Kansas State University

Discussion

Obviously, it is important to determine whether an equation looks like an Euler equation in the neighborhood of a singular point. If it does, we say the singular point is a regular singular point. Otherwise the singular point is an irregular singular point. The precise definition is the following.

Definition: A point $x_0$ is a regular singular point for the linear differential equation $p(x)y''+q(x)y'+r(x)y=0$ if $p(x)$, $q(x)$ and $r(x)$ are all analytic in a neighborhood of $x_0$ and the following limits exist: $$ \lim_{x\to x_0}\frac{(x-x_0)q(x)}{p(x)}{\buildrel\text{def}\over =} \alpha_0\qquad\text{and}\qquad \lim_{x\to x_0}\frac{(x-x_0)^2r(x)}{p(x)}{\buildrel\text{def}\over =} \beta_0 $$ If either limit fails to exist, $x_0$ is an irregular singular point.

As before, don't worry about functions being "analytic." Any polynomial is analytic as is any rational, exponential, logarithmic or trigonometric function away from their singularities. The $\buildrel\text{def}\over =$ symbol means that $\alpha_0$ and $\beta_0$ are defined to be the values of those limits (if they exist). We will use $\alpha_0$ and $\beta_0$ in what follows.

The rationale behind this definition (other than that it is the one that makes the theorems we have later true) is the following. Divide through the equation $p(x)y''+q(x)y'+r(x)y=0$ by $p(x)$ and multiply by $(x-x_0)^2$ and you obtain $$ (x-x_0)^2y''+(x-x_0)\left(\frac{(x-x_0)q(x)}{p(x)}\right)y'+ \left(\frac{(x-x_0)^2r(x)}{p(x)}\right)y=0. $$ Now assuming $x_0$ is a regular singular point, we make the substitution $u=x-x_0$ and then can approximate this equation by the associated Euler equation $$ u^2y''+\alpha_0uy'+\beta_0y=0. $$ So the differential equation does indeed look like an Euler equation in the neighborhood of a regular singular point as promised. We will discuss how to use this information to obtain a series solution about a regular singular point in the next section.

Paradigm

Step 1: Find the singular points.

Since all the coefficients are polynomials, the singular points are just the roots of the leading coefficient, in this case $x^3-3x^2$. So the singular points are $0$ and $3$.

Step 2: For each singular point, classify it as regular or irregular by computing the appropriate limits.

For $x=0$, $$ \lim_{x\to0}\frac{(x-0)1}{x^3-3x^2}=\lim_{x\to0}\frac{1}{x^2-3x} \text{ is undefined.} $$ so $x=0$ is an irregular singular point.

For $x=3$, $$ \eqalign { &\lim_{x\to3}\frac{(x-3)1}{x^3-3x^2}=\lim_{t\to3}\frac{1}{x^2}=\frac19 \cr &\lim_{x\to3}\frac{(x-3)^22}{x^3-3x^2}=\lim_{x\to3}\frac{2(x-3)}{x^2}=0 \cr} $$ so both limits exist and $x=3$ is a regular singular point.