kostoglotov said:
Homework Statement
I have the second order diff eq:
Solving by Laplace transform gets me to:
I could use the inverse laplace transform that takes me back to e^{at}cos(bt) 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?
Homework Equations
The Attempt at a Solution
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##.
