2.4.2 circular membrane, no circular symmetry, example 1

   2.4.2.1 Animations

pict

Using polar coordinates. Disk has radius \(a\). Wave displacement is \(u\equiv u\left ( r,\theta ,t\right ) \) (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}+\frac{1}{r^{2}}\frac{\partial ^{2}u}{\partial \theta ^{2}}\right ) \\ 0 & <r<a\\ -\pi & <\theta <\pi \end{align*}

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

And boundary conditions on \(\theta \)\begin{align*} u\left ( r,-\pi ,t\right ) & =u\left ( r,\pi ,t\right ) \\ \frac{\partial u\left ( r,-\pi ,t\right ) }{\partial \theta } & =\frac{\partial u\left ( r,\pi ,t\right ) }{\partial \theta } \end{align*}

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

Solution

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

The time ODE is\[ T^{\prime \prime }+c^{2}\lambda T=0 \] Now we separate again the space ODE’s (remember to move the \(\lambda \) with the \(R\) and not the \(\Theta \))\begin{align*} \frac{R^{\prime \prime }}{R}+\frac{1}{r}\frac{R^{\prime }}{R}+\lambda & =-\frac{1}{r^{2}}\frac{\Theta ^{\prime \prime }}{\Theta }\\ r^{2}\frac{R^{\prime \prime }}{R}+r\frac{R^{\prime }}{R}+r^{2}\lambda & =-\frac{\Theta ^{\prime \prime }}{\Theta } \end{align*}

Let separation constant be \(\mu \), therefore\begin{align*} -\frac{\Theta ^{\prime \prime }}{\Theta } & =\mu \\ \Theta ^{\prime \prime }+\mu \Theta & =0 \end{align*}

With periodic boundary conditions and\begin{align*} r^{2}\frac{R^{\prime \prime }}{R}+r\frac{R^{\prime }}{R}+r^{2}\lambda & =\mu \\ r^{2}R^{\prime \prime }+rR^{\prime }+\lambda r^{2}R-\mu R & =0\\ rR^{\prime \prime }+R^{\prime }-\frac{\mu }{r}R & =-\lambda rR \end{align*}

Now it is in SL form, where \(p=r,q=-\frac{\mu }{r},\sigma =r\). This is singular SL. Can be written as\[ R^{\prime \prime }+\frac{1}{r}R^{\prime }+\left ( \lambda -\frac{\mu }{r^{2}}\right ) R=0 \] Before we solve the above \(R\) ODE, we solve the \(\Theta ^{\prime \prime }+\mu \Theta =0\) to find \(\mu \) Eigenvalues. The solution is\[ \Theta =A\cos \left ( \sqrt{\mu }\theta \right ) +B\sin \left ( \sqrt{\mu }\theta \right ) \] With B.C \(\Theta \left ( -\pi \right ) =\Theta \left ( \pi \right ) \) and \(\Theta ^{\prime }\left ( -\pi \right ) =\Theta ^{\prime }\left ( \pi \right ) \). From first B.C. we obtain\begin{align} A\cos \left ( \sqrt{\mu }\pi \right ) -B\sin \left ( \sqrt{\mu }\pi \right ) & =A\cos \left ( \sqrt{\mu }\pi \right ) +B\sin \left ( \sqrt{\mu }\pi \right ) \nonumber \\ 2B\sin \left ( \sqrt{\mu }\pi \right ) & =0 \tag{1} \end{align}

Looking at second B.C. \(\Theta ^{\prime }\left ( -\pi \right ) =\Theta ^{\prime }\left ( \pi \right ) \)\[ \Theta ^{\prime }\left ( \theta \right ) =-A\sqrt{\mu }\sin \left ( \sqrt{\mu }\theta \right ) +\sqrt{\mu }B\cos \left ( \sqrt{\mu }\theta \right ) \] Hence\begin{align} A\sqrt{\mu }\sin \left ( \sqrt{\mu }\pi \right ) +\sqrt{\mu }B\cos \left ( \sqrt{\mu }\pi \right ) & =-A\sqrt{\mu }\sin \left ( \sqrt{\mu }\pi \right ) +\sqrt{\mu }B\cos \left ( \sqrt{\mu }\pi \right ) \nonumber \\ A\sqrt{\mu }\sin \left ( \sqrt{\mu }\pi \right ) & =-A\sqrt{\mu }\sin \left ( \sqrt{\mu }\pi \right ) \nonumber \\ 2A\sin \left ( \sqrt{\mu }\pi \right ) & =0 \tag{2} \end{align}

