differentiation we obtain (and uniformly) convergent integrals, which easily follows from. condition (1 6) The function a2(x, 77, p) is an integral with fimte L and ita differentiation does not occasion any difficulties. Differentiating (3.9) with respect to and using (2.4) (cf. footnote on p. 439), we obtain
The functions a1(x, 17;
445
§4. FURTHER PROPERTIES OF THE GREEN'S FUNCTION
g(x,
ft).
p.),
which
coincides with (3.7), and the assertion is proved. §4. Further properties of the function G(x,
1.
G(x,
properties which were established above, the function
Besides
still other important properties, which will be derived
has
in this section.
x the function G(x,
LEIi&MA 4.1. For equation
satisfies
the differential
a2G/o,,2=$
(4.1)
PROOF. By (3.8) the equation (3.4) can be written in the form 'ç;p.)
q
(x)}
g
f)
—
g
(x
i)
{q
—q
(x))
G
from which it follows that
± p}g—çg(x,
g
this equation with equation (1.3), we conclude on the basis uniqueness of solutions that p.),
coincides with (4.1). This proves the lemma.
446
XII. ASYMPTOTIC NUMBER OF EIGENVALUES
is symmetric in the variables x
LEMMA 4.2. The function G(x,y; y, i.e. (4.2)
PRooF. Let x and y be two distinct real numbers, and suppose definiteness that x
i whose existence follows from the fact that G(x, E 82G(x, which we mentioned in the preceding s E
Put p)— G(x,
= Il + '2 Since
ii)— G(x,
G(y, E;
p.)], we have
= G (y, x; p.)
12=G(y,
G (x,
— G (x, x; p.)
p.)
y;
G (y,
p.)
G(x, p.)
G (x,
p.)
p.)
(ii'
Adding these equalities and taking (2.5) into account, we — G(x, y; az), which proves (4.2) since it follows
I = G(y, x;
(4.1)that 1=0.
+
2. If the function q(x) is bounded from below and q then the Sturm-Liouville operator has a discrete spectrum i
§5. DIFFERENTIATION OF THE GREEN'S FUNCTION
unique Green's function for all outside the spectrum (cf. Chapter 4,
447 §1).
It was shown above that the function G(x, satisfies all the conditions of the Green's function of the Sturm-Liouville operator. Consequently, by the uniqueness of the Green's function, G(x, is the Green's function. We know from Chapter 2 that the Green's function 0) and can be expressed in terms of the solutions (x, E (— —* cx) which are for the case q(x)—* as lxi E + essentially uniquely determined.
From the integral representation of the resolvent (cf. Chapter 2, a)) §3) it follows that for any function f(x) E a)
a)
(4.3)
G (x, y;
where
/ (g) dg
1
=
±
J.°.
the are the eigenvalues, and the (x) corresponding eigenfunctions of the Sturm-Liouville operator, i.e. Here
§5.
Differentiation of the Green's function with respect to
1. It follows from Lemmas 1.1 and 1.3 that if q3"2(x) E then the Green's function G(x,
,i) belongs
a)),
to the space X, i.e.
a) (c
(5.1)
dx
a)), then (5.1) may fail to hold. q3"2(x) Let us assume that for some integer r > 0 If
(5.2)
E ..V'(— a), a)).
(x In this case it is natural to consider the iteration z) of the Green's functions, and one may hope that this iterate will be a HilbertSchmidt kernel, i e will satisfy the condition (5 1) From the integral representation of the Green's function (cf Chapter 2,
*3) it follows that the iterated Green's functions coincide up to sign with respect to We will study in detail which corresponds to the first The case of higher denvatives (iterates) can be studied similarly, we will restnct ourselves in this case to simply stating the final results
From equation (1.3) follows
448
XII. ASYMPTOTIC NUMBER OF EIGENVALUES
(g (x,
— q (x)1 G
[(J
(5.3)
If q7"2(x) E ..V'( — co, co), then it follows from Lemma 1 6 thi g(x, and consequently by Lemma 1.1 G(x, belongs to space
i.e.
(54)
Since the function G(x, be written in the form
symmetric, the condition (5.4)
is
(55) Moreover, it follo*s from Lemma 1.1 (and subsequent
that x;
G2
{q
+
dx}
{ (5.6)
=o
g2
(x,
fq
+
dx).
Put
=
(x,
(5.7)
[q
LEMMA 5.1. Let the function q(x) 1. for Ix—El
— q (x)] G
satisfy the
conditions:
Iq(E) where 2.
A>O and O
for Ix—EI B,
where
B is
3. for
Ix
a constant;
>1 q2(E)exp{ —
Ix
449
§5. DIFFERENTIATION OF THE GREEN'S FUNCTION
where C is a constant; 4. q7"2(x) E .V'(— Then the function a (x,
aD).
defined by (5.7), belongs to the space X, i.e.
(57) PROOF.
a (x,
Let us write the function
,i)
Estimating this integral analogously to what was done in the proof of Lemma 1.1, we can prove (5.7'). Moreover, arguing as in the proof of Lemma 1 1 and usmg the estimate (5 6), it is not hard to obtain the estimate ==o(IIg(x, 'i; 2.
It can be verified directly that if q712(x) E
the function og(x,
IL) /ô1i
—
ca), then
belongs to the space X, i.e. a
Taking account of (5 7), the mtegral equation (5 3) can be rewritten in the form
(53')
(x,
—
g (x,
[q
— q (x)}
Solving this equation by iteration, we obtain (59)
ÔG
ôg
it follows from Lemma 1.1 that as IL (510)
Since
G
450
XII. ASYMPTOTIC NUMBER OF EIGENVALUESO
(5.10) can be written in the form Ij[3(x,
(5.11) We
will show that (x,
(x,
(5.12)
This estimate will follow from the estimate (5 8), if we can show as!4—3cx)
(5.13)
II
g (x,
=°
+
[g
g (x,
A direct computation yields (5.14)
g (x,
C
{q (x)±
C is a constant. Further, using the condition (1.20) we obtain
where
=
g2(x, {
+
g2 (x,
dx
p.) [q ('n) + J
(5.15) g2 (x,
p.) [q (x) + p.r2
+ ciI{IexP(—u\Iq(x)+
{q (x)±
+
C4
dx
dx
(x)± p.) dx
§5. DIFFERENTIATION OF THE GREEN'S FUNCTION
451
Obviously (5.13) follows from (5.14) and (5.15). Consequently we have
proved (5.12). Now from the estimates (5.11) and (5.12) we obtain (5.16)
li13(x,
From equation (5.9) there follows, by virtue of (5.12) and (5.16),
+0(1)).
(5.17)
now return to equation (5.3'), which we rewrite in the form
We
E;
Taking the norm of both sides and estimating the norm of the integral by means of Lemma 1.1, we obtain 3G
'\
from which
(1.+o(1)).
(5.18)
From the estimates (5 17) and (5 18) we conclude that (518')
J.G(x,
7b
+ o(t)),
urn 3. We wifi now formulate without proof the results for the general ase, which covers iterates of higher order.
THEOREM 5 1 Suppose that q(x) satisfies conditions 1 and 2 of Lemma .1 and, moreover, the conditions:
3. for some integer r
0
q3/'221(x) E .V'(— 4.
for
here
C is a constant.
__________________
452
XII. ASYMPTOTIC NUMBER OF EIGENVALUES
belongs to the space X, i.e.
Then the function 0) (0)
Içrat
2
-CO
and as '
II
Ot
(5.19)
hx
(1.+o(1)).
Since by (4.3) {G (x,
5
(—1 )t
ft))
-0)
it follows from the Parseval equality that 0)
(5.20)
5
0)
" (x,
=
—th
+
from which 0) (0) 5
5
0)
[/G (x,
EL)]
)2
dx= n
Therefore
'.'l
the asymptotic formula (5.19) can be written in the 0) (0)
)]2d}d
(5.21) 5
-0)
5
Let us calculate the integral in the right side of (5.21). Putting =ic = [q(x) r = Ix —ui in the well-known eq 0)
I
-0)
we obtain (cf. (1.2)) 0)
g(x,
I I
/
2it
5
-0)
a2
Differentiating this relation r times with respect to 0) I
{a2-fq(x)
EL)
-0)
we obtaz
453
§6. ASYMPTOTIC DISTRIBUTION
Therefore it follows from the Parseval equality that
from which, on the basis of the preceding equality and (2.1), follows 12
c
(—1
I
From the asymptotic formula (5.21) and the above equality we finally
obtain, for (5.20')
S_—
V
I
dx
ç 3
{q(x)
*6. Asymptotic distribution of the eigenvaiues
1. In this section, using a Tauberian theorem of (cf. Chapter 14, Theorem 4.3), we wifi obtain from the asymptotic formula (5.20')
an asymptotic formula for the distribution of the eigenvalues. This formula leads to the following theorem. THEOREM 6.1. Let q(x) satisfy the conditions of Theorem 5.1. introduce the monotonic function ti(X) = mes$ q(x)
0•) = (X —
(X).
Let us assume that there exist positive constants a and
such that for
sufficiently large A we have the inequality (6.1)
9#(A).
Then for A
we
N(X)
I
(62)
have the asymptotic formula {? 1
q(x)<X
0
PltooE. Let
=mes{q(x)
We
_q(x)}'/sdx
454
XII. ASYMPTOTIC NUMBER OF EIGENVALUES
dx
(63) 3
We
{q (x)
da(k)
ç
—I
_L
wifi show that dG (X)
ç
(6.4)
+
1
0
T
I' ± 2) (2v + 3/2 r (1/2)
dX
ç
(X —
+
3
0
(v)
0
In fact, changing the order of integration, we obtain (6.5)
=
(?.
dX}.
(v)
Further, ç
12
—
3 (X
(t
3
1
dt
+
ç
(I ±
Hence, by well-known formulas for Euler integrals we obtain dX—--
I
+2)
which together with (6.5) proves (6.4). =foA(X —v)"2du(v). Then Let us put (v) dA.
(X —
(X)
Therefore (6.3) can be written in the form ç
6.
do (X)
—
2r
+ 2)
f
± 3/2)r(1/2)3
From the asymptotic formula (5 20') and from (6 3) it follows, virtue of (6.6), that
s
(4c±l)!!
ç
3/2)T(1/2)
3 0
the values of the Euler functions F(p), we obtain (67)
s
3
0
(X
455
§6. ASYMPTOTIC DISTRIBUTION
By definition C,
+ we can write S. in the form
Putting N(A)
s
ç
—
dN(X)
3
0
Therefore from (6.7) we finally obtain the asymptotic formula dN (?.)
(6 8)
I
3
0
0
By virtue of the condition (6.1) the above-mentioned Tauberian theorem of totic formula
is applicable to (6.8), from which follows the asympco) N(A) 'S.' or in expanded form
1 —4 (A_' o
-q(x)
and the theorem is proved. 2. In this section we wifi obtain a formula from which, on the one hand, (6.2) will follow as a special case, and on the other hand, using this formula we will prove in the next section a theorem on the convergence and summabifity of expansions and differential expansions, in the eigenfunctions of Sturm-Liouvile operators, of functions which grow at infinity like polynomials. THEOREM 6.2. Let r and p be positive integers. Assume that
4'. for
1
PROOF. We will prove the asymptotic formula (6.9) for the simple case r = 0, p = 1. The proof for the general case is completely analogous,
but involves somewhat more cumbersome calculations.
456
XII. ASYMPTOTIC NUMBER OF EIGENVALUES
From the integral equation (5.3) follows (multiplying by q(x)) = q(x)
q (x)
—
q (x)
{g (x,
[q
— q (x)j G
(6.10)
Put
Arguing as in the proof of Lemma 1.1, one can show that lix =o(ll G(x,fl;1L) iix).
