Mathematical Tools for Physics - Department of Physics - University

10—Partial Differential Equations 260

α=n^2 > 0

d^2 g

dφ^2

=−n^2 g =⇒ g(φ) =Acosnφ+Bsinnφ

r^2

d^2 f

dr^2

+r

df

dr

−n^2 f= 0 =⇒ f(r) =Crn+Dr−n

There’s not yet a restriction thatnis an integer, though that’s often the case. Verifying the last

solution forfis easy.

A general solution that is the sum of all these terms is

V(r,φ) = (C 0 +D 0 lnr)(A 0 +B 0 φ) +

∑

n

(Cnrn+Dnr−n)(Ancosnφ+Bnsinnφ) (10.51)

Some of these terms should be familiar:

C 0 A 0 is just a constant potential.

D 0 A 0 lnr is the potential of a uniform line charge;dlnr/dr= 1/r, and that is the way that the

electric field drops off with distance from the axis.

C 1 A 1 rcosφ is the potential of a uniform field (as is thersinφ term). Write this in the form

C 1 A 1 rcosφ=C 1 A 1 x, and the gradient of this isC 1 A 1 ˆx. The sine givesyˆ.

See problem10.24.

Example

A conducting wire, radiusR, is placed in a uniform electric fieldE~ 0 , and perpendicular to it. Put the

wire along thez-axis and call the positivex-axis the direction that the field points. That’sφ= 0.

In the absence of the wire, the potential for the uniform field isV =−E 0 x=−E 0 rcosφ, because

−∇V =E 0 xˆ. The total solution will be in the form of Eq. (10.51).

Now turn the general form of the solution into a particular one for this problem. The entire range

ofφfrom 0 to 2 πappears here; you can go all the way around the origin and come back to where you

started. The potential is a function, meaning that it’s single valued, and that eliminatesB 0 φ. It also

implies that all thenare integers. The applied field has a potential that is even inφ. That means that

you should expect the solution to be even inφtoo. Does it really? You also have to note that the

physical system and its attendant boundary conditions are even inφ— it’s a cylinder. Then too, the

first few times that you do this sort of problem you should see what happens to the odd terms; what

makes them vanish? I won’t eliminate thesinφterms from the start, but I’ll calculate them and show

that they do vanish.

V(r,φ) = (C 0 +D 0 lnr)B 0 +

∑∞

n=1

(Cnrn+Dnr−n)(Ancosnφ+Bnsinnφ)

Carrying along all these products of (still unknown) factors such asDnAnis awkward. It makes it look

neater and it is easier to follow if I combine and rename some of them.

V(r,φ) =C 0 +D 0 lnr+

∑∞

n=1

(Cnrn+Dnr−n) cosnφ+

∑∞

n=1

(Cn′rn+D′nr−n) sinnφ (10.52)

Asr→∞, the potential looks like−E 0 rcosφ. That implies thatCn= 0forn > 1 , and thatCn′ = 0

for alln, and thatC 1 =−E 0.

Now use the boundary conditions on the cylinder. It is a conductor, so in this static case the
potential is a constant all through it, in particular on the surface. I may as well take that constant to
be zero, though it doesn’t really matter.

V(R,φ) = 0 =C 0 +D 0 lnR+

∑∞

n=1

(CnRn+DnR−n) cosnφ+

∑∞

n=1

(Cn′Rn+D′nR−n) sinnφ