From (1,2), we see that both are satisfied if \begin{align*} \sqrt{\mu }\pi & =n\pi \qquad n=1,2,3,\ldots \\ \mu & =n^{2} \end{align*}

Hence \[ \Theta _{n}=A_{n}\cos \left ( n\theta \right ) +B_{n}\sin \left ( n\theta \right ) \] There is another solution for \(\mu =0\) which is constant (that is why one of the sums below starts from \(n=0\)). We can combine the zero eigenvalue with the above and write\[ \Theta _{n}=A_{n}\cos \left ( n\theta \right ) +B_{n}\sin \left ( n\theta \right ) \qquad n=0,1,2,3,\ldots \] Since at \(n=0\) the above reduces to constant \(A_{0}\).

Now that we know \(\mu _{n}=n^{2}\), from solving the \(\theta \) part, we go and solve the \(r\) ODE. For each \(n\), the solution to the \(r\) (Bessel) ode

\[ R^{\prime \prime }+\frac{1}{r}R^{\prime }+\left ( \lambda -\frac{n^{2}}{r^{2}}\right ) R=0 \] The solution turns out to be  \[ R_{nm}\left ( r\right ) =J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) \qquad m=1,2,3,\cdots \] Where \(\lambda _{nm}\) is found from roots of \(0=J_{n}\left ( \sqrt{\lambda _{nm}}a\right ) \) giving the eigenvalues. Now the time ODE is solved\begin{align*} T_{nm}^{\prime \prime }+c^{2}\lambda _{nm}T_{nm} & =0\\ T_{nm} & =C_{nm}\cos \left ( c\sqrt{\lambda _{nm}}t\right ) +D_{nm}\sin \left ( c\sqrt{\lambda _{nm}}t\right ) \qquad n=0,1,2,3,\ldots ,m=1,2,3,\cdots \end{align*}

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

We now break this sum as follows\begin{align*} u\left ( r,\theta ,t\right ) & =\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }\left ( C_{nm}\cos \left ( c\sqrt{\lambda _{nm}}t\right ) +D_{nm}\sin \left ( c\sqrt{\lambda _{nm}}t\right ) \right ) J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) A_{n}\cos \left ( n\theta \right ) \\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }\left ( C_{nm}\cos \left ( c\sqrt{\lambda _{nm}}t\right ) +D_{nm}\sin \left ( c\sqrt{\lambda _{nm}}t\right ) \right ) J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) B_{n}\sin \left ( n\theta \right ) \end{align*}

Or\begin{align*} u\left ( r,\theta ,t\right ) & =\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }C_{nm}\cos \left ( c\sqrt{\lambda _{nm}}t\right ) J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) A_{n}\cos \left ( n\theta \right ) +D_{nm}\sin \left ( c\sqrt{\lambda _{nm}}t\right ) J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) A_{n}\cos \left ( n\theta \right ) \\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }C_{nm}\cos \left ( c\sqrt{\lambda _{nm}}t\right ) J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) B_{n}\sin \left ( n\theta \right ) +D_{nm}\sin \left ( c\sqrt{\lambda _{nm}}t\right ) J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) B_{n}\sin \left ( n\theta \right ) \end{align*}

Then we break the above into 4 sums\begin{align*} u\left ( r,\theta ,t\right ) & =\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }C_{nm}\cos \left ( c\sqrt{\lambda _{nm}}t\right ) J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) A_{n}\cos \left ( n\theta \right ) \\ & +\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }D_{nm}\sin \left ( c\sqrt{\lambda _{nm}}t\right ) J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) A_{n}\cos \left ( n\theta \right ) \\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }C_{nm}\cos \left ( c\sqrt{\lambda _{nm}}t\right ) J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) B_{n}\sin \left ( n\theta \right ) \\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }D_{nm}\sin \left ( c\sqrt{\lambda _{nm}}t\right ) J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) B_{n}\sin \left ( n\theta \right ) \end{align*}

