Runge-Kutta 4th order method (RK4) for second order Diff Eq

[Helmholtz]

Homework Statement

Hello, maybe this is due to my lack of understand of RK4, but I have an equation: x'' + b^2*x=0 (derivatives with respect to variable t) and I need to use RK4 to find the solution on an interval. I can readily find solutions analytically, but my understanding of RK4 is that I need the problem in a first order diff eq, which I do not accomplish when I use characteristic equations. Initial conditions are x(t)=0 and x'(0)=0. The intervals is 0 to 4$\pi$ and step sizes of h.

Homework Equations

Find diff eq, such that y'=f(t,y)

RK4 equations:
y_n+1= y_n + h/6*(k₁ + 2*k₂ + 2*k₃ + k₄)

k₁=f(t_n,y_n)

k₂=f(t_n+h/2,y_n+k₁*h/2)

k₃=f(t_n+h/2,y_n+k₂*h/2)

k₄=f(t_n+h,y_n+k₃*h)

The Attempt at a Solution

My attempt to break the problem to first order is a substitution with x'=v

And it follows that that: v'= -b^2*x = f(t,x).

Now I have two first order differential equations. I can use RK4 to get values for v and x, but it seems like my values of v depend on x and vice verse? Here is what I have so far, but it is coupled so I think it's wrong.

First I solved for the higher ordered one, aka v. So:

v_n+1 = v_n + 1/6*(l₁ + 2*l₂ + 2*l₃ +l₄).

Where:

l₁= -b^2*x_n*h
l₂= -b^2*x_n*h*(1+h/2)
l₃= -b^2*x_n*h*(1+h/2+h²/4)
l₄= -b^2*x_n*h*(1+h/2+h²/2+h³/4)

Now I solved for x with the following:

x_n+1 = x_n + 1/6*(k₁ + 2*k₂ + 2*k₃ + k₄)

Where:
k₁= v_n*h
k₂= v_n*h*(1+h/2)
k₃= v_n*h*(1+h/2+h²/4)
k₄= v_n*h*(1+h/2+h²/2+h³/4)

Is this right? Since I have initial conditions I think I can build this up to the full solution since I know v₀ and x₀.

 

[DrClaude]

Sorry to say, but you got it all wrong! The point is to solve both equations at the same time, and use the intermediate solutions to get a better guess for the next intermediate solution, and when you have all 4 of them you take an "average" to actually advance one time step.

Check this thread for more details:
https://www.physicsforums.com/showthread.php?t=671526

 

[Helmholtz]

I have figured it out, thanks DrClaude.