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,
.
We wish to impose an orthogonality argument so that we can isolate each individual coefficient,
. Recall that if we have a problem in Sturm-Liouville form,
![{\displaystyle {\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}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/dbc463d38d9eeb2e5c8423cfae7c607d1e9a9b01)
then the eigenfunctions
have the orthogonality relationship
![{\displaystyle \int _{a}^{b}\psi _{n}\psi _{m}\sigma (r){\textrm {d}}r=0,\quad {}n\neq {m}.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/2b3762d3ab379db12347c918d994b9a25ca70b7e)
Our eigenfunction problem for r was
![{\displaystyle \displaystyle {}r^{2}R''+rR'+\left({\frac {n\pi }{\ln {2}}}\right)^{2}R=0}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/21f3eb006572faa12e3270a5a4c0d9e367ffbc27)
which we can write in Sturm-Liouville form to get
![{\displaystyle {\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}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/90f2b1fdf4bbf506d692c3c67f2014f7a9776c47)
so our orthogonality relation is
![{\displaystyle {\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}}.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/7cc10804d6c733704a096dc6384f71b9f4dcc203)
If we compute the integrals we can use a substitution of
![{\displaystyle {\begin{aligned}x&={\frac {\ln {r}}{\ln {2}}}\\{\textrm {d}}x&={\frac {1}{\ln {2}}}{\frac {1}{r}}{\textrm {d}}r\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/00778fc38db71cafb37ea56970414d75cc1ccf49)
to get
![{\displaystyle \ln {2}\int _{0}^{1}\sin \left(n\pi {x}\right)\sin \left(m\pi {x}\right){\textrm {d}}x.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/66546e1f700e62ad3d7c5e99f92f8e775845be17)
We immediately recognize this new integral as the standard sine orthogonality relation and so we get
![{\displaystyle {\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}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/e99f99b72ddbc90f5f68e161e80e9c75a634cccf)
Returning to
![{\displaystyle \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)}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/49febf55693e6146885c2c7dba95513b8be4a192)
we can use our orthogonality relation to get
![{\displaystyle {\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}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/4109dab7a79c32fdf3ace943f67d44676063ffbb)
Therefore we now have the constants
for any
and therefore we have the solution to the partial differential equation,
![{\displaystyle 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).}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/231cb392966e0be23ef2f8553297a18f6d4c9a24)