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On the resolvent of the Pekeris operator with a Neumann condition
Journal of Sound and Vibration (1978) 56(l),
ON THE
RESOLVENT
87-95
OF THE
NEUMANN
PEKERIS
OPERATOR
WITH
A
CONDITION
D. HABAULT and P. J. T. FILIPPI Laboratoire de Mtkanique et d’Acoustique, Centre National de la Recherche Scientijque, 13214 Marseille Cedex 2, France (Received 8 June 1977, and in revisedform 30 August 1977)
A constant thickness plane layer of homogeneous isotropic medium is limited, on one side, by a perfectly rigid plane, and, on the other side, by a second homogeneous isotropic medium occupying the half-space. The sound pressure due to a harmonic point spherical source is studied. An exact representation of the solution is given for both positions of the source: outside of or within the layer. In the first case, this representation comprises an image source radiation and layer potentials, the densities of which are expressed in terms of the well-known modes. In the second case, the field within the layer is expressed as layer potentials, the densities of which are developed in terms of the modes. The result is valid for conservative media or dispersive ones as well.
1. INTRODUCTION In a recent work, Wilcox [I] has studied the spectrum of the Pekeris operator with a Dirichlet condition in the case of conservative media. Though this theory provides a complete theoretical result (propagation of transient waves or harmonic ones can be obtained), it seems difficult to use his solution for practical applications; furthermore the assumption that the media are conservative ones (self-adjointness of the Pekeris operator) is fundamental. The aim of this paper is more restricted in the sense that simple harmonic waves, only, are considered, but the restriction to conservative media is not necessary. Furthermore, the representation of the solution avoids the branch integrals which usually appear, and has a form which can provide analytic approximations and easy numerical computations. The choice of the Neumann condition at the back of the layer is often appropriate in both room and outdoor acoustics : indeed, in many cases, an absorbing boundary is composed of a porous medium layer on a hard obstacle. Furthermore, in outdoor acoustics, many kinds of ground can be approximated by such a model. In the second section the statement of the problem is presented. The third section is devoted to the calculation of the pressure field in the unbounded domain when the source is in this domain. In the fourth section, the case of a source within the layer is considered, and the pressure field is calculated there. The pressure transmitted into the source-free region has not been calculated : the technique developed in sections 3 and 4 could easily be applied.
2. STATEMENT
OF THE PROBLEM
Let Q, be the half-space z > h > 0, and Q2 be the domain 0 < z < h. In Qi(i = 1,2) a fluid or a porous medium is characterized by a mass density pi (real or complex) and a sound speed cI (real or complex); the plane z = 0 is assumed to be perfectly rigid. A point spherical source, generating a simple harmonic (e-‘“0’) signal is located at the point S(O,O,s). The problem is to find the sound pressuresp, in Q, andp, in Q, satisfying the following equations 87
88
D. HABAULT AND P. J. T. FILIPPI
and boundary conditions : (A + w$cf)pl
= 6,
in
Q1,
(1)
(A + &/1)P2
= 6s
in
Oz,
(2)
Trp, = Trpz,
Tr arp2 = 0
on
z=h,
on
Tr %P~/P~ = Tr aZp2/p2
(3,4)
z=O.
(5)
The uniqueness of the solution is ensured by taking (p1,p2) as the limit of the bounded solution of the system of equations (1) to (5) when o,, is replaced by o0 + is(c > 0,~ --f 0); in the following, it is assumed that kf and k: are defined by kf = (coo + is)2/c:,
(6)
with the assumption Im(k:) > 0 (absorption condition for porous media 12, 31). The final results will be valid even ifs = 0. We will solve separately the cases Sin a, and Sin Qz; in each case we will be essentially interested in p1 or p2, respectively. The application of the first cor@uration is to the reflection of a sound wave by a layer of an absorbing medium on a hard plane. The second configuration is very close to the problem of the propagation of sound waves in shallow water, the Neumann condition having to be replaced by a Dirichlet one.
3. THE SOURCE
IN THE
HALF-SPACE
z > h
In this configuration, the right-hand side of equation (2) is identically zero. Upon using the two-dimensional Fourier transform with respect to the variables x and y, and taking into account the cylindrical symmetry of the problem, equations (1) to (5) become (d2/dz2 + K:)/?, = as (d2/dz2 + K:)p2 = 0 Tr@, = Trj?,,
z > h,
(7)
0~ Z-Ch,
(8)
in in
on
Tr dzb2 = 0
z=h,
on
TrdZP1lpl = T&02/~, z = 0,
(9,10>
(11)
with Kf = -47~’C2+ kf,
(12)
m ,Ci = 2~
s
r = (x2 + y2)‘j2.
Jo(27t{r)p,(r, z) r dr,
(13)
0
The solution of equations (7) to (11) is taken in the form &K,(s+z) eiKl Is-2I A
=
___
-
2iK,
(14)
’
2iK,
fA
(15)
The functionj, obviously satisfies the condition (11); the continuity conditions (9) and (10) lead to the linear system eiK2h
eiK,Cs+h) ----_A
----APl
eiK,Cs-h)
e-iK,h B=-----
K2
K1 eiK,(s+h)
+
_
eiK2h
_
K, e-iKZh
B=-----. P2
’
&K,(s-h)
Pl
(16)
89
RESOLVENT OF THE PEKERIS OPERATOR
An easy calculation shows that A is given by A
=
e-ZiK,h
p2 Kl cos K2 h + ip, K2 sin K2 h p2 K, cos Kz h - ip, K, sin K2 h
=e
_ZiK,hp$ Kf cos2 K2 h - p: Kg sin’ Kz h + 2ip, p2 Kl K2 sin K2 h cos K2 h pt K: cos2 K2 h + pf K: sin2 K, h
.