Finally, we merge constants in the above as follows\begin{align*} A_{n}C_{nm} & \equiv A_{nm}\\ A_{n}D_{nm} & \equiv B_{nm}\\ B_{n}C_{nm} & \equiv C_{nm}\\ B_{n}D_{nm} & \equiv D_{nm} \end{align*}

Hence the final solution now becomes\begin{align} u\left ( r,\theta ,t\right ) & =\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }A_{nm}\cos \left ( c\sqrt{\lambda _{nm}}t\right ) J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) \cos \left ( n\theta \right ) \nonumber \\ & +\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }B_{nm}\sin \left ( c\sqrt{\lambda _{nm}}t\right ) J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) \cos \left ( n\theta \right ) \nonumber \\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }C_{nm}\cos \left ( c\sqrt{\lambda _{nm}}t\right ) J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) \sin \left ( n\theta \right ) \nonumber \\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }D_{nm}\sin \left ( c\sqrt{\lambda _{nm}}t\right ) J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) \sin \left ( n\theta \right ) \tag{3} \end{align}

Now initial conditions \(u\left ( r,\theta ,0\right ) =f\left ( r,\theta \right ) \) is used to find \(A_{nm},C_{nm}\,\)using orthogonality. At \(t=0\) the solution simplifies to (all terms with \(\sin \left ( c\sqrt{\lambda _{nm}}t\right ) \) vanish giving\begin{align*} u\left ( r,\theta ,t\right ) & =\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }A_{nm}J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) \cos \left ( n\theta \right ) \\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }C_{nm}J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) \sin \left ( n\theta \right ) \end{align*}

Hence\begin{equation} f\left ( r,\theta \right ) =\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }A_{nm}J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) \cos \left ( n\theta \right ) +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }C_{nm}J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) \sin \left ( n\theta \right ) \tag{4} \end{equation} When iterating over \(m\) index, the terms \(\cos \left ( n\theta \right ) \) and \(\sin \left ( n\theta \right ) \) will be constant. So for each \(n\), we have \(\sum _{m=1}^{\infty }A_{nm}J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) \) and \(\sum _{m=1}^{\infty }C_{nm}J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) \). So orthogonality is carried out on the \(m\) index on the Bessel functions. Multiplying (4) by \(J_{n}\left ( \sqrt{\lambda _{nk}}r\right ) \) and integrating\begin{align*} \int _{0}^{a}f\left ( r,\theta \right ) J_{n}\left ( \sqrt{\lambda _{nk}}r\right ) rdr & =\sum _{n=0}^{\infty }\left ( \int _{0}^{a}\sum _{m=1}^{\infty }A_{nm}J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) J_{n}\left ( \sqrt{\lambda _{nk}}r\right ) rdr\right ) \cos \left ( n\theta \right ) \\ & +\sum _{n=1}^{\infty }\left ( \int _{0}^{a}\sum _{m=1}^{\infty }C_{nm}J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) J_{n}\left ( \sqrt{\lambda _{nk}}r\right ) rdr\right ) \sin \left ( n\theta \right ) \end{align*}

Or\begin{align*} \int _{0}^{a}f\left ( r,\theta \right ) J_{n}\left ( \sqrt{\lambda _{nk}}r\right ) rdr & =\sum _{n=0}^{\infty }A_{nk}\left ( \int _{0}^{a}J_{n}^{2}\left ( \sqrt{\lambda _{nk}}r\right ) rdr\right ) \cos \left ( n\theta \right ) \\ & +\sum _{n=1}^{\infty }C_{nk}\left ( \int _{0}^{a}J_{n}^{2}\left ( \sqrt{\lambda _{nk}}r\right ) rdr\right ) \sin \left ( n\theta \right ) \end{align*}

Replacing \(k\) back with \(m\), the above becomes\begin{align} \int _{0}^{a}f\left ( r,\theta \right ) J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) rdr & =\sum _{n=0}^{\infty }A_{nm}\left ( \int _{0}^{a}J_{n}^{2}\left ( \sqrt{\lambda _{nm}}r\right ) rdr\right ) \cos \left ( n\theta \right ) \nonumber \\ & +\sum _{n=1}^{\infty }C_{nm}\left ( \int _{0}^{a}J_{n}^{2}\left ( \sqrt{\lambda _{nm}}r\right ) rdr\right ) \sin \left ( n\theta \right ) \tag{5} \end{align}

