2.4.3 circular membrane, no circular symmetry, speciﬁc problem from exam

This solves wave PDE on disk also as above. This was problem from exam, moved here. Math 322. Spring 2018. Solve

$u_{tt}=u_{xx}+u_{yy}$ On the unit disk $$x^{2}+y^{2}\leq 1$$ with with boundary conditions$u\left ( x,y\right ) =0\qquad \text{if }x^{2}+y^{2}=1$ And initial conditions\begin{align*} u\left ( x,y,0\right ) & =0\\ u_{t}\left ( x,y,0\right ) & =\left \{ \begin{array} [c]{ccc}\frac{1}{\pi \epsilon ^{2}} & & \text{if }\sqrt{x^{2}+y^{2}}\leq \epsilon \\ 0 & & \text{otherwise}\end{array} \right . \end{align*}

Where $$0<\epsilon <1$$. Plot the solution $$u\left ( r,t\right )$$ for $$\epsilon =\frac{1}{2}$$

Solution

The PDE and initial and boundary conditions are converted to polar coordinates to become$$u_{tt}=u_{rr}+\frac{1}{r}u_{r}+\frac{1}{r^{2}}u_{\theta \theta } \tag{1}$$ On the unit disk with radius $$1$$. The boundary conditions are\begin{align*} u\left ( 1,\theta ,t\right ) & =0\\ u\left ( 0,\theta ,t\right ) & <\infty \end{align*}

Where $$u\left ( 0,\theta ,t\right ) <\infty$$ means the solution is bounded at center of disk $$r=0$$.  The boundary conditions on $$\theta$$ are the standard periodic boundary conditions\begin{align*} u\left ( r,-\pi ,t\right ) & =u\left ( r,\pi ,t\right ) \\ u_{\theta }\left ( r,-\pi ,t\right ) & =u_{\theta }\left ( r,\pi ,t\right ) \end{align*}

And initial conditions are\begin{align*} u\left ( r,\theta ,0\right ) & =0\\ u_{t}\left ( r,\theta ,0\right ) & =\left \{ \begin{array} [c]{ccc}\frac{1}{\pi \epsilon ^{2}} & & \text{if }r\leq \epsilon \\ 0 & & \text{otherwise}\end{array} \right . \end{align*}

The above PDE is solved by separation of variables.  Let $$u\left ( r,\theta ,t\right ) =T\left ( t\right ) R\left ( r\right ) \Theta \left ( \theta \right )$$. Substituting this in the PDE (1) gives$T^{\prime \prime }R\Theta =R^{\prime \prime }T\Theta +\frac{1}{r}R^{\prime }T\Theta +\frac{1}{r^{2}}\Theta ^{\prime \prime }RT$ Dividing by $$RT\Theta$$$\frac{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 }=-\lambda ^{2}$ Where $$\lambda$$ is the ﬁrst separation variable. This results in two equations\begin{align} \frac{T^{\prime \prime }}{T} & =-\lambda ^{2}\tag{1}\\ \frac{R^{\prime \prime }}{R}+\frac{1}{r}\frac{R^{\prime }}{R}+\frac{1}{r^{2}}\frac{\Theta ^{\prime \prime }}{\Theta } & =-\lambda ^{2} \tag{2} \end{align}

The time ODE (1) is$$T^{\prime \prime }+\lambda ^{2}T=0 \tag{1A}$$ Multiplying (2) by $$r^{2}$$ and rearranging$r^{2}\frac{R^{\prime \prime }}{R}+r\frac{R^{\prime }}{R}+r^{2}\lambda ^{2}=-\frac{\Theta ^{\prime \prime }}{\Theta }=\mu ^{2}$ Where $$\mu$$ is the second separation constant. This gives the $$R$$ ODE as$$r^{2}R^{\prime \prime }+rR^{\prime }+\left ( r^{2}\lambda ^{2}-\mu ^{2}\right ) R=0 \tag{3}$$ And the $$\Theta$$ ODE as$$\Theta ^{\prime \prime }+\mu ^{2}\Theta =0 \tag{4}$$ The eigenvalues for (4) determine the Bessel equation (3) order. Therefore (4) needs to be solved ﬁrst to determined the order. The ODE boundary conditions for (4) are periodic\begin{align*} \Theta \left ( -\pi \right ) & =\Theta \left ( \pi \right ) \\ \Theta ^{\prime }\left ( -\pi \right ) & =\Theta ^{\prime }\left ( \pi \right ) \end{align*}

