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2.11 HW 11

2.11.1 HW 11 questions

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2.11.2 Problem 1

Find the normal modes of a rectangular drum with sides of length $L_{x}$ and $L_{y}$

solution

The geometry of the problem is

Figure 2.37:Problem to solve

Using Cartesian coordinates. Wave displacement is $u\equiv u\left ( x,y,t\right )$ (out of page).

$$\begin{align*} \frac{\partial ^{2}u\left ( x,y,t\right ) }{\partial t^{2}} & =c^{2}\left ( \frac{\partial ^{2}u}{\partial x^{2}}+\frac{\partial ^{2}u}{\partial y^{2}}\right ) \\ 0 & <x<L_{x}\\ 0 & <y<L_{y} \end{align*}$$

Boundary conditions on $x$

$$\begin{align*} u\left ( 0,y,t\right ) & =0\\ u\left ( L_{x},y,t\right ) & =0 \end{align*}$$

And boundary conditions on $y$

$$\begin{align*} u\left ( x,0,t\right ) & =0\\ u\left ( x,L_{y},t\right ) & =0 \end{align*}$$

Solution

Let $u=X\left ( x\right ) Y\left ( y\right ) T\left ( t\right )$. Substituting into the PDE gives

$$\begin{align*} \frac{1}{c^{2}}T^{\prime \prime }XY & =X^{\prime \prime }YT+Y^{\prime \prime }XT\\ \frac{1}{c^{2}}\frac{T^{\prime \prime }}{T} & =\frac{X^{\prime \prime }}{X}+\frac{Y^{\prime \prime }}{Y} \end{align*}$$

Hence, using $\lambda$ as first separation constant we obtain

$$\begin{align*} \frac{1}{c^{2}}\frac{T^{\prime \prime }}{T} & =-\lambda \\ \frac{X^{\prime \prime }}{X}+\frac{Y^{\prime \prime }}{Y} & =-\lambda \end{align*}$$

The time ODE becomes

$$T^{\prime \prime }+c^{2}\lambda T=0$$ And the space ODE becomes $$\frac{X^{\prime \prime }}{X}+\frac{Y^{\prime \prime }}{Y}=-\lambda$$ Separating the space ODE again $$\frac{X^{\prime \prime }}{X}=-\lambda -\frac{Y^{\prime \prime }}{Y}=-\mu$$ Where $\mu$ is the new separation variable. This gives two new separate ODE’s $$\begin{align*} \frac{X^{\prime \prime }}{X} & =-\mu \\ -\lambda -\frac{Y^{\prime \prime }}{Y} & =-\mu \end{align*}$$

Or

$$\begin{align*} X^{\prime \prime }+\mu X & =0\\ Y^{\prime \prime }+Y\left ( \lambda -\mu \right ) & =0 \end{align*}$$

Solving for $X$ ODE first, and knowing that $\mu >0$ from nature of boundary conditions, we obtain

$$X\left ( x\right ) =A\cos \left ( \sqrt{\mu }x\right ) +B\sin \left ( \sqrt{\mu }x\right )$$ Applying B.C. at $x=0$ $$0=A$$ Hence $X\left ( x\right ) =B\sin \left ( \sqrt{\mu }x\right )$. Applying B.C. at $x=L_{x}$ $$0=B\sin \left ( \sqrt{\mu }L_{x}\right )$$ Hence $$\begin{align} \sqrt{\mu }L_{x} & =n\pi \nonumber \\ \mu _{n} & =\left ( \frac{n\pi }{L_{x}}\right ) ^{2}\qquad n=1,2,3,\cdots \tag{1} \end{align}$$

Therefore the $X_{n}\left ( x\right )$ solution is

$$\begin{equation} X_{n}\left ( x\right ) =B_{n}\sin \left ( \frac{n\pi }{L_{x}}x\right ) \qquad n=1,2,3,\cdots \tag{2} \end{equation}$$ Solving the $Y\left ( y\right )$ ODE using the same eigenvalues found above $$Y^{\prime \prime }+Y\left ( \lambda -\left ( \frac{n\pi }{L_{x}}\right ) ^{2}\right ) =0$$ The solution is $$Y\left ( y\right ) =C\cos \left ( \sqrt{\lambda -\left ( \frac{n\pi }{L_{x}}\right ) ^{2}}y\right ) +D\sin \left ( \sqrt{\lambda -\left ( \frac{n\pi }{L_{x}}\right ) ^{2}}y\right )$$ Applying first B.C. $Y\left ( 0\right ) =0$ gives $$0=C$$ Hence $$Y\left ( y\right ) =D\sin \left ( \sqrt{\lambda -\left ( \frac{n\pi }{L_{x}}\right ) ^{2}}y\right )$$ Applying second B.C. $Y\left ( L_{y}\right ) =0$ $$0=D\sin \left ( \sqrt{\lambda -\left ( \frac{n\pi }{L_{x}}\right ) ^{2}}L_{y}\right )$$ Hence $$\begin{align*} \sqrt{\lambda -\left ( \frac{n\pi }{L_{x}}\right ) ^{2}}L_{y} & =m\pi \qquad m=1,2,3,\cdots \\ \lambda _{nm}-\left ( \frac{n\pi }{L_{x}}\right ) ^{2} & =\left ( \frac{m\pi }{L_{y}}\right ) ^{2}\\ \lambda _{nm} & =\left ( \frac{m\pi }{L_{y}}\right ) ^{2}+\left ( \frac{n\pi }{L_{x}}\right ) ^{2}\qquad n=1,2,3,\cdots ,m=1,2,3,\cdots \end{align*}$$

Hence the $Y_{nm}$ solution is

$$Y_{nm}=D_{nm}\sin \left ( \frac{m\pi }{L_{y}}y\right ) \qquad n=1,2,3,\cdots ,m=1,2,3,\cdots$$ We notice that $X_{n}\left ( x\right )$ solution depends on $n$ only, while $Y_{nm}\left ( y\right )$ solution depends on $n$ and $m.$ Now that we found $\lambda$ we can we solve the time $T\left ( t\right )$ ode $$\begin{align*} T_{nm}^{\prime \prime }+c^{2}\lambda _{nm}T_{nm} & =0\\ T_{nm}\left ( t\right ) & =E_{nm}\cos \left ( c\sqrt{\lambda _{nm}}t\right ) +F_{nm}\sin \left ( c\sqrt{\lambda _{nm}}t\right ) \end{align*}$$

