Math 555: Differential Equations

5.4 Regular Singular Points, Part I

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5.4.1 Euler Equations

In this chapter, we've been studying second-order linear differential equations with variable coefficients such as

$$P(x)y'' + Q(x)y' + R(x)y = G(x).$$
If $P(x)=0$ for all $x$ then we have a first-order differential equation, so we usually divide both sides by the function $P(x)$ to obtain the form

$$y'' + p(x)y' + q(x)y = g(x),$$
where $p(x) = \dfrac{Q(x)}{P(x)}$, $q(x) = \dfrac{R(x)}{P(x)}$ and $g(x) = \dfrac{G(x)}{P(x)}$. So the singular points are the points where $P(x)=0$.

An Euler Differential Equation is of the form

$$ax^2y'' + bxy' + cy = 0,$$
where $a$, $b$ and $c$ are constants. We have that $0$ is a singular point because $P(x) = ax^2 = 0$ when $x=0$. It is a regular singular point because we have two finite, well-defined limits.

$$\begin{align*} \displaystyle\lim_{x\to 0}\ (x-0)\dfrac{bx}{ax^2} &= \displaystyle\lim_{x\to 0}\ \dfrac{b}{a} = \dfrac{b}{a} \\ \\ \displaystyle\lim_{x\to 0}\ (x-0)^2\dfrac{c}{ax^2} &= \displaystyle\lim_{x\to 0}\ \dfrac{c}{a} = \dfrac{c}{a} \\ \end{align*}$$

This differential equation can be transformed into a second-order linear differential equations with constant coefficients using the substitution

$$t = \ln(x).$$
If we compute our new derivatives of $y$ with respect to $t$ we obtain

$$\begin{align*} \dfrac{dt}{dx} &= \dfrac{1}{x} \\ \\ \dfrac{dy}{dx} &= \dfrac{dy}{dt}\dfrac{dt}{dx} = \dfrac{1}{x}\dfrac{dy}{dt} \\ \\ \dfrac{d^2y}{dx^2} &= \dfrac{d}{dx}\dfrac{dy}{dx} = \dfrac{d}{dx}\left(\dfrac{1}{x}\dfrac{dy}{dt}\right) \\ \\ &= -\dfrac{1}{x^2}\frac{dy}{dt} + \dfrac{1}{x}\left(\dfrac{d}{dx}\dfrac{dy}{dt}\right) \\ \\ &= -\dfrac{1}{x^2}\frac{dy}{dt} + \dfrac{1}{x}\left(\dfrac{d^2y}{dt^2}\dfrac{dt}{dx}\right) \\ \\ &= \dfrac{1}{x^2}\left(\dfrac{d^2y}{dt^2} - \frac{dy}{dt}\right). \end{align*}$$

Substituting these expressions into our differential equation we have

$$\begin{align*} ax^2y'' + bxy' + cy &= ax^2\left(\dfrac{1}{x^2}\left(\dfrac{d^2y}{dt^2} - \frac{dy}{dt}\right)\right) + bx\left(\dfrac{1}{x}\dfrac{dy}{dt}\right) + cy \\ \\ &= a\dfrac{d^2y}{dt^2} - a\dfrac{dy}{dt} + b\dfrac{dy}{dt} + cy \\ \\ &= a\dfrac{d^2y}{dt^2} + (b-a)\dfrac{dy}{dt} + cy = 0. \end{align*}$$
This new 2nd-order homogeneous linear differential equation with constant coefficients can be easily be solved using the techniques of Chapter 3 .

Recall we will find two roots; either real distinct roots, complex conjugate roots, or repeated roots, $r_1$ and $r_2$. We can write the solution to the homogeneous equation

Solutions to Homogeneous Second-Order Equations with Constant Coefficients

1. Real Distinct Roots $\quad r_1 \neq r_2$

$$y(t) = c_1e^{r_1t} + c_2e^{r_2t},$$
2. Complex Conjugate Roots $\quad r = \mu \pm i\delta$

$$y(t) = e^{\mu t}\left(c_1\cos(\delta t) + c_2\sin(\delta t)\right),$$ 3. Repeated Roots $\quad r_1 = r_2$

$$y(t) = e^{rt}\left(c_1 + c_2t\right).$$

Recall that we are looking for the solution $y(x)$, so we must substitute $t = \ln|x|$ and we have the solutions

