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Half-space Poisson extension of a plane wave

Example

Assume Countable Choice and n≥3. Write en for the last canonical basis vector en−1. Fix ξ∈Rn−1 and let g(x′)=exp⁡(2πi ξ⋅x′) on ∂H=Rn−1. Then the half-space Poisson integral of g is Ug(x′,t)=exp⁡(−2π∣ξ∣t)exp⁡(2πi ξ⋅x′),(x′,t)∈H, including the case ξ=0, where the extension is the constant 1. Consequently ∂tUg(x′,0)=−2π∣ξ∣ g(x′) and the outward normal derivative at the boundary is +2π∣ξ∣ g, so for a single spatial frequency the Dirichlet-to-Neumann map is multiplication by 2π∣ξ∣.

Facts & Assumptions

Given: Countable Choice, an integer n≥3, a frequency ξ∈Rn−1 and the datum g(x′)=exp⁡(2πi ξ⋅x′) on ∂H.

[F1]

For bounded continuous g the half-space Poisson integral Ug is the unique bounded harmonic function on H, continuous on H‾, with trace g; the Poisson kernel is PH((x′,t),z)=2t/(ωn−1(∣x′−z∣2+t2)n/2) (Poisson kernel and bounded Dirichlet problem on a half-space).

[F2]

Laplacian and partial derivatives: Δf=∑i∂i∂if, and for the exponential exp⁡(2πi ξ⋅x′) the tangential derivatives give Δx′exp⁡(2πi ξ⋅x′)=−4π2∣ξ∣2exp⁡(2πi ξ⋅x′) by the chain and product rules, while ∂t2exp⁡(−2π∣ξ∣t)=4π2∣ξ∣2exp⁡(−2π∣ξ∣t) (The Laplacian of a C2 function and of a C2 vector field, Directional derivatives and partial derivatives of a map U⊆Rm→Rn, Sums, scalar multiples, products and quotients: (f+g)′(c)=f′(c)+g′(c), (αf)′(c)=αf′(c), (fg)′(c)=f′(c)g(c)+f(c)g′(c), and (f/g)′(c)=(f′(c)g(c)−f(c)g′(c))/g(c)2 when g(c)≠0, The exponential function is smooth and (exp⁡)′=exp⁡, The complex exponential is entire and its complex derivative is itself).

[F3]

In the negative-sign 2π normalisation the Fourier transform of the plane wave x′↦e2πiξ⋅x′ is δξ (Fourier transform of delta constants plane waves and polynomials).

[F4]

Countable Choice ACω is the standing hypothesis (The Axiom of Countable Choice (ACω)).

Verification

technique · direct
1.1givenF4algebra

Work under [F4] and define V(x′,t):=exp⁡(−2π∣ξ∣t)exp⁡(2πi ξ⋅x′) on H‾. Since ∣V(x′,t)∣=exp⁡(−2π∣ξ∣t)≤1 and V(x′,0)=g(x′), the function V is bounded and continuous on H‾ with trace g.

2.1step 1.1F2

By [F2], ΔV=Δx′V+∂t2V=−4π2∣ξ∣2V+4π2∣ξ∣2V=0 on H; so V is harmonic (all derivatives exist and are continuous, being those of an exponential).

3.1step 1.1step 2.1F1

Applying uniqueness in [F1] to V and to the Poisson integral Ug of the bounded continuous datum g gives Ug=V, which is the displayed formula; for ξ=0 this reads Ug≡1.

4.1step 3.1algebra

Differentiating the formula at t=0 gives ∂tUg(x′,0)=−2π∣ξ∣ g(x′); the outward unit normal of H at the boundary plane is −en, so the outward normal derivative is −∂tUg(x′,0)=+2π∣ξ∣ g(x′). This is the single-mode Dirichlet-to-Neumann computation: the half-space Poisson multiplier e−2π∣ξ∣t differentiates to the boundary multiplier 2π∣ξ∣ in the outward normal.

5.1step 3.1F3algebra∎

The same multiplier is visible in the Fourier description: [F3] says the datum g has Fourier transform δξ, and the extension multiplies that mode by the factor e−2π∣ξ∣t; the constant mode ξ=0 is fixed and does not decay.

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