case $$\mu =0$$. This leads to solution \begin{align*} \Theta & =c_{1}\theta +c_{2}\\ \Theta ^{\prime } & =c_{1} \end{align*}

First BC gives\begin{align*} -c_{1}\pi +c_{2} & =c_{1}\pi +c_{2}\\ c_{1} & =0 \end{align*}

And since second BC $$\Theta ^{\prime }=c_{1}$$, this implies $$\Theta \left ( \theta \right )$$ is constant. So $$\mu =0$$ is an eigenvalue, with $$\Theta _{0}\left ( \theta \right ) =1$$ being the eigenfunction.

Case $$\mu >0$$ The solution to (4) becomes$\Theta \left ( \theta \right ) =A\cos \left ( \mu \theta \right ) +B\sin \left ( \mu \theta \right )$ To satisfy the periodic boundary conditions, $$\mu$$ must be an integer, and since $$\mu >0$$, then $$\mu =n$$ for $$n=1,2,3,\cdots$$. Therefore\begin{align} \Theta _{0}\left ( \theta \right ) & =1\qquad n=0\tag{5A}\\ \Theta _{n}\left ( \theta \right ) & =A_{n}\cos n\theta +B_{n}\sin n\theta \qquad n=1,2,3,\cdots \tag{5B} \end{align}

The above solution can be combined to one$$\Theta _{n}\left ( \theta \right ) =A_{n}\cos n\theta +B_{n}\sin n\theta \qquad n=0,1,2,\cdots \tag{5}$$ Because when $$n=0$$ the above solution gives $$\Theta _{0}\left ( \theta \right ) =A_{0}$$ which is the constant eigenfunction. Now that $$\mu$$ is found, Bessel ODE (3) can be solved.\begin{align} r^{2}R^{\prime \prime }\left ( r\right ) +rR^{\prime }\left ( r\right ) +\left ( r^{2}\lambda ^{2}-n^{2}\right ) R\left ( r\right ) & =0\qquad n=0,1,2,3,\cdots \tag{5C}\\ R\left ( 1\right ) & =0\nonumber \\ R\left ( 0\right ) & <\infty \nonumber \end{align}

$$\lambda =0$$ is not a possible eigenvalue. This can be shown as follows. When $$\lambda =0$$ equation (5C) becomes the Euler ODE$r^{2}R^{\prime \prime }\left ( r\right ) +rR^{\prime }\left ( r\right ) +n^{2}R\left ( r\right ) =0\qquad n=0,1,2,3,\cdots$ Now, when $$n=0$$, then the ODE becomes $$r^{2}R^{\prime \prime }\left ( r\right ) +rR^{\prime }\left ( r\right ) =0$$ whose solution is $$R\left ( r\right ) =c_{1}+c_{2}\ln \left ( r\right )$$. Since solution is bounded at $$r=0$$, then $$R\left ( r\right ) =c_{1}$$. And since $$R\left ( 1\right ) =0$$ then $$c_{1}=0$$ also, leading to trivial solution. When $$n>0$$, the ODE becomes $$r^{2}R^{\prime \prime }\left ( r\right ) +rR^{\prime }\left ( r\right ) +n^{2}R\left ( r\right ) =0$$ whose solution is $$R\left ( r\right ) =c_{1}r^{n}+c_{2}\frac{1}{r^{n}}$$. Since solution is bounded at $$r=0$$, then $$c_{2}=0$$ and the solution now becomes $$R\left ( r\right ) =c_{1}r^{n}$$. Using BC $$R\left ( 1\right ) =0$$ gives $$c_{1}=0$$ leading again to trivial solution. This shows that $$\lambda =0$$ is not eigenvalue.

