4.9 HW 9

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4.9.1 Chapter 13, problem 6.1 Mary Boas. Second edition

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4.9.2 chapter 13, problem 4.1. Mary Boas, second edition

Complete the plucked string problem to get equation 4.0

Solution

Here we start with the solution given in 4.8

$$y_0={\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{b}_n\sin \left(\frac{n\pi x}{L}\right)=f(x)(1)$$

Where $f(x)$ represents the initial position (shape) of the string.

Now need to find $b_n$

First need to define $f(x)$, from diagram we see that from $x=0$ to $x=L/2$ the slope is $\frac{h}{L/2}=\frac{2h}{L}$ hence from equation of line we get $y=\frac{2h}{L}x$

From $x=L/2$ to $x=L$, slope is $-\frac{2h}{L}$, so $y=h-\frac{2h}{L}(x-\frac{L}{2})=h-\frac{2h}{L}x+h=2h-\frac{2h}{L}x=2(h-\frac{hx}{L})$

so we have $$f(x)=\{\begin{array}{lll}\frac{2h}{L}x& & 0\le x\le \frac{L}{2}\\ 2(h-\frac{hx}{L})& & \frac{L}{2}<x\le L\end{array}$$

so from $(1)$ we get, after applying inner product w.r.t. $\sin \left(\frac{n\pi x}{L}\right)$

$$\begin{array}{rl}{b}_n{\displaystyle \underset{0}{\overset{L}{\int }}}{\sin }^2\left(\frac{n\pi x}{L}\right)& ={\displaystyle \underset{0}{\overset{L}{\int }}}f(x)\sin \left(\frac{n\pi x}{L}\right)dx\\ b_n\frac{L}{2}& ={\displaystyle \underset{0}{\overset{\frac{L}{2}}{\int }}}f(x)\sin \left(\frac{n\pi x}{L}\right)dx+{\displaystyle \underset{\frac{L}{2}}{\overset{L}{\int }}}f(x)\sin \left(\frac{n\pi x}{L}\right)dx\\ b_n\frac{L}{2}& ={\displaystyle \underset{0}{\overset{\frac{L}{2}}{\int }}}\frac{2h}{L}x\sin \left(\frac{n\pi x}{L}\right)dx+{\displaystyle \underset{\frac{L}{2}}{\overset{L}{\int }}}2(h-\frac{hx}{L})\sin \left(\frac{n\pi x}{L}\right)dx\\ b_n\frac{L}{2}& =\frac{2h}{L}{\displaystyle \underset{0}{\overset{\frac{L}{2}}{\int }}}x\sin \left(\frac{n\pi x}{L}\right)dx+{\displaystyle \underset{\frac{L}{2}}{\overset{L}{\int }}}2h\sin \left(\frac{n\pi x}{L}\right)dx-{\displaystyle \underset{\frac{L}{2}}{\overset{L}{\int }}}\frac{2hx}{L}\sin \left(\frac{n\pi x}{L}\right)dx\\ b_n\frac{L}{2}& =\frac{2h}{L}{\displaystyle \underset{0}{\overset{\frac{L}{2}}{\int }}}x\sin \left(\frac{n\pi x}{L}\right)dx+2h{\displaystyle \underset{\frac{L}{2}}{\overset{L}{\int }}}\sin \left(\frac{n\pi x}{L}\right)dx-\frac{2h}{L}{\displaystyle \underset{\frac{L}{2}}{\overset{L}{\int }}}x\sin \left(\frac{n\pi x}{L}\right)dx\\ b_n\frac{L}{2}& =\frac{16hL\cos \left(\frac{n\pi }{4}\right)\sin {\left(\frac{n\pi }{4}\right)}^3}{n^2{\pi }^2}\\ b_n& =\frac{32h\cos \left(\frac{n\pi }{4}\right)\sin {\left(\frac{n\pi }{4}\right)}^3}{n^2{\pi }^2}\end{array}$$

so

$$b_n=\frac{32hL\cos \left(\frac{n\pi }{4}\right)\sin {\left(\frac{n\pi }{4}\right)}^3}{n^2{\pi }^2}$$

Looking at few values of n to see the pattern

$$\begin{array}{rl}{b}_n& =\frac{32h\cos \left(\frac{\pi }{4}\right)\sin {\left(\frac{\pi }{4}\right)}^3}{1^2{\pi }^2},\frac{32h\cos \left(\frac{2\pi }{4}\right)\sin {\left(\frac{2\pi }{4}\right)}^3}{2^2{\pi }^2},\frac{32h\cos \left(\frac{3\pi }{4}\right)\sin {\left(\frac{3\pi }{4}\right)}^3}{3^2{\pi }^2},...\\ & =\frac{8h}{{\pi }^2},0,-\frac{8h}{9{\pi }^2},0,\frac{8h}{25{\pi }^2},\dots \\ & =\frac{8h}{{\pi }^2}(1,0,-\frac{1}{9},0,\frac{1}{25},\dots )\end{array}$$

Notice that we have terms for only odd n.

Now, substituting the above in the general solution given in equation 4.7 in book, which is

$$y={\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{b}_n\sin \left(\frac{n\pi x}{L}\right)\cos \left(\frac{n\pi vt}{L}\right)$$

Gives

$$\begin{array}{rl}y& =\frac{8h}{{\pi }^2}(\sin \left(\frac{\pi x}{L}\right)\cos \left(\frac{\pi vt}{L}\right)+0+-\frac{1}{9}\sin \left(\frac{3\pi x}{L}\right)\cos \left(\frac{3\pi vt}{L}\right)+0+\frac{1}{25}\sin \left(\frac{5\pi x}{L}\right)\cos \left(\frac{5\pi vt}{L}\right)+...)\\ y& =\frac{8h}{{\pi }^2}(\sin \left(\frac{\pi x}{L}\right)\cos \left(\frac{\pi vt}{L}\right)-\frac{1}{9}\sin \left(\frac{3\pi x}{L}\right)\cos \left(\frac{3\pi vt}{L}\right)+\frac{1}{25}\sin \left(\frac{5\pi x}{L}\right)\cos \left(\frac{5\pi vt}{L}\right)+...)\end{array}$$

The above is the result we are asked to show.

4.9.3 chapter 13, problem 4.2. Mary Boas, second edition

A string of length L has zero initial velocity and a displacement $y_0\left(x\right)$ as shown. Find the displacement as a function of x and t.

Solution

The PDE that governs this problem is the wave equation ${\mathrm{\nabla }}^{2}y=\frac{1}{v^2}\frac{{\partial }^{2}y}{\partial t^2}$

The candidate solutions are

$$y=\{\begin{array}{l}\sin (kx)\sin (\omega t)\\ \sin (kx)\cos (\omega t)\\ \cos (kx)\sin (\omega t)\\ \cos (kx)\cos (\omega t)\end{array}$$

where $\omega =kv$ and $k=\frac{2\pi }{\lambda }$ where $\lambda$ is the wave length

Now we discard solutions that contains $\cos kx$ since the string is fixed at $x=0$.

So we are left with

$$y=\{\begin{array}{l}\sin (kx)\sin (\omega t)\\ \sin (kx)\cos (\omega t)\end{array}$$

Now, $y=0$ at $x=L$ then from $\sin kx=0$ or $\sin kL=0$we need $k=\frac{n\pi }{L}$

Hence solutions become

$$y=\{\begin{array}{l}\sin (\frac{n\pi }{L}x)\sin (\frac{n\pi }{L}vt)\\ \sin (\frac{n\pi }{L}x)\cos (\frac{n\pi }{L}vt)\end{array}$$

Applying initial conditions, which says that at time $t=0$, velocity is zero.

