Results 1 to 5 of 5

Math Help - inverse Laplace transform by contour integration

  1. #1
    Member
    Joined
    Apr 2008
    Posts
    204

    inverse Laplace transform by contour integration

    use contour integration to find the inverse Laplace transform of
    \frac{1}{s^3 + a^3}

    i thought i should first split the fraction, so I got
    \frac{1}{s^3 + a^3} = \frac{1}{(s+a)(s^2 - as + a^2)}
    but i tried to use partial fractions to split this fraction and it didn't work!
    the inverse Laplace transform formula is
    f(x) = \frac{1}{2 \pi i}\int^{\gamma + i\infty}_{\gamma - i\infty} \frac{1}{s^3 + a^3}e^{sx} ds
    i'm not sure where to go from here!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,677
    Thanks
    1499

    Re: inverse Laplace transform by contour integration

    Quote Originally Posted by wik_chick88 View Post
    use contour integration to find the inverse Laplace transform of
    \frac{1}{s^3 + a^3}

    i thought i should first split the fraction, so I got
    \frac{1}{s^3 + a^3} = \frac{1}{(s+a)(s^2 - as + a^2)}
    but i tried to use partial fractions to split this fraction and it didn't work!
    the inverse Laplace transform formula is
    f(x) = \frac{1}{2 \pi i}\int^{\gamma + i\infty}_{\gamma - i\infty} \frac{1}{s^3 + a^3}e^{sx} ds
    i'm not sure where to go from here!
    Why should you have to use Contour Integration? Partial Fraction should have put this into a form where you could recognise the Inverse Laplace Transform, or find it on your table...

    What did you do with your Partial Fractions?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Apr 2008
    Posts
    204

    Re: inverse Laplace transform by contour integration

    well when i did partial fractions i got
    \frac{1}{(s + a)(s^2 - as + a^2)} = \frac{A}{s + a} + \frac{Bs + C}{s^2 - as + a^2}
    equating the coefficients i got
    0 = A + B
    A = -B

    0 = -Aa + Ba + C
    0 = -2Aa + C
    2Aa = C
    Aa = \frac{C}{2}

    1 = Aa^2 + Ca
    1 = \frac{Ca}{2} + \frac{2Ca}{2}
    1 = \frac{3Ca}{2}
    C = \frac{2}{3a}

    so Aa = \frac{C}{2} = \frac{2}{6a} = \frac{1}{3a}
    A = \frac{1}{3a^2}

    so B = -A = -\frac{1}{3a^2}

    so \frac{1}{(s + a)(s^2 - as + a^2)} = \frac{1}{3a^2(s+a)} + \frac{-\frac{1}{3a^2}s+\frac{2}{3a}}{s^2 - as + a^2}

    is this right so far? from here it just seems to get very complicated and messy, and i am still left with the s^2!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,677
    Thanks
    1499

    Re: inverse Laplace transform by contour integration

    Quote Originally Posted by wik_chick88 View Post
    well when i did partial fractions i got
    \frac{1}{(s + a)(s^2 - as + a^2)} = \frac{A}{s + a} + \frac{Bs + C}{s^2 - as + a^2}
    equating the coefficients i got
    0 = A + B
    A = -B

    0 = -Aa + Ba + C
    0 = -2Aa + C
    2Aa = C
    Aa = \frac{C}{2}

    1 = Aa^2 + Ca
    1 = \frac{Ca}{2} + \frac{2Ca}{2}
    1 = \frac{3Ca}{2}
    C = \frac{2}{3a}

    so Aa = \frac{C}{2} = \frac{2}{6a} = \frac{1}{3a}
    A = \frac{1}{3a^2}

    so B = -A = -\frac{1}{3a^2}

    so \frac{1}{(s + a)(s^2 - as + a^2)} = \frac{1}{3a^2(s+a)} + \frac{-\frac{1}{3a^2}s+\frac{2}{3a}}{s^2 - as + a^2}

    is this right so far? from here it just seems to get very complicated and messy, and i am still left with the s^2!
    \displaystyle \begin{align*} \frac{1}{(s + a)(s^2 - as + a^2)} &= \frac{A}{s + a} + \frac{Bs + C}{s^2 - as + a^2} \\ \frac{1}{(s +a)(s^2 - as + a^2)} &= \frac{A(s^2 - as + a^2) + (Bs + C)(s + a)}{(s + a)(s^2 - as + a^2)} \\ 1 &= A(s^2 - as + a^2) + (Bs + C)(s + a) \\ 1 &= As^2 - Aas + Aa^2 + Bs^2 + Bas + Cs + Ca \\ 0s^2 + 0s + 1 &= (A + B)s^2 + (Ba - Aa + C)s + Aa^2 + Ca  \end{align*}

    So \displaystyle A + B = 0, Ba - Aa + C = 0, Aa^2 + Ca = 1

    From the first equation, we get \displaystyle A = -B, and substituting into the other two equations gives

    \displaystyle Ba + Ba + C = 0, -Ba^2 + Ca = 1

    \displaystyle 2Ba + C = 0, (-Ba + C)a = 1

    \displaystyle 3Ba - Ba + C = 0, -Ba + C = \frac{1}{a}

    Substituting the second of these into the first gives

    \displaystyle \begin{align*} 3Ba + \frac{1}{a} &= 0 \\ 3Ba &= -\frac{1}{a} \\ B &= -\frac{1}{3a^2} \end{align*}

    Substituting into \displaystyle -Ba + C = \frac{1}{a} gives

    \displaystyle \begin{align*} -\left(-\frac{1}{3a^2}\right)a + C &= \frac{1}{a} \\ \frac{1}{3a} + C &= \frac{1}{a} \\ C &= \frac{1}{a} - \frac{1}{3a} \\ C &= \frac{2}{3a} \end{align*}

    And substituting into \displaystyle A + B = 0 we find

    \displaystyle \begin{align*} A - \frac{1}{3a^2} &= 0 \\ A &= \frac{1}{3a^2} \end{align*}


    Therefore

    \displaystyle \frac{1}{(s + a)(s^2 - as + a^2)} = \frac{\frac{1}{3a^2}}{s + a} + \frac{-\frac{1}{3a^2}s + \frac{2}{3a}}{s^2 - as + a^2} = \frac{1}{3a^2(s + a)} + \frac{2a - s}{3a^2(s^2 - as + a^2)}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5

    Re: inverse Laplace transform by contour integration

    Quote Originally Posted by wik_chick88 View Post
    use contour integration to find the inverse Laplace transform of
    \frac{1}{s^3 + a^3}

    i thought i should first split the fraction, so I got
    \frac{1}{s^3 + a^3} = \frac{1}{(s+a)(s^2 - as + a^2)}
    but i tried to use partial fractions to split this fraction and it didn't work!
    the inverse Laplace transform formula is
    f(x) = \frac{1}{2 \pi i}\int^{\gamma + i\infty}_{\gamma - i\infty} \frac{1}{s^3 + a^3}e^{sx} ds
    i'm not sure where to go from here!
    The procedure for finding the inverse Laplace Tranform via a contour integration was found by the Britisch mathematician Thomas John l'Anson Bromwich about a centiry ago. The details are in...

    The Complex inversion formula. Bromwich contour.

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: November 7th 2011, 02:04 AM
  2. Inverse Laplace Transform
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: May 23rd 2011, 10:57 AM
  3. Inverse Laplace Transform
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 25th 2010, 10:14 PM
  4. Inverse Laplace Transform
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: June 9th 2009, 03:48 AM
  5. Replies: 0
    Last Post: May 5th 2009, 12:30 PM

Search Tags


/mathhelpforum @mathhelpforum