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Fredholm solvability and residual stability for volume scattering

Statement

Assume AC and the hypotheses of The three-dimensional volume integral scattering model with 0<∣D∣<∞. If I−Kk is injective, then for every f∈L2(D;C) there is a unique u∈L2(D;C) solving (I−Kk)u=f, and Ak:=I−Kk has a bounded inverse. Writing Rk:=∥Ak−1∥, every v∈L2(D;C) satisfies ∥u−v∥2≤Rk∥f−(I−Kk)v∥2. This is an operator-theoretic Fredholm branch; no physical uniqueness statement for the exterior PDE is asserted.

Facts & Assumptions

Given: AC, D,q,k, compact Kk, and injectivity of I−Kk.

[A1]

AC supplies Countable Choice; complex L2(D) is complete under Countable Choice (The Axiom of Choice, The Axiom of Countable Choice (ACω), Complex Lp completeness and almost-everywhere subsequences). Thus it is a Banach space (Banach space).

[F1]

The operator Kk is compact (The volume integral scattering operator is compact).

[F2]

On a Banach space, if K is compact and I−K is injective, then I−K is surjective with bounded inverse (Fredholm alternative for identity minus compact); this result explicitly assumes AC.

[F3]

A compact linear operator is bounded, so Ak=I−Kk is bounded (Compact linear operator, A bounded linear operator between normed spaces).

Proof

Given: AC and the data in the Statement.

1.1givenA1F1F2

By [A1], L2(D;C) is Banach. By [F1] Kk is compact, and the assumed injectivity permits [F2] to be applied to Ak=I−Kk. Therefore Ak is surjective and has a bounded inverse; for each f the unique solution is u=Ak−1f.

2.1givenF3step 1.1algebra∎

For arbitrary v∈L2(D;C), linearity gives Ak(u−v)=Aku−Akv=f−Akv. Applying the bounded inverse from step 1.1 yields u−v=Ak−1(f−Akv), so ∥u−v∥2≤∥Ak−1∥ ∥f−Akv∥2=Rk∥f−(I−Kk)v∥2.

Depends on

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Sources