4.2. GREEN’S REPRESENTATION THEOREM 57

i.e., the normal velocity on the boundary is proportional to the excess pressure onthe boundary. The coefficient χ is called the acoustic impedance of the obstacle D,and is, in general, a space dependent function defined on the boundary ∂D. Thisimpedance condition leas to a boundary-value problem for the velocity potential uof the form

∂nu+ iλu = 0,

where λ = χρ(ω + iγ).

4.2 Green’s representation theorem

We begin our analysis by establishing the basic property that any solution to theHelmholtz equation can be represented as the combination of a single- and a double-layer acoustic surface potential. It is easily verified that the function

G(x, y) =1

4π

eiκ|x−y|

|x− y|, x, y ∈ R3, x = y,

is a solution to the Helmholtz equation

∆G(x, y) + κ2G(x, y) = 0

with respect to x for any fixed y. Because of its polelike singularity at x = y, thefunction G is called a fundamental solution to the Helmholtz equation, i.e.,

∆G(x, y) + κ2G(x, y) = −δ(x− y).

Given a function ϕ ∈ C(∂D), the function

(4.1) u(x) =

∫∂D

G(x, y)ϕ(y)dsy, x ∈ R2 \ ∂D,

is called the acoustic single-layer potential with density ϕ. Since for x ∈ R3 \∂D, wecan differentiate under the integral sign, we see that u is a solution of the Helmholtzequation.

Given a function ψ ∈ C(∂D), the function

(4.2) v(x) =

∫∂D

∂nyG(x, y)ψ(y)dsy, x ∈ R3 \ ∂D

is called the acoustic double-layer potential with density ψ. We assume unit normaln to be directed into the exterior domain R3 \ D. We note that the double-layer

58 CHAPTER 4. OBSTACLE SCATTERING

potential v is also a solution to the Helmholtz equation. In the following, we shalldistinguish by indices + and − the limits obtained by approaching the boundary∂D from inside R3 \D and D, respectively, i.e.,

v+(x) = limy→x, y∈R3\D

v(y), v−(x) = limy→x, y∈D

v(y), x ∈ ∂D.

For any domain Ω with boundary ∂Ω of class C2, we introduce the linearspace X(G) of all complex-valued function u ∈ C2(G)∩C(G) for which the normalderivative on the boundary exists in the sense that the limit

∂nu(x) = limh→0, h>0

⟨n(x),∇u(x− hn(x))⟩, x ∈ ∂G,

exists uniformly on ∂G. We note that the assumption u, v ∈ X(G) suffices toguarantee the validity of the first Green’s theorem

(4.3)

∫G

u∆vdx =

∫∂G

u∂nvds−∫G

∇u · ∇vdx

and the second Green’s theorem

(4.4)

∫G

(u∆v − v∆u)dx =

∫∂G

(u∂nv − v∂nu)ds

for a bounded domain G with C2 boundary ∂G.

We denote by D a bounded region in R3 with a boundary ∂D consisting of afinite number of disjoint, closed, bounded surfaces belonging to the class C2. Theexterior R3 \D is assumed to the connected, whereas D itself may have more thanone component. We assume the normal n to ∂D to be directed into the exterior ofD.

Theorem 4.2.1. Let u ∈ X(D) be a solution to the Helmholtz equation

∆u+ κ2u = 0 in D.

Then ∫∂D

[u(y)∂nyG(x, y)− ∂nyu(y)G(x, y)

]dsy =

−u(x), x ∈ D,

0, x ∈ R3 \D.

Proof. We choose an arbitrary fixed point x ∈ D and circumscribe it with a sphereBx,r := y ∈ R3, |x − y| = r. We assume the radius r to be small enough suchthat Bx,r ⊂ D and direct the unit normal ν to Bx,r into the interior of Bx,r, as seen

4.2. GREEN’S REPRESENTATION THEOREM 59

x

D

Bx,r

ν

Figure 4.1: Domain D and ball Bx,r.

in Figure 4.1. Now we apply the second Green’s theorem (4.4) to the function u(y)and G(x, y) in the region y ∈ D, |x− y| > r to obtain∫

∂D+Bx,r

[u(y)

∂G(x, y)

∂νy− ∂u

∂ν(y)G(x, y)

]ds(y) = 0.

Since on Bx,r, we have

G(x, y) =eiκr

4πr, ∇yG(x, y) =

(1

r− iκ

)eiκr

4πrν(y).

Simple calculation, using the mean value theorem, shows that

limr→0

∫Bx,r

[u(y)

∂G(x, y)

∂νy− ∂u

∂ν(y)G(x, y)

]ds(y) = u(x).

The statement for x ∈ R3 \D readily follows from Green’s theorem applied tothe function u(y) and G(x, y) in the region D.