Hence from above, after taking $\frac{\partial y}{\partial t}$, we get

$$\frac{\partial y}{\partial t}=\{\begin{array}{l}\frac{n\pi v}{L}\sin (\frac{n\pi }{L}x)\cos (\frac{n\pi v}{L}t)\\ -\frac{n\pi v}{L}\sin (\frac{n\pi }{L}x)\sin (\frac{n\pi v}{L}t)\end{array}$$

For the above to be zero at $t=0$ then we discard first solution above with $\cos t$ in it. Hence final general solution is now

$$y=\{\begin{array}{l}\sin (\frac{n\pi }{L}x)\cos (\frac{n\pi }{L}vt)\end{array}$$

A general solution is a linear combination of the above solutions, hence

$$y={\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{b}_n\sin \left(\frac{n\pi }{L}x\right)\cos \left(\frac{n\pi }{L}vt\right)(1)$$

To find $b_n$, we apply the second initial condition, which is $y=y_0=f(x)$

(Notice that we use two initial conditions, i.e. at time t=0 we are looking at speed and position, this is because we started with a PDE with $\frac{{\partial }^{2}y}{\partial t^2}$ in it, which is a second order in t.)

At t=0, (1) becomes

$$y={\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{b}_n\sin \left(\frac{n\pi }{L}x\right)=f(x)(2)$$

To find $f(x)$ from diagram, we see that for $0\le x\le \frac{L}{4}$, $y=x\frac{h}{L/4}=\frac{4h}{L}x$

For $\frac{L}{4}<x\le \frac{L}{2}$, $y=-(x-\frac{L}{4})\frac{h}{L/4}+h=-(x-\frac{L}{4})\frac{4h}{L}+h=-x\frac{4h}{L}+\frac{L}{4}\frac{4h}{L}+h=-x\frac{4h}{L}+2h$

For $\frac{L}{2}<x\le L$, $y=0$

Hence

$$y=\{\begin{array}{lll}\frac{4h}{L}x& & 0\le x\le \frac{L}{4}\\ 2h-x\frac{4h}{L}& & \frac{L}{4}<x\le \frac{L}{2}\\ 0& & \frac{L}{2}<x\le L\end{array}$$

Do the inner product on both sides of equation (2) w.r.t. $\sin \frac{n\pi }{L}x$

$$\begin{array}{rl}{b}_n{\int }_0^L{\sin }^2\frac{n\pi }{L}xdx& ={\int }_0^{L}f\left(x\right)\sin \frac{n\pi }{L}xdx\\ b_n\frac{L}{2}& ={\int }_0^{\frac{L}{4}}f(x)\sin \frac{n\pi }{L}xdx+{\int }_{\frac{L}{4}}^{\frac{L}{2}}f(x)\sin \frac{n\pi }{L}xdx+{\int }_{\frac{L}{2}}^{L}f(x)\sin \frac{n\pi }{L}xdx\\ & ={\int }_0^{\frac{L}{4}}\frac{4h}{L}x\sin \frac{n\pi }{L}xdx+{\int }_{\frac{L}{4}}^{\frac{L}{2}}(2h-x\frac{4h}{L})\sin \frac{n\pi }{L}xdx+{\int }_{\frac{L}{2}}^{L}0\sin \frac{n\pi }{L}xdx\\ & ={\int }_0^{\frac{L}{4}}\frac{4h}{L}x\sin \frac{n\pi }{L}xdx+{\int }_{\frac{L}{4}}^{\frac{L}{2}}2h\sin \left(\frac{n\pi }{L}x\right)-x\frac{4h}{L}\sin \left(\frac{n\pi }{L}x\right)dx\\ & ={\int }_0^{\frac{L}{4}}\frac{4h}{L}x\sin \frac{n\pi }{L}xdx+{\int }_{\frac{L}{4}}^{\frac{L}{2}}2h\sin \left(\frac{n\pi }{L}x\right)dx-{\int }_{\frac{L}{4}}^{\frac{L}{2}}x\frac{4h}{L}\sin \left(\frac{n\pi }{L}x\right)dx\\ & =\frac{4h}{L}{\int }_0^{\frac{L}{4}}x\sin \frac{n\pi }{L}xdx+2h{\int }_{\frac{L}{4}}^{\frac{L}{2}}\sin \left(\frac{n\pi }{L}x\right)dx-\frac{4h}{L}{\int }_{\frac{L}{4}}^{\frac{L}{2}}x\sin \left(\frac{n\pi }{L}x\right)dx\\ b_n& =\frac{8h}{n^2{\pi }^2}(2\sin \left(\frac{n\pi }{4}\right)-\sin \frac{n\pi }{2})\end{array}$$

Looking at few values of $b_n$

$$\begin{array}{rl}{b}_n& =\frac{8h}{1^2{\pi }^2}(2\sin \left(\frac{\pi }{4}\right)-\sin \frac{\pi }{2}),\frac{8h}{2^2{\pi }^2}(2\sin \left(\frac{2\pi }{4}\right)-\sin \frac{2\pi }{2}),\frac{8h}{3^2{\pi }^2}(2\sin \left(\frac{3\pi }{4}\right)-\sin \frac{3\pi }{2}),\dots \\ & =\frac{8h}{{\pi }^2}[(2\sin \left(\frac{\pi }{4}\right)-\sin \frac{\pi }{2}),\frac{1}{2^2}(2\sin \frac{2\pi }{4}-\sin \frac{2\pi }{2}),\frac{1}{3^2}(2\sin \frac{3\pi }{4}-\sin \frac{3\pi }{2}),\dots ]\\ & =\frac{8h}{{\pi }^2}\left[\frac{1}{n^2}\{(2\sin \left(\frac{\pi }{4}\right)-\sin \frac{\pi }{2}),(2\sin \frac{2\pi }{4}-\sin \frac{2\pi }{2}),(2\sin \frac{3\pi }{4}-\sin \frac{3\pi }{2}),\dots \}\right]\\ & =\frac{8h}{{\pi }^2}\left[\frac{1}{n^2}(2\sin \left(\frac{n\pi }{4}\right)-\sin \frac{n\pi }{2})\right]\end{array}$$

Hence from equation (1) above, we get

$$\begin{array}{rl}y& ={\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{b}_n\sin \frac{n\pi }{L}x\cos \frac{n\pi }{L}vt\\ & ={\displaystyle \sum _{n=1}^{\mathrm{\infty }}}\frac{8h}{{\pi }^2}\left[\frac{1}{n^2}(2\sin \left(\frac{n\pi }{4}\right)-\sin \frac{n\pi }{2})\right]\sin \frac{n\pi }{L}x\cos \frac{n\pi }{L}vt\\ & =\frac{8h}{{\pi }^2}{\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{B}_n\sin \frac{n\pi }{L}x\cos \frac{n\pi }{L}vt\end{array}$$

Where $$B_n=\frac{1}{n^2}(2\sin \left(\frac{n\pi }{4}\right)-\sin \frac{n\pi }{2})$$

The above is the result required to show.

4.9.4 chapter 13, problem 4.6. Mary Boas, second edition

A string of length L is initially stretched straight, its ends are fixed for all time t. At time t=0 its points are given the velocity $V(x)={\left(\frac{\partial y}{\partial t}\right)}_{t=0}$as shown in diagram below. Determine the shape of the string at time t.

Solution

The PDE that governs this problem is the wave equation ${\mathrm{\nabla }}^{2}y=\frac{1}{v^2}\frac{{\partial }^{2}y}{\partial t^2}$

The candidate solutions are

$$y=\{\begin{array}{l}\sin (kx)\sin (\omega t)\\ \sin (kx)\cos (\omega t)\\ \cos (kx)\sin (\omega t)\\ \cos (kx)\cos (\omega t)\end{array}$$

Where $\omega =kv$ and $k=\frac{2\pi }{\lambda }$ where $\lambda$ is the wave length

Now we discard solutions that contains $\cos kx$ since the string is fixed at $x=0$.

