A Perfect Absorbing Layer for High-Order Simulation of Wave Scattering Problems

Abstract

We report a novel approach to design artificial absorbing layers for spectral-element discretisation of wave scattering problems with bounded scatterers. It is essentially built upon two techniques: (i) a complex compression coordinate transformation that compresses all outgoing waves in the open space into the artificial layer, and then forces them to be attenuated and decay exponentially; (ii) a substitution (for the unknown) that removes the singularity induced by the transformation, and diminishes the oscillations near the inner boundary of the layer. As a result, the solution in the absorbing layer has no oscillation and is well-behaved for arbitrary high wavenumber and very thin layer. It is therefore well-suited and perfect for high-order simulations of scattering problems.

Appendices

Appendix 1. Proof of Theorem 1

Proof

Given the transformation (14), (8) becomes

$$\displaystyle{ \mathbf{J} = \frac{\partial (x,y)} {\partial (\tilde{x},\tilde{y})} = \frac{\partial (x,y)} {\partial (r,\theta )} \frac{\partial (r,\theta )} {\partial (\tilde{\rho },\theta )} \frac{\partial (\tilde{\rho },\theta )} {\partial (\tilde{x},\tilde{y})}. }$$

(54)

With $\tilde{\rho }$ in place of ρ in (9)– (10), we have

$$\displaystyle{ \mathbf{J} = \mathbf{R}\,\mathbf{J}_{\!1}\mathbf{R}^{t}\quad \mathrm{with}\;\;\;\mathbf{J}_{\! 1} = \left [\begin{array}{*{10}c} dr/d\tilde{\rho }& 0 \\ 0 &r/\tilde{\rho } \end{array} \right ], }$$

(55)

and

$$\displaystyle{ \mathbf{C} = \mathbf{R}\left [\begin{array}{*{10}c} c& 0\\ 0 &1/c \end{array} \right ]\mathbf{R}^{t},\quad n = \frac{\tilde{\rho }} {r} \frac{d\tilde{\rho }} {dr} =\alpha \beta \frac{\rho } {r} \frac{d\rho } {dr},\quad c:= \frac{\tilde{\rho }} {r} \frac{dr} {d\tilde{\rho }} = \frac{\beta } {\alpha } \frac{\rho } {r} \frac{dr} {d\rho }. }$$

(56)

Then we can work out the explicit expressions of n, c in (18) as (13).

Note that the asymptotic boundary condition at r = b is transformed from the Sommerfeld radiation condition in (1b).

We now derive the estimate (19). For this purpose, we expand the solution and data in Fourier series:

$$\displaystyle{ \{u,\varPsi \}=\sum _{ \vert m\vert =0}^{\infty }\{\hat{u}_{ m}(r),\hat{\psi }_{m}(r)\}e^{\mathrm{i}m\theta }, }$$

(57)

where $\big\{\hat{u}_{m}(r),\hat{\psi }_{m}(r)\big\}$ are the Fourier coefficients. Then we can reduce the problem (16)– (17) to

$$\displaystyle{ \frac{1} {r}\big(rc\,\hat{u}_{m}^{{\prime}}\big)' -\frac{m^{2}} {r^{2}c}\hat{u}_{m} + k^{2}n\;\hat{u}_{ m} = 0,\;\;\;r \in [a,b),\;\;\vert m\vert = 0,1,2,\cdots \,, }$$

(58)

$$\displaystyle{ \hat{u}_{m} =\hat{\psi } _{m}\;\;\mathrm{at}\;\;r = a;\quad \frac{1} {\alpha } \frac{dr} {d\rho } \hat{u}_{m}^{{\prime}}-\mathrm{ i}ku = o(\vert \tilde{\rho }\vert ^{-1/2})\;\;\mathrm{as}\;\;r \rightarrow b^{-}. }$$

(59)

One can verify by using the Bessel equation of Hankel function (cf. [1]):

$$\displaystyle{r^{2}y'' + ry' + (r^{2} - m^{2})y = 0,\quad y = H_{ m}^{(1)}(r),}$$

that the unique solution of (16)– (17) is

$$\displaystyle{ u =\sum _{ \vert m\vert =0}^{\infty }\hat{u}_{ m}(r)e^{\mathrm{i}m\theta }\;\;\mathrm{with}\;\;\hat{u}_{ m}(r) =\hat{\psi } _{m} \frac{H_{m}^{(1)}(k\tilde{\rho })} {H_{m}^{(1)}(ka)}. }$$

