A Kansa-Radial Basis Function Method for Elliptic Boundary Value Problems in Annular Domains

Abstract

We employ a Kansa-radial basis function (RBF) method for the numerical solution of elliptic boundary value problems in annular domains. This discretization leads, with an appropriate selection of collocation points and for any choice of RBF, to linear systems in which the matrices possess block circulant structures. These linear systems can be solved efficiently using matrix decomposition algorithms and fast Fourier transforms. A suitable value for the shape parameter in the various RBFs used is found using the leave-one-out cross validation algorithm. In particular, we consider problems governed by the Poisson equation, the inhomogeneous biharmonic equation and the inhomogeneous Cauchy–Navier equations of elasticity. In addition to its simplicity, the proposed method can both achieve high accuracy and solve large-scale problems. The feasibility of the proposed techniques is illustrated by several numerical examples.

Appendix

We consider the RBFs $\phi _{mn}, \, m=1, \ldots , M, n=1, \ldots , N$. These satisfy, see, e.g., [7, Appendix D],

$$\begin{aligned} \phi _{mn}(x,y)= \Phi (r_{mn}), \quad r_{mn}^2(x,y)=(x-x_{mn})^2+(y-y_{mn})^2. \end{aligned}$$

(6.1)

It can be easily seen that

$$\begin{aligned} \dfrac{\partial \phi _{mn}}{\partial x} (x,y)= & {} \dfrac{\Phi ^\prime (r_{mn})}{r_{mn}} (x-x_{mn}), \quad \dfrac{\partial \phi _{mn}}{\partial y} (x,y) = \dfrac{\Phi ^\prime (r_{mn})}{r_{mn}}(y-y_{mn}),\end{aligned}$$

(6.2)

$$\begin{aligned} \frac{\partial ^2 \phi _{mn}}{\partial x^2} (x,y)= & {} \dfrac{\Phi ^\prime (r_{mn})}{r_{mn}} + \left( \dfrac{r_{mn}\Phi ^{\prime \prime }(r_{mn})-\Phi ^\prime (r_{mn})}{r_{mn}^3}\right) (x-x_{mn})^2,\end{aligned}$$

(6.3)

$$\begin{aligned} \frac{\partial ^2 \phi _{mn}}{\partial y^2} (x,y)= & {} \dfrac{\Phi ^\prime (r_{mn})}{r_{mn}} + \left( \dfrac{r_{mn}\Phi ^{\prime \prime }(r_{mn})-\Phi ^\prime (r_{mn})}{r_{mn}^3}\right) (y-y_{mn})^2, \end{aligned}$$

(6.4)

and

$$\begin{aligned} \frac{\partial ^2 \phi _{mn}}{\partial x \partial y} (x,y) = \left( \dfrac{r_{mn}\Phi ^{\prime \prime }(r_{mn})-\Phi ^\prime (r_{mn})}{r_{mn}^3}\right) (x-x_{mn}) (y-y_{mn}). \end{aligned}$$

(6.5)

Also,

$$\begin{aligned} \Delta \phi _{mn} (x,y) = \Phi ^{\prime \prime }(r_{mn}) +\dfrac{\Phi ^\prime (r_{mn})}{r_{mn}} \end{aligned}$$

(6.6)

and

$$\begin{aligned} \Delta ^2 \phi _{mn} (x,y) = \Phi ^{\prime \prime \prime \prime }(r_{mn}) +\dfrac{2\Phi ^{\prime \prime \prime }(r_{mn})}{r_{mn}}- \dfrac{\Phi ^{\prime \prime }(r_{mn})}{r_{mn}^2} +\dfrac{\Phi ^{\prime }(r_{mn})}{r_{mn}^3}. \end{aligned}$$

(6.7)

In the sequel, we shall be using the notation (cf. (2.8))

$$\begin{aligned} \delta _{m_1,m_2} x= & {} x_{m_1,n_1}-x_{m_2,n_2} = \mathsf{r}_{n_1} \cos ({\vartheta }_{m_1} + \frac{2\pi \alpha _{n_1}}{N})-\mathsf{r}_{n_2} \cos ({\vartheta }_{m_2} + \frac{2\pi \alpha _{n_2}}{N}),\nonumber \\ \delta _{m_1,m_2} y= & {} y_{m_1,n_1}-y_{m_2,n_2} = \mathsf{r}_{n_1} \sin ({\vartheta }_{m_1} + \frac{2\pi \alpha _{n_1}}{N})-\mathsf{r}_{n_2} \sin ({\vartheta }_{m_2} + \frac{2\pi \alpha _{n_2}}{N}). \end{aligned}$$

(6.8)

We shall also need the normal derivatives on the boundary $\partial \Omega _1$ which are given by

$$\begin{aligned} (\mathrm{n}_x, \mathrm{n}_y) (x_{m_1,1},y_{m_1,1}) = -\left( \cos ({\vartheta }_{m_1} + \frac{2\pi \alpha _1}{N}),\; \sin ({\vartheta }_{m_1} + \frac{2\pi \alpha _1}{N})\right) . \end{aligned}$$

(6.9)

We next state and prove the following lemmata:

Lemma 1

For any radial function, each submatrix $\left( {A}_{n_1,n_2}\right)$ in (2.12) possesses a circulant structure.

