5Distributions
IB Methods
5.2 Green’s functions
One of the main uses of the
δ
function is the Green’s function. Suppose we wish
to solve the 2nd order ordinary differential equation
Ly
=
f
, where
f
(
x
) is a
bounded forcing term, and L is a bounded differential operator, say
L = α(x)
∂
2
∂x
2
+ β(x)
∂
∂x
+ γ(x).
As a first step, we try to find the Green’s function for
L
, which we shall write as
G(x, ξ), which obeys
LG(x, ξ) = δ(x − ξ).
Given G(x, ξ), we can then define
y(x) =
Z
b
a
G(x, ξ)f(ξ) dξ.
Then we have
Ly =
Z
b
a
LG(x, ξ)f(ξ) dξ =
Z
b
a
δ(x − ξ)f(ξ) dξ = f(x).
We can formally say
y
=
L
−1
f
, and the Green’s function shows that
L
−1
means
an integral.
Hence if we can solve for Green’s function, then we have solved the differential
equation, at least in terms of an integral.
To find the Green’s function G(x, ξ) obeying the boundary conditions
G(a, ξ) = G(b, ξ) = 0,
first note that
LG
= 0 for all
x ∈
[
a, ξ
)
∪
(
ξ, b
], i.e. everywhere except
ξ
itself.
Thus we must be able to expand it in a basis of solutions in these two regions.
Suppose
{y
1
(
x
)
, y
2
(
x
)
}
are a basis of solutions to
LG
= 0 everywhere on
[a, b], with boundary conditions y
1
(a) = 0, y
2
(b) = 0. Then we must have
G(x, ξ) =
(
A(ξ)y
1
(x) a ≤ x < ξ
B(ξ)y
2
(x) ξ < x ≤ b
So we have a whole family of solutions. To fix the coefficients, we must decide
how to join these solutions together over x = ξ.
If
G
(
x, ξ
) were discontinuous at
x
=
ξ
, then
∂
x
G|
x=ξ
would involve a
δ
function, while
∂
2
x
G|
x=ξ
would involve the derivative of the
δ
function. This is
not good, since nothing in
LG
=
δ
(
x − ξ
) would balance a
δ
0
. So
G
(
x, ξ
) must
be everywhere continuous. Hence we require
A(ξ)y
1
(ξ) = B(ξ)y
2
(ξ). (∗)
Now integrate over a small region (ξ − ε, ξ + ε) surrounding ξ. Then we have
Z
ξ+ε
ξ−ε
α(x)
d
2
G
dx
2
+ β(x)
dG
dx
+ γ(x)G
dx =
Z
ξ+ε
ξ−ε
δ(x − ξ) dx = 1.
By continuity of
G
, we know that the
γG
term does not contribute. While
G
0
is
discontinuous, it is still finite. So the
βG
0
term also does not contribute. So we
have
lim
ε→0
Z
ξ+ε
ξ−ε
αG
00
dx = 1.
We now integrate by parts to obtain
lim
ε→0
[αG
0
]
ξ+ε
ξ−ε
+
Z
ξ+ε
ξ−ε
α
0
G
0
dx = 1.
Again, by finiteness of G
0
, the integral does not contribute. So we know that
α(ξ)
∂G
∂x
ξ
+
−
∂G
∂x
ξ
−
!
= 1
Hence we obtain
B(ξ)y
0
2
(ξ) − A(ξ)y
0
1
(ξ) =
1
α(ξ)
.
Together with (∗), we know that
A(ξ) =
y
2
(ξ)
α(ξ)W (ξ)
, B(ξ) =
y
1
(ξ)
α(ξ)W (ξ)
,
where W is the Wronskian
W = y
1
y
0
2
− y
2
y
0
1
.
Hence, we know that
G(x, ξ) =
1
α(ξ)W (ξ)
(
y
2
(ξ)y
1
(x) a ≤ x ≤ ξ
y
1
(ξ)y
2
(x) ξ < x ≤ b.
Using the step function Θ, we can write this as
G(x, ξ) =
1
α(ξ)W (ξ)
[Θ(ξ − x)y
2
(ξ)y
1
(x) + Θ(x − ξ)y
1
(ξ)y
2
(x)].
So our general solution is
y(x) =
Z
b
a
G(x, ξ)f(ξ) dξ
=
Z
b
x
f(ξ)
α(ξ)W (ξ)
y
2
(ξ)y
1
(x) dξ +
Z
x
a
f(ξ)
α(ξ)W (ξ)
y
1
(ξ)y
2
(x) dξ.
Example. Consider
Ly = −y
00
− y = f
for
x ∈
(0
,
1) with
y
(0) =
y
(1) = 0. We choose our basis solution to satisfy
y
00
= −y as {sin x, sin(1 − x)}. Then we can compute the Wronskian
W (x) = −sin x cos(1 − x) − sin(1 − x) cos x = −sin 1.
Our Green’s function is
G(x, ξ) =
1
sin 1
[Θ(ξ − x) sin(1 − ξ) sin x + Θ(x − ξ) sin ξ sin(1 −x)].
Hence we get
y(x) = sin x
Z
1
x
sin(1 − ξ)
sin 1
f(ξ) dξ + sin(1 − x)
Z
x
0
sin ξ
sin 1
f(ξ) dξ.
Example.
Suppose we have a string with mass per unit length
µ
(
x
) and held
under a tension T . In the presence of gravity, Newton’s second law gives
µ
∂
2
y
∂t
2
= T
∂
2
y
∂x
2
+ µg.
We look for the steady state solution
˙y
= 0 shape of the string, assuming
y(0, t) = y(L, t) = 0. So we get
∂
2
y
∂x
2
= −
µ(x)
T
g.
We look for a Green’s function obeying
∂
2
y
∂x
2
= −δ(x − ξ).
This can be interpreted as the contribution of a pointlike mass located at
x
=
ξ
.
0 L
ξ
y
The homogeneous equation y
00
= 0 gives
y = Ax + B(x − L).
So we get
G(x, ξ) =
(
A(ξx) 0 ≤ x < ξ
B(ξ)(x − L) ξ < x ≤ L.
Continuity at x = ξ gives
A(ξ)ξ = B(ξ)(ξ − L).
The jump condition on the derivative gives
A(ξ) − B(ξ) = 1.
We can solve these to get
A(ξ) =
ξ − L
L
, B(ξ) =
ξ
L
.
Hence the Green’s function is
G(x, ξ) =
ξ − L
L
Θ(ξ − x) +
ξ
L
(x − L)Θ(x − ξ).
Notice that
ξ
is always less that
L
. So the first term has a negative slope; while
ξ is always positive, so the second term has positive slope.
For the general case, we can just substitute this into the integral. We can
give this integral a physical interpretation. We can think that we have many
pointlike particles m
i
at small separations ∆x along the string. So we get
g(x) =
n
X
i=1
G(x, x
i
)
m
i
g
T
,
and in the limit ∆x → 0 and n → ∞, we get
g(x) →
g
T
Z
L
0
G(x, ξ)µ(ξ) dξ.