Journal of Lie Theory Volume 13 (2003) 167–188 \begin{matrix}c \\ ◯ \end{matrix} 2003 Heldermann Verlag

C^{∞}−Symmetries and Reduction of Equations Without Lie Point Symmetries

C. Muriel and J. L. Romero

Communicated by P. Olver

Abstract. It is proved that several usual methods of reduction for ordinary differential equations, that do not come from the Lie theory, are derived from the existence of C^{∞} -symmetries. This kind of symmetries is also applied to obtain two successive reductions of an equation that lacks Lie point symmetries but is a reduced equation of another one with a three dimensional Lie algebra of point symmetries. Some relations between C^{∞} -symmetries and potential symmetries are also studied.

1. Introduction

Let us consider the nth-order ordinary differential equation

Δ(x, u^{(n)}) = 0.  (1)

In the literature there appear several methods of reduction for (1). One of the most important is based on the existence of Lie point symmetries of the equation. However, there are also equations that lack Lie point symmetries but can be reduced. This is the case, for instance, when by means of

y = y(x, u), v = f(x, u, u_{x}),  (2)

equation (1) transforms into

Δ_{1}(y, v^{(n−1)}) = 0,  (3)

or when (1) can be written in the form

D_{x}(Δ_{2}(x, u^{(n−1)})) = 0,  (4)

where D_{x} denotes the total derivative with respect to the independent variable x. There are also many examples of integrable equations that lack Lie point symmetries ([2, 8, 9, 10, 11, 17]). In this paper we extend the concept of C^{∞}-symmetry, that appears in [17], and we prove that these classes of reductions are particular

ISSN 0949–5932 / $2.50 \begin{matrix}c \\ ◯ \end{matrix} Heldermann Verlag cases of the algorithm of reduction derived from the existence of extended C^{∞}-sy-mmetries of the equation. These C^{∞}-symmetries can be found by a well-defined algorithm, somewhat similar to the Lie algorithm.

It may also happen that (1) has no Lie point symmetries but, by means of a Bäcklund transformation

u = f(x, v, v_{x}),  (5)

equation (1) is transformed into an equation of the form

Δ̃(x, v^{(n+1)}) = 0  (6)

that has a non-trivial Lie algebra of point symmetries G. This case can happen when (1) is the reduced equation of (6) after using a generator of G: it may occur that the unused generators are not inheritable to the reduced equation. In the literature, these lost symmetries are called type I hidden symmetries (the term type II hidden symmetries refers to the symmetries that are gained after an order reduction). It may be said that the origin of the theory of hidden symmetries is in the concept of exponential vector fields ([20]), that provides order reductions but are not local vector fields. Many recent studies about lost symmetries have been done (see [1]-[3],[10]-[12], [16] and references therein). In [13], the concept of solvable structure ([6]) is applied to study hidden symmetries of type II. In particular, the authors show how the hidden symmetries of type II appearing in [12] are related to a solvable structure for the unreduced equations.

Let us observe that when (5) does not depend on v then X = \frac{∂}{∂v} is a Lie point symmetry of (6), (1) is the corresponding reduced equation and, if n ≥ 3, the order of the original equation can be reduced by two. By using some results that appear in [18], in this paper we prove that if an equation of the form (6) admits a three dimensional Lie algebra of point symmetries then the order of (6) can successively be reduced by three: if any of the generators of G is used to reduce the order then the remaining generators are inheritable, at some stage of the reduction process, as C^{∞}−symmetries of the reduced equations. We also show, through an example, that these C^{∞}−symmetries (derived from type I hidden symmetries) can be used to construct a solvable structure of the reduced equations.

It may also happen that, for some function f, (6) can be written in the conserved form

D_{x}(Δ_{3}(x, v^{(n)})) = 0,  (7)

for some function Δ_{3}. In this case we also have the trivial reduction

Δ_{3}(x, v^{(n)}) = 0,  (8)

which, as we prove in this paper, corresponds to a C^{∞}-symmetry of (6). Then, we could use the symmetries of (8) to obtain solutions of the original equation (1). This is the way followed by Bluman ([7]): these symmetries are called potential symmetries of (1). Potential symmetries are not, in general, either contact or Lie-Bäcklund symmetries of (1) because v, as defined by (5), cannot be expressed in terms of x, u and derivatives of u with respect to x to some finite order. Since Lie point symmetries of (8) can be used to reduce its order, potential symmetries are useful to find solutions of equation (1), because if v(x) = ϕ(x) solves equation (8) then u(x) = f(x, ϕ(x), ϕ_{x}(x)) solves equation (1). Let us observe that Lie point symmetries of any reduced equation

Δ_{3}(x, v^{(n)}) = C,  (9)

where C ∈ ℝ is an arbitrary constant, lead to a similar process for equation (1). Therefore, in this paper we will understand a potential symmetry of equation (1) as a Lie point symmetry of equation (8), for some C ∈ ℝ. In practice, Bäcklund transformations (5) that let us write equation (1) in conserved form (7) are difficult to find, if there is one, because this form is too restrictive. In this paper we prove that some special potential symmetries of (1), that are here called superpotential symmetries, can be considered as C^{∞}-symmetries of (1) and, therefore, two procedures to obtain solutions of (1) are available. This is illustrated through an example and both methods are compared.

2. Notations and preliminary results

Let us consider an nth-order ordinary differential equation

Δ(x, u^{(n)}) = 0,  (10)

with (x, u) ∈ M, for some open subset M ⊂ X × U ≃ ℝ^{2}. We denote by M^{(k)} the corresponding k−jet space M^{(k)} ⊂ X × U^{(k)}, for k ∈ ℕ. Their elements are (x, u^{(k)}) = (x, u, u_{1}, ⋯, u_{k}), where, for 1 ≤ i ≤ k, u_{i} denotes the derivative of order i of u with respect to x. We assume that the implicit function theorem can be applied to equation (10), and, as a consequence, that this equation can locally be written in the explicit form

u_{n} = Ψ(x, u^{(n−1)}).  (11)

The vector field

A_{(x,u)} = \frac{∂}{∂x} + u_{1} \frac{∂}{∂u} + ⋯ + Ψ(x, u^{(n−1)})\frac{∂}{∂u_{n−1}}  (12)

will be called the vector field associated with equation (11).

It is well-known ([23]) that a vector field X on M is a Lie point symmetry of equation (11) if and only if there exists a function ρ ∈ C^{∞}(M^{(1)}) such that

[X^{(n−1)}, A_{(x,u)}] = ρA_{(x,u)},  (13)

where X^{(n−1)} denotes the usual (n − 1)th prolongation of the vector field X. The generalized Lie symmetries ([22]) are vector fields Y defined on M^{(n−1)} that satisfy [Y, A_{(x,u)}] = ρA_{(x,u)}, for some function ρ ∈ C^{∞}(M^{(j)}).

A Lie point symmetry X can be used to reduce the order of the equation by one: we introduce a change of variables {y = y(x, u), α = α(x, u)} such that the vector field X can be written as X = \frac{∂}{∂α}, in some open set of variables (y, α), that will also be denoted by M. Since X is a Lie point symmetry of the equation,

(11) can be written in terms of variables (y, α^{(n)}) of M^{(n)} in the form

α_{n} = Φ(y, α_{1}, α_{2}, ⋯, α_{n−1}).  (14)

If we set w = α_{1} in (14) we obtain a reduced equation

w_{n−1} = Φ(y, w, w_{1}, ⋯, w_{n−2}),  (15)

where (y, w) are in some open set M_{1} ⊂ ℝ^{2}.

