Science:Math Exam Resources/Courses/MATH257/December 2011/Question 05 (a)/Solution 1

To use the method of separation of variables, we write $u$ as

$u(r,\theta )=R(r)\Theta (\theta ).\,$

We compute that

$u_{r}(r,\theta )=R'(r)\Theta (\theta ),\;u_{rr}(r,\theta )=R''(r)\Theta (\theta ),$

and

$u_{\theta \theta }(r,\theta )=R(r)\Theta ''(\theta ).\,$

Substitute the above into our differential equation

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

we get that

$R''\Theta +{\frac {1}{r}}R'\Theta +{\frac {1}{r^{2}}}R\Theta ''=0.$

Next, we would like to group all the $R$ terms onto one side and all the $\Theta$ terms onto another, so divide the entire equation by $R\Theta$ to get

${\frac {R''\Theta }{R\Theta }}+{\frac {1}{r}}{\frac {R'\Theta }{R\Theta }}+{\frac {1}{r^{2}}}{\frac {R\Theta ''}{R\Theta }}=0$

which gives

${\frac {R''}{R}}+{\frac {1}{r}}{\frac {R'}{R}}+{\frac {1}{r^{2}}}{\frac {\Theta ''}{\Theta }}=0.$

To get rid of the factor of ${\frac {1}{r^{2}}}$ in the last term on the left, we will multiply the equation by $r^{2}$ to get

$r^{2}{\frac {R''}{R}}+r{\frac {R'}{R}}+{\frac {\Theta ''}{\Theta }}=0$

which gives

$r^{2}{\frac {R''}{R}}+r{\frac {R'}{R}}=-{\frac {\Theta ''}{\Theta }}.$

Here, $r$ and $\theta$ are independent of each other. However, the right hand side of the equation is a function of $r$ while the right hand side a function of $\theta$, so both sides of the equation must be equal to a constant which we will denote $\mu$. In other words, we have

$r^{2}{\frac {R''}{R}}+r{\frac {R'}{R}}=-{\frac {\Theta ''}{\Theta }}=\mu .$

This gives us two equations

$r^{2}R''+rR'=\mu R\;\;{\text{and}}\;\;-\Theta ''=\mu \Theta .$

We will first solve the equation for $R$. Before we go ahead to do so, we would like to figure out the boundary conditions for $R$. The boundary conditions

$u(1,\theta )=0\;\;{\text{and}}\;\;u(2,\theta )=0$

imply that

$R(1)\Theta (\theta )=0\;\;{\text{and}}\;\;R(2)\Theta (\theta )=0.$

Since we do not want $\Theta$ to be $0$ since this will give the trivial solution (i.e. $u\equiv 0$) which is not of interest to us. So the boundary conditions give us

$R(1)=0\;\;{\text{and}}\;\;R(2)=0.$

We will now solve the equation for $R$ which says

$r^{2}R''+rR'-\mu R=0.\,$

This is the Euler's equation and we recall that we should look for solutions in the form

$R(r)=r^{\alpha }\,$

for an unknown constant $\alpha$ to be determined. We compute that

$R'=\alpha r^{\alpha -1}\;\;{\text{and}}\;\;R''=\alpha (\alpha -1)r^{\alpha -2}.$

Substituting these into

$r^{2}R''+rR'-\mu R=0\,$

we get that

$\alpha (\alpha -1)+\alpha =\mu \,$

which after simplifying gives

$\alpha ^{2}=\mu .\,$

Here, the sign of $\mu$ will affect the form of the solution, so we have three cases to consider. They are

case 1: $\mu =-\lambda ^{2}<0$

case 2: $\mu =0$

case 3: $\mu =\lambda ^{2}>0$

Further calculations show that case 2 and case 3 only give trivial solutions. If you are not sure, you should try verifying that yourself. For completeness of the solution, I will also include the computation in the appendix of this solution.

