Boundary Problem for Electrostatic Potential

Question

I have been working on a exercise that asks me to resolve the 2nd order differential equation for a electrostatiic problem. Here it is the exercise statement:

Letting u be the electrostatic potential between $2$ concentric metal spheres, with $R_1<R_2$ and if we have a ODE:

$$\dfrac{d^2u}{dr^2} + \dfrac{2}{r}\dfrac{du}{dr} = 0$$

with the boundary conditions as $u(R_1) = V$ and $u(R_2) = 0$, where $V$ is the potential of the sphere inside.

I have tried to solve this problem in the following way:

1º Since I don't have the u term in the equation I just assumed a order reduction using a new variable:

$$\dfrac{du}{dr} = v \Rightarrow \dfrac{dv}{dr} + \dfrac{2}{r}v = 0$$

And by solving this equation I got that my anwser is

$$ u(r) = \dfrac{1}{r^2}K$$

with $K$ being the integration constant.

However, this solution doesn't satisfy my problem's boundary conditions. I have done some research and found out that for this boundary problems there are methods of separating the boundary problem in two initial value problems. Despite that I have not being able to come uo with any satisfying results. Am I missing something here? If you could provide steps for tackling this problem (at least the essential parts) I would apreciate it a lot.

Thanks in advance!!

Answer 1

HINT:

Note that we have

$$\frac1{r^2}\frac{\partial }{\partial r} \left(r^2\frac{\partial u}{\partial r}\right)=\frac{\partial^2 u}{\partial r^2}+\frac 2r \frac{\partial u}{\partial r}$$

Can you find a solution to the differencial equation

$$\frac1{r^2}\frac{\partial }{\partial r} \left(r^2\frac{\partial u}{\partial r}\right) =0?\tag 1$$

The solution to $(1)$ should be $u=A/r+B$.

Answer 2

You get $v=Kr^{-2}$, not $u$. You now need to use your knowledge of $v$ to find $u$.

(Related: as a sanity check, with a $n$th order linear ODE you should have $n$ coefficients appearing as constants of integration before you apply any boundary conditions. So you could have used this fact to conclude that $Kr^{-2}$ couldn't be the general solution to the original ODE.)