It can easily be checked that the vector field associated with equation (14), written in the new variables, is

A_{(y,α)} = \frac{1}{D_{x}(y(x, u))}A_{(x,u)}.  (16)

The vector field associated with the reduced equation (15) can be constructed as follows. Let π_{X}^{(k)} : M^{(k)} → M_{1}^{(k−1)} be the projection (y, α, α_{1}, ⋯, α_{k}) ↦ (y, w, ⋯, w_{k−1}) = (y, α_{1}, ⋯, α_{k}), for k ∈ ℕ. A vector field V on M^{(k)} will be called π_{X}^{(k)}−projectable if

[X^{(k)}, V] = fX^{(k)},  (17)

for some function f ∈ C^{∞}(M^{(k)}). This implies that V , in the variables (y, α^{(k)}), must take the following form

V = ξ(y, α_{1}, ⋯, α_{k})\frac{∂}{∂y} + η(y, α, α_{1}, ⋯, α_{k})\frac{∂}{∂α} + ∑_{i=1}^{k} η_{i}(y, α_{1}, ⋯, α_{k})\frac{∂}{∂α_{i}} . (18)

The π_{X}^{(k)}−projection of V on M_{1}^{(k−1)} is the vector field

(π_{X}^{(k)})_{∗}(V) = ξ(y, w, ⋯, w_{k−1})\frac{∂}{∂y} + ∑_{i=1}^{k} η_{i}(y, w, ⋯, w_{k−1})\frac{∂}{∂w_{i−1}} . (19)

With this definition, it can be checked that the vector field A_{(y,α)} is π_{X}^{(n−1)}−projectable and its projection is the vector field A_{(y,w)} associated with the reduced equation (15).

The concept of Lie point symmetry for an ordinary differential equation can be generalized in several ways: conditional symmetries, Lie-Bäcklund symmetries, etc. ([5],[4],[20], [21]). In [17], we have introduced the concept of C^{∞}−symmetry. This concept is somewhat similar to the concept of Lie point symmetry, but it is based on a different way to prolong vector fields. The following prolongation method generalizes the method that appears in [17].

Definition 2.1. Generalized prolongation formula

Let X = ξ(x, u)\frac{∂}{∂x} + η(x, u)\frac{∂}{∂u} be a vector field defined on M, and let λ ∈ C^{∞}(M^{(k)}) be an arbitrary function. The λ−prolongation of order n of X, denoted by X^{[λ,(n)]}, is the vector field defined on M^{(n+k−1)} by

X^{[λ,(n)]} = ξ(x, u)\frac{∂}{∂x} + ∑_{i=0}^{n}η^{[λ,(i)]}(x, u^{(i+k−1)})\frac{∂}{∂u_{i}}  (20)

where η^{[λ,(0)]}(x, u) = η(x, u) and

\begin{aligned}η^{[λ,(i)]}(x, u^{(i+k−1)}) = & D_{x} (η^{[λ,(i−1)]}(x, u^{(i+k−2)})) − D_{x}(ξ(x, u))u_{i} \\ & +λ (η^{[λ,(i−1)]}(x, u^{(i+k−2)}) − ξ(x, u)u_{i}) , \end{aligned}  (21)

for 1 ≤ i ≤ n.

Let us observe that, if λ = 0, the λ−prolongation of order n of X is the usual nth prolongation of X. If Q = η(x, u) − ξ(x, u)u_{1} is the characteristic of X = ξ(x, u)\frac{∂}{∂x} + η(x, u)\frac{∂}{∂u} then

X^{[λ,(n)]} = X_{Q}^{[λ,(n)]} + ξ(x, u)D_{x},  (22)

where

X_{Q}^{[λ,(n)]} = ∑_{i=1}^{n}(D_{x} + λ)^{i}(Q)\frac{∂}{∂u_{i}} .  (23)

Definition 2.2. Let Δ(x, u^{(n)}) = 0 be an nth-order ordinary differential equation. We will say that a vector field X, defined on M, is a C^{∞}(M^{(k)})−symmetry of the equation, 1 ≤ k < n, if there exists a function λ ∈ C^{∞}(M^{(k)}) such that

X^{[λ,(n)]}(Δ(x, u^{(n)})) = 0, when Δ(x, u^{(n)}) = 0.  (24)

In this case we will also say that X is a λ−symmetry or a C^{∞}-symmetry, if there is no place for confusion.

By a straightforward generalization of a result that appears in [17], it can be checked that a vector field X on M is a C^{∞}(M^{(k)})−symmetry of the equation (11) if and only if there exist two functions, λ, ρ ∈ C^{∞}(M^{(k)}), such that

[X^{[λ,(n−1)]}, A_{(x,u)}] = λX^{[λ,(n−1)]} + ρA_{(x,u)}.  (25)

Let us observe that if X is a Lie point symmetry then, [X^{(n−1)}, A_{(x,u)}] = ρA_{(x,u)} for some ρ ∈ C^{∞}(M^{(1)}), and for any function f ∈ C^{∞}(M), fX^{(n−1)} satisfies

[fX^{(n−1)}, A_{(x,u)}] = −A_{(x,u)}(f)X^{(n−1)} + fρA_{(x,u)}.  (26)

Therefore, fX is a C^{∞}(M^{(1)})−symmetry for λ = −\frac{A_{(x,u)}(f)}{f}.

Conversely, if X is a λ−symmetry, for λ ∈ C^{∞}(M^{(k)}), and f ∈ C^{∞}(M^{(j)}), then, by using (25), we have

[fX^{[λ,(n−1)]}, A_{(x,u)}] = (fλ − A_{(x,u)}(f))X^{[λ,(n−1)]} + fρA_{(x,u)}.  (27)

If we choose f such that A_{(x,u)}(f) = fλ then fX^{[λ,(n−1)]} becomes a Lie symmetry in the generalized sense.

Example 2.3. It can be proved ([17]) that the second order equation:

u_{xx} + \frac{x^{2}}{4u^{3}} + u + \frac{1}{2u} = 0  (28)

has no Lie point symmetries. The vector field X = u\frac{∂}{∂u} is a λ−symmetry, for

λ = \frac{x}{u^{2}} , of equation (28). In this case, A_{(x,u)} = \frac{∂}{∂x} + u_{x} \frac{∂}{∂u} − (\frac{x^{2}}{4u^{3}} + u + \frac{1}{2u}) \frac{∂}{∂u_{x}}

and, by (21), X^{[λ,(1)]} = u\frac{∂}{∂u} + (u_{x} + \frac{x}{u})\frac{∂}{∂u_{x}} . The vector field Y = X^{[λ,(1)]} verifies formula (25), because

[Y, A_{(x,u)}] = [X^{[λ,(1)]}, A_{(x,u)}] = \frac{x}{u} \frac{∂}{∂u} + (\frac{xu_{x}}{u^{2}} + \frac{x^{2}}{u^{3}} ) \frac{∂}{∂u_{x}} = λY.  (29)

However, there is no function ρ such that (13) is satisfied. Therefore, [Y, A_{(x,u)}] ≠ ρA_{(x,u)} for any function ρ; i.e. Y is not a Lie symmetry, in the generalized sense, of equation (28). In order to find a function f such that fY is a generalized Lie symmetry, we must solve the following partial differential equation:

f_{x} + f_{u}u_{x} − f_{u_{x}} (\frac{x^{2}}{4u^{3}} + u + \frac{1}{2u}) = \frac{x}{u^{2}} f.  (30)

Therefore, it seems that it is easier to calculate C^{∞}−symmetries than the associated generalized Lie symmetries.

An algorithm to determine the C^{∞}−symmetries of an equation follows from (24): this equation generates a system of equations for the infinitesimals of the C^{∞}−symmetry X, in which λ is also an unknown function. This gives, with respect to Lie method, a higher level of freedom in the resolution of these determining equations. In particular, it may happen that the determining equations for Lie point symmetries only admit the trivial solution but the corresponding equations for C^{∞}−symmetries have non-trivial solutions, as in Example 2.3.

