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\begin{equation} u_{tt}=u_{rr}+\frac{1}{r}u_{r}+\frac{1}{r^{2}}u_{\theta \theta } \tag{1} \end{equation} 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\begin{equation} T^{\prime \prime }+\lambda ^{2}T=0 \tag{1A} \end{equation} 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\begin{equation} r^{2}R^{\prime \prime }+rR^{\prime }+\left ( r^{2}\lambda ^{2}-\mu ^{2}\right ) R=0 \tag{3} \end{equation} And the \(\Theta \) ODE as\begin{equation} \Theta ^{\prime \prime }+\mu ^{2}\Theta =0 \tag{4} \end{equation} 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\begin{equation} \Theta _{n}\left ( \theta \right ) =A_{n}\cos n\theta +B_{n}\sin n\theta \qquad n=0,1,2,\cdots \tag{5} \end{equation} 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\begin{equation} 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 \end{equation} 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\begin{equation} 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} \end{equation} 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\begin{equation} \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} \end{equation} 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\begin{equation} 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} \end{equation} 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\begin{equation} rJ_{1}\left ( r\right ) =\int rJ_{0}\left ( r\right ) dr \tag{10A} \end{equation} 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\begin{equation} \int _{0}^{1}rJ_{0}^{2}\left ( \lambda _{0m}r\right ) dr=\frac{1}{2}J_{1}^{2}\left ( \lambda _{0m}\right ) \tag{10C} \end{equation} 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\begin{equation} 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} \end{equation}

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