On Recursion Operators
Jan A. Sanders & Jing Ping Wang
Vrije Universiteit, Amsterdam
Faculty of Sciences
Division of Mathematics & Computer Science
De Boelelaan 1081a
1081 HV Amsterdam
The Netherlands
Abstract
We observe that application of a recursion operator of Burgers equation does not
produce the expected symmetries. This is explained by the incorrect assumption
that Dx−1 Dx = 1. We then proceed to give a method to compute the symmetries
using the recursion operator as a first order approximation.
Key words: Recursion operator, cylindrical Korteweg-de Vries equation (cKdV),
(co)symmetries
1
Introduction
If we look at the historical development of the theory of integrable systems
(with the KdV equation as the most famous example), we see that equations
and symmetries, which did not explicitly depend on the time or space variables, were first studied. Based on this assumption a theoretical framework
was developed. When later on examples were found of both equations and
symmetries with explicit time or space dependency, the theoretical framework
was slightly adapted to cover the new situation, but by that time the way
things were calculated was already somewhat fixed, and these new features
were not analyzed in any fundamental way.
In this paper we first describe some of the difficulties one has applying the results in the literature to explicitly time-dependent situations. These difficulties
1
J. P. Wang gratefully acknowledges the support from NWO for this research.
Preprint submitted to Elsevier Preprint
3 October 2000
came as quite a surprise to us, since they appear at such an elementary level,
but they seem to be common to problems related to nonlocal variables (for
details see [OSW00]). We then proceed to lay the foundations of a recursion
operator formalism in the t-dependent case.
The goal of this paper is not to compute the symmetries of time-dependent
equations; for instance one can derive the symmetries of cKdV from those of
KdV using the transformation given in [Fuc93]. The point is that we want
foolproof algorithms so that we can compute symmetries given any equation
and its recursion operator.
To illustrate the problem we consider the well-known (Cf [Olv93, p.315]) Burgers equation
Ã
ut = K = u2 + uu1
∂u
ut =
,
∂t
!
∂ iu
ui =
,
∂xi
which has a time-dependent recursion operator
1
1
1
1
< = t(Dx + u + u1 Dx−1 ) + x + Dx−1 .
2
2
2
2
Applying this operator repeatedly to a symmetry like 1 + tu1 , it is supposed to
produce a hierarchy of symmetries. There is, however, one problem with this
elementary fact: while <(1 + tu1 ) is indeed a symmetry, <2 (1 + tu1 ) is not!
Let LK denote the Lie derivative. If Q is an evolutionary vectorfield, LK Q can
be defined as
LK Q =
∂Q
+ DQ K − DK Q,
∂t
and we say that Q is a symmetry if LK Q = 0. Since by definition, (LK <)Q =
LK (<Q) − <(LK Q), a sufficient condition for an operator to map a symmetry
to a new symmetry is that LK < = 0. Thus one often defines a recursion
operator by the property that LK < = 0. It is easily checked that the above
operator satisfies this condition for the Burgers equation. This seems a fairly
paradoxical situation until one realizes how the proof that LK < = 0 works:
in the computation one uses the relation Dx−1 Dx = 1. But this relation is
obviously not true 2 , since it is wrong on ker Dx , e.g., Dx−1 Dx t = 0 6= t.
Instead, we should read Dx−1 Dx = 1 − Π, where Π is the projection on ker Dx
(For example Π(t + t2 u1 ) = t).
2
A similar problem was observed by Guthrie [Gut94]. He states that if one only
considers constants when applying Dx−1 without allowing time–dependent functions,
this may lead to what he calls ‘bogus symmetries’.
2
Let us now compute LK < for the Burgers equation. We find that
LK < =
∂<
1
+ D< [K] − DK < + <DK = − (1 + tu1 )ΠDx .
