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

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MATH257 December 2011

• Q1 • Q2 (a) • Q2 (b) • Q3 (a) • Q3 (b) • Q4 (a) • Q4 (b) • Q5 (a) • Q5 (b) •

Other MATH257 Exams

• 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 $f(r)$ .

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