Selçuk J. Appl. Math. Selçuk Journal of Vol. 6. No. 1. pp. 3-12, 2005 Applied Mathematics
On Compact Integral Equation for the Acoustic Scattering Problem in a Penetrable Inhomogeneous Medium
Adil M¬s¬r
Gazi University, Faculty of Arts and Sciences, Department of Mathematics 06500 Teknikokullar Ankara, Turkey;
e-mail:adilm @ gazi.edu.tr
Received: January 12, 2005
Summary.In this paper, the scattering problem corresponding to time-harmonic acoustic waves in an inhomogeneous object in which %i(x) and ki(x)
continu-ously di¤erentiable density and wave numbers respectively, has been investi-gated. First, the scattering problem has been reduced to an integral equation. Then, compactness of the integral operator corresponding to this integral equa-tion has been shown under some restricequa-tion.
Key words: Scattering theory, acoustic waves, transmission problem 1. Introduction
Recently, acoustic scattering problem in an inhomogeneous medium has been investigated by a lot of mathematicians [1,3,4,5,7]. Colton and Wendland [5], Colton [1], Colton and Monk [3], Kleinman and Martin [7] have investigated acoustic scattering problem with di¤erent boundary conditions. In this paper, we are interested in such a problem.
Let S be a simply connected closed piecewise Lyapuno¤ surface in R3 [9].
Let Vibe the region interior to S (Vi= Vi[ S) and Vebe the region exterior to
S (Ve= Ve[ S), where the bar denotes the closure of the domain.
Erect a coordinate system with the orijin in Vi. The unit vector normal
to the surface at x will be denoted by n, and will always be taken as directed from S into Ve. Di¤erentation in the direction of n will be denoted by
@ @n. Furthermore, u+(x) and @u
+(x)
@n will denote the limiting values of u and @u @n as x ! x 2 S from Ve and similarly u (x) and
@u (x)
@n will denote the limiting values of u and @u
Let %; k : R3! C be continuously di¤erentiable local density and local wave number functions such that when x 2 Vi %(x) = %i(x) and k(x) = ki(x), when
x 2 Ve %(x) = %e and k(x) = ke, where %e and ke are complex constants.
The problem we consider is that of …nding the total …eld at all points in R3 given a time-harmonic incident …eld ui(x). Time dependence exp( iwt) is
assumed. It is convenient to represent the total …eld u(x) as a sum of incident and scattered …elds for x 2 Ve, i.e., u(x) = ui(x) + us(x), x 2 Ve and the
problem then is that of determining u(x) such that, u(x) 2 C2(V
i) \ C1(Vi), us(x) 2 C2(V e) \ C1(Ve) (1) M +k2(x) u(x) = 0; x 2 R3 (2) M +k2(x) us(x) = 0; x 2 Ve (3) %eu+(x) %i(x)u (x) = 0; x 2 S (4) @u +(x) @n @u (x) @n = 0; x 2 S (5) lim r!1r @us(x) @r ikeu s(x) = 0 in all directions
whereM is laplacian operator, i =p 1 and r = jxj. Theorem 1 (uniqueness theorem) Let Im(ke) > 0 and
(6) m(x) = ke
%e%i(x) If
(7) Im(m(x)) < 0; Im k2
i(x)m(x) 0; Vi
and let u(x) 2 C2(V
i) \ C1(Vi) be such that (8) Im 8 > < > :ke Z Vi u(x)r v(x) ke :ru(x)dx 9 > = > ; 0
then the scattering problem (1)-(5) has at most solution, where the bar denotes the complex conjugate of a function.