Solutions to Euler Equations

1. Real Distinct Roots $\quad r_1 \neq r_2$

$$\begin{align*} y(t) &= c_1e^{r_1\ln|x|} + c_2e^{r_2\ln|x|}, \\ \\ &= c_1|x|^{r_1} + c_2|x|^{r_2}, \end{align*}$$
2. Complex Conjugate Roots $\quad r = \mu \pm i\delta$

$$\begin{align*} y(t) &= e^{\mu\ln|x|}\left(c_1\cos(\delta\ln|x|) + c_2\sin(\delta\ln|x|)\right) \\ \\ &= |x|^{\mu}\left(c_1\cos\left(\ln\left(|x|^{\delta}\right)\right) + c_2\sin\left(\ln\left(|x|^{\delta}\right)\right)\right), \end{align*}$$
3. Repeated Roots $\quad r_1 = r_2$

$$\begin{align*} y(t) &= e^{r\ln|x|}\left(c_1 + c_2\ln|x|\right) \\ \\ &= |x|^r\left(c_1 + c_2\ln|x|\right).\end{align*}$$

Notice that the independent variable $y = e^{rt} = x^r$ are the fundamental solutions. If we use this substitution

$$y = x^r,\qquad y' = rx^{r-1},\qquad y'' = r(r-1)x^{r-2},$$
then our differential equation becomes

$$\begin{align*} ax^2y'' + bxy' + cy &= ax^2\left(r(r-1)x^{r-2}\right) + bx\left(rx^{r-1}\right) + cx^r \\ \\ &= a\left(r^2 - r\right)x^r + brx^r + cx^r \\ \\ &= \left(ar^2 + (b-a)r + c\right)x^r = 0. \end{align*}$$
Since $x=0$ is a singular point of the differential equation, $x^r\neq 0$ for any $x$ in the domain of our solution. Recall the domain of our solution must either be $(-\infty,0)$ or $(0,\infty)$. The gives us the indicial equation because the solution will give us the exponent or index of our independent variable $x$.

$$ar^2 + (b-a)r + c = 0.$$
The solutions of this quadratic equation is precisely the solution of our characteristic equation after the substitution $t = \ln(x)$, so we have our general solution and the fundamental solutions to the differential equation.

5.4.2 Practice Solving Euler Equations

The process for solving Euler equations is very similar to solving homogeneous second-order equations. We begin by finding the indicial equation , which is analogous to the characteristic equation, and determine the values of the index $r$ so that we can write down our solution using the form corresponding to the values of $r$.

Exercise 5.4.1

Find fundamental solutions and a general solution to the Euler equation

$$2x^2y'' + 3xy' - y = 0,\qquad x > 0.$$

View Solution

Substituting $y = x^r$ into our differential equation results in

$$\begin{align*} 2x^2r(r-1)x^{r-2} + 3xrx^{r-1} - x^r &= 0 \\ \\ \left(2(r^2 - r) + 3r - 1\right)x^r &= 0 \\ \\ 2r^2 + r - 1 &= 0 \\ \\ (2r - 1)(r + 1) &= 0 \\ \\ r = \dfrac{1}{2},\qquad r &= -1 \\ \\ y_1(x) = x^{1/2},\qquad &y_2(x) = x^{-1} \\ \\ y(x) = c_1x^{1/2} + c_2x^{-1},\qquad &x \gt 0. \\ \end{align*}$$

Exercise 5.4.2

Find the solution to the given initial value problem, plot the solution and describe the behavior of the solution as $x\rightarrow 0$.

$$x^2y'' - 3xy' + 4y = 0,\qquad y(-1) = 2,\ y'(-1) = 3.$$

View Solution

Substituting $y = x^r$ into our differential equation results in

$$\begin{align*} x^2r(r-1)x^{r-2} - 3xrx^{r-1} + 4x^r &= 0 \\ \\ \left(r^2-r\right)x^r - 3rx^r + 4x^r &= 0 \\ \\ r^2 - 4r + 4 &= 0 \\ \\ (r - 2)^2 &= 0 \\ \\ y(t) &= e^{2t}\left(c_1 + c_2t\right) \\ \\ y(x) &= x^2\left(c_1 + c_2\ln|x|\right) \end{align*}$$
Here we move to finding the values of $c_1$ and $c_2$ from the initial conditions.

$$\begin{align*} y(-1) &= (-1)^2\left(c_1 + c_2\ln|-1|\right) \\ \\ 2 &= c_1 \\ \\ y'(x) &= 2x\left(2 + c_2\ln|x|\right) - x^2\left(\dfrac{c_2}{|x|}\right) \\ \\ y'(-1) &= 2(-1)\left(2 + c_2\ln|-1|\right) - (-1)^2\left(\dfrac{c_2}{|-1|}\right) \\ \\ 3 &= -4 - c_2 \\ \\ \Rightarrow c_2 &= -7 \\ \\ y(x) &= x^2\left(2 - 7\ln|x|\right),\qquad x \lt 0. \end{align*}$$

