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I have the problem:

$$x^2y''+(1/4)y=\lambda y, 1<x<e$$ $$y(1)=y(e)=0$$

And I'm trying to find a weighting function for its solutions (I already calculated the solutions and eigenvalues). If we plug $x=e^t$ we can see the equation becomes:

$$Y''-Y'+(1/4-\lambda)Y=0, 0<x<1$$

This seems like something I should be able to transform into Sturm-Liouville form, but since the coefficients of Y'' and $\lambda Y$ are different it doesn't seem feasible.

Any suggestions?

Thanks!

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@JohnD: Will do, once I get home. –  add Feb 1 '13 at 21:46
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2 Answers

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Since the eigenvalues are negative, let $\lambda\to -\lambda$. Divide by $-x^2$ so $$-y'' - \frac{1}{4x^2}y = \frac{\lambda}{x^2} y.$$ We are looking for an ODE of the form $$-(p y')' + q y = \lambda w y.$$ (We follow the notation used here.) Note that if $p=1$, $(p y')' = y''$. Thus, we can simply read off $w$ from the right hand side. It is a good exercise to check that the eigenfunctions are orthogonal with respect to this weight function.

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The equation $$a_2(x)y''(x)+a_1(x)y'(x)+a_0(x)y(x)+\lambda y=0, \quad a<x<b,$$ can be brought into the so-called self-adjoint form given by $${1\over w(x)}[(p(x)y')'+q(x)y]+\lambda y=0,$$ via $$ w(x)=\exp\left(\int {a_1(x)-a_2'(x)\over a_2(x)}\,dx\right), \quad p(x)=a_2(x)w(x), \quad q(x)=a_0(x)w(x). $$ (It's a good exercise to show why this is true.)

The $w(x)$ here is the weight function.

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