Wave diffraction from the truncated hollow wedge: analytical regularization and Wiener–Hopf analysis

Abstract

The electromagnetic wave diffraction from perfectly conducting truncated wedges is considered on the rigorous level in cylindrical coordinates. An analytical regularization method is developed to obtain mathematically accurate problem solutions. The solution method is based on the unknown field representation through the principal value Kontorovich–Lebedev integral and the eigenfunctions series. We analyze the scattering from the semi-infinite truncated wedge, which consists of two non-parallel and non-intersecting perfectly conducting and infinitely thin half-planes, and develop this technique for analysis of more complicated problems of wave diffraction from the truncated wedge of finite length. The problems are reduced to the infinite systems of linear algebraic equations (ISLAE) of the first kind. The convolution type operators and their inverse ones are used to reduce them to the ISLAE of the second kind applied to the analytical regularization procedure. Two versions of the procedure, such as left- and right-sides regularization, are considered. The developed technique is compared with the Wiener–Hopf method. The numerical examples of wave scattering from the truncated wedge, including its well-known geometries as the semi-plane and the slit in the infinite plane, are analyzed.

Appendices

Appendix A

Let us consider the series representation (2) through the principal value integral (5) using the following definition

$$\begin{aligned} \dfrac{1}{2\pi \mathrm {i}} P\int \limits _{-i\infty }^{+\mathrm {i}\infty }\left\langle \cdot \right\rangle \mathrm {d}\nu \equiv \dfrac{1}{2\pi \mathrm {i}} {\mathop {\lim }\limits _{\begin{array}{c} N\rightarrow \infty \\ \upsilon _{1} \rightarrow 0 \end{array}}} \left[ \int \limits _{-\mathrm {i}N}^{-\mathrm {i}\upsilon _{1}}\left\langle \cdot \right\rangle d\nu + \int \limits _{\mathrm {i}\upsilon _{1} }^{\mathrm {i}N}\left\langle \cdot \right\rangle \mathrm {d}\nu \right] =\dfrac{1}{2\pi \mathrm {i}} {\mathop {\lim }\limits _{\begin{array}{c} N\rightarrow \infty \\ \upsilon _{1} \rightarrow 0 \end{array}}} \left[ \int \limits _{\Gamma _{\upsilon _{1} } }\left\langle \cdot \right\rangle \mathrm {d}\nu - \int \limits _{C_{\upsilon _{1}}}\left\langle \cdot \right\rangle \mathrm {d}\nu \right] \end{aligned}$$

(79)

with

$$\begin{aligned} \dfrac{1}{2\pi \mathrm {i}} {\mathop {\lim }\limits _{\begin{array}{c} N\rightarrow \infty \\ \upsilon _{1} \rightarrow 0 \end{array}}} \int \limits _{\Gamma _{\upsilon _{1}}}\left\langle \cdot \right\rangle \mathrm {d}\nu =-\sum \limits _{(\xi _{n} )}res\left\langle \cdot \right\rangle _{\begin{array}{c} \xi _{n}\\ \mathrm {Re}\xi _{n}>0) \end{array}}-\dfrac{1}{2\pi \mathrm {i}} {\mathop {\lim }\limits _{R_{N} \rightarrow \infty }} \int \limits _{C_{R_{N} } }\left\langle \cdot \right\rangle \mathrm {d}\nu , \end{aligned}$$

(80)

where $\left\langle \cdot \right\rangle$ indicates the integrand; the integration path $\Gamma _{\upsilon _{1}}$ runs along the imaginary axis $(-\mathrm {i}N,-\mathrm {i}\upsilon _{1})\bigcup$ $(\mathrm {i}\upsilon _{1},\mathrm {i}N)$ and bypasses the simple pole at $\nu =0$ embracing this point from the right side by the semi-circle $C_{\upsilon _{1}}$ of the infinitely small radius $\upsilon _{1}$ (see Fig. 9). The singularities of this integrand take the form of simple poles at $\nu =n$, $n=1,2,3,\ldots$ that lie at the right-hand semi-plane of the complex plane $\nu$; $C_{R_{N}}$ is the semi-circle in right semi-plane with radius $R_{N}$.

Let us use the asymptotic estimations of the Macdonald and modified Bessel functions for the large indexes and find that on the imaginary axis $\mathrm {arg}(\nu )=\pm \pi /2$, and for the real wave number parameter $s=-\mathrm {i}k$, $\mathrm {arg}(s)=0$ the product $K_{\nu }(sr_{>})I_{\nu }(sr_{<})\sim |\nu |^{-1}$, if $|\nu |\rightarrow \infty$. This product decays exponentially on the contour $C_{R_{N}}$ for $-\pi /2<\mathrm {arg}(\nu )<\pi /2$, if $R_{N}\rightarrow \infty$.

Fig. 9

Contour for integration in the expression (79)

Considering this, we find that the integral (5) becomes absolutely convergent and, if the parameter s is real, can be evaluated applying the residues theorem. The equivalence of the representations (2) and (5) directly follows from this evaluation.

