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

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

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.

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

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

$\displaystyle \underline{SOLUTION}$:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

For $\displaystyle \underline{r = 0}$:
$\displaystyle F(0+1) = F(1) = 1(1-1) + \frac{3}{4}(1) = \frac{3}{4}$

$\displaystyle F(0+2) = F(2) = 2(2-1) + \frac{3}{4}(2) = \frac{7}{2}$
$\displaystyle F(0+3) = F(3) = 3(3-1) + \frac{3}{4}(3) = \frac{33}{4}$

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

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

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

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

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

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

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

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

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

The recurrence relation is given by

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

For $\displaystyle \underline{r = 0}$

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

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

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

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

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

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

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

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

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

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

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

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

The solutions are given by:

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

and

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

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

Regards,
Fractalus

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

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

Regards,
Fractalus

3. ## Need my method to be checked!

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.