∂t
2
The combination ΠDx is only nonzero on symmetries that contain a term of
the form g(t)x. If we now compute <(1 + tu1 ), we see that it equals
t2 u2 + t2 uu1 + tu + tu1 x + x
and there is indeed a term of the form g(t)x, namely x. This explains why
<2 (1 + tu1 ) is not a symmetry.
Having located the source of the problem, the next question is: how to adapt
the definition of recursion operator so that it will also handle this case correctly? After giving the necessary definitions in section 2, we answer this question in section 3. In section 4 we explicitly compute the u-independent term of
all the symmetries of the cKdV equation generated by the corrected recursion
operator.
2
Recursion operators
In order to understand the behavior of recursion operators, it helps to have
a good formal setup so that we know clearly in which spaces the objects
we study live. Our approach is based on [GD79,Dor93] and is described in
detail in [Wan98]. We only use one variable u to keep the notation simple, the
generalization to systems being fairly obvious.
Let C be the space of C ∞ -functions of t and P the space of local functions,
that is, formal power series x, u, u1 , · · · uk , for some k < ∞ with coefficients in
C. We now define the representation space A as P divided out by the image
of Dx , that is, P/im Dx , where
Dx =
∞
X
∂
∂
+
uk+1
.
∂x k=0
∂uk
The equivalence
class of a local function g ∈ P is called a functional and
R
denoted by g, the idea being that one can do partial integration, since
Z
0≡
Z
Dx (gh) =
3
Z
hDx g +
gDx h.
Next the Lie algebra h is defined as follows: Consider the space of vertical
P
∂
vectorfields of the form ∞
k=0 fk ∂u , fk ∈ P. Divide out the image of the Lie
k
derivative of Dx , where the Lie derivative is given by the commutator of the
vectorfields. An immediate consequence of this construction is that the eleP
∂
k
ments in h commute with Dx and that if ∞
k=0 fk ∂u ∈ h, then fk = Dx f0 .
k
This implies that the elements in the Lie algebra h are determined by f0 and
we simply write this as f0 ∈ h. Under this identification the Lie bracket now
looks like
[k0 , f0 ] = Df0 [k0 ] − Dk0 [f0 ],
where Df0 [k0 ] is the Fréchet derivative
Df0 [k0 ] =
∞
X
∂f0
k=0 ∂uk
Dxk k0 .
One associates the evolution equation
ut = K 0
∂
with the vectorfield K = ∂t
+ K0 , where K0 ∈ h. The Lie derivative of the
evolution equation on h is given by the commutator. One verifies that this
defines an action LK : h → h and
LK S =
∂S
+ DS [K0 ] − DK0 [S],
∂t
S ∈ h.
(1)
Note that the equation does not live in the same space as its symmetries. One
does not see this difference when the symmetries are time–independent.
For any c(t) ∈ C and k0 ∈ h, we have c(t)k0 ∈ h since Dx c(t) = 0. Therefore C
is the natural ring of coefficients for h. We now have a ring of coefficients C,
a Lie algebra h and a representation space A, with the Lie derivative as the
representation. From here on we can construct a Lie algebra complex. Let us
give the first steps here, since we do not need the general theory.
We denote the space of C-multilinear n-forms, taking their values in A, by Ωn .
In particular, Ω0 = A. Since LK : h → h, the action of the Lie derivative of
LK : Ωn → Ωn makes sense.
R
R
R³
´
∂g
+ Dg [K0 ] . We now
For any functional g ∈ Ω0 , we define LK g =
∂t
1
define and compute the action of LK on Ω . Taking ω ∈ Ω1 , for any f0 ∈ h we
have
4
(LK ω)(f0 ) = LK ω(f0 ) − ω(LK f0 )
!
∂(ω · f0 )
∂f0
=
+ Dω·f0 [K0 ] − ω · (
+ Df0 [K0 ] − DK0 [f0 ])
∂t
∂t
!
Z Ã
∂ω
?
