8136-1-T THE UNIVERSITY OF MICHIGAN COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRICAL ENGINEERING Radiation Laboratory ON THE SOLUTION SPACES FOR CERTAIN EQUATIONS OF ACOUSTICAL SCATTERING THEORY by EBLPH,E A R RALPH E. KLEINMAN January 1967 Grant GP-6140 8136-1-T = RL-2174 Contract With:Natioal Science Foundation Applied Mathematics and Statistics Wahington, D. C. 20550 Administered through: OFFICE OF RESEARCH ADMINISTRATION * ANN ARBOR

THE UNIVERSITY OF MICHIGAN 8136-1-T ABSTRACT The certain equations arising in the theory of acoustical scattering theory are discussed. The Neumann problem is rigorously solved. i

THE UNIVERSITY OF MICHIGAN 8136-1-T TABLE OF CONTENTS ABSTRACT i I INTRODUCTION 1 nI AN ALTERNATIVE SPACE FOR ITERATION 4 III REFERENCES 25 ii

THE UNIVERSITY OF MICHIGAN 8136-1-T I INTRODUCTION We consider a spherical coordinate system p = (r, 0, p) erected with origin interior to a smooth, closed and bounded surface B, and let V denote the exterior volume. Then if G is the static Dirichlet Green's o function for the surface B, and if Gk is the Dirichlet Green's function for the Helmholtz equation then it was shown by Kleinman [3] that G (p,p) ~ -ikr u(p,po) = - 2ik j dv1 G ar [rle u(P,,P)] or1 ar 1 1 uk(PlPo V + -ikrB + ikR(pB po) B B a + d e (pp 4dr | B R(PBP) an o PB) B -ikr Here uk is the regular part of Gk, and u = e uk, R(p, p) denotes the distance between points p and p1, dv1 the volume element, dac the surface element, a/an the normal derivative directed out of V, and k is the complex wave number. An integral equation for the corresponding Neumann problem was given by Ar and Kleinman [2, that is, if G is the potential Neumann Green's function for surface B and if Gk (with the regular part uk) is the Neumann Green's function for the Helmholtz equation then 1

THE UNIVERSITY OF MICHIGAN 8136-1-T -V/ u p = -2ik V Go(P' P1) a dv1 -.r ar [rI u(p1) + ik LB 9 duB Go(P, PB) B dkB Go(p, PB) e nrB u(pB) -ikrB a u(pi an with the same notation used above.. We write the above equation for the Neumann problem in the operator form '(p) = Lo.u+ u(~) with L = kL = kO + kO1, where (A) -4 0*(A = - 2i i GO(P, P) dv -1 a ar1 [r o (p)] i C) -4 0 ~ ~0 = daB G (p, pB) r (p ) and B -ikrB a u(p ) doB G (p, pB) e an B oB 2

THE UNIVERSITY OF MICHIGAN 8136-1-T With this notation it was shown by Ar [i] that above equation can be solved iteratively in the following function space W consisting of functions w: V -> E such that (a) w C2 (V), C (, (b) w is analytic on the closed unit disc, in the complex z = 1/r plane, having the expansion a) = 7 (n, '), I n+ 1, nO (c) f (0 ) = Ym (0,) where Y is an m order spherical n ' ' m m=n harmonic, i.e. Y (e,) A Pm (cos ) e e =m tin m with the norm Ilw| = max |(p) + max | a(0,0, l/z) pfV \7<1 for 0 <<_ r, O<< 2,r. In what follows we shall define another norm on the space W. Then we shall show that the operator L is bounded in this norm. The rest of the analysis for solving the main problem is the same as given in [1. We shall only do the analysis for the more complicated Neumann problem. The analysis for the Dirichlet problem is essentially the same. Applications and the consequences of the method have been indicated in [1], [2] and 3] and therefore will not be repeated here. 3

