Sinc-Galerkin solution to the clamped plate eigenvalue problem

Abstract

We propose an accurate and computationally efficient numerical technique for solving the biharmonic eigenvalue problem. The technique is based on the sinc-Galerkin approximation method to solve the clamped plate problem. Numerical experiments for plates with various aspect ratios are reported, and comparisons are made with other methods in literature. The calculated results accord well with those published earlier, which proves the accuracy and validity of the proposed method.

Appendix 1: Proof of Theorem 3.1

For $U_{xxxx}$, the inner product with sinc basis element is given by

$$\begin{aligned} \left<\Phi ^{4} U_{xxxx},S_k S_l\right>=\Phi ^{4} \int _{0}^1 \int _0^1 U_{xxxx}(x,y)S_k(x)S_l(y)w(x)v(y)dx dy. \end{aligned}$$

Integrating by parts to remove the fourth order derivatives from the dependent variable U leads to the equality.

$$\begin{aligned} \left<\Phi ^{4} U_{xxxx},S_k S_l\right>= B_{x}+\Phi ^{4} \int _{0}^1 \int _0^1 U(x,y)\left[ S_k(x)S_l(y)w(x)v(y)\right] _{xxxx}dx dy. \end{aligned}$$

(5.1)

where the boundary term is

$$\begin{aligned} B_x=\Phi ^{4} \int _0^1\left[ U_{xxx}(S_kS_lwv)-U_{xx}(S_kS_lwv)_x +U_x(S_kS_lwv)_{xx}-U(S_kS_lwv)_{xxx}\right] _0^1 dy. \end{aligned}$$

The boundary terms in Eq. (5.1) vanished. Continuing only with the remaining integral in (5.1) and expanding the derivative results in

$$\begin{aligned} \left<\Phi ^{4} U_{xxxx},S_k S_l\right>=\Phi ^{4} \int _{0}^1 \int _0^1 \sum _{i=0}^4U(x,y)\left[ S_k^{(i)}\mu _i\right] S_l v(y) dx dy. \end{aligned}$$

(5.2)

where $S_k^{(i)}$ denotes the ith derivative of $S_k$ with respect to the $\phi _1$ and

$$\begin{aligned} \mu _4= & {} (\phi _1')^{4}w, \\ \mu _3= & {} 6(\phi '_1)^2\phi ''_1w+4(\phi '_1)^3w', \\ \mu _2= & {} 3(\phi ''_1)^2w+4\phi '_1\phi '''_1w+12\phi _1'\phi _1''w'+6(\phi _1')^2w'', \\ \mu _1= & {} \phi _1''''w+4\phi _1'''w'+6\phi _1''w''+4\phi _1'w'''. \end{aligned}$$

and

$$\begin{aligned} \mu _0=w'''' \end{aligned}$$

Applying the sinc quadrature in the x-domain and y-domain to Eq. (5.2) yields Eq. (3.10).

The inner product for $U_{yyyy}$ may be handled in a similar manner. This gives the expression (3.11) where

$$\begin{aligned} \eta _4= & {} (\phi _2')^{4}v, \\ \eta _3= & {} 6(\phi '_2)^2\phi ''_2v+4(\phi '_2)^3v', \\ \eta _2= & {} 3(\phi ''_2)^2v+4\phi '_2\phi '''_2v+12\phi _2'\phi _2''v'+6(\phi _2')^2v'', \\ \eta _1= & {} \phi _2''''v+4\phi _2'''v'+6\phi _2''v''+4\phi _2'v'''. \end{aligned}$$

and

$$\begin{aligned} \eta _0=v'''' \end{aligned}$$

For $U_{xxyy}$, the inner product with sinc basis element is given by

$$\begin{aligned} \left<2\,\Phi ^{2} u_{xxyy},S_k\, S_l\right>=2\,\Phi ^{2} \int _{0}^1 \int _0^1 \,U_{xxyy}(x,y)\,S_k(x)\,S_l(y)\,w(x)\,v(y)\,dx \,dy. \end{aligned}$$

Integrating by parts to remove the fourth derivatives from the dependent variable U leads to the equality

$$\begin{aligned} \left<2\,\Phi ^{2} U_{xxyy},S_k \,S_l\right> = B_{xy}+ \Phi ^{2} \int _{0}^1 \int _0^1 U(x,y)\,\left[ 2\,S_k(x)\,S_l(y)\,w(x)\,v(y)\right] _{xxyy}\,dx\, dy. \end{aligned}$$

(5.3)

where the boundary term is

$$\begin{aligned} B_{xy}= & {} \Phi ^{2}\int _0^1\left[ U_{xyy}(S_k\,S_l\,w\,v)-U_{yy}(S_k\,S_l\,w\,v)_x\right] _0^1 dy \\&+ \; \int _0^1\left[ U_{y}(S_k\,S_l\,w\,v)_{xx}- U(S_k\,S_l\,w\,v)_{xxy}\right] _0^1 dx=0 \end{aligned}$$

Continuing with the remaining integral in (5.3) and expanding the derivative result in

$$\begin{aligned} \left<2\,\Phi ^{2} U_{xxyy},S_k S_l\right>=2\,\Phi ^{2} \int _{0}^1\int _0^1 \sum _{r=0}^2\,\,\sum _{p=0}^2\,\,U(x,y)S_k^{(r)}\,S_l^{(p)}\,\,\tau _r\,\xi _p\, \,dx\, dy. \end{aligned}$$

(5.4)

where

$$\begin{aligned} \tau _2=(\phi _1')^2\,w,\quad \tau _1=2\phi _1'w'+\phi _1''w,\quad \tau _0=w'', \end{aligned}$$

and

$$\begin{aligned} \xi _2=(\phi _2')^2v,\quad \xi _1=2\phi _2'v'+\phi _2''v,\quad \xi _0=v'' \end{aligned}$$

Applying the sinc quadrature in the x-domain and y-domain to the Eq. (5.4) yields Eq. (3.12).

For $\Phi ^{4} \lambda ^2 U(x,y)$, the inner product is

$$\begin{aligned} \left<\Phi ^{4} \lambda ^2 U ,S_k S_l\right>=\Phi ^{4} \int _0^1\int _0^1 \lambda ^2 U(x,y)S_k S_lw(x)v(y)dx dy \end{aligned}$$

(5.5)

Applying the sinc quadrature to (5.5) yields

$$\begin{aligned} \left<\Phi ^{4} \lambda ^2 U ,S_k S_l\right>\approx \Phi ^{4} \lambda ^2 h_x h_y\frac{w(x_k)U(x_k,y_l)v(y_l)}{\phi _1'(x_k)\phi _2'(y_l)} \end{aligned}$$

(5.6)

as given in Eq. (3.13).