1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Staff Emeritus
    Science Advisor

    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

    User Avatar
    Homework Helper
    Gold Member

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Coupled ODEs
  1. Coupled ODEs (Replies: 3)

  2. Help: Coupled ODE (Replies: 2)

Loading...