THE UNIVERSITY OF MICHIGAN 8136-1-T II AN ALTERNATIVE SPACE FOR ITERATION We recall that if w is in our space W, then @ = n+i. v r> 6-= 1 n== r f = Y (), y mP (cos O)e n m m m m f.=C011 n) () tm where Implicit in this definition is the fact that the series converge absolutely and uniformly and derivatives with respect to 0 and 0 also converge absolutely and uniformly. Thus, there exists a constant M such that O (n) 7 Y) < M for all 8 and. m=n Note, here and all that follows Y(n) denote a m order spherical m harmonic (depending on n). With this in mind, we define the following function mapping W into E. u -|| u|| = max fmax |u(p)I, = max max ul PEV 00 00, max / O0< < X n=O m=n r>b 00 00 max A= X 9,0 n=O m=i n+l a b (X). ) 4

THE UNIVERSITY OF MICHIGAN 8136-1-T where b is a constant such that b > 6 = 1. Lemma 1 Function defined by (1) is a norm. Proof Since nax (max u, X) pVuf unless u- 0, it follows that > max| uI for all X, and since max lul > 0 peV pEV I||u| > 0 unless u-0. (2) We have Ia u =I a l ul for any complex number max. a u =[ a maxl u pcV pcV we also have oa. Therefore, (3) - n+1 n=0 m=n r = IaL m= n=O m=n r Therefore, max, 6, n=O a y(n) m = Ial bn+l b max, 0 nD man (n) mn I n+1 b (4) From (3) and (4), it follows that (n) max l max ma u, max = = [C max max Iu|, max / Z pEV 0,0 bn+l or I aulI = 11 llul (5) 5

TURI UWIWIVftLS1Ty @F URCU1GA 8136 -1-T For u and v in W let n=O m=n ii=O m=n YU" (O, 0) m n + 1 r (n) 80) m n+ 1 r pr> 6= 1 r> 6=1I Since Ilu+ Vl:~lulI+ IlVI max lu+vI:~ p EV max pcV Jul + max JvJ pEY (6) Also, OD m mj OD m=n (7) From (7) it follows that. max < max 0,0 n=O M=n1 In)I ~m bn+ 1 + + max OJ O 4 jn) (8) + 1 From (6) and (8) it follows that To justify the last step we must show that if A,, B,, C., D,, are non-negative (9) 6

THE UNIVERSITY OF MICHIGAN 8136-1-T constants then max (A + B, C + D) < max (A, C) + max(B, D). Let max (A + B, C + D) = A + B. Then since (10) A < max (A, C) and B < max (B, D), we have A + B < pa (A, C) + max (B, D) This is sufficient to prove (10). The validity of (10) justifies the last step (9). With (2), (5) and (9) it follows that (1) defines a norm,proving the lemma. Lemma 2 If u e W then u - - < C u for all p e V, where f is tht first r - ' * o r coefficient in the expansion for u, and C a positive constant. Proof We have for i e W U = i n=O f n n+l r r> 6 = 1; also, max 8,b n=0 r>b If I n+1 r max7 D n 1 max n+ 1 O'. n=O b D Y(n) < max O, m=n nm l bn+l b+ 7

THE UNIVERSITY OF MICHIGAN 8136-1-T Hence, with the definition of our norm (1) it follows that max,. in" O bi Lb I lul (11) If r A b, then since |f| o <|ubyIb (11) I u Idu -j^ <.< br llul+llulH I II - II II I '- r" r1 (12) = - ( +b) II||u r I1 If r > b, then f r = n=1 f n+ 1 r 1 2 r b2 2 r IfI n-1 r flb nI bn+l 1 < I r 2 < 2 r [f.[ n-l b - IfL bn+l (I3) 2 < Irlu, by(ll) r From (12) and (13) we conclude that f r I < C2 I u r for all p V for all p ~ V (14) S where C is a positive constant. Thus proving the lemma. 8