=
+ Dω [K0 ] + DK0 [ω] · f0 ,
∂t
Z Ã
?
where DK
is the dual operator of DK0 defined by
0
Z
Z
ω · DK0 [f0 ] =
?
[ω] · f0 ,
DK
0
ω ∈ Ω1 ,
f0 ∈ h.
Since the pairing Ω1 × h → A is nondegenerate (Cf. [Dor93]), this leads to
LK ω =
∂ω
?
+ Dω [K0 ] + DK
[ω]
0
∂t
(2)
Definition 1 Let ut = K0 be an evolution equation. If a vectorfield S ∈ h (a
∂
1-form ω ∈ Ω1 ) is invariant under K = ∂t
+ K0 , K0 ∈ h, i.e., in the kernel of
LK , we call S (ω) a symmetry (cosymmetry) of the equation.
An operator < : h → h is called a recursion operator if it maps symmetries
to new symmetries. If <S is well-defined, we know it is a symmetry when
S ∈ h is a symmetry and (LK <)S = 0 due to the definition of LK <, i.e.,
LK (<S) = (LK <)S + <LK S.
Similarly, an operator = : Ω1 → Ω1 is called a conjugate recursion operator if it maps cosymmetries to new cosymmetries.
3
The corrected recursion operator
ˆ : h → h a weak recursion operator of the equation
We call an operator <
ˆ = ∂ <ˆ + D ˆ [K0 ] − DK <
ˆ + <D
ˆ K = 0 using the
ut = K0 if it satisfies that L̂K <
0
0
<
∂t
−1
rule that Dx Dx = 1. In this section, we give the corrected recursion operator
ˆ
< based on a weak recursion operator <.
Definition 2 We define a projection Π : P → C as follows. Given any
f (t, x, u, · · · , uk ) we put (Πf )(t) = f (t, 0, 0, · · · , 0).
For example, let f = t cos(u), then Πf = t, while with f = t sin(u), Πf = 0.
Definition 3 We denote all the terms with the projection operator Π in an
operator · by [[·]]. For example, we have [[Dx−1 Dx ]] = −Π.
5
Lemma 4 For any local function P ∈ im Dx and any K0 ∈ h, we have
DDx−1 P [K0 ] = Dx−1 DP [K0 ] − [[Dx−1 DP ]]K0 .
PROOF. Let Dx F = P . We know DDx F [K0 ] = Dx DF [K0 ]. Then
DDx−1 P [K0 ] = D(1−Π)F [K0 ] = DF [K0 ] = Dx−1 DP [K0 ] + ΠDF [K0 ]
= Dx−1 DP [K0 ] − [[Dx−1 DP ]]K0 ,
using Dx−1 Dx = 1 − Π and DΠF [K0 ] = 0.
This lemma implies that D<S [K0 ] = D< [K0 ]S + <DS [K0 ] − [[<DS ]]K0 if <S
is well defined, where the operator < : h → h and S, K0 ∈ h.
Lemma 5 Assume that <S is well defined, where < : h → h is an operator
∂
and S ∈ h. Then, for K = ∂t
+ K0 , K0 ∈ h, we have
Ã
(LK <)S =
!
∂<
+ D< [K0 ] − DK0 < + <DK0 S − [[<DS ]]K0 .
∂t
(3)
PROOF. By the definition of the Lie derivative and formula (1), we have
(LK <)S = LK (<S) − <LK S
∂(<S)
∂S
=
+ D<S [K0 ] − DK0 [<S] − <(
+ DS [K0 ] − DK0 [S])
∂t
∂t
Ã
!
∂<
=
+ D< [K0 ] − DK0 < + <DK0 S − [[<DS ]]K0 .
∂t
Note that the Lie derivative of an operator depends also on the object it acts
on, i.e., it can not be written as some (differential) operator.
