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

  1. Apr 29, 2013 #1
    1. The problem statement, all variables and given/known data

    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[itex]\pi[/itex] and step sizes of h.

    2. Relevant equations
    Find diff eq, such that y'=f(t,y)

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

    k1=f(tn,yn)

    k2=f(tn+h/2,yn+k1*h/2)

    k3=f(tn+h/2,yn+k2*h/2)

    k4=f(tn+h,yn+k3*h)


    3. 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:

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

    Where:

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

    Now I solved for x with the following:

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

    Where:
    k1= vn*h
    k2= vn*h*(1+h/2)
    k3= vn*h*(1+h/2+h2/4)
    k4= vn*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.
     
    Last edited: Apr 29, 2013
  2. jcsd
  3. May 2, 2013 #2

    DrClaude

    User Avatar

    Staff: Mentor

    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
     
  4. May 8, 2013 #3
    I have figured it out, thanks DrClaude.
     
    Last edited: May 8, 2013
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