1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Step Function Laplace w/ piecewise

  1. Jul 25, 2015 #1
    1. The problem statement, all variables and given/known data

    ##y'' + 2y' + 2y = h(t); y(0)=0, y'(0)=1,
    h(t) = \begin{cases} t &, \pi \leq t < 2\pi \\
    0 &, 3\pi \leq t < \infty \\
    \end{cases}##



    2. Relevant equations




    3. The attempt at a solution

    Take the Laplace of both sides:


    ##\mathcal{L}(y'' + 2y' + 2y) = \mathcal{L}(h(t)) ##

    ##s^2 F(s) - sf(0) - f'(0) + 2sF(s) - 2f(0) + 2F(s) = \mathcal{L}(h(t)) ##

    ##(s^2 + 2s +2)F(s) - 1 = \mathcal{L}(h(t)) = \int_{\pi}^{2\pi} te^{-st}##

    ##(s^2 + 2s +2)F(s) - 1 =\frac {\pi e^{-\pi s}} {s} + \frac {e^{-\pi s}} {s^2} - \frac {2\pi e^{-2\pi s}} {s} - \frac {e^{2\pi s}} {s^2} ##

    Solve for F(s):

    This is where I always screw up, If I even dare to add the 1 to the other side and divide by ##(s^2 + 2s + 2)## I get a ridiculously long expansion, that I don't even think is plausible for a practice problem, or any other problem.

    I have no idea where to continue once I get to this point.
     
  2. jcsd
  3. Jul 25, 2015 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    It is actually not so bad, once you simplify things a bit. Start with
    [tex] F(s) = \frac{e^{-\pi s}}{D_2(s)} + \frac{\pi e^{-\pi s}}{D_1(s)} - \frac{e^{-2 \pi s}}{D_2(s)} - \frac{2 \pi e^{-2 \pi s}}{D_1(s)} + \frac{1}{D_0(s)},[/tex]
    where ##D_2(s) = s^2 (s^2 + 2 s + 2), \: D_1(s) = s (s^2 + 2 s + 2), \; D_0(s) = s^2 + 2s + 2##.

    It is convenient to first take the inverse Laplace transforms of the ##1/D_k(s)## functions:
    [tex] \frac{1}{D_0(s)} \leftrightarrow e^{-t} \sin(t) \equiv e_0(t) \\
    \frac{1}{D_1(s)} = \frac{1}{s} \frac{1}{D_0(s)} \leftrightarrow \int_0^t e_0(\tau) \, d \tau = \frac{1}{2} - \frac{1}{2} e^{-t} (\sin(t) + \cos(t)) \equiv e_1(t) \\
    \frac{1}{D_2(s)} = \frac{1}{s} \frac{1}{D_1(s)} \leftrightarrow \int_0^t e_1 (\tau) \, d \tau = \frac{1}{2}(e^{-t} \cos(t) + t-1) \equiv e_2(t) [/tex]

    Now we have
    [tex] F(t) = [e_2(t-\pi)+\pi e_1(t-\pi)]u(t-\pi) -[e_2(t-2 \pi) + 2 \pi e_1(t - 2 \pi)]u(t - 2 \pi) + e_0(t) [/tex]

    These are obtained using the following two standard properties, wherein below we have ##f(t) \leftrightarrow g(s)##:
    [tex] \begin{array}{rl}1. & \int_0^t f(\tau) d \tau \leftrightarrow \frac{1}{s} g(s) \\
    2. & f(t-a)u(t-a) \leftrightarrow e^{-as} g(s) \end{array} [/tex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted