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0 there exists a linear combination of (p'ks, such that <£. fc=i a(u-uh,ip-Ilip)
1,*
Proof. Since V is separable, V, as a subset of V, is also separable. Let {fk} be a basis in V of functions from V. We orthonormalise {<^fc} in V by the Schmidt procedure, to obtain another denumerable set {
(v-Vu,u) = 0,
Vu,wG(C£°(ft)) 3 .
(u'V«,to) = - ( » - V » , « ) ,
(1.85)
(1-86)
Proof. Property (1.86) is a direct consequence of (1.85) when we replace the u in it with u + w. To prove (1.85) we have / {v-Vu)-udx=
in
-
I v • V | u | 2 dx = 0.
2 JQ
Theorem 29. T/ie problem (1.84) admits at least one solution.
•
46
Numerical Methods for Exterior
Problems
Proof. We use the Galerkin method to prove existence. Let {
where Cfc satisfy (Vum,V
=< f,
Vfc = l , - - - , m .
(1-87)
To prove the existence of Cfc, we make some estimates first. Multiplying (1.87) by Cfc and summing up with respect to A;, we obtain
) =< f,um > .
(Vu m , V u m ) + (u m • Vu
By (1.85) the second term on the left hand side vanishes. The right hand side is bounded by ||/||i/'|l u m||v- Let M be a constant such that M > H/llv, then we get (Vu m , V u m ) + (u m • Vu m , um) >< f,um>,
V|u m |i = M.
Owing to the Corollary 1 of the Brouwer theorem, the solution to (1.87) exists, and \um\i < \\f\\vLetting m —> oo, we can extract a subsequence of um, converging weakly in V, and by Lemma 5 the subsequence converges strongly in Lfoc(fl). Let u be the limit. For the second term on the left hand side of (1.87) we have (u • Vu, tpk) - {um • Vu m , ipk) - (u • V(u - um), tpk) + ((u - um) • Vum, ipk). Noting that (fk is compactly supported, we find the limit of the above expression is zero. Then we obtain (Vu, V(fk) + (u • Vu, tpk) = < f,
V/c = 1, • • • , oo.
Since the linear hull of {(fik} is dense in V, the conclusion follows.
•
We turn now to the inhomogeneous boundary conditions, lim u = Uoo,
u\9a = g,
(1.88)
| a: | —*oo
and assume that g € (ii'1/2(<9fi)) . To prove the existence of (1.80),(1.81), (1.88) we need the Sobolev inequality. Let us prove it for a special case, the inequality for three dimensional domains. Lemma 28. For all u € C~(ft)
Ho,| < AHV«||o,i-
(1-89)
Exterior Problems of Partial Differential
Proof.
47
Equations
We have
1 fa
du dxi
dxx =
-F1(x2,x3)
and similar estimates for x2 and X3. With the obvious meaning of the symbols we then deduce \2u{x)\i <
{F1(x2,X3)F2{x1,xa)F3(x1,X2))i.
Integrating over x\, and using the Schwarz inequality, oo
\2u(x)\% dxi /
-OO
< {Fi(x2,x3))i
I /
F2(x1,x3)dx1)
I /
F3(xi,x2)dx1J
.
Integrating it successively over x2 and x3 and applying the same procedure, we find Ho,
n/..
du dxi
dx
<
aS^
du dxi
dx,
which, in turn, after employing the inequality (ai + a 2 + a3)2 < 3(af + a\ + a\) gives (1.89).
D
L e m m a 29. For all u G Cg°(n) 2 ||w||o,6 < 7/|l u li-
(1.90)
Proof. Replacing u with uA in (1.89) and then estimate the right hand side by the Holder inequality. We have
l<6<^=ll«" 30,6 |V«||0. Then (1.90) follows.
•
By the inverse trace theorem of Sobolev spaces there exists u\ € (H1^)) with a compact support on Q, such that ||ui||i < \\g — Woo||i> and u\ = g — u^ on dQ. Let R be large enough such that B(0,R) D supp Mi, and let fix = Q n B(0, R). We apply Lemma 20 to get a function u2 G ( H ^ f t i ) ) 3 such that V • u2 = - V • ui and ||w2||i < C||V • iti||o- The Lemma 20 is about an exterior domain, however it can be verified that it is
48
Numerical Methods for Exterior
Problems
also applied to bounded domains. Let us extend the function ui to zero to the exterior of B(0, R), then u-i € (HQ (&)) • We define UQ = u\ + ui + u^,, then uo satisfies the boundary condition, and V • UQ = 0, and we notice that the derivatives of UQ are compactly supported. Let w = u — wo, then the weak formulation for (1.80),(1.81),(1.88) is: Find u £ V + u0, such that (1.84) holds. Theorem 30. There is a positive constants > 0, such that if\\g — Woo||i < S, then the weak solution to (1.80), (1.81), (1.88) exists. Proof. Being analogous to Theorem 29, we set the approximation to w = u — UQ to be m
wm = y^c fc y fc , then Cfc satisfy (Vio m , Vv*;) + (wm • Vw m , (fk) + (wm • Vu 0 ,
+ (u0-'Vuo,
= -(Vu 0 ,V(^fc)+,
(1-91)
V/c = 1, ••• ,m.
Some terms in (1.91) are the same as those in (1.87). Let us estimate the other ones. By the Holder inequality, < ||Wm||o,6l|VUo||0,§-
\(wm-Vu0,Wm)\
Since Vuo is compactly supported we get \(wm • Vu0,wm)\
< C|«; m |f||Vuo||o,fii,
where we have applied Lemma 29. Being the same |(u 0 - Vu0,wm)\
< C||u 0 ||i,ni||Vuo||o|^m|i,
where we have applied the Sobolev embedding theorem to get ||«o||o,6,ni < C||wo||i,ni- By (1.85) we have (u0 •Vwm,wm)
=0.
Finally we have |(Vuo,Vu> m )| < |wo|i,f)i|w m |i. By the definition of UQ, there is a constant C\, such that C||Vuo||o,ni < C\\\g - Woo||I. We take 5 and M such that 5 < 1/Ci and (1 - 8CX)M > ll/llv' + C||wo||i,fii ||Vuo||o + |«o|i,ni- By the assumption of the theorem, llff —«oo 111 < S, then if |u>m|i = M the combination of the above inequalities
Exterior Problems of Partial Differential
49
Equations
can be summarized as (Vw m , Vw m ) + (wm • Wwm,wm)
+ (wm • Vu0, wm)
+(u0 • Vwm, wm)(u0 • V«o, wm) > - ( V u o , V w m ) + < / , wm > . The remaining part of the proof is the same as that in Theorem 29
•
The condition for g can be reduced to | fg^g • vds\ < 5, see [Galdi, G.P. (1994)]. Then for a particular case, j9ng • vds = 0, there is no restriction on the size of data. Generally speaking the solutions are not unique. An example is given in [Temam, R. (1984)]. Some results for the two dimensional case can be found in [Galdi, G.P. (1994)]. 1.9
Heat equation
To investigate evolution problems, some results of the theory of linear operators are needed. Let us introduce some definitions and some results first. Let H be a Hilbert space and A : D(A) —> H be a linear operator with domain D(A) C H. Let D(A) = H. Then the adjoint operator A* is defined by D(A*) = {ueH;3he
H, (u, Av) = (h,v),\/v
€ D(A)},
A*u = h.
A is symmetric if and only if A C A*. A is self-adjoint if and only if A = A*. Let A be a closed symmetric operator, then the Cayley transform of A is defined by UA = (A-iI)(A
+
iI)-\
where / is the identity. A bounded linear operator A : H —> H is isometric, if and only if A keeps the inner product invariant: (Au,Av) = (u,v),
\/u,v£H.
In particular if the range of A, R(A) = H, then A is called a unitary operator. A family of orthogonal projection operators E(X), —oo < A < oo, in a Hilbert space H is called a resolution of the identity if E(\)E(v)=E(mm(\,iJ,)), E{-oo) = 0,E(+oo) = I,
50
Numerical
Methods
for Exterior
Problems
where E(-oo)u
— s-lim E(X)u,
E(+oo)u = s-lim E(X)u,
A—» — oo
£(A + 0) = £(A),
A—>+oo
where E(X + 0)u = s-lim
E{y)u,
H—»A+0
and s-lim is the strong limit. Proposition 2. For ant/ u,v € H, (E(X)u,v) variation with respect to X.
is a function of bounded
Proposition 3. / / /(A) is a complex continuous function defined on (—oo, oo), and u £ H, then for — oo < a < (3 < oo we can define f{X)dE(X)u,
/i a
J a Riemann sum which is the strong limit of the
^2f(Xj)E(Xj,Xj+i]u,
a = Xi < X2 < • • • < Xn = fl,Xj e (Aj,A j+ i]
3
as max|Aj + i — Aj| —> 0, where E(Xj,Xj+i]
= E(Xj+\)
— E(Xj).
Theorem 3 1 . For a given u € H the following three conditions are equivalent:
(i) iZo fw dEWu exists> Wf^lfWfdlEMufKoo, (3) F(v) = f™ f(X)d(E(X)u,v)
defines a bounded linear functional.
Theorem 32. If f(X) is a real continuous function, then by oo
f(X)d(E(X)u,v)
/
-oo
determines a self-adjoint operator A with D(A) = D, where
D = UJ°° |/(A)|2d|£(AH2
/
Ad{E(X)u, v),
Vu G D(A), v e H,
-oo
which is denoted by oo
/
XdE(X). -oo
It is called the spectral resolution of the self-adjoint operator A.
Exterior Problems of Partial Differential
Equations
51
Theorem 33. A self-adjoint operator in a Hilbert space possesses a uniquely determined spectral resolution. Theorem 34. For a symmetric operator A there exists a closed symmetric extension A**. A closed symmetric operator A possesses a uniquely determined spectral resolution if and only if A is self-adjoint, and A is self-adjoint if and only if its Cayley transform is unitary. Let A : D{A) C H —> H be a linear operator. The resolvent set of A is defined by p(A) = {A G C; R{A - XI) — H,(A- AJ)one - to - one, (A - XI)_1 continuous}. The spectrum of A is denned by o~{A) = C \ p(A), which is further divided into three subsets. The point spectrum is o-p(A) = { A s o-(A); A — XI is not one — to — one}. The continuous spectrum is ac(A) ={A G cr(A); R(A - XI) = H, (A - A7)one - to - one, (A — A / ) - 1 is not continuous}. The residual spectrum is ar(A) = {A G a{A); R(A - XI) ^ H,{A- XI)one - to - one}. Theorem 35. Let A be a self-adjoint operator. Then X G o~p{A) if and only if E(X) — E(X — 0) 7^ 0, crr(A) is a null set, and the necessary and sufficient condition for X G cr(A) is that Ve > 0, E(I*) ^ 0, where J» = (A — £, A + e). Let us now investigate the initial boundary value problem of the heat equation. We remark that although the above definitions and results are all for complex numbers, we will assume the functions are real in the remaining part of this section, and it is easy to see that the following result is valid for both real and complex cases. The problem reads — = Au + f, u = 0,
xedCl,
(1.92) (1.93)
u(x,0) = u0(x). (1.94) We assume that dim(Q) = 3, but the results are true for two and one dimensional problems as well. We define H = L2(Q), and operator A = — A with domain D(A) = { « £ # o ( ^ ) ; A " G H}. Lemma 30. A : D(A) —> H is a self-adjoint operator .
52
Numerical Methods for Exterior
Problems
Proof. Clearly A is closed and symmetric. We consider the Cayley transform UA- Let v = (A + il)~lu, then by Theorem 17 u —> v and u —» Av are bounded in H, therefore UA is bounded in H. Besides we have UAVA = I, so UA keeps inner product invariant. From U*A = UA we get R(UA) = D(UAX) = D(UA) = H, so UA is a unitary operator. Then A is self-adjoint. • We assume that u0 G L2(fl), and / e L 2 ((0,T); L2(fl)) for T > 0. Let us write the problem (1.92)-(1.94) in the operator form: Find u : (0,T] —> .0(^4), such that dxi — +Au = f, limu(t)=«o. (1-95) at t-t+o T h e o r e m 36. The problem (1.95) admits a unique solution u{t), and ^ £ L 2 ((0,T),L 2 (fi)). Proof.
Formally we get u(t) = e~Atu0 + [ e-MeATf{r) Jo
dr.
By Theorem 17 we have cr{A) C [0, oo). Then in the notations of spectral resolution the solution to (1.95) can be expressed as / e-xtdE{X)u0+ / / e - A ( t - r ) dE{\) / ( r ) dr. Jo Jo Jo Let us verify it is the desired solution. Differential of it gives u=
J
9
/-OO
/"OO
ft
Xe Xt
= - /
~ dE(X)u0
+ /(*)"/
Ae- A ('- T ) dE(X) f(r) dr.
/
Here we notice that Ae~A* is bounded for positive t, so the above expression makes sense. For the first term we see that /»00
A/ Jo
/-OO
e-xtdE{X)u0=
/•oo
Jo
f\
Jo
fidllE(X)E(n)u0
poo
Xe~xt dE(X)u0. Jo Jo Jo The second term can be deduced in the same way. Thus u satisfies the equation. It is obvious that the initial condition is satisfied. To prove uniqueness we consider a final boundary value problem =
e~xtdx
/'OO
e~xtdx
fidllE(n)u0=
dv -— + Av — g, dt
lim v(t) = 0. t-r-o
Exterior Problems of Partial Differential
Equations
53
Then by the above argument v exists. Let u be the solution to (1.95) associated with / = 0 and UQ = 0. We have (u(t),g{t))dt=l
(u,--^+Av)dt =j
(^+Au,v)dt-
(v(T),u(T))
+ ( U (0),v(0)) = 0.
Since g is arbitrary, we obtain u(t) = 0. Thus the proof is complete.
1.10
•
Wave equation
We investigate the following initial boundary value problem of the wave equation (1.24) in this section. The initial and boundary conditions are u = 0, u(x,0) = uo(x), Let uo,ui £ L2(Q) = H. C([0, oo); H), which satisfies
L
xedfl,
(1.96)
—(x,0)=u1(x).
(1.97)
We are looking for a weak solution u €
JQu{w-Av)dxdt+lUoix)irt{x>0)dx -
/ ui(x)v(x,0)dx
= 0,
(1.98) Vw € V,
./n where V = Co([0, oo); D(A)) n C 2 ([0, oo); H), and the operator A is the same as that in the previous section. Theorem 37. The problem (1.98) admits a unique solution. Proof.
The equation (1.24) can be expressed as d2u
A
The solution to the problem can be formally expressed as u(t) = cos(A1/2t)u0
+ A~1/2
sm(A1/2t)Ul.
54
Numerical Methods for Exterior
Problems
Then in the notations of spectral resolution the solution to (1.98) can be expressed as u(t)=
[ Jo
cos(VXt) dE{X)u0 + [ Jo
sin
^ A ^ dE(X)Ul. vA
(1.99)
Let us verify that u(t), defined by (1.99), is the desired weak solution. u(t) G H is well defined because 3in(v/At)
|cos(\/A)| < 1,
In order to prove u is continuous in t, we estimate \\u(t)-u(t0)\\2 <2 ( J
| cos(v/At) - cos(\/Ato)| 2 d\\E(X)u0\\
r°° i + / Jo
- | sin(\/At) - sin(\/At 0 )| 2 d\\E{X)Ul ^
<2(x0\t-t0\2
f ° d\\E(X)uo\\2
/>oo
/-Ao
\
+ 4 / d||£(A) Uo || 2 + |t - *o|2 / d\\E(X)Ulf J\0 Jo J 2 2 2 2 <2|t - io| (A 0 ||«o|| + ||wi|| ) + 8(|| W0 || - ||E(A0)«oH2). We first take Ao large enough, then take \t — io| small. Then \\u(t) — u(to)\\ is arbitrary small, which gives u £ C([0,oo);H). To verify the weak formulation (1.98) we take v 6 V, then use Lebesgue's dominated convergence theorem to obtain /*
-
S-i
/ uo(x)~(x,0)dx
POO
=- J f00 d =J
Q
d(E(X)u0,-£(;0))
- J
= j™ 1^
f°°
r-
dv
Cos{^t)d{E{X)uo,-Q-t{;t))dt
Uos(VXt) d(E(X)u0, ^ ( - , t))
- V\sm{y/Xt)d(E(\)uo,-£(;t)))
dt,
Exterior Problems of Partial Differential
55
Equations
and analogously /•OO
0=
Jo
= / Jo
J
— dt
/>00
1 /Asin(v
Jo
/ Jo
/
At)rf( J B(A)uo,w(-^))di
{Xcos{VXt)d(E(X)u0,v{;t))
+\/Asin(\/Ai) d(E(X)u0, -^(-, t))) dt. Thus we obtain /.oo
I
/.oo
I
r
Q2V
cos(Vxt)d{E(X)u0,-^(-,t)+Av(-,t))dt
= -
Qv
uo{x) —
(x,0)dx.
Similarly we get
/°° r
sm
!/x' t) ^ E ^ ) u i > §^ <•> *)+Av(-' *)) * = - / "i^w 1 - ° ) ^
The proof of existence is thus complete. To prove uniqueness we take an arbitrary g £ CQ°(Q. X (0,OO)) and solve the equation d2v
„
-d¥~Av = 9> to get v satisfying the homogeneous boundary condition and vanishing for sufficiently large t. Let u be a solution to the problem (1.98) with homogeneous initial data, then by (1.98)
f°° f
/ Jo
(d2v
\
f°° f
I u\ -7T7 — &v dxdt — Ja \dt2 J Jo
/ uqdxdt = 0. Ja
Since CQ° is dense in L2, we obtain u = 0, which implies uniqueness. To prove the existence of v, we write down the formal expression oo
/
A~1/2
sin(A1/2{t
- s))g(s) ds.
We define a cut-off function C G C°°(R), such that C > 0, £(t) = 1 for t > 0, and £(£) = 0 for t < T, T < 0 fixed. Then we set
v(t) = ~C(t) f
J™ S l n ( V ^
S))
d(E(X)g)(; s) ds.
(1.100)
Let us verify that v(t) expressed by (1.100) is the desired solution. Following the same lines we can see v(t) is the weak solution, so the only problem is
56
Numerical Methods for Exterior
Problems
the regularity of v. For a bounded interval I C [0, oo), AE{I) is a bounded operator and AE(I)v(t)
= -C(t) j°° I MdE{j£) ULLH/y/J,) ^ I Jt J^ei Jo
ssm(y/\(t i n
( ^ VA
- s)) *» d(E(X)g)(; s) ds.
Since E(/j,)E(X) = S(min(/x, A)), we have r 0 • dE(\) = 0, A < fi, dE(fi)dE(X) = I dE((i), A = /x, [ 0 • dE(fx) =0,\>fi. Therefore AE(I)v(t)
= -((t)
r
S m (
f
^
S))
d(E(v)Ag)(;
s) ds.
Letting / —> [0,oo), one gets oo
/
\t-s\\\Ag(;s)\\0ds
so v(t) £ D(A). Moreover \\Av(t) -
Av(t0)\\0 /•OO
POO
+ C(M
\\Ag(;s)\\0ds+\at)-ato)\
/ Jt
\t-s\\\Ag(;s)\\0ds
s\\\Ag(-,s)\\0ds Jt0
which implies v £ Co([0, oo);D(A)). v is also second order continuously differentiable with respect to t, because
= -g(-,t)-Aj
J
V
ll
^d(£(A)As)(-,s)
Therefore v E V.
1.11
Maxwell equations
We work with complex-valued functions in this section.
•
Exterior Problems of Partial Differential
Equations
57
The Maxwell equations in vacuum are: div£=^,
(1.101)
an
curl£ = - — ,
(1.102)
div£ = 0,
(1.103)
cuilB = p (e^
+j \,
(1.104)
where E = (E1,E2,E3)T is the electric field, B = (B1,B2,B3)T is the magnetic flux density, e is the dielectric constant, (i is the permeability, p is the electric charge density, and j is the current density. The relation between p and j is | ? + d i v j = 0.
(1.105)
Let H = (H\,H2,H3)T be the magnetic field, then B = p,H. For given p and j there are six unknowns, E and B, in the system (1.101)-(1.104). It seems the system is over determined with eight equations. However (1.101) and (1.103) are not independent. If (1.103) is satisfied at time t = 0, then applying the operator div to the both sides of (1.102), we find that (1.103) is satisfied at all time t. Being the same, applying div to the both sides of (1.104) and noting (1.105) we can get d_
di
(divE
-?\=0.
Therefore if (1.101) is satisfied at the initial time, then it is satisfied at all times. Let us consider the total reflection boundary condition, E x j/|an = 0, -gjB • V\QQ = 0, and the initial conditions E\t=o = Eo, H\t=o = Ho. To investigate the initial boundary value problem, we introduce some Hilbert spaces first. Define ff (curl; fi) = {v G (L2{n)f; curl v £ (L 2 (fi)) 3 }, equipped with norm \\v\\H(curW) =
{\\v\\20+\\cm\v\\l}1'2.
58
Numerical Methods for Exterior
Problems
Lemma 31. The mapping v —> v x v\en, defined on (C°°(Sl))3 can be extended by continuity to a linear continuous mapping from i7(curl; SI) onto ( J ff- 1 /2(sn))3. Then we define #o(curl; Si) = {v € ff (curl; SI); v x i/\da = 0}. We choose H=
{L2{Sl))3x(L2(Sl))3
with the weight matrix
\0 pi J i
such that
(U,V)H
= {u,Mv) and ||u||# = (u,u)fj.
«-(!)• —
We define
'(;T).
with domain D(A) = H0(cuil; fi) x /f(curl; fi). The initial boundary value problem can be written in the form of - ^ + iAu = / ,
u(0) = uo-
(1.106)
Lemma 32. A : -D(^4) —* H is a self-adjoint operator. Proof,
clearly A is closed and symmetric. By definition D(A*) = {v£ H;3g € H,(Au,v)H
= (u,fl)j/,Vu e D(A)}.
We denote v = (vi,V2) T and g = (
V£ £
ff0(curl;fi).
Therefore g\ = ificurl vi and the domain for v2 is H(cui\;Sl). letting E = 0, we have V - X ( V x H,Vl)
= (H,g2),
Similarly
Vtf £ tf (curl; ft).
Therefore g% — —iecurl v\ and the domain for v\ is i^o(curl; SI). Hence D{A*) = ff0(curl; SI) x ff (curl; SI) = D(A).
