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616 CHAPTER 17 The Heat Equation

Case 1: λ = 0
We get X(x) = cx + d,so X(0) = d = 0. Then X(x) = cx,so

X (L) + AX(L) = c + cL = 0
and this forces c = 0. This case yields only the trivial solution, so 0 is not an eigenvalue of this
problem.

Case 2: λ< 0

2
2

Write λ =−α with α> 0. Then X − α X = 0, with solutions
αx
X(x) = ce + de −αx .
Now X(0) = c + d = 0 implies that d =−c,so
αx
X(x) = c(e − e −αx ).
From the radiation condition at L,
X (L) = αc(e αL + e −αL ) =−AX(L) =−Ac(e αL − e −αL ).

If c = 0, then

αc(e αL + e −αL )> 0 and − Ac e αL − e −αL
< 0,
contradicting the preceding line. We conclude that c = 0, so this case also has only the trivial
solution. This problem has no negative eigenvalue.

Case 3: λ> 0
2
2

Set λ = α with α> 0. Now X + α X = 0, so
X(x) = c cos(αx) + d sin(αx).
Then X(0) = c = 0, so X(x) = d sin(αx).Next,

X (L) = αd cos(αL) =−AX(L) =−Ad sin(αL).
If d = 0, we have only the trivial solution. If d = 0, then α cos(αL) =−A sin(αL),so
α
tan(αL) =− . (17.3)
A
We have a nontrivial solution for X only if α is chosen to satisfy this transcendental equa-
tion, which we cannot solve algebraically for α. However, let z = αL. Then equation (17.3) is
tan(z) =−z/AL. Part of the graphs of y = tan(z) and y =−z/AL are shown in Figure 17.1,
and they have infinitely many points of intersection for z > 0. The z-coordinates of these points
are solutions of tan(z) =−z/AL. Let these z-coordinates be z 1 , z 2 ,··· in increasing order. Since
α = z/L, the eigenvalues of this problem for X are
z 2
2 n
λ n = α = .
n 2
L
The eigenfunctions are functions X n (x) = sin(α n x) = sin(z n x/L).
With these eigenvalues, the problem for T is
2
z k
n
T + T = 0
L 2
with solutions that are constant multiples of
2
T n (t) = e −z n kt/L 2 .

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October 14, 2010 15:25 THM/NEIL Page-616 27410_17_ch17_p611-640