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Existence and uniqueness for the spherical exact-DtN variational problem

Statement

Fix 0<a<R, k>0, and smooth f on S2. The exact-DtN variational problem of Exact-DtN truncated modal variational formulation has exactly one solution u∈Vf. It is the restriction to a≤r≤R of the outgoing modal field u(r,ω)=∑ℓ=0∞hℓ(1)(kr)hℓ(1)(ka)fℓ(ω), where fℓ is the degree-ℓ projection of f.

Facts & Assumptions

Given: 0<a<R, k>0, smooth f, and the variational definition [F1].

[F1]

Exact-DtN truncated modal variational formulation: the trial trace at a is fℓ, tests have zero trace at a, and the form subtracts R2τℓ at the outer boundary.

[F2]

Modal trace bound and continuity of the exact-DtN variational form: the form and traces are continuous on the coefficient energy space; finite-degree tests are dense by truncating the defining Hilbert sum.

[F3]

The spherical modal DtN map is well-defined and exact for smooth data: the outgoing modal series for smooth data at any positive sphere is classical, solves the Helmholtz equation, has the prescribed trace, and satisfies uℓ′(R)=τℓuℓ(R).

[F4]

Spherical harmonic modes are orthogonal, complete for continuous data, and have polynomial evaluation bounds: ∥fℓ∥2 decays faster than every inverse power of 1+ℓ and the degree-ℓ angular energy is ℓ(ℓ+1) times the squared L2 norm.

[F5]

Outgoing spherical Hankel modes have positive flux and no positive-real zeros and Spherical Hankel modulus and uniform radial bounds: hℓ(1)(ka)≠0, the outgoing/incoming Wronskian is nonzero, ∣hℓ(kr)/hℓ(ka)∣≤1 for r≥a, and ∣hℓ′∣≤(1+(ℓ+1)/(kr))∣hℓ∣.

Proof

Given: The data and suppliers [F1]–[F4].

1.1F1F3F4F5algebra

By [F3] applied at the inner radius a, the proposed series has inner trace f. Its degree-ℓ radial ratio has modulus at most one for r≥a by [F5], and its radial derivative is bounded by k+(ℓ+1)/r times that ratio. The angular energy of degree ℓ is ℓ(ℓ+1) times its squared L2 norm. Consequently its total V energy is at most Ca,R,k∑ℓ(1+ℓ)2∥fℓ∥22, which is finite by [F4]. Thus it belongs to Vf.

1.2F1F2algebra

Let w∈V0 solve the homogeneous variational equation. Testing against functions supported in a single degree and compactly supported in (a,R) shows distributionally that (r2wℓ′)′+(k2r2−ℓ(ℓ+1))wℓ=0. Since the coefficients are smooth and r>0, this one-dimensional equation implies wℓ∈C2([a,R];Hℓ). Integrating by parts and allowing arbitrary outer test trace yields wℓ′(R)=τℓwℓ(R); the inner trace is wℓ(a)=0.

2.1F1F2F3step 1.1algebra

Each modal coefficient satisfies (r2uℓ′)′+(k2r2−ℓ(ℓ+1))uℓ=0 and uℓ′(R)=τℓuℓ(R) by [F3]. Integration by parts against a finite-degree test with zero inner trace makes its modal volume term equal R2uℓ′(R)vℓ(R)‾; the outer term in [F1] subtracts exactly R2τℓuℓ(R)vℓ(R)‾. Hence the form vanishes for every finite-degree test. Finite-degree truncations converge in V, and [F2] extends the identity to every test in V0. This proves existence.

3.1F1F5step 1.2algebra∎

The radial ODE solution is wℓ(r)=Aℓhℓ(1)(kr)+Bℓhℓ(2)(kr): the two solutions have nonzero Wronskian by [F5], and matching value and derivative at one radius followed by uniqueness of the linear initial-value equation spans all solutions. At R, the condition from step 1.2 gives Bℓk((hℓ(2))′(kR)−[(hℓ(1))′(kR)/hℓ(1)(kR)]hℓ(2)(kR))=0. The parenthesized factor is the nonzero Wronskian divided by hℓ(1)(kR), so Bℓ=0. The inner condition and [F5] then give Aℓ=0. Every modal projection vanishes, hence w=0 in V. This proves uniqueness.

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