Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Inverse laplace transform

  1. Oct 14, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the Inverse Laplace Transform of


    3. The attempt at a solution

    for this question i found the singularities to be at 0 and when s = 1. (as the sqrt of 1 is ± 1) there is also a branch point that runs from 0→-∞. so if you take a contour that runs vertically to the right of all singularities. then arcs down towards the axis then along the branch point around the singularity at s=0, then back along the branch point and arc back towards the the vertical part, you should avoid crossing any branch points. can you take the contour integral over each part of the contour and then use the residue theorem to account for the singularity at s=1?

    the other bit i am stuck on is the integral that runs either side of the branch point. i get:

    - [itex]\int[/itex] e-xt* ((√x)i-1/ x(√x)+1) dx

    and im not sure if this is right or how to integrate from ∞ to 0

    thanks for any help
  2. jcsd
  3. Oct 15, 2011 #2
    If you saw that other one I commented on, I may have created some confussion: the point s=1 is a pole only for the branch:


    but that's not the branch you're integrating on (I think) so it's not a singular point for this problem. Here's what I think you should do: first get the answer in Mathematica:

    Code (Text):

    myinverse = InverseLaplaceTransform[
       (Sqrt[s] - 1)/(s*(Sqrt[s] + 1)), s, t]

    -1 + E^t - E^t*Erf[Sqrt[t]] + E^t*Erfc[Sqrt[t]]
    notice how that answer has the erf function. That's a non-elementary integral so that means when you do the contour analysis, you'll encounter integrals you won't be able to express in simple form.

    The next thing to do is to just guess at the answer and compute it numerically to see if it agrees with the actual answer. So just guess (for now), that the answer is just the two horizontal legs of that key-hole contour. Now, can you combine both legs and arrive at the expression:

    [tex]\int_{\infty}^0 \frac{e^{-rt}}{r}\left(\frac{4i \sqrt{r}}{1+r}\right)dr[/tex]

    Now, suppose that is the only contribution to the answer. Then the transform would actually be:

    [tex]-\frac{1}{2\pi i}\int_{\infty}^0 \frac{e^{-rt}}{r}\left(\frac{4i \sqrt{r}}{1+r}\right)dr[/tex]


    Now, compute that numerically for say t=3. Does that agree with the actual inverse transform at t=3?

    If it doesn't then maybe there are other parts of the contour contributing to the answer. Get it working first numerically, then work towards expressing the numerical solution analytically.
    Last edited: Oct 15, 2011
  4. Oct 15, 2011 #3
    thanks jackmell. that helps a heap.

    i have done the analysis of all the other contours and unless i have made a mathematical mistake im pretty confident they are all 0. but your suggestion is a great way to check it.

    the only thing im confused about is how you arrived at your expression for when you combine the two legs? the result i get for one of the horizontal legs of the key hole is in my previous post? have i got that wrong? or am i missing something on how to combine the two expressions.

    thanks heaps for you help
  5. Oct 16, 2011 #4
    How about the one around the origin? If we let [itex]s=\rho e^{it}[/itex], I get:

    [tex]\lim_{\rho\to 0} \int_{\pi}^{-\pi} \frac{e^{t\rho e^{it}}}{\rho e^{it}}\frac{\sqrt{\rho e^{it}}-1}{\sqrt{\rho e^{it}}+1} \rho i e^{it}dt[/tex]

    What's that one?

    Ok, on the contour above the negative real axis, we let [itex]s=re^{\pi i}[/itex]. Make that substituion, simplify, and I get:
    [tex]\int_{\infty}^0 \frac{e^{-rt}}{r} \frac{i\sqrt{r}-1}{i\sqrt{r}+1}dr[/tex]

    On the one below the negative axis, let [itex]s=re^{-\pi i}[/itex] and that one is:

    [tex]\int_{\infty}^0 \frac{e^{-rt}}{r} \frac{i \sqrt{r}+1}{1-i\sqrt{r}}dr[/tex]

    Also, it's a little ambigious to say "above the axis" and "below it". Actually, the two contours are exactly on the real axis but to meet unreasonable analytic requirements, we treat them just infinitely close to it.
    Last edited: Oct 16, 2011
  6. Oct 17, 2011 #5
    yeah i realized after i wrote it that it is a little ambiguous to say it like that. but i felt like you would know what i meant.

    so i think when you combine those two integrals (by addition) i come out with something like this...

    [itex]\int[/itex] e-rt*[itex]\frac{1}{r}[/itex]*(2 - [itex]\frac{4}{r+1}[/itex])dr

    which is an integral that looks a lot like a laplace transform? i went through and got a final result of

    -2 + 4etEi(-t) which looks waaaaaaay off from what mathematica gave you as a result? have i gone down the wrong track?

    thanks again for all your help with this.
  7. Oct 17, 2011 #6
    The two integrals along the branch-cut reduces to:

    [tex]\int_{\infty}^{0} \frac{e^{-rt}}{r}\left(\frac{i\sqrt{r}-1}{\sqrt{r}+1}+\frac{i\sqrt{r}+1}{1-i\sqrt{r}}\right)dr=\int_{\infty}^0 \frac{e^{-rt}}{r}\left(\frac{4i \sqrt{r}}{1+r}\right)dr[/tex]

    That limit around the origin as rho goes to zero is 2pi i. The integral over the large circular arcs should tend to zero and we're left with:

    [tex]\int_{\infty}^0 \frac{e^{-rt}}{r}\left(\frac{4i \sqrt{r}}{1+r}\right)dr+2\pi i+\int_{\sigma-i\infty}^{\sigma+i\infty} \frac{e^{st}}{s}\frac{\sqrt{s}-1}{\sqrt{s}+1}ds=0[/tex]

    So, we get that into the inverse transform and write:

    [tex]\mathcal{L}^{-1}\left\{\frac{\sqrt{s}-1}{s(\sqrt{s}+1}\right\}=-\frac{1}{2\pi i}\left(\int_{\infty}^0 \frac{e^{-rt}}{r}\left(\frac{4i \sqrt{r}}{1+r}\right)dr+2\pi i\right)[/tex]

    That agrees with Mathematica:

    Code (Text):

    tval = 5;

    N[(-(1/(2*Pi*I)))*(Integrate[(Exp[(-r)*tval]/r)*((4*I*Sqrt[r])/(1 + r)), {r, Infinity, 0}] + 2*Pi*I)]

    myinverse = InverseLaplaceTransform[(Sqrt[s] - 1)/(s*(Sqrt[s] + 1)), s, t]

    N[myinverse /. t -> tval]

    -0.5353474112470704 + 0.*I

    -1 + E^t - E^t*Erf[Sqrt[t]] + E^t*Erfc[Sqrt[t]]

    Last edited: Oct 17, 2011
  8. Oct 17, 2011 #7
    oh whoops, stupid error by me.

    that gets me to the result though so that great! thanks heaps for all your help with this question.
  9. Oct 19, 2011 #8
    Hi guys, when writing the Bromwich contour for this integral, as is done in "http://en.wikipedia.org/wiki/Bromwich_integral", [Broken] would the restriction on [itex]\gamma[/itex] be [itex]\gamma > 0[/itex] or [itex]\gamma > 1[/itex] ?
    Last edited by a moderator: May 5, 2017
  10. Oct 20, 2011 #9
    [itex]\gamma>0[/itex] for this problem but if you were integrating over the branch which had the pole at x=1, gamma would have to be larger than one.
    Last edited by a moderator: May 5, 2017
  11. Oct 23, 2011 #10
    Thanks a lot jackmell :)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook