Determining Weight function in Sturm Liouville problem

Question

By choosing the proper weight function $\sigma (x) $ solve the Sturm-Liouville problem and determine its eigenvalues and eigenfunctions.

$$ \frac{d}{dx}\left[x\frac{dy(x)}{dx}\right] + \frac{2}{x}y(x) +\lambda \sigma (x)y(x)=0,\; y'(1)=y'(2)=0,\; 1 \leq x \leq 2. $$

I don't understand what it means to "choose" the proper weight function. I tried to rewrite the problem in this form.

$$\frac{1}{\sigma(x)}\left[\frac{d}{dx}\left[x\frac{dy(x)}{dx} + \frac{2}{x}y(x)\right] +\lambda\sigma(x)=0\right], $$

then calculate it by setting $p(x)=A(x)\sigma (x), p'(x)=B(x)\sigma(x)$ and using this formula:

$$\sigma(x)=e^{\int \frac{A-B'}{B}\,dX}, $$ but it doesn't get me anywhere; solving this gives you just $1=1.$

I tried extracting information about the weight function from the boundary condition but i am failing at that too and i tried solving the differential equation using an infinite series but that won't work either because of the unknown weight function. Any tips?

Answer 1

For generic 2nd order (homogeneous) ODE:

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

To transform it to Sturm Liouville form you need:

$ a(x)>0$ and $a,b,c$ to be on continuous on the interval of definition.

Now the weight function is defined as follows:

$w(x) = \dfrac{1}{a(x)}e^{\int \frac{b(x)}{a(x)}dx}$

We now set:

$ p(x) = w(x)a(x)\\ q(x)=w(x)c(x)$

To do a sanity check find $p'(x)$ which is $b(x)w(x)$

so now you can rewrite the DE to:

$\dfrac{1}{w(x)} \left[ w(x)a(x)y'' + w(x)b(x)y' + w(x)c(x)y \right] =0\\ \dfrac{1}{w(x)} \left[ p(x)y'' + p'(x)y' + q(x) \right] = 0\\ \dfrac{1}{w(x)} \left[ (p(x)y')' + q(x) \right] = 0 $

That's how you reformulate to SL form. Hope this helps and you can work the details for your problem.

Answer 2

I think the problem might be a tad easier than you're making it out to be; most of the time, you can simply read off what the weight function should be. See this link for a type-up of some notes I got on SL problems, including a downloadable pdf.

For your problem, you need to massage the equation into the right form, from which you can simply read it off: \begin{align*} \frac{d}{dx}\left[x\frac{dy(x)}{dx}\right] + \frac{2}{x}y(x) +\lambda \sigma (x)y(x)&=0\\ \frac{d}{dx}\left[x\frac{dy(x)}{dx}\right] + \frac{2}{x}y(x)&=-\lambda \sigma (x)y(x); \end{align*} then you just need a positive weight function. The hope is that some combination of $\sigma(x)$ or $\lambda\sigma(x)$ or $-\lambda\sigma(x)$ or $-\sigma(x)$ is positive.

Alternatively, you could view $1/x$ as the weighting function by writing as $$\frac{d}{dx}\left[x\frac{dy(x)}{dx}\right]+\lambda \sigma (x)y(x)=- \frac{2}{x}y(x).$$ Depending on how $\sigma(x)$ behaves, this may be the only option, since $1/x$ does not change sign on your interval.

Answer 3

Your equation may be written as $$ -xy''-y'+\frac{2}{x}y=\lambda\sigma(x)y \\ -x^2y''-xy'+2y=\lambda \sigma(x)xy. $$ If you are allowed to choose $\sigma(x)$, I would choose $\sigma(x)x=1$. Then you have Euler's equation, which you can explicitly solve: $$ x^2y''+xy'+(2-\lambda)y=0,\;\;\; y'(1)=y'(2)=0. $$ The solutions where $y'(1)=0$ can be normalized by an added condition such as $y(1)=1$, for example. Then you can solve for $\lambda$ for which $y'(2)=0$, and that determines the eigenvalues $\lambda$.