Exterior Problems of Partial Differential
59
Equations
We assume that / £ C([0, oo); H). The formulation of a weak solution to (1.106) reads: Find u £ C([0, oo); H) such that
/
/ \U \ ~df~
+ iA
^)
dxdt
~ ^M
~ I Mx)v>(x,0)dx
= 0, (1.107)
for all ip £ C o ([0, oo); D(A)) n C 1 ^ , oo); ff). The weak formulation (1.107) is equivalent to the classical formulation (1.101)-(1.105) with the total reflection boundary condition and the initial condition. Just like the previous argument claiming that (1.101) and (1.103) are not independent, we take
/ 1U V T T ) ~ ^ f
/
dxdt
~
/ u0(x)<j>(x,0) dx = 0.
Since
\\u(t)\\H < \\U0\\H + / Wf(T)\\HdT Jo
(1.108)
holds. Proof.
Formally we get u(t) = e-iAtu0
e-iA^-T^f(T)dT.
+ [ Jo
In the notations of spectral resolution the solution can be expressed as pt
OO
/
e-iXtdE{X)u0+ -oo
/
/>00
/
Jo J —oo
e-
iA(
*- T > dE{\) / ( r ) dr.
£
Numerical Methods for Exterior
60
Problems
Let us verify that it is the desired solution. We have oo
d{E(X)uo,
-oo -oo />oo
= -J
e-iXtd{E{X)uQ,V{t))Hdt
jtJ
/»oo
/*oo
= /
iXe-iXtd(E(X)u0,f(t))Hdt
/
-^oof
r°° /"
/•OO
JO
fj-iXtd(E(X)u0,^-)Hdt
J- OO /•OO OO
ie-iXtd(E(X)u0,A
pC pOO
,-i\tMi?f\\..
Jo
<*¥>(*) d{E{\)uQ,^j^) Hdt.
J-t
Similarly we have /•OO
(/(r), ^(r))„ = / JT
/*00
/
ie-W-^
d(E(X)f(r),
A
J — OO
' FII
e iX{t T) WA)/(T)
~
~
' ^r)H c
Then (1.107) follows by noting f£° dr J^° dt = f£° dtfQ dr. It is easy to see that u is continuous with respect to t. Then by (1.107) ^ exists in the sense of distributions and (1.106) holds, therefore u is continuously differentiable. Taking the inner product of (1.106) with u we get
l^L+i(Au,u)H
= (f,u)H.
For the real part we have ld\\ufH 2 dt
= Re(/,«) ff
<||/||HHH.
Then (1.108) follows, which implies uniqueness.
•
Up to now we have investigated three kinds of evolution problems. Applying some similar argument, the results for some other problems, for example the evolution elasticity problem, can also be obtained.
Exterior Problems of Partial Differential
1.12
61
Equations
Darwin model
The Darwin model is a simplified model [Hewett, D.W. and Nielson, C. (1978)] [Li, T.-t. and Qin, T. (1997)] for the Maxwell equations (1.101)(1.105). It was shown in [Degond, P. and Raviart, P.A. (1992)] that the Darwin model approximates the Maxwell equations up to the second order for B and to the third order for E, provided 77 = v/c is small, where v is a characteristic velocity, and c is the light velocity. Let us introduce the Darwin model for bounded domains first. For a domain Qo let I \ , 0 < i < m, be the connected components of the boundary dCto, and TQ be the outer boundary. Note that the electric field E can be written as the sum of a transverse component ET and a longitudinal component EL, such that E = ET + EL, and V • ET = 0,
V x EL = 0.
The equation (1.104) becomes / 3EL
dET
\
In order to get the Darwin model, we neglect the term for ET and get cuxW = n ( e ^
+j \.
(1.109)
Then the system (1.101)-(1.103),(1.109) with the total reflection boundary condition and the initial condition will be reduced to (a) EL = — V, and <j> satisfies
(/>|r; = oii,
0 < i < m.
a — {ai} satisfies „
d a
1 f • „
a(0) = a 0 ,
,
62
Numerical Methods for Exterior
Problems
where OJO depends on Eo, C = {cy} is the capacitance matrix. Here Cyi r ~~dv ^ s ' anc ^ X = {Xi} i s t n e solution of Axi = 0, Xi|r\,
=Sij.
(b) 5 satisfies - A B = fiW x j ,
(1.110)
V - 5 = 0,
(1-111)
B-v\dn0=B0-v\dn0,
(1.112)
(V x B) x ^|an 0 = /A7 x Han 0 ,
(1.113)
AET = T T V x 5 ,
(1-114)
V - £ r = 0,
(1.115)
£ r x Han 0 = 0,
(1.116)
where B 0 = HHQ. (c) £ r satisfies
I ET-vds
= 0,
\
(1-117)
Now we turn to the exterior domain $7 and assume that dim(fi) = 2. We investigate problem (c) as an example, the method to problem (b) is similar. In order to figure out the suitable functional setting for the problem, we introducing some spaces. The space #(curl,div; Q 0 ) = {u£ (L 2 (fi 0 )) 2 ; V • u, V x u e L 2 (fto)}, provided with the norm ||u||o,curl,div = ( H o + IIV • U||g + || V X «||g)* ,
Exterior Problems of Partial Differential
63
Equations
is a Hilbert space. We take a cut-off function £ £ C°°(f2), which satisfies: C = 1 near the boundary dfl, £ = 0 near the infinity and 0 < C < 1. Let V be the space of distributions. Define tfoc(ft) = { « £ V; ||Cu||o,curi,div < oo, (1 - 0 « e (#o'*( f i )) 2 >«
x
Han = 0},
and ||u||, = {||V x u||§ + ||V • u||g+ < u • i/, 1 > I n } ^ However || • ||* does not provide a norm in Hoc(Q), and actually we have the following result: Lemma 33. Let V0 = {u £ Hoc(ty', ||u||* = 0}, then Vo ={u = ( U I , U 2 ) ; U I = -~—,u 2 = - - — ,
ox 2
oxi
4> = a(x2 + f(x)) + b{Xl + g(x)), (a, b) £ M2}, where f,g £ H1'*^)
satisfy - A V' = 0, dlj) ~dv~
dQ
iell,
dx dv
x
where x{^\-,^'i) — %i, \ separately. Proof.
We take any u £ Vo, then it will satisfy V - u = 0,
(1.118)
V x u = 0,
(1.119)
u x i/| fin = 0,
(1.120)
/ u-vds = 0. (1.121) Jan From (1.119)-(1.121), J r —v,2dx\ + U\ dxi = 0 is true for any closed curve r in Q.. Thus we can define
4>(x) = f (-«2 dxi
+ u\ dx2),
Jxn
where Xo can be any but fixed point in fi. Notice now ui = J^jf-, «2 = — J^~-
64
Numerical Methods for Exterior
Problems
From (1.119),(1.120), we can get
A
^ * ' ^ - f l = v * « = °-
and d
x Han = u x Han = 0. 0x2' dx\ da Thus starting from every u G Vo, we will end up with a problem: find a function <j>, satisfying ~dv
A
xeQ (1.122)
=0.
du
an Besides, u is a bounded harmonic function, which can be developed in a neighborhood of the infinity as follows:
E
ZOk
fc=0
2
w h e r e zofc = Cfc + ibk, (dk, bk) G M , z = x + iy = re%e. T h a t is / ^ > a,k cos k6 + bk sin fc# ^ ^
(«1,«2)= n
^
afc
sin k6 — bk cos k6
,2^
\fc=0
^
fc=0
/
From (1.118),(1.121) we know ai = 0, so m —iu2=
[a0 + \
bi sin k9 r '
b\ cos A;# \ r J
^ ^
z
ok zk
Hence the asymptotic expansion of (p(x) near the infinity will be cj){x) -» a0{x2-X2o)+bo(x1-xw)
+ b1(lnr~lnr0)
+ 0(-)
^
h(x)+c+0(-),
where h(x) = 0,0X2 + boxi + bi lnr. Introduce a new function ip, such that (/> = I/J + h. Then ip will satisfy A v> =0, chp_ dv
iefl dh
an Notice that it suffices to consider the solution in H1'*(Q.)/M. now.
(1.123)
Exterior Problems of Partial Differential
65
Equations
Knowing that the well-posedness of the problem (1.123) is equivalent to
/. so let flr = flfl B(0,r), /
then by the Green's formula we get
— ds = \
Jandv
9h , — ds = 0,
Ahdx — /
Jnr
—— ds = — 2itb\.
JdD(r) °v
Therefore b\ = 0 is sufficient and necessary in order that (1.123) admits a unique solution in # 1 -*(ft)/R. Once bi = 0, there exist f,g£ i7 1 '*(n)/M, such that i> = a0f(x)
- b0g{x),
where / , g are the functions to (1.123) with h(x) = x-i,x\ Consequently
respectively.
4>(x) = a0(x2 + f(x)) + b0(xi + g(x)). On the other hand it is easy to see that the function u given in the Lemma for all a, b are in Vo. • Let V = {v e Ho c (ft); < v • u, 1 > a n = 0}, and let the closure of the quotient space V/VQ with respect to the norm || • ||* be W. Besides, let Q = {p G L2(Q); supp p C C il}, equipped with norm IIPIIB = (||PIIO + I ^ P ^ I
:
Then we take closure to obtain a Hilbert space Q. lip £ Q and limpn = p, pn S Q, then we define f^pdx = lim fnpndx. Notice this can be seen as a generalized integral. The subspace of Q, {p £ Q; j„p dx — 0} is denoted byQoWe define bilinear forms on W x W and W x Qo, d(u, v) = / (V x u) • (V x v) dx+ / (V • u)(V • v) dx
Jn
Jn
+ < u • v, 1 > a n < v-v,l
b(u,p) — \ (V • u)pdx,
>an,
u € W,p e Q0.
u,v
€W,
66
Numerical Methods for Exterior
Problems
Now the variational formulation for the problem will be: Find u G W, p 6 Q o ) such that d( u,v)+b(v,p)=
I S - ( V xv)dx, Jn b(u,q)=Q,
VveW,
VgeQo,
(1.124) (1.125)
In order to prove that there is a unique solution to the problem (1.124),(1.125), we introduce one result on the inf-sup condition, which appears in Section 1.7, and will be used later on in some different problems. [Girault, V. and Raviart, P.A. (1988)] Let W, Q be two Hilbert spaces, and b(u,p) be a bounded linear form defined on W x Q. Define Z={w&
W;b(w,q) = 0,Wq € Q},
z° = {/ eW-,< f,w >=oyw e z}, Z± = {we W; {w,v)w
=
0,Vv€Z},
where (•, -)w is the inner product in W. Define two operators B : W —• Q' and B' : Q —» W, satisfying < Bw,q >= b(w,q),
\/w
£W,q£Q,
< B'q,w>=b(w,q),
Vw
£W,q£Q,
then ker B = Z. Theorem 39. The three following properties are equivalent: (a) there exists a constant f3 > 0, such that inf
sup
^ V > / 3 ;
(1-126)
(b) the operator B' is an isomorphism from Q onto Z° and \\B'q\\w> > P\\q\\Q,
V96Q;
(c) the operator B is an isomorphism from Z \\Bw\\Q. > P\\w\\w,
(1.127)
onto Q' and
Vw e W.
(1.128)
Exterior Problems of Partial Differential
Proof, to
Equations
67
(i) Let us show that (a)<=>(b). The statement (a) is equivalent
sup
< B'q,w > ^ ... .. ——i: > p\\q\\Q,
TIT
P
_ q£Q,
Iff
that is, (1.126) is equivalent to (1.127). It remains to prove that B' is an isomorphism. (1.127) implies that B' is a one-to-one operator from Q onto its range R(B'). Moreover, it also implies that the inverse of B' is continuous. Hence B' is an isomorphism from Q onto R(B'). Therefore R(B') is a closed subspace of W, since B' is an isomorphism. We can apply the closed range theorem of Banach spaces (see [Yosida, K. (1974)]) which says that R(B') = (ker(B)) 0 = Z°.
(ii) (b)«=Kc). We observe that Z° can be identified with (Z-1)'. Indeed, for w € W let w1 denote the orthogonal projection of w on Z1-. Then with each / G {Z1-)1 we associate the element / of W defined by < f,w >=< f,w1
>,
Vw e W.
Obviously / e Z° and it is easy to check that the correspondence / —> / maps isometrically (Z1)' onto Z°. This permits to identity (Z^)' and Z°. As a consequence, statements (b) and (c) are equivalent. • Besides, we need an auxiliary lemma. Lemma 34. For a given q £ Qo there is a (yq + Vo) £ W such that V • vg = q,
V x vq = 0,
vq x
V\9Q
= 0,
consequently
\\vq\\l = h\\l = h\\l Proof. For a given e > 0 we have qe £ Q so that \\q — qe\\$ < e. Then consider the following problem: Find
By Lax-Milgram theorem there is a unique solution, and
HIMTake a cut-off function ( G C^°(M2) such that C = l f o r | x | < l , C = 0 for |x| > 2, and 0 < £ < 1. Define Ca(^) = C(x/a) a n d ^a = <^Ca- Let qea =
68
Numerical Methods for Exterior
Problems
Acf>a, then q£a £ Q. When a is large enough, we have C a A0 = Acj> = qe therefore for sufficiently large a it holds that \\Qea - qe\\t =\\&a ~ AJWf = \\ACa
2
|| ACa4> + 2VCa • W | | g + | / ( A C a 0 + 2VCa • V
Jn (1.129) Let us estimate the right hand side of (1.129). Since \x\2 In2 \x\ < |a;|3 < 8a 3 in the domain D = {x; a < \x\ < 2a}, we have
||AC*||§ <4l|AC||^ )00 / \4>(x)\2dx
JD
JD \X\ \D.
\x\
QS~t
< —l|AC||3, O ollW||3, n ->0, a
(a-»oo).
Analogously we have l|2VCa-V^||g<^||VC||§ i O O ||V0||g i n -»O. For the second term of (1.129) we notice that / A^dx Jn
= - / VCaV^cb, Jn
therefore / ( A C o ^ + 2VCa-V0)da:| Jn in <J
|VCa • V<£| dx < |||VC||o,n||V^|| 0 ,D < C\\V
To conclude there is a ao such that ||g ea — <7e||j < £ for a > ao. We fix such an a£. Let vq£ = V
|V • vq£ - a||o + \
:
>9n - / qdx\ Jn
- q\\i < \We - q\% + ||fca. - 9 e | | | <
2e
-
Exterior Problems of Partial Differential
Equations
69
Therefore vqE converges as £ —> 0. Let vq be the limit, then vq + Vo is the element in W we are looking for. • Now we are ready to prove the well-posedness. Theorem 40. The problem (1.124), (1.125) admits a unique solution and p = 0. Proof.
(u,p),
Because
d(u, u) = / |V x u\2 dx+ / |V • u\2 dx = \\u\\l, Jo. Jo. the bilinear form d(-, •) is coercive. By Lemma 34 SUP v€W
> "^
INI*
[j— -
-TT—rp- -
IKII*
g
•
(1.13U)
Ikllit
Therefore the inf-sup condition with /3 = 1 holds. We define W0 = {u G W;6(p, M) = 0,Vp G QO}, then being analogous to Theorem 26, the problem: find u G WQ, such that d(u, w) = f B • (V x v) dx,
VuGWo,
(1.131)
admits a unique solution. To prove the existence and uniqueness of p, we note that the equation (1.124) defines a linear operator F(v) = -d(u, v)+
/ B • (V x v) dx Ja.
on W, and by the equation (1.131) it is in the space Z°. Then by Theorem 39, p is determined by the inverse of B' uniquely. Finally let us show p = 0. According to Lemma 34 there is vp corresponding to the solution p. We take v = vp in (1.124), then we have d(u,vp) — / (V • u)(V • vp)dx — / (V -u)pdx = b(u,p) = 0, Ju Jo. I B • (V x vp) dx = 0, Jn and b(vP,p) = (P,p)-
70
Numerical Methods for Exterior
Problems
Equation (1.124) implies (p,p) = 0, which gives p = 0.
D
Remark 1. The solutions are not unique. They may differ from a function in Vo- It seems the most natural way to make the solution unique is to impose a boundary condition at the infinity, u||,,.|=00 = 0, because a function in Vo satisfying this condition is zero. Unfortunately we still can not prove the existence if this boundary condition is imposed.
Chapter 2
Boundary Element Method
The boundary element method is based on a reduction of a boundary value problem on a domain to an equivalent problem defined on the boundary. The dimension of the problem is thus deducted by one, and it does not care very much if the underlying domain is interior or exterior, so it is suitable to deal with exterior problems. The potential theory is classical. We will derive the boundary equations by using the potential theory. On the other hand we will introduce the approach by Feng and his students, using the theory of singular integrations. The advantage of this approach is that the energy norm of it is exactly the same as that of the original boundary value problem. For inhomogeneous equations a Newton potential is usually applied to reduce it to homogeneous ones, and a numerical scheme of integration can be applied to the Newton potential. Therefore we consider homogeneous equations only in this chapter.
2.1
Some typical domains
For some typical domains the solutions to exterior problems can be expressed explicitly. We work with complex-valued functions in this section. 2.1.1
Harmonic
equation
Let us begin with the exterior domain fi of the unit disk B(0,1) Dirichlet boundary value problem of the harmonic equation. In polar coordinates the problem is
iiL (
du
l d2u
\
-o
r2 d82
r dr \ dr J 71
and the
72
Numerical
Methods for Exterior
Problems
« | r = l = 9(0)-
The solution can be expressed in terms of Fourier series, oo
Pn(r)ein9-
u = Y, n = —oo
Plug it into the equation and obtain
y
n2pn\
fid fdPn\
^
1 r dr \ dr J v
n= — oo
N
ine_Q
2
r J
'
'
Therefore each term vanishes: \d_( d£n\ _ n2pn _ r dr \ dr J r2 The general solutions to these ordinary differential equations are: p n = anr~n + bnrn, Vn ^ 0.
Po — a0 + &o lnr,
Since u £ H1'*(Q), we get the solution oo
u=
anr-^ein9.
Y, n= — oo
By the boundary condition we can determine an:
^j\-in*g{
an = Therefore the solution is
u
r \ S r-Me*0-Ag(
=t J
®
l,n= — oo
(2.1
J
The power series in (2.1) are equal to: oo
1
E
X
n
in(e-
Finally we get n=0
- 1 V r-|nLi"(0-¥>) = _
~~ l-.r-lei(6-V>)'
Z ^ n = —oo
(-27T
U =
1 /""r 2^y0 r2
+
r^ - 1 l-2rcos(0-
r-lp-i(9-ip)
!
l_r-le-t(fl-v)'
(2.2)
Boundary
Element
73
Method
This is the Poisson formula for harmonic equations. If we replace the boundary condition by u\r=R = g(0), and solve the problem exterior to B(0,R), let v(^,0) = u(r,0). Replacing r by -^ in the Poisson formula we get for the general case, 27r
2Wo
2 2 ~ r - R " 2
r2 +
g(
R -2rRcos{6-
(2.3)
Under a conformal mapping a harmonic function is transformed to a new harmonic function. Applying this property one can get some more solutions for other exterior domains. For example, if we define w = az H—, a > 6 > 0, z where z = x\ + ix^ = re10, w = £1 + i& = pezlf', then the exterior of the unite circle r = 1 is transformed to fii, the exterior of an ellipse, dQ.\ = {(£1162); £1 = (a + b)cos#,£ 2 = (a — b)sm6}. The function v(p,ip) =u{r,6) can be expressed in terms of the Poisson formula, 1 r2ir r2 - 1 (P,
Av = o,
b pezv = are16 + — ^ , re
(6,6)efii,
vlecit =g(0)There is another approach to derive the Poisson formula (2.2), which is the use of the potential theory. Let us investigate the integral equation (1.11). Since 917 is the unit circle, we have d
1
duy
_ c o s ( ( x - y),vy)
\x-y\
_
\x-y\
so the equation is
1 If a x o ( ) ~ TZ \ 2
47T Jdn
a
(v) dsy = g(x).
1 2'
74
Numerical Methods for Exterior
Problems
We replace the boundary value g(x) by g(x) = g(x) — •— j d n g(y) dsy, then the integral equation 1
1 /*
admits a solution cr(x) = 2g(x). The double layer potential V{x) =
^La(y)
\x-y\
dS
«
satisfies (1.1),(1.4), and V\QQ = g. Therefore If 27r
cos({x-y),vy)
1 A 27r
J do.
F ~ 2/1 ./an 2\x - y\ cos((a: - y), vy) + \x- y\2 |2 27r7aan F-J/l' ;
9{y)asy
Noting that \x — y\2 = r2 + 1 — 2r cos(# — y>) and 2|a; — y\ cos((a; — y), vy) -f \x — y\2 = r2 — 1, we derive the Poisson formula (2.2) again. The above approach can be applied to derive the three dimensional Poisson formula of the Dirichlet problem. Because the boundary dQ, is the unite sphere S3 = dB(0,1), the integral equation (1.11) is
r{x)~iL3a{y)\^v\dSy=9{x)We replace the boundary value g(x) by If ~g{x) = g{x) - — J^
1 g(y)j——-^dsy,
then <J{X) = 2g(x) is the solution. We notice that the difference between g and g is nothing but a single layer potential, so the solution u is I f , . cos((x — yy),vv) u(x)=—
=J- / 4TT JS3
,
1 f , . +~ g(y)4TT Jan
zi» - yl«»((» - y). ^») +1» - yl2
( }
1 -dsy \x-y\
d,
\x - y\A
Hence we get the Poisson formula l [ r2 - 1 U X ( ) = JZ 77--; 1 7l9{y)dsy, 47r ^Sa (r 2 + 1 - 2r cos 95) 2
Boundary
Element
75
Method
where r = \x\ and ip is the angle between the vectors x and y & S3. If we replace the boundary condition by u\r=R = g(x), and solve the problem exterior to B(0, R), let v(^, 9) = u(r,6). Replacing r by ^ in the Poisson formula we get for the general case, 1 /• r2 - R2 «(*) = 7 z / ! 2 ^ m 7i9(y)Rdsv. 0 p 47T Js3 (r 2 + R2 - 2rR cos 9?) 2 2.1.2
Bi-harmonic
(2.4)
equation
Let us solve the bi-harmonic equation next. The problem is: 1 d ( 8_\ r dr \ dr)
u\r=i = gi(0),
1 d2 \ r2 d82 J
—
2
= 92(0). r=l
The solution is expressed in terms of Fourier series, 00
u=
Y,
Pn(r)ein9.
Then we get 1 d ( d\ r dr \ dr I
n2s2 r2
Therefore we have I d f d , --r [r-r- ) po = ao + bo In r, r dr \ dr '
Solving these equations again and noting that u € H2'*(Q), we get Po = ao + M l + 21nr),
pn = anr~M
+ bnr-^+2,
Vn ^ 0.
76
Numerical Methods for Exterior
Problems
Taking the boundary conditions into account we solve the coefficients as follows: $1 (0) = J > „ + bn)ein9 + (ao + b0), njtO
- (\n\ ~ 2)bn)einB + 260.