(17)
To obtain the inverse Fourier transform of the second term of expression (14) (the scattered field), use will be made of the following inversion formula which can be deduced from equation (13): f(r) = 7t 7 Hb1)(2?c
(18)
--m with the definition
Hbl’(-x) = Hy)(x e’“).
If~(4rc2~2) has non-real poles only and no branch points, the integral (18) is obtained by the residue method, by using a contour of integration along the semi-circle r at infinity in the upper complex plane : ,f(r) = 2irc2x Res{Hf’(2x&-) f(4rr2 5’) 5} - rr s Hb”(2z
(19) r Furthermore, the scattered field will be decomposed into three terms: one corresponding to the image of the source with respect to the interface plane, and two layer potentials; this procedure will avoid the computation of the branch integrals which usually are involved. 3.1.
THE POLES OF A
The poles of A are given by the equation D(47r2r2) = p$ Kf cos2 K2h + pf K$ sin2 K2 h = 0.
(20)
A first remark is that if (2&) is a solution of equation (20), (-2~~5,) is a solution too, because (20) is an even function; furthermore, because of the definition of k:, the 5, have non-zero imaginary parts; it is thus ensured that there exist as many poles in the upper as in the lower complex half-plane, and none on the real axis. We will make use of the following notation : A, = 2lr5”
with
Im(,?,) > 0.
The classical mode equation is tan K2 h = -i p2 K,lp, K2.
(21)
The second remark is that, though equation (20) has the same solutions as equation (21), it is obvious that it can have other solutions: namely, those of tan Kz h = i p2 K,/p, K2.
(21’)
But if pl, cl, p2 and c2 are all real, it can be easily seen that equation (21’) will introduce no additional modes. Nevertheless, the fact that equation (20) can have more solutions than equation (21) does not introduce any difficulty. Note that, for simplicity, we assume that all the zeros of D(4rc25’) are simple ones. Finally, we will denote by D’(;i2) the function dD(A2)
D’(A2) = -
W’)
= -
p:cos2K2h+p:sin2K,h+h
sin K2 h cos K2 h
which is the derivative of D with respect to A2= 4n2 t2.
K,
D. HABAULT AND P. J. T. FILIPPI
90
3.2.
THE FIRST TWO TERMS OF THE SCATTERED FIELD
The first two terms of the scattered field are given by the inverse Fourier transform q1 of the function PgKf c~s2 K2h _ p:K:~in2K2heiK~(~+Z-2h) ^ eiKI(S+z-h) (T1= (23) pf Kf cos2 K2 h + pf K$ sin’ K2 h 2iKi = ‘I 2iK, ’ The function P1 cannot be inverted by formula (18) because the Fourier integral (18) is not convergent. It is necessary to write it in the form 3 I = (-4n2 5’ + a2) i,/(-479 t2 + LY2) = (-47~~5’ + a’) h,(4r2 r’)/[(-47r2 t2 + cr’) D(47c2[‘)I
V,cc,
with
Ima#O,
(24)
which leads to
(A(,, being the two-dimensional Laplace operator). No branch integral occurs because P1is an even function of Kl and K,; the integral of h, D-‘(-4n2 t2 + u2)-’ Hkl)r alongthe real axis is convergent because of the term (-479c2 + cr2)-‘; the integral along r of this function is zero because of the exponential decrease of the Hankel function. Using the well-known property of the Hankel function (Dirac measure in rW’>, one obtains i
W:) Hb”(J.”p) + aC2, D’(A;)
r, = -
4c
(26) c
The coefficient of a(,, can be calculated by remarking that it is the sum of the residues in the upper half plane of the function -(4n/i) h,(4n2 5’)
(27)
This function being odd in 5, its integral along the real axis is zero; to evaluate the integral of expression (27) on I’, it is necessary to remark that Kz ~-4~2 R2 e2ie
(5 = Re”),
cos’ K2 h E ${exp(-4rrReie h) + 2 + exp(47cRe” h)},
sin” K2h z -&{exp (-4nReie h) - 2 + exp(4nReie h)}. Either one or the other of the exponential functions will give a contribution, the sign of case. Nevertheless, in each quadrant, one has hi/D N (P: + PMP:
depending on
- ~3.
and so --
4n Rh,(4n2 R2 e2je)
RZ $io
i s D(47r2 R2 ezie) -47~’ RZ ezie + a2
dO=-----. _
P2 + PI
P: - Pt
(28)
0
Finally, q1 will involve a wave emitted by the image source S’ = (O,O,2h - s), and a simple layer potential due to a source distribution in the plane containing the image source and
RESOLVENT
91
OF THE PEKERIS OPERATOR
parallel to the interface : pf + pf 41 =-
i
eik,r(S’sX)
h,(2)
-
p: - p: 47rr(S’,X) - 4
D’(#J
eik,rW') Ha ‘[A, p(X’)]
z’=2h-s
x = (XT y. 3.3.
