7.4: Regular Singular Points Euler Equations

In the next three sections we’ll continue to study equations of the form

$$\label{eq:7.4.1} P_0(x)y''+P_1(x)y'+P_2(x)y=0$$

where $P_0$, $P_1$, and $P_2$ are polynomials, but the emphasis will be different from that of Sections 7.2 and 7.3, where we obtained solutions of Equation $\ref{eq:7.4.1}$ near an ordinary point $x_0$ in the form of power series in $x-x_0$. If $x_0$ is a singular point of Equation $\ref{eq:7.4.1}$ (that is, if $P(x_0)=0$), the solutions can’t in general be represented by power series in $x-x_0$. Nevertheless, it is often necessary in physical applications to study the behavior of solutions of Equation $\ref{eq:7.4.1}$ near a singular point. Although this can be difficult in the absence of some sort of assumption on the nature of the singular point, equations that satisfy the requirements of the next definition can be solved by series methods discussed in the next three sections. Fortunately, many equations arising in applications satisfy these requirements.

Definition 7.5.1 : Regular Singular Points

Let $P_0$, $P_1$, and $P_2$ be polynomials with no common factor and suppose $P_0(x_0)=0$. Then $x_0$ is a regular singular point of the equation

$$\label{eq:7.4.2} P_0(x)y''+P_1(x)y'+P_2(x)y=0$$

if $\eqref{eq:7.4.2}$ can be written as

$$\label{eq:7.4.3} (x-x_0)^2A(x)y''+(x-x_0)B(x)y'+C(x)y=0$$

where $A$, $B$, and $C$ are polynomials and $A(x_0)\ne0$; otherwise, $x_0$ is an irregular singular point of Equation $\ref{eq:7.4.2}$.

Example 7.5.1 : Bessel’s equation

Bessel’s equation,

$$\label{eq:7.4.4} x^2y''+xy'+(x^2-\nu^2)y=0,$$

has the singular point $x_0=0$. Since this equation is of the form Equation $\ref{eq:7.4.3}$ with $x_0=0$, $A(x)=1$, $B(x)=1$, and $C(x)=x^2-\nu^2$, it follows that $x_0=0$ is a regular singular point of Equation $\ref{eq:7.4.4}$.

Example 7.5.2 : Legendre’s equation

Legendre’s equation,

$$\label{eq:7.4.5} (1-x^2)y''-2xy'+\alpha(\alpha+1)y=0,$$

has the singular points $x_0=\pm1$. Mutiplying through by $1-x$ yields

$$(x-1)^2(x+1)y''+2x(x-1)y'-\alpha(\alpha+1)(x-1)y=0, \nonumber$$

which is of the form Equation $\ref{eq:7.4.3}$ with $x_0=1$, $A(x)=x+1$, $B(x)=2x$, and $C(x)=-\alpha(\alpha+1)(x-1)$. Therefore $x_0=1$ is a regular singular point of Equation $\ref{eq:7.4.5}$. We leave it to you to show that $x_0=-1$ is also a regular singular point of Equation $\ref{eq:7.4.5}$.

Example 7.5.3

The equation

$$x^3y''+xy'+y=0 \nonumber$$

has an irregular singular point at $x_0=0$. (Verify.)

For convenience we restrict our attention to the case where $x_0=0$ is a regular singular point of Equation $\ref{eq:7.4.2}$. This isn’t really a restriction, since if $x_0\ne0$ is a regular singular point of Equation $\ref{eq:7.4.2}$ then introducing the new independent variable $t=x-x_0$ and the new unknown $Y(t)=y(t+x_0)$ leads to a differential equation with polynomial coefficients that has a regular singular point at $t_0=0$. This is illustrated in Exercise 7.4.22 for Legendre’s equation, and in Exercise 7.4.23 for the general case.

Euler Equations

The simplest kind of equation with a regular singular point at $x_0=0$ is the Euler equation, defined as follows.

Definition 7.5.2 : Euler Equations

An Euler equation is an equation that can be written in the form

$$\label{eq:7.4.6} ax^2y''+bxy'+cy=0,$$

where $a,b$, and $c$ are real constants and $a\ne0$.