We now apply orthogonality on \(n\) using the \(\cos \left ( n\theta \right ) \) results in\[ \int _{-\pi }^{\pi }\left ( \int _{0}^{a}f\left ( r,\theta \right ) J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) r\ dr\right ) \cos \left ( n\theta \right ) d\theta =A_{nm}\int _{-\pi }^{\pi }\left ( \int _{0}^{a}J_{n}^{2}\left ( \sqrt{\lambda _{nm}}r\right ) rdr\right ) \cos ^{2}\left ( n\theta \right ) d\theta \] But \(\int _{0}^{a}J_{n}^{2}\left ( \sqrt{\lambda _{nm}}r\right ) rdr\) does not depend on \(\theta \), therefore the above becomes\begin{align*} \int _{-\pi }^{\pi }\left ( \int _{0}^{a}f\left ( r,\theta \right ) J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) r\ dr\right ) \cos \left ( n\theta \right ) d\theta & =A_{nm}\left ( \int _{0}^{a}J_{n}^{2}\left ( \sqrt{\lambda _{nm}}r\right ) rdr\right ) \int _{-\pi }^{\pi }\cos ^{2}\left ( n\theta \right ) d\theta \\ & =A_{nm}\pi \int _{0}^{a}J_{n}^{2}\left ( \sqrt{\lambda _{nm}}r\right ) rdr \end{align*}

Therefore\[ A_{nm}=\frac{\int _{-\pi }^{\pi }\left ( \int _{0}^{a}f\left ( r,\theta \right ) J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) r\ dr\right ) \cos \left ( n\theta \right ) d\theta }{\pi \int _{0}^{a}J_{n}^{2}\left ( \sqrt{\lambda _{nm}}r\right ) rdr}\] Similarly for \(\sin \left ( n\theta \right ) \), which gives\[ C_{nm}=\frac{\int _{-\pi }^{\pi }\left ( \int _{0}^{a}f\left ( r,\theta \right ) J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) r\ dr\right ) \sin \left ( n\theta \right ) d\theta }{\pi \int _{0}^{a}J_{n}^{2}\left ( \sqrt{\lambda _{nm}}r\right ) rdr}\] Now we will look at the second initial conditions \(\frac{\partial u}{\partial t}\left ( r,\theta ,0\right ) =g\left ( r,\theta \right ) .\) Taking derivative w.r.t. time \(t\) of the solution in (3) gives\begin{align*} \frac{\partial u}{\partial t}\left ( r,\theta ,t\right ) & =\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }-c\sqrt{\lambda _{nm}}A_{nm}\sin \left ( c\sqrt{\lambda _{nm}}t\right ) J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) \cos \left ( n\theta \right ) \\ & +\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }c\sqrt{\lambda _{nm}}B_{nm}\cos \left ( c\sqrt{\lambda _{nm}}t\right ) J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) \cos \left ( n\theta \right ) \\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }-c\sqrt{\lambda _{nm}}C_{nm}\sin \left ( c\sqrt{\lambda _{nm}}t\right ) J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) \sin \left ( n\theta \right ) \\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }c\sqrt{\lambda _{nm}}D_{nm}\cos \left ( c\sqrt{\lambda _{nm}}t\right ) J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) \sin \left ( n\theta \right ) \end{align*}

At time \(t=0\) the above becomes (all terms with \(\sin \left ( c\sqrt{\lambda _{nm}}t\right ) \) vanish).\begin{align*} g\left ( r,\theta \right ) = & \sum _{n=0}^{\infty }\sum _{m=1}^{\infty }c\sqrt{\lambda _{nm}}B_{nm}\cos \left ( c\sqrt{\lambda _{nm}}t\right ) J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) \cos \left ( n\theta \right ) \\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }c\sqrt{\lambda _{nm}}D_{nm}\cos \left ( c\sqrt{\lambda _{nm}}t\right ) J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) \sin \left ( n\theta \right ) \end{align*}

Now orthogonality is used. At \(t=0\) the above becomes\begin{align*} g\left ( r,\theta \right ) & =\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }c\sqrt{\lambda _{nm}}B_{nm}J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) \cos \left ( n\theta \right ) \\ & +\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }c\sqrt{\lambda _{nm}}D_{nm}J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) \sin \left ( n\theta \right ) \end{align*}