Proof. Assume that the function u1(x) and u2(x) are solutions of the problems
(1)-(5). Set w(x) = u1(x) u2(x). Then w satis…es the problem:
(9) 4 + k2(x) w(x) = 0 , x 2 R3 (10) 4 + ke2(x) ws(x) = 0 , x 2 Ve (11) %ew+(x) %i(x)w (x) = 0, x 2 S (12) @w +(x) @n @w (x) @n = 0, x 2 S (13) lim r!1 @ws(x) @r ikew s(x) = 0 in all directions
Let R= x 2 R3: jxj = R and assume that for su¢ ciently large R, S lies in
the sphere R. Let = x 2 R3: jxj = R nVi. From Green’s …rst identity, we
get Z w(x)M w(x) + jrw(x)j2 dx = Z S w+(y)@w +(y) @n dS(y) + (14) Z R w(y)@w +(y) @n dS(y) If we use conditions (11) and (12) in equation (14), we obtain
Z R w(y)@w(y) @R dS(y) = Z S %i(x) %e w (y)@w (y) @n dS(y) (15) + Z k2 ejw(x)j 2 dx Z jrw(x)j2dx
¬f we use Green’s …rst identy for the …rst integral of the right hand side of equation (15), we obtain
Z R w(y)@w(y) @R dS(y) = Z k2 ejwj 2 dx Z % i(x) %e jrwj 2 dx (16) Z Vi ki2(x)%i(x) %e jwj 2 dx + Z Vi %i(x) %e jrwj 2 dx + Z Vi w(x)r %i(x)% e :rw(x)dx
Multiplying equation (16) by keand taking its imaginary part use the fact that
lim R!1Im 0 @ke Z R w(y)@w(y) @R dS(y) 1 A = 0
( Because of Im ke> 0 and w decays exponentially) we get for R ! 1
(17) Im keke2 Z Ve jw(x)j2dx Im(ke) Z Ve jrw(x)j2dx Z Vi Im(ki(x)m(x)) jw(x)j2dx + Z Vi Im(m(x)) jrw(x)j2dx+ Im 0 @ke Z Vi w(x)r m(x)ke rw(x)dx 1 A = 0
Since Im(ke) > 0 it follows Im keke2 < 0 and if we use the conditions (7) and
(8) in equation (17) we get w 0: 2. Integral Representation Let (18) G(x; y) = e ik(x)jx yj jx yj ; x 6= y
and x(S) be the solid angle subtended by the surface S at the point x [8]. Then
(19) 1 4 R S us+(y)@G(x;y) @n(y) G(x; y) @us+(y) @n dS(y) = 8 > < > : us(x) ; x 2 V e x(S10) 4 u s+(x) ; x 2 S 0 ; x 2 Vi and (20) 1 4 R S ui+
(y)@G(x;y)@n(y) G(x; y)@ui+@n(y) dS(y)
= 8 > < > : ui(x) ; x 2 V e x(S01) 4 u i+ (x) ; x 2 S 0 ; x 2 Vi where S0
1= fy 2 Ve: jx yj = 1 ; x 2 Sg. Adding (19) and (20), we obtain
(21) 1 4 R S u+(y)@G(x;y) @n(y) G(x; y) @u+(y) @n dS(y) = 8 > < > : u(x) ; x 2 Ve x(S10) 4 u +(x) ; x 2 S 0 ; x 2 Vi
If we use (3) and (4) in equation (21), we get
(22) 1 4 R S %i(x) %e u (x) @G(x;y) @n(y) G(x; y) @u @n (y) dS(y) = u(x) ; x 2 Ve (x)u (x) ; x 2 S where (23) (x) = x(S10) 4 %e %i(x)
If we apply the divergence theorem o the left hand side of equation (22), we get
(24) 1 4 ke R Vi m(x)M G(x; y)u(x)dx +4 k1 e R Vi (rm(x)) (rG(x; y)) u(x)dx + 1 4 R Vi k2 i(x)G(x; y)u(x)dx +41 R Vi (rG(x; y) ru(x)) m(x)ke 1 dx = u(x) ; x 2 Ve (x)u (x) ; x 2 S
Hence, we have the following theorem.