92(6) = ^2(-\n\an Then i
r* e-m<egi{
an + bn = —l
-\n\an
- (\n\ - 2)bn = i - f
e'^g2{V)
dtp.
Then the solution is <.27r
-J_ /i u=—
in
^-f)
e
• {(51 - \92 + \92T2)r-^ 1
+
2n
f2n
- M S 1 ( 1 - r*)r-W} dtp
(9i+92^r)d(p.
After some calculation we finally get r2T
r* + 1 -2r Je(0 -
=i l 1 +
/-2,r 2r2 - r(r 2 + 1) cos(0 -
2^y0 1
+
Ji
27r/
/"27r
a
(r2 + l - 2 , C o s ( / - y ) ) 2 ( 1 - r ) g l ( y ) ^ 1
^lnr+ 2 ~
1
2r^92^d
^
Boundary Element
77
Method
If we replace the boundary condition by 011
= 92(0). r=R
and solve the problem exterior to B(0, R), let v(^,6) the boundary condition, v\r=i = gi{6),
dv dr
= u(r, 6). v satisfies
= RtoW). r=l
Then for this problem the solution is ,2TT
_j_ r ~2Wo +
r
2 _
^2
r» + R* - 2rRcoS(6 I ^ j <*<*> " f *<*> + h .
1 f2n 2R2r2 - Rr(r2 + R2) cos(0 - cp) M 2^J0 (r 2 + R? - 2rRcos(9 - &)* ( 1 " 2n 1 f r 1 1 r2
r2 ,
9
,
* ^ *P
., *"
&
)9l(ip)
(2.5) 2.1.3
Stokes
equation
For the exterior problem of the Stokes equation, - A u + Vp = 0,
(2.6) (2.7)
u\r=R = g{6),
(2.8)
we introduce the stream function 1p(x) = <j> {-U2
78
Numerical Methods for Exterior
Problems
Applying the operator x T
d
d
dx2'
dx\
to the equation (2.6) we get A2ip = 0, so the stream function ip is the solution to the bi-harmonic equation. We assume
I
2TT
g • v d6 = 0
first, then in the polar coordinate ip(R,2n) = tp(R,0). By this assumption ip is single valued. Let (ur, ug) be the components of u in polar coordinates, that is ur\ ug J
( cos6 s i n 8 \ (u\ \ — sin 9 cos 9 J \ u^
Under the polar coordinates it holds that /(r,e) ip(r,9) = (b (-uedr + urrd9), •Wo) The boundary condition (2.8) becomes Ur\r=R = 9r{9),
1 dip ur = - — , r 06
dip ue = - — . Or
U6\r=R = ge(0).
The boundary condition for ip is lp\r=R =R r=R =
9r(v)dr),
=
-fr
-9e(9).
r=R
Applying the expression for the solutions to the bi-harmonic equation we obtain r2 - R2 ^=W 0 ri + R>-2rRcos(9-v){RJ0 R 1 \ d
+
f2"
(
[* *<*>*>
J_ r2" 2R2r2 - Rr{r2 + R2) cos(9 - tp) r2 2 2 2 ( 2n J0 (r + R - 2rRcos(9 - tp)) R?' 2 2 rv> 1 f* r 1 1 r
(RJ
gr(r))dr})dtp - — J
(\n- +
-~-^)Rg9{ip)d
Boundary
Element
79
Method
Then after some calculation we obtain the solution u: 1
f2"
_ J_ f2n
+ 3i? 2 )sin(6>-y)
[*
(r 2 - R2)2 sin(0 - ip) 2Tg9(tp)dip (r * {I- + -f R n~2-2rRcos(6 — zrrtcos^c- - y>)) sin(6> - p) (r2 + R2-2rRcos(6-
2Wo 27r
+
R2(-r2
2^7o
(2.9)
r2 fv (1-^2)/ gr(v)dvd
^r2w 2(r2-iZ2)(r-J?cos(fl-y)),n ^
,
2^J0 (ri + R2-2rRcoS(0-v))2iRJo R 1 + -^geiv) - ^9e(
1 f(-27T
+ 2^J0
ri
2r R -2rRcoS(6-lp){RJ0 2
+
^
fv ^
^
) d n
R 1 + -^9e{9) - ^99(
_2
1 r 2TT 7
_
^2
. - 2ri? cos((9 -
r r2 + # 2 ip) H^ _ J _ f ^ ( 2 i ? V - Rr(r2 + R2) cos(6 - ip))(r - Rcos(6 - ip)) 2n J0 (r 2 + R? - 2rRcos(6 - ip)f 2 r [* • ( 1 _ ^ 2 " ) - R / 9r(v)dr]dip 0 2
+
1 /-^ <±R3r - 4Rr3 + (5r 4 - R4) cos(6> - ip) fv 2^/o (r-2+^-2ri?cos(0-„))2 /0 ^
1 f2n R
,
x
, , < ^
r
-2^1 ( 7 - ^ M ^ . (2.10) If /•27T
1 /"^ — y gr(6)d0 = gjLO,
80
Numerical Methods for Exterior
Problems
we define a particular solution
to the equations (2.6),(2.7). Then /•27T
/ Jo
(u- u0)\r=R • vdQ = 0.
u — UQ can be expressed in term of the above expressions. 2.1.4
Plane
elasticity
For the plane elasticity problem, —fi A u — (A + /i)grad divu = 0, UllN = l =flli«2||a:| = l =
iefi.
92,
we introduce the complex expression for the displacement u = (u\,U2)T first. Let f(z) be an analytic function, / = p + iq, z — x + iy. Let L — —jU A — (A + ^i)grad div. By the Cauchy-Riemann equation, we have
""(A)-0-
div
C)-0'
and Ap = Aq = 0, therefore
For the function zf(z) = (xp + yq) + i{yp - xq), we have
div^
+ y
\yp-xqj
g r addivf^
+
^)=2f|V
\yp-xq)
and
-(rs)-(i-i)-
y-^J
Boundary
Element
Method
81
Therefore
£
+
£-S)~°* *"(i)
Let af(z) - zf'(z) = U + iV, then U L( )
= -2a(\ + n) (^]+2(X
We set a = ^ f ,
then L (^
+ 2^ ( j f M .
) = 0.
Let us return to the Lame equation(1.35). If Ui+iu2 = aip(z) — z(p'(z) — tp{z), where
00
i>{z) = J2
n=0
bnr-ne-in9.
n=0
Thus we get the expression, 00
00
«! + iu2 = a J2 anr-ne-inf> n=0
+^
00
na^r-neiin+2'>9
n=0
- J2
Kr~nein9.
n—0
Applying the boundary conditions we get 00
00
9 = 9x + i92 = a Y^ "ne-™9 + £ n=0
00
na^e^+V9
- £
n=0
KeinB'.
n=0
The coefficients can be evaluated with 60 = 0, 1
1 /"27r e in ~2^J ~ ^(4>)d
—
bn =
1
bn =
~2^J
C2w
/- 2 7 r
77 — 2
e
~
in
*
9(<W#+
~2^~/
n = l,2,
/" 27r
e-^-WgWty,
n = 3, • • • .
82
Numerical Methods for Exterior
Problems
Finally we get /"27r f °°
1
•'°
°°
ln=0
n=0
+ J2r-nein{e-*)g(
—n
- J2 n=3
e^e^-W-Vgit)
a
I
+
\l_r-le-i(0-¥>)
a (1 _ r - i e ^ - v ) ) 2 27r
2TT JO 2W0
- —
' r2 - 1 I r 2 + 1 - 2r cos(0 - ip) ff(¥>)
r ( r _ e i ( e -, ) ) 2 3(
<**•
If we replace the boundary condition by Ul\\x\=R
= 9l,U2\\x\
= R = 32,
then 1 Ul+lU2=
2^J0
/2w ( {ri +
r2-R2 Ri-2rRCos(e-V)gM
e2i9 ei(9-V)R(r2
a 2.1.5
Helmholtz
_ #2)
ReW-v))2
r(r -
(2-11)
-g(4>) } dc/>.
equation
In order to solve the exterior problem of the Helmholtz equation let us introduce the definition and some basic properties of the Bessel functions. The Bessel equation is dx2
x dx
x2 J
\
One solution of it is the Bessel function of the first kind of order s, which can be developed into a series, ^
,
1
/a:\ 2 f c + s
'•M-B-U'rjrnE+yk) fc=0
'
Boundary
Element
83
Method
where T is the Gamma function. The asymptotic behavior of it as x —» oo is S
Js(x) =
]ffxCos(x-
^-^+0(x^).
Another linear independent solution of the Bessel equation is usually denoted by Ys(x), the Bessel function of the second kind of order s, which possesses a singular point x = 0. We omit its expression, but give the asymptotic expression of it here:
y.W.yX*,(x-f-J) + 0(x-n Moreover the Bessel functions of the third kind of order s are called the Hankel functions of the first kind of order s and the second kind of order s. The definitions of them are: Hi1\x) = Js(x)+iYs(x), H?\x) = Js(x)-iYs{x). Therefore the asymptotic behavior of them is the following: HM(x) = A /A e <*-¥-i) + 0(x-V2), V ixx
(2.12)
H™(x) = . / l e - ' ( - ¥ - ? ) + 0(x-*/2).
(2.13)
and
V TTX
Multiplying them by a factor e~luJt, we find Hs (x)e~"lu't represents an outgoing plane wave and Hs (x)e~ZUJt represents an incoming plane wave. For the Dirichlet boundary value problem of the homogeneous Helmholtz equation (A + w2)u = 0, u = g,
zefi,
xedfl,
(2.14) (2.15)
84
Numerical Methods for Exterior
Problems
we consider the case of two dimension first and let <9f2 = {x; r = R}. The solution can be expressed in terms of Fourier series, oo
u(r,0)=
J2
anH^(ujr)ein9.
(2.16)
n= — oo
Taking into account of the boundary condition we get oo
g(6)=
anHil\uR)ein9,
£
(2.17)
n = —oo
therefore
2Wo
g(6)e-in9de/Hi1\u;R).
Lemma 35. If g £ #2(dft), the series (2.16) converges and solves the problem (2.14),(2.15). Proof. By (2.17) the boundary condition (2.15) is satisfied, provided the series converges. Making use of separation of variables it is easy to see that each term of (2.16) is a solution to the equation (2.14). It represents an outgoing sphere wave, so the Sommerfield radiation condition is satisfied. It remains to prove convergence. We recall the estimate H I v < Cll/llo,,,
(2.18)
where / is the right hand side of the equation (1.25) and the support of / is bounded. For the problem (2.14),(2.15) we apply the inverse trace theorem of Sobolev spaces to take a function UQ in H2(Q) with a compact support, so that t*o|an = g, and ||ito||2 < C\\g\\ 3. Let u\ = u — u0, then it is the solution to (A + w 2 ) Wl = - ( A + w 2 )u 0 . By the estimate (2.18) we have \Hv<
||«o||v + | | « i | | v < C | | s | | | .
Therefore the convergence of the series (2.17) in H^(dil) vergence of (2.16) in V.
implies the con•
Boundary Element
85
Method
The treatment for three dimensional case is analogous. The solution can be represented as a series of spherical functions: oo
n
u(r,9) = ^2
E
^nm^ll2(u>r)Ynm(e,
(2.19)
n=0 m=—n
where Cgi(z) = y ^ < + i W .
P m)
n (x)
2.2
PlT\cos6)eim*,
Ynm{9,tp) =
(1 — x2)^ dn+m = dx^+m^2 2"n
~
1) -
"
General domains
Let us consider the harmonic equation first. The integral equations (1.11) and (1.13) can be regarded as a boundary reduction, so the corresponding boundary element method is based on the approximation to these integral equations. However for Dirichlet boundary value problems we have shown that the integral equations are not in fact equivalent to the original problems, which causes some inconvenience. Another choice is representing the solutions of Dirichlet problems in terms of single layer potentials. We start from the two dimensional case and consider the problem (1.1),(1.2),(1.4). Let the solution u be expressed as
<x) = rrif
"(y)ln T-^—\ dsy + c -
( 2 - 2 °)
to Jan F - V\ then the boundary condition (1.2) leads to a Predholm integral equation of the first type: g{x) = -!- / w(y) In — ! — dsy + C, x€ dQ, to Jan F - V\ where the density w and the constant C are to be determined.
(2.21)
Theorem 41. If g e C2(dQ) then the equation (2.21) admits a unique solution (w,C) satisfying f u(y)dsy Jan
= 0.
(2.22)
86
Numerical Methods for Exterior Problems
Proof. Owing to the theory of elliptic equations the solution u to the problem (1.1),(1.2) satisfies u £ C 1 (fi) under the hypothesis of this theorem. Therefore f^ e C(dfi). By Theorem 7 u can be expressed in terms of a double layer potential plus a constant. It can be derived from (1.7) that Vu = 0(|:r|~ 2 ) for large |a;|, then we notice J n Audx = 0 and make use of the Green's formula to obtain fm | ^ ds = 0. Let us regard | ^ as a boundary value, then by Theorem 8 u can be expressed in terms of a single layer potential plus one constant and the density satisfies f9n uj(y) dsy = 0, which proves the existence. To prove uniqueness, we assume that g = 0 in (2.21), then define u by (2.20). By (2.21) u is a constant on the boundary dQ. By uniqueness of the Dirichlet problem u is identity to a constant on the entire plane. Then by Lemma 3w = 0. Then the equation (2.21) implies C = 0. • To consider more general g's, we need to study weak solutions and the variational formulation. We will see that the variational formulation is convenient to implementation by using the finite element method. Letting X 6 C(d£l) and fm x{x) dsx = 0, we multiply the equation (2.21) by x a n d integrate the equation on dQ to obtain b(u>,x) = — 2?r
/
u;{y)x{x)\n-
JOQ J9Q
\X
-dsydsx= - y\
g{x)x{x)dsx. Jdn
(2.23) To study the equation (2.23), we take v 6 Co°(R 2 ), and consider the bilinear form a(u,v) = / Wu-Wvdx= Jv
h»
V f — / ui(y)\n-. \27r Jan \x-y\
rdsv
+ C ) • Vvdx. J
By exchanging the order of integrations we get a(u,v) = / ijj(y)dsy \ V x — h i : 27r Jan hi \x-y\
:-Vxvdx.
Let / = — Av, then by the property of Newton potential (see Section 1.2) f 1 1 v(y) —In: -f(x)dx= Ky = / > JR22ir \x-y\JK '
f 1 1 / V ^ — In:S7v(x)dx. Jw x2n \x - y\
Therefore a(u,v)=
f u(y)v(y)dsy. Jan
(2.24)
Boundary Element
87
Method
Now we set v G F 1 '*(E 2 ) and take a series {vn} c Co°(IR2) tending to v. Passing to the limit in (2.24) we find that (2.24) holds for all v G Hl'*(R2). Letting it be an equation for u, we will prove it is well-posed. By the trace theorem of Sobolev spaces, v G H1^2 on the boundary dft, so we may assume that w G H~1^2(dfl), the dual space of H1/2(dQ,). The equation (2.24) is thus written in the form of distributions: Find u G H1'*($?), such that a(u, v) =< u, v >an,
Vv G
ff1-*(R2).
(2.25)
We define W = {w G H~1/2{dVl)\ < w, 1 >dn= 0}, then we have Lemma 36. The problem (2.25) admits a solution for all w G W, and the difference of any two solutions is a constant. Proof. We define V = {u e Hl>*(M.2); JB(0 Ro)udx and consider the problem: Find u G V, such that a(u,v) =< UJ,V >an,
= 0} with R0 > 0,
V « e V.
Since |• 11 is a norm of V, it admits a unique solution u. For any v G H1'*(M.2) we can divide it into two parts: v = v\ + C, where v\ G V and C is a constant. Noting that < w, 1 >aa= 0, we find u is also a solution to the problem (2.25). If u is a solution to the homogeneous problem, then because a(u, u) = 0, u is a constant. • We have shown that a classical solution u can be expressed in terms of a single layer potential (2.20), and satisfies (2.25). Now we set w G W, then solve (2.25) to get u G H1'*(M2). The mapping from w to u is continuous. Therefore we can take the closure in (2.20) and define a generalized single layer potential u(x) = < 57I11 urzq,w >an, which defines a continuous operator from W to .ff 1 '*(R 2 ). Let us return to the problem (2.23). b(uj,x) is n o w ready to be defined OTIW xW. The problem is: For a given g G H1/2(dCl), find UJ €W, such that b{u, x) =< 9, X >dn,
VX G W.
(2.26)
Theorem 42. The problem (2.26) admits a unique solution. Proof. We consider a single layer potential v(x) =< ^ In T^ITT, X >9n, then b(uj,x) =< v,v >sn- By (2.25) we get b(w,x) = a(u,v). For a given v G H1'2(dD.), we consider the solutions to the interior and exterior
88
Numerical Methods for Exterior
Problems
problems with Dirichlet boundary value v, still denoted by v, then v £ / ^ ' " ( R 2 ) , and ||V||I,*,R2 < M||w|| 1 / 2 ,an with a constant M. Hence for all UJ £ W it holds that ||w||_l/2=
SUp
-r—r
vem/2 <M
<M
SUp
\\v\\i/2,esi <w,v>«o dU
sup
—rr-r
ugjfV! ,, =M
IMIi,*,n a(u,v)
sup
| |
, ,, , < M|M|I.
We get b(u>,u>) > -j^j\W\\^-i/2- Owing to the Lax-Milgram theorem uniqueness and existence follows. • Using the boundary element method to solve (1.1),(1.2),(1.4), we first solve (2.26) for a given g, then evaluate u(x) —< <^ln urrq,w >aa +C, where the constant C is determined by g(x) = < ^ In T^T,W >an + C , x £ dil. It is easy to verify that u is the desired solution. For three dimensional problems the situation is similar. The solution to (1.1),(1.2),(1.4) can be expressed in terms of a single layer potential,
<^ = -k I ^y)u^7\dsv 4?r Jan
( 2 - 27 )
\x - y\
The kernel u> satisfies the problem: Find UJ £ H~1^2(dQ), b(oj, x) =< 9, X >an,
VX £ H-V2(dn),
such that (2.28)
where &(w,x) = T " / / UJ(y)x(x)] -dsydsx. 47r Jen Jan \x ~ V\ Because u,x € H~1^2(dD,), the bilinear form &(•,•) should be understood in the sense of distributions. Being the same, (2.27) is also in the sense of distributions. For a given g £ H1^2(dCl) the problem (2.28) admits a unique solution u. It is easy to verify that u defined by (2.27) is the desired solution. For the Neumann problem (1.1),(1.3),(1.4) if we express the solution in terms of a double layer potential, then u x
()
= TT" / 2TT J9n
<J
(2 / ) ~a~
ln
1
1
dvv \x - y\ for two dimensional case. Then the kernel a satisfies Lai case. Then the kernel a satisfies V
'
^Jan
Ky
'dvxdvy
\x-y\
yi
ds
y
Boundary
Element
89
Method
The kernel is hyper-singular. We will discuss this kind of integral equations later on. Next let us consider the Dirichlet boundary value problem of the biharmonic equation, (1.53). The fundamental solution K(r) = -^r2lnr, r = | a; |, satisfies A 2 u = 5{x), and K depends on r only. The solution u to the problem (1.53) is expressed in terms of a potential form, u(x) =
w(y)K(x-y)dsy+ Jan
Jan
a(y)—K(x-y)dsy+p(x), °vv
p 6 ?i(n).
(2.29) Being analogous to (2.22) we require that the densities w, a satisfy the condition,
J ("(y)v(y) + °(y)^r)
ds
v = °.
Vv e p
i•
(2-3°)
We verify that under the condition (2.30) the function u determined by (2.29) is in the space H2'*{Q). Then we derive the variational formulations for u and (w,«r). Finally we verify that u is the solution to (1.53). L e m m a 37. If u> and a are continuous functions and satisfy (2.30), then u G H2'*{Sl). Proof.
Let XQ £ Q c , then by the condition (2.30) we have
/ (w(y)(K(x - x0)+VK(x Jan +VK(x
- xQ) • (x0 - y)) + a(y) — (K(x - x0) CVy - XQ) • (XQ - y))) dsy = 0.
Then we get u(x) = / u(y){K(x Jan + / Jan
av
y
-y)-
K(x - a;0) - VK(x - x0) • {x0 - y)} dsy o-{y)-—{K(x-y)-K(x-xo)-VK(x-x0)-(x0-y)}dsy+p(x).
Applying Taylor's expansion we find K(x -y)-
K(x - x0) - VK{x - x0) • (x0 - y) = 0(\D2K\
• \x0 - y\2),
90
Numerical Methods for Exterior
Problems
as i —> oo, with xo being a fixed point and y bounded. expression of K we easily find u2
Then by the
f
/ -A dx < oo. Jn |x| 4 ln 4 |a:/i?o| The other two terms in (1.52) can be verified in the same way. • 2 We substitute (2.29) into (1.47) with v G C^°(R ). Let us evaluate the first term as the following: f vA\ f w(y)K{x-y)dsy JVL2
KJdQ
+ /
er(y)~—K(x - y) dsv +p(x) > • Av(x) dx
JdCl
"Vy
J
= / u(y) dsy / 2 uAxK(x Jan Jw + /
er(2/)-— /
J do.
Wy
— y)Av(x)
i/AxK(x-y)
dx
•
Av(x)dxdsy
JW-
= / ijj{y) dsy I vK{x — y)A v(x) dx Jan JM2 + / cr(y)^— / 2 i>K(x— y) • Jan vvy Jm.
A2v(x)dxdsy.
By the property of the Newton potential JR2 K(x — y)A2v(x) the first term of (1.47) is
dx = v(y), so
i> oj{yHy)+aiy)
L{
^')dsy
The second term can be evaluated in the same way, and we get
a(U,V) =
Ian (^{y)v^
+a
^^~)
ds
v
( 2 - 31 )
Lemma 38. Let w £ F " 3 / 2 ( 5 f i ) , u G H^^idCl), and (2.30) be satisfied, 2 2 then the solution u G # -*(R ) to the equation (2.31) for all v G H2'*(R2) exists, and the difference of any two solutions belongs to P\. Proof. We define V = {u G H2'*{R2); JR2 updx = 0,Vp G P i } , then follow the same line as the proof of Lemma 36 to obtain the result. • We remark that some integrals are understood in the sense of distributions. Since there is no misunderstanding we will not distinguish them with normal integrals.
Boundary Element Method
91
From the boundary conditions of (1.53) and the expression (2.29) we get gi(x) = Jan
uj{y)K{x-y)dsy+
a{y)—K{x-y)dsy+p{x),
ov
Jan
y
x£dQ, and 92{x)=-—\l
u{y)K{x-y)dsy
+ /
cr
(y)-z—K(x-y)dsv+p(x)\>
Jan
xedn.
