Representation of Solutions

11 Representation of Solutions

In this chapter we derive representation formulae for the solution to the transmission problem 1 ∇ · ( ∇u) + ω 2 εδ u = 0 in Ω , (11.1) µδ similar to (2.17) and (2.19).

11.1 Preliminary Results We begin this chapter by deriving the outgoing fundamental solution to the Helmholtz equation. We refer to the books of Colton and Kress [90], [91] and the one of N´ed´elec [224] for detailed treatments of the Helmholtz equation, emphasizing existence and uniqueness results for the exterior problem. Let Ω be a bounded domain in IRd , d = 2 or 3, with a connected Lipschitz boundary, a permeability equal to µ0 > 0, and a permittivity equal to ε0 > 0. Consider a bounded domain D ⊂⊂ Ω with a connected Lipschitz boundary, a permeability 0 < µ
= µ0 < +∞, and a permittivity 0 < ε
= ε0 < +∞. Let √ √ k0 := ω ε0 µ0 and k := ω εµ, where ω > 0 is a given frequency. A fundamental solution Γk (x) to the Helmholtz operator ∆ + k 2 in IRd is a solution (in the sense of distributions) of (∆ + k 2 )Γk = δ0 ,

(11.2)

where δ0 is the Dirac mass at 0. Solutions are not unique, since we can add to a solution any plane wave (of the form eikθ·x , θ ∈ IRd : |θ| = 1) or any combination of such plane waves. We need to specify the behavior of the solutions at infinity. It is natural to look for radial solutions of the form Γk (x) = wk (r) that is subject to the extra Sommerfeld radiation condition or outgoing wave condition





dwk −(d+1)/2



− ikw at infinity. (11.3) k ≤ Cr

dr H. Ammari and H. Kang: LNM 1846, pp. 185–195, 2004. c Springer-Verlag Berlin Heidelberg 2004


186

11 Representation of Solutions

If d = 3, equation (11.2) becomes 1 d 2 dwk r + k 2 wk = 0, r2 dr dr

r>0,

whose solution is

eikr e−ikr + c2 . r r It is easy to check that the Sommerfeld radiation condition (11.3) leads to c2 = 0 and then (11.2) leads to c1 = −1/(4π). If d = 2, equation (11.2) becomes wk (r) = c1

1 d dwk r + k 2 wk = 0, r dr dr

r>0.

This is a Bessel equation whose solutions are not elementary functions. It is known that the Hankel functions of the first and second kind of order (1) (2) 0, H0 (kr) and H0 (kr), form a basis for the solution space. At infinity (1) (r → +∞), only H0 (kr) satisfies the outgoing radiation condition (11.3). (1) (1) At the origin (r → 0), H0 (kr) behaves like H0 (kr) ∼ (2i/π) log(r). The following lemma holds. Lemma 11.1 The outgoing fundamental solution Γk (x) to the operator ∆ + k 2 is given by  i 1    − H0 (k|x|) , d = 2 , 4 Γk (x) = eik|x|   , d=3, − 4π|x| for x
= 0, where H01 is the Hankel function of the first kind of order 0. Let for x
= 0  1  log |x| , d = 2 ,  2π Γ0 (x) := Γ (x) = 1  − , d=3. 4π|x| k k For a bounded domain D in IRd and k > 0 let SD and DD be the single and double layer potentials defined by Γk , that is,  k SD ϕ(x) = Γk (x − y)ϕ(y) dσ(y) , x ∈ IRd , ∂D ∂Γk (x − y) k DD ϕ(x) = ϕ(y) dσ(y) , x ∈ IRd \ ∂D , ∂ν y ∂D

for ϕ ∈ L2 (∂D). Because Γk − Γ0 is a smooth function, we can easily prove from (2.12) and (2.13) that

11.1 Preliminary Results

 k ∂(SD ϕ)