THE UNIVERSITY OF 8136-1-T MICHIGAN with Bounding the Operator L. We recall that L = kL = kO + kO ) -- O&~ = - 2i -v w - 0 a = i1 d JIB (15) dv1 Go(P, P1) r1 a ar1 1 W(p1] (16) and aB G (p, p) n' 9B(p ) (17) From (16) integrating by parts with respect to r1 once we obtain O0 = 2i JB da Go(P, p) [(p ) - B B B B + (18) Vf + 2i [ dv FrG$p.pl Fw(p)- 1 Irl rL rj 1 1 = 02- W+ 03, thus splitting the operator O into two parts with and -0 2 o = 2if - 3B = y > 30* = 21 Jv daBGo(p, PB) [(p B (19) dv - 1 rl a ar1 [rGo(P (P ] LrlG(p,piL w(pI)-2] (20) 9

THE UNIVERSITY OF MICHIGAN 8136-1-T We now proceed to show in the following lemmas that the operators 0 1 02 and 03 ild A e L1 + 1 W0 + 02 + 0+ 3 is bounded in 2 3 1 1 2 3 norm (1). Lemma 3 The operator 03 (Eq. (20) is bounded. Proof From the Eq. (20) and the estimate of the Lamma 2, it follows that for some constant C > 0, 3 - r 1 2 ar 1 ( 1 |0 CI} dv L )-kr1G. (21) The integrand is of 0 ( R(p p ) as R - 0, am of ( - )as r --- D>, hence for some constant C > 0.1 max [03 * < Cj||I. (22) Next, in order to show that 1I03' ~|j < C JIJ|| for some constant C > 0 we must show (see definition of norm (1)) that if O^ g (o,0) 3 =, r >6 = 1 (23) n=0 r and 7 (n) (24) m=n 10

THE UNIVERSITY OF MICHIGAN 8136-1-T then OD Z |(n) m 2* II c Ii, =0 n+ 1 C211W n=m=n b max e,9 (25) for some constant C2 > 0 (b> 6 = 1). Note that 03o o has the expansion tidicated by (23), (24) and (25) because we have already shown that L and all the other operators involved map the space W into itself. Now we proceed to carry out the quite tedious task of showing (25). First, neglecting the constants 03- = fv d1 r ar 1 rG 0 rl1 (26).il 1 f0 a I I o a r1 1 f r< v r r uv 1 1J1 r1 Or1r = d~ 11{tO-l' Er -- where u is the regular part of the first integral. For r > 6 = 1 static Green's function. Consider the C f 0 a dv -( - - )- (r u ) c 1l r1 ) (r u)1 " int ext v, r, 0 ) dv (1 - 1 r1 rI a F ar, 1L n + 1 n=0 r + (27) o a f ar (; 01 01) r r (rrl)n n=0 1 11

THE UNIVERSITY OF MICHIGAN 8136-1-T where Vint is the volume exerior to the surface B and interior to the sphere of radius 6 = 1, V is the volume exterior to that sphere. The expansions ext in spherical harmonics for the Static Green's function are given previously. Since the integration with respect to p1 does not change the fact that th order spherical harmonics, we must show that the integrals are still n order spherical harmonics, we must show that max iv ei OD 1 b i vint f dv 1 0r(a -) - 1 1 r Sr1 1 Kt i 1 1 f 0 n rl n+l Ya''r;e 1) 1 (2) = max Oe, OD I11 n+ 1 n= b m=n m=n <K1 I |w, some constant K1 > 0. t;' -,iri.Clearly, n+1 n= b 1 dv1 (4 ext Vint f 1 o a [rlYn(O ~; pl)~ dv (WrY(6,FJ; p 1 r1 r1 ar L L 1 1 1 fo n Yn.;9..P.1) 1 to - ).. < r 1, r rn+l -- bn+1 1 1 r b 1 - (29) L Vint 1 f a' dv --- - (rY) + r1 r1 ext dv,;,. i i r, 12