Combining all solution , and merging all constants into two, we find

$$\begin{align*} u_{nm}\left ( x,y,t\right ) & =X_{n}\left ( x\right ) Y_{nm}\left ( y\right ) T_{nm}\left ( t\right ) \\ & =\left ( B_{n}X_{n}\right ) \left ( D_{nm}\sin \left ( \frac{m\pi }{L_{y}}y\right ) \right ) \left ( E_{nm}\cos \left ( c\sqrt{\lambda _{nm}}t\right ) +F_{nm}\sin \left ( c\sqrt{\lambda _{nm}}t\right ) \right ) \\ & =B_{n}X_{n}\sin \left ( \frac{m\pi }{L_{y}}y\right ) \left ( E_{nm}^{\prime }\cos \left ( c\sqrt{\lambda _{nm}}t\right ) +F_{nm}^{\prime }\sin \left ( c\sqrt{\lambda _{nm}}t\right ) \right ) \\ & =X_{n}\sin \left ( \frac{m\pi }{L_{y}}y\right ) \left ( E_{nm}^{\prime \prime }\cos \left ( c\sqrt{\lambda _{nm}}t\right ) +F_{nm}^{\prime \prime }\sin \left ( c\sqrt{\lambda _{nm}}t\right ) \right ) \end{align*}$$

Where $E_{nm}^{\prime \prime },F_{nm}^{\prime \prime }$ are the new constants after merging them with the other constants. Renaming $E_{nm}^{\prime \prime }=A_{nm},F_{nm}^{\prime \prime }=B_{nm}$ the above solution can be written as

$$\begin{align} u\left ( x,y,t\right ) & =\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }X_{n}\left ( x\right ) Y_{mn}\left ( y\right ) T_{mn}\left ( t\right ) \nonumber \\ & =\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }A_{nm}\sin \left ( \frac{n\pi }{L_{x}}x\right ) \sin \left ( \frac{m\pi }{L_{y}}y\right ) \cos \left ( c\sqrt{\lambda _{nm}}t\right ) \nonumber \\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }B_{nm}\sin \left ( \frac{n\pi }{L_{x}}x\right ) \sin \left ( \frac{m\pi }{L_{y}}y\right ) \sin \left ( c\sqrt{\lambda _{nm}}t\right ) \tag{3} \end{align}$$

To solve this completely, we apply initial conditions to find $A_{nm},B_{nm}$. But the problem is just asking for the normal modes. These are given by $X_{n}\left ( x\right ) Y_{mn}\left ( y\right )$. Therefore for $n=1$, we have the modes $\sin \left ( \frac{\pi }{L_{x}}x\right ) \sin \left ( \frac{\pi }{L_{y}}y\right ) ,\sin \left ( \frac{\pi }{L_{x}}x\right ) \sin \left ( \frac{2\pi }{L_{y}}y\right ) ,\sin \left ( \frac{\pi }{L_{x}}x\right ) \sin \left ( \frac{3\pi }{L_{y}}y\right ) ,\cdots$ and for $n=2$ we have $\sin \left ( \frac{2\pi }{L_{x}}x\right ) \sin \left ( \frac{\pi }{L_{y}}y\right ) ,\sin \left ( \frac{2\pi }{L_{x}}x\right ) \sin \left ( \frac{2\pi }{L_{y}}y\right ) ,\sin \left ( \frac{2\pi }{L_{x}}x\right ) \sin \left ( \frac{3\pi }{L_{y}}y\right ) ,\cdots$ and so on.

$n$	$m=1$	$2$	$3$	$4$
$1$	$\sin \left ( \frac{\pi }{L_{x}}x\right ) \sin \left ( \frac{\pi }{L_{y}}y\right )$	$\sin \left ( \frac{\pi }{L_{x}}x\right ) \sin \left ( \frac{2\pi }{L_{y}}y\right )$	$\sin \left ( \frac{\pi }{L_{x}}x\right ) \sin \left ( \frac{3\pi }{L_{y}}y\right )$	$\cdots$
$2$	$\sin \left ( \frac{2\pi }{L_{x}}x\right ) \sin \left ( \frac{\pi }{L_{y}}y\right )$	$\sin \left ( \frac{2\pi }{L_{x}}x\right ) \sin \left ( \frac{2\pi }{L_{y}}y\right )$	$\sin \left ( \frac{2\pi }{L_{x}}x\right ) \sin \left ( \frac{3\pi }{L_{y}}y\right )$	$\cdots$
$3$	$\sin \left ( \frac{3\pi }{L_{x}}x\right ) \sin \left ( \frac{\pi }{L_{y}}y\right )$	$\sin \left ( \frac{3\pi }{L_{x}}x\right ) \sin \left ( \frac{2\pi }{L_{y}}y\right )$	$\sin \left ( \frac{3\pi }{L_{x}}x\right ) \sin \left ( \frac{3\pi }{L_{y}}y\right )$	$\cdots$
$\vdots$	$\vdots$	$\vdots$	$\vdots$	$\vdots$

To draw these modes, let us assume that $L_{x}=1,L_{y}=1$. This gives

$n$	$m=1$	$2$	$3$	$4$
$1$	$\sin \left ( \pi x\right ) \sin \left ( \pi y\right )$	$\sin \left ( \pi x\right ) \sin \left ( 2\pi y\right )$	$\sin \left ( \pi x\right ) \sin \left ( 3\pi y\right )$	$\cdots$
$2$	$\sin \left ( 2\pi x\right ) \sin \left ( \pi y\right )$	$\sin \left ( 2\pi x\right ) \sin \left ( 2\pi y\right )$	$\sin \left ( 2\pi x\right ) \sin \left ( 3\pi y\right )$	$\cdots$
$3$	$\sin \left ( 3\pi x\right ) \sin \left ( \pi y\right )$	$\sin \left ( 3\pi x\right ) \sin \left ( 2\pi y\right )$	$\sin \left ( 3\pi x\right ) \sin \left ( 3\pi y\right )$	$\cdots$
$\vdots$	$\vdots$	$\vdots$	$\vdots$	$\vdots$

The following is a plot of the above modes for illustrations with the code used to generate these plots.

Figure 2.38:Modes using $L_x=1,L_y=1$

Figure 2.39:Code used to draw above plot

The following is 3D view of the above modes.

Figure 2.40:3D view of the modes using $L_x=1,L_y=1$

Figure 2.41:Code used to draw above plot

2.11.3 Problem 2

Find the normal modes of an acoustic waves in a hollow sphere of radius $R$. The wave equation is

$$\nabla ^{2}\psi \left ( r,\theta ,\phi ,t\right ) =\frac{1}{c^{2}}\psi _{tt}$$ With boundary conditions $\psi _{r}=0$ at $r=0$ and at $r=r_{0}$. (I used $r_{0}$ in place of $R$ because wanted to use $R\left ( r\right )$ for separation of variables).