Similarly, the formula for the Lie derivative of an operator = : Ω1 → Ω1 is as
follows:
Lemma 6 Assume that =ω is well defined, where = : Ω1 → Ω1 is an operator
∂
+ K0 , K0 ∈ h, we have
and ω ∈ Ω1 . Then, for K = ∂t
Ã
(LK =)ω =
!
∂=
∗
∗
+ D= [K0 ] + DK
= − =DK
ω − [[=Dω ]]K0 .
0
0
∂t
6
(4)
If ΠDxm K0 = 0 for all m ∈ N, i.e., the expression K0 contains no terms of the
form of c(t)xj , j ∈ N, we have the usual formula for the Lie derivative:
∂<
+ D< [K0 ] − DK0 < + <DK0 ,
∂t
∂=
∗
∗
LK = =
+ D= [K0 ] + DK
= − =DK
,
0
0
∂t
LK < =
< : h → h.
(5)
= : Ω1 → Ω1 .
(6)
Theorem 7 Consider the equation ut = K0 . Assume that it has a weak reˆ of the form Pn f (i) Di + Pj∈Γ h(j) ⊗ D−1 ξ (j) , where Γ is
cursion operator <
i=0
x
x
some index set such that the h(j) ∈ h are symmetries and the ξ (j) ∈ Ω1 are
cosymmetries of the equation. Then
ˆ K ]] = −
[[<D
0
XX
h(j) ΠDxl−1 ξ (j)
j∈Γ l≥1
ˆ ∗ D∗ ]] =
[[<
K0
XX
∂K0
,
∂ul
(−1)l ξ (j) ΠDxl−1 h(j)
j∈Γ l≥1
∂K0
.
∂ul
If S is a symmetry of the equation, the next symmetry is
¯
µ
¯
¶
¯
ˆ − [[<D
ˆ K ]] ¯¯ R t
ˆ
¯ R
<
0
¯Π= Π S + [[<DS ]] ¯Π= t Π K0 .
(7)
If ω is a cosymmetry of the equation, the next cosymmetry is
¯
µ
¯
¶
¯
ˆ ∗ + [[<
ˆ ∗ D∗ ]] ¯¯ R t
ˆ∗
¯ R
<
K0 ¯Π= Π ω + [[< Dω ]] ¯Π= t Π K0 .
(8)
¯
¯
By the notation ¯¯Π=R t Π we mean the following: we insert before the Π the
integration with respect to explicit time t. And we do not suppose that
Dx−1 f (t) = f (t)Dx−1 . If we would have allowed Dx−1 to commute with elements
in ker Dx , the result would depend on where we would have put these
elements
Rt
−1
relative to Π. For instance, if we have
tDx , this would lead to t dτ Π, but
R
if we start with Dx−1 t this leads to t τ dτ Π.
Corollary 8 If, moreover, ΠDxm K0 = 0 for all m ∈ N, the recursion operator
reads
ˆ+
<=<
XX
¯
Zt
(j)
h
ΠDxl−1 ξ (j)
j∈Γ l≥1
7
∂K0
ˆ ¯¯ R t .
ˆ − LK <
=<
¯Π= Π
∂ul
and the conjugate recursion operator reads
∗
ˆ∗
< =< +
XX
¯
Zt
l (j)
(−1) ξ
ΠDxl−1 h(j)
j∈Γ l≥1
∂K0
ˆ ∗ − LK <
ˆ ∗ ¯¯ R t .
=<
¯Π= Π
∂ul
Note that the correction preserves Dx−1 terms. This implies that the usual
ˆ are still valid, such as the structure of
results of the “recursion operators” <
ˆ k S are local if L̂S <
ˆ = 0, cf.
their non-local terms and the statement that <
[Wan98,Bil93].
This approach to correct recursion operators removes Guthrie’s objection “the
‘obvious’ definitions of Dx−1 (Q) do not survive simple coordinate changes”. We
remark that the example in [Gut94] does not lead to any obstructions, only
illustrates the freedom of choice.
