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• V•
“ “
V
„E






E← VE VE
5. DEFORMATIONS OF THE BURGERS EQUATION 221

V V
where „E and „N are the Cartan coverings of the equation and its covering
respectively, V• is naturally constructed by •, while the mapping V„¦ is
de¬ned by V„¦ (v) = v „¦, v ∈ V N . The pair (V• , V„¦ ) is the B¨cklund
a
transformation corresponding to the recursion operator de¬ned by „¦.
This interpretation is another way to understand why shadows of recur-
sion operators take symmetries to shadows of symmetries (see Section 8 of
Chapter 3).

5. Deformations of the Burgers equation
Deformations of the Burgers equation
ut = uu1 + u2 (5.83)
will be discussed from the point of view of the theory of deformations in
coverings. We start with the following theorem (see Theorem 5.19 above):
Theorem 5.26. The only solution of the deformation equation
(1)
E („¦) =0
for the Burgers equation (5.83) is ω = ±ω0 where ± is a constant and
ω0 = du ’ u1 dx ’ (uu1 + u2 ) dt
i.e., Cartan form associated to u. This leads to the trivial deformation of
UE for (5.83).
In order to ¬nd nontrivial deformations for the Burgers equation, we
have to discuss them in the nonlocal setting. So in order to arrive at an
augmented system, a situation similar to that one for the construction of
nonlocal symmetries (see Section 4 of Chapter 3), we ¬rst have to construct
conservation laws for the Burgers equation and from this we have to intro-
duce nonlocal variables.
The only conservation law for the Burgers equation is given by
1
Dt (u) = Dx u2 + u1 , (5.84)
2
which is just the Burgers equation itself.
In (5.84), the total derivative operators Dx and Dt are given in local
coordinates on E, x, t, u, u1 , u2 , . . . , by
‚ ‚ ‚
Dx = + u1 + u2 + ...,
‚x ‚u ‚u1
‚ ‚ ‚
Dt = + ut + u1t + ... (5.85)
‚t ‚u ‚u1
The conservation law (5.84) to the introduction of the new nonlocal variable
y, which satis¬es formally the additional partial di¬erential equations
yx = u,
1
yt = u 2 + u 1 . (5.86)
2
222 5. DEFORMATIONS AND RECURSION OPERATORS

We now start from the covering E 1 = E ∞ —R, where the Cartan distribution,
or equivalently the total derivative operators Dx , Dt , is given by

Dx = D x + u ,
‚y
12 ‚
Dt = D t + u + u1 , (5.87)
2 ‚y
x
where y is the (formal) nonlocal variable y = u dx with the associated
Cartan form ω’1 de¬ned by
12
ω’1 = dy ’ u dx ’ u + u1 dt. (5.88)
2
Local coordinates in E 1 are given by
(x, t, y, u, u1 , . . . ).
We now demonstrate the calculations involved in the computations of defor-
mations of a partial di¬erential equation or a system of di¬erential equations.
In order to construct deformations of the Burgers equation (5.83)

i
Dx („¦) —
U= , (5.89)
‚ui
we start at the generating form
„¦ = F 0 ω0 + F 1 ω1 + F 2 ω2 + F 3 ω3 + F ’1 ω’1 , (5.90)
where F i , i = ’1, . . . , 3, are functions dependent on u, u1 , u2 , u3 , u4 , u5 , y.
The Cartan forms ω’1 , . . . , ω3 are given by
ω0 = du ’ u1 dx ’ (uu1 + u2 ) dt,
ω1 = du1 ’ u2 dx ’ (u2 + uu2 + u3 ) dt,
1
ω2 = du2 ’ u3 dx ’ (uu3 + 3u1 u2 + u4 ) dt,
ω3 = du3 ’ u4 dx ’ (uu4 + 4u1 u3 + 3u2 + u5 ) dt,
2
12
ω’1 = dy ’ u dx ’ u + u1 dt, (5.91)
2
and it is a straightforward computation to show that
Dx (ωi ) = ωi+1 ,
Dt (ω’1 ) = uω0 + ω1 ,
Dt (ω0 ) = u1 ω0 + uω1 + ω2 ,
Dt (ω1 ) = u2 ω0 + 2u1 ω1 + uω2 + ω3 ,
Dt (ω2 ) = u3 ω0 + 3u2 ω1 + 3u1 ω2 + uω3 + ω4 ,
Dt (ω3 ) = u4 ω0 + 4u3 ω1 + 6u2 ω2 + 4u1 ω3 + uω4 + ω5 , (5.92)
where i = ’1, 0, . . .
5. DEFORMATIONS OF THE BURGERS EQUATION 223

