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# 3.4: Regular Singular Points: Euler Equations

• • William F. Trench
• Andrew G. Cowles Distinguished Professor Emeritus (Mathematics) at Trinity University
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## Regular Singular Points

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

\begin{equation}\label{eq:3.4.1}
P_0(x)y''+P_1(x)y'+P_2(x)y=0
\end{equation}

where $$P_0$$, $$P_1$$, and $$P_2$$ are polynomials, but the emphasis will be different from that of Sections 3.2 and 3.3, where we obtained solutions of \eqref{eq:3.4.1} near an ordinary point $$x_0$$ in the form of power series in $$x-x_0$$. If $$x_0$$ is a singular point of \eqref{eq:3.4.1} (that is, if $$P(x_0)=0$$), the solutions can't in general be represented by power series in $$x-x_0$$. Nevertheless, it's often necessary in physical applications to study the behavior of solutions of \eqref{eq:3.4.1} 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 $$\PageIndex{1}$$:

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 $$\textcolor{blue}{\mbox{regular singular point}}$$ of the equation

\begin{equation} \label{eq:3.4.2}
P_0(x)y''+P_1(x)y'+P_2(x)y=0
\end{equation}

if \eqref{eq:3.4.2} can be written as

\begin{equation} \label{eq:3.4.3}
(x-x_0)^2A(x)y''+(x-x_0)B(x)y'+C(x)y=0
\end{equation}

where $$A$$, $$B$$, and $$C$$ are polynomials and $$A(x_0)\ne0$$; otherwise, $$x_0$$ is an $$\textcolor{blue}{\mbox{irregular}}$$ singular point of \eqref{eq:3.4.2}.

Example $$\PageIndex{1}$$:

Bessel's equation,

\begin{equation} \label{eq:3.4.4}
x^2y''+xy'+(x^2-\nu^2)y=0,
\end{equation}

has the singular point $$x_0=0$$. Since this equation is of the form \eqref{eq:3.4.3} 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 \eqref{eq:3.4.4}.

Example $$\PageIndex{2}$$:

Legendre's equation,

\begin{equation} \label{eq:3.4.5}
(1-x^2)y''-2xy'+\alpha(\alpha+1)y=0,
\end{equation}

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

\begin{eqnarray*}
(x-1)^2(x+1)y''+2x(x-1)y'-\alpha(\alpha+1)(x-1)y=0,
\end{eqnarray*}

which is of the form \eqref{eq:3.4.3} 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 \eqref{eq:3.4.5}. We leave it to you to show that $$x_0=-1$$ is also a regular singular point of \eqref{eq:3.4.5}.

Example $$\PageIndex{3}$$:

The equation

\begin{eqnarray*}
x^3y''+xy'+y=0
\end{eqnarray*}

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 \eqref{eq:3.4.2}. This isn't really a restriction, since if $$x_0\ne0$$ is a regular singular point of \eqref{eq:3.4.2} 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 $$(3.4E.22)$$ for Legendre's equation, and in Exercise $$(3.4E.23)$$ 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 $$\PageIndex{2}$$:

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

\begin{equation} \label{eq:3.4.6}
ax^2y''+bxy'+cy=0,
\end{equation}

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

Theorem $$(2.1.1)$$ implies that \eqref{eq:3.4.6} has solutions defined on $$(0,\infty)$$ and $$(-\infty,0)$$, since \eqref{eq:3.4.6} can be rewritten as

\begin{eqnarray*}
ay''+{b\over x}y'+{c\over x^2}y=0.
\end{eqnarray*}

For convenience we'll restrict our attention to the interval $$(0,\infty)$$. (Exercise $$(3.4E.19)$$ deals with solutions of \eqref{eq:3.4.6} 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 \eqref{eq:3.4.6} produces

\begin{equation} \label{eq:3.4.7}
\begin{array}{lcl}
ax^2(x^r)''+bx(x^r)'+cx^r&=&ax^2r(r-1)x^{r-2}+bxrx^{r-1}+cx^r\\
&=&[ar(r-1)+br+c]x^r.
\end{array}
\end{equation}

The polynomial

\begin{eqnarray*}
p(r)=ar(r-1)+br+c
\end{eqnarray*}

is called the $$\textcolor{blue}{\mbox{indicial polynomia}}$$ of \eqref{eq:3.4.6}, and $$p(r)=0$$ is its $$\textcolor{blue}{\mbox{indicial equation}}$$. From \eqref{eq:3.4.7} we can see that $$y=x^r$$ is a solution of \eqref{eq:3.4.6} 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 \eqref{eq:3.4.6} on $$(0,\infty)$$, since $$y_2/y_1=x^{r_2-r_1}$$ is nonconstant. In this case

\begin{eqnarray*}
y=c_1x^{r_1}+c_2x^{r_2}
\end{eqnarray*}

is the general solution of \eqref{eq:3.4.6} on $$(0,\infty)$$.

Example $$\PageIndex{4}$$

Find the general solution of

\begin{equation} \label{eq:3.4.8}
x^2y''-xy'-8y=0
\end{equation}

on $$(0,\infty)$$.

The indicial polynomial of \eqref{eq:3.4.8} is

\begin{eqnarray*}
p(r)=r(r-1)-r-8=(r-4)(r+2).
\end{eqnarray*}

Therefore $$y_1=x^4$$ and $$y_2=x^{-2}$$ are solutions of \eqref{eq:3.4.8} on $$(0,\infty)$$, and its general solution on $$(0,\infty)$$ is

\begin{eqnarray*}
y=c_1x^4+{c_2\over x^2}.
\end{eqnarray*}

Example $$\PageIndex{5}$$

Find the general solution of

\begin{equation} \label{eq:3.4.9}
6x^2y''+5xy'-y=0
\end{equation}

on $$(0,\infty)$$.

