solving a PDE with with coordinates First i need to find eigenfunctions in spherical coordinates with source term.
I am having trouble with this . Can someone help guide me
i also have 2 b.c.'s.
 A: The eigenfunctions are $\sin \left( {\frac {\beta_{{n}}r}{R}} \right)$ where $\beta_{{n}}$ are the roots of
$\sin \left( \beta_{{n}} \right) =\beta_{{n}}\cos \left( \beta_{{n}}
 \right) 
$;
We look for a solution of
${\frac {\partial }{\partial t}}w \left( r,t \right) =\alpha\,{\frac {
\partial ^{2}}{\partial {r}^{2}}}w \left( r,t \right) +rg \left( r,t
 \right) 
$
with the form
$w \left( r,t \right) =\sum _{n=1}^{\infty }A_{{n}} \left( t \right) 
\sin \left( {\frac {\beta_{{n}}r}{R}} \right) 
$
$rg \left( r,t \right) =\sum _{n=1}^{\infty }B_{{n}} \left( t \right) 
\sin \left( {\frac {\beta_{{n}}r}{R}} \right) 
$
Then we obtain
${\frac {d}{dt}}A_{{n}} \left( t \right) =-{\frac {\alpha\,A_{{n}}
 \left( t \right) {\beta_{{n}}}^{2}-B_{{n}} \left( t \right) {R}^{2}}{
{R}^{2}}}
$
Using the method of the integrating factor the last equation is rewritten as
${{\rm e}^{-{\frac {\alpha\,{\beta_{{n}}}^{2}t}{{R}^{2}}}}}{\frac {
\partial }{\partial t}} \left( A_{{n}} \left( t \right) {{\rm e}^{{
\frac {\alpha\,{\beta_{{n}}}^{2}t}{{R}^{2}}}}} \right) =B_{{n}}
 \left( t \right) 
$
or as
${{\rm e}^{-{\frac {\alpha\,{\beta_{{n}}}^{2}\tau}{{R}^{2}}}}}{\frac {
\partial }{\partial \tau}} \left( A_{{n}} \left( \tau \right) {{\rm e}
^{{\frac {\alpha\,{\beta_{{n}}}^{2}\tau}{{R}^{2}}}}} \right) =B_{{n}}
 \left( \tau \right) 
$
From this we deduce that
${\frac {\partial }{\partial \tau}} \left( A_{{n}} \left( \tau \right) 
{{\rm e}^{{\frac {\alpha\,{\beta_{{n}}}^{2}\tau}{{R}^{2}}}}} \right) =
B_{{n}} \left( \tau \right) {{\rm e}^{{\frac {\alpha\,{\beta_{{n}}}^{2
}\tau}{{R}^{2}}}}}
$
integrating both sides we have
$\int _{0}^{t}\!{\frac {\partial }{\partial \tau}} \left( A_{{n}}
 \left( \tau \right) {{\rm e}^{{\frac {\alpha\,{\beta_{{n}}}^{2}\tau}{
{R}^{2}}}}} \right) {d\tau}=\int _{0}^{t}\!B_{{n}} \left( \tau
 \right) {{\rm e}^{{\frac {\alpha\,{\beta_{{n}}}^{2}\tau}{{R}^{2}}}}}{
d\tau}
$
and then we obtain
$A_{{n}} \left( t \right) {{\rm e}^{{\frac {\alpha\,{\beta_{{n}}}^{2}t}
{{R}^{2}}}}}-A_{{n}} \left( 0 \right) =\int _{0}^{t}\!B_{{n}} \left( 
\tau \right) {{\rm e}^{{\frac {\alpha\,{\beta_{{n}}}^{2}\tau}{{R}^{2}}
}}}{d\tau}
$
it is to say
$A_{{n}} \left( t \right) = \left( \int _{0}^{t}\!B_{{n}} \left( \tau
 \right) {{\rm e}^{{\frac {\alpha\,{\beta_{{n}}}^{2}\tau}{{R}^{2}}}}}{
d\tau}+A_{{n,0}} \right) {{\rm e}^{-{\frac {\alpha\,{\beta_{{n}}}^{2}t
}{{R}^{2}}}}}
$
where $A_{n,0}=A_{n}(0)$
Then we have
$w \left( r,t \right) =\sum _{n=1}^{\infty } \left(  \left( \int _{0}^{
t}\!B_{{n}} \left( \tau \right) {{\rm e}^{{\frac {\alpha\,{\beta_{{n}}
}^{2}\tau}{{R}^{2}}}}}{d\tau}+A_{{n,0}} \right) {{\rm e}^{-{\frac {
\alpha\,{\beta_{{n}}}^{2}t}{{R}^{2}}}}}\sin \left( {\frac {\beta_{{n}}
r}{R}} \right)  \right) 
$
and 
$F \left( r,t \right) =\sum _{n=1}^{\infty } \left(  \left( \int _{0}^{
t}\!B_{{n}} \left( \tau \right) {{\rm e}^{{\frac {\alpha\,{\beta_{{n}}
}^{2}\tau}{{R}^{2}}}}}{d\tau}+A_{{n,0}} \right) {{\rm e}^{-{\frac {
\alpha\,{\beta_{{n}}}^{2}t}{{R}^{2}}}}}\sin \left( {\frac {\beta_{{n}}
r}{R}} \right)  \right) {r}^{-1}
$
