An effective method to solve ODE in terms of power series with regular singular point

**Fractalus** · July 19th 2011, 04:33 AM

The goal of this post is to see if what I did is correct for an exercice but also to show an effective method to use when you need to find power series of an ODE with regular singular point. The method is probably well known but I think that having a detailed example of it can help other students.

$\underline{QUESTION}$ : Find around the origin, the solutions in power series the following equation.

$4xy'' + 3y' + 3y = 0$

$\underline{SOLUTION}$ :

$\underline{Step 1}$ : Find the singular points
$P(x)= 4x, Q(x)=3, R(x)=3$

The only x such that $P(x)=0$ is $x=0$ . Therefore, $x=0$ is the only singular point and all other real points are ordinary points.

$\underline{Step 2}$ : Determine if the singular points are regular or irregular

$\lim_{x\to 0} xp(x)=\lim_{x\to 0} x\frac{Q(x)}{P(x)} = \lim_{x\to 0} x\frac{3}{4x}= \frac{3}{4}$ which exists.

$\lim_{x\to 0} x^2p(x)=\lim_{x\to 0} x^2\frac{R(x)}{P(x)} = \lim_{x\to 0} x^2\frac{3}{4x}= \lim_{x\to 0} x\frac{3}{4} = 0$ which exists.

Therefore, $x=0$ is a regular singular point.

$\underline{Step 3: Find \sum_{n=0}^\infty p_nx^n and \sum_{n=0}^\infty q_n x^n}$
Because $x = 0$ is a regular point, this means that $xQ(x)/P(x) = xp(x)$ and $x^2R(x)/P(x) = x^2q(x)$ have infinite limits when $x \to 0$ and are analytics in $x = 0$ . Therefore, they have a converging power series expansion of the form:

$xp(x) = \sum_{n=0}^\infty p_nx^n, x^2q(x)= \sum_{n=0}^\infty q_n x^n$

in a neighbourhoud \ $\mid x \mid < \rho$ around the origin, where $\rho > 0$

In this example,
$xp(x) = 3/4$ and $x^2q(x) = \frac{3}{4}t$ which means that
$p_0 = \frac{3}{4}, p_1 = \cdots = p_n = \cdots = 0$ and
$q_1 = \frac{3}{4}, q_0 = q_2 = \cdots = q_n = \cdots = 0$

$\underline{Step 4}$ : : Find the roots of $F(r)$ .

$F(r) = r(r-1) + p_0r + q_0 = 0$
Here, we have $F(r) = r(r-1) + \frac{3}{4}r = r^2 - \frac{1}{4}r = r(r - \frac{1}{4}) = 0$ . Therefore, $r_1 = 0$ and $r_2 = \frac{1}{4}$ .

$\underline{Step 5}$ : Find the $F(r+n)$ for the first root and the second root.

We first have to compute $F(r+n)$ for some n (let say five), and for $r = 0$ . We are going to do the same thing for $r = \frac{1}{4}$ .

For $\underline{r = 0}$ :
$<br /> F(0+1) = F(1) = 1(1-1) + \frac{3}{4}(1) = \frac{3}{4}$

$F(0+2) = F(2) = 2(2-1) + \frac{3}{4}(2) = \frac{7}{2}$
$<br /> F(0+3) = F(3) = 3(3-1) + \frac{3}{4}(3) = \frac{33}{4}$

$F(0+4) = F(4) = 4(4-1) + \frac{3}{4}(4) = 15$

$F(0+5) = F(5) = 5(5-1) + \frac{3}{4}(5) = \frac{95}{4}$

For $\underline{r = \frac{1}{4}}$ :

$F(\frac{1}{4}+1) = F(\frac{5}{4}) = \frac{5}{4}(\frac{5}{4}-1) + \frac{3}{4}(\frac{5}{4}) = \frac{5}{4}$

$F(\frac{1}{4}+2) = F(\frac{9}{4}) = \frac{9}{4}(\frac{9}{4}-1) + \frac{3}{4}(\frac{9}{4}) = \frac{9}{2}$

$F(\frac{1}{4}+3) = F(\frac{13}{4}) = \frac{13}{4}(\frac{13}{4}-1) + \frac{3}{4}(\frac{13}{4}) = \frac{39}{4}$

$F(\frac{1}{4}+4) = F(\frac{17}{4}) = \frac{17}{4}(\frac{17}{4}-1) + \frac{3}{4}(\frac{17}{4}) = 17$

$F(\frac{1}{4}+5) = F(\frac{21}{4}) = \frac{21}{4}(\frac{21}{4}-1) + \frac{3}{4}(\frac{21}{4}) = \frac{105}{4}$

$\underline{Step 6}$ : Find the coefficients for the first root and the second root.

The recurrence relation is given by

$a_n = - \frac{\sum_{k=0}^{n-1} a_k[kp_{n-k}+q_{n-k}]}{F(r+n)}$ $n \geq 1$ and $a_0 = 1$

For $\underline{r = 0}$

$a_1 = - \frac{a_0(0\cdot p_1 + q_1)}{F(1)} = \frac{-3/4}{3/4} = -1$

$a_2 = - \frac{a_1q_1)}{F(2)} = \frac{-(-1)(3/4)}{7/2} = 3/14$

$a_3 = - \frac{a_2(q_1)}{F(3)} = \frac{-(3/14)(3/4)}{33/4} = -3/154$

$a_4 = - \frac{a_3q_1}{F(4)} = \frac{-(-3/154)(3/4)}{15} = 3/3080$

$a_5 = - \frac{a_4q_1}{F(5)} = \frac{(3/3080)(3/4)}{95/4} = -9/292800$

For $\underline{r = \frac{1}{4}}$

$a_1 = - \frac{a_0q_1}{F(1)} = \frac{(-3/4)}{5/4} = -3/5$

$a_2 = - \frac{a_1q_1}{F(2)} = \frac{-(-3/5)(3/4)}{9/2} = 1/10$

$a_3 = - \frac{a_2q_1}{F(3)} = \frac{-(1/10)(3/4)}{39/4} = -1/130$

$a_4 = - \frac{a_3q_1}{F(4)} = \frac{-(-1/130)(3/4)}{17} = 3/8840$

$a_5 = - \frac{a_4q_1}{F(5)} = \frac{-(3/8840)(3/4)}{105/4} = -9/928200$

$\underline{Step 7}$ : Write the solutions

The solutions are given by:

$y_1(x) = 1 - x + \frac{3}{14}x^2 - \frac{3}{154}x^3 + \frac{3}{3080}x^4 -\frac{9}{292600}x^5 + \cdots$

and

$y_2(x) = \mid x \mid^{-1/4} \left[1 - \frac{3}{5}x + \frac{1}{10}x^2 - \frac{1}{130}x^3 + \frac{3}{8840}x^4 -\frac{9}{928200}x^5 + \cdots \right]$

And that's all! Did I made a mistake somewhere? Can I find a better solution than that?

Regards,
Fractalus

**Fractalus** · July 19th 2011, 04:12 PM

Anyone knows if what I did is correct? This was the main point of my post.

Regards,
Fractalus

**Fractalus** · July 20th 2011, 01:41 PM

I would really appreciate if someone tells if my solution is correct or not and why. I gave a detailed solution to help people finding my errors.

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An effective method to solve ODE in terms of power series with regular singular point

Re: An effective method to solve ODE in terms of power series with regular singular p

Need my method to be checked!

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