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Laplace Transform Question

  1. Dec 31, 2013 #1
    1. The problem statement, all variables and given/known data

    Find

    $$ L^{-1} \left[ \frac{1}{ (p^2 + a^2)^2} \right] $$

    2. Relevant equations

    $$ L [ x \cos ax ] = \frac{p^2 - a^2} { (p^2 + a^2) ^2 } $$

    3. The attempt at a solution

    I have no idea. Any thoughts?
     
  2. jcsd
  3. Dec 31, 2013 #2

    LCKurtz

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    Look at this thread:

    https://www.physicsforums.com/showthread.php?t=725295
     
  4. Dec 31, 2013 #3

    berkeman

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    Thread closed temporarily for Moderation...

    @piercebeatz -- Please check your PMs. You must always show some effort before we can offer tutorial help.

    Thread is re-opened.
     
  5. Dec 31, 2013 #4
    I'm terribly sorry for that. Here's where I got (I'm supposed to use the above identity in the problem, and convolutions is in the next chapter):

    $$ L^ {-1} \left[ \frac{1}{(p^2 + a^2)^2} \right] = L^{-1} \left[ \frac{1}{p^2 - a^2} \frac{p^2 - a^2}{(p^2 + a^2)^2} \right] $$

    By the theorem

    $$ L^{-1} [ F(p) G(p) ] = \int_0^x f(x-t)g(t) \, dt $$

    (where F(p) = L(f(t)) etc.), this becomes

    $$ L^{-1} \left[ \frac{1}{p^2 - a^2} \frac{p^2 - a^2}{(p^2 + a^2)^2} \right] = \int_0^x \frac{1}{a} \sinh (a (x-t)) \cos(at) \, dt $$

    By the substitution t = iy, this becomes

    $$ i/a \int_0^{x/i} \sinh( a( x-iy)) \cosh ay \, dy $$

    $$ = i/2a \int_0^{x/i} \left[ \sinh (a (x + y - iy)) + \sinh(a( x- y - iy) ) \right] \, dy $$

    $$ = i/2a \left[ \frac{\cosh( a ( x+ y - iy)) }{a(1 - i)} + \frac{ \cosh(a (x - y -iy))}{a(-1-i)} \right]_0^{x/i} $$

    $$ = i/2a \left[ \frac{\cosh( -aix)}{a(1-i)} + \frac{\cosh(aix)}{a(-1-i)} \right] $$

    $$ = \frac{1}{a} \cos ax \cdot \frac{i}{2} \left[ \frac{1}{a(1-i)} - \frac{1}{a(1+i)} \right] $$

    $$ = \frac{1}{a} \cos ax \cdot \frac{i}{2} \left[ \frac{ a( -1 - i + 1 - i )}{-2 a^2 } \right] $$

    $$ = - \frac{ \cos ax}{2a^2} $$

    That's the only way I could think of doing it
     
    Last edited: Dec 31, 2013
  6. Dec 31, 2013 #5

    LCKurtz

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    That is the convolution theorem stated in inverse form.

    Why would you break it up that way? Why not $$
    \mathcal L^{-1}\left(\frac {1}{p^2+a^2}\cdot \frac {1}{p^2+a^2}\right)$$I haven't checked the rest of your work, but, at least according to Maple, your answer isn't correct.
     
  7. Jan 1, 2014 #6

    vanhees71

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    Why don't you use complex integration and the theorem of residues to invert the Laplace transform?
     
  8. Jan 1, 2014 #7
    I don't know that theorem. Does anyone see any error in the above calculation? I can't find one
     
  9. Jan 1, 2014 #8

    LCKurtz

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    You don't know what theorem? Please quote the message to which you are replying.
     
  10. Jan 1, 2014 #9

    Sorry, I meant the "theorem of residues"
     
  11. Jan 1, 2014 #10

    vela

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    You didn't take the inverse Laplace transform of the two pieces correctly. According to Mathematica, the inverse Laplace transform of ##\frac{p^2-a^2}{(p^2+a^2)^2}## is ##t \cos at##. You forgot the ##t## out front.

    In any case, I suggest you take LCKurtz's suggestion for how to break up the original transform to use with the convolution theorem.
     
  12. Jan 1, 2014 #11
    Right, I forgot the t. Putting it in and doing integration by parts yields the answer, which I have obtained on paper. Thank you all for your time.
     
  13. Jan 2, 2014 #12

    vanhees71

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    Sometimes, it's easier to directly use the back-transformation formula
    [tex]f(t)=\frac{1}{2 \pi \mathrm{i}} \int_{C} \mathrm{d} p \exp(p t) F(p),[/tex]
    where [itex]C[/itex] is a straight line parallel to the imaginary axis in the complex [itex]p[/itex] line.

    The integral can be evaluated with help of the theorem of residues very easily in this case. The image function is
    [tex]F(p)=\frac{p^2-a^2}{(p^2+a^2)^2}.[/tex]
    For simplicity let's assume that [itex]a>0[/itex]. Then the poles of [itex]F[/itex] are on the imaginary axis, [itex]p_{1,2}=\pm \mathrm{i} a[/itex]. For the real part of the integration path we can thus choose any positive value, and thus we can close this path by a semi-circle with infinite radius to the left [itex]p[/itex] plane (since in the back-transformation formula above we always tacitly assume [itex]t>0[/itex].

    Thus the integral is just given by the sum over the residues of the function [itex]F(p)\exp(p t)[/itex]. Since the poles are of second order, the residues are evaluated as
    [tex]\text{res}_{\pm \mathrm{i} a}[F \exp(p t)]=\lim_{p \rightarrow \mathrm{i} a} \frac{\mathrm{d}}{\mathrm{d} p}[(p \mp \mathrm{i}a)^2 F(p) \exp(p t)] = \frac{t}{2} \exp(\pm \mathrm{i} a t).[/tex]
    Thus the original function is
    [tex]f(t)=\text{res}_{\mathrm{i} a}[F \exp(p t)] + \text{res}_{-\mathrm{i} a}[F \exp(p t)]=t \cos(a t).[/tex]
     
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