Exterior waves, absorbing boundaries and PML

For the stated mode, ${\mathrm{\partial }}_{x}u=iqs(x)u$, so $s^{-1}{\mathrm{\partial }}_{x}u=iqu$ and $s^{-1}{\mathrm{\partial }}_x(s^{-1}{\mathrm{\partial }}_{x}u)=s^{-1}{\mathrm{\partial }}_x(iqu)=-q^{2}u$. Also ${\mathrm{\Delta }}_{y}u=-\mathrm{\mid }\eta {\mathrm{\mid }}^{2}u$. Since $q^2+\mathrm{\mid }\eta {\mathrm{\mid }}^2=k^2$, substitution into [F1] gives zero.

At $x=0$, $S(0)=0$ and $\sigma (0)=0$ by continuity and the definition, so $X(0)=0$, $s(0)=1$, and the stretched mode and its normal flux equal $e^{i\eta \cdot y}$ and $iq{e}^{i\eta \cdot y}$, respectively, exactly as in the physical outgoing mode. Since $\mathrm{\mid }{e}^{iq(x+iS)+i\eta \cdot y}\mathrm{\mid }=e^{-qS}$, the bounds follow from $q\ge q_0$ and, when assumed, $S(x)\ge {\sigma }_{0}x$. As $q\downarrow 0$, the attenuation bound tends to one, so this proof gives no uniform grazing decay.

Direct differentiation of $h_0=-i{e}^{iz}\mathrm{/}z$ gives the finite expression $h_{\mathrm{\ell }}(z)=(-i{)}^{\mathrm{\ell }+1}{e}^{iz}{\sum }_{m=0}^{\mathrm{\ell }}{i}^m{A}_{\mathrm{\ell }m}{z}^{-m-1},A_{\mathrm{\ell }m}=\frac{(\mathrm{\ell }+m)!}{m!(\mathrm{\ell }-m)!2^m}\mathrm{.}$ Indeed it holds at $\mathrm{\ell }=0$; substituting this expression in $h_{\mathrm{\ell }+1}=\mathrm{\ell }{h}_{\mathrm{\ell }}\mathrm{/}z-h_{\mathrm{\ell }}^{\mathrm{\prime }}$ gives $A_{\mathrm{\ell }+1,m}=A_{\mathrm{\ell }m}+(\mathrm{\ell }+m)A_{\mathrm{\ell },m-1}$ (out-of-range coefficients are zero), and the displayed factorial formula satisfies that recurrence and the $m=0$ base.

Let $F_{\mathrm{\ell }}=\mathrm{\mid }{h}_{\mathrm{\ell }}{\mathrm{\mid }}^2$. The radial recurrence gives $h_{\mathrm{\ell }+1}=(2\mathrm{\ell }+1)h_{\mathrm{\ell }}\mathrm{/}z-h_{\mathrm{\ell }-1}$. Also $F_{\mathrm{\ell }}^{\mathrm{\prime }}=2\mathrm{Re}({\bar{h}}_{\mathrm{\ell }}{h}_{\mathrm{\ell }-1})-2(\mathrm{\ell }+1)F_{\mathrm{\ell }}\mathrm{/}z$ by [F2]. Expanding the squared modulus and eliminating the real cross term yields $F_{\mathrm{\ell }+1}=F_{\mathrm{\ell }-1}-(2\mathrm{\ell }+1)F_{\mathrm{\ell }}^{\mathrm{\prime }}\mathrm{/}z-(2\mathrm{\ell }+1)F_{\mathrm{\ell }}\mathrm{/}{z}^2\mathrm{.}$ From $F_0=z^{-2}$ and $F_1=z^{-2}+z^{-4}$, coefficient comparison yields $S_{\mathrm{\ell }+1,0}=S_{\mathrm{\ell }-1,0}=1$ and, for $1\le q\le \mathrm{\ell }$, $S_{\mathrm{\ell }+1,q}=S_{\mathrm{\ell }-1,q}+(2\mathrm{\ell }+1)(2q-1)S_{\mathrm{\ell },q-1}$. The displayed factorial coefficients satisfy this: the ratios are $S_{\mathrm{\ell }-1,q}\mathrm{/}{S}_{\mathrm{\ell }+1,q}=(\mathrm{\ell }-q)(\mathrm{\ell }-q+1)\mathrm{/}[(\mathrm{\ell }+q)(\mathrm{\ell }+q+1)]$ and $S_{\mathrm{\ell }+1,q}\mathrm{/}{S}_{\mathrm{\ell },q-1}=(2q-1)(\mathrm{\ell }+q)(\mathrm{\ell }+q+1)\mathrm{/}(2q)$, whose difference gives the recurrence. At the top index $q=\mathrm{\ell }+1$, $S_{\mathrm{\ell }+1,\mathrm{\ell }+1}=(2\mathrm{\ell }+1{)}^2{S}_{\mathrm{\ell },\mathrm{\ell }}$; thus induction proves the stated sum at every degree.

