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6.3 Solutions About Singular Points: Class Notes

6.3a

A singular point for the equation $a_2(x)y″+a_1(x)y\prime +a_0(x)y=0(*)$ is a point $x=x_0$ such that $a_2(x_0)=0$. For our work with solving differential equations, we will only consider $x_0=0$. Standard form for $(*)$ is $y″+P(x)y\prime +Q(x)y=0$. Multiplying by $x^2$ (or by $(x-x_0{)}^2$ if $x_0\ne 0$) yields $x^{2}y″+x^{2}P(x)y\prime +x^{2}Q(x)y=0$. By relabeling $p(x)=xP(x)$, $q(x)=x^{2}Q(x)$, we get $x^{2}y″+xp(x)y\prime +q(x)y=0$.

Definition

$x=x_0$ is a regular singular point if $p(x)=(x-x_0)P(x)$ and $q(x)=(x-x_0{)}^{2}Q(x)$ are analytic at $x=x_0$ and an irregular singular point if either $p(x)$ or $q(x)$ is not analytic at $x=x_0$.

Example

Determine the singular points of the given differential equation. Classify each singular point as regular or irregular.

$$(x^2-9{)}^{2}y″+(x+3)y\prime +2y=0$$

Solution

The leading coefficient $(x^2-9{)}^2$ is zero at $x=3$ and $x=-3$, so $x=3,-3$ are singular points.

Divide by $(x^2-9{)}^2=(x+3{)}^2(x-3{)}^2$ to get standard form:

$${\displaystyle y″+\frac{1}{(x+3)(x-3{)}^2}y\prime +\frac{2}{(x+3{)}^2(x-3{)}^2}y=0}$$

$${\displaystyle P(x)=\frac{1}{(x+3)(x-3{)}^2},Q(x)=\frac{2}{(x+3{)}^2(x-3{)}^2}}$$

$x=-3$:

$${\displaystyle (x-(-3))P(x)=(x+3)P(x)=\frac{1}{(x-3{)}^2},(x+3{)}^{2}Q(x)=\frac{2}{(x-3{)}^2}}$$

Both are analytic at $x=-3$ (their denominators are not zero there), so

$x=3$:

$${\displaystyle (x-3)P(x)=\frac{1}{(x+3)(x-3)}}$$

This is not analytic at $x=3$ (its denominator is zero there), so

Theorem: (Frobenius’ Theorem)

If $x=x_0$ is a regular singular point of $(*)$, then there is at least one solution of the form $y=(x-x_0{)}^r\sum _{n=0}^{\infty }{c}_n(x-x_0{)}^n$ $=\sum _{n=0}^{\infty }{c}_n(x-x_0{)}^{n+r}$, where the number $r$ is a constant to be determined.

For us, $x_0=0$, so the form is

The number $r$ can be an integer or a fraction. We use this form for equations with terms like $xy″$ or $x^{2}y″$.

6.3b

Example

$x=0$ is a regular singular point of the given differential equation. Show that the indicial roots of the singularity do not differ by an integer. Use the method of Frobenius to obtain two linearly independent solutions about $x=0$. Form the general solution on $(0,\infty )$.

$$2xy″+5y\prime +xy=0$$

Solution

The solutions can include fractional powers, such as $c_0\sqrt{x}+c_1{x}^{3/2}+\cdots$. Write $y$, $y\prime$, and $y″$ in Frobenius form:

$$y=\sum _{n=0}^{\infty }{c}_n{x}^{n+r}\Rightarrow y\prime =\sum _{n=0}^{\infty }(n+r)c_n{x}^{n+r-1},y″=\sum _{n=0}^{\infty }(n+r)(n+r-1)c_n{x}^{n+r-2}$$

Tip: The starting value for the index does not change with this method: every series starts at $n=0$.

