Question

Question: A square membrane of side l is distorted into the shape

$\mathbf{f}\mathbf{}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\mathbf{x}\mathbf{}\mathbf{y}\mathbf{}\mathbf{(}\mathbf{l}\mathbf{-}\mathbf{x}\mathbf{)}\mathbf{(}\mathbf{l}\mathbf{-}\mathbf{y}\mathbf{)}$

and released. Express its shape at subsequent times as an infinite series. Hint: Use a double Fourier series as in Problem 5.9.

Short answer

The solution if the membrane is square is given below.
$\mathrm{z}(\mathrm{x},\mathrm{y}.\mathrm{t})=\sum _{\mathrm{n}=1,3,5...}^{\infty }\sum _{\mathrm{n}=1,3,5...}^{\infty }\frac{64{\mathrm{l}}^4}{{\mathrm{\pi }}^6{\mathrm{n}}^3}\sin \left(\frac{\mathrm{n\pi }}{\mathrm{l}}\mathrm{x}\right)\sin \left(\frac{\mathrm{m\pi }}{\mathrm{l}}\mathrm{y}\right)\cos \left(\frac{\mathrm{t\pi v}}{\mathrm{l}}\sqrt{{\mathrm{n}}^2+{\mathrm{m}}^2}\right)$

Step-by-step solution

Given Information.

The function of distorted shape is as given below.

$\mathrm{f}(\mathrm{x},\mathrm{y})=\mathrm{x}\mathrm{y}(1-\mathrm{x})(1-\mathrm{y})$.

Definition of Laplace’ equation.

The total of the second-order partial derivatives of z, the unknown function, with respect to the Cartesian coordinates equals , according to Laplace's equation.

The wave equation is ${\mathbf{\nabla }}^{\mathbf{2}}\mathbf{z}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{v}}^{\mathbf{2}}}\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{z}}{\mathbf{\partial }{\mathbf{t}}^{\mathbf{2}}}$.

Use wave equation.

Start from a wave equation.
${\mathbf{\nabla }}^{\mathbf{2}}\mathbf{z}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{v}}^{\mathbf{2}}}\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{z}}{\mathbf{\partial }{\mathbf{t}}^{\mathbf{2}}}$

Put a solution of the form mentioned below in the above equation.

$\mathrm{Z}\left(\mathrm{z}\right)=\mathrm{F}(\mathrm{x},\mathrm{y})\mathrm{T}\left(\mathrm{t}\right)$

Dividing by $\mathrm{F}(\mathrm{x},\mathrm{y})\mathrm{T}\left(\mathrm{t}\right)$.

$\frac{{\nabla }^2\mathrm{F}}{\mathrm{F}}=\frac{1}{V^{2}T}\frac{{\partial }^{2}T}{\partial t^2}=-{\mathrm{K}}^2$

The both sides are a function of a different variable and they must be equal to some constant if they are to be equal.

Write the derived two equation.
$$\begin{gathered} {\nabla }^2\mathrm{F}+{\mathrm{K}}^2\mathrm{F}=0; \\ \frac{{\partial }^2\mathrm{T}}{\partial {\mathrm{t}}^2}+{\mathrm{K}}^2{\mathrm{V}}^2\mathrm{T}=0 \end{gathered}$$

Write the solution of the time equation.
$\mathrm{T}\left(\mathrm{t}\right)=\left\{\begin{array}{l}\sin \left(\mathrm{Kvt}\right)\\ \cos \left(\mathrm{Kvt}\right)\end{array}\right.$

Put $\mathrm{F}(\mathrm{x},\mathrm{y})=\mathrm{X}\left(\mathrm{x}\right)\mathrm{Y}\left(\mathrm{y}\right)$and divide by$\mathrm{X}\left(\mathrm{x}\right)\mathrm{Y}\left(\mathrm{y}\right)$ to further separate the space equation.

$\frac{1}{\mathrm{X}}\frac{d^2\mathrm{X}}{{\mathrm{dx}}^2}+\frac{1}{\mathrm{Y}}\frac{{\mathrm{d}}^2\mathrm{Y}}{{\mathrm{dy}}^2}+{\mathrm{K}}^2=0$

Present the constant.

