# Infinite coupled ODE's

1. Nov 18, 2009

### dirk_mec1

1. The problem statement, all variables and given/known data

$$\dot{x}_i = \sum_{ n=1}^{\infty} a(n,i) x_n +b(n,i) y_n$$

$$\dot{y}_i = \sum_{ n=1}^{\infty} c(n,i) x_n +d(n,i) y_n$$

$$\forall i \in \mathbb{N}$$

2. Relevant equations
a,b,c and d are constants (though dependent on the constants n and i).

3. The attempt at a solution
I want to know how I can solve such an infinite large coupled system of ODE's. Can someone help me?

2. Nov 18, 2009

### Nick Bruno

well, you have multiple unknowns and 2 equations. u need as many equations as you have unknowns. Maybe you can choose i such that you have enough equations to solve?

3. Nov 18, 2009

### dirk_mec1

Maybe I wasn't clear enough, i runs from 1 till infinity, in the first post I've denoted that i is part of the natural numbers so

i = 1,2,3,...,inf.

4. Nov 18, 2009

### D H

Staff Emeritus
He has an infinite number of equations there, Nick.

To the OP: Convert this problem to a set of equations of the form

$$\dot u_i = \sum_{n=1}^{\infty}e_{n,i} u_n$$

with the following:

\aligned u_{2n-1} &= x_n \\ u_{2n} &= y_n \\ e_{2n-1,2m-1} &= a_{n,m} \\ e_{2n-1,2m} &= b_{n,m} \\ e_{2n,2m-1} &= c_{n,m} \\ e_{2n,2m} &= d_{n,m} \endaligned

Now you have a problem in one infinite-dimensioned vector (u) rather than two (x and y). See if you can take it from there.

5. Nov 19, 2009

### dirk_mec1

Very smart: creating one infinite dimensional vector but shouldn't there be another summation sign for m?

6. Nov 19, 2009

### D H

Staff Emeritus
No.

Suppose that instead of an infinite number of x and y, we only have two:

\aligned \dot x_1 &= a_{1,1} x_1 + b_{1,1} y_1 + a_{2,1} x_2 + b_{2,1} y_2 \\ \dot y_1 &= c_{1,1} x_1 + d_{1,1} y_1 + c_{2,1} x_2 + d_{2,1} y_2 \\ \dot x_2 &= a_{1,2} x_1 + b_{1,2} y_1 + a_{2,2} x_2 + b_{2,2} y_2 \\ \dot y_2 &= c_{1,2} x_1 + d_{1,2} y_1 + c_{2,2} x_2 + d_{2,2} y_2 \endaligned

Define the four-vector $\vec u = [u_1, u_2, u_3, u_4]^T = [x_1, y_1, x_2, y_2]^T$. The above becomes

\aligned \dot u_1 &= a_{1,1} u_1 + b_{1,1} u_2 + a_{2,1} u_3 + b_{2,1} u_4 \\ \dot u_2 &= c_{1,1} u_1 + d_{1,1} u_2 + c_{2,1} u_3 + d_{2,1} u_4 \\ \dot u_3 &= a_{1,2} u_1 + b_{1,2} u_2 + a_{2,2} u_3 + b_{2,2} u_4 \\ \dot u_4 &= c_{1,2} u_1 + d_{1,2} u_2 + c_{2,2} u_3 + d_{2,2} u_4 \endaligned

This is just a matrix-vector equation: $\dot{\vec u} = \mathbf A \vec u$ if you treat u as a column vector (or $\dot{\vec u} = vec u \mathbf A^T$ if you use row vectors). Each of those a, b, c, and d elements maps to exactly one of the elements of the state matrix A.

This won't change when you go to infinite dimensional space.

7. Nov 19, 2009

### dirk_mec1

D_H, thanks for clarifying that. Now with this mapping the equation we've got is:

$$\dot{ u} = \mathbf A u$$

Normally I would determine eigenvalues and eigenvectors to get an explicit solution because the general solution is:

$$u(t) = e^{At}$$

But now I've got an infinite large matrix. I've thought about this and there's isn't an exact solution that can be found, right?

8. Nov 20, 2009

### D H

Staff Emeritus
Infinite state matrices are the subjects of many books. Long books. Long books with lots of hairy math.

The best I can do is refer you to some. Here is one: http://books.google.com/books?id=G_x-F-l2V2UC

Google the term "infinite dimensional linear system" and you will find many more.