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I have the ODE $$-(1-x^2) \frac{d^2 f(x)}{dx^2} + x \frac{df(x)}{dx}+g(x)f(x)=\lambda f(x)$$

and I want to reduce it to Sturm-liouville form.

The problem is that we don't have $2x$ but just $x$. otherwise it would be similar to the Legendre differential equation.

Could anybody help me with that? By the way, does this mean that the Sturm-Liouville operator is not self-adjoint?

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3 Answers 3

up vote 1 down vote accepted

I believe the correct approach is to divide both sides by $-(1-x^2)$ then use the approach of integrating factor.

In this particular case, you need to multiply the equation by $$\exp\big(\int-\frac{x}{1-x^2}dx\bigg)$$ then you should be able to do it. You should replace $x$ by $2x$ in the integrating factor (as in Lengendre polynomial) to see what happens.

A good example on the wikipedia page:

This operator is, of course, self-adjoint.

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Here is an approach. Assume we have the ode

$$ a(x)y''+b(x)y'+c(x)y = 0 \longrightarrow (*)$$

and we want to have the Sturm Liouville Form. Multiply $(*)$ by the function $\mu(x)$ as

$$ \mu\,a y''+\mu b y'+\mu c y = 0 .$$

To determine $\mu(x)$ we have

$$(\mu a y' )' +\mu c y = 0 $$

$$ \implies (\mu a)' = \mu b $$

$$ \implies \mu(x) = e^{\int \frac{b(x)-a'(x)}{a(x)}}. $$

Now you can advance.

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Don't change your answer. An alternative substitution is $y = (1-x^{2})^{1/4} f$. This gives the alternative form $$ -\frac{d}{dx}(1-x^{2})\frac{df}{dx} +\left(\frac{1}{4(1-x^{2})}+\frac{1}{4}+g\right)f =\lambda f $$ This form is in symmetric form for $f$, but not for $y$.

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