Proof

We first consider the submatrices $\left( {A}_{n_1,n_2}\right) , \; n_1=2, \ldots , N-1, \; n_2=1, \ldots , N,$ which are defined by (2.13a). From (6.6), $\Delta \phi _{m_2,n_2}$ is an RBF, i.e. it is a function of $r_{m_2,n_2}$ (cf. (6.1)). Moreover,

$$\begin{aligned}&r_{m_2,n_2}^2(x_{m_1,n_1},y_{m_1,n_1})= (\delta _{m_1,m_2} x)^2+(\delta _{m_1,m_2} y)^2=\mathsf{r}_{n_1}^2+\mathsf{r}_{n_2}^2\\&\quad - 2\mathsf{r}_{n_1}\mathsf{r}_{n_2} \cos \left( {\vartheta }_{m_1}-{\vartheta }_{m_2} + \frac{2\pi (\alpha _{n_1}-\alpha _{n_2}) }{N}\right) , \end{aligned}$$

shows that for fixed $n_1, n_2$, the quantity $r_{m_2,n_2}(x_{m_1,n_1},y_{m_1,n_1})$ only depends on $({\vartheta }_{m_1}-{\vartheta }_{m_2})$, i.e. $({m_1}-{m_2})$ and therefore the submatrices are circulant.

The proof that the submatrices $\left( {A}_{n_1,n_2}\right) , \; n_1=1, N, \; n_2=1, \ldots , N$, defined by (2.13b)–(2.13c), are circulant in the case of Dirichlet boundary conditions is identical since their elements involve the RBFs $\phi _{m_2,n_2}$, which only depend on $r_{m_2,n_2}$.

In the case of a Neumann boundary condition on $\partial \Omega _1$, however, we have that the submatrices $\left( {A}_{1,n}\right) , \; n=1, \ldots , N$, are defined by (2.15), or

$$\begin{aligned} \left( A_{1,n_2}\right) _{m_1,m_2}= & {} \dfrac{\partial \phi _{m_2,n_2}}{\partial x} (x_{m_1,1},y_{m_1,1}) \mathrm{n}_x (x_{m_1,1},y_{m_1,1})\\&+ \,\dfrac{\partial \phi _{m_2,n_2}}{\partial y} (x_{m_1,1},y_{m_1,1}) \mathrm{n}_y (x_{m_1,1},y_{m_1,1}), \end{aligned}$$

and from (6.2) and (6.9) we obtain

$$\begin{aligned}&= - \dfrac{\Phi ^\prime (r_{m_2n_2})(x_{m_1,1},y_{m_1,1})}{r_{m_2n_2}(x_{m_1,1},y_{m_1,1}))} \left[ (x_{m_1,1}-x_{m_2,n_2}) \cos \left( {\vartheta }_{m_1} + \frac{2\pi \alpha _1}{N}\right) \right. \nonumber \\&\quad \left. +\, (y_{m_1,1}-y_{m_2,n_2}) \sin \left( {\vartheta }_{m_1} + \frac{2\pi \alpha _1}{N}\right) \right] . \end{aligned}$$

Ignoring the radial part, we examine

$$\begin{aligned}&(x_{m_1,1}-x_{m_2,n_2}) \cos \left( {\vartheta }_{m_1} + \frac{2\pi \alpha _1}{N}\right) + (y_{m_1,1}-y_{m_2,n_2}) \sin \left( {\vartheta }_{m_1} + \frac{2\pi \alpha _1}{N}\right) \nonumber \\&\quad =\left( \mathsf{r}_{1} \cos \left( {\vartheta }_{m_1} + \frac{2\pi \alpha _1}{N}\right) - \mathsf{r}_{n_2} \cos \left( {\vartheta }_{m_2} + \frac{2\pi \alpha _{n_2}}{N}\right) \right) \cos \left( {\vartheta }_{m_1} + \frac{2\pi \alpha _1}{N}\right) \nonumber \\&\qquad +\left( \mathsf{r}_{1} \sin \left( {\vartheta }_{m_1} + \frac{2\pi \alpha _1}{N}\right) - \mathsf{r}_{n_2} \sin \left( {\vartheta }_{m_2} + \frac{2\pi \alpha _{n_2}}{N}\right) \right) \sin \left( {\vartheta }_{m_1} + \frac{2\pi \alpha _1}{N}\right) \nonumber \\&\quad =\mathsf{r}_{1}-\mathsf{r}_{n_2} \cos \left( {\vartheta }_{m_2}-{\vartheta }_{m_1} + \frac{2\pi \left( \alpha _{n_2}-\alpha _1\right) }{N}\right) , \end{aligned}$$