To compute the limit
$$\begin{align*} \displaystyle\lim_{x\to0^-} 2x^2 - 7x^2\ln\left(|x|\right) &= 0 - 7\displaystyle\lim_{x\to 0^-} x^2\ln\left(|x|\right) \\ \\ &= -7\displaystyle\lim_{x\to0^-}\dfrac{\ln\left(|x|\right)}{x^{-2}} \\ \\ &= -7\displaystyle\lim_{x\to0^-}\dfrac{-\dfrac{1}{|x|}}{-2x^{-3}} \\ \\ &= -7\displaystyle\lim_{x\to0^-}\dfrac{x^3}{|x|} \\ \\ &= 7\displaystyle\lim_{x\to0^-} x^2 = 0. \end{align*}$$

Exercise 5.4.3

Determine the general solution of the differential equation

$$x^2y'' - 3xy' + 4y = 0,$$ in any interval not including the singular point.

View Solution

$$\begin{align*} x^2r(r-1)x^{r-2} + 4xrx^{r-1} + \frac{25}{4}x^r &= 0 \\ \\ \left(r^2 + 3r + \frac{25}{4}\right)x^r &= 0 \\ \\ r^2 + 3r + \frac{25}{4} &= 0 \\ \\ r &= \dfrac{-3\pm\sqrt{9-25}}{2} \\ \\ r &= -\dfrac{3}{2} \pm 2i \\ \\ y(t) &= e^{-3t/2}\left(c_1\cos(2t) + c_2\sin(2t)\right) \\ \\ y(x) &= x^{-3/2}\left(c_1\cos\left(2\ln|x|\right) + c_2\sin\left(2\ln|x|\right)\right) \\ \\ &= x^{-3/2}\left(c_1\cos\left(\ln\left(x^2\right)\right) + c_2\sin\left(\ln\left(x^2\right)\right)\right) \\ \end{align*}$$


$y(x) = x^{-3/2}\left(\cos\left(\ln\left(x^2\right)\right) + \sin\left(\ln\left(x^2\right)\right)\right)$.
If you play the video for a couple of seconds you will see the plot window zoom into the positive $x$-axis toward the origin. The natural logarithm function grows without bound toward minus infinity as $x\to 0$. The cosine and sine functions create a periodic curve that oscillates an infinite number of times along their negative domain to minus infinity. Thus the curve oscillates an infinite number of times as $x$ approaches zero. The cosine and sine function also achieve the maximum amplitude an infinite number of times as $x$ approaches zero. Since our solution divides the cosine and sine function by $x^{3/2}$, the amplitude gets large without bound to an asymptote all while oscillating faster and faster as $x$ approaches zero.

5.4.3 Discontinuities

In calculus, we discussed a difference between removable discontinuites and nonremovable discontinuities . Recall the definition of continuity

Definition of Continuity

A function $f$ defined on an interval of the real line $I=[c,d]\subset\mathbb{R}$ is continuous at $x=a$ if and only if

$$\displaystyle\lim_{x\to a^+} f(x) = \displaystyle\lim_{x\to a^-} f(x) = f(a) = L < \infty.$$

If $f$ is not continuous at $x=a$, then $f$ is said to have a discontinuity at $a$. There are degrees of discontinuity.

Definition of Removable Discontinuities

A function $f$ defined on an interval of the real line $I=[c,a)\cup(a,d]\subset\mathbb{R}$ has a removable discontinuity at $x=a$ if and only if
$$\displaystyle\lim_{x\to a^+} f(x) = \displaystyle\lim_{x\to a^-} f(x) = L < \infty,$$
but $f$ is not well-defined at $x=a$; that is $f(a)$ does not exist or $x=a$ is not in the domain of $f$.

Removable Discontinuity at $x=3$.

We call this weak discontinuity removable because we can fill the discontinuity by defining a new function $g:I\to\mathbb{R}$

$$g(x) = \left\{\begin{array}{lcl} f(x) & \qquad & x\neq a \\ L = \displaystyle\lim_{x\to a} f(x) & \qquad & x=a \end{array}\right..$$

A more serious discontinuity occurs when the two-sided limit does not exist; either because the limit from the left and the limit from the right are different finite values, or because one or both of the one-sided limits is infinite.