)
°vy
We define v(y) = / x(x)K(x-y)dsxJan
I Jan
T(X)-—K{x-y)dsx,
OXJ
x
with dq{x) I ( X(x)q{x) + T{X)dvx Jan \
dsx = 0,
VgePi,
then we consider the dual products <9i,X> + / Jan +
+ < 92,T>=
\ Jan vJan
w{y)K{x-y)dsy
a
°vy
I 4 r \ ( Jan avx i J an
(.y)-^—K(x-y)dsy+p{x)\x(x)dsx )
(2.32)
»(v)K{x-y)dsv
+ / a(y)^— K{x-y)dsy+p{x) Jan My
J
}T{x)dsx.
By the expression of v we have < 5 i , X > + <92,T
>=
+
.
(2.33)
92
Numerical Methods for Exterior
Problems
With regard to (2.32) we define a bilinear form on (H~3/2(dft) H-^2(dn)) x (H~3/2{dfl) x H-l'2{dQ)) as follows:
x
b((co,a),(X,r)) =
\ u(y)K(x-y)dsy+ Jen Udn +
Jan
-S—\ u(y)K(x-y)dsy+ Jan °vx Udn
a(y)-—K{x °vy Jan
- y) dsy\ x{x) dsx )
a(y)-—K(x 0v y
- y)dsy\
)
r(x)dsx
We define a space
w = {(u,a) e(H-3/2(dn) x H-^2(dn))<w,v>dn
dv +
Theorem 43. TTie problem: find (w,cr) G W, suc/i i/iat b((w,a),(x,T))=
+ <92,T>,
V(X,T)eW,
(2.34)
admits a unique solution. Proof. By (2.31) we see that b((u),a), (x> r )) = a ( u i y )> then following the same lines as the proof of Theorem 42 we get the result. D With the above argument the problem (1.53) is reduced to a boundary equation (2.34). Then we use (W,
Boundary
2.3
Element
93
Method
Subdivision of the domain
If the domain Q is in a complicated shape, the simplest way to deal with exterior problems is to truncate the domain. Let us consider the problem Au = 0,
u = g,
(2.35)
(2.36)
x G d£l,
in three dimension first. Let f2o be a domain with a simple shape, and Jlo 3 fic- We define Qi = Qo H fi. The domain ft is replaced by fii, and some artificial boundary condition is imposed on the artificial boundary d£l0.
•X
X
The most self evident attempt is to impose u = 0,
x G 3f2o.
(2.37)
Let Uh be the solution to (2.35),(2.36),(2.37) on £l\. It is indeed an approximation, because we have the following estimate.
94
Numerical Methods for Exterior
Problems
Lemma 39. We suppose that dim (fi) = 3. Let R = min{|a;|;a; e then \u-uh\
dflo},
=O
Proof. By the expression in Section 1.1 we get u = 0(1/R) on dttoThen applying the maximum principle for u — Uh the conclusion follows.• For two dimensional problems the situation is worse, because the solution u does not tend to zero at the infinity. However we still have the following: Lemma 40. We suppose that dim(Q) = 2. Let g £ Hl/2(dfl). Then the approximate solution Uh converges to u on any bounded domains weakly in H1 and strongly in L2, as R —> oo. Proof. Owing to the inverse trace theorem of Sobolev spaces, we can take a function UQ e iJ 1 (fi) with a compact support, such that uo|sn = g and ||«o||i < C||g||i. Let R be large enough, so that uo = 0 on dfto. Let Mi = Uh — UQ, then by the weak formulation (1.20) it holds that a(ui,v) = -a(uo,v),
Vv £ ffo(fli).
Let v = Mi, then |wi|f < |uo|il u i|ii which gives |ui|i < |wo|i < C||fl||ii where the constant C is independent of the domain Q\. We extend Uh by zero on D.2 = K2 \ ^ , then Uh G Hl'*{Q). Uh satisfies
KHi,.
Vveflo1^!),
a(u,v) = 0,
Wei^fti).
to obtain
Since R —> oo, u is the solution to the problem (2.35),(2.36). Because u is unique, Uh converges to u as R —» oo. • Although the above strategy generates convergent approximate solutions, the rate is very slow. The artificial boundary condition (2.37) should be improved. Using the theory in Section 1.1 and Section 2.2 the solution
Boundary
Element
Method
95
u on the domain SI2 can be expressed in terms of single or double layer potentials. For definiteness, let us consider two dimensional domains. By (2.20)
"(y)ln 7-^—r dsv + c '
«(*) = TT I 2
/
u
ds
(y)
(2-38)
v = °.
x
^ Jen2 \ ~y\ Jdn2 so we can impose a boundary condition on dflo instead of (2.37), u x
()
=
7T / 27r
w
(2/) m 1
7afi 2
F dsv + C, x € 8Q.Q,
F - 2/1
Here w and C are unknown, so (1.9) 1 du(x0) = - - w ( x/ 0 )x + dv+ 2
/
uj(y) dsy = 0.
JdQ0
(2.39) we need another boundary condition. By
1 /f w(j/)^— , x d ,In, 1 rdsy, — 2TT Jano dvx \XQ - y\ for XQ S 9J7O- Therefore we get another boundary condition, du dv+
W -nu(x) + 7T (y)-S— l n l \dsv x€dn0. (2. 40) w 2 2TT 7 a n o 5z/x |x-y| Theorem 44. Let g £ Hll2{dSl). The problem (2.35), (2.S6), (2.39), (2.40) with unknowns U,OJ,C on the domain fii is equivalent to the problem (2.35), (2.36) on the domain fi.
Proof. Let u be the solution to (2.35),(2.36), then u is a harmonic function, so it is analytic. By Theorem 42 u can be expressed by (2.38) on SI2, so u is the solution to (2.35),(2.36),(2.39),(2.40). Conversely, if u is a solution to (2.35),(2.36),(2.39),(2.40), then we can define u on ^2 by (2.38). Then u £ Hl,*{ST), u is a harmonic function on Q,\ and Q.2 respectively, and u satisfies the boundary condition (2.36). We only need to show that u is a harmonic function on the entire domain SI. The weak formulations are: /
Vu- Vvdx=<
--u+
— /
u>(y)—\nVveH1(n1),v\dil
;dsy,v>dQo, = 0.
and / Vu- Vvdx=< 7n2
-w- — / w(y)— In, r dsj,,v >an 0 , 2 2TT J a n o Si/ I - —2/|
96
Numerical Methods for Exterior
Problems
Now we take v G H0'*(Q) and add them together, then we get a(u, v) = I VM • Vv dx + Vu • Vv dx = 0. Jsix Jn2 Therefore u is a harmonic function on Q,.
•
The artificial boundary conditions (2.39) (2.40) are exact, so they yield better results.
2.4
Dirichlet to Neumann operator
The above mapping U\QQ0 —• (w,C) -+ | ^ | e n defines an operator if : Hl'2{dQ,o) -> l/'- 1 / 2 (ano), which is called a "Dirichlet to Neumann operator" . Using the operator an exact artificial boundary condition is designed to get a problem equivalent to (2.35)(2.36): Aw = 0, u = g, du — = Ku, av
x G fli, x £ dfl, x G d£l0.
The operator K depends only on the the equation and the artificial boundary dtto- To obtain K, one needs to solve an integral equation for ui. However the choosing of flo is quite flexible. If on some occasions fio = B(0, R), then using the results in Section 2.1 K can be expressed explicitly. This idea is due to Feng and named after "natural boundary reduction". See [ Feng, K. (1982)] [ Feng, K. (1983)] [Feng, K. (1984)] [Feng, K. (1994)] [Feng, K. and Yu, D. (1983)], also see [Han, H. and Ying, L. (1980)]. Let us consider the exterior problem of harmonic functions first. Differential of (2.3) gives du _ 1 f ~dr ~ 2TT J0
27r
ArR2 - 2R(r2 + R2) cos(6 -
Let r —> R, then we get the following formula formally, du dr
= T^ r 4nR J0
—Te^9^)dV. 2
Sm
^
(2.41)
Boundary
Element
Method
97
Let v denote the outward unit normal vector along the boundary, then we get an explicit expression for K:
For the Helmholtz equation differential of (2.16) gives oo
ur(r,0)=
Y,
anuHV'(ujr)eine,
(2.42)
a
(2.43)
n— — oo
then formally we get oo
ur(R,6)=
n"Hiiy'{wR)einf>'.
J2 n = —oo
Let v be directed to the exterior of the domain Clc, we define tr{u){6) = u(R,6) = g(0), tr{uv)(6) = uv{R,6), then (2.16) and (2.43) together defines an operator K, such that tr{uv) =
K(tr(u)),
or explicitly (•27T
-ur(R,6)
= K(6) *u(R,6)
= / Jo
K{6 - 6')u(R,0')d8',
(2.44)
where the function K is
The integral in (2.44) is hyper-singular. K is in fact a pseudo-differential operator of order 1 on the boundary manifold Oil, and defines a linear continuous map
K
:Hs{dn)-^Hs-1{dn).
Parallel to the association of the operator A + w2 in the domain D, with the bilinear functional a(u,v) = / ( y u • y v + to2uv) dx, Jn
98
Numerical
Methods
for Exterior
Problems
there is also an association of the operator K on the boundary dd with the bilinear functional /•27T
b(
Jo
/»27T
tp(9) dB / Jo
K{9 - 0')v(6')
dO',
which is inherently related t o a(-, •) by t h e equality, a(u,v) =
b{tr(u),tr(v)).
Being t h e same, for three dimensional case differential of (2.19) gives oo
n
ur{r,0) = J2
Yl
uanm^i±{ur)Ynm{6,
(2.45)
n = 0 m= — n
t h e n (2.19) a n d (2.45) together define a n operator K. We will provide a rigorous mathematical setting of t h e operator K in t h e next section.
2.5
Finite part of divergent integrals
Sometimes some integrals in t h e expressions of some D t N operators are divergent in t h e sense of Riemann integral. In fact they are singular integrals in t h e sense of Cauchy or H a d a m a r d . To provide a rigorous meaning of these integrals, let us introduce some basic definitions. [ Feng, K. (1982)] Let xm, x>0, 0, x < 0, ' where m is a complex number. Letting
/
/-OO
xr£
/ x^ip(x)dx, (2.46) •oo Jo which converges in t h e sense of Riemann if R e ( m ) > —1. Therefore x™ is a distribution, denoted by [x™]. It is easy t o see t h a t )
dx^\T{m
r rn-\-p-i I J
+ p+l)[X+
J
^
r ml
T(m+l)[X+1,
for positive p . Let us define distributions for R e ( m ) < — 1. T h e dual p r o d u c t ([x™], tp) defines a n analytic function for Re(m) > — 1. T h e analytic continuation can
Boundary Element
99
Method
be defined as the following: For m ^ negative integers, and a non-negative integer p, satisfying Re(m) +p > — 1, define a distribution, PfTm J
1
dP
=
+
[
Wm+P]
{m + p)---(m+l)dxP
+
J
T(m + 1) d? T(m + p+l)dxP[
=
+
J
+
'
The above definition is independent of the choosing of the parameter p, and Pfx™ = [x^] provided Re(m) > - 1 . By the definition of distributions one gets a finite part of (2.46), (Pfxy,
(2.47)
where ip^ = ^ £ . It is easy to check that (Pfx™, ip) is an analytic function with first order poles m = negative integers. One explicit expression of (2.47) is derived as follows. For e > 0, we consider J°° xmtp(x) dx. Let a > e be a fixed number, and
(fc)
^ =£^E ^(oLrfe^ fc=0
then /»oo
/ Je
pa
xm
pa
/ xm(ip(x)-tpp(x))dx+ Je
/•oo
xmcpp(x)dx+ Je
If rn is not a negative integer, then
, .. , _
k=o
xmV{x)dx. Ja
k\
m+ k+1
fc=o
m P(x) dx = If /m xis a
0
^ ^
v(fc)(0)
,,
fc!
a™+fc+i
;—7 1+
m + fc + 1
^ ( 0 ) £ m + f c +! k\ m + k + 1
w(-" , - 1 )(0) ( - m - 1 ) ! In a (/ p(-
m 1
- )(0) (-m-1)!
\i k ^ —m — 1, the limit of em+k+x is either zero or infinity as e —+ 0. Naturally the finite part of the divergent integral is defined by: If m is not
100
Numerical Methods for Exterior
Problems
a negative integer, then xmcp{x)dx=
Pf
xm(ip(x)-
J0
xm
I
J0
Ja (2 48)
A y W ( Q ) a"+fc+i ^ k\ m + k + 1'
|
-
fc=0
If m is a negative integer, then /»oo
Pf / Jo
/>a
/»oo
xm(ip(x) -
x"V(:r) cte = / Jo
y> +
^
xm
Ja y(fc)(Q) am+k+l
fc!
m +fc+ 1
y(-m-l)(Q)
+
(-m-1)!
n
°'
(2.49)
Lemma 41. 7/m zs not a negative integer, then /•OO
xmcp(x)dx = (Pfx™,lf).
Pf Proof.
(2.50)
It is easy to see that
I
xmtp(x)dx = ([x+l],
It remains to prove that the definition of (2.48)(2.49) is independent of the choosing of parameters p and a. If m is not a negative integer, the assertion is already contained in the previous lemma. Here we give another proof which is valid for all complex numbers m. Lemma 42. The definition of (2.48), (2.49) is independent of p and a. Proof. Letfc*be the smallest non-negative integer p satisfying Re(m) + p + 1 > 0. If k > k* and e —> 0, then
m+k + 1 ~~*
Boundary
Element
101
Method
By the definition, if m is not a negative integer, Pf
7 /
V
/.oo
xmip(x) dx = lim
/
£-•0
Je
= lim
/ A
Jo
e-tO
<^fc)(0) em+k+1 k\ m + k + 1 fc=0
xm
£^(fc)(0)
xmip(x) dx + y^k=0
k\
m+k+1 £
m + k+1
The right hand side is independent of p and a. Noting that if m is a negative integer, k* = —m — 1, then we get /*oo
Pf /
z " M x)dx
Jo
— lim £->0
/»oo j /s
fc* y <^fc)(0) e^+fc+i x cp{x) dx+^2 A;! m + k + l fc=o m
+
kJ
which is also independent of p and a.
lne
a
We have derived (2.41) formally in the previous section. Now we are ready to prove it rigorously. Let us assume that g € Cco(dQ) first. By (2.1) we get
dr ~
2ir J0
I ^
""
|n
r
-|n|-lem(9-^)l5^-)^
)
^ n = —oo
where for simplicity we take R= I. We define a function 1
°°
(2.51)
then for a fixed r € (1, oo) the right hand side can be expressed in terms of a convolution product, or a dual product on the space Cco(dQ): du or
—
=
- K
r
* g ••
For this boundary condition the solution u to the exterior problem belongs to C°°(fi), Therefore | ^ converges as r —> 1. The functions Kr is thus convergent weakly with respect to r in the dual space C°°(fi)'. Let the limit be K £ C°°(d)', which is a distribution by definition. We obtain the derivative at r = 1,
fr=-
(2.52)
102
Numerical
Methods
for Exterior
Problems
Now let us derive the expression of K. Let us define 1
°°
r-l
n
l-1ei™e
(2.53)
n = —oo
and
**<«> = -2ns n =<
(2.54)
ra^O then fl#ir 06
Kr
OK-2r "
=
06
Kir,
for r £ ( l , o o ) . Consequently < tfr(0 - •), = < A"lr(« " O.fl' > = < ^ 2 r ( 0 " O.fl" > Letting r —> 1, by (2.51) .ft^r converges in L 2 (9fi) t o 1 YimK2r{6) r-+l
=
°°
1
==
- Y—
27T
In 2 s i n -
2
7T
n—~ oo
\-K t-^ Ira which is a function in the classical sense. Therefore
<9u Or
p27T
i r Wo
In 2 sin- - V
g"{
(2.55)
This is the expression of the distribution —K. Taking classical derivatives on 6 £ (0, 27r) we get
A/I In d0 \ ^
2sin2
27rCOt2'
d f 1 6»1 —I —cot- } =
i6\2TT d0
2J
1 47rsir 2
e
Noting (2.53),(2.54) we can define some distributions as the limit of some divergent series, K =
Pf-
47r sin 2
e
'Alt 2rc
\n\e
Boundary
Element
103
Method
and
K1
=
-.
f 1 9\ Pf \ 7 - cot - } \2TT 2J
^^
^ 27U
I
I
^ n n = -co
>.
The derivative of u is the finite part of a hyper-singular integral, du dr
1
2TT
Pf- 4TT JO
sin2 6-^
;9{
(2.56)
For general R, the formulas are, ne
in&
and du dr
2.6
p, r=R
iv)dr
{-ssr^'
.
Numerical approximation
We investigate the implementation of the boundary element method in this section. Some meshes on the boundaries are required in the numerical computation of the above integral equations, and if the exterior domains are truncated, some meshes on some bounded domains are required as well. First of all let us consider the integral equation (1.11) for two dimensional problems as an example. Let us fix a point on dSl and let s = 0 at this point, s £ [0, L] is denned by the arc length along anticlockwise direction. The curve dfl is divided into some arc segments by some nodes s\ = 0, s%, • • • , sjv, and the elements, Ci, i = 1,2, •, N, are the arc segments s{s2, S2S3, • • • , srfsi. Let the approximate solution be
Si
—crh(Si+i) H
Si+i
r
-
104
Numerical Methods for Exterior
Problems
on SiS'i+i, where hi is the arc length of this element. Usually there are two methods to solve a^. 1. Collocation method Let us write the function K by K(s,t), s G [0,L],t e [0,L]. We assume that the equation (1.11) is satisfied at nodes, 2crh(si)+
ah(t)K{si,t)dt
= g(si),
i = l,2,---,N,
(2.57)
which is an algebraic system with N equations and N unknowns ah(si). We can define a finite dimensional space S(dQ) = {ah G C([o,L]);ah(0)
= ah(L),ah\ei
G Pi(eO,t = l , 2 , - - - ,N},
where Pi is the set of linear functions. Then we can define a set of basis functions in Sh'. UeSidU),
Li(si) = l,Li(sj)=0,j^i,
i=
l,2,---,N.
The function ah can be developed in terms of the basis functions: N
Vh(s) = ^2Msi)Li(s).
(2.58)
Then the equation (2.57) can be written as N
1
a
( fL
\
2 h{si) + X ! I /
Lj(t)K(si,t)dt
3=
J ah(sj) = g(si) t),
i=
l,2,---,N.
2. Galerkin method Multiply the equation by r G C(dQ.) and integrate the equation over dfi,, then we have - /
a(x)r(x)dx+
* Jdo.
/
a(y)K(x,y)T(x)dsydsx=
Jan J an
g(x)r(x)dsx.
Jan
The Galerkin scheme is: Find ah G S(dfl), such that - / ah(x)Th(x) * Jon = f g(x)Th{x)dsx Jan
dx + / / Jen i s n
ah(y)K(x,y)Th(x)dsydsx C2
KQ\
\/Th£S(dn).
Since S(dQ) is a iV-dimensional linear space, the equation (2.59) is a N x N algebraic system. Let the unknown function ah be developed in terms of
Boundary
Element
105
Method
the basis functions as (2.58), and N T
h(s) = y"Vh(Si)Li(s), i=l
then the equation (2.59) can be written as N
Y^kijah{sj)
= gi,
i = 1,2, ••• ,N,
3=1
where -i
/ȣ
pL
kij = - / Lj(s)Li(s)ds+ 2 Jo
pL
/ Vo Vo
Lj(t)K(s,t)Li(s)dtds,
and /
./o
9 ( 5 ) ^ ( 5 )Ids. ,
The strategy for solving the integral equation (1.11) for three dimensional problems is similar. However, it is difficult to construct a continuous and piecewise polynomial function on a closed smooth surface dQ, so usually the boundary dQ, is approximated by an surface dflh, which causes an additional error. Next let us consider the problem (2.26). Let us define a subspace of S(dfi): S0(d£l) = I u € S(dtl); I
u{s) ds = 0 L ,
then the approximation to (2.26) is: Find u>h & So(dil), such that & K , Xh) =< g, Xh >aa,
VX G So{dSl).
The bilinear form b(-, •) defined in (2.23), b
/ w(y)x{x)\n-dsydsx, 2TT Jan Jan F - V\ is symmetric, so the equation (2.60) is equivalent to (u,x)
= 7T
2—
1 2b(UJh,UJh">~
< 9 Wh
'
>m
I
=
°'
(2.60)
106
Numerical Methods for Exterior
Problems
With respect to the constraint fQ Uh(s) ds = 0, we introduce a Lagrangian multiplier A and define a function -b(uh,wh)2
< g,tOh >dn +A /
u!h(s)ds.
The problem (2.60) is equivalent to: Find (uh, -M G S(d£l) x R, such that H"h,Xh) + *
Jo
Xh(s)ds=
Jo
g(s)Xh(s)ds,
VX € S(dfi),
and fL / u>h(s) ds =0. Jo This is a (N + 1) x (N + 1) algebraic system. Finally let us consider the exterior problem of the Poisson equation, -Au
= f,
u = 0,
rrefi, x€dQ
(2.61) (2.62)
in two dimensional space, where / G Co(fi). For simplicity of the domain partition, let us assume that Clc is a polygon. Let f2o be a disk B(0, R) with R large enough, so that supp / C B{0,R). Let fii = flo f~l fi, then by the argument in Section 2.4 the problem is equivalent to -Aw = /, u = 0,
i£f!i, z G dCl,
— - Ku, x£ dfto, a;/ where K is the Dirichlet to Neumann operator. Let a(u,v) = /
(2.63) (2.64) (2.65)
Vu-Vvdx,
JQI
then the variational formulation of the problem is: Find u G Hl{£l\), that U\QQ = 0 and a(u,v) - (Ku,v)da0
= (f,v)ni,
W G tf^fii^Hafi = °-
such
Boundary Element
107
Method
Applying the finite element method to solve this variational problem, we set a mesh. For example, the domain fii is divided into triangular elements {ej}, and linear interpolation is designed on each element with respect to the vertices. Since dflo is a circle, each element neighboring the circle possesses an arc edge. The finite element space is denned by, S(fii) = {u € ^ ( O i ) ; ^ , G Pi(eO,Vidian = 0}. The finite element formulation is: Find Uh € S(D.i), such that a(uh, v) - (Kuh, v)dno
= (/, v)Ql,
Vu £ S^).
(2.66)
The evaluation of a{uh, v) and (/, v) is routine. However the evaluation of (Kuh, V)SQ0 requires some special consideration. There are two kinds of approaches: 1. Truncation r-2-K
p
(Kuh,v)dno=j
•,
(
[-2TT
f \ - ^ l
°°
"I
\n\ein^-^uh(R,lp)diP\v(R,6)de.
