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Which inverse Laplace form can I use?

  1. Oct 17, 2016 #1
    1. The problem statement, all variables and given/known data

    I have the second order diff eq: CAK3UGm.gif

    Solving by Laplace transform gets me to: sUwM9Mn.gif

    I could use the inverse laplace transform that takes me back to [itex]e^{at}cos(bt)[/itex] with b=0, but that only solves for the homogeneous (complementary) part of the equation, it won't reproduce the dirac delta function.

    What should I manipulate the Y(s) into? Should I tackle it with partial fraction decomposition?

    2. Relevant equations


    3. The attempt at a solution
     
  2. jcsd
  3. Oct 17, 2016 #2

    Mark44

    Staff: Mentor

    I didn't verify your work.
    ##Y(s) =\frac{}{} = \frac{\alpha s}{(s - 4)^2} + \frac{\kappa - 8\alpha + \beta}{(s - 4)^2}##
    Take the inverse Laplace transform of each of the two parts on the right. The numerator on the right is just a constant. A table of inverse Laplace transforms should have both of these.
     
  4. Oct 18, 2016 #3

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    If you want to have ##y(x) = 0## for ##x < 0##, then the DE + initial conditions are compatible only if ##\alpha = 0## and ##\kappa = \beta##. To see this, look at the left-hand-side near ##x=0##, and interpret the derivatives in terms of generalized functions (##\delta (x)##, etc.). If ##\alpha \neq 0## we have ##y'(x) = \alpha \delta (x) + \; \text{smooth terms}## and ##y''(x) = \alpha \delta'(x) + \beta \delta (x) + \: \text{smooth terms}##, so we would need
    $$\alpha \delta'(x) +\beta \delta(x) - 8 \alpha \delta (x) = \kappa \delta(x). $$
    This fails because ##\delta## and ##\delta'## are very different as generalize functions. On the other hand, if ##y(x)## is continuous at ##x = 0##, so that ##\alpha = 0##, we have ##\beta \delta (x) = \kappa \delta (x)##, and this works when ##\kappa = \beta##.

    If ##\alpha, \beta## and ##\kappa## are three different parameters, we cannot have ##y(x) = 0## for ##x < 0##, and so we cannot solve the problem using Laplace transforms. We could try a Fourier transform instead, but by far the easiest way would be the direct approach where we apply different valuations of the homogeneous solution for ##x > 0## and for ##x < 0##, then match up the boundary conditions at ##x = 0##. By this I mean: use homogeneous solution ##y(x) = a_1\, y_1(x) + a_2 \, y_2(x)## for ##x < 0## and ##y(x) = b_1\, y_1(x) + b_2 \,y_2(x)## for ##x > 0## (with the same functions ##y_1(x)## and ##y_2(x)## in both). Assuming ##y## continuous at ##x = 0## we get ##y(0-) = \alpha## and ##y(0+) = \alpha##. Next: integrate the DE from ##x = -\epsilon## to ##x = +\epsilon## and then look at the limit as ##\epsilon \to 0##. That gives ##y'(\epsilon)- y'(-\epsilon) + O(\epsilon) = \kappa##, so the condition is ##y'(0+) - y'(0-) = \kappa##. Thus, ##y'(0+) = \beta## and ##y'(0-) = \beta - \kappa##. Those four conditions allow determination of ##a_1,a_2, b_1,b_2##.
     
    Last edited: Oct 18, 2016
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