THE UNIVERSITY OF MICHIGAN 8136-1-T Note the Y n Green's function. in both integrals arise from the regular part of the static That is OD n=0 Yn(w, O;Pl) n+r r for r > 6 = 1 (30) O Di-, Yn(9,;O 11) dA (rr ) 1 n=0 1 for r,r1 > 6 = 1. The fact that the series converges absolutely and remains absolutely convergent after multiplying by r1 and differentiating with respect to r means that r [rYn( < A 6n+ A for some constant A independent of n, 0, 0, and p1. Similarly, there is a constant B such that 2n Yn(e. i1, 6 1) < B62 = B. Now using the estimate in the Lemma 2, Eqs. (31) and (32), we obtain o | (r Y2 dvl < C {11 ' int (31) (32) (33) * A 1 sin 0 dr r1 1 1 int dO1 d01 < D 11 1, 13

THE UNIVERSITY OF MICHIGAN 8136-1-T for some constant D > 0. And, similarly V ext nf fO J Y(e,0;1, 1 L -'- n+1 dv1 < C r rI n+l - (34) * B v V ext n+4dvl < E [ tll rt for some constant E > 0. With the help of (33) and (34) we obtain max 9, n= n=0 1 1 fo a bn+l rI 'xi ar1LL1n( l int -[ dvaOn C 3 11w11 dv ( - rl ' 1 r n+l n'11 bn+" ext (35) - el — l b - 1 4 III11.I C3, C4 are appropriate positive constants; we also recall that b > 6 = 1. This completes the first part (for the regular part of static Green's function). Next we go on to the singular part of the static Green's fucntion (second integral in (26)). 14

THE UNIVERSITYOFMCIA OF MICHIGAN 8136-1-T 1 4ir cv r1 f 0 a Orl -) dv = (36) 1 4i 1 f a (w- -24 -.. rr1 OrI r I r1 A~ r n+1 P (c Os'$ dv 1 where cos y = cos 0coos61 + sinG9 sin01 Cos (O -~ ). For r> 6 = 1 this becomes 1 r, - A v int 6 - f aL ( 0 Oc Or18 n=O ii+1 P Cos$ + n+1 r + ~1; r dr 1 * sine0 2wr 1 I do, 1 1 n+1I OD r1 Pn(cosi#)}+ (37) +jL50 driS o r 2w0 dO r sine0(- )1 1 1 r8 Or n r n n r 1 15

THE UNIVERSITY OF MICHIGAN 8136 -1-T For r > 6 = 1., however, f 0 OD n + 1 121MX (37) may be written as 1 41 00 n+1 n+1 r 1 f0 -(w.-)rnP (cos)) r1 r in Vint r 4wr 2w it dr1 0so0 x x sino c1dffdo m=1 n=O n n (n+1)P n(anB)4 + rn (38) +?IT dr1S sine Id6ldol 4, 7 m= 1 m 00 L LLn (Cos'-h. n = n+1n Recall that so that fm (6l J 0 ) = 2wr do, I 400x y (M) (61Io, d1 si 1fm(6i' 01) P coo y) = 0, n <in. -4wr y (M) n (0, 0), n:,> m. 16

THE UNIVERSITY OF MICHIGAN 8136-1-T Hence from (38) we have 1 4z n=0 (n+ 1) dv I(( n+l int r int f -— ~) rnP (coos r i n I drI x 00 n=m (2n+1) n+l r1 n+1 r (n+ 1) - r dr1 ) Y(m)(O p) m=l n=m (39) n nr x n+m+1 r1 41 4wr uOD n+l dv n+ 1 1 r int f (o - i) rn P(cos i) + 01 m=l n=m n=m 20n (, ) 2n + 1 n+1 (rn-m+ 1-1) n+1 (n-m+1) r n=m Y(m (e, pnrn (2n+1)(n+m)nm Now we reorder the last two sums to put them into the "right" tar-. These sums may be written as m=l n=m m=l (O.) an-m+1 m+ rm(2n+ 1) m+ (40) n=m n=m y(m)(, p)(n + l) r 1(2n + 1)(n-m + 1) The first term can be written, changing the roles of m and n and starting the first sum at zero as 17

