2Partial Differential Equations
II Integrable Systems
2.2 Sine–Gordon equation
We next look at another equation that again has soliton solutions, known as the
sine–Gordon equation.
Definition (Sine–Gordon equation). The sine–Gordon equation is given by
u
tt
− u
xx
+ sin u = 0.
This is known as the sine–Gordon equation, because there is a famous
equation in physics known as the Klein–Gordon equation, given by
u
tt
− u
xx
+ u = 0.
Since we have a sine instead of a u, we call it a sine-Gordon equation!
There are a few ways we can motive the sine-Gordon equation. We will use
one from physics. Suppose we have a chain of pendulums of length
`
with masses
m:
m m m m m m
∆x
`
The pendulum will be allowed to rotate about the vertical plane, i.e. the plane
with normal along the horizontal line, and we specify the angle by
θ
i
(
t
). Since
we want to eventually take the limit as ∆
x →
0, we imagine
θ
is a function of
both space and time, and write this as θ
i
(t) = θ(i∆x, t).
Since gravity exists, each pendulum has a torque of
−m`g sin θ
i
.
We now introduce an interaction between the different pendulum. We imagine
the masses are connected by some springs, so that the
i
th pendulum gets a
torque of
K(θ
i+1
− θ
i
)
∆x
,
K(θ
i−1
− θ
i
)
∆x
.
By Newton’s laws, the equations of motion is
m`
2
d
2
θ
i
dt
2
= −mg` sin θ
i
+
K(θ
i+1
− 2θ
i
+ θ
i−1
)
∆x
.
We divide everything by ∆
x
, and take the limit as ∆
x →
0, with
M
=
m
∆x
held
constant. We then end up with
M`
2
∂
2
θ
∂t
2
= −Mg` sin θ + K
∂
2
θ
∂x
2
.
Making some simple coordinate scaling, this becomes
u
tt
− u
xx
+ sin u = 0.
There is also another motivation for this from differential geometry. It turns out
solutions to the sine-Gordon equation correspond to pseudospherical surfaces in
R
3
, namely the surfaces that have constant negative curvature.
If we pick so-called “light cone coordinates”
ξ
=
1
2
(
x − t
) and
τ
=
1
2
(
x
+
t
),
then the sine-Gordon equations become
∂
2
u
∂ξ∂τ
= sin u,
and often this is the form of the sine-Gordon equations we will encounter.
This also admits soliton solutions
u(x, t) = 4 tan
−1
exp
x − vt
√
1 − v
2
.
We can check that this is indeed a solution for this non-linear PDE.
This solution looks like
2π
Now remember that
θ
was an angle. So 2
π
is just the same as 0! If we think of
the value of
u
as living in the circle
S
1
, then this satisfies the boundary condition
u → 0 as x → ±∞:
If we view it this way, it is absolutely obvious that no matter how this solution
evolves in time, it will never become, or even approach the “trivial” solution
u = 0, even though both satisfy the boundary condition u → 0 as x → ±∞.