Page 527 - Advanced engineering mathematics

15.1 Eigenfunction Expansions
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Case 1: λ = 0
Then y(x) = cx + d for some constants c and d.But y(0) = 0 = d and then y(L) = cL = 0
requires that c =0, so all solutions are trivial when λ=0. Therefore λ is not an eigenvalue of this
problem.
Case 2: λ< 0
2
2
Say λ =−k for k > 0. The general solution of y − k y = 0is
kx
y(x) = c 1 e + c 2 e −kx .
But y(0) = c 1 + c 2 = 0 means that c 2 =−c 1 ,so
kx −kx
y(x) = c 1 e − e .
Then
kL −kL
y(L) = c 1 e − e = 0.
If e kL − e −kL = 0 we would have e 2kL = 1 and then 2kL = 0, impossible if k > 0 and L > 0.
Therefore c 1 = 0, so c 2 = 0alsoand y is the trivial solution. This problem has no negative
eigenvalue.
Case 3: λ> 0
2
2
Say λ = k for k > 0. Now y + k y = 0 has general solution
y(x) = c 1 cos(kx) + c 2 sin(kx).
Since y(0) = c 1 = 0, then y = c 2 sin(kx). Then
y(L) = c 2 sin(kL) = 0.
To have a nontrivial solution we cannot have c 2 vanish, so we must choose k so that sin(kL) = 0.
Then kL = nπ for n any positive integer, so, using n as an index,
nπ
2
2
λ n = k = .
L
These are the eigenvalues of this problem, for each positive integer n. Corresponding to each
such eigenvalue is the eigenfunction
nπx
y n (x) = sin .
L
Any nonzero constant multiple of y n is also an eigenfunction corresponding to λ n .
EXAMPLE 15.2 A Periodic Sturm-Liouville Problem
The problem
y + λy = 0; y(−L) = y(L), y (−L) = y (L)
is periodic on [−L, L]. If we compare y + λy = 0 with the Sturm-Liouville equation we have
r(x) = 1, so r(−L) =r(L), as is required for a periodic problem on [−L, L]. We will also solve
this problem by taking cases on λ.
Case 1: λ = 0
Then y(x) = cx + d.Now
y(−L) =−cL + d = y(L) = cL + d
implies that c = 0. The constant function y(x) = d satisfies both boundary conditions. Thus 0 is
an eigenvalue of this problem with constant (nonzero) functions as eigenfunctions.
October 14, 2010 15:20 THM/NEIL Page-507 27410_15_ch15_p505-562