(6.10)

which, again, only depends on $(m_2-m_1)$. Hence, this part is also radial and the corresponding submatrices are circulant. $\square$

Following identical arguments we also arrive at the corresponding result for the biharmonic case.

Lemma 2

For any radial function, each submatrix $\left( {A}_{n_1,n_2}\right)$ in (2.12) corresponding to collocation scheme (3.2), for both the first and the second biharmonic problem, possesses a circulant structure.

Finally, we consider the corresponding result for the Cauchy–Navier equations of elasticity.

Lemma 3

For any radial function, each submatrix $\left( \tilde{A}_{n_1,n_2}\right)$ in (4.9) possesses a $2 \times 2$ block circulant structure.

Proof

In order to prove the lemma we need to show that, see, e.g. [23],

$$\begin{aligned} R_{{\vartheta }_{m_1}} \left( A_{n_1,n_2}\right) _{m_1,m_2} R_{{\vartheta }_{m_2}}= & {} R_{{\vartheta }_{m_1+m}} \left( A_{n_1,n_2}\right) _{m_1+m,m_2+m} R_{{\vartheta }_{m_2+m}}, \nonumber \\&m_1,m_2=1,\ldots ,M, \; n_1, n_2=1, \ldots , N, \end{aligned}$$

(6.11)

provided $m_1+m, m_2+m \le M$. In case $m_1+m > M$, $m_1+m$ is replaced by $m_1+m-M$ and in case $m_2+m > M$, $m_2+m$ is replaced by $m_2+m-M$.

Since $R_{{\vartheta }_k}^2=I_2$ proving (6.11) is equivalent to proving

$$\begin{aligned}&R_{{\vartheta }_{m_1+m}} R_{{\vartheta }_{m_1}} \left( A_{n_1,n_2}\right) _{m_1,m_2} R_{{\vartheta }_{m_2}}R_{{\vartheta }_{m_1+m}}=\left( A_{n_1,n_2}\right) _{m_1+m,m_2+m},\nonumber \\&m_1,m_2=1,\ldots ,M, \; n_1, n_2=1, \ldots , N. \end{aligned}$$

(6.12)

However,

$$\begin{aligned} R_{{\vartheta }_{m_1+m}} R_{{\vartheta }_{m_1}}= & {} \left( \begin{array}{cc} \cos {\vartheta }_{m_1+m} &{} \sin {\vartheta }_{m_1+m} \\ \sin {\vartheta }_{m_1+m} &{} -\!\cos {\vartheta }_{m_1+m} \end{array} \right) \! \left( \begin{array}{cc} \cos {\vartheta }_{m_1} &{} \sin {\vartheta }_{m_1} \\ \sin {\vartheta }_{m_1} &{} -\!\cos {\vartheta }_{m_1} \end{array} \right) \! \\= & {} \left( \begin{array}{cc} \cos {\vartheta }_m &{} -\sin {\vartheta }_m \\ \sin {\vartheta }_m &{} \!\cos {\vartheta }_m \end{array} \right) \! = W_{{\vartheta }_m}, \end{aligned}$$

and

$$\begin{aligned} R_{{\vartheta }_{m_1}} R_{{\vartheta }_{m_1+m}}= & {} \left( \begin{array}{cc} \cos {\vartheta }_{m_1} &{} \sin {\vartheta }_{m_1} \\ \sin {\vartheta }_{m_1} &{} -\!\cos {\vartheta }_{m_1} \end{array} \right) \! \left( \begin{array}{cc} \cos {\vartheta }_{m_1+m} &{} \sin {\vartheta }_{m_1+m} \\ \sin {\vartheta }_{m_1+m} &{} -\!\cos {\vartheta }_{m_1+m} \end{array} \right) \! \\= & {} \left( \begin{array}{cc} \cos {\vartheta }_m &{} \sin {\vartheta }_m \\ -\sin {\vartheta }_m &{} \!\cos {\vartheta }_m \end{array} \right) \! = W_{{\vartheta }_m}^{-1}. \end{aligned}$$

