5Distributions

IB Methods 5.3 Green’s functions for initial value problems
Consider
Ly
=
f
(
t
) subject to
y
(
t
=
t
0
) = 0 and
y
0
(
t
=
t
0
) = 0. So instead of
having two boundaries, we have one boundary and restrict both the value and
the derivative at this boundary.
As before, let
y
1
(
t
)
, y
2
(
t
) be any basis of solutions to
Ly
= 0. The Green’s
function obeys
L(G) = δ(t τ ).
We can write our Green’s function as
G(t, τ) =
(
A(τ)y
1
(t) + B(τ)y
2
(t) t
0
t < τ
C(τ )y
1
(t) + D(τ)y
2
(t) t > τ.
Our initial conditions require that
y
1
(t
0
) y
2
(t
0
)
y
0
1
(t
0
) y
0
2
(t
0
)
A
B
=
0
0
However, we know that the matrix is non-singular (by definition of
y
1
, y
2
being a
basis). So we must have
A, B
= 0 (which makes sense, since
G
= 0 is obviously
a solution for t
0
t < τ).
Our continuity and jump conditions now require
0 = C(τ )t
1
(t) + D(τ)y
2
(t)
1
α(τ)
= C(τ )y
0
1
(τ) + D(τ )y
0
2
(τ).
We can solve this to get C(τ) and D(τ). Then we get
y(t) =
Z
t
0
G(t, τ)f(τ) dτ =
Z
t
t
0
G(t, τ)f(τ) dτ,
since we know that when
τ > t
, the Green’s function
G
(
t, τ
) = 0. Thus the
solution
y
(
t
) depends on the forcing term term
f
only for times
< t
. This
expresses causality!
Example.
For example, suppose we have
¨y
+
y
=
f
(
t
) with
f
(0) =
˙y
(0) = 0.
Then we have
G(t, τ) = Θ(t τ )[C(τ ) cos(t τ) + D(τ ) sin(t τ)]
for some
C
(
τ
)
, D
(
τ
). The continuity and jump conditions gives
D
(
τ
) = 1
, C
(
τ
) =
0. So we get
G(t, τ) = Θ(t τ ) sin t(τ).
So we get
y(t) =
Z
t
0
sin(t τ)f(τ) dτ.