2Partial Differential Equations
II Integrable Systems
2.3 B¨acklund transformations
For a linear partial differential equation, we have the principle of superposition
— if we have two solutions, then we can add them to get a third solution. This is
no longer true in non-linear PDE’s.
One way we can find ourselves a new solution is through a B¨acklund trans-
formation. This originally came from geometry, where we wanted to transform a
surface to another, but we will only consider the applications to PDE’s.
The actual definition of the B¨acklund transformation is complicated. So we
start with an example.
Example. Consider the Cauchy-Riemann equation
u
x
= v
u
, u
y
= −v
x
.
We know that the pair
u, v
satisfies the Cauchy-Riemann equations, if and only
if both u, v are harmonic, i.e. u
xx
+ u
yy
= 0 etc.
Now suppose we have managed to find a harmonic function
v
=
v
(
x, y
). Then
we can try to solve the Cauchy-Riemann equations, and we would get another
harmonic function u = u(x, y).
For example, if v = 2xy, then we get the partial differential equations
u
x
= 2x, u
y
= −2y.
So we obtain
u(x, y) = x
2
− y
2
+ C
for some constant
C
, and this function
u
is guaranteed to be a solution to
Laplace’s equations.
So the Cauchy-Riemann equation generates new solutions to Laplace’s equa-
tion from old ones. This is an example of an (auto-)B¨acklund transformation for
Laplace’s equation.
In general, we have the following definition:
Definition
(B¨acklund transformation)
.
A B¨acklund transformation is a system
of equations that relate the solutions of some PDE’s to
(i) A solution to some other PDE; or
(ii) Another solution to the same PDE.
In the second case, we call it an auto-B¨acklund transformation.
Example.
The equation
u
xt
=
e
u
is related to the equation
v
xt
= 0 via the
B¨acklund transformation
u
x
+ v
x
=
√
2 exp
u − v
2
, u
t
− v
t
=
√
2 exp
u + v
2
.
The verification is left as an exercise on the first example sheet. Since
v
xt
= 0 is
an easier equation to solve, this gives us a method to solve u
xt
= e
u
.
We also have examples of auto-B¨acklund transformations:
Example. For any non-zero constant ε, consider
∂
∂ξ
(ϕ
1
− ϕ
2
) = 2ε sin
ϕ
1
+ ϕ
2
2
∂
∂τ
(ϕ
1
+ ϕ
2
) =
2
ε
sin
ϕ
1
− ϕ
2
2
.
These equations come from geometry, and we will not go into details motivating
these. We can compute
∂
2
∂ξ∂τ
(ϕ
1
− ϕ
2
) =
∂
∂τ
2ε sin
ϕ
1
+ ϕ
2
2
= 2ε cos
ϕ
1
+ ϕ
2
2
∂
∂τ
ϕ
1
+ ϕ
2
2
= 2ε cos
ϕ
1
+ ϕ
2
2
1
2
·
2
ε
sin
ϕ
1
− ϕ
2
2
= 2 cos
ϕ
1
+ ϕ
2
2
sin
ϕ
1
− ϕ
2
2
= sin ϕ
1
− sin ϕ
2
.
It then follows that
∂
2
ϕ
2
∂ξ∂τ
= sin ϕ
2
⇐⇒
∂
2
ϕ
1
∂ξ∂τ
= sin ϕ
1
.
In other words,
ϕ
1
solves the sine-Gordon equations in light cone coordinates,
if and only if
ϕ
2
does. So this gives an auto-B¨acklund transformation for the
sine-Gordon equation. Moreover, since we had a free parameter
ε
, we actually
have a family of auto-B¨acklund transforms.
For example, we already know a solution to the sine-Gordon equation, namely
ϕ
1
= 0. Using this, the equations say we need to solve
∂ϕ
∂ξ
= 2ε sin
ϕ
2
∂ϕ
∂τ
= −
2
ε
sin
ϕ
2
.
We see this equation has some sort of symmetry between
ξ
and
τ
. So we use an
ansatz
ϕ(ξ, τ ) = 2χ(εξ − ε
−1
τ).
Then both equations tell us
dχ
dx
= sin χ.
We can separate this into
csc χ dχ = dx.
Integrating this gives us
log tan
χ
2
= x + C.
So we find
χ(x) = 2 tan
−1
(Ae
x
).
So it follows that
ϕ(ξ, τ ) = 2 tan
−1
(A(εξ + ε
−1
τ)),
where
A
and
ε
are free parameters. After a bit more work, this was the 1-soliton
solution we previously found.
Applying the B¨acklund transform again to this new solution produces multi-
soliton solutions.