Jeffrey Hellrung
Wednesday, December 06, 2006
Math 266A, Homework 07
1. For the operator and boundary conditions
Lu =u′′ + cos xu′;
u(0) = u(1) = 0;
find a weighted inner product
(u, v)q=Z1
0
uvqdx
such that Lis self-adjoint, i.e.,
(u, Lv)q= (Lu, v)q.
Hint: Look for q=er.
Solution
Using integration by parts and the boundary conditions on uand v,
(u, Lv)q=Z1
0
u(Lv)qdx
=Z1
0
uqv′′ dx +Z1
0
uq cos xv′dx
=uqv′|1
0−Z1
0
(uq)′v′dx +uq cos xv|1
0−Z1
0
(uq cos x)′vdx
=−(uq)′v
1
0+Z1
0
(uq)′′ vdx −Z1
0
(uq cos x)′vdx
=Z1
0(uq)′′ −(uq cos x)′vdx.
Equating this with (Lu, v)q, and noting the equality can only hold for all vif the integrands are
identical, we find that
(uq)′′ −(uq cos x)′=Luq
for all u. Expanding each side yields
qu′′ + (2q′−cos xq)u′+ (q′′ −(cos xq)′)u=qu′′ + cos xq u′.
Equating the coefficients on u′and ugives, in both cases,
q′−cos xq = 0,
which has solution
q(x) = cesin x.
For (·,·)qto be an inner product, we’d require c > 0, e.g., c= 1.
2. If Lis self-adjoint in a weighted inner product (·,·)qwith q > 0, show that the eigenvalues are real and
that eigenfunctions with distinct eigenvalues are orthogonal.
Solution
Suppose (λ, u) is an eigenvalue / eigenfunction pair. Then
(u, Lu)q= (u, λu)q=λ(u, u)q=λkuk2
q;
1