Further, from this same lemma and the integral equation (1 3) follows ii
G(x, n; ii) lix —
g(x,
II x).
Therefore
lla(x,n;,z) lix =o(iig(x,n;,L) lix). By a direct calculation it is not hard to establish the estimate flg(x, which with the preceding estimates leads to the estimate
this estimate and the integral equation (6.10) there fo analogously to the derivatkrn of the asymptotic equality (5.18), From
+ o(1)), which
proves the assertion of the theorem for r = 0 and p = 1,
THEOREM 6.3. Suppose that the function q(x) satisfies the of Theorem 6.2. We introduce the monotonic function i(X) = and put
457
§6. ASYMPTOTIC DISTRIBUTION
(A)
(6.11)
5(X —
(v)
satisfies the inequalities
and assume that the function
Then as a,(,P)
(6.12)
4
5 (X
— v)'/2v2Pdc
q2P 5
(x) {X —
q
(x)}1/ dx,
where
(6.13)
5
and con(x) is an eigenfunction of the operator (1.1) corresponding to the eigenvalue
PROOF. Replacing the number r in (5.20) by r +p, we obtain )}2
G(
{
=
+ p)
I
If we now multiply both sides of this equality by q2"(x) and integrate to + and then use the asymptotic formula (6.9) and equality (6.12), we obtain for
with respect to x from —
± p)! }2
±
5
dx.
g (x,
(x) {s
From this asymptotic equality, by virtue of (5.22) (putting there +p in place r) we arrive at the equality n=1
(6.14) 22
(2't + 2P + 1)!
çq
(z)
+
dx
Proceeding in the same way as we did to obtain the asymptotic equality
(6.8) from (5.20'), for the case under consideration we arrive from (6.14) at the equality (6 15)
c 3
(X)
(X) 1
458
XII. ASYMPTOTIC NUMBER OF EIGENVALUES
is defined by (6.11). and the function Then the asymptotic formula (6.12) follows from (6.15) by theorem. This proves the theorem. Formula (6.2) is obviously a special case of (6.12) (corresponding to the case p =0). where
THE CASE OF THE HALFLINE. Similar results can be obtained for the case of the halihine [0, cx'). Suppose that q (x) is defined on the haliline [0, u) and satisfies the conditions (1.5), (1.6) and (1.20).
We wifi consider the operator
l(y)=—y"+q(x)y
(6.16)
y'(0) —hy(0) =0,
(6.17)
where h is an arbitrary fixed number. Let us first clarify what the analog of the function g(x, this case. Let ic2 =q(x) >0), X(x) = chKx +hK1shKx. We consider the function (618)
I
g+(x,
+ h)' X
is for
(ii
It is not hard to verify that the function = and the conditions tion
satisfies the equa-
—o —
g
we can construct an With the help of the function equation analogous to equation (1.3), and then prove the existence for the operator (6.16) —(6.17). a Green's function To ultimately see this, we only have to see how to estimate the ii). From (6.18) follows tion
(6.19)
+ (1 +
= 4 4
Since
(6.20)
=o(1) as
e
(1
+
e
(1
+0
{1 + 0 V1))
it follows from (6.19) that +0(1)
+0(1)
______
459
*6. ASYMPTOTIC DISTRIBUTION
The representation (6.20) makes it easy to extend the results obtained earlier to the case of the halfline.
We remark only that for the case of the halfline the coefficient in front of the integral in the asymptotic formulas (6.2) and (6.12) has to be replaced by 1/2ir To conclude this section we will indicate various conditions which imply the inequality (6.1). Suppose that q(x) is monotone increasing, has a first and second derivative and is concave. Then q' (x) > 0, q" (x) > 0. Further, suppose that the condition q(x)q"(x)
(6.21)
is satisfied, where C is some finite nonzero constant. Let x = p (A) denote the function inverse to q(x) = A. From consideration of the graph of q (x) it is evident that p (A) is monotone increasing, has a first and second derivative and is convex, i.e. p' (A) > 0, p" (A) <0. It is easily seen that from (6.21) follows the inequality
p' (A) >
(6.22) In
— C'Ap" (A).
fact, by differentiating q(x) = A we obtain dX
I
I
dX
/1\
d
q
d(
(dx
I
\dX J
Inserting these values of q(x), q' (x) and q" (x) in (6.21), we obtain
(
—X
p' (X) > —-C'Ap" (X),
C
i.e. the condition (6.22).
Let us further express the function of the function p (A). It is obvious that from the definition of the function
2
ç
(A) follows
x
x
I' (X)—_!
q(x)
<.A} =mes{x
=mes{q(x)
Therefore
=
('u)
3 (X_v)V2
!
ç
2
3
(v) dv
(X_v)'/2
(A)
=
(A))
460
XII. ASYMPTOTIC NUMBER OF EIGENVALUES
Integrating by parts and taking into account that p' (X) > 0 and p" (A) <0, we obtain (A) = —t
(A
— v)h12
dv +
(v)
(A
[_pU (v)] dv
—
(6.23) vP (A
— v)Vi [—p" (v)J dv.
On the other hand, by (6.22) we have (A)
(A — v)'/2 p' (v) dv> C-1A
(A —
v)V8
[—p" (v)] dv.
From this inequality there follows, by (6.23), i.e. second inequality in (6.1). The first inequality in (6.1) is trivial. §7. Eigenfunction expansions for an unbounded increasing potential
We wifi consider the following eigenvalue problem: (7.1) (7.2)
y"+$A—q(x)}y=0 y(0)cosa +y'(O)sina =0.
ce),
Suppose that q(x) satisfies the conditions of Theorem 6 2, and moreove that for large x it satisfies the inequality (7.3)
q(x)
axe,
where a and ö are certain positive constants. It is known that if the condition (7.3) is satisfied (more prec then the eigenfunctions of the prol if q(x) + as x + (7.1) + (7.2) decrease rapidly. Thus, for example, if the function has polynomial growth at infinity, then the eigenfunction more rapidly "than e_CN, where C> 0 is any constant. Therefore
any s> 0 the' integral (7.4)
exists, and to construct the Fourier series of a function we do not to require that it be square integrable: In this section, using the asymptotic formula (6.14), we will several theorems concerning the convergence and sunimabL.
__________
461
§7. UNBOUNDED INCREASING POTENTIAL
expansions and differentiated expansions, in the eigenfunctions of the problem (7.1) + (7.2), of functions which have polynomial growth at infinity.
We wifi first prove several lemmas. LEMMA 7.1. Suppose that the function q(x) satisfies the conditions of Theorem 6.2 and the condition (7.3). Then for ,j we have the estimate
=
(7.5)
where we have put = A,, (the A,, are the eigenualues of the problem (7.1) — (7.2)), and the numbers are defined by (7.4).
PROOF. By virtue of (7.3) we have (7.6)
u(A)
=
=mes{q(x)
=
Put q,(A) = 10X v'd0r(v), where s is some fixed number. Then, using the
notation of the preceding section, and replacing r +p +1 in (6.14) by p, and 2p by s, we can write (6.14) in the
3 (A. ±
do,
ç
—c
c
a, (A.) di.
,) (A. +
(A. +
From (7.6) we obtain the following bound for Therefore from (7.7) follows the inequality (A.)
ç
)8+112dX
— —t
J_
C
ii,(A)
—
ç
(I +
dt —
and so a fortiori (p.+1
(7• 8' /
3
)
2
d'i (At f u2)2P
(A
'
C EL
>
{x8 [(EL
On the other hand, (79)
+ 1)2] — x, (ps)).
From the bounds (7.8) and (7.9) follows the bound To be correct, we wifi use a formula analogous to (6.14) for the case of the haifline [0, a). These formulas differ only by a constant factor.
462
XII. ASYMPTOTIC NUMBER OF EIGENVALUES
+
/8 as
/8 (ij.2) =
1)2]
ag')
<
was to be proved.
Suppose that the function 1(x) is suxnmable over every finite a positive number s such that (7.10)
Put
S(x, where
c,=5 f(x) satisfies the condition (7.10), and q(i we have the satisfies the hypothesis of Lemma 7.1, then for —* }L+1
V {S (x, p.)) = 0
(7.11)
IL
By the definition of the function S(x,ii) we have
PROOF.
IL+1
V /
B
p 2 (x)
-.
/ Further, applymg the Cauchy-Bunjakovskii mequahty, by virtue ( (7.10) we obtain
= 5 1(x)
(x) dx
=
5
()
/2(x) q8 (x) ds)
(x) qS/2 (x) dx
(s
(x) q1 (x)
dx)
Therefore
V
{S (x, p.))
(x)r. ( ( The lemma folkvs from this inequality by virtue of (7.5) and IL
Chapter 7.
C
§7. UNBOUNDED INCREASING POTENTIAL
463
In the very same way, there follows from (7.5) and (1.21), Chapter 8: lEMMA 7.3. Suppose that the hypothesis of Lemma 7.2 is satisfied, and moreover that q(x) has a kth derivative which is summable over every finite interval. Then for we have the estimate (x,
(7.13)
—'0
Using the estimates (7.11) and (7.13) and the methods of Chapters 7 and 8, it is easy to prove the following theorems. THEOREM 7.1. Suppose that the function q(x) satisfies the hypotheses of Theorem 6.2 as well as the condition (7.3), and suppose that f(x) is summable over every finite interval and there exists a positive number s such that /2(x)
Then for
>s +1/6
(x) dx < co.
we have, at every point of continuity of f(x),
lim
f(x) i.e. the Riesz mean (of order > s + = of the expansion of f(x) in the eigenfunctions of the Sturm-Liouville operator tends to f(x) at every point of continuity of f(x). where
THEOREM 7.2. Let f(x) and q(x) satisfy all the hypotheses of the preceding theorem, and suppose moreover that q(x) has a (k — 1)th derivative, and f(x) a kth derivative, which derivatives are summable over every finite
interval. Then for r> s +1/6 +f +k we have, at every point of continuity of
i e the Riesz mean of order > S + 1/6 + + k of the kth derivative of the expansion of f(x) in the eigenfunctions of the Sturm Liouville operator tends at every point of continuity of f (x) to the value (x)
ThEOREM 7.3. Suppose that q(x) satisfies the hypothesis of Theorem
7.1. Suppose further that there exist operators 2' °f = f(x), —q(x)f, conditions: 1.
.
.
2'f = f"
22kf on the interval [0, cx) which satisfy the following
XII. ASYMPTOTIC NUMBER OF
464
2.
o {exp
3.
= o {exp
for x—*
dx]
for
x —*
(i = 0,1,2,
.
. ., k
(i = 0, 1,2,
.
. ., k
—
}
4.
There exists a positive number s such that
the equality
Then
/ (x) =
(x)
holds uniformly in every finite interval. THEOREM 7 4 Suppose that all the conditions of the preceding are satisfied, and moreover that f(x) and q(x) have mth derivatives are summable over every finite interval. Then if k> [s/2 + 1/2ô + 1/4 +m/2], we have at every point of of (x)
=
(x).
§8. Asymptotic behaviour of the number of eigenvalues of an operator of order 2n
We will consider the
ordinary differential operator L o
the entire line
Ly =
+po(x)y + L1(x, d/dx)y
(—
L0
where L1(x, d/dx) is an operator of order not greater than 2n —2. wifi assume that the characteristic polynomial P(s,x,A)
+p0(x) +A
is positive for all sufficiently large x and A> 0 mid all real s we. will assume that the coefficients of L and the roots P(s, x; A) are such that: 1.
465
*8. OPERATOR OF ORDER 2n 2.
I
<
I
x
for some
I
constant c0> 0 which depends upon the characteristic polynomial P(s,x; A).
for Jx 1 and some c>O. < for some 1> 0.