Now that $$\lambda$$ is is shown not to be zero, the Bessel ODE (5C) is solved . The ﬁrst step is to convert the ODE to a Bessel ODE in the classical form in order to use the standard solution. Let $$t=\lambda r$$, then $$R^{\prime }\left ( r\right ) =R^{\prime }\left ( t\right ) \lambda$$ and $$R^{\prime \prime }\left ( r\right ) =R^{\prime \prime }\left ( t\right ) \lambda ^{2}$$. ODE (5C) becomes\begin{align*} \frac{t^{2}}{\lambda ^{2}}\lambda ^{2}R^{\prime \prime }\left ( t\right ) +\frac{t}{\lambda }\lambda R^{\prime }\left ( t\right ) +\left ( \frac{t^{2}}{\lambda ^{2}}\lambda ^{2}-n^{2}\right ) R\left ( t\right ) & =0\\ t^{2}R^{\prime \prime }\left ( t\right ) +tR^{\prime }\left ( t\right ) +\left ( t^{2}-n^{2}\right ) R\left ( t\right ) & =0 \end{align*}

This is now in standard Bessel ODE form. This is of order $$n$$, where $$n$$ is $$n=0,1,2,3,\cdots$$. Since the order is integer, then the solution is given by$R_{n}\left ( t\right ) =C_{n}J_{n}\left ( t\right ) +D_{n}Y_{n}\left ( t\right )$ Where $$J_{n}\left ( t\right )$$ is the Bessel function of order $$n$$ and $$Y_{n}\left ( t\right )$$ is the Bessel function of second kind of order $$n$$. In terms of $$r$$ the above solution becomes$R_{n}\left ( r\right ) =C_{n}J_{n}\left ( \lambda r\right ) +D_{n}Y_{n}\left ( \lambda r\right )$ Because the solution is bounded at $$r=0$$ and since $$Y_{n}\left ( 0\right )$$ blows up, then $$D_{n}=0$$. The above solution simpliﬁes to$R_{n}\left ( r\right ) =C_{n}J_{n}\left ( \lambda r\right )$ Applying the second boundary conditions, when $$r=1$$ then$0=C_{n}J_{n}\left ( \lambda \right )$ For non-trivial solution $$J_{n}\left ( \lambda \right ) =0$$. Hence $$\lambda$$ are the positive zeros of $$J_{n}\left ( z\right )$$. Let the positive zeros of $$J_{n}\left ( z\right )$$ be $$j_{nm}$$. For $$m=1,2,3,\cdots$$. Therefore $\lambda _{nm}=j_{nm}\qquad n=0,1,2,\cdots ,m=1,2,3,\cdots$ This means that $$j_{nm}$$ is the $$m^{th}$$ eigenvalue for the $$n^{th}$$ order Bessel function $$J_{n}\left ( z\right )$$. So there are two indices to handle in this problem. The order of the Bessel function is determined from the $$\Theta _{n}\left ( \theta \right )$$ eigenvalues, and then once this order $$n$$ is ﬁxed, the second eigenvalue $$\lambda _{nm}$$ is determined from the zeros of the Bessel function $$J_{n}\left ( z\right )$$. Hence the $$R_{nm}\left ( r\right )$$ solution is$R_{nm}\left ( r\right ) =C_{nm}J_{n}\left ( \lambda _{nm}r\right ) \qquad n=0,1,2,3,\cdots ,m=1,2,3,\cdots$ Now that $$\lambda _{nm}$$ is known, the time ODE (1) can be solved\begin{align*} T_{nm}^{\prime \prime }+\lambda _{nm}^{2}T_{nm} & =0\\ T_{nm} & =A_{nm}\cos \left ( \lambda _{nm}t\right ) +B_{nm}\sin \left ( \lambda _{nm}t\right ) \qquad n=0,1,2,3,\cdots ,m=1,2,3,\cdots \end{align*}

