General examples of Sturm-Liouville operators The topic:
My question pertains to examples of Sturm-Liouville operators in the context of a technical research paper on functional determinants of differential operators : http://arxiv.org/abs/0711.1178.
My background reading on Sturm-Liouville operators:
I have been reading up on Wikipedia on the definition of a Sturm-Liouville operator and I understand that all second-order linear ordinary differential operators can be expressed in Sturm-Liouville form. Does that imply that the most general operator which can be recast in Sturm-Liouville form must contain a first-order derivative term?
Actual question:
In the beginning of Section 3, the paper explains that functional determinants of operators of the form $$-\frac{d^{2}}{dx^{2}} + V(x)$$ can be found using a so-called Gel'fand-Yaglom theorem. The paper goes on to state the theorem and give some examples. That's all good and well.
Now, at the end of Section 3, the author mentions that the Gel'fand-Yaglom theorem is, in fact, applicable for all Sturm-Liouville problems (which means that the Gel'fand-Yaglom theorem is applicable for all second-order differential operators?) The author then goes on to use $$-\frac{d^{2}}{dr^{2}}+\frac{(l+\frac{d-3}{2})(l+\frac{d-1}{2})}{r^{2}}+V(r)$$ as an example of a Sturm-Liouville operator, the determinant of which he calculates.
Now, I don't see how the operator $-\frac{d^{2}}{dr^{2}}+\frac{(l+\frac{d-3}{2})(l+\frac{d-1}{2})}{r^{2}}+V(r)$ is any different from the operator $-\frac{d^{2}}{dx^{2}} + V(x)$, as both do not have a first-order derivative term, and so the second differential operator is not really a fitting example of a very general Sturm-Liouville operator, is it?
 A: The general Sturm Liouville eigenvalue problem is
$$
          -\frac{d}{dx}\left(p\frac{df}{dx}\right)+qf = \lambda wf.
$$
Here it is assumed that $w > 0$, $p > 0$. In this context, one must use the weighted $L_w^2$ space with inner product
$$
                (f,g)_{w}=\int_{I} f \overline{g}w dx.
$$
The actual operator which is symmetric in this case is
$$
               Lf = \frac{1}{w}\left[-\frac{d}{dx}\left(p\frac{df}{dx}\right)+qf\right],
$$
and the eigenvalue problem is $Lf=\lambda f$.
If you take $w=1$ and $p=1$, then you get standard Cartesian potential form:
$$
                       Lf = -\frac{d^{2}f}{dx^{2}}+qf.
$$
There are multiple choices of changes of variable that will transform
$$
                    -a(x)f''(x)+b(x)f'(x)+c(x)f(x) = \lambda d(x)f(x)
$$
into standard Sturm-Liouville form, assuming $a > 0$ and $d > 0$. One way is to multiply by an integrating factor for the first two terms, which is $e^{-\int (b/a) dx}$, and you get
$$
               -\frac{d}{dx}\left(e^{-\int (b/a)dx}\frac{df}{dx}\right)
    +\frac{c}{a}e^{-\int(b/a)dx}f = \lambda \frac{d}{a}e^{-\int(b/a)dx} f.
$$
There are other ways to do this by letting $f = \rho g$ and choosing $\rho$ to get the form $-\frac{d}{dx}( h\frac{df}{dx})$ for the first two terms. Both of these approaches have merit; the second approach requires more differentiability for the coefficients, but the second approach is the one most often used in classical equations of Physics/Engineering. Combining the two (assuming differentiability of the coefficients) and a change of independent variable, you can even reduce to the form
$$
                 -\frac{d^{2}f}{dx^{2}}+qf = \lambda w f.
$$
