1
Physics 384 2016 Assignment 8 SOLUTIONS
Problem I
Consider the inhomogeneous ODE
d
2
y
(
x
)
dx
2

k
2
y
(
x
) =
f
(
x
)
where the boundary conditions are that
y
(
±∞
) =
finite. Use the Fourier transform method
to show that the Green’s function
G
(
x
;
x
0
)
for this problem, which satisfies
d
2
G
(
x
;
x
0
)
dx
2

k
2
G
(
x
;
x
0
) =
δ
(
x

x
0
)
(1)
can be expressed as
G
(
x
;
x
0
) =

1
2
π
∞
ˆ
∞
e
ik
0
(
x

x
0
)
k
0
2
+
k
2
dk
0
Evaluate the integral and show that
G
(
x
;
x
0
) =

1
2
k
e

k

x

x
0

∀
x
Solution
To get started, it’s handy to use the fact that the Green’s function should depend only on
the separation
x

x
0
, since there is no preferred orgin in Eq. (1), to write the ODE as
d
2
G
(
x
)
dx
2

k
2
G
(
x
) =
δ
(
x
)
(2)
We write
G
(
x
)
as a Fourier transform
G
(
x
) =
1
2
π
∞
ˆ
∞
˜
G
(
k
0
)
e
ik
0
x
dk
0
apply the differential operator, and identify the result with the deltafunction in terms of its
own Fourier transform
d
2
dx
2

k
2
G
(
x
)
=
1
2
π
∞
ˆ
∞
(

k
0
2

k
2
)
˜
G
(
k
0
)
e
ik
0
x
dk
0
=
1
2
π
∞
ˆ
∞
e
ik
0
x
dk
0
2
Equating the integrands gives
˜
G
(
k
0
) =

1
k
0
2
+
k
2
Now it’s time to use complex analysis to explicitly evalute the F.T. for
G
(
x
)
G
(
x
) =

1
2
π
∞
ˆ
∞
1
k
0
2
+
k
2
e
ik
0
x
dk
0
The integrand has poles at
k
0
=
±
ik
As usual, to get damped exponentials at complex infinity, we close the contour above the
real line for
x >
0
, and below the real line for
x <
0
, hence
G
(
x
) =

1
2
π
×
2
πi
×
e
i
(
k
0
=
ik
)
x
2
ik
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 Winter '20
 Physics, Fourier Series