The C^{∞}(M^{(k)})-symmetries can be used to obtain reduction processes. The corresponding method is described in Theorem 2.5. In order to prove this theorem, we need a preliminary result.

Theorem 2.4. Let X be a vector field defined on M ⊂ X × U and let λ ∈ C^{∞}(M^{(k)}). If α = α(x, u^{(j)}), β = β(x, u^{(j)}) are functions in C^{∞}(M^{(j)}) such that

X^{[λ,(j)]}(α(x, u^{(j)})) = X^{[λ,(j)]}(β(x, u^{(j)})) = 0,  (31)

then

X^{[λ,(j+1)]} (\frac{D_{x}α(x, u^{(j)})}{D_{x}β(x, u^{(j)})}) = 0.  (32)

Proof. It is clear that

[X^{[λ,(j+1)]}, D_{x}] = λX^{[λ,(j+1)]} + μD_{x},  (33)

where μ = −D_{x}(X(x)) − λX(x) ∈ C^{∞}(M^{(k)}). Therefore,

\begin{matrix}X^{[λ,(j+1)]} (\frac{D_{x}α}{D_{x}β}) = \frac{1}{(D_{x}β)^{2}} (D_{x}β · X^{[λ,(j+1)]}(D_{x}α) − D_{x}α · X^{[λ,(j+1)]}(D_{x}β)) \\ = \frac{1}{(D_{x}β)^{2}} (D_{x}β · [X^{[λ,(j+1)]}, D_{x}](α) − D_{x}α · [X^{[λ,(j+1)]}, D_{x}](β)) \\ = \frac{1}{(D_{x}β)^{2}} (D_{x}β · (μ · D_{x}α) − D_{x}α · (μ · D_{x}β)) = 0. \end{matrix} (34)

This proves the theorem.

The following theorem, and its proof, gives a method to reduce an equation that admits a C^{∞}-symmetry.

Theorem 2.5. Let X be a λ-symmetry, with λ ∈ C^{∞}(M^{(k)}), of the equation Δ(x, u^{(n)}) = 0. Let y = y(x, u) and w = w(x, u, u_{1}, ⋯, u_{k}) be two functionally independent invariants of X^{[λ,(n)]}. The general solution of the equation can be obtained by solving a reduced equation of the form Δ_{r}(y, w^{(n−k)}) = 0 and an auxiliary kth-order equation w = w(x, u, u_{1}, ⋯, u_{k}).

Proof. Let y = y(x, u) and w = w(x, u, u_{1}, ⋯, u_{k}) be two functionally independent invariants of X^{[λ,(k)]} such that w depends on u_{k}. By Theorem 2.4,

w_{1} = \frac{D_{x}w(x, u, u_{1}, ⋯, u_{k})}{D_{x}y(x, u)}  (35)

is an invariant for X^{[λ,(k+1)]}. The set {y, w, w_{1}} is functionally independent, because w_{1} depends on u_{k+1}. From w_{1} and y we can obtain, by derivation, a (k + 2)th-order invariant for X^{[λ,(n)]} and so on. Therefore, the set

{y, w, w_{1}(x, u, u_{1}, ⋯, u_{k+1}), ⋯, w_{n−k}((x, u, u_{1}, ⋯, u_{n})}  (36)

is a set of functionally independent invariants of X^{[λ,(n)]}. Since X is, by hypothesis, a C^{∞}(M^{(k)})-symmetry of Δ(x, u^{(n)}) = 0, it can be checked, by Definition 2.1, that this equation can be written in terms of (36). The resulting equation is a (n−k)thorder equation of the form

Δ_{r}(y, w^{(n−k)}) = 0.  (37)

We can recover the general solution of Δ(x, u^{(n)}) = 0 from the general solution of (37) and the corresponding kth-order auxiliary equation:

w = w(x, u, u_{1}, ⋯, u_{k}).  (38)

3. C^{∞}−Symmetries and order reductions

In this section we show that many of the known reduction processes for ordinary differential equations can be obtained, through the former method, as a consequence of the existence of C^{∞}−symmetries of the given equations.

Theorem 3.1. Let

Δ_{1}(x, u^{(n)}) = 0  (39)

be an nth-order ordinary differential equation. Let us suppose that there exists a transformation

\begin{matrix}y = y(x, u), \\ w = w(x, u, u_{1}), \end{matrix} \}  (40)

where w_{u_{1}} ≠ 0, such that (39) can be written, in terms of variables (y, w), in the form

Δ_{2}(y, w^{(n−1)}) = 0.  (41)

There exists a C^{∞}−symmetry X of equation (39) such that (41) is the corresponding reduced equation.

Proof. Let α ∈ C^{∞}(M) be such that the functions y, α are functionally independent. We denote α_{1} = \frac{dα}{dy} = \frac{D_{x}(α(x,u))}{D_{x}(y(x,u))} ∈ C^{∞}(M^{(1)}). We consider on M^{(1)} the local coordinates (y, α, α_{1}). We determine a vector field of the form X = ξ(y, α)\frac{∂}{∂y} + η(y, α)\frac{∂}{∂α} and a function λ(y, α, α_{1}) ∈ C^{∞}(M^{(1)}) such that X is a λ−symmetry of the equation and the functions y, w in (40) are invariants of X^{[λ,(1)]}.

We set ξ = 0 and η = 1; by Definition 2.1, X^{[λ,(1)]} = \frac{∂}{∂α} + λ\frac{∂}{∂α_{1}} . We determine λ with the condition X^{[λ,(1)]}(w) = 0 and we find that λ = −\frac{w_{α}}{w_{α_{1}}} .

1. Let us prove that the vector field X = \frac{∂}{∂α} is a λ−symmetry of the equation for the function λ = −\frac{w_{α}}{w_{α_{1}}} . We denote w_{i} = \frac{d^{i)}w}{dy^{i)}}, for 1 ≤ i ≤ n−1. It is clear that the set {y, α, w, ⋯, w_{n−1}} is a system of coordinates in M^{(n)}. By the construction of X and λ, we have that {y, w, ⋯, w_{n−1}} are invariants for the vector field X^{[λ,(n)]}. Therefore, in the new local coordinates, X^{[λ,(n)]} = \frac{∂}{∂α}. Since, by hypothesis, equation (39) can be written in these local coordinates as equation (41), we obtain

X^{[λ,(n)]}(Δ_{2}(y, w^{(n−1)})) = \frac{∂}{∂α}(Δ_{2}(y, w^{(n−1)})) = 0.  (42)

This proves that X is a λ−symmetry of the equation.

2. In order to check that (41) is the reduced equation that, by Theorem 2.5, corresponds to the λ-symmetry, it is sufficient to observe that the reduced equation can be obtained by writing the equation in terms of the complete system {y, w, ⋯, w_{n−1}} of invariants of X^{[λ,(n)]}.

This proves the theorem.

Theorem 3.2. Let

D_{x}(Δ(x, u^{(n−1)})) = 0,  (43)

be an nth-order ordinary differential equation such that Δ is an analytical function of its arguments. There exists a function λ ∈ C^{∞}(M^{(k)}), k ≤ n − 1, such that the vector field X = \frac{∂}{∂u} is a λ−symmetry of the equation. The trivial order reduction

Δ(x, u^{(n−1)}) = C, C ∈ ℝ,  (44)

admitted by the equation, can be obtained as the auxiliary equation that corresponds to the reduction process, by means of X, that appears in Theorem 2.5.

