**3. Numerical solution**

We use the Nyström method based on weighted trigonometric interpolation quadratures as the numerical method used to approximate the solution to the integral equations (24), (26) and (31) with a mesh of 2*n* points. We parameterize the boundary *∂D* as

$$\mathbf{x}(t) = \left(\mathbf{x}\_1(t), \mathbf{x}\_2(t)\right), \qquad t \in \left[0, 2\pi\right]. \tag{36}$$

So for x, y ∈ *∂D*, we let

$$x = x(t) = \left(\mathbf{x}\_1(t), \mathbf{x}\_2(t)\right), \qquad t \in \left[0, 2\pi\right],\tag{37}$$

$$y = x(\tau) = (\mathbf{x}\_1(\tau), \mathbf{x}\_2(\tau)), \qquad \tau \in [0, 2\pi] \,. \tag{38}$$

The outward pointing unit normal at x(*τ*) is

$$m(x(\tau)) = \frac{\left(\mathbf{x}\_2'\left(\tau\right), -\mathbf{x}\_1'\left(\tau\right)\right)}{J(\tau)},\tag{39}$$

where *J*(*τ*) is the Jacobian factor

$$J(\tau) = \sqrt{\left(\mathbf{x}\_1' \left(\tau\right)\right)^2 + \left(\mathbf{x}\_2' \left(\tau\right)\right)^2}.\tag{40}$$

The operators (25), (27) and (32) may then be expressed as

$$(\mathcal{K}\phi)(x(t)) = \int\_0^{2\pi} \mathcal{K}\_0(t,\tau)\phi(\tau)d\tau,\tag{41}$$

$$(\mathcal{S}\phi)(x(t)) = \int\_0^{2\pi} \mathcal{S}\_0(t,\tau)\phi(\tau)d\tau,\tag{42}$$

$$(\mathcal{K}'\phi)(x(t)) = \int\_0^{2\pi} K\_0'(t,\tau)\phi(\tau)d\tau,\tag{43}$$

where *φ* (*τ*) = *φ* (*x* (*τ*)), and the associated kernels

This solution is unique provided that *k* is not an interior Dirichlet eigenvalue [3]. Uniqueness is guaranteed by considering a suitable combination of single- and double-layer potentials,

where *η* = 0 such that *η* Re *k* ≥ 0, solves the exterior impedance problem uniquely provided

We use the Nyström method based on weighted trigonometric interpolation quadratures as the numerical method used to approximate the solution to the integral equations (24), (26)

K + *iη*T + *iηλ*K + *λ*S

x(*t*) = (*x*1(*t*), *x*2(*t*)), *t* ∈ [0, 2*π*] . (36)

x = x(*t*) = (*x*1(*t*), *x*2(*t*)), *t* ∈ [0, 2*π*] , (37) y = x(*τ*) = (*x*1(*τ*), *x*2(*τ*)), *τ* ∈ [0, 2*π*] . (38)

> <sup>1</sup> (*τ*)

*<sup>J</sup>*(*τ*) , (39)

*K*0(*t*, *τ*)*φ*(*τ*)*dτ*, (41)

*S*0(*t*, *τ*)*φ*(*τ*)*dτ*, (42)

<sup>0</sup>(*t*, *τ*)*φ*(*τ*)*dτ*, (43)

. (40)

*<sup>φ</sup>*(y)*ds*(y), <sup>x</sup> <sup>∈</sup> **<sup>R</sup>**2\*D*¯ , (34)

*φ* = −2*m*. (35)

ie the combined potential

8 Advanced Electromagnetic Waves

**3. Numerical solution**

So for x, y ∈ *∂D*, we let

*usc*(x) =

*∂G*(x, y)

that the density *φ*(x) ∈ *∂D* is a solution of the integral equation [3]

and (31) with a mesh of 2*n* points. We parameterize the boundary *∂D* as

n(x(*τ*)) =

 *x* <sup>1</sup> (*τ*)