What is the lowest frequency?

solution

Let

$$\psi \left ( r,\theta ,\phi ,t\right ) =u\left ( r,\theta ,\phi \right ) e^{-i\omega t}$$ Substituting this back in the original PDE gives $$\nabla ^{2}u\left ( r,\theta ,\phi \right ) +\frac{\omega ^{2}}{c^{2}}u\left ( r,\theta ,\phi \right ) =0$$ Let $k=\frac{\omega }{c}$ (wave number) and the above becomes $$\begin{equation} \nabla ^{2}u+k^{2}u=0 \tag{1} \end{equation}$$ The above is called the Helmholtz PDE. In spherical coordinates it becomes $$\overset{\text{Radial part}}{\overbrace{u_{rr}+\frac{2}{r}u_{r}}}+\overset{\text{Angular part}}{\overbrace{\frac{1}{r^{2}}\left ( \frac{\cos \theta }{\sin \theta }u_{\theta }+u_{\theta \theta }\right ) +\frac{1}{r^{2}\sin ^{2}\theta }u_{\phi \phi }}}+k^{2}u=0$$ Let $u\left ( r,\theta ,\phi \right ) =R\left ( r\right ) \Theta \left ( \theta \right ) \Phi \left ( \phi \right )$ and the above becomes $$R^{\prime \prime }T\Theta \Phi +\frac{2}{r}R^{\prime }T\Theta \Phi +\frac{1}{r^{2}}\left ( \frac{\cos \theta }{\sin \theta }\Theta ^{\prime }RT\Phi +\Theta ^{\prime \prime }RT\Phi \right ) +\frac{1}{r^{2}\sin ^{2}\theta }\Phi ^{\prime \prime }R\Theta T+k^{2}R\Theta T=0$$ Dividing by $R\Theta \Phi \neq 0$ gives $$\begin{align*} \frac{R^{\prime \prime }}{R}+\frac{2}{r}\frac{R^{\prime }}{R}+\frac{1}{r^{2}}\left ( \frac{\cos \theta }{\sin \theta }\frac{\Theta ^{\prime }}{\Theta }+\frac{\Theta ^{\prime \prime }}{\Theta }\right ) +\frac{1}{r^{2}\sin ^{2}\theta }\frac{\Phi ^{\prime \prime }}{\Phi }+k^{2} & =0\\ r^{2}\sin ^{2}\theta \frac{R^{\prime \prime }}{R}+r^{2}\sin ^{2}\theta \frac{2}{r}\frac{R^{\prime }}{R}+\sin ^{2}\theta \left ( \frac{\cos \theta }{\sin \theta }\frac{\Theta ^{\prime }}{\Theta }+\frac{\Theta ^{\prime \prime }}{\Theta }\right ) +k^{2}r^{2}\sin ^{2}\theta & =-\frac{\Phi ^{\prime \prime }}{\Phi } \end{align*}$$

The left side depends only on $r,\theta$ and the right side depends only on $\phi$. Let the second separation constant be $m^{2}$ and the above becomes

$$\begin{equation} r^{2}\sin ^{2}\theta \frac{R^{\prime \prime }}{R}+r^{2}\sin ^{2}\theta \frac{2}{r}\frac{R^{\prime }}{R}+\sin ^{2}\theta \left ( \frac{\cos \theta }{\sin \theta }\frac{\Theta ^{\prime }}{\Theta }+\frac{\Theta ^{\prime \prime }}{\Theta }\right ) +k^{2}r^{2}\sin ^{2}\theta =-\frac{\Phi ^{\prime \prime }}{\Phi }=m^{2} \tag{2} \end{equation}$$ Which gives the first angular ODE as $$\begin{equation} \Phi ^{\prime \prime }+m^{2}\Phi =0 \tag{2A} \end{equation}$$ We now go back to (2) to obtain the rest of the solutions. We now have $$\begin{align*} r^{2}\sin ^{2}\theta \frac{R^{\prime \prime }}{R}+r^{2}\sin ^{2}\theta \frac{2}{r}\frac{R^{\prime }}{R}+\sin ^{2}\theta \left ( \frac{\cos \theta }{\sin \theta }\frac{\Theta ^{\prime }}{\Theta }+\frac{\Theta ^{\prime \prime }}{\Theta }\right ) +k^{2}r^{2}\sin ^{2}\theta & =m^{2}\\ k^{2}r^{2}+r^{2}\left ( \frac{R^{\prime \prime }}{R}+\frac{2}{r}\frac{R^{\prime }}{R}\right ) +\left ( \frac{\cos \theta }{\sin \theta }\frac{\Theta ^{\prime }}{\Theta }+\frac{\Theta ^{\prime \prime }}{\Theta }\right ) & =\frac{m^{2}}{\sin ^{2}\theta }\\ k^{2}r^{2}+r^{2}\left ( \frac{R^{\prime \prime }}{R}+\frac{2}{r}\frac{R^{\prime }}{R}\right ) & =-\left ( \frac{\cos \theta }{\sin \theta }\frac{\Theta ^{\prime }}{\Theta }+\frac{\Theta ^{\prime \prime }}{\Theta }\right ) +\frac{m^{2}}{\sin ^{2}\theta } \end{align*}$$

The left side depends on $r$ and the right side depends on $\theta$ only. Let the separation constant be $l\left ( l+1\right )$ where $l$ is integer which results in

$$\begin{equation} k^{2}r^{2}+r^{2}\left ( \frac{R^{\prime \prime }}{R}+\frac{2}{r}\frac{R^{\prime }}{R}\right ) =-\left ( \frac{\cos \theta }{\sin \theta }\frac{\Theta ^{\prime }}{\Theta }+\frac{\Theta ^{\prime \prime }}{\Theta }\right ) +\frac{m^{2}}{\sin ^{2}\theta }=l\left ( l+1\right ) \tag{3} \end{equation}$$ Therefore the next angular ODE is $$\begin{align} -\left ( \frac{\cos \theta }{\sin \theta }\frac{\Theta ^{\prime }}{\Theta }+\frac{\Theta ^{\prime \prime }}{\Theta }\right ) +\frac{m^{2}}{\sin ^{2}\theta } & =l\left ( l+1\right ) \nonumber \\ -\left ( \frac{\cos \theta }{\sin \theta }\frac{\Theta ^{\prime }}{\Theta }+\frac{\Theta ^{\prime \prime }}{\Theta }\right ) +\frac{m^{2}}{\sin ^{2}\theta }-l\left ( l+1\right ) & =0\nonumber \\ \left ( \frac{\cos \theta }{\sin \theta }\frac{\Theta ^{\prime }}{\Theta }+\frac{\Theta ^{\prime \prime }}{\Theta }\right ) -\frac{m^{2}}{\sin ^{2}\theta }+l\left ( l+1\right ) & =0\nonumber \\ \Theta ^{\prime \prime }+\frac{\cos \theta }{\sin \theta }\Theta ^{\prime }+\left ( l\left ( l+1\right ) -\frac{m^{2}}{\sin ^{2}\theta }\right ) \Theta & =0 \tag{4} \end{align}$$