THE UNIVERSITY OF MICHIGAN 8136-1-T O O y(n+ 1)(e.) n2 m ( r2m + 1)r+1 IitQ1 Im=A+1 m+ 1 nm ) m -1 n+m+ (41) n=0 m=n+l Y( (0, )(n+ 1) r -n)(n + 1) r (m-n)(n + m + 1) Next, the second oo oo m=l n=m term in (40) can be handled Y(n(, )(n + ) oD r (2n+ 1)(n-m + 1) n= 1 similarly; thus n-1 Y(n-m)( n r Y ( ) (n+l) 0 r(2n+)(m+l) m=0 (42) Thus, we can write the integral (36) involving the singular part of the static Green's function, with the help of (41) and (42) as 4 r I 8r 1r1 1 1,i1 rfo 1 ~I: 4 dv r (w-r-) R 5V n=O 1 n+1 r ' ' I dv1 x 4V t Vint 1o n,:,, x (t — l)rl P1(cos) +2 (m - n)(n + m + 1) m=b+1 (43) -2" m11=0 (n-m) Y( ), 0)(n + 1) (2n + l)(m+ 1) I where the last term is identically zero for n = 0. Our task is to show that for some constant C > 0, 5 18

THE UNIVERSITYOFMCIA OF MICHIGAN 8136 -1 -T max 9, fi ri=0 1 1 + m=n I M (9,p 0) <0c5 I1ca)II (44) lp where 11n 4 C 1 r 4w*Vint1r f n (o — i r P (coo ~-h i11n 2u+l n-i y(n-rn) (6,p m=0 ( andi (n)= z= m (n + 1), )n+1 m (m - n)(m +n + 1) m>n+ 1 Clearly OD n=0 1 ~n+i m=n (n) = m n=0 ~n1 m=n+i z (n) mH.'7I -- K fiiw-h* - - 7r,,, < n+i n=0 4irb f dv ((o —2-)r n P (Cos _$ 1 rI in int + (45) n +1 (2n+l)b '~1 n-i -M n(m)( 0) m+i + n=0 1 4, 7 m =n+i (m - n) (m +ati) 19

THE UNIVERSITY OF MICHIGAN 8136-1-T Consider the three terms on the right hand side of the inequality (45) separately. Firstly, f (wo- r)r Pn(cos rin int y)dv < Vit int f o n cO-1 r P (cos y) dvl r 1 I n 1 But f r 1 <C 2 rI ifwI, fP (cos 4 < 1, O<r < 6 = 1. So 1dv dv rl(w 1 r f 0 ) n(cos 6Q1 1 ',II (46) where Q1 is a positive constant independent of w. Notice that, with b> 6 = 1, n=0 - + 4Q7T 11 = 4nb+ Q4 1Ilq 47rb n=0 n+1 bn bn1 (47) Q1 [ill d 47rb d(1) b =0 n=O 1 n+1 b Q414k 47rb b2 (l-b)2 < Q2 II 1, some Q2 > 0. Thus, with (47), (46) we have n=0 n+1 4,Tbn 3 f dv -(w — ) r P (cos$ y r1 r 1 n 1 1 Q2 11II 11 (48)' A -A. int 20