(K*φ*)(x(*t*)) =

(S*φ*)(x(*t*)) =

*φ*)(x(*t*)) =

(K

*J*(*τ*) =

The operators (25), (27) and (32) may then be expressed as

 *x*

<sup>2</sup> (*τ*), −*x*

<sup>2</sup> + *x* <sup>2</sup> (*τ*) 2

 2*π*

0

 2*π*

0

 2*π*

0 *K*

(*<sup>I</sup>* <sup>−</sup> *<sup>i</sup>ηλ*) *<sup>φ</sup>* <sup>−</sup>

*<sup>∂</sup>*n(y) <sup>−</sup> *<sup>i</sup>ηG*(x, <sup>y</sup>)

*∂D*

The outward pointing unit normal at x(*τ*) is

where *J*(*τ*) is the Jacobian factor

$$S\_0(t,\tau) = \mathcal{Z}G(x(t), x(\tau))f(\tau),\tag{44}$$

$$K\_0(t, \tau) = 2 \frac{\partial G(x(t), x(\tau))}{\partial n\_\perp(\tau)} J(\tau),\tag{45}$$

$$K\_0'(t,\tau) = 2\frac{\partial G(x(t), x(\tau))}{\partial n\,(t)} f(\tau),\tag{46}$$

each have a logarithmic singularity at *t* = *τ*. Thus we transform the integral operator formulation (25) of the exterior Dirichlet problem into the parametric form

$$\phi(t) + \int\_0^{2\pi} \left\{ \mathcal{K}\_0(t, \tau) - i\eta \mathcal{S}\_0(t, \tau) \right\} \phi(\tau) d\tau = g(t), \quad 0 \le t \le 2\pi,\tag{47}$$

the integral operator formulation (27) of the exterior Neumann problem into the parametric form

$$-\left.\phi(t) + \int\_0^{2\pi} \mathcal{K}\_0'(t, \tau)\phi(\tau)d\tau = h(t), \quad 0 \le t \le 2\pi,\tag{48}$$

and the integral operator formulation (32) of the exterior impedance problem into the parametric form

$$-\phi(t) + \int\_0^{2\pi} \left\{ K\_0'(t, \tau) + ik\lambda S\_0(t, \tau) \right\} \phi(\tau) d\tau = m(t), \quad 0 \le t \le 2\pi. \tag{49}$$

A method developed by Martensen and Kussmaul [4] for the logarithmic singularities arising in (41), (42) and (43) was employed. The singular parts of the kernels (44), (45) and (46) are isolated in the following manner so that

$$K\_0(t, \tau) = K\_1(t, \tau) \ln\left(4\sin^2\frac{t-\tau}{2}\right) + K\_2(t, \tau),\tag{50}$$

$$S\_0(t, \tau) = S\_1(t, \tau) \ln\left(4\sin^2\frac{t-\tau}{2}\right) + S\_2(t, \tau),\tag{51}$$

$$K\_0'(t, \tau) = K\_1'(t, \tau) \ln\left(4\sin^2\frac{t-\tau}{2}\right) + K\_2'(t, \tau),\tag{52}$$

where *K*1, *K*2, *S*1, *S*2, *K* <sup>1</sup>, *K* <sup>2</sup> are analytic. The smooth components of the kernel *K*0(*t*, *τ*) are evaluated using the trapezoidal rule to approximate

$$\int\_0^{2\pi} K\_2(t,\tau)\phi(\tau)d\tau \approx \frac{\pi}{n} \sum\_{j=0}^{2n-1} K\_2(t,t\_j)\phi(\tau\_j)dt.\tag{53}$$

An identical rule was applied for

$$\int\_{0}^{2\pi} \mathbf{S}\_2(t, \tau) \phi(\tau) d\tau,\tag{54}$$

and

$$\int\_0^{2\pi} \mathbf{K}\_2'(t, \tau) \phi(\tau) d\tau. \tag{55}$$

A different quadrature rule is used to estimate the singular part of the kernel *K*0(*t*, *τ*) which replaces the integrand by its trigonometric interpolation polynomial and integrates this interpolant exactly. We apply the following quadrature rule

$$\int\_0^{2\pi} \ln\left(4\sin^2\frac{t-\tau}{2}\right) K\_1(t,\tau)\phi\left(\tau\right)d\tau \approx \sum\_{j=0}^{2n-1} \mathbb{R}\_j^{(n)}(t)\mathcal{K}\_1(t,t\_j)\phi\left(t\_j\right), \qquad \text{for } 0 \le t \le 2\pi,\tag{56}$$

to approximate the integral of the logarithmic part of the kernel *K*0(*t*, *τ*). The quadrature weights *R*(*n*) *<sup>j</sup>* are given by

$$R\_j^{(n)}(t) = -\frac{2\pi}{n} \sum\_{m=1}^{n-1} \frac{1}{m} \cos m(t - t\_j) - \frac{\pi}{n^2} \cos n(t - t\_j), \qquad \text{for } j = 0, \ldots, 2n - 1. \tag{57}$$