THE THIRD
47W(X , X’)
z),
do’.
da’ = dx’ dp’.
TERM OF THE SCATTERED
(29:1
FIELD
The third term of the scattered field is given by the inverse Fourier transform q2 of the function eiK,Cs+z-2hl 2p, p2 K, sin K2 h cos K, h 42
=
pi Kf cos2 K2 h + pi Ki sin’ K, h
=
F,iK,
eik’,b’+r-2h)
h2(drc2
2iK,
(2)
= D(4n2r2)
iK,
2iK,
eiK,(s+r-2h)
iK
2iK,
1
(30)
’
Here again, PZis even with respect to K1 and K,, and no branch integral will occur. It can be easily seen that the integral of P2HP) { along the real axis is convergent: indeed e2inlCl P2Hd” 5 E m x constant,
for
151-+ a,
which is integrable at infinity. Furthermore, the integral along r is zero because of the exponential decrease of the Hankel function; one thus obtains i
r2 = -
4c
h2(X) H::‘(A, p). D’(1.i)
-
(31)
The coefficient iK, in C&can be interpreted as the derivative factor of eiKl(s+Z-Zh) with respect to z’ = -2h + s and q2 is represented by a double layer potential :
g
elkv(X,X')
WV-ndJ31 n
I&
x,)
4~r(X
do’.
(32)
r’=2h-s
3.4.
REMARKS
ON THE FINAL RESULT
The total pressure field pi in the unbounded domain Q, is finally given by eikIr(S.X)
PIG% X) = -
pi
+
pf
eiklrG:X)
-___ p: - pf 4nr(S’. X) -4 4nr(S, X)
i
h,(X) r’=2h-s
eik,r’X.X”
’
4m( X, X’)
SC
M%) -Hb”[A,p(X’)] D’(X)
S’ = (0, 0, -s),
x = (x3 Y, z).
da’ + f
D’(X)
H;“[i, p( A”)] x a
eiklr(X.X’)
-
az’ 4nr(X, x’)
d,s’,
(33)
where
.s= a p(Y)
0, s>,
= (x’2 + y’2)1’2,
X’ = (x’, y’. z’),
da’ = dx’ dy’,
D’(A2), h,(A’) and h2(A2) are given by equations (22), (24) and (30), respectively, and the A, are solutions of D(A$J= 0, with Im(A,) > 0. Formula (33) can be compared to formula (20) given in reference [3] in the case of the reflexion of a spherical wave by a half space (Sz, = z < 0) : the amplitude of the field due to the image source is the same in both cases; as a first approximation (highly absorbing medium in
92
D. HABAULT
AND P. J. T. FILIPPI
Q,), the scattered field is that of the image source, and it does not depend on the thickness of the layer.
4. THE SOURCE IN THE LAYER 0 < z < h When the source is in the layer, the right-hand side of equation (1) is identically zero. By a Fourier transform, bne again obtains equations (7) to (1 l), the only difference being that equation (7) has a zero second member, while that of equation (8) is 6,. The solution ($&) may be taken in the form 0, = AeiKl’/2iK, P2
=
(34) eiK2(s+z)
eiK21s-zl
e
2iK2 +--- 2iK,
iK,z
+
e-iK2z
(35)
2iK,
fB
The coefficients A and B are solutions of the linear system: e iK,h
--A
KI
eiKlh
+
eiK,” --A+
eiKICh-sl
e-iKzh
+
eiKZ(h+s)
g=--
+
K2 eiK,h
_
Kz
e-iK2h
eiK,(h-s,
+
’ eiK2(h+s)
B=-
(36) Pz
Pz
Pl
which gives B=(l/D){-K:pzcosK,h+
iK:p?sinK,h+
+ K1 p, K2 p2 cos K2 h - iK1 p, K2 p2 sin K2 h} cos K2 seiKzh.
(37)
The procedure for invertingfi, is much the same as in section 3, and will lead to layer potentials due to source distributions in the planes z = h and z = -h. 4.1.
FIRST COMPONENT
q1 OF pz
The first component, ql, is the inverse of 4, =--
K: p$ cos K2 h cos K2 seiKlr + e-1K2zeiK,h
=-p:KfD
2iK2
D g, e”‘zz + e-iK>
2 eiKlh
2iK2
’
The Fourier integral of Kfg,/D is convergent: indeed, for large positive t it has the following asymptotic behaviour : (_4$ 52eZnC(s+h))/(_4TC2 (2 ednr”), (39) which decreases exponentially because h > s. On r, this function is of the form -47r2 R2 ezie exp{f2nRe”(s
+ h)}/[-4n2 R2 ezioexp {-&47rRele h}],
which is, for the same reason, exponentially decreasing when R --f cc. The function q1 is thus given by
(40)
93
RESOLVENT OF THE PEKERIS OPERATOR
with X = (x, y, z),
p(X”) = (x”2 + y”2)‘/2,
p(Y)
X” = (x”, y”, zq,
X’ = (x’, y’, z’),
= (x2 + y’2)1’z,
da” = dx”dy”.
da’ = dx’ dy’,
4.2. SECOND COMPONENT q2 OF& q2, is given by the inverse of
The second component,
q2 = iK2pf
K2 sin K2 h cos Kz s eiKIr + emiK2’
eiKZh
D g2eiKzz
=
+
2iK2 e-iK2z eiK2h,
iKZpi-
(42)
2iK,
D
It is easily seen that a double layer potential
of q2 can be obtained
representation a
g2m
eikzr(X,X’)
H;“[& p( X’)] -
-
azl 47cr(~, x’)
D’( j.:)
:
da’ -
(43)
4.3. THIRD COMPONENT q3 0FP2 The function
g3 is cos K2 h cos K2 s
1
43=--K:ip1P2
D
K,
eiKzz+e-iK,z eiK,h.
iK2
(44)
2iK2
As shown in section 4.1, $75-1K2
cosK,hcosK,s 1
i
D
=4
g,(X) c
(G
-
A3
m
n
W,‘V.
