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

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,

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

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,

$\frac{\text{d}}{\text{d}r}(p(r)\frac{\text{d}\psi }{\text{d}r})-q(r)\psi +\sigma (r)\mu \psi =0$

then the eigenfunctions $\psi$ have the orthogonality relationship

$\underset{a}{\overset{b}{\int }}{\psi }_n{\psi }_m\sigma (r)\text{d}r=0,n\ne m.$

Our eigenfunction problem for r was

${\displaystyle }{r}^2{R}^{″}+r{R}^{\prime }+{\left(\frac{n\pi }{\ln 2}\right)}^{2}R=0$

which we can write in Sturm-Liouville form to get

$\frac{\text{d}}{\text{d}r}\left(r\frac{\text{d}R}{\text{d}r}\right)+\frac{1}{r}{\left(\frac{n\pi }{\ln 2}\right)}^{2}R=0$

so our orthogonality relation is

$\begin{array}{rl}& \underset{a}{\overset{b}{\int }}\frac{R_n{R}_m}{r}\text{d}r\\ & =\underset{1}{\overset{2}{\int }}\sin (\frac{n\pi }{\ln 2}\ln r)\sin (\frac{m\pi }{\ln 2}\ln r)\frac{1}{r}\text{d}r\end{array}.$

If we compute the integrals we can use a substitution of

$\begin{array}{rl}x& =\frac{\ln r}{\ln 2}\\ \text{d}x& =\frac{1}{\ln 2}\frac{1}{r}\text{d}r\end{array}$

to get

$\ln 2\underset{0}{\overset{1}{\int }}\sin \left(n\pi x\right)\sin \left(m\pi x\right)\text{d}x.$

We immediately recognize this new integral as the standard sine orthogonality relation and so we get

$\begin{array}{rl}& \underset{1}{\overset{2}{\int }}\sin (\frac{n\pi }{\ln 2}\ln r)\sin (\frac{m\pi }{\ln 2}\ln r)\frac{1}{r}\text{d}r\\ & =\ln 2\underset{0}{\overset{1}{\int }}\sin \left(n\pi x\right)\sin \left(m\pi x\right)\text{d}x=\{\begin{array}{ll}0& ,n\ne m\\ \frac{\ln 2}{2}& ,n=m\end{array}.\end{array}$

Returning to

$\sum _{n=1}^{\mathrm{\infty }}{B}_n\sin (\frac{n\pi }{\ln (2)}\ln (r))\sinh \left({\lambda }_n\frac{\pi }{2}\right)=f(r)$

we can use our orthogonality relation to get

$\begin{array}{rl}& B_n\sinh \left({\lambda }_n\frac{\pi }{2}\right)\frac{\ln 2}{2}=\underset{1}{\overset{2}{\int }}f(r)\sin (\frac{n\pi }{\ln (2)}\ln (r))\frac{1}{r}\text{d}r\\ & B_n=\frac{2}{\ln 2}\frac{1}{\sinh \left({\lambda }_n\frac{\pi }{2}\right)}\underset{1}{\overset{2}{\int }}f(r)\sin (\frac{n\pi }{\ln (2)}\ln (r))\frac{1}{r}\text{d}r.\end{array}$

Therefore we now have the constants $B_m$ for any $f(r)$ and therefore we have the solution to the partial differential equation,

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