(x) = ± ∂ν ±



k

(DD ϕ) (x) = ∓ ±

 1 k ∗ I + (KD ) ϕ(x) a.e. x ∈ ∂D , 2  1 k I + KD ϕ(x) a.e. x ∈ ∂D , 2

k is the operator defined by for ϕ ∈ L2 (∂D), where KD  ∂Γk (x − y) k KD ϕ(x) = p.v. ϕ(y)dσ(y) , ∂νy ∂D k ∗ k ) is the L2 -adjoint of KD , that is, and (KD  ∂Γk (x − y) k ∗ ) ϕ(x) = p.v. ϕ(y)dσ(y) . (KD ∂νx ∂D

187

(11.4) (11.5)

(11.6)

(11.7)

k k ∗ The singular integral operators KD and (KD ) are bounded on L2 (∂D). We will need the following important result from the theory of the Helmholtz equation. For its proof we refer to [91] (Lemma 2.11).

Lemma 11.2 (Rellich’s lemma) Let R0 > 0 and BR (0) = {|x| < R}. Let u satisfy the Helmholtz equation ∆u + k02 u = 0 for |x| > R0 . Assume, furthermore, that  lim

R→+∞

∂BR (0)

|u(x)|2 dσ(x) = 0 .

Then, u ≡ 0 for |x| > R0 . Note that the assertion of this lemma does not hold if k0 is imaginary or k0 = 0. Now we can prove the following uniqueness result. Lemma 11.3 Suppose d = 2 or 3. Let D be a bounded Lipschitz domain in 1,2 IRd . Let u ∈ Wloc (IRd \ D) satisfy   ∆u + k02 u = 0 in IRd \ D ,  



   

x 

∂u −(d+1)/2

, as |x| → +∞ uniformly in

∂|x| − ik0 u = O |x| |x|     ∂u   dσ = 0 . u  ∂D ∂ν Then, u ≡ 0 in IRd \ D. Proof. Let BR (0) = {|x| < R}. For R large enough, D ⊂ BR (0). Notice first that by multiplying ∆u + k02 u = 0 by u and integrating by parts over BR (0) \ D we arrive at  ∂u dσ = 0 ,  u ∂ν ∂BR (0)

188

11 Representation of Solutions



since 

u ∂D

∂u dσ = 0 . ∂ν

   ∂u − ik0 u dσ = −k0 u |u|2 . ∂ν ∂BR (0) ∂BR (0)



But 

Applying the Cauchy–Schwarz inequality,

2 1/2

 1/2     

∂u



∂u 2





 ≤ −ik −ik u |u| 0 u dσ

0 u dσ



∂ν ∂BR (0) ∂BR (0) ∂BR (0) ∂ν and using the Sommerfeld radiation condition



 

∂u

−(d+1)/2



as |x| → +∞ ,

∂|x| − ik0 u = O |x| we get

 

1/2 

C ∂u − ik0 u dσ

≤ u |u|2 , ∂ν R ∂BR (0) ∂BR (0)





for some positive constant C independent of R. Consequently, we obtain that 1/2  C , |u|2 ≤ R ∂BR (0) which indicates by the Rellich’s Lemma that u ≡ 0 in IRd \ BR (0). Hence, by the unique continuation property for ∆ + k02 , we can conclude that u ≡ 0 up to the boundary ∂D. This finishes the proof.  

11.2 Representation Formulae We now present two representations of the solution of (11.8) similar to the representation formula (2.39) for the transmission problem for the harmonic equation. Let f ∈ W 12 (∂Ω), and let u and U denote the solutions to the 2 Helmholtz equations   ∇ · ( 1 ∇u) + ω 2 ε u = 0 in Ω , δ µδ (11.8)  u = f on ∂Ω , and



∆U + ω 2 ε0 µ0 U = 0 in Ω , U = f on ∂Ω .

(11.9)

The following theorem is of importance to us for establishing our representation formulae.

