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Unformatted text preview: n is just the projection of h in the fn direction.
Example. We consider the operator
L=x d
dx
109 x d
dx , which acts on the space V0 of functions f : [1, eπ ] → R which vanish at the
endpoints of the interval [1, eπ ]. To solve the eigenvalue problem, we need to
ﬁnd the nontrivial solutions to
x d
dx x df
(x)
dx f (1) = 0 = f (eπ ). = λf (x), (4.32) We could ﬁnd these nontrivial solutions by using the techniques we have learned
for treating CauchyEuler equations.
However, there is a simpler approach, based upon the technique of substitution. Namely, we make a change of variables x = ez and note that since
dx = ez dz, 1d
d
=z
dx
e dz Thus if we set and hence x d
d
1d
= ez z
=
.
dx
e dz
dz ˜
f (z ) = f (x) = f (ez ), the eigenvalue problem (4.32) becomes
˜
d2 f
˜
(z ) = λf (z ),
2
dz ˜
˜
f (0) = 0 = f (π ), a problem which we have already solved. The nontrivial solutions are
λn = −n2 , ˜
fn (z ) = sin nz, where n = 1, 2, . . . . Thus the eigenvalues for our original problem are
λn = −n2 , for n = 1, 2, . . . , and as corresponding eigenfunctions we can take
fn (x) = sin(n...
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This document was uploaded on 01/12/2014.
 Winter '14
 Equations

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