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.


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) $


$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} $

  • $\begingroup$ Ok, the only last part i am confused on is integrating factor part..i do not see how you came to your answer with integrating factor. $\endgroup$ – Jackson Hart Dec 8 '14 at 21:18
  • $\begingroup$ Wow, thanks a lot! $\endgroup$ – Jackson Hart Dec 8 '14 at 21:56
  • 1
    $\begingroup$ Is this answer equivalent to other with the Bromwich integral? $\endgroup$ – Jackson Hart Dec 8 '14 at 21:57
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    $\begingroup$ You are welcome. Please note the advantage of the Laplace transform method that it does not need the integrating factor but the dis-advantage is the Laplace transform needs the Bromwich integral. $\endgroup$ – Juan Ospina Dec 8 '14 at 21:58
  • $\begingroup$ any ideas here? math.stackexchange.com/questions/1055041/… $\endgroup$ – Jackson Hart Dec 8 '14 at 22:08

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