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Integral of matrix exponential

  1. Mar 23, 2013 #1
    1. The problem statement, all variables and given/known data
    [itex]\vec{N} = \frac{1}{\tau}(\int_0^\infty exp(t' \textbf{T}) dt') \vec{n}_0[/itex]

    Show that [itex]\vec{N} = -\frac{\textbf{T}^{-1}}{\tau}\vec{n}_0[/itex] where [itex]\textbf{T}^{-1}[/itex] is the inverse matrix of [itex]\textbf{T}[/itex].

    It is part of a small project called Gambler's Ruin in which we investigate the random walk of a gambler on a "money axis" where he has a probability for moving right and left. So the matrix T is the Laplacian matrix for a 1d graph (1)--(2)--(3)--(4)--(5)--(6)--(M).


    2. Relevant equations
    I have two main questions:

    1. How to integrate the matrix
    2. That [itex]\int_0^\infty exp(t) dt = \infty[/itex] as I expect this will be a part of the solution.


    3. The attempt at a solution
    First of all the matrix exponential is just the Taylor series of the exponential with the matrix as exponent:

    [itex]exp(t\textbf{T}) = \sum_{k=0}^\infty \frac{1}{k!} (t\textbf{T})^k[/itex]

    If the matrix T is diagonal then the exp(tT) will just be an matrix with exponential exp(t) along it's diagonal. If it is not diagonal all elementes will be proportinal to exp(xt).

    So if this matrix T, the laplacian for a 1d graph, is diagonal (which I hope it is or can be made to be) I am stuck when I have to do the integral of a matrix with diagonal entries exp(t).

    [itex]\frac{1}{\tau}(\int_0^\infty
    \left[\begin{array}{ccc}
    exp(t) & 0 & 0 \\
    0 & exp(t) & 0 \\
    0 & 0 & exp(t) \end{array}\right] dt') \vec{n}_0[/itex]

    I'd be glad for any advice on how to proceed or hints to where I can find out about this.

    Best regards
    Wuhtzu
     
  2. jcsd
  3. Mar 24, 2013 #2
    The best thing I've been able to find is that the integral of a matrix is the integral of the individual elements (just as the derivative of a matrix is a matrix containing the derivatives of the original matrix).

    If that is the case then (and if my T was diagonal):

    [itex] \int_0^\infty \left [
    \begin{array}{ccc}
    exp(t) & 0 & 0 \\
    0 & exp(t) & 0\\
    0 & 0 & exp(t) \end{array}
    \right ] dt =
    \left [
    \begin{array}{ccc}
    \int_0^\infty exp(t) dt& \int_0^\infty 0 dt & \int_0^\infty 0 dt \\
    \int_0^\infty 0 dt & \int_0^\infty exp(t) dt & \int_0^\infty 0 dt\\
    \int_0^\infty 0 dt & \int_0^\infty 0 dt & \int_0^\infty exp(t) dt \end{array}
    \right ] =
    \left [
    \begin{array}{ccc}
    \int_0^\infty exp(t) dt & c & c\\
    c & \int_0^\infty exp(t) dt & c\\
    c & c & \int_0^\infty exp(t) dt \end{array}
    \right ]
    [/itex]

    where c is a constant. Is that correct?
     
  4. Mar 24, 2013 #3

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Getting rid of constants, you want to show that [itex]\textbf{T}^{-1} = \int_0^\infty exp(t' \textbf{T}) dt'[/itex]
    As t' is scalar, ##exp(t' \textbf{T}) = \sum_0^\infty \frac{1}{k!} t'^k\, \textbf{T}^k##
    Every summand can be integrated, and we get ##\sum_0^\infty \frac{1}{(k+1)!} t'^{k+1}\, \textbf{T}^{k+1} \textbf{T}^{-1}## assuming T is invertible.
    Now you can re-construct your exponential and get the result. You just have to make sure that the exchange of integration and summation is valid.
     
  5. Mar 25, 2013 #4
    So you are right, aprt from constants I need to show that

    [itex]\textbf{T}^{-1} = \int_0^\infty exp(t'\textbf{T}) dt' [/itex]

    With your suggestion this is what I can do:

    [itex]\int_0^\infty exp(t'\textbf{T}) dt' = \int_0^\infty \sum_{k=0}^\infty \frac{1}{k!} t'^k \textbf{T}^k dt[/itex]


    [itex]= \sum_{k=0}^\infty \int_0^\infty \frac{1}{k!} t'^k \textbf{T}^k dt'[/itex]
    [itex]= \sum_{k=0}^\infty \frac{1}{k!} \textbf{T}^k \int_0^\infty t'^k dt'[/itex]
    [itex]= \sum_{k=0}^\infty \frac{1}{k!} \textbf{T}^k \left[ \frac{1}{k+1} t'^{k+1} \right]_0^\infty[/itex]
    [itex]= \sum_{k=0}^\infty \frac{1}{k!} \textbf{T}^k \frac{1}{k+1} \left[ t'^{k+1} \right]_0^\infty[/itex]
    [itex]= \sum_{k=0}^\infty \frac{1}{k!(k+1)} \textbf{T}^{-1} \textbf{T}^{k+1} \left[ t'^{k+1} \right]_0^\infty[/itex]

    I can imagine your idea of maybe recognizing the exponential taylor series again (only with k+1 this time), but before getting into that (and what may come from it) I cannot see how I avoid multiplying each summand by [itex]\infty[/itex] since [itex]\left[ t'^{k+1} \right]_0^\infty=\infty[/itex]
     
  6. Mar 25, 2013 #5

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Don't plug in your integration limits before you pack it into an exponential function again. Alternatively, split the integral in two parts (0 to 1 and 1 to infinity), and use clever sign prefactors to get finite expressions.
     
  7. Mar 25, 2013 #6
    Per se you do not need to perform the integration... just leave the integral, and do the splitting of ##\mathbf{T}^k## just as before. Then bring all except ##\mathbf{T}^{-1}## inside the integral and you will have again the exponential... just check with the limits for the sum if they are correct... Only thing I don't see where you can go this way... I don't think you can get the result you need...
     
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