So we are left with

$$y=\{\begin{array}{l}\sin (kx)\sin (\omega t)\\ \sin (kx)\cos (\omega t)\end{array}$$

Now, $y=0$ at $x=L$ then from $\sin kx=0$ or $\sin kL=0$we need $k=\frac{n\pi }{L}$

Hence solutions become

$$y=\{\begin{array}{lll}\sin \frac{n\pi }{L}x\sin \frac{n\pi }{L}vt& & \\ \sin \frac{n\pi }{L}x\cos \frac{n\pi }{L}vt& & \end{array}$$

Applying initial conditions, which says that at time $t=0$, velocity is given by $V(x)$

Hence from above, after taking $\frac{\partial y}{\partial t}$, we get

$$\frac{\partial y}{\partial t}=\{\begin{array}{lll}\frac{n\pi v}{L}\sin \frac{n\pi }{L}x\cos \frac{n\pi v}{L}t& & \\ -\frac{n\pi v}{L}\sin \frac{n\pi }{L}x\sin \frac{n\pi v}{L}t& & \end{array}$$

For the above we discard velocity solution above with $\sin t$ in it since that will give zero velocity at time t=0, which is not the case here. Hence we discard y solution with $\cos t$ in it, then the final general solution for y is now

$$y=\begin{array}{lll}\sin \frac{n\pi }{L}x\sin \frac{n\pi }{L}vt& & \end{array}$$

A general solution is a linear combination of the above solutions, hence

$$y={\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{b}_n\sin \frac{n\pi x}{L}\sin \frac{n\pi vt}{L}(1)$$

To find $b_n$, we apply the velocity initial condition. Hence differentiate equation (1) and set t=0, we have

$$\frac{\partial y}{\partial t}={\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{b}_n\frac{n\pi v}{L}\sin \frac{n\pi x}{L}\cos \frac{n\pi vt}{L}$$

Setting t=0

$$\frac{\partial y}{\partial t}={\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{b}_n\frac{n\pi v}{L}\sin \frac{n\pi x}{L}=V_{t=0}(2)$$

Now to find $V_{t=0}$. From diagram, we see that for $0\le x\le \frac{L}{2}-w$, $V_{t=0}=0$

For $\frac{L}{2}-w<x\le \frac{L}{2}+w$,$V_{t=0}=h$

For $\frac{L}{2}+w<x\le L$, $V_{t=0}=0$

Hence

$$V_{t=0}=\{\begin{array}{lll}0& & 0\le x\le \frac{L}{2}-w\\ h& & \frac{L}{2}-w<x\le \frac{L}{2}+w\\ 0& & \frac{L}{2}+w<x\le L\end{array}$$

Do the inner product on both sides of equation (2) w.r.t. $\sin \frac{n\pi }{L}x$

$$\begin{array}{rl}{b}_n\frac{n\pi v}{L}{\int }_0^L{\sin }^2\frac{n\pi }{L}xdx& ={\int }_0^{L}V(x)\sin \frac{n\pi x}{L}dx\\ b_n\frac{n\pi v}{2}& ={\int }_0^{\frac{L}{2}-w}0\sin \frac{n\pi }{L}xdx+{\int }_{\frac{L}{2}-w}^{\frac{L}{2}+w}h\sin \frac{n\pi x}{L}dx+{\int }_{\frac{L}{2}+w}^{L}0\sin \frac{n\pi }{L}xdx\\ b_n\frac{n\pi v}{2}& ={\int }_{\frac{L}{2}-w}^{\frac{L}{2}+w}h\sin \frac{n\pi x}{L}dx\\ b_n\frac{n\pi v}{2}& =-h\frac{L}{n\pi }{[\cos \frac{n\pi x}{L}]}_{\frac{L}{2}-w}^{\frac{L}{2}+w}\\ b_n\frac{n\pi v}{2}& =-h\frac{L}{n\pi }[\cos \frac{n\pi (\frac{L}{2}+w)}{L}-\cos \frac{n\pi (\frac{L}{2}-w)}{L}]\\ b_n\frac{n\pi v}{2}& =-h\frac{L}{n\pi }[\cos (\frac{n\pi }{2}+\frac{n\pi w}{L})-\cos (\frac{n\pi }{2}-\frac{n\pi w}{L})]\\ b_n& =-\frac{2hL}{n^2{\pi }^{2}v}[\cos (\frac{n\pi }{2}+\frac{n\pi w}{L})-\cos (\frac{n\pi }{2}-\frac{n\pi w}{L})]\end{array}$$

But $\cos (a+b)=\cos (a)\cos (b)-\sin (a)\sin (b)$

and $\cos (a-b)=\cos (a)\cos (b)+\sin (a)\sin (b)$

Let $a=\frac{n\pi }{2},b=\frac{n\pi w}{L}$

Hence $b_n$ becomes

$$\begin{array}{rl}{b}_n& =-\frac{2hL}{n^2{\pi }^{2}v}[\cos (a+b)-\cos (a-b)]\\ & =-\frac{2hL}{n^2{\pi }^{2}v}[\cos (a)\cos \left(b\right)-\sin (a)\sin (b)-\{\cos \left(a\right)\cos (b)+\sin (a)\sin (b)\}]\\ & =-\frac{2hL}{n^2{\pi }^{2}v}[\cos (a)\cos \left(b\right)-\sin (a)\sin (b)-\cos (a)\cos (b)-\sin (a)\sin (b)]\\ & =-\frac{2hL}{n^2{\pi }^{2}v}[-\sin (a)\sin \left(b\right)-\sin (a)\sin (b)]\\ & =\frac{4hL}{n^2{\pi }^{2}v}\sin (a)\sin (b)\\ & =\frac{4hL}{n^2{\pi }^{2}v}\sin \left(\frac{n\pi }{2}\right)\sin \left(\frac{n\pi w}{L}\right)\end{array}$$

For even $n$, the term $\sin \left(\frac{n\pi }{2}\right)$ is zero. For $n$ odd $\sin \left(\frac{n\pi }{2}\right)=1$ when $n=1,5,9,\dots$ and $\sin \left(\frac{n\pi }{2}\right)=-1$ when $n=3,7,11,\dots$Hence

$$b_n=A(n)\frac{4hL}{n^2{\pi }^{2}v}\sin \left(\frac{n\pi w}{L}\right)n=1,3,5,7,\dots$$

And $A(n)$ is a function which returns $1$ when $n=1,5,9,..$ and returns $-1$ when $n=3,7,11,\dots$

Hence now we have $b_n$ we can substitute in (1)

$$\begin{array}{rl}y& ={\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{b}_n\sin \frac{n\pi x}{L}\sin \frac{n\pi vt}{L}\\ y& ={\displaystyle \sum _{nodd}^{\mathrm{\infty }}}A(n)\frac{4hL}{n^2{\pi }^{2}v}\sin \left(\frac{n\pi w}{L}\right)[\sin \frac{n\pi x}{L}\sin \frac{n\pi vt}{L}]\\ y& =\frac{4hL}{{\pi }^{2}v}{\displaystyle \sum _{nodd}^{\mathrm{\infty }}}A(n)\frac{1}{n^2}\sin \left(\frac{n\pi w}{L}\right)[\sin \frac{n\pi x}{L}\sin \frac{n\pi vt}{L}]\end{array}$$

Which is the general solution. Looking at few expanded terms in the series we get

$y=\frac{4hL}{{\pi }^{2}v}\{\sin \left(\frac{\pi w}{L}\right)\sin \frac{\pi x}{L}\sin \frac{\pi vt}{L}-\frac{1}{9}\sin \left(\frac{3\pi w}{L}\right)\sin \frac{3\pi x}{L}\sin \frac{3\pi vt}{L}+\frac{1}{25}\sin \left(\frac{5\pi w}{L}\right)\sin \frac{5\pi x}{L}\sin \frac{n\pi vt}{L}\}$

Which is the result required.