Similarly to the above we now find \(B_{nm}\) and \(D_{nm}\). The only difference, is that now we have extra \(c\sqrt{\lambda _{nm}}\) terms that show up. The final result will be\[ B_{nm}=\frac{\int _{-\pi }^{\pi }\int _{0}^{a}g\left ( r,\theta \right ) J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) \cos \left ( n\theta \right ) r\ d\theta dr}{c\pi \sqrt{\lambda _{nm}}\int _{0}^{a}J_{n}^{2}\left ( \sqrt{\lambda _{nm}}r\right ) r\ dr}\] And\[ D_{nm}=\frac{\int _{-\pi }^{\pi }\int _{0}^{a}g\left ( r,\theta \right ) J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) \sin \left ( n\theta \right ) r\ d\theta dr}{c\pi \sqrt{\lambda _{nm}}\int _{0}^{a}J_{n}^{2}\left ( \sqrt{\lambda _{nm}}r\right ) r\ dr}\] Summary of solution\begin{align*} u\left ( r,\theta ,t\right ) & =\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }A_{nm}\cos \left ( c\sqrt{\lambda _{nm}}t\right ) J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) \cos \left ( n\theta \right ) \\ & +\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }B_{nm}\sin \left ( c\sqrt{\lambda _{nm}}t\right ) J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) \cos \left ( n\theta \right ) \\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }C_{nm}\cos \left ( c\sqrt{\lambda _{nm}}t\right ) J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) \sin \left ( n\theta \right ) \\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }D_{nm}\sin \left ( c\sqrt{\lambda _{nm}}t\right ) J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) \sin \left ( n\theta \right ) \end{align*}

\[ A_{nm}=\frac{\int _{-\pi }^{\pi }\left ( \int _{0}^{a}f\left ( r,\theta \right ) J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) r\ dr\right ) \cos \left ( n\theta \right ) d\theta }{\pi \int _{0}^{a}J_{n}^{2}\left ( \sqrt{\lambda _{nm}}r\right ) rdr}\]\[ C_{nm}=\frac{\int _{-\pi }^{\pi }\left ( \int _{0}^{a}f\left ( r,\theta \right ) J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) r\ dr\right ) \sin \left ( n\theta \right ) d\theta }{\pi \int _{0}^{a}J_{n}^{2}\left ( \sqrt{\lambda _{nm}}r\right ) rdr}\]\[ B_{nm}=\frac{\int _{-\pi }^{\pi }\int _{0}^{a}g\left ( r,\theta \right ) J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) \cos \left ( n\theta \right ) r\ d\theta dr}{c\pi \sqrt{\lambda _{nm}}\int _{0}^{a}J_{n}^{2}\left ( \sqrt{\lambda _{nm}}r\right ) r\ dr}\]\[ D_{nm}=\frac{\int _{-\pi }^{\pi }\int _{0}^{a}g\left ( r,\theta \right ) J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) \sin \left ( n\theta \right ) r\ d\theta dr}{c\pi \sqrt{\lambda _{nm}}\int _{0}^{a}J_{n}^{2}\left ( \sqrt{\lambda _{nm}}r\right ) r\ dr}\] With \(\lambda _{nm}\) being the solutions for \(0=J_{n}\left ( \sqrt{\lambda _{nm}}a\right ) \). For each \(n\), we find \(\lambda _{n,1},\lambda _{n,2},\lambda _{n,3},\cdots \), which are the zeros of the Bessel \(J_{n}\left ( x\right ) \) function. So for each \(n\), we have infinite number of zeros. This generates all the needed \(\lambda _{nm}\). Hence \(\sqrt{\lambda _{nm}}a=BesselJZero\left ( n,m\right ) \), therefore \(\sqrt{\lambda _{nm}}=\frac{a}{BesselJZero\left ( n,m\right ) }\)

2.4.2.1 Animations

The following animations run for 80 seconds. They are for different \(n,m\,\ \)modes. All have zero for initial velocity, which means \(g\left ( r,\theta \right ) =0\). Radius used is \(a=1\) and \(c=0.2\), initial position used is \(f\left ( r,\theta \right ) =r\theta \).

Cases for \(n=0\)

Cases for \(n=1\)

Cases for \(n=2\)

Cases for \(n=3\)

Source code for all the above animations