Let $z=\cos \theta$, then $\frac{d\Theta }{d\theta }=\frac{d\Theta }{dz}\frac{dz}{d\theta }=-\frac{d\Theta }{dz}\sin \theta$ and

$$\begin{align*} \frac{d^{2}\Theta }{d\theta ^{2}} & =\frac{d}{d\theta }\left ( -\frac{d\Theta }{dz}\sin \theta \right ) \\ & =-\frac{d^{2}\Theta }{dz^{2}}\frac{dz}{d\theta }\sin \theta -\frac{d\Theta }{dz}\cos \theta \\ & =\frac{d^{2}\Theta }{dz^{2}}\sin ^{2}\theta -\frac{d\Theta }{dz}\cos \theta \end{align*}$$

But $\sin ^{2}\theta =1-\cos ^{2}\theta =1-z^{2}$ and the above becomes

$$\frac{d^{2}\Theta }{d\theta ^{2}}=\frac{d^{2}\Theta }{dz^{2}}\left ( 1-z^{2}\right ) -\frac{d\Theta }{dz}z$$ Using these in (4) gives $$\begin{align} \frac{d^{2}\Theta }{dz^{2}}\left ( 1-z^{2}\right ) -\frac{d\Theta }{dz}z+\frac{z}{\sin \theta }\left ( -\frac{d\Theta }{dz}\sin \theta \right ) +\left ( l\left ( l+1\right ) -\frac{m^{2}}{1-z^{2}}\right ) \Theta \left ( z\right ) & =0\nonumber \\ \left ( 1-z^{2}\right ) \Theta ^{\prime \prime }-2z\Theta ^{\prime }+\left ( l\left ( l+1\right ) -\frac{m^{2}}{1-z^{2}}\right ) \Theta \left ( z\right ) & =0 \tag{3A} \end{align}$$

And finally, we obtain the final ODE, which is the radial ODE from (3)

$$\begin{align} k^{2}r^{2}+r^{2}\left ( \frac{R^{\prime \prime }}{R}+\frac{2}{r}\frac{R^{\prime }}{R}\right ) & =l\left ( l+1\right ) \nonumber \\ k^{2}r^{2}R+r^{2}\left ( R^{\prime \prime }+\frac{2}{r}R^{\prime }\right ) -l\left ( l+1\right ) R & =0\nonumber \\ r^{2}R^{\prime \prime }+2rR^{\prime }+\left ( k^{2}r^{2}-l\left ( l+1\right ) \right ) R & =0\nonumber \\ R^{\prime \prime }+\frac{2}{r}R^{\prime }+\left ( k^{2}-\frac{l\left ( l+1\right ) }{r^{2}}\right ) R & =0 \tag{4A} \end{align}$$

In summary we have obtained the following 4 ODE’s to solve (1A,2A,3A,4A)

$$\begin{align} \Phi ^{\prime \prime }+m^{2}\Phi & =0\tag{2A}\\ \left ( 1-z^{2}\right ) \Theta ^{\prime \prime }-2z\Theta ^{\prime }+\left ( l\left ( l+1\right ) -\frac{m^{2}}{1-z^{2}}\right ) \Theta \left ( z\right ) & =0\tag{3A}\\ R^{\prime \prime }+\frac{2}{r}R^{\prime }+\left ( k^{2}-\frac{l\left ( l+1\right ) }{r^{2}}\right ) R & =0 \tag{4A} \end{align}$$

Solution to (2A) requires $m$ to be integer due to periodicity requirements of solution. The solution is $\Phi \left ( \phi \right ) =e^{\pm im\phi }$. Equation (3A) is the associated Legendre ODE. Since we are taking $l$ as integer then the solution is known to be $\Theta \left ( z\right ) =P_{l}^{m}\left ( z\right ) +Q_{l}^{m}\left ( z\right )$ where $P_{l}^{m}\left ( z\right )$ is called the associated Legendre polynomial and $Q_{l}^{m}$ is the Legendre function of the second kind. Finally (4A) can be converted to Bessel ODE as shown in class notes using the transformation $R\left ( r\right ) =\frac{u\left ( r\right ) }{\sqrt{r}}$ which results in

$$u^{\prime \prime }+\frac{1}{r}u^{\prime }+\left ( k^{2}-\frac{\left ( l+\frac{1}{2}\right ) ^{2}}{r^{2}}\right ) u=0$$ Which has solution $J_{l+\frac{1}{2}}\left ( kr\right )$. The second solution $J_{-\left ( l+\frac{1}{2}\right ) }\left ( kr\right )$ is rejected since it is not finite at zero and hence makes the solution blow up at center of sphere. Therefore solution to (4A) is $$\begin{align*} R\left ( r\right ) & =C\sqrt{\frac{\pi }{2kr}}J_{l+\frac{1}{2}}\left ( kr\right ) \\ & =Cj_{l}\left ( kr\right ) \end{align*}$$

Where $C$ is arbitrary constant. Putting all the above together, then the final solution is

$$\psi \left ( r,\theta ,\phi ,t\right ) =\left \{ \begin{array} [c]{c}e^{-i\omega t}\end{array} \right . \left \{ \begin{array} [c]{c}e^{im\phi }\\ e^{-im\phi }\end{array} \right . \left \{ \begin{array} [c]{c}P_{l}^{m}\left ( \cos \theta \right ) \\ Q_{l}^{m}\left ( \cos \theta \right ) \end{array} \right . \left \{ \begin{array} [c]{c}j_{l}\left ( kr\right ) \end{array} \right .$$ Where $j_{l}\left ( kr\right )$ are the spherical Bessel functions. Now we need to satisfy the boundary conditions. Since only $j_{l}\left ( kr\right )$ depends on $r$, then $\psi _{r}=0$ at $r=0$ and at $r=r_{0}$ are equivalent to looking at $R^{\prime }\left ( r\right ) =0$ at $r=0$ and $r=r_{0}$.  Therefore we need to find the smallest $l,k$ which satisfy both conditions. This will give the lowest frequency.