Now, case 1: says that

$\alpha ^{2}=\mu =-\lambda ^{2}.\,$

This gives

$\alpha =\pm \lambda i.$

Recall that for the Euler's equation, if $\alpha$ is in the form of $\alpha =a\pm bi$, the solution of the Euler's equation can be written as

$R(r)=C_{1}r^{a}\cos(b\ln(r))+C_{2}r^{a}\sin(b\ln(r))\,$

for some arbitrary constants $C_{1}$ and $C_{2}$. For us, we have $a=0$ and $b=\lambda$, so our solution for $R$ is

$R(r)=C_{1}\cos(\lambda \ln(r))+C_{2}\sin(\lambda \ln(r)).\,$

Next, we need to match the boundary conditions $R(1)=0\;\;{\text{and}}\;\;R(2)=0.\,$

$R(1)=0\Rightarrow 0=C_{1}\cos(\lambda \underbrace {\ln(1)} _{=0})+C_{2}\sin(\lambda \underbrace {\ln(1)} _{=0})=C_{1}\underbrace {\cos(0)} _{=1}+C_{2}\underbrace {\sin(0)} _{=0}=C_{1}.$

Hence, $C_{1}=0$ and

$R(r)=C_{2}\sin(\lambda \ln(r)).\,$

Now,

$R(2)=0\Rightarrow C_{2}\sin(\lambda \ln(2))=0.$

If $C_{2}=0$, we will have the trivial solution. To get non-trivial solutions, we need

$\sin(\lambda \ln(2))=0\Rightarrow \lambda \ln(2)=n\pi \Rightarrow \lambda ={\frac {n\pi }{\ln(2)}}.$

In other words, we have an infinite numbers of eigenvalues $\lambda _{n}$ give by

$\lambda _{n}={\frac {n\pi }{\ln(2)}}\;\;{\text{for}}\;\;n={\color {red}1},2,3,\cdots$

with the corresponding eigenfunctions

$R_{n}(r)=C_{n}\sin(\lambda _{n}\ln(r)).\,$

Now, we will go back to solve the $\theta$ equation. Recall that we have $\mu =-\lambda ^{2}$ so the $\theta$ equation $-\Theta ''=\mu \Theta$ becomes

${\begin{aligned}-\Theta ''&=-\lambda ^{2}\Theta \\\Theta ''&=\lambda ^{2}\Theta \end{aligned}}$

which gives solutions of the form

$\Theta =A\cosh(\lambda \theta )+B\sinh(\lambda \theta )\,$

for some arbitrary constants $A$ and $B$. Since we found that an infinite number of $\lambda _{n}={\frac {n\pi }{\ln(2)}}$, we have an infinite number of $\Theta _{n}$ given by

$\Theta _{n}=A_{n}\cosh(\lambda _{n}\theta )+B_{n}\sinh(\lambda _{n}\theta ).\,$

Putting everything together, we can express $u$ as

$u(r,t)=\sum _{n=1}^{\infty }C_{n}\sin(\lambda _{n}\ln(r))\left(A_{n}\cosh(\lambda _{n}\theta )+B_{n}\sinh(\lambda _{n}\theta )\right).$

We can absorb the arbitrary constant $C_{n}$ into $A_{n}$ and $B_{n}$ to get

$u(r,t)=\sum _{n=1}^{\infty }\sin(\lambda _{n}\ln(r))\left(A_{n}\cosh(\lambda _{n}\theta )+B_{n}\sinh(\lambda _{n}\theta )\right).$

Finally, we will use the other two boundary conditions $u(r,0)=0$ and $u(r,{\frac {\pi }{2}})=f(r)$ to find $A_{n}$ and $B_{n}$.

${\begin{aligned}&u(r,0)=0\,\\&\sum _{n=1}^{\infty }\sin(\lambda _{n}\ln(r))(A_{n}\underbrace {\cosh(0)} _{=1}+B_{n}\underbrace {\sinh(0)} _{=0})=0\\&\sum _{n=1}^{\infty }A_{n}\sin(\lambda _{n}\ln(r))=0.\\&A_{n}=0\;\;{\text{for each}}\;\;n=1,2,3,\cdots \end{aligned}}$

Hence

${\begin{aligned}u(r,t)&=\sum _{n=1}^{\infty }B_{n}\sin(\lambda _{n}\ln(r))\sinh(\lambda _{n}\theta )\\&=\sum _{n=1}^{\infty }B_{n}\sin \left({\frac {n\pi }{\ln(2)}}\ln(r)\right)\sinh \left(\lambda _{n}\theta \right)\end{aligned}}.$

Finally, to find $B_{n}$, we use the last boundary condition

$u\left(r,{\frac {\pi }{2}}\right)=\sin \left({\frac {2\pi }{\ln(2)}}\ln(r)\right)$.