Proof. 1. We try to find λ ∈ C^{∞}(M^{(k)}), k ≤ n − 1, with the condition

X^{[λ,(n−1)]}(Δ(x, u^{(n−1)})) = 0 when D_{x}(Δ(x, u^{(n−1)})) = 0.  (45)

In terms of the characteristic Q ≡ 1 of X, we have

X^{[λ,(n−1)]} = ∑_{i=0}^{n−1}(D_{x} + λ)^{i}(1)\frac{∂}{∂u_{i}} .  (46)

Hence, the equation X^{[λ,(n−1)]}(Δ(x, u^{(n−1)})) = 0 can be written as

(D_{x} + λ)^{n−1}(1)\frac{∂Δ}{∂u_{n−1}} = − ∑_{i=0}^{n−2}(D_{x} + λ)^{i}(1)\frac{∂Δ}{∂u_{i}}.  (47)

Let us observe that, since the set of analytical functions is closed under differentiation, if λ is an analytical function in M^{(k)} then, for 1 ≤ i ≤ n − 1, the function defined by (D_{x} + λ)^{i}(1) is analytical in M^{(k+i−1)} and in the partial derivatives of λ with respect to all its arguments up to the order i − 1.

Since the order of (43) is n, we have Δ_{u_{n−1}}(x, u^{(n−1)}) ≠ 0, in some open set of M^{(n−1)}. Therefore, the implicit function theorem for analytical functions ([15]) let us, locally, write (43) in the form u_{n} = F (x, u^{(n−1)}), where F is an analytical function on its arguments.

In (47), we must replace u_{n} by F and u_{n+h}, h ≥ 1, by the corresponding derivatives. The resulting equation is defined by functions that depend analytically on their arguments. It is easy to see that, in (47), the derivative

\frac{∂^{n−2)}λ}{∂x^{n−2)}}  (48)

does only appear in the first member and its coefficient is Δ_{u_{n−1}} ≠ 0. Equation (47) can be solved for \frac{∂^{n−2)}λ}{∂x^{n−2)}} and the resulting partial differential equation for λ can be written in the form

\frac{∂^{n−2)}λ}{∂x^{n−2)}} = G(x, u^{(n−1)}, λ^{(n−2)}),  (49)

where λ^{(n−2)} denotes the partial derivatives of λ with respect to its arguments, of orders ≤ n − 2, and the function G is analytical on its arguments and does not depend on \frac{∂^{n−2)}λ}{∂x^{n−2)}}.

With these conditions, Cauchy-Kovalevsky Theorem ([15]) ensures the existence of analytical solutions, λ(x, u^{(n−1)}), to equation (49).

Next, we prove that, if λ is a solution of (49), the vector field X = \frac{∂}{∂u} is a λ-symmetry of the equation.

By Definition 2.2, it can be checked, that

[X^{[λ,(n−1)]}, D_{x}] = λX^{[λ,(n−1)]}.  (50)

By applying both members of this expression to Δ(x, u^{(n−1)}) we get

\begin{matrix}X^{[λ,(n−1)]}(D_{x}(Δ(x, u^{(n−1)}))) − D_{x}(X^{[λ,(n−1)]}(Δ(x, u^{(n−1)}))) \\ = λX^{[λ,(n−1)]}(Δ(x, u^{(n−1)})). \end{matrix}  (51)

Since λ is such that X^{[λ,(n−1)]}(Δ(x, u^{(n−1)})) = 0 when D_{x}(Δ(x, u^{(n−1)})) = 0, we get

X^{[λ,(n−1)]}(D_{x}(Δ(x, u^{(n−1)}))) = 0, when D_{x}(Δ(x, u^{(n−1)})) = 0,  (52)

i.e. X is a λ-symmetry of the equation.

2. The algorithm to reduce the order of the equation, by using the λ-sy-mmetry X, leads to the first order ordinary differential equation

w_{y} = 0,  (53)

where y = x and w = Δ(x, u^{(n−1)}) are two functionally independent invariants of the vector field X^{[λ,(n−1)]}. The general solution of equation w_{y} = 0 is w = C, C ∈ ℝ. Therefore, the general solution the original equation is obtained by solving the ordinary differential equation (44).

4. Reduction of equations without Lie point symmetries.

Let us suppose that the ordinary differential equation

Δ(x, u^{(n)}) = 0  (54)

has no Lie point symmetries. We will also suppose that by means of u = f(x, v_{x}) equation (54) is transformed into the (n + 1)th-order equation

Δ̃(x, v^{(n+1)}) = 0,  (55)

and that this equation has a three dimensional Lie algebra of point symmetries G. By Theorem 3.1, the order reduction of equation (55) to equation (54) corresponds to the use of the Lie point symmetry \frac{∂}{∂v}. By the classification of three dimensional Lie algebras that appears in [14], the structure of the Lie algebra G generated by X_{1}, X_{2} and X_{3} corresponds, by means of some linear combination of the generators, to some of the types enumerated in the following table:

Let us analyze the reduction of (54) by using the three dimensional symmetry algebra G of (55). We denote X = \frac{∂}{∂v}, the vector field that lets reduce (55) to (54). Here X may be any of the generators X_{i}, 1 ≤ i ≤ 3 and, depending on i, several ways of step by step reduction of (54) can be followed.

Table 1: Three-dimensional solvable algebras

I  II  III  IV
[X_{1}, X_{2}] = 0  [X_{1}, X_{2}] = 0  [X_{1}, X_{2}] = X_{3}  [X_{1}, X_{2}] = 0
[X_{1}, X_{3}] = 0  [X_{1}, X_{3}] = X_{3}  [X_{1}, X_{3}] = 0  [X_{1}, X_{3}] = aX_{1} + bX_{2}
[X_{2}, X_{3}] = 0  [X_{2}, X_{3}] = 0  [X_{2}, X_{3}] = 0  [X_{2}, X_{3}] = cX_{1} + dX_{2}

Table 2: Three-dimensional non-solvable algebras

\begin{matrix}V & VI \\ [X_{1}, X_{2}] = 2X_{3} & [X_{1}, X_{2}] = X_{3} \\ [X_{1}, X_{3}] = X_{1} & [X_{1}, X_{3}] = −X_{2} \\ [X_{2}, X_{3}] = −X_{2} & [X_{2}, X_{3}] = X_{1} \end{matrix}

1. In cases I to III, the kernel Z(G) = {X ∈ G : [X, Y] = 0, Y ∈ G} is not trivial. If we use any of the generators to perform a first reduction then the corresponding reduced equation conserves at least one Lie point symmetry. Since, by hypothesis, (54) has no Lie point symmetries, X = \frac{∂}{∂v} cannot be in G. This contradiction proves that, with our hypotheses, these three cases cannot happen.

2. In case IV, the following chains of normal subalgebras in G hold:

\begin{matrix}< X_{1} > ▷ < X_{1}, X_{2} > ▷ < X_{1}, X_{2}, X_{3} >, \\ < X_{2} > ▷ < X_{1}, X_{2} > ▷ < X_{1}, X_{2}, X_{3} > . \end{matrix}

If we first reduce with X_{i}, 1 ≤ i ≤ 2, the reduced equation always inherits a Lie point symmetry. Since (54) has no Lie point symmetries, necessarily the first reduction is performed by using X = X_{3} = \frac{∂}{∂v}. Let us study how the symmetries X_{1} and X_{2} can be used to reduce, successively, the order of (54) by two.

