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Inverse Laplace Transform Equation

  1. Sep 10, 2011 #1
    Good day to all,

    I encounter this expression in analyzing my equation after transform it using Laplace Transform, to get the answer I have to invert it back, I have no idea on how to find its inversion.

    [text]-\text{Cosh}\left[\sqrt{2 s+s^2} x_0\right]+\text{Cosh}\left[s h_0+\sqrt{2 s+s^2} x_0\right]+\text{Sinh}\left[\sqrt{2 s+s^2} x_0\right]-\text{Sinh}\left[s h_0+\sqrt{2 s+s^2} x_0\right][\text]

    After looking at the table of Laplace Transforms I just could find the expression of [text]\text{Sinh}[\sqrt{s}][\text] or [text]\text{Cosh}[\sqrt{s}][\text], in my equation there are terms of $\sqrt{2 s+s^2}$, i think because of that make it more difficult, if not i could use convolution theorem and utilize special Laplace transforms properties available in the table.

    I do appreciate if someone could give me some advice.... thank you in advance...
     
    Last edited: Sep 10, 2011
  2. jcsd
  3. Sep 10, 2011 #2
    This is what you have:


    [tex]-\text{Cosh}\left[\sqrt{2 s+s^2} x_0\right]+\text{Cosh}\left[s h_0+\sqrt{2 s+s^2} x_0\right]+\text{Sinh}\left[\sqrt{2 s+s^2} x_0\right]-\text{Sinh}\left[s h_0+\sqrt{2 s+s^2} x_0\right][/tex]

    It's not professional-looking to use those capital letters as a formal math syntax. Probably though you got that from Mathematica which by convention capitalizes all it's functions. So obviously Mathematica's InverseLaplaceTransform function doesn't return an answer maybe because of the multi-valued terms. However the multi-valued transform may still exist since all that's required is for the Bromwich integral to converge except you might have to express it as a complex integral expression which I'm pretty sure Mathematica won't do. So for starters, since I'm assuming you're using Mathematica, is to just see if it looks like it might converge numerically but I'd just start with one or a pair:

    Code (Text):
    myTable =
      Table[Im[NIntegrate[
         I (Sinh[Sqrt[2 s + s^2]] - Cosh[Sqrt[2 s + s^2]]) Exp[2 s] /.
          s -> 3 + I y, {y, -yval, yval}]], {yval, 5, 150, 5}];
    ListPlot[myTable, Joined -> True]
    Just experiment with that (check Im and Re) to see if enlarging the integration range shows that it's tending to some limit. That though may not be helpful. Sometimes it's not. Finally, we can do it the old-fashion way via contour integration I guess starting with a double key-hole contour around the two branch points and maybe through that analysis, you'll find that the transform actually does not exist for any values of the parameter x.
     
  4. Sep 11, 2011 #3
    Know what, after looking at this problem, I'd like to make the following working conjecture:

    The inverse transform:

    [tex]\text{L}^{-1}\left\{e^f\right\}[/tex]

    does not exist if:

    [tex]e^f\geq O(e^s)[/tex]

    and since this problem can be written as:

    [tex](1-e^s)e^{-s-\sqrt{2s+s^2}}=O(e^s)[/tex]

    I suspect it does not have an inverse transform and believe if the original expression resulted from a real problem that should have an answer, then I suspect some other error could have been made in the calculation.

    Anyone here knows for sure?
     
    Last edited: Sep 11, 2011
  5. Sep 11, 2011 #4
    Thanks Jackmell... thanks for the advice and the coding. I'm not familiar with the numerical coding above as I am working it out analytically. I'm pretty sure that someone (familiar with mathematica) at my school will enlighten me on this.

    As for the conjecture you made, I am going to discuss it and will be back to post the result later.

    Thank you so much...
     
  6. Sep 12, 2011 #5

    hunt_mat

    User Avatar
    Homework Helper

    There is a standard result for the inverse Laplace transform of
    [tex]
    \frac{e^{b\sqrt{s^{2}\pm a^{2}}}}{\sqrt{s^{2}\pm a^{2}}}
    [/tex]
    Which are:
    [tex]
    J_{0}(a\sqrt{x^{2}-b^{2}}),\quad I_{0}(a\sqrt{x^{2}-b^{2}})
    [/tex]
    From here you can just differentiate to get the required integrand.
     
  7. Sep 12, 2011 #6
    Ok. I can begin to see the general outline of the procedure Hunt is suggesting but haven't worked it completely through. Looks like it may involve a DiracDelta and a convolution but not sure ok. Just suggesting that and don't want to get in more trouble. Just work it out and see what it actually is. I'll do so too eventually and if you guys don't post anything, I'll post my results as well as a numeric check if that's ok. Sorry for leading you wrong above Suga.

    Thanks Hunt.
     
    Last edited: Sep 12, 2011
  8. Sep 12, 2011 #7

    hunt_mat

    User Avatar
    Homework Helper

    I think that most of the functions can be put into the form:
    [tex]
    e^{as+b\sqrt{(s-c)^{2}\pm d^{2}}}
    [/tex]
    Then this will be a product of transforms:
    [tex]
    e^{as},\quad e^{b\sqrt{(s-c)^{2}\pm d^{2}}}
    [/tex]
    Which if you do a co-ordinate transformation will get to look like the first post. You will need to differentiate w.r.t b under the integral sign to get rid of the square root. The inverse transform will be some kind of convolution.