Definition of Jump Discontinuity

A function $f$ defined on an interval of the real line $I=[c,d]\subset\mathbb{R}$ has a jump discontinuity at $x=a$ if and only if

$$\displaystyle\lim_{x\to a^-} f(x) = L \neq M = \displaystyle\lim_{x\to a+} f(x).$$

The most serious discontinuity occurs where the graph of a function encounters a vertical asymptote.

Definition of Vertical Asymptote

A function $f$ defined on an interval of the real line $I=[c,a)\cup(a,d]\subset\mathbb{R}$ has a vertical asymptote at $x=a$ if either

$$\begin{align*} \lim_{x\to a^-} f(x) &= \pm\infty, \\ \\ \displaystyle\lim_{x\to a^+} f(x) &= \pm\infty, \end{align*}$$
or both.

5.4.3 Singular Points of a Differential Equation

Recall the first equation in Section 5.4.1

Definition of Singular Point

For a second-order linear differential equation given by

$$P(x)y'' + Q(x)y' + R(x)y = G(x),$$
any point $x=a$ such that $P(a)=0$ is a singular point of the differential equation.

In standard form our differential equation takes the form

$$y'' + p(x)y' + q(x)y = g(x),$$
where $p(x) = \dfrac{Q(x)}{P(x)}$, $q(x) = \dfrac{R(x)}{P(x)}$ and $g(x) = \dfrac{G(x)}{P(x)}$. Like discontinuities, there are different types of singular points of a differential equation. If the singular point is similar to the singular point for Euler's Equation then we would like to find a similar solution

$$y(x) = x^r\phi(x),$$
where $\phi(x)$ is an analytic function on some interval of convergence. If our singular point is $x=0$ and $\phi(x)$ is analytic on some circle of radius $\rho$ centered at $(0,0)$, then it has an absolutely convergent Maclaurin series

$$\phi(x) = \displaystyle\sum_{n=0}^{\infty} a_nx^n.$$
That means our solution takes the form

$$y(x) = x^r\displaystyle\sum_{n=0}^{\infty} a_nx^n = \displaystyle\sum_{n=0}^{\infty} a_nx^{n+r}.$$
Notice that if $0 < r < 1$, then the solution may not be differentiable at $x=0$. To see this let us look at the graph of $\phi(x) = |x|^{\frac{1}{2}}\cos(2x)$.


if $r < 0$, then $\displaystyle\lim_{x\to 0} \phi(x)=\pm\infty$. Too see this let us look at the graph of $\phi(x) = |x|^{-\frac{1}{2}}\cos(2x)$.

5.4.4 Solving a Differential Equation with a Singular Point

Before discussing more theoretical information, let us do an example to connect our method to solve Euler's equations to solving equations with a "well-behaved" singular point.

Example 5.4.1

Solve the differential equation

$$2x^2y'' - xy' + (1 + x)y = 0.$$

Solution

We will assume that this differential equation has such a well-behaved singularity at $x=0$, just like Euler's equation, and proceed to find a pair of fundamental solutions.

$$y = x^r\displaystyle\sum_{n=0}^{\infty} a_nx^n = \displaystyle\sum_{n=0}^{\infty} a_nx^{n+r}.$$
Differentiating we twice we obtain

$$y'(x) = \displaystyle\sum_{n=0}^{\infty} a_n(n+r)x^{n+r-1}\text{ and }y''(x) = \displaystyle\sum_{n=0}^{\infty} a_n(n+r)(n+r-1)x^{n+r-2}.$$
Substituting these expression into our differential equation results in

$$2x^2\displaystyle\sum_{n=0}^{\infty} a_n(n+r)(n+r-1)x^{n+r-2} - x\displaystyle\sum_{n=0}^{\infty} a_n(n+r)x^{n+r-1} + (1 + x)\displaystyle\sum_{n=0}^{\infty} a_nx^{n+r} = 0.$$
Distributing multiplication over addition yields

$$\displaystyle\sum_{n=0}^{\infty} 2a_n(n+r)(n+r-1)x^{n+r} - \displaystyle\sum_{n=0}^{\infty} a_n(n+r)x^{n+r} + \displaystyle\sum_{n=0}^{\infty} a_nx^{n+r} + \displaystyle\sum_{n=0}^{\infty} a_nx^{n+r+1} = 0.$$
We need to adjust the index of the last sum by substituting $n-1$ for $n$ to obtain

