The convolution theorem is very useful, for example to solve certain types of integral equa
tions.
f
(
x
) =
g
(
x
) +
λ
ˆ
∞
∞
d
x
0
h
(
x

x
0
)
f
(
x
0
)
(26a)
F
(
k
) =
G
(
k
) +
√
2
πλH
(
k
)
F
(
k
)
.
(26b)
Assuming that
g
and
h
are known, and we can calculate
G
and
H
.
F
(
k
) =
G
(
k
)
1

√
2
πλH
(
k
)
(27a)
f
(
x
) =
1
√
2
π
ˆ
∞
∞
d
k
G
(
k
)
1

√
2
πλH
(
k
)
e
ikx
.
(27b)
7
Wave equation on infinite string
∂
2
u
(
x, t
)
∂x
2

1
c
2
∂
2
u
(
x, t
)
∂t
2
= 0
.
(28)
Inspired by the complex exponential form of the Fourier series, we can combine our previously
found trigonometric solutions into complex exponential solutions
u
(
x, t
) =
e
ik
(
x
±
ct
)
. An infinite
string only has initial conditions, no boundary conditions, and thus no string lengthscale to
select certain normal modes, so any
k
is valid:
u
(
x, t
) =
1
√
2
π
ˆ
∞
∞
d
k
[
A
(
k
)
e
ik
(
x

ct
)
+
B
(
k
)
e
ik
(
x
+
ct
)
]
.
(29)
We scaled
A
(
k
)
, B
(
k
) are arbitrary (so far), so we can pull out a factor of 1
/
√
2
π
, no problem.
So the initial conditions are
φ
(
x
) =
u
(
x,
0)
(30a)
=
1
√
2
π
ˆ
∞
∞
d
k
[
A
(
k
) +
B
(
k
)]
e
ikx
(30b)
ψ
(
x
) =
∂u
∂t
t
=0
(30c)
=

ic
√
2
π
ˆ
∞
∞
d
k k
[
A
(
k
)

B
(
k
)]
e
ikx
.
(30d)
We can invert the inverse Fourier transforms above to find relations between the Fourier coef
6
ficients and the initial conditions.
A
(
k
) +
B
(
k
) =
1
√
2
π
ˆ
∞
∞
d
x φ
(
x
)
e

ikx
(31a)
= Φ(
k
)
(31b)
A
(
k
)

B
(
k
) =
i
√
2
πck
ˆ
∞
∞
d
x ψ
(
x
)
e

ikx
(31c)
=
i
ck
Ψ(
k
)
(31d)
A
(
k
) =
1
2
Φ(
k
) +
i
ck
Ψ(
k
)
(31e)
B
(
k
) =
1
2
Φ(
k
)

i
ck
Ψ(
k
)
.
(31f)
A
(
k
)
, B
(
k
) determine everything.
8
Summary
•
FT of Lorentzian:
F
1
x
2
+
a
2
=
r
π
2
e

k

a
a
.
(32)
•
Convolution theorem:
F
ˆ
∞
∞
d
y f
(
x

y
)
g
(
y
)
=
√
2
πF
(
k
)
G
(
k
)
.
(33)
•
FT solution of infinite string: coefficients are functions of FTs of ICs.
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 Fall '18
 Theoretical Physics, dx, dk