Mathematics · Ordinary Differential Equations · Linear Ordinary Differential Equations

Boundary Value Problems

Last edited 9/18/2026, 7:27:50 PM

An ODE with conditions at two points, rather than initial conditions at one, has nonzero solutions only for special values of the constant: the eigenvalues. For

X'' + \lambda X = 0

with both ends pinned to zero, the general solution is

A\cos\sqrt{\lambda}x + B\sin\sqrt{\lambda}x

; the condition at

x = 0

kills the cosine, and the condition at

x = L

forces

\sqrt{\lambda}L = n\pi

. Two things to hold onto: the eigenvalue is the square of

n\pi/L

, and it is the sine that vanishes at both ends. A cosine would need the derivative to vanish instead.

$X'' + \lambda X = 0, \qquad X(0) = X(L) = 0$
$X(0) = 0 \Rightarrow A = 0, \qquad X(L) = 0 \Rightarrow \sin\sqrt{\lambda}L = 0 \Rightarrow \sqrt{\lambda}L = n\pi$
$\lambda_n = \left(\frac{n\pi}{L}\right)^2, \qquad X_n(x) = \sin\frac{n\pi x}{L}, \qquad n = 1, 2, 3, \dots$