The fundamental solution is therefore$u_{nm}\left ( r,\theta ,t\right ) =\Theta _{n}\left ( \theta \right ) T_{nm}\left ( t\right ) R_{nm}\left ( r\right )$ The complete solution is the superposition of the fundamental solutions given by \begin{align*} u\left ( r,\theta ,t\right ) & =\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }\Theta _{n}\left ( \theta \right ) T_{nm}\left ( t\right ) R_{nm}\left ( r\right ) \\ & =\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }\left ( A_{n}\cos n\theta +B_{n}\sin n\theta \right ) \left \{ A_{nm}\cos \left ( \lambda _{nm}t\right ) +B_{nm}\sin \left ( \lambda _{nm}t\right ) \right \} C_{nm}J_{n}\left ( \lambda _{nm}r\right ) \end{align*}

The above can now be written as\begin{align*} u\left ( r,\theta ,t\right ) & =\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }A_{n}\cos n\theta \left ( \left ( A_{nm}\cos \left ( \lambda _{nm}t\right ) +B_{nm}\sin \left ( \lambda _{nm}t\right ) \right ) C_{nm}J_{n}\left ( \lambda _{nm}r\right ) \right ) \\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }B_{n}\sin n\theta \left ( \left ( A_{nm}\cos \left ( \lambda _{nm}t\right ) +B_{nm}\sin \left ( \lambda _{nm}t\right ) \right ) C_{nm}J_{n}\left ( \lambda _{nm}r\right ) \right ) \end{align*}

Or\begin{align} u\left ( r,\theta ,t\right ) & =\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }A_{n}\cos n\theta A_{nm}\cos \left ( \lambda _{nm}t\right ) C_{nm}J_{n}\left ( \lambda _{nm}r\right ) \nonumber \\ & +\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }A_{n}\cos n\theta B_{nm}\sin \left ( \lambda _{nm}t\right ) C_{nm}J_{n}\left ( \lambda _{nm}r\right ) \nonumber \\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }B_{n}\sin n\theta A_{nm}\cos \left ( \lambda _{nm}t\right ) C_{nm}J_{n}\left ( \lambda _{nm}r\right ) \nonumber \\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }B_{n}\sin n\theta B_{nm}\sin \left ( \lambda _{nm}t\right ) C_{nm}J_{n}\left ( \lambda _{nm}r\right ) \tag{6} \end{align}

Constants are now merged and renamed as follows in order to simplify the rest of the solution. Let\begin{align*} A_{n}A_{nm}C_{nm} & =\bar{A}_{nm}\\ A_{n}B_{nm}C_{nm} & =\bar{B}_{nm}\\ B_{n}A_{nm}C_{nm} & =\bar{C}_{nm}\\ B_{n}B_{nm}C_{nm} & =\bar{D}_{nm} \end{align*}

Equation (6) can now be written as\begin{align} u\left ( r,\theta ,t\right ) & =\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }\bar{A}_{nm}\cos \left ( n\theta \right ) \cos \left ( \lambda _{nm}t\right ) J_{n}\left ( \lambda _{nm}r\right ) \nonumber \\ & +\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }\bar{B}_{nm}\cos \left ( n\theta \right ) \sin \left ( \lambda _{nm}t\right ) J_{n}\left ( \lambda _{nm}r\right ) \nonumber \\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }\bar{C}_{nm}\sin \left ( n\theta \right ) \cos \left ( \lambda _{nm}t\right ) J_{n}\left ( \lambda _{nm}r\right ) \nonumber \\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }\bar{D}_{nm}\sin \left ( n\theta \right ) \sin \left ( \lambda _{nm}t\right ) J_{n}\left ( \lambda _{nm}r\right ) \tag{7} \end{align}

