10—Partial Differential Equations 264I picked the axis forφ= 0pointing toward the split between the cylinders. No particular reason, but
you have to make a choice. I make the approximation that the cylinder is infinitely long so thatz
dependence doesn’t enter. Also, the two halves of the cylinder almost touch so I’m neglecting the
distance between them.
(b) What is the electric field,−∇V on the central axis? Is this answer more or less what you would
estimate before solving the problem? Ans: (b)E= 4V 0 /πR.
10.17 Solve the preceding problemoutside the cylinder. The integerncan be either positive or
negative, and this time you’ll need the negative values. (Andwhymustnbe an integer?)
Ans:(4V 0 /π)
∑
nodd(1/n)(R/r)
nsinnφ
10.18 In the split cylinder of problem10.16, insert a coaxial wire of radiusR 1 < R. It is at zero
potential. Now what is the potential in the domainR 1 < r < R? You will needboththe positive and
negativenvalues,
∑
(Anrn+Bnr−n) sinnφ
Ans:(4V 0 /π)
∑
moddsinmφ[−R
−m1 r
m+Rm
1 r
−m]/m[R−mRm
1 −R
−m1 R
m]10.19 Fill in the missing steps in deriving Eq. (10.48).
10.20 Analyze how rapidly the solution Eq. (10.48) approaches a constant aszincreases from zero.
Compare Eq. (10.44).
10.21 A broad class of second order linear homogeneous differential equations can, with some manip-
ulation, be put into the form (Sturm-Liouville)
(p(x)u′)′+q(x)u=λw(x)u
Assume that the functionsp,q, andware real, and use manipulations much like those that led to
the identity Eq. (5.15). Derive the analogous identity for this new differential equation. When you use
separation of variables on equations involving the Laplacian you will commonly come to an ordinary
differential equation of exactly this form. The precise details will depend on the coordinate system you
are using as well as other aspects of the PDE.
10.22 Carry on from Eq. (10.31) and deduce the separated solution that satisfies these boundary
condition. Show that it is equivalent to Eq. (10.32).
10.23 The Laplacian in cylindrical coordinates is Eq. (10.49). Separate variables for the equation
∇^2 V = 0and you will see that the equations inzandφare familiar. The equation in thervariable
is less so, but you’ve seen it (almost) in Eqs. (4.18) and (4.22). Make a change of variables in the
r-differential equation,r=kr′, and turn it into exactly the form described there.
10.24 In the preceding problem suppose that there’s noz-dependence. Look at the case where the
separation constant is zero for both therandφfunctions, finally assembling the product of the two
for another solution of the whole equation.
These results provide four different solutions, a constant, a function ofralone, a function ofφalone,
and a function of both. In each of these cases, assume that these functions are potentialsV and that
E~ =−∇V is the electric field from each potential. Sketch equipotentials for each case, then sketch
the corresponding vector fields that they produce (a lot of arrows).