Derivation of Asymptotic Formulae

12 Derivation of Asymptotic Formulae

Suppose that the domain D is of the form D = δB + z, and let u be the solution of (11.8). The function U is the background solution as before. In this chapter we derive an asymptotic expansion of ∂u/∂ν on ∂Ω as δ → 0 in terms of the background solution U . The leading-order term in this asymptotic formula has been derived in by Vogelius and Volkov in [259] (see also [25] where the second-order term in the asymptotic expansions of solutions to the Helmholtz equation is obtained). The proof of our asymptotic expansion is radically different from the variational ones in [259, 25]. It is based on layer potential techniques and the decomposition formula (11.14) of the solution to the Helmholtz equation. For simplicity, although the asymptotic expansions are valid in the two-dimensional case we only consider d = 3 in what follows.

12.1 Asymptotic Expansion We first derive an estimate of the form (11.15) with a constant C independent of δ. Proposition 12.1 Let D = δB + z and (ϕ, ψ) ∈ L2 (∂D) × L2 (∂D) be the unique solution of (11.13). There exists δ0 > 0 such that for all δ ≤ δ0 , there exists a constant C independent of δ such that   −1

ϕ L2 (∂D) + ψ L2 (∂D) ≤ C δ H L2(∂D) + ∇H L2 (∂D) . (12.1) Proof. After the scaling x = z + δy, (11.13) takes the form  1 k0 δ kδ   SB ϕδ − SB ψδ = δ Hδ



on ∂B , k0 δ kδ 1 ∂(SB ϕδ )

ψδ )

1 ∂(SB 1 ∂Hδ   − = 



µ ∂ν µ0 ∂ν δµ0 ∂ν − + H. Ammari and H. Kang: LNM 1846, pp. 197–205, 2004. c Springer-Verlag Berlin Heidelberg 2004


198

12 Derivation of Asymptotic Formulae

kδ where ϕδ (y) = ϕ(z + δy), y ∈ ∂B, etc, and the single layer potentials SB and k0 δ SB are defined by the fundamental solutions Γkδ and Γk0 δ , respectively. It then follows from Theorem 11.4 that for δ small enough the following estimate holds:

ϕδ L2 (∂B) + ψδ L2 (∂B) ≤ Cδ −1 Hδ W12 (∂B) ,

for some constant C independent of δ. By scaling back, we obtain (12.1).   Let H be the function defined in (11.12). Fix n ∈ IN, define Hn (x) =

n ∂ i H(z) (x − z)i , i!

|i|=0

and let (ϕn , ψn ) be the unique solution of  k k0  SD ϕn − SD ψn = Hn+1



k0 k ψn )

1 ∂(SD ϕn )

1 ∂(SD 1 ∂Hn+1  

− µ0

= µ0 ∂ν µ ∂ν ∂ν − +

on ∂D .

(12.2)

Then (ϕ − ϕn , ψ − ψn ) is the unique solution of (12.2) with the right-hand sides defined by H − Hn+1 . Therefore, by (12.1), we get

ϕ − ϕn L2 (∂D) + ψ − ψn L2 (∂D)   ≤ C δ −1 H − Hn+1 L2 (∂D) + ∇(H − Hn+1 ) L2 (∂D) .

(12.3)

By the definition of Hn+1 , we have

H − Hn+1 L2 (∂D) ≤ C|∂D|1/2 H − Hn+1 L∞ (∂D) ≤ C|∂D|1/2 δ n+2 H C n+2 (D) , and

∇(H − Hn+1 ) L2 (∂D) ≤ C|∂D|1/2 δ n+1 H C n+1(D) .

It then follows from (12.3) and Proposition 11.6 that

ϕ − ϕn L2 (∂D) + ψ − ψn L2 (∂D) ≤ C|∂D|1/2 δ n+1 .

