Geometric approach for differential equations
Raouf Dridi
Michel Petitot
Sylvain Neut
L.I.F.L. Univ. Lille I
L.I.F.L. Univ. Lille I
L.I.F.L. Univ. Lille I
Abstract
Present differential equations solver are often based on a list of equations the solutions of which one knows (e.g. as listed in [9]). Each equation of this list and its
solution are ordered in a table. Significant progress would be made if it was possible
to compute in advance the differential invariants that allow to decide if one equation to
solve is equivalent to one of the list by a change of coordinates. We will show that, for
the computation of these invariants, a geometrical approach offers advantages over non
geometrical approaches (e.g. such as that of Riquier, Ritt, Kolchin etc.)
The sought change of coordinates is solution of certain PDE system. Differential
algebra allows to compute the integrability conditions of this PDE system. Although,
in practice one is often confronted with computer output consisting of several pages
of intricate formulae. Even so, for more complicated equivalence problems with higher
complexity such those coming from biology, physics, etc. differential algebra is not
efficient due to expression swell.
In his equivalence method, E. Cartan [5] formulated the PDE system in term of
linear Pfaffian system. Then the required conditions are computed with a process
called absorption of torsion leading to sparse structure equations. In addition, this
computation is done by separately and symmetrically treating the linear Pfaffian system
and this has the effect of dividing the number of variables by two. Cartan’s method has
a wide range of applications to problems arising in classical invariant theory, differential
equation classification, exact solution of PDEs, general relativity and many others.
Cartan’s equivalence method is complementary to Lie symmetry method. In the
case when the equation which we want to solve admits the symmetry group of finite
cardinality, we prove that the sought change of coordinates can be computed without
integrating differential equation.
keywords : Cartan’s method; differential algebra.
1
Introduction
Two systems of differential equations Ef and Ef¯ are said to be equivalent under a pseudogroup of transformations Φ (we write Ef ∼Φ Ef¯) if and only if there exist a local diffeomorphism φ ∈ Φ wich map the solutions of Ef to the solutions of Ef¯. The change of coordinates
φ ∈ Φ is solution of a PDE system that we can compute by an algorithm. A program like
diffalg or rif can compute the integrability conditions of such a PDE system, then the
existence of φ is decidable.
Example Consider two ordinary differential equations (Ef ) and (Ef¯) of the form
dy
d2 ȳ
dȳ
d2 y
¯
= f x, y,
and
= f x̄, ȳ,
(1)
dx2
dx
dx̄2
dx̄
and Φ the pseudo-group of local diffeomorphisms φ : R2 → R2 defined by the Lie equations
∂ x̄
∂ x̄
∂ ȳ
= 1,
= 0,
6= 0.
∂x
∂y
∂y
1
This gives (x̄, ȳ) = φ(x, y) = (x + C, η(x, y)) where C is a constant and η(x, y) is an
arbitrary function.
2
Equivalence problems and differential elimination
We say that an equivalence problem is an asymmetrical equivalence problem if we know f¯
and our task is to find explicit conditions on f such that Ef ∼Φ Ef¯. For the equivalence (1)
the jet space J2 (R, R) has coordinates x = (x, y, p = y 0 , q = y 00 ). Thus the asymmetrical
equivalence problem reads
∃ φ ∈ Φ,
φ∗ (q̄ − f¯(x̄, ȳ, p̄)) = 0
mod q − f (x, y, p).
(2)
The prolongation formulae [14] of φ are
φ∗ x̄ = x + C,
φ∗ ȳ = η(x, y),
φ∗ p̄ = Dx η,
φ∗ q̄ = Dx2 η
(3)
∂
∂
∂
where Dx = ∂x
+ p ∂y
+ f (x, y, p) ∂p
is the total derivative. Such problems can be solved
by elimination methods using diffalg or rif.
Example 1 On the above example, let us suppose that f¯ is identically null. In this case
∗
φ p̄ = ηx + ηy p and φ∗ q̄ = ηxx + 2ηxy p + ηyy p2 + ηy q. The equations (2) take the form of a
polynomial PDE system
ηxx + 2ηxy p + ηyy p2 + ηy f = 0,
ηp = 0,
ηy 6= 0.
