The Hybrid Method for the Dirichlet Case

In this section we will consider the hybrid method for the numerical solution to the in-verse obstacle scattering problem 2.15 with Dirichlet boundary condition. This means the solutionu^sto the direct problem satisfies

∆u^s+k²u^s = 0, x ∈R^m\D, u= 0, x∈Γ,

r→∞lim r^m−12 ∂ u^s

∂r −iku^s

= 0,

where again the total fieldu is the sum of the given incident field uⁱ and the scattered fieldu^s. The goal is to recover the domain D from the knowledge of the far-field pat-ternu∞for one incident plane waveuⁱ(x) =e^ikx·d, |d|= 1.

38 3.3. The Hybrid Method for the Dirichlet Case

We will consider that the solution to the inverse problem can be parameterized, that is,

Γ ={z∗(t) : t∈X}

where the parameterization z∗ ∈ C²(X) is such that Γ is C²–smooth (in R² we will assume thatz∗ ∈ C²([0,2π])is2π–periodic and is counter-clockwise oriented) and has the following mapping properties

( z:X →R², X := [0,2π], m= 2, z:X →R³, X := [0, π]×[0,2π], m= 3.

At the n–iteration of the hybrid method we will consider that the current approxima-tionγnto the correct boundaryΓis given by

γ_n ={z_n(t) : t∈X}.

Later on we will assume that z_n ∈ U, whereU is a finite dimensional subspace of the previous parameterization space, that is

U ⊂ {z : z ∈C²(X), z isX–periodic}.

Following the comments on the solvability of (3.3), we will now state an assumption that will be of crucial importance throughout this work.

Assumption 3.2(Analytic Continuation Principle). The solutionu^sto the direct problem of scattering byDcan be analytically extended as a solution to the Helmholtz equation in a neighbourhood of the boundaryΓofD.

Remark3.3. If the boundaryΓis analytic, then assumption 3.2 holds (see [13]).

Assuming thatγ_nis sufficiently close toΓin a way that the solutionu^s to the direct problem of scattering by Dcan be analytically extended up to γ_nand assuming that k² is not an interior Dirichlet eigenvalue of the negative Laplacian for the interior ofγ_n, we can represent the scattered fieldu^sas a single layer potential overγ_n(see [8, thm. 3.30]), that is

u^s(x) = Z

γn

Φ(x, y)ϕ(y)ds(y), x∈R^m\γ_n with densityϕ∈C(γ_n).

Remark3.4. We choose a single-layer representation because it leads to less complexity later on in the implementation. However, the condition on the wave numberkis needed and can not be guaranteeda priori for the successive approximationsγn. We also note that for the Neumann and Robin cases we will use a combined single-and double-layer potential in order to illustrate the differences in the implementation.

Due to analyticity of the single layer potential inR^m\γ_n, it is clear that if the analytic continuation principle 3.2 does not hold, then the previous representation would not make sense for the cases whereγn∩D6=∅. One knows that the exterior trace to the boundary of the previous potential is given by

u^s(x) = S_γnϕ

(x), x∈γ_n

whereS_γn : C(γ_n) → C(γ_n) is the single-layer operator (2.18). By the asymptotics of the single layer potential the far-field must satisfy the equation

S_γn,∞ϕ=u∞inΩ_m (3.8)

with the far-field operatorS_γ,∞:C(γ)→C(Ω_m)given by Sγ,∞ϕ =%_m

Z

γ

e^−ikˆx·yϕ(y)ds(y), xˆ∈Ω_m,

with %_m given as in (2.14). As referred in the previous section, the previous operator is compact (since it has a continuous kernel) so (3.8) must be replaced by a regularized equation. In order to show that a regularization scheme is applicable one needs to show that the operatorS_γn,∞is injective.

Theorem 3.5. Assume that k² is not an interior Dirichlet eigenvalue of the negative Laplacian with respect to the open bounded domain D_γ with boundary γ. Then the operatorSγ,∞ :L²(γ)→L²(Ωm)is injective and has dense range.

Proof. Assume thatψ ∈L²(γ)satisfiesS_γ,∞ψ = 0. Then the single-layer potential v(x) =

Z

γ

Φ(x, y)ψ(y)ds(y), x∈R^m\γ

has a vanishing far-field. By Rellich’s lemma and analyticity we conclude thatvvanishes inR^m\D_γ. By continuity up to the boundary we conclude that S_γψ = 0overγ. Ask² is not an interior Dirichlet eigenvalue with respect to the open bounded domainD_γ we know that S_γ is injective (see [8, thm.3.30]) and so we conclude that ψ = 0, proving injectivity ofS_γ.

To prove denseness one shows by a similar argument that the adjoint operatorS_γ^∗ is injective and concludes thatS_γis therefore dense, since for a linear bounded operatorA the closure of the range ofA is the orthogonal complement of the nullspace ofA^∗ (for details see [9, thm.5.17]).

