An Explicit Formula for the Nonstationary Diffracted Wave Scattered on a NN-Wedge

Abstract

We consider two dimensional nonstationary scattering of plane waves by a NN-wedge. We prove the existence and uniqueness of a solution to the corresponding mixed problem and we give an explicit formula for the solution. Also the Limiting Amplitude Principle is proved and a rate of convergence to the limiting amplitude is obtained.

Appendices

Appendix 1

Lemma 11.1

The function H _N (β), defined by (11), has the following properties

• (i) H _N (−β+iπ)=−H _N (β), for any $\beta\in \mathbb{C}$.

• (ii) H _N (β+2iΦ)=H _N (β), for any $\beta\in \mathbb{C}$.

Proof

It follows directly from (11). □

Lemma 11.2

The function $\widehat{u}_{s}$ given in (20) satisfies “stationary” NN-problem (19).

Proof

It is easy to check that $f_{1}(\rho,\theta,\omega) := - \widehat{g}(\omega) e^{i\rho\omega\cos(\theta-\alpha)}$ satisfies system (19). Therefore by (20), to prove (19) for $\widehat{u}_{s}$ it suffices to prove that

$$\begin{aligned} f_2(\rho,\theta,\omega) := \int _{\mathcal{C}} e^{-\rho\omega\sinh\beta} H_N(\beta+i\theta)\,\mathrm{d}\beta \end{aligned}$$

(91)

satisfies for any ρ>0

$$ \begin{cases} (\Delta+ \omega^2)\ f_2(\rho,\theta,\omega) = 0,&\theta\in[\phi ,2\pi] \\ \frac{\partial}{\partial y_2} f_2(\rho,\theta,\omega) = 0,&\theta = 2\pi\\ \frac{\partial}{\partial\mathbf{n}_2}f_2(\rho,\theta,\omega) = 0,&\theta= \phi \end{cases} $$

(92)

The Helmholtz equation in (92) follows by differentiation of the integral (91) after the change of variable β↦β′−iθ, since (Δ+ω ²)e ^{−ρωsinh(β−iθ)}=0. Moreover, the integral in (91) converges absolutely after the differentiation for any $\omega\in\mathbb{C}^{+}$ by (21), the condition $\omega\in\mathbb{C}^{+}$ and (11).

Let us prove that f ₂ satisfies the second equality of (92). Since $\frac{\partial}{\partial y_{2} } |_{\theta=2\pi}=\frac{1}{\rho }\cdot\frac{\partial}{\partial\theta} |_{\theta=2\pi}$ it suffices to prove that

$$ \left.\frac{\partial}{\partial\theta}\, f_2(\rho,\theta ,\omega) \right|_{\theta=2\pi} = 0. $$

(93)

Since $\frac{\partial}{\partial\theta} = i\, \frac{\mathrm{d}}{\mathrm {d}\beta}$ and the integral $\int_{\mathcal{C}} e^{-\rho\omega\sinh\beta}\frac {\partial}{\partial\theta} H_{N}(\beta+i\theta) \,\mathrm{d} \beta$ converges uniformly with respect to θ (by (21), $\omega \in\mathbb{C}^{+}$ and (11)) to prove (93) it suffices to prove that

$$ \int_{\mathcal{C}}e^{-\rho\omega\sinh\beta }\frac{\mathrm{d}}{\mathrm{d}\beta} H_N(\beta+2i\pi)\,\mathrm{d}\beta=0. $$

(94)

The function $e^{-\rho\omega\sinh\beta}\frac{\mathrm{d} }{\mathrm{d}\beta} H_{N}(\beta+2i\pi)$ is invariant with respect to $\mathcal{S}(\beta)=-\beta-3i\pi$, for any $\beta\in\mathcal {C}$. It follows from Lemma 11.1, i) and the fact that $\frac{\mathrm{d} }{\mathrm{d}\beta}\, H_{N}(\beta+2i\pi)$ is invariant with respect to $\mathcal{S}(\beta)$ for any $\beta\in\mathcal{C}$. Hence (94) holds by the symmetry of $\mathcal{C}$ given in (21) with respect to $-i\frac{3\pi}{2}$, see Fig. 4.