(4)
By eliminating η in (4) using diffalg with the ranking η f , we obtain the characteristic
set
1
1
1
ηxx = −ηy f + pηy fp − pηy fpp , ηxy = − ηy fp + pηy fpp ,
2
2
2
1
(5)
ηyy = − ηy fpp , ηp = 0,
2
1
fppp = 0, fxp = −fpp f + 2fy + fp2 − pfyp .
2
00
0
Conditions on f so that the equation y = f (x, y, y ) is equivalent to ȳ 00 = 0 are given by
the two last equations. It follows
Theorem 1. The equation y 00 = f (x, y, y 0 ) is reduced to the equation ȳ 00 = 0 by a transformation φ if and only if
fppp = 0,
1
fxp = −fpp f + 2fy + fp2 − pfyp .
2
(6)
To have the change of coordinates φ we have to solve the PDE system given by the first
four equations of (5). In the favorable cases, we can determine the change of coordinates
without any integration.
Example 2 Suppose that the target equation is the Painlevé first equation (P1) :
ȳ 00 = 6ȳ 2 + x̄ . The problem formulation is as above and in this case DiffAlg gives the
2
following characteristic set
η(x, y) = 1/12 fy − 1/24 fx,p − 1/24 fp,p f + 1/48 fp 2 − 1/24 pfy,p ,
(7)
2
C = −x + 1/16 pfp fy,p,p f + 1/16 p fp,p fp fy,p + 1/16 pfp fp,p fy
..
.
+1/96 fx,p fp 2 − 1/96 fp,p 2 f 2 ,
fx,x,x,x,p = −24 + 5/2 fpfp,p fp fy,p − 4 fx fx,y,p + 2 fx,x,x,y − 2 pfy,p fp fx,p
..
.
+fp fx,x,x,p + fy fx,x,p + 5/2 fx,p fx,x,p + 1/4 fx,x,p fp 2 − 2 pfx,x,y,y ,
fx,x,y,p = −pfp,p fy,y + 3 pfy,p,p fy + fp,p p2 fy,y,p − 3/2 pfp fy,y,p
..
.
+pfp,p fx,y,p + pfp,p 2 fy − pfp,p fp fy,p − ffy,y,p − fy fy,p ,
fx,y,y,p = 2 fy,y,y + fy,p 2 − pfy,y,y,p − fy,y,p,p f − 2 fy,p,p f y −
1/2 fp,p fy,p,p f + fp fy,y,p − 1/2 fp,p fx,y,p − 1/2 fp,p pfy,y,p −
1/2 fp,p 2 fy + 1/2 fp,p fp fy,p ,
fx,p,p = fy,p − pfy,p,p ,
fp,p,p = 0.
These formulae show that the change of variable φ is obtained without integrating differential equation and is unique. We will see in section 6 that this results from the fact that
the group of symmetries of (P1) is reduced to the identity. The others equations gives the
requested conditions on f .
As the reader has noticed, such explicit formulae consisting of several lines (pages) prove
quite useless for practical application. More dramatically, this method is rarely efficient
due to expression swell involved by the use of commutative derivations. Indeed, the study
of the equivalence of the 3rd order differential equation y 000 = f (x, y, p, q) with ȳ 000 = 0
under contact transformations (x̄, ȳ) = (ξ(x, y, p), η(x, y, p)) , requires the prolongation to
J3 = (x, y, p = yx , q = yxx , r = yxxx ) with
p̄ =
Dx η
,
Dx ξ
q̄ =
Dx 2 ηDx ξ − Dx ηDx 2 ξ
,
Dx ξ 3
2
r̄ =
Dx 3 ηDx ξ 2 − 3Dx 2 ηDx 2 ξDx ξ + 3Dx ηDx 2 ξ − Dx ηDx 3 ξDx ξ
Dx ξ 5
Therefore, the change of coordinates satisfy the PDE system {r̄ = 0, ∂∂qp̄ = 0, ξq = 0, ηq = 0}
where one will have replaced r by f . In commutative derivations, the equations blow up
and take more than one hundreds of lines
r̄ = −(−3pηxxy ξx 3 + 12ηxx p2 ξxy ξy ξx + 6p3 ηy ξxxy ξy ξx − 3pηy ξxx 2 ξx
+6ηxx ξxx ξx pξy − 12p3 ηy ξxx ξxy ξy − 12ηx ξxx p2 ξxy ξy + 6p3 ηxy ξxx ξy 2
..
.
100 lines of differential polynomials
..
.
−3ηx ξxx 2 ξx − q 2 ηxpp ξx 3 + ηx ξxxx ξx 2 + 3ηxx ξxx ξx 2 )
/((ξx + pξy + qξp )3 (ξx + pξy )3 )
3
and calculation (treatment by DiffAlg) do not finish !
First optimization is given by the use cross derivations guided by the ranking of elim∂
∂ ∂
ination ξ η f such in [8] using the non commutative derivations {Dx , ∂y
, ∂p
, ∂q }. A
better alternative is to use the associated invariant derivations discussed later on.
3
Cartan’s method of equivalence
E. Cartan recast the problem of local equivalence Ef ∼Φ Ef¯ into to the calculation of the
integrability conditions of a linear Pfaffian system :
i
θ̄ (ā, x̄) = θi (a, x), (1 ≤ i ≤ m)
(8)
θ1 ∧ θ2 ∧ θ3 6= 0.
The change of coordinates φ ∈ Φ is solution of the linear Pfaffian system where two set of
variables (ā, x̄) and (a, x) play a symmetrical role. The problem is reduced to a symmetrical
calculation of the integrability conditions. These conditions symmetrically presented under
the form
Ī(ā, x̄) = I(a, x)
(9)
are called fundamental invariants. Every algorithm like diffalg based on the notion of
“ranking” breaks this symmetry. Cartan’s algorithm computes the integrability conditions
of (9) by separately and symmetrically treating the invariant 1-forms θ and θ̄. This divides
the number of variables by two.
To fix the ideas, let return to the equivalence problem (1). The differential equation
y 00 = f (x, y, p) is equivalent to the Pfaffian system (Σf )
{dp − f (x, y, p) dx = 0,
dy − p dx = 0}.
(10)
In the same way, the equation ȳ 00 = f¯(x̄, ȳ, p̄) is equivalent to the Pfaffian system (Σf¯).
Since the first prolongation preserves the module of the contact forms then (Σf ) and (Σf¯)
are equivalent under φ ∈ Φ if there exist functions a1 (x), a2 (x) and a3 (x) such that

 


dp̄ − f¯(x̄) dx̄
a1 (x) a2 (x) 0
dp − f (x) dx
 dȳ − p̄ dx̄  =  0
a3 (x) 0  dy − p dx  ,
dx̄
0
0
1
dx
this in turn can be rewritten in the ”symmetrical” form


 


ā1 (x̄) ā2 (x̄) 0
dp̄ − f¯(x̄) dx̄
a1 (x) a2 (x) 0
dp − f (x) dx
 0
ā3 (x̄) 0  dȳ − p̄ dx̄  =  0
a3 (x) 0  dy − p dx  .
0
0
1
dx̄
0
0
1
dx
{z
}|
{z
} |
{z
}|
{z
}
|
ḡ(ā)
ω̄
g(a)
ω
According to Cartan we define θ = g(a) ω and the equivalence problem takes the form (8).
We lunch Cartan’s method as explained in [15]. For brevity, we don’t present here the steps
and details of calculations but we shall only give the final involutive structure equations
[13, Page 59]
dθ1 = −θ1 ∧ θ4 + I1 θ2 ∧ θ3 ,
dθ2 = −θ1 ∧ θ3 − θ2 ∧ θ4 ,
dθ3 = 0,
dθ4 = I2 θ1 ∧ θ2 + I3 θ2 ∧ θ3 .
4
involving the fundamental invariants
1
1
I1 (a, x) = − (fp )2 − fy + Dx fp ,
4
2
I2 (a, x) =
fppp
,
2a3 2
I3 (a, x) =
fyp − Dx fpp
2a3
and
1
1
θ = a3 ( fp (x, y, p)p − f (x, y, p))dx − fp (x, y, p)dy + dp , θ2 = a3 (dy − pdx) ,
2
2
1
1
1
1
θ3 = dx, θ4 = ( fp (x, y, p) − fpp (x, y, p)p)dx + fpp (x, y, p)dy + da3 .
