MATH257 December 2011
• Q1 • Q2 (a) • Q2 (b) • Q3 (a) • Q3 (b) • Q4 (a) • Q4 (b) • Q5 (a) • Q5 (b) •
Question 05 (b)

Consider the following problem involving Laplace's equation in an annular region:
 $u_{rr}+{\frac {1}{r}}u_{r}+{\frac {1}{r^{2}}}u_{\theta \theta }=0,\quad 1<r<2,\quad 0<\theta <\pi /2.$
 $u(1,\theta )=0,\quad u(2,\theta )=0,\quad u(r,0)=0,\quad u(r,\pi /2)=f(r).$
Find the solution for a general function $f(r)$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint

If we want to find a solution for all f(r), we're going to need to find an orthogonality relation. How can we use SturmLiouville theory to find it?

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work.
 If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work.
 If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies.
In part (a) we found a solution for a specific f(r) that happened to be one of the original eigenfunctions and so we were able to compare just by analyzing the coefficient of one term. However now we have a general f(r) and if we continue from part (a) we have,
 $u\left(r,{\frac {\pi }{2}}\right)=\sum _{n=1}^{\infty }B_{n}\sin \left({\frac {n\pi }{\ln(2)}}\ln(r)\right)\sinh \left(\lambda _{n}{\frac {\pi }{2}}\right)=f(r)$.
We wish to impose an orthogonality argument so that we can isolate each individual coefficient, $B_{n}$. Recall that if we have a problem in SturmLiouville form,
 ${\frac {\textrm {d}}{{\textrm {d}}r}}\left(p(r){\frac {{\textrm {d}}\psi }{{\textrm {d}}r}}\right)q(r)\psi +\sigma (r)\mu \psi =0$
then the eigenfunctions $\psi$ have the orthogonality relationship
 $\int _{a}^{b}\psi _{n}\psi _{m}\sigma (r){\textrm {d}}r=0,\quad {}n\neq {m}.$
Our eigenfunction problem for r was
 $\displaystyle {}r^{2}R''+rR'+\left({\frac {n\pi }{\ln {2}}}\right)^{2}R=0$
which we can write in SturmLiouville form to get
 ${\frac {\textrm {d}}{{\textrm {d}}r}}\left(r{\frac {{\textrm {d}}R}{{\textrm {d}}r}}\right)+{\frac {1}{\color {blue}r}}\left({\frac {n\pi }{\ln {2}}}\right)^{2}R=0$
so our orthogonality relation is
 ${\begin{aligned}&\int _{a}^{b}{\frac {R_{n}R_{m}}{\color {blue}r}}{\textrm {d}}r\\&=\int _{1}^{2}\sin \left({\frac {n\pi }{\ln {2}}}\ln {r}\right)\sin \left({\frac {m\pi }{\ln {2}}}\ln {r}\right){\frac {1}{r}}{\textrm {d}}r\end{aligned}}.$
If we compute the integrals we can use a substitution of
 ${\begin{aligned}x&={\frac {\ln {r}}{\ln {2}}}\\{\textrm {d}}x&={\frac {1}{\ln {2}}}{\frac {1}{r}}{\textrm {d}}r\end{aligned}}$
to get
 $\ln {2}\int _{0}^{1}\sin \left(n\pi {x}\right)\sin \left(m\pi {x}\right){\textrm {d}}x.$
We immediately recognize this new integral as the standard sine orthogonality relation and so we get
 ${\begin{aligned}&\int _{1}^{2}\sin \left({\frac {n\pi }{\ln {2}}}\ln {r}\right)\sin \left({\frac {m\pi }{\ln {2}}}\ln {r}\right){\frac {1}{r}}{\textrm {d}}r\\&=\ln {2}\int _{0}^{1}\sin \left(n\pi {x}\right)\sin \left(m\pi {x}\right){\textrm {d}}x={\begin{cases}0&,n\neq {m}\\{\frac {\ln {2}}{2}}&,n=m\end{cases}}.\end{aligned}}$
Returning to
 $\sum _{n=1}^{\infty }B_{n}\sin \left({\frac {n\pi }{\ln(2)}}\ln(r)\right)\sinh \left(\lambda _{n}{\frac {\pi }{2}}\right)=f(r)$
we can use our orthogonality relation to get
 ${\begin{aligned}&B_{n}\sinh \left(\lambda _{n}{\frac {\pi }{2}}\right){\frac {\ln {2}}{2}}=\int _{1}^{2}f(r)\sin \left({\frac {n\pi }{\ln(2)}}\ln(r)\right){\frac {1}{r}}{\textrm {d}}r\\&B_{n}={\frac {2}{\ln {2}}}{\frac {1}{\sinh \left(\lambda _{n}{\frac {\pi }{2}}\right)}}\int _{1}^{2}f(r)\sin \left({\frac {n\pi }{\ln(2)}}\ln(r)\right){\frac {1}{r}}{\textrm {d}}r.\end{aligned}}$
Therefore we now have the constants $B_{m}$ for any $f(r)$ and therefore we have the solution to the partial differential equation,
 $u(r,\theta )=\sum _{n=1}^{\infty }B_{n}\sin \left({\frac {n\pi }{\ln(2)}}\ln(r)\right)\sinh \left(\lambda _{n}\theta \right).$
