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1 ™

ψ = Bul + (lu0 B + Bx + b)ul’1 + O[l ’ 2]
2
and let us compute (2.68) for these functions temporary denoting ku0 A +
™ ™ ¯ ¯
A + a and lu0 B + B + b by A and B respectively. Then we have:

1¯ 1 l’1 1¯ ¯
l
{•, ψ} = Dx (Auk + Auk’1 )B + Dx (Auk + Auk’1 )B
2 2 2
1¯ 1 k’1 1¯ ¯
k
’ Dx (Bul + Bul’1 )A ’ Dx (Buk + Bul’1 )A + O[k + l ’ 1]
2 2 2
1 ¯1 1¯
¯ ¯
= (lDx (A)uk+l’2 + Auk+l’1 )B + (Auk+l’1 + Auk+l’2 )B
2 2 2
1 ¯1 1¯
¯ ¯
’ (kDx (B)uk+l’2 + Buk+l’1 )A ’ (Buk+l’1 + Buk+l’2 )A +
2 2 2
O[k + l ’ 3],
or in short,
1™ ™
{•, ψ} = (lAB ’ K BA)uk+l’2 + O[k + l ’ 3]. (2.69)
2
3. THE BURGERS EQUATION 83

3.4. Low order symmetries. These computations were done already
in Section 3 of Chapter 1 (see equation (1.61)). They can also be done
independently taking k = 2 and solving equation (2.64) directly. Then one
obtains ¬ve independent solutions which are
•0 = u 1 ,
1
•1 = tu1 + 1,
1
•0 = u 2 + u 0 u 1 ,
2
1 1
•1 = tu2 + (tu0 + x)u1 + u0 ,
2
2 2
2 2 2
•2 = t u2 + (t u0 + tx)u1 + tu0 + x. (2.70)

3.5. Action of low order symmetries. Let us compute the action
def
Tij = {•j , •} = ’
•j •j
i i i


of symmetries •j on other symmetries of the equation E.
i
0 one has
For •1
‚ ‚
0
’ ’ Dx = ’ .
T1 = = ui+1
u1 u1
‚ui ‚x
i≥0

Hence, if • = Auk + O[k ’ 1] is a function of the form (2.67), then we obtain
1™
0
T1 • = ’ Auk’1 + O[k ’ 2].
2
Consequently, if • is a symmetry, then, since sym(E) is closed under the
Jacobi bracket,
k’1
dk’1 A
1
(T1 )k’1 •
0

= u1 + O[0]
dtk’1
2
is a symmetry as well. But from (2.70) one sees that ¬rst-order symmetries
are linear in t. Thus, we have the following result:
Proposition 2.25. If • = Auk + O[k ’ 1] is a symmetry of the Burgers
equation, then A is a k-th degree polynomial in t.
3.6. Final description. Note that direct computations show that the
equation E possesses a third-order symmetry of the form
3 3 3
•0 = u 3 + u 0 u 2 + u 2 + u 2 u 1 .
3
20 40
2
2 0
Using the actions T2 and T3 , one can see that
k’1
k!(k ’ 1)!
3
((T2 )i
2 0
T2 )k’1 )u1
2
—¦ —¦ ’ uk + O[k ’ 1]
(T3 = (2.71)
(k ’ i)!
2
is a symmetry, since u1 is the one.
84 2. HIGHER SYMMETRIES AND CONSERVATION LAWS

Theorem 2.26. The symmetry algebra sym(E) for the Burgers equation
E = {ut = uux + uxx }, as a vector space, is generated by elements of the
form
•i = ti uk + O[k ’ 1], k ≥ 1, i = 0, . . . , k,
k

which are polynomial in all variables. For the Jacobi bracket one has
1
{•i , •j } = (li ’ kj)•i+j’1 + O[k + l ’ 3]. (2.72)
k l k+l’2
2
The Lie algebra sym(E) is simple and has •0 , •2 , and •0 as its generators.
1 2 3

Proof. It only remains to prove that all •i are polynomials and that
k
sym(E) is a simple Lie algebra. The ¬rst fact follows from (2.71) and from
2 0
the obvious observation that coe¬cients of both T2 and T3 are polynomials.
Let us prove that sym(E) is a simple Lie algebra. To do this, let us
introduce an order in the set {•i } de¬ning
k

def
¦ k(k+1) +i = •i .
k
2


Then any symmetry may be represented as s »± ¦± , » ∈ R.
±=1
Let I ‚ sym(E) be an ideal and ¦ = ¦s + s’1 »± ¦± be its element.
±=1
i for some k ≥ 1 and i ¤ k.
Assume that ¦s = •k
Note now that
‚ ‚ ‚
1 ±
’ tDx = ’t
T1 = Dx (tu1 + 1)
‚u± ‚u0 ‚x
±≥0

and
‚ ‚
0 ± 2
’ D x ’ u 0 Dx ’ u 1 = ’ .
T2 = Dx (u2 + u0 u1 )
‚u± ‚t
±≥0

Therefore,
((T1 )k’1 —¦ (T2 )i )¦ = c•0 ,
1 0
1

where the coe¬cient c does not vanish. Hence, I contains the function •0 . 1
But due to (2.71) the latter, together with the functions •2 and •0 , generates
2 3
the whole algebra.

