Simple Partial Differential Equation

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jaydnul
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Homework Statement


This is actually an electromagnetism problem but all the physics is done, I just don't remember how to solve the PDE:
[itex]\frac{d^2V}{dr^2}=-\frac{2}{r}\frac{dV}{dr}[/itex]
The d's should be del's, just don't know how to do that...

Homework Equations



Not sure.

The Attempt at a Solution


[/B]
Don't know where to start.
 
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There are no other independent variables in the problem besides r. Assume V can be written as a product of functions of all the independent variables. I.e. it's separable. V(r)=u(r)*v(theta)*w(phi). The treat it as an ODE. Do you remember those? I would try putting dV/dr=u(r) and separating variables.
 
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Ahh I think that's just what I needed. So

[itex]V_{rr}=-k^2u(r)[/itex] where k=sqrt{2/r}
[itex]u(r)=Ae^{ikr}+Be^{-ikr}[/itex]
[itex]V(r)=\frac{-iA}{k}e^{ikr}+\frac{iB}{k}e^{-ikr}[/itex]
then sub in for k

Is that correct?
 
Jd0g33 said:
Ahh I think that's just what I needed. So

[itex]V_{rr}=-k^2u(r)[/itex] where k=sqrt{2/r}
[itex]u(r)=Ae^{ikr}+Be^{-ikr}[/itex]
[itex]V(r)=\frac{-iA}{k}e^{ikr}+\frac{iB}{k}e^{-ikr}[/itex]
then sub in for k

Is that correct?

Not really. If u(r)=dV/dr then the equation becomes du/dr=(-2/r)u. Doesn't it? Solve that.
 
Ahh, I'm an idiot. [itex]u=r^{-2}[/itex]

Thanks Dick!
 
Jd0g33 said:

Homework Statement


This is actually an electromagnetism problem but all the physics is done, I just don't remember how to solve the PDE:
[itex]\frac{d^2V}{dr^2}=-\frac{2}{r}\frac{dV}{dr}[/itex]
The d's should be del's, just don't know how to do that...

Homework Equations



Not sure.

The Attempt at a Solution


[/B]
Don't know where to start.

Note that
[tex]\frac{d}{dr}\left( r^k \frac{dF(r)}{dr} \right) = r^k \frac{d^2F(r)}{dr^2}+ k r^{k-1} \frac{dF(r)}{dr}[/tex]
Can you see how that fact helps?