11.2 Representation Formulae

189

Theorem 11.4 Suppose that k02 is not a Dirichlet eigenvalue for −∆ on D. For each (F, G) ∈ W12 (∂D) × L2 (∂D), there exists a unique solution (f, g) ∈ L2 (∂D) × L2 (∂D) to the system of integral equations  k k0  SD f − SD g = F



k0 k on ∂D . (11.10) g)

f )

1 ∂(SD 1 ∂(SD  − =G 



µ ∂ν µ0 ∂ν −

+

Furthermore, there exists a constant C independent of F and G such that   (11.11)

f L2 (∂D) + g L2 (∂D) ≤ C F W12 (∂D) + G L2 (∂D) , where in the three-dimensional case the constant C can be chosen independently of k0 and k if k0 and k go to zero. Proof. We only give the proof for d = 3 and µ0
= µ leaving the general case to the reader. Let X := L2 (∂D) × L2 (∂D) and Y := W12 (∂D) × L2 (∂D), and define the operator T : X → Y by

  k0 k f )

g)

1 ∂(SD 1 ∂(SD k0 k T (f, g) := SD f − SD g, − . µ ∂ν − µ0 ∂ν + We also define T0 by T0 (f, g) :=



  0 0 f )

g)

1 ∂(SD 1 ∂(SD 0 0 SD f − SD g, − . µ ∂ν − µ0 ∂ν +

k0 0 − SD : L2 (∂D) → W12 (∂D) is a compact operator, We can easily see that SD k0 ∂ ∂ 0 |± : L2 (∂D) → L2 (∂D). Therefore, T − T0 is a and so is ∂ν SD |± − ∂ν SD compact operator from X into Y . It can be proved that T0 : X → Y is invertible. In fact, a solution (f, g) of the equation T0 (f, g) = (F, G) is given by 0 −1 f = g + (SD ) (F )

g=

  µ0 µ 1 1 0 ∗ −1 0 ∗ 0 −1 (λI + (KD ) ) ) )((SD ) (F )) , G + ( I − (KD µ0 − µ µ 2

0 where λ = (µ + µ0 )/(2(µ − µ0 )). Recall now that the invertibility of SD and 0 ∗ λI+(KD ) was proved in Theorems 2.8 and 2.13. Thus we see, by the Fredholm alternative, that it is enough to prove that T is injective. Suppose that T (f, g) = 0. Then the function u defined by  k0 SD g(x) if x ∈ IRd \ D , u(x) := k SD f (x) if x ∈ D ,

190

11 Representation of Solutions

is the unique solution of the transmission problem (11.8) with Ω replaced by IRd and the Dirichlet boundary condition replaced by the Sommerfeld radiation condition



∂u

−(d+1)/2



) , |x| → ∞ .

∂|x| (x) − ik0 u(x) = O(|x| Using the fact that



   µ0 µ0 ∂u

∂u

u ¯ dσ = u ¯ dσ = (|∇u|2 − k 2 |u|2 ) dx ,



∂ν µ ∂ν µ ∂D ∂D D + − 

we find that 

∂D

∂u

u ¯ dσ = 0 , ∂ν +

which gives, by applying Lemma 11.3, that u ≡ 0 in IRd \ D. Now u satisfies (∆ + k 2 )u = 0 in D and u = ∂u/∂ν = 0 on ∂D. By the unique continuation property of ∆ + k 2 , we readily get u ≡ 0 in D, and hence in IRd . In particular, k0 k0 SD g = 0 on ∂D. Since (∆ + k02 )SD g = 0 in D and k02 is not a Dirichlet k0 eigenvalue for −∆ on D, we have SD g = 0 in D, and hence in IRd . It then follows from the jump relation (11.4) that



k0 k0 g)

g)

∂(SD ∂(SD − =0 g= ∂ν + ∂ν −

on ∂D .

k k k On the other hand, SD f satisfies (∆ + k 2 )SD f = 0 in IRd \ D and SD f =0 on ∂D. It then follows from Lemma 11.3 (see also Theorem 3.7 of [91]), that k f = 0. Then, in the same way as above, we can conclude that f = 0. This SD finishes the proof of solvability of (11.10). The estimate (11.11) is an easy consequence of solvability and the closed graph theorem. Finally, it can be easily proved in the three-dimensional case that if k0 and k go to zero, then the constant C in (11.11) can be chosen independently of k0 and k. We leave the details to the reader.   The following representation formula holds.