I found from DLMF that the series expansion of $j_{l}\left ( kr\right )$ is

$$\begin{equation} j_{l}\left ( kr\right ) =\frac{\left ( kr\right ) ^{l}}{\left ( 2l+1\right ) !!}\left ( 1-\frac{\left ( kr\right ) ^{2}}{2\left ( 2l+3\right ) }+\frac{\left ( kr\right ) ^{4}}{8\left ( 2l+5\right ) \left ( 2l+3\right ) }+\cdots \right ) \tag{5} \end{equation}$$ Hence for $r\rightarrow 0$, we can approximate the above as the following by ignoring all higher order terms $$\lim _{r\rightarrow 0}j_{l}\left ( kr\right ) =\frac{\left ( kr\right ) ^{l}}{\left ( 2l+1\right ) !!}$$ Which means for small $r$, the derivative is $$\frac{d}{dr}j_{l}\left ( kr\right ) =\frac{l\left ( kr\right ) ^{l-1}}{\left ( 2l+1\right ) !!}$$ At $r=0$ then setting $\left [ \frac{d}{dr}j_{l}\left ( kr\right ) \right ] _{r\rightarrow 0}=0$ is satisfied for all $l$. Now taking derivative of (5) gives

$$\frac{d}{dr}j_{l}\left ( kr\right ) =\frac{l\left ( kr\right ) ^{l-1}}{\left ( 2l+1\right ) !!}\left ( 1-\frac{\left ( kr\right ) ^{2}}{2\left ( 2l+3\right ) }+\frac{\left ( kr\right ) ^{4}}{8\left ( 2l+5\right ) \left ( 2l+3\right ) }+\cdots \right ) +\frac{\left ( kr\right ) ^{l}}{\left ( 2l+1\right ) !!}\left ( 1-\frac{2\left ( kr\right ) }{2\left ( 2l+3\right ) }+\frac{4\left ( kr\right ) ^{3}}{8\left ( 2l+5\right ) \left ( 2l+3\right ) }+\cdots \right )$$

At $r=r_{0}$ the above becomes  

$$\left [ \frac{d}{dr}j_{l}\left ( kr\right ) \right ] _{r\rightarrow r_{0}}=\frac{l\left ( kr_{0}\right ) ^{l-1}}{\left ( 2l+1\right ) !!}\left ( 1-\frac{\left ( kr_{0}\right ) ^{2}}{2\left ( 2l+3\right ) }+\frac{\left ( kr_{0}\right ) ^{4}}{8\left ( 2l+5\right ) \left ( 2l+3\right ) }+\cdots \right ) +\frac{\left ( kr_{0}\right ) ^{l}}{\left ( 2l+1\right ) !!}\left ( 1-\frac{2\left ( kr_{0}\right ) }{2\left ( 2l+3\right ) }+\frac{4\left ( kr_{0}\right ) ^{3}}{8\left ( 2l+5\right ) \left ( 2l+3\right ) }+\cdots \right )$$

Now we ask, for which values of $l$ is the above zero? If we let $l\rightarrow \infty$ then we obtain

$$\begin{align*} \left [ \frac{d}{dr}j_{l}\left ( kr\right ) \right ] _{\substack{r\rightarrow r_{0}\\l\rightarrow \infty }} & =\lim _{l\rightarrow \infty }\frac{l\left ( kr_{0}\right ) ^{l-1}}{\left ( 2l+1\right ) !!}+\frac{\left ( kr_{0}\right ) ^{l}}{\left ( 2l+1\right ) !!}\\ & =0 \end{align*}$$

Therefore, to satisfy both $\left [ \frac{d}{dr}j_{l}\left ( kr\right ) \right ] _{r\rightarrow 0}=0$ and $\left [ \frac{d}{dr}j_{l}\left ( kr\right ) \right ] _{r\rightarrow r_{0}}=0$ we need $l\rightarrow \infty$. In other words, a very large integer. The larger $l$ is, the lower the radial frequency. In addition, increasing $k$ while keeping $l$ fixed will increase the frequency. And decreasing $k$ while keeping $l$ fixed decreases the frequency. And for fixed $k$, increasing $l$ decreases the frequency.

2.11.4 Problem 3

A sphere of radius $R$ is at temperature $u=0$. At time $t=0$ it is immersed in a heat bath of temperature $u_{0}$. What is the temperature distribution $u\left ( r,t\right )$ as function of time?

solution

Note: I Used $u\left ( r,t\right )$ instead of $T\left ( r,t\right )$ as the dependent variable to allow using $T\left ( t\right )$ for separation of variables without confusing it with the original $T\left ( r,t\right )$.

The PDE specification is, solve for $u\left ( r,t\right )$

$$u_{t}=k\nabla ^{2}u\qquad t>0,0<r<R$$ With initial conditions $$u\left ( r,0\right ) =0$$ And boundary conditions $$\begin{align*} u\left ( R,t\right ) & =u_{0}\\ \left \vert u\left ( 0,t\right ) \right \vert & <\infty \end{align*}$$

Where the second B.C. above means the temperature $u$ is bounded at origin (center of sphere). In spherical coordinates, the PDE becomes (There are no dependency on $\theta ,\phi$ due to symmetry), and only radial dependency.

$$\begin{equation} \frac{1}{k}u_{t}=\frac{1}{r}\left ( ru\right ) _{rr}\tag{1} \end{equation}$$ To simplify the solution, let $$U\left ( r,t\right ) =ru\left ( r,t\right )$$ And we obtain a new PDE $$\begin{equation} \frac{1}{k}U_{t}=U_{rr}\tag{2} \end{equation}$$ And the boundary conditions $u\left ( R,t\right ) =u_{0}$ becomes $U\left ( R,t\right ) =Ru_{0}$ and the initial conditions becomes $U\left ( r,0\right ) =0$. So we will solve (2) and not (1). But since the boundary conditions are not homogenous, we can not use separation of variables. We introduce a reference function $w\left ( r\right )$ which need to satisfy the nonhomogeneous boundary conditions only. Let $w\left ( r\right ) =Br$. When $r=R$ then $Ru_{0}=BR$ or $B=u_{0}$ When $r=0$ then $w=0$ which is bounded. Hence $$w\left ( r\right ) =u_{0}r$$ Therefore, the solution now can be written as $$\begin{equation} U\left ( r,t\right ) =v\left ( r,t\right ) +u_{0}r\tag{3} \end{equation}$$ Where $v\left ( r,t\right )$ now satisfies the PDE but with homogenous B.C.  Substituting (3) into (2) gives $$\begin{align} v_{t} & =k\frac{\partial ^{2}}{\partial r^{2}}\left ( v\left ( r,t\right ) +u_{0}r\right ) \nonumber \\ v_{t} & =kv_{rr}\left ( r,t\right ) \tag{4} \end{align}$$

