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System of Coupled Differential Equations

  1. Apr 4, 2009 #1
    I'm not sure if I'm using the correct terminology, but just to be sure let me explain what I understand to be coupled DE's (or CDEs); A set of DEs with the all derivatives taken with respect to a single parameter t and each DE depending on the functions f1,f2,...,fn. If you guys would like to correct me on nomenclature, I'd be glad to hear it.

    Now that definitions are out of the way, I can begin to ask my questions. You don't have to answer all of them, though that would be great as well =).

    1)Where can I find some good resources on methods to solve CDEs (websites, books, etc..)

    2)How can I solve this system:

    ax''+bx+cy=d
    ex''+fy''=g

    Where a,b,c,d,e,f,g are constants and x and y is what were looking for. A general solution is ideal, but if that is impossible here is some initial conditions.

    x(0)=0; x'(0)=0
    y(0)=0; y'(0)=0

    From what I understand, this could possibly be solved with Laplace Transform, however my math is a little lacking in that department so while your reading this question, I may be brushing up on the subject.

    Anyways, thanks in advance people.
     
  2. jcsd
  3. Apr 4, 2009 #2
    A technique I've seen used is to integrate / differentiate the equations until some substitution is possible.

    So..

    ax''+bx+cy=d
    ex''+fy''=g
    =>
    ax'''+bx''+cy''=0
    ex''+fy''=g
    =>
    ax''' + bx'' + (c/f)(g-ex'') = 0
    =>
    ax''' + bx'' - (ce/f)x'' = -(cg/f)
    =>
    ax''' + (b - ce/f)x'' = -(cg/f)
    =>
    w' + (b/a - ce/af)w = -(cg/af)
    =>
    w' + Mw = N

    Which can easily be solved, yielding x''. From that, you should be able to get the form of x' and x. Then... well, y'', y', and y should be solvable as well. Applying initial conditions should handle the four arbitrary constants introduced by x''->x', x'->x, y''->y', y'->y, but I'm not sure I remember off-hand how to handle the constant introduced by x'''->x''. There must be a way, though.
     
  4. Apr 5, 2009 #3
    Hi!

    Since the equations a linear, you could try following:

    Any ODE of order 2 can be rewritten as a system of 2 first-order ODEs, e.g.:

    x=x(t), v=v(t)

    x''+x'+x=0

    v=x'
    v'+v+x=0

    x'=v
    v'=-v-x

    [tex]\frac{d}{dt}\left(\begin{array}{c}x\\v\end{array}\right)=\left(\begin{array}{cc}0&1\\-1&-1\end{array}\right)\left(\begin{array}{c}x\\v\end{array}\right)[/tex]

    In your case each equation will produce another 2, so you'll end up with a system of 4 coupled linear first-oder ODEs.(one usually says of such system to have the order 4)


    Then you separate in the LHSs the derivatives, and everything else in the RHSs. You have to then once again rewrite the system in terms of matrices and vectors, to be able to perform a basis transformation on the matrix.

    This transformation is usually a DIAGONALISATION. However, some matrices are not diagonalisable where you have to transform them to their Jordan normal form.

    After all this algebraic stuff you have achieved the following very important and useful thing:
    Your equations have been (fully, in case the matrix is diagonalisable) decoupled and are independably solvable :)

    hope you can follow it :)

    best regards,
    marin
     
  5. Apr 18, 2009 #4
    All right thanks guys, you've all given me great insights on how to solve these problems in the future. And if anyone wants to take a shot at it, how would one solve a coupled partial differential equation (I've taken a course in PDE's so don't worry bout that part). Thanks guys.
     
  6. Mar 10, 2011 #5
    [tex]x^2\sqrt{x}[\tex]
     
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