Hence proving (6.12) is equivalent to proving that

$$\begin{aligned} W_{{\vartheta }_m} \left( A_{n_1,n_2}\right) _{m_1,m_2} W_{{\vartheta }_m}^{-1} =\left( A_{n_1,n_2}\right) _{m_1+m,m_2+m}. \end{aligned}$$

(6.13)

From (4.4a), for $m_1,m_2=1,\ldots ,M, \; n_1=2, \ldots , N-1, \; n_2=1, \ldots , N$, we can write

$$\begin{aligned}&\left( A_{n_1,n_2}\right) _{m_1,m_2}=\mu \Delta \phi _{m_2,n_2}(x_{m_1,n_1},y_{m_1,n_1}) I_2\nonumber \\&\quad +\, \dfrac{\mu }{1-2\nu } \left( \begin{array}{c c} \dfrac{\partial ^2 \phi _{m_2,n_2}}{\partial x^2} (x_{m_1,n_1},y_{m_1,n_1}) &{} \dfrac{\partial ^2 \phi _{m_2,n_2}}{\partial x \partial y} (x_{m_1,n_1},y_{m_1,n_1}) \\ \dfrac{\partial ^2 \phi _{m_2,n_2}}{\partial x \partial y} (x_{m_1,n_1},y_{m_1,n_1}) &{} \dfrac{\partial ^2 \phi _{m_2,n_2}}{\partial y^2} (x_{m_1,n_1},y_{m_1,n_1}) \end{array} \right) . \end{aligned}$$

(6.14)

We first consider the first term in (6.14), namely $\mu \Delta \phi _{m_2,n_2}(x_{m_1,n_1},y_{m_1,n_1}) I_2$.

We clearly have that

$$\begin{aligned} \mu \Delta \phi _{m_2,n_2}(x_{m_1,n_1},y_{m_1,n_1}) W_{{\vartheta }_m} I_2 W_{{\vartheta }_m}^{-1}= & {} \mu \Delta \phi _{m_2,n_2}(x_{m_1,n_1},y_{m_1,n_1}) I_2\\= & {} \mu \Delta \phi _{m_2+m,n_2}(x_{m_1+m,n_1},y_{m_1+m,n_1}) I_2, \end{aligned}$$

since the Laplacian of $\phi _{m_2,n_2}(x,y)$ is radial from (6.6) and

$$\begin{aligned}&(\delta _{m_1,m_2} x)^2+(\delta _{m_1,m_2} y)^2=\mathsf{r}_{n_1}^2+\mathsf{r}_{n_2}^2- 2\mathsf{r}_{n_1}\mathsf{r}_{n_2} \cos \left( {\vartheta }_{m_1}-{\vartheta }_{m_2} + \frac{2\pi (\alpha _{n_1}-\alpha _{n_2}) }{N}\right) \nonumber \\&\quad =(\delta _{m_1+m,m_2+m} x)^2+(\delta _{m_1+m,m_2+m} y)^2. \end{aligned}$$

(6.15)

Therefore, the first term in (6.14) satisfies (6.13).

For the second term in (6.14), from (6.3)–(6.5) and ignoring the radial parts we only need show that

$$\begin{aligned}&\left( \begin{array}{ll} \cos {\vartheta }_m &{} -\sin {\vartheta }_m \\ \sin {\vartheta }_m &{} \!\cos {\vartheta }_m \end{array} \right) \! \left( \begin{array}{c c} (\delta _{m_1,m_2} x)^2 &{} (\delta _{m_1,m_2} x)(\delta _{m_1,m_2} y)\\ (\delta _{m_1,m_2} x)(\delta _{m_1,m_2} y) &{} (\delta _{m_1,m_2} y)^2 \end{array} \right) \left( \begin{array}{ll} \cos {\vartheta }_m &{} \sin {\vartheta }_m \\ -\sin {\vartheta }_m &{} \!\cos {\vartheta }_m \end{array} \right) \!\nonumber \\&\quad = \left( \begin{array}{c c} (\delta _{m_1+m,m_2+m} x)^2 &{} (\delta _{m_1+m,m_2+m} x)(\delta _{m_1+m,m_2+m} y)\\ (\delta _{m_1+m,m_2+m} x)(\delta _{m_1+m,m_2+m} y) &{} (\delta _{m_1+m,m_2+m} y)^2 \end{array} \right) . \end{aligned}$$

(6.16)