(60)

We next resort to a uniform estimate of Hankel functions first derived in [7, Lemma 2.2]: For any complex z with Re(z), Im(z) ≥ 0, and for any real Θ such that 0 < Θ ≤ | z |, we have for any real order ν, 

$$\displaystyle{ \vert H_{\nu }^{(1)}(z)\vert \leq e^{-\mathrm{Im}(z)\big(1- \frac{\varTheta ^{2}} {\vert z\vert ^{2}} \big)^{1/2} }\vert H_{\nu }^{(1)}(\varTheta )\vert, }$$

(61)

which implies

$$\displaystyle{ \max _{\vert m\vert \geq 0}\bigg\vert \frac{H_{m}^{(1)}(k\tilde{\rho })} {H_{m}^{(1)}(ka)}\bigg\vert \leq \mathrm{ exp}\Big\{-k\sigma _{0}(\rho -a)\Big(1 - \frac{a^{2}} {k^{2}\rho ^{2} + k^{2}\sigma _{0}^{2}(\rho -a)^{2}}\Big)^{1/2}\Big\},\;\;\rho> a. }$$

(62)

Therefore, we can derive (19) by using the Parseval’s identity of Fourier series and (62). □

Appendix 2. Proof of Theorem 2

Proof

We first deal with the boundary term $\langle \mathbf{C}\nabla u \cdot \mathbf{n},\phi \rangle _{\varGamma _{\!b}}$ in (24). By a direct calculation and (56), we have

$$\displaystyle{ (\partial _{r}u,r^{-1}\partial _{\theta }u)^{t} = \mathbf{R}^{t}\nabla u,\quad \mathbf{C}\nabla u \cdot \mathbf{n} = \mathbf{R}\,\mathrm{diag}(c,c^{-1})\,\mathbf{R}^{t}\,\nabla u \cdot \mathbf{n} = c\,\partial _{ r}u. }$$

(63)

Thus, using (56) and the substitutions: ϕ = wψ and u = wv, we can write

$$\displaystyle{ \begin{array}{rl} &\langle \mathbf{C}\nabla u \cdot \mathbf{n},\phi \rangle _{\varGamma _{\!b}} =\langle cu_{r},\phi \rangle _{\varGamma _{\!b}} =\langle c\bar{w}u_{r},\psi \rangle _{\varGamma _{\!b}} = a^{3/2}\bigg\langle \frac{\beta } {r}\sqrt{\frac{b-r} {s(r)}}e^{-\mathrm{i}k(\rho -a)}\frac{1} {\alpha } \frac{dr} {d\rho } u_{r},\psi \bigg\rangle _{\varGamma _{\!b}} \\ & = a^{3/2}\bigg\langle \frac{\beta } {r}\sqrt{\frac{b-r} {s(r)}}e^{-\mathrm{i}k(\rho -a)}\Big(\frac{1} {\alpha } \frac{dr} {d\rho } u_{r} -\mathrm{ i}ku\Big),\psi \bigg\rangle _{\varGamma _{\!b}} +\mathrm{ i}ka^{3/2}\bigg\langle \frac{\beta } {r}\sqrt{\frac{b-r} {s(r)}}e^{-\mathrm{i}k(\rho -a)}u,\psi \bigg\rangle _{\varGamma _{\! b}} \\ & = a^{3/2}\bigg\langle \frac{\beta } {r}\sqrt{\frac{b-r} {s(r)}}e^{-\mathrm{i}k(\rho -a)}\Big(\frac{1} {\alpha } \frac{dr} {d\rho } u_{r} -\mathrm{ i}ku\Big),\psi \bigg\rangle _{\varGamma _{\!b}} +\mathrm{ i}ka^{3}\bigg\langle \frac{\beta } {r} \frac{(b-r)^{2}} {s^{2}(r)} v,\psi \bigg\rangle _{\varGamma _{\!b}}. \end{array} }$$

Noting that the integral along Γ _​b is in θ, we obtain from the transformed Sommerfeld radiation condition (17) that $\langle \mathbf{C}\nabla u \cdot \mathbf{n},\phi \rangle _{\varGamma _{\!b}} \rightarrow 0$ as r → b ⁻. 