Straightforward calculations show that(x

|x|,∇xG(x, y)

)− iκG(x, y) = O

(1

|x|2

), |x| → ∞

and (x

|x|,∇x

∂G(x, y)

∂ν(y)

)− iκ

∂G(x, y)

∂ν(y)= O

(1

|x|2

), |x| → ∞

uniformly for all directions x/|x| and uniformly for all y contained in the boundedset ∂D. From this we conclude the following.

60 CHAPTER 4. OBSTACLE SCATTERING

Theorem 4.2.2. Both the single-layer acoustic potential defined by (4.1) and thedouble-layer acoustic potential defined by (4.2) satisfy the Sommerfeld radiation con-dition (

x

|x|,∇u(x)

)− iκu(x) = o

(1

|x|

), |x| → ∞

uniformly for all directions x/|x|.

As we shall soon see, the Sommerfeld radiation condition completely charac-terizes the behavior of solutions to the Helmholtz equation at infinity.

Theorem 4.2.3. Let u ∈ X(R3 \D) be a solution to the Helmholtz equation

∆u+ κ2u = 0 in R3 \D

satisfying the Sommerfeld radiation condition(x

|x|,∇u(x)

)− iκu(x) = o

(1

|x|

), |x| → ∞

uniformly for all directions x/|x|. Then∫∂D

[u(y)∂nyG(x, y)− ∂nyu(y)G(x, y)

]dsy =

0, x ∈ D,

u(x), x ∈ R3 \D.

Proof. We first show that

(4.5)

∫|y|=R

|u|2ds = O(1), R → ∞.

To accomplish this, we first observe that from the radiation condition it follows that

0 = limR→∞

∫|y|=R

∣∣∣∂u∂ν

− iκu∣∣∣2ds

= limR→∞

∫|y|=R

∣∣∣∂u∂ν

∣∣∣2 + |κ|2|u|2 + 2Im

(κu∂u

∂ν

)ds(4.6)

where ν denotes the outward unit normal to the sphere BR := y ∈ R3, |y| = R.We take R large enough so that BR ⊂ R3 \D and apply the first Green’s theoremin the domain DR := y ∈ R3 \D, |y| < R to obtain

κ

∫|y|=R

u∂u

∂νds = κ

∫∂D

u∂u

∂νds− κ|κ|2

∫DR

|u|2dy + κ

∫DR

|∇u|2dy.

4.2. GREEN’S REPRESENTATION THEOREM 61

Now we substitute the imaginary part of the above equation into (4.6) and find that

limR→∞

∫|y|=R

∣∣∣∂u∂ν

∣∣∣2 + |κ|2|u|2ds+ 2Im

∫DR

|κ|2|u|2 + |∇u|2dy

= −2Imκ

∫∂D

u∂u

∂νds.(4.7)

All four terms on the left-hand side of above equation are nonnegative since Imκ ≥ 0.Hence these terms must be individually bounded as R → ∞ since their sum tendsto a finite limit. Equation (4.5) follows immediately.

We now note the identity∫|y|=R

[u(y)

∂G(x, y)

∂ν(y)− ∂u

∂ν(y)G(x, y)

]ds(y)

=

∫|y|=R

u(y)

[∂G(x, y)

∂ν(y)− iκG(x, y)

]ds(y)

−∫|y|=R

G(x, y)

[∂u

∂ν(y)− iκu(y)

]ds(y) =: I1 + I2

and apply Schwartz’s inequality to each of the integrals I1 and I2. From the radiationcondition

∂G(x, y)

∂ν(y)− iκG(x, y) = O

(1

R2

), y ∈ BR,

for the fundamental solution and (4.5) we see that I1 = O(1/R) as R → ∞. Theradiation condition and G(x, y) = O(1/R), y ∈ BR, yield I2 = o(1) for R → ∞.Hence

limR→∞

∫|y|=R

[u(y)

∂G(x, y)

∂ν(y)− ∂u

∂ν(y)G(x, y)

]ds(y) = 0.

The proof is now completed as in Theorem 4.2.1 by applying the second Green’stheorem in the domain y ∈ DR, |x− y| > r if x ∈ R3 \D or DR if x ∈ D.

Remark 4.2.4. It is obvious that any solution of the Helmholtz equation satisfyingthe Somerfeld radiation condition automatically satisfies

u(x) = O

(1

|x|

), |x| → ∞

uniformly for all directions x/|x|.

Note that it is not necessary to impose this additional condition for the repre-sentation theorem to be valid. Physically, the fundamental solutionG(x, y) describesan outgoing spherical wave of the form

ei(κ|x−y|−ωt)

4π|x− y|.