${\mathrm{K}}^2={\mathrm{k}}_{\mathrm{x}}^2+{\mathrm{k}}_2$

Use the boundary condition.

Write the equation.

$\left(\frac{1}{\mathrm{X}}\frac{{\mathrm{d}}^2\mathrm{X}}{{\mathrm{dx}}^2}+{\mathrm{k}}_{\mathrm{x}}^2\right)+\left(\frac{1}{\mathrm{Y}}\frac{d^{2}Y}{d{y}^2}+k_y^2\right)=0$

The solutions are the trigonometric solutions.
$$\begin{gathered} \mathrm{X}\left(\mathrm{x}\right)=\left\{\begin{array}{l}\cos \left({\mathrm{k}}_{\mathrm{x}}\mathrm{x}\right)\\ \sin \left({\mathrm{k}}_{\mathrm{x}}\mathrm{x}\right)\end{array}\right. \\ \mathrm{Y}\left(\mathrm{y}\right)=\left\{\begin{array}{l}\cos \left({\mathrm{k}}_{\mathrm{y}}\mathrm{y}\right)\\ \sin \left({\mathrm{k}}_{\mathrm{y}}\mathrm{y}\right)\end{array}\right. \end{gathered}$$

On the boundary so use the boundary condition.
$$\begin{gathered} \mathrm{Z}(0,\mathrm{y},\mathrm{t})=0 \\ \mathrm{Z}(\mathrm{l},\mathrm{y},\mathrm{t})=0 \\ \mathrm{Z}(\mathrm{x},0,\mathrm{t})=0 \\ \mathrm{Z}(\mathrm{x},\mathrm{l},\mathrm{t})=0 \end{gathered}$$

$$\begin{gathered} \mathrm{X}\left(0\right)=0 \\ \Rightarrow 0=\mathrm{Csin}\left({\mathrm{k}}_{\mathrm{x}}0\right)+\mathrm{D}\cos \left({\mathrm{k}}_{\mathrm{x}}0\right) \end{gathered}$$

$\mathrm{D}=0$
$$\begin{gathered} \mathrm{X}\left(\mathrm{a}\right)=0 \\ =\sin \left({\mathrm{k}}_{\mathrm{x}}\mathrm{a}\right) \end{gathered}$$

$$\begin{gathered} {\mathrm{k}}_{\mathrm{x}}\mathrm{l}=\mathrm{n\pi } \\ {\mathrm{k}}_{\mathrm{x}}=\frac{\mathrm{n\pi }}{\mathrm{l}} \end{gathered}$$

Solve further.

Repeat the same.

