We just solved x'' + x = f(t), where f(t) = u(t) - u_pi(t), x(0) = 0, x'(0) = 0, using Laplace transform. I want to point out that this is also easy to solve using undetermined coefficients. In the domain x less than pi we have x'' + x = 1; this has general solution x = 1 + a cos t + b sin t . The initial condition x(0) = 0, x'(0) = 0, force a = -1, b = 0. Then x' = sin(t), and we obtain the values x(pi) = 2, x'(pi) = 0, which we use as initial values for the ODE x'' + x = 0 valid for t greater than pi. The general solution is a cos(t) + b sin(t), and the initial conditions force a = -2, b = 0, so for t greater than pi, x(t) = -2 cos(t).
Let's return to the same system but impose a periodic "crenelation" or square wave:
f = u - u_pi + u_{2pi} - u_{3pi} + ... = sum_{n=0}^infinity (-1)^n u_{n pi}.
With rest of the initial conditions we obtain
(s^2+1) X(s) = (1/s)(1 - e^{-pi s} + e^{-2 pi x} - ...
This can be rewritten using the geometric series
sum_{n=0}^infinity x^n = 1/(1-x) , |x| < 1 ,
as
(s^2+1) X(s) = (1/s)(1/(1 + e^{-pi s}).
This is pretty but is not particularly useful for us, since we're going to have to take it apart again when we form the inverse transform.
If we write G(s) = 1/s(s^2+1) as we did before, then
X(s) = (1 - e^{-pi x} + e^{-2 pi s} - ...) G(s)
Using the t-translation rule, this is the Laplace tranform of
x(t) = g(t) - g_pi(t) + g_{2 pi}(t) - ...
where as before g(t) = 1 - cos(t). Then
g_pi(t) = 1 + cos(t) for t greater than pi
g_{2pi}(t) = 1 - cos(t) for t greater than 2 pi, ...
and so
x(t) = 1 - cos(t) for t less than pi
= -2 cos(t) for t between pi and 2 pi
= 1 -3 cos(t) for t between 2 pi and 3 pi ....
The graph of this shows the effect of resonance.
At this point we should step back and try to see what the Laplace transform is telling us. If we assume rest initial conditions, x(0) = 0 = x'(0), then the ODE
p(D) x = f(t)
has as its Laplace transform
p(s) X(s) = F(s)
or
X(s) = W(s)F(s)
where W(s) = 1/p(s), the reciprocal of the characteristic polynomial, is the "transfer function" of the system. In the s-plane, the system multiplies the (transformed) input function by the transfer function. We know very well by now that the roots of the characteristic polynomial control the behavior of solutions to the ODE. These roots correspond to places where the transfer function becomes infinite. Since the roots may be complex, we had better allow the variable s to be complex. We have the table of corresponsdences between three different chapters in this course:
linear second order equations: roots of the characteristic polynomial
linear first order systems: eigenvalues
Laplace transform: poles of the weight function
For example, suppose x'' + 2x' + 2x = 0 with initial condition x(0) = 0, x'(0) = 1. (This should obviously be done using our earlier methods, but let's use it to learn about the Laplace transform now.) Then
X(s) = (F(s) + x(0))W(s) = 1/(s^2 + 2x + 2).
To find the inverse Laplace transform of this, complete the square and use the s-shift theorem and the fact that L(sin(t);s) = 1/(s^2+1):
X(s) = 1/((s+1)^2 + 1) = :(e^{-t}sin(t);s),
so x(t) = e^{-t}sin(t).
The roots of the characteristic polynomial, and the poles of the transfer function, are -1 + i and -1 - i. In this case these coincide with the poles of X(s). The fact that the largest real part among the poles X(s) is -1 tells us that x(t) decays like e^{-t}. The fact that there are poles with imaginary parts 1, -1, tells us that x(t) exhibits oscillatory (but not periodic) behavior with angular frequency 1. These are general comments; the location of the poles of X(s) tells us a great deal about the function x(t).