Further details on the structure of sym(E) one can ¬nd in [60].


4. The Hilbert“Cartan equation
We compute here classical and higher symmetries of the Hilbert“Cartan
equation [2]. Since higher symmetries happen to depend on arbitrary func-
tions, we consider some special choices of these functions [38].
4. THE HILBERT“CARTAN EQUATION 85

4.1. Classical symmetries. The Hilbert“Cartan equation is in e¬ect
an underdetermined system of ordinary di¬erential equations in the sense of
De¬nition 1.10 of Subsection 2.1 in Chapter 1. The number of independent
variables, n, is one while the number of dependent variables, m, is two. Local
coordinates are given by x, u, v in J 0 (π), while the order of the equations is
two, i.e.,
2
ux = vxx (2.73)
The representative morphism (see De¬nition 1.6 on p. 6) ¦ is given by
2
¦∆ (x, u, v, ux , vx , uxx , vxx ) = ux ’ vxx . (2.74)
The total derivative operator Dx is given by the formula
‚ ‚ ‚ ‚ ‚
+ ···
D = Dx = + ux + vx + uxx + vxx (2.75)
‚x ‚u ‚v ‚ux ‚vx
To construct classical symmetries for (2.73), we start from the vector ¬eld
X, given by
‚ ‚ ‚
X = X(x, u, v) + U0 (x, u, v) + V0 (x, u, v)
‚x ‚u ‚v
‚ ‚
+ U1 (x, u, x, ux , vx ) + V1 (x, u, x, ux , vx )
‚ux ‚vx
‚ ‚
+ U2 (x, u, x, ux , vx , uxx , vxx ) + V2 (x, u, x, ux , vx , uxx , vxx ) .
‚uxx ‚vxx
The de¬ning relations (1.34) (see p. 26) for U1 , V1 , U2 , V2 are
U1 = D(U0 ) ’ ux D(X) = D(U0 ’ ux X) + uxx X,
V1 = D(V0 ) ’ vx D(X) = D(V0 ’ vx X) + vxx X,
U2 = D(U1 ) ’ uxx D(X) = D 2 (U0 ’ ux X) + uxxx X,
V2 = D(V1 ) ’ vxx D(X) = D 2 (V0 ’ vx X) + vxxx X. (2.76)
From (2.76) we derive the following explicit expressions for U1 , V1 , U2 , V2 :
U1 = U0,x + U0,u ux + U0,v vx ’ ux (X0,x + X0,u ux + X0,v vx ),
V1 = V0,x + V0,u ux + V0,v vx ’ ux (X0,x + X0,u ux + X0,v vx ),
U2 = U0,xx + 2U0,xu ux + 2U0,xv vx + U0,uu u2 + 2U0,uv ux vx + U0,u uxx
x
2
+ U0,vv vx + U0,v vxx ’ 2uxx (X0,x + X0,u ux + X0,v vx )
’ ux (X0,xx + 2X0,xu ux + 2X0,xv vx
+ X0,uu u2 + 2X0,uv ux vx + X0,u uxx + X0,vv vx + X0,v vxx ),
2
x
V2 = V0,xx + 2V0,xu ux + 2V0,xv vx + V0,uu u2 + 2V0,uv ux vx + V0,u uxx
x
2
+ V0,vv vx + V0,v vxx ’ 2uxx (X0,x + X0,u ux + X0,v vx )
’ vx (X0,xx + 2X0,xu ux + 2X0,xv vx + X0,uu u2
x
2
+ 2X0,uv ux vx + X0,u uxx + X0,vv vx + X0,v vxx ). (2.77)
86 2. HIGHER SYMMETRIES AND CONSERVATION LAWS

Now the symmetry-condition X(¦∆ ) |E = 0 results in
2
U1 ’ 2vxx V2 = »(ux ’ vxx ) (2.78)
which is equivalent to
1
U1 ’ 2(ux ) 2 V2 = 0 mod ¦∆ = 0, (2.79)
which results in
U0,x + U0,u ux + U0,v vx ’ ux (X0,x + X0,u ux + X0,v vx )
’ V0,xx + 2V0,xu ux + 2V0,xv vx + V0,uu u2 + 2V0,uv ux vx + V0,u uxx
x
2
+ V0,vv vx + V0,v vxx ’ 2uxx (X0,x + X0,u ux + X0,v vx )
’ vx (X0,xx + 2X0,xu ux + 2X0,xv vx + X0,uu u2 + 2X0,uv ux vx
x

+ X0,u uxx + X0,vv vx + X0,v vxx ) · 2(ux )1/2 = 0.
2
(2.80)