For example, consider the cylindrical KdV equation (cKdV)
ut = K0 = u3 + uu1 −
u
.
2t
(9)
√
√
1
−1
ˆ = t(D2 + 2 u) + 1 x + ( tu1 + √
ˆ =0
The operator <
)D
t satisfies L̂K <
x
x
3
3
3
6 t
ˆ ∗ = 0, cf. [ZC86]. Since ΠDm K0 = 0 for all m ∈ N, by formula (5)
and L̂K <
x
and (6), we have
√
√
√
1
1
ˆ ∗ = tΠD2 √
ˆ = −( tu1 + √
)ΠDx2 t;
LK <
.
LK <
x
3
6 t
6 t
¯
¯
Using the notation ¯¯Π=R t Π , the recursion operator now reads
√
t
√
tu1
1 Z
ˆ
+ √ ) dτ τ ΠDx2
<=<+(
3
6 t
and the conjugate recursion operator now reads
√ Zt
1
< = < − t dτ √ ΠDx2 .
6 τ
∗
ˆ∗
When S is any symmetry of the equation, we see that
√
Ã
! Zt
¯
¶
√
¯
tu1
1
ˆ
ˆ
¯
R
LK < ¯Π= t Π S = (LK <)S − LK (
+ √ )
dτ τ ΠDx2 S.
3
6 t
µ
LK
8
Table 1
Symmetries of Burgers: ut = u2 + uu1
0
tu1 + 1
1
t2 (u2 + uu1 ) + t(u + u1 x) + x
2
t3 (u3 + 32 u21 + 32 uu2 + 34 u1 u2 ) + t2 ( 32 xu2 + 32 xuu1 + 3u1 + 34 u2 )
+t( 32 xu + 34 x2 u1 + 32 ) + 43 x2
3
t4 (u4 + 2u3 u + 32 u2 u2 + 5u1 u2 + 3u21 u + 12 u1 u3 )
+t3 (2xu3 + 3xuu2 + 6u1 u + 12 u3 + 32 xu1 u2 + 3xu21 + 5u2 )
+t2 ( 32 x2 u2 + 32 x2 uu1 + 6xu1 + 32 xu2 + 3u)
+t( 12 x3 u1 + 3x + 32 x2 u) + 12 x3
Table 2
Symmetries of cKdV: ut = u3 + uu1 −
√
1
0
tu1 + 2√
t
1
t3/2 (u3 + uu1 ) +
2
t5/2 (u5 + 53 uu3 +
3
√
t
2 (u
u
2t
x
√
4 t
u2 )
+ xu1 ) +
10
3 u1 u2
+ 56 u1
√ 5
5 2
5 2
+t3/2 ( 65 xu3 + 53 u2 + 56 xuu1 + 12
u ) + t( 12
xu + 24
x u1 ) +
¡
35
70
2
t7/2 u7 + 73 uu5 + 7u1 u4 + 35
18 u3 u + 3 u3 u2 + 9 uu1 u2
¢
¡
35
3
3 + t5/2 7 xu + 7 u + 35 xuu + 35 xu u
+ 35
5
3
1 2
18 u1 + 54 u1 u
6
2 4
18
9
¢
35 2
35
35 3
2
+ 35
9 uu2 + 12 u1 + 36 xu1 u + 108 u
35 2
+t3/2 ( 72
x u3 +
√ 35 3
+ t( 432 x u1 +
35
18 xu2
+
35 2
144 x u
35 2
72 x uu1
+
35
144 )
+
+
35
24 u1
+
5x√2
48 t
35
2
72 xu )
35 √
x3
864 t
√
1
1
Since tu
+ 6√
is a symmetry, this ensures that (LK <)S = 0, and thus < is
3
t
the recursion operator. We list some of the lower order symmetries of Burgers
and cKdV in table 1 and 2, respectively.