By performing the multiplications on the left hand side of (6.16) we obtain that it is equal to

$$\begin{aligned} \left( \begin{array}{cc} (\cos {\vartheta }_m\delta _{m_1,m_2} x-\sin {\vartheta }_m\delta _{m_1,m_2}y)^2 &{} (\cos {\vartheta }_m\delta _{m_1,m_2} x-\sin {\vartheta }_m\delta _{m_1,m_2}y)(\sin {\vartheta }_m\delta _{m_1,m_2} x +\cos {\vartheta }_m\delta _{m_1,m_2}y) \\ (\cos {\vartheta }_m\delta _{m_1,m_2} x-\sin {\vartheta }_m\delta _{m_1,m_2}y)(\sin {\vartheta }_m\delta _{m_1,m_2} x +\cos {\vartheta }_m\delta _{m_1,m_2}y) &{} (\sin {\vartheta }_m\delta _{m_1,m_2} x+\cos {\vartheta }_m\delta _{m_1,m_2}y)^2 \end{array} \right) \!. \end{aligned}$$

Moreover, it can be easily shown that

$$\begin{aligned} \cos {\vartheta }_m\delta _{m_1,m_2} x-\sin {\vartheta }_m\delta _{m_1,m_2}y= & {} \mathsf{r}_{n_1} \cos \left( {\vartheta }_{m_1+m} + \frac{2\pi \alpha _{n_1}}{N}\right) \\&\quad -\mathsf{r}_{n_2} \cos \left( {\vartheta }_{m_2+m} + \frac{2\pi \alpha _{n_2}}{N}\right) \end{aligned}$$

and

$$\begin{aligned} \sin {\vartheta }_m\delta _{m_1,m_2} x\!+\!\cos {\vartheta }_m\delta _{m_1,m_2}y \!=\!\mathsf{r}_{n_1} \sin \left( {\vartheta }_{m_1+m} \!+ \!\frac{2\pi \alpha _{n_1}}{N}\right) \!-\!\mathsf{r}_{n_2} \sin \left( {\vartheta }_{m_2+m} \!+ \!\frac{2\pi \alpha _{n_2}}{N}\right) , \end{aligned}$$

from which (6.16) follows. Hence, the second term in (6.14) also satisfies (6.13).

We next need to prove that

$$\begin{aligned}&W_{{\vartheta }_m} \left( A_{1,n_2}\right) _{m_1,m_2} W_{{\vartheta }_m}^{-1} =\left( A_{1,n_2}\right) _{m_1+m,m_2+m} \quad \text{ and } \quad \nonumber \\&W_{{\vartheta }_m} \left( A_{N,n_2}\right) _{m_1,m_2} W_{{\vartheta }_m}^{-1} =\left( A_{N,n_2}\right) _{m_1+m,m_2+m}. \end{aligned}$$

(6.17)

Since from (4.4b) we can write

$$\begin{aligned} \left( A_{1,n_2}\right) _{m_1,m_2}=\phi _{m_2,n_2}(x_{m_1,1},y_{m_1,1}) I_2 \quad \text{ and } \quad \left( A_{N,n_2}\right) _{m_1,m_2}=\phi _{m_2,n_2}(x_{m_1,N},y_{m_1,N}) I_2, \end{aligned}$$

from (6.15), following a similar argument to the one used for the first term of (6.14), (6.17) follows.

Finally, we show that if instead of the displacements $u_1, u_2$ we have that the tractions $t_1, t_2$ are prescribed on, say, the boundary $\partial \Omega _1$, cf. Sect. 4.2.1, then

$$\begin{aligned} W_{{\vartheta }_m} \left( A_{1,n_2}\right) _{m_1,m_2} W_{{\vartheta }_m}^{-1} =\left( A_{1,n_2}\right) _{m_1+m,m_2+m} \end{aligned}$$

(6.18)

is still true.

We write

$$\begin{aligned} \left( A_{1,n}\right) _{m_1,m_2}=\left( \begin{array}{cc} T_{11} &{} T_{12} \\ T_{21} &{} T_{22} \end{array}\right) , \end{aligned}$$

(6.19)

where $T_{ij}, i,j=1,2,$ are the appropriate quantities in (4.6).

We can easily show that

$$\begin{aligned}&W_{{\vartheta }_m} \left( \begin{array}{c c} T_{11} &{} T_{12} \\ T_{21} &{} T_{22} \end{array}\right) W_{{\vartheta }_m}^{-1}\\&\quad = \left( \begin{array}{c c} \cos ^2 {\vartheta }_m T_{11} -\cos {\vartheta }_m\sin {\vartheta }_m (T_{21} + T_{12}) + \sin ^2 {\vartheta }_m T_{22}&{} \cos ^2 {\vartheta }_m T_{12} +\cos {\vartheta }_m\sin {\vartheta }_m (T_{11} - T_{22}) - \sin ^2 {\vartheta }_m T_{21} \\ \cos ^2 {\vartheta }_m T_{21} +\cos {\vartheta }_m\sin {\vartheta }_m (T_{11} - T_{22}) - \sin ^2 {\vartheta }_m T_{12} &{} \sin ^2 {\vartheta }_m T_{11} +\cos {\vartheta }_m\sin {\vartheta }_m (T_{21} + T_{12}) + \cos ^2 {\vartheta }_m T_{22} \end{array}\right) . \end{aligned}$$