We need to solve the above but with homogenous boundary conditions

$$\begin{align*} v\left ( R,t\right ) & =0\\ \left \vert v\left ( 0,t\right ) \right \vert & <\infty \end{align*}$$

This is standard PDE, who can be solved by separation of variables. let $v=F\left ( r\right ) T\left ( t\right )$, hence (4) becomes

$$\begin{align*} T^{\prime }F & =kF^{\prime \prime }T\\ k\frac{T^{\prime }}{T} & =\frac{F^{\prime \prime }}{F}=-\lambda ^{2} \end{align*}$$

Which gives

$$F^{\prime \prime }+\lambda ^{2}F=0$$ Due to boundary conditions only $\lambda >0$ is eigenvalues. Hence solution is $$F\left ( r\right ) =A\cos \left ( \lambda r\right ) +B\sin \left ( \lambda r\right )$$ At $r=0$, since bounded, say $0$, then we can take $A=0$, leaving the solution $$F\left ( r\right ) =B\sin \left ( \lambda r\right )$$ At $r=R$ $$0=B\sin \left ( \lambda R\right )$$ For nontrivial solution $$\begin{align*} \lambda R & =n\pi \qquad n=1,2,3,\cdots \\ \lambda _{n} & =\frac{n\pi }{R} \end{align*}$$

Hence eigenfunctions are

$$F_{n}\left ( r\right ) =\sin \left ( \frac{n\pi }{R}r\right ) \qquad n=1,2,3,\cdots$$ The time ODE is therefore $T^{\prime }+\lambda ^{2}kT=0$ with solution $T_{n}\left ( t\right ) =A_{n}e^{-\left ( \frac{n\pi }{R}\right ) ^{2}kt}$. Hence the solution to (4) is $$v\left ( r,t\right ) =\sum _{n=1}^{\infty }A_{n}e^{-\left ( \frac{n\pi }{R}\right ) ^{2}kt}\sin \left ( \frac{n\pi }{R}r\right )$$ Therefore from (3) $$U\left ( r,t\right ) =\left ( \sum _{n=1}^{\infty }A_{n}e^{-\left ( \frac{n\pi }{R}\right ) ^{2}kt}\sin \left ( \frac{n\pi }{R}r\right ) \right ) +u_{0}r$$ But $U\left ( r,t\right ) =ru\left ( r,t\right )$, hence $$\begin{equation} u\left ( r,t\right ) =\left ( \frac{1}{r}\sum _{n=1}^{\infty }A_{n}e^{-\left ( \frac{n\pi }{R}\right ) ^{2}kt}\sin \left ( \frac{n\pi }{R}r\right ) \right ) +u_{0}\tag{5} \end{equation}$$ Now we find $A_{n}$ from initial conditions. At $t=0$ $$\begin{align*} 0 & =u_{0}+\frac{1}{r}\sum _{n=1}^{\infty }A_{n}\sin \left ( \frac{n\pi }{R}r\right ) \\ -ru_{0} & =\sum _{n=1}^{\infty }A_{n}\sin \left ( \frac{n\pi }{R}r\right ) \end{align*}$$

Therefore $A_{n}$ are the Fourier series coefficients of $-ru_{0}$

$$\begin{align*} \frac{R}{2}A_{n} & =-\int _{0}^{R}ru_{0}\sin \left ( \frac{n\pi }{R}r\right ) dr\\ A_{n} & =-\frac{2u_{0}}{R}\int _{0}^{R}r\sin \left ( \frac{n\pi }{R}r\right ) dr\\ & =-\frac{2u_{0}}{R}\left ( -1\right ) ^{n+1}\frac{R^{2}}{n\pi }\\ & =\left ( -1\right ) ^{n}\frac{2R}{n\pi }u_{0} \end{align*}$$

Hence the solution (5) becomes

$$\begin{align} u\left ( r,t\right ) & =u_{0}+u_{0}\frac{2R}{r\pi }\sum _{n=1}^{\infty }\left ( -1\right ) ^{n}\frac{1}{n}e^{-k\left ( \frac{n\pi }{R}\right ) ^{2}t}\sin \left ( \frac{n\pi }{R}r\right ) \nonumber \\ & =u_{0}\left ( 1+\frac{2R}{r\pi }\sum _{n=1}^{\infty }\left ( -1\right ) ^{n}\frac{1}{n}e^{-k\left ( \frac{n\pi }{R}\right ) ^{2}t}\sin \left ( \frac{n\pi }{R}r\right ) \right ) \tag{7} \end{align}$$

Verification of solution

Verification that (7) satisfies the PDE $u_{t}=k\nabla ^{2}u$. Taking time derivative of (7) gives

$$\begin{equation} u_{t}=-u_{0}\frac{2R}{r\pi }k\sum _{n=1}^{\infty }\left ( -1\right ) ^{n}\frac{1}{n}\left ( \frac{n\pi }{R}\right ) ^{2}e^{-k\left ( \frac{n\pi }{R}\right ) ^{2}t}\sin \left ( \frac{n\pi }{R}r\right ) \tag{8} \end{equation}$$ And taking space derivatives of (7) gives $$\begin{align*} u_{x} & =u_{0}\frac{2R}{r\pi }\sum _{n=1}^{\infty }\left ( -1\right ) ^{n}\frac{1}{n}e^{-k\left ( \frac{n\pi }{R}\right ) ^{2}t}\frac{n\pi }{R}\cos \left ( \frac{n\pi }{R}r\right ) \\ u_{xx} & =-u_{0}\frac{2R}{r\pi }\sum _{n=1}^{\infty }\left ( -1\right ) ^{n}\frac{1}{n}e^{-k\left ( \frac{n\pi }{R}\right ) ^{2}t}\left ( \frac{n\pi }{R}\right ) ^{2}\sin \left ( \frac{n\pi }{R}r\right ) \end{align*}$$

Hence $ku_{xx}$ becomes

$$\begin{equation} ku_{xx}=-u_{0}\frac{2R}{r\pi }k\sum _{n=1}^{\infty }\left ( -1\right ) ^{n}\frac{1}{n}e^{-k\left ( \frac{n\pi }{R}\right ) ^{2}t}\left ( \frac{n\pi }{R}\right ) ^{2}\sin \left ( \frac{n\pi }{R}r\right ) \tag{9} \end{equation}$$ Comparing (8) and (9) shows they are the same expressions.

