Exterior waves, absorbing boundaries and PML — Examples

For a propagating mode, substitute $q=k\cos \theta$ into [F2]. At $\theta =0$ the factor is $e^{-kS}$; as $\theta \uparrow \pi \mathrm{/}2$, $\cos \theta \to 0$, so it tends to $1$.

For $q=i\alpha$, direct substitution into [F1] gives $e^{iqX}=e^{i(i\alpha )(x+iS)}=e^{-\alpha x-i\alpha S}$, whose modulus is $e^{-\alpha x}$. The added stretch contributes only a phase, confirming that the propagating attenuation theorem does not cover evanescent modes.

Divide numerator and denominator of [F1] by $k>0$ to obtain $\mathcal{R}=(\cos \theta -1)\mathrm{/}(\cos \theta +1)$. At $\theta =0$, this is $0$; at $\theta =\pi \mathrm{/}3$, $\cos \theta =1\mathrm{/}2$ and it is $(-1\mathrm{/}2)\mathrm{/}(3\mathrm{/}2)=-1\mathrm{/}3$.

As $\theta \uparrow \pi \mathrm{/}2$, continuity of cosine gives $\cos \theta \to 0$ and the denominator tends to $1$, so $\mathcal{R}\to -1$ and $\mathrm{\mid }\mathcal{R}\mathrm{\mid }\to 1$. The endpoint is a limit; the propagating theorem assumed $q>0$.

Parameterize the stretched contour by the real physical coordinate $x\in [0,L]$. For a source at $x^{\mathrm{\prime }}$, set $x_{<}=\min (x,x^{\mathrm{\prime }})$, $x_{>}=\max (x,x^{\mathrm{\prime }})$, and $X_{<}=X(x_{<})$, $X_{>}=X(x_{>})$. The functions $y_1(X)=\sin (kX)$ and $y_2(X)=\sin (k(X_L-X))$ solve the homogeneous equation and vanish at the respective endpoints; their Wronskian in the order $y_1{y}_2^{\mathrm{\prime }}-y_1^{\mathrm{\prime }}{y}_2$ is the constant $-k\sin (k{X}_L)$. Thus $G(x,x^{\mathrm{\prime }})=\frac{\sin (k{X}_{<})\sin (k(X_L-X_{>}))}{k\sin (k{X}_L)}$ has the endpoint conditions and solves the homogeneous equation away from $x=x^{\mathrm{\prime }}$. The real parameter orders the two sides of the contour, and the one-sided derivatives in the contour coordinate satisfy ${{\mathrm{\partial }}_{X}G\mid }_{x\to x^{\mathrm{\prime }+}}-{{\mathrm{\partial }}_{X}G\mid }_{x\to x^{\mathrm{\prime }-}}=-1$, as required by [A1].

At $X^{\mathrm{\prime }}=x_0=L\mathrm{/}2$, $X_L=L+iS$, and $kL=(2m+1)\pi$, the numerator magnitude is $\cosh (kS)$ and the denominator magnitude is $k\sinh (kS)$; this follows from $\sin ((m+1\mathrm{/}2)\pi )=(-1{)}^m$ and $\mathrm{\mid }\sin (n\pi +it)\mathrm{\mid }=\sinh (t)$ for integer $n$ and $t>0$. Hence $\mathrm{\mid }G(x_0,x_0)\mathrm{\mid }=\coth (kS)\mathrm{/}k$. Since $\sinh t\sim t$ and $\cosh t\to 1$ as $t\downarrow 0$, this is asymptotic to $1\mathrm{/}(k^{2}S)$. At $S=0$, $\sin (kL)=0$ and $\sin (kx)$ satisfies both homogeneous Dirichlet conditions, proving resonance. The same damping parameter in [F2] gives round-trip reflected amplitude $e^{-2kS}\to 1$, so the growing cavity response accompanies vanishing attenuation.

Substituting [F2] into the Hankel ratio cancels the factor $-i\mathrm{/}k$ and gives $u(r)=(a\mathrm{/}r)e^{ik(r-a)}$, so $u(a)=1$. Direct differentiation yields $u^{\mathrm{\prime }}=(ik-r^{-1})u$ and $u^{\mathrm{\prime }\mathrm{\prime }}=(-k^2-2ik\mathrm{/}r+2\mathrm{/}{r}^2)u$. In three dimensions a radial function satisfies $\mathrm{\Delta }u=u^{\mathrm{\prime }\mathrm{\prime }}+2u^{\mathrm{\prime }}\mathrm{/}r$; hence $u^{\mathrm{\prime }\mathrm{\prime }}+2u^{\mathrm{\prime }}\mathrm{/}r+k^{2}u=0$ by cancellation of each coefficient, in agreement with [F3] after setting $z=kr$.

The same derivative identity gives $r(u^{\mathrm{\prime }}-iku)=-u(r)$ and $\mathrm{\mid }u(r)\mathrm{\mid }=a\mathrm{/}r\to 0$. Thus the outgoing condition holds, and the displayed mode is an outgoing monopole solution.

If $w$ is any other classical outgoing solution with boundary value $1$, then $w-u$ is a classical outgoing solution with zero Dirichlet trace on $r=a$. By [F4], $w-u=0$. Hence the displayed field is the unique outgoing monopole solution.

By [F3], $h_0(z)=-i{e}^{iz}\mathrm{/}z$, so $h_0^{\mathrm{\prime }}(z)=(-i)e^{iz}(i\mathrm{/}z-1\mathrm{/}{z}^2)=h_0(z)(i-1\mathrm{/}z)$.

As $h_0(kR)\mathrm{\ne }0$ by [F3], substitute $z=kR$ into step 1.1 and multiply by $k$: ${\tau }_0=k(i-1\mathrm{/}(kR))=ik-1\mathrm{/}R$. Theorem [F2] identifies this multiplier with the exact outgoing normal derivative.