Equation (2.80) is a polynomial in the “variables” (ux )1/2 , vx , uxx , the
coe¬cients of which should vanish. From this observation we obtain the
following system of equations:
1: U0,x = 0,
u1/2 : ’2V0,xx = 0,
x

u1/2 vx : ’4V0,xv + 2X0,xx = 0,
x

u1/2 uxx : ’2V0,u = 0,
x

u1/2 uxx vx : 2X0,u = 0,
x

u1/2 vx :
2
’2V0,vv + 4X0,xv = 0,
x

u1/2 vx :
3
2X0,vv = 0,
x
U0,u ’ X0,x ’ 2V0,v + 4X0,x = 0,
ux :
’X0,v + 4X0,v + 2X0,v = 0,
u x vx :
u2 : ’X0,u + 4X0,u = 0,
x

u3/2 : ’4v0,xu = 0,
x

u3/2 vx : ’4V0,uv + 4X0,xu = 0,
x

u3/2 vx :
2
4X0,uv = 0,
x

u5/2 : ’2V0,uu = 0,
x

u5/2 vx : 2X0,uu = 0,
x
vx : U0,v = 0. (2.81)
From system (2.81) we ¬rst derive:
X0,u = X0,v = 0, V0,uu = V0,uv = V0,vv = 0 = V0,u = V0,xx ,
4. THE HILBERT“CARTAN EQUATION 87

[Ai , Aj ] A1 A2 A3 A4 A5 A6
A1 0 0 0 0 A1 A3
0 2A2 ’3A2
A2 0 0
A3 0 A3 0 0
’A6
A4 0 0
A5 0 A6
A6 0

Figure 2.1. Commutator table for classical symmetries of
the Hilbert“Cartan equation

which results in the equality X(x, u, v) = H(x) and in the fact that V0 is
independent of u, being of degree 1 in v and of degree 1 in x, i.e.,
X(x, u, v) = H(x), V0 = a0 + a1 x + a2 v + a3 xv.
1/2
Now from the equation labeled by ux vx in (2.81) we derive
H(x) = a3 x2 + a4 x + a5 . (2.82)
From the equations U0,v = 0 and U0,u + 3X0,x ’ 2V0,v = 0 we get
U0 = ’(4a3 x + 2a2 ’ 3a5 )u + G(x). (2.83)
Finally from U0,x = 0 we arrive at a3 = 0, G(x) = a6 , from which the general
solution is obtained as
U0 = (2a2 ’ 3a4 )u + a6 ,
X = a4 x + a5 , V0 = a0 + a1 x + a2 v.
This results in a 6-dimensional Lie algebra, the generators of which are given
by

A1 = ,
‚x

A2 = ,
‚u

A3 = ,
‚v
‚ ‚
A4 = 2u +v ,
‚u ‚v
‚ ‚
’ 3u ,
A5 = x
‚x ‚u

A6 = x ,
‚v
while the commutator table is given on Fig. 2.1.

4.2. Higher symmetries. As a very interesting and completely com-
putable application of the theory of higher symmetries developed in Subsec-
tion 2.1, we construct in this section the algebra of higher symmetries for
88 2. HIGHER SYMMETRIES AND CONSERVATION LAWS

the Hilbert“Cartan equation E
2
ux ’ vxx = 0. (2.84)
First of all, note that E ∞ is given by the system of equations:
Di (ux ’ vxx ) = 0,
2
i = 0, 1, . . . (2.85)
where D is de¬ned by
∞ ∞
‚ ‚ ‚
D= + uk+1 + vk+1 , (2.86)
‚x ‚uk ‚vk
k=0 k=0
and uk = ux . . . x . So from (2.84) we have
k times

D1 F = u2 ’ 2v2 v3 = 0,
D2 F = u3 ’ 2v3 ’ 2v2 v4 = 0,
2

i
i
i
D F = u1+i ’ v2+l v2+i’l = 0,
l
l=0
2
i = 3, . . . , with F (x, u, v, u1 , v1 , u2 , v2 ) = u1 ’ v2 = 0.
In order to construct higher symmetries of (2.84), we introduce internal
coordinates on E ∞ which are
x, u, v, v1 , v2 , v3 , · · · (2.87)
The total derivative operator restricted to E ∞ , again denoted by D, is given
by the following expression
‚ 2‚ ‚ ‚
D= + v2 + v1 + vi+1 ,
‚x ‚u ‚v ‚vi
i>0
n
‚ 2‚ ‚ ‚
D(n) = + v2 + v1 + vi+1 . (2.88)
‚x ‚u ‚v ‚vi
i>0
Suppose that a vertical vector ¬eld V = with the generating function ¦,
¦

¦ = f u (x, u, v, v1 , . . . , vn ), f (x, u, v, v1 , . . . , vn ) ,
v
(2.89)
is a higher symmetry of E. We introduce the notation
f [vk ] = f (x, u, v, v1 , . . . , vk ). (2.90)
Since the vertical vector ¬eld V is formally given by
‚ ‚ ‚ ‚

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