As usual we consider the cases
,
and
separately. Consider the
equation first, since this has the most
restrictive explicit boundary conditions (
8.10
).
We have to solve
(8.11) |
which has as a solution
(8.12) |
Applying the boundary conditions, we get
If we eliminate one of the coefficients from the equation, we get
(8.15) |
which leads to
(8.16) |
which in turn shows
(8.17) |
and thus we only have a non-zero solution for , an integer. We have found
(8.18) |
We have
(8.19) |
This implies that
(8.20) |
The boundary conditions are satisfied for ,
(8.21) |
The solution (hyperbolic sines and cosines) cannot satisfy the boundary conditions.
___________________________
Now let me look at the solution of the equation for each of the two cases (they can be treated as one),
(8.22) |
Let us attempt a power-series solution (this method will be discussed in great detail in a future lecture)
(8.23) |
We find the equation
(8.24) |
If we thus have two independent solutions (as should be)
(8.25) |
The term with the negative power of diverges as goes to zero. This is not acceptable for a physical quantity (like the temperature). We keep the regular solution,
(8.26) |
For we find only one solution, but it is not very hard to show (e.g., by substitution) that the general solution is
(8.27) |
We reject the logarithm since it diverges at .