E
Let N be a finite number, then approximately 2TT J"
/
I
p2TT
\ - ^ j
N
1
\n\ein{9-'fi)uh(R,
S
2. Integration by parts If Uh and v are infinitely differentiable, we denote g(6)=uh(R,9),1(e)=v(R,6), and
W ) = -^ln
«2 s •i n -e 2
then by (2.55) it holds that Kuh{6) =
-
Numerical Methods for Exterior
108
Problems
Consequently we have PZTT
(Kuh,v)ano=-
/ Jo
< K2{6 - •),
i(0)RM
/•2ir
:/ Jo
I
Jo Jo
that is r2n />Z7T
-i
(Kuh,v)dQo
= - / "• 7o
2ir r/*Z7T
/ Jo
Q
In 2 sin •
_
^
g'(
(2.68)
The definition of S(fii) implies that U\QQ0 S iJ^dfio), so applying the fact that C°°(d£lo) is dense in H1(dQo) we can show that (2.68) holds for Uh, v € S'(fii). The integrals in (2.68) are convergent, and can be evaluated using a numerical scheme of integration. 2.7
Error estimates
Let us prove convergence and estimate errors to some of the above schemes. The first example is (2.60). We assume that the exact solution w £ H2(dd), and let II denote the interpolation operator. Let h be the maximum length of all elements. Then owing to the error estimates of interpolation: [Ciarlet, P.G. (1978)] ||IL; — w||_i/ 2 < C||na> — w||o < Ch2\u>\2Let eh =< HLJ, 1 > / / af2 ds, then | < IIu; - <J, 1 > |
|ino;-^l|_1/2l[l||1/2 ~ < Oft \u\2? Ja-n ds Jan ds Let u>i = IIw — eh, then u>j £ So(dtt), and \eh\ =
< C
\\ui -u}\\-i/2
<
Ch2\u>\2.
The following error estimate is routine: a
w w - Uh\\_i/
2
-ujh) — b(u> -u>h,oJ — wj)
<||w - w^||-i/ 2 ||w - w/H-i/2-
Boundary Element
109
Method
Consequently we have the following: Theorem 45. It holds that < Ch2\u\2.
\\u-Uh\\-i/2
The second example is the scheme (2.66). We assume that the partition is regular, that is the interior angles of all elements possess a common lower bound 9Q > 0. First of all let us show the existence and uniqueness of the solution to the problem (2.66). We extend a function Uh € S(fli) to D,2, the exterior of dfi,o, by solving a boundary value problem of the harmonic equation on $72 with the boundary value u/i|an 0 i then Uh G -ff1'*(fi), and the application of Green's formula leads to a(uh, v) - (Kuh, v)an0 = / Vuh • Vv dx, Vv e H1'*^). Jn Owing to the estimate for exterior problems of the harmonic equation and the trace theorem of Sobolev spaces, we obtain |"h|i,n 2 ^ c\\uh\\i/2,dn0
< C'lluhUi,^
Consequently it holds that \a(uh,v) - (Kuh,v)9n0\
< Clluhlli^illulli.fi!,
Vuh,v G 5(fii),
and \a(uh,uh)
- (Kuh,uh)da0\
> a\\uh\\jni,
Vuh G S(Cli),
with a > 0. Owing to the Lax-Milgram theorem there is a unique solution. Let u be the exact solution to (2.61),(2.62), then a standard argument leads to the following: (see [Ciarlet, P.G. (1978)]) Theorem 46. It holds that ||w-"/i||i,ni ^ Ch\u\2,niFinally let us consider the error of the truncation (2.67). Let u be the solution to (2.63)-(2.65), and UN be the solution to —AUJV = / , UJV
= 0,
x £
fli,
x G dQ,
110
Numerical Methods for Exterior
KNUN,
~&7
Problems
x e d£l0,
where JV
,2TT
KNuN
= -—
27r
\n\ein(e-^uN{R,V)dip.
V
Jo
n=-JV Jr'K,
The variational formulation is: Find UJV G - ^ ( ^ l ) ) such that wjv|an = 0, and bN{uN,v)
= (f,v),
W G ^ ( n ^ . u l a n = 0,
(2.69)
where /•27T
i>N(uN,v) = a(uN,v) — / Jo
KNU^(0)v(R,0)Rd6
and a(uN,v)
= = /
Vujv • Vv da:,
where we notice that although the problem and functions are real, the Fourier series are expressed in terms of complex functions. Define b(u,v) = a(u,v) -
(Ku,v)dno,
then the variational formulation of the problem (2.63)-(2.65) is: Find u G i7 1 (Oi) such that U\QQ, = 0, and W G H1^),
b(u, v) = (/, v),
(2.70)
v\dn = 0.
Let us define a norm: \\WN\\\
=bN(uN,uN)* r2n
= \ l"Ar|i,ni +
N
2^
2
inv
h= i^ E \/W\e- uN(R,ip)dV 27r
Jo
„=_JV
then \bN(uN,v)\
< \\\uN\\\ • \\\v\\
Owing to the Lax-Milgram theorem the problem (2.69) admits a unique solution.
Boundary Element
111
Method
We turn now to prove the convergence of (2.69). We have bjv(u — UN, u — UN) =6JV(U, u — UJV) — b(u,u
— UJV) + b(u,u
— UN) . — 6JV(UJV; w — UN)
=6JV(W, u — ujv) — b{u, u — UN)-
By the definition of b and 6JV, 6JV(W,W —
«jv) - b(u,u — UN) = (Ku -
KNU,U
- UN)an0-
We apply the Schwarz inequality to get \bN(u,u - uN) - b(u,u - uN)\ < \\Ku - KNu\\0taa0\\u
- UN\\o,ano- ( 2 - 71 )
Let us estimate \\u — Ujv||o,9n0- Let RQ be the maximum positive constant, such that B(0, RQ) C fic, Ri be the minimum constant, such that supp (/) C B(0,Ri), and let the constant R satisfy R> R\. Extend the function u — UN by zero on Clc, then we get R
(u-uN)(R,6)=
d(u — dr
f JRo
UN)
dr.
By the Schwarz inequality we obtain ((u-uN)(R,0))2
p2ir
T)
((u-uN)(R,e))2d6<—
d(u — dr
UN)
dr.
JRo
Hence we have / Jo
f
-fto
pR
flir
rdr JR0
JO
d(u — dr
R
UN)
d6 <
-£-\u-uN\ln
that is u
N\\lidn0
R2 < -5~\u - uN\ln
R2 <
-p-\\\u-uN\
(2.72)
Next let us estimate \\Ku—K Nu\\otan0- The solution u can be developed into a series u=
J^
anr-{nleine,
with DI«I
1 2TT
r2ir
/ Jo
e-^uiR^^dtp,
112
Numerical
Methods
for Exterior
Problems
for r > R\. Therefore du dr
Kv
Y^ |n|ani?-l"l-1e™e, r=R
and ^ InlanR-W-1^inO
Ku-KNu=-
\n\>N
The norm is equal to \\Ku-KNu\\ldUo=
\Ku-KNu\2ds=
f JdQ.0
\Ku-KNu\2R dO
[ JO
=2nR J2 | n | 2 K | 2 i T 2 H - 2 |n|>AT
(2.73)
Integrating by parts gives \n\
r2n
R[;
2TT ni
|„|2|_
Jo
|2 _ p2|"l 2TT
\n\
(2.74)
dip,
dip
\an\
J0
dip
— rtj
2
Plugging it into (2.73) gives
\-^MR^)d^ dip
\n\>N -2(N+1)
2
*
<±R-> (2•2TT \RI
n\>N 2(JV+1)
x
<-R~ -2?r
_inv,du(RUip) dip d
oo
r2n
y
(# \Ri
f\-^
f r ^ Jo
-2(N+1)
9p
r2TT
a
R_y*"-° Ri
f
I
(du{RUip)
(•
ds
ds.
Boundary
Element
113
Method
Therefore \\Ku-KNu\\0,da0
< ( ^-J
Mi.aBCo.flO-
(2-75)
The combination of (2.71),(2.72),(2.75) gives the following: Theorem 47. It holds that |||U-UJV||| < - r f ^ - j
\u\i,aFHo,R1y
R§ Moreover, the solution u is infinitely differentiable on Q. If we replace (2.74) by integrating by parts for k times, then the result is generalized to Theorem 48. It holds that
|||«-^|||<^-(]^T) 2.8
(§)
+2
\u\k,aBio,Rl),
Vfc = l , 2 , . . . .
Domain decomposition
The stiffness matrices of the finite element method are usually sparse, while the matrices of the natural boundary element method are usually circulant. It is known that effective numerical methods to solve these two kinds of matrices are different. For the coupling of these two methods one strategy to separate these two matrices is using an iterative procedure. For example the problem (2.66) can be solved by « K , v) - (K\n, v)dno
= (/, u)m,
to
e S(fii),
A"+1 = enunh + (i - en)xn, where n = 1,2, •••, and 6n £ (0,1) is the relaxation factor. This is the conventional Dirichlet-Neumann alternating scheme in the non-overlapping domain decomposition method. Another approach is the overlapping domain decomposition method. Again for the problem (2.66) let R > Ri, B(0,R±) D supp (/), and define fii = Qf]B{0,R), n2 = {x;r > Rr}. The scheme reads: u? h G S(fii), n —'1 uih\r=R = IIu22hh \r=R, and a « h ) v) = (/, v)ni,
Vv e 5(fii), v\r=R = 0,
Numerical Methods for Exterior
114
Problems
and U2h is given by the Poisson formula on Q.2'^
=
1 t o j
f2v
r2 - R2 Ri-2rR11coS(0-v)U^R^)dif>
r> +
0
where n = 1, 2, • • •, and II is the interpolation operator to the finite element space. Convergence proof is essentially standard. For details, see [Yu, D. (2002)]. 2.9
Boundary perturbation
For two dimensional case if the boundary is a small perturbation of a circle: d£l = {(r,6);r = Ro + £7(0)}, then the solution to the exterior problem can be developed into a power series in terms of the parameter e. Following the argument by [Bruno, O.P. and Reitich, F. (1996)] let us introduce this approach for the exterior boundary value problem of the Helmholtz equation: (A + w 2 )u = 0,
ieO,
(2.76)
with the boundary conditions, u = g,
x£dQ,
(2.77)
£-iu;« =0(l). Let B(0,R) D JF, and U\ = {(r,9);R0 developed as
(2.78)
< r < R}. The solution u can be
00
u=Y/Mr,0)en.
(2.79)
n=0
The functions un satisfy {A + LU2)un = Q,
un = gn, dun —— = Kun, av
XGQ.I,
r=
RQ,
r — R,
Boundary
Element
Method
115
where K is the Dirichlet to Neumann operator given by (2.44), and gn are equal to gn{9) - 6n0g(9) - ^ 1=0
1—— v
^ - j
.
'
It is observed that the above algorithm suffers from ill conditioning due to significant cancelations [Nicholis, D.P. and Reitich, F. (2004a)][Nicholis, D.P. and Reitich, F. (2004b)]. There is another approach. First of all a mapping is denned to transfer the domain $7 Q B(0, R) to fii: ,
(R - R0)r - Re^e) (R - Ro) - e 7 (0)
& = 6.
The system is modified, upon dropping primes, to du \ r-— dr Vr dr )
d2u — 2 +r'iLj2u = F(r,6,u), d6
R0 < r < R,
Ro, 3n — =Ku + J{6),
,r = R,
where F and J are some known functions. Then following the same steps u is developed into a series (2.79), and un are solved by recursion. A spectralGalerkin method is given to solve this problem. For details, see [Nicholis, D.P. and Shen, J.].
Chapter 3
Infinite Element Method
Using an infinite number of degree of freedom, one can make a direct discretization of the original exterior problem. The problem is the implementation of the approximate problem which contains infinite unknowns. The introducing of a structured mesh can solve the problem. It is possible to analyze the discrete problem and eliminate infinite number of unknowns in advance. Then only a very small amount of unknowns are needed to be taken into account in the real computation to result in a solution with infinite unknowns exactly. We will present this approach and study the related mathematical problems in this chapter.
3.1 3.1.1
Harmonic equation-two dimensional problems Infinite
element
formulation
We start with the problem (1.1),(1.2) and (1.4). We assume that flc is a polygon in a plane, and it is in star-shape with respect to the point O, that is, the line segment linking each point on dfl with the point O is included entirely in the interior of dQ. Applying the approach in Section 1.3, we have the variational formulation: For a given g € H1/2(dil) find u G iJ 1 '*(fi) such that u\on = 9, and Vveffo'*( fi )>
a{u,v)=0, where a(u, v) = I
Jn 117
S/u-Vvdx.
C3-1)
118
Numerical Methods for Exterior
Problems
To solve (3.1) fl is divided into an infinite number of triangular elements. Let us denote dd by r 0 . Taking a constant £ > 1, we draw the similar curves of To with center O and the constants of proportionality £>£2>' •' i£ fc >''' i which are denoted by T i ^ , • • • ,Tfc, • • • respectively. The domain between two polygons is named a "layer". Afterwards each layer is further divided into elements. It may be proceeded in such a way: some points on To are selected as nodes, where the vertices of To must be nodes and some other nodes on the line segments can be properly selected according to our requirement, then rays are drawn from the origin to the nodes, consequently every layer is divided into some quadrilaterals which are similar to each other among the layers, finally each quadrilateral is divided into two triangles such that the manner of partition for each layer is the same. The triangular elements are denoted by e,, i = 1,2, • • •. We denote £kQ. the exterior of Tk and Q.k = (£ fc-1 fi) \ (£fcft).
• * i
Fig. 4
Sometimes the domain Qc may not be in star-shape, and the curve To may be a polygon in very complicated shape. Then we may decompose
Infinite Element
Method
119
the exterior domain fi to an unbounded domain with regular shape and a bounded domain with complicated shape like Section 2.3. We still denote by n the unbounded domain, which can be divided in the above way, while the bounded domain can be divided into finite elements in conventional way. We need the two styles of partition to be conform to each other, that is, the nodes and line segments of the elements are coincide to each other along the interior boundary. We still first analyze the domain fl for this situation. We will state an algorithm for calculating the combined stiffness matrix of 0,, which is a matrix with finite order, and which is applied either to solve a boundary value problem directly, or to solve a boundary value problem for a domain with complicated shape by assembling it with the stiffness matrices of other elements. For the later case 0, is treated as one element. we consider the infinite element subspaces of i71'*(f2). Let the nodes be the vertices of all triangular elements, and linear interpolation is applied in all elements. Then let the interpolating functions belong to the space iJ 1 '*(Q). The infinite element spaces is defined as the following: S(Q) = {ue
H^ffl;
u\et G P^eO, t = 1,2, • • • } ,
So(n) = {ueS(n);u\dQ
S(rk)
=
= 0},
{u\rk;ueS(Q)}.
Let gj G S(To) be an approximation of g. The formulation of the infinite element scheme is: find Uh G S(Q) such that U\QQ = gj, and a{uh,v)=0,
VD e 5 0 (0).
(3.2)
L e m m a 43. The problem (3.2) admits a unique solution. Proof. We define a function «o £ S(Q), such that uo\an = 9i and uo = 0 for all x G £Q, which can be obtained by interpolation with respect to all nodes. Let Wh = Uh — «oi then u>h is the solution to: find Wh G SQ(Q), such that a(wh, v) = -a(u0, v),
Vv G So(fi).
Owing to the Lax-Milgram theorem, existence and uniqueness follows.
•
120
3.1.2
Numerical Methods for Exterior
Tranfer
Problems
matrix
We now turn to the implementation of the above scheme. In order to carry out the real computation, we deduce an equivalent and finite form of it. We consider the polygon Tk, and starting from one node, we arrange the nodes of Tk in an order according to the anticlockwise direction. Let the number of nodes be n. The nodal values of the solution of (3.2) are also arranged in an order, which are denoted by yk ,yk , ••• ,yk , and they form a n-dimensional column vecter yk = (yk ,yk ,••• ,yk ) T .Let us denote yk = BkUh, where Bk is the trace operator. We will identify yk & K n with u\„ rather than distinguish them. ' >- k
We consider the fc-th layer Clk. The element stiffness matrix of each triangular element can be evaluated by conventional way (for instance, see [Ciarlet, P.G. (1978)]). Upon assembling the element stiffness matrices by the nodes in the A;-th layer, we obtain a stiffness matrix of one layer, which is denoted by
that is,
/„^-vw"("K-r^)(r> where zk-i = Bk-\v and zk = Bkv, and KQ, K0, and A are n x n matrices. It is known by the fundamental theory of the finite element method that (3.3) is a symmetric matrix, therefore AT and A are the transpose of each other, and KQ,KQ are symmetric. By the equation (1.1) and the similarity of the partition it is easy to see that the stiffness matrices for all layers are the same. Upon assembling the layer stiffness matrices by the nodes, we obtain an infinite by infinite total stiffness matrix. Noting the equation and the boundary condition, we get an algebraic system of equations of infinite order. Let K = KQ + KQ, then it is -Ay0 + Kyi - ATy2 = 0, -Ayk-i
+ Kyk - ATyk+x
= 0,
Infinite Element
121
Method
which takes the form of a block tri-diagonal Toeplitz matrix. If the boundary condition is the Neumann boundary condition (1.3), there is another equation for yo, K0yo - ATVl
= /o,
(3.5)
where /o is the "load vecter", which can be evaluated from the boundary value g by conventional finite element method. Now our aim is giving the algorithm of the combined stiffness matrix on ft, thus we ignore the difference between these two boundary conditions at this occasion. In order to reduce the algebraic system of equations of infinite order, we need a careful study of the properties of the solutions. Lemma 44. Ifu £ S(fl), yk = BkU, then there is a constant C independent of u, such that .
oo
.
2
oo
2
/ \Vu\ dx
-J2\yk-yk-i\ < fc=l
^
n
(3.6)
fc=0
Proof. Let ej C fl be an arbitrary triangular element. Let the length of the side pointing to the origin O be L, and the values of u at the both ends of it be yk and Uk_v Let measej denote the area of the triangle ej, and p the least height of e-j. Then
ly^-Vk-il L
^ , „ i ^ \yk\ + \yk-i\
< |Vu| <
Therefore it holds that iJ
L2
-\v{?-VkU2
|V«|2dx< = £ ^ 1 + ^ 1 ) 2 . Jet
P
By similarity me^ei a n d me^S63' depend only on the position of ej in fifc and are independent of k. Summing up the above inequalities with respect to all triangles ej we obtain (3.6). • Lemma 45. KQ and K'0 are positive definite matrices. Proof. We take an arbitrary yo ^ 0 and set y\ = 0. The interpolating function u of them is not a constant, hence a(u,u) = /
in,
|VM| 2 dx > 0.
122
Numerical Methods for Exterior
Problems
On the other hand a(u,u)
-AT K'n
K0 -A
( « )
T
Vo T
Thus VoKoyo > 0,
Vy0 ^ 0,
so Ko is a positive definite matrix. The argument for K'Q is similar.
•
By Lemma 43 the problem (3.2) admits a unique solution Uh for any yo G K™ such that B$Uh = 2/o- Let yk = BkUh, then 2/1 is determined by yo uniquely. It is easy to verify that the mapping 2/0 —• 2/1 is linear. Therefore there is a real matrix X such that 2/1 = Xy 0 Let tu(:r) = Ufe(£fc:r), then w G S,(fi), Bow = 2/fc and w satisfies the problem (3.2), hence 2/fc+i =
Xyk.
We obtain by induction that yk = Xky0,
k = l,2,---.
(3.7)
The matrix X is the transfer matrix for the problem. We will make use of the complex function solutions to the Laplace equation in the remaining part of this section. The above argument can be applied for this case without any modification except we regard that 2/i, 2/2, • • •, yk, • • • are vectors over the field of complex numbers. We observe that the transfer matrix X is real as we have proved. Substituting (3.7) into (3.4) we obtain {-A + KX-
ATX2)y0
= 0.
2/o is an arbitrary vector, hence X satisfies the equation ATX2 -KX
+ A = 0.
(3.8)
The solutions to (3.8) are certainly not unique. We need the following lemmas to identify a unique solution. L e m m a 46. 7/A is an eigenvalue of the transfer matrix X and |A| > 1, then A = 1.
Infinite Element Method
123
Proof. Let g be t h e eigenvector associated with A. Taking yo = g as t h e b o u n d a r y value of problem (3.2), we get j / i , 2/2> •• •, 2/fc, •• •, a n d by (3.7) yk = Xkg = Xkg. Hence oo
£ \yk - yk^
oo
oo
= £ \Xk - \*-*\*\g\* = £ |A|2fc-2|A - 1|2|5|2.
fc=i fe=i fc=i By L e m m a 44, t h e series on the right h a n d side converges. We have \g\ ^ 0, hence A = 1. • L e m m a 4 7 . The elementary divisor corresponding of the transfer matrix X is linear.
to an eigenvalue
X—1
Proof. Suppose it were not t h e case, then there would be nonvanishing vectors g a n d go such t h a t (X-
7)3o=0,
(X-
I)g = g0,
where 7 is a unit matrix, i.e. Xg = go + gBy induction we would get Xkg
= kg0 + g.
Taking yo = 9 as the b o u n d a r y value of the problem (3.2), we obtain j / i , 2/2,--•, 2/fc,..-, then by Vk = kg0 + g we would get oo
oo
2
X ] 12/fc - 2/fc-i I = X ] l^ol2 = +°°> fe=l fc=l which contradicts u € 771'*(J7) according t o L e m m a 44. L e m m a 4 8 . There is an eigenvalue X = 1 of the transfer the associated eigenvector is g\ = ( 1 , 1 , . . . , 1)T.
• matrix X , and
124
Numerical Methods for Exterior
Problems
Proof. We take yo =
Xgi=gi.
n
Lemma 49. As k —+ oo, Xkyo —> j/oo for any complex vector yo, where 2/oo = agi, and a is a constant. Proof. On the basis of Lemma 46 and Lemma 47, we conclude that the k limit of X exists as k —> oo. Denote by J/QQ the limit of Xkyo, which is (j/oo , J/oo , • • •, I/oo1 ) T in component form. It will suffice to prove yio' — yio = • • • = y£ . If it were not true, we might assume that y^ / J/oo , which correspond to the i-th. node and the (i + l)-th node respectively. There must be one element ej in fifc such that the connecting line segment of the above two nodes is one side of ej. Let the length of this side be L, then we have .„
/
,
mease^. a\
fi+ii.o
| V ul9| 2 c f c > — ^ y W - y J T 1 ^ 2 .
J en
By similarity me^ei [s independent of k. Summing up the above inequality with respect to k, we obtain / fc=i
The general term of the series on the right hand side would take \y,O(0 O ylo | 2 7^ 0 as its limit as k —» oo, hence 2
/ |Vu| dx = oo, in JQ
which leads to a contradiction.
