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 $x$?
Best Answer
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!