Theorem 2 Let m(x); G(x; y) and (x) be given as (6), (18) and (23) respec-tively. Then we have integral representation (24) for u(x).
Now, introduce the integral operators:
L1: Cp1 Vi ! C Vi and Lj : C Vi ! C Vi ; j = 2; 3; 4 de…ned by (25) (L1f )(x) = 1 4 Z Vi (y)rG(x; y) rf(y)dy (26) (L2f )(x) = 1 4 ke Z Vi m(y)M G(x; y)f(y)dy (27) (L3f )(x) = 1 4 ke Z Vi
[rm(y) rG(x; y)] f(y)dy
and (28) (L4f )(x) = 1 4 Z Vi
k2i(y)G(x; y)f (y)dy
where C1
p Vi = f 2 C1 Vi : krfk p kfk , k k denotes the supremum norm
and
(29) (x) = m(x)
ke
1
Thus the integral equation (24) can be writen in the following operator form. (30) (L1u)(x) + (L2u)(x) + (L3u)(x) + (L4u)(x)= u(x); x 2 Ve
(x)u (x); x 2 S
Lemma 1 L1 is a uniform continuous operator on Cp1(Vi).
Proof. Let x12 Vi and x = x1+ h such that 0 < jhj <2:
j(L1f )(x1+ h) ((L1f )(x1)j p k k kfk 4 Z Vi jryG(x1+ h; y) ryG(x1; y)j dy if we de…ne A(t; y) by
(31) A(t; y) = eik(y)jt yj i jt yj rk(y) +(t y) jt yj(1 ik(y)) then (32) j(L1f )(x1+h) (L1f )(x1)j pk kkf k 4 R Vi A(x1+h;y) jx1+h yj2 A(x1;y) jx1 yj2 dy pk kkf k 4 ( R jx1 yj< A(x1+h;y) jx1+h yj2 A(x1;y) jx1 yj2 dy +R 1 A(x1+h;y) jx1+h yj2 A(x1;y) jx1 yj2 dy ) pk kkf k 4 R jx1 yj< krkk jx1+h yj2 L jx1 yj2 dy+ R jx1 yj< krkk jx1 yj2 L jx1 yj2 dy + R 1 A(x1+h;y) jx1+h yj2 A(x1;y) jx1 yj2 dy ! where krkk = max x2Vi jrk(x)j , L = max x2Vi j1 ik(x)j and 1= Vi\fjx1 yj g : Let (33) I1= Z jx1 yj< krkk jx1+ h yj2 + L jx1+ h yj2 ! dy and (34) I2= Z jx1 yj< krkk jx1 yj2 + L jx1 yj2 ! dy
Since jx1+ h yj jx1 yj + jhj < 32 and the ball jx1+ h yj < 32 covers
the ball jx1 yj < ; I1 Z jx1+h yj<32 krkk jx1+ h yj2 + L jx1+ h yj2 ! dy: We kn¬w from [8] that Z r<" d r = Z S 8 < : " Z 0 rm 1dr 9 = ;dS1= jS1j "m m
where m = 3 ( because we are in R3), jS1j is area of unit sphere and 0 < < m: So that (35) I1 9 2 krkk + 6 L (36) I1 (2 krkk + 4 L)
We prescribe a number " and choose su¢ ciently small, so that
(37) p k k kfk 4 13 2 krkk + 10 L < 1 2" Then it follows from (32)-(37) that
(38) j(L1f )(x1+ h) ((L1f )(x1)j "2+pk kkf k4 Z 1 A(x1+h;y) jx1+h yj2 A(x1;y) jx1 yj2 dy
We now …x the number . In the domain 1, the inequality jx1 yj is
satis…ed, so that the function A(x1;y)
jx1 yj2 is uniformly continuous with respect to
the total of points x1 and y. We can therefore choose an arbitrary small h0,
such that for jhj < h0, we have
(39) A(x1+ h; y) jx1+ h yj2 A(x1; y) jx1 yj2 < " 2 pk kkf k4 jVij
where jVij is the measure of Vi. Then by (38) we have
(40) j(L1f )(x1+ h) ((L1f )(x1)j 1 2" + " j 1j 2 pk kkf k4 jVij < "; jhj < h0
where j 1j is the measure of 1.The quantity h0 depends only on ". It depends
neither on the point x nor the function f , and the lemma is proven. Lemma 2 L1 is bounded on Cp1(Vi). Proof. kL1f k = sup x2Vi 1 4 Z Vi (y)ryG(x; y)rf(y)dy (41) p k k kfk 4 Z Vi ry eik(y)jx yj jx yj dy
From [8], we have (42) ry eik(y)jx yj jx yj = A(x1; y) jx1 yj2 H krkk + L jx1 yj2 where H is diameter of Vi.