Initial conditions are used to determine the $$4$$ new constants above. Using initial condition at $$t=0,u\left ( r,\theta ,0\right ) =0$$ the above equation becomes$$0=\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }\bar{A}_{nm}\cos \left ( n\theta \right ) J_{n}\left ( \lambda _{nm}r\right ) +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }\bar{C}_{nm}\sin \left ( n\theta \right ) J_{n}\left ( \lambda _{nm}r\right ) \nonumber$$ Applying orthogonality on $$\cos \left ( n\theta \right )$$ and $$\sin \left ( n\theta \right )$$ in turn shows that $$\bar{A}_{nm}=0$$ and $$\bar{C}_{nm}=0$$. Therefore the solution (7) reduces to the following two sums only$$u\left ( r,\theta ,t\right ) =\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }\bar{B}_{nm}\cos \left ( n\theta \right ) \sin \left ( \lambda _{nm}t\right ) J_{n}\left ( \lambda _{nm}r\right ) +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }\bar{D}_{nm}\sin \left ( n\theta \right ) \sin \left ( \lambda _{nm}t\right ) J_{n}\left ( \lambda _{nm}r\right ) \tag{8}$$ Taking time derivative gives$u_{t}\left ( r,\theta ,t\right ) =\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }\bar{B}_{nm}\cos \left ( n\theta \right ) \lambda _{nm}\cos \left ( \lambda _{nm}t\right ) J_{n}\left ( \lambda _{nm}r\right ) +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }\bar{D}_{nm}\sin \left ( n\theta \right ) \lambda _{nm}\cos \left ( \lambda _{nm}t\right ) J_{n}\left ( \lambda _{nm}r\right )$ Applying the second initial condition at $$t=0$$ gives$$\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }\bar{B}_{nm}\cos \left ( n\theta \right ) \lambda _{nm}J_{n}\left ( \lambda _{nm}r\right ) +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }\bar{D}_{nm}\sin \left ( n\theta \right ) \lambda _{nm}J_{n}\left ( \lambda _{nm}r\right ) =\left \{ \begin{array} [c]{ccc}\frac{1}{\pi \epsilon ^{2}} & & \text{if }r\leq \epsilon \\ 0 & & \text{otherwise}\end{array} \right . \tag{9}$$ Case $$n=0$$ (9) becomes$\sum _{m=1}^{\infty }\bar{B}_{0m}\lambda _{0m}J_{0}\left ( \lambda _{0m}r\right ) =\left \{ \begin{array} [c]{ccc}\frac{1}{\pi \epsilon ^{2}} & & \text{if }r\leq \epsilon \\ 0 & & \text{otherwise}\end{array} \right .$ Applying orthogonality on $$J_{0}\left ( \lambda _{0m}r\right )$$ results in\begin{align} \bar{B}_{0m}\lambda _{0m}\int _{0}^{1}rJ_{0}^{2}\left ( \lambda _{0m}r\right ) dr & =\frac{1}{\pi \epsilon ^{2}}\int _{0}^{\epsilon }rJ_{0}\left ( \lambda _{0m}r\right ) dr\nonumber \\ \bar{B}_{0m} & =\frac{1}{\pi \epsilon ^{2}\lambda _{0m}}\frac{\int _{0}^{\epsilon }rJ_{0}\left ( \lambda _{0m}r\right ) dr}{\int _{0}^{1}rJ_{0}^{2}\left ( \lambda _{0m}r\right ) dr} \tag{9A} \end{align}

Case $$n>1$$ Applying orthogonality on $$\cos \left ( n\theta \right ) ,$$ equation (9) becomes\begin{align*} \sum _{m=1}^{\infty }\bar{B}_{nm}\left ( \int _{-\pi }^{\pi }\cos ^{2}\left ( n\theta \right ) d\theta \right ) \lambda _{nm}J_{n}\left ( \lambda _{nm}r\right ) & =\left \{ \begin{array} [c]{ccc}\frac{1}{\pi \epsilon ^{2}}\int _{-\pi }^{\pi }\cos \left ( n\theta \right ) d\theta & & \text{if }r^{2}\leq \epsilon \\ 0 & & \text{otherwise}\end{array} \right . \\ \sum _{m=1}^{\infty }\pi \bar{B}_{nm}\lambda _{nm}J_{n}\left ( \lambda _{nm}r\right ) & =\left \{ \begin{array} [c]{ccc}0 & & \text{if }r^{2}\leq \epsilon \\ 0 & & \text{otherwise}\end{array} \right . \end{align*}

Hence $$\bar{B}_{nm}=0$$ for all $$n>0$$.