$$\displaystyle\sum_{n=0}^{\infty} 2a_n(n+r)(n+r-1)x^{n+r} - \displaystyle\sum_{n=0}^{\infty} a_n(n+r)x^{n+r} + \displaystyle\sum_{n=0}^{\infty} a_nx^{n+r} + \displaystyle\sum_{n=1}^{\infty} a_{n-1}x^{n+r} = 0.$$
The first three sums start at $n=0$, while the last sum starts at $n=1$ so we will re-write the first three sums

$$ \begin{align*} 2a_0(r)(r-1)x^r &- a_0(r)x^r + a_0x^r + \sum_{n=1}^{\infty} 2a_n(n+r)(n+r-1)x^{n+r} + \ldots \\ &\sum_{n=1}^{\infty} a_n(n+r)x^{n+r} + \sum_{n=1}^{\infty} a_nx^{n+r} + \sum_{n=1}^{\infty} a_{n-1}x^{n+r} = 0. \end{align*}$$
or

$$ a_0\left[2r(r-1) - r + 1\right]x^r + \displaystyle\sum_{n=1}^{\infty} \left[\left(2(n+r)(n+r-1) - (n+r) + 1\right)a_n + a_{n-1}\right]x^{n+r} = 0.$$
Since all of our coefficients must be equal to zero we have two equations; an indicial equation for $r$

$$2r(r-1) - r + 1 = 2r^2 - 3r + 1 = (2r-1)(r-1) = 0,$$
and a recurrence equation

$$\left(2(n+r)(n+r-1) - (n+r) + 1\right)a_n + a_{n-1} = 0,\qquad n\ge 1.$$
This gives us two distinct exponents at the singularity $r=\frac{1}{2}$ and $r=1$. Using the recurrence equation we solve for $a_n$ to get the recurrence formula

$$ a_n = -\dfrac{a_{n-1}}{2(n+r)^2 - 3(n+r) + 1} = -\dfrac{a_{n-1}}{\left(2(n+r)-1\right)\left(n+r-1\right)},\qquad n\ge 1.$$
Notice that we get our two fundamental solutions from the two indices like in the case of Euler's equation instead of obtaining a recurrence formula that separates into two separate solutions. For $r=1$, the recurrence formula becomes

$$a_n = -\dfrac{a_{n-1}}{(2n+1)n},\qquad n\ge 1.$$

The first few coefficients for the first fundamental series

$$\begin{align*} a_1 &= -\dfrac{a_0}{3\cdot 1}, \\ \\ a_2 &= -\dfrac{a_1}{5\cdot 2} = \dfrac{a_0}{(3\cdot 5)\cdot (1\cdot 2)}, \\ \\ a_3 &= -\dfrac{a_2}{7\cdot 3} = \dfrac{a_0}{(3\cdot 5\cdot 7)\cdot (1\cdot 2\cdot 3)}, \\ \end{align*}$$

give us the pattern for the equation of the general $n^{th}$ term

$$\begin{align*} a_n &= (-1)^n\dfrac{a_0}{(1\cdot 3\cdots (2n+1))n!} \\ \\ &= \dfrac{(-1)^n \color{#ec008c}{(2\cdot 4\cdots 2n)}}{1\cdot \color{#ec008c}{2}\cdot 3\cdot \color{#ec008c}{4}\cdots \color{#ec008c}{(2n)}(2n+1))n!}a_0 \\ \\ &= \dfrac{(-1)^n\color{#307fe2}{2^n}\color{#ec008c}{(1\cdot 2\cdots n)}}{(2n+1)!n!}a_0 \\ \\ &= \dfrac{(-1)^n2^nn!}{(2n+1)!n!}a_0 = \dfrac{(-1)^n2^n}{(2n+1)!}a_0. \end{align*}$$
This makes our series solution

$$\begin{align*} y_1(x) &= \displaystyle\sum_{n=0}^{\infty} a_nx^{n+1} \\ \\ &= a_0x + \displaystyle\sum_{n=1}^{\infty} \dfrac{(-1)^n2^n}{(2n+1)!}a_0x^{n+1} \\ \\ &= a_0\left(x + \displaystyle\sum_{n=1}^{\infty} \dfrac{(-1)^n2^n}{(2n+1)!}x^{n+1}\right) \\ \\ &= a_0x\left(1 + \displaystyle\sum_{n=1}^{\infty} \dfrac{(-1)^n2^n}{(2n+1)!}x^n\right) \\ \end{align*}$$
Thus our first fundamental solution is given by

$$y_1(x) = x\left(1 + \displaystyle\sum_{n=1}^{\infty} \dfrac{(-1)^n2^n}{(2n+1)!}x^n\right).$$