If we substitute (42) into (41) and use again [8], we get
(43) kL1f k p k k H2krkk + HL kfk
if we use same procedure for operators L2; L3and L4, we have
(44) kL2f k kmk 4 jkejk k kfk (45) kL3f k krmk jkej H2krkk + HL kfk and (46) kL4f k 1 2M H 2 kfk where for …x y, R = jx yj and
(x) = G(x; y) jrRj2 2 R 2ik(x) R k 2(x) R2 jrk(x)j2 +2Rk(x)rk(x):rR + iR k(x) + 2 R ik(x)R 1 R and M = max x2Vi k2 i(x):
Somewhat, since the kernels of the operators Lj, j = 2; 3; 4 are weakly
singular, that operators are compact operators on C1
p Vi [2].
Now introduce the operator T : C1
p Vi ! C Vi de…ned by
(47) T u = L1u + L2u + L3u + L4u
Theorem 2 The operator T de…ned by (47) is a bounded operator on C1 p(Vi):
Proof. kT fk = sup
x2Vi
j(T f)(x)j kL1f k + + kL2f k + kL3f k + kL4f k : From
(48) kT fk C kfk where C = p k k +krmk jkej H2krkk + HL +kmk k k 4 jkej +1 2M H 2 Conclusion
T is a contraction operator if C < 1 [6]. Then from Banach …x point theorem (T u)(x) = u(x) has only one solution if there exist. We can arrive to this solution by successive aproximation. Let uo(x) be given, then we construct
a (un(x)) sequence such that lim
n!1un(x) = u(x) and u(x) solve the problem
(1)-(5) [6], where un+1(x) = T (un(x)); n = 0; 1; 2; : : :.
References
1. Colton, D. (1998;): Dense sets and far-…eld patterns for acoustic waves in an inhomogeneous medium, Proceeding of the Edinburg Mathematical Society, Vol.31. 2. Colton, D., Kress, R. (1983): Integral Equation Methods in Scattering Theory, John Wiley.
3. Colton, D., Monk, P. (1998): The inverse scattering problem for time-harmonic acoustic waves in an inhomogenous medium, Q. J. Mech. Appl. Math., Vol.41. 4. Colton, D., Monk, P. (1992): A comparision of two methods for solving the in-verse scattering problem for acoustic waves in an inhomogenous medium, Journal of Computational and Appl. Math., Vol.42.
5. Colton, D., Wendland, W. (1975/76): Constructive methods for solving the exterior neumann problem for the reduced wave equation in a spherical symmetric medium, Proceeding of the Royal Society of Edinburg, 75A, Vol.8.
6. Kreyszic, E. (1978): Introductory Functional AnalysisWith Applications, John Wi-ley and Sons.
7. Kleinman, R.E., Martin, P.A. (1988): On single integral equations for the trans-mission problem of acoustic, Siam J. Appl. Math., Vol.48.
8. Mikhlin, S.G. (1970): Mathematical Physics An Advenced Course, Nort-Holland Publishing Co.
