Well-Posedness

10 Well-Posedness

In this chapter, by using the theory of collectively compact operators, we prove existence and uniqueness of a solution to the Helmholtz equation  (∆ + ω 2 ε0 µ0 )u = 0 in Ω \ ∪m  s=1 Ds ,     2  (∆ + ω εs µs )u = 0 in Ds , s = 1, . . . , m ,  



 1 ∂u

1 ∂u

(10.1) − = 0 on ∂Ds , s = 1, . . . , m ,  µs ∂ν − µ0 ∂ν +   



   u − − u + = 0 on ∂Ds , s = 1, . . . , m ,    u = f on ∂Ω , assuming that ω 2 ε0 µ0 is not an eigenvalue for the operator − ∆ in L2 (Ω) with homogeneous Dirichlet boundary conditions.

(10.2)

10.1 Existence and Uniqueness of a Solution In order to define the natural weak formulation of the problem (10.1), let aδ denote the sesquilinear form   1 aδ (u, v) = ∇u · ∇v − ω 2 εδ uv , (10.3) Ω µδ Ω defined on W01,2 (Ω) × W01,2 (Ω). Let b be a given conjugate-linear functional on W01,2 (Ω). Our assumption (10.2) is that the variational problem a0 (u, v) = b(v) for all v ∈ W01,2 (Ω) has a unique solution. The following lemma from [259] shows that the assumption (10.2) also leads to the unique solvability of (10.1).

H. Ammari and H. Kang: LNM 1846, pp. 179–183, 2004. c Springer-Verlag Berlin Heidelberg 2004


180

10 Well-Posedness

Lemma 10.1 Suppose (10.2) is satisfied, and let aδ , 0 ≤ δ, be the sesquilinear forms introduced by (10.3). There exists a constant 0 < δ0 , such that given any 0 ≤ δ < δ0 , and any bounded, conjugate-linear functional, b, on W01,2 (Ω), there is a unique u ∈ W01,2 (Ω) satisfying aδ (u, v) = b(v) for all v ∈ W01,2 (Ω). Furthermore, there exists a constant C, independent of δ and b, such that ||u||W 1,2 (Ω) ≤ C

sup v∈W01,2 (Ω),||v||W 1,2 (Ω) =1

|b(v)| .

Proof. In order to prove this lemma it is convenient to introduce a decomposition of aδ . Pick a fixed positive constant, λ, with λ > ω 2 , and write aδ as aδ = Aδ + Bδ , where   1 2 ∇u · ∇v + (λ − ω ) εδ uv Aδ (u, v) = Ω µδ Ω 

and Bδ (u, v) = −λ

εδ uv . Ω

Suppose (10.2) is satisfied. Then the sesquilinear form Aδ is uniformly continuous and uniformly coercive on W01,2 (Ω) × W01,2 (Ω). It is also convenient to introduce a family of bounded linear operators Kδ : W01,2 (Ω) → W01,2 (Ω) by  Aδ (Kδ u, v) = Bδ (u, v) = −λ εδ uv , Ω

for all u and v in W01,2 (Ω). Let δn be a sequence converging to zero. We first show that the linear operators {Kδn } are compact and Kδn converges pointwise to K0 as δn approaches 0. We remind the reader that the operators {Kδn } are collectively compact iff the set {Kδn (u) : n ≥ 1, u ∈ W01,2 (Ω), ||u||W 1,2 (Ω) ≤ 1} is relatively compact (its closure is compact) in W01,2 (Ω). Fix u ∈ W01,2 (Ω), then Aδn ((Kδn − K0 )u, v) = Bδn (u, v) − B0 (u, v) + A0 (K0 u, v) − Aδn (K0 u, v) , for all v ∈ W01,2 (Ω). We easily see that |B0 (u, v) − Bδn (u, v)| → 0 ,

(10.4)

|A0 (K0 u, v) − Aδn (K0 u, v)| → 0 ,

(10.5)

sup v∈W01,2 (Ω),||v||W 1,2 (Ω) =1

as δn → 0. It is also clear that sup

v∈W01,2 (Ω),||v||W 1,2 (Ω) =1

as δn → 0. A combination of (10.4) and (10.5) yields

10.1 Existence and Uniqueness of a Solution

sup v∈W01,2 (Ω),||v||W 1,2 (Ω) =1

181

|Aδn ((Kδn − K0 )u, v)| → 0 ,

as δn → 0. Since Aδn is uniformly coercive on W01,2 (Ω) × W01,2 (Ω), it follows now that ||(Kδn − K0 )u||W 1,2 (Ω) → 0 , as δn → 0. This verifies the pointwise convergence of the operators {Kδn }. Let Kτm (um ) be any sequence from the set

1,2 Kδn (u) : n ≥ 1, u ∈ W0 (Ω), ||u||W 1,2 (Ω) ≤ 1 . In order to verify the collective compactness of the operators {Kδn } we need to show that the sequence Kτm (um ) contains a convergent subsequence. By extraction of a subsequence (still referred to as Kτm (um )) we may assume that either: (1) τm = τ is constant (i.e., independent of m) or: (2) τm → 0 as m → +∞. We may also assume that um converges weakly to some u∞ ∈ W01,2 (Ω). We introduce the sequence u
m = um − u∞ . Clearly ||u
m ||W 1,2 (Ω) ≤ 2 and u
m converges weakly to zero. Since the imbedding W01,2 (Ω) → L2 (Ω) is compact, this gives that u
m has a subsequence (still referred to as u
m ) converging strongly to zero in L2 (Ω). From the definition of Kτm it follows immediately that sup v∈W01,2 (Ω),||v||W 1,2 (Ω) =1

|Aτm (Kτm u
m , v)| =

sup v∈W01,2 (Ω),||v||W 1,2 (Ω) =1

|Bτm (u
m , v)|

≤ C||u
m ||L2 (Ω) . Since Aτm is uniformly coercive and since ||u
m ||L2 (Ω) → 0, we conclude from the above estimate that ||Kτm u
m ||W 1,2 (Ω) → 0, which is exactly what we are aiming at. We want now to solve the variational problem: Find u ∈ W01,2 (Ω) such that aδ (u, v) = b(v) for all v ∈ W01,2 (Ω) .

