I recalled our little table of values of the LT, ending with
L(t^{a-1);s) = Gamma(a)/s^a, a positive.
Gamma(a) = integral_0^infinity u^{a-1} e^{-u} du
enjoys some neat properties. On Friday I noted that Gamma(n) = (n-1)! for n = 0,1,2,.... In particular Gamma(1) = 1. I remark without proof that Gamma(a+1) = a Gamma(a) and that Gamma(1/2) = sqrt(pi).
I also recalled the properties we've obtained so far:
0. L(f) is unchanged if we alter any single value of f.
1. L is linear: L(f+g) = L(f) + L(g), L(cf) = cL(f).
2. L(f) essentially determines f: if f is piecewise continuous and L(f) = 0, then f(a) = 0 wherever f is continuous.
3. L preserves real and imaginary parts.
4. s-derivative rule: -(d/ds)L(f(t);s) = L(tf(t);s).
Shifting rules: Here is some important notation: if f(t) is a function on [0,infinity), and a is a nonnegative number, then
f_a(t) = f(t-a) if t is bigger than or equal to a.
= 0 if t is less than a.
f_a is f shifted to the right by a units, with the gap filled in by 0. This is consistent with the u_a notation we already have. Compute:
L(f_a(t);s) = integral_0^infinity f_a(t) e^{-st} dt
= integral_a^infinity f(t-a) e^{-st} dt
= integral_0^infinity f(u) e^{-s(a+u)} du using u = t-a
= e^{-as} L(f(t);s):
5. t-shift rule: L(f_a(t);s) = e^{-as} L(f(t);s).
Application: L(u_a(t);s) = e^{-at}/s.
What about the s-shift? Compute:
L(f(t);s-a) = integral_0^infinity f(t) e^{-(s-a)t} dt = integral_0^infinity e^{at}f(t) e^{-st} dt = L(e^{at}f(t);s):
6. s-shift rule: L(f(t);s-a) = L(e^{at}f(t);s).
Application: L(e^{kt}t^{a-1};s) = Gamma(a)/(s-k)^a.
Of paramount importance to us will be the Laplace transform of a derivative:
L(f'(t);s) = integral_0^infinity e^{-st} f'(t) dt integrate by parts:
= e^{-st}f(t)|_0^infinity - integral_0^infinity (-s e^{-st}) f(t) dt
= -f(0) + s L(f(t);s):
7. t-derivative rule: L(f'(t);s) = s L(f(t);s) - f(0).
All the applications of the LT to ODEs depend upon this rule. We can apply it to f'(t):
L(f''(t);s) = s L(f'(t);s) - f'(0) = s^2 L(f(t);s) - s f(0) - f'(0).
and so on:
L(f^{(n)}(t);s) = s^n L(f(t);s) - s^{(n-1)} f(0) - ... - f^{(n-1)}(0).
Example: we have an undamped oscillator controlled by x'' + x = f(t) and initial conditions x(0) and x'(0) given.
Step 1: Apply L to the equation:
s^2 X(s) - s x(0) - x'(0) + X(s) = F(s)
Step 2: Solve for X(s):
X(s) = (F(s) + s x(0) + x'(0))/(s^2 + 1)
(NB: the denominator is the characteristic polynomial of the differential operator, evaluated at s.)
Step 3: Find the function x with this Laplace transform. This is the hard step. Like integration, it is more of an art than a science. Among the basic tools is the method of partial fractions.
At this point let's take f(t) = u_0 - u_pi : so the pendulum is being pushed steadily for pi time units and then allowed to swing freely. Also, suppose the system is at rest at t = 0 : x(0) = x'(0) = 0. Then
F(s) = (1 - e^{- pi s})/s, X(s) = (1 - e^{- pi s}) / s(s^2 + 1).
Deal separately with the two terms in the numerator. Partial fractions enjoins us to write
G(s) = 1/s(s^2 + 1) = a/s + (bs + c)/(s^2 + 1) .
One way to find a, b, c, is to cross multiply and equate coefficients. Here is another way, called the "coverup method" by its friends. To find a , multiply through by s and then set s = 0:
1/(0^2 + 1) = a + 0 : a = 1.
This works very simply for linear factors in the denominator. Now that we know about complex numbers we can make it work for quadratic factors too: to find b and c, multiply through by a factor of s^2 + 1 and then set that factor to zero. Let's use s - i; s^2 + 1 = (s-i)(s+i):
1/s(s+i) = a(s-i)/s + (bs+c)/(s+i)
Now set s = i: 1/i(2i) = 0 + (bi+c)/2i
so - i = bi + c, and hence b = -1, c = 0. We've found
G(s) = 1/s(s^2 + 1) = 1/s - s/(s^2 + 1)
and it's easy to check this work. It's also easy to see what function has this as its Laplace transform: g(t) = 1 - cos(t)
The other term is e^{-pi s}G(s), which by the t-shift rule is the Laplace transform of g_pi(t).
g_pi(t) = 1 - cos(t - pi) = 1 + cos(t) if t is at least pi, = 0 if t < pi.
x(t) = g(t) - g_pi(t),
and it's easy to graph the result. This same result could also be obtained pretty easily by undetermined coefficients; first you find the solution to x'' + x = 1 with initial conditions x(0) = x'(0) = 0. You compute that x(pi) = 2 and x'(pi) = 0, and then use these as initial conditions for the equation x'' + x = 0 that holds for the rest of time.