4—Differential Equations 96What isc(x)^2 +s(x)^2? Differentiate this expression to getd
dx[c(x)^2 +s(x)^2 ] = 2c(x)c′(x) + 2s(x)s′(x) =− 2 c(x)s(x) + 2s(x)c(x) = 0This combination is therefore a constant. What constant? Just evaluate it atx= 0and you see that it is one.
There are many more such results that you can derive, but that’s left for the exercises.
4.5 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 I can figure
out the result of an impact by one molecule, I 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.
m ̈x+kx=Fext(t)Suppose that everything is at rest 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′)The amplitude will depend on the strength of the kick. A constant forceFapplied 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 won’t
have had a chance to move very far before the force is turned off, so from then on it’s subject only to the−kx
force. This kick givesma velocityF∆t′/m, and that’s what determines the unknown constantA.
Just aftert=t′,vx=Aω 0 =F∆t′/m, so the position ofmis
x(t) ={F∆t′
mω 0 sin(
ω 0 (t−t′))
(t > t′)
0 (t≤t′)