Solving Coupled ODEs with Boundary Conditions

[Madz]

Hi,

Can anyone please 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.

 

[Pseudo Statistic]

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]

 

[Madz]

Yep,thats right:)

 

[HallsofIvy]

Wicked! That can be reduced to a single 6th order equation or a 6 by 6 first order system.

 

[Madz]

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:)

 

[siddharth]

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:)

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.