Now the equation for nonlocal deformations is (4.65), see p. 185,
1
E 1 („¦) = 0.
Since this one amounts to
2
Dt („¦) ’ u1 „¦ ’ uDx („¦) ’ Dx („¦) = 0, (5.93)
we are led to an overdetermined system of partial di¬erential equations for
the functions F ’1 , . . . , F 3 , by equating coe¬cients of ω’1 , . . . , ω4 to zero,
i.e.,
0 = ’2Dx (F 3 ),
ω4 :
0 = ’2Dx (F 2 ) + (Dt ’ uDx ’ u1 ’ Dx )(F 3 ) + 4u1 F 3 ,
2
ω3 :
0 = ’2Dx (F 1 ) + (Dt ’ uDx ’ u1 ’ Dx )(F 2 ) + 6u2 F 3 + 3u1 F 2 ,
2
ω2 :
0 = ’2Dx (F 0 ) + (Dt ’ uDx ’ u1 ’ Dx )(F 1 ) + 4u3 F 3 + 3u2 F 2
2
ω1 :
+ 2u1 F 1 ,
0 = ’2Dx (F ’1 ) + (Dt ’ uDx ’ u1 ’ Dx )(F 0 ) + u4 F 3 + u3 F 2
2
ω0 :
+ u2 F 1 + u1 F 0 ,
0 = (Dt ’ uDx ’ u1 ’ Dx )(F ’1 ).
2
ω’1 : (5.94)
Note that in each coe¬cient related to ω’1 , . . . , ω3 there is always a number
of terms which together are just
Dt ’ uDx ’ u1 ’ Dx (F i ),
2
i = ’1, 0, 1, 2, 3, (5.95)
which arise by action of 1 1 on the coe¬cient F i of the term F i ωi in „¦,
E
(5.90). From these equations we obtain the solution by solving the system
in the order as given by the equations in (5.94).
This leads to the following solutions
F 3 = c1 ,
3
F 2 = c1 u + c 2 ,
2
3
F 1 = c1 u2 + 3u1 + c2 u + c3 ,
4
1 9 1 3 1
F 0 = c1 u3 + uu1 + 2u2 + c2 u2 + u1 + c3 u + c4 ,
8 4 4 2 2
3 3 3 1 1 1 1
F ’1 = c1 u2 u1 + uu2 + u2 + u3 + c2 uu1 + u2 + c3 u1 + c5 .
41 2
8 4 2 2 2
(5.96)
Combination of (5.90) and (5.96)) leads to the following independent solu-
tions
W 1 = ω0 ,
W 2 = u1 ω’1 + u0 ω0 + 2ω1 ,
224 5. DEFORMATIONS AND RECURSION OPERATORS

W 4 = 2(uu1 + u2 )ω’1 + (u2 + 6u1 )ω0 + 4uω1 + 4ω2 ,
W 7 = (3u2 u1 + 6uu2 + 6u2 + 4u3 )ω’1 + (u3 + 18uu1 + 16u2 )ω0
1
+ 6(u2 + 4u1 )ω1 + 12uω2 + 8ω3 . (5.97)
1

In case we start from functions F i , i = ’1, . . . , 2, in (5.90), dependent on
x, t, u, u1 , u2 , u3 , u4 , u5 , y, and taking F 3 = 0, and solving the system of
equations (5.96) in a straightforward way, we arrive to
F 2 = c1 (t),
1
F 1 = c1 (t)x + c1 (t)u + c2 (t),
2
1
F 0 = (c1 (t)x2 + 2c1 (t)xu + 2c1 (t)u2 + 12c1 (t)u1 + 4c2 (t)x
8
+ 4c2 (t)u + c3 (t)),
1
F ’1 = (c1 (t)x3 ’ 6c1 (t)x + 12c1 (t)u + 12c1 (t)xu1
48
+ 24c1 (t)(uu1 + u2 ) + 6c2 (t)x2 + 24c2 (t)u1 + 24c3 (t)x + 48c4 (t)).
(5.98)
Finally, from the last equation in (5.94) we arrive at
c1 (t) = ±1 + ±2 t + ±3 t2 ,
c2 (t) = ±4 + ±5 t,
3
c3 (t) = ±6 + t,
2
1
c4 (t) = ’ c5 , (5.99)
2
which leading to the six independent solutions
W 1 = ω0 ,
±6 :
W 2 = u1 ω’1 + u0 ω0 + 2ω1 ,
±4 :
W 3 = (tu1 + 1)ω’1 + (tu + x)ω0 + 2tω1 ,
±5 :
W 4 = 2(uu1 + u2 )ω’1 + (u2 + 6u1 )ω0 + 4uω1 + 4ω2 ,
±1 :
W 5 = (2tuu1 + 2tu2 + xu1 + u)ω’1 ,
±2 :
+ (tu2 + 6tu1 + xu)ω0 + (4tu + 2x)ω1 + 4tω2 ,
W 6 = (2t2 (uu1 + u2 ) + 2txu1 + 2tu + 2x)ω’1 ,
±3 :
+ (t2 (uu2 + 6u1 ) + 2txu + 6t + x2 )ω0 ,
+ (4t2 u + 4tx)ω1 + 4t2 ω2 . (5.100)
If we choose the term F 3 in (5.90) to be dependent of x, t, u, u1 , u2 ,
u3 , u4 , u5 , y too, the general solution of the deformation equation (5.93),
or equivalently the resulting overdetermined system of partial di¬erential
5. DEFORMATIONS OF THE BURGERS EQUATION 225

