H2.4-4

Statement: X'' + lambda X = 0, X'=0 at x=0 and x=L has only solution X=0 for lambda negative.

Details: Write lambda = -a^2 with a positive. Then the differential equation can be solved for general solution

X = c1 exp(ax) + c2 exp(-ax)

Compute X' = a c1 exp(ax) - a c2 exp(-ax), then translate X'=0 at x=0 and x=L into two equations for c1, c2
(symbol a appears also, a positive constant).


  a c1 - a c2 = 0,  a c1 exp(aL) - a c2 exp(-aL) = 0.

Cancel symbol a, because it is positive. Solve for c2 in terms of c1. Substitute into the second equation to
get

  c1 (exp(aL) - exp(-aL)) = 0

Then either c1 = 0 or else exp(aL) - exp(-aL) = 0. The latter cannot happen (why? Try an example like a=2, L=10). Then
c1 = 0. Conclude that also c2 = 0. Then X=0 is the only solution.