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Frobenius Method (Regular singular point)

For the 2nd-order linear differential equation

$\displaystyle y^{\prime\prime} + P(x)y^{\prime} + Q(x)y = R(x) $

if $x=a$ is not an ordinary point, then $x=a$ is called a singular point. For $x=a$ singular point and

$\displaystyle (x - a)P(x), \ (x- a)^{2}Q(x) $

are analytic. Then $x=a$ is called a regular singular point. Otherwise, irregular singular point.

Example 4..3   Find the singular point of the following differential equation

$\displaystyle (x-1)^{2}x^{2}(x-2)y^{\prime\prime} + 5x^{2}y^{\prime} - (x-2)y = 0 .$

SOLUTION Since

$\displaystyle P(x) = \frac{5x^{2}}{(x-1)^{2}x^{2}(x-2)}, \ Q(x) = \frac{-(x-2)}{(x-1)^{2}x^{2}(x-2)} $

$x = 1, 0, 2$ are singular points. Now

$\displaystyle (x-1)P(x) = \frac{5x^{2}}{(x-1)x^{2}(x-2)} $

implies that $x = 1$ is irregular singular point and the rest is regular singular point. $\ \blacksquare$

Theorem 4..7   For the 2nd-order differential equation

$\displaystyle y^{\prime\prime} + P(x)y^{\prime} + Q(x)y = 0 $

if $x=a$ is the regular singular point, then there exists a solution around $x=a$ of the form

$\displaystyle y(x) = \sum_{n=0}^{\infty}c_{n}(x - a)^{n+r}, \ c_{0} \neq 0 $

The power series $\sum_{n=0}^{\infty}c_{n}(x - a)^{n+r}$ is called Frobinius series. Using the Frobinious series to find $c_{n}$ is called Frobenius method.

Example 4..4   Find the power series solution of the follwing differential equation around $x = 0$.

$\displaystyle L(y) = $

SOLUTION Write $L(y) = x^{2}y^{\prime\prime} + xy^{\prime} + (x^2 -4)y = 0$ as the standard form

$\displaystyle y^{\prime\prime} + \frac{1}{x}y^{\prime} + \frac{x^2 - 4}{x^2}y = 0. $

Then $P(x) = \frac{1}{x}, \ Q(x) = \frac{x^2 - 4}{x^2}$. Thus $x = 0$ is the regular singular point. Now we set a solution of the differential equation as $y = \sum_{n=0}^{\infty}c_{n}x^{n+r}, \ c_{0} \neq 0$. Differentiate to get

$\displaystyle y^{\prime} = \sum_{n=0}^{\infty}(n+r)c_{n}x^{n+r-1}, \ y^{\prime\prime} = \sum_{n=0}^{\infty}(n+r)(n+r-1)c_{n}x^{n+r-2} $

Substitute these into $L(y) = 0$ to obtain

$\displaystyle L(y) = \sum_{n=0}^{\infty}(n+r)(n+r-1)c_{n}x^{n+r} + \sum_{n=0}^{\infty}(n+r)c_{n}x^{n+r} + (x^2 - 4)\sum_{n=0}^{\infty}c_{n}x^{n+r} = 0$

Now let the power of $x$ be the least number $n+r$. Then

$\displaystyle \sum_{n=0}^{\infty}(n+r)(n+r)c_{n}x^{n+r} - \sum_{n=2}^{\infty}c_{n-2}x^{n+r} - 4\sum_{n=0}^{\infty}c_{n}x^{n+r} = 0 .$

Next we let the index be the largest number $n=2$
$\displaystyle \underbrace{(r^2 -4)c_{0}x^{r}}_{n = 0} + \underbrace{((r+1)^2 -4... ...x^{r+1}}_{n = 1} + \sum_{n=2}^{\infty}[((n+r)^2 - 4)c_{n} + c_{n-2}]x^{n+r} = 0$      

For $n=0$, we have the indicial equation $r^2 - 4 = 0$ which implies $r = \pm 2$.

We find a solution for $r=2$. Note that the coefficients of $x^{n+r}$ are 0. Thus

$\displaystyle ((r+1)^2 - 4)c_{1} = 0$

and

$\displaystyle ((n+r)^2 - 4)c_{n} + c_{n-2} = 0, \ n \geq 2$

The for $r=2$, we have $c_1 = 0$. Similarly, for $r=2$, we have

$\displaystyle c_{n} = -\frac{c_{n-2}}{n(n+4)} $


$\displaystyle c_{2n}$ $\displaystyle =$ $\displaystyle \frac{c_{2n}}{c_{2n-2}}\frac{c_{2n-2}}{c_{2n-4}}\cdots\frac{c_{2}}{c_{0}}c_{0}$  
  $\displaystyle =$ $\displaystyle \frac{-1}{2n(2n+4)}\frac{-1}{(2n-2)(2n+2)}\cdots \frac{-1}{2\cdot 6}c_{0}$  
  $\displaystyle =$ $\displaystyle \frac{(-1)^{n}c_{0}}{2^{n}n!2^{n-1}(n+2)!} = \frac{(-1)^{n}c_{0}}{2^{2n-1}n!(n+2)!}$  

Thus,

$\displaystyle y_{1}(x) = \sum_{n=0}^{\infty}\frac{(-1)^{n}c_{0}}{2^{2n-1}n!(n+2)!}x^{2n+2}\ensuremath{\ \blacksquare}$

Example 4..5   Find a solution of the following differential equation around $x = 0$.

