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Infinite coupled ODE's

  1. Nov 18, 2009 #1
    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

    [tex] \forall i \in \mathbb{N} [/tex]

    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. jcsd
  3. Nov 18, 2009 #2
    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?
  4. Nov 18, 2009 #3
    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.
  5. Nov 18, 2009 #4

    D H

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    He has an infinite number of equations there, Nick.

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

    [tex]\dot u_i = \sum_{n=1}^{\infty}e_{n,i} u_n[/tex]

    with the following:

    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}

    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.
  6. Nov 19, 2009 #5
    Very smart: creating one infinite dimensional vector but shouldn't there be another summation sign for m?
  7. Nov 19, 2009 #6

    D H

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

    \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

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

    \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

    This is just a matrix-vector equation: [itex]\dot{\vec u} = \mathbf A \vec u[/itex] if you treat u as a column vector (or [itex]\dot{\vec u} = vec u \mathbf A^T[/itex] 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.
  8. Nov 19, 2009 #7
    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:

    [tex] u(t) = e^{At} [/tex]

    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?
  9. Nov 20, 2009 #8

    D H

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    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.
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