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