#### 2.4.4 circular membrane, with circular symmetry

In this case, there is no $$\theta$$ dependency in boundary conditions or in initial conditions. Using polar coordinates. Disk has radius $$a$$. Wave displacement is $$u\equiv u\left ( r,t\right )$$ (out of page). \begin{align*} \frac{\partial ^{2}u\left ( r,\theta ,t\right ) }{\partial t^{2}} & =c^{2}\left ( \frac{\partial ^{2}u}{\partial r^{2}}+\frac{1}{r}\frac{\partial u}{\partial r}\right ) \\ 0 & <r<a \end{align*}

Boundary conditions on $$r$$\begin{align*} \left \vert u\left ( 0,t\right ) \right \vert & <\infty \\ u\left ( a,t\right ) & =0 \end{align*}

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

Solution

Let $$u=T\left ( t\right ) R\left ( r\right )$$. Plug in the PDE$\frac{1}{c^{2}}T^{\prime \prime }R=R^{\prime \prime }T+\frac{1}{r}R^{\prime }T$ Dividing by $$RT$$$\frac{1}{c^{2}}\frac{T^{\prime \prime }}{T}=\frac{R^{\prime \prime }}{R}+\frac{1}{r}\frac{R^{\prime }}{R}$ Hence\begin{align*} \frac{1}{c^{2}}\frac{T^{\prime \prime }}{T} & =-\lambda \\ \frac{R^{\prime \prime }}{R}+\frac{1}{r}\frac{R^{\prime }}{R} & =-\lambda \end{align*}

The time ODE is$T^{\prime \prime }+c^{2}\lambda T=0$ And the $$r$$ ODE is (Sturm-Liouville)

$rR^{\prime \prime }+R^{\prime }+\lambda rR=0$ Where $$p=r,q=0,\sigma =r$$. This is singular SL.  The solution turns out to be  $R_{n}\left ( r\right ) =A_{n}J_{0}\left ( \sqrt{\lambda _{n}}r\right ) \qquad n=1,2,3,\cdots$ Where $$\lambda _{n}$$ is found from roots of $$0=J_{n}\left ( \sqrt{\lambda _{n}}a\right )$$ giving the eigenvalues. Now the time ODE is solved\begin{align*} T_{n}^{\prime \prime }+c^{2}\lambda _{n}T_{n} & =0\\ T_{n} & =B_{n}\cos \left ( c\sqrt{\lambda _{n}}t\right ) +C_{n}\sin \left ( c\sqrt{\lambda _{n}}t\right ) \qquad n=1,2,3,\ldots , \end{align*}

Hence the solution is\begin{align} u\left ( r,t\right ) & =\sum _{n=1}^{\infty }T_{n}R_{n}\nonumber \\ & =\sum _{n=1}^{\infty }A_{n}\cos \left ( c\sqrt{\lambda _{n}}t\right ) J_{0}\left ( \sqrt{\lambda _{n}}r\right ) +B_{n}\sin \left ( c\sqrt{\lambda _{n}}t\right ) J_{0}\left ( \sqrt{\lambda _{n}}r\right ) \tag{1} \end{align}

Now initial conditions $$u\left ( r,0\right ) =f\left ( r\right )$$ is used to ﬁnd $$A_{n}\,$$using orthogonality. At $$t=0$$ the solution simpliﬁes to $u\left ( r,0\right ) =\sum _{n=1}^{\infty }A_{n}J_{0}\left ( \sqrt{\lambda _{n}}r\right )$ Hence\begin{align*} f\left ( r\right ) & =\sum _{n=1}^{\infty }A_{n}J_{0}\left ( \sqrt{\lambda _{n}}r\right ) \\ \int _{0}^{a}f\left ( r\right ) J_{0}\left ( \sqrt{\lambda _{n}}r\right ) rdr & =A_{n}\int _{0}^{a}J_{0}^{2}\left ( \sqrt{\lambda _{n}}r\right ) rdr\\ A_{n} & =\frac{\int _{0}^{a}f\left ( r\right ) J_{0}\left ( \sqrt{\lambda _{n}}r\right ) rdr}{\int _{0}^{a}J_{0}^{2}\left ( \sqrt{\lambda _{n}}r\right ) rdr} \end{align*}

Now we will look at the second initial conditions $$\frac{\partial u}{\partial t}\left ( r,0\right ) =g\left ( r\right ) .$$ Taking derivative w.r.t. time $$t$$ of the solution in (1) gives$\frac{\partial u}{\partial t}\left ( r,t\right ) =\sum _{n=1}^{\infty }-c\sqrt{\lambda _{n}}A_{n}\sin \left ( c\sqrt{\lambda _{n}}t\right ) J_{0}\left ( \sqrt{\lambda _{n}}r\right ) +B_{n}c\sqrt{\lambda _{n}}\cos \left ( c\sqrt{\lambda _{n}}t\right ) J_{0}\left ( \sqrt{\lambda _{n}}r\right )$ At time $$t=0$$ the above becomes $g\left ( r\right ) =\sum _{n=1}^{\infty }B_{n}c\sqrt{\lambda _{n}}J_{0}\left ( \sqrt{\lambda _{n}}r\right )$ Now orthogonality is used. The above becomes$B_{n}=\frac{\int _{0}^{a}g\left ( r\right ) J_{0}\left ( \sqrt{\lambda _{n}}r\right ) rdr}{c\sqrt{\lambda _{n}}\int _{0}^{a}J_{0}^{2}\left ( \sqrt{\lambda _{n}}r\right ) rdr}$ Summary of solution$u\left ( r,t\right ) =\sum _{n=1}^{\infty }A_{n}\cos \left ( c\sqrt{\lambda _{n}}t\right ) J_{0}\left ( \sqrt{\lambda _{n}}r\right ) +B_{n}\sin \left ( c\sqrt{\lambda _{n}}t\right ) J_{0}\left ( \sqrt{\lambda _{n}}r\right )$$A_{n}=\frac{\int _{0}^{a}f\left ( r\right ) J_{0}\left ( \sqrt{\lambda _{n}}r\right ) rdr}{\int _{0}^{a}J_{0}^{2}\left ( \sqrt{\lambda _{n}}r\right ) rdr}$$B_{n}=\frac{\int _{0}^{a}g\left ( r\right ) J_{0}\left ( \sqrt{\lambda _{n}}r\right ) rdr}{c\sqrt{\lambda _{n}}\int _{0}^{a}J_{0}^{2}\left ( \sqrt{\lambda _{n}}r\right ) rdr}$ With $$\lambda _{n}$$ being the solutions for $$0=J_{0}\left ( \sqrt{\lambda _{n}}a\right )$$. We have inﬁnite number of zeros. This generates all the needed $$\lambda _{n}$$. Hence $$\sqrt{\lambda _{n}}a=BesselJZero\left ( 0,n\right )$$, therefore $$\sqrt{\lambda _{n}}=\frac{a}{BesselJZero\left ( 0,n\right ) }$$

Animation

This animation runs for 40 seconds. Using radius $$a=1$$ and zero initial velocity. Initial position is $$u\left ( r,0\right ) =f\left ( r\right ) =r.$$ And $$c=0.2$$. Number of terms in the sum used was $$30$$. Which means $$u\left ( r,t\right ) =\sum _{n=1}^{30}A_{n}\cos \left ( c\sqrt{\lambda _{n}}t\right ) J_{0}\left ( \sqrt{\lambda _{n}}r\right )$$. The $$B_{n}$$ terms are all zero since initial velocity $$g\left ( r\right )$$ was zero.

Source code for all the above animation