# For example to nd the explicit solution to the heat

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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 Cauchy-Euler 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.

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