4.9.5 chapter 13, problem 5.1. Mary Boas, second edition

Compute numerically the coefficients $c_m=\frac{200}{k_m{J}_1\left(k_m\right)}$ for the first 3 terms of the series $u={\displaystyle \sum _{m=1}^{\mathrm{\infty }}}{c}_m{J}_0\left(k_{m}r\right)e^{-k_{m}z}$ for the steady state temp. in a solid semi-infinite cylinder when $u=0$ at $r=1$ and $u=100$ at $z=0.$ find $u$ at $r=1/2,z=1$

Solution

Here, we are looking at the solution for temp. inside a semi-infinite cylinder. This solution is for the case of a uniform temp. distribution on the boundary $z=0$ is given by $u$ equation shown above. Note that in the expression $c_m=\frac{200}{k_m{J}_1\left(k_m\right)}$, the $k_m$ are the zeros of $J_0$ not $J_1.$

Need to find $c_1,c_2,c_3$ where $c_1=\frac{200}{k_1{J}_1\left(k_1\right)}$

To find $k_1$ and $J_1\left(k_1\right)$ I used mathematica.

I plotted $J_0(x)$ to see where the zeros are located first

So I see there is a zero near 2,5, and 9. I use mathematica to find these:

Now I need to find $J_1\left(k_m\right)$. This is the result for 3 terms:

Hence, now the $c_m$ terms can be found:

$$\begin{array}{rl}{c}_1& =\frac{200}{k_1{J}_1\left(k_1\right)}=\frac{200}{\left(2.404\right)\left(0.519\right)}=160.30\\ c_2& =\frac{200}{k_2{J}_1\left(k_2\right)}=\frac{200}{\left(5.52\right)(-0.34)}=-106.56\\ c_3& =\frac{200}{k_3{J}_1\left(k_3\right)}=\frac{200}{\left(8.65\right)\left(0.27\right)}=85.635\end{array}$$

Evaluating $u={\displaystyle \sum _{m=1}^{\mathrm{\infty }}}{c}_m{J}_0\left(k_{m}r\right)e^{-k_{m}z}$ for the first 3 terms when $r=1/2,z=1$

$$\begin{array}{rl}u& =c_1{J}_0\left(k_{1}r\right)e^{-k_{1}z}+c_2{J}_0\left(k_{2}r\right)e^{-k_{2}z}+c_3{J}_0\left(k_{3}r\right)e^{-k_{3}z}\\ & =c_1{J}_0\left(k_1\frac{1}{2}\right)e^{-k_1}+c_2{J}_0\left(k_2\frac{1}{2}\right)e^{-k_2}+c_3{J}_0\left(k_3\frac{1}{2}\right)e^{-k_3}\\ & =\left(160.30\right)J_0\left(2.404\frac{1}{2}\right)e^{-2.404}-\left(106.56\right)J_0\left(5.52\frac{1}{2}\right)e^{-5.52}+\left(85.635\right)J_0\left(8.65\frac{1}{2}\right)e^{-8.65}\\ & =\left(160.30\right)J_0\left(1.202\right)e^{-2.404}-\left(106.56\right)J_0\left(2.76\right)e^{-5.52}+\left(85.635\right)J_0\left(4.325\right)e^{-8.65}\end{array}$$

Mathematica was used to evaluate $J_0$ values above.

Hence

$$\begin{array}{rl}u& =\left(160.30\right)\left(0.67\right)e^{-2.404}-\left(106.56\right)(-0.168)e^{{}^{-5.52}}+\left(85.635\right)(-0.356)e^{-8.65}\\ u& =9.7043+7.1713\times {10}^{-2}-5.3389\times {10}^{-3}\\ u& =9.7707\text{degrees}\end{array}$$

4.9.6 chapter 13, problem 5.2. Mary Boas, second edition

Find the solution for the steady state temp. distribution in a solid semi-infinite cylinder if the boundary temp. are $u=0$ at $r=1$ and $u=y=r\sin \theta$ at $z=0$.

Solution

The candidate solutions are given by the solution to the Laplace equation in cylindrical coordinates which are

$$u=\{\begin{array}{lll}{J}_n\left(kr\right)\sin \left(n\theta \right)e^{-kz}& & (1)\\ & & \\ J_n\left(kr\right)\cos \left(n\theta \right)e^{-kz}& & (2)\end{array}$$

Where $k$ is a zero of $J_n$ (This is because we have used the B.C. of $u=0$ at $r=1$ to determine that the $k^{\prime }s$ have to be the zeros of $J_n$) when deriving the above solutions. See book page 560.

From boundary conditions we want $u=r\sin \theta$ when $z=0$, hence we need to keep the solution (1) above, with $n=1$. Hence a solution is $$u=J_1\left(kr\right)\sin \left(\theta \right)e^{-kz}(3)$$

A general solution is a linear series combinations (eigenfunctions) of (3), each eigenfunction for each of the zeros of $J_1$. Call these zeros $k_m$

$$u=\sum _{m=1}^{\mathrm{\infty }}{c}_m{J}_1\left(k_{m}r\right)\sin \left(\theta \right)e^{-k_{m}z}(4)$$

We now apply B.C. at $z=0$ to find $c_m$. From (4) when $z=0$ $$r\sin \theta =\sum _{m=1}^{\mathrm{\infty }}{c}_m{J}_1\left(k_{m}r\right)\sin \left(\theta \right)(5)$$

We use (5) to find $c_m$ and then substitute into (4) to obtain the final solution.

To find $c_m$ from (5), take the inner product of each side with respect to $r{J}_1\left(k_{u}r\right)$ from $r=0$ to $r=1$

$$\begin{array}{rl}{\int }_0^{1}r\sin \theta \left[r{J}_1\left(k_{u}r\right)\right]dr& =\sum _{m=1}^{\mathrm{\infty }}{c}_m({\int }_0^1{J}_1\left(k_{m}r\right)\sin \left(\theta \right)\left[r{J}_1\left(k_{u}r\right)\right]dr)\\ \sin \theta {\int }_0^1{r}^2{J}_1\left(k_{u}r\right)dr& =\sum _{m=1}^{\mathrm{\infty }}{c}_m\sin \left(\theta \right)\left({\int }_0^1{J}_1\left(k_{m}r\right)\left[r{J}_1\left(k_{u}r\right)\right]dr\right)\end{array}$$

Dividing each side by $\sin \theta$

$${\int }_0^1{r}^2{J}_1\left(k_{u}r\right)dr=\sum _{m=1}^{\mathrm{\infty }}{c}_m\left({\int }_0^1{J}_1\left(k_{m}r\right)\left[r{J}_1\left(k_{u}r\right)\right]dr\right)$$

From orthogonality of Bessel function, we know that

$${\int }_0^1{J}_p\left(k_{m}r\right)r{J}_p\left(k_{u}r\right)dr=0$$

If $m\ne u$. Hence in above equation all terms on the right drop except for one when $u=m$. We get

$${\int }_0^1{r}^2{J}_1\left(k_{m}r\right)dr=c_m{\int }_0^{1}r{J}_1\left(k_{m}r\right)J_1\left(k_{m}r\right)dr$$

Or

$$c_m=\frac{{\int }_0^1{r}^2{J}_1\left(k_{m}r\right)dr}{{\int }_0^{1}r{J}_1\left(k_{m}r\right)J_1\left(k_{m}r\right)dr}(6)$$

The integral in the denominator above is found from equation 19.10 in text on page 523 which gives $${\int }_0^{1}r{J}_1\left(k_{m}r\right)J_1\left(k_{m}r\right)dr=\frac{1}{2}{\left[J_2\left(k_m\right)\right]}^2(7)$$

Now, we need to find the integral of the numerator in equation (6).

