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Reduction of differential equations then numerical methods

  1. Jun 7, 2009 #1
    1. The problem statement, all variables and given/known data

    I have a function that attempts to imitate the internal forces of the cord used for bungee jumping. I have been told to resolve it using Runge-Kutta 1st order numerical aproximation, therefore i have to reduce it to a system of equations. The preffered Runge-Kutta method is euler forward: y1 = y0 + k1 where k1 = h*f(y0). Plus I have to do this in a spreadsheet.

    2. Relevant equations

    M*q''(t)+K*q(t)=b(t)

    Where M and K are 3x3 matrix and k and b are vectors with 3 components

    3. The attempt at a solution

    If I make q'(t) = v(t)
    then v'(t) = M^(-1)*[b(t)-K*q(t)]

    i can divide v'(t) = M^(-1)*b(t)+[-M^(-1)*K*q(t)]
    and then obtain a vector plus matriz to solve it. My problem is that i dont know how to take the data from the matriz and apply it.

    Thanks
     
  2. jcsd
  3. Jun 8, 2009 #2
    Your differential system (written as a system of first order differential equations) is:

    [tex]
    \left[ \begin{array}{1c}
    q' \\
    v' \\
    \end{array} \right] = \left[ \begin{array}{1c} v(t) \\ M^{-1}b(t)-M^{-1}Kq(t) \\\end{array} \right]

    [/tex]


    And so your Runge-Kutta numerical method is:

    [tex]
    \left[ \begin{array}{1c}
    q_{n+1} \\
    v_{n+1} \\
    \end{array} \right] = \left[ \begin{array}{1c} v_{n} \\ M^{-1}b(t_{n})-M^{-1}Kq_{n} \\\end{array} \right]

    [/tex]

    supposedly you have the intial values vo, qo, so you go on generating the solution at the mesh points ti=i*h, where h is your step-size. The matrix M is constant so all you have to do is invert it and use it as I've indicated above to generate your numerical solution . . . in your spreadsheet go on generating the solution for each time step in a new row of cells based on the value of the previous time step which you take from the previous row of cells and the values of the components of K, M^(-1) and b which you store in a separate set of cells (you can evaluate b(tn) in a seperate cell at each time step) , . . . um, I'm not sure if I understand your question altogether lol.
     
  4. Jun 8, 2009 #3
    Yeah actually i read my message back, and it was kind of funny, my problem was that the graph didn't end up the way i was told it was supposed to look like. Ended up being a problem of both the method selected and the step i had chosen, fixed those and it fell into place.

    I'll attach it when i get home in case anyone else was pondering the problem, thanks for responding
     
  5. Jun 8, 2009 #4
    Oh I see, okay:) yes, the numerical method you chose is an explicit one and has a certain region of absolute stability that limits the step-size you can use for it to work well . . .
    glad you sorted it out :)
     
  6. Jun 8, 2009 #5
    yeah that was exactly the problem, function went haywire
    thanks for the support
     
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