Using Tikhonov regularization, as referred in the previous section, we need to re-place (3.8) by

α_nI+S_γ^∗_n,∞S_γn,∞

ϕ⁽ⁿ⁾ =S_γ^∗_n,∞u∞ (3.9)

40 3.3. The Hybrid Method for the Dirichlet Case

solving it with respect to ϕ⁽ⁿ⁾ for some regularization parameter α_n > 0 decreasing withn. The scattered fieldu^scan now be approximated by

u^s_n(x) = Z

γn

Φ(x, y)ϕ⁽ⁿ⁾ds(y), x∈R^m\γn. (3.10) and using the jump relations we also get the approximations tou^sand its exterior normal derivative∂u^s/∂ν onγngiven by

u^s_n(x) = (S_γnϕ⁽ⁿ⁾)(x), x∈γ_n, (3.11)

∂ u^s_n

∂ν (x) = −1

2ϕ⁽ⁿ⁾+ (K_γ^∗_nϕ⁽ⁿ⁾)(x), x∈γ_n, (3.12) respectively.

For a fixed analytic fieldu, we now define the operatorG_D that maps the parameter-izationz of the contourγ to the exterior trace of the Dirichlet boundary condition of that fielduoverγ, that is,

G_D :z 7→u◦z.

If the field u is the total field, then in order to find the position of the boundary of the obstacleDas the location where the boundary condition is satisfied, we want to find the solution to

G_D(z) = 0.

In the spirit of a Newton method we now linearize the previous equation aroundz_n and solve the linearized equation

G_D(z_n) +G⁰_D(z_n)h= 0 inX (3.13) with respect to the shift h. In the next theorem we characterize the Fr´echet derivative ofGD.

Theorem 3.6. The operator G_D : C²(X) → C(X) is Fr´echet differentiable and the Fr´echet derivative is given by

G⁰_D(z)h= (gradu◦z)·h.

Proof. By the Taylor formula, the Fr´echet differentiability ofG_D is a direct consequence of the analyticity ofuand theC²–smoothness ofz. Moreover, from the Taylor formula foruone gets for eacht ∈Xthat

u z(t) +h(t)

=u z(t)

+ gradu z(t)

·h(t) +O |h(t)|² , as||h||∞ →0.Therefore by definition ofG_D we have

||G_D(z+h)−G_D(z)−(gradu◦z)·h||_∞ =O(||h||²_∞) as||h||∞ →0and by definition of the Fr´echet derivative one has the result.

With this characterization, equation (3.13) can be rewritten in the following way (grad u◦z_n)·h =−u◦z_ninX. (3.14) In this way, at each iterationnwe approximateu^s byu^s_nobtained in the first step of the iteration and solve

(grad u^s_n+ grad uⁱ)◦zn

·h=− u^s_n+uⁱ

◦zn inX

with respect tohin a least squares sense, obtaining the new approximationγ_n+1 param-eterized by z_n+1 = z_n+ h. Note that we use the jump relations (3.11) and (3.12) to compute the terms involved, through the decomposition

gradu|_γn = ν ∂ u

∂ν γn

+∇_tu, (3.15)

where∇_turepresents the surface gradient ofu, which inR² reduces itself to the tangen-tial derivative times the tangentangen-tial unit vector. We now repeat the two steps until some stopping criteria is fulfilled. The details on the numerical implementation will be given in sections 5.1.2 and 5.2.2 forR²andR³, respectively.

Remark3.7. This method can also be extended to recover sound-soft cracks. The main difference is that one needs to choose the right or left hand side normal derivative in (3.15) in an appropriate way. A further penalty term on the length of the crack is also needed to prevent it from degenerating to a point. For details we refer to [42].

Remark3.8. Note that to show solvability of (3.13) we would need to show that G⁰_D(z_n) is injective. However this can only be done if we are over the correct boundaryΓ, ifuis considered to be the exact total field and if there exists an open set ofΓwhereh·ν 6= 0.

In this case, by the boundary condition we have that if

0 =G⁰_D(z∗)h=h·gradu◦z∗ =h·ν ∂ u

∂ν

thenh= 0everywhere. By contradiction, assume than this is not the case. Then, as there exists an open subset ofΓ where h·ν 6= 0, the normal derivative of u would need to vanish on that open subset ofΓ. By the boundary condition and Holmgren’s theorem we conclude that the total fieldu = 0 inR^m\D, which can not happen since the scattered field goes to zero at infinity and the incident field does not. A similar result can be shown for the Neumann and Robin boundary condition (see [59, thm.5]).

Im Dokument A Hybrid Method for Inverse Obstacle Scattering Problems (Seite 45-50)