Using equation 15.1 in text, page 514, which says

$$\frac{d}{dx}[x^p{J}_p(x)]=x^p{J}_{p-1}\left(x\right)$$

Putting $p=2$ above, and letting $x=k_{m}r$ gives

$$\begin{array}{rl}\frac{1}{k_m}\frac{d}{dr}\left[{\left(k_{m}r\right)}^2{J}_2\left(k_{m}r\right)\right]& ={\left(k_{m}r\right)}^2{J}_1\left(k_{m}r\right)\\ \frac{1}{k_m}\frac{d}{dr}\left[k_m^2{r}^2{J}_2\left(k_{m}r\right)\right]& =k_m^2{r}^2{J}_1\left(k_{m}r\right)\\ \frac{1}{k_m}\frac{d}{dr}\left[r^2{J}_2\left(k_{m}r\right)\right]& =r^2{J}_1\left(k_{m}r\right)\end{array}$$

Integrating each side w.r.t $r$ from $0\dots 1$

$$\begin{array}{rl}\frac{1}{k_m}{\int }_0^1\frac{d}{dr}\left[r^2{J}_2\left(k_{m}r\right)\right]dr& ={\int }_0^1{r}^2{J}_1\left(k_{m}r\right)dr\\ \frac{1}{k_m}{\left[r^2{J}_2\left(k_{m}r\right)\right]}_0^1& ={\int }_0^1{r}^2{J}_1\left(k_{m}r\right)dr\\ \frac{1}{k_m}[J_2\left(k_m\right)-0]& ={\int }_0^1{r}^2{J}_1\left(k_{m}r\right)dr\\ \frac{1}{k_m}{J}_2\left(k_m\right)& ={\int }_0^1{r}^2{J}_1\left(k_{m}r\right)dr(8)\end{array}$$

Substituting (7) and (8) into (6)

$$\begin{array}{rl}{c}_m& =\frac{{\int }_0^1{r}^2{J}_1\left(k_{m}r\right)dr}{{\int }_0^{1}r{J}_1\left(k_{m}r\right)J_1\left(k_{m}r\right)dr}\\ & =\frac{\frac{1}{k_m}{J}_2\left(k_m\right)}{\frac{1}{2}{\left[J_2\left(k_m\right)\right]}^2}\\ & =\frac{2}{k_m{J}_2\left(k_m\right)}\end{array}$$

Substituting this into (4) above, we get

$$\begin{array}{rl}u& =\sum _{m=1}^{\mathrm{\infty }}{c}_m{J}_1\left(k_{m}r\right)\sin \left(\theta \right)e^{-k_{m}z}\\ u& =\sum _{m=1}^{\mathrm{\infty }}\frac{2}{k_m{J}_2\left(k_m\right)}{J}_1\left(k_{m}r\right)\sin \left(\theta \right)e^{-k_{m}z}\end{array}$$

where $k_m$ are zeros of $J_1$

The above is the result we are asked to show.

4.9.7 chapter 13, problem 5.4. Mary Boass, second edition

A flat circular plate of radius 1 is initially at temp. 100${}^0$. From $t=0$ on, the circumference of the plate is held at 0${}^0.$ Find the time-dependent temp distribution $u(r,\theta ,t)$

Solution

First convert heat equation from Cartesian coordinates to polar.

heat equation in 2D Cartesian is $${\mathrm{\nabla }}^{2}u=\frac{{\partial }^2}{\partial x^2}u+\frac{{\partial }^2}{\partial x^2}u=\frac{1}{{\alpha }^2}\frac{\partial }{\partial t}u$$

First need to express Laplacian operator ${\mathrm{\nabla }}^2$ in polar coordinates: $$\begin{array}{rl}x& =r\cos \theta \\ y& =r\sin \theta \end{array}$$

Hence

$$\begin{array}{rl}\text{(A)}& \frac{\partial x}{\partial r}& =\cos \theta \frac{\partial y}{\partial r}& =\sin \theta \end{array}$$

And

$$\begin{array}{rl}\text{(B)}& \frac{\partial x}{\partial \theta }& =-r\sin \theta \frac{\partial y}{\partial \theta }& =r\cos \theta \end{array}$$

From geometry, we also know that $$\begin{array}{rl}r& =\sqrt{x^2+y^2}\\ \theta & =\arctan \frac{y}{x}\end{array}$$

The above 2 relations imply

$\frac{\partial }{\partial r}=\frac{\partial x}{\partial r}\frac{\partial }{\partial x}+\frac{\partial y}{\partial r}\frac{\partial }{\partial y}$ and   $\frac{\partial }{\partial \theta }=\frac{\partial x}{\partial \theta }\frac{\partial }{\partial x}+\frac{\partial y}{\partial \theta }\frac{\partial }{\partial y}$

Hence we can express the above, using equations (A) and (B) as follows:

$$\begin{array}{rl}\frac{\partial }{\partial r}& =\frac{\partial x}{\partial r}\frac{\partial }{\partial x}+\frac{\partial y}{\partial r}\frac{\partial }{\partial y}\\ & =\cos \theta \frac{\partial }{\partial x}+\sin \theta \frac{\partial }{\partial y}\end{array}$$

Multiply each side by $r$

$$\begin{array}{rl}r\frac{\partial }{\partial r}& =r\cos \theta \frac{\partial }{\partial x}+r\sin \theta \frac{\partial }{\partial y}\\ \text{(1)}& & =x\frac{\partial }{\partial x}+y\frac{\partial }{\partial y}\end{array}$$

Squaring each sides of (1) gives

$$\begin{array}{rl}r\frac{\partial }{\partial r}\left(r\frac{\partial }{\partial r}\right)& ={(x\frac{\partial }{\partial x}+y\frac{\partial }{\partial y})}^2\\ r(r\frac{{\partial }^2}{\partial r^2}+\frac{\partial }{\partial r})& =x\frac{\partial }{\partial x}x\frac{\partial }{\partial x}+y\frac{\partial }{\partial y}y\frac{\partial }{\partial y}+2x\frac{\partial }{\partial x}y\frac{\partial }{\partial y}\\ r^2\frac{{\partial }^2}{\partial r^2}+r\frac{\partial }{\partial r}& =x\frac{\partial }{\partial x}\left(x\frac{\partial }{\partial x}\right)+y\frac{\partial }{\partial y}\left(y\frac{\partial }{\partial y}\right)+2x\frac{\partial }{\partial x}\left(y\frac{\partial }{\partial y}\right)\\ & =x(x\frac{{\partial }^2}{\partial x^2}+\stackrel{=1}{\overbrace{\frac{\partial }{\partial x}x}}\frac{\partial }{\partial x})+y(y\frac{{\partial }^2}{\partial y^2}+\stackrel{=1}{\overbrace{\frac{\partial }{\partial y}y}}\frac{\partial }{\partial y})+2x(y\frac{{\partial }^2}{\partial x\partial y}+\stackrel{=0}{\overbrace{\frac{\partial }{\partial x}y}}\frac{\partial }{\partial y})\\ \text{(2)}& & =x^2\frac{{\partial }^2}{\partial x^2}+x\frac{\partial }{\partial x}+y^2\frac{{\partial }^2}{\partial y^2}+y\frac{\partial }{\partial y}+2xy\frac{{\partial }^2}{\partial x\partial y}\end{array}$$

Notice that when manipulating of differential operators, $x\frac{\partial }{\partial x}\ne \frac{\partial }{\partial x}x$. Similarly

$$\begin{array}{rl}\frac{\partial }{\partial \theta }& =\frac{\partial x}{\partial \theta }\frac{\partial }{\partial x}+\frac{\partial y}{\partial \theta }\frac{\partial }{\partial y}\\ & =-r\sin \theta \frac{\partial }{\partial x}+r\cos \theta \frac{\partial }{\partial y}\\ \text{(3)}& & =-y\frac{\partial }{\partial x}+x\frac{\partial }{\partial y}\end{array}$$

