Results 1 to 6 of 6

Math Help - Laplace inverse of a function

  1. #1
    Newbie
    Joined
    Jun 2008
    Posts
    20

    Laplace inverse of a function

    Hi,
    Can anybody help me to find inverse laplace transform of one of these functions.
    Thanks,

     <br />
f(s)=\frac{1}{s+a+b\sqrt{s+c}}<br />

     <br />
f(s)=\frac{1}{s(s+a+b\sqrt{s+c})}<br />
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,111
    Thanks
    2
    Currently, I am of the opinion that you will not find a convenient expression for those. It may be time for the fundamental definition and figuring out how to deal with the integral expressions.

    However, for the first, you may wish to consider a transformation of variables, maybe:

    w = \sqrt{s+c}+\frac{b}{2}

    v = \sqrt{a-c-\frac{b^{2}}{4}}

    Then consider the transform with respect to "w", rather than 's', and see if that leads to anything even close to

    \frac{\sin(v \cdot t)}{v}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Aug 2008
    Posts
    903
    Hi. Those look like a challenge. Isn't someone in here taking Honors Complex Analysis? Hummmm . . . Anyway, Mathematica can't even compute \mathcal{L}^{-1}\left\{\frac{1}{s+\sqrt{s-1}}\right\}. Doesn't mean there isn't one of course. Even though I'm not certain the integral converges, I found it interesting that I could (apparently) get a reasonably accurate numerical approximation to what I think is the inverse transform by computing numerically \frac{1}{2\pi i}\int_{2-i\infty}^{2+i\infty}\frac{e^{st}}{s+\sqrt{s-1}}ds over the interval t\in (0,4). The first plot is the data points I calculated and the blue line is a fit to the equation a+bt+ct^2. Granted that's probably not exactly the inverse transform. But I'm encouraged by next taking the Laplace transform of my fit:

    \mathcal{L}\left\{a+bt+ct^2\right\}=\frac{a}{s}+\f  rac{b}{s^2}+\frac{2c}{s^3} (in blue) against \frac{1}{s+\sqrt{s-1}} (in red) which is the second plot. Granted it's not perfect, but the agreement is surprisingly close in the range (1.5,4).

    I'm surprised I could get even this close with numerical methods and I'm optimistic that the numerical result is approximately the inverse transform in this range of t.
    Attached Thumbnails Attached Thumbnails Laplace inverse of a function-inverse-laplace-transform.jpg   Laplace inverse of a function-transform-pair.jpg  
    Last edited by shawsend; September 16th 2008 at 12:18 PM. Reason: Not sure about convergence of integral
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jun 2008
    Posts
    20

    The Problem Can be Solved This Way

    Hi, Thanks for your help and interest to solve this problem, but I could solve it this way.


     <br />
f(s)=\frac{1}{s+a+b\sqrt{s+c}}=\frac{1}{(\sqrt{s+c  }+B_1)(\sqrt{s+c}+B_2)}<br />

     <br />
=\frac{1}{B_1-B_2}(\frac{1}{\sqrt{s+c}+B_2}-\frac{1}{\sqrt{s+c}+B_1})<br />

    where

     <br />
B_1,B_2=\frac{1}{2}[b\pm\sqrt{\Delta}]<br />

    and

    \Delta=b^2-4(a-c)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Aug 2008
    Posts
    903
    you da' man Ehsan!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Aug 2008
    Posts
    903
    I was curious how close I came with the numerical approximation. So using you expression, I used Mathematica and calculated the inverse transform of \frac{1}{s+\sqrt{s-1}}:

    f(t)=\frac{1}{b_1-b2}\left[e^t\left(\frac{1}{\sqrt{\pi t}}-b_2e^{b_2^2 t}\textbf{Erfc}(b_2\sqrt{t})\right)-e^t\left(\frac{1}{\sqrt{\pi t}}-b_1e^{b_1^2 t}\textbf{Erfc}(b_1\sqrt{t})\right)\right]

    The plot below shows f(t) above in red, and the numerical fit I calculated in an earlier post still in blue with the blue data points. I'm really surprised I guess the agreement is pretty good outside the end points.
    Attached Thumbnails Attached Thumbnails Laplace inverse of a function-lt-comparisons.jpg  
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Laplace Inverse of polynomial function?
    Posted in the Calculus Forum
    Replies: 0
    Last Post: August 1st 2011, 06:32 AM
  2. Inverse Laplace of this function
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: May 8th 2011, 02:46 PM
  3. Inverse Laplace transform of a function
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: April 15th 2011, 04:28 PM
  4. laplace inverse
    Posted in the Differential Equations Forum
    Replies: 7
    Last Post: October 9th 2010, 10:15 PM
  5. Laplace/Inverse Laplace Questions
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: August 14th 2010, 12:29 PM

Search Tags


/mathhelpforum @mathhelpforum