Let us prove that f ₂ satisfies the third equality of (92). Since (91), $\frac{\partial}{\partial\mathbf{n}_{2} } |_{\theta=\phi} = -\frac{1}{\rho}\cdot\frac{\partial}{\partial \theta} |_{\theta=\phi}$, $\frac{\partial}{\partial\theta} = i\, \frac{\mathrm{d}}{\mathrm {d}\beta}$, and the fact that the integral $\int_{\mathcal{C}} e^{-\rho\omega\sinh\beta}\frac {\partial}{\partial\theta} H_{N}(\beta+i\theta) \,\mathrm{d} \beta$ converges uniformly with respect to θ (by (21), $\omega \in\mathbb{C}^{+}$ and (11)), it suffices to prove that $\int_{\mathcal{C}} e^{-\rho\omega\sinh\beta} \frac {\mathrm{d}\ }{\mathrm{d}\beta} H_{N}(\beta+i\phi) \,\mathrm{d} \beta= 0$. Integrating by parts and using that for any $\beta\in\mathcal{C}$, H _N (β+iϕ) is a bounded function and e ^{−ρωsinhβ}⟶0 when $\vert\operatorname{Re} \beta\vert \rightarrow+\infty$ we obtain $\int_{\mathcal{C}} e^{-\rho\omega\sinh\beta} \frac {\mathrm{d}}{\mathrm{d}\beta} H_{N}(\beta+i\phi) \,\mathrm{d} \beta = \rho\omega\int_{\mathcal{C}} e^{-\rho\omega\sinh\beta} \cosh \beta\, H_{N}(\beta+i\phi)\, \mathrm{d}\beta$. The last integral is equal to 0, because of the invariance of the integrand with respect to −β−3iπ, for any $\beta\in \mathcal{C}$ and by the symmetry of $\mathcal{C}$ with respect to $-i\frac{3\pi}{2}$. □

Appendix 2

Proof of Lemma 4.3

(i) The analytic continuation of $\widehat{g}(\omega_{1})$ to $\mathbb{C}^{+}$ and (37) follow from the Paley-Wiener type theorem for convex cones ([21, Theorem I.5.2]) since suppf⊂[0,∞) by (2). The estimate (38) follows from (2) since $\omega_{0}\in \mathbb{R}$.

(ii) $f'\in C_{0}^{\infty}(\mathbb{R})$ since supp(f′) is a compact set by (2) and $f\in C^{\infty}(\mathbb{R})$. Hence existence of the analytic continuation of $\widehat{g}_{1}(\omega_{1})$ to $\mathbb{C}$ and (40) follow from the Classic Paley-Wiener Theorem [32]. It is easy to check that $\widehat{g}_{1}(\omega)=(\omega-\omega _{0})\widehat{g}(\omega)$, for any $\omega\in\mathbb{C}^{+}$ by (34), (35) and the Analytic Continuation Principle. This implies the first identity in (41). Hence, the second identity in (41) follows from (37) and (40). The statement (42) follows from the first identity in (41) and (39). □

Appendix 3

Definition 13.1

For a function h(s), we denote the jump of h(s), at a point $s=s^{*}\in\mathbb{R}$ as $\mathfrak{J}(h,s^{*}) := \lim_{\varepsilon\rightarrow0+} h(s^{*} + i \varepsilon) - \lim_{\varepsilon\rightarrow0+} h(s^{*} - i \varepsilon)$, if the limits exist.

Lemma 13.2

Let $f\in\mathrm{C}_{0}(\mathbb{R})$ and for |ε|≤1 let $F(\varepsilon) := \frac{1}{2i\pi}\int_{-\infty }^{\infty} f(s) \coth(qs+i\varepsilon)\,\mathrm{d}s$. Then there exist the limits lim _ε→0+ F(ε), lim _ε→0− F(ε) and $\mathfrak{J}(F,0) = -\frac{1}{q}\ f(0)$.

Proof

It follows by the Sokhotsky-Plemelj theorem. □

Lemma 13.3

Let u _r , u _d be the functions given by (17) and (10), respectively. Then

$$\begin{aligned} \mathfrak{J} \biggl( \frac{\partial ^{(k)}u_r}{\partial\theta^k}, \theta_l \biggr) = - \mathfrak{J} \biggl( \frac{\partial^{(k)}u_d}{\partial \theta^k}, \theta_l \biggr), \quad k\in\mathbb{N}_0, \ l=1,2. \end{aligned}$$

(95)

Proof

We consider the case θ=θ ₁. The case θ=θ ₂ is analyzed similarly.

