5Symmetry methods in PDEs
II Integrable Systems
5.3 Symmetries of differential equations
So far we’ve just been talking about Lie groups in general. We now try to apply
this to differential equations. We will want to know when a one-parameter group
of transformations is a symmetry of a differential equation.
We denote a general (ordinary) differential equation by
∆[x, u, u
x
, u
xx
···] = 0.
Note that in general, ∆ can be a vector, so that we can have a system of equations.
We say
u
=
u
(
x
) is a solution to the differential equation if it satisfies the above
equation.
Suppose
g
ε
be a 1-parameter group of transformations generated by a vector
field V , and consider the new coordinates
(
˜
x, ˜u) = g
ε
(x, u).
Note that we transform both the domain
x
and the codomain
u
of the function
u(x), and we are allowed to mix them together.
We call g
ε
a Lie point symmetry of ∆ if
∆[x, u, u
x
, ···] = 0 =⇒ ∆[
˜
x,
˜
u, ˜u
˜
x
, ···] = 0
In other words, it takes solutions to solutions.
We say this Lie point symmetry is generated by V .
Example. Consider the KdV equation
∆ = u
t
+ u
xxx
− 6uu
x
= 0.
Then translation in the t direction given by
g
ε
(x, t, u) = (x, t + ε, u)
is a Lie point symmetry. This is generated by
V =
∂
∂t
.
Indeed, by the chain rule, we have
∂˜u
∂
˜
t
=
∂u
∂
˜
t
=
∂t
∂
˜
t
∂u
∂t
+
∂x
∂
˜
t
∂u
∂x
=
∂u
∂t
.
Similarly, we have
˜u
˜x
= u
x
, ˜u
˜x˜x˜x
= u
xxx
.
So if
∆[x, t, u] = 0,
then we also have
∆[˜x,
˜
t, ˜u] = ∆[x, t, u] = 0.
In other words, the vector field
V
=
∂
∂t
generates a Lie point symmetry of the
KdV equation.
Obviously Lie point symmetries give us new solutions from old ones. More
importantly, we can use it to solve equations!
Example. Consider the ODE
du
dx
= F
u
x
.
We see that there are things that look like
u/x
on both sides. So it is not too
hard to see that this admits a Lie-point symmetry
g
ε
(x, u) = (e
ε
x, e
ε
u).
This Lie point symmetry is generated by
V = x
∂
∂x
+ t
∂
∂t
.
The trick is to find coordinates (
s, t
) such that
V
(
s
) = 0 and
V
(
t
) = 1. We call
these “invariant coordinates”. Then since
V
is still a symmetry of the equation,
this suggests that
t
should not appear explicitly in the differential equation, and
this will in general make our lives easier. Of course, terms like
t
s
can still appear
because translating t by a constant does not change t
s
.
We pick
s =
u
x
, t = log |x|,
which does indeed satisfy V (s) = 0, V (t) = 1. We can invert these to get
x = e
t
, u = se
t
.
With respect to the (s, t) coordinates, the ODE becomes
dt
ds
=
1
F (s) − s
,
at least for
F
(
s
)
6
=
s
. As promised, this does not have an explicit
t
dependence.
So we can actually integrate this thing up. We can write the solution as
t = C +
Z
s
ds
0
F (s
0
) − s
0
.
Going back to the original coordinates, we know
log |x| = C
Z
u/x
ds
F (s) − s
.
If we actually had an expression for
F
and did the integral, we could potentially
restore this to get an expression of
u
in terms of
x
. So the knowledge of the Lie
point symmetry allowed us to integrate up our ODE.
In general, for an nth order ODE
∆[x, u, u
0
, ··· , u
(n)
] = 0
admitting a Lie point symmetry generated by
V = ξ(x, u)
∂
∂x
+ η(x, u)
∂
∂u
,
we introduce coordinates
s = s(u, x), t = t(u, x)
such that in the new coordinates, we have
V =
∂
∂t
.
This means that in the new coordinates, the ODE has the form
∆[s, t
0
, ··· , t
(n)
] = 0.
Note that there is no explicit
t
! We can now set
r
=
t
0
, so we get an (
n −
1)th
order ODE
∆[s, r, r
0
, ··· , r
(n−1)
],
i.e. we have reduced the order of the ODE by 1. Now rinse and repeat.