Substitute. Multiplying by $2x$ and by $x$ raises the exponents:

$$\sum _{n=0}^{\infty }2(n+r)(n+r-1)c_n{x}^{n+r-1}+\sum _{n=0}^{\infty }5(n+r)c_n{x}^{n+r-1}+\sum _{n=0}^{\infty }{c}_n{x}^{n+r+1}=0$$

Let $k=n$ in the first two series and $k=n+2$ in the third, so every exponent is $x^{k+r-1}$:

$$\sum _{k=0}^{\infty }[2(k+r)(k+r-1)+5(k+r)]c_k{x}^{k+r-1}+\sum _{k=2}^{\infty }{c}_{k-2}{x}^{k+r-1}=0$$

Factor out $k+r$: $(k+r)(2k+2r-2+5)=(k+r)(2k+2r+3)$

$$\sum _{k=0}^{\infty }(k+r)(2k+2r+3)c_k{x}^{k+r-1}+\sum _{k=2}^{\infty }{c}_{k-2}{x}^{k+r-1}=0$$

Write out the $k=0$ and $k=1$ terms of the first series:

$$r(2r+3)c_0{x}^{r-1}+(r+1)(2r+5)c_1{x}^r+\sum _{k=2}^{\infty }[(k+r)(2k+2r+3)c_k+c_{k-2}]x^{k+r-1}=0$$

Each coefficient is zero. The term with the smallest power, $x^{r-1}$, gives $r(2r+3)c_0=0$. The first coefficient $c_0$ is not zero, so

$$r(2r+3)=0\text{(indicial equation)}$$

$${\displaystyle r=0,-\frac{3}{2}\text{(indicial roots)}}$$

The roots differ by ${\displaystyle \frac{3}{2}}$, which is not an integer. The $x^r$ term gives $(r+1)(2r+5)c_1=0$. For these roots $(r+1)(2r+5)\ne 0$, so $c_1=0$.

From the remaining series, $(k+r)(2k+2r+3)c_k+c_{k-2}=0$. Solve for $c_k$:

$r=0$: ${\displaystyle c_k=-\frac{1}{k(2k+3)}{c}_{k-2}}$ ($k=2,3,4,\dots$)

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle k=2:c_2}& {\displaystyle =-\frac{1}{2(7)}{c}_0=-\frac{1}{14}{c}_0}\hfill \\ \hfill {\displaystyle k=3:c_3}& {\displaystyle =(\dots )c_1=0}\hfill \\ \hfill {\displaystyle k=4:c_4}& {\displaystyle =-\frac{1}{44}{c}_2=\frac{1}{616}{c}_0}\hfill \end{array}}$$

${\displaystyle r=-\frac{3}{2}}$: ${\displaystyle c_k=-\frac{1}{(k-\frac{3}{2})(2k)}{c}_{k-2}=-\frac{1}{(2k-3)k}{c}_{k-2}}$

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle k=2:c_2}& {\displaystyle =-\frac{1}{2}{c}_0}\hfill \\ \hfill {\displaystyle k=3:c_3}& {\displaystyle =(\dots )c_1=0}\hfill \\ \hfill {\displaystyle k=4:c_4}& {\displaystyle =-\frac{1}{20}{c}_2=\frac{1}{40}{c}_0}\hfill \end{array}}$$

Each root gives a solution with its own $c_0$; call it $C_1=c_0$ for $r=0$ and $C_2=c_0$ for ${\displaystyle r=-\frac{3}{2}}$. The general solution is

$${\displaystyle y=C_1(1-\frac{1}{14}{x}^2+\frac{1}{616}{x}^4+\cdots )+C_2{x}^{-3/2}(1-\frac{1}{2}{x}^2+\frac{1}{40}{x}^4+\cdots )}$$

The factor $x^{-3/2}$ is the $x^r$ from the Frobenius form.

Two linearly independent solutions:

6.3c

The equation $r(2r+3)=0$ is an indicial equation and can be generalized as follows:

$x=0$ is a regular singular point of $x^{2}y″+xp(x)y\prime +q(x)y=0$ if $p$ and $q$ are analytic at $x=0$. That is,

$$p(x)=a_0+a_{1}x+\cdots ,q(x)=b_0+b_{1}x+\cdots$$

Substitute the Frobenius series:

$$\begin{array}{cc}\hfill {\displaystyle \sum _{n=0}^{\infty }(n+r)(n+r-1)c_n{x}^{n+r}+(a_0+a_{1}x+\cdots )\sum _{n=0}^{\infty }(n+r)c_n{x}^{n+r}}& {\displaystyle }\hfill \\ \hfill {\displaystyle +(b_0+b_{1}x+\cdots )\sum _{n=0}^{\infty }{c}_n{x}^{n+r}}& {\displaystyle =0}\hfill \end{array}$$

We want the term with the smallest power on $x$. That is $x^r$, from $n=0$:

$$r(r-1)c_0{x}^r+a_{0}r{c}_0{x}^r+b_0{c}_0{x}^r=0$$

Divide by $c_0{x}^r$:

$$r(r-1)+a_{0}r+b_0=0\text{(indicial equation)}$$

where $a_0$ is the constant of $p(x)$ and $b_0$ is the constant of $q(x)$.