An identical rule was applied for

$$\int\_0^{2\pi} \ln\left(4\sin^2\frac{t-\tau}{2}\right) S\_1(t,\tau) \phi\left(\tau\right) d\tau,\tag{58}$$

and

$$\int\_0^{2\pi} \ln\left(4\sin^2\frac{t-\tau}{2}\right) K\_1'(t,\tau)\phi\left(\tau\right)d\tau. \tag{59}$$

Three different spacings of the 2*n* mesh points were used. For smooth scatterers we used a mesh of 2*<sup>n</sup>* uniformly spaced points *tj* <sup>=</sup> *<sup>π</sup><sup>j</sup> <sup>n</sup>* , for *j* = 0, 1, ..., 2*n* − 1, in the parameterisation (36). However, for domains with corners, the solutions to (25), (27) and (32) have singularities in the derivatives in the corners. To deal with these singularities, the uniform mesh is replaced by a non-uniform graded mesh. This is achieved by substituting a new variable such that the derivatives of the transformed integrand vanish up to a certain order at the corners [4]. The previous quadrature rules (Martensen-Kussmaul and trapezoidal) are then modified as follows. For any function *f*(*t*), its definite integral over [0, 2*π*] is evaluated by the trapezoidal quadrature rule after the substitution *t* = *w*(*s*) by an appropriately chosen function *w*(*s*):

$$\int\_0^{2\pi} f\left(t\right) dt = \int\_0^{2\pi} f\left(w\left(s\right)\right) w'\left(s\right) ds \approx \frac{\pi}{n} \sum\_{j=1}^{2n-1} a\_j f\left(s\_j\right),\tag{60}$$

with weights *aj* = *w tj* and mesh points *sj* = *w tj* .

For a domain with a single corner, the scatterer boundary *∂D* is defined as having one corner at the point <sup>x</sup><sup>0</sup> and *<sup>∂</sup>D*\ {x0} is assumed to be *<sup>C</sup>*<sup>2</sup> and piecewise analytic. The angle *<sup>γ</sup>* at the corner is assumed to be 0 < *γ* < 2*π*. A suitable choice of the function *w*(*s*) is [4]

$$w\left(s\right) = 2\pi \frac{[v\left(s\right)]^p}{[v\left(s\right)]^p + [v\left(2\pi - s\right)]^p}, \qquad 0 \le s \le 2\pi,\tag{61}$$

where

The smooth components of the kernel *K*0(*t*, *τ*) are evaluated using the trapezoidal rule to

*n*

A different quadrature rule is used to estimate the singular part of the kernel *K*0(*t*, *τ*) which replaces the integrand by its trigonometric interpolation polynomial and integrates

*R*(*n*)

to approximate the integral of the logarithmic part of the kernel *K*0(*t*, *τ*). The quadrature

 *K*

*<sup>j</sup>* (*t*)*K*1(*t*, *tj*)*φ*

 *tj* 

2*n*−1 ∑ *j*=0

2*n*−1 ∑ *j*=0

*K*2(*t*, *tj*)*φ*(*τj*)*dt*. (53)

*S*2(*t*, *τ*)*φ*(*τ*)*dτ*, (54)

<sup>2</sup>(*t*, *τ*)*φ*(*τ*)*dτ*. (55)

*<sup>n</sup>*<sup>2</sup> cos *<sup>n</sup>*(*<sup>t</sup>* <sup>−</sup> *tj*), for *<sup>j</sup>* <sup>=</sup> 0, ..., 2*<sup>n</sup>* <sup>−</sup> 1. (57)

*S*1(*t*, *τ*)*φ* (*τ*) *dτ*, (58)

<sup>1</sup>(*t*, *τ*)*φ* (*τ*) *dτ*. (59)

, for 0 ≤ *t* ≤ 2*π*, (56)