We can find the second fundamental solution using the second root of the indicial equation $r=\dfrac{1}{2}$. The recurrence formula becomes

$$a_n = -\dfrac{a_{n-1}}{\left(2\left(n+\frac{1}{2}\right)-1\right)\left(n+\frac{1}{2}-1\right)} = -\frac{a_{n-1}}{2n\left(n - \frac{1}{2}\right)} = -\dfrac{a_{n-1}}{n(2n-1)},\qquad n\ge 1.$$
The first few coefficients of the second fundamental solution

$$\begin{align*} a_1 &= -\dfrac{a_0}{1\cdot 1}, \\ \\ a_2 &= -\dfrac{a_1}{2\cdot 3} = \dfrac{a_0}{(1\cdot 2)\cdot(1\cdot 3)}, \\ \\ a_3 &= -\dfrac{a_2}{3\cdot 5} = -\dfrac{a_0}{(1\cdot 2\cdot 3)\cdot(1\cdot 3\cdot 5)}, \\ \\ a_4 &= -\dfrac{a_3}{4\cdot 7} = \dfrac{a_0}{(1\cdot 2\cdot 3\cdot 4)\cdot(1\cdot 3\cdot 5\cdot 7)}, \end{align*}$$
give us the pattern for the equation of the general $n^{th}$ term

$$\begin{align*} a_n &= (-1)^n\dfrac{a_0}{n!\left(1\cdot 3\cdot 5\cdots (2n-1)\right)} \\ \\ &= \dfrac{(-1)^n\color{#ec008c}{2\cdot 4\cdots 2n}}{n!\left(1\cdot \color{#ec008c}{2}\cdot 3\cdot \color{#ec008c}{4}\cdot 5\cdots(2n-1)\color{#ec008c}{(2n)}\right)}a_0 \\ \\ &= \dfrac{(-1)^n2^nn!}{n!(2n)!}a_0 = \dfrac{(-1)^n2^n}{(2n)!}a_0 \end{align*}$$
This makes our second fundamental solution

$$\begin{align*} y_2(x) &= \displaystyle\sum_{n=0}^{\infty} a_nx^{n+\frac{1}{2}} \\ \\ &= a_0x^{1/2} + \displaystyle\sum_{n=1}^{\infty}\dfrac{(-1)^n2^n}{(2n)!}a_0x^{n+\frac{1}{2}} \\ \\ &= a_0\left(x^{1/2} + \displaystyle\sum_{n=1}^{\infty}\dfrac{(-1)^n2^n}{(2n)!}x^{n+\frac{1}{2}}\right) \\ \\ &= a_0x^{1/2}\left(1 + \displaystyle\sum_{n=1}^{\infty}\dfrac{(-1)^n2^n}{(2n)!}x^n\right) \\ \end{align*}$$
Thus our second fundamental solution is given by

$$y_2(x) = x^{1/2}\left(1 + \displaystyle\sum_{n=1}^{\infty}\dfrac{(-1)^n2^n}{(2n)!}x^n\right),\qquad x>0.$$

5.4.5 Regular Singular Points

The question we need to answer is

What condition(s) must a second-order linear differential equation

$$P(x)y'' + Q(x)y' + R(x)y = 0$$
satisfy to have an absolutely convergent solution of the form

$$y = \displaystyle\sum_{n=0}^{\infty} a_nx^{n+r}$$
on some interval $(-\rho,0)$ or $(0,\rho)$.

If we write our differential equation as we did in 5.4.1 ,

$$y'' + p(x)y' + q(x)y = 0.$$
or multiplying both sides by $x^2$

$$x^2y'' + x\left(xp(x)\right)y' + \left(x^2q(x)\right)y = 0,$$
where $xp(x)$ and $x^2q(x)$ are analytic functions on some interval of radius $\rho$ centered at $0$. That is each has an absolutely convergent power series

$$xp(x) = \color{#307fe2}{\displaystyle\sum_{n=0}^{\infty} p_nx^n}$$
and

$$x^2q(x) = \color{#ec008c}{\displaystyle\sum_{n=0}^{\infty} q_nx^n}.$$
Thus we can compute two limits

$$\displaystyle\lim_{x\to 0} x\dfrac{x^2p(x)}{x^2} = \displaystyle\lim_{x\to 0} xp(x) = p_0 + \displaystyle\sum_{n=1}^{\infty} p_n0^n = p_0,$$ and $$\displaystyle\lim_{x\to 0} x^2\dfrac{x^2q(x)}{x^2} = \displaystyle\lim_{x\to 0} x^2q(x) = q_0 + \displaystyle\sum_{n=1}^{\infty} q_n0^n = q_0.$$
These are finite limits and thus this condition we need to satisfy becomes

Definition of Regular Singular Point

If the differential equation

$$y'' + p(x)y' + q(x)y = 0$$
has a singular point at $x=x_0$ and the two limits

$$ \lim_{x\to x_0} (x-x_0)p(x) = p_0\qquad\qquad\qquad \displaystyle\lim_{x\to x_0} (x-x_0)^2q(x) = q_0$$
are both well-defined finite limits then the singular point is called a regular singular point .