(a) We may assume that b = 0: if this is not the case, we can use a linear change of coordinates (possibly with complex coefficients) to get b = 0. Let f_{1} ∈ C^{∞}(M) be a function such that X_{3}(f_{1}) = af_{1}. Then

[f_{1}X_{1}^{(k)}, X_{3}^{(k)}] = f_{1}(aX_{1}^{(k)}) − X_{3}(f_{1})X_{1}^{(k)} = 0 (k ∈ ℕ). (56)

By (17), f_{1}X_{1}^{(k)} is a π_{X_{3}}^{(k)}−projectable vector field. Since [X_{1}^{(k)}, D_{x}] = −D_{x}(X_{1}(x))D_{x}, we get

[f_{1}X_{1}^{(k)}, D_{x}] = −\frac{D_{x}(f_{1})}{f_{1}}f_{1}X_{1}^{(k)} − f_{1}D_{x}(X_{1}(x))D_{x},  (57)

and, by taking (33) into account, it follows that

f_{1} · X_{1}^{(k)} = (f_{1}X_{1})^{[λ_{1},(k)]}, for λ_{1} = −\frac{D_{x}(f_{1})}{f_{1}}.  (58)

Let us denote Y_{1} = (π_{X_{3}}^{(1)})_{∗}(f_{1}X_{1}^{(1)}). The vector field Y_{1} is a C^{∞}−symmetry of equation (54), for the function λ_{1} given, in coordinates (x, v), by:

λ_{1} = −\frac{D_{x}(f_{1})}{f_{1}}.  (59)

(b) By Theorem 2.5, we use the λ_{1}−symmetry Y_{1} to reduce the order of equation (54). Let {y = y(x, u), w = w(x, u, u_{1})} be two functionally independent invariants of Y_{1}^{[λ_{1},(1)]}. We denote by

Δ̂(y, w^{(n−1)}) = 0  (60)

the corresponding reduced equation. Let β = β(x, u) be such that Y_{1}(β) = 1. We denote π^{(k)} = π_{Y_{1}}^{(k−1)} ∘ φ ∘ π_{X_{3}}^{(k)}, where φ stands for the change of variables {x, u^{(k)}} ↔ {y, β, w^{(k−1)}}. Let f_{2} be a function such that X_{3}(f_{2}) = df_{2} and X_{1}(f_{2}) = 0. Since, for 2 ≤ k ≤ n − 1,

(i) [f_{1}X_{1}^{(k)}, f_{2}X_{2}^{(k)}] = f_{1}X_{1}^{(k)}(f_{2})X_{2}^{(k)} − f_{2}X_{2}^{(k)}(f_{1})X_{1}^{(k)} = f_{12}^{1}f_{1}X_{1}^{(k)}, where f_{12}^{1} = −\frac{f_{2}}{f_{1}} X_{2}^{(k)}(f_{1}),

(ii) [f_{2}X_{2}^{(k)}, X_{3}^{(k)}] = f_{2}(cX_{1}^{(k)} + dX_{2}^{(k)}) − X_{3}^{(k)}(f_{2})X_{2}^{(k)} = f_{23}^{2}f_{1}X_{1}^{(k)}, where f_{13}^{2} = \frac{f_{1}}{f_{2}} c,

the vector field f_{2}X_{2}^{(k)} is π^{(k)}−projectable. Since, by hypothesis, the vector field X_{2} is a Lie point symmetry of the equation (55), we can write

[f_{2}X_{2}^{(n−1)}, A_{(x,v)}] = λ_{2}f_{2}X_{2}^{(n−1)} + μA_{(x,v)}

for some functions λ_{2}, μ ∈ C^{∞}(M^{(1)}). By (i) and (ii), the Jacobi identity for the vector fields

{f_{1}X_{1}^{(n−1)}, f_{2}X_{2}^{(n−1)}, μA_{(x,v)}} and {f_{2}X_{2}^{(n−1)}, X_{3}^{(n−1)}, μA_{(x,v)}}

let us prove that the functions λ_{2} and μ are both f_{1}X_{1}^{(n−1)}-invariant and X_{3}^{(n−1)}-invariant. Hence,

[(π^{(n−2)})_{∗}(f_{2}X_{2}^{(n−1)}), A_{(y,w)}] = λ̃_{2}(π^{(n−2)})_{∗}(f_{2}X_{2}^{(n−1)}) + μ̃A_{(y,w)}, (61)

where (π^{(n−2)})^{∗}(λ_{2}) = λ̃_{2} and (π^{(n−2)})^{∗}(μ) = μ̃. Let us denote Z_{2} = (π^{(2)})_{∗}(f_{2}X_{2}^{(2)}). Clearly, (61) shows that Z_{2} is a C^{∞}-symmetry of the reduced equation, and Z_{2} can be used to reduce again the order.

3. In case V we have

< X_{1} > ▷ < X_{1}, X_{3} >, < X_{2} > ▷ < X_{2}, X_{3} > .  (62)

It can be assumed, as before, that vector field X is X_{3}. By proceeding as for Case IV, it can be checked ([18]) that if f_{1}, f_{2} ∈ C^{∞}(M) are such that

X_{3}(f_{1}) = f_{1}, X_{3}(f_{2}) = −f_{2},  (63)

then the vector field f_{1}X_{1}^{(1)} and f_{2}X_{2}^{(1)} are π_{X_{3}}^{(1)}−projectable. The projections Y_{1} = (π_{X_{3}}^{(1)})_{∗}(f_{1}X_{1}^{(1)}) and Y_{2} = (π_{X_{3}}^{(1)})_{∗}(f_{2}X_{2}^{(1)}), are C^{∞}−symmetries of equation (54). Any of these two C^{∞}−symmetries can be used to reduce the order of equation (54). The unused C^{∞}−symmetry can also be recovered as a C^{∞}−symmetry of the corresponding reduced equation ([18]).

4. In case VI the vector field \frac{∂}{∂v} can be any of the generators and the main results are also valid. We have developed an procedure that allows us to recover the unused generators for the algebra as C^{∞}−symmetries for the reduced equations. Theoretical results about this situation have been developed in [19].

The former results prove that if (54) has no Lie point symmetries but (55) has a three dimensional Lie algebra of point symmetries G such that \frac{∂}{∂v} ∈ G then, by using the generators of G, the order of (54) can successively be reduced by two.

In order to illustrate these ideas, let us consider the following second order differential equation:

\begin{matrix}8(ux + 1)u_{xx} − 24xu_{x}^{2} − 2(u^{2}x^{2} + 2ux + 24u + 1)u_{x} \\ +x^{3}u^{5} + (5x^{2} + 8x)u^{4} + (7x + 32)u^{3} + 3u^{2} = 0. \end{matrix}  (64)

In Appendix A we prove that this equation has no Lie point symmetries. We are going to transform (64) into a third order equation that admits a (non-solvable) three-dimensional algebra of symmetry. We show how this algebra allows us to recover two of its point symmetries as C^{∞}−symmetries of equation (64). As a consequence, equation (64) can be solved through two first order equations: one of them is a Ricatti equation, and the other one can be solved by a quadrature, because it is a linear equation.

By means of the transformation u = v_{x}, equation (64) becomes the third order equation

\begin{matrix}8(v_{x}x + 1)v_{xxx} − 24xv_{xx}^{2} − 2(v_{x}^{2}x^{2} + 2v_{x}x + 24v_{x} + 1)v_{xx} \\ +x^{3}v_{x}^{5} + (5x^{2} + 8x)v_{x}^{4} + (7x + 32)v_{x}^{3} + 3v_{x}^{2} = 0. \end{matrix}  (65)

It can be checked that this equation admits a three-dimensional Lie algebra generated by

X_{1} = e^{−v} \frac{∂}{∂x}, X_{2} = −e^{v}x^{2} \frac{∂}{∂x} + 2e^{v}x\frac{∂}{∂v} , X_{3} = \frac{∂}{∂v} .  (66)

Since

[X_{1}, X_{2}] = 2X_{3}, [X_{1}, X_{3}] = X_{1}, [X_{2}, X_{3}] = −X_{2},  (67)

the symmetry algebra of equation (65) is the non-solvable Lie algebra sl(2, ℝ) associated to the unimodular group, and corresponds to case V. Equation (64) can be considered as the reduced equation that corresponds to the reduction derived from the use of the Lie point symmetry X_{3}. Symmetries X_{1} and X_{2} are not inheritable, as Lie point symmetries, to the reduced equation (64). However, these lost symmetries (hidden symmetries of type I) can be recovered ([18]) as C^{∞}−symmetries of equation (64). Let us choose f_{1} = e^{v} and f_{2} = e^{−v}. The vector fields f_{1}X_{1}^{(1)} and f_{2}X_{2}^{(1)} are π_{X_{3}}^{(1)}−projectable and the projections

Y_{1} = \frac{∂}{∂x} + u^{2} \frac{∂}{∂u}, Y_{2} = −x^{2} \frac{∂}{∂x} + (2 + 4xu + x^{2}u^{2})\frac{∂}{∂u}  (68)

are C^{∞}−symmetries of equation (64), for λ_{1} = −u and λ_{2} = u, respectively (see (59)). Let us observe that the C^{∞}−symmetries Y_{1} and Y_{2} could also be found by solving the determining equations for the C^{∞}−symmetries of equation (64).