~1.
Furthermore, 1/K,) = -2i eikIP/4np.
P-‘( The function
q3 can be represented
a 1 aelk2rW.X”) 1(X”)-
by double layer potentials
:
eiklr(X,X’)
r(k:-~,2,)gl(E.:)H~l)(r3.P):2)e’k’P D’(j.$ - L;) g;$
Hb”(& p)t,
es
n where the symbol & stands for the convolution f(d2~
g(p)
=
product
af 4Xrr(~, x~)
do”,
(45)
in the plane:
Y’,o)lgW’> 01 dg’,
jfb(
Y' = (x’, y’),
y = (x9Y),
4np
dt+= dx’ dy’.
4.4. FOURTH COMPONENT q4 OFP, The last component,
q4, is obtained
4 =-LKZ’ l’PlP2 Kl
by inverting
Kz sin K2 h cos K, s eiKzr + eciKzz D
2iK,
eiKth ’
(46)
D. HABAULT
94
AND P. J. T. FILIPPI
Upon using the results of sections 4.2 and 4.1, the function q4 is readily seen to be elkzr(X,X’)
47cr(X, X’)
do’ +
eik2rW.X”)
4zr(X, X’) 4.5.
da”.
(47)
THE FINAL RESULT
The total pressure field pz in the layer Q, is finally given by
g1m (x> + $p; 1 (k: - 2:)D’. n HW, P) [b=, + k4,]
eik2r
p2 = -{d, + S,,} * -
4nr
* ;;
(x) +
e% r
HW, P)i&h - &=-,I * 4nr
-
g,
W:- 2) ~‘0
(x) -
fz
m W,% P):z, 4np
n
[S,,, - S,,_,] efktr
e’k’ [6z = ph + d,,_,] *j$X) W’(~.P):,,~ where * indicates the convolution product in R3, or, more explicitly, elk&,X)
PI6
Jo = -
elkzr
i
(S’.X)
47cr(S, X) - 4nr(S’, X) elkzr
(
2 (k: - A:) g
+-P:
4
a
Hb%
P(X’)I a
&W -
D’tX)
WV.
4nr(X
x”) 9
r”=-h g2G)
-_h
[A,p(X)] x eik2r’X.X”’
p(X”)]
da’ +
D’(X)
&‘=
Hi"
X,X’)
’ 47cr(X, X’)
-
II
G
elkzr’X.X”
4nr(X
7
x,) cb’ -
eikzr(X,X”)
P(X”)IF 4xrcX x,,j da” + 3
da”
95
RESOLVENTOFTHEPEKERISOPERATOR
where
s = (0, 0, 4, da’ = dx’dy’,
S’= (0, 0, -s), da” = dx” dy”.
x= (X,Y,Z),
X’ = (x’, y’, z’),
X” = (xP, y”, z”),
D’Q’), gl(l’) and gz(n2) are given by expressions (22) (38) and (42), respectively, and the 2, are solutions of D(,Q = 0, with Im(&) > 0. 5. CONCLUDING REMARKS The representation of the sound pressure field given by formulas (33) and (48) has several advantages. First of all, it does not involve branch integrals, which usually appear in the literature. The second advantage is that more or less simple analytic approximations of the integrals can be established, depending on the absorption rate of sound waves in the different media which governs the decrease of the layer densities; furthermore, numerical computations can easily be performed. The third advantage is that the contributions of each mode appear as layer potentials: that is, as functions which are analytic everywhere in the propagation domain; a global inversion of the Fourier transforms a1 or d2 would have given products of functions of the z variable by Hankel functions, the singularity of which could introduce difficulties in the computation of the near field. The last point of interest about these results is that they are valid whether the media are conservative ones (pl, cl, p2 and c2 all real parameters) or not (pl, cl, p2 and c2 real or complex). This shows that the spectral representation of the resolvent of the Pekeris operator proposed in reference [l] for the real case applies equally for the complex one. It would be interesting to see if the spectral theory can be applied to solve transient wave propagation problems in absorbing media, just as it can be for conservative systems. REFERENCES 1. C. WILCOX 1976Archives of Rational Mechanics and Analysis 60,259-300.
Spectral analysis of the Pekeris operator in the theory of acoustic wave propagation in shallow water.
2. TH. LEVY and E. SANCHEZ-PALENCIA (to appear) Journal of Mathematical Analysis and Applications. Equations and interface conditions for acoustic phenomena in porous media. 3. P. J. T. FILIPPIand D. HABAULT 1978Journal of Sound and Vibration 56,97-103. Reflexion of a spherical wave by the plane interface between a perfect fluid and a porous medium.
