Mathematical Tools for Physics

10—Partial Differential Equations 309

10.7 In the analysis leading to Eq. ( 23 ) the temperature aty =b was set toT 0. If instead, you have the
temperature atx=aset toT 0 with all the other sides at zero, write down the answer for the temperature within
the rod. Now use the fact that Eq. ( 18 ) is linear to write down the solution ifboththe sides aty=bandx=a
are set toT 0.

10.8 In leading up to Eq. ( 22 ) I didn’t examine the third possibility for the separation constant, that it’s zero.
Do so.

10.9 Look at the boundary condition of Eq. ( 19 ) again. Another way to solve this problem is to use the solution
for which the separation constant is zero, and to use it to satisfy the conditions aty = 0 andy =b. You
will then have one term in the separated solution that is T 0 y/b, and that means that in Eq. ( 20 ) you will
have to choose the separation variable to be positive instead of negative. Why? Because now all the rest of
the terms in the sum over separated solutions must vanish aty= 0andy=b. You’ve already satisfied the
boundary conditions on those surfaces by using theT 0 y/bterm. Now you have to satisfy the boundary conditions
onx= 0andx=a because the total temperature there must be zero. That in turn means that the sum
over all the rest of the separated terms must add to−T 0 y/batx= 0andx=a. When you analyze this
solution in the same spirit as the analysis of Eq. ( 23 ), compare the convergence properties of that solution to
your new one. In particular, look ata banda bto see which version converges better in each case.
Ans:T 0 y/b+ (2T 0 /π)

∑∞

1

[

(−1)n/n

]

sin(nπy/b) cosh

(

nπ(x−a/2)/b

)/

cosh(nπa/ 2 b)

10.10 Finish the reanalysis of the electrostatic boundary value problem Eq. ( 42 ) starting from Eq. ( 43 ). This
will get the potential forz 6 = 0with perhaps less work than before.

10.11 Examine the solution Eq. ( 39 ) atz= 0in the light of problem5.11.

10.12 A thick slab of material is alternately heated and cooled at its surface so the its surface temperature
oscillates as

T(0,t) =

{

T 1 ( 0 < t < t 0 )

−T 1 (t 0 < t < 2 t 0 )

T(0,t+ 2t 0 ) =T(0,t)

That is, the period of the oscillation is 2 t 0. Find the temperature inside the material, forx > 0. How does this
behavior differ from the solution in Eq. ( 17 )?