Science:Math Exam Resources/Courses/MATH257/December 2011/Question 05 (b)

MATH257 December 2011

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  •  December 2011  •  

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 Sturm-Liouville theory to find it?

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