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The boundary conditions (8.24) for u imply
dX dX
u x (0, t) = (0)T(t) = 0, u x (L, t) = (L)T(t) = 0.
dx dx
SInce u is nontrivial it follows that

dX dX
(0) = (L) = 0.
dx dx
Therefore, the function X should be a solution of the eigenvalue problem

2
d X
+ λX = 0, 0 < x < L, (8.25)
dx 2
dX dX
(0) = (L) = 0. (8.26)
dx dx

This eigenvalue problem is also called the Neumann problem. We have already written
the general solution of the ODE (8.25):
√ √
1) if λ < 0, then X(x) = α cosh( −λx) + β sinh( −λx),

2) if λ = 0, then X(x) = α + βx,
√ √
3) if λ > 0, then X(x) = α cos( λx) + β sin( λx),

where α, β are arbitrary real numbers.
Negative eigenvalue (λ < 0) The first boundary condition (dX/dx)(0) = 0 implies
√
that β = 0. Then (dX/dx)(L) = 0 implies that sinh( −λL) = 0. Therefore, X(x) ≡ 0
and the eigenvalue problem (8.25)-(8.26) does not admit negative eigenvalues.
Zero eigenvalue (λ = 0) The general solution is a linear function X(x) = α + βx.
Substituting this solution into the boundary conditions (8.26) implies that λ = 0 is an
eigenvalue with a unique eigenfunction X 0 (x) ≡ 1 (the eigenfunction is unique up to a
multiplicative factor).
Positive eigenvalue (λ > 0) The general solution for λ > 0 has the form
√ √
X(x) = α cos( λx) + β sin( λx). (8.27)

Substituting it in dX (0) = 0, we obtain β = 0. The boundary condition dX (L) = 0
dx √ √ dx
implies now that sin( λL) = 0. Thus λL = πn, where n ∈ N. Consequently, λ > 0 is
an eigenvalue if and only if:
πn
2
λ = ( ) , n = 1, 2, 3, . . .
L
The associated eigenfunction is
πnx
X(x) = cos ,
L

and it is uniquely determined up to a multiplicative factor.

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