We shall first show that the first term in (4.6) is radial. Using the notation of (6.8) with $n_1=1$, we have from (6.2) that

$$\begin{aligned}&\dfrac{\partial \phi _{m_2,n}}{\partial x}(x_{m_1,1},y_{m_1,1})\mathrm{n}_x(x_{m_1,1},y_{m_1,1}) + \dfrac{\partial \phi _{m_2,n}}{\partial y}(x_{m_1,1},y_{m_1,1})\mathrm{n}_y (x_{m_1,1},y_{m_1,1})\\&\quad =\dfrac{\Phi ^\prime (r_{m_2n})}{r_{m_2n}} \left[ \delta _{m_1,m_2}x \, \mathrm{n}_x(x_{m_1,1},y_{m_1,1})+\delta _{m_1,m_2}y \, \mathrm{n}_y(x_{m_1,1},y_{m_1,1})\right] , \end{aligned}$$

and ignoring the radial factor, we have, using (6.10) that

$$\begin{aligned}&\delta _{m_1,m_2}x \, \mathrm{n}_x(x_{m_1,1},y_{m_1,1})+\delta _{m_1,m_2}y \, \mathrm{n}_y(x_{m_1,1},y_{m_1,1})\\&\quad =\mathsf{r}_{1}-\mathsf{r}_{n_2} \cos ({\vartheta }_{m_2}-{\vartheta }_{m_1} + \frac{2\pi (\alpha _{n_2}-\alpha _1)}{N}), \end{aligned}$$

which only depends on $(m_2-m_1)$, hence it is radial.

We next consider the second term in (4.6). Dropping the multiplying constants and the obviously radial parts, and with the appropriate notation, we consider the quantities

$$\begin{aligned} \tilde{T}_{11}(m_1,m_2)= & {} \delta _{m_1,m_2}x \, \mathrm{n}_x(x_{m_1,1},y_{m_1,1})= \mathsf{r}_{1} \cos ^2 \left( {\vartheta }_{m_1} + \frac{2\pi \alpha _1}{N}\right) \\&-\,\mathsf{r}_{n_2} \cos \left( {\vartheta }_{m_2} + \frac{2\pi \alpha _{n_2}}{N}\right) \cos \left( {\vartheta }_{m_1} + \frac{2\pi \alpha _1}{N}\right) ,\\ \tilde{T}_{12}(m_1,m_2)= & {} 2\nu \delta _{m_1,m_2}y \, \mathrm{n}_x(x_{m_1,1},y_{m_1,1}) +(1-2\nu ) \delta _{m_1,m_2}x \, \mathrm{n}_y(x_{m_1,1},y_{m_1,1})\\= & {} \mathsf{r}_{1} \sin \left( {\vartheta }_{m_1} + \frac{2\pi \alpha _1}{N}\right) \cos \left( {\vartheta }_{m_1} + \frac{2\pi \alpha _1}{N}\right) \\&-\, \mathsf{r}_{n_2} \left( 2 \nu \cos \left( {\vartheta }_{m_1} + \frac{2\pi \alpha _1}{N}\right) \sin \left( {\vartheta }_{m_2} + \frac{2\pi \alpha _{n_2}}{N}\right) \right. \\&+\left. (1-2\nu ) \sin \left( {\vartheta }_{m_1} + \frac{2\pi \alpha _1}{N}\right) \cos \left( {\vartheta }_{m_2} + \frac{2\pi \alpha _{n_2}}{N}\right) \right) \\= & {} \mathsf{r}_{1} \sin \left( {\vartheta }_{m_1} + \frac{2\pi \alpha _1}{N}\right) \cos \left( {\vartheta }_{m_1} + \frac{2\pi \alpha _1}{N}\right) \\&-\, \mathsf{r}_{n_2} \left( 2 \nu \sin \left( {\vartheta }_{m_2}-{\vartheta }_{m_1} + \frac{2\pi (\alpha _{n_2}-\alpha _1)}{N}\right) \right. \\&\left. +\sin \left( {\vartheta }_{m_1} + \frac{2\pi \alpha _1}{N}\right) \cos \left( {\vartheta }_{m_2} + \frac{2\pi \alpha _{n_2}}{N}\right) \right) ,\\&\tilde{T}_{21}(m_1,m_2)= (1-2\nu ) \delta _{m_1,m_2}y \, \mathrm{n}_x(x_{m_1,1},y_{m_1,1}) \\&\quad +\, 2\nu \delta _{m_1,m_2}x \, \mathrm{n}_y(x_{m_1,1},y_{m_1,1})\\= & {} \mathsf{r}_{1} \sin \left( {\vartheta }_{m_1} + \frac{2\pi \alpha _1}{N}\right) \cos \left( {\vartheta }_{m_1} + \frac{2\pi \alpha _1}{N}\right) \\&-\, \mathsf{r}_{n_2} \left( (1-2 \nu ) \cos \left( {\vartheta }_{m_1} + \frac{2\pi \alpha _1}{N}\right) \sin \left( {\vartheta }_{m_2} + \frac{2\pi \alpha _{n_2}}{N}\right) \right. \\&+\left. 2 \nu \sin \left( {\vartheta }_{m_1} + \frac{2\pi \alpha _1}{N}\right) \cos ({\vartheta }_{m_2} + \frac{2\pi \alpha _{n_2}}{N}) \right) \\= & {} \mathsf{r}_{1} \sin \left( {\vartheta }_{m_1} + \frac{2\pi \alpha _1}{N}\right) \cos \left( {\vartheta }_{m_1} + \frac{2\pi \alpha _1}{N}\right) \\&-\, \mathsf{r}_{n_2} \left( 2 \nu \sin \left( {\vartheta }_{m_1}-{\vartheta }_{m_2} + \frac{2\pi (\alpha _1-\alpha _{n_2})}{N}\right) \right. \\&+\left. \cos \left( {\vartheta }_{m_1} + \frac{2\pi \alpha _1}{N}\right) \sin \left( {\vartheta }_{m_2} + \frac{2\pi \alpha _{n_2}}{N}\right) \right) , \\ \tilde{T}_{22}(m_1,m_2)= & {} \delta _{m_1,m_2}y \, \mathrm{n}_y(x_{m_1,1},y_{m_1,1})= \mathsf{r}_{1} \sin ^2 \left( {\vartheta }_{m_1} + \frac{2\pi \alpha _1}{N}\right) \\&-\,\mathsf{r}_{n_2} \sin \left( {\vartheta }_{m_2} + \frac{2\pi \alpha _{n_2}}{N}\right) \sin \left( {\vartheta }_{m_1} + \frac{2\pi \alpha _1}{N}\right) . \end{aligned}$$