Verification that (7) satisfies the boundary condition.

When $r=R$, therefore (7) gives, when replacing $r$ by $R$

$$\begin{align*} u\left ( R,t\right ) & =u_{0}\left ( 1+\frac{2R}{R\pi }\sum _{n=1}^{\infty }\left ( -1\right ) ^{n}\frac{1}{n}e^{-k\left ( \frac{n\pi }{R}\right ) ^{2}t}\sin \left ( \frac{n\pi }{R}R\right ) \right ) \\ & =u_{0}\left ( 1+\frac{2R}{R\pi }\sum _{n=1}^{\infty }\left ( -1\right ) ^{n}\frac{1}{n}e^{-k\left ( \frac{n\pi }{R}\right ) ^{2}t}\sin \left ( n\pi \right ) \right ) \\ & =u_{0}\left ( 1+0\right ) \\ & =u_{0} \end{align*}$$

But $n$ is integer. Hence $\sin \left ( n\pi \right ) =0$ for all $n$. And the above becomes

$$\begin{align*} u\left ( R,t\right ) & =u_{0}\left ( 1+0\right ) \\ & =u_{0} \end{align*}$$

Verified.

Verification that (7) satisfies the initial conditions $u\left ( r,0\right ) =0$ for $r<R$.

At $t=0$ (7) becomes

$$\begin{align*} u\left ( r,0\right ) & =u_{0}\left ( 1+\frac{2R}{r\pi }\sum _{n=1}^{\infty }\left ( -1\right ) ^{n}\frac{1}{n}\sin \left ( \frac{n\pi }{R}r\right ) \right ) \\ & =u_{0}+\frac{2R}{r\pi }u_{0}\sum _{n=1}^{\infty }\frac{\left ( -1\right ) ^{n}}{n}\sin \left ( \frac{n\pi }{R}r\right ) \\ & =u_{0}+\frac{2R}{r\pi }u_{0}\left ( -\sin \left ( \frac{\pi }{R}r\right ) +\frac{1}{2}\sin \left ( \frac{2\pi }{R}r\right ) -\frac{1}{3}\sin \left ( \frac{3\pi }{R}r\right ) +\frac{1}{4}\sin \left ( \frac{4\pi }{R}r\right ) -\cdots \right ) \end{align*}$$

I could not simplify the above by hand, but using the computer, I verified numerically it is zero for $0<r<R$ for a given $R$ and given $u_{0}$.

Figure 2.42:Obtaining the sum using the computer

2.11.5 Problem 4

Consider the Helmholtz equation

$$\begin{equation} \nabla ^{2}u\left ( r,\theta \right ) +k^{2}u\left ( r,\theta \right ) =0 \tag{1} \end{equation}$$

inside the circle $r=r_{0}$ with the boundary condition $u\left ( r_{0},\theta \right ) =f\left ( \theta \right )$. The solution can be written in the form $u\left ( r,\theta \right ) =\int _{0}^{2\pi }f\left ( \theta ^{\prime }\right ) G\left ( r,\theta ;\theta ^{\prime }\right ) d\theta ^{\prime }$. Find the Green function $G$.

solution

I will solve (1) directly and then compare the solution obtain to $u\left ( r,\theta \right ) =\int _{0}^{2\pi }f\left ( \theta ^{\prime }\right ) G\left ( r,\theta ;\theta ^{\prime }\right ) d\theta ^{\prime }$ in order to read off the Green function expression.  (1) in polar coordinates becomes

$$u_{rr}+\frac{1}{r}u_{r}+\frac{1}{r^{2}}u_{\theta \theta }+k^{2}u=0$$

Writing $u\left ( r,\theta \right ) =R\left ( r\right ) \Theta \left ( \theta \right )$, the above PDE becomes

$$\begin{align*} R^{\prime \prime }\Theta +\frac{1}{r}R^{\prime }\Theta +\frac{1}{r^{2}}\Theta ^{\prime \prime }R+k^{2}R\Theta & =0\\ \frac{R^{\prime \prime }}{R}+\frac{1}{r}\frac{R^{\prime }}{R}+\frac{1}{r^{2}}\frac{\Theta ^{\prime \prime }}{\Theta }+k^{2} & =0\\ r^{2}\frac{R^{\prime \prime }}{R}+r\frac{R^{\prime }}{R}+r^{2}k^{2} & =-\frac{\Theta ^{\prime \prime }}{\Theta }=m \end{align*}$$

Where $m$ is the separation constant. The eigenvalue problem is taken as

$$\Theta ^{\prime \prime }+m\Theta =0$$ Due to periodicity of the solution on the disk, then $\Theta \left ( -\pi \right ) =\Theta \left ( \pi \right )$ and $\Theta ^{\prime }\left ( -\pi \right ) =\Theta ^{\prime }\left ( \pi \right )$. These boundary conditions restrict $m$ to only positive integer values. Hence let $m=n^{2}$ and the solution to the above becomes $$\Theta _{\alpha }\left ( \theta \right ) =A_{n}\cos \left ( n\theta \right ) +B_{n}\sin \left ( n\theta \right )$$ Now the radial ODE is $$\begin{align*} r^{2}\frac{R^{\prime \prime }}{R}+r\frac{R^{\prime }}{R}+r^{2}k^{2} & =\alpha ^{2}\\ r^{2}R^{\prime \prime }+rR^{\prime }+\left ( r^{2}k^{2}-n^{2}\right ) R & =0\\ R^{\prime \prime }+\frac{1}{r}R^{\prime }+\left ( k^{2}-\frac{n^{2}}{r^{2}}\right ) R & =0 \end{align*}$$

