A function which has a convergent power series expansion about a point is called analytic at that point. A function may not be analytic at some points but analytic every where else. This means that the function can be expanded about every point other than the analytic point.

Assuming that the above is correct, consider the application of power series to solve a second order ODE. Regular power series solutions are not applicable around singular points (not considering Frobenius method). This is understood. My question is this: Is it possible to define a convergent solution of ODE using Power series expanded about points other than the singular point? For example, consider, a certain ODE has a singular point at $x=-1,1$. Is it possible to have a convergent power series solution in the domain $x<-1$ and $x>1$ using the regular power series?


1 Answer 1


Of course it´s possible, although the series won´t be convergent for any interval containing x=-1, 1. The interest in developing series in singular points lies in the fact that singular points are often related to special points in the geometry of the problem where the equation comes from, such as corners or singularities in the system of coordinates.

Maybe it´s helpful to notice that the need of distinguishing between ordinary and singular points arise from the application of uniqueness and existence of solutions theorem for ordinary differential equations. Let me explain it with second order ones, considering no inhomogeneities:


Theis theorem says that this equation has a solution given that it´s defined within certain invertal $I$, you have initial conditions for $y$ and $y'$ inside it and coeficients $p$, $q$ and $r$ are continuous.

But equations to be solved by series often are presented like that:


So you have to redefine coeficients in order to apply the theorem:



Remember that a point is singular when $p(t)=0$. If there´re no singular points within the interval of the problem, you can apply the theorem, but if there´s some you can´t apply it building a series around that point because the coefficient $P$ and $Q$ is not defined, although if it´s regular you can transform it to an ode very similar to Euler´s equation and therefore prove the existence and uniqueness of the solution.

I´ve told you all this stuff because I wan´t you to understand that the singularity of one point in those problems only implies two things, that you can´t develop correctly any series around that point and that you can´t be sure that the solution exists and is only one. If the singular point is also regular, you can be sure that the solution exists because of the similarities with Euler´s equation, but you have to calculate following Frobenius path. If you develop a series in your example around any ordinary point you could arrive to no possible solutions, more that one solution, or maybe just one solution, but you´ve no previous information from any theorem, you just have to check it in that particular problem.


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