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Coupled ODEs

  1. Jul 14, 2007 #1
    Hi,

    Can anyone plz tell me how to go about solving this system of coupled ODEs.?

    1) (-)(lambda) + vH''' = -2HH' +(H')^2 - G^2
    2) vG'' = 2H'G - 2G'H

    lambda and v are constants.
    And the boundary conditions given are
    H(0) = H(d) = 0
    H'(0) = omega * ( c1 * H''(0) + c2 * H'''(0) )
    G(0) = omega * ( 1 + c1*G'(0) + c2*G''(0) )
    H'(d) = omega * ( c3 * H''(d) + c4 * H'''(d) )
    G(d) = omega * ( c3 * G'(d) + c4 * G''(d) )
    ... c1,c2,c3,c4,omega are constants

    -Maddy.
     
  2. jcsd
  3. Jul 14, 2007 #2
    Ouch. This should probably be tidied up..
    Is this what you meant? :D
    [tex]-\lambda + v\frac{d^3 H}{dt^3} = -2H\frac{dH}{dt} + (\frac{dH}{dt})^2 - G^2[/tex]
    [tex]v\frac{d^2 G}{dt^2} = 2\frac{dH}{dt}G - 2\frac{dG}{dt}H[/tex]
     
  4. Jul 14, 2007 #3
    Yep,thats right:)
     
  5. Jul 14, 2007 #4

    HallsofIvy

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    Wicked! That can be reduced to a single 6th order equation or a 6 by 6 first order system.
     
  6. Jul 16, 2007 #5
    Oh..How do we do it? It would also be great if you could suggest me some good book that would help me with such problems.Thanks:)
     
  7. Jul 16, 2007 #6

    siddharth

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    Just substitute the value for [tex]\frac{dH}{dt}[/tex] and then [tex]\frac{d^3H}{dt^3}[/tex] (by diff twice) from the second equation to the first.
     
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