The same is now done to ﬁnd $$\bar{D}_{nm}$$. Applying orthogonality on $$\sin \left ( n\theta \right ) ,$$ equation (9) becomes\begin{align*} \sum _{m=1}^{\infty }\bar{D}_{nm}\left ( \int _{-\pi }^{\pi }\sin ^{2}\left ( n\theta \right ) d\theta \right ) \lambda _{nm}J_{n}\left ( \lambda _{nm}r\right ) & =\left \{ \begin{array} [c]{ccc}\frac{1}{\pi \epsilon ^{2}}\int _{-\pi }^{\pi }\sin \left ( n\theta \right ) d\theta & & \text{if }r^{2}\leq \epsilon \\ 0 & & \text{otherwise}\end{array} \right . \\ \sum _{m=1}^{\infty }\bar{D}_{nm}\left ( \int _{-\pi }^{\pi }\sin ^{2}\left ( n\theta \right ) d\theta \right ) \lambda _{nm}J_{n}\left ( \lambda _{nm}r\right ) & =\left \{ \begin{array} [c]{ccc}0 & & \text{if }r^{2}\leq \epsilon \\ 0 & & \text{otherwise}\end{array} \right . \end{align*}

Hence all $$\bar{D}_{nm}=0$$ for all $$n>0$$.

Therefore the solution (8) reduces to only using $$n=0,m=1,2,3,\cdots$$. The solution can now be written as$$u\left ( r,\theta ,t\right ) =\sum _{m=1}^{\infty }\bar{B}_{0m}\sin \left ( \lambda _{0m}t\right ) J_{0}\left ( \lambda _{0m}r\right ) \tag{10}$$ Where $$\bar{B}_{0m}=\frac{1}{\pi \epsilon ^{2}\lambda _{0m}}\frac{\int _{0}^{\epsilon }rJ_{0}\left ( \lambda _{0m}r\right ) dr}{\int _{0}^{1}rJ_{0}^{2}\left ( \lambda _{0m}r\right ) dr}$$ And $$\lambda _{0m}$$ are all the positive zeros of $$J_{0}\left ( z\right )$$, $$m=1,2,3,\cdots$$.

$$\bar{B}_{0m}$$ is now simpliﬁed more. Considering ﬁrst the numerator of $$\bar{B}_{0m}$$ which is $$\int _{0}^{\epsilon }rJ_{0}\left ( \lambda _{0m}r\right ) dr$$. The hint given says that$\frac{d}{dr}\left ( rJ_{1}\left ( r\right ) \right ) =rJ_{0}\left ( r\right )$ This is the same as saying$$rJ_{1}\left ( r\right ) =\int rJ_{0}\left ( r\right ) dr \tag{10A}$$ However the integral in $$\bar{B}_{0m}$$ is $$\int rJ_{0}\left ( \lambda _{0m}r\right ) dr$$ and not $$\int rJ_{0}\left ( r\right ) dr$$. To transform it so that the hint can be used, let $$\lambda _{0m}r=\bar{r}$$, then $$\frac{dr}{d\bar{r}}=\frac{1}{\lambda _{0m}}$$ or $$dr=\frac{d\bar{r}}{\lambda _{0m}}$$. Now $$\int rJ_{0}\left ( \lambda _{0m}r\right ) dr$$ becomes $$\int \frac{\bar{r}}{\lambda _{0m}}J_{0}\left ( \bar{r}\right ) \frac{d\bar{r}}{\lambda _{0m}}$$ or $$\frac{1}{\lambda _{0m}^{2}}\int \bar{r}J_{0}\left ( \bar{r}\right ) d\bar{r}$$ and now the hint (10A) can be used on this integral giving$\frac{1}{\lambda _{0m}^{2}}\left ( \int \bar{r}J_{0}\left ( \bar{r}\right ) d\bar{r}\right ) =\frac{1}{\lambda _{0m}^{2}}\left ( \bar{r}J_{1}\left ( \bar{r}\right ) \right )$ Replacing $$\bar{r}$$ back by $$\lambda _{0m}r$$, gives the result needed\begin{align*} \frac{1}{\lambda _{0m}^{2}}\left ( \bar{r}J_{1}\left ( \bar{r}\right ) \right ) & =\frac{1}{\lambda _{0m}^{2}}\left ( \lambda _{0m}rJ_{1}\left ( \lambda _{0m}r\right ) \right ) \\ & =\frac{1}{\lambda _{0m}}rJ_{1}\left ( \lambda _{0m}r\right ) \end{align*}