Example: the indicial equation from the general form

Consider $2xy″+5y\prime +xy=0$. Multiply by ${\displaystyle \frac{x}{2}}$ to put it in the form $x^{2}y″+xp(x)y\prime +q(x)y=0$:

$${\displaystyle x^{2}y″+\frac{5}{2}xy\prime +\frac{1}{2}{x}^{2}y=0}$$

$${\displaystyle p(x)=\frac{5}{2},q(x)=\frac{1}{2}{x}^2\Rightarrow a_0=\frac{5}{2},b_0=0}$$

$${\displaystyle r(r-1)+\frac{5}{2}r+0=0}$$

$$2r(r-1)+5r=0$$

$$2r^2+3r=0$$

$$r(2r+3)=0$$

6.3d

Example

Use the method of Frobenius to obtain two linearly independent solutions about the regular singular point $x=0$.

$$9x^{2}y″+9x^{2}y\prime +2y=0$$

Solution

Indicial roots. Divide by 9:

$${\displaystyle x^{2}y″+x^{2}y\prime +\frac{2}{9}y=0}$$

Matching $xp(x)y\prime$ and $q(x)y$:

$${\displaystyle p(x)=x\Rightarrow a_0=0,q(x)=\frac{2}{9}\Rightarrow b_0=\frac{2}{9}}$$

Indicial equation:

$${\displaystyle r(r-1)+0r+\frac{2}{9}=0}$$

$$9r^2-9r+2=0$$

$$(3r-1)(3r-2)=0$$

$${\displaystyle r=\frac{1}{3},\frac{2}{3}\text{(these don’t differ by an integer)}}$$

Back to the equation. Substitute $y=\sum _{n=0}^{\infty }{c}_n{x}^{n+r}$:

$$9x^{2}y″+9x^{2}y\prime +2y=0$$

$$\sum _{n=0}^{\infty }9(n+r)(n+r-1)c_n{x}^{n+r}+\sum _{n=0}^{\infty }9(n+r)c_n{x}^{n+r+1}+\sum _{n=0}^{\infty }2c_n{x}^{n+r}=0$$

Exponents: let $k=n$ in the first and third series and $k=n+1$ in the second:

$$\sum _{k=0}^{\infty }[9(k+r)(k+r-1)+2]c_k{x}^{k+r}+\sum _{k=1}^{\infty }9(k+r-1)c_{k-1}{x}^{k+r}=0$$

Indices: the $k=0$ term gives $[9r(r-1)+2]c_0=0$, so ${\displaystyle r=\frac{1}{3},\frac{2}{3}}$, the same roots as above. For $k\ge 1$:

$$(9(k+r)(k+r-1)+2)c_k+9(k+r-1)c_{k-1}=0$$

${\displaystyle r=\frac{1}{3}}$: ${\displaystyle c_k=-\frac{3k-2}{k(3k-1)}{c}_{k-1}}$

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle k=1:c_1}& {\displaystyle =-\frac{1}{2}{c}_0}\hfill \\ \hfill {\displaystyle k=2:c_2}& {\displaystyle =\frac{1}{5}{c}_0}\hfill \end{array}}$$

${\displaystyle r=\frac{2}{3}}$: ${\displaystyle c_k=-\frac{3k-1}{k(3k+1)}{c}_{k-1}}$

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle k=1:c_1}& {\displaystyle =-\frac{1}{2}{c}_0}\hfill \\ \hfill {\displaystyle k=2:c_2}& {\displaystyle =\frac{5}{28}{c}_0}\hfill \end{array}}$$

With $C_1=c_0$ for the first root and $C_2=c_0$ for the second:

$${\displaystyle y=C_1{x}^{1/3}(1-\frac{1}{2}x+\frac{1}{5}{x}^2+\cdots )+C_2{x}^{2/3}(1-\frac{1}{2}x+\frac{5}{28}{x}^2+\cdots )}$$

6.3e

If $r_1$ and $r_2$ differ by an integer–that is, if $r_2-r_1\in \mathbb{Z}$–we can use $r_1$ to find $y_1$ and then reduction of order to find $y_2$.