We next deal with the other two terms in (24). Using the basic differentiation rules

$$\displaystyle{ \nabla u = w\,\nabla v + v\,\nabla w,\quad \nabla \bar{\phi } =\bar{ w}\,\nabla \bar{\psi } +\bar{\psi }\, \nabla \bar{w}, }$$

we derive from (24) and a direct calculation that

$$\displaystyle{ \begin{array}{rl} \mathbb{B}_{_{\varOmega _{\mathrm{ ab}}}}(u,\phi ) =&\big(\vert w\vert ^{2}\mathbf{C}\nabla v,\nabla \psi \big)_{\varOmega _{\mathrm{ ab}}} +\big (w\,\mathbf{C}\nabla v \cdot \nabla \bar{w},\psi \big)_{\varOmega _{\mathrm{ab}}} +\big (v\,\bar{w}\mathbf{C}\,\nabla w,\nabla \psi \big)_{\varOmega _{\mathrm{ab}}} \\ & +\big (\mathbf{C}\,\nabla w \cdot \nabla \bar{w}\,v,\psi \big)_{\varOmega _{\mathrm{ab}}} - k^{2}\big(\vert w\vert ^{2}\,n\,v,\psi \big)_{\varOmega _{\mathrm{ab}}}.\end{array} }$$

(64)

As C is symmetric, one verifies readily that for any vectors a and b with two components, we have (Ca) ⋅ b = (Cb) ⋅ a. Thus, we can rewrite

$$\displaystyle{ \begin{array}{rl} &\big(w\,\mathbf{C}\nabla v \cdot \nabla \bar{w},\psi \big)_{\varOmega _{\mathrm{ab}}} =\big (w\,\mathbf{C}\nabla \bar{w} \cdot \nabla v,\psi \big)_{\varOmega _{\mathrm{ab}}}. \end{array} }$$

(65)

As w is independent of θ, we immediately get $\nabla w = \frac{dw} {dr} \mathbf{n}.$ Then by (56),

$$\displaystyle{ \mathbf{C}\nabla w = \frac{dw} {dr} \mathbf{R}\,\mathrm{diag}(c,c^{-1})\,\mathbf{R}^{t}\,\mathbf{n} = c\frac{dw} {dr} \mathbf{n}. }$$

(66)

Thus, we have

$$\displaystyle{ w\mathbf{C}\nabla \bar{w} = c\,w\frac{d\bar{w}} {dr} \mathbf{n},\quad \bar{w}\mathbf{C}\nabla w = c\,\bar{w}\frac{dw} {dr} \mathbf{n},\quad \mathbf{C}\,\nabla w \cdot \nabla \bar{w} = c\Big\vert \frac{dw} {dr} \Big\vert ^{2}. }$$

(67)

Introducing

$$\displaystyle{ \varpi _{1} = \vert w\vert ^{2},\quad \varpi _{ 2} = c\,\bar{w}\frac{\alpha } {\beta } \frac{dw} {dr},\quad \varpi _{3} = c\Big\vert \frac{dw} {dr} \Big\vert ^{2} - k^{2}\vert w\vert ^{2}n,\quad \partial _{\mathbf{ n}} = \mathbf{n} \cdot \nabla, }$$

(68)

we can derive (25) from (64)– (65) and (67)– (68). By (20),

$$\displaystyle{ \frac{dw} {dr} = w \frac{d\rho } {dr}\Big(-\frac{3} {2\rho } +\mathrm{ i}k\Big). }$$

(69)