(12.4)

By (11.18), we obtain ∂(GkD0 ψn ) ∂(GkD0 (ψ − ψn )) ∂U ∂u (x) = (x) + (x) + (x) , ∂ν ∂ν ∂ν ∂ν Since dist(D, ∂Ω) ≥ c0 , we get



∂Gk0



sup (x, y)

≤ C

∂ν x∈∂Ω, y∈∂D

x ∈ ∂Ω .

12.1 Asymptotic Expansion

199

for some C. Hence, for each x ∈ ∂Ω, we have from (12.4)

#

$1/2

2

∂(Gk0 (ψ − ψ ))



(x, y) ∂G



n k 0 D

dσ(y) (x) ≤

ψ − ψn L2 (∂D)







∂ν ∂νx ∂D ≤ C|∂D|1/2 |∂D|1/2 δ n+1 ≤ C
δ n+d , where C and C
are independent of x ∈ ∂Ω and δ. Thus we conclude that ∂u ∂U ∂(GkD0 ψn ) (x) = (x) + (x) + O(δ n+d ) , ∂ν ∂ν ∂ν

uniformly in x ∈ ∂Ω . (12.5)

For each multi-index i, define (ϕi , ψi ) to be the unique solution to  kδ k0 δ i  SB ϕi − SB ψi = x



k0 δ kδ on ∂B . (12.6) 1 ∂(SB ϕi )

ψi )

1 ∂(SB 1 ∂xi  − = 



µ ∂ν µ0 ∂ν µ0 ∂ν − + Then, we claim that ϕn (x) =

n+1

δ |i|−1

∂ i H(z) ϕi (δ −1 (x − z)) , i!

δ |i|−1

∂ i H(z) ψi (δ −1 (x − z)) . i!

|i|=0

ψn (x) =

n+1 |i|=0

In fact, the expansions follow from the uniqueness of the solution to the integral equation (11.10) and the relation   n+1 i k0  |i|−1 ∂ H(z) −1 ϕi (δ (· − z)) (x) δ SD i! |i|=0

=

n+1

δ |i|

|i|=0

∂ i H(z) k0 δ (SB ϕi )(δ −1 (x − z)) , i!

for x ∈ ∂D. It then follows from (12.5) that n+1 ∂u ∂U ∂ i H(z) ∂ k0 (x) = (x) + G (ψi (δ −1 (· − z)))(x) δ |i|−1 ∂ν ∂ν i! ∂ν D |i|=0

+O(δ n+d ) , uniformly in x ∈ ∂Ω. Note that

(12.7)

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12 Derivation of Asymptotic Formulae

GkD0 (ψi (δ −1 (· − z)))(x) =



Gk0 (x, y)ψi (δ −1 (y − z)) dσ(y)  d−1 =δ Gk0 (x, δw + z)ψi (w) dσ(w) . ∂D

∂B

Moreover, for x near ∂Ω, z ∈ Ω, w ∈ ∂B, and sufficiently small δ, we have ∞ δ |j| j ∂ Gk (x, z)wj . j! z 0

Gk0 (x, δw + z) =

|j|=0

Therefore, we get GkD0 (ψi (δ −1 (· − z)))(x) =

 ∞ δ |j|+d−1 j ∂z Gk0 (x, z) wj ψi (w) dσ(w) . j! ∂B

|j|=0

Define, for multi-indices i and j in INd ,  Wij := wj ψi (w) dσ(w) .

(12.8)

∂B

Then we obtain the following theorem from (12.7). Theorem 12.2 The following pointwise asymptotic expansion on ∂Ω holds: n+1 ∂u ∂U d−2 (x) = (x) + δ ∂ν ∂ν



n−|j|+1

|j|=0

+O(δ

n+d

|i|=0

δ |i|+|j| i ∂∂ j Gk (x, z) ∂ H(z) z 0 Wij i!j! ∂νx

(12.9)