Appendix B

Here, we show the schemes briefly for the derivation of the series Eq. (24). Let us substitute the expression (20) into Eq. (18). Then using Theorem 1, we represent this equation in the form which keeps the integrals as

$$\begin{aligned} U_{n}^{(l)} (sr)=\dfrac{1}{\pi \mathrm {i}} P\int \limits _{-\mathrm {i}\infty }^{\mathrm {i}\infty }\nu M_{l} (\nu )\int \limits _{c}^{\infty }K_{z_{n}^{(l)} } (sr')\left\{ \begin{array}{l} {{\mathop {I_{\nu } (sr')K_{\nu } (sr)}\limits _{r\ge r'}} } \\ {{\mathop {K_{\nu } (sr')I_{\mathrm{\nu }} (sr)}\limits _{r\le r'}} } \end{array}\right\} \dfrac{\mathrm {d}r'}{r'}\mathrm {d}\nu . \end{aligned}$$

(81)

Here, $r\in (c,\infty )$, $n=1,2,3, \ldots$. Let us evaluate the internal integral (81) using the formula:

$$\begin{aligned} \int \limits _{c}^{\infty }K_{z_{n}^{(l)}} (sr')\left\{ \begin{array}{l} {{\mathop {I_{\nu } (sr')K_{\nu } (sr)}\limits _{r\ge r'}} } \\ {{\mathop {K_{\nu } (sr')I_{\nu } (sr)}\limits _{r\le r'}} } \end{array}\right\} \dfrac{\mathrm {d}r'}{r'} = \dfrac{K_{z_{n}^{(l)}} (sr)}{\nu ^{2}-(z_{n}^{(l)} )^{2}}-K_{\nu } (sr)\dfrac{scW[K_{z_{n}^{(l)} } I_{\nu }]_{sc}}{\nu ^{2} -(z_{n}^{(l)})^{2}}. \end{aligned}$$

(82)

Then, according to the definition (79) and representation (81), the first term in (82) forms a couple of integrals for each $n=1,2,\, 3,\ldots$ and $l=1,2$ as

$$\begin{aligned} J_{n}^{(l)} =\dfrac{1}{\pi \mathrm {i}} {\mathop {\lim }\limits _{ \begin{array}{c} N\rightarrow \infty \\ \upsilon _{1} \rightarrow 0 \end{array}}} \int \limits _{\Gamma _{\upsilon _{1} } }\dfrac{\nu M_{l} (\nu )}{\nu ^{2} -(z_{n}^{(l)} )^{2} } \mathrm {d}\nu -\dfrac{\delta _{l}^{1} }{\pi \mathrm {i}} {\mathop {\lim }\limits _{\upsilon _{1} \rightarrow 0}}\int \limits _{C_{\upsilon _{1} } }\dfrac{\nu M_{l} (\nu )}{\nu ^{2}-(z_{n}^{(l)})^{2} }\mathrm {d}\nu . \end{aligned}$$

(83)

Let us formulate the further result as Theorem B: The integral $J_{n}^{(l)}$ is equal to zero for any $z_{n}^{(l)}$.

Proving: Taking into account that $M_{l} (\nu )=M_{l} (-\nu )$, let us rewrite the first integral (83) as

$$\begin{aligned} {\mathop {J}\limits ^{\frown }}_{n}^{(l)} =\dfrac{1}{\pi \mathrm {i}} {\mathop {\lim }\limits _{\begin{array}{c} N\rightarrow \infty \\ \upsilon _{1} \rightarrow 0 \end{array}}} \int \limits _{\Gamma _{\upsilon _{1} } }\dfrac{\nu M_{l} (\nu )}{\nu ^{2} -(z_{n}^{(l)} )^{2} } d\nu =\dfrac{1}{2\pi \mathrm {i}} {\mathop {\lim }\limits _{\begin{array}{c} N\rightarrow \infty \\ \upsilon _{1} \rightarrow 0 \end{array}}} \left[ \int \limits _{\uparrow \Gamma _{\upsilon _{1} } + \delta }\dfrac{M_{l} (\nu )}{\nu -z_{n}^{(l)} } \mathrm {d}\nu +\int \limits _{\downarrow \Gamma _{\upsilon _{1}} -\delta }\dfrac{M_{l} (\nu )}{\nu - z_{n}^{(l)} }\mathrm {d}\nu \right] . \end{aligned}$$

(84)

Here, $\uparrow \Gamma _{\upsilon _{1} } +\delta$ is the integration path formed from the contour $\Gamma _{\upsilon _{1} }$ that is determined in “Appendix A” by shifting it to the right of the imaginary axes by the value $0<\delta <1$ and $\downarrow \Gamma _{\upsilon _{1} } -\delta$ is the integration path formed from the contour $\Gamma _{\upsilon _{1} }$ by shifting it to the left by the value $0<\delta <1$ and from the $\uparrow \Gamma _{\upsilon _{1} } +\delta$ by changing $\nu \in (\uparrow \Gamma _{\upsilon _{1} } +\delta )$ $\rightarrow -\nu$; these contours are oppositely oriented, the arrows $\uparrow ,\downarrow$ show the directions of integration. Considering that the integrands decay as $O(\nu ^{-2} )$, if $|\nu |\rightarrow \infty$, we reduce ${\mathop {J}\limits ^{\frown }}_{n}^{(l)}$ to the integration along the closed contour ${\mathop {C}\limits ^{\frown }}_{\upsilon _{1} }$ (see Fig. 10), which encompasses the second-order pole of the function $M_{l} (\nu )$ at $\nu =0$, if $l=1$ and the closed contour ${\mathop {C}\limits ^{\frown }}_{\upsilon _{1} }$ encompasses the regular function, if $l=2$. Therefore, applying the residues theorem, we find that