ON THE
RESOLVENT
87-95
OF THE
NEUMANN
PEKERIS
OPERATOR
WITH
A
CONDITION
D. HABAULT and P. J. T. FILIPPI Laboratoire de Mtkanique et d’Acoustique, Centre National de la Recherche Scientijque, 13214 Marseille Cedex 2, France (Received 8 June 1977, and in revisedform 30 August 1977)
A constant thickness plane layer of homogeneous isotropic medium is limited, on one side, by a perfectly rigid plane, and, on the other side, by a second homogeneous isotropic medium occupying the half-space. The sound pressure due to a harmonic point spherical source is studied. An exact representation of the solution is given for both positions of the source: outside of or within the layer. In the first case, this representation comprises an image source radiation and layer potentials, the densities of which are expressed in terms of the well-known modes. In the second case, the field within the layer is expressed as layer potentials, the densities of which are developed in terms of the modes. The result is valid for conservative media or dispersive ones as well.
1. INTRODUCTION In a recent work, Wilcox [I] has studied the spectrum of the Pekeris operator with a Dirichlet condition in the case of conservative media. Though this theory provides a complete theoretical result (propagation of transient waves or harmonic ones can be obtained), it seems difficult to use his solution for practical applications; furthermore the assumption that the media are conservative ones (self-adjointness of the Pekeris operator) is fundamental. The aim of this paper is more restricted in the sense that simple harmonic waves, only, are considered, but the restriction to conservative media is not necessary. Furthermore, the representation of the solution avoids the branch integrals which usually appear, and has a form which can provide analytic approximations and easy numerical computations. The choice of the Neumann condition at the back of the layer is often appropriate in both room and outdoor acoustics : indeed, in many cases, an absorbing boundary is composed of a porous medium layer on a hard obstacle. Furthermore, in outdoor acoustics, many kinds of ground can be approximated by such a model. In the second section the statement of the problem is presented. The third section is devoted to the calculation of the pressure field in the unbounded domain when the source is in this domain. In the fourth section, the case of a source within the layer is considered, and the pressure field is calculated there. The pressure transmitted into the source-free region has not been calculated : the technique developed in sections 3 and 4 could easily be applied.
2. STATEMENT
OF THE PROBLEM
Let Q, be the half-space z > h > 0, and Q2 be the domain 0 < z < h. In Qi(i = 1,2) a fluid or a porous medium is characterized by a mass density pi (real or complex) and a sound speed cI (real or complex); the plane z = 0 is assumed to be perfectly rigid. A point spherical source, generating a simple harmonic (e-‘“0’) signal is located at the point S(O,O,s). The problem is to find the sound pressuresp, in Q, andp, in Q, satisfying the following equations 87
88
D. HABAULT AND P. J. T. FILIPPI
and boundary conditions : (A + w$cf)pl
= 6,
in
Q1,
(1)
(A + &/1)P2
= 6s
in
Oz,
(2)
Trp, = Trpz,
Tr arp2 = 0
on
z=h,
on
Tr %P~/P~ = Tr aZp2/p2
(3,4)
z=O.
(5)
The uniqueness of the solution is ensured by taking (p1,p2) as the limit of the bounded solution of the system of equations (1) to (5) when o,, is replaced by o0 + is(c > 0,~ --f 0); in the following, it is assumed that kf and k: are defined by kf = (coo + is)2/c:,
(6)
with the assumption Im(k:) > 0 (absorption condition for porous media 12, 31). The final results will be valid even ifs = 0. We will solve separately the cases Sin a, and Sin Qz; in each case we will be essentially interested in p1 or p2, respectively. The application of the first cor@uration is to the reflection of a sound wave by a layer of an absorbing medium on a hard plane. The second configuration is very close to the problem of the propagation of sound waves in shallow water, the Neumann condition having to be replaced by a Dirichlet one.
3. THE SOURCE
IN THE
HALF-SPACE
z > h
In this configuration, the right-hand side of equation (2) is identically zero. Upon using the two-dimensional Fourier transform with respect to the variables x and y, and taking into account the cylindrical symmetry of the problem, equations (1) to (5) become (d2/dz2 + K:)/?, = as (d2/dz2 + K:)p2 = 0 Tr@, = Trj?,,
z > h,
(7)
0~ Z-Ch,
(8)
in in
on
Tr dzb2 = 0
z=h,
on
TrdZP1lpl = T&02/~, z = 0,
(9,10>
(11)
with Kf = -47~’C2+ kf,
(12)
m ,Ci = 2~
s
r = (x2 + y2)‘j2.
Jo(27t{r)p,(r, z) r dr,
(13)
0
The solution of equations (7) to (11) is taken in the form &K,(s+z) eiKl Is-2I A
=
___
-
2iK,
(14)
’
2iK,
fA
(15)
The functionj, obviously satisfies the condition (11); the continuity conditions (9) and (10) lead to the linear system eiK2h
eiK,Cs+h) ----_A
----APl
eiK,Cs-h)
e-iK,h B=-----
K2
K1 eiK,(s+h)
+
_
eiK2h
_
K, e-iKZh
B=-----. P2
’
&K,(s-h)
Pl
(16)
89
RESOLVENT OF THE PEKERIS OPERATOR
An easy calculation shows that A is given by A
=
e-ZiK,h
p2 Kl cos K2 h + ip, K2 sin K2 h p2 K, cos Kz h - ip, K, sin K2 h
=e
_ZiK,hp$ Kf cos2 K2 h - p: Kg sin’ Kz h + 2ip, p2 Kl K2 sin K2 h cos K2 h pt K: cos2 K2 h + pf K: sin2 K, h
.
