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A Inverse Laplace transform of a piecewise defined function

  1. Feb 17, 2017 #1
    I understand the conditions for the existence of the inverse Laplace transforms are
    $$\lim_{s\to\infty}F(s) = 0$$
    and
    $$
    \lim_{s\to\infty}(sF(s))<\infty.
    $$
    I am interested in finding the inverse Laplace transform of a piecewise defined function defined, such as

    $$F(s) =\begin{cases} 1-s &\text{ if }0\le s\le1 \text{ and}\\
    0&\text{ if } s>1
    \end{cases}$$
    Clearly the limits above do satisfy the existence of the inverse condition, but I'm not sure how to determine the inverse.

    I'm not sure whether the Bromwich integral method can be applied, since it would appear that if I choose gamma (the Browmich integral integration limits: gamma - i*Inf to gamma + i*Inf) between 0 and 1 the function to integrate is (1-s), whereas if I choose gamma > 1 then the Bromwich integral is obviously 0. I'm also not sure whether Post's inversion formula can be used since I'm not sure I understand how to evaluate high-order derivatives of a function which is not differentiable at s = 1. Clearly for a finite k, the kth order derivative of F exists for all s except 1, but how about as k -> Inf?

    Finally, just wondering if the two conditions I listed initially (the two limits) are sufficient for the inverse Laplace transform of $F(s)$ to exist.
     
  2. jcsd
  3. Feb 18, 2017 #2
    I'm still not sure about how to arrive analytically at the inverse Fourier transform, but I have played around trying to fit a sum of exponentials to F(s), using F(s) values for s from 0 to 10 with a step size of 1e-3 and using 600 bins for the t from 0.005 to 30. The f(t) determined by least-squares fitting looks something like this (log10 scale for the x-axis):
    inverse Laplace transform.jpg
    Clearly this looks a lot like a simple wavelet, and presumably summing up to infinity would converge to F(s) exactly. The overlay of the fit to F(s) is below:
    iLt fit.jpg
     
  4. Feb 19, 2017 #3

    jasonRF

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    As I posted in your other thread
    https://www.physicsforums.com/threads/inverse-laplace-transform-of-f-s-exp-as-as-delta-t-a.904385
    (I may ask the moderators to combine these threads as they are highly related), the Laplace transform will be analytic in a right-half of the complex plane. I have never ever worked with piecewise Laplace transforms (am not even sure they can possibly represent a valid Laplace transform). If we blindly assume that your Laplace transform is valid, for your case it is analytic for ##\Re(s)>1## (since it is not analytic for, say ##\Re(s)>1/2##). So your Bromwich integral must be along a vertical contour with ##\Re(s)>1##; your function is zero there, so the answer you get is zero. Indeed, any ##F(s)## that is zero for a right-half plane has an inverse transform that is zero (for a reference see Chapter 6 of "Mathematics for the Physical Sciences" by Schwartz, hopefully in a library you have access to). I am starting to think that your ##F(s)## probably cannot represent the Laplace transform of either a regular or generalized function, but I may be wrong at this point.

    jason
     
  5. Feb 19, 2017 #4

    jasonRF

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    Wait - in your example above you have ##f(t)## that is not zero for ##t<0##. So are both of your threads using a bilateral Laplace transform?

    If so, then the Laplace transform should be analytic in (have a region of convergence of) a strip ##\sigma_0 < \Re(s) < \sigma_1##. There is not unique inverse transform without specifying the strip in which the Laplace transform should be considered analytic. A classic textbook example is ##F(s) = -3 / ( (s+2)(s-1)##. IF the region of convergence is ##-2 < \Re(s) < 1## then ##f(t) = u(t)e^{-2t} + u(-t)e^{t}##; if the region of convergence is ##\Re(s)>1## then ##f(t) = u(t) (e^{-2t} - e^t)##; if the region of convergence is ##\Re(s)<-2## then ##f(t) = u(-t) (-e^{-2t} + e^t)##. Of course, this example has an ##F(s)## that is analytic everywhere in the complex ##s## plane except at the 2 poles, so falls under that standard types of functions for which we use the Bromwich integral approach. With these standard ##F(s)## examples, in the original problem formulation you should probably be able to deduce the region of convergence. Or, you can try all options and then pick the solution that makes physics sense given the problem you are trying to solve.

    I really think you need to post your real problem that you are trying to solve. I have probably helped all I can without some real understanding of what you are actually trying to do.

    jason
     
    Last edited: Feb 19, 2017
  6. Feb 19, 2017 #5
    Hi Jason, thanks again for the detailed comments, I'll go in greater length tomorrow about the other issues that you have raised, but regarding the sign of f(t), the plot above is probably confusing because it shows log10(t) on the x axis and the pre-exponential factor in the sum of exponentials I've used to do the fit. So f(t) is zero for t < 0, but can take positive or negative values for t>0 (where the log10 is real).

    In other words my simulation seems to suggest that the inverse Laplace transform of F(s) exists and is given by f(t) = A(t) sin (w*ln(t)) where A(t) is the envelope of the wavelet above, and w is the frequency of the wave.
     
    Last edited: Feb 19, 2017
  7. Mar 1, 2017 #6
    I have thought about the problem, pursuing the sum of exponentials method above, and have realized that adding the faster and slower exponential terms towards log(t) -> ±∞ would not actually significantly affect the behavior of the sum of exponentials around s = 1. The very fast exponentials would only make a difference at very low s, whereas the very slow ones would only make a difference at very high s, therefore the smoothness of the sum of exponentials (red line in plot above) around s = 1 cannot be made into the kink that F(s) displays there no matter how many exponential terms one adds and therefore no sum of exponentials will converge to F(s).

    That is to say, as Jason points out, F(s) cannot be expressed as a sum of exponentials or a Laplace transform because it not differentiable (and therefore analytic) at s = 1.
     
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