# 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

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:)

6. Jul 16, 2007

### siddharth

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.