$\displaystyle L(y) = 4x^2 y^{\prime\prime} + (3x + 1)y = 0. $

SOLUTION Write in the standard form. Then since $Q(x) = \frac{3x + 1}{4x^{2}}$, $x = 0$ is a regular singular point. We let $y = \sum_{n=0}^{\infty}c_{n}x^{n+r}$.Then

$\displaystyle L(\sum c_{n}x^{n+r}) = \sum_{n=0}^{\infty}4(n+r)(n+r-1)c_{n}x^{n+r} + \sum_{n=0}^{\infty}3c_{n}x^{n+r+1} + \sum_{n=0}^{\infty}c_{n}x^{n+r} = 0 $

Now let the power of $x$ be the least number $x^{n+r}$. Then

$\displaystyle \sum_{n=0}^{\infty}[4(n+r)(n+r-1) + 1]c_{n}x^{n+r} + \sum_{n=1}^{\infty}3c_{n-1}x^{n+r} = 0. $

Thus,

$\displaystyle \underbrace{(4r(r-1) + 1)c_{0}x^{r}}_{n = 0} + \sum_{n=1}^{\infty}\{[4(n+r)(n+r-1) + 1]c_{n} + 3c_{n-1}\}x^{n+r} = 0 $

Now the indicial equation is $4r^2 - 4r + 1 = 0$. Thus $r = \frac{1}{2}$ and the recurrence relation

$\displaystyle c_{n} = \frac{-3c_{n-1}}{4(n+r)(n+r-1) + 1} $

can be written as

$\displaystyle c_{n} = \frac{-3c_{n-1}}{4(n+\frac{1}{2})(n - \frac{1}{2}) + 1} = \frac{-3c_{n-1}}{4n^2}. $

Thus a solution is

$\displaystyle y_{1}(x) = \vert x\vert^{1/2}\sum_{n=0}^{\infty}c_{n}x^{n}, \ $   ただし$\displaystyle \ c_{n} = \frac{-3c_{n-1}}{4n^2 }, \ n \geq 1.\ensuremath{\ \blacksquare} $

Theorem 4..8   In the 2nd-order linear differential equation

$\displaystyle y^{\prime\prime} + P(x)y^{\prime} + Q(x)y = 0, $

if $x = 0$ is the regular singular point and the root $r$ of the indicial equation is multiple root, then the linearly independent solutions $y_{1}$ and $y_{2}$ are given by the following forms.
$\displaystyle y_{1}$ $\displaystyle =$ $\displaystyle \vert x\vert^{r}\sum_{n=0}^{\infty}c_{n}x^{n} \ (c_{n} \neq 0),$  
$\displaystyle y_{2}$ $\displaystyle =$ $\displaystyle y_{1}\log{\vert x\vert} + \vert x\vert^{r}\sum_{n=0}^{\infty}C_{n}x^{n} .$  

Now we find $y_{2}$. By the theorem, we find a solution of the form

$\displaystyle y_{2} = y_{1}\log{\vert x\vert} + \vert x\vert^{3/2}\sum_{n=0}^{\infty}C_{n}x^{n}$

For the matter of calculation we set $x > 0, \ c_{0} = 1$. Substitute $y_2$ into $L(y) = 0$.
$\displaystyle L(y_{2})$ $\displaystyle =$ $\displaystyle 4x^{2} (y_{1}^{\prime\prime}\log{x} + \frac{2y^{\prime}}{x} - \frac{y_{1}}{x^2})$  
  $\displaystyle +$ $\displaystyle \sum_{n=0}^{\infty}4(n+\frac{3}{2})(n + \frac{1}{2})C_{n}x^{n+3/2} + (3x+1)y_{1}\log{x} + \sum_{n=0}^{\infty}3C_{n}x^{n+5/2}$  
  $\displaystyle +$ $\displaystyle \sum_{n=0}^{\infty}C_{n}x^{n+3/2}$  
  $\displaystyle =$ $\displaystyle \log{x}L(y_{1}) + 8xy^{\prime} - 4y_{1} + 3C_{0}x^{3/2} + C_{0}x^{3/2}$  
  $\displaystyle +$ $\displaystyle \sum_{n=1}^{\infty}[4(n + \frac{3}{2})(n + \frac{1}{2})C_{n} + C_{n} + 3C_{n-1}]x^{n+3/2} = 0$  

Now express a few terms of $y_{1}$.