*<sup>K</sup>*2(*t*, *<sup>τ</sup>*)*φ*(*τ*)*d<sup>τ</sup>* <sup>≈</sup> *<sup>π</sup>*

 2*π*

0

 2*π*

0 *K*

cos *<sup>m</sup>*(*<sup>t</sup>* <sup>−</sup> *tj*) <sup>−</sup> *<sup>π</sup>*

4 sin<sup>2</sup> *<sup>t</sup>* <sup>−</sup> *<sup>τ</sup>* 2

4 sin2 *<sup>t</sup>* <sup>−</sup> *<sup>τ</sup>* 2

this interpolant exactly. We apply the following quadrature rule

*K*1(*t*, *τ*)*φ* (*τ*) *dτ* ≈

approximate

10 Advanced Electromagnetic Waves

and

 2*π*

4 sin<sup>2</sup> *<sup>t</sup>* <sup>−</sup> *<sup>τ</sup>* 2

*<sup>j</sup>* are given by

*<sup>j</sup>* (*t*) = <sup>−</sup>2*<sup>π</sup>*

An identical rule was applied for

*n*

*n*−1 ∑ *m*=1

1 *m*

> 2*π*

0 ln 

 2*π*

0 ln 

0 ln 

and

weights *R*(*n*)

*R*(*n*)

 2*π*

0

An identical rule was applied for

$$w(s) = \left(\frac{1}{p} - \frac{1}{2}\right) \left(\frac{\pi - s}{\pi}\right)^3 + \frac{1}{p} \frac{s - \pi}{\pi} + \frac{1}{2},\tag{62}$$

and the integer *p* is chosen to be at least 2. The function *w* (*s*) is strictly monotonically increasing and the derivatives at the end points *s* = 0 and 2*π* vanish up to order *p*. This choice of substitution ensures that approximately half of the quadrature points are uniformly distributed around the surface of the scatterer and that the other half are concentrated at the corner end points *s* = 0 and 2*π*. In this study we set *p* = 8. Use of this particular function *w*(*s*) (61) requires that the parameterisation of the surface (36) is such that the corner x<sup>0</sup> occurs at *t* = 0.

The required substitution is applied to the discretization of (41) by setting *t* = *w* (*s*) and *τ* = *w* (*σ*) to obtain

$$\int\_0^{2\pi} \mathbf{K}\_0(t,\tau)\phi(\tau)d\tau = \int\_0^{2\pi} \mathbf{K}\_0(w\left(s\right),w\left(\sigma\right))\phi\left(w\left(\sigma\right)\right)w'\left(\sigma\right)d\sigma,\tag{63}$$

and decomposing

$$K\_0\left(w\left(s\right), w\left(\sigma\right)\right) = K\_1(s, \sigma) \ln\left(4\sin^2\frac{s-\sigma}{2}\right) + K\_2(s, \sigma),\tag{64}$$

where

$$K\_1(\mathbf{s}, \sigma) = K\_1(w\left(\mathbf{s}\right), w\left(\sigma\right)), \tag{65}$$

and

$$K\_2(s, \sigma) = K\_0\left(w\left(s\right), w\left(\sigma\right)\right) - K\_1(s, \sigma) \ln\left(4\sin^2\frac{s-\sigma}{2}\right), \qquad s \neq \sigma. \tag{66}$$

The kernels *K*1(*s*, *σ*) and *K*2(*s*, *σ*) are analytic. The operator is now discretized using the points *sj* = *w tj* and weights *aj* = *w* (*tj*) . Fuller details are in [4]. The same discretization procedure is applied to discretize (42) and (43).

For a domain with two corners, the scatterer boundary *∂D* is defined as having a corner at the point <sup>x</sup><sup>0</sup> and a second at the point <sup>x</sup>*<sup>π</sup>* and *<sup>∂</sup>D*\ {x<sup>0</sup> <sup>∪</sup> <sup>x</sup>*π*} is assumed to be *<sup>C</sup>*<sup>2</sup> and piecewise analytic. The angle *γ* at the corners is assumed to satisfy 0 < *γ* < 2*π*. Our choice of the function *w*(*s*) is

$$w(s) = s - \frac{3}{4}\sin 2s + \frac{3}{20}\sin 4s - \frac{1}{60}\sin 6s, \qquad 0 \le s \le 2\pi. \tag{67}$$

The function *w* (*s*) is strictly monotonically increasing between the corners and the derivatives at the corner points *s* = 0, *π* and 2*π* vanish. This choice (67) of substitution ensures that approximately half of the quadrature points are uniformly distributed around the surface of the scatterer between the two corners and that the other half is concentrated at the corner end points *s* = 0, *π* and 2*π*. Use of this particular function *w*(*s*) (67) requires that the parameterisation of the surface (36) is such that the corner x<sup>0</sup> occurs at *t* = 0 and that the corner x*<sup>π</sup>* occurs at *t* = *π*.