•
Lemma 50. There is a unique eigenvalue X = 1 of the transfer matrix X, and the absolute values of all other eigenvalues are less than 1. Proof. If it were not true, then by Lemma 46, Lemma 47 and Lemma 48, there would be another eigenvector g' associated with the eigenvalue A = 1. According to Lemma 49 we have Xkg' —> ag\ (k —> oo), but Xkg' — g', hence g' = ag\, which contradicts the fact that g\ and g' are linearly independent. •
Infinite Element
125
Method
Let
(3 9)
* = l'/"o")- * - ( ? ? ' ) •
'
where I is the unit matrix. The equation (3.4) can be written as Rl(
y* )
= R
KVk-\J
Jy^).
(3.10)
\ Vk
Let A, g be a couple of eigenvalue and eigenvector, then Xg = Xg.
(3.11)
Taking k = 1 and setting y0 = g in (3.10), and noting (3.7),(3.11) we get
Therefore A and I 1 are the generalized eigenvalue and generalized eigenV 9 J vector of the matrices bundle R\ — XR2. To find matrix X, we first solve the above generalized eigenvalue problem to get all 2n eigenvalues and eigenvectors, then pick out the n eigenvalues and eigenvectors of X. We will always denote by det the determinant of a matrix. Now we evaluate det(i?i - \R2) = det .
'K-\AT
-A
This matrix is split into blocks. We add the second column by the product of the first column and A, then obtain , ,n ^r,•, , (K-XAT det(Ri - XR2) = det (
-A +
= ( - l ) " d e t ( - A + XK-
XK-\2AT\ j X2AT).
K is symmetric, so this expression is a symmetric polynomial with the independent variable A. Therefore if A ^ 0 is an eigenvalue, then so is 1/A. Therefore we only need to pick out the eigenvalues, which satisfy |A| < 1, of the matrices bundle Ri — XR2, then supply a A = 1. Let the eigenvalues be Ai, A2, • • • , An, and the corresponding eigenvectors be g\, g2, • • • ,gn, and we set T = (gi,g2, • • • , gn), A = diag(Ai, A2, • • • , A n ), where diag means a diagonal matrix or a block diagonal matrix. By (3.11) we have
XT = TA,
126
Numerical Methods for Exterior
Problems
that is X = TAT'1.
(3.13)
(3.13) is the formula for evaluating the transfer matrix X. Once X is obtained, then (3.7) gives the solution on the entire domain ft. If A is invertible, then the above generalized eigenvalue problem can be reduced to a conventional eigenvalue problem. Multiplying R^1 on the left of (3.12), we get
( A A
rr- T x ;)-e; which is a conventional eigenvalue problem. Eigenvalue A may be a complex number, thus the above calculation may encounter complex numbers. To prevent from this inconvenience, we can vary (3.13) slightly. Eigenvalues and eigenvectors must appear in conjugate pairs. Let A = a ± i(3, g = p ± iq be a pair of eigenvalues and eigenvectors, then by (3.11) we have X(p ± iq) = (a ± i(3)(p ± iq). Separating the real and imaginary parts we obtain Xp = ap - (3q, Xq = /3p + aq, that is
*<»>«)-(")(",!! We substitute the corresponding two columns in matrices T and A, and let Ti
=
(•••,?,
*.-""(••••(-"„£).•••)• It is easy to see X = TiAiTf 1 . After substituting with respect to all complex eigenvalues, the calculation of the transfer matrix X becomes real.
Infinite Element Method
3.1.3
Further discussion
127
for the transfer
matrix
We need to prove that the above conditions for X are not only necessary but sufficient. The argument is the following. Let
+ C2(f2 H
h Ciipi
oftp'iS, then So(iY) C So(£l). Each function in So(fi) has a bounded support. Theorem 49. So(Q) is dense in SQ(£1). Proof. We take an arbitrary u £ S0 (fi). Since S0(il) C Hi\'*(n), there exists MI G C Q ° ( Q ) , such that ||M — MI||I,* < £ for any £ > 0. We consider the following problem: Find u2 G So(fi), s u c n that a(u2,v) = a(ui,v),
Vv G S0(il).
(3.14)
We can prove that the problem (3.14) admits a unique solution as we did for the conclusion of Lemma 43. By the positive definiteness of the following quadratic functional we obtain -a(u2,u2)-a(ui,U2)=
min
\-a(v,v)
-
a(ui,v)\,
hence -a(u2,u2)
-a{ui,u2)
< -a(u,u)
-a(ui,u).
By adding |a(ui,Mi) to the both sides of the above inequality we obtain a(u2 — Mi,t*2 — Mi) < a(u — MI,M — Mi), thus a(u2
— Mi, u2 — Mi) < £.
Applying the inequality (1.19) we get ||"2 - W l | | l , * < 5£.
128
Numerical Methods for Exterior
Problems
But «i € Co°(f2), so there exists a natural number fco such that u\ = 0 on the domain £fc°fi. The equation (3.14) imply that VvG5o(^on),
a(u2,u)=0,
where we accept that every function in So(£k°Q) is extended to zero on ft, \ £fc°fi. According to the conclusion of Lemma 49, u 2 tends to a constant a as |a;| —> oo, therefore u 2 is a bounded function. We construct truncated functions «2,fc 6 So(Q), fc = 1,2,... such that _ J u 2 , for x e fi \ £fcfi, ' \ 0 , for re G £ fc+1 n.
U2 fc_
Let yfc = BkU2, then by Lemma 44 we have [ \Vu2M\2dx<
Jn
f
Jn
\Vu2\2dx+hyk\2.
°
Therefore {u2,k} is bounded in So(fi). Taking any v e So(^) and making use of the Schwarz inequality we obtain / Vv • V(u 2 Jn
f < ( / \Jnk+1
u2,k)dx
Vt; V(u 2 - u2,k)dx + \Wv\2dx) J
f
Vv Vw
2
cfa;
([ |V(u2-w2,fc)|2^] \JQk+1 J
+
/ -V+ifi
Wv-Vu2dx
Because both u2 and u2]fe are bounded functions, JQ |V(u 2 ~ w2,fc)|2 dx are uniformly bounded with respect to fc. And on account of v, u2 6 So(Q,), /
\Vv\2dx->0,
/
Vv-Vu 2 ote
as fc —> oo, therefore /
Wv • V(u 2 - u2,k) dx —y 0
as fc —* oo, which means u2,fc weakly converges to u2. By the Mazur Theorem [Yosida, K. (1974)], there is a linear combination, U 3 = Ci 1*2,1 + C 2 U 2 ,2 "I
1"
Infinite Element
129
Method
such that ||u 3 - it 2 ||i,* < £• Thus ||u - U3II1,* < 7e. But u 3 £ S0(Q), which leads to the conclusion. • Now we continue to discuss the Laplace equation. We give another formulation for the infinite element method: Find u G S(Cl), such that u h\an = 9i &nd a(uh,
i = l,2, •••.
(3.15)
Theorem 50. The formulation (3.15) is equivalent to (3.2). Proof. It will suffice to prove that (3.15) implies (3.2). Let Uh be the solution to (3.15), then Uh satisfies a(uh,v) = 0,
Vv € 5o(fi),
which yields (3.2 by noting Theorem 49.
•
Expressing (3.15) in terms of matrices, we get the system of equations (3.4), and the equation (3.8) for the transfer matrix X. We should bear in mind that the solutions to (3.8) are not unique. Theorem 51. The necessary and sufficient conditions to determine the transfer matrix X are (a) X satisfies (3.8); (b) X admits an eigenvalue A = 1 which corresponds to a linear elementary divisor and the eigenvector g\; (c) The absolute values of all other eigenvalues of X are less than 1. Proof. The lemmas from Lemma 46 to Lemma 50 and the equation (3.8) show that those conditions are necessary. It remains to prove they are sufficient. If a matrix X satisfies (a),(b),(c), let us prove it is the transfer matrix. We take an arbitrary g £ W1, and set yk = Xkg, k = 0 , 1 , . . . , then yk satisfy (3.4). Let Uh be the interpolating function. If we can prove Uh 6 S(Q), then owing to Theorem 50, Uh is the solution to the problem (3.2), which means X is the desired one. By (b) and (c), the limit of Xkyo exists as k —> 00. Let it be y^. We have Xyoo = X lim Xky0 k—>oo
= lim Xk+1y0
= yx,
k—>oo
therefore y ^ = agi. Let y'k = yk — y^, then the interpolating function of y'k is just u' = Uh — a. We only need to show u' £ S(fl).
130
Numerical Methods for Exterior
Problems
We express X in the Jordan canonical form, X = TJT~\
(3.16)
where 1 where J\ consists of those Jordan blocks, the absolute values of the eigenvalues corresponding to which are less than 1. By the definition of yk, Xy'k_x = X(yk-i
- t/oo) = Vk - ?/oo = y'k-
(3-17)
Let zk = T~ly'k, then (3.16) and (3.17) yield zk = Jzk-i,
k = 1,2,-•• .
By induction we obtain Zk = J
Z0,
that is
Zfc=
0./x fc ) ZO '
(3 18)
'
Because y'k —> 0 as k —> oo, we have Zk —> 0. (3.18) implies that the first component of ZQ is zero, otherwise the limit of zk would not be zero. Thus
Let || • || be the spectral norm of a matrix. Denote by p\ = p(J\) the spectral radius of the matrix J\, and p the highest order among the Jordan blocks of J\ whose spectral radius are equal to p\, then [ Varga, R.S. (1962)]
Pi fc U
(3.19)
where k > p — 1, c(p) is a constant depending only on p, and Ck~ combinational number. We have oo
oo
2
oo
fc=p—1
2
2
p+1
2
E I^I ^ E II^II M < E c (p){crvr } M 2 . k=p— 1
2
is the
fc=p—1
Infinite Element
131
Method
It is known that p\ < 1, hence the series on the right hand side converges. Therefore oo
oo
oo
fc=0 fc=0 fc=0
converges too. On account of Lemma 44 /
Vu'| 2 dx < +oo. n
1, Applying Lemma 7 we have u' G i?n,* *(fi).
D
We remark that using the Jordan canonical form (3.16) and (3.7) we can get the limit of the solution at the infinity, yoo=Xooy0-T('10)r-1.
3.1.4
Combined
stiffness
matrix
The combined stiffness matrix is the Dirichlet to Neumann operator in discrete form. Substituting (3.7) into (3.5) we obtain (K0 - ATX)y0 T
We define Kz = KQ — A X. role of this matrix.
= /o.
The following theorem indicates the important
Theorem 52. Let Uh be the solution to the problem (3.2), then for an arbitrary v £ S(Q) it holds that a(uh,v)
= ZoKzy0,
yo = B0uh,z0
= B0v.
(3.20)
Proof. We decompose v = v\ + v2 in S(D,), where BoVi = BQV = z 0 , BkV\ = 0, k = 1,2, • • •, and B0V2 = 0, BkV2 = B^v, k = 1,2, • • •. Then V2 S So(n), and by (3.2) we get a(u^,v 2 ) = 0. By (3.3) we have a(uh,v) =a(uh,vi)
= /
Vuh • Vvi dx = (j/
Since the stiffness matrix is symmetric, we have
*»•»>- w")(-VO(i)which gives (3.20).
D
132
Numerical Methods for Exterior
Problems
Therefore Kz is defined as a combined stiffness matrix. We should note that the combined stiffness matrix is different from the total stiffness matrix. At this circumstance, the total stiffness matrix is an infinite by infinite matrix, by which we can get the strain energy arising from any nodal values, but the combined stiffness matrix is only a n x n matrix, by which we can only get the strain energy when the nodal values yo, y\, • • • satisfy the equations (3.4). The procedure from the the total stiffness matrix to the combined stiffness matrix can be viewed as a procedure of elimination. By this procedure, y\, 1/2, • • • are all eliminated. Lemma 51. The eigenvectors of the transfer matrix X associated with the eigenvalue A = 1 are necessarily the null eigenvectors of the combined stiffness matrix Kz. Proof. Let g be such an eigenvector. Since it corresponds to a constant solution Uh, we have
Here I fore I
,
I is a symmetric and semi-positive definite matrix, there-
I is its null eigenvector, that is
-A
K'Q )
\g)
We obtain from the first row that (K0 - AT)g = 0, then by substituting g = Xg into it we get Kzg = (K0 - ATX)g = 0. Theorem 53. Kz is a symmetric and semi-positive definite matrix. Proof. a(-,-) is symmetric, hence by (3.20) we have ZQKzy0 = y%Kzz0 provided v is a solution to the problem (3.2). yo and zo are arbitrary, so Kz is symmetric. Because a(u,u) > 0 for any u € S(Q), Kz is semi-positive definite. D
Infinite Element
133
Method
If the domain ftc is not in star-shape, or the curve To is in complicated shape, we can decompose the exterior domain ft to an unbounded domain with regular shape and a bounded domain with complicated shape, as we have stated previously. Then we can find the combined stiffness matrix for the unbounded domain, which can be regarded as a single element, owing to the conclusion of Theorem 52. Upon assembling the combined stiffness matrix with the stiffness matrices of the other elements, we get an algebraic system of equations. Then the problem can be solved.
3.2
General elements
The pattern of partition in Section 3.1 can be generalized. On the one hand the elements are not restricted to be linear triangular elements, on the other hand each layer may not be divided by rays. We should notice that we have not made any other assumption to the stiffness matrices in the preceding argument except that they are the same for all layers. Therefore the previous argument is applicable for more general partitions. The triangular elements can be of higher order, and quadrilateral elements can also be applied. There are only two restrictions, namely, one to one correspondence of the nodes between Tk-i and Tfc and the geometry of the layer mesh is similar to each other. It should be noticed that there may be some nodes which do not belong to Tfc or Tfc_i. It is needed to eliminate those nodal values from the systems and obtain the combined stiffness matrix for one layer. Let yfe_ j. be a column vector consisting of the interior nodal values, then we get the matrix expression for the general cases: oo (Kn Ki2Ki3\ a(u,v) = £ (Vk-i Vk-i Vk) K2i K22 K23 fc=1
2
\K31K32K33J
f
z
k-i\ zk_,_ = 0,
(3.21)
\ zk )
in terms of the total stiffness matrix, v is arbitrary in So (ft). Let it vanish on all sub-domains except one ftfc for a fixed k. Then
(yl-ivLxvl)
I K22 \zk_h = 0.
134
Numerical Methods for Exterior
Besides, zk_i
is arbitrary. Therefore J/fc-1^12 + yl-1^22
yk_i
Problems
+ VkKz2 = 0.
can be solved explicitly: 2/fc-i = -K^iK^yk-i
+
Kj2yk).
Upon substituting it into (3.21), and rearranging the terms, we can get
«->=£(*.*) (_*r£)(t')This expression is the same as that of the linear triangular elements, except the stiffness matrices may be different. Then all above algorithm can be applied here without any change. .
3.3
Harmonic equation-three dimensional problems
The discussion for the Laplace equation in three space dimension is similar to that for the two dimensional case. Let To be a closed convex surface in a three dimensional space, it is stitched up with space quadrilaterals, and the point O is in the interior of I V Taking £ > 1, we draw the similar surfaces to To with center O and the constants of proportionality £, £ 2 , • • • , £fc, • • •, which are denoted by Ti, I V • • • , Tk, • • • respectively. The domain between Tfc_i and Tk may be divided into many sorts of elements, for instance eight nodes hexahedron isoparametric elements. Let
(K0-AT\ \-A K0 j be the stiffness matrix of the layer between To and T\, where the degree of freedom with respect to yi has already been eliminated, as we have done in Section 3.2. Then the stiffness matrices of the other layers are
« " ( - ! " « ) •
* - ' • » • • • • •
Let K = £,*Ko + £~?Kb, then the equation corresponding to (3.4) is -Ayk-i
+ ^Kyk
- £ATyk+i
= 0.
Infinite Element Method
135
Hence the transfer matrix X satisfies the equation
£ATX2-^KX
+ A = 0.
If we set £ 5 X = Y, then Y satisfies the equation ATY2 - KY + A = 0, the form of which is the same as (3.8). The solutions are not unique, so it is necessary to study the properties of X. Lemma 52. IfuG S(Cl), yk = BkU, then there is a constant C independent of u, such that 1
„
oo
^n
fc=i
oo
fc=o
The proof of it is analogous to that of Lemma 44, thus omitted. Being analogous to two dimensional cases, KQ and K'0 are symmetric and positive definite matrices. Moreover we have Lemma 53. Each eigenvalue A of the transfer matrix X satisfies |A| < £~ i . Proof. Let g be the eigenvector associated with A. We take yo = 9 as the boundary value of the problem (3.2) (three dimensional case), and get 2/1,2/2, • • •, Vk, • • •, then yk = Xkg. Hence oo
oo
fc=l fc=l By Lemma 52 the above series converges. If A ^ 1, then £fc|A|2fc < 1, t h a t is, |A| < £~ 2. If A = 1, then y0 = y1 = •••= yk = •••=: g. We have
Jnk
^4d >ce, \x X
thus lull,* = oo, which is unsuited to present needs. Therefore A = 1 is not an eigenvalue. • By this result all eigenvalues A of the matrix Y satisfy | A| < 1. We define matrices R\, i?2 by (3.9), then we can evaluate the general eigenvalues and eigenvectors of the matrices bundle R\ — \Ri- Finally we get the matrix Y by the approach in Section 3.1.
Numerical Methods for Exterior
136
Problems
Theorem 54. The necessary and sufficient conditions to determine the transfer matrix X are (a) X satisfies the equation ZATX2-^KX
+ A = 0;
(b) The absolute values of all eigenvalues of X are less than £ - 5 .
3.4
Inhomogeneous equations
We study the infinite element method for the Poisson equation - A u = /,
x £fl,
(3.22)
in two space dimension in this section. We assume that / is compactly supported, otherwise the computing procedure can not be terminated in finite steps. The variational formulation for the Dirichlet problem is: For given / G L2{Q) and g G ^^(dfl) find u G F x '*(fi) such that u\an = g, and a(u, v) = (/, v),
Vw G Ho1*(CI).
(3.23)
For simplicity we still use the linear triangular elements and the same partition as that in Section 3.1. The infinite element formulation is: find Uh G S(il) such that Uh\dQ =
= (f,v),
(3-24)
VUGSO(Q).
Being analogous to (3.3) there are vectors fk,gk such that /
Vuh • Vw dx — / /
/ • vdx TX
/
/
N
(3-25)
Let {y£} be a solution of the infinite element scheme to the equation, but it does not necessary satisfy the boundary conditions. Let jjk = Vk — J/)t, then yk+i = Xjjk- We have 2/o =yo + Vo, 2/1=^1 + 2/1= Xy0 + 2/i = X(y0 - y*) + y* = Xy0 + qlt
Infinite Element
Method
Vk = Xyk-i
+qk,
137
Let W
«= E
\\ I \^h\2dx-
2 /iff =fc+ii 1 ^
f f-uhdx\
• / n«
J
= \ylKzyk + ylhk. (3.26) 2
On the other hand
(3.27)
+ (»* ^M-i) ( ^ J
+ Ivk+iK.yk+i + yl+ihk+ij
We take the partial derivatives with respect to yk+i, then let it be zero to obtain (K'0 + Kz)yk+i
- Ayk + gk+i + hk+i = 0.
We solve yk+i from this equation then substitute it into (3.27) and compare (3.27) with (3.26). It is deduced that hk = AT(K'0 + Kzr\gk+i
+ hk+i) + fk+i,
(3.28)
and we have qk = -(K^ + Kzy1(gk
+ hk).
(3.29)
Thus the procedure to solve inhomogeneous equations is: (1) evaluate hk from the recurrence formula (3.28); (2) evaluate qk from the formula (3.29); (3) then get yk. The recurrence is backward in the first step. Since / is compactly supported, hk+x = 0 for k large enough, then it is the initial point.
Numerical Methods for Exterior
138
3.5
Problems
Plane elasticity
We consider plane stress or plane strain here. The Lame equation for the displacement u = (wi,W2)T is -fiAu-(X
+ /i)grad divu = / ,
x G
fi.
(3.30)
The only difference for these two cases lies in the constants. We consider the Dirichlet boundary condition, u = 0,
xedfl,
(3.31)
as an example, and the approach for the Neumann problem is similar. Inhomogeneous problems can be dealt with applying the approach in Section 3.4, so we consider homogeneous problems only in this section. We will work in the spaces V = (.ff1,*(fi)) , and Vo = {u G V;U\QQ = 0}, and the corresponding bilinear form is
f
2
a(u,v) = i 2_, crij(u)€ij{u)
dx 2
• /
^ A div u div v + 2/u Y J eij(u)eij(v) > dx,
/n Ja
i,j = l
where the strains are
and the stresses are <Tij(u) = o-ji{u) = A Y^e^(u)
J Sij + 2fj,eij(u).
The variational formulation of the problem (3.30),(3.31) is: For a given g G Hll2(d£l), find u G V, such that U\QQ = g, and a(u,v)=0,
VVGV0.
(3.32)
For simplicity we consider linear triangular elements here. The infinite element formulation is: Find u G (S(Q))2, such that U\QQ = gi, and a(u, v) = 0,
Vv G (S0(fl))2.
(3.33)
Infinite Element
Method
139
Let the number of nodes on Tfc be n. According to the anticlockwise direction the displacements on the nodes are u\k', u2k,u\k\u2k>''' >uifc >u2fc successively, and they form a (2n)- dimensional column vector y^. Now we can evaluate the stiffness matrix and get
a(u,v) = g (l/ZLi Vl) (K°A ~£, ) ( ^ the form of which is the same as that for the Laplace equation. In parallel with the propositions in Section 3.1, we establish the following lemmas and theorems. The proof of some propositions is the same as the previous one, thus omitted. Lemma 54. KQ and K0 are positive definite matrices. Proof. We take j / o ^ 0 and yk = 0, k = 1, 2, • • •, then the corresponding interpolating function is not a rigid body motion, hence we get a(u, u) > 0. It follows that KQ is a positive definite matrix. The proof for K0 is the same. • Lemma 55. / / A is an eigenvalue of the transfer matrix X and |A| > 1, then A = 1. Lemma 56. The elementary divisor associated with an eigenvalue A = 1 of the transfer matrix X is linear. Lemma 57. As k —> oo, Xkyo —> y ^ for any complex vector yo, where 2/oo = agi + /3g2, where g\ = (1,0,1,0, •• • , 1,0) T , g2 = (0,1,0,1, • • • , 0 , 1 ) T , and a, (3 are constants. Lemma 58. If an eigenvalue A of the transfer matrix X is equal to 1, then the associated eigenvector of it is a linear combination of g\ and g2 • Theorem 55. The necessary and sufficient conditions to determine the transfer matrix X are (a) X satisfies the equation, ATX2 -KX
+ A = 0;
(b) There are two eigenvalues A = 1 of X, the associated elementary divisors are linear, and the corresponding eigenvectors are g\, g2; (c) The absolute values of all other eigenvalues of X are less than 1.