4. 5.
I
P' [sk (x, A)J
6.
JX+po(x)I for
2n
all x and A> 0, k=1,2,...,2n.
We now introduce the function p
(X)
Ls:(x, A)1'
where the Sk (x, A) are the characteristic roots tor which Im Si (x, A) > 0. As is evident, p (A) is an analytic function in the A = + ir plane slit
along the positive semiaxis. Now let r(t) be defined by
o
The following holds. THEOREM 8 1 Suppose that the operator L satisfies the conditions 1—6, and that (for each t) the coefficient q,(x) of the ith derivative in the operator L1 (x, d/dx) is subordinate to Po (x) in the sense that I
q,(x)
> 0 Further, let the function a (t) defined above satisfy the Tauberian condition ta' (t) au(t) (a > 0)
for some
Then for A—* we have the asymptotic formula L where the are the eigenvalues of the
1
N(A) '—j
This theorem can be proved m accordance with the classical scheme of T Carleman first one finds the asymptotic behaviour of the Green's function K(x, A), the kernel of the resolvent of L, for large A <0 Then one apphes the Tauberian theorem of We note that for
466
XII. ASYMPTOTIC NUMBER OF EIGENVALUES
our purposes we need the asymptotic behaviour of A (x,
A)
with respect to x over the entire space. The asymptotic behaviour sought by the "parametrics" method, i.e. one writes down in form the kernel of the resolvent of the principal part of the operatoi with "frozen" coefficients, i.e. of the operator —
L0 = (—1 )"
d2"
+
d2"2 (E)
+
+ P0 (s).
This kernel can be found by means of the Fourier transform. One 1
constructs an integral equation for the kernel of the resolvent of original operator L in the usual way. One can prove that it can solved by the method of successive approximations, if the hypothes of the theorem are satisfied. We note two special cases. I. Let 0 (k =1,2, . . ., 2n — 2). Then the calculation of function tr(t) can easily be carried through. In this case q
II. 2n = 4. We write the operator L in the form Ly =y" —
+p0(x)y +L1(x, d/dx)y.
it is not hard to verify that the conditions 1, 3 and 6 wifi be satisfie if
c[po(x) +x0] for some A0 and c <1. In this case also t
calculation of the function q(t) can be carried through. Namely {X—p0(x))dx. p,(x)<X
This same result can be obtained by studying the Green's func G(x, E; t) of a parabolic equation for small t. By this route one can c......
results concerning the asymptotic behaviour of N(X) for systems ordinary differential equations, and also for singular elliptic operat with a discrete spectrum. Bibliographical references
The first to obtain the asymptotic behaviour of N(X) were de and :Mandl [1], who used a variational, theorem of Courant. [2] applied Carleman's method to this problem and obtained an asr totic formula for N(X) under less restrictive conditions on the pots
REFERENCES
467
function q(x) (in the book cited the case of two variables was considered; however the method of this book is also applicable to the case of a single variable).
The complete series for the Green's function was used by Levitan [21]. The application of estimates for the weighted trace of the Green's function to questions concerning eigenfunction expansions with unbounded increasing potentials is due to Levitan (cf. also the authors' joint paper [3]).
The formula for N(X) for the case of an nth order operator, given in §8, is due to [3,4].
CHAPTER 13
ELEMENTS OF THE SPECTRAL THEORY OF LINEAR OPERATORS IN HILBERT SPACE. RELATION TO DIFFERENTIAL OPERATORS
For a deeper understanding of many questions which have discussed in the previous chapters of this book, it is useful to con these questions from the point of view of the general spectral of linear selfadjoint operators in Hilbert space.
At the present time there are many excellent texts on the abstra theory of linear operators in Hubert space; however in applying general theory to differential operators a number of analytic diffici arise and, unfortunately, not everything as yet appears as smooth
definitive in this topic as in the general theory. Possibly this from an incompleteness of the theory, in particular when spe about partial differential operators. In this chapter we give a brief presentation of some basic topics f the spectral theory of linear selfadjoint operators in an abstract space. As we have already mentioned, at the present time this t has been brought to a high degree of completenes, and can be studie
from many texts. Therefore this part of the chapter will in be presented without proofs, and can serve as a first introduction abstract spectral theory in Hilbert space (the reader who is acquain with this theory to the extent, example, of the books by Glazman [1], Ljusternik and Sobolev [1], or Smirnov [1] can the corresponding sections of this chapter). The end of this chapter and 7) is devoted to various applica of the abstract theory to differential operators. and
§1. Abstract Hilbert space
1. Definition of an abstract Hilbert space. Let H be a complex 1 space with elements x, y, z,.... We assume that with every pair X, H there is associated a complex number denoted by ( called the scalar product of these elements and satisfying the foil conditions (axioms): 468
§1. ABSTRACT HILBERT SPACE
469
a) (x,y)=(y,x), b) c)
(x1+x2,y)
=(x1,y) +(x2,y),
(Ax, y) = X(x, y) for any complex number A,
d) (x, x) 0, and (x, x) = 0 if and only if x = 0 (the zero element of the space H). The number II x = (x, x)1"2 is called the norm of the element x. From
the axioms of the scalar product one can deduce that this quantity satisfies all the conditions of the norm in a linear normed space. If H is complete relative to the metric p(x,y) =11 x —yII and is infinite dimensional, then it is called a complex (unitary) Hubert space. REMARK. A real Hubert space is defined analogously. For this case the scalar product is assumed to be a real number. We give two important examples of Hilbert space. 10. The space 12. The elements of this space are infinite sequences y = (fli, of complex numbers such that = (si,
The scalar product is defined by (x,
One can show that 12 satisfies all the conditions of a unitary Hubert space. 20. The space 2(a, b). The elements of this space are measurable
complex-valued functions f(x) ,g(x),... defined on the interval [a, bJ and having square-integrable modulus. The scalar product is defined by (/ (a,), g (x))
f(a,)
dx.
We have essentially been dealing with this space throughout the book.
Namely, we have studied the spectral theory of a Stiirm-Liouville co) and differential operatorin the spaces .5/ 2(0,?r), connection with Parseval's formula we were led to consider the space In
(a, b), which also is a Hilbert space. Finally, in connection with the spectral theory of Dirac-type systems of equations we considered a Hilbert space whose elements are vector-
functions 2. Orthogonality. Two elements x, y E H are said to be orthogonal, if (x, y) = 0 In this case one writes x y An element x is said to be orthogonal to a subspace
we write x I .5/
H if x
is orthogonal to every
yE ..V. In this case
XIII. LINEAROPERATORSIN HILBERT SPACE
470
One has the following important result. THEOREM 1 1 Let be a closed subspace of the space H Then every x. E H has a unique decomposition
x=y±z,
(1.1) where
y m the decomposition (1 1) is called the projection of the element x on .V. We remark that the set M of all elements in H which are orthogonal to is itself a closed subspace Therefore the element z from (1 1) is the projection of x on M The subspace M is called the orthogonal complement of and is denoted by H is the orthogonal sum of and M, written ai
-f-M 3. Orthogonal systems. A system e1, e2,••• of elements of the space is said to be orthonormal if —
Let
if i U, if i =j. ço,
be the closed subspace generated by an orthonormal
(i.e., the closure in the norm of H of the set of all
IL.
combinations of the form where the are complex or numbers according as H is a complex or real Hilbert space). Obviow for any element x E and any > 0 there exists a linear combinati such
that
(L2)
A simple computation shows that
x_kek) (13)
=(x,
e
ek)—
i—i
=11z112+ where ck = (x, ek).
The numbers ck are called the Fourier coeffic of x (relative to the system e1, e2, ...). It follows from the identity (1.3) that the difference x
ABSTRACT HILBERT SPACE
471
has minimum norm when the numbers equal the corresponding Fourier coefficients: = Ck. In this case we obtain from (1.2) and (1.3) (1.4)
x
and since
=11 x112
—
—
J
J2
<€2,
can be chosen arbitrarily small, it follows from (1.4) that
x = urn
Ckék
N-*a
From (1.4) also follows the convergence of the series
j2
and
that 0x112=Ek=11ckl2
If now x is an arbitrary element of H, we can put x = y + z, where Then
=
Cj:
=
= (y,
ek) + (z, ek) = (x.
ek),
(1.5)
Since
x 112 = II y +z 112 = II y2
+ ft
2,
it follows that
(16)
From (1 5) and (1 6) follows (17)
eb)
The mequahty (1 7) is called Bessel's inequality 4 Closed orthononnal systems. Let e, } be an orthonormal system m a Hubert space H DEFINITION 1 1 The orthonormal system e, us said to be complete if there exists no vector x E H other than zero which is orthogonal to all the elements of the system e, DEFINITION 1 2 The orthonormal system { e, is said to be closed if the generated by this system coincides with H closed subspace
Obviously the Fourier series of an element x, corresponding to a closed orthonormal system, converges to x for every x H, and one the Parseval equahty A closed orthonormal system is also called an orthonormal basis of the
H THEOREM 1 2 The concepts of a complete and a closed orthonormal
system coincide
XIII. LINEAR OPERATORS IN HILBERT SPACE
472 PROOF
If an orthonormal system is complete, then there exists no
nonzero element which is orthogonal to the closed subspace
generated
by this system But then z = 0 m the decomposition (1 1), and consequently x = y Conversely, if the orthonormal system { e, } is closed, then =H 12 and so for every element x E H we have II x 112 = Therefore if x .1. e, (i = 1,2, = 0, i e x =0, which says that the ), then I
II
e, is complete REMARK The considerations of §1 4 are concerned with so called separable Hilbert spaces These spaces are characterized by the fact that they contam a countable complete orthonormal system There exist, however, nonseparable Hubert spaces, for example the space of almost periodic functions. In these spaces every complete orthonormal system is uncountable. In this book we do not encounter nonseparable Hubert spaces.
system
*2. Linear functionals and bounded linear operators
1. Linear functionais and linear operators. Let D be some set of elements of a Hubert space H. DEFINITION 2.1. If with each element fED we associate some number +(f), then we say that we have defined a functional $ in H with domain D. If with each element f E D we associate some element Tf H, then T is called an operator in H with domain D. The set = { Tf; f E D is called the range of T The notation is also used for the domain of and for DT and
the domain and range of T. From now on we will in this section consider only so-called linea functionals and linear operators, and we will assume that D = H. DEFINITION 2 2 A functional $ (operator T) is said to be and homogeneous if for any complex numbers a, fi and any
f,gEHwe have $(ag+$g)
a$(f)
aTf-l--(3Tg).
DEFINITION 2.3. A functional $ (operator T) is said to bounded if sup j
(sup I! Tf
(/)1
DEFINITION 2.4. A linear functional (operator) is a functional which is homogeneous, additive and bounded.
The norm H TII of a linear operator T is defined by T 1= sup Till. i/i
§2. LINEAR FUNCTIONALS
473
The norm can also be defined by fell
A bounded linear operator is continuous.
If a homogeneous and additive operator is continuous at one point, then it is bounded. The analogous property holds for linear functionals. 2. Bilinear functionals. If with every ordered pair of elements f,g E H we associate some (generally speaking, complex) number in such a way that
+
Q
(2.1)
g)
b)
(1,
+
c)
sup
Q (/, g)
= =
(ti, g) (1, gi)
+ +
(12'
g),
(I' g2),
CU,
we say that we have defined a bilinear functional in H. in c) is called the norm of the bilinear functional c2 and is denoted by
then
The
THEOREM 2.1. Every bilinear functional in H can be represented in the form = (Af,g), where A is a bounded operator in H which furthermore A = is uniquely determined by the functional II
3.
