TITLE: Intuition for the Frobenius method
QUESTION [19 upvotes]: I'm teaching a differential equations class now and I am hoping to give a reason for the Frobenius series method beyond simply "we guess these solutions".  Now, for the Euler equation
$$t^n x^{(n)}(t) + a_{n - 1} t^{n - 1} x^{(n - 1)}(t) + \dots + a_0 x(t) = 0$$
there is a good, easy explanation for why the fundamental solutions are of the form $x(t) = t^r$, where $r$ solves the indicial equation and repeated roots are handled by multiplying by powers of $\ln t$: just make the change of variables $s = \ln t$ and verify that this makes $t^n x^{(n)}(t)$ a constant-coefficient linear combination of the $x^{(k)}(s)$'s, and copy down the solutions to a constant-coefficient linear equation: $s^k e^{rs}$, with $k$ less than the multiplicity of $r$ as a root of the characteristic polynomial, which turns out here to be exactly the indicial polynomial.
But there isn't an apparent generalization of this analogy to arbitrary differential equations with regular singular points,
$$t^n x^{(n)}(t) + a_{n - 1}(t) t^{n - 1} x^{(n - 1)}(t) + \dots + a_0(t) x(t) = 0,$$
with the $a_i(t)$ analytic around $t = 0$.  You can make the same change of variables and render the equation non-singular, but it will:

still have variable coefficients;
even if you got the solutions as power series, the substitution $s = \ln t$ would make them into series in $\ln t$, which is not desirable;
when two roots differ by an integer then one of the solutions won't even be of the desired form;
when there's a repeated root, the second solution looks like
$$x_2(t) = x_1(t) \ln t + x^r v(t),$$
where $v(t)$ is some other power series, anyway, which is not what you'd get from the change of variables in any obvious way.  Now, it is true that there is the following relationship between $x_1(t)$ and $v(t)$:
$$x_1(t) = \sum_{i = 0}^\infty b_i(r) t^{n + r} \qquad
    v(t) = \sum_{i = 0}^\infty b_i'(r) t^{n + r}$$
where we differentiate the coefficients with respect to $r$, considered somehow as a continuous variable.

So, my question:

Is there some connection, via a transform, change of variables, or approximation, that produces the Frobenius method by analogy with non-singular equations?  Perhaps just when the roots of the indicial polynomial do not differ by integers?

REPLY [4 votes]: As you note, Ryan, it is quite sensible to try power functions for the Euler equation. Now the general equation with regular singular points is an analytic deformation of the Euler equation (near each of those singular points), in that its coefficients are analytic functions which, when evaluated at 0, give you an Euler equation. It is therefore very natural to look for solutions which are deformations of the solutions of the Euler equation. That is precisely what the Frobenius method does.
And it works marvels. For example, the indicial equation of the general equation is precisely the same as that for the corresponding Euler equation, so that the behaviour of solutions is qualitatively the same as for the Euler equation.