$$\begin{aligned} {\mathop {J}\limits ^{\frown }}_{n}^{(l)} =\dfrac{1}{\pi \mathrm {i}} {\mathop {\lim }\limits _{\begin{array}{c} N\rightarrow \infty \\ \upsilon _{1} \rightarrow 0 \end{array}}} \int \limits _{\Gamma _{\upsilon _{1}}}\dfrac{\nu M_{l} (\nu )}{\nu ^{2} -(z_{n}^{(l)} )^{2} } \mathrm {d}\nu =\dfrac{1}{2\pi \mathrm {i}} {\mathop {\lim }\limits _{\begin{array}{c} N\rightarrow \infty \\ \upsilon _{1} \rightarrow 0 \end{array}}} \int \limits _{{\mathop {C}\limits ^{\frown }}_{\upsilon _{1} } }\dfrac{M_{l} (\nu )}{\nu -z_{n}^{(l)} } \mathrm {d}\nu ={\left\{ \begin{array}{ll} -1/[\pi (z_{n}^{(l)} )^{2}],&{} \text{ if } l=1, \\ 0, &{} \text{ if } l=2. \end{array}\right. } \end{aligned}$$

(85)

Then taking into account the fact that the integral along the semi-circle $C_{\upsilon _{1} }$ in (83) accepts the same value, we find that $J_{n}^{(l)}$ is equal to zero for any $z_{n}^{(l)}$. $\square$

We evaluate the integral formed from (81) by the second term in the right-hand side of the expression (82) using the residue theorem. Considering the definition (79), (80) we evaluate the integral along the contour $\Gamma _{\upsilon _{1} }$embracing the integrand singularity in the right semi-plane by the semi-circle $C_{R_{N}}$. Similarly, we transform the known part of Eq. (18) and arrive at the series equation (24a). The integration along the semi-circle $C_{\upsilon _{1}}$ takes into account the singularity at $\nu =0$ and leads to the equation (24b).

Fig. 10

Integration contour for evaluation of the integral $J_{n}^{(l)}$

Appendix C

Let us consider the series

$$\begin{aligned} \Upsilon _{qn}^{(l)}=\sum \limits _{p=1}^{\infty }\dfrac{\tau _{qp}^{(l)}}{\xi _{p}^{(l)}+z_{n}^{(l)}}=\dfrac{1}{\left[ M_{l}^{-}(z_{q}^{(l)})\right] '}\sum \limits _{p=1}^{\infty }\dfrac{1}{\left\{ [M_{l}^{-}{(\xi _{p}^{(l)})]^{-1}}\right\} '\, (z_{q}^{(l)}-\xi _{p}^{(l)})(\xi _{p}^{(l)}+z_{n}^{(l)})}. \end{aligned}$$

(86)

In order to obtain the summation formula for the series (51), let us introduce the integral as

$$\begin{aligned} J_{qn}^{(l)}=\dfrac{1}{2\pi \mathrm {i}{[M_{l}^{-}(z_{q}^{(l)})]' }}\int \limits _{C_{R}}{\dfrac{M_{l}^{-}(t)\mathrm {d}t}{(z_{q}^{(l)}-t)(t+z_{n}^{(l)})}}. \end{aligned}$$

(87)

Here, $C_{R}$ is the circular integration path in the complex plane t, the points $t=0$ and R are the center and the radius of the circle respectively; $C_{R}$ outline encompasses the simple poles of the integrand at $t=\xi _{q}^{(l)}$ ($p=1,2,3,\ldots$) and $t=-z_{n}^{(l)}$. For $|t|\rightarrow \infty$ the integrand as a function of t tends to zero not slower than $t^{-5/2}$, therefore, $J_{qn}^{(l)}\rightarrow 0$, if $R\rightarrow \infty$. Then, applying the residues theorem, we arrive at the equality as

$$\begin{aligned} \dfrac{1}{[M_{l}^{-}(z_{q}^{(l)})]'}\sum \limits _{p=1}^{\infty }\dfrac{1}{\left\{ [M_{l}^{-}(\xi _{p}^{(l)})]^{-1}\right\} '(z_{q}^{(l)}-\xi _{p}^{(l)})(\xi _{p}^{(l)}+z_{n}^{(l)})}=-\dfrac{M_{l}^{+}(z_{n}^{(l)})}{[M_{l}^{-}(z_{q}^{(l)})]'(z_{q}^{(l)}+z_{n}^{(l)})}. \end{aligned}$$

(88)

This formula proves representation (51).