1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Laplace transform of a matrix exponential

  1. Jan 18, 2012 #1

    A_B

    User Avatar

    1. The problem statement, all variables and given/known data
    show that the Laplace transform of e^(At) = (sI - A)^(-1)

    [tex]
    \mathcal{L}\left\{ e^{At} \right\}(s) = \left(sI - A \right)^{-1}
    [/tex]

    3. The attempt at a solution

    I find
    [tex]
    \left( e^{At} \right)_{ij} = \sum_{k=0}^{\infty} \frac{(A^k)_{ij}t^k}{k!}
    [/tex]

    and since
    [tex]
    \mathcal{L}\left\{ (A^k)_{ij}t^k \right\}(s) = \frac{k!}{s^{k+1}} (A^k)_{ij}
    [/tex]

    we have
    [tex]
    \mathcal{L}\left\{\left( e^{At} \right)_{ij}\right\}(s) = \sum_{k=0}^{\infty} \frac{(A^k)_{ij}}{s^{k+1}}
    [/tex]

    and there I'm stuck.

    Thanks
    A_B
     
    Last edited: Jan 18, 2012
  2. jcsd
  3. Jan 18, 2012 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    First show it's true for a diagonal matrix D. Then show it's still true if A is diagonalizable, i.e. PAP^(-1)=D for some invertible matrix P.
     
  4. Jan 18, 2012 #3

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    [tex](sI-A)^{-1}=
    \frac{1}{s}(I-\frac{1}{s}A)^{-1}.[/tex]

    RGV
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Laplace transform of a matrix exponential
  1. Matrix Exponential (Replies: 4)

  2. Matrix exponential (Replies: 4)

  3. Matrix exponential (Replies: 4)

  4. Matrix Exponentials (Replies: 2)

Loading...