Every $S_{\mathrm{\ell }m}$ is positive, so differentiating the finite sum gives $F_{\mathrm{\ell }}^{\mathrm{\prime }}(z)<0$. For $0\le m\le \mathrm{\ell }-1$, $S_{\mathrm{\ell }m}\mathrm{/}{S}_{\mathrm{\ell }-1,m}=(\mathrm{\ell }+m)\mathrm{/}(\mathrm{\ell }-m)\ge 1$, and the degree-$\mathrm{\ell }$ sum has one additional positive term. Hence $\mathrm{\mid }{h}_{\mathrm{\ell }}\mathrm{\mid }$ decreases with $z$ and $\mathrm{\mid }{h}_{\mathrm{\ell }-1}\mathrm{\mid }\le \mathrm{\mid }{h}_{\mathrm{\ell }}\mathrm{\mid }$. For $\mathrm{\ell }\ge 1$, [F2] now implies $\mathrm{\mid }{h}_{\mathrm{\ell }}^{\mathrm{\prime }}\mathrm{\mid }\le \mathrm{\mid }{h}_{\mathrm{\ell }-1}\mathrm{\mid }+(\mathrm{\ell }+1)\mathrm{\mid }{h}_{\mathrm{\ell }}\mathrm{\mid }\mathrm{/}z\le (1+(\mathrm{\ell }+1)\mathrm{/}z)\mathrm{\mid }{h}_{\mathrm{\ell }}\mathrm{\mid }$; for $\mathrm{\ell }=0$ the same bound follows by differentiating $h_0=-i{e}^{iz}\mathrm{/}z$.

Differentiating the finite sum in step 1.1 cancels its $i{e}^{iz}$ derivative in $h_{\mathrm{\ell }}^{\mathrm{\prime }}-i{h}_{\mathrm{\ell }}$, so $\mathrm{\mid }z(h_{\mathrm{\ell }}^{\mathrm{\prime }}-i{h}_{\mathrm{\ell }})\mathrm{\mid }\le {\sum }_{m=0}^{\mathrm{\ell }}(m+1)A_{\mathrm{\ell }m}{z}^{-m-1}$. The lowest-power term of the positive modulus sum gives $\mathrm{\mid }{h}_{\mathrm{\ell }}(z_0)\mathrm{\mid }\ge A_{\mathrm{\ell }\mathrm{\ell }}{z}_0^{-\mathrm{\ell }-1}$. For $j=\mathrm{\ell }-m$, factorial cancellation gives $\frac{A_{\mathrm{\ell },\mathrm{\ell }-j}}{A_{\mathrm{\ell }\mathrm{\ell }}}=\frac{2^j}{j!}{\prod }_{s=0}^{j-1}\frac{\mathrm{\ell }-s}{2\mathrm{\ell }-s}\le \frac{2^j}{j!}\mathrm{.}$ Using $z\ge z_0$ and $m+1\le \mathrm{\ell }+1$, the quotient in the claim is at most $(\mathrm{\ell }+1){\sum }_{j=0}^{\mathrm{\ell }}(2z_0{)}^j\mathrm{/}j!\le (\mathrm{\ell }+1)e^{2z_0}$.