By using (6.9), we first consider the term

$$\begin{aligned}&\cos ^2 {\vartheta }_m \tilde{T}_{11} -\cos {\vartheta }_m\sin {\vartheta }_m (\tilde{T}_{21} + \tilde{T}_{12}) + \sin ^2 {\vartheta }_m \tilde{T}_{22}\\&\quad = \cos ^2 {\vartheta }_m \delta _{m_1,m_2}x \, \mathrm{n}_x -\cos {\vartheta }_m\sin {\vartheta }_m \left( \delta _{m_1,m_2}y \, \mathrm{n}_x+ \delta _{m_1,m_2}x \, \mathrm{n}_y \right) + \sin ^2 {\vartheta }_m \delta _{m_1,m_2}y \, \mathrm{n}_y\\&\quad =\mathsf{r}_{1} \cos ^2 \left( {\vartheta }_{m_1}\!+\! {\vartheta }_{m} \!+\! \frac{2\pi \alpha _1}{N}\right) -\mathsf{r}_{n_2} \cos \left( {\vartheta }_{m_2} + {\vartheta }_{m} \!+\! \frac{2\pi \alpha _{n_2}}{N}\right) \cos \left( {\vartheta }_{m_1}+ {\vartheta }_{m} \!+\! \frac{2\pi \alpha _1}{N}\right) \\&\quad =\tilde{T}_{11}(m_1+m,m_2+m). \end{aligned}$$