),

where the remainder O(δ d+n ) is dominated by Cδ d+n f W 21 (∂Ω) for some C 2

independent of x ∈ ∂Ω. In view of (11.18), we obtain the following expansion: n+1 ∂(GkD0 ψ) (x) = δ d−2 ∂ν

|j|=0



n−|j|+1

|i|=0

δ |i|+|j| i ∂∂ j Gk (x, z) ∂ H(z) z 0 Wij i!j! ∂νx

(12.10)

+O(δ n+d ) . Observe that ψi , and hence, Wij depends on δ, and so does H. Thus the formula (12.9) is not a genuine asymptotic formula. However, since it is simple and has some potential applicability in solving the inverse problem for the Helmholtz equation, we made a record of it as a theorem. Observe that by the definition (12.6) of ψi , ψi L2 (∂B) is bounded, and hence |Wij | ≤ Cij , ∀ i, j ,

12.1 Asymptotic Expansion

201

where the constant Cij is independent of δ. Since δ is small, we can derive an asymptotic expansion of (ϕi , ψi ) using their definition (12.6). Let us briefly explain this. Let   k0 δ kδ f − SB g SB  

 f  k0 δ kδ :=  1 ∂(SB Tδ g)

 on ∂B , f )

1 ∂(SB g −

µ ∂ν

µ0 ∂ν −

+

and let T0 be the operator when δ = 0. Then the solution (ϕi , ψi ) of the integral equation (12.6) is given by      −1 xi ϕi (12.11) T0−1  1 ∂xi  . = I + T0−1 (Tδ − T0 ) ψi µ0 ∂ν By expanding Tδ − T0 in a power series of δ, we can derive the expansions )i , ψ)i ) be the leading-order term in the of ψi and Wij . Let, for i, j ∈ INd , (ϕ ) expansion of (ϕi , ψi ). Then (ϕ )i , ψi ) is the solution of the system of the integral equations  0 0 ) )i − SB ψi = xi  SB ϕ



0 0 )

on ∂B . (12.12) 1 ∂(SB ϕ )i )

ψi )

1 ∂(SB 1 ∂xi  − =  µ ∂ν − µ0 ∂ν + µ0 ∂ν As a simplest case, let us now take n = 1 in (12.9) to find the leading-order term in the asymptotic expansion of ∂u/∂ν|∂Ω as δ → 0. We first investigate the dependence of Wij on δ for |i| ≤ 1 and |j| ≤ 1. If |i| ≤ 1, then both sides of the first equation in (12.12) are harmonic in B, and hence 0 0 ) ϕ )i − SB SB ψi = xi

in B .

Therefore we get



0 0 )

∂(SB ϕ )i )

ψi )

∂(SB ∂xi − =



∂ν ∂ν ∂ν − −

on ∂B .

This identity together with the second equation in (12.12) yields



  0 )

0 )

µ ∂(SB ψi )

ψi )

∂(SB µ ∂xi . − = 1 − µ0 ∂ν + ∂ν − µ0 ∂ν In view of the relation (11.4), we have       µ 1 1 µ ∂xi ∗ ∗ ) ) I + KB ψi − − I + KB ψi = 1 − , µ0 2 2 µ0 ∂ν

202

12 Derivation of Asymptotic Formulae

∗ where KB is the operator defined in (11.6) when k = 0. Therefore, we have  i 

∗ −1 ∂x

) , (12.13) ψ)i = (λI − KB ∂ν ∂B

where λ :=

µ µ0

+1

2(1 −

µ µ0 )

=

µ0 µ +1 2( µµ0 − 1)

.

(12.14)

∗ Here we have used the fact from Theorem 2.8 that the operator λI − KB on 2 0 ) )i = 1. Hence L (∂B) is invertible. Observe that if |i| = 0, then ψi = 0 and SB ϕ kδ kδ ϕi = 1 + O(δ). Moreover, since SB ϕi depends we obtain ψi = O(δ) and SB 2 2 kδ on δ analytically and (∆ + k δ )SB ϕi = 0 in B, we conclude that

ψi = O(δ)

kδ and SB ϕi = 1 + O(δ 2 ) ,

|i| = 0 .