We can work out {ϖ _j } _{j = 1} ³ by using (12), (56) and (69). □

Appendix 3. Proof of Theorem 3

Proof

We take v = ψ in (25). By (56) and (63), we have

$$\displaystyle{ \begin{array}{rl} &\mathrm{Re}\big(\varpi _{1}\mathbf{C}\nabla v,\nabla v\big)_{\varOmega _{\mathrm{ab}}} =\mathrm{ Re}\int _{0}^{2\pi }\!\!\!\!\int _{a}^{b}\Big\{c\vert v_{r}\vert ^{2} + \frac{1} {cr^{2}} \vert v_{\theta }\vert ^{2}\Big\}\varpi _{1}\,rdrd\theta \\ &\qquad =\int _{ 0}^{2\pi }\!\!\!\!\int _{a}^{b}\Big\{\mathrm{Re}\Big(\frac{\beta }{\alpha }\Big)\frac{a^{3}} {r\rho ^{2}} \frac{dr} {d\rho } \Big\}\vert v_{r}\vert ^{2}rdrd\theta +\int _{ 0}^{2\pi }\!\!\!\!\int _{ a}^{b}\Big\{\mathrm{Re}\Big(\frac{\alpha }{\beta }\Big)\frac{a^{3}} {r\rho ^{4}} \frac{d\rho } {dr}\Big\}\vert v_{\theta }\vert ^{2}rdrd\theta. \end{array} }$$

(70)

Using (26) and integration by parts leads to

$$\displaystyle{ \begin{array}{rl} &\mathrm{Re}\Big\{\frac{1} {\alpha } \big(\beta D_{\mathbf{n}}v,v\varpi _{2}\big)_{\varOmega _{\mathrm{ab}}} + \frac{1} {\alpha } \big(\beta v\varpi _{2},D_{\mathbf{n}}v\big)_{\varOmega _{\mathrm{ab}}}\Big\} = 2\int _{0}^{2\pi }\!\!\!\!\int _{a}^{b}\!\mathrm{Re}\Big(\frac{\beta }{\alpha }\Big)\,\mathrm{Re}(\varpi _{2}v\bar{v}_{r})rdrd\theta \\ & =\int _{ 0}^{2\pi }\!\!\!\!\int _{a}^{b}\mathrm{Re}\Big(\frac{\beta }{\alpha }\Big)\,\mathrm{Re}(\varpi _{2})(\partial _{r}\vert v\vert ^{2})rdrd\theta - 2\int _{0}^{2\pi }\!\!\!\!\int _{a}^{b}\mathrm{Re}\Big(\frac{\beta }{\alpha }\Big)\,\mathrm{Im}(\varpi _{2})\mathrm{Im}(v\bar{v}_{r})rdrd\theta \\ & = \frac{3} {2} \frac{1} {1+\sigma _{0}^{2}} \|v(a,\cdot )\|_{L^{2}(0,2\pi )}^{2} + \frac{3} {2}\int _{0}^{2\pi }\!\!\!\!\int _{ a}^{b}\Big\{\frac{a^{3}} {r\rho ^{4}} \Big( \frac{4\sigma _{0}^{2}} {1+\sigma _{0}^{2}} \frac{a} {\rho } - 3\Big) \frac{d\rho } {dr}\Big\}\vert v\vert ^{2}rdrd\theta \\ &\quad - 2k\int _{0}^{2\pi }\!\!\!\!\int _{a}^{b}\mathrm{Re}\Big(\frac{\beta }{\alpha }\Big)\frac{a^{3}} {\rho ^{2}} \mathrm{Im}(v\bar{v}_{r})drd\theta.\end{array} }$$

(71)

It is evident that

$$\displaystyle{ \mathrm{Re}\big(\varpi _{3}v,v\big)_{\varOmega _{\mathrm{ab}}} =\int _{ 0}^{2\pi }\!\!\!\!\int _{ a}^{b}\mathrm{Re}(\varpi _{ 3})\vert v\vert ^{2}rdrd\theta. }$$

(72)

Note from (15) that

$$\displaystyle{ \mathrm{Re}\Big(\frac{\beta } {\alpha }\Big) = 1 - \frac{\sigma _{0}^{2}} {1 +\sigma _{ 0}^{2}} \frac{a} {\rho }> \frac{1} {1 +\sigma _{ 0}^{2}},\quad \mathrm{Re}\Big(\frac{\alpha } {\beta }\Big) = 1 + \frac{\sigma _{0}(1 - a/\rho )} {1 +\sigma _{ 0}^{2}(1 - a/\rho )^{2}} \frac{a} {\rho }> 1. }$$

(73)

