# Coupled ODEs

1. Jul 14, 2007

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

2. Jul 14, 2007

### Pseudo Statistic

Ouch. This should probably be tidied up..
Is this what you meant? :D
$$-\lambda + v\frac{d^3 H}{dt^3} = -2H\frac{dH}{dt} + (\frac{dH}{dt})^2 - G^2$$
$$v\frac{d^2 G}{dt^2} = 2\frac{dH}{dt}G - 2\frac{dG}{dt}H$$

3. Jul 14, 2007

Yep,thats right:)

4. Jul 14, 2007

### HallsofIvy

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

5. Jul 16, 2007

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