At $x=R$, the bracket in the definition of $u$ is $1+\mathcal{R}$ and its $x$ derivative is $iq(1-\mathcal{R})$. Substitution into [F1], after cancelling the nonzero tangential exponential, gives $q(1-\mathcal{R})-k(1+\mathcal{R})=0$.

Since $q+k>0$, solving the equation in step 1.1 gives $\mathcal{R}=(q-k)\mathrm{/}(q+k)$. The numerator is nonpositive for $q\le k$, so $\mathrm{\mid }\mathcal{R}\mathrm{\mid }=(k-q)\mathrm{/}(k+q)$; this is zero at $q=k$ and tends to $1$ as $q\downarrow 0$.

The condition $u(L)=0$ reads $A{e}^{iqX(L)}+B{e}^{-iqX(L)}=0$. Since exponentials never vanish, if $A\mathrm{\ne }0$ division gives $B\mathrm{/}A=-e^{2iqX(L)}$.

With $X(L)=L+iS(L)$, $\mathrm{\mid }{e}^{2iqX(L)}\mathrm{\mid }=\mathrm{\mid }{e}^{2iqL-2qS(L)}\mathrm{\mid }=e^{-2qS(L)}$. The result measures the reflected-to-incident coefficient at $x=0$, where $X(0)=0$; it is the two-way attenuation through the continuous layer.

Let $z=kR$. The boundary condition, divided by $k$, is $A((h_{\mathrm{\ell }}^{(1)}{)}^{\mathrm{\prime }}(z)-i{h}_{\mathrm{\ell }}^{(1)}(z))+B((h_{\mathrm{\ell }}^{(2)}{)}^{\mathrm{\prime }}(z)-i{h}_{\mathrm{\ell }}^{(2)}(z))=0.$ If the incoming factor in parentheses were zero, then $(h_{\mathrm{\ell }}^{(2)}{)}^{\mathrm{\prime }}=i{h}_{\mathrm{\ell }}^{(2)}$, so $\mathrm{Im}(\overline{h_{\mathrm{\ell }}^{(2)}}(h_{\mathrm{\ell }}^{(2)}{)}^{\mathrm{\prime }})=\mathrm{\mid }{h}_{\mathrm{\ell }}^{(2)}{\mathrm{\mid }}^2>0$ by nonvanishing, contradicting its negative flux from [F2]. Hence that factor is nonzero.

Divide the equation in step 1.1 by its nonzero incoming factor to obtain the displayed coefficient equality for every $A,B$. For $A\mathrm{\ne }0$, division by $A$ gives the displayed ratio. If $A=0$, the coefficient equality forces $B=0$, so the boundary condition has no nonzero purely incoming mode and the zero mode carries no ratio.

By [F4], for every degree $\mathrm{\ell }$ the projection of $u$ has the form $u_{\mathrm{\ell }}(r,\cdot )=c_{\mathrm{\ell }}{h}_{\mathrm{\ell }}^{(1)}(kr)$ with $c_{\mathrm{\ell }}$ independent of $r$. Taking the degree-$\mathrm{\ell }$ projection of the zero trace at $r=a$ gives $0=c_{\mathrm{\ell }}{h}_{\mathrm{\ell }}^{(1)}(ka)$. The nonvanishing in [F3] implies $c_{\mathrm{\ell }}=0$ for every $\mathrm{\ell }$.

Thus every harmonic projection of $u(r,\cdot )$ vanishes for each $r>a$. By the uniform density in [F2], choose for each $\epsilon >0$ a finite sum $p$ of harmonic modes with $\mathrm{\parallel }u(r,\cdot )-p{\mathrm{\parallel }}_{\mathrm{\infty }}<\epsilon$. Orthogonality gives $⟨u,p⟩=0$, so $\mathrm{\parallel }u(r,\cdot ){\mathrm{\parallel }}_2^2=\mathrm{\mid }⟨u,u-p⟩\mathrm{\mid }\le \mathrm{\parallel }u(r,\cdot ){\mathrm{\parallel }}_2\mathrm{\parallel }u(r,\cdot )-p{\mathrm{\parallel }}_2\le \sqrt{4\pi }\epsilon \mathrm{\parallel }u(r,\cdot ){\mathrm{\parallel }}_2$. Letting $\epsilon \downarrow 0$ gives $u(r,\cdot )=0$ in $L^2$, and continuity makes it zero pointwise. Since $r>a$ was arbitrary, $u\equiv 0$.