Theorem 11.5 Suppose that k02 is not a Dirichlet eigenvalue for −∆ on D. Let u be the solution of (11.8) and g := ∂u ∂ν |∂Ω . Define k0 k0 (g)(x) + DΩ (f )(x) , H(x) := −SΩ

x ∈ IRd \ ∂Ω ,

and let (ϕ, ψ) ∈ L2 (∂D) × L2 (∂D) be the unique solution of  k k0  SD ϕ − SD ψ = H



k0 k on ∂D .



 1 ∂(SD ϕ) − 1 ∂(SD ψ) = 1 ∂H 



µ ∂ν µ0 ∂ν µ0 ∂ν − +

(11.12)

(11.13)

11.2 Representation Formulae

Then u can be represented as  k0 H(x) + SD ψ(x) , x ∈ Ω \ D , u(x) = k SD ϕ(x) , x ∈ D . Moreover, there exists C > 0 independent of H such that  

ϕ L2 (∂D) + ψ L2 (∂D) ≤ C H L2 (∂D) + ∇H L2 (∂D) .

191

(11.14)

(11.15)

Proof. Note that u defined by (11.14) satisfies the differential equations and the transmission condition on ∂D in (11.8). Thus in order to prove (11.14), it suffices to prove that ∂u/∂ν = g on ∂Ω. Let f := u|∂Ω and consider the following transmission problem:   (∆ + k02 )v = 0 in (Ω \ D) ∪ (IRd \ Ω) ,      (∆ + k 2 )v = 0 in D ,   



  1 ∂v

1 ∂v

  v|− − v|+ = 0 , − = 0 on ∂D , µ ∂ν − µ0 ∂ν + (11.16)







 ∂v ∂v 

−

= g on ∂Ω ,  v|− − v|+ = f,   ∂ν − ∂ν +   



  ∂v

 



(x) − ik0 v(x)

= O(|x|−(d+1)/2 ) , |x| → ∞ . ∂|x| We claim that (11.16) has a unique solution. In fact, if f = g = 0, then we can show as before that v = 0 in IRd \ D. Thus

∂v

v= = 0 on ∂D . ∂ν − By the unique continuation for the operator ∆ + k 2 , we have v = 0 in D, and hence v ≡ 0 in IRd . Note that vp , p = 1, 2, defined by  v1 (x) =

u(x) , 0,

x∈Ω,

x ∈ IR \ Ω , d

 v2 (x) =

k0 ψ(x) , H(x) + SD k SD ϕ(x)

,

x∈Ω\D ,

x∈D,

are two solutions of (11.16), and hence v1 ≡ v2 . This finishes the proof.   Proposition 11.6 For each n ∈ IN there exists Cn independent of D such that

H C n(D) ≤ Cn f W 21 (∂Ω) . 2

192

11 Representation of Solutions

Proof. Let g := ∂u/∂ν|∂Ω . By the definition (11.12), it is easy to see that  

H C n (D) ≤ C g W 2 1 (∂Ω) + f W 21 (∂Ω) , −

2

2

where the constant C depends only on n and dist(D, ∂Ω). Therefore it is enough to show that

g W 2 1 (∂Ω) ≤ C f W 21 (∂Ω) −

2

2

for some C independent of D. Let ϕ be a C ∞ -cutoff function which is 0 in a neighborhood of D and 1 in a neighborhood of ∂Ω. Let v ∈ W 12 (∂Ω) and define v ∈ W 1,2 (Ω) to be the 2 unique solution to ∆ v = 0 in Ω and v = v on ∂Ω. Let  , 12 ,− 12 denote the W 12 − W−2 1 pairing on ∂Ω. Then 2

2



v, g 12 ,− 12 =

 ∇(ϕu) · ∇ v dx

∆(ϕu) v dx + Ω

=

Ω ∆ϕu v dx + 2



Ω

Ω

∇ϕ · ∇u v dx − k02

 ϕu v dx Ω

∇(ϕu) · ∇ v dx .

