Question 3.24: Solve Laplace’s equation by separation of variables in cylin...

\frac{1}{s} \frac{\partial}{\partial s}\left(s \frac{\partial V}{\partial s}\right)+\frac{1}{s^{2}} \frac{\partial^{2} V}{\partial \phi^{2}}=0 .

Look for solutions of the form V(s, \phi)=S(s) \Phi(\phi) :

\frac{1}{s} \Phi \frac{d}{d s}\left(s \frac{d S}{d s}\right)+\frac{1}{s^{2}} S \frac{d^{2} \Phi}{d \phi^{2}}=0 .

Multiply by s^{2} \text { and divide by } V=S \Phi :

\frac{s}{S} \frac{d}{d s}\left(s \frac{d S}{d s}\right)+\frac{1}{\Phi} \frac{d^{2} \Phi}{d \phi^{2}}=0 .

Since the first term involves s only, and the second \phi only, each is a constant:

\frac{s}{S} \frac{d}{d s}\left(s \frac{d S}{d s}\right)=C_{1}, \quad \frac{1}{\Phi} \frac{d^{2} \Phi}{d \phi^{2}}=C_{2}, \quad \text { with } C_{1}+C_{2}=0 .

Now C_{2} must be negative (else we get exponentials for \Phi, which do not return to their original value—as geometrically they must— when \phi is increased by 2π).

C_{2}=-k^{2} . \quad \text { Then } \frac{d^{2} \Phi}{d \phi^{2}}=-k^{2} \Phi \Rightarrow \Phi=A \cos k \phi+B \sin k \phi .

Moreover, since \Phi(\phi+2 \pi)=\Phi(\phi) , k must be an integer: k = 0, 1, 2, 3, . . . (negative integers are just repeats, but k = 0 must be included, since \Phi=A (a constant) is OK).

s \frac{d}{d s}\left(s \frac{d S}{d s}\right)=k^{2} S can be solved by S=s^{n} , provided n is chosen right:

s \frac{d}{d s}\left(s n s^{n-1}\right)=n s \frac{d}{d s}\left(s^{n}\right)=n^{2} s s^{n-1}=n^{2} s^{n}=k^{2} S \Rightarrow n=\pm k .

Evidently the general solution is S (s)=C s^{k}+D s^{-k} , unless k = 0, in which case we have only one solution to a second-order equation—namely, S = constant. So we must treat k = 0 separately. One solution is a constant—but what’s the other? Go back to the diferential equation for S, and put in k = 0:

s \frac{d}{d s}\left(s \frac{d S}{d s}\right)=0 \Rightarrow s \frac{d S}{d s}=\text { constant }=C \Rightarrow \frac{d S}{d s}=\frac{C}{s} \Rightarrow d S=C \frac{d s}{s} \Rightarrow S=C \ln s+D (another constant).

So the second solution in this case is ln s. [How about \Phi ? That too reduces to a single solution, \Phi=A , in the case k = 0. What’s the second solution here? Well, putting k = 0 into the \Phi equation:

\frac{d^{2} \Phi}{d \phi^{2}}=0 \Rightarrow \frac{d \Phi}{d \phi}=\text { constant }=B \Rightarrow \Phi=B \phi+A .

But a term of the form B\phi is unacceptable, since it does not return to its initial value when \phi is augmented by 2π.] Conclusion: The general solution with cylindrical symmetry is

V(s, \phi)=a_{0}+b_{0} \ln s+\sum\limits_{k=1}^{\infty }{} \left[s^{k}\left(a_{k} \cos k \phi+b_{k} \sin k \phi\right)+s^{-k}\left(c_{k} \cos k \phi+d_{k} \sin k \phi\right)\right] .

Yes: the potential of a line charge goes like ln s, which is included.