Squaring each side of (3) gives

$$\begin{array}{rl}{\left(\frac{\partial }{\partial \theta }\right)}^2& ={(-y\frac{\partial }{\partial x}+x\frac{\partial }{\partial y})}^2\\ \frac{\partial }{\partial \theta }\frac{\partial }{\partial \theta }& =-y\frac{\partial }{\partial x}(-y\frac{\partial }{\partial x})+x\frac{\partial }{\partial y}\left(x\frac{\partial }{\partial y}\right)-y\frac{\partial }{\partial x}\left(x\frac{\partial }{\partial y}\right)+x\frac{\partial }{\partial y}(-y\frac{\partial }{\partial x})\\ \frac{{\partial }^2}{\partial {\theta }^2}& =-y(-y\frac{{\partial }^2}{\partial x^2}+\stackrel{=0}{\overbrace{\frac{\partial }{\partial x}(-y)}}\frac{\partial }{\partial x})+x(x\frac{{\partial }^2}{\partial y^2}+\stackrel{=0}{\overbrace{\frac{\partial }{\partial y}(x)}}\frac{\partial }{\partial y})\\ & -y(x\frac{{\partial }^2}{\partial y\partial x}+\stackrel{=1}{\overbrace{\frac{\partial }{\partial x}x}}\frac{\partial }{\partial y})+x(-y\frac{{\partial }^2}{\partial x\partial y}+\stackrel{=1}{\overbrace{\frac{\partial }{\partial y}y}}\frac{\partial }{\partial x})\\ & =y^2\frac{{\partial }^2}{\partial x^2}+x^2\frac{{\partial }^2}{\partial y^2}-yx\frac{{\partial }^2}{\partial y\partial x}-y\frac{\partial }{\partial y}-xy\frac{{\partial }^2}{\partial x\partial y}-x\frac{\partial }{\partial x}\\ \text{(4)}& & =y^2\frac{{\partial }^2}{\partial x^2}+x^2\frac{{\partial }^2}{\partial y^2}-2yx\frac{{\partial }^2}{\partial y\partial x}-y\frac{\partial }{\partial y}-x\frac{\partial }{\partial x}\end{array}$$

Adding equation (2) and (4) and carry cancellations

$r^2\frac{{\partial }^2}{\partial r^2}+r\frac{\partial }{\partial r}+\frac{{\partial }^2}{\partial {\theta }^2}=(x^2\frac{{\partial }^2}{\partial x^2}+x\frac{\partial }{\partial x}+y^2\frac{{\partial }^2}{\partial y^2}+y\frac{\partial }{\partial y}+2xy\frac{{\partial }^2}{\partial x\partial y})+(y^2\frac{{\partial }^2}{\partial x^2}+x^2\frac{{\partial }^2}{\partial y^2}-2yx\frac{{\partial }^2}{\partial y\partial x}-y\frac{\partial }{\partial y}-x\frac{\partial }{\partial x})$

$r^2\frac{{\partial }^2}{\partial r^2}+r\frac{\partial }{\partial r}+\frac{{\partial }^2}{\partial {\theta }^2}=(x^2\frac{{\partial }^2}{\partial x^2}+y^2\frac{{\partial }^2}{\partial y^2})+(y^2\frac{{\partial }^2}{\partial x^2}+x^2\frac{{\partial }^2}{\partial y^2})$

Hence we get

$$\begin{array}{rl}{r}^2\frac{{\partial }^2}{\partial r^2}+r\frac{\partial }{\partial r}+\frac{{\partial }^2}{\partial {\theta }^2}& =x^2\frac{{\partial }^2}{\partial x^2}+y^2\frac{{\partial }^2}{\partial y^2}+y^2\frac{{\partial }^2}{\partial x^2}+x^2\frac{{\partial }^2}{\partial y^2}\\ & =(x^2+y^2)(\frac{{\partial }^2}{\partial x^2}+\frac{{\partial }^2}{\partial y^2})\end{array}$$

Dividing by $r^2$

$$\frac{{\partial }^2}{\partial r^2}+\frac{1}{r}\frac{\partial }{\partial r}+\frac{1}{r^2}\frac{{\partial }^2}{\partial {\theta }^2}=\frac{(x^2+y^2)}{r^2}(\frac{{\partial }^2}{\partial x^2}+\frac{{\partial }^2}{\partial y^2})$$

But $r^2=x^2+y^2$ hence

$$\frac{{\partial }^2}{\partial r^2}+\frac{1}{r}\frac{\partial }{\partial r}+\frac{1}{r^2}\frac{{\partial }^2}{\partial {\theta }^2}=(\frac{{\partial }^2}{\partial x^2}+\frac{{\partial }^2}{\partial y^2})={\mathrm{\nabla }}^2$$

Now that we have the Laplacian in polar coordinates, we can solve the problem by applying separation of variables on the heat PDE expressed in polar coordinates.

$$\begin{array}{}\text{(5)}& \frac{{\partial }^2}{\partial r^2}u+\frac{1}{r}\frac{\partial }{\partial r}u+\frac{1}{r^2}\frac{{\partial }^2}{\partial {\theta }^2}u=\frac{1}{{\alpha }^2}\frac{\partial }{\partial t}u\end{array}$$

Let solution $u(r,\theta ,t)$ be a linear combination of functions each depends on only $r,\theta ,$ or $t$

$$\begin{array}{}\text{(6)}& u(r,\theta ,t)=R(r)\mathrm{\Theta }\left(\theta \right)T(t)\end{array}$$

Substitute (6) in (5). First evaluate the various derivatives:

$$\frac{\partial }{\partial r}u=\mathrm{\Theta }\left(\theta \right)T(t)\frac{\partial }{\partial r}R(r)$$

$$\frac{{\partial }^2}{\partial r^2}u=\mathrm{\Theta }\left(\theta \right)T\left(t\right)\frac{{\partial }^2}{\partial r^2}R(r)$$

$$\frac{\partial }{\partial \theta }u=R(r)T(t)\frac{\partial }{\partial \theta }\mathrm{\Theta }\left(\theta \right)$$

$$\frac{{\partial }^2}{\partial {\theta }^2}u=R(r)T(t)\frac{{\partial }^2}{\partial {\theta }^2}\mathrm{\Theta }\left(\theta \right)$$

$$\frac{\partial }{\partial t}u=R(r)\mathrm{\Theta }\left(\theta \right)\frac{\partial }{\partial t}T(t)$$

Hence equation (5) becomes

$$\begin{array}{rl}\frac{{\partial }^2}{\partial r^2}u+\frac{1}{r}\frac{\partial }{\partial r}u+\frac{1}{r^2}\frac{{\partial }^2}{\partial {\theta }^2}u& =\frac{1}{{\alpha }^2}\frac{\partial }{\partial t}u\\ \mathrm{\Theta }\left(\theta \right)T(t)\frac{d^2}{d{r}^2}R\left(r\right)+\frac{1}{r}\mathrm{\Theta }\left(\theta \right)T(t)\frac{d}{dr}R(r)+\frac{1}{r^2}R(r)T\left(t\right)\frac{d^2}{d{\theta }^2}\mathrm{\Theta }\left(\theta \right)& =\frac{1}{{\alpha }^2}R(r)\mathrm{\Theta }\left(\theta \right)\frac{d}{dt}T(t)\end{array}$$

Divide by $R(r)\mathrm{\Theta }\left(\theta \right)T(t)$

$$\begin{array}{rl}\frac{1}{R(r)}\frac{d^2}{d{r}^2}R(r)+\frac{1}{r}\frac{1}{R(r)}\frac{d}{dr}R(r)+\frac{1}{r^2}\frac{1}{\mathrm{\Theta }\left(\theta \right)}\frac{d^2}{d{\theta }^2}\mathrm{\Theta }\left(\theta \right)& =\frac{1}{{\alpha }^2}\frac{1}{T(t)}\frac{d}{dt}T(t)\\ \frac{1}{R(r)}[\frac{d^2}{d{r}^2}R(r)+\frac{1}{r}\frac{d}{dr}R(r)]+\frac{1}{r^2}\frac{1}{\mathrm{\Theta }\left(\theta \right)}\frac{d^2}{d{\theta }^2}\mathrm{\Theta }\left(\theta \right)& =\frac{1}{{\alpha }^2}\frac{1}{T(t)}\frac{d}{dt}T(t)\end{array}$$

We notice that the RHS depends only on $t$ and the LHS depends only on $r,\theta$ and they equal to each others, hence they both must be constant. Let this constant be $-k^2$