The adjoint operator. Let A be a bounded linear operator defined
everywhere in H. The expression (f, Ag) is obviously a bilinear functional in H with norm All. Therefore it follows from the preceding theorem
that there exists a unique (bounded) linear operator A* for which =(A*f,g), (2.2) and that IA *11 = II A
The operator A*, satisfying (2.2) for all f,g E H, is called the adjoint of A It is easily seen that (A*) * = A If A* = A, then A is said to be a selfadjoint operator A bounded selfadjoint operator A is said to be positive if (Af f) 0
every fEH 4 Projection operators Let H be a Hilbert space and G a closed sub : space of H The operator P which associates with every element f E H its projection g on G is called the projection operator or projector (on G), [and is denoted by A projection operator is obviously additive and homogeneous More over, it is bounded, and its norm equals unity We give without proofs the basic properties of projection operators
474
XIII. LINEAR OPERATORS IN HILBERT SPACE
a) In order that an operator P, defined everywhere in H, be a projector,
it is necessary and sufficient that 1) P2 = P, 2) P* P02 is a projector if and only if b) The product of two projectors they commute, i.e. = PG2PG1, and if this condition is satisfied, then P01P02 = P0. where G = G1 fl G2.
Q (n < co) is a proare pairjector if and only if POkPGJ = 0 (j k), i e if the subspaces wise orthogonal. In this case Q P0. where G = G1 + G2 + d) The difference of two projectors, P01 — PG2, is a projector if and
ë) A sum of projectors, P01 +
+
only if G2_CG1. e) If Ph } (k =1,2,...) is an infinite monotonic sequence of projectors,, converges strongly to a projector P. i e the projector than as k for every fEH we have Pkf=Pf 5 Unitary operators A linear operator U, defined on all of the space H, and having as its range all of H, is called a unitary operator if
(2 3)
(Uf, Ug) = (f, g)
for all f,gEH We state, again without proofs, the following simple properties of unitary operators exists and is a) If U is a unitary operator, then its inverse unitary I (the identity operator) b) U" = U', i e 6 Isometric mappings Suppose that we are given two Hubert H1 and H2, let us agree to differentiate between the scalar products these spaces by the subscripts 1 and 2 1 DEFINITIoN 25 A linear operator V. defined on all of H1 and mapping H1 onto all of H2, is said to be an isometric mapping (of H1 onto H2) (2 4)
(Vf, Vg)2 = (f, g)1
for all f,gEH1 If the range v of V does not coincide with all of H2 (but (2 4) s holds), then V is said to be partially isometric By virtue of these definitions Theorem 1 1 of Chapter 2 can regarded as proving that the indicated mapping of the space ..V 2(0 (— co, co) is partially isometric, and Theorem 4 into the space
of Chapter 2, that the indicated mapping between these spaces isometric 7 Isometric operators
DEFINITION 26 A linear operator V defined on a Hubert space H,
475
§3. GENERAL CONCEPTS
said to be isometric if it preserves the norm of every element, i.e. if
VxII
for all x E H. = Like any linear operator, V is defined on all of H, but it is not required that V shall map H onto all of H. Therefore an isometric operator need not be unitary, which is evident from the following simple example. Let e1, e2,..• be an orthonormal basis in H, and let x be an arbitrary element of H. Then x xiei, where Xi = (x, ek). The operator V defined by Vx is isometric and maps all of H onto the subspace of H which is orthogonal to e1. x
§3. Some general concepts of the theory of linear operators
In this section we wifi consider some general concepts of the theory
of linear but, as a rule, unbounded operators. Such operators are generally defined not on the entire space, but only on some dense subset
of it Thus, for example, differential operators are defined on differ entiable functions which form a dense set m the space 2(a, b) Two operators A and B are regarded as equal or as coinciding if DA —DB and Ax=Bx for every XEDA If DACDB and Ax =Bx for every XEDA, then B is said to be an extension of A andA is said to bea re striction of B In this case we write A c B 1 The concept of a closed operator DEFINITION 3 1 An operator T (not necessarily linear) is said to be closed, if from the existence of the limits
with
it follows that fE DT and g = Tf Thus every continuous operator (with closed domain) is obviously closed, but the converse is false, since for a discontmuous operator the
does not necessarily imply that of existence of lirn,,... T is not closed, but whenever two sequences { Tg,,
}
and
and in Dr,' converge to the same limit, then the sequences either both diverge or both converge to the same limit, then T
has closed extensions Among these extensions we distinguish the minimal
closed extension, which is a restnction of every other closed extension of T This operator is usually denoted by T and is called the closure of (the closure all those elements f T To obtam T we adjoin to DT which for which there exists at least one sequence f1, f2, of converges to f, and for which the sequence { Th converges, we then }
put
476
XIII. LINEAR OPERATORS IN HILBERT SPACE
2. The general definition of the adjoint operator. Let T be an arbitrary linear whose domain DT is dense in H. We consider the scalai product (Tf,g), fEDT. There exist pairs of elements g, g* for which (Tf,g) (f,g*) (3.1)
holds for every DT. For example, (3.1) holds for g =g* =0. If DT is dense in H (which we are assuming), then the element g* uniquely determined by the element g. We put g* = T*g. By definition, T* is called the adjoint of T. Its domain DT* is the set of all those g for which there exists a g* satisfying The followmg assertions follow easily from the (3 1) for every f E
definition of the adjomt operator
a) T* is linear b) IfS_CT, then S*DT* The operator T* is closed even when T is not closed d) If T has a closure T, then (T) * = T* c)
If the operator T** exists, then T T** 3 Synunetnc and selfadjomt operators A linear operator A is said e)
to be symmetric if a) its domain DA is dense in H,
b) for every pair of elements f,g E DA one has (Af,g) = (f, Ag) If there exists a number y> — such that (Af,f) 'y(f,f), fE 1)4 then A is said to be lower semibounded Obviously for a symmetric operator A we have A * A, from whk it follows that a symmetric operator always has a closure. If B is a symmetnc extension of A, then B ç A * An operator A is said to be selfadjoint if it coincides with its adjoin
i.e.A*=A.
We present without proofs the followmg properties of operators.
a) If a selfadjoint operator has an inverse, then this inverse is adjoint (bounded or unbounded). b) The eigenvalues of a selfadjoint operator are real. c) Two eigenvectors f1, f2 of a selfadjoint operator corresponding, distinct eigenvalues A1, A2 are orthogonal 1)
In the theory of unbounded operators the term "linear" stands for "homogen and additive".
477
§4. SPECTRAL ANALYSIS
4. The spectrum of a linear operator. Let T be a closed linear operator defined on a dense subset DT of H. We consider the equation Tf — Af =g,
I E DT, g E H, with a complex parameter A. Let us denote by the range of the operator T — Al. If T — Al establishes a one-to-one then T — Al has an inverse correspondence between DT and (T—AIY' with domain and range DT. DEFINITION 3.2. The values of the parameter A for which the inverse operator (T — Al)-' exists, is defined everywhere in H (i.e. and is bounded, is called a regular value (or regular point) of the operator T. The remaining points of the complex A-plane form the spectrum of T.
One can show that the spectrum of a selfadjoint operator lies entirely on the real axis. 5. The resolvent. Let T be a closed linear operator with domain which is dense in H. The resoluent of T is defined as the operator-valued function of A,
defined on the set of all those values of A for which it exists and for which its domain, i.e. is dense in H. At every regular point of T the
resolvent RA is a bounded operator defined on all of H. One can show that for a selfadjoint operator A we have is a regular point of A.
=
if A
§4. Spectral analysis of selfad joint operators
1. DEFINITION 4.1. By a spectral function is meant an operator-valued function E, of a real parameter A satisfying the following conditions: a) E, is a projector for every value of A, for A (i.e. EM — E, is a nonnegative operator), b) E,
c) E_ = 0,
E
for every x the vector-function E,x is continuous from the right, i e
llniIlE,+,x—E,xII =0.
It easily follows from property b) that the limits lim
,
x-
=
_0x,
Urn
=
= EM This condition E H Condition d) says that is not essential and represents only a specific normalization of the spectral function
XIII. LINEAR OPERATORS IN JULEERT SPACE
478
We wifi use the following notation: if denotes one of the intervals (a, [a, (a, fi] or [a, then wifi denote, respectively,
=
—
—
—
— —
=—
—
It easily follows from property b) that is a projector for any •, 4, are pairwise It is also easy to verify that if the intervals disjomt, then = 0 for j 2 The integral with respect to a spectral function By means of an arbitrary spectral function one can construct an mtegral analogous to the Stieltjes mtegral Let f be an arbitrary numerical function, defined and continuous c an interval [a, b]. We consider an arbitrary partition of [a, b] by into parts of certain points a=X0
II
(4.1)
From the existence of the integral (4.1) there obviously follows existence of the integral (4.2)
f (X)
x
H,
which is defined as the limit of the sums If in the equality
=Sx.
479
§4. SPECTRAL ANALYSIS
we pass, to the limit, we obtain (4.4)
The
f(A)
x).
right side of this equality is an ordinary Stieltjes integral. If
the function f is continuous for all finite A, then one can define the integral / (A)
(45)
b—÷ + By virtue of the Cauchy criterion and formula (4.4), the integral (4.5) exists if and only if the ordinary Stieltjes integral
as the limit of the integral (4.2) as a—+ —
x)
exists. We then have x),
d
/ (A)
which is obtained from (4.4) by passing to the limit.
3. The basic spectral theorem. THEOREM
4i.
For every selfadjoint operator A there exists a unique
having the following properties: spectral function a) a given vector x belongs to DA if and only if lXI2d(EAx,x) b) when this condition is satisfied, then (4.6)
Ax =
d
A
x).
Conversely, every operator defined by a) and b) by means of some spectral
function EA is a selfadjoint operator. For a proof of this theorem see, for example, Ahiezer and Glazman [1]. Theorem 1.1 of Chapter 2 is easily seen to be a special case of Theorem 4.1, for which A)dp(A),
F(X)
(t, A) dt.
From this same theorem it follows that to every point of Weyl's limit circle there corresponds a selfadjoint extension of the Sturm-
480
XIII. LINEAE OPERATORS IN HILBERT SPACE
Liouvile operator (consequently, in the case of a limit point there exists
only one selfadjoint extension. See also the following section of this chapter). 4. Reducibility. Let M be a closed subspace of H, and P the projector
onM DEFINITION 4.2. The subspace M reduces an operator A, if x E D4 lmplles that Px E DA and APx = PAx THEOREM 4.2. A closed subspace M reduces a selfadjoint operator A i and only if the projector PM commutes with the spectral function of for every value of A.
For the proof see Ahiezer and Glazman [1]. 5. Description of the spectrum of a selfadjoint operator by means of I spectral decomposition. Recall that a number A is called a regular = (A — Al)' exists, is defined of an operator A if the operator is the entire space H and is bounded. In this case the operator the resolvent of the operator A. The set of all nonregular points of A is its spectrum. Obviously every eigenvalue A of A belongs to its spectrum
since in this case the operator (A — A I)' does not exist. The set of a' eigenualues of A is called its discrete spectrum. All other points of spectrum, if such exist, are called points of the continuous spectrum, the set of these points is called the continuous spectrum of the operator. A is selfadjoint, then eigenvectors corresponding to distinct eigenva
of A are mutually orthogonal, and therefore A can have only a or countable set of eigenvalues (in a separable space there cannot an uncountable set of pairwise orthogonal vectors). Thus the dis spectrum of a selfadjoint operator (in a separable Hilbert space) is a finite or countable set of real numbers. We wifi show that every nonreal number A is a regular point of
selfadjoint operator A, and that (4.7)
= (A — Al)—' =
where E,, is the spectral function of A (compare with Theorems 3.1
7.1 of Chapter 2). Indeed, the integral in the right side of (4.7) exists (see of this section). Furthermore, for any x E H we have
§5. EXTENSION OF SYMMETRIC OPERATORS
(4.8)
dE S
2
dE
481
I
Hence this integral is a bounded operator, and it is easy to verify that it is equal to RA. 9 1/ Im A. Moreover, It follows from the inequality (4.8) that it follows from (4.7) that (RA) * The following theorem gives a certain characterization of the spectrum
of a selfadjoint operator A. be its THEOREM 4.3. Let A be a selfadjoint operator in H, and let spectral function. Then a) A real number A0 is a regular point of A if and only if the function EA is constant in some neighborhood of this point. — 0; b) A real number Ao is an eigenvalue of A if and only if in this case
to the
is the projector on the eigenspace of A corresponding
— A0.
