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Solve for x, a and b in matrix equation aAx + bBx = C

  1. Jun 20, 2012 #1
    Hello everybody

    I recently encountered the following equation [itex]C(t) = a\int_0^t{x(\tau)d\tau} + b\int_0^t{\int_0^\tau{x(\tau')d\tau'd\tau}}[/itex], where C, a, b and x are greater or equal to zero. C and x are vectors - in my case around 3500 long - and a and b are constants.

    If we take sufficiently small steps we can replace the integrals with summations:
    [itex]C(t) = a\Sigma_0^t{x} + b\Sigma_0^t\Sigma_0^t{x}[/itex].
    Such a summation can also be written as a matrix of the form [1 0 0; 1 1 0; 1 1 1] etc. using Matlab notation and [1 0 0; 2 1 0; 3 2 1] etc. for det double summation.

    Now we have a system: [itex]C = aAx + bBx[/itex] where C and x are Nx1 matrices, a and b are constants, and A and B are NxN matrices. I wish to solve it in some least norm sense for x, a and b, with the constraints that x, a and b should be equal to or greater than zero.

    I have tried to solve the first equation using some of the nonlinear optimization tools in Matlab with poor results. I hoped it would be easier to solve when rewritten as a linear system, but I cannot see how.

    Any suggestions would be most welcome.
     
  2. jcsd
  3. Jun 20, 2012 #2
    I suppose τ' is a different τ? Like saying τ1 and τ2?

    If so, then the inner integral is equal to x(τ2), which yields:

    [tex]C(t)= a\Sigma_0^t{x} + b\Sigma_0^t{x(τ_2)}[/tex]

    Edit: Nevermind, I thought it was x'(τ')
     
  4. Jun 20, 2012 #3
    Ok, I found what was bugging me:

    You replaced the double integral with two sums from 0-t. However, the inner sum should be from τ'=0 to τ'=τ, and the outer sum should be τ=0 to τ=t
     
  5. Jun 21, 2012 #4
    Thank you. You are correct. That is what I meant. Equivalent to using two 'cumsum' in Matlab.
     
  6. Jun 21, 2012 #5

    I like Serena

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    Welcome to PF, Panteren! :smile:

    I take it your system is actually the following?
    $$C(t_i) = aAx(t_i) + bBx(t_i)$$

    In that case the solution in a least-norm-sense is given by a least-squares solution.

    What you'd do is minimize ##\sum_i (C(t_i) - aAx(t_i) + bBx(t_i))^2##, which is the sum-squared deviation given a certain a and b.
    To solve it you'd calculate the partial derivatives to a and also to b and set them to zero.

    You'll find the system of equations:
    $$a \sum_i (Ax(t_i))^2 + b \sum_i Ax(t_i) \cdot Bx(t_i) = \sum_i C(t_i) \cdot Ax(t_i)$$
    $$a \sum_i Ax(t_i) \cdot Bx(t_i) + b \sum_i (Bx(t_i))^2 = \sum_i C(t_i) \cdot Bx(t_i)$$

    Its solution (for a and b) appears to be what you want.
     
  7. Jun 21, 2012 #6
    Thank you. :smile: I have read the forums for quite some time. Lots of interesting stuff and insight to be found.


    Yes, but I am not sure I understand the distinction between that and what I wrote? Please elaborate what I have misunderstood or stated unclear :-/?

    Thank you, but the problem is, that [itex]x[/itex] is also unknown. Perhaps it is obvious how to get that in addition to [itex]a[/itex] and [itex]b[/itex] from the system of equations, but I do not follow :-(
    If I had [itex]x[/itex] I could just turn it into a standard linear regression problem and likewise if I had the constants [itex]a[/itex] and [itex]b[/itex], but when I only have [itex]A[/itex], [itex]B[/itex], [itex]C[/itex] and the non-negativity constraints on [itex]a[/itex], [itex]b[/itex] and [itex]x[/itex] ... ?
     
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