I know this looks like a physics thing, but it's purely the math that I'm stuck on. I'm attempting to write Laplace's equation, $\nabla^2V=0$, in cylindrical coordinates for a potential, $V(r,\phi,z)$, independent of $z$.
By definition (in cylindrical),
$$
\nabla^2V=\frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial V}{\partial r}\right)+\frac{1}{r^2}\frac{\partial^2 V}{\partial \phi^2}+\frac{\partial^2 V}{\partial z^2}
$$
in cylindrical coordinates. $\frac{\partial^2 V}{\partial z^2}=0$ and $V(r,\phi) = R(r)\Phi(\phi)$. Plugging in $V(r,\phi)$ and using separation of variables:
$$
\frac{r}{R}\frac{d}{dr}\left(r\frac{dR}{dr}\right)+\frac{1}{\Phi}\frac{d^2\Phi}{d\phi^2}=0
$$
Now I have two ordinary differential equations. Using a separation constant $k^2$, I can solve for $\Phi(\phi)$ easily:
$$
\frac{d^2\Phi}{d\phi^2}=-\Phi k^2 \quad \Rightarrow \quad \Phi(\phi)=A\sin(k\phi)+B\cos(k\phi)
$$
Here's where I'm stuck, solving for $R(r)$
$$
\frac{r}{R}\frac{d}{dr}\left(r\frac{dR}{dr}\right)=k^2
$$
I tried applying the product rule and writing it out like this:
$$
rR''+R'-\frac{k^2}{r}R=0
$$
but I don't see how to get past this. I'm eventually supposed to end up with something in this form:
$$
V(r, \phi)=A_0+B_0\ln(r)+\sum^\infty_{k=1}[r^k(A_k\cos(k\phi)+B_k\sin(k\phi))+r^{-k}(C_k\cos(k\phi)+D_k\sin(k\phi)]
$$
Some of the constants might be different but this is what I'm going for. A Fourier thing happened somewhere.
[Math] Solving Laplace’s equation in cylindrical coordinates (ODE)
ordinary differential equations
Best Answer
The radial equation is
$$ r^2R'' + rR' - k^2R = 0 $$
If you didn't know, this is called a Cauchy-Euler equation, with well-known solutions of the form $x^n$. Plugging in this form results in the characteristic equation
$$ m(m-1) + m - k^2 = 0 \implies m^2-k^2 = 0 \implies m = \pm k $$
If $k\ne0$, the general solution is
$$ R_k(r) = A_kr^k + B_kr^{-k} $$
If $k=0$, we have a double root and the solution has the form
$$ R_0(r) = A_0 + B_0\ln r $$
You may verify this by directly solving
$$ r^2R'' + rR' = 0 \implies \frac{R''}{R'} = -\frac{1}{r} \implies \ln (R') = -\ln r + c \implies R' = \frac{B}{r} \implies \dots $$