5Symmetry methods in PDEs
II Integrable Systems
5.5 Painlev´e test and integrability
We end with a section on the Painlev´e test. If someone just gave us a PDE,
how can we figure out if it is integrable? It turns out there are some necessary
conditions for integrability we can check.
Recall the following definition.
Definition
(Singularirty)
.
A singularity of a complex-valued function
w
=
w
(
z
)
is a place at which it loses analyticity.
These can be poles, branch points, essential singularities etc.
Suppose we had an ODE of the form
d
2
w
dz
2
+ p(z)
dw
dz
+ q(z)w = 0,
and we want to know if the solutions have singularities. It turns out that
any singularity of a solution
w
=
w
(
z
) must be inherited from the functions
p
(
z
)
, q
(
z
). In particular, the locations of the singularities will not depend on
initial conditions w(z
0
), w
0
(z
0
).
This is not the case for non-linear ODE’s. For example, the equation
dw
dz
+ w
2
= 0
gives us
w(z) =
1
z − z
0
.
The location of this singularity changes, and it depends on the initial condition.
We say it is movable.
This leads to the following definition:
Definition (Painlev´e property). We will say that an ODE of the form
d
n
w
dz
n
= F
d
n−1
w
dz
n−1
, ··· , w, z
has the Painlev´e property if the movable singularities of its solutions are at worst
poles.
Example. We have
dw
dz
+ w
2
= 0.
has a solution
w(z) =
1
z − z
0
.
Since this movable singularity is a pole, this has the Painleve´e property.
Example. Consider the equation
dw
dz
+ w
3
= 0.
Then the solution is
w(z) =
1
p
2(z − z
0
)
,
whose singularity is not a pole.
In the olden days, Painlev´e wanted to classify all ODE’s of the form
d
2
w
dz
2
= F
dw
dz
, w, z
,
where F is a rational function, that had the Painlev´e property.
He managed to show that there are fifty such equations (up to simple
coordinate transformations). The interesting thing is that 44 of these can be
solved in terms of well-known functions, e.g. Jacobi elliptic functions, Weierstrass
℘ functions, Bessel functions etc.
The remaining six gave way to solutions that were genuinely new functions,
called the six Painlev´e transcendents. The six differential equations are
(PI)
d
2
w
dz
2
= 6w
2
+ z
(PII)
d
2
w
dz
2
= 2x
3
+ zw + α
(PIII)
d
2
w
dz
2
=
1
w
dw
dz
2
+
1
z
−
dw
dz
+ αw
2
+ β
+ γw
3
+
δ
w
(PIV)
d
2
w
dz
2
=
1
2w
dw
dz
2
+
3w
3
2
+ 4zw
2
+ 2(z
2
− α)w +
β
w
(PV)
d
2
w
dz
2
=
1
2w
+
1
w −1
dw
dz
2
−
1
z
dw
dz
+
(w −1)
2
z
2
αw +
β
w
+
γw
z
+
δw(w + 1)
w −1
(PVI)
d
2
w
dz
2
=
1
2
1
w
+
1
w −1
+
1
w −z
dw
dz
2
−
1
z
+
1
z − 1
+
1
w −z
dw
dz
+
w(w − 1)(w − z)
z
2
(z − 1)
2
α +
βz
w
2
+
γ(z − 1)
(w −1)
2
+
δz(z − 1)
(w −z)
2
.
Fun fact: Painlev´e served as the prime minister of France twice, for 9 weeks and
7 months respectively.
This is all good, but what has this got to do with integrability of PDE’s?
Conjecture
(Ablowitz-Ramani-Segur conjecture (1980))
.
Every ODE reduction
(explained later) of an integrable PDE has the Painlev´e property.
This is still a conjecture since, as we’ve previously mentioned, we don’t
really have a definition of integrability. However, we have proved this conjecture
for certain special cases, where we have managed to pin down some specific
definitions.
What do we mean by ODE reduction? Vaguely speaking, if we have a Lie
point symmetry of a PDE, then we can use it to introduce coordinates that are
invariant and then form subsequence ODE’s in these coordinates. We can look
at some concrete examples:
Example.
In the wave equation, we can try a solution of the form
u
(
x, t
) =
f
(
x − ct
), and then the wave equation gives us an ODE (or lack of) in terms of
f.
Example. Consider the sine–Gordon equation in light cone coordinates
u
xt
= sin u.
This equation admits a Lie-point symmetry
g
ε
: (x, t, u) 7→ (e
ε
x, e
−ε
t, u),
which is generated by
V = x
∂
∂x
− t
∂
∂t
.
We should now introduce a variable invariant under this Lie-point symmetry.
Clearly z = xt is invariant, since
V (z) = xt − tx = 0.
What we should do, then is to look for a solution that depends on z, say
u(x, t) = F (z).
Setting
w = e
iF
,
the sine–Gordon equation becomes
d
2
w
dz
2
=
1
w
dw
dz
2
−
1
z
dw
dz
−
1
2z
.
This is equivalent to PIII, i.e. this ODE reduction has the Painlev´e property.
Example. Consider the KdV equation
u
t
+ u
xxx
− 6uu
x
= 0.
This admits a not-so-obvious Lie-point symmetry
g
ε
(x, t, u) =
x + εt +
1
2
ε
2
, t + ε, u −
1
6
ε
This is generated by
V = t
∂
∂x
+
∂
∂t
−
1
6
∂
∂u
.
We then have invariant coordinates
z = x −
1
2
t
2
, w =
1
6
t + u.
To get an ODE for w, we write the second equation as
u(x, t) = −
1
6
t + w(z).
Then we have
u
t
= −
1
5
− tw
0
(z), u
x
= w
0
(z), u
xx
= w
00
(z), u
xxx
= w
000
(z).
So KdV becomes
0 = u
t
+ u
xxx
− 6uu
x
= −
1
6
+ w
000
(z) − 6ww
0
(z).
We would have had some problems if the
t
’s didn’t get away, because we wouldn’t
have an ODE in
w
. But since we constructed these coordinates, such that
w
and
z
are invariant under the Lie point symmetry but
t
is not, we are guaranteed
that there will be no t left in the equation.
Integrating this equation once, we get an equation
w
00
(z) − 3w(z)
2
−
1
5
z + z
0
= 0,
which is is PI. So this ODE reduction of KdV has the Painlev´e property.
In summary, the Painlev´e test of integrability is as follows:
(i) Find all Lie point symmetries of the PDE.
(ii) Find all corresponding ODE reductions.
(iii) Test each ODE for Painlev´e property.
We can then see if our PDE is not integrable. Unfortunately, there is no real
test for the converse.