Power series solution : can we have a unique series solution(around $x=0$) for an IVP having $x=0$ a singular point?

Question

Studying power series and Frobenius method , I found a theorem stating that there is a unique Maclaurin series $y(x)$ satisfying the IVP $$y''+a(x)y'+b(x)y=0 ,\ \ y(0)=\alpha \ \ ,y'(0)=\beta$$ provided $a(x)$ and $b(x)$ are analytic at $x=0$ .

So does this mean that if we have $x=0$ is a regular singular point , we do not have always a unique solution for the IVP ? or not always , we may have a unique solution or not and we just do not guarantee the unique solution ?

For example , the problem $$xy''-xy'+y=e^{x}\ \ \ ,y(0)=1\ \ \ ,y'(0)=2$$ The general solution ( particular and homogeneous solutions) is $$y(x)=(e^x-x)+c_1x+c_2(-1+x\ln(x)+\frac{x^2}{2}+\frac{x^3}{12}+...)$$ (Note : the details of the solution is here Solving this non homogeneous IVP using power series) Applying initial conditions , we find that $c_1=2$ , $c_2=0$ (unique values for the constants although $x=0$ is singular point!)

However , another problem $$x^2y''-3xy'+3y=0,\ \ y(0)=0\ \ ,y'(0)=1$$ has $x=0$ a regular singular point , and has solution $$y=c_1x+c_2x^3$$ Applying intial conditions , we find that $c_1=1$ but $c_2$ has infinite values.

Answer 1

A unique solution to a differential equation is not always guaranteed, so a differential equation might have infinite solutions (same applies for a system of differential equations).

If you have an IVP of the form $x'=f(x,t), x(0)=x_0$, one way to check uniqueness is by checking if the function of the RHS $f(x,t)$ is Lipschitz or in other words, if its derivative is bounded.

Other than that, as you found out by two different examples, a solution may be unique or not.

The method of power series generates a general solution to the problem and gives you the ability to calculate constants by assuming solutions of the form :

$$y(x) = \sum_{n=0}^\infty a_n(x-x_0)^n$$

which is a series solution around the ordinary point $x_0$.

Uniqueness of a solution around a singular point that happens to be part of the initial value (such as your example), is again not guaranteed and can be determined via usage of theorems/lemmas, as mentioned.

You may read more about uniqueness and existence, here.