Example 5.4.2

Consider the differential equation

$$2xy'' + y' + xy = 0.$$
Show that $x=0$ is a regular singular point and find the general solution.

Solution

Show that $x=0$ is a regular singular point.
The standard form of the second-order linear differential equation is

$$y'' + \dfrac{1}{2x}y' + \dfrac{1}{2}y = 0,$$
and $x=0$ is a singular point since $P(0) = 2(0) = 0$. If we compute the two limits

$$\begin{align*} \displaystyle\lim_{x\to 0} xp(x) &= \displaystyle\lim_{x\to 0} x\dfrac{1}{2x} = \displaystyle\lim_{x\to 0} \dfrac{1}{2} = \dfrac{1}{2} \\ \\ \displaystyle\lim_{x\to 0} x^2q(x) &= \displaystyle\lim_{x\to 0} x^2\dfrac{1}{2} = 0, \end{align*}$$
and we conclude that $x=0$ is a regular singular point.

Determine the indicial equation and the roots of the indicial equation.
First we need to multiply both sides of our differential equation by $x$ to get

$$2x^2y'' + xy' + x^2y = 0.$$

$$y(x) = \displaystyle\sum_{n=0}^{\infty} a_nx^{n+r}$$
We obtain the series form of our differential operator $L[y]$

$$\begin{align*} &= 2x^2\displaystyle\sum_{n=0}^{\infty} a_n(n+r)(n+r-1)x^{n+r-2} + x\displaystyle\sum_{n=0}^{\infty} a_n(n+r)x^{n+r-1} + x^2\displaystyle\sum_{n=0}^{\infty} a_nx^{n+r} \\ \\ &= \displaystyle\sum_{n=0}^{\infty} 2a_n(n+r)(n+r-1)x^{n+r} + \displaystyle\sum_{n=0}^{\infty} a_n(n+r)x^{n+r} + \displaystyle\sum_{n=0}^{\infty} a_nx^{n+r+2} \\ \\ &= \displaystyle\sum_{n=0}^{\infty} 2a_n(n+r)(n+r-1)x^{n+r} + \displaystyle\sum_{n=0}^{\infty} a_n(n+r)x^{n+r} + \displaystyle\sum_{n=2}^{\infty} a_{n-2}x^{n+r} \\ \\ &= 2a_0(r)(r-1)x^r + 2a_1(r+1)rx^{r+1} + a_0rx^r + a_1(r+1)x^{r+1} \\ &\qquad\ + \displaystyle\sum_{n=2}^{\infty} 2a_n(n+r)(n+r-1)x^{n+r} + \displaystyle\sum_{n=x}^{\infty} a_n(n+r)x^{n+r} + \displaystyle\sum_{n=2}^{\infty} a_{n-2}x^{n+r} \\ \\ &= a_0\left(2(r^2-r) + r\right)x^r + a_1\left(2(r^2+r) + r + 1\right)x^{r+1} \\ &\qquad\ + \displaystyle\sum_{n=2}^{\infty} \left[2a_n(n+r)(n+r-1) + a_n(n+r) + a_{n-2}\right]x^{n+r} \\ \\ &= a_0\left(2r^2-r\right)x^r + a_1\left(2r^2 + 3r + 1\right)x^{r+1} \\ &\qquad\ + \displaystyle\sum_{n=2}^{\infty} \left[a_n\left(2(n+r)(n+r-1) + (n+r)\right) + a_{n-2}\right]x^{n+r} \end{align*}$$
Since all the coefficients of the series must be zero, yet $a_0\neq 0$ for a non-trivial solution, the indicial equation is given by

$$2r^2-r = r(2r-1) = 0.$$
So the roots of the indicial equation are $r=0$ and $r=\frac{1}{2}$.

Determine the recurrence relation.