In terms of variables

\{z = \frac{1}{u} + x, β = x, μ = \frac{β_{z}}{β − z} = \frac{u^{3}}{u_{x} − u^{2}} \} ,  (69)

the vector field Y_{1}^{[λ_{1},(1)]} is simply expressed as Y_{1}^{[λ_{1},(1)]} = \frac{∂}{∂β} . The corresponding reduced equation is

24μ + 8μ_{z}z + μ^{3}z^{3} − 2μ^{2}z(4 + z) = 0.  (70)

This first order equation inherits Y_{2} as the C^{∞}−symmetry

Z_{2} = −2z^{2} \frac{∂}{∂z} − 2μz(−3 + μz)\frac{∂}{∂μ}, λ̃_{2} = −μ.  (71)

In the system of coordinates \{s = \frac{μz−2}{z^{3}μ}, r = \frac{1}{2z}\} the vector field Z_{2} can be written as Z_{2} = \frac{∂}{∂r}. Therefore, in these coordinates, equation (70) takes the form of the Ricatti equation

r_{s} = \frac{4r^{2}}{s} − 1.  (72)

When the general solution of equation (72) is expressed, in terms of {z, μ}, as μ = H(z, C_{1}), the auxiliary first order equation that let us recover the solution of equation (70) is the linear equation β_{z} = H(z, C_{1})(β − z), which can be solved by quadrature.

Next we show how the C^{∞}−symmetries (68) can be used to construct a solvable structure for equation (64). The C^{∞}−symmetries Y_{1} and Y_{2} are in involution because [Y_{1}^{[λ_{1},(1)]}, Y_{2}^{[λ_{2},(1)]}] = −2xY_{1}^{[λ_{1},(1)]}. Let g_{1} be any function such that A_{(x,u)}(g_{1}) = λ_{1}g_{1}. Then g_{1}Y_{1}^{[λ_{1},(1)]} is a generalized Lie symmetry of equation (64), that is

[g_{1}Y_{1}^{[λ_{1},(1)]}, A_{(x,u)}] = ρ_{1}A_{(x,u)},  (73)

for some function ρ_{1}. Let g_{2} be a function such that

A_{(x,u)}(g_{2}) = λ_{2}g_{2} and Y_{1}^{[λ_{1},(1)]}(g_{2}) = 0.  (74)

A function g_{2} can be found as follows: since {z, μ} in (69) are invariants of Y_{1}^{[λ_{1},(1)]} = \frac{∂}{∂β} , we can choose any function g_{2} = g_{2}(z(x, u), μ(x, u, u_{x})) such that A_{(z,μ)}(g_{2}) = λ̃_{2}g_{2} = −μg_{2}, where A_{(z,μ)} is the vector field associated to equation (70). If g_{2} verifies (74), then

[g_{2}Y_{2}^{[λ_{2},(1)]}, A_{(x,u)}] = ρ_{2}A_{(x,u)}, and [g_{2}Y_{2}^{[λ_{2},(1)]}, g_{1}Y_{1}^{[λ_{1},(1)]}] = gY_{1}^{[λ_{1},(1)]}, (75)

for some function ρ_{2} and g = 2xg_{1}g_{2} + g_{2}Y_{2}^{[λ_{2},(1)]}(g_{1}). By (73) and (75), we deduce that {A_{(x,u)}, g_{1}Y_{1}^{[λ_{1},(1)]}, g_{2}Y_{2}^{[λ_{2},(1)]}} constitutes a solvable structure of equation (64).

The same construction can be done for any equation with symmetry algebra of case V, because the C^{∞}−symmetries Y_{1} and Y_{2} are always in involution (see Theorem 4 in [18]).

5. C^{∞}-symmetries and potential symmetries.

In this section we study some relationships between potential symmetries, introduced by Bluman ([7]), and C^{∞}-symmetries. Although we only consider 2nd-order equations, the ideas included in this section may directly be generalized to equations of greater order.

Let us assume that equation

Δ(x, u^{(2)}) = 0  (76)

has no Lie point symmetries and that by means of the Bäcklund transformation u = f(x, v_{x}) equation (76) can be written in conserved form

D_{x}(Δ̃(x, v^{(2)})) = 0.  (77)

Equation (77) can be reduced to (76) by means of the Lie point symmetry \frac{∂}{∂v}. By other hand, (77) can trivially be reduced to equation

Δ̃(x, v^{(2)}) = C,  (78)

where C is an arbitrary constant. By Theorem 3.2, this reduction is also associated to the existence of a C^{∞}-symmetry of (77); in general, this reduction does not come from the existence of a Lie point symmetry.

If, for some C ∈ ℝ, X is a Lie point symmetry of (78) then X is not necessarily a point symmetry of equation Δ̃(x, v^{(2)}) = C′, for C′ ≠ C, and is not a point symmetry of (77). In this paper, any point symmetry of equation (78), for some C ∈ ℝ, will be called potential symmetry of equation (76). Let us recall that Bluman ([7]) considered the concept of potential symmetry only for C = 0 and, in this case, it may happen that the general solution of (76) cannot be obtain from one of the equations of type (78). This occurs in some trivial cases. For instance, if equation (76) is u_{xx} = 0 then, by writing u = v_{x}, equation (77) is v_{xxx} = 0 and equation (78) is v_{xx} = 0. The general solution of this equation takes the form v = ax + b, with a, b ∈ ℝ. Clearly u = v_{x} = a does not give the general solution of u_{xx} = 0.

If X is a vector field, in variables (x, v), that is a point symmetry of every equation of the form (78) (and does not depend on C) then X will be called a super-potential symmetry of equation (76). We prove in this section that superpotential symmetries can be recovered as C^{∞}-symmetries of (76).

Let us assume that the vector field X is a Lie point symmetry of equation (78), for every C ∈ ℝ; i.e. X is a super-potential symmetry of equation (76). In this case Δ̃(x, v^{(2)}) is a X^{(2)}-invariant function. If ϕ = ϕ(x) is an arbitrary solution of (77) then D_{x}(Δ̃(x, ϕ^{(2)}(x))) = 0 and there exists a constant C ∈ ℝ such that Δ̃(x, ϕ^{(2)}(x)) = C. Hence ϕ is a solution of equation (78) and X transforms ϕ into another solution of the same equation (78). This proves that the transformed solution is also a solution of (77). Therefore X is also a Lie point symmetry of (77). This fact can also be proved by another procedure: since [X^{(3)}, D_{x}] = μD_{x}, for some function μ, it follows that μD_{x}Δ̃ = X^{(3)}(D_{x}Δ̃) − D_{x}(X^{(3)}Δ̃) = X^{(3)}(D_{x}Δ̃) and X^{(3)}(D_{x}Δ̃) = 0 when D_{x}Δ̃ = 0.

