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Perfect matching and attenuation for non-grazing propagating modes

Statement

Let k>0, η∈R2 satisfy ∣η∣<k, and let q=k2−∣η∣2>0 be the outgoing normal branch. In the stretch of Cartesian complex-stretch PML for a homogeneous Helmholtz mode, u(x,y)=eiqX(x)+iη⋅y solves the pulled-back Helmholtz equation. At x=0, its value and normal flux (1/s)∂xu agree with those of the physical mode eiqx+iη⋅y. Its modulus is e−qS(x). If a family of modes satisfies q≥q0>0, then ∣u(x,y)∣≤e−q0S(x) uniformly; whenever S(x)≥σ0x for some σ0>0, this is at most e−q0σ0x. No uniform bound is asserted as q↓0.

Facts & Assumptions

Given: The constant-coefficient Cartesian operator and stretch data in [F1].

[F1]

Cartesian complex-stretch PML for a homogeneous Helmholtz mode: X=x+iS, s=X′=1+iσ, and the operator is s−1∂x(s−1∂x)+Δy+k2.

Proof

Given: k>0, ∣η∣<k, q=k2−∣η∣2, and [F1].

1.1givenF1algebra

For the stated mode, ∂xu=iqs(x)u, so s−1∂xu=iqu and s−1∂x(s−1∂xu)=s−1∂x(iqu)=−q2u. Also Δyu=−∣η∣2u. Since q2+∣η∣2=k2, substitution into [F1] gives zero.

2.1F1step 1.1algebra∎

At x=0, S(0)=0 and σ(0)=0 by continuity and the definition, so X(0)=0, s(0)=1, and the stretched mode and its normal flux equal eiη⋅y and iqeiη⋅y, respectively, exactly as in the physical outgoing mode. Since ∣eiq(x+iS)+iη⋅y∣=e−qS, the bounds follow from q≥q0 and, when assumed, S(x)≥σ0x. As q↓0, the attenuation bound tends to one, so this proof gives no uniform grazing decay.

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