2.4.1 Rectangular membrane. Fixed on all edges

pict

Using Cartesian coordinates. Wave displacement is \(u\equiv u\left ( x,y,t\right ) \) (out of page).\begin{align*} \frac{\partial ^{2}u\left ( x,y,t\right ) }{\partial t^{2}} & =c^{2}\left ( \frac{\partial ^{2}u}{\partial x^{2}}+\frac{\partial ^{2}u}{\partial y^{2}}\right ) \\ 0 & <x<L\\ 0 & <y<H \end{align*}

Boundary conditions on \(x\)\begin{align*} u\left ( 0,y,t\right ) & =0\\ u\left ( L,y,t\right ) & =0 \end{align*}

And boundary conditions on \(y\)\begin{align*} u\left ( x,0,t\right ) & =0\\ u\left ( x,H,t\right ) & =0 \end{align*}

Initial conditions\begin{align*} u\left ( x,y,0\right ) & =f\left ( x,y\right ) \\ \frac{\partial u}{\partial t}\left ( x,y,0\right ) & =g\left ( x,y\right ) \end{align*}

Solution

Let \(u=X\left ( x\right ) Y\left ( y\right ) T\left ( t\right ) \). Plug in the PDE\begin{align*} \frac{1}{c^{2}}T^{\prime \prime }XY & =X^{\prime \prime }YT+Y^{\prime \prime }XT\\ \frac{1}{c^{2}}\frac{T^{\prime \prime }}{T} & =\frac{X^{\prime \prime }}{X}+\frac{Y^{\prime \prime }}{Y} \end{align*}

Hence\begin{align*} \frac{1}{c^{2}}\frac{T^{\prime \prime }}{T} & =-\lambda \\ \frac{X^{\prime \prime }}{X}+\frac{Y^{\prime \prime }}{Y} & =-\lambda \end{align*}

Hence time ODE becomes\[ T^{\prime \prime }+c^{2}\lambda T=0 \] And the space ODE is\[ \frac{X^{\prime \prime }}{X}+\frac{Y^{\prime \prime }}{Y}=-\lambda \] Separating this again\[ \frac{X^{\prime \prime }}{X}=-\lambda -\frac{Y^{\prime \prime }}{Y}\] Using new separation variable \(\mu \), we obtain two new ODE’s\begin{align*} \frac{X^{\prime \prime }}{X} & =-\mu \\ -\lambda -\frac{Y^{\prime \prime }}{Y} & =-\mu \end{align*}

Or\begin{align*} X^{\prime \prime }+\mu X & =0\\ Y^{\prime \prime }+Y\left ( \lambda -\mu \right ) & =0 \end{align*}

Solving for \(X\) ODE, and knowing that \(\mu >0\) from earlier, we obtain\[ X=A\cos \left ( \sqrt{\mu }x\right ) +B\sin \left ( \sqrt{\mu }x\right ) \] Applying B.C. at \(x=0\)\[ 0=A \] Hence \(X\left ( x\right ) =B\sin \left ( \sqrt{\mu }x\right ) \). Applying B.C. at \(x=L\) \[ 0=B\sin \left ( \sqrt{\mu }L\right ) \] Hence \begin{align*} \sqrt{\mu }L & =n\pi \\ \mu & =\left ( \frac{n\pi }{L}\right ) ^{2}\qquad n=1,2,3,\cdots \end{align*}

