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### Series Solutions About Regular Singular Points

#### Discussion

We are finally ready to work out series solutions about regular singular points. About ordinary points we guess the form of the solution is $\sum_{n=0}^{\infty}a_n(x-x_0)^n$, plug that into the equation and solve for the $a_n$. About regular singular points we do pretty much the same thing, except we must pick a different form for the solution. We pick the form in analogy with Euler equations. If $x_0$ is a regular singular point for the equation $p(x)y''+q(x)y'+r(x)y=0$ with $\lim_{x\to x_0}(x-x_0)q(x)/p(x)=\alpha_0$ and $\lim_{x\to x_0}(x-x_0)^2r(x)/p(x)=\beta_0$, then near the point $x_0$ the equation $p(x)y''+q(x)y'+r(x)y=0$ should behave similarly to the equation $(x-x_0)^2y''+\alpha_0(x-x_0)y'+\beta_0y=0$, called the associated Euler equation. We can solve this Euler equation and find it has solutions that look like $cx^r$ where $r$ is a root of the**indicial equation**$r^2+(\alpha_0-1)r+\beta_0=0$ and $c$ is a constant (remember that $r$ as used in this sentence is a constant, not to be confused with the function $r(x)$ that appears in the original equation). So near the point $x_0$, the solution to $p(x)y''+q(x)y'+r(x)y=0$ should look like $cx^r$. Now the series $\sum_{n=0}^{\infty}a_n(x-x_0)^n$ looks like the constant $a_0$ near $x_0$, since all the other terms drop out. If we want to make it look like a constant times $x^r$, the obvious guess is to multiply through by $x^r$. And that is exactly the right guess to make. In the paradigm that follows we will assume that the roots of the associated Euler equation are real. If they are complex, we can still find a solution, but the situation becomes messier than I care to think about. If you are curious about the complex valued case, stop by my office and I'll go over it with you.

#### Paradigm:

Solve $2x^2y''-xy'+(x+1)y=0$ about $x_0=0$.
*Step 1:* Check that $x_0$ is a regular singular point. (If it
isn't, give up and call a mathematician for help.)
$$ \eqalign {
&\lim_{x\to0}\frac{(x-0)(-x)}{2x^2}=-\frac12 \cr
&\lim_{x\to0}\frac{(x-0)^2(x+1)}{2x^2}=\frac12 \cr}
$$
Both limits exist so $x_0=0$ is a regular singular point.

*Step 2:* Find the roots of the indicial equation,
$r^2+(\alpha_0-1)r+\beta_0=0$.

The indicial equation in this case is $r^2-3/2r+1/2=0$ and it has roots $r=1$ and $r=1/2$.

*Step 3:* Guess that a solution takes the form
$(x-x_0)^r\sum_{n=0}^{\infty}a_n(x-x_0)^n$ with $a_0=1$ using the
*larger* $r$ found in step 2. (This assumes the roots are real).
Then plug in and solve for the $a_n$.

The process of carrying out step 3 is very similar to the process of finding the series solution about an ordinary point.

SubStep 1: Make the guess and compute all the pieces. $$ \eqalign { y&=x\sum_{n=0}^{\infty}a_nx^n=\sum_{n=0}^{\infty}a_nx^{n+1} \cr y'&=\sum_{n=0}^{\infty}(n+1)a_nx^n \cr y''&=\sum_{n=0}^{\infty}(n+1)na_nx^{n-1} \cr} $$ and from these we get $$ \eqalign { (x+1)y&=\sum_{n=0}^{\infty}a_nx^{n+2}+\sum_{n=0}^{\infty}a_nx^{n+1} \cr -xy'&=\sum_{n=0}^{\infty}-(n+1)a_nx^{n+1} \cr 2x^2y''&=\sum_{n=0}^{\infty}2(n+1)na_nx^{n+1} \cr} $$ SubStep 2: Change all terms to the original form of $y$.