The indicial polynomial of \eqref{eq:3.4.9} is

\begin{eqnarray*}
p(r)=6r(r-1)+5r-1=(2r-1)(3r+1).
\end{eqnarray*}

Therefore the general solution of \eqref{eq:3.4.9} on $$(0,\infty)$$ is

\begin{eqnarray*}
y=c_1x^{1/2}+c_2x^{-1/3}.
\end{eqnarray*}

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

\begin{equation} \label{eq:3.4.10}
ax^2y''+bxy'+cy=0,
\end{equation}

on $$(0,\infty)$$, but \eqref{eq:3.4.10} has no other solution of the form $$y=x^r$$. If the indicial equation has complex conjugate zeros then \eqref{eq:3.4.10} has no real-valued solutions of the form $$y=x^r$$. Fortunately we can use the results of Section 3.2 for constant coefficient equations to solve \eqref{eq:3.4.10} in any case.

Theorem $$\PageIndex{3}$$

Suppose the roots of the indicial equation

\begin{equation} \label{eq:3.4.11}
ar(r-1)+br+c=0
\end{equation}

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

\begin{equation} \label{eq:3.4.12}
ax^2y''+bxy'+cy=0
\end{equation}

on $$(0,\infty)$$ is

\begin{eqnarray*}
y&=&c_1x^{r_1}+c_2x^{r_2}\mbox{ if $$r_1$$ and $$r_2$$ are distinct real numbers };\\
y&=&x^{r_1}(c_1+c_2\ln x)\mbox{ if $$r_1=r_2$$ };\\
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{eqnarray*}

Proof

We first show that $$y=y(x)$$ satisfies \eqref{eq:3.4.12} on $$(0,\infty)$$ if and only if $$Y(t)=y(e^t)$$ satisfies the constant coefficient equation

\begin{equation} \label{eq:3.4.13}
a{d^2Y\over dt^2}+(b-a){dY\over dt}+cY=0
\end{equation}

on $$(-\infty,\infty)$$. To do this, it's convenient to write $$x=e^t$$, or, equivalently, $$t=\ln x$$; thus, $$Y(t)=y(x)$$, where $$x=e^t$$. From the chain rule,

\begin{eqnarray*}
{dY\over dt}={dy\over dx}{dx\over dt}
\end{eqnarray*}

and, since

\begin{eqnarray*}
{dx\over dt}=e^t=x,
\end{eqnarray*}

it follows that

\begin{equation} \label{eq:3.4.14}
{dY\over dt}=x{dy\over dx}.
\end{equation}

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

\begin{eqnarray*}
{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)\\
&=&{dx\over dt}{dy\over dx}+x{d^2y\over dx^2}{dx\over dt}\\
&=&x{dy\over dx}+x^2{d^2y\over dx^2}\quad\left(\mbox{ since } {dx\over dt}=x\right).
\end{eqnarray*}

From this and \eqref{eq:3.4.14},

\begin{eqnarray*}
x^2{d^2y\over dx^2}={d^2Y\over dt^2}-{dY\over dt}.
\end{eqnarray*}

Substituting this and \eqref{eq:3.4.14} into \eqref{eq:3.4.12} yields \eqref{eq:3.4.13}. Since \eqref{eq:3.4.11} is the characteristic equation of \eqref{eq:3.4.13}, Theorem $$(3.2.1)$$ implies that the general solution of \eqref{eq:3.4.13} on $$(-\infty,\infty)$$ is

\begin{eqnarray*}
Y(t)&=&c_1e^{r_1t}+c_2e^{r_2t}\mbox{ if $$r_1$$ and $$r_2$$ are distinct real numbers; }\\
Y(t)&=&e^{r_1t}(c_1+c_2t)\mbox{ if $$r_1=r_2$$; }\\
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{eqnarray*}

Since $$Y(t)=y(e^t)$$, substituting $$t=\ln x$$ in the last three equations shows that the general solution of \eqref{eq:3.4.12} on $$(0,\infty)$$ has the form stated in the theorem.

Example $$\PageIndex{6}$$

Find the general solution of

\begin{equation} \label{eq:3.4.15}
x^2y''-5xy'+9y=0
\end{equation}

on $$(0,\infty)$$.

The indicial polynomial of \eqref{eq:3.4.15} is

\begin{eqnarray*}
p(r)=r(r-1)-5r+9=(r-3)^2.
\end{eqnarray*}

Therefore the general solution of \eqref{eq:3.4.15} on $$(0,\infty)$$ is

\begin{eqnarray*}
y=x^3(c_1+c_2 \ln x).
\end{eqnarray*}

Example $$\PageIndex{7}$$

Find the general solution of

\begin{equation} \label{eq:3.4.16}
x^2y''+3xy'+2y=0
\end{equation}

on $$(0,\infty)$$.

The indicial polynomial of \eqref{eq:3.4.16} is

\begin{eqnarray*}
p(r)=r(r-1)+3r+2=(r+1)^2+1.
\end{eqnarray*}

The roots of the indicial equation are $$r=-1 \pm i$$ and the general solution of \eqref{eq:3.4.16} on $$(0,\infty)$$ is

\begin{eqnarray*}
y={1\over x}\left[c_1\cos(\ln x)+c_2\sin(\ln x)\right].
\end{eqnarray*}