Therefore the \(X\) solution is\[ X_{n}\left ( x\right ) =B_{n}\sin \left ( \left ( \frac{n\pi }{L}\right ) x\right ) \qquad n=1,2,3,\cdots \] Solving the \(Y\) ODE\[ Y^{\prime \prime }+Y\left ( \lambda -\left ( \frac{n\pi }{L}\right ) ^{2}\right ) =0 \] The solution is\[ Y=A\cos \left ( \sqrt{\lambda -\left ( \frac{n\pi }{L}\right ) ^{2}}y\right ) +B\sin \left ( \sqrt{\lambda -\left ( \frac{n\pi }{L}\right ) ^{2}}y\right ) \] Applying first B.C. \[ 0=A \] Hence\[ Y=B\sin \left ( \sqrt{\lambda -\left ( \frac{n\pi }{L}\right ) ^{2}}y\right ) \] Applying second B.C.\[ 0=B\sin \left ( \sqrt{\lambda -\left ( \frac{n\pi }{L}\right ) ^{2}}H\right ) \] Hence\begin{align*} \sqrt{\lambda -\left ( \frac{n\pi }{L}\right ) ^{2}}H & =m\pi \qquad m=1,2,3,\cdots \\ \lambda _{nm}-\left ( \frac{n\pi }{L}\right ) ^{2} & =\left ( \frac{m\pi }{H}\right ) ^{2}\\ \lambda _{nm} & =\left ( \frac{m\pi }{H}\right ) ^{2}+\left ( \frac{n\pi }{L}\right ) ^{2}\qquad n=1,2,3,\cdots ,m=1,2,3,\cdots \end{align*}

Hence the \(Y\) solution is\[ Y_{nm}=B_{nm}\sin \left ( \frac{m\pi }{H}y\right ) \qquad n=1,2,3,\cdots ,m=1,2,3,\cdots \] Now we solve the time \(T\) ode\begin{align*} T_{nm}^{\prime \prime }+c^{2}\lambda _{nm}T_{nm} & =0\\ T_{nm}\left ( t\right ) & =A_{nm}\cos \left ( c\sqrt{\lambda _{nm}}t\right ) +B_{nm}\sin \left ( c\sqrt{\lambda _{nm}}t\right ) \end{align*}

Combining all solution , and merging all constants into two, we find\begin{align} u_{nm}\left ( x,y,t\right ) & =X_{n}\left ( x\right ) Y_{nm}\left ( y\right ) T_{nm}\left ( t\right ) \nonumber \\ u\left ( x,y,t\right ) & =\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }X_{m}\left ( x\right ) Y_{mn}\left ( y\right ) T_{mn}\left ( t\right ) \nonumber \\ & =\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }A_{nm}\sin \left ( \left ( \frac{n\pi }{L}\right ) x\right ) \sin \left ( \frac{m\pi }{H}y\right ) \cos \left ( c\sqrt{\lambda _{nm}}t\right ) \nonumber \\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }B_{nm}\sin \left ( \left ( \frac{n\pi }{L}\right ) x\right ) \sin \left ( \frac{m\pi }{H}y\right ) \sin \left ( c\sqrt{\lambda _{nm}}t\right ) \tag{1} \end{align}

We now use initial conditions to find \(A_{nm},B_{nm}\). At \(t=0\)\begin{align*} u\left ( x,y,0\right ) & =f\left ( x,y\right ) \\ \frac{\partial u}{\partial t}\left ( x,y,0\right ) & =g\left ( x,y\right ) \end{align*}

Applying first initial condition to (1)  gives\[ f\left ( x,y\right ) =\sum _{n=1}^{\infty }\left ( \sum _{m=1}^{\infty }A_{nm}\sin \left ( \frac{m\pi }{H}y\right ) \right ) \sin \left ( \left ( \frac{n\pi }{L}\right ) x\right ) \] Applying 2D orthogonality gives\begin{align*} \int _{0}^{L}\int _{0}^{H}f\left ( x,y\right ) \sin \left ( \left ( \frac{n\pi }{L}\right ) x\right ) \sin \left ( \frac{m\pi }{H}y\right ) \ dxdy & =A_{nm}\left ( \frac{L}{2}\right ) \left ( \frac{H}{2}\right ) \\ A_{nm} & =\frac{4}{LH}\int _{0}^{L}\int _{0}^{H}f\left ( x,y\right ) \sin \left ( \left ( \frac{n\pi }{L}\right ) x\right ) \sin \left ( \frac{m\pi }{H}y\right ) \ dxdy \end{align*}