For a proof see Naimark [iJ, §12.5 or Ahiezer and Glazman [1], §82. §5. The extension of symmetric operators
1. The deficiency subspaces and deficiency indices. One of the basic problems of the abstract theory of symmetric operators is the description of all symmetric extensions of a given symmetric operator A. A particularly important special case of this problem is that of clarifying conditions under which a given symmetric operator has selfadjoint extensions, and descrIbing in this case all such extensions. If B is a symmetric extension of a symmetric operator A, i.e. A ç B,
then B*
A*. But B is a symmetric operator; therefore B
B*
and hence (5.1)
ACBCB*CA*.
It follows from these inclusion relations that every symmetric extension of A is a restriction of the operator A*. DEFINITION 5.1. A symmetric operator A is said to be maximal if it has no symmetric extensions. It follows from this definition that a selfadjoint operator A is maximal.
In fact, in this case A* =A and (5.1) shows that B =A, i.e. every symmetric extension B of A coincides with A.
482
XHL LINEAR OPERATORS IN HILBERT SPACE
Let us denote by the subspace consistmg of all elements of the form y = (A + ii) x, x E DA, where i = and by the subspace of all elements of the form (A — ii) x, x E DA LEMMA 5.1.
PRooF. Simple and therefore
and
are
calculations
(5.2)
II
closed subspaces, if A is closed. yield
(A ± ii) x
II
if
(A ± ii) x 112 = Ax 112 +11 x 112,
x if.
and that
Let us now assume that y,, = (A ±
Then
and consequently by (5.2) 0. Since H is — Xm —* x0 E H. Thus x0, complete, it follows that E DA, A is a closed operator, we have x0 E DA and Ax0 ix0. It follows that as was to be proved. We denote by the orthogonal complement of II
yn — yrn if
0,
if
if
DEFINITION 5.2. The are called the deficiency subspaces of the operator A Their dimensions are called the deficiency indices of A REMARK 1. Let A denote an arbitrary nonreal number. Similarly to the foregoing one can introduce the two subspaces and prove
that they are closed, define the subspaces N,, and and define the deficiency mdices as the dimensions of the subspaces N,, and The subspaces N,, and depend upon A, but one can prove that theff dimensions m each of the halfspaces Im A >0 and Im A <0 do not depend upon A (for a proof see, for example, Ahiezer and Glazman [1]) Therefore the deficiency mdices of a symmetric operator do not depend upon the choice of the (nonreal) number A REMARK 2. If for a Sturm-Liouvifie operator 1(y)
—y" ±q(x)y,
0 x<
with the boundary condition y'(O)cosa +y(0)sina =0 the limit-point case holds, then the deficiency indices are (0,0), whil m the bnut circle case they are (1, 1) 2 Description of the domain of the adjoint operator An elemer z His orthogonal to the subspace if and only if, for every x E Th one has (z, Ax — ix) = 0 or (Ax, z) = (x, iz), i e when z E DA an A = iz. Thus the orthogonal complement of which we introduce above and denoted by is the subspace of eigenvectors of A * corr sponding to the eigenvalue i A similar result holds for and N..
483
§5. EXTENSION OF SYMMETRIC OPERATORS
THEOREM 5.1. The domain DA. of A* (the adjoint of a closed symmetric operator) is the direct sum 2) of DA, N, and N_1, i.e. (5.3)
PROOF. Let y be an arbitrary element in DA. and consider the N_i, we have
element A *y — iy. Since H (5.4)
Further, Ax =A*x and A*zj =
—
Hence y — x — iio = z EN1. Putting
Therefore (5.4)
i10
=
we
implies that
obtain
y=x-f-z-f-i,
(5.5)
which yields the desired decomposition (5.3). We shall show that this decomposition is unique. Suppose that along with (5.5) we have another Then decomposition of the element y: y = x1 + z1 (x—xj) +(z—z1)
(5.6)
=0.
Applying A* to both sides of (5.6), we obtain =0. A(x—x1)+i(z—zi) (5.7) Multiplying (5.6) by i and subtracting from (5.7), we have A(x—x1) —i(x—x1) =0. (5.8)
I
it follows from (5.8) that A(x—x1) —i(x—-x1) =0, =0.
Since A (x — x1)
—
i(x
— x1)
—
Similarly = z, and therefore x1 = x. If instead of i we choose an arbitrary nonreal number X, we obtain
Therefore
the decomposition (5.9)
be subspaces of E. If every element = x1 + k = 1, .. + and writes E = then one says that E is the direct sum of the Let E be a linear space and let
.
. .,
484
XIII. LINEAR OPERATORS IN HILBERT SPACE
lithe deficiency indices of A are (0,0), i.e. the corresponding deficiency subspaces and are the zero subspace, it follows from (5.5) that DA. = DA,
and so A is selfadjoint.
3. Let us denote by F the set of those elements y E DA. for which the quadratic form (A*y,y) assumes real values. THEOREM 5.2. In order that an element y E DA. belong to the set F, it is necessary and sufficient that in (5.5) one have lizil = PROOF. If y =x +z then (A*y,y)
(5.10)
=(Ax,x) +(x,A*(z =A(x,x) +2Re(x,A*(z+z)) =A(x,x) +2Re(x,A*(z+z))
—i(i,z)
Since (Ax, x) and i — i (1, z) are real numbers, it follows from 2 (5.10) that Im (A *y, y) = lIz 112 — from which the theorem follows. 4. ThEOREM 5.3. To every symmetric extension B of a closed symmetric operator A there correspond two subspaces T, C and and an isometric operator U mapping T, onto such that a) the domain DB consists of all elements of the form
ç
(511)
y=x+z+Uz,
where XEDA and zET,, b) for every y of the form (5 11) (512)
By=Ax+iz—iUz
Conversely, given two subspaces T, c N, and T_, N, and an metric operator U which maps T, onto then the operator B on the set of elements of the form (5 11) by formula (5 12) is a extension of A The operator B is closed if and only if the subspaces T, and T_, are
iso..4
PROOF Let B be a symmetric extension of A and let y E DB As have seen, DB c DA and therefore accordmg to (5 5) y = x + z where, smce y E F, we have by Theorem 5 2
(513)
=flzfl
If y runs through all of DB, then z runs through some subspace
1
§5. EXTENSION OF SYMMETRIC OPERATORS
485
and runs through some subspace We shall show that with each element z E one can associate only one element E In fact, if Yi
Xi
+ 21 +21,
Y2
= X2 + Z1 + Z2,
Yi , Y2 E DB,
then Yi —Y2
=x1
Therefore from (5.13) follows 1111 =0, i.e. Associating with each element z E T1 the unique element corresponding to it from (5.5), we obtain an isomorphic and isometric mapping of onto Denoting this mapping by U, we arrive at (5.11), and then at
By__A*yA*(x+z+Uz) =Ax+iz—iUz, which is (5.12).
ç be two subspaces and let U let T, and The operator B defined be an isometric operator mapping T1 onto on the elements of the form (5.11) by the formula (5.12) is a symmetric extension of A, since the subspace DB of element of the form (5.11) satisfies the condition DR ç F fl DA., and By = A *y, YE DB. It remains to prove the last part of the theorem. To this end we note that for B to be closed it is necessary and sufficient that the subspace of elements of the form (B + iI)y, y E DB, be closed. The necessity was proved in Lemma 5.1 of this section. We prove its sufficiency. Let us assume that is closed, but B is not closed. Taking the of B, we would have to add new limit elements to which would contradict being closed. Further, for any y E DR Conversely,
(B+iI)y=(B+iI)(x+z+z)
=(A +iI)x+2iz.
Therefore (5.14)
the set of elements of the form (A + ii) x, x E DA. Since is closed if and only if T, is. This proves sufficiency. 5. Let us assume that we have extended a symmetric operator A, by the method indicated in Theorem 5.3, to a symmetric operator B. The question arises, what are the deficiency subspaces and the deficiency indices of this extension? The answer is given by the following theorem. where
is is closed,
Every symmetric operator has a closure;
cf. §3.
486
xiii. LINEAR OPERATORS IN Hi LBERT SPACE
THEOREM 5.4. Let B be a closed symmetric extension of a closed symmetric
operator A, with domain DB =
DA
+ (I + U)
Let (m1, m2) be the deficiency indices of A: m1 =
dim N_i,
and (mc, mc) the deficiency indices of B: mc = dim N[, mc = dim Then N, =N( T, dim = 1, we have m1 mc + 1, m2 = mc + 1. ®N1, we have by (5.14) H= Therefore = N[ ® Similarly = one proves that The relations between the deficiency mdices follow at once from these relations Since
By means of Theorem 5 3 and 5 4 one can give a complete description of all closed symmetric extensions of a given closed symmethc operator A One can also state conditions for the existence of selfadjomt extensions and describe all of these extensions For this and other topics in abstract
spectral theory the reader can turn to the monographs of Plesner [iJ and Ahiezer and Glazman [1] *6 A sufficient condition for the selfadjomtness of a Sturm-Liouville operator on the real line or haffime
1. It follows directly from the definition of the adjoint and of a adjoint operator that to prove the selfadjointness of a symmetric a A it is sufficient to show that if some element x is in DA., then x E
i.e. that (6.1)
DA.—DA.
In order to prove the equality (6.1), we must have a detailed descriptioi of DA.. For a Sturm-Liouville operator (6.2)
1(y) = —y" ±q(x)y
this can be accomplished, and on the basis of such a description can give a sufficient condition for the selfadjointness of the o 1 (the closure of the operator 1) in terms of the function q(x). The can be accomplished for first-order Dirac-type systems, and wifi done in the following section. In the subsequent discussion an important role is played by following result.
6.1. Let f(x) (a x b) satisfy the following conditions: a) it has an absolutely continuous first derivative in the interval (..,
§6. SELFADJOINTNESS OF A STURM-LIOUVILLE OPERATOR
b) f"(x) c)
f(a)
487
/
—f'(a) =f(b) =f'(b) =0.
Let h(x) be an arbitrary fixed function from 2(a, b). If for every function f(x) satisfying conditions a), b) and c) we have f"(x)h(x)dx =0,
(6.3)
then h(x) =a +$x almost
where a and $ are
everywhere,
constants.
PROOF. First we give a characterization of the class of functions satisfying conditions a), b) and c). Let f" (t) = X (t). Then
f'(x) =
5X
(t) dt,
/
(x)
=
5 (x — t) X (t) dt.
It therefore follows from condition c) that
5X(t)dt=0,
(6.4)
5tX(t)dt=zO.
