Difficult non-linear differential equation

[matteo86bo]

Homework Statement

I need to solve this differential equation numerically. I hope you can help me, it's not homework!

Homework Equations

This is the equation to be solved

$x''=\dfrac{\alpha}{x}-\beta\dfrac{(x'-\gamma x)^2}{x}-\delta x - \dfrac{\varepsilon}{x^2}$

The Attempt at a Solution

I've tried with a simple discretization, but it doesn't seem to be stable.

$\dfrac{x(i)-2x(i-1)+x(i-2)}{h^2}=\dfrac{\alpha}{x(i-1)}-\beta\dfrac{\left(\dfrac{x(i)-x(i-1)}{h}-\gamma x(i-1)\right)^2}{x(i-1)}-\delta x(i-1) - \dfrac{\varepsilon}{x(i-1)^2}$

I think it's better to use Runge-Kutta IV but I have no idea how to apply it on such a complicated equation.

 

[Filip Larsen]

You can reduce a second order ODE like that to a set of first order ODE by introducing an extra state variable. For instance, in your case you could set v = x' and get the following first order equations

v' = α/x - β (v-γx)²/x - δx - ε/x², and
x' = v

Thus, (x,v) is the state of the system and the field that you have to provide to methods like RK4 is given by the right hand side of the equations.

 

[matteo86bo]

Thanks that helped but I have still some problem in computing v.

According to what you wrote and the runge Kutta IV method definition the k should be:

$k1=hf(t_{i},x_{i})$
$k2=hf(t_{i+1/2},x_{i}+k1/2)$
$k3=hf(t_{i+1/2},x_{i}+k2/2)$
$k4=hf(t_{i+1},x_{i}+k3)$

How is v for every k?

 

[Filip Larsen]

No, your state consist of two variables, not just one, meaning that in the context of the RK4 method the state variable (not to be confused with the x in your equations), the field f and the k-values are all two-dimensional. For instance, when applied to your state variables the stage k₁ = h f(t_i,y_i) from the RK4 method (where I used y = (x,v) to denote the complete state) really means k₁ = (k_x1, k_v1) = h f(t_i,x_i,v_i).

 

[matteo86bo]

Ah ok, thanks. So in my case, since I don't have an explicit time dependency the ks should be


$k1=hf(x_{i},v_{i})$
$k2=hf(x_{i}+k1/2,v_{i}+k1/2,)$
$k3=hf(x_{i}+k2/2,v_{i}+k2/2)$
$k4=hfx_{i}+k3,v_{i}+k3)$

Is that right?

 

[Filip Larsen]

Correct, but remember as was mentioned, that the field f also evaluates to two values, one for x' and one for v' so you need to treat the k-values as vectors also when you combine them to a new y value.

 

[matteo86bo]

Sorry but I don't understand how to compute the Ks considering them as arrays. I tried googling it but I couldn't find anything. Can please you make an example?

 

[Filip Larsen]

You should treat them as vectors with two elements. If we define the field as the (column) vector

f(x,v) = ( f_x(x,v), f_v(x,v) )^T ,

that is, a vector with the two elements

f_x(x, v) = v and
f_v(x, v) = α/x - β (v-γx)2/x - δx - ε/x2,

then from the definition of the RK4 method we have

k₁ = h f(x_i, v_i) = ( h f_x(x_i,v_i), h f_v(x_i,v_i) )^T

that is, the two elements of k₁ is

h f_x(x_i, v_i) and
h f_v(x_i, v_i).

If you are still confused by this I suggest that you read up on vector algebra.

 

[matteo86bo]

Ok, now I got it, thanks a lot for you time!