THE UNIVERSITY OF MICHIGAN 8136-1-T Secondly, oo n=1 (n+1) (2n+l)b"+' n-1 (n-m) n / m+1 m=O on n=l n-1 (n+1) y(n-m) = (2 b (2n+l)bn+l(m+l) rn=O < y 1 - bn n=l oo bn= nO oo m=n m=n (replacing n by n - 1) (49) < b ||Ilw (by definition of our norn(. ThiRdly, 1 bn+1 b m=n+l -{ n+l) | ( t) oo -m-'., < v (m-n)(m+n+l) - n=O,n+l m - m=n+l (50) < b Jwlj ( since (-n)(m+n+ < 1 (m-n)(m+n+) - ) as above. Combining (47), (49) and (50) we obtain the inequality (44) which completes the analyris for the singular part. Equation (44) with the inequalityl for the regular part, (35), yield (25). Finally, combining (25) witht (22) we obtain lIa s wll c joiw where C is an appropriate constant, proving the Lemma 3. (51) 21

THE UNIVERSITY OF MICHIGAN 8136-1-T Lemma 4 The operator 02, Eq. (19), is bounded. Proof From the Eq. (19) it follows, with the estimate of Lemma 2, that for some constant C > 0, Go I (PllP D, pB C Oe < < Cclo| duB C x B B (52) x d[uoP + B l r B d d BB 4srrBR(PP') (B + IOP'PB) doB + The first term on the right is the potential of a single layer distribution of density 1/4xrrB. Since rB f 0 (the origin was taken with B) and the surface is smooth, closed and finite, this density is uniformly Holder continuous which means that the potential is continuously differentiable for all points pEV. The second term on the right hand side of (52) is the integral of a bounded function over a finite surface and, hence, is bounded. Thus, there exists a constant N1 > 0 such that max 0 2w < N1I| (53) pEV If 02~w n= r> 6=1 n=O r and oo n m= m=n 22

THE UNIVERSITY OF MICHIGAN 8136-1-T then by exactly the same long procedure of Lemma 3 it can be shown that 0o max ' pV O2 m=n z(n) m < N2 tI2ll bn+1 - 2 for some constant N2 > 0. From (53), (54) and the definition of our norm, it now follows that 1102o w < N J Jf for some constant N > 0, proving the lemma. Lemma 5 The operator 01, Eq. (17), is bounded. Proof We have with Eq. (17) [1o O j < _ dB Go(p, B)j Ij rB I W(PB)I B By definition I|(PB)| < IWJ Also, in and '? are unit vectors, B fi<B1 I, 1 (54) (55) (56) (57) (58) thus, 1o " <WI 11P11 I daB B (59) 23

THE UNIVERSITY OF MICHIGAN 8136-1-T The surface integral is bounded by the same argument given in the first part of Lemma 4 (Eq. (52) ). Hence, for some constant A > 0 max O-1 w < A1[i| I F per Furthermore, if (60) and O gn(O, ) 01~ ~=: ' n+l r n=0 r (n) g(e, 0 ) L7 zm m=n r> 6 = 1 (61) (62) then, mGin by exactly the Aoo OD max n=0 m=n same argument of Lemma 3, it follows that z(n) < A2 I< n+l - 2 1101i b for some constant A2 > 0. From (60), (63) and the definition of the norm, it follows that |[10 i||>- A [IJ)U for some constant A > 0, proving the lemma. It now follows immediately from the Lemmas 3, 4 and 5 that: Corollary. The operator L = kL1 (Eq. (15)) is bounded; and there exists a complex number k > 0 such that for |k < |ko|, | LJ < 1. O I 1'lii l (63) (64) 24

THE UNIVERSITY OF MICHIGAN i8136-1-T III REFERENCES [i] Ar, Ergun (1966), "On the Helmholtz Equation for an Acoustically Rigid Scatterer, " The University of Michigan Radiation Laboratory Report 7359-1-T, NSF Grant GP-4581 12] Ar, Ergun and R. E. Kleinman (1966), "The Exterior Neumann Problem for the Three-dimensional Helmholtz Equation," Arch. Rat. Mech. Anal. 23, pp. 218-236. [3] Kleinman, R. E. (1965), "The Dirichlet Problem for the Helmholtz Equation," Arch. Rat. Mech. Anal. 18, No. 3, pp. 205-229. 25