By [F3], $\mathrm{\mid }{\tau }_{\mathrm{\ell }}\mathrm{\mid }\le k+(\mathrm{\ell }+1)\mathrm{/}R$. The $C^p$ estimate in [F5] and the super-polynomial decay in [F2] show that for every $p$, ${\sum }_{\mathrm{\ell }}\mathrm{\parallel }{\tau }_{\mathrm{\ell }}{g}_{\mathrm{\ell }}{\mathrm{\parallel }}_{C^p}<\mathrm{\infty }$. Thus the defining series converges in every smooth norm to a smooth function $T_{k,R}g$.

For $r\ge R$, the positive coefficient formula for $\mathrm{\mid }{h}_{\mathrm{\ell }}(z){\mathrm{\mid }}^2$ in [F3] makes $\mathrm{\mid }{h}_{\mathrm{\ell }}(kr)\mathrm{/}{h}_{\mathrm{\ell }}(kR)\mathrm{\mid }\le 1$. The derivative estimate in [F3] bounds one radial derivative of the ratio by $k+(\mathrm{\ell }+1)\mathrm{/}r$ times this ratio; repeated radial derivatives are bounded by polynomials in $1+\mathrm{\ell }$ on each compact annulus by differentiating the radial ODE in [F5]. Angular derivatives likewise contribute only polynomial factors. Thus [F2] and the evaluation bound make the extension series and its derivatives through order two uniformly convergent on $R\le r\le R_1$. Termwise differentiation is valid; each summand solves the Helmholtz equation by [F5], its trace is $g_{\mathrm{\ell }}$, and its radial derivative at $R$ is ${\tau }_{\mathrm{\ell }}{g}_{\mathrm{\ell }}$. Summing proves the trace, derivative, and PDE claims.

For each fixed $\mathrm{\ell }$, the finite formula in [F3] gives $r({\mathrm{\partial }}_r-ik)(h_{\mathrm{\ell }}(kr)\mathrm{/}{h}_{\mathrm{\ell }}(kR))\to 0$. The last bound in [F3], multiplied by $\mathrm{\parallel }{g}_{\mathrm{\ell }}{\mathrm{\parallel }}_{\mathrm{\infty }}$, dominates the absolute value of this degree's radiation residual by $(\mathrm{\ell }+1)e^{2kR}\sqrt{(2\mathrm{\ell }+1)\mathrm{/}(4\pi )}\mathrm{\parallel }{g}_{\mathrm{\ell }}{\mathrm{\parallel }}_2$, a summable sequence by [F2]. Dominated convergence for this series proves uniform convergence of the full radiation residual to zero.

Let $w$ be any other classical outgoing extension of $g$. By [F4], its degree-$\mathrm{\ell }$ projection is $d_{\mathrm{\ell }}{h}_{\mathrm{\ell }}^{(1)}(kr)$ for a fixed $d_{\mathrm{\ell }}\in {\mathcal{H}}_{\mathrm{\ell }}$. Its trace at $R$ is $g_{\mathrm{\ell }}$, so $d_{\mathrm{\ell }}{h}_{\mathrm{\ell }}^{(1)}(kR)=g_{\mathrm{\ell }}$. Nonvanishing in [F3] forces $d_{\mathrm{\ell }}=g_{\mathrm{\ell }}\mathrm{/}{h}_{\mathrm{\ell }}^{(1)}(kR)$, the same coefficient as the constructed extension in [F1]. Their difference has every harmonic projection zero and is continuous; density in [F2] implies the difference vanishes. This proves uniqueness and the well-definedness claimed in [F1].