(17)
To obtain the inverse Fourier transform of the second term of expression (14) (the scattered field), use will be made of the following inversion formula which can be deduced from equation (13): f(r) = 7t 7 Hb1)(2?c
(18)
--m with the definition
Hbl’(-x) = Hy)(x e’“).
If~(4rc2~2) has non-real poles only and no branch points, the integral (18) is obtained by the residue method, by using a contour of integration along the semi-circle r at infinity in the upper complex plane : ,f(r) = 2irc2x Res{Hf’(2x&-) f(4rr2 5’) 5} - rr s Hb”(2z
(19) r Furthermore, the scattered field will be decomposed into three terms: one corresponding to the image of the source with respect to the interface plane, and two layer potentials; this procedure will avoid the computation of the branch integrals which usually are involved. 3.1.
THE POLES OF A
The poles of A are given by the equation D(47r2r2) = p$ Kf cos2 K2h + pf K$ sin2 K2 h = 0.
(20)
A first remark is that if (2&) is a solution of equation (20), (-2~~5,) is a solution too, because (20) is an even function; furthermore, because of the definition of k:, the 5, have non-zero imaginary parts; it is thus ensured that there exist as many poles in the upper as in the lower complex half-plane, and none on the real axis. We will make use of the following notation : A, = 2lr5”
with
Im(,?,) > 0.
The classical mode equation is tan K2 h = -i p2 K,lp, K2.
(21)
The second remark is that, though equation (20) has the same solutions as equation (21), it is obvious that it can have other solutions: namely, those of tan Kz h = i p2 K,/p, K2.
(21’)
But if pl, cl, p2 and c2 are all real, it can be easily seen that equation (21’) will introduce no additional modes. Nevertheless, the fact that equation (20) can have more solutions than equation (21) does not introduce any difficulty. Note that, for simplicity, we assume that all the zeros of D(4rc25’) are simple ones. Finally, we will denote by D’(;i2) the function dD(A2)
D’(A2) = -
W’)
= -
p:cos2K2h+p:sin2K,h+h
sin K2 h cos K2 h
which is the derivative of D with respect to A2= 4n2 t2.
K,
D. HABAULT AND P. J. T. FILIPPI
90
3.2.
THE FIRST TWO TERMS OF THE SCATTERED FIELD
The first two terms of the scattered field are given by the inverse Fourier transform q1 of the function PgKf c~s2 K2h _ p:K:~in2K2heiK~(~+Z-2h) ^ eiKI(S+z-h) (T1= (23) pf Kf cos2 K2 h + pf K$ sin’ K2 h 2iKi = ‘I 2iK, ’ The function P1 cannot be inverted by formula (18) because the Fourier integral (18) is not convergent. It is necessary to write it in the form 3 I = (-4n2 5’ + a2) i,/(-479 t2 + LY2) = (-47~~5’ + a’) h,(4r2 r’)/[(-47r2 t2 + cr’) D(47c2[‘)I
V,cc,
with
Ima#O,
(24)
which leads to
(A(,, being the two-dimensional Laplace operator). No branch integral occurs because P1is an even function of Kl and K,; the integral of h, D-‘(-4n2 t2 + u2)-’ Hkl)r alongthe real axis is convergent because of the term (-479c2 + cr2)-‘; the integral along r of this function is zero because of the exponential decrease of the Hankel function. Using the well-known property of the Hankel function (Dirac measure in rW’>, one obtains i
W:) Hb”(J.”p) + aC2, D’(A;)
r, = -
4c
(26) c
The coefficient of a(,, can be calculated by remarking that it is the sum of the residues in the upper half plane of the function -(4n/i) h,(4n2 5’)
(27)
This function being odd in 5, its integral along the real axis is zero; to evaluate the integral of expression (27) on I’, it is necessary to remark that Kz ~-4~2 R2 e2ie
(5 = Re”),
cos’ K2 h E ${exp(-4rrReie h) + 2 + exp(47cRe” h)},
sin” K2h z -&{exp (-4nReie h) - 2 + exp(4nReie h)}. Either one or the other of the exponential functions will give a contribution, the sign of case. Nevertheless, in each quadrant, one has hi/D N (P: + PMP:
depending on
- ~3.
and so --
4n Rh,(4n2 R2 e2je)
RZ $io
i s D(47r2 R2 ezie) -47~’ RZ ezie + a2
dO=-----. _
P2 + PI
P: - Pt
(28)
0
Finally, q1 will involve a wave emitted by the image source S’ = (O,O,2h - s), and a simple layer potential due to a source distribution in the plane containing the image source and
RESOLVENT
91
OF THE PEKERIS OPERATOR
parallel to the interface : pf + pf 41 =-
i
eik,r(S’sX)
h,(2)
-
p: - p: 47rr(S’,X) - 4
D’(#J
eik,rW') Ha ‘[A, p(X’)]
z’=2h-s
x = (XT y. 3.3.