With any of the above quadrature rules evaluated at the 2*n* points *tj* we have obtained a system of 2*n* linear equations for the boundary values *φ tj* for *j* = 0, 1, ..., 2*n* − 1 that is a discretization of the integral equations (25), (27) and (32). The solutions are obtained by the usual Gaussian elimination procedure.

Implementation of the graded mesh ensures an exponentially fast convergence rate (as a function of *n*) for scatterers with one or two corners with the Neumann and impedance boundary conditions. In the case where these scatterers have a Dirichlet boundary condition further modifications are necessary to achieve comparable convergence rates. For these domains the kernel of (24) is no longer weakly singular at the corner.

The modification for domains with a single corner at x<sup>0</sup> and the Dirichlet boundary condition [4], uses the fundamental solution

$$G\_0\left(x, y\right) = \frac{1}{2\pi} \ln \frac{1}{|x - y|}, \qquad x \neq y\_\prime \tag{68}$$

to the Laplace equation in **R**<sup>2</sup> to subtract a vanishing term. This transforms (24) into

$$u^{\mathcal{K}}(x) = \int\_{\partial D} \left\{ \left\{ \frac{\partial G(x, y)}{\partial n(y)} - i\eta G(x, y) \right\} \phi(y) - \frac{\partial G\_0(x, y)}{\partial n(y)} \phi(x\_0) \right\} ds(y), \qquad x \in \mathbb{R}^2 \backslash \bar{D} \tag{69}$$

and the associated boundary equation (25) is reformulated as

$$\begin{split} \left\{ \phi(\boldsymbol{x}) - \phi(\boldsymbol{x}\_{0}) + 2 \int\_{\partial D} \left\{ \frac{\partial G(\boldsymbol{x}, \boldsymbol{y})}{\partial n(\boldsymbol{y})} - i\eta G(\boldsymbol{x}, \boldsymbol{y}) \right\} \phi(\boldsymbol{y}) d\boldsymbol{s}(\boldsymbol{y}) \\ - 2 \int\_{\partial D} \frac{\partial G\_{0}(\boldsymbol{x}, \boldsymbol{y})}{\partial n(\boldsymbol{y})} \phi(\boldsymbol{x}\_{0}) d\boldsymbol{s}(\boldsymbol{y}) = -2\boldsymbol{u}^{\text{inc}}(\boldsymbol{x}), \qquad \boldsymbol{x} \in \partial D. \end{split} \tag{70}$$

An analysis showing the existence of a solution to (70) is provided in [4]. The integral equation (70) is rewritten in parameterised form

*φ* (*t*) − *φ* (0) − 2*π* 0 *K*ˆ(*t*, *τ*)*φ* (*τ*) *dτ* − 2*π* 0 *H*(*t*, *τ*)*φ* (0) *dτ* = *g* (*t*), 0 ≤ *t* ≤ 2*π*, (71)

where

and decomposing

12 Advanced Electromagnetic Waves

where

and

points *sj* = *w*

 *tj* 

of the function *w*(*s*) is

the corner x*<sup>π</sup>* occurs at *t* = *π*.

usual Gaussian elimination procedure.

*K*<sup>0</sup> (*w* (*s*), *w* (*σ*)) = *K*1(*s*, *σ*)ln

*K*2(*s*, *σ*) = *K*<sup>0</sup> (*w* (*s*), *w* (*σ*)) − *K*1(*s*, *σ*)ln

<sup>4</sup> sin 2*<sup>s</sup>* <sup>+</sup>

system of 2*n* linear equations for the boundary values *φ*

domains the kernel of (24) is no longer weakly singular at the corner.

3

and weights *aj* = *w*

procedure is applied to discretize (42) and (43).