Each $H^1((a,R);{\mathcal{H}}_{\mathrm{\ell }})$ is complete because ${\mathcal{H}}_{\mathrm{\ell }}$ is finite dimensional. A Cauchy sequence in the weighted square-sum norm is Cauchy in each coordinate; take its coordinatewise $H^1$ limits. For every finite number of coordinates the squared norm of the limit is bounded by the liminf of the sequence norms, so the coordinatewise limit has finite total norm; applying the same argument to differences shows convergence in the sum norm. Tails of the defining convergent nonnegative series tend to zero, so finite-degree truncations are dense. For any scalar or finite-dimensional Hilbert-valued $w\in H^1(a,R)$, set $A=\mathrm{\parallel }w{\mathrm{\parallel }}_{L^2(a,R)}$ and $B=\mathrm{\parallel }{w}^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^2(a,R)}$. Absolute continuity and $\mathrm{\mid }w(R){\mathrm{\mid }}^2-\mathrm{\mid }w(t){\mathrm{\mid }}^2=2\mathrm{Re}{\int }_t^R⟨w^{\mathrm{\prime }}(s),w(s)⟩ds$ give, after averaging $t$ over $[a,R]$ and applying Cauchy–Schwarz, $\mathrm{\parallel }w(R){\mathrm{\parallel }}^2\le (R-a{)}^{-1}{A}^2+2AB$. Multiplying by $1+\mathrm{\ell }$ and using $2(1+\mathrm{\ell })AB\le (1+\mathrm{\ell }{)}^2{A}^2+B^2$, $(1+\mathrm{\ell }{)}^2\le 2(1+\mathrm{\ell }(\mathrm{\ell }+1))$, and $B^2\le a^{-2}{\int }_a^R{r}^2\mathrm{\parallel }{w}^{\mathrm{\prime }}{\mathrm{\parallel }}^{2}dr$ gives $(1+\mathrm{\ell })\mathrm{\parallel }w(R){\mathrm{\parallel }}^2\le C_{a,R}{\int }_a^R(r^2\mathrm{\parallel }{w}^{\mathrm{\prime }}{\mathrm{\parallel }}^2+(1+\mathrm{\ell }(\mathrm{\ell }+1))\mathrm{\parallel }w{\mathrm{\parallel }}^2)dr$. Replacing $r$ by $a+R-r$ proves the same inequality at the inner endpoint. Summing with $w=u_{\mathrm{\ell }}$ proves the weighted trace estimate.

The volume terms of $\mathfrak{a}$ are bounded by a constant times $\mathrm{\parallel }u{\mathrm{\parallel }}_{\mathcal{V}}\mathrm{\parallel }v{\mathrm{\parallel }}_{\mathcal{V}}$ because $r$ lies in $[a,R]$ and $k$ is fixed. By [F2], $\mathrm{\mid }{\tau }_{\mathrm{\ell }}\mathrm{\mid }\le C_{k,R}(1+\mathrm{\ell })$. Cauchy–Schwarz over degrees and the trace estimate from step 1.1 give ${\sum }_{\mathrm{\ell }}\mathrm{\mid }{\tau }_{\mathrm{\ell }}⟨u_{\mathrm{\ell }}(R),v_{\mathrm{\ell }}(R)⟩\mathrm{\mid }\le C{({\sum }_{\mathrm{\ell }}(1+\mathrm{\ell })\mathrm{\parallel }{u}_{\mathrm{\ell }}(R){\mathrm{\parallel }}_2^2)}^{1\mathrm{/}2}{({\sum }_{\mathrm{\ell }}(1+\mathrm{\ell })\mathrm{\parallel }{v}_{\mathrm{\ell }}(R){\mathrm{\parallel }}_2^2)}^{1\mathrm{/}2}\le C\mathrm{\parallel }u{\mathrm{\parallel }}_{\mathcal{V}}\mathrm{\parallel }v{\mathrm{\parallel }}_{\mathcal{V}}\mathrm{.}$ Thus the boundary series is absolutely convergent and the whole form is continuous.