THE THIRD
47W(X , X’)
z),
do’.
da’ = dx’ dp’.
TERM OF THE SCATTERED
(29:1
FIELD
The third term of the scattered field is given by the inverse Fourier transform q2 of the function eiK,Cs+z-2hl 2p, p2 K, sin K2 h cos K, h 42
=
pi Kf cos2 K2 h + pi Ki sin’ K, h
=
F,iK,
eik’,b’+r-2h)
h2(drc2
2iK,
(2)
= D(4n2r2)
iK,
2iK,
eiK,(s+r-2h)
iK
2iK,
1
(30)
’
Here again, PZis even with respect to K1 and K,, and no branch integral will occur. It can be easily seen that the integral of P2HP) { along the real axis is convergent: indeed e2inlCl P2Hd” 5 E m x constant,
for
151-+ a,
which is integrable at infinity. Furthermore, the integral along r is zero because of the exponential decrease of the Hankel function; one thus obtains i
r2 = -
4c
h2(X) H::‘(A, p). D’(1.i)
-
(31)
The coefficient iK, in C&can be interpreted as the derivative factor of eiKl(s+Z-Zh) with respect to z’ = -2h + s and q2 is represented by a double layer potential :
g
elkv(X,X')
WV-ndJ31 n
I&
x,)
4~r(X
do’.
(32)
r’=2h-s
3.4.
REMARKS
ON THE FINAL RESULT
The total pressure field pi in the unbounded domain Q, is finally given by eikIr(S.X)
PIG% X) = -
pi
+
pf
eiklrG:X)
-___ p: - pf 4nr(S’. X) -4 4nr(S, X)
i
h,(X) r’=2h-s
eik,r’X.X”
’
4m( X, X’)
SC
M%) -Hb”[A,p(X’)] D’(X)
S’ = (0, 0, -s),
x = (x3 Y, z).
da’ + f
D’(X)
H;“[i, p( A”)] x a
eiklr(X.X’)
-
az’ 4nr(X, x’)
d,s’,
(33)
where
.s= a p(Y)
0, s>,
= (x’2 + y’2)1’2,
X’ = (x’, y’. z’),
da’ = dx’ dy’,
D’(A2), h,(A’) and h2(A2) are given by equations (22), (24) and (30), respectively, and the A, are solutions of D(A$J= 0, with Im(A,) > 0. Formula (33) can be compared to formula (20) given in reference [3] in the case of the reflexion of a spherical wave by a half space (Sz, = z < 0) : the amplitude of the field due to the image source is the same in both cases; as a first approximation (highly absorbing medium in
92
D. HABAULT
AND P. J. T. FILIPPI
Q,), the scattered field is that of the image source, and it does not depend on the thickness of the layer.
4. THE SOURCE IN THE LAYER 0 < z < h When the source is in the layer, the right-hand side of equation (1) is identically zero. By a Fourier transform, bne again obtains equations (7) to (1 l), the only difference being that equation (7) has a zero second member, while that of equation (8) is 6,. The solution ($&) may be taken in the form 0, = AeiKl’/2iK, P2
=
(34) eiK2(s+z)
eiK21s-zl
e
2iK2 +--- 2iK,
iK,z
+
e-iK2z
(35)
2iK,
fB
The coefficients A and B are solutions of the linear system: e iK,h
--A
KI
eiKlh
+
eiK,” --A+
eiKICh-sl
e-iKzh
+
eiKZ(h+s)
g=--
+
K2 eiK,h
_
Kz
e-iK2h
eiK,(h-s,
+
’ eiK2(h+s)
B=-
(36) Pz
Pz
Pl
which gives B=(l/D){-K:pzcosK,h+
iK:p?sinK,h+
+ K1 p, K2 p2 cos K2 h - iK1 p, K2 p2 sin K2 h} cos K2 seiKzh.
(37)
The procedure for invertingfi, is much the same as in section 3, and will lead to layer potentials due to source distributions in the planes z = h and z = -h. 4.1.
FIRST COMPONENT
q1 OF pz
The first component, ql, is the inverse of 4, =--
K: p$ cos K2 h cos K2 seiKlr + e-1K2zeiK,h
=-p:KfD
2iK2
D g, e”‘zz + e-iK>
2 eiKlh
2iK2
’
The Fourier integral of Kfg,/D is convergent: indeed, for large positive t it has the following asymptotic behaviour : (_4$ 52eZnC(s+h))/(_4TC2 (2 ednr”), (39) which decreases exponentially because h > s. On r, this function is of the form -47r2 R2 ezie exp{f2nRe”(s
+ h)}/[-4n2 R2 ezioexp {-&47rRele h}],
which is, for the same reason, exponentially decreasing when R --f cc. The function q1 is thus given by
(40)
93
RESOLVENT OF THE PEKERIS OPERATOR
with X = (x, y, z),
p(X”) = (x”2 + y”2)‘/2,
p(Y)
X” = (x”, y”, zq,
X’ = (x’, y’, z’),
= (x2 + y’2)1’z,
da” = dx”dy”.
da’ = dx’ dy’,
4.2. SECOND COMPONENT q2 OF& q2, is given by the inverse of
The second component,
q2 = iK2pf
K2 sin K2 h cos Kz s eiKIr + emiK2’
eiKZh
D g2eiKzz
=
+
2iK2 e-iK2z eiK2h,
iKZpi-
(42)
2iK,
D
It is easily seen that a double layer potential
of q2 can be obtained
representation a
g2m
eikzr(X,X’)
H;“[& p( X’)] -
-
azl 47cr(~, x’)
D’( j.:)
:
da’ -
(43)
4.3. THIRD COMPONENT q3 0FP2 The function
g3 is cos K2 h cos K2 s
1
43=--K:ip1P2
D
K,
eiKzz+e-iK,z eiK,h.
iK2
(44)
2iK2
As shown in section 4.1, $75-1K2
cosK,hcosK,s 1
i
D
=4
g,(X) c
(G
-
A3
m
n
W,‘V.