Conversely, if a function X(t) from (6.4), then putting / (x)
=
2(a,
b)
satisfies
the conditions
(x — t) A (t) dt,
we obtain a function f(x) satisfying conditions a), b) and c). If now a function h(t) E 2(a, b) does not satisfy the conditions (6.4), we can choose constants a and [3 such that the function h1(t) =h(t) —a —$t
(6.5)
will satisfy these conditions. Indeed, to determine a
and j3 we have
two equations h1 5
(t)
dt
5
h (t) dt —
(b — a)
—4- (b — a)2 =0,
(6.6)
5th1 (t) dt
= 5th
(t)
dt
j. (b — a)2 —4. (b—a)3=0,
from which a and [3 are uniquely determined.
Let us now turn to equation (6.3). It can be written in the form
488
XIII. LINEAR OPERATORS IN HILBERT SPACE
X(x)h(xydx—_O,
(6.7)
Using (6.4), we rewrite this last equation in the form (6.8)
where h1(x) has the form (6.5), with constants a and satisfying the equations (6.6). Putting, in (6.8), A (x) = h1 (x) (which we can do since h1(x) satisfies the conditions (6.4)), we obtain =0, from which it follows that h (x) = a + almost everywhere, as was to be proved.
2. The case of the entire line. Let 1(y) = —y" ±q(x)y (— cx, <x < a,), and let q(x) be everywhere continuous. We denote by D(l) the set of all twice continuously differentiable functions having compact supports. First of all we shall give a descnption of D(l *) According to the definition
of the adjoint of an operator, if yE D(l) and z E D(l*), then there exists a function E 2 — a,, a,) such that (6.9)
y. f dx.
(—u" + q (x) y) z dx
Integrating by parts, we obtain
f dx =
y".
(x) dx,
Y(x)dx, where Z* (x)
(x — t) ? (t) dt,
Y (x) =
(x — t) q (t) z (t) dt.
Therefore (6.9) can be rewritten in the form (6.10)
(x)
+ V (x) —
On the basis of Lemma 6.1 we obtain
(x)] dx =0.
489
§6. SELFADJOINTNESS OF A STURM-LIOUVILLE OPERATOR
From this equality there follows a complete description of the functions z (x) belonging to D (1*), which is given in the following lemma. LEMMA 6.2. The set D(l*) consists of all functions z(x) satisfying the following conditions: a)
z(x) is continuous and has a first derivative which is absolutely
continuous in every finite interval, c) —z" +q(x)z =z*(x)
With the help of Lemma 6.2 we can prove the following theorem. THEOREM 6.1. If the function q(x) is bounded from below, then the operator 1 (the closure of 1) is selfadjoint.
We will first prove another lemma, which contains the main part of the proof of Theorem 6.1. LEMMA 6.3. If the function q(x) is bounded from below, then from
a, cx,), l(y) = —y"
cx, cx,)
follows
(6.11) PROOF.
(6.12)
Integrating by parts, we have y'2dx
+
yy"dx — y' (a) y (a) — y' (—a) y (—a).
ccx) From and —y" cx,, the existence, as a —f + of a finite limit for the sum
(6.13)
yy" dx
+
cx>, ccx)
follows
(x) y2 dx.
Since by hypothesis the function q(x) is bounded from below, it follows from the boundedness of the sum (6.13) that the first term of this sum either tends to a finite limit or tends to — cx• But then the left side of (6.12), and therefore the quantity y'(a)y(a) —y'(—a)y(—-a), either tends, as a + >cx, to a finite limit or to + cx,• If we can show that the case (6.14)
urn [y'(a)y(a) —y'(—a)y(—a)} = +
is not possible, then from (6.12) we would obtain
490
XIII. LINEAR OPERATORS IN HILBERT SPACE
Urn a -*
—a
which would prove the lemma. To prove that (6.14) is not possible,
consider the function 1(a) = y2(a) +y2( — a) and its derivative I' (a:) =2y'(a)y(a) —2y'(—a)y(—a). If (6.14) holds, then I' (a) —*
contradicts the fact that y(x) E PROOF
a fortiori
and 2
1(a) —*
which
cx').
(—
OF THEOREM 6 1 We will show that the operator 1 * is symmetric
on its domam It will follow that (615) On the other hand, for a symmetric operator 1 we always have the inclusion 4
(6.16)
It follows from (6.15) and (6.16) that l** = 1*, i.e., the operator 1* i selfadjoint. Further, for_any symmetric operator A (with a domain) A* * = A, where A is the closure of A (cf. Ahiezer and Glaz
and 51). Therefore
[1],
and hence 1 is selfadjoint. 'it
1
it remains for us to prove that i * is symmetric. Let y, z E D(l It f cx') from Lemma 6.3 that y',z' and therefore y'Z,) cx'). Consequently there exist sequences —p
+
such
that 0,
jim y (an) z' (afl) = 0,
0,
Urn y (ba) z' (b,,) = 0.
urn y' (a,,) z (a,,) = (6.17) Passing
z (ba) =
bm y'
to the limit in Green's identity (z) — zl* (y)} dx
—•
(zy' —
and using (6.17), we obtain (z) — zl* (y)} dx =0,
which proves that i *
is
symmetric on its domain D(l
The case of a haifline. We now consider the Sturm-Liouville opei 1(y) = —y" +q(x)y on the halffine (0, cx'), assuming as before 3.
q(x) is a continuous function. We take for D(l) the set of all
§6. SELFADJOINTNESS OF A STURM-LIOUVILLE OPERATOR
491
continuously differentiable functions with compact supports satisfying the boundary condition
y'(O)cosa +y(O)sina =0.
(6.18)
The following lemma gives a description of D(l LEMMA 6.4. The domain D(l*) of 1* consists of those functions z(x) satisfying the following conditions: a) z(x)
is
continuous and has a first derivative which is absolutely
continuous in every finite interval, c)
d) z'(0)cosa +z(0)sina =0. The proof is based upon Lemma 6.1. Assume first that y(x) satisfies the conditions y(0) =y'(O) =0. Applying Lemma 6.1, we obtain z (x) =
+ + (x — t) [q (t) z (t) —
(t)J dt.
It follows from this representation that z(x) satisfies conditions b) and c), and also that the limits z( + 0) = a, z' ( + 0) = exist.
We now consider an arbitrary function satisfying the condition (6.18) and write down the equality which defines the adjoint operator: (6.19)
Integrating by parts, we obtain
dx = —y' (0) Therefore
z
(0) + y (O) z' (0)
—
y (—z") dx.
(6.19) can be written in the form
{—y" + q (x) y) z dx (6.20)
+
Since, as was proved, l*(z) = (6.21)
—z"
+q(x)z, it follows from (6.20) that
—y'(O)z(O) +y(0)z'(0) =0.
492
XIII. LINEAR OPERATORS IN HILBERT SPACE
By (6.18), y'(O)/y(O) = —tga. Inserting this expression in (6.21),we obtain condition d) for z(x). Using Lemma 6.4, we can obtain the following theorem, which is analogous to Theorem 6.1. THEOREM 6.2. If q(x) below, then the operator I was
(0 (the
at the beginning of
given
x < a)
and bounded from closure of the operator 1, whose definition is contihuous
this subsection) is selfadjoint.
REMARK Another proof of Theorem 62 was given m Chapter 2 A substantially more general theorem (Titchmarsh's theorem) was also proved in Chapter 2
§7 The selfadjomtness of a first-order system
We consider a first order operator, defined on vector-functions y(x)
(0
=
x < a))
(7.1)
d
(7.2)
We assume the coefficients a, b and c to be continuous functions, an* take to be the set of all continuous vector-functions with
supports having continuous first derivative, and satisfying the condi tion (7.2).
Let us find the domam D(.V *) of the adjornt of
Instead 0
Lemma 6 1 we will use the followmg lemma, whose proof is analogous to that of Lemma 6 1 LEMMA 7 1
a) f(x)
Let f(x)
be a
continuous function satisfying the conditioni interval (a, b),
is absolutely continuous in the
b) f'(x) c)
f(a)
=f(b) =0.
h(x)
= const almost
If
everywhere.
§7. SELFADJOINTNESS OF A FIRST-ORDER SYSTEM
493
From the definition of the adjoint operator we obtain
+ b(x)y1 + c(x)y2Jz2}dx
+b(x)112]zi
+
(7.3)
dx,
where z2
Putting successively Y2 =0, Yi =0 and using Lemma 7.1, it is not hard to obtain from (7.3) the following lemma (cf. the proof of Lemma 6.4). LEMMA 7.2. The domain D(fI' *) consists of all vector-functions \z2(x)
satisfying the following conditions: a)
b) z1(x) and z2(x) are absolutely continuous in every finite interval, cx),
c)
i.e. d)
co).
(z) z1 (0) cos
a
+ z2 (0) sin a
0.
From Lemma 7.2 we deduce the following remarkable theorem. THEOREM 7.1. If the coefficients of the system (7.1) are continuous, then the operator (the closure of the operator is selfadjoint. PROOF. It is sufficient (cf. Theorem 6.1). Let
to prove the symmetry of the operator
*
y=(Yi) and z=C) belong to
By Lemma 7.2 we have y1z2,
there exists a sequence (7.4)
such
{
=0,
y2zi
E
2(0
that =0.
). Therefore
XIII. LINEAR OPERATORS IN HILBERT SPACE
494
Further, on the basis of Green's identity we have (again using the result of Lemma 7.2) (y, 2* (z)) — (2* (y), z9)_4—(zi, y2)]dx
*(y), z) *(z)) = Passing to the limit and using (7.4), we obtain (y, *), for any y, z E D(.V as was to be proved. Theorem 7.1 shows that for a first-order system whose coefficients are continuous, the limit circle case cannot occur (i.e. the deficiency indices
are (0,0)). Bibliographical
references
—5. In these sections we have given a brief discussion of the foundations of the theory of symmetric and selfadj omt operators in Hubert space The reader can find a detailed discussion of this theory m the books by Ahiezer and Glazman [1] and Riesz and Sz -Nagy [1], for example
§6 Theorem 6 1 is due to Weyl [1] In presentmg the results of this section we have used the book by Glazman [1]
§7 Theorem 7 1 is due to Levitan A similar theorem was obtained independently by Martynov [1]
CHAPTER 14
SOME THEOREMS OF ANALYSIS *1. The Riemann-Lebesgue lemmas
Throughout this book we have frequently referred to lemmas on the of trigonometric integrals of the form behaviour, as A —* sin Ax dx,
(1.1)
cos Ax dx,
where (a, b) is a finite or infinite interval, and f(x) E (a, b). These lemmas are well known and are usually called the Riemann-Lebesgue lemmas. Less known are lemmas on the behaviour of the integrals (1.1) when the function 1(x) depends upon a parameter. (a, b). Then LEMMA 1.1. Let (a, b) be a finite interval and f(x) E (1.2)
sin Axdx=O.
urn a
PROOF. Let us first assume that f(x) is the characteristic function and f(x) = 0 of an interval (a, C (a, b), i.e. f(x) =1 for x E (a,
for
Then b
1(x) sin Ax dx =
c
.
sin Ax dx
from which (1.1) follows immediately.
Now assume that f(x) is a step function, i.e. the interval (a, b) can be subdivided into a finite number of subintervals (ak_i, a,,), k r1, . . ., n, such that f(x) is equal to a constant d,, throughout the interval (a,,_i, ak),
k=1,...,n.Then b
,,
liiii
/(x) sin Axdx=
urn a
sin Ax dx = 0.
k_—i
If f(x) is an arbitrary function from Lusin's theorem; see, for example,
(a, b), then it is [1].
495
that
496
XIV. THEOREMS OF ANALYSIS
for any > 0 one can find a step function
(x)
such that
(1.3)
It follows from the foregoing that for fixed so large that for H > we have
we can choose Ao =
(1.4)
It therefore follows from (1.3) and (1.4) that for xj > I
1(x) sin Ax dx
[1(x) —
(x)} sin Ax dx +
A0
(x) sin Ax dx
whence the lemma follows by the arbitrariness of REMARK
1. One can similarly prove that Jim a
The same remark holds for the subsequent lemmas of this We will therefore state them only for the first of the integrals (1 1YI
LEMMA 1 2 Let (a, b) be an infinite (on one or both sides) interval an4J let f(x) E (a, b). Then (1.2) holds.
