Solving an ODE using shooting method

1. Dec 29, 2014

spaghetti3451

Hi, I am trying to solve the following ODE for my maths project:

$y'' = \frac{\alpha}{2}y^3 - \frac{3}{2}y^2 + y - \frac{3}{x} y'$

under the following boundary conditions:

$y'(0) = 0$
$y(x) \rightarrow y \_ \equiv 0\ \text{as}\ x \rightarrow \infty$

As a first step, I converted this problem into a set of coupled ODEs:

$\frac{dy}{dx} = z$
$\frac{dz}{dx} = \frac{\alpha}{2}y^3 - \frac{3}{2}y^2 + y - \frac{3}{x} z$

under the following boundary conditions:

$z(0) = 0$
$y(x) \rightarrow y \_ \equiv 0\ \text{as}\ x \rightarrow \infty$

Next, my source tells me to use the shooting method to convert the BVP into an IVP, which means that I have to use two initial guesses of $y(0)$ to be able to use the secant method to find the correct value of $y(0)$.

Now, my question is, according to my source, I can avoid the singularity at x = 0 using Taylor expansion as follows:

$y(r_{0}) = y_{0} + \frac{1}{16} r_{0}^{2} (2y_{0} - 3y_{0}^{2} + \alpha y_{0}^{3})$

I see how you can estimate $y(r_{0})$ where $r_{0}$ is a tiny distance away from the origin, but I don't really see how they dervied this expression. Could anyone help me out?

2. Dec 29, 2014

HomogenousCow

Just set y(0)'=0 and y(0)=y_0 in the original DE.

EDIT: Tried it myself, something weird is going on. The numerical factors come out different for me.
Maybe an error? I don't see where the factor of 1/16 comes from. It comes out as 1/4 for me.

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Last edited: Dec 29, 2014
3. Dec 29, 2014

Staff: Mentor

Start out by representing y as a function of x at small values of y by y(x) = y(0)+bx2. Then

y' = 2bx
y''=2b

3y'/x = 6b

So $8b=\frac{\alpha}{2}y(0)^3-\frac{3}{2}y(0)^2+y(0)$

So, $y(r_0)=y(0)+br_0^2=y(0)+\frac{1}{8}r_0^2(\frac{\alpha}{2}y(0)^3-\frac{3}{2}y(0)^2+y(0))$

Chet

4. Jan 2, 2015

spaghetti3451

Thanks for the solution. I also came to the conclusion, but I used $\frac{y''(0)}{2}$ in place of $b$. I have written the code that solves the ODE in my original post, but it's giving me unexpected results. I was wondering if anyone in Physics Forums might want to have a look at it.