When solving a differential equation using series expansion method, if it has the following form :


; where $p$ and $q$ are analytic at $x_0$; if we want to find the solution in a series fprm expanded around $x_0$, it can't be solved using the regular method of series expansion and we must use Frobenius method to find the series around $x_0$.

Why do we need to expand around the singular point? why we don't solve the equation by expanding $y$ around a regular, nonsingular point (I mean a point different than $x_0$) and without using Frobenius method ? Is it a matter of convergence of the solution series at the singularity point?


First of all expansion around singular points are important/interesting for applications. For example, Bessel equation $$x^2y''+xy'+(x^2-\nu^2)y=0$$ often arises in problems with axial symmetry. Its singular point $x=0$ corresponds precisely to approaching the symmetry axis.

Second, expanding around an arbitrarily chosen point, in general, does not allow to obtain all expansion coefficients in a nice closed form - one only finds a recursion formula they satisfy. Whereas the recursion relation for the corresponding expansion coefficients around a singularity is often solvable. Example: the very same Bessel equation has a nice series solution $$J_{\nu}(x)=\sum_{k=0}^{\infty}\frac{(-1)^k}{k!\,\Gamma(k+\nu+1)}\left(\frac{x}{2}\right)^{2k+\nu}.$$

A more philosophical viewpoint: any 2nd order differential equation can be written as a rank 2 1st order linear system. Suppose we have its $2\times 2$ fundamental matrix solution $\Phi(x)$. Being analytically continued along a closed path on the Riemann sphere, $\Phi$ will be multiplied by a constant matrix, called monodromy matrix, which depends only on the homotopy class of the path.

If we choose $\Phi$ arbitrarily, the monodromy matrix for the loop encircling a particular singular point is nothing special. However, when $\Phi$ is constructed from the solutions obtained by the Frobenius method, the monodromy matrix is special (e.g. diagonal in the Bessel case). This is to say that such solutions are rather distinguished from the complex-analytic point of view. Yet in other words, Frobenius method suggests a particular "good" basis in the solution space. For instance, this choice of the basis is largely responsible for the existence of differentiation and recursion formulas for Bessel functions.

  • $\begingroup$ So, in general, "we can't expand $y$ at an analytical point and use that series solution for singular points too ( and vice versa)", because it doesn't necessarily converge at singular points too. Am I right? $\endgroup$ – Zorich Jul 3 '13 at 15:32
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    $\begingroup$ @Monopole Sorry, I do not understand your question. Expansion at non-singular point in general will have a non-zero radius of convergence but it will almost surely diverge at singular point (unless your equation accidentally has solutions regular at this singular point and you have accidentally picked such solution). The expansion at singular point is more delicate to deal with: it will be convergent in some neighbourhood of it if you have a regular singularity (like $x=0$ in Bessel equation), but it will be at best asymptotic if the singularity is irregular (like $x=\infty$ for Bessel). $\endgroup$ – Start wearing purple Jul 3 '13 at 18:26

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