17—Densities and Distributions 414Green’s functions
In the discussion of the Green’s function solution to a differential equation in section4.6, I started with
the differential equation
mx ̈+kx=F(t)
and found a general solution in Eq. (4.34). This approach pictured the external force as a series of
small impulses and added the results from each.
x
t′
impulset
x(t) =
∫∞
−∞dt′G(t−t′)F(t′) where G(t) =
{ 1
mω 0 sinω^0 t (t≥^0 )
0 (t < 0 )
(17.15)
I wrote it a little differently there, but it’s the same. Can you verify that this is a solution to the stated
differential equation? Simply plug in, do a couple of derivatives and see what happens. One derivative
at a time:
dx
dt
=
∫∞
−∞dt′G ̇(t−t′)F(t′) where G ̇(t) =
{ 1
mcosω^0 t (t≥^0 )
0 (t < 0 )
(17.16)
Now for the second derivative.Oops.I can’t differentiateG ̇. It has a step att= 0.
This looks like something you saw in a few paragraphs back, where there was a step in thefunctionθ. I’m going to handle this difficulty now by something of a kludge. In the next section you’ll
see the notation that makes this manipulation easy and transparent. For now I will subtract and add
the discontinuity fromG ̇ by using the same step functionθ.
G ̇(t) =
{ 1
m[
cosω 0 t−1 + 1
]
(t≥ 0 )
0 (t < 0 )
=
{ 1
m[
cosω 0 t− 1
]
(t≥ 0 )
0 (t < 0 )
}
+
1
m
θ(t) =G ̇ 0 (t) +
1
m
θ(t) (17.17)
The (temporary) notation here is thatG ̇ 0 is the part ofG ̇that doesn’t have the discontinuity att= 0.
That part is differentiable. The expression fordx/dtnow has two terms, one from theG ̇ 0 and one
from theθ. Do the first one:
d
dt
∫∞
−∞dt′G ̇ 0 (t−t′)F(t′) =
∫∞
−∞dt′
d
dt
G ̇ 0 (t−t′)F(t′)
andd
dt
G ̇ 0 (t) =
{ 1
m[−ω^0 sinω^0 t] (t≥^0 )
0 (t < 0 )
The original differential equation involvedm ̈x+kx. TheG ̇ 0 part of this is
m
∫∞
−∞dt′
{ 1
m[−ω^0 sinω^0 (t−t
′)] (t≥t′)
0 (t < t′)
}
F(t′)
+k
∫∞
−∞