PROOF. For definiteness let us consider an interval of the for! >0 one can find A
(a, + co), where a is finite. For arbitrary
such that (1.5)
By Lemma 1.1, for fixed A we can choose A0
so large that
(16)
for lxi
>x0 It follows from (15) and (16) that for lxi
>x0 one
§1. RIEMANN-LEBESGUE LEMMAS
497
sin
f(x)sin
1(x) dx
from which the lemma follows. The following lemma gives us information on the behaviour of integrals
of the form (1.1) when the function f(x) depends upon a parameter (or parameters). LEMMA 1.3. Suppose that we have an infinite set E = f(x) of functions which satisfy the following conditions: 10. 1(x) where the interval (a,b) can be finite as well as }
infinite.
2°. The set E is relatively compact in the metric space (a, b). 0 one can find Then the relation (1.2) holds uniformly, i.e. for any such that for xl > A0 the inequality = (1.7)
holds for all f(x) E E. PROOF. Let E > 0. From a well-known criterion for the relative compactness of a set in a metric space it follows that one can find a
finite E-set for E, i.e. there exists a finite set of functions E, = }
fk(X)
c E such that for each function f(x) E E there exists a function
E E, such that
(1.8)
For thed we can obviously find A0 =
>0 such that the inequality
sin Axdx
(1.9)
holds for xi > A0 and for all k = 1, . .. n. The lemma follows from (1.8) and (1.9). REMARK 2. If the interval (a, b) is finite and the functions f(x) of the family E are continuous, then for the relative compactness of E in I
498
XIV.THEOREMS OF ANALYSIS
the metric of .1' (a, b) it is sufficient that E be relatively compact in the metric of the space C(a, b), and for this it is sufficient that E be uniformly bounded and equicontmuous 2) If the interval (a, b) is mfinite, then for the relative compactness of E it is sufficient to require that for any > 0 there exist a fimte interval
(a', b') C (a b) such that
for every function f(x) E E, and that the restrictions of the functions in E to the interval (a', b') form a relatively compact family in t (a', b'). In fact, denoting the set of all restrictions mention c
metric of
above by E', it is clear that if the restrictions to (a', b') of f1(x), . . .,f,(x) E E form an -set in E', then these functions will form a 3-set in E; hence E is relatively compact since is arbitrary
A case of interest is that in which the family E is generated by suminable function ={f(x±t)}, xE(a,b), (a,b) finite interval and t belongs to some closed bounded set M on the Let tx+t}E(A,B) and 1(x) E9'(A,B). Then the set E is compac
In fact, let
be an infinite sequence of numbers belonging to J set M If we choose a subsequence { of which converges to number 4, then }
jf(x+
—/(x+ t0)l dx
b+to
a+t,
B
A
§2. Helly's theorems
1. Letuj (X), u2(X),... be an infinite sequence of nondecreasing bouni finite: closed interval [a,b]. FOr defhi functioiis defined on
let us assume that all the functions of this sequence are from the left: —0) = FIRST THEOREM. If the functions
are uniformly
then we eon find a monotone function u(X) and a subsequence which converges to cr(X) at every point of continuity of a(X) 2)
a,,
This is the so called Arzelà Ascoh lemma see for example Ljustermk and Sobol
§2. HELLY'S THEOREMS
499
PROOF. Let Ai, A2,... be a countable everywhere dense subset of the interval [a, b]. By means of the diagonal procedure (see, for example, Smirnov [1], §12) one can find a subsequence which conr,,,(A,) = r,. verges at each of the points Ai, A2, .... Put If
= sup t,,
=4 {a(X + 0) + a(X —0)).
is a nondecreasing function. We wifi now show that at every point of continuity of
Obviously (2.1)
one has
or(X).
Let A be any point from the interval [a, b]. Since for A
( inf
'c,
a (A
+ 0).
In the same way one can prove that Thus J
At every point of continuity o o- we have fore the preceding inequalities imply that
—0) = u(A + 0). Therecr,,,(A) = i(A),
as was
to be shown. HELLY'S SECOND THEOREM. Let [a, b] be a finite of a continuous function on [a, b]. Suppose that a sequence nondecreasing functions defined on [a, b] converges to a nondecreasing function defined on [a, b], at all points of continuity of If, moreover, (2.2)
urn
a (a),
Jim
a (b),
500
XIV. THEORRMS OF ANALYSIS
then (2.3)
da (A).
urn 5 / (A) a
PROOF.
such that
Let a that
a
Let >0 be arbitrary. We choose a positive number ô <5 implies Jf(x') —f(X")I A0
A1 — A1_i
< ...
A], we obtain b
5/(X)
/
(A)
—
(Ak—
p
(2.4)
5
—
5
xi_I
b
p 5
/ (A) —
/
(A) <
s
(A) 5
= {a,,(b) where M is chosen to satisfy the condition way we can prove the inequality
<2ME,
I <M In the
b
51(X) da (A)
(2.5)
/ (pt.) [a (As) — a
—
<2 ME.
We now choose the points A1, , to be pomts of continuity of It is obvious that, holding these points fixed, we can choose n so 1
that the inequality p
p
(2.6)
/ (p.) [a (As) —
a
1=]
<S
will hold. Since is arbitrary, (2.4), (2.5) and (2.6) imply (2.3). 2. Helly's second theorem admits the following generalization infinite interval: Suppose that a sequence { of nondecreasing cx)) converges to a nondecreasing function defined on (— at all of continuity of r, and suppose that f is a continuous function defiru (—
ccx). Assume
that for any E >0 there exists a number A
such that for all a,b> A and all n we have (2.7)
1(A)
I
<
51 f(A)
<e.
501
§2. HELLY'S THEOREMS
Then urn —Ca
PROOF. We wifi first show that f is absolutely integrable with respect (A) <€. to the function r. Indeed, if b, c> A, then by hypothesis f(x) I
Holding b and c fixed (we can assume that b and c are points of continuity of Letting c
and letting n —*
o-)
we
we
obtain
f(A) dcr(X) <E. I
obtain
(2.8)
In the same way we can prove that (2.9)
These inequalities imply the absolute integrabiity of f. By the previous theorem, if — a and b are points of continuity of u, then lim
(2.10)
(X)
(X) = / (A)
(A).
Further, by virtue of the inequalities (2.7) we have tim
(2.11)
—a,
-
It
follows from (2.8) —(2.11) that n-*a,
—a,
—a,
—a,
"-'a,
—a
+ Since
—a,
—a
I
+
is arbitrary, the theorem is proved.
1(X) dc(X)
502
XIV.
OF ANALYSIS
§3. The Stieltjes inversion formula
Let
+ io-2(X) be a complex function of bounded variation
on the entire real We put
m
The Stieltjes inversion formula gives us a means of expressmg cr(X) terms of (z)
We put (with z = + Zr) rp(z)__p(z)
t)=
(31)
ç —
THEOREM 3 1
(32)
—
3
(i
If the points a, b are points of continuity of
then
a(b)—a(a)=hm We may clearly assume without loss of generality that; Integrating (3.1) by parts, we obtain
coj) = 0.
-4 Integrating this equality with respect to
from a to b, we obtain
b)—I(; a). We begin by considering the first integral. We have
I(r, b) =
d0 (X) dX =!
That is, the real functions
and °2 are
a
(X + b)
dX.
of bounded vanation on the real
503
§3. STIELTJES INVERSION FORMULA
Since
dX=1, it follows that
I(; b) —a
{a(X +
b)
—a(b)} dx
{o(X+b)—a(b)}dX
—7
+4We pick > 0 and then choose so that the inequality or(X +b) — or(b)I <s / 2 holds for XI o. This is possible, since by assumption b is a point of continuity of o-(X). Having chosen o in this way, we obtain I
I
s
J ii L.—
ç 3
—a)
Let M> 0 be such that
[tj +
X2
lb
I <M Then
I
4-). For fixed a we can choose r so small that the last expression will be
I(r, b) = is arbitrary, we have proved that In the same way we can prove that I(r, a) = o-(a). Therefore
<s/2. Since
(3.2) follows from (3.3). are two
THEOREM 3.2 (UNIQUENESS THEOREM). If ori(X) and complex-valued
functions of bounded variation on (— a)
a'), and
a)
ç
do1 (X)
ç
(A)
3
z—A —
3
z—x
—a)
for all nonreal z, then
a).
—01
PROOF. By Theorem 3.1,
—a)
= const. o-1(b) — orj(a)
whenever a, b are points of continuity of both
and and
—
a2(a)
coincide
Hence (assuming
that all ftthctions of bounded variation are taken to be continuous from the left) it follows that 01 and u2 can only REMARK. If
in this case
is
a real function, then
a constant. =
Therefore
504
XIV.THEOREMS OF ANALYSIS
=
p(z)
Im
{cp
(z)}.
For r> 0 the Stieltjes inversion formula assumes the form 1'
(b) — o
(a) = urn — — Im
(z)) da.
a
*4. Tauberian theorems
1. Levitan's Tauberian theorems for Fourier integrals. THEOREM 4.1. Let the function o-(v) iatisfy the following conditions: is of bounded variation in every finite interval. a) (4.1')
c) For every function which has a bounded derivative of order r + 2 and which vanishes outside the interval (— A, A), h (v) da (v) = 0,
Then for
—*
where it (v)
= 4—
(t)
have the estimate
we IL
—
)8 da (v)
(4.1)
= o (pr_a)
(s
0).
PROOF. Let us consider the integral 1
(4.2)
s)
— v2)s
(v).
The proof wifi consist in giving another representation for this integral,
from which the estimate (4.1) follows easily. We put v,
(s>0),
f
s)
By the Fourier mversion formula we have
s)
§4. TAUBERIAN THEOREMS
p
v,
h (t,
s)
505
s)
Inserting this expression in (4.2), we obtain (4.2')
1
s)
=
Ii (t,
da (v)
Let n denote the integer part of s: n = (4.3) by parts n times, we obtain h (t,
(4.4)
s)
[s].
(ii,
s)
Integrating the integral in
v, s)
We will consider the case of integer and noninteger s separately. Suppose
first that s is an integer: s =n, with n ? 1 (the case s =0 will be considered separately). Integrating the right side of (4.4) by parts once more, we obtain h (t,
s)
=
—
v, s)
v,
s)
(4.5) p(n+1)
v, s)
dv.
Let #(t) be a function which has a bounded derivative of order n + 2, and which equals 1 for
t
and zero outside the interval (—1, 1).
Put #A(t) =#(t/A). The function equals zero outside the interval (— A, A). Therefore condition c) of the theorem implies that da (v)
(t) h (t,
s)
o.
Subtracting this relation from (4.2), we obtain
I
s)
=
If we replace here J(M,s) where
(v)
—
(t)] h (t,
s)
by the expression (4.5), we obtain
XIV. THEOREMS OF ANALYSIS
506
I—
a)
'1,
=
2t
$
do
(t)
(v) { .1
Co
=
a)
(EL,
{I
$ do (v)
2it
.-a)
a)
— (_j)fl+1
do (v)
—
f
I
—a)
(t)
S
p(fl+l)
(t) etdt1,
—
J
s)
dt
L
We now put Co
ç I
(4.6)
2ic
J
(t) e
(tt)
—a)
It is easily seéñ that this integral converges absolutely for s> 1, and converges for s =1. Moreover, integrating by parts, it is not hard to show that for v one has the estimate cx
A9(v) =
(4.7)
Interchanging the order of integration in J (,z, s), which is permissible by virtue of the convergence of the integral (4.6), we obtain (4.8)
J(3) (p., s) =
(v)
s)
(p.,
(v +
Further, using the notation (4.6), we obtain a)
(4.9)
s)=
p(8) (EL, v, a)
d,o (v + EL),
$ Aa
2it
—a) a)
(4.10)
(#)
v, a) L=_L
J(2) (p., s) —
A
(v)
a (v — p.).