+ Ω

Therefore, it follows from the Cauchy–Schwartz inequality that |v, g 12 ,− 12 | ≤ C u W 1,2 (Ω\D)  v W 1,2 (Ω) ≤ C u W 1,2 (Ω\D) v W 21 (∂Ω) . 2

2

Since v ∈ W 1 (∂Ω) is arbitrary, we get 2

g W 2 1 (∂Ω) ≤ C u W 1,2 (Ω\D) . −

(11.17)

2

Note that the constant C depends only on dist(D, ∂Ω). On the other hand, since k02 is not a Dirichlet eigenvalue for the Helmholtz equation (11.8) in Ω we can prove that

u W 1,2 (Ω) ≤ C f W 21 (∂Ω) , 2

where C depends only on ω 2 , µ0 , µ, ε0 , and ε. It then follows from (11.17) that

g W 2 1 (∂Ω) ≤ C f W 21 (∂Ω) . −

2

2

This completes the proof.   We now transform the representation formula (11.14) into the one using the Green’s function and the background solution U , that is, the solution of (11.9).

11.2 Representation Formulae

193

Let Gk0 (x, y) be the Dirichlet Green’s function for ∆ + k02 in Ω, i.e., for each y ∈ Ω, G is the solution of  (∆ + k02 )Gk0 (x, y) = δy (x) , x ∈ Ω , Gk0 (x, y) = 0 , x ∈ ∂Ω . Then,

 U (x) = ∂Ω

∂Gk0 (x, y) f (y)dσ(y) , ∂νy

x∈Ω.

Introduce one more notation. For a Lipschitz domain D ⊂⊂ Ω and ϕ ∈ L2 (∂D), let  Gk0 (x, y)ϕ(y) dσ(y) , x ∈ Ω . GkD0 ϕ(x) := ∂D

Our second representation formula is the following. Theorem 11.7 Let ψ be the function defined in (11.13). Then ∂(GkD0 ψ) ∂u ∂U (x) = (x) + (x) , ∂ν ∂ν ∂ν

x ∈ ∂Ω .

(11.18)

To prove Theorem 11.7 we first observe an easy identity. If x ∈ IRd \ Ω and z ∈ Ω then

 ∂Gk0 (z, y)

Γk0 (x − y) (11.19)

dσ(y) = Γk0 (x − z) . ∂νy ∂Ω ∂Ω As a consequence of (11.19), we have

   1 ∂Gk0 (z, ·)

∂Γk0 (x − z) k0 ∗ I + (KΩ ) , (x) =

2 ∂νy ∂νx ∂Ω

(11.20)

for all x ∈ ∂Ω and z ∈ Ω. Our second observation is the following. Lemma 11.8 If k02 is not a Dirichlet eigenvalue for −∆ on Ω, then (1/2) I + k0 ∗ ) : L2 (∂Ω) → L2 (∂Ω) is injective. (KΩ   k0 ∗ Proof. Suppose that ϕ ∈ L (∂Ω) and (1/2) I + (KΩ ) ϕ = 0. Define 2

k0 ϕ(x) , x ∈ IRd \ Ω . u(x) := SΩ

Then u is a solution of (∆ + k02 )u = 0 in IRd \ Ω, and satisfies the Sommerfeld radiation condition



 

∂u

−(d+1)/2



− ik u as |x| → +∞ , = O |x| 0

∂|x|

194

11 Representation of Solutions

and the Neumann boundary condition

  1 ∂u

k0 ∗ I + (KΩ ) ϕ = 0 . = ∂ν ∂Ω 2 k0 Therefore, by Lemma 11.3, we obtain SΩ ϕ(x) = 0, x ∈ IRd \ Ω. Since k02 is not a Dirichlet eigenvalue for −∆ on Ω, we can prove that ϕ ≡ 0 in the same way as before. This completes the proof.  