First we find $\mathfrak{J} (\frac{\partial ^{(k)}u_{r}}{\partial\theta^{k}}, \theta_{1} )$. From (17) it follows that

$$\begin{aligned} \mathfrak{J} \biggl(\frac{\partial ^{(k)}u_r}{\partial\theta^k}, \theta_1 \biggr) = -\frac{\partial^{(k)}}{\partial\theta^k} u_{r,1}(\rho,\theta_1,t), \quad k\in\mathbb{N}_0. \end{aligned}$$

(96)

Using polar coordinates y=(ρcosθ,ρsinθ) in u _r,1 and making the change of variable β=−i(θ−θ ₁) we obtain $u_{r,1}(\rho,\theta_{1},t) = e^{-i\omega_{0}(t-\rho\cos(\theta-\theta_{1}))} f(t-\rho\cos (\theta-\theta_{1})) = A(\beta,\rho,t)$, where

$$\begin{aligned} A(\beta,\rho,t) :=& e^{-i\omega_0(t-\rho\cosh\beta )} f(t-\rho\cosh\beta). \end{aligned}$$

(97)

Then $\frac{\partial^{(k)} }{\partial\theta^{k}} u_{r,1}(\rho,\theta,t) = (-i)^{k} \frac{\partial^{(k)} }{\partial\beta^{k}} A(\beta,\rho ,t)$, $k\in\mathbb{N}_{0}$. Hence by (96) we obtain

$$\begin{aligned} \mathfrak{J} \biggl(\frac{\partial ^{(k)}u_r}{\partial\theta^k}, \theta_1 \biggr) = -(-i)^k \frac{\partial^{(k)}}{\partial\beta^k} A(0,\rho,t),\quad k\in \mathbb{N}_0. \end{aligned}$$

(98)

Now we find $\mathfrak{J} (\frac{\partial^{(k)} u_{d}}{\partial\theta^{k}} ,\theta_{1} )$. From (18) and (11) it follows that Z _N (ρ,θ,t)=coth[q(β+iθ−iθ ₁)]+coth[q(β+iθ−iθ ₂)]−coth[q(β+iθ−2iπ−iα)]−coth[q(β+iθ−iα)]. Since function coth[q(β+iθ−iθ ₁)] is discontinuous in θ=θ ₁, then Z _N (ρ,θ,t) is also discontinuous in θ=θ ₁. Hence from (51) we have that

$$\begin{aligned} \mathfrak{J} \biggl(\frac{\partial^{(k)} u_d}{\partial\theta^k} ,\theta_1 \biggr) = \mathfrak{J} \biggl(\frac{\partial^{(k)} u_1}{\partial \theta^k} ,\theta_1 \biggr), \end{aligned}$$

(99)

where $u_{1}(\rho,\theta,t) = -\frac{q}{2i\pi} \int_{-\infty}^{+\infty} A(\beta ,\rho,t) \coth[ q(\beta+ i\theta- i\theta_{1} ) ] \,\mathrm{d}\beta$ and A(β,ρ) is given by (97). Using $\frac{\partial }{\partial\theta} \coth[ q(\beta+ i\theta- i\theta_{1} ) ] = i\frac{\partial }{\partial\beta} \coth[ q(\beta+ i\theta- i\theta_{1} ) ]$, integrating by parts k times (for k=0 we do not integrate $\frac {\partial^{(k)} u_{1}}{\partial\theta^{k}}$), and using that by (2) the integration is realized on a compact interval, we obtain $\frac{\partial^{(k)} }{\partial\theta^{k}} u_{1}(\rho,\theta,t) = (-1)^{k-1}\frac{i^{k} q}{2i\pi} \int_{-\infty }^{+\infty} \frac{\partial^{(k)} }{\partial\beta^{k}} A(\beta,\rho,t) \cdot \coth[ q(\beta+ i\theta- i\theta_{1}) ] \,\mathrm{d}\beta$. Applying Lemma 13.2 with $f(\beta)=(-1)^{k-1} i^{k} q \frac {\partial^{(k)} }{\partial\beta^{k}} A(\beta,\rho,t)$ we obtain

$$\begin{aligned} \mathfrak{J} \biggl(\frac{\partial^{(k)} u_1}{\partial \theta^k} , \theta_1 \biggr) = (-i)^k\, \frac{\partial^{(k)} }{\partial\theta^k} A(0,\rho,t). \end{aligned}$$

(100)

Therefore (95) follows from (99), (100) and (98). □

Appendix 4

Lemma 14.1

The function $[\frac{t}{t+\sqrt{t^{2} - \rho^{2}}} ]^{m}$ admits the following asymptotic behavior

$$\begin{aligned} \biggl[\frac{t}{t+\sqrt{t^2 - \rho^2}} \biggr]^m =& \biggl( \frac{1}{2} \biggr)^m + m \biggl(\frac{1}{2} \biggr)^{m-1} \frac{1}{8}\cdot\frac{\rho^2}{t^2} + O \biggl( \frac{1}{t^4} \biggr), \quad t\rightarrow\infty. \end{aligned}$$

(101)