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yersinia29
25th September 2004, 01:45 PM
I need to solve a system of 3 differential equations that are all coupled to each other. I think matrices would be the best way to do it, but i need some help.

Lets say that the 3 variables are Mx, My, and Mz, and the diffeq setup is:

dMx/dt = A*Mx + B*My + C*Mz

dMy/dt = D*Mx + E*My + F*Mz

dMz/dt = G*Mx + H*My + J*Mz

So the matrix would be:

dM/dt = [A B C
D E F
G H J] M + [K L M]

Can somebody give me a step by step of how to solve this system of ODEs?

TillEulenspiegel
25th September 2004, 02:07 PM
http://www.physicsforums.com/

Benguin
25th September 2004, 02:07 PM
My calculus at that level is at best rusty, and it was never that good to start with.

I think the problem is intended to guide you towards LaGrange ... have you covered that yet?

SGT
28th September 2004, 10:29 AM
Call x the vector [Mx My Mz]<sup>T</sup> ,
F the matrix [A B C
D E F
G H J]
and b the vector [K L M]]<sup>T</sup>.

dx /dt = Fx + b

if b = 0 the solution is
x = x(0)e<sup>F t</sup>
where x(0) is the value of your vector x at t = 0.

if b ~= 0

the solution will be:
x = x(0)e<sup>F t </sup>+ e<sup>F t</sup>x(0) integral e<sup>F (t-&tau; )</sup>b d&tau;,
The integral is calculated between 0 and t.

The matrix e<sup>F t</sup> is defined as:

e<sup>F t</sup> = I + F t + F<sup>2</sup> t<sup>2</sup>/2! + F<sup>3</sup> t<sup>3</sup>/3! + ...

yersinia29
29th September 2004, 01:30 PM
SGT, I've got another question.

I know the general solution for 2x2 ODE matrix is of the form Ce^At + Ce^Bt + ....

However, I've seen some 3x3 ODE matrices where the solution is not of this form, but is instead of the form

(Ce^At)*(Ce^Bt)*(Ce^Dt).....

This seems like a fundamentally different solution set than the one above.

I guess my real question is, what determines the nature of additive exponentials vs multiplicative exponentials in teh solution set?

yersinia29
29th September 2004, 04:45 PM
Here's another way of asking my question:

Suppose you have the following diffeq:

dF/dt = (A + B)F

Now from what I understand, the solution to this is:

F = (e^At*e^Bt)F(0)

My question is why/how is that the case?

Vorticity
29th September 2004, 08:10 PM
Originally posted by yersinia29
Here's another way of asking my question:

Suppose you have the following diffeq:

dF/dt = (A + B)F

Now from what I understand, the solution to this is:

F = (e^At*e^Bt)F(0)
No, what you have there is not quite right. Assuming that you mean A and B to be matrices, the correct solution would be:

F(t)=e^[(A+B)t] F(0)

This is not the same as what you have unless A and B commute. (If they commute, that means that AB=BA.)

Suppose you have a system of ODEs of the form:

dx/dt = Ax + b

where x is an n-dimensional vector, A is a constant n by n matrix, and b is a constant n-dimensional vector. We can start off by defining a new variable y, which satisfies:

x=y - A^-1 b

Here "A^-1" means the inverse of A. Substituting in this expression for x simplifies our equation to:

dy/dt = Ay

Now, to find the general solution of this homogeneous n-dimensional ODE, we assume that A has eigenvectors

v1, v2, ... , vn

with corresponding eigenvalues

e1, e2, ... , en

This means that for each i, we have A vi = ei vi. (For more about eigenvalues/vectors, see http://mathworld.wolfram.com/Eigenvalue.html )

A little fiddling around will show that the general solution to the equation for y(t) is:

y(t) = C1 v1 e^( e1 t ) + C2 v2 e^( e2 t ) + ... + Cn vn e^( en t )

(try substituting this back into the equation for dy/dt to see that it is a solution.) Here C1, C2, ... , Cn are arbitrary constants. If we happen to have an initial condition for each of the n components of x, then we also have initial conditions for all of the components of y. We can then find the C's by setting t=0 in the above expression for y(t):

y(0) = C1 v1 + C2 v2 + ... + Cn vn

which is a set of n linear algebraic equations for the n unknown C's. Solve it and you're done.