(10.6)

This problem can be rewritten as Aδ (u, v) + Bδ (u, v) = b(v) for all v ∈ W01,2 (Ω) , or as

Aδ ((I + Kδ )uδ , v) = b(v) for all v ∈ W01,2 (Ω) .

(10.7)

W01,2 (Ω),

it now follows Since Aδ is uniformly continuous and coercive on that the variational problem (10.7) is equivalent to the problem of finding u ∈ W01,2 (Ω) such that (I + Kδ )u = Fδ . Here the function Fδ ∈ W01,2 (Ω) is defined by Aδ (Fδ , v) = b(v) for all v ∈ W01,2 (Ω), and therefore satisfies

182

10 Well-Posedness

||Fδ ||W 1,2 (Ω) ≤ C

sup v∈W01,2 (Ω),||v||W 1,2 (Ω) =1

|b(v)| .

(10.8)

By the same arguments as we just went through earlier in this proof, the original variational problem to find U ∈ W01,2 (Ω) such that a0 (U, v) = b(v) for all v ∈ W01,2 (Ω) is thus equivalent to the problem to find U ∈ W01,2 (Ω) such that (I + K0 )U = F0 with F0 ∈ W01,2 (Ω) defined by A0 (F0 , v) = b(v) for all v ∈ W01,2 (Ω). The fact that this problem has a unique solution (assumption (10.2)) implies that I + K0 is an invertible operator. For any sequence δn converging to zero we have already verified that the operators {Kδn } are collectively compact and converge pointwise to K0 . From the theory of collectively compact operators [35] (see Theorem A.4 in Appendix A.3) it follows that there exists a constant 0 < δ0 , such that given any 0 ≤ δ < δ0 , the operator I + Kδ is invertible with ||(I + Kδ )−1 Fδ ||W 1,2 (Ω) ≤ C||Fδ ||W 1,2 (Ω)

(10.9)

for some constant C, independent of δ. It then follows from (10.8) and (10.9) that the variational problem (10.6) has a unique solution u ∈ W01,2 (Ω) satisfying sup |b(v)| . ||u||W 1,2 (Ω) ≤ C v∈W01,2 (Ω),||v||W 1,2 (Ω) =1

Thus the proof of Lemma 10.1 is complete.   Suppose that there exists a constant c0 > 0 such that |zs − zs
| ≥ 2c0 > 0 , ∀ s
= s


and dist(zs , ∂Ω) ≥ 2c0 > 0, ∀ s .

(10.10)

Using the above lemma we can show as in [259] that the following holds. Proposition 10.2 Suppose (10.10) and (10.2) are satisfied. There exists 0 < δ0 such that, given an arbitrary f ∈ W 12 (∂Ω), and any 0 < δ < δ0 , the 2

boundary value problem (10.1) has a unique weak solution u in W 1,2 (Ω). The m constant δ0 depends on the domains {Bs }m s=1 , Ω, the constants {µs , εs }s=0 , and c0 , but is otherwise independent of the points {zs }m . Moreover, let U s=1 denote the unique weak solution to the boundary value problem:  (∆ + ω 2 ε0 µ0 )U = 0 in Ω , U = f on ∂Ω . There exists a constant C, independent of δ and f , such that d

||u − U ||W 1,2 (Ω) ≤ Cδ 2 ||f ||W 21 (∂Ω) . 2

{Bs }m s=1 , Ω,

the constants {µs , εs }m The constant C depends on the domains s=0 , and c0 , but is otherwise independent of the points {zs }m . s=1

10.1 Existence and Uniqueness of a Solution

183

Proof. The function u − U is in W01,2 (Ω), and for any v ∈ W01,2 (Ω)   1 ∇(u − U ) · ∇v − ω 2 εδ (u − U )v aδ (u − U, v) = Ω µδ Ω   m  1 1 − )∇U · ∇v + ω 2 (εs − ε0 )U v . = ( µ0 µs s=1 δBs +zs







Next



δBs +zs



1 1 2 − )∇U · ∇v + ω (εs − ε0 )U v

( µ0 µs

is bounded by C||U ||W 1,2 (δBs +zs ) ||v||W 1,2 (Ω) . Since the inclusions are away from the boundary ∂Ω, standard elliptic regularity results give that ||U ||W 1,∞ (δBs +zs ) ≤ C||U ||W 1,2 (Ω) ≤ C||f ||W 21 (∂Ω) , 2

and so d

1

d

||U ||W 1,2 (δBs +zs ) ≤ ||U ||W 1,∞ (δBs +zs ) δ 2 |Bs | 2 ≤ Cδ 2 ||f ||W 21 (∂Ω) . 2

From Lemma 10.1 it then follows immediately that d

||u − U ||W 1,2 (Ω) ≤ Cδ 2 ||f ||W 21 (∂Ω) , 2

exactly as desired.