*<sup>w</sup>* (*s*) <sup>=</sup> *<sup>s</sup>* <sup>−</sup> <sup>3</sup>

4 sin<sup>2</sup> *<sup>s</sup>* <sup>−</sup> *<sup>σ</sup>* 2

The kernels *K*1(*s*, *σ*) and *K*2(*s*, *σ*) are analytic. The operator is now discretized using the

For a domain with two corners, the scatterer boundary *∂D* is defined as having a corner at the point <sup>x</sup><sup>0</sup> and a second at the point <sup>x</sup>*<sup>π</sup>* and *<sup>∂</sup>D*\ {x<sup>0</sup> <sup>∪</sup> <sup>x</sup>*π*} is assumed to be *<sup>C</sup>*<sup>2</sup> and piecewise analytic. The angle *γ* at the corners is assumed to satisfy 0 < *γ* < 2*π*. Our choice

<sup>20</sup> sin 4*<sup>s</sup>* <sup>−</sup> <sup>1</sup>

The function *w* (*s*) is strictly monotonically increasing between the corners and the derivatives at the corner points *s* = 0, *π* and 2*π* vanish. This choice (67) of substitution ensures that approximately half of the quadrature points are uniformly distributed around the surface of the scatterer between the two corners and that the other half is concentrated at the corner end points *s* = 0, *π* and 2*π*. Use of this particular function *w*(*s*) (67) requires that the parameterisation of the surface (36) is such that the corner x<sup>0</sup> occurs at *t* = 0 and that

With any of the above quadrature rules evaluated at the 2*n* points *tj* we have obtained a

discretization of the integral equations (25), (27) and (32). The solutions are obtained by the

Implementation of the graded mesh ensures an exponentially fast convergence rate (as a function of *n*) for scatterers with one or two corners with the Neumann and impedance boundary conditions. In the case where these scatterers have a Dirichlet boundary condition further modifications are necessary to achieve comparable convergence rates. For these

 *tj* 

*K*1(*s*, *σ*) = *K*1(*w* (*s*), *w* (*σ*)), (65)

4 sin2 *<sup>s</sup>* <sup>−</sup> *<sup>σ</sup>* 2

(*tj*) . Fuller details are in [4]. The same discretization

<sup>60</sup> sin 6*s*, 0 <sup>≤</sup> *<sup>s</sup>* <sup>≤</sup> <sup>2</sup>*π*. (67)

for *j* = 0, 1, ..., 2*n* − 1 that is a

+ *K*2(*s*, *σ*), (64)

, *s* = *σ*. (66)

$$H(t) = \begin{cases} \frac{1}{\pi} \frac{\mathbf{x}\_2'(\tau)[\mathbf{x}\_1(t) - \mathbf{x}\_1(\tau)] - \mathbf{x}\_1'(\tau)[\mathbf{x}\_2(t) - \mathbf{x}\_2(\tau)]}{|\mathbf{x}(t) - \mathbf{x}(\tau)|^2}, & t \neq \tau, \\\\ \frac{1}{2\pi} \frac{\mathbf{x}\_2'(t)\mathbf{x}\_1''(t) - \mathbf{x}\_1'(t)\mathbf{x}\_2''(t)}{|\mathbf{x}'(t)|^2}, & t = \tau, t \neq 0, 2\pi, \end{cases} \tag{72}$$

and

$$
\hat{K}(t,\tau) = K(t,\tau) - i\eta S(t,\tau), \qquad 0 \le t \le 2\pi. \tag{73}
$$

We now apply the substitution (60) to (71) and obtain

$$\begin{split} \int\_{0}^{2\pi} \hat{K}(t,\tau)\phi\left(\tau\right)d\tau - \int\_{0}^{2\pi} H(t,\tau)\phi\left(0\right)d\tau \\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \frac{2\pi}{\int\_{0}^{2\pi}} \hat{K}(w\left(s\right),w\left(\sigma\right))w'\left(\sigma\right)\phi\left(w\left(\sigma\right)\right)d\sigma \\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \frac{2\pi}{\int\_{0}^{2\pi}} \hat{H}(w\left(s\right),w\left(\sigma\right))w'\left(\sigma\right)\phi\left(0\right)d\sigma. \tag{74} \end{split} \tag{75}$$