By [F3] applied at the inner radius $a$, the proposed series has inner trace $f$. Its degree-$\mathrm{\ell }$ radial ratio has modulus at most one for $r\ge a$ by [F5], and its radial derivative is bounded by $k+(\mathrm{\ell }+1)\mathrm{/}r$ times that ratio. The angular energy of degree $\mathrm{\ell }$ is $\mathrm{\ell }(\mathrm{\ell }+1)$ times its squared $L^2$ norm. Consequently its total $\mathcal{V}$ energy is at most $C_{a,R,k}{\sum }_{\mathrm{\ell }}(1+\mathrm{\ell }{)}^2\mathrm{\parallel }{f}_{\mathrm{\ell }}{\mathrm{\parallel }}_2^2$, which is finite by [F4]. Thus it belongs to ${\mathcal{V}}_f$.

Let $w\in {\mathcal{V}}_0$ solve the homogeneous variational equation. Testing against functions supported in a single degree and compactly supported in $(a,R)$ shows distributionally that $(r^2{w}_{\mathrm{\ell }}^{\mathrm{\prime }}{)}^{\mathrm{\prime }}+(k^2{r}^2-\mathrm{\ell }(\mathrm{\ell }+1))w_{\mathrm{\ell }}=0$. Since the coefficients are smooth and $r>0$, this one-dimensional equation implies $w_{\mathrm{\ell }}\in C^2([a,R];{\mathcal{H}}_{\mathrm{\ell }})$. Integrating by parts and allowing arbitrary outer test trace yields $w_{\mathrm{\ell }}^{\mathrm{\prime }}(R)={\tau }_{\mathrm{\ell }}{w}_{\mathrm{\ell }}(R)$; the inner trace is $w_{\mathrm{\ell }}(a)=0$.

Each modal coefficient satisfies $(r^2{u}_{\mathrm{\ell }}^{\mathrm{\prime }}{)}^{\mathrm{\prime }}+(k^2{r}^2-\mathrm{\ell }(\mathrm{\ell }+1))u_{\mathrm{\ell }}=0$ and $u_{\mathrm{\ell }}^{\mathrm{\prime }}(R)={\tau }_{\mathrm{\ell }}{u}_{\mathrm{\ell }}(R)$ by [F3]. Integration by parts against a finite-degree test with zero inner trace makes its modal volume term equal $R^2{u}_{\mathrm{\ell }}^{\mathrm{\prime }}(R)\overline{v_{\mathrm{\ell }}(R)}$; the outer term in [F1] subtracts exactly $R^2{\tau }_{\mathrm{\ell }}{u}_{\mathrm{\ell }}(R)\overline{v_{\mathrm{\ell }}(R)}$. Hence the form vanishes for every finite-degree test. Finite-degree truncations converge in $\mathcal{V}$, and [F2] extends the identity to every test in ${\mathcal{V}}_0$. This proves existence.

The radial ODE solution is $w_{\mathrm{\ell }}(r)=A_{\mathrm{\ell }}{h}_{\mathrm{\ell }}^{(1)}(kr)+B_{\mathrm{\ell }}{h}_{\mathrm{\ell }}^{(2)}(kr)$: 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 $B_{\mathrm{\ell }}k((h_{\mathrm{\ell }}^{(2)}{)}^{\mathrm{\prime }}(kR)-[(h_{\mathrm{\ell }}^{(1)}{)}^{\mathrm{\prime }}(kR)\mathrm{/}{h}_{\mathrm{\ell }}^{(1)}(kR)]h_{\mathrm{\ell }}^{(2)}(kR))=0$. The parenthesized factor is the nonzero Wronskian divided by $h_{\mathrm{\ell }}^{(1)}(kR)$, so $B_{\mathrm{\ell }}=0$. The inner condition and [F5] then give $A_{\mathrm{\ell }}=0$. Every modal projection vanishes, hence $w=0$ in $\mathcal{V}$. This proves uniqueness.