We next consider

$$\begin{aligned}&\cos ^2 {\vartheta }_m \tilde{T}_{12} +\cos {\vartheta }_m\sin {\vartheta }_m (\tilde{T}_{11} - \tilde{T}_{22}) - \sin ^2 {\vartheta }_m \tilde{T}_{21}\\&\quad =\cos ^2 {\vartheta }_m \left( 2\nu \delta _{m_1,m_2}y \, \mathrm{n}_x \!+\!(1\!-\!2\nu ) \delta _{m_1,m_2}x \, \mathrm{n}_y\right) \!+\!\cos {\vartheta }_m\sin {\vartheta }_m (\delta _{m_1,m_2}x \, \mathrm{n}_x \!- \!\delta _{m_1,m_2}y \, \mathrm{n}_y)\\&\quad \quad - \sin ^2 {\vartheta }_m \left( (1-2\nu ) \delta _{m_1,m_2}y \, \mathrm{n}_x + 2\nu \delta _{m_1,m_2}x \, \mathrm{n}_y \right) =-2\nu \mathsf{r}_{n_2} \sin ({\vartheta }_{m_2}-{\vartheta }_{m_1} \\&\qquad + \frac{2\pi (\alpha _{n_2}-\alpha _1)}{N}) +\mathsf{r}_{1} \sin \left( {\vartheta }_{m_1} +{\vartheta }_m + \frac{2\pi \alpha _1}{N}\right) \cos \left( {\vartheta }_{m_1} +{\vartheta }_m + \frac{2\pi \alpha _1}{N}\right) \\&\quad \quad - \mathsf{r}_{n_2} \sin \left( {\vartheta }_{m_1} +{\vartheta }_m + \frac{2\pi \alpha _1}{N}\right) \cos \left( {\vartheta }_{m_2} +{\vartheta }_m + \frac{2\pi \alpha _{n_2}}{N}\right) =\tilde{T}_{12}(m_1+m,m_2+m). \end{aligned}$$

Similarly, we have that

$$\begin{aligned}&\cos ^2 {\vartheta }_m \tilde{T}_{21} +\cos {\vartheta }_m\sin {\vartheta }_m (\tilde{T}_{11} - \tilde{T}_{22}) - \sin ^2 {\vartheta }_m \tilde{T}_{12}=\cos ^2 {\vartheta }_m \\&\qquad \times \left( (1-2\nu ) \delta _{m_1,m_2}y \, \mathrm{n}_x + 2\nu \delta _{m_1,m_2}x \, \mathrm{n}_y\right) +\cos {\vartheta }_m\sin {\vartheta }_m (\delta _{m_1,m_2}x \, \mathrm{n}_x - \delta _{m_1,m_2}y \, \mathrm{n}_y)\\&\qquad - \sin ^2 {\vartheta }_m \left( 2\nu \delta _{m_1,m_2}y \, \mathrm{n}_x + (1-2\nu ) \delta _{m_1,m_2}x \, \mathrm{n}_y \right) =-2\nu \mathsf{r}_{n_2} \sin \left( {\vartheta }_{m_1}-{\vartheta }_{m_2} \right. \\&\qquad \left. + \,\frac{2\pi (\alpha _1-\alpha _{n_2})}{N}\right) +\mathsf{r}_{1} \sin \left( {\vartheta }_{m_1} +{\vartheta }_m + \frac{2\pi \alpha _1}{N}\right) \cos \left( {\vartheta }_{m_1} +{\vartheta }_m + \frac{2\pi \alpha _1}{N}\right) \\&\qquad - \,\mathsf{r}_{n_2} \cos \left( {\vartheta }_{m_1} +{\vartheta }_m + \frac{2\pi \alpha _1}{N}\right) \sin \left( {\vartheta }_{m_2} +{\vartheta }_m + \frac{2\pi \alpha _{n_2}}{N}\right) =\tilde{T}_{21}(m_1+m,m_2+m). \end{aligned}$$

Finally,

$$\begin{aligned}&\sin ^2 {\vartheta }_m \tilde{T}_{11} +\cos {\vartheta }_m\sin {\vartheta }_m (\tilde{T}_{21} + \tilde{T}_{12}) + \cos ^2 {\vartheta }_m \tilde{T}_{22}\\&\quad = \sin ^2 {\vartheta }_m \delta _{m_1,m_2}x \, \mathrm{n}_x +\cos {\vartheta }_m\sin {\vartheta }_m \left( \delta _{m_1,m_2}y \, \mathrm{n}_x+ \delta _{m_1,m_2}x \, \mathrm{n}_y \right) + \cos ^2 {\vartheta }_m \delta _{m_1,m_2}y \, \mathrm{n}_y.\\&\quad =\mathsf{r}_{1} \sin ^2 \left( {\vartheta }_{m_1}\!+\! {\vartheta }_{m} \!+\! \frac{2\pi \alpha _1}{N}\right) \!-\!\mathsf{r}_{n_2} \sin \left( {\vartheta }_{m_2} + {\vartheta }_{m} \!+\! \frac{2\pi \alpha _{n_2}}{N}\right) \sin \left( {\vartheta }_{m_1}\!+\! {\vartheta }_{m} \!+\! \frac{2\pi \alpha _1}{N}\right) \\&\quad =\tilde{T}_{22}(m_1+m,m_2+m). \end{aligned}$$

Therefore (6.18) is satisfied and the proof is complete. $\square$