The logarithmic singularity present in the kernel *K*ˆ(*t*, *τ*) remains to be accounted for. This is done in the same manner as (64). Using the quadrature rules (53) and (56) to discretize the kernel, and the trapezoidal rule to discretize the kernel *H*(*t*, *τ*) and *φ*<sup>0</sup> = *φ* (0) at the corner *s*<sup>0</sup> = 0 gives

$$\begin{aligned} \phi\_i &= \phi\_0 + \sum\_{j=1}^{2n-1} \left[ \mathcal{R}\_{[i-j]}(t) \left\{ \mathcal{K}\_1(w\left(s\_i\right), w\left(s\_j\right)\right) - i\eta \mathcal{S}\_1(w\left(s\_i\right), w\left(s\_j\right)) \right\} \\ &+ \frac{\pi}{n} \left\{ \mathcal{K}\_2(w\left(s\_i\right), w\left(s\_j\right)\right) - i\eta \mathcal{S}\_2(w\left(s\_i\right), w\left(s\_j\right)) \right\} ] a\_j \phi\_j \\ &- \sum\_{j=1}^{2n-1} \frac{\pi}{n} H(w\left(s\_i\right), w\left(s\_j\right)) a\_j \phi\_0 = g\left(w\left(s\_i\right)\right), \quad \text{for } i = 0, \ldots, 2n - 1. \tag{75} \end{aligned} \tag{76}$$

We have obtained a system of 2*n* − 1 linear equations for the boundary values *φ tj* , for *j* = 1, 2, ..., 2*n* −1, that is a discretization of the integral equation (70). The solution is obtained by the usual Gaussian elimination procedure.

The described modification (70) applied to (25) ensures that exponentially fast convergence is achieved for scatterers with the Dirichlet boundary condition and a single corner on *∂D*.

This modification needs to be extended when the scatterer has two corners on *∂D*. There are now two points in the domain with singularities in the derivatives: at *t* = 0 and *t* = *π*. Each of these singularities have a contributing effect that needs to be accounted for. We use the fundamental solution to the Laplace equation in **R**<sup>2</sup> (68) to subtract vanishing terms. To reflect these combined contributions (69) is reformulated as

$$\begin{split} u^{\rm sc}(x) &= \int\_{\partial D} \left\{ \left\{ \frac{\partial \mathcal{G}(x,y)}{\partial n(y)} - i\eta \mathcal{G}(x,y) \right\} \phi(y) \\ &- \cos^{2} \frac{t}{2} \frac{(x)}{\partial n(y)} \frac{\partial \mathcal{G}\_{0}(x,y)}{\partial n(y)} \phi(x\_{0}) - \sin^{2} \frac{t}{2} \frac{(x)}{\partial n(y)} \frac{\partial \mathcal{G}\_{0}(x,y)}{\partial n(y)} \phi(x\_{\pi}) \right\} ds(y), \qquad x \in \mathbb{R}^{2} \backslash D, \tag{76} \end{split}$$

where x<sup>0</sup> and x*<sup>π</sup>* are the two corner points and *t*(x) denotes the parameter value of point x. The associated boundary equation (70) is now

$$\begin{split} \phi(x) - \left(\cos^{2}\frac{t(x)}{2}\phi(x\_{0}) + \sin^{2}\frac{t(x)}{2}\phi(x\_{\pi})\right) \\ &+ 2\int\_{\partial D} \left\{\frac{\partial G(x,y)}{\partial n(y)} - i\eta G(x,y)\right\} \phi(y) ds(y) \\ - 2\int\_{\partial D} \left(\cos^{2}\frac{t(x)}{2}\frac{\partial G\_{0}(x,y)}{\partial n(y)}\phi(x\_{0}) + \sin^{2}\frac{t(x)}{2}\frac{\partial G\_{0}(x,y)}{\partial n(y)}\phi(x\_{\pi})\right) ds(y) \\ &= -2u^{inc}(x), \qquad x \in \partial D, \quad \text{(77)} \end{split}$$

which in parameterised form is

We now apply the substitution (60) to (71) and obtain

*<sup>K</sup>*ˆ(*t*, *<sup>τ</sup>*)*<sup>φ</sup>* (*τ*) *<sup>d</sup><sup>τ</sup>* <sup>−</sup>