If X_{1}, X_{2} are super-potential symmetries of (76) and X_{1}, X_{2}, X_{3}, with X_{3} = \frac{∂}{∂v}, are the generators of a Lie algebra, then we have proved, in section 4., that X_{1}, and X_{2}, can be recovered as C^{∞}−symmetries of equation (76). Since C^{∞}−symmetries can be calculated by a well-defined algorithm, and can be used to reduce the order, we could solve the original equation without knowing the associated conserved form (77), which is needed to calculate potential symmetries. In this case two methods of reduction can be used, that are illustrated through the following example.

Let us consider the following second order differential equation:

u^{5} + e^{2(\frac{1}{u} +x)} (u^{4} + u^{5} − 3 u_{1}^{2} + u u_{2}) = 0.  (79)

It can be checked (see Appendix B) that this equation has no Lie point symmetries.

Method A. By means of the transformation u = v_{x} = v_{1} equation (79) becomes the third order differential equation:

v_{1}^{5} + e^{2 ( \frac{1}{v_{1}} +x)} (v_{1}^{4} + v_{1}^{5} − 3 v_{2}^{2} + v_{1} v_{3}) = 0.  (80)

Equation (80) can be expressed in conserved form as

D_{x} (e^{−2 v} (e^{−2 ( \frac{1}{v_{1}} +x)} + (1 + \frac{1}{v_{1}} − \frac{v_{2}}{v_{1}^{3}} )^{2})) = 0.  (81)

Let us consider any second order equation associated to equation (81) or (80):

e^{−2 ( \frac{1}{v_{1}} +x)} + (1 + \frac{1}{v_{1}} − \frac{v_{2}}{v_{1}^{3}} )^{2} = Ce^{2 v},  (82)

where C is an arbitrary constant. It can be checked that equation (82) admits

X_{1} = e^{−v} \frac{∂}{∂x}, X_{2} = −e^{−v}(v + x + 1)\frac{∂}{∂x} + e^{−v} \frac{∂}{∂v}  (83)

as Lie point symmetries. Hence, X_{1} and X_{2} are super-potential symmetries of equation (79). Since [X_{1}, X_{2}] = 0, any of these two Lie point symmetries can be used to reduce the order of equation (82), in such a way that the reduced equation inherits the unused symmetry as a Lie point symmetry. As a consequence, equation (82) can be solved by quadratures.

Next, we use the Lie point symmetry X_{1} to reduce the order of equation (82). Let us introduce coordinates {y = v, α = e^{v}x} in some open set M ⊂ ℝ^{2}. In variables \{y, α\}, the vector field X_{1} can be written as \frac{∂}{∂α}. We consider the corresponding system of coordinates {y, α^{(2)}} in M^{(2)}, and the map π_{X_{1}}^{(2)} : M^{(2)} → M_{1}^{(1)} defined by (y, α^{(2)}) ↦ (y, w, w_{1}), where w = α_{1}. In terms of {y, w^{(1)}} equation (82) takes the form of the first order reduced equation:

e^{−2 (we^{−y}−y)} + (w_{1} + e^{y} − w)^{2} = e^{4 y} C.  (84)

It can be checked that the vector field (π_{X_{1}}^{(1)})_{∗}X_{2}^{(1)}, that will be denoted by X̃_{2}, can be written in terms of {y, w} as

X̃_{2} = e^{−y} \frac{∂}{∂y} + (−1 + e^{−y}w)\frac{∂}{∂w}.  (85)

The vector field X̃_{2} is a Lie point symmetry of equation (84), and therefore, it can be used to integrate the equation. By means of the change of variables \{z = e^{−y}w + y, β = e^{y}\}, for which X̃_{2} = \frac{∂}{∂β} , equation (84) takes the form

β_{z} = ±\frac{e^{z}}{\sqrt{−1 + e^{2 z} C}} .  (86)

The general solution of (86) can be obtained by a quadrature:

C_{1}β = ± ln(C_{1} e^{z} + \sqrt{−1 + C_{1}^{2} e^{2 z}}) + C_{2},  (87)

where C_{1}^{2} = C and C_{2} is an arbitrary constant. This solution can be written in the simple form C_{1}e^{z} = cosh(C_{1}β − C_{2}). Since z = e^{−y}w + y and β = e^{y}, the general solution of equation (84) can be expressed as

w = −e^{y} (C_{2} + ln(2C_{1}) − C_{1} e^{y} + y − ln(1 + e^{2 (C_{2}− C_{1} e^{y})})) .  (88)

By integration with respect to y we get :

α = e^{y} (\frac{C_{1}}{2} e^{y} − (−1 + C_{2} + ln(2C_{1}) + y)) − \frac{1}{2C_{1}} ∫ \frac{ln(1 + t)}{t}dt + C_{3}, (89)

where t = e^{2 (C_{2}− C_{1} e^{y})} and C_{3} is an arbitrary constant. The solution of equation (81) can be expressed as

x = (\frac{C_{1}}{2} e^{v} − (−1 + C_{2} + ln(2C_{1}) + v)) − e^{−v} \frac{1}{2C_{1}} ∫ \frac{ln(1 + t)}{t}dt + C_{3}e^{−v}, (90)

where t = e^{2 (C_{2}− C_{1} e^{v})}. Since u = v_{x} = \frac{e^{y}}{w−α}, the general solution of equation (79) is given by

u^{−1} = ln(1 + e^{2 (C_{2}− C_{1} e^{v})}) − 1 + \frac{C_{1} e^{v}}{2} + \frac{e^{−v}}{2C_{1}} ∫ \frac{ln(1 + t)}{t}dt − C_{3}e^{−v}.  (91)

Method B. Let us observe that the Lie point symmetries X_{1} and X_{2} of equation (82), that are super-potential symmetries of equation (79), are also Lie point symmetries of equation (80). Equation (80) does also admit the vector field X_{3} = \frac{∂}{∂v} as Lie point symmetry. It can be checked that the following relations hold:

[X_{1}, X_{2}] = 0, [X_{1}, X_{3}] = X_{1}, [X_{2}, X_{3}] = X_{1} + X_{2}.  (92)

Therefore, equation (80) admits a three-dimensional solvable algebra, G generated by {X_{1}, X_{2}, X_{3}}, that corresponds to case IV (with a = 1, b = 0, c = d = 1), and equation (79) is the X_{3}−reduced equation.

In what sequel, we show that the point symmetries X_{1}, and X_{2}, used above as super-potential symmetries, can be recovered as C^{∞}−symmetries of equation (79). Since C^{∞}−symmetries can be calculated by an algorithm, and can be used to reduce the order, we could solve the original equation without knowing the associated conserved form (81) needed to calculate potential symmetries.

A function f_{1} such that X_{3}(f_{1}) = f_{1} is given by f_{1} = e^{v}. Clearly, the vector field f_{1}X_{1}^{(1)} is π_{X_{3}}^{(1)}−projectable and the vector field

Y_{1} = (π_{X_{3}}^{(1)})_{∗}(f_{1}X_{1}^{(1)}) = \frac{∂}{∂x} + u^{2} \frac{∂}{∂u}  (93)

is a C^{∞}−symmetry of equation (79), for the function

λ_{1} = −\frac{D_{x}(e^{v})}{e^{v}} = −v_{x} = −u.  (94)

By Theorem 2.5, the order of equation (79) can be reduced by using the λ_{1}−symmetry Y_{1}. It can be checked that in terms of variables

\{z = \frac{1}{u} + x, β = x, μ = \frac{β − z}{β_{z}} = \frac{u_{x}}{u^{3}} − \frac{1}{u}\}  (95)

the vector field Y_{1}^{[λ_{1},(1)]} is simply expressed as Y_{1}^{[λ_{1},(1)]} = \frac{∂}{∂β} and the corresponding reduced equation is

1 − e^{2 z} (−1 + μ + μ μ_{z}) = 0.  (96)

The unused Lie point symmetry X_{2} can also be recovered as a C^{∞}−symmetry of equation (96). By taking f_{2} = e^{v}, the vector field f_{2}X_{2} is projectable to the space of variables {z, μ}; its projection is given by

Y_{2} = −\frac{∂}{∂z} + (−1 + μ)\frac{∂}{∂μ}.  (97)

The vector field Y_{2} is a C^{∞}−symmetry of equation (96) for the function λ_{2} = −\frac{D_{z}(e^{v})}{e^{v}} = −\frac{D_{x}(v)}{D_{x}(z)} = \frac{1}{μ}.