    I had to do one of these recently, it wasn't pleasant.
     
  9. Sep 15, 2011 #8
    Just wish to follow-up with this:

    Given:
    [tex]\mathcal{L}^{-1}\left\{\frac{e^{b\sqrt{s^2-a^2}}}{\sqrt{s^2-a^2}}\right\}=I_0(a\sqrt{x^2-b^2})[/tex]
    find:
    [tex]\mathcal{L}^{-1}\left\{(1-e^{s h_0})e^{-sh_0-x_0\sqrt{2s+s^2}}\right\}[/tex]
    Splitting up the calculation, and completing the square:
    [tex]
    \begin{align*}
    \mathcal{L}^{-1}\left\{(1-e^{s h_0})e^{-sh_0-x_0\sqrt{2s+s^2}}\right\}&=
    \mathcal{L}^{-1}\left\{e^{-sh_0-x_0\sqrt{(s+1)^2-1}}\right\}-\mathcal{L}^{-1}\left\{e^{-x_0\sqrt{(s+1)^2-1}}\right\}\\
    &=\mathcal{L}^{-1}\left\{e^{-sh_0} e^{-x_0\sqrt{(s+1)^2-1}}\right\}-\mathcal{L}^{-1}\left\{e^{-x_0\sqrt{(s+1)^2-1}}\right\}
    \end{align*}
    [/tex]
    Now using the exponential shifting theorem:
    [tex]\mathcal{L}^{-1}\left\{f(s-a)\right\}=e^{at}\mathcal{L}^{-1}\left\{f(s)\right\}[/tex]
    and Hunt's suggestion about differentiation under the integral sign,
    [tex]\mathcal{L}^{-1}\left\{e^{-x_0\sqrt{s^2-1}}\right\}=\frac{x_0 I_1(\sqrt{x^2-x_0^2})}{\sqrt{x^2-x_0^2}}[/tex]
    and finally the Convolution theorem I obtain:
    [tex]\mathcal{L}^{-1}\left\{(1-e^{s h_0})e^{-sh_0-x_0\sqrt{2s+s^2}}\right\}=\int_0^t \frac{e^{-\beta}x_0 I_1(\sqrt{\beta^2-x_0^2})}{\sqrt{\beta^2-x_0^2}}\delta(t-\beta-h_0)d\beta-\frac{e^{-t}x_0 I_1(\sqrt{t^2-x_0^2})}{\sqrt{t^2-x_0^2}}
    [/tex]

    However I'm not confident of that answer until I could check it with some real data. May follow up on that as well.
     
  10. Sep 16, 2011 #9
    Using properties of the delta-function, this can be written as:

    [tex]
    \begin{align*}
    \mathcal{L}^{-1}\left\{(1-e^{s h_0})e^{-sh_0-x_0\sqrt{2s+s^2}}\right\}&=\int_0^t \frac{e^{-\beta}x_0 I_1(\sqrt{\beta^2-x_0^2})}{\sqrt{\beta^2-x_0^2}}\delta(t-\beta-h_0)d\beta-\frac{e^{-t}x_0 I_1(\sqrt{t^2-x_0^2})}{\sqrt{t^2-x_0^2}}\\
    &=\frac{e^{-(t-h_0)}x_0 I_1(\sqrt{(t-h_0)^2-x_0^2})}{\sqrt{(t-h_0)^2-x_0^2}}-\frac{e^{-t}x_0 I_1(\sqrt{t^2-x_0^2})}{\sqrt{t^2-x_0^2}}
    \end{align*}
    [/tex]

    and in this particular case, direct integration over the Bromwich path cannot be used to check these results because of oscillation problems however, we can equate the Bromwich integral to an integral over a dumb-bell contour around the branch points and integrate that path numerically:

    Code (Text):

    xval = 3;
    Subscript[x, 0] = 1.2;
    Subscript[h, 0] = 2;

    NIntegrate[(1/(2*Pi*I))*((1 - Exp[y*Subscript[h, 0]])*Exp[(-y)*Subscript[h, 0] - Subscript[x, 0]*Sqrt[(y + 1)^2 - 1]] -
        (1 - Exp[y*Subscript[h, 0]])*Exp[(-y)*Subscript[h, 0] + Subscript[x, 0]*Sqrt[(y + 1)^2 - 1]])*Exp[y*xval], {y, 0, -2}]

    g[x_, x0_, h0_] := N[(Exp[-(x - h0)]*x0*BesselI[1, Sqrt[(x - h0)^2 - x0^2]])/Sqrt[(x - h0)^2 - x0^2]] -
        N[(Exp[-x]*x0*BesselI[1, Sqrt[x^2 - x0^2]])/Sqrt[x^2 - x0^2]];

    g[xval, Subscript[x, 0], Subscript[h, 0]]

    Out[17]=
    0.1402728521842268 + 0.*I

    Out[19]=
    0.14027285270349493 + 0.*I
     
    Because of these agreements, I now have high confidence this is the correct inverse transform.
     
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