A weakly damped truncated layer can produce a resonant cavity response

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.