~1.
Furthermore, 1/K,) = -2i eikIP/4np.
P-‘( The function
q3 can be represented
a 1 aelk2rW.X”) 1(X”)-
by double layer potentials
:
eiklr(X,X’)
r(k:-~,2,)gl(E.:)H~l)(r3.P):2)e’k’P D’(j.$ - L;) g;$
Hb”(& p)t,
es
n where the symbol & stands for the convolution f(d2~
g(p)
=
product
af 4Xrr(~, x~)
do”,
(45)
in the plane:
Y’,o)lgW’> 01 dg’,
jfb(
Y' = (x’, y’),
y = (x9Y),
4np
dt+= dx’ dy’.
4.4. FOURTH COMPONENT q4 OFP, The last component,
q4, is obtained
4 =-LKZ’ l’PlP2 Kl
by inverting
Kz sin K2 h cos K, s eiKzr + eciKzz D
2iK,
eiKth ’
(46)
D. HABAULT
94
AND P. J. T. FILIPPI
Upon using the results of sections 4.2 and 4.1, the function q4 is readily seen to be elkzr(X,X’)
47cr(X, X’)
do’ +
eik2rW.X”)
4zr(X, X’) 4.5.
da”.
(47)
THE FINAL RESULT
The total pressure field pz in the layer Q, is finally given by
g1m (x> + $p; 1 (k: - 2:)D’. n HW, P) [b=, + k4,]
eik2r
p2 = -{d, + S,,} * -
4nr
* ;;
(x) +
e% r
HW, P)i&h - &=-,I * 4nr
-
g,
W:- 2) ~‘0
(x) -
fz
m W,% P):z, 4np
n
[S,,, - S,,_,] efktr
e’k’ [6z = ph + d,,_,] *j$X) W’(~.P):,,~ where * indicates the convolution product in R3, or, more explicitly, elk&,X)
PI6
Jo = -
elkzr
i
(S’.X)
47cr(S, X) - 4nr(S’, X) elkzr
(
2 (k: - A:) g
+-P:
4
a
Hb%
P(X’)I a
&W -
D’tX)
WV.
4nr(X
x”) 9
r”=-h g2G)
-_h
[A,p(X)] x eik2r’X.X”’
p(X”)]
da’ +
D’(X)
&‘=
Hi"
X,X’)
’ 47cr(X, X’)
-
II
G
elkzr’X.X”
4nr(X
7
x,) cb’ -
eikzr(X,X”)
P(X”)IF 4xrcX x,,j da” + 3
da”
95
RESOLVENTOFTHEPEKERISOPERATOR
where
s = (0, 0, 4, da’ = dx’dy’,
S’= (0, 0, -s), da” = dx” dy”.
x= (X,Y,Z),
X’ = (x’, y’, z’),
X” = (xP, y”, z”),
D’Q’), gl(l’) and gz(n2) are given by expressions (22) (38) and (42), respectively, and the 2, are solutions of D(,Q = 0, with Im(&) > 0. 5. CONCLUDING REMARKS The representation of the sound pressure field given by formulas (33) and (48) has several advantages. First of all, it does not involve branch integrals, which usually appear in the literature. The second advantage is that more or less simple analytic approximations of the integrals can be established, depending on the absorption rate of sound waves in the different media which governs the decrease of the layer densities; furthermore, numerical computations can easily be performed. The third advantage is that the contributions of each mode appear as layer potentials: that is, as functions which are analytic everywhere in the propagation domain; a global inversion of the Fourier transforms a1 or d2 would have given products of functions of the z variable by Hankel functions, the singularity of which could introduce difficulties in the computation of the near field. The last point of interest about these results is that they are valid whether the media are conservative ones (pl, cl, p2 and c2 all real parameters) or not (pl, cl, p2 and c2 real or complex). This shows that the spectral representation of the resolvent of the Pekeris operator proposed in reference [l] for the real case applies equally for the complex one. It would be interesting to see if the spectral theory can be applied to solve transient wave propagation problems in absorbing media, just as it can be for conservative systems. REFERENCES 1. C. WILCOX 1976Archives of Rational Mechanics and Analysis 60,259-300.
Spectral analysis of the Pekeris operator in the theory of acoustic wave propagation in shallow water.
2. TH. LEVY and E. SANCHEZ-PALENCIA (to appear) Journal of Mathematical Analysis and Applications. Equations and interface conditions for acoustic phenomena in porous media. 3. P. J. T. FILIPPIand D. HABAULT 1978Journal of Sound and Vibration 56,97-103. Reflexion of a spherical wave by the plane interface between a perfect fluid and a porous medium.