 2*π*

*H*(*t*, *τ*)*φ* (0) *dτ*

*K*ˆ(*w* (*s*), *w* (*σ*))*w* (*σ*) *φ* (*w* (*σ*)) *dσ*

*H*(*w* (*s*), *w* (*σ*))*w* (*σ*) *φ* (0) *dσ*. (74)

(75)

 *tj* , for

0

0

− 2*π*

0

The logarithmic singularity present in the kernel *K*ˆ(*t*, *τ*) remains to be accounted for. This is done in the same manner as (64). Using the quadrature rules (53) and (56) to discretize the kernel, and the trapezoidal rule to discretize the kernel *H*(*t*, *τ*) and *φ*<sup>0</sup> = *φ* (0) at the corner

*K*1(*w* (*si*), *w*(*sj*)) − *iηS*1(*w* (*si*), *w*(*sj*))

We have obtained a system of 2*n* − 1 linear equations for the boundary values *φ*

 *φ*(y)

2

*∂G*0(x, y) *<sup>∂</sup>*n(y) *<sup>φ</sup>*(x*π*)

*ds*(y), <sup>x</sup> <sup>∈</sup> **<sup>R</sup>**2\*D*¯ , (76)

*j* = 1, 2, ..., 2*n* −1, that is a discretization of the integral equation (70). The solution is obtained

The described modification (70) applied to (25) ensures that exponentially fast convergence is achieved for scatterers with the Dirichlet boundary condition and a single corner on *∂D*. This modification needs to be extended when the scatterer has two corners on *∂D*. There are now two points in the domain with singularities in the derivatives: at *t* = 0 and *t* = *π*. Each of these singularities have a contributing effect that needs to be accounted for. We use the fundamental solution to the Laplace equation in **R**<sup>2</sup> (68) to subtract vanishing terms. To

 ]*ajφ<sup>j</sup>*

*H*(*w* (*si*), *w*(*sj*))*ajφ*<sup>0</sup> = *g* (*w* (*si*)), for *i* = 0, ..., 2*n* − 1.

= 2*π*

*K*2(*w* (*si*), *w*(*sj*)) − *iηS*2(*w* (*si*), *w*(*sj*))

 2*π*

14 Advanced Electromagnetic Waves

0

*s*<sup>0</sup> = 0 gives

*φ<sup>i</sup>* − *φ*<sup>0</sup> +

*usc*(x) =

*∂D*

<sup>−</sup> cos<sup>2</sup> *<sup>t</sup>*(x) 2

+ *π n* 

− 2*n*−1 ∑ *j*=1

2*n*−1 ∑ *j*=1

[*R*|*i*−*j*|(*t*)

by the usual Gaussian elimination procedure.

*<sup>∂</sup>G*(x, <sup>y</sup>)

*∂G*0(x, y)

reflect these combined contributions (69) is reformulated as

*<sup>∂</sup>*n(y) <sup>−</sup> *<sup>i</sup>ηG*(x, <sup>y</sup>)

*<sup>∂</sup>*n(y) *<sup>φ</sup>*(x0) <sup>−</sup> sin2 *<sup>t</sup>*(x)

*π n* 
$$\begin{aligned} \left(\phi\left(t\right) - \left(\cos^2\frac{t}{2}\phi\left(0\right) + \sin^2\frac{t}{2}\phi\left(\pi\right)\right) - \int\_0^{2\pi} \hat{\mathbb{K}}(t,\tau)\phi\left(\tau\right)d\tau \\ - \cos^2\frac{t}{2}\int\_0^t H(t,\tau)\phi\left(0\right)d\tau - \sin^2\frac{t}{2}\int\_0^t H(t,\tau)\phi\left(\pi\right)d\tau = \mathbf{g}\left(t\right), \qquad 0 \le t \le 2\pi, \end{aligned} \tag{78}$$

where *<sup>H</sup>*(*t*) is as (72) except that for *<sup>t</sup>* <sup>=</sup> *<sup>τ</sup>*, *<sup>t</sup>* <sup>=</sup> 0, *<sup>π</sup>*, 2*<sup>π</sup>* and *<sup>K</sup>*ˆ(*t*, *<sup>τ</sup>*) is as in (73). We then apply the substitution (60) as in the case for the single corner domain with graded mesh (67) and discretize in the same fashion.