140
Numerical Methods for Exterior
Problems
For three dimensional elastic problems the approach in Section 3.3 can be applied. The algorithm is similar.
3.6
Bi-harmonic equations
According to Section 1.6 the bi-harmonic equations are A2u = /,
(3.34)
and the corresponding bilinear form is a(u,v) = / DAu-
Jn
Avdx
fn _ - l f — — JQ \dx\dx\
d2^ d2v dx\dx2dxidx2
d2ud2v\ dx\dx\) (3.35)
We consider homogeneous equation here. For inhomogeneous the algorithm in Section 3.4 can be applied here without change. Then variational formulation of the Dirichlet problem G H3/2(dty, and g2 € H^2(dn), find u G # 2 ' * ( 0 ) , gi u\,dn = gi,$\aa =92, and a{u,v) = 0,
equations significant is: Given such that
VvGffo'*(°)-
( 3 - 36 )
For definiteness we use the Morley triangular elements here [Ciarlet, P.G. (1978)]. The values of u are given at the vertices and the normal derivatives of u are given at the middle points. Quadratic functions are applied as the interpolation functions and six coefficients are determined uniquely by the six nodal values. Here it should be noticed that at each middle point we must fix one normal direction and all layers must be similar, including the direction. Let the six nodes of an element e* be -X.- , j = 1,2,3, and xff,
1 < j < I < 3. Let the set of all xf> s be S i , and let the
set of all Xjl s be £2- Then the infinite element space is denned as S(Q) = { « £ P2(ei)>Vi; u continuous on S i , Vu continuous on S2, 2 , Iwl2,ei < °°}-
Infinite Element
So(fi) = {uG S{tt);u(x)
141
Method
= 0,Va; £ To n S i , ^ ^ - = 0,Va; e T 0 D E 2 } .
The corresponding bilinear form is a,h(u,v) =2_\ /
7n
vAu-Avdx
\<9a;2 5a;2
dx\dx2dx\dx2
L e m m a 59. Let ||u||s(0) = (J2i Mi.eJ 2 > then ^
dx\dx\)
J5 a norm
over
dx.
SoCty-
Proof. If ||u||s(n) = 0, then u £ Pi(ej) for all i. By the continuity of u and Vu at nodes, u e Pi(fi). Then by the zero boundary condition on To, u = 0. D Let g\j and g2J be the approximation of g\ and <72- The infinite element formulation is: Find UH € S(Q), such that u(a;) = 31/(2;), Vx £ To n S i , ^i=52(x),Va;eronS2,and ah(u,v)=0,
VVGS'O(O).
(3.37)
Theorem 56. The problem (3.37) admits a unique solution. Proof. It is routine to transfer the problem into a problem with homogeneous boundary condition. Applying the Lax-Milgram theorem we need to verify that a/»(-, •) is coercive. Clearly a,h(u,u) is equivalent to ||u||s(n)) thus the conclusion follows. • We turn now to the implementation. We consider the fc-th layer between Tk-i and Tfc. If the number of triangular elements is N, then there are N nodes on Tk-i (or Tk) and inside the A;-th layer there are other iV nodes. Let yk-i (or yk) be a vector with components of A or | ^ at the nodes on Tfc-i (or Tfc), and J/J._I be a vector with components of | ^ at the interior nodes. Notice that we use the value of nr then by (3.35) the total stiffness matrices of all layers are just the same. Upon introducing some matrices, we have the following, which is in the same form as (3.21), 00
ah(u,v)
= ^ fc
=i
(Kn
( j / L i 2/Li Vl)
Ki2
K13\
K21 K22 K2z
\K31 K32K33)
(
z
*>-i\
z fc _i
= 0. (3.38)
\ zk J
Then following the same lines as those of Section 3.2, y ^ - i nated. Then there is a transfer matrix X.
can
be elimi-
142
Numerical Methods for Exterior
Problems
Lemma 60. All eigenvalues of X which are not equal to 1 have a magnitude less than 1. When A = 1, A is a double eigenvalue and its eigenvectors correspond to u = x and u = y. Proof. Let A be an eigenvalue of X with an associated eigenvector g. suppose that |A| > 1, then yk = Xkyo- 2/fc-i a r e then derived, which depend linearly on yk and yk-i- If . /KuK12K13\
Wl = (yl y\ y[)
(yo\
K21 K22 K2Z U i U 0, V ^ i K$2 K33J \ 2/1 /
then ah(uh,uh)
= ^Afc
1
Wi=oo,
fc=i
which is impossible. If W\ — 0, then ||M/I||S(Q) = 0, which implies that Uh = x, or Uh = y, 01 Uh = 1, or their linear combination. It can be seen that Uh = 1 corresponds to A = j < 1 by noting that the ratio of |x| between Tk and Tk-i is £ and A is the variable defined on nodes. If Uh is not the linear combination of x, y, then the magnitude of A has to be less than 1. This completes the proof. D
3.7
Stokes equation
We study two dimensional problems first. As before we consider homogeneous equations only. The system of equations is: - A u + Vp = 0,
(3.39)
V • u = 0,
(3.40)
with the boundary condition, u = g,
xedfl.
The corresponding bilinear forms are: a(u,v) = / Vw : Vvdx,
b(u,p) = (p, VM).
(3.41)
Infinite Element
143
Method
The corresponding variational problem is: For a given g G (tf 1/2 (ft)) ,find u G (H1,*{$!)') and p G L2(fl), such that U\SQ = g, and a(u,v)-b(v,p)=0,
6(u,g) = 0,
toe
(FQ1'*^))2,
VgGL 2 (fi).
(3.42)
(3.43)
The problem is essentially equivalent to (1.64) and (1.67) with a slightly difference, where the equation is inhomogeneous and the boundary condition is homogeneous. In Section 1.7 it has been explained how a problem with inhomogeneous boundary condition can be reduced to a problem with homogeneous boundary condition. Therefore the problem (3.42),(3.43) admits a unique solution. We consider the infinite element method for the problem (3.42),(3.43). We use P2 — Po elements as an example, where u is approximated by piecewise quadratic polynomials and p is approximated by piecewise constants. The stability and convergence proof of this kind of elements for interior problems can be found in [Girault, V. and Raviart, P.A. (1988)]. The infinite element spaces are S(Q) = {ue
(ff x '*(fi)) 2 ; u\ei G (P 2 (e0) 2 ,Vt} ,
M(Q) = {pe L2(n);p\ei
G Po(ei), V*} ,
5„(fl) = { « e S ( f i ) ; U | f f i = 0},
S(rk) =
{u\rk;ues(n)}.
Let gi G S(To) be the approximation of g. The infinite element formulation is: Find Uh G S(fl) and ph G M(Q), such that u/j|en = gi, and a(uh, v) - b(v,ph) = 0, b(uh,q) = 0,
Vv G So(fi),
VgGM(fi).
(3.44) (3.45)
To prove the well-posedness of (3.44),(3.45) we need to verify the inf-sup condition for this problem.
144
Numerical Methods for Exterior
Problems
Lemma 61. There exists a constant (3 > 0 independent ofh such that inf sup -^PL->p. peM(n) lP ^o„ eS . o(n):U ^o IMIi,*l|Pl|o
(3.46)
Proof. By Lemma 20, for any p G M(Cl), there exists u G (H0'*(Cl)) , such that V • u = p, and ||w||i,* < C||p||o with a constant C depending on fi. We apply the Clement interpolation operator Ph [Clement, P. (1975)] on u. P^u satisfies the following: Let e^ be an arbitrary element, and Ei be a "macro element" consisting of e, and all elements neighboring e^, then \U - Phu\ltei
+
ff_1||u
- Phu\\o>et
<
C\U\hEi,
where H is the longest length of diameters of elements in Ei. Let Xj, j = 1, 2,3 be the vertices of e, and Xji, 1 < j < I < 3, be the middle points of the edges Sji. One interpolation u —» w is defined (see [Girault, V. and Raviart, P.A. (1988)]) such that w £ S{Q) and w(Xj) =Phu(Xj),j = 1,2,3,
/ (u-w)ds = 0,1 < j < I < 3. Jsjl
Clearly / V • wdx = I i/ei
w • uds = /
J dei
u • v ds = / V • u ete.
Jdei
J ei
It is proved in [Girault, V. and Raviart, P.A. (1988)] that \PhU~w\1<ei
< C ( / l _ 1 | | u - P f t u | | o , e i + Iw-PfeWll.eJ,
where h is the diameter of e*. Then |« - w|i, ei < C(l + — )|w|l,£ i . The constant C depends on the shape of elements. However the number of elements in one layer is finite, and the elements are similar to each other on different layers, so there is a uniform constant C for all elements. Certainly C depends on the mesh. Being the same j - is bounded by a uniform constant. Hence |u;|i < C|w|i, and b{w,p)
IHIi,*lbllo which gives (3.46).
_
b(u,p)
IIHIi,*INIo
>
1_
c" H
Infinite Element
Method
145
By Theorem 39, we obtain that Theorem 57. The problem (3.44), (3-45) admits a unique solution. We consider the k-th layer flk- If the number of triangular elements is N, then there are N nodes on Tfc_i (or Tfc). Inside the A;-th layer there are other N nodes. The total number of nodes on one layer is 3iV. There are 2N degrees of freedom of the velocity components on Fk-i (or Tk), which are arranged as a 2N dimensional vector yk-i (or yk). There is an analogous 2N dimensional vector t/j._i associated with the interior nodes. The values of pressure in the k-th layer form a N dimensional vector, which is denoted by q. We impose an additional constraint for the implementation of (3.44),(3.45): [ gj-i>ds = 0. Jan
(3.47)
The reason is interpreted as follows. Let fio be a sub-domain consisting of a finite number of elements. We take q = 1 in f2o and q = 0 outside fioi then by (3.45) Jn V • Uhdx = 0, and then by the Green's formula Ian uh • v ds = 0. That is, the total flux on dflo vanishes and the nodal values are constrained by a linear relationship. Therefore the basis functions in V do not have local supports. In order to get a block tri-diagonal Toeplitz matrix like (3.4) the property of local supports of the basis functions is essential. That is the difficulty. On the other hand, if (3.47) is satisfied, then one can see that there is a vector 7 such that 7Tj/o = 0. We take a particular function q e M(fl) at the equation (3.45) as follows: q = 0 on £fcf2, and q = 1 on fi\£fcfi, then we get lTVk = 0.
(3.48)
We normalize 7 to a unit vector, then construct an orthogonal matrix Tr = (7G), and set yk = GTyk, then there is an one to one correspondence between yk and yk- There is no restriction between yk-i and yk, so it is easy to construct stiffness matrices on Qk and get a block tri-diagonal Toeplitz matrix with respect to ykGiven j/o and j/i, satisfying (3.48), we solve the finite element approximation of the Stokes equation on Q\ with boundary data yo,Vi and the given mesh. Let the approximate solution be u/,, then there are matrices
146
KO,KQ,
Numerical Methods for Exterior
Problems
and A, such that
where Zk = GTZk, and Zfc is associated with v S V. The above expression is valid for all layers Qfc, hence the infinite element scheme is deduced to the a system of infinite equations, which takes the same form as (3.4). Then the algorithm is the same as that for the Laplace equation. The transfer matrix X and the combined stiffness matrix Kz can be evaluated. If Jdn g • v ds = 0 then it is easy to get an approximation gi to satisfy Ion 9i • vds = 0. If Jdn g • v ds ^ 0, we can set Mi = u+ — V In |x| / 2TT Jen
g-vds,
then u\ satisfies the same equation and the condition (3.47). The link between the combined elements and other conventional elements is as follows: If the conventional elements occupy a domain Q*, then we can define a(u, v) = I
Vu : Vvdx +
y^Kzzo,
and b(u,p) = /
in*
pV • udx,
where yo = GTyo and ZQ = GTZQ, and To is the boundary of the domain Q. Define spaces S{W) = ( u e {H\n*))2;u\ei M ( S T ) =
€ (P 2 (ei)) 2 ,Vi},
G Po(e<),Vt, f
pdx = 0
Let the union of Q and fi* be Q<x,. The numerical computation for the problem, — A u + Vp = 0, V - u = 0,
x e Hoo,
i6fl(
Infinite Element
u = g,
147
Method
x £ dCloo,
is reduced to a conventional finite element scheme on fi*. The above algorithm can be extended to three dimensional problems and the axial symmetric flow. For the later the system of equations is 1 „,
„
.
1
dp
OC-t
0X~i
V(ziVui) + - j u i + ~
X\
-—V(iiVu2)
= 0,
+ ^ - = 0, OX2
Xi
d . . d , . - — (XiUl) + -z—{XiU2) = 0. 8x\ OX2
The corresponding bilinear forms are a(u, v) =
xi ( Vui • Vvi + V«2 • Vt/2 H
6(w,p) =
P[ g—(xiui)
j - 1
+ -g—{x1u2) J dx.
The corresponding Hilbert spaces are equipped with norms lbllo= I
xlP2(x)dx,
JA
IMI? = jf*i(|V« a (*)| a + ^ ) c f a : 1
For details, see [Ying, L. (1986)]. 3.8
Darwin model
Let us recall some definitions in Section 1.12. ff(curl,div;fi0)
= { « € (L 2 (ft 0 )) 2 ; V • «, V x « e L 2 (fi 0 )},
||«||0,curl,div = ( N l o + IIV • «||§ + || V X U|| 2 ) * ,
148
Numerical Methods for Exterior
Problems
H0c(n) = {UG V; ||C«||o,curi,div < oo, (1 - ()u e (fi"o'* ( n ) ) 2 . u
x
"Ian = 0},
ll«ll* = {||Vx u||§ + ||V • u\\20+ < u • i/, 1 > I f i } i V = {v £ # 0 c (ft); < « • i/, 1 > a n = 0} Let the closure of the quotient space V/VQ with respect to the norm || • ||„ be W. Let Q = {p £ L 2 (fi);suppp CC fi}, equipped with norm
blllt=(lbll§+I^P^|2)a. Then take closure to obtain a Hilbert space Q. If p £ Q and limpn = P,Pn £ Q, then define JQpdx = limjnpndx. The subspace of Q, {p £ Q; Jnpdx = 0} is denoted by Qo. The bilinear forms are d(u, v) = / (V x u) • (V x v) dx + / (V • w)(V • v) dx Jo, Jo. + < u • v, 1 >aa< v-i/,1 >an,
u,v
eW,
and 6(u,p) = / (V • u)pdx,
u £ W,p £ Q 0 .
The problem: Find w £ W, p £ Qo, such that d(u, v) + b{v,P) = f 5 • (V xv)dx, &(«,
Vv £ W,
(3.50)
V?€Q0,
(3.51)
admits a unique solution. We consider the infinite element method for (3.50),(3.51). Being the same as the Stokes problem, Pi — PQ elements are applied for u and p. The mesh restricted on a bounded domain 0 \ £kQ is also a mesh with finite number of elements. In general let fio be a union of element, finite or infinite, we define E(n0) = {uG (C(Th))\u\et
£ (P 2 ( e i )) 2 ,V e i C fio}.
Let D{n\£kfL)
= lu£E(n\£kn);ux
v\ro = 0 , J u-vds
=0
Infinite Element
149
Method
and let u
ll.,n\«*n = ( / \Jn\£kn
\V-u\2dx+
J Jn\?>=n
Lemma 62. || • ||*,n\^fcn *s a norm on D(D. Proof.
\Vxu\2dx >
\£kfl).
If ||u||^tQ\^kQ = 0, then Au = 0, so u is a harmonic function.
u G P2 on individual elements, hence u G P 2 on Q \ £fcfi. We extend u to the interior of To analytically. Then we define the stream function tp, such that u = (Q£-, — gjf-) , then Aip = 0. By the boundary condition of u, -g^lr0 = 0 , so ip is a constant in the interior of To. Therefore u = 0.
•
We return now to the exterior domain Q. Lemma 63. The space Wh(Q) = {u G E(Q); v • u, V x u G L 2 (fi),u x ^|an = 0, < M • v, 1 >afi= 0} is a Hilbert space under the norm \\ • ||*. Proof, because VoC\E(n) = {0}, || • ||* is a norm on Wh(Q). Let us prove it is complete. Let {«„} be a Cauchy sequence with limit u. We are going to prove u G E(Q). {un} is also a Cauchy sequence on D(fl \ £kfl). By Lemma 62, u G D(fl \ £fcfi). Since A; is arbitrary, u G J5(fi). • Lemma 64. Wh{n) c ^ ' " ( f i ) . Proo/. Let u G Wh(fl). We define a function < G C°°(ft), such that £ = 0 near To, C = 1 n e a r the infinity, and 0 < C < 1- We take an arbitrary
Jn
Jn
(CM))
dx + [ (V x
K«|i =
sup
Jn
, ,
v w
< ||c«||„
therefore £u G # o ' * ( n ) > t h a t is, u S H 1 ' * ^ ^ ) for large k. Then by u G F 1 on any bounded domain we have u G i7 1, *(fi). • Lemma 65. The norm || • ||* is equivalent to || • ||i t , on Wh(Q). Proof. It is the direct consequence of the closed graph theorem and Lemma 64. •
150
Numerical Methods for Exterior
Problems
We assume an inhomogeneous boundary condition, u x V\QQ = g, and let gi be an approximation of g, then the formulation of the infinite element method is: Find Uh G E(Q.)nH1'*(Cl), ph G M(fi), such that uh x v\r-0 = gi, Jr Uh • uds = 0, and d(uh,v) + b(v,Ph)=
I B-(Vxv)dx,
V« e W h (n),
(3.52)
Jn b(uh, q) = 0,
Vq G M(fi).
(3.53)
Theorem 58. T7ie problem (3.52), (3.53) admits a unique solution. Proof. d(-, •) is coercive. To verify the inf-sup condition we apply the inf-sup condition of the infinite element method for the Stokes equation, sup ^ ^ v e So(fi) , v | 1 Now Wh(fi) D E(Q) n fl'o'*^). sup
»e^)
* ^ |w|1
> P\\p\\0,
Vp G M(Q).
thus
> /3||P||o,
V^ G M(Q).
Moreover we have ||t>||ij| < 2|u|i, thus sup v€Wh(il)
b IHI
-^Pl>±\\p\\0, * ^
VpGM(fi).
Then the conclusion follows, and the proof is the same as the that in Theorem 40. • The relationship between the solutions of the infinite element method to the Stokes equation and to the Darwin model is the following: Theorem 59. The solution to (3.52),(3.53) is a solution to (3.44),(3.45) with appropriate boundary value and inhomogeneous term. Proof. We take v in (3.52) such that v\r0 = 0, then v G H0'*(n). Let a series {vn} C Co°(fi) tend to v in HQ'*^), then by the Green's formula, d(uh,vn) = a(uh,vn). Letting n - » o o w e verify that Uk,Ph satisfy (3.44),(3.45). •
Infinite Element
151
Method
Now let us go to the algorithm for the Darwin model. For the formulation (3.52),(3.53), we take a boundary value 7/j for the Stokes equation such that 7/j x v = gi, and / r ^h-vds — 0. Then solve the following problem: Find Wh £ S(Q), rh € M(fl), such that Wh\r0 = 7h, a n d a(wh,v) + b(v,rh)=
[ B-{Vxv)dx,
6(«; h ,9) = 0,
V«eS0(fi),
VgeM(fi).
(3. 54) (3.55)
The solutions to (3.52),(3.53) also satisfy a(uh, v) + b(v, rh)=
/ 5 • (V x v) dx,
b{uh,q) = 0,
W e S'o(fi),
V?eM(fi).
(3.56)
(3.57)
Let Uh — Wh = Uh and Ph — fh — Ph, then we have a(Uh,v) + b(v,Ph) = 0,
b(Uh,q) = 0,
VveS0{n),
V9eM(Q).
(3.58)
(3.59)
Let the vectors yk = BkUh and z/t = GTyk, then it holds that Zfc+i = X^fc, which yields yk+1 = G I G r | / , .
(3.60)
Let the function form of (3.60) be Uh\vx = YUh\r0i where F is an operator. So far the problem (3.52),(3.53) is deduced to a finite element approximation on a bounded domain fti for the following system of equations: - A « = VxB, V • u = 0, u x
xefii, i e fii,
Hr 0 = 3 h ,
/ u • v ds = 0, •/r0 (u - w/i)|rx = Y(u - Wh)|r0,
152
Numerical Methods for Exterior
Problems
where Wh is given. We solve the finite element problem on fii with the same mesh, and get the solution Uh to (3.52),(3.53) on Sl\. Then we use the formula (3.60) and the solution Wh to (3.54),(3.55) to get the solution Uh on the entire domain Q. Finally we need to show that (3.52),(3.53) with homogeneous boundary condition is an approximation to (3.50),(3.51). If the solution to (3.50),(3.51) satisfies u G i7 1, *(fi), then we can modify the formulation to: Find u G # 1 , *(fi) nW,pG M(fi), such that d{u,v) + b(v,p)=
[ B-(V
Jo,
xv)dx,
b(u,q) = 0,
W€#1,*(fi)nW)
Vg G M(fi).
(3.61)
(3.62)
This is because the solution u satisfies V • u = 0, b(u,p) = 0 for all p G L 2 (fi)., Because H1'*^) n W is dense in W, uniqueness holds for the problem (3.61),(3.62). The spaces in (3.52),(3.53) are subspaces of those in (3.61),(3.62), so the solutions to (3.52),(3.53) are indeed approximate solutions.
3.9 3.9.1
Elliptic equations with variable coefficients A homogeneous
equation
Let us consider an equation, V • (fc(z)Vu) = 0
(3.63)
on the domain f2. Let A > 1 be a parameter. We make a similarity transformation Ax —> x. Under the transformation the equation (3.63) becomes V • (Jfc(Ax)Vu) = 0,
(3.64)
which is still considered on Q. The bilinear form associated with (3.64) is thus defined by a\(u,v)
= / k(\x)VuVvdx.
(3.65)
We assume that (3.63) is elliptic, that is, there is a constant fco > 0 such that k(x) > k0 for all x G fi. In addition we assume that k(x) is continuous
Infinite Element
153
Method
and approaches to a constant as x —> oo. It has no harm in assuming Hindoo k(x) = 1. We still use linear triangular elements for the infinite element mesh. The total stiffness matrix on f2i is defined by
(K0-AT\ \-A K'Q ) ' that is, k(Xx)Vw • Vv dx = ( y% yj) (K0
/
\-A
JO.!
-A'\
K'Q
z0\
\zj>
where u,v £ S(Q), and the space S(Q) is defined in Section 3.1. The infinite formulation for the equation (3.64) is: for a given gi G S(To), find UH G S(Q), such that ax(uh, v) = 0,
Vv G So(fi).