—a)
It is easily seen that as z — (4.11)
is,
s) I
=
Let us now estimate the integrals (4.9) and (4.10). We will consider (4.9); the estimation of (4.10) is similar. Let N be a positive integer and put
507
§4. TAUBERIAN THEOREMS
Given an arbitrary
> 0, for
>
and
N we will have
'I
j+1
V Therefore (4.12)
Further, (4.13) and a similar estimate holds -for i3:
(4.13')
If we first choose N, and then and (4.13') we obtain for M >
then from the estimates (4.12), (4.13)
(4.14) It
follows from (4.11) that for
L
(4.15)
J J
for
s)
for large
As before, let N be a positive
integer. We put
Just as before, (4.17)
Further, s)
12+ 13)
508
XIV. THEOREMS OF ANALYSIS
It follows from (4.17) that (4.18)
Iifl +
E,
s) (j
jr
+ 1)dE
N±i
since
Let us now consider s)
Let I
v
We have A, (v)
(v
+
be an arbitrary positive number and choose a so large that for I > a one has
I
j+l
V J
We then obtain
Therefore
I
Hence
I
<
(4.19)
Let be an arbitrary positive number. We first choose N so that <e/3. Finally, 2/(N + 1) 3. We then choose a so large that we choose
3.
so large that
Then from (4.19) we obtain
the estimate (4.19')
J3(L,S)
From the estimates (4.16) and (4.19') it follows that
509
§4. TAUBERIAN THEOREMS
(4.20)
Dividing by
we
obtain
as was to be proved. Let us now consider the case s =0. In this case (cf. the notation (4.3)) h (t,
e
0)
e
Therefore
/
0)
=
f
da (v)
da (v) (A0 (v
=
A0
[1 —
(t)]
e ipi — e —ipi
— — A0 (v + (v +
—
A0 (v)
(v
—
Ao(v) has all the properties of the function A8(u), we can carry out the previous estimates, as a consequence of which we obtain Since
which proves the theorem for s = 0. Now suppose that the number s = n + a (a this case we have to estimate the integral J
s)
—
0)
is noninteger. In
(v).
We will transform this integral by means of integration by parts, and reduce its estimation to the case of integer powers of s. Integrating by parts n +1 times, we obtain 1
n
+
=C
—
v
— t2Y da (t)]
dv =
XIV. THEOREMS OF ANALYSIS
510
=
n)dv,
where C is some constant. Further, put J
n+
—
(v, n)
dv = J' + J".
Then the estimate (4.20) implies that (4.21)
I
1"
=
—
I
Further, integrating by parts once more, we obtain —
1'
1)
n + 1) dv.
Therefore, applying the estimate (4.20) again, we obtain (4.22)
P=o
+
s) = O(,LT+8) for —* Thus It follows from (4.21) and (4.22) that the proof of the theorem is complete. REMARK. One can similarly prove that in the estimates (4.1) and
(4.1') one can replace o by 0. For the case of an odd
(precisely this case is applied in this book) the functions A2(v) and h (t, ,i, s) can
be assumed even, i.e. one can consider the cosine Fourier transform of the even function which equals zero outside the interval (— A, A) For this case the basic estimate assumes the form v2)8
(v) =
or
(1— Theorem
da (v) = o
4.1 in fact gives us the asymptotic behaviour of the odd
component of the function x(v). To obtain the asymptotic behaviour of u(v) itself, one has to consider a kernel of a more general kind than
511
§4. TAUBERIAN THEOREMS
theorems in this section) or else consider,
v; s) (cf.
together with the kernel (ii, v; s), some odd kernel as well, which would enable us to clarify the asymptotic behaviour of the even component of r.
In the following theorem this is done in one of the simplest THEOREM 4.1'. Let us assume that the function conditions:
a) It
has
satisfies the following
bounded variation in every finite interval.
b) For a+1
c)
and
For every function is equal to zero outside
h
(v)
which has a continuous second derivative the interval (—A,A),
=0,.
h (v)
=
(t)
d) The point v = 0 is a point of continuity of r(u). Then for A—4
+o(—A) = 0(1).
(4.23)
REMARK. We note that the condition d) is not a restriction, as there always exists a number v0 such that the function = o(v +v0) will will be continuous at v =0. At the same time, it is easy to see that satisfy the conditions a), b) and c) if does, and that u will satisfy (4.23) ii does (in view of satisfying b)). We assume in (4.23) that the points A and — A are points of continuity
of o(v).
PROOF OF THEOREM 4.1'. For A >0 we put
(
Ofor
u<—A,
for
v—X,
—1 for —X
for for ior 0 for 0 1
This theorem is due to Levitan and
is
v0,
0
vA,
v>A, published here for the first time.
512
XIV. THEOREMS OF ANALYSIS
I —cos At
K(t, A)
It follows from the Fourier inversion formula that (4.24)
x (A,
v)z..
COS
We put, by definition —0
f(v)da(v)zlim f(v)da(v),
/(v)da(v)lim
—x
(4.25)
It is not hard to see that (A) + (—A) =
(0) +
x (A, v)
(v),
or, inserting for x(A, v) the expression (4.24), =
(4.26)
Let be the same as in Theorem 4.1. It follows from condition c) of the theorem that (4.27)
da
(v) {_t
—
(t)
At e_iYtdt}
= 0.
Since tl(v) and (t)
—
At
are continuous at the point v = 0, the outer integral in (4.27) can be understood in the sense of (4 25) Subtractmg (4 27) from (4 26), we obtam
§4. TAUBERIAN THEOREMS
(X) + (—k)
(0)
+
{j
da 5
(t) (1
— cos Xt)
(v) {s cx:,
5
1—
'
= 2o (0) ±L 5
—
513
('i)
t—y (t)
'dt
5
(4.28) I—
(t)
5
— 5 A1(v+X)da(v),
where Ai(v) is the same as in Theorem 4.1. Since Ai(v) I = O(1/v2) for large it, the estimate (4.23) can be obtained from the integral representation (4.28) by means of estimates similar to those which were carried out in the proof of Theorem 4.1. 2. Tauberian theorems for Fourier integrals. THEOREM 4.2. Suppose that for some integer n 0 the nondecreasing and left-continuous functions p (A) and o(X) (—
I. On the set of all infinitely differentiable functions f(x) which vanish outside a given interval (—h,h) one has the identity (X) d {p (X) —
(X))
5
=
5
(x) G (x) dx,
where G(x) is some function, defined and summable over (—h,h), and (A)
=
5
1(x)
514
XIV. THEOREMS OF ANALYSIS
II. One of the functions p(A) and r—
cr(X), say (7(A), is
such that
a(a+Ao)_a(a)*(X)<
for some A0> 0. Further, suppose that the kernel T(N, A) (0 satisfies the following conditions:
A. For every N the function T(N, A)
is
N<
N is a parameter)
of bounded variation in
A
A)ldX< B. For every N the function T(N, A) has k 0 derivatives with respect (m=0,1,...,k). The A, and IT(m)(N,A)l =o(1A1"2) for kth derivative T (N, A) is of bounded variation (as a function of A) and to
X)I
S
±
(N, X)
(the number n in A, B and C is the same as in conditions I and II). Let us denote by co(x) an even infinitely differentiable function such that
p(x)=l,
p
Then
(N, X)
§4. TAUBERIAN THEOREMS
—
lim
$ T (N, X) 4 (p(X) _a(A)) — $ [f1(N,
515
G(x)dx
(4)
X)J
the constant C, which depends only upon
can be chosen to
C <8. ipecial case of Theorem 4.2. A special case of Theorem 4.2 has been
led in this book, namely: 1,
lxI< N,
T(N,x)= 0,
k = n = 0,
the function G(x) is of bounded variation in some neighborhood ro.
en if the hypotheses of Theorem 4.2 are satisfied for any h, and (+ 0) = 0 (for example, if (X) = A), then for N —
N)—p(—N)=
4-{G(+O)+G(_O)}±o(t).
Suppose that the functions p(X) and a(X) satisfy 'ions I and II of Theorem 4.2 with n = 0. Let g(x) be an arbitrary EOREM
on with compact support which coincides with G(x) in some neighd of zero, and let
(A) =
g (ii.)
g
d'i..
there exists an absolute constant C such that Sg(A)I <
—p(—X)} —p(-—X)} —{o(X)
—S'(A)
Ca*(1/h),
Cq*(1/h),
A is some constant. Theorem 4.2' is due to
and is published here for the first time.
XIV. THEOREMS OF ANALYSIS
516
Let us remark, first of all, that if g1 (x) is another function with compact support which coincides with G(x) in a neighborhood of zero, then by a well-known localization theorem _Sg1(A)} =0,
and
lim{S(X) exists.
—Sj(X)}
Therefore it suffices to prove the theorem for one function g(x)
with compact support which coincides with G(x) in a neighborhood of zero.
We also remark that the function p (A) is defined up to an additive constant; therefore we can always assume it to be normalized in such
a way that A =0. PROOF. The assertion I) is a corollary of Theorem 4.2, if we take the kernel T(N, A) = 1 for XI
I
We precede the proof of the assertion II) by the following remark: if the function a(A), XE (0, + aD), is such that for some ö ? 0 we have, dN = the inequality for any sequences dN }, { bN }, bN> dN, (4.29)
ixnla(dN) —a(bN)l
then there exists a number A such that (4.30) V2
Indeed, it is obvious from (4.29) that
Iimja(X)J
Therefore all the limit values of a (A), A
+ a,
lie in the rectangle
limRea(X) ? Rez liniRea(A), limlma(A) Imz > liinlma(A),
o. It is clear that if the complex number A is taken as the center of this rectangle, then the inequality (4.30) whose sides are of length will hold.
Let us now verify that the function
________
517
§4. TAUBERIAN THEOREMS
(4.31)
+i(—X) } —S(X)
a(A) ==$p(A) +p(—X) } —
satisfies the condition (4.29), if g(x) =ç(x/h)G(x) and o = Co.*(1/h). For this we put for lxi E(dN,bN), for (dN, bN).
T(N A) —
— to
kernel satisfies the conditions A, B and C (for k = 0) of Theorem 4.2, and therefore Theorem 4.2 is applicable to it. Taking note of the inequality proved in that theorem, we see that the function a (A) defined There follows by (4.31) satisfies the condition (4.29) with ö = the existence of A for which (4.30), and hence condition II), holds, which was to be proved. This
If the conditions of Theorem 4.2 are satisfied for any h, and ( + 0) = 0, while G(x) satisfies conditions in the neighborhood of zero which guarantee the convergence of its Fourier series and conjugate series, then for A ± one has the following asymptotic formula: COROLLARY.
+A + o(1).
p
3. A Taubenan theorem of and be positive increasing functions, defined for x> 0, where (x) is differentiable and satisfies the conditions = and for x> a > 0, where a and <x4,'(x) are positive numbers,° while =0 for x
a)
/ (x) — —
ç
dp(1)
g
—
m is the integer part of
If f(x)/g(x)—1 for x—* — urn
6)
— 0
0
where
ç
—
cx,
then
p(x)
—=1.
proved his theorem under the assumption that the numbers a and satisfy the inequality 0 <5
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