Now the limits are applied, using the fundamental theory of calculus\begin{align} \int _{0}^{\epsilon }rJ_{0}\left ( \lambda _{0m}r\right ) dr & =\frac{1}{\lambda _{0m}}\left [ rJ_{1}\left ( \lambda _{0m}r\right ) \right ] _{0}^{\epsilon }\nonumber \\ & =\frac{\epsilon }{\lambda _{0m}}J_{1}\left ( \lambda _{0m}\epsilon \right ) \tag{10B} \end{align}

This completes ﬁnding the numerator integral in $$\bar{B}_{0m}$$. The denominator integral in $$\bar{B}_{0m}$$ is $$\int _{0}^{1}rJ_{0}^{2}\left ( \lambda _{0m}r\right ) dr$$. This was found in HW4, from problem 3, which is $\int _{0}^{1}rJ_{0}^{2}\left ( \lambda _{0m}r\right ) dr=\frac{1}{2}\left [ J_{0}^{\prime }\left ( \lambda _{0m}\right ) \right ] ^{2}$ But $$J_{0}^{\prime }\left ( \lambda _{0m}\right ) =-J_{1}\left ( \lambda _{0m}\right )$$, hence the above becomes$$\int _{0}^{1}rJ_{0}^{2}\left ( \lambda _{0m}r\right ) dr=\frac{1}{2}J_{1}^{2}\left ( \lambda _{0m}\right ) \tag{10C}$$ Applying (10B) and (10C), $$\bar{B}_{0m}$$ simpliﬁes to the following expression\begin{align*} \bar{B}_{0m} & =\frac{1}{\pi \epsilon ^{2}\lambda _{0m}}\frac{\frac{\epsilon }{\lambda _{0m}}J_{1}\left ( \lambda _{0m}\epsilon \right ) }{\frac{1}{2}J_{1}^{2}\left ( \lambda _{0m}\right ) }\\ & =\frac{2}{\pi \epsilon \lambda _{0m}^{2}}\frac{J_{1}\left ( \lambda _{0m}\epsilon \right ) }{J_{1}^{2}\left ( \lambda _{0m}\right ) } \end{align*}

Therefore the ﬁnal solution becomes\begin{align} u\left ( r,\theta ,t\right ) & =\sum _{m=1}^{\infty }\bar{B}_{0m}\sin \left ( \lambda _{0m}t\right ) J_{0}\left ( \lambda _{0m}r\right ) \nonumber \\ u\left ( r,\theta ,t\right ) & =\frac{2}{\pi \epsilon }\sum _{m=1}^{\infty }\frac{1}{\lambda _{0m}^{2}}\frac{J_{1}\left ( \lambda _{0m}\epsilon \right ) }{J_{1}^{2}\left ( \lambda _{0m}\right ) }J_{0}\left ( \lambda _{0m}r\right ) \sin \left ( \lambda _{0m}t\right ) \tag{11} \end{align}

When $$\epsilon =\frac{1}{2}$$, the above solution (11) becomes$$u\left ( r,\theta ,t\right ) =\frac{4}{\pi }\sum _{m=1}^{\infty }\frac{1}{\lambda _{0m}^{2}}\frac{J_{1}\left ( \frac{1}{2}\lambda _{0m}\right ) }{J_{1}^{2}\left ( \lambda _{0m}\right ) }J_{0}\left ( \lambda _{0m}r\right ) \sin \left ( \lambda _{0m}t\right ) \tag{11A}$$

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