equations (5.94) for the coe¬cients F i , i = ’1, . . . , 3, is a linear combination
of the following ten solutions

W 1 = ω0 ,
W 2 = u1 ω’1 + u0 ω0 + 2ω1 ,
W 3 = (tu1 + 1)ω’1 + (tu + x)ω0 + 2tω1 ,
W 4 = 2(uu1 + u2 )ω’1 + (u2 + 6u1 )ω0 + 4uω1 + 4ω2 ,
W 5 = (2tuu1 + 2tu2 + xu1 + u)ω’1
+ (tu2 + 6tu1 + xu)ω0 + (4tu + 2x)ω1 + 4tω2 ,
W 6 = (2t2 (uu1 + u2 ) + 2txu1 + 2tu + 2x)ω’1
+ (t2 (u2 + 6u1 ) + 2txu + 6t + x2 )ω0
+ (4t2 u + 4tx)ω1 + 4t2 ω2 ,
W 7 = (3u2 u1 + 6uu2 + 6u2 + 4u3 )ω’1 + (u3 + 18uu1 + 16u2 )ω0
1
+ 6(u2 + 4u1 )ω1 + 12uω2 + 8ω3 ,
W 8 = (t(3u2 u1 + 6uu2 + 6u2 + 4u3 ) + x(2uu1 + 2u2 ) + u2 )ω’1
1
+ (t(u3 + 18uu1 + 16u2 ) + x(u2 + 6u1 ) + 2u)ω0
+ (t(6u2 + 24u1 ) + x(4u))ω1
+ (12tu + 4x)ω2
+ 8tω3 ,
W 9 = (t2 (3u2 u1 + 6uu2 + 6u2 + 4u3 ) + tx(4uu1 + 4u2 ) + x2 (u1 )
1
+ 2tu2 + 2xu ’ 6)ω’1 + (t2 (u3 + 18uu1 + 16u2 ) + tx(2u2 + 12u1 )
+ x2 u + 4tu ’ 2x)ω0 + (t2 (6u2 + 24u1 ) + 8txu + 2x2 )ω1
+ (12t2 u + 8tx)ω2 + (8t2 )ω3 ,
W 10 = (t3 (3u2 u1 + 6uu2 + 6u2 + 4u3 ) + t2 x(6uu1 + 6u2 ) + 3tx2 u1
1
+ t2 (3u2 + 12u1 ) + 6txu + 3x2 + 6t)ω’1 + (t3 (u3 + 18uu1 + 16u2 )
+ t2 x(3u2 + 18u1 ) + 3tx2 u + x3 + 18t2 u + 18tx)ω0 + (t3 (6u2 + 24u1 )
+ 12t2 xu + 6tx2 + 24t2 )ω1 + (12t3 u + 12t2 x)ω2 + (8t3 )ω3 . (5.101)

In order to compute the classical recursion operators for symmetries
resulting from the deformations constructed in (5.100) induced by the char-
acteristic functions W1 , W2 , . . . , we use Proposition 4.29. Suppose we start
at a (nonlocal) symmetry X of the Burgers equation; its presentation is

‚ ‚
i
= X’1 + Dx (X) . (5.102)
X
‚y ‚ui
i
226 5. DEFORMATIONS AND RECURSION OPERATORS

The nonlocal component X’1 is obtained from the invariance of the equa-
tions, cf. (5.87)
yx = u,
1
yt = u 2 + u 1 ,
2
i.e.,
Dx (X’1 ) = X, (5.103)
from which we have
’1
X’1 = Dx (X). (5.104)
Theorem 4.30, stating that U1 is a symmetry, yields for the component
X
‚/‚u,
’1
W 2 = u1 X’1 + uX + 2Dx X = u1 Dx + u + 2Dx X (5.105)
X

and similar for W 3

W 3 = (tu1 + 1)X’1 + (tu + x)X + 2tDx X
X
’1
= (tu1 + 1)Dx + (tu + x) + 2tDx X. (5.106)
From formulas (5.105) and (5.106) together with similar results with respect
to W4 , . . . , W7 we arrive in a straightforward way at the recursion operators
R1 = id,
’1
R2 = u1 Dx + u + 2Dx ,
’1 ’1
R3 = t(u1 Dx + u + 2Dx ) + x + Dx ,
’1
R4 = 2(uu1 + u2 )Dx + (u2 + 6u1 ) + 4uDx + 4Dx ,
2

