#### 2.4.2 circular membrane, no circular symmetry, example 1

2.4.2.1 Animations

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$$f\left ( r,\theta \right ) =\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }A_{nm}J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) \cos \left ( n\theta \right ) +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }C_{nm}J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) \sin \left ( n\theta \right ) \tag{4}$$ 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