In this case the original form involves $x^{n+1}$. So we want all the
pieces to be sums of $x^{n+1}$ or $x^{j+1}$ or some such term. So let $j=n+1$.
Then we have
$$ \eqalign {
(x+1)y&=\sum_{j=1}^{\infty}a_{j-1}x^{j+1}+
\sum_{n=0}^{\infty}a_nx^{n+1} \cr
-xy'&=\sum_{n=0}^{\infty}-(n+1)a_nx^{n+1} \cr
2x^2y''&=\sum_{n=0}^{\infty}2(n+1)na_nx^{n+1} \cr}
$$
SubStep 3: Change all indices to the same letter (I use $m$) and
plug into the equation.
$$
2x^2y''-xy'+(x+1)y=
\eqalign {
&\sum_{m=0}^{\infty}2(m+1)ma_mx^{m+1}+
\sum_{m=0}^{\infty}-(m+1)a_mx^{m+1} \cr
&+\sum_{m=1}^{\infty}a_{m-1}x^{m+1} +
\sum_{m=0}^{\infty}a_mx^{m+1} \cr }
=0
$$
SubStep 4: Collect like terms.
$$
(0a_0-1a_0+1a_0)x^1 +
\sum_{m=1}^{\infty}(2(m+1)ma_m-(m+1)a_m+a_{m-1}+a_m)x^{m+1} = 0
$$
SubStep 5: Equate coefficients to 0 and solve for the recurrence
relation.
$$ \eqalign {
0&=0\qquad(m=0\text{ term}) \cr
(2m^2+m)a_m+a_{m-1}&=0\qquad(\text{general term}) \cr }
$$
Solving the last equation for $a_m$, the highest order term, we find
the recurrence relation
$$
a_m=-\frac{a_{m-1}}{2m^2+m}.
$$
SubStep 6: Plug in $a_0=1$ to find the solution.
$$ \eqalign {
a_0&=1 \cr
a_1&=-\frac{1}{3} \cr
a_2&=-\frac{-1/3}{10}=\frac{1}{30} \cr
a_3&=-\frac{1/30}{21}=-\frac{1}{630} \cr
&\vdots \cr }
$$
So the solution is
$$
y_1(x)=x-\frac13x^2+\frac{1}{30}x^3-\frac{1}{630}x^4+\ldots
$$
*Step 4:* If the smaller root of the indicial equation doesn't
differ from the larger root by an integer, find a second solution
using the smaller root. (If it does differ by an integer, the process
of finding a second solution is much harder. Stop by my office for
details if you are interested. Fortunately, in applications the
solution corresponding to the larger root is usually more interesting
if the two roots differ by an integer. There is a good reason for
this, and again you can stop by my office if you are curious.)

The substeps here are the same as for step 3 and so we will just run through them quickly and without comment.

SubStep 1:
$$ \eqalign {
y&=x^{1/2}\sum_{n=0}^{\infty}a_nx^n=\sum_{n=0}^{\infty}a_nx^{n+1/2} \cr
y'&=\sum_{n=0}^{\infty}(n+1/2)a_nx^{n-1/2} \cr
y''&=\sum_{n=0}^{\infty}(n+1/2)(n-1/2)a_nx^{n-3/2} \cr }
$$
So we get
$$ \eqalign {
(x+1)y&=\sum_{n=0}^{\infty}a_nx^{n+3/2}+\sum_{n=0}^{\infty}a_nx^{n+1/2} \cr
-xy'&=\sum_{n=0}^{\infty}-(n+1/2)a_nx^{n+1/2} \cr
2x^2y''&=\sum_{n=0}^{\infty}2(n+1/2)(n-1/2)a_nx^{n+1/2} \cr }
$$
SubStep 2: Let $j=n+1$.
$$ \eqalign {
(x+1)y&=\sum_{j=1}^{\infty}a_{j-1}x^{j+1/2}
+\sum_{n=0}^{\infty}a_nx^{n+1/2} \cr
-xy'&=\sum_{n=0}^{\infty}-(n+1/2)a_nx^{n+1/2} \cr
2x^2y''&=\sum_{n=0}^{\infty}2(n+1/2)(n-1/2)a_nx^{n+1/2} \cr }
$$
SubStep 3:
$$ \eqalign {
2x^2y''-xy'+(x+1)y&=
\sum_{m=0}^{\infty}2(m+1/2)(m-1/2)a_mx^{m+1/2} \cr
&\qquad+\sum_{m=0}^{\infty}-(m+1/2)a_mx^{m+1/2} \cr
&\qquad+\sum_{m=1}^{\infty}a_{m-1}x^{m+1/2} \cr
&\qquad+\sum_{m=0}^{\infty}a_mx^{m+1/2} \cr
&=0 \cr}
$$
SubStep 4:
$$ \eqalign {
&(-1/2a_0-1/2a_0+a_0)x^{1/2} \cr
&+\sum_{m=1}^{\infty}(2(m+1/2)(m-1/2)a_m-(m+1/2)a_m+a_{m-1}+a_m)x^{m+1/2}
\cr }
=0
$$
SubStep 5:
$$ \eqalign {
0&=0\qquad(m=0\text{ term}) \cr
(2m^2-m)a_m+a_{m-1}&=0 \qquad\text{(general term)} \cr }
$$
We solve the last equation for the recurrence relation
$$
a_m=-\frac{a_{m-1}}{2m^2-m}
$$
SubStep 6:
$$ \eqalign {
a_0&=1 \cr
a_1&=-\frac{1}{2-1}=-1 \cr
a_2&=-\frac{-1}{8-2}=\frac16 \cr
a_3&=-\frac{1/6}{18-3}=-\frac{1}{90} \cr
&\vdots \cr}
$$
So the solution is
$$
y_2(x)=x^{1/2}-x^{3/2}+\frac16x^{5/2}-\frac{1}{90}x^{7/2}+\ldots
$$
*Step 5* The general solution is $c_1y_1(x)+c_2y_2(x)$.