Example

$x=0$ is a regular singular point of the given differential equation. Show that the indicial roots of the singularity differ by an integer. Use the method of Frobenius to obtain at least one series solution about $x=0$. Use reduction of order and a CAS to find a second solution. Form the general solution on $(0,\infty )$.

$${\displaystyle y″+\frac{3}{x}y\prime -2y=0}$$

Solution

Indicial roots. Multiply by $x^2$:

$$x^{2}y″+3xy\prime -2x^{2}y=0$$

$$p(x)=3,q(x)=-2x^2\Rightarrow a_0=3,b_0=0$$

$$r(r-1)+3r=0$$

$$r^2+2r=0\Rightarrow r=-2,0$$

The roots differ by 2, an integer.

Substitute $y=\sum _{n=0}^{\infty }{c}_n{x}^{n+r}$ into the original equation:

$$\sum _{n=0}^{\infty }(n+r)(n+r-1)c_n{x}^{n+r-2}+\sum _{n=0}^{\infty }3(n+r)c_n{x}^{n+r-2}-\sum _{n=0}^{\infty }2c_n{x}^{n+r}=0$$

Combine the first two series, using

$$(n+r)(n+r-1)+3(n+r)=(n+r)(n+r+2)$$

and let $k=n$ there and $k=n+2$ in the third:

$$\sum _{k=0}^{\infty }(k+r)(k+r+2)c_k{x}^{k+r-2}-\sum _{k=2}^{\infty }2c_{k-2}{x}^{k+r-2}=0$$

The $k=0$ and $k=1$ terms:

$$r(r+2)c_0=0\Rightarrow r=0,-2,(r+1)(r+3)c_1=0\Rightarrow c_1=0$$

For $k\ge 2$:

$$(k+r)(k+r+2)c_k-2c_{k-2}=0,k=2,3,4,\dots$$

$r=0$: ${\displaystyle c_k=\frac{2}{k(k+2)}{c}_{k-2}}$

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle k=2:c_2}& {\displaystyle =\frac{1}{4}{c}_0}\hfill \\ \hfill {\displaystyle k=3:c_3}& {\displaystyle =\frac{2}{15}{c}_1=0}\hfill \\ \hfill {\displaystyle k=4:c_4}& {\displaystyle =\frac{1}{48}{c}_0}\hfill \end{array}}$$

$r=-2$: ${\displaystyle c_k=\frac{2}{k(k-2)}{c}_{k-2}}$. At $k=2$ the denominator $k(k-2)$ is zero, so this root does not give a second series solution.

From $r=0$ (with $c_0=1$):

$${\displaystyle y_1=1+\frac{1}{4}{x}^2+\frac{1}{48}{x}^4+\cdots }$$

This is the “at least one” power series solution from Frobenius’ Theorem.

Here ${\displaystyle P(x)=\frac{3}{x}}$:

$${\displaystyle e^{-\int P(x)dx}=e^{-\int \frac{3}{x}dx}=e^{-3\ln x}=e^{\ln x^{-3}}=\frac{1}{x^3}}$$

Tip: Move the $-3$ into the exponent of $x$ (a property of logarithms), so that $e^{\ln u}=u$ applies.

$${\displaystyle y_2=y_1\int \frac{1}{x^3{(1+\frac{1}{4}{x}^2+\frac{1}{48}{x}^4+\cdots )}^2}dx}$$

Use a CAS such as WolframAlpha.com. As a series, the integrand is ${\displaystyle \frac{1}{x^3}-\frac{1}{2x}+\frac{7x}{48}+\cdots }$, and

$${\displaystyle \int (\frac{1}{x^3}-\frac{1}{2x}+\frac{7x}{48})dx=\frac{7x^2}{96}-\frac{1}{2x^2}-\frac{\ln x}{2}}$$

(WolframAlpha writes $\ln x$ as log(x).) So

$${\displaystyle y_2=y_1(\frac{7}{96}{x}^2-\frac{1}{2x^2}-\frac{1}{2}\ln x+\cdots ),y_1=1+\frac{1}{4}{x}^2+\frac{1}{48}{x}^4+\cdots }$$