Similarly, we prove that <∗ is the conjugate recursion operator. We list some
of the lower order cosymmetries cKdV in table 3.
Remark 9 Assume S0 ∈ h is the starting symmetry of the recursive procedure
ˆ is explicitly time–dependent and
(7). The correction term only appears when <
m
m
either ΠDx S0 6= 0 or ΠDx K0 6= 0 for some m. This explains why it does not
happen so often though surprisingly it does for the Burgers equation.
There exist integrable equations that do not satisfy the condition ΠDxm K0 = 0
for all m ∈ N in Corollary 8. In this case, we have to compute their symmetries
and cosymmetries one by one according to formula (7) and (8) in Theorem 7.
9
Table 3
u
Cosymmetries of cKdV: ut = u3 + uu1 − 2t
√
0
t
√
1 t3/2 u + x2 t
2√
2 t5/2 (u2 + 12 u2 ) + x2 ut3/2 + x8 t
¡
¢
5 3
3 t7/2 u4 + 53 uu2 + 56 u21 + 18
u + 65 t5/2 (xu2 + u1 + 21 xu2 )
√
5 3/2 2
5 3
+ 24
t (x u + 1) + 144
x t
¡
¢
7 2
35 2
35
35 4
2
4 t9/2 u6 + 73 uu4 + 14
3 u1 u3 + 2 u2 + 18 u u2 + 18 uu1 + 216 u
¡
¢
35
35
35
2
3
+t7/2 76 xu4 + 73 u3 + 35
18 xuu2 + 36 xu1 + 18 uu1 + 108 xu
35 2
+t5/2 ( 72
x u2 +
√
35 4
+ 3456
x t
35
36 xu1
+
35
72 u
+
35 2 2
144 x u )
35 3
+ t3/2 ( 432
x u+
Table 4
Symmetries of the equation ut + u3 + 6uu1 +
3u
t
−
5x
2t2
35
144 x)
=0
0
t3 u1 − 12 t2
1
2
t9 (u3 + 6uu1 ) − t8 (3xu1 + 3u) + 23 t7 x
¢
¡
¢
¡
t15 u5 + 10uu3 + 20u1 u2 + 30u2 u1 − t14 5xu3 + 10u2 + 30xuu1 + 15u2
3
15 12 2
2
+t13 ( 15
2 x u1 + 15xu) − 4 t x
¢
¡
t21 u7 + 14uu5 + 42u1 u4 + 70u2 u3 + 70u2 u3 + 280uu1 u2 + 70u31 + 140u3 u1
−t20 (7xu5 + 21u4 + 70xuu3 + 140xu1 u2 + 140uu2
¢
+105u21 + 210xu2 u1 + 70u3
¡
¢
105
2
2
2
+t19 35
2 x u3 + 70xu2 + 105x uu1 + 2 u1 + 105xu
¡
¢ 35 17 3
105 2
35
3
−t18 35
2 x u1 + 2 x u + 4 + 4 t x
The generalized Korteweg–de Vries equation [Cho87]
ut + u3 + 6uu1 + 6f u − x(f 0 + 12f 2 ) = 0,
f ∈C
is integrable with the weak recursion operator
´
³
ˆ = g 2 D2 + 4(u − xf ) + 2g(u1 − f )D−1 g,
<
x
x
Rt
1
where g = e 6f (τ )dτ . We have ΠDx K0 6= 0 when f 0 + 12f 2 6= 0, i.e., f 6= C+12t
.
1
3u
5x
Let us take f = 2t . Then the equation becomes ut + u3 + 6uu1 + t − 2t2 = 0
ˆ = t6 D2 + 4t6 u − 2xt5 + (2t3 u1 − t2 )D−1 t3 . We list some of lower order
and <
x
x
symmetries and cosymmetries of this equation in table 4 and 5, respectively.