This is Bessel ODE whose solutions are (since $n$ are integers) is

$$R_{\alpha }\left ( r\right ) =C_{n}J_{n}\left ( kr\right ) +E_{n}Y_{n}\left ( kr\right )$$ But $Y_{n}\left ( kr\right )$ blows up at $r=0$, hence it is rejected leaving solution $R_{n}\left ( r\right ) =C_{n}J_{n}\left ( kr\right )$. Hence the final solution is $$\begin{equation} u\left ( r,\theta \right ) =\sum _{m=1}^{\infty }\left ( A_{n}\cos \left ( n\theta \right ) +B_{n}\sin \left ( n\theta \right ) \right ) J_{n}\left ( kr\right ) \tag{2} \end{equation}$$ Where the constant $C_{n}$ is merged with the other two constants.  Now, at $r=r_{0}$ we are told that $u\left ( r_{0},\theta \right ) =f\left ( \theta \right )$. Hence the above becomes $$f\left ( \theta \right ) =\sum _{m=1}^{\infty }\left ( A_{n}\cos \left ( n\theta \right ) +B_{n}\sin \left ( n\theta \right ) \right ) J_{n}\left ( kr_{0}\right )$$ By orthogonality of $\cos \left ( n\theta \right ) ,\sin \left ( n\theta \right )$ we find the Fourier cosine and Fourier sine coefficients $A_{n},B_{n}$ as $$\begin{align*} A_{n}J_{n}\left ( kr_{0}\right ) \frac{1}{\pi } & =\int _{0}^{2\pi }f\left ( \theta \right ) \cos \left ( n\theta \right ) d\theta \\ B_{n}J_{n}\left ( kr_{0}\right ) \frac{1}{\pi } & =\int _{0}^{2\pi }f\left ( \theta \right ) \sin \left ( n\theta \right ) d\theta \end{align*}$$

Substituting the above back into the solution found in (2) results in

$$\begin{align} u\left ( r,\theta \right ) & =\sum _{m=1}^{\infty }\left [ \left ( \frac{\pi }{J_{n}\left ( kr_{0}\right ) }\int _{0}^{2\pi }f\left ( \theta ^{\prime }\right ) \cos \left ( n\theta ^{\prime }\right ) d\theta ^{\prime }\right ) \cos \left ( n\theta \right ) +\left ( \frac{\pi }{J_{n}\left ( kr_{0}\right ) }\int _{0}^{2\pi }f\left ( \theta ^{\prime }\right ) \sin \left ( n\theta ^{\prime }\right ) d\theta ^{\prime }\right ) \sin \left ( n\theta \right ) \right ] J_{n}\left ( kr\right ) \nonumber \\ & =\sum _{m=1}^{\infty }\frac{\pi }{J_{n}\left ( kr_{0}\right ) }\left ( \int _{0}^{2\pi }f\left ( \theta ^{\prime }\right ) \cos \left ( n\theta ^{\prime }\right ) \cos \left ( n\theta \right ) d\theta ^{\prime }+\int _{0}^{2\pi }f\left ( \theta ^{\prime }\right ) \sin \left ( n\theta ^{\prime }\right ) \sin \left ( n\theta \right ) d\theta ^{\prime }\right ) J_{n}\left ( kr\right ) \tag{3} \end{align}$$

Using trig relations

$$\begin{align*} \cos A\cos B & =\frac{1}{2}\left ( \cos \left ( A+B\right ) +\cos \left ( A-B\right ) \right ) \\ \sin A\sin B & =\frac{1}{2}\left ( \cos \left ( A-B\right ) -\cos \left ( A+B\right ) \right ) \end{align*}$$

Then (3) becomes

$$u\left ( r,\theta \right ) =\sum _{m=1}^{\infty }\frac{\pi }{2J_{n}\left ( kr_{0}\right ) }\left ( \int _{0}^{2\pi }f\left ( \theta ^{\prime }\right ) \left ( \cos \left ( n\left ( \theta ^{\prime }+\theta \right ) \right ) +\cos \left ( n\left ( \theta ^{\prime }-\theta \right ) \right ) \right ) d\theta ^{\prime }+\int _{0}^{2\pi }f\left ( \theta ^{\prime }\right ) \left ( \cos \left ( n\left ( \theta ^{\prime }-\theta \right ) \right ) -\cos \left ( n\left ( \theta ^{\prime }+\theta \right ) \right ) \right ) d\theta ^{\prime }\right ) J_{n}\left ( kr\right )$$

Which is simplified to, after combining both integrals to one

$$\begin{align*} u\left ( r,\theta \right ) & =\sum _{m=1}^{\infty }\frac{\pi }{2J_{n}\left ( kr_{0}\right ) }\left ( \int _{0}^{2\pi }f\left ( \theta ^{\prime }\right ) \left ( \cos \left ( n\left ( \theta ^{\prime }+\theta \right ) \right ) +\cos \left ( n\left ( \theta ^{\prime }-\theta \right ) \right ) +\cos \left ( n\left ( \theta ^{\prime }-\theta \right ) \right ) -\cos n\left ( \theta ^{\prime }+\theta \right ) \right ) d\theta ^{\prime }\right ) J_{n}\left ( kr\right ) \\ & =\sum _{m=1}^{\infty }\frac{\pi }{2J_{n}\left ( kr_{0}\right ) }\left [ \int _{0}^{2\pi }f\left ( \theta ^{\prime }\right ) 2\cos \left ( \theta ^{\prime }-\theta \right ) d\theta ^{\prime }\right ] J_{n}\left ( kr\right ) \\ & =\sum _{m=1}^{\infty }\left [ \int _{0}^{2\pi }f\left ( \theta ^{\prime }\right ) \frac{\pi }{J_{n}\left ( kr_{0}\right ) }\cos \left ( \theta ^{\prime }-\theta \right ) d\theta ^{\prime }\right ] J_{n}\left ( kr\right ) \end{align*}$$

Exchanging integration with summation gives

$$u\left ( r,\theta \right ) =\int _{0}^{2\pi }f\left ( \theta ^{\prime }\right ) \left ( \sum _{m=1}^{\infty }\frac{\pi }{J_{n}\left ( kr_{0}\right ) }\cos \left ( \theta ^{\prime }-\theta \right ) J_{n}\left ( kr\right ) \right ) d\theta ^{\prime }$$ Comparing the above to $$u\left ( r,\theta \right ) =\int _{0}^{2\pi }f\left ( \theta ^{\prime }\right ) G\left ( r,\theta ;\theta ^{\prime }\right ) d\theta ^{\prime }$$ Shows that Green function is $$G\left ( r,\theta ;\theta ^{\prime }\right ) =\sum _{m=1}^{\infty }\frac{\pi }{J_{n}\left ( kr_{0}\right ) }\cos \left ( \theta ^{\prime }-\theta \right ) J_{n}\left ( kr\right )$$ Where $r_{0}$ is radius of disk. It is symmetric in $\theta$ as expected.

2.11.6 Key solution for HW 11

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