With these two observations available we are now ready to prove Theorem 11.7. Proof of Theorem 11.7. Let g := ∂u/∂ν and g0 := ∂U /∂ν on ∂Ω for convenience. By the divergence theorem, we get k0 k0 (g0 )(x) + DΩ (f )(x) , U (x) = −SΩ

x∈Ω.

It then follows from (11.12) that k0 k0 H(x) = −SΩ (g)(x) + SΩ (g0 )(x) + U (x) ,

x∈Ω.

Consequently, substituting (11.14) into the above equation, we see that for x∈Ω

  k0

ψ)

∂(SD k0 ∂H

k0 + (g0 )(x) + U (x) . H(x) = −SΩ (x) + SΩ ∂ν ∂Ω ∂ν ∂Ω Therefore the jump formula (11.4) yields

   k0 ∂(SD ψ)

∂H

1 ∂H k0 ∗ = − − I + (KΩ ) + ∂ν 2 ∂ν ∂Ω ∂ν ∂Ω   1 k0 ∗ + I + (KΩ ) (g0 ) on ∂Ω . 2

(11.21)

By (11.20), we have for x ∈ ∂Ω 

∂Γk0 (x − y) ψ(y) dσ(y) ∂νx ∂D

  1 ∂(GkD0 ψ)

k0 ∗ = I + (KΩ ) 2 ∂ν

k0 ∂(SD ψ) (x) = ∂ν

 (x) .

(11.22)

∂Ω

Thus we obtain

   k0 ∂(SD ψ)

1 k0 ∗ ) − I + (KΩ 2 ∂ν ∂Ω 

     1 ∂(GkD0 ψ)

1 k0 ∗ k0 ∗ I + (KΩ ) = − I + (KΩ ) 2 2 ∂ν ∂Ω

on ∂Ω .

11.2 Representation Formulae

195

It then follows from (11.21) that 

     1 ∂(GkD0 ψ)

1 ∂H

k0 ∗ k0 ∗ I + (KΩ ) + − I + (KΩ ) − g0 = 0 2 ∂ν ∂Ω 2 ∂ν ∂Ω on ∂Ω and hence, by Lemma 11.8, we arrive at

   1 ∂H

∂(GkD0 ψ)

k0 ∗ + − I + (KΩ ) − g0 = 0 ∂ν ∂Ω 2 ∂ν ∂Ω

on ∂Ω .

(11.23)

By substituting this equation into (11.14), we get ∂u ∂U = − ∂ν ∂ν



1 k0 ∗ − I + (KΩ ) 2



 k0 ∂(GkD0 ψ)

ψ) ∂(SD +

∂ν ∂ν ∂Ω

on ∂Ω .

Finally, using (11.22) we conclude that (11.18) holds and the proof is then complete.   Observe that, by (11.4), (11.23) is equivalent to  

  k0



∂ ∂(G ψ) k0 D

H + SΩ − U

= 0 ∂ν ∂ν ∂Ω −

on ∂Ω .

On the other hand, by (11.19),  k0 SΩ

 ∂(GkD0 ψ)

k0 ψ(x) , (x) = SD ∂ν ∂Ω

x ∈ ∂Ω .

Thus, by (11.14), we obtain  k0 H(x) + SΩ

 ∂(GkD0 ψ)

(x) − U (x) = 0 , ∂ν ∂Ω

x ∈ ∂Ω .

Then, by the unique continuation for ∆ + k02 , we obtain the following Lemma. Lemma 11.9 We have  k0 H(x) = U (x) − SΩ

 ∂(GkD0 ψ)

(x) , ∂ν ∂Ω

x∈Ω.

(11.24)