(3.66)
The problem (3.66) admits a unique solution. Then a combined stiffness matrix KZ(X) is defined by y^Kz(X)zo = a\(v,h,v). Kz{\) is real and symmetric. The transfer matrix X(X) is defined by y\ = X(X)yo. Let the domain Q be regarded as a union of Q\ and £fi. The bilinear form a\ can also be written as ax(uh,v)
= (yl $X{XF)
(f° ^
)
(*)
+!#X(A)3X(AO*1
Since z\ is arbitrary, we have yl{-A
+ X{X)TK'0 + X(X)KZ(X£))
= 0.
yo is arbitrary too, so X(X) = (K^ +
Kz(X0)-1A.
On the other hand, being analogous to Theorem 52 it holds that KZ{X) =K0-
ATX{X).
Therefore KZ(X) = K0-
AT{K'Q + tf^AO)-1 A
(3.67)
154
Numerical Methods for Exterior
Problems
To solve an equation with variable coefficients like (3.64), it is convenient to consider the critical case £ = 1 + . We set y\ = j/o + (£ — 1)771, then
K0 -AT\ fy0 -A K'Q ) U K0 - A - AT + K'Q - ( £ - 1)(AT - K'Q) \(y0 -{£-\){A-K'0) (d-l)2K>0 ){m
L0 -#r\ fy0 -B L'0 ) W By (3.67) we get
KZ(X) = Kz(XO + L0 '
BT
- Kz(XO) (—^-,
+ Kz(XO)
(-^-
- Kz(XO
Dividing it by £ — 1, letting £ —> 1 and defining LQ
= lim - — - , £-i£-l'
B = lim - — - , C-U-l'
L'0 = lim ° ° C-U-l'
to obtain A
^A^
=
"L° +
(i?T
" K*(X))(L'o)~l(B ~ K*W)'
(3-68)
which is the ordinary differential equation for KZ(X). Letting A —> 00, the equation (3.64) approaches to the Laplace equation, Aw = 0 asymptotically. We denote by KZQ the combined stiffness matrix associated with the Laplace equation, which is the initial data of (3.68) at A = 00. If we know more properties of the function k{x) near |a;| = 00 we can get more accurate asymptotic expression for KZ(X). Let us assume that k{x) = l + ^ + o ( ± ) , \x\ \\x\J
0= tan-^, X!
then t(A
,)
= 1 +
^ )
A
- .+
0
(^
Infinite Element
Method
155
accordingly we have L0 = Loo + LoiX'1 + o(A~1), L'0 = L'00 + B^Bo
L'01X~l+o(X~l), + oiX-1).
+ BxX^
Formally we set KZ(X) = Kzo + KziX~l we obtain a linear equation for Kz±: Kzl - K^L'^iBo
and substitute it into (3.68), then
=Loi - (5 0 T - K^iL'wr'L'M^iBo - B^{L'mr\Bo
Kzo){L'o0rlKzl
- Kz0) - (B% -
- Kz0) T
(3.69)
1
- Kzo) - (5 0 - ^ 0 ) ( ^ o ) - B i .
We can take a constant f slightly larger than 1, then get an approximation of Kzo- Using (3.69) we get Kz\. KZO + KZ\XQ1 is taken as an approximation to Kz{Xo) for a large Ao. Then we solve (3.68) backward to get KZ{X) for all A G [1, Ao]. KZ{1) is our desired combined stiffness matrix for the equation (3.63). 3.9.2
An inhomogeneous
equation
Next let us consider an equation, - V • (Jfc(a;)V«) = f{x).
(3.70) 2
To guarantee well-posedness it is assumed that /(x)|a;| log \x\ is bounded for large \x\. Under the transformation Arc —• x it becomes - V • {k(\x)Vu)
= X2f(Xx).
Fx(v) = X2 [ Ja
f(Xx)vdx,
(3.71)
We define
and J\(u,v)
= a\(u,v)
-F\(v).
The infinite element formulation of the Dirichlet boundary value problem to (3.71) is: Find Uh G S(fi) for given gi G S(To) such that uh\ro = 91 and Jx(uh,v)=0,
Vv€S0(fl).
(3.72)
156
Numerical Methods for Exterior
Problems
If vi,v2 € S(fl) and u i | r o = v 2 |r 0 = 2 o, then v = vi — v2 € S0(fl). (3.72) implies that J\(u,vi) = J\(u,V2). Therefore as a functional on the space S(fi), J\{u,-) depends on zQ only. We write J\ as the sum of a bilinear part and a linear part: Jx(u,v)
= y%Kz(X)z0 - H{X)z0,
(3.73)
where H(X) is a row vector. Because (3.73) is valid for all homogeneous equations, KZ(X) is just the combined stiffness matrix associated with (3.64). Let us derive the equation for H(X). We divide the domain fi into two parts: fii and £fi for given £ > 1, then we have y%Kz(X)z0 - H(X)z0 = y?X(A£)zi
+ (vi vi) ( 5 l ~KI) ( Z ) ~ H(x^Zl' A2 £ f{Xx)vdxThe last term depends on (
) linearly. We define
($o(A) $i(A)) (Z°)
= A2 J
f(Xx)vdx.
Noting (3.67) we get y%(K0 - AT(K'Q + Kz(XO)-1A)z0
+
- H(X)z0 =
y^K^X^
^^)(-rO(*)- F(A0 * i - ( *° (A) * i(A)) (*)(3.74)
Since Uh is the solution, (3.74) holds for all z\, we have yjKz(XO
- yTQA + y?K'0 - H(\Q - $j(A) = 0,
which gives yi
= {K + K^XOr'iAyo
+ HT(X£) + *f(A)).
(3.75)
Infinite Element
157
Method
We substitute (3.75) into (3.74) to obtain -H(X)z0
= - H(\£)(K'0
+
K^XOT'Azo
- *i(A)(tf£ + K^XOT'Azo
- $o(A)z 0 .
Since ZQ is arbitrary, it holds that -H(X)
= -(H(XO
+ *i(\))(K'0
+ K^XOr'A
- $ 0 (A).
We set z\ = ZQ + (£ — l)Ci then (*o(A) *i(A)) ( j )
= ($ 0 (A) + ^ ( A ) ( ^ - 1 ) $ ! ( A ) ) (*?
We define *o(A) - *„(A) + $!(A),
*i(A) = (£ - l)*i(A),
then it holds that
ff (A) = H(\£) +
*i(A)\
Z-l
^+177^+^(^)1 f-^T-^(AoU+*o v ^_1)2 *v ^ y ^ _ ! "yJ (
*'
C-l (3.76)
Let V>o (A) = lim ^ ,
Vi (A) = lim 7JZTJ2 •
We divide (3.76) by £ — 1 and let £ —• 1 to obtain A
^ T
=
-HWiL'^iB
- KZ(X)) - Vo(A),
which is the ordinary differential equation for H(X). If f(x) = / o (0)|z|- 3 +o(|:r|- 3 ), then Vo(A) = ^oX*1+o{X~1). we set H(X) = H0 + HiX'1, then by (3.77) we have
(3.77)
Formally
-H0(L'oo)-1(B0-KzO)=0, -Hi = - H0(L'00)~
L'0l(L'00)~ 1
(B0 -
Kz0)
- F 1 ( ^ 0 ) - ( B o - KzQ) - H0{L'M)-l{Bi
- KzX) - V>oo,
158
Numerical Methods for Exterior
Problems
which gives
I \-lfD. ri(B V-K M iPoo{I-(L' oo 0 z0)}
H1 = 3.9.3
General
multiply
connected
l
domains
Let Dj c Q, j = 1, • • • , N, where Dj are simply-connected domains with boundary jj. We assume that Dj are disjoint to each other and set do = n \ U j = i A/> thenfio is a multiply connected domain. Let us consider the equation (3.70) on fioFirst of all we consider the Neumann boundary condition: du
fa
= 0,
j = l,;N.
(3.78)
1i
The boundary condition on To is flexible as usual. If the boundaries jj are suitable regular, say, satisfying the Lipschitz condition, then functions in Hloc(Q0) can be extended into Dj , still in H1. Therefore if we assume w
"\o,
xen\n0,
to-{™'*
efi0,
and consider the equation - V • (fc(x)Vu) = f(x) on Q, then it is equivalent to (3.70) on fio- We define a seminorm Hull!,. = ([ k(x) (\Vu(x)\2 \Jn V
+
f{x) ) dx) |z| log \x\J J 2
on tf1 •*(£}) and let T = {uG/r 1 '*(n);||u||i,» = 0} )
(3.79)
Infinite Element
Method
159
then for a given function UQ on To we consider the variational problem: find 1, u e i7 *(fi)/T, such that u|r 0 = gi, and i>*/ VveHo'*{Q).
Jx(u,v)=0,
The problem is thus defined on fi, and the approaches in the previous section can be applied here. We provide more details about the mathematical formulations in the next subsection. Secondly we consider the Dirichlet boundary conditions u|7i=0,
j = l,---,N.
(3.80)
We use penalty method and consider the problem: find u £ such that u|r 0 = gi and
H1'*(Q)/T
N
f k{x)Vu -VvdxJn
f f(x)vdx Jn
+ ~ ^ / uvds = 0, e j=i I'J
Vv £ #o'*(fi),
where e > 0 is a small constant. We denote by A - 1 7j the image of 7,- under the transformation Xx —> x, then define N
J\(u,v)
2
= / fc(Aa;)Vw • Vvdx — A / f(Xx)vdx Jn Jn
-\—YJ / £ ~[J(\-i-iJ)r\n
uvds,
(3.81) where we notice that (A - 1 7,) fl ft may not be a closed curve. Then we use the approach in the previous section to solve this problem. Concerning the last term of (3.81) the scheme to evaluate KZ{X) should be modified. Let us consider one domain Dj as an example. Let T\ be the similar figures of To with the center O and the constant of proportionality A > 1. We assume that 7j consists of some parts, some of which lie in Tx1 and r^ 2 , and some are transversal to T\ (Fig. 5). The matrix KZ(X) is discontinuous at A = Aj and A2: y g X ( A 2 - 0)z0 = j / 2 X ( A 2 + 0)z 0 + — f uvds, £ ./(AjSiJnro
j/gX(Ai - 0)^o = s/2X(Ai + 0)z0 + ^
£
/
uvds.
1 (A-'T^nro Jixr
The other parts of jj have some contribution to the matrix LQ. Let us consider "element stiffness matrices".
160
Numerical Methods for Exterior
Fig. 5
rAl
r\2
Problems
°
Fig. 6
Suppose 1 — 2 — 3 — 4 is a combination of two elements, the intersection of A _1 7j with it is approximated by a line segment 7' with an oblique angle #.(Fig 6) 7'- divides the line segment 1 — 2 into two parts with lengths li, I2 respectively. Let
(^3 and y>4 have no contribution to the stiffness matrices, and the length of 7J is r(£ — l ) / s i n # , therefore
lim
1 / vf „, _ hra _i_ f fc y ^ >
€-»i € - 1 /y;.
€ - i £ - l V'l+^y
=
sine
' f_A_y. sinflVji+W
The other entries can be evaluated in the same way. We get
esinfl I hh \ (h+h)2
( h \2 I ' \h+h] )
We assemble all matrices K of one layer then add the sum to Lo in the equation (3.68).
Infinite Element
3.9.4
Transfer
161
Method
matrices
To solve a boundary value problem after getting Kz{\) matrices X(Ao,A) and Xi(Ao,A) by yx=X(\o,X)yx0+X1(X0,\),
we define transfer
A > A0,
(3.82)
where y\ = w|rx- If Ux0 is given, then we regard it as a boundary data. By the well-posedness of the problem y\ is determined uniquely, hence the matrix X and X\ are well defined. By some calculation we get from (3.75) that
* = < 7 + ((^52 + *«(*>) (^k - K>w1 > y° L
'°
+KZ(XO)
1
(HT(XO + *I{X)
Therefore lim ^ | ^
= ( L i ) - 1 ^ - ^ ( A ) ) y 0 + (Lj,)" 1 ff r (A).
We define an infinitesimal transfer matrix Y(X) = (L'0)~1(B — KZ(X)) and an infinitesimal transfer vector Y\(X) = (L'Q)"1 HT(X). We then take Ao > 1, AAo > 0, and make a similarity transformation Xox —» x, then (Ao + AA0):r -> (1 + ^)x. Let £ = 1 + ^ , then approximately 2AO+AAO
= yx0 + ^-l){Y(X0)yXo+Y1{X0))
= yXo +
—l(Y(X0)yx0+Y1(X0)),
and yx=X(X0,X)yx0+X1(X0,X) =X(X0 + AA0, X)yx0+AX0 + *i(Ao + AA0, A) =X(X0 + AAo, A) (yXo + ^(Y(X0)yXo
+
^(AQ)))
+ Xi(A 0 + AA0, A).
Since yx0 is arbitrary, we get approximately X(X0, A) = X(A 0 + AAo, A) (i + \
*i(Ao, A ) = ^X(X0
^Y(X0) A0
+ AAo, A)yi(Ao) + XX(A0 + AA0, A).
Dividing them by AAo and letting AAo —> 0 we get -£-X(X0, CAo
X) = -A 0 - 1 X(A 0 , A)Y(Ao),
(3.83)
162
Numerical Methods for Exterior
Problems
— X i ( A 0 , A ) = -Ao 1 X(A 0 ,A)y 1 (A 0 ).
(3.84)
We have initial value X(X,X) = I,
X1(X,X) = 0,
so we can solve (3.83),(3.84) to get X and Xi, then get the solution from a given boundary condition. 3.10
Convergence
All schemes in this chapter can be proved convergent. We investigate one of them for example. Let us consider the Poisson equation, -Aw = /,
ie(l,
(3.85)
with homogeneous boundary condition, u\dn = 0,
(3.86)
where dim (Cl) = 2. We assume that / G Hm~1(Cl) for a positive integer m, and / is compactly supported. Moreover we assume that u G H^l(Cl). The partition of the domain CI is described in Section 3.1. Let h be a reference length of the mesh, for example we can set h to be the maximum length of the diameters of all elements in fii. Some assumptions are made on the mesh: (A) The partition is regular, that is the interior angles of all elements possess a common lower bound 8o > 0. (B) O $ Si, Vi. And let d(0,ei) be the distance from e, to the point O, then there is a constant x, such that meas e, < \h2(d(0,
e,)) 2 ,
Vi.
For instance the similar partition given in Section 3.1 satisfies the condition (B). Let m-th order elements are employed in the scheme. Then the infinite element subspaces are S(Cl) = { « £ H1'*^);
M * G P m (ei), Vt},
and S0(Cl) = {u G S(Cl);u\an
= 0}.
Infinite Element
Method
163
The infinite element formulation of the problem is: Find Uh £ that a(uh,v)
= (f,v),
such
SQ(Q,),
VveS0(Q).
(3.87)
By the expressions of the solutions of the Laplace equation in Section 1.1, we find the weighted semi-norm |u|m+l(m) "
/n J is bounded, where D u denotes the (m + l)-th order derivatives of u. Let II be the interpolating operator with respect to S{Q), then we have m+1
Theorem 60. If the conditions (A),(B) hold, u G #,7 c +1 (f2), \u\m+i{m) < oo, then we have the interpolating estimate, |u-nu|i
U
f j^(u(x)-Uu(x))2dx\2
and (3.89)
< Chm+l\u\m+1{m).
(3.90)
Proof. By the embedding theorem of Sobolev spaces, u is continuous, so II u makes sense. According to the interpolating estimate on triangular elements [Ciarlet, P.G. (1978)], |u-n«|.,e.
s = 0,l,
(3.91)
where hi is the diameter of e,. By the condition (A) we have meas e, > h\ sin 3 9Q COS 9Q , then by the condition (A),(B) we have 2
> 2 (d(Q,e,)) 2 sin t'ocosfc'o
Substituting it into (3.91) we get \U - Uu\s,ei < Chm+1-S(d(0, We have d(0,ei)
ei))
m+1 S
- |«U+1,e,.
< \x\ on the triangle e,, therefore
(d(0,ei))m+1-s\u\m+1,ez
< (^O.ei))1-^!^!^),^.
We obtain
(d(o, ei )) s -V-nu| S;ei <
chm+l-s\u\m+l{mUi.
Upon summing them up with respect to i we get (3.89) and (3.90).
D
164
Numerical Methods for Exterior
Problems
We turn to estimate the error u — it/,. Theorem 61. / / the conditions (A),(B) hold, and the solution u to the problem (3.85),(3.86) satisfies |u| m +i( m ) < oo, then |«-Ufc|i
(3.92)
The proof is routine. We take an arbitrary v € S(£l), then a(u — Uh,v) = 0.
Thus a(u — Uh,u — Uh) = a(u — ith,u — Uu), which gives \u - Uh\f <\u-
uh\\\u~
Ylu\i.
(3.92) follows by applying Theorem 60.
•
Thus the error estimates are optimal. Using Nitsche's trick, we can prove L 2 -norm estimate. However, because the exterior domain Q can not be convex, and the boundary dfi, is not smooth, the result is not optimal. There are some approaches to deal with reentrant corners, but it is another story. Let R > 0 be large enough, such that B(0,R) D flc. We have the theorem: Theorem 62. Under the assumptions of Theorem 61 it holds that II" - Uh\\o,ar\B(o,R) < Chs\u - uh\i,
(3.93)
where 0 < s < 1, depending on Q, and the constant C depends on Q, and R. Proof.
We set ( u - Uh, \x\ < R, ;
[0,
ja;j>i2,
and consider an auxiliary problem: find ip S H0'*(Q), such that a(v,
veH^*(Q).
Let us estimate ip first. By Theorem 14
IMIi,.
(3.94)
Infinite Element
165
Method
If xo is a corner point on dfl, we consider two small domains Q^ = Q fl B(x0,S), fl2s = O n B(x0,2S) with S > 0. Then by the separation of variables, the solution (p in Q-28 can be expressed explicitly, and it can be estimated that IMIi+ s ,n 4 < C(||/|| 0 ,n, + IM|o,n2J> where 0 < s < 1, depending on the angle. Let f2i = fl n B(0, R + 1), and fl2 = {z;]-Xl > # + ! } > then using the interior estimate of elliptic equations [ Gilbarg, D. and Trudinger, N.S. (1977)], we obtain
IMIi+.,n 1
Jn The second inequality is due to the embedding theorem of Sobolev spaces. Thus we obtain
!MI1+..0! +
(J
\x\2\D^{x)\2 dx\ ' < C||/||o.
Letting v = u — Uh in (3.94) and noting that a(u - uh,Uf)
= 0,
we get \u- uh\i\
=
(f,u-Uh).
Using the normal estimate for interpolating operators on fii and Theorem 60 on Q2> we have
lv-n v |i
we
obtain finally
H/llo^Cfc'lu-Ufclx, and the proof is complete. Other kinds of error estimates can be found in [Ying, L. (1995)].
HI
Chapter 4
Artificial Boundary Conditions
Some artificial boundary conditions, exact or approximate, have been given in the previous chapters, for example, the application of "Dirichlet to Neumann" operator or the combined stiffness matrix. In this chapter we will investigate more artificial boundary conditions. A pseudo-differential operator is introduced for the wave equation to reduce the boundary condition at the infinity to an interface. Since this condition is global in space, some works have been done to localize the boundary condition on the analytic level. The problem of stability for some approximations is still open, because there are some higher order derivatives in the conditions. In this chapter we will also show some works on the artificial boundary conditions of the Navier-Stokes equations.
4.1
Absorbing boundary conditions
Absorbing boundary conditions are applied for wave problems. Let Q be an exterior domain, and let CIQ be a bounded domain such that do ^> ^ c The truncated domain is fix = fl D fio and dilo is the artificial boundary. Suppose there is no incoming waves from the exterior of CIQ. The artificial boundary condition on dflo is denned to absorb all outgoing waves from Qi, so that the waves do not reflect from dflo. It is named as an "absorbing boundary condition". We start from one dimensional case and consider the wave equation,
^
= ^ .
at2
dx2 167
(41) {
'
168
Numerical Methods for Exterior
Problems
Suppose the domain is (0, oo), and the initial and boundary conditions are du ^7
u\t=o = wo,
ui,
x > 0,
(4.2)
ot w|x=o = 0,
t > 0,
(4.3)
where M0(0) = 0 , m(0) = 0 . The equation (4.1) can be factorized to
, d !:l«-o\dt - ! : dxJ ) ( \dtl + dx
<«)
It yields two equations
5-S-°.
<"'
du
,, „. (4 6)
and du
+
m o-x=°-
-
Obviously the solutions to (4.5),(4.6) are solutions to (4.1). Conversely, if we make a transform of variables £ = x — t and r\ = x +1, then the equation (4.1) is transformed to d2u _ Q d£dr] and the general solution is u = /(£) + g(rj) = f(x — t) + g(x + t), where / , g are arbitrary functions. Therefore the general solution to (4.1) is the linear combination of the solutions to (4.5) and (4.6). Using the expression of general solutions one can get the solution to the initial-boundary value problem (4.1),(4.2),(4.3): \{ua(x -t) + u0(x + *)) + ! / * + / u i ( 0 d£ | ( - u 0 ( * -x)+
u0(x + *)) + § / t ! i * « i ( 0 C i < *.
If wo and u\ vanish for x > a, then on the domain {(x,t);x solution is ,,,„. ts _ / 5"o(z - 0 + | C t MO « , 1
x>t,
> a} the
a: > *,
' j " \ -§«o(t - a:) + I /;_, «i(0 d6, x < t.
Thus only outgoing waves appear in this domain. The solution u satisfies (4.5) on the line x = a. (4.5) is the exact absorbing boundary condition
Artificial Boundary
169
Conditions
on x = a. With this boundary condition the exterior problem is reduced to an initial boundary value problem on the interval (0,a), and the DtN (Dirichlet to Neumann) operator is thus K = J^. We turn now to investigate two dimensional problems and consider the equation = A«-
W
(4-7)
Let the plane x\ — a be the artificial boundary, and there is no incoming waves from xi > a. Like (4.4), formally we can make the factorization:
/
V
d dxi
e2 d
Qx\
0.
di\ i - aF
(4.8)
Then the absorbing boundary condition at x = a is o
I
~&*~\ (4.9)
and the DtN operator is K
a2
d_ dt\
82
aF In order to interpret (4.9) precisely, we introduce some basic definitions of pseudo-differential operators briefly.[Taylor, M. (1974)] The Fourier transform is defined on the spaces: S(Rn) = {
lim xaD^if(x)
= 0,Va,/3},
|x|—>oo
where a, (3 are multiple indices, and C£°(R n ) C <S(R") C C°°(K n ). The dual space of C£° (E"), «S(Rn), and C°°(E n ) are denote by X>'(E