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Asymptotic evaluation of Laplace inverse transform

  1. Mar 27, 2007 #1

    tpm

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    If we have integrals of the form:

    [tex] \oint_{C} ds f(s) e^{st} [/tex] for [tex] t\sim \infty [/tex]

    or [tex] \int_{-\infty}^{\infty}dxf(a+ix)e^{ixt} [/tex]

    In both cases i would like to know some techniques to evaluate 'asymptotically' the integrals given above for big t using only a few residues or other methods... thanks
     
    Last edited: Mar 27, 2007
  2. jcsd
  3. Mar 30, 2007 #2
    What you are looking for is the method of stationary phase. It is called this way because if one consider the integral

    [tex]I(x)=\int_a^b e^{ixp(t)}q(t)dt,[/tex]

    then the integrand oscillates rapidly in such way that the oscillations cancel out, but near stationary points ([itex]p'(t_0)=0[/itex]), it oscillates more slowly, hence contributing more to the integral. To obtain the expansion, we consider the contributions given by this stationary points, and the edges of the interval.

    Near [itex]t=a[/itex], the integrand is approximately

    [tex]\exp[ix\{p(a)+(t-a)p'(a)\}]q(a).[/tex]

    An indefinite integral of this function is

    [tex]\frac{\exp[ix\{p(a)+(t-a)p'(a)\}]q(a)}{ixp'(a)},[/tex]

    provided that [itex]p'(a)\ne 0[/itex]. The lower limit [itex]t=a[/itex] asympototically contributes

    [tex]-\frac{e^{ixp(a)}q(a)}{ixp'(a)}[/tex]

    to [itex]I(x)[/itex]. In similar fashion, the point [itex]t=b[/itex] contributes asymptotically

    [tex]\frac{e^{ixp(b)}q(b)}{ixp'(b)}.[/tex]

    Next, if [itex]t_0 \in (a,b)[/itex] is a stationary point of [itex]p(t)[/itex], then near this point the integrand is approximately

    [tex]\exp[ix\{p(t_0)+(t-t_0)^2p''(t_0)\}]q(t_0),[/tex]

    provided that [itex]p''(t_0)[/itex] and [itex]q(t_0)[/itex] are not zero. Since we believe that only a neighborhood of [itex]t_0[/itex] contributes, we extend the domain to the whole real axis, so we can explicitly evaluate the integral obtained from such reasoning

    [tex]\int_{-\infty}^\infty \exp[\pm iy t^2]dt=e^{\pm i\pi/4}\left(\frac{\pi}{y}\right)^{1/2}\qquad (y>0).[/tex]

    So the contribution to [itex]I(x)[/itex] from the neighborhood of [itex]t_0[/itex] is expected to be

    [tex]e^{\pm i \pi/4}q(t_0)\exp\{ixp(t_0)\}\left|\frac{2\pi}{xp''(t_0)}\right|^{1/2},[/tex]

    where the upper and lower sign is taken accordingly as [itex]xp''(t_0)[/itex] is positive or negative.

    The asymptotic expression for [itex]I(x)[/itex] will be the sum of all contributions from stationary points along with the ones of the upper and lower limits of the integral.

    You can find more terms considering higher stationary points.

    Full exposition of this method can be found in Olver's great book Asymptotics and Special Functions, wich include complex variables and contour integration.
     
    Last edited: Mar 30, 2007
  4. Apr 29, 2007 #3

    tpm

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    Thanks 'AIRAVATA' the problem is that Laplace transform uses the integral..

    [tex] \oint _{C} ds f(s)e^{st} [/tex]

    as you can see , even for big 't' the function st has no 'extremal' except when s=-oo or s=oo, hence saddle point or steepest descent method can't be applied, i'm downloading from E-mule the book you pointed to me , to see if i get a clearer result, thankx..
     
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