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How to abstractly prove a Laplace transform identity?

  1. Apr 8, 2017 #1
    1. The problem statement, all variables and given/known data
    "Suppose that ##F(s) = L[f(t)]## exists for ##s > a ≥ 0##.
    (a) Show that if c is a positive constant, then
    ##L[f(ct)]=\frac{1}{c}F(\frac{s}{c})##

    2. Relevant equations
    ##L[f(t)]=\int_0^\infty f(t)e^{-st}dt##

    3. The attempt at a solution
    ##L[f(ct)]=\int_0^\infty f(ct)e^{-st}dt##

    D: ##e^{-st}## I: ##f(ct)##
    D: ##-e^{-st}## I: ##\frac{1}{c}F(ct)##
    I: ##F(ct)##

    ##L[f(ct)]=\int_0^\infty f(ct)e^{-st}dt=(\frac{1}{c}e^{-st}F(ct))+\frac{1}{c}\int_0^\infty e^{-st}F(ct)dt##

    D: ##F(ct)## I: ##e^{-st}##
    D: ##cf(ct)## I: ##-\frac{1}{s}e^{-st}##

    ##L[f(ct)]=\int_0^\infty f(t)e^{-st}dt=(\frac{1}{c}e^{-st}F(t))+\frac{1}{c}(-\frac{1}{s}e^{-st}F(ct))+\frac{c}{s}\int_0^\infty e^{-st}f(ct)dt##
    ##\frac{s-c}{s}\int_0^\infty f(ct)e^{-st}dt=(\frac{1}{c}e^{-st}F(ct))+(-\frac{1}{cs}e^{-st}F(ct))##
    ##L[f(ct)]=\frac{s}{s-c}(\frac{1}{c}(e^{-st}F(ct)-\frac{1}{s}e^{-st}F(ct))=\frac{s}{s-c}(e^{-st}F(ct))(\frac{1}{c}-\frac{1}{s})##
    ##L[f(ct)]=\frac{s}{s-c}(e^{-st}F(ct))(\frac{s}{cs}-\frac{c}{cs})=\frac{s}{c}e^{-st}F(ct)##

    Can anyone tell me what it is I'm doing wrong?
     
  2. jcsd
  3. Apr 8, 2017 #2

    FactChecker

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    Try ∫0f(ct)e-stdt = ∫0f(ct)e-s/c⋅ctdt = 1/c∫0f(h)e-s/c⋅hdh, where h=ct.

    Note: If c is negative, you have to take into account the change of limits of the integral. Since it was stated that c>0, that is not a concern here.
     
    Last edited: Apr 14, 2017
  4. Apr 8, 2017 #3

    pasmith

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    Homework Helper

    This is a straightforward consequence of integration by substitution. First, to avoid confusion, define [itex]g(t) = f(ct)[/itex]. Then [tex]L[g(t)] = G(s) = \int_0^\infty g(t)e^{-st}\,dt = \int_0^\infty f(ct)e^{-st}\,dt.[/tex] Now set [itex]ct = u[/itex] and compare to [tex]
    F(p) = \int_0^\infty f(u)e^{-pu}\,du.[/tex]
     
  5. Apr 11, 2017 #4
    Thanks for the help, everyone. I couldn't have finished the homework without it.
     
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