Theorem 5.1.1 implies that Equation $\ref{eq:7.4.6}$ has solutions defined on $(0,\infty)$ and $(-\infty,0)$, since Equation $\ref{eq:7.4.6}$ can be rewritten as

$$ay''+{b\over x}y'+{c\over x^2}y=0. \nonumber$$

For convenience we’ll restrict our attention to the interval $(0,\infty)$. (Exercise 7.4.19 deals with solutions of Equation $\ref{eq:7.4.6}$ on $(-\infty,0)$.) The key to finding solutions on $(0,\infty)$ is that if $x>0$ then $x^r$ is defined as a real-valued function on $(0,\infty)$ for all values of $r$, and substituting $y=x^r$ into Equation $\ref{eq:7.4.6}$ produces

$$\label{eq:7.4.7} \begin{array}{lcl} ax^2(x^r)''+bx(x^r)'+cx^r&= ax^2r(r-1)x^{r-2}+bxrx^{r-1}+cx^r\\[4pt] &= [ar(r-1)+br+c]x^r. \end{array}$$

The polynomial

$$p(r)=ar(r-1)+br+c\nonumber$$

is called the indicial polynomial of Equation $\ref{eq:7.4.6}$, and $p(r)=0$ is its indicial equation. From Equation $\ref{eq:7.4.7}$ we can see that $y=x^r$ is a solution of Equation $\ref{eq:7.4.6}$ on $(0,\infty)$ if and only if $p(r)=0$. Therefore, if the indicial equation has distinct real roots $r_1$ and $r_2$ then $y_1=x^{r_1}$ and $y_2=x^{r_2}$ form a fundamental set of solutions of Equation $\ref{eq:7.4.6}$ on $(0,\infty)$, since $y_2/y_1=x^{r_2-r_1}$ is nonconstant. In this case

$$y=c_1x^{r_1}+c_2x^{r_2}\nonumber$$

is the general solution of Equation $\ref{eq:7.4.6}$ on $(0,\infty)$.

Example 7.5.4

Find the general solution of

$$\label{eq:7.4.8} x^2y''-xy'-8y=0$$

on $(0,\infty)$.

Solution

The indicial polynomial of Equation $\ref{eq:7.4.8}$ is

$$p(r)=r(r-1)-r-8=(r-4)(r+2). \nonumber$$

Therefore $y_1=x^4$ and $y_2=x^{-2}$ are solutions of Equation $\ref{eq:7.4.8}$ on $(0,\infty)$, and its general solution on $(0,\infty)$ is

$$y=c_1x^4+{c_2\over x^2}.\nonumber$$

Example 7.5.5

Find the general solution of

$$\label{eq:7.4.9} 6x^2y''+5xy'-y=0$$

on $(0,\infty)$.

Solution

The indicial polynomial of Equation $\ref{eq:7.4.9}$ is

$$p(r)=6r(r-1)+5r-1=(2r-1)(3r+1).\nonumber$$

Therefore the general solution of Equation $\ref{eq:7.4.9}$ on $(0,\infty)$ is

$$y=c_1x^{1/2}+c_2x^{-1/3}. \nonumber$$

If the indicial equation has a repeated root $r_1$, then $y_1=x^{r_1}$ is a solution of

$$\label{eq:7.4.10} ax^2y''+bxy'+cy=0,$$

on $(0,\infty)$, but Equation $\ref{eq:7.4.10}$ has no other solution of the form $y=x^r$. If the indicial equation has complex conjugate zeros then Equation $\ref{eq:7.4.10}$ has no real–valued solutions of the form $y=x^r$. Fortunately we can use the results of Section 5.2 for constant coefficient equations to solve Equation $\ref{eq:7.4.10}$ in any case.