First notice the other term outside of the series must also be zero,

$$a_1\left(2r^2 + 3r + 1\right) = 0$$
If $r=0$, then $a_1(2\cdot 0^2 + 3\cdot 0 + 1) = a_1 = 0$. And

if $r=\frac{1}{2}$, then $a_1\left(2\left(\frac{1}{4}\right) + 3\left(\frac{1}{2}\right) + 1\right) = 3a_1 = 0$. In either case $a_1=0$.

For the indices greater than one we still need each of the coefficients to be equal to zero, so the recurrence equation is

$$\begin{align*} a_n\left(2(n+r)(n+r-1) + (n+r)\right) + a_{n-2} &= 0 \\ \\ a_n(n+r)\left(2(n+r-1) + 1\right) &= -a_{n-2} \\ \\ a_n(n+r)(2n + 2r - 1) &= -a_{n-2}, \end{align*}$$
which we solve for $a_n$ to find

$$a_n = -\dfrac{a_{n-2}}{(n+r)(2n + 2r - 1)}.$$

Find the fundamental series solution for $x>0$ corresponding to the larger root.

For $r=\frac{1}{2}$, the recurrence formula becomes

$$\begin{align*} a_n &= -\dfrac{a_{n-2}}{\left(n+\frac{1}{2}\right)\left(2n + 2\frac{1}{2} - 1\right)} \\ \\ &= -\dfrac{a_{n-2}}{\left(n+\frac{1}{2}\right)(2n)} \\ \\ &= -\dfrac{a_{n-2}}{(2n+1)n},\qquad n\ge 2. \end{align*}$$
From the first few coefficients

$$\begin{align*} a_1 &= 0 \\ a_2 &= -\dfrac{a_0}{5\cdot 2} \\ a_3 &= -\dfrac{a_1}{7\cdot 3} = 0 \\ a_4 &= -\dfrac{a_2}{9\cdot 4} = \dfrac{a_0}{(5\cdot 9)\cdot (2\cdot 4)} \\ a_5 &= -\dfrac{a_3}{11\cdot 5} = 0 \\ a_6 &= -\dfrac{a_4}{13\cdot 6} = -\dfrac{a_0}{(5\cdot 9\cdot 13)\cdot (2\cdot 4\cdot 6)} \\ &\ddots \\ a_{2n-1} &= -\dfrac{a_{2n-3}}{(4n-1)(2n-1)} = 0 \\ a_{2n} &= -\dfrac{a_{2n-2}}{(4n+1)(2n)} = (-1)^n\dfrac{a_0}{\left(5\cdot 9\cdots (4n+1)\right)(2\cdot 4\cdots 2n)} \\ &= \dfrac{(-1)^n}{\left(5\cdot 9\cdots (4n+1)\right)2^nn!}a_0, \\ \\ \end{align*}$$
the first fundamental solution of our differential equation is

$$y_1(x) = x^{1/2}\left(1 + \displaystyle\sum_{n=1}^{\infty} (-1)^n\dfrac{x^{2n}}{\left(5\cdot 9\cdots (4n+1)\right)2^nn!}\right).$$

Determine the series solution of the other indicial root.

For $r=0$, the recurrence formula becomes

$$a_n = -\dfrac{a_{n-2}}{n(2n - 1)}.$$
From the first few coefficients

$$\begin{align*} a_1 &= 0 \\ a_2 &= -\dfrac{a_0}{2(3)} \\ a_3 &= -\dfrac{a_1}{3(5)} = 0 \\ a_4 &= -\dfrac{a_2}{4\cdot 7} = \dfrac{a_0}{(2\cdot 4)\cdot(3\cdot 7)} \\ a_5 &= -\dfrac{a_3}{5\cdot 9} = 0 \\ a_6 &= -\dfrac{a_4}{6\cdot 11} = -\dfrac{a_0}{(2\cdot 4\cdot 6)\cdot(3\cdot 7\cdot 11)} \\ &\ddots \\ a_{2n-1} &= -\dfrac{a_{2n-3}}{(2n-1)\cdot(4n-3)} = 0 \\ a_{2n} &= -\dfrac{a_{2n-2}}{2n\cdot(4n-1)} = (-1)^n\dfrac{a_0}{(2\cdot 4\cdots(2n))\cdot(3\cdot 7\cdots(4n-1))} \\ &= \dfrac{(-1)^n}{2^nn!(3\cdot 7\cdots(4n-1))}a_0, \end{align*}$$
the second fundamental solution of our series is

$$y_2(x) = 1 + \displaystyle\sum_{n=1}^{\infty}(-1)^n\dfrac{x^{2n}}{2^nn!(3\cdot 7\cdots(4n-1))}.$$

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