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Laplace transform on a DE

  1. Feb 21, 2012 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    I'd like to solve a DE using Laplace transform.
    [itex]\ddot y (t)+\omega ^2 y(t)=f(t)[/itex] for all t>0.
    Initial conditions: [itex]y(0)=\dot y (0)=0[/itex]. The dot denotes the derivative with respect to t.

    2. Relevant equations

    [itex]\mathbb{L}( \dot y )=s\mathbb{L}( y )-y(0)[/itex].
    Convolution: if [itex]h=f*g[/itex] then [itex]H(s)=G(s)F(s)[/itex].
    3. The attempt at a solution
    I apply the LT on the DE: [itex]s^2 Y(s)-sy(0)-\dot y(0)+\omega ^2 Y(s)=F(s)[/itex]. Therefore [itex]Y(s)=\frac{F(s)}{s^2+\omega ^2 }[/itex]. Now I can use the convolution property: [itex]y(t)=f(t)*\mathbb{L ^{-1}} \left ( \frac{1}{s^2+\omega ^2} \right )[/itex].
    Unfortunately I do not find the inverse Laplace transform of [itex]\left ( \frac{1}{s^2+\omega ^2} \right )[/itex] in any table.
    So it means I must perform the integral [itex]\frac{1}{2\pi i }\int _{\sigma -i \infty}^{\sigma + i \infty }e^{st}\frac{1}{s^2+\omega ^2 }ds[/itex]. Any help for this is appreciated.

    Edit: Let [itex]f(z)=\frac{e^{zt}}{z^2+\omega ^2}[/itex]. I need to employ the residue theorem. The denominator can be rewritten into [itex](z-i \omega )(z+i \omega)[/itex]. So the residue at [itex]z=i\omega[/itex] is worth [itex]\lim _{z \to i \omega } \frac{e^{zt}}{z+i \omega }=\frac{e^{i \omega t }}{2i \omega}[/itex]. While the residue at [itex]z=-i \omega[/itex] is worth [itex]-\frac{e^{-i \omega t}}{2i \omega }[/itex].
     
    Last edited: Feb 21, 2012
  2. jcsd
  3. Feb 21, 2012 #2

    vela

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    It is in the table. Look for ##\frac{\omega}{s^2+\omega^2}##. Remember ω is just a constant here.
    So what do you get when you sum those and simplify?
     
  4. Feb 21, 2012 #3

    fluidistic

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    Ok thanks! I'd rather look when I'm done, just for the surprise of getting a good result.
    What I had done is [itex]\mathbb{L ^{-1}} \left ( \frac{1}{s^2+\omega ^2} \right )=\frac{1}{2\pi i }\int _{\sigma -i \infty}^{\sigma + i \infty }e^{st}\frac{1}{s^2+\omega ^2 }ds[/itex]. When performing the complex integral, I had that it's worth [itex]2\pi i \sum _i res f(z)[/itex]. So that the [itex]2 \pi i[/itex]'s terms cancel out. So that the inverse LT of [itex]\frac{1}{s^2+\omega^2}[/itex] should be just what you said, the sum of the residues. This gave me [itex]\frac{i}{2\omega } (e^{i\omega t }-e^{-i \omega t })=\frac{i \sin (\omega t)}{\omega}[/itex].

    Therefore [itex]y(t)=f(t)*\frac{i\sin (\omega t )}{\omega}=\int _0 ^t f(t-\tau )i \frac{\sin (\omega \tau)}{\omega}d\tau[/itex].
    Does this look possible? I'm going to check out the table.

    Edit: I'm wrong it seems. I should have found [itex]\frac{\sin (\omega t )}{\omega}[/itex]?
     
    Last edited: Feb 21, 2012
  5. Feb 21, 2012 #4

    vela

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    Yeah, sine has an i in the denominator, so you should have just left it down there. Other than that extra factor of i, your results look good.
     
  6. Feb 21, 2012 #5

    fluidistic

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    Thank you vela, I found out the mistake.
     
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