Taking time derivative of (1)\begin{align*} \frac{\partial u}{\partial t}\left ( x,y,t\right ) & =\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }-c\sqrt{\lambda _{nm}}A_{nm}\sin \left ( \left ( \frac{n\pi }{L}\right ) x\right ) \sin \left ( \frac{m\pi }{H}y\right ) \sin \left ( c\sqrt{\lambda _{nm}}t\right ) \\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }c\sqrt{\lambda _{nm}}B_{nm}\sin \left ( \left ( \frac{n\pi }{L}\right ) x\right ) \sin \left ( \frac{m\pi }{H}y\right ) \cos \left ( c\sqrt{\lambda _{nm}}t\right ) \end{align*}

AT \(t=0\) the above becomes\[ g\left ( x,y\right ) =\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }c\sqrt{\lambda _{nm}}B_{nm}\sin \left ( \left ( \frac{n\pi }{L}\right ) x\right ) \sin \left ( \frac{m\pi }{H}y\right ) \] Applying 2D orthogonality gives\begin{align*} \int _{0}^{L}\int _{0}^{H}g\left ( x,y\right ) \sin \left ( \left ( \frac{n\pi }{L}\right ) x\right ) \sin \left ( \frac{m\pi }{H}y\right ) \ dxdy & =B_{nm}\left ( \frac{L}{2}\right ) \left ( \frac{H}{2}\right ) \\ B_{nm} & =\frac{4}{LH}\int _{0}^{L}\int _{0}^{H}g\left ( x,y\right ) \sin \left ( \left ( \frac{n\pi }{L}\right ) x\right ) \sin \left ( \frac{m\pi }{H}y\right ) \ dxdy \end{align*}

Summary of solution\begin{align*} u\left ( x,y,t\right ) & =\sum _{n=1}^{\infty }\left ( \sum _{m=1}^{\infty }A_{nm}\sin \left ( \frac{m\pi }{H}y\right ) \cos \left ( c\sqrt{\lambda _{nm}}t\right ) \right ) \sin \left ( \left ( \frac{n\pi }{L}\right ) x\right ) \\ & +\left ( \sum _{n=1}^{\infty }\sum _{m=1}^{\infty }B_{nm}\sin \left ( \frac{m\pi }{H}y\right ) \sin \left ( c\sqrt{\lambda _{nm}}t\right ) \right ) \sin \left ( \left ( \frac{n\pi }{L}\right ) x\right ) \\ A_{nm} & =\frac{4}{LH}\int _{0}^{L}\int _{0}^{H}f\left ( x,y\right ) \sin \left ( \left ( \frac{n\pi }{L}\right ) x\right ) \sin \left ( \frac{m\pi }{H}y\right ) \ dxdy\\ B_{nm} & =\frac{4}{LH}\int _{0}^{L}\int _{0}^{H}g\left ( x,y\right ) \sin \left ( \left ( \frac{n\pi }{L}\right ) x\right ) \sin \left ( \frac{m\pi }{H}y\right ) \ dxdy\\ \lambda _{nm} & =\left ( \frac{m\pi }{H}\right ) ^{2}+\left ( \frac{n\pi }{L}\right ) ^{2} \end{align*}

Animation

This first animation was done for sum first \(15\) terms in the double series above and using \(L=1,H=2,c=0.1\). Using \(u\left ( x,y,0\right ) =f\left ( x,y\right ) =x\cos \left ( y\right ) \) and \(\frac{\partial }{\partial t}u\left ( x,y,0\right ) =g\left ( x,y\right ) =0\). This one is done in Mathematica

Here is the Mathematica source code for the above

The following is the animation done in Maple

Here is the Maple source code for the above

The following additional animations, all using \(L=1,H=2,c=0.1\). All using Using \(u\left ( x,y,0\right ) =f\left ( x,y\right ) =x\cos \left ( y\right ) \) and \(\frac{\partial }{\partial t}u\left ( x,y,0\right ) =g\left ( x,y\right ) =0\). For different number of terms in the series. Number of terms, or the mode \(n,m\) is given below each animation. Boundary conditions zero on all 4 edges.

Source code for all the above animations