2
2
2
a3
1
Dual the coframe {θi }i=1,4 is the frame of invariant derivations {Xi }i=1,4
1 ∂
1 ∂
1 fp (x, y, p) ∂
1
∂
,
, X2 =
+
− fpp (x, y, p)
a3 ∂p
a3 ∂y 2
a3
∂p 2
∂a3
∂
∂
∂
1
∂
∂
X3 =
, X4 = a3
.
+p
+ f (x, y, p)
− fp (x, y, p)a3
∂x
∂y
∂p 2
∂a3
∂a3
P
The differential of a function H can re-expressed as dH = 4i=1 Xi (H) θi . Performing a
similar computation for the barred system Σf¯ leads to analogous structure equations (but
with bars!).
X1 =
4
Differential relations between the fundamental invariants
The invariants algebra attached to a given equation (for a given group of transformation
Φ) is a differential algebra generated by fundamental invariants and closed by the so called
invariant derivations. These derivations, which generally do not commute, allow to compute
a complete system of invariants from fundamental invariants. One obtains most of the
differential relations between the fundamental invariants using Poincaré lemma d ◦ d =
0 where d denotes the exterior derivation. This low cost computation does not require
the expression of the invariants in local coordinates or any elimination process. This is
particularly useful since the memory space taken up by the invariants can reach 1.1 M
bytes for some examples treated by our software.
For the problem (1), the relations provided by Poincaré lemma are
X1 I1 = −I3 ,
X4 I1 = 0,
X4 I2 = −2I2 ,
X1 I3 = −X3 I2 ,
X4 I3 = −I3 .
Clearly I1 and I2 form a basis of the differential algebra generated by the three invariants.
For the equivalence problem (1), where f¯ = 0 the invariants vanish. According to (9), the
corresponding invariants of y 00 = f (x, y, p) must vanish too. But we have already proved
that the vanishing of I1 and I2 is enough. One finds the same conditions of Theorem 1.
5
Change of coordinates calculation
Even when one has proved the existence of a change of coordinates φ ∈ Φ, the explicit computation of φ requires the solution of a Pfaffian system satisfying the Frobenius condition.
One obtains the transformation φ without integrating any differential equation when the
equations (E) and (Ē) have a symmetry group S ⊂ Φ with finite cardinality. Indeed, if the
function φ is the general solution of a differential system, it depend on at least one arbitrary
constant, and then the symmetry group S is continuous
5
In the other case, the transformation φ is algebraic and is obtained by solving in the
system of equations of the form (9) where appear the fundamental invariants and a finite
number of derivated invariants.
Example Let’s go back to the equivalence problem (1) with the first Painlevé equation
(P1). Suppose that (P1) is the target equation (that with bars) then the corresponding
invariants are
Ī1 (ā, x̄) = 12ȳ, Ī2 (ā, x̄) = Ī3 (ā, x̄) = 0.
According to the equality of the invariants (9) on the one hand I2 (a, x) = I3 (a, x) = 0,
giving the two last equations of (7). On the other hand, we have
1
1
12ȳ = Ī1 (ā, x̄) = I1 (a, x) = − fp 2 − fy + Dx fp
4
2
and this gives the sought change of coordinates ȳ = η(x, y). To find x̄ = x + C, we have
X̄32 Ī1 (ā, x̄) = −72ȳ 2 − 12x̄
and according again to the equality of the invariants we obtain
C = x̄ − x = −
1
1
I1 (a, x)2 − X32 I1 (a, x) − x.
24
12
If one wants to find same the conditions (7) on the function f , one differentiates C according to the invariant derivations and taking account of the relations provided by Poincaré.
The long and cumbersome formulas (7) take more compact forms expressed in terms of
differential invariants
X1 X32 I1 (a, x) = 0,
I1 (a, x) X2 I1 (a, x) + X2 X32 I1 (a, x) = 0,
I1 (a, x) X3 I1 (a, x) + X33 I1 (a, x) − 1 = 0
Remark If the target equation (P 1) is changed, the differential invariants provides
by Cartan’s method will not change and can be used afterwards. This is not the case of
elimination algorithms where one is brought for each target equation to remake all the
calculations.
Our software is available at www.lifl.fr/~neut/logiciels.
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