EXAMPLE: Find one solution of Bessel's equation of order 0, $x^2y''+xy'+x^2y=0$ about $x_0=0$.

Step 1: $$ \eqalign { \lim_{x\to0}\frac{x\cdot x}{x^2}&=1=\alpha_0 \cr \lim_{x\to0}\frac{x^2\cdot x^2}{x^2}&=0=\beta_0 \cr } $$ So $x_0=0$ is a regular point.

Step 2: $r^2+(1-1)r+0=0$ has a double root at $r=0$.

Step 3: Since the largest (and only) root of the indicial equation is 0, we try $y=x^0\sum_{n=0}^{\infty}a_nx^n$.

SubStep 1: $$ \eqalign { y&=\sum_{n=0}^{\infty}a_nx^n \cr y'&=\sum_{n=0}^{\infty}na_nx^{n-1} \cr y''&=\sum_{n=0}^{\infty}n(n-1)a_nx^{n-2}. \cr } $$ Plugging these in we get $$ \eqalign { x^2y&=\sum_{n=0}^{\infty}a_nx^{n+2} \cr xy'&=\sum_{n=0}^{\infty}na_nx^n \cr x^2y''&=\sum_{n=0}^{\infty}n(n-1)a_nx^n. \cr } $$ SubStep 2: Let $j=n+2$ so $$ \eqalign { x^2y&=\sum_{j=2}^{\infty}a_{j-2}x^j \cr xy'&=\sum_{n=0}^{\infty}na_nx^n \cr x^2y''&=\sum_{n=0}^{\infty}n(n-1)a_nx^n. \cr } $$ SubStep 3: $$ x^2y''+xy'+x^2y= \sum_{m=0}^{\infty}m(m-1)a_mx^m+\sum_{m=0}^{\infty}ma_mx^m +\sum_{m=2}^{\infty}a_{m-2}x^m=0 $$ SubStep 4: $$ 0a_0x^0+1a_1x^1+\sum_{m=2}^{\infty} \bigl(m(m-1)a_m+ma_m+a_{m-2}\bigl)x^m=0 $$ SubStep 5: $$ \eqalign { 0&=0 \cr a_1&=0 \cr m^2a_m+a_{m-2}&=0,\qquad m\ge2 \cr } $$ The recurrence relation is $$ a_m=-\frac{a_{m-2}}{m^2},\qquad m\ge2 $$ SubStep 6: $$ \eqalign { a_0&=1 \cr a_1&=0 \cr a_2&=-\frac14 \cr a_3&=0 \cr a_4&=\frac{1}{64} \cr &\vdots \cr } $$ $$ J_0(x)=1-\frac14x^2+\frac{1}{64}x^4+\cdots $$ Step 4: Since the indicial equation has a double root, the two roots differ by an integer, 0, and so we stop here.

No, you can't claim 1 point for a misprint for my writing $J_0(x)$ instead
of $y_1(x)$ in the last example. The equation $x^2y''+xy'+(x^2-\nu^2)y=0$
is called Bessel's equation of order $\nu$. It arises in solving Laplace's
equation in cylindrical coordinates. $J_0(x)$ is the standard notation for
the Bessel function of the first kind of order 0, which is the solution we
just computed. A second linearly independent solution is $Y_0(x)$, the
Bessel function of the second kind of degree 0. Since when we have a
double root in an Euler equation we tack on a $\log$ term, it shouldn't be
too surprising that $Y_0(x)$ has a logarithmic singularity at $x=0$.
Accordingly, if we want a bounded solution near $x=0$, we want $J_0$ which
is why it is accorded the privilege of being of the *first* kind.
Bessel functions fall just after the trigonometric functions in their
importance in practical applications. You
can read much much more about Bessel functions in *A Treatise on Bessel
Functions* by Watson.

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©2010, 2014 Andrew G. Bennett