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### Regular Singular Points

#### Discussion

We have seen earlier in this chapter how to find series solutions about ordinary points. While those techniques don't work in general around singular points, there are different techniques that will work around some singular points. The key is whether the differential equation looks like an Euler equation near the singular point. If it does, then we can guess the correct form of the series solution by considering an Euler equation. If it doesn't look like an Euler equation, we are probably out of luck. There are no general techniques for that case.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

Find and classify the singular points of $(x^3-3x^2)y''+y'+2y=0$.
*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.

If you have any problems with this page, please contact bennett@math.ksu.edu.

©2010, 2014 Andrew G. Bennett