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Thread: Characteristic relaxation in nonhomogenous differential equations

  1. #1 Characteristic relaxation in nonhomogenous differential equations 
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    Hey all,

    I think I have a dumb question, but here we go. I have a stochastic simulation of a biochemical reaction system for which I want to know the characteristic relaxation time. The way to figure this out, I understand, is to take the mean/macroscropic differential equations for the system and calculate the eigenvalues. Then the inverse of the smallest eigenvalue gives you the characteristic relaxation time.

    Question: how do you solve for eigenvalues if the system of DEs is nonhomogenous? Or does the nonhomogeneous part not contribute to the relaxation time? My system is a bit more complicated than this, but my question only concerns what to do with the constant 'k' in the first equation.




    This is sort of a physics question I suppose, let me know if it's better suited for that forum--and thanks for your time!


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    So if you have some functions A_n dependent on only one variable t and there is a matrix B_nk: dA_n/dt = B_nk A_k (where there is a summation in index k) if B_nk is a constant matrix then knowing its eigenvalues will give
    an easy solution to the system. Say B has eigenvectors x,y and eigenvalues ex, ey. Take a function u(t) and multiply by x to get something two-dimensional. Bx*u(t)=ex*x*u(t) so plugging it into the differential equation will give x*u'(t)=x*ex*u(t) which clearly has a solution u=exp(ex*t)*K1. Since the system is linear sums of solutions are solutions. If the eigenvalues are different the solutions exp(t ey)y*Ky and exp(t ex)x*Kx are linearly independent and so the system has a complete solution.

    The eigenvalues here have units inverse time and if they are negative i guess they could be interpreted as a relaxation time.

    If the t0, t1 and stuffs are just constants then this applies right away. If they are functions of t for instance then you'd have to linearize at some point in order to use this method.


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  4. #3  
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    Quote Originally Posted by mathman89 View Post
    So if you have some functions A_n dependent on only one variable t and there is a matrix B_nk: dA_n/dt = B_nk A_k (where there is a summation in index k) if B_nk is a constant matrix then knowing its eigenvalues will give
    an easy solution to the system. Say B has eigenvectors x,y and eigenvalues ex, ey. Take a function u(t) and multiply by x to get something two-dimensional. Bx*u(t)=ex*x*u(t) so plugging it into the differential equation will give x*u'(t)=x*ex*u(t) which clearly has a solution u=exp(ex*t)*K1. Since the system is linear sums of solutions are solutions. If the eigenvalues are different the solutions exp(t ey)y*Ky and exp(t ex)x*Kx are linearly independent and so the system has a complete solution.

    The eigenvalues here have units inverse time and if they are negative i guess they could be interpreted as a relaxation time.

    If the t0, t1 and stuffs are just constants then this applies right away. If they are functions of t for instance then you'd have to linearize at some point in order to use this method.
    I totally didn't see your response, for some reason I didnt get an email despite subscribing to the thread. Sorry! And thank you. Some follow-up:

    To make sure I understand your response, the short answer is you can ignore the constant 'k' in my equation and look at the matrix A only. All of the terms are constants in matrix A, so I can easily calculate the eigenvalues (no need to linearize).

    When I do this, the eigenvalues are unique and negative. Why do you say "i guess they could be interpreted as a relaxation time"? To be honest, this assertion (1/(smallest eigenvalue)=characteristic relaxation time) was made to me by a senior professor, but I have not been able to corroborate it anywhere. And this method I'm working on depends on the assertion.
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    It is a theorem of linear algebra that any linear transformation can be represented by a matrix. So you'll note that for the vector

    and the matrix,
    .
    Then we may rewrite your equation as,


    As you can see the transformation, that is the differentiation of the vector with respect to time, is not represented as a matrix (instead we have a matrix and a vector) so the transformation is non-linear. Depending on whether you seek to minimise the relative or absolute error you may be able to get away with using the approximation, k = 0, in which case you take the smallest eigenvalue from the characteristic polynomial below.


    I'm not exactly sure how one would proceed if the assumption i mentioned was not valid or yields an inaccurate result.
    Last edited by wallaby; February 23rd, 2012 at 07:43 AM. Reason: Original post was rushed
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