Hence

$$\begin{array}{}\text{(7)}& \frac{1}{{\alpha }^2}\frac{1}{T(t)}\frac{d}{dt}T\left(t\right)& =-k^2\text{(8)}& \frac{1}{R(r)}[\frac{d^2}{d{r}^2}R(r)+\frac{1}{r}\frac{d}{dr}R(r)]+\frac{1}{r^2}\frac{1}{\mathrm{\Theta }\left(\theta \right)}\frac{d^2}{d{\theta }^2}\mathrm{\Theta }\left(\theta \right)& =-k^2\end{array}$$

equation (7) is a linear first order ODE with constant coeff. $\frac{d}{dt}T(t)=-{\alpha }^{2}T(t)k^2$ or $\frac{dT(t)}{T(t)}=-{\alpha }^2{k}^{2}dt$

Integrating to solve gives $$\begin{array}{rl}\int \frac{dT(t)}{T(t)}& =\int -{\alpha }^2{k}^{2}dt\\ \ln T(t)& =-{\alpha }^2{k}^{2}t\end{array}$$

or$$\begin{array}{}\text{(9)}& T(t)=e^{-{\alpha }^2{k}^{2}t}\end{array}$$ Looking at equation (8). Multiply each sides by $r^2$ we get$$\begin{array}{rl}\frac{r^2}{R(r)}[\frac{d^2}{d{r}^2}R\left(r\right)+\frac{1}{r}\frac{d}{dr}R(r)]+\frac{1}{\mathrm{\Theta }\left(\theta \right)}\frac{d^2}{d{\theta }^2}\mathrm{\Theta }\left(\theta \right)& =-r^2{k}^2\\ \frac{r^2}{R(r)}[\frac{d^2}{d{r}^2}R\left(r\right)+\frac{1}{r}\frac{d}{dr}R(r)]+r^{2}k+\frac{1}{\mathrm{\Theta }\left(\theta \right)}\frac{d^2}{d{\theta }^2}\mathrm{\Theta }\left(\theta \right)& =0\\ \text{(10)}& r^2(\frac{1}{R(r)}[\frac{d^2}{d{r}^2}R(r)+\frac{1}{r}\frac{d}{dr}R(r)]+k^2)+\frac{1}{\mathrm{\Theta }\left(\theta \right)}\frac{d^2}{d{\theta }^2}\mathrm{\Theta }\left(\theta \right)& =0\end{array}$$

The second term depends only on $\theta$ and the first term depends only on $r$ and they are equal, hence they must be both constant. Let this constant be $-n^2$ hence

$$\begin{array}{rl}\frac{1}{\mathrm{\Theta }\left(\theta \right)}\frac{d^2}{d{\theta }^2}\mathrm{\Theta }\left(\theta \right)& =-n^2\\ \frac{d^2}{d{\theta }^2}\mathrm{\Theta }\left(\theta \right)& =-n^2\mathrm{\Theta }\left(\theta \right)\end{array}$$

This is a second order linear ODE with constant coeff. Solution is

$$\begin{array}{}\text{(11)}& \mathrm{\Theta }\left(\theta \right)=\{\begin{array}{lll}\sin n\theta & & \\ & & \\ \cos n\theta & & \end{array}\end{array}$$

From (10) we now have

$$\begin{array}{rl}{r}^2(\frac{1}{R(r)}[\frac{d^2}{d{r}^2}R(r)+\frac{1}{r}\frac{d}{dr}R(r)]+k^2)-n^2& =0\\ \frac{r^2}{R(r)}[\frac{d^2}{d{r}^2}R\left(r\right)+\frac{1}{r}\frac{d}{dr}R(r)]+r^2{k}^2-n^2& =0\\ r^2[\frac{d^2}{d{r}^2}R(r)+\frac{1}{r}\frac{d}{dr}R(r)]+(r^2{k}^2-n^2)R\left(r\right)& =0\\ \text{(12)}& r^2\frac{d^2}{d{r}^2}R(r)+r\frac{d}{dr}R(r)+(r^2{k}^2-n^2)R(r)& =0\end{array}$$

Equation (12) is the Bessel D.E., its solutions are $J_n\left(kr\right)$ and $N_n\left(kr\right)$ . As described on book on page 560, we can not use the $N_n\left(kr\right)$ solution since plate contains the origin and $N_n(0)$ is not defined. So we use solution $R\left(r\right)=J_n\left(kr\right).$ From boundary conditions, we want solution to be zero at $r=1$, hence we want $J_n(k)=0$, hence the k’s are the zeros of $J_n$

Putting these solutions together, we get from (6)

$$\begin{array}{rl}u(r,\theta ,t)& =R(r)\mathrm{\Theta }\left(\theta \right)T(t)\\ & =\{\begin{array}{lll}{J}_n\left(kr\right)\sin n\theta e^{-{\alpha }^2{k}^{2}t}& & \\ & & \\ J_n\left(kr\right)\cos n\theta e^{-{\alpha }^2{k}^{2}t}& & \end{array}\end{array}$$

From symmetry of plate, the solution can not depend on the angle $\theta$, hence let $n=0$ and so as not to get $u=0$, we must pick the solution with $\cos n\theta$ term. Hence our solution now is

$$u(r,t)=J_0\left(kr\right)e^{-{\alpha }^2{k}^{2}t}$$

Where $k$ is a zero of $J_0$

The general solution is a linear combination of this eigenfunction for all zeros of $J_0$, hence

$$\begin{array}{}\text{(13)}& u(r,t)=\sum _{m=1}^{\mathrm{\infty }}{c}_m{J}_0\left(k_{m}r\right)e^{-{\alpha }^2{k}_m^{2}t}\end{array}$$

We find $c_m$ by using initial condition. When $t=0$ $,$ temp. was 100${}^0$ hence

$$100=\sum _{m=1}^{\mathrm{\infty }}{c}_m{J}_0\left(k_{m}r\right)$$

Applying inner product w.r.t. $r{J}_0\left(k_{u}r\right)$ from $0\dots 1$

$$\begin{array}{rl}{\int }_0^{1}100r{J}_0\left(k_{u}r\right)dr& ={\int }_0^1\left(\sum _{m=1}^{\mathrm{\infty }}{c}_m{J}_0\left(k_{m}r\right)\right)r{J}_0\left(k_{u}r\right)dr\\ 100{\int }_0^{1}r{J}_0\left(k_{u}r\right)dr& =\sum _{m=1}^{\mathrm{\infty }}{c}_m{\int }_0^1{J}_0\left(k_{m}r\right)r{J}_0\left(k_{u}r\right)dr\end{array}$$

From orthogonality of $J_0\left(k_{m}r\right)$and $J_0\left(k_{u}r\right)$, all terms drop expect when $m=u$

$$100{\int }_0^{1}r{J}_0\left(k_{u}r\right)dr=c_u{\int }_0^{1}r{\left[J_0\left(k_{u}r\right)\right]}^{2}dr$$

From here we can follow the book on page 561 to get $$c_m=\frac{200}{k_m{J}_1\left(k_m\right)}$$

Substitute this in equation 13

$$\begin{array}{rl}u(r,t)& =\sum _{m=1}^{\mathrm{\infty }}{c}_m{J}_0\left(k_{m}r\right)e^{-{\alpha }^2{k}_m^{2}t}\\ & =\sum _{m=1}^{\mathrm{\infty }}\frac{200}{k_m{J}_1\left(k_m\right)}{J}_0\left(k_{m}r\right)e^{-{\alpha }^2{k}_m^{2}t}\\ & =200\sum _{m=1}^{\mathrm{\infty }}\frac{1}{k_m{J}_1\left(k_m\right)}{J}_0\left(k_{m}r\right)e^{-{\alpha }^2{k}_m^{2}t}\end{array}$$

Where $k_m$ are zeros of J${}_0$

Notice that final solution does not depend on $\theta$

4.9.8 chapter 13, problem 5.11. Mary Boas, second edition

Solve $$\begin{array}{rl}r\frac{d}{dr}\left(r\frac{dR}{dr}\right)& =n^{2}R\\ \frac{d}{dr}\left(r^2\frac{dR}{dr}\right)& =l(l+1)R\end{array}$$