’1
R5 = t (2uu1 + 2u2 )Dx + (u2 + 6u1 ) + 4uDx + 4Dx
2

’1 ’1
+ x u1 Dx + u + 2Dx + uDx ,
’1
R6 = t2 (2uu1 + 2u2 )Dx + (u2 + 6u1 ) + 4uDx + 4Dx
2

’1
+ 2tx u1 Dx + u + 2Dx + x2
’1 ’1
+ t(2uDx + 6) + 2xDx ,
’1
R7 = (3u2 u1 + 6uu2 + 6u2 + 4u3 )Dx + (u3 + 18uu1 + 16u2 )
1
+ 6(u2 + 4u1 )Dx + 12uDx + 8Dx .
2 3
(5.107)
The operator R1 is just the identity operator while R2 is the ¬rst classical
recursion operator for the Burgers equation.
This application shows that from the deformations of the Burgers equa-
tion one arrives in a straightforward way at the recursion operators for
symmetries. It will be shown in forthcoming sections that the representa-
tion of recursion operators for symmetries in terms of deformations of the
di¬erential equation is more favorable, while it is in e¬ect a more condensed
6. DEFORMATIONS OF THE KDV EQUATION 227

presentation of this recursion operator. Moreover the appearance of formal
integrals in these operators is clari¬ed by their derivation.
The deformation of an equation is a geometrical object, as is enlightened
in Chapter 6: it is a symmetry in a new type of covering.

6. Deformations of the KdV equation
Motivated by the results obtained for the Burgers equation, we search
for deformations in coverings of the KdV equation. In order to do this, we
¬rst have to construct conservation laws for the KdV equation
ut = uu1 + u3 , (5.108)
i.e., we have to ¬nd functions F x , F t , depending on x, t, u, u1 , . . . such that
on E ∞ one has
Dt (F x ) = Dx (F t ), (5.109)
where Dx , Dt are total derivative operators, which in local coordinates x, t,
u, u1 , u2 , u3 , . . . on E ∞ have the following presentation
‚ ‚ ‚ ‚
+ ··· ,
Dx = + u1 + u2 + u3
‚x ‚u ‚u1 ‚u2
‚ ‚ ‚ ‚
+ ···
Dt = + ut + ut1 + ut2 (5.110)
‚t ‚u ‚u1 ‚u2
Since the KdV equation is graded,
deg(x) = ’1, deg(t) = ’3
deg(u) = 2, deg(u1 ) = 3, . . . , (5.111)
F x , F t will be graded too being of degree k and k + 2 respectively.
In order to avoid trivialities in the construction of these conservation
laws, we start at a function F triv which is of degree k ’ 1 and remove in the
expression F x ’ Dx (F triv ) special terms by choosing coe¬cients in F triv in
an appropriate way, since the pair (Dx (F triv ), Dt (F triv )) leads to a trivial
conservation law.
After this, we restrict ourselves to conservation laws of the type (F x ’
Dx (F triv ), F t ’ Dt (F triv )). Searching for conservation laws satisfying the
condition deg(F x ) ¤ 6, we ¬nd the following three conservation laws
12
x t
F1 = u, F1 = u + u2 ,
2
1 13 12
F2 = u2 ,
x t
u ’ u1 + uu2 ,
F2 =
2 3 2
34
F3 = u3 ’ 3u2 , F3 =
x t
u + 3u2 u2 ’ 6uu2 ’ 6u1 u3 + 3u2 . (5.112)
1 1 2
4
We now introduce the new nonlocal variables y1 , y2 , y3 by the following
system of partial di¬erential equations
(y1 )x = u,
228 5. DEFORMATIONS AND RECURSION OPERATORS

1
(y1 )t = u2 + u2 ,
2
1
(y2 )x = u2 ,
2
1 1
(y2 )t = u3 ’ u2 + uu2
21
3
(y3 )x = u3 ’ 3u2 ,
1
3
(y3 )t = u4 + 3u2 u2 ’ 6uu2 ’ 6u1 u3 + 3u2 . (5.113)
1 2
4
The compatibility conditions for these equations (5.113) are satis¬ed because
of (5.109).
If we now repeat the construction of ¬nding conservation laws on E ∞ —
R3 , where local variables are given by x, t, u, y1 , y2 , y3 , u1 , u2 , . . . and
where the system of partial di¬erential equations is given for u, y1 , y2 , y3
by (5.113), we ¬nd yet another conservation law

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