4—Differential Equations 784.6 Green’s Functions
Is there a general way to find the solution to the whole harmonic oscillator inhomogeneous differential
equation? One that does not require guessing the form of the solution and applying initial conditions?
Yes there is. It’s called the method of Green’s functions. The idea behind it is that you can think of
any force as a sequence of short, small kicks. In fact, because of the atomic nature of matter, that’s
not so far from the truth. If you can figure out the result of an impact by one molecule, you can add
the results of many such kicks to get the answer for 1023 molecules.
I’ll start with the simpler case where there’s no damping, b = 0in the harmonic oscillator
equation.
mx ̈+kx=Fext(t) (4.30)
Suppose that everything is at rest at the origin and then at timet′the external force provides a small
impulse. The motion from that point on will be a sine function starting att′,
Asin
(
ω 0 (t−t′)
)
(t > t′) (4.31)
The amplitude will depend on the strength of the kick. A constant forceF applied for a very short
time,∆t′, will change the momentum of the mass bym∆vx=F∆t′. If this time interval is short
enough the mass doesn’t have a chance to move very far before the force is turned off, then from that
time on it’s subject only to the−kxforce. This kick givesma velocityF∆t′/m, and that’s what
determines the unknown constantA.
Just aftert=t′,vx=Aω 0 =F∆t′/m. This determinesA, so the position ofmis
x(t) =
{F∆t′
mω 0 sin(
ω 0 (t−t′)
)
(t > t′)
0 (t≤t′)
(4.32)
x
t′
impulset
When the external force is the sum of two terms, the total solution is the sum of the solutions
for the individual forces. If an impulse at one time gives a solution Eq. (4.32), an impulse at a later
time gives a solution that starts its motion at that later time. The key fact about the equation that
you’re trying to solve is that it is linear, so you can get the solution for two impulses simply by adding
the two simpler solutions.
m
d^2 x 1
dt^2
+kx 1 =F 1 (t) and m
d^2 x 2
dt^2
+kx 2 =F 2 (t)
then