Theorem 7.5.3

Suppose the roots of the indicial equation

$$\label{eq:7.4.11} ar(r-1)+br+c=0$$

are $r_1$ and $r_2$. Then the general solution of the Euler equation

$$\label{eq:7.4.12} ax^2y''+bxy'+cy=0$$

on $(0,\infty)$ is

$$\begin{aligned} y&= c_1x^{r_1}+c_2x^{r_2}\mbox{ if $r_1$ and $r_2$ are distinct real numbers }; \\[4pt] y&= x^{r_1}(c_1+c_2\ln x)\mbox{ if $r_1=r_2$ }; \\[4pt] y&= x^{\lambda}\left[c_1\cos\left(\omega\ln x\right)+ c_2\sin\left(\omega\ln x \right)\right]\mbox{ if $r_1,r_2=\lambda\pm i\omega$ with $\omega>0$}.\end{aligned} \nonumber$$

Proof

We first show that $y=y(x)$ satisfies Equation $\ref{eq:7.4.12}$ on $(0,\infty)$ if and only if $Y(t)=y(e^t)$ satisfies the constant coefficient equation

$$\label{eq:7.4.13} a{d^2Y\over dt^2}+(b-a){dY\over dt}+cY=0$$

on $(-\infty,\infty)$. To do this, it is convenient to write $x=e^t$, or, equivalently, $t=\ln x$; thus, $Y(t)=y(x)$, where $x=e^t$. From the chain rule,

$${dY\over dt}={dy\over dx}{dx\over dt}\nonumber$$

and, since

$${dx\over dt}=e^t=x,\nonumber$$

it follows that

$$\label{eq:7.4.14} {dY\over dt}=x{dy\over dx}.$$

Differentiating this with respect to $t$ and using the chain rule again yields

$$\begin{aligned} {d^2Y\over dt^2}&= {d\phantom{t}\over dt}\left(dY\over dt\right)={d\phantom{t}\over dt}\left(x{dy\over dx}\right)\\[4pt] &= {dx\over dt}{dy\over dx}+x{d^2y\over dx^2}{dx\over dt}\\[4pt] &= x{dy\over dx}+x^2{d^2y\over dx^2}\quad\left(\mbox{ since } {dx\over dt}=x\right).\end{aligned}\nonumber$$

From this and Equation $\ref{eq:7.4.14}$,

$$x^2{d^2y\over dx^2}={d^2Y\over dt^2}-{dY\over dt}.\nonumber$$

Substituting this and Equation $\ref{eq:7.4.14}$ into Equation $\ref{eq:7.4.12}$ yields Equation $\ref{eq:7.4.13}$. Since Equation $\ref{eq:7.4.11}$ is the characteristic equation of Equation $\ref{eq:7.4.13}$, Theorem 5.2.1 implies that the general solution of Equation $\ref{eq:7.4.13}$ on $(-\infty,\infty)$ is

$$\begin{aligned} Y(t)&= c_1e^{r_1t}+c_2e^{r_2t}\mbox{ if $r_1$ and $r_2$ are distinct real numbers; }\\[4pt] Y(t)&= e^{r_1t}(c_1+c_2t)\mbox{ if $r_1=r_2$; }\\[4pt] Y(t)&= e^{\lambda t }\left(c_1\cos\omega t+c_2\sin\omega t \right)\mbox{ if $r_1,r_2=\lambda\pm i\omega$ with $\omega\ne0$}.\end{aligned}\nonumber$$

Since $Y(t)=y(e^t)$, substituting $t=\ln x$ in the last three equations shows that the general solution of Equation $\ref{eq:7.4.12}$ on $(0,\infty)$ has the form stated in the theorem.

Example 7.5.6

Find the general solution of

$$\label{eq:7.4.15} x^2y''-5xy'+9y=0$$

on $(0,\infty)$.

Solution

The indicial polynomial of Equation $\ref{eq:7.4.15}$ is

$$p(r)=r(r-1)-5r+9=(r-3)^2.\nonumber$$

Therefore the general solution of Equation $\ref{eq:7.4.15}$ on $(0,\infty)$ is

$$y=x^3(c_1+c_2 \ln x).\nonumber$$

Example 7.5.7

Find the general solution of

$$\label{eq:7.4.16} x^2y''+3xy'+2y=0$$

on $(0,\infty)$.

Solution

The indicial polynomial of Equation $\ref{eq:7.4.16}$ is

$$p(r)=r(r-1)+3r+2=(r+1)^2+1.\nonumber$$

The roots of the indicial equation are $r=-1 \pm i$ and the general solution of Equation $\ref{eq:7.4.16}$ on $(0,\infty)$ is

$$y={1\over x}\left[c_1\cos(\ln x)+c_2\sin(\ln x)\right].\nonumber$$