Second order linear ODE, self adjoint (Sturm-Liouville) form. Orthogonality of solutions - confused about the weight factor.

Question

If I have an ODE of the form $$a(x)y''+b(x)y'+c(x)y= \lambda y$$

Such that $b=a'$, then it is equivalent to:

$$(a(x)y')'+c(x)y= \lambda y$$

So the solutions corresponding to two different eigenvalues (suppose they are indexed by integers) are orthogonal (suppose $x\in[0,1]$):

$$\int_0^1 y_n(x)y_m(x)dx=\delta_{nm}||y_n||.$$

But now if $b \neq a'$, I want to multiply the equation through a weight $w(x)$ so that $wb =(wa)'$, therefore $$w(x)=\frac{1}{a} \exp{\int \frac{b}{a}}.$$

After this, the solutions are orthogonal, this time with the weight factor added in:

$$\int_0^1 y_n(x)y_m(x) w(x)dx=\delta_{nm}||y_n||.$$

Is the above correct? If so, consider the Bessel equation:

$$x^2y''+xy'+(x^2-n^2)y$$

The weight is $w(x)=\frac{1}{x}$, so why on earth does it say here that:

$$\int_0^1 xJ_\alpha (xu_{\alpha,n})J_\alpha (xu_{\alpha,m})dx = \delta_{nm}(\text{stuff})$$

Are those 'scaled' Bessel functions solution to some different differential equation with weight $x$?

Answer 1

I think you might have confused the term "weight function":

If you have $y'' + \frac{1}{x} y' + \frac{(x^2-n^2)}{x} y =0$ and you want to write this as a self-adjoint equation, then it must be:

$$\mathcal{L} = (p(x) y')' + (q(x) - \lambda \omega(x) ) y = 0,$$

where $\omega(x)$ is the weight function, $p(x)$ is a $C^1_x$ function and $q(x)$ is a $C^0_x$ function. Expanding this and identifying with the original ode, it yields:

$$\frac{p'}{p} = \frac{1}{x} \Rightarrow p(x) = x,$$

so it allows you to identify $q(x)$ and $\omega(x)$, if you put $\lambda = -n^2$.

Cheers!

Answer 2

Firstly, the version with the $1/x$ weight is also given in the Wikipedia page you linked to. It reads $$ \int_0^\infty J_\alpha J_\beta \frac{\mathrm{d}x}{x} = \frac2\pi \frac{\sin(\frac\pi2(\alpha - \beta))}{\alpha^2 - \beta^2} $$ Note the endpoints of the integral: boundary conditions are important!

Now, for the rescaled version, let me sketch the derivation here. The basic idea is to show that the functions $J_{\alpha}(u_{\alpha,n} x)$ is the eigenfunction over $[0,1]$ of some second order ODE with eigenvalue dictated by $u_{\alpha,n}$. The key is that $x^2 y'' + x y'$ is homogeneous. So if we let $s = \lambda x$ and $\frac{dy}{ds} = \dot{y}$, we have that $x^2 y'' + xy' = s^2 \ddot{y} + s \dot{y}$. So Bessel's equation is equivalent to, in this $s$ variable, $$ s^2 \ddot{y} + s \dot{y} - \alpha^2 y = - \frac{1}{\lambda^2} s^2 y \tag{*} $$

Here $\frac{1}{\lambda^2}$ plays the role of the eigenvalue. Observe that for $\lambda^{-1} = u_{\alpha,n}$ the roots of $J_\alpha$, this function now lives over $[0,1]$ and has the right boundary conditions. This means that the different "eigenvalues" are precisely those roots.

Now doing the "$1/x$" weight to bring it into self-adjoint form we have that the equation is now

$$ \frac{d}{ds}(s \dot{y}) - \frac{n^2}{s} y = - \frac{1}{\lambda^2} s y $$

and so we see that for the orthogonality argument to work, we have to use the weight coming in from the right hand side which is $s^1$.

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To further convince you, note that as an eigenvalue problem the "rescaled" equation (*) is in fact $$ \ddot{y} + \frac{1}{s} \dot{y} - \frac{n^2}{s^2} y = \mu y $$ where $\mu = - \frac{1}{\lambda^2}$ are the eigenvalues. This brings the equation to the form you are most familiar with in your original post, and you see here now that the correct weight function, in your language, should be $w(s) = s$.