Using the Cauchy-Schwarz inequality, we obtain

$$\displaystyle{ \begin{array}{rl} 2k\int _{0}^{2\pi }\!\!\!\!\int _{a}^{b}\mathrm{Re}\Big(\frac{\beta }{\alpha }\Big)\frac{a^{3}} {\rho ^{2}} \mathrm{Im}(v\bar{v}_{r})drd\theta & \leq \frac{1} {\epsilon } \int _{0}^{2\pi }\!\!\!\!\int _{ a}^{b}\Big\{\mathrm{Re}\Big(\frac{\beta }{\alpha }\Big)\frac{a^{3}} {r\rho ^{2}} \frac{dr} {d\rho } \Big\}\vert v_{r}\vert ^{2}rdrd\theta \\ & +\epsilon k^{2}\int _{0}^{2\pi }\!\!\!\!\int _{a}^{b}\Big\{\mathrm{Re}\Big(\frac{\beta }{\alpha }\Big)\frac{a^{3}} {r\rho ^{2}} \frac{d\rho } {dr}\Big\}\vert v\vert ^{2}rdrd\theta,\end{array} }$$

(74)

where ε is a positive constant independent of k. Thus, by (25), (70)– (74) and collecting the terms, we obtain

$$\displaystyle{ \begin{array}{rl} \mathrm{Re}\big\{&\mathbb{B}_{\varOmega _{\mathrm{ab}}}(u,u)\big\} \geq \Big (1 -\frac{1} {\epsilon } \Big)\int _{0}^{2\pi }\!\!\!\!\int _{ a}^{b}\Big\{\mathrm{Re}\Big(\frac{\beta }{\alpha }\Big)\frac{a^{3}} {r\rho ^{2}} \frac{dr} {d\rho } \Big\}\vert v_{r}\vert ^{2}rdrd\theta \\ & +\int _{ 0}^{2\pi }\!\!\!\!\int _{a}^{b}\Big\{\mathrm{Re}\Big(\frac{\alpha }{\beta }\Big)\frac{a^{3}} {r\rho ^{4}} \frac{d\rho } {dr}\Big\}\vert v_{\theta }\vert ^{2}rdrd\theta + \frac{3} {2} \frac{1} {1+\sigma _{0}^{2}} \|v(a,\cdot )\|_{L^{2}(0,2\pi )}^{2} \\ & +\int _{ 0}^{2\pi }\!\!\!\!\int _{a}^{b}\bigg\{\mathrm{Re}(\varpi _{3}) +\bigg (\frac{3} {2} \frac{1} {\rho ^{2}} \Big( \frac{4\sigma _{0}^{2}} {1+\sigma _{0}^{2}} \frac{a} {\rho } - 3\Big) -\epsilon k^{2}\mathrm{Re}\Big(\frac{\beta }{\alpha }\Big)\bigg)\frac{a^{3}} {r\rho ^{2}} \frac{d\rho } {dr}\bigg\}\vert v\vert ^{2}rdrd\theta. \end{array} }$$

(75)

We next work out and estimate the functions in the brackets. We have

$$\displaystyle{ s(r) = a^{2} + r(b - 2a) \leq \vert I\vert \bar{c},\;\;\;\bar{c}:=\mathrm{ max}\{a,\vert I\vert \},\;\;\vert I\vert:= b - a. }$$

(76)

By (12), (26) and (76),

$$\displaystyle{ \frac{a^{3}} {r\rho ^{2}} \frac{dr} {d\rho } = \frac{a^{3}} {\vert I\vert ^{2}} \frac{(b - r)^{4}} {rs^{2}(r)} \geq \frac{a^{3}} {b\,\bar{c}^{2}\vert I\vert ^{4}}(b-r)^{4};\;\; \frac{a^{3}} {r\rho ^{4}} \frac{d\rho } {dr} = \frac{a^{3}\vert I\vert ^{2}} {r} \frac{(b - r)^{2}} {s^{4}(r)} \geq \frac{a^{3}} {b\,\bar{c}^{4}\vert I\vert ^{2}}(b-r)^{2}, }$$

(77)

so we can obtain the lower bounds of the first two terms.

With a careful calculation, we can work out the summation in the curly brackets of the last term in (75) by using (12), (15) and (26). □