$\displaystyle y_{1} = x^{1/2} - \frac{3}{4}x^{3/2} + \frac{9}{64}x^{5/2} - \frac{27}{(36)(64)}x^{7/2} + \cdots . $

Note that $L(y_{1}) \equiv 0$. Then

$\displaystyle (4C_{0} - 6)x^{3/2} + (16C_{1} + 3C_{0} + \frac{9}{4})x^{5/2} + (36C_{2} + 3C_{1} - \frac{9}{32})x^{7/2} + \cdots = 0 $

From this $C_{0} = \frac{3}{2}, C_{1} = - \frac{27}{64}, C_{2} = \frac{11}{256}, \ldots $. Thus

$\displaystyle y_{2} = \log{x}(x^{1/2} - \frac{3}{4}x^{3/2} + \frac{9}{64}x^{5/2... ...{27}{64}x^{5/2} + \frac{11}{256}x^{7/2} + \cdots . \ensuremath{\ \blacksquare} $

Example 4..6   Find a solution of the following differential equation

$\displaystyle L(y) = x^2 y^{\prime\prime} + (x^2 - 3x)y^{\prime} + 3y = 0. $

SOLUTION Since $P(x) = \frac{x^2 - 3x}{x^2}, \ Q(x) = \frac{3}{x^2}$, $x = 0$ is a regular singular point. Then let $y = \sum_{n=0}^{\infty}c_{n}x^{n+r}$.

$\displaystyle L(\sum_{n=0}^{\infty}c_{n}x^{n+r})$ $\displaystyle =$ $\displaystyle \sum_{n=0}^{\infty}(n+r)(n+r-1)c_{n}x^{n+r} + \sum_{n=0}^{\infty}(n+r)c_{n}x^{n+r-1}$  
  $\displaystyle -$ $\displaystyle \sum_{n=0}^{\infty}3(n+r)c_{n}x^{n+r} + \sum_{n=0}^{\infty}3c_{n}x^{n+r} = 0 .$  

Let the power of $x$ be put together with $x^{n+r}$. Then

$\displaystyle \sum_{n=0}^{\infty}[(n+r)(n+r-1) - 3(n+r) + 3]c_{n}x^{n+r} + \sum_{n=1}^{\infty}(n+r-1)c_{n-1}x^{n+r} = 0. $

From this
    $\displaystyle \underbrace{(r(r-1) -3r + 3)c_{0}x^{r}}_{n=0}$  
  $\displaystyle +$ $\displaystyle \sum_{n=1}^{\infty}\{[(n+r)(n+r-1) - 3(n+r) + 3]c_{n} + (n+r-1)c_{n-1}\}x^{n+r} = 0$  

Now the indicial equation $r^2 - 4r + 3 = 0$ implies that $r = 1, 3$．Then we have two solutions.

$\displaystyle y_{1}(x) = \sum_{n=0}^{\infty}c_{n}x^{n+1}, \ y_{2}(x) = \sum_{n=0}^{\infty}C_{n}x^{n+3} $

Put the coefficients of $x^{n+r}$ equal to 0. Then we have the following recurrence equation

$\displaystyle c_{n} = \frac{-(n+r-1)c_{n-1}}{(n+r)(n+r-1) - 3(n+r) + 3} $

For $r = 1$, we have

$\displaystyle c_{n} = \frac{-nc_{n-1}}{(n+1)n - 3(n+1) + 3} = \frac{-nc_{n-1}}{n^2 - 2n} $

For $r = 3$, we have

$\displaystyle C_{n} = \frac{-(n+2)C_{n-1}}{(n+3)(n+2) - 3(n+3) + 3} = \frac{-(n+2)C_{n-1}}{n^2 + 2n}. $

Now note that $c_{n+2} = C_{n}$．Thus $y_{1} = y_{2}$. Then we need to find the independent solution．

Theorem 4..9   For the 2nd-order linear differential equation

$\displaystyle y^{\prime\prime} + P(x)y^{\prime} + Q(x)y = 0, $

if $x = 0$ is a regular singular point and the difference of the roots $r_{1},r_{2}$ is positive integer, then the linearly independent solutions $y_{1},y_{2}$ are given by the following:
$\displaystyle y_{1}$ $\displaystyle =$ $\displaystyle \vert x\vert^{r_{1}}\sum_{n=0}^{\infty}c_{n}x^{n} \ (c_{n} \neq 0),$  
$\displaystyle y_{2}$ $\displaystyle =$ $\displaystyle cy_{1}\log{\vert x\vert} + \vert x\vert^{r_{2}}\sum_{n=0}^{\infty}C_{n}x^{n} \ (C_{0} \neq 0) .$  

Thus $y_{2} = cy_{1}\log{x} + x\sum_{n=0}^{\infty}C_{n}x^{n}$ can be found by substituting $y_2$ into $L(y) = 4x^2 y^{\prime\prime} + (3x + 1)y = 0$. $\ \blacksquare$


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