Solution

First equation, use power series method.

$$\begin{array}{rl}r\frac{d}{dr}\left(r\frac{dR}{dr}\right)& =n^{2}R\\ r(r\frac{d^{2}R}{d{r}^2}+\frac{dR}{dr})-n^{2}R& =0\\ r^2\frac{d^{2}R}{d{r}^2}+r\frac{dR}{dr}-n^{2}R& =0\end{array}$$

Let $R=a_0{r}^s+a_1{r}^{s+1}+a_2{r}^{s+2}+a_3{r}^{s+3}+a_4{r}^{s+4}+\cdots$ then

$$\begin{array}{rl}R& =a_0{r}^s+a_1{r}^{s+1}+a_2{r}^{s+2}+a_3{r}^{s+3}+a_4{r}^{s+4}+\cdots \\ -n^{2}R& =-n^2{a}_0{r}^s-n^2{a}_1{r}^{s+1}-n^2{a}_2{r}^{s+2}-n^2{a}_3{r}^{s+3}-n^2{a}_4{r}^{s+4}-\cdots \\ \frac{dR}{dr}& =s{a}_0{r}^{s-1}+(s+1)a_1{r}^s+(s+2)a_2{r}^{s+1}+(s+3)a_3{r}^{s+2}+\cdots \\ r\frac{dR}{dr}& =s{a}_0{r}^s+(s+1)a_1{r}^{s+1}+(s+2)a_2{r}^{s+2}+(s+3)a_3{r}^{s+3}+\cdots \\ \frac{d^{2}R}{d{r}^2}& =(s-1)s{a}_0{r}^{s-2}+s(s+1)a_1{r}^{s-1}+(s+1)(s+2)a_2{r}^s+(s+2)(s+3)a_3{r}^{s+1}+\cdots \\ r^2\frac{d^{2}R}{d{r}^2}& =(s-1)s{a}_0{r}^s+s(s+1)a_1{r}^{s+1}+(s+1)(s+2)a_2{r}^{s+2}+(s+2)(s+3)a_3{r}^{s+3}+\cdots \end{array}$$

Table is

	$r^s$	$r^{s+1}$	$r^{s+2}$	$r^{s+m}$
$-n^{2}R$	$-n^2{a}_0$	$-n^2{a}_1$	$-n^2{a}_2$	$-n^2{a}_m$
$r\frac{dR}{dr}$	$s{a}_0$	$(s+1)a_1$	$(s+2)a_2$	$(s+m)a_m$
$r^2\frac{d^{2}R}{d{r}^2}$	$(s-1)s{a}_0$	$s(s+1)a_1$	$(s+1)(s+2)a_2$	$(s+m-1)(s+m)a_m$

Hence, from first column we see , and since $a_0$ $\ne 0$ we solve for $s$

$$\begin{array}{rl}-n^2{a}_0+s{a}_0+(s-1)s{a}_0& =0\\ a_0(-n^2+s+(s-1)s)& =0\\ -n^2+s+(s-1)s& =0\\ -n^2+s^2& =0\\ s& =\pm n\end{array}$$

We see from second column, $a_1(-n^2+(s+1)+s^2+s)=0$ or $a_1(-s^2+2s+1+s^2)=0$, hence $a_1(2s+1)=0$

For $a_1\ne 0$ then $s=-\frac{1}{2}$, this means $n$ is not an integer since $s=\pm n$. hence $a_1$ must be zero.

The same applies to all $a_m$ , $m>0$ Hence solution contains only $a_0$

$$\begin{array}{rl}R& =a_0{r}^{\pm n}\\ R& =\{\begin{array}{lll}{a}_0{r}^{-n}& & \\ & & \\ a_0{r}^{+n}& & \end{array}\end{array}$$

For some constant $a_0$. This solution is when $n\ne 0$

If $n=0$, table is

	$r^s$	$r^{s+1}$	$r^{s+2}$	$r^{s+m}$
$-n^{2}R$	$0$	$0$	$0$	$0$
$r\frac{dR}{dr}$	$s{a}_0$	$(s+1)a_1$	$(s+2)a_2$	$(s+m)a_m$
$r^2\frac{d^{2}R}{d{r}^2}$	$(s-1)s{a}_0$	$s(s+1)a_1$	$(s+1)(s+2)a_2$	$(s+m-1)(s+m)a_m$

From first column:

$$\begin{array}{rl}s{a}_0+s^2{a}_0-s{a}_0& =0\\ a_0(s+s^2-s)& =0\\ s^2& =0\\ s& =0\end{array}$$

And all other $a^{\prime }s$ are zero. Hence $R=a_0$ or $R$ is constant.

Now for the second ODE

$$\begin{array}{rl}\frac{d}{dr}\left(r^2\frac{dR}{dr}\right)& =l(l+1)R\\ r^2\frac{d^{2}R}{d{r}^2}+2r\frac{dR}{dr}-l(l+1)R& =0\end{array}$$

Let $R=a_0{r}^s+a_1{r}^{s+1}+a_2{r}^{s+2}+a_3{r}^{s+3}+a_4{r}^{s+4}+\cdots$ then

$$\begin{array}{rl}R& =a_0{r}^s+a_1{r}^{s+1}+a_2{r}^{s+2}+a_3{r}^{s+3}+a_4{r}^{s+4}+\cdots \\ -l(l+1)R& =-l(l+1)a_0{r}^s-l(l+1)a_1{r}^{s+1}-l(l+1)a_2{r}^{s+2}-l(l+1)a_3{r}^{s+3}-l(l+1)a_4{r}^{s+4}-\cdots \\ \frac{dR}{dr}& =s{a}_0{r}^{s-1}+(s+1)a_1{r}^s+(s+2)a_2{r}^{s+1}+(s+3)a_3{r}^{s+2}+\cdots \\ 2r\frac{dR}{dr}& =2s{a}_0{r}^s+2(s+1)a_1{r}^{s+1}+2(s+2)a_2{r}^{s+2}+2(s+3)a_3{r}^{s+3}+\cdots \\ \frac{d^{2}R}{d{r}^2}& =(s-1)s{a}_0{r}^{s-2}+s(s+1)a_1{r}^{s-1}+(s+1)(s+2)a_2{r}^s+(s+2)(s+3)a_3{r}^{s+1}+\cdots \\ r^2\frac{d^{2}R}{d{r}^2}& =(s-1)s{a}_0{r}^s+s(s+1)a_1{r}^{s+1}+(s+1)(s+2)a_2{r}^{s+2}+(s+2)(s+3)a_3{r}^{s+3}+\cdots \end{array}$$

Table is

	$r^s$	$r^{s+1}$	$r^{s+2}$	$r^{s+m}$
$-n^{2}R$	$-l(l+1)a_0$	$-l(l+1)a_1$	$-l(l+1)a_2$	$-l(l+1)a_m$
$2r\frac{dR}{dr}$	$2s{a}_0$	$2(s+1)a_1$	$2(s+2)a_2$	$2(s+m)a_m$
$r^2\frac{d^{2}R}{d{r}^2}$	$(s-1)s{a}_0$	$s(s+1)a_1$	$(s+1)(s+2)a_2$	$(s+m-1)(s+m)a_m$

From first column: $$\begin{array}{rl}-l(l+1)a_0+2s{a}_0+(s-1)s{a}_0& =0\\ a_0(-l(l+1)+2s+(s-1)s)& =0\\ -l(l+1)+2s+(s-1)s& =0\\ -l(l+1)+s+s^2& =0\\ (s-l)(s-(-l-1))& =0\end{array}$$

Hence $s=l$ or $s=-l-1$.

We also see that all other $a^{\prime }s$ will be zero, since recursive formula has only $a_m$ in it and no other $a$. Hence

$$\begin{array}{rl}R& =a_0{r}^s\\ R& =\{\begin{array}{lll}{a}_0{r}^l& & \\ & & \\ a_0{r}^{-l-1}& & \end{array}\end{array}$$

For some constant $a_0$