$$\begin{gathered} \mathrm{Y}\left(0\right)=0 \\ \Rightarrow 0=\mathrm{E}\sin \left({\mathrm{k}}_{\mathrm{y}}0\right)+\mathrm{F}\cos \left({\mathrm{k}}_{\mathrm{y}}0\right) \end{gathered}$$
$$\begin{gathered} \mathrm{Y}\left(\mathrm{b}\right)=0 \\ =\sin \left({\mathrm{k}}_{\mathrm{y}}\mathrm{b}\right) \end{gathered}$$
$$\begin{gathered} {\mathrm{k}}_{\mathrm{y}}\mathrm{l}=\mathrm{m\pi } \\ {\mathrm{k}}_{\mathrm{y}}=\frac{\mathrm{m\pi }}{\mathrm{l}} \end{gathered}$$
$$\begin{gathered} {\mathrm{K}}_{\mathrm{nm}}^2={\mathrm{k}}_{\mathrm{x}}^2+{\mathrm{k}}_{\mathrm{y}}^2 \\ ={\mathrm{\pi }}^2\left(\frac{{\mathrm{n}}^2}{{\mathrm{l}}^2}+\frac{{\mathrm{m}}^2}{{\mathrm{l}}^2}\right) \\ {\mathrm{\phi }}_{\mathrm{nm}}={\mathrm{K}}_{\mathrm{nm}}\mathrm{v} \\ =\frac{\mathrm{v\pi }}{\mathrm{l}}\sqrt{{\mathrm{n}}^2+{\mathrm{m}}^2} \end{gathered}$$
Write the solution.
$\mathrm{Z}(\mathrm{x},\mathrm{y},\mathrm{t})=\sum _{\mathrm{n}-1}^{\infty }\sum _{\mathrm{m}-1}^{\infty }\sin \left(\frac{\mathrm{m\pi }}{\mathrm{l}}\mathrm{y}\right)\sin \left(\frac{\mathrm{m\pi }}{\mathrm{l}}\mathrm{y}\right)\left({\mathrm{A}}_{\mathrm{nm}}\sin \left(\frac{\mathrm{t\pi v}}{\mathrm{l}}\sqrt{{\mathrm{n}}^2+{\mathrm{m}}^2}\right)+{\mathrm{B}}_{\mathrm{nm}}\cos \left(\frac{\mathrm{t\pi v}}{\mathrm{l}}\sqrt{{\mathrm{n}}^2+{\mathrm{m}}^2}\right)\right)$
At $\mathrm{t}=0$
$$\begin{gathered} \mathrm{Z}(\mathrm{x},\mathrm{y},0)=\mathrm{f}(\mathrm{x},\mathrm{y}) \\ =\mathrm{xy}(\mathrm{l}-\mathrm{x})(\mathrm{l}-\mathrm{y}) \\ =\sum _{\mathrm{n}=1}^{\infty }\sum _{\mathrm{m}=1}^{\infty }\sin \left(\frac{\mathrm{n\pi }}{\mathrm{l}}\mathrm{x}\right)\sin \left(\frac{\mathrm{m\pi }}{\mathrm{l}}\mathrm{y}\right){\mathrm{B}}_{\mathrm{nm}} \end{gathered}$$

It can be seen that so only the cosine terms remain.
There is a double Fourier series.|
The${\mathrm{A}}_{\mathrm{nm}}$ coefficient are calculated almost the same as in the one dimensional case.
$$\begin{gathered} {\mathrm{A}}_{\mathrm{nm}}=\frac{2}{\mathrm{l}}\frac{2}{\mathrm{l}}{\int }_0^{\mathrm{l}}\sin \left(\frac{\mathrm{n\pi }}{\mathrm{l}}\mathrm{x}\right)\mathrm{dx}{\int }_0^{\mathrm{l}}\mathrm{f}(\mathrm{x},\mathrm{y})\sin \left(\frac{\mathrm{m\pi }}{\mathrm{l}}\mathrm{y}\right)\mathrm{dy} \\ =\frac{4}{{\mathrm{l}}^2}{\int }_0^{\mathrm{l}}\mathrm{x}(\mathrm{x}-\mathrm{l})\sin \left(\frac{\mathrm{n\pi }}{\mathrm{l}}\mathrm{x}\right)\mathrm{dx}{\int }_0^{\mathrm{l}}\mathrm{y}(\mathrm{y}-\mathrm{l})\sin \left(\frac{\mathrm{m\pi }}{\mathrm{l}}\mathrm{y}\right)\mathrm{dy} \\ {\mathrm{A}}_{\mathrm{nm}}=\frac{64{\mathrm{l}}^4}{{\mathrm{\pi }}^6{\mathrm{n}}^3{\mathrm{m}}^3};\mathrm{n},\mathrm{m}=1,3,5...... \end{gathered}$$

Hence the final result.
$\mathrm{Z}(\mathrm{x},\mathrm{y},\mathrm{t})=\sum _{\mathrm{n}=1,3,5...}^{\infty }\sum _{\mathrm{m}=1,3,5...}^{\infty }\frac{64{\mathrm{l}}^4}{{\mathrm{\pi }}^6{\mathrm{n}}^3{\mathrm{m}}^3}\sin \left(\frac{\mathrm{n\pi }}{\mathrm{l}}\mathrm{x}\right)\sin \left(\frac{\mathrm{m\pi }}{\mathrm{l}}\mathrm{y}\right)\cos \left(\frac{\mathrm{t\pi v}}{\mathrm{l}}\sqrt{{\mathrm{n}}^2+{\mathrm{m}}^2}\right)$