10
Table 5
Cosymmetries of the equation ut + u3 + 6uu1 +
3u
t
−
5x
2t2
=0
0
t3
1
t9 u − 12 t8 x
2
t15 (u2 + 3u2 ) − 3t14 xu + 34 t13 x2
¡
¢
¡
¢
t21 u4 + 10uu2 + 5u21 + 10u3 − t20 5xu2 + 5u1 + 15xu2
3
4
4
5
5 18 3
2
+t19 ( 15
2 x u + 4) − 4t x
¡
¢
t27 u6 + 14uu4 + 28u1 u3 + 21u22 + 70u2 u2 + 70uu21 + 35u4
¡
¢
−t26 7xu4 + 14u3 + 70xuu2 + 35xu21 + 70uu1 + 70xu3
¢
¢
¡
¡
2 u + 35xu + 105 x2 u2 + 35 u − t24 35 x3 u + 35 x +
+t25 35
x
2
1
2
2
2
2
4
35 23 4
16 t x
More details on symmetries of cKdV
In this section we want to derive the u-independent part of all symmetries.
If by K00 we denote the linear part of the equation ut = K0 , and by f the
u-independent part of the symmetry, then f obeys
ft = DK00 f.
So for cKdV this implies that
ft = f3 −
f
.
2t
We find that if we try the homogeneous expression fλ,j =
j = 0, 1, 2, this gives us the recursion relation
(λ +
P∞
p=0
3p+j λ−p
aλ,j
t ,
p x
1
λ,j
− p)aλ,j
p = (3p + j + 1)(3p + j + 2)(3p + j + 3)ap+1 .
2
This results in a divergent sequence 3 unless we make it finite by taking λ =
n − 21 , n ∈ N. We then obtain:
³ ´
fn,j =
n
X
p=0
³
n
p
´
3p+j
3p
p! 3p+j n−p− 1
2b
x
t
n,j ,
(3p)!
3
Observe that this is one of the very rare instances that topology plays a role in the
otherwise completely formal and algebraic theory of integrable systems. It would
be interesting to see whether this is a general phenomenon, that we usually miss
because we assume expressions to be polynomial (or formal power series).
11
n− 12 ,j
where we put bn,j = a0
. Thus we obtain
fn,j
j + 1 j + 2 j + 3 −x3
,
,
;
]
= xj tn−1/2 2 F3 [−n, 1;
bn,j
3
3
3
27 t
where 2 F3 is a generalized hypergeometric function [Olv74]. Then
√
¯
Zt
√
¯
tu1
1
ˆ
¯
R
LK <fn,j ¯Π= t Π = −(
+ √ )Π Dx2 τ fn,j dτ =
3
6 t
³ ´
√
t
n
Z
n
X
1
tu1
p! 3p+j n−p
p
2
³
´
= −(
+ √ )Π Dx
x
τ bn,j dτ
3p+j (3p)!
3
6 t
p=0
3p
³ ´
√
n
Zt X
n
tu1
p! 3p+j−2 n−p
1
p
³
= −(
+ √ )Π
j(j − 1) 3p+j−2´
x
τ bn,j dτ
3
(3p)!
6 t
p=0
3p
√
2
tu1
1
=−
δj,2 (
+ √ )tn+1 bn,2 .
n+1
3
6 t
Observe that only the first term in the expression for fn,j is used in the answer.
This allows us to compute explicitly the correction terms for all orders. The
only problem we have is that we usually normalize the symmetries such that
the term with the highest derivative and the same homogeneity degree as fn,j
looks like
1
t3n+j+ 2 u6n+2j+1 .
Since the recursion operator does not change the coefficient of this term, we
can compute bn,j be applying < to fn,j , j = 0, 1, 2. We find that
fn,j+1
fn,j n + 13 j + 12
=
ΠDxj+1
, j = 0, 1
bn,j
j+1
bn,j+1
1 1
fn,2
fn+1,0
= (2 +
, j = 2.
