Math292 Textbook Examples of Regular Singular Points

Section 5.2 Examples of Regular Singular Points

Matching this up with

P (x) y^{″} + Q (x) y^{'} + R (x) y = 0,

we see that

P (x) = x^{2},

Q (x) = x,

and

R (x) = x^{2} .

Note that all of these are (technically) power series in

x .

Next, we find the singular points of the ODE, which starts by setting

P (x) = x^{2} = 0 .

Thus, the only possible singular point is

x = 0 .

Next, we check if

x = 0

is regular. As in the method, define

q (x) = Q (x) / P (x) = x / x^{2} = 1 / x

and

r (x) = R (x) / P (x) = x^{2} / x^{2} = 1 .

Then, we'll compute the following limits:

\begin{matrix} lim_{x \to 0} x q (x) = lim_{x \to 0} x \cdot \frac{1}{x} = 1, \end{matrix}

and

\begin{matrix} lim_{x \to 0} x^{2} r (x) = lim_{x \to 0} x^{2} \cdot 1 = 0. \end{matrix}

Since both of these limits are finite,

x = 0

is a regular singular point.

Try this one yourself! We have

P (x) = (x - 1)^{2} (x + 1),

Q (x) = x,

and

R (x) = x - 1

in this case.

Start by setting $P(x) = 0$ and solving for $x$ to find the possible singular points of the ODE. You'll see that $x = 1, -1\text{.}$

Next, find $q(x) = Q(x)/P(x)$ and $r(x) = R(x)/P(x)\text{.}$ Calculate the two limits

\begin{gather*} \lim_{x \to 1} (x - 1)q(x) \qquad \text{and} \qquad \lim_{x \to 1} (x-1)^2 r(x), \end{gather*}

and the two limits

\begin{gather*} \lim_{x \to -1} (x + 1)q(x) \qquad \text{and} \qquad \lim_{x \to -1} (x+1)^2 r(x). \end{gather*}

Which of these limits are finite?

$x = -1$ is a regular singular point and $x = 1$ is not.