A function g such that gY_{2} is a Lie point symmetry of (96) is given by

g = \frac{e^{z}}{\sqrt{1 + e^{2z} (1 − μ)^{2}}} .  (98)

In coordinates \{s = (1 − μ)e^{z}, r = \sqrt{1 + e^{2 z} (−1 + μ)^{2}}e^{−z}\} the vector field gY_{2} can be written as \frac{∂}{∂r} and equation (96) can be integrated by a quadrature:

r_{s} = \frac{1}{\sqrt{1 + s^{2}}} .  (99)

Let us compare the two methods of reduction we have used. When method A is applied to equation (79), we have to construct the conserved form (81). The order of the integrated equation (82) is reduced by means of the potential symmetries (83). The solution (90) of equation (82) is obtained by quadratures. To recover solutions of equation (79), we must solve equation (90) for v and then u can be obtained by derivation (u = v_{x}).

When method B (C^{∞}−symmetries) is applied, we construct directly the reduced equation (99). The solutions of equation (79) can be obtained as follows. We first solve equation (99) in the form r = H(s, C_{1}) and obtain μ in the form μ = G(z, C_{1}). Since μ = \frac{β−z}{β_{z}}, we must solve the linear equation β = z + β_{z}G(z, C_{1}). From the solution β = J(z, C_{1}, C_{2}) of this equation we directly obtain the solution of (79) in the implicit form x = J(x + \frac{1}{u}, C_{1}, C_{2}, C_{3}).

6. Conclusions

The introduction of the concept of C^{∞}(M^{(1)})−symmetry ([17]) was motivated by the existence of ordinary differential equations that can be reduced or integrated but lack Lie point symmetries. In this paper, that concept is extended to define the C^{∞}(M^{(k)})−symmetries and it is proved that several classes of order reduction methods come from of the existence of C^{∞}(M^{(k)})−symmetries but not from the existence of Lie point symmetries.

We have also studied another equations, without Lie point symmetries, that can be obtained by reduction of equations with a three-dimensional Lie algebra of point symmetries G, that may be solvable or not. The elements of G are hidden symmetries of the original equation, can be recovered as C^{∞}−symmetries of the reduced equation and can be used to get new order reductions. This is illustrated through an example, that corresponds to a non-solvable algebra. We have also shown how these C^{∞}−symmetries can be used to construct a solvable structure of the reduced equation.

We have also proved that a class of potential symmetries, that are here called super-potential symmetries, can be recovered as C^{∞}−symmetries of the original equation. Since these C^{∞}−symmetries can be calculated by a well-defined algorithm, we do not have to determine the Bäcklund transformations needed to write the equation in conserved form, that in practice are difficult to find.

7. Appendix A: Equation (64) has no Lie point symmetries.

If a vector field X = p(x, u)\frac{∂}{∂x} + r(x, u)\frac{∂}{∂u} is a Lie point symmetry of equation (64), the infinitesimal p(x, u) must satisfy the equation (ux + 1)\frac{∂^{2}p}{∂u^{2}} + 3x\frac{∂p}{∂u} = 0. Therefore,

p(x, u) = \frac{p_{1}(x)u(ux + 2) + 2p_{2}(x)}{(u x + 1)^{2}},  (100)

where p_{1} and p_{2} depend only on x. The infinitesimal r(x, u) must satisfy

\begin{matrix}(ux + 1)^{4} \frac{∂^{2}r}{∂u^{2}} − 3x(ux + 1)^{3} \frac{∂r}{∂u} + 3x^{2}(ux + 1)^{2}r + 2 p_{2} u^{2} x^{3} \\ −p_{1} u^{2} x^{2} + 8 p′_{2} u x^{2} + 4 p_{2} u x^{2} − 3 p_{1} u^{2} x + 32 p_{2} u x \\ −4 p′_{1} u x − 2 p_{1} u x + 8 p′_{2} x + 2 p_{2} x − 18 p_{1} u + 2 p_{2} − 4 p′_{1} − p_{1} = 0. \end{matrix}  (101)

It can be checked that the general solution of equation (101) is given by

\begin{aligned}r(x, u) = & \frac{1}{6x^{3}(ux+1)^{2}} (6r_{1}u^{5}x^{5} + 30r_{1}u^{4}x^{4} + 6x^{3}(r_{2}x^{2} + 10r_{1})u^{3} \\ & −2x^{2}(2p_{2}x^{2} − 9r_{2}x^{2} − p_{1}x − 3p_{1} − 30r_{1})u^{2} − (6p′_{2}x^{2} \\ & +8p_{2}x^{2} − 18r_{2}x^{2} + 24xp_{2} − 3p′_{1}x − 4p_{1}x − 21p_{1} − 30r_{1})xu \\ & +6r_{2}x^{2} − 6p′_{2}x^{2} − 4p_{2}x^{2} − 12p_{2}x + 3p′_{1}x + 2p_{1}x + 9p_{1} + 6r_{1}) , \end{aligned} (102)

where r_{1} and r_{2} are arbitrary functions depending on x. If, in the determining equations, we cancel out the coefficients of powers of u then r_{1}(x) = 0, r_{2}(x) = \frac{−2p′_{1}(x)}{3x}, p_{2}(x) = \frac{2p′_{1}(x)}{x^{2}} + \frac{(x^{2}−12)p_{1}(x)}{2x^{3}} and p_{1}(x) must satisfy:

(x^{2} + 36)xp′_{1}(x) − 108p_{1}(x) = 0.  (103)

By solving this equation, we get p_{1}(x) = \frac{a x^{3}}{(x^{2}+36)^{3/2}} , for a ∈ ℝ, and hence p_{2}(x) =

\frac{a x^{4}+24 a x^{2}}{2(x^{2}+36)^{5/2}} .

With these values, we evaluate again the determining equations and we find a = 0. Therefore p_{1} = p_{2} = 0. By (100), p = 0 and, by (102), r = 0. Equation (64) has no Lie point symmetries.

8. Appendix B: Equation (79) has no Lie point symmetries

Next, we prove that equation (79) has no Lie point symmetries. If a vector field X = p(x, u)\frac{∂}{∂x} + r(x, u)\frac{∂}{∂u} is a Lie point symmetry of equation (79) the infinitesimals p and r must satisfy the following determining equations:

\begin{aligned}e1 : & \frac{∂^{2}p}{∂u^{2}} u + 3 \frac{∂p}{∂u} = 0, \\ e2 : & (\frac{∂^{2}r}{∂u^{2}} − 2 \frac{∂^{2}p}{∂u∂x}) u^{2} − 3 \frac{∂r}{∂u} u + 3 r = 0, \\ e3 : & 3 \frac{∂p}{∂u} u^{5} e^{−2 x− \frac{2}{u}} + 3 \frac{∂p}{∂u} u^{5} + 3 \frac{∂p}{∂u} u^{4} + 2 \frac{∂^{2}r}{∂u∂x} u − \frac{∂^{2}p}{∂x^{2}} u − 6 \frac{∂r}{∂x} = 0, \\ e4 : & (−\frac{∂r}{∂u} u^{4} + 2 \frac{∂p}{