This boundary condition gives us that

$u\left(r,{\frac {\pi }{2}}\right)=\sum _{n=1}^{\infty }B_{n}\sin \left({\frac {n\pi }{\ln(2)}}\ln(r)\right)\sinh \left(\lambda _{n}{\frac {\pi }{2}}\right)=\sin \left({\frac {2\pi }{\ln(2)}}\ln(r)\right).$

If we expand out the summation, we see that

${\begin{aligned}&B_{1}\sin \left({\frac {\pi }{\ln(2)}}\ln(r)\right)\sinh \left(\lambda _{1}{\frac {\pi }{2}}\right)+{\color {blue}B_{2}\sin \left({\frac {2\pi }{\ln(2)}}\ln(r)\right)\sinh \left(\lambda _{2}{\frac {\pi }{2}}\right)}\\&+B_{3}\sin \left({\frac {3\pi }{\ln(2)}}\ln(r)\right)\sinh \left(\lambda _{3}{\frac {\pi }{2}}\right)+\cdots ={\color {blue}\sin \left({\frac {2\pi }{\ln(2)}}\ln(r)\right)}\end{aligned}}.$

Here, we see that $n=2$ is the only term on the left hand side that matches with the right hand side. Hence, we have

$B_{n}\sinh \left(\lambda _{n}{\frac {\pi }{2}}\right)=0\;\;{\text{if}}\;\;n\not =2$

and

$B_{2}\sinh \left(\lambda _{2}{\frac {\pi }{2}}\right)=1$

$\Rightarrow B_{2}={\frac {1}{\sinh(\lambda _{2}{\frac {\pi }{2}})}}={\frac {1}{\sinh({\frac {2\pi }{\ln(2)}}{\frac {\pi }{2}})}}.$

In other words, the solution is

$u(r,\theta )={\frac {1}{\sinh \left({\frac {2\pi }{\ln(2)}}{\frac {\pi }{2}}\right)}}\sin \left({\frac {2\pi }{\ln(2)}}\ln(r)\right)\sinh \left({\frac {2\pi }{\ln(2)}}\theta \right).$

Appendix:

We will address two final questions in this appendix.

Question 1: Refer to the solution above, why doesn't $\mu =0$ lead to non-trivial solutions for the Euler's equation?

Answer to question 1: Suppose $\mu =0$, then

$\alpha ^{2}=\mu \Rightarrow \alpha =0.$

The solution to the Euler's equation is

$R(r)=C_{1}\ln(r)+C_{2}\,$

for some arbitrary constants $C_{1}$ and $C_{2}$. The boundary condition $R(1)=0$ implies

$R(1)=C_{1}\underbrace {\ln(1)} _{=0}+C_{2}=0\Rightarrow C_{2}=0.$

So

$R(r)=C_{1}\ln(r).\,$

The boundary condition $R(2)=0$ implies

$R(2)=C_{1}\ln(2)=0\Rightarrow C_{1}=0.$

Hence, we only get the trivial solution.

Question 2: Refer to the solution above, why doesn't $\mu =\lambda ^{2}>0$ lead to non-trivial solutions for the Euler's equation?

Answer to question 2: Suppose $\mu =\lambda ^{2}>0$, then

$\alpha ^{2}=\mu \Rightarrow \alpha =\pm \lambda .$

The solution to the Euler's equation is

$R(r)=C_{1}r^{\lambda }+C_{2}r^{-\lambda }\,$

for some arbitrary constants $C_{1}$ and $C_{2}$. The boundary condition $R(1)=0$ implies

$R(1)=C_{1}+C_{2}=0\Rightarrow C_{1}=-C_{2}.$

So

$R(r)=C_{1}r^{\lambda }-C_{1}r^{-\lambda }.\,$

The boundary condition $R(2)=0$ implies

$R(2)=C_{1}2^{\lambda }-C_{1}2^{-\lambda }=0.\,$

If $C_{1}=0$, we get a trivial solution, so suppose $C_{1}\not =0$, then we must have

$2^{\lambda }=2^{-\lambda }\Rightarrow 2^{2\lambda }=1\Rightarrow \lambda =0.$

However, this contradicts with our original assumption that $\lambda ^{2}>0$.