Π<
)Π
bn,2
3 n + 1 bn+1,0
ΠDxj+1 <
1
In the second expression the 13 n+1
comes from the correction term. Thus we
should define the bn,j by the recursion relation
n + 31 j + 21
bn,j , j = 0, 1
j+1
n + 23
bn+1,0 = 2
bn,2 , j = 2.
n+1
bn,j+1 =
12
We have b0,0 = 12 , since the first order symmetry is
5
35
b0,2 = 48
, b1,0 = 144
and in general, since
bn+1,j =
(n +
3+2j
)(n
6
+ 5+2j
)(n +
6
n+1
√
1
tu1 + 2√
. We find b0,1 = 14 ,
t
7+2j
)
6
bn,j ,
j = 0, 1, 2,
we find that
bn,j = (
3 + 2j 5 + 2j 7 + 2j b0,j
) (
) (
)
.
6
6
6
n
n
n n!
where (a)n is the Pochhammer symbol
f1,j =
=
√
√
³³
´
3
3+j
+ x6t
3
³ ´
3+j
3
Γ(a+n)
.
Γ(a)
Thus we have
´
tx
b
j 1,j
tx
(3 + 2j)(5 + 2j)(7 + 2j)
j
63
³
3+j
3
´
³³
3+j
3
´
+
x3
6t
´
b0,j
and we see that
Ã
f1,0
!
√ 35
x3
1+
.
= t
144
6t
7
This agrees with the symmetry with leading term t 2 u7 as given in table 3.
The methods in this section can be used for other equations, and to do similar
computations for cosymmetries.
5
Concluding remarks
We have seen the theory of recursion operators is not quite as well settled as
we may have believed, but we have shown how the correction can be obtained
by a combination of abstract methods and concrete calculations. Although the
time–dependent symmetries and recursion operators are not all that common,
this has clarified the way things should be defined.
13
References
[Bil93]
Ayşe Hümeyra Bilge. On the equivalence of linearization and formal
symmetries as integrability tests for evolution equations. J. Phys. A,
26(24):7511–7519, 1993.
[Cho87] T. Chou. Symmetries and a hierarchy of the general KdV equation.
Journal of Physics A, 20:359–366, 1987.
[Dor93] Irene Dorfman. Dirac structures and integrability of nonlinear evolution
equations. John Wiley & Sons Ltd., Chichester, 1993.
[Fuc93]
Benno Fuchssteiner. Integrable nonlinear evolution equations with timedependent coefficients. J. Math. Phys., 34(11):5140–5158, 1993.
[GD79]
I. M. Gel’fand and I. Ja. Dorfman. Hamiltonian operators and algebraic
structures associated with them. Funktsional. Anal. i Prilozhen., 13(4):13–
30, 96, 1979.
[Gut94] Graeme A. Guthrie. Recursion operators and non-local symmetries. Proc.
Roy. Soc. London Ser. A, 446(1926):107–114, 1994.
[Olv74]
F. W. J. Olver. Asymptotics and special functions. Academic Press [A
subsidiary of Harcourt Brace Jovanovich, Publishers], New York-London,
1974. Computer Science and Applied Mathematics.
[Olv93]
Peter J. Olver. Applications of Lie groups to differential equations, volume
107 of Graduate Texts in Mathematics. Springer-Verlag, New York, second
edition, 1993.
[OSW00] Peter J. Olver, Jan A. Sanders, and Jing Ping Wang. Nonlocal symmetries
and ghosts. Technical report, Vrije Universiteit Amsterdam, 2000. In
preparation.
[Wan98] Jing Ping Wang. Symmetries and Conservation Laws of Evolution
Equations. PhD thesis, Vrije Universiteit, Amsterdam, 1998.
[ZC86]
G.C. Zhu and H.H. Chen. Symmetries and integrability of the cylindrical
Korteweg–de Vries equation. Journal of Mathematical Physics, 27(1):100–
103, 1986.
14