It also follows from (12.13) that if |i| = |j| = 1, then  i  

j ∗ −1 ∂y

x (λI − KB ) (x) dσ(x) + O(δ) . Wij = ∂ν ∂B ∂B

(12.15)

(12.16)

The first quantity in the right-hand side of (12.16) is the polarization tensor Mij as defined in (3.1). In summary, we obtained that Wij = Mij + O(δ) ,

|i| = |j| = 1 .

(12.17)

Suppose that either i = 0 or j = 0. By (11.4) and (12.6), we have



k0 δ k0 δ ∂(SB ∂(SB ψi )

ψi )

ψi = −



∂ν ∂ν + −



k0 δ kδ ∂(SB ψi )

ϕi )

µ0 ∂(SB ∂xi = (12.18)

− ∂ν −

. µ ∂ν ∂ν − − It then follows from the divergence theorem that    µ0 k0 δ kδ xj ψi dσ = −k 2 δ 2 xj SB ϕi dx + k02 δ 2 xj SB ψi dx µ ∂B B B   ∂xj kδ ∂xj k0 δ µ0 SB ϕi dσ − SB ψi dσ . + µ ∂B ∂ν ∂B ∂ν

(12.19)

From (12.19), we can observe the following. Wij = −k 2 δ 2

µ0 |B| + O(δ 3 ) = −δ 2 ω 2 εµ0 |B| + O(δ 3 ) , µ

|i| = |j| = 0 , (12.20)

2

|i| = 1 , |j| = 0 ,

(12.21)

2

|i| = 0 , |j| = 1 .

(12.22)

Wij = O(δ ) , Wij = O(δ ) ,

12.1 Asymptotic Expansion

203

In fact, (12.20) and (12.22) follow from (12.15) and (12.19), and (12.21) immediately follows from (12.19). As a consequence of (12.20), (12.21), (12.22), and (12.10), we obtain ∂(GkD0 ψ) (x) = O(δ d ) , ∂ν

uniformly on x ∈ ∂Ω .

Since the center z is apart from ∂Ω, it follows from (11.24) that |H(z) − U (z)| + |∇H(z) − ∇U (z)| = O(δ d ) . We now consider the case |i| = 2 and |j| = 0. In this case, one can show using (12.18) that   ψi dσ = − ∆xi dx + O(δ 2 ) . ∂B

B

Therefore, if |j| = 0, then 1 ∂ i H(z)Wij = −∆H(z)|B| + O(δ 2 ) = k02 H(z)|B| + O(δ 2 ) . (12.23) i!j!

|i|=2

So (12.9) together with (12.17)-(12.23) yields the following expansion formula of Vogelius-Volkov [259]. In fact, in [259], the formula is expressed in terms of the free space Green’s function Γk instead of the Green’s function Gk0 . However, these two formulae are the same, as we can see using the relation (11.20). Theorem 12.3 For any x ∈ ∂Ω, ∂U ∂u (x) = (x) ∂ν  ∂ν  ∂∇z Gk0 (x, z) ∂Gk0 (x, z) d 2 + δ ∇U (z)M (12.24) + ω µ0 (ε − ε0 )|B|U (z) ∂νx ∂νx + O(δ d+1 ) , where M is the polarization tensor defined in (3.1) with λ given by (12.14). Before returning to (12.9) let us make the following important remark. The tensors Wij play the same role as the generalized polarization tensors. As defined in Chap. 3 the GPT’s are given for i, j ∈ INd by  wj ψ)i (w) dσ(w) , Mij := ∂B

where ψ)i is defined by (12.12). The following result makes the connection between Wij and Mij . Its proof is immediate.

204

12 Derivation of Asymptotic Formulae

Lemma 12.4 Suppose that ai are constants such that



ai wi is a harmonic

i

polynomial. Then

ai Wij →

i



as δ → 0 .

ai Mij

i

Observing now that the formula (12.9) still contains ∂ i H factors, the remaining task is to convert (12.9) to a formula given solely by U and its derivatives. Substitution of (12.10) into (11.24) yields that, for any x ∈ Ω, H(x) = U (x) −δ d−2

n+1 n+1−|j| |j|=0

|i|=0

j δ |i|+|j| i k0 ∂∂z Gk0 (x, z) ∂ H(z)SΩ ( )Wij i!j! ∂νx

(12.25)

+O(δ n+d ) . In (12.25) the remainder O(δ n+d ) is uniform in the C n -norm on any compact subset of Ω for any n ∈ IN and therefore (∂ γ H)(z) + δ d−2

n+1 n+1−|j| |j|=0

δ |i|+|j| ∂ i H(z)Pijγ = (∂ γ U )(z) + O(δ d+n ) ,

|i|=0

for all γ ∈ INd with |γ| ≤ n + 1 where Pijγ

j

1 γ k0 ∂∂z Gk0 (·, z)

Wij ∂ SΩ ( = )

. i!j! ∂νx x=z

Define the operator Pδ by Pδ : (wγ )γ∈INd ,|γ|≤n →

 n+1 wγ + δ d−2 |j|=0



n+1−|j|

δ |i|+|j| wi Pijγ



|i|=0

. γ∈INd ,|γ|≤n

Observe from (12.11) that Pδ can be written as Pδ = I + δ d P1 + . . . + δ n+d−1 Pn−1 + O(δ n+d ) . Defining as in (4.17) Qp , p = 1, . . . , n − 1, by (I + δ d P1 + . . . + δ n+d−1 Pn−1 )−1 = I + δ d Q1 + . . . + δ n+d−1 Qn−1 + O(δ n+d ) , we finally obtain that ((∂ i H)(z))i∈INd ,|i|≤n+1 = (I+

n

δ d+p−1 Qp )((∂ i U )(z))i∈INd ,|i|≤n+1 +O(δ d+n ) ,

p=1

which yields the main result of this chapter.

12.1 Asymptotic Expansion

205

Theorem 12.5 The following pointwise asymptotic expansion on ∂Ω holds: n+1 ∂u ∂U (x) = (x) + δ d−2 ∂ν ∂ν



n+1−|j|

|j|=0

|i|=0

δ |i|+|j| × i!j!

   n+2−|i|−|j|−d ∂∂zj Gk0 (x, z) d+p−1 γ δ Qp )(∂ U (z)) Wij (I + ∂νx i p=1 +O(δ n+d ) , where the remainder O(δ d+n ) is dominated by Cδ d+n f W 21 (∂Ω) for some C 2

independent of x ∈ ∂Ω. When n = d, we have a simpler formula d+1 ∂u ∂U (x) = (x) + δ d−2 ∂ν ∂ν

|j|=0



d+1−|j|

|i|=0

δ |i|+|j| i ∂∂ j Gk (x, z) ∂ U (z) z 0 Wij i!j! ∂νx (12.26)

2d

+O(δ ) . Let us now consider the case when there are several well separated inclusions. The inclusion D takes the form ∪m s=1 (δBs + zs ). The magnetic permeability and electric permittivity of the inclusion δBs + zs are µs and εs , s = 1, . . . , m. By iterating the formula (12.26) we can derive the following theorem. Theorem 12.6 The following pointwise asymptotic expansion on ∂Ω holds: ∂u ∂U (x) = (x) ∂ν ∂ν m d+1 d+1−|j| δ |i|+|j| ∂∂ j Gk (x, z) s +δ d−2 ∂ i U (z) z 0 Wij + O(δ 2d ) . i!j! ∂ν x s=1 |j|=0

(12.27)

|i|=0

Here Wijs is defined by (12.8) with B, µ, ε replaced by Bs , µs , εs . We conclude this chapter by making one final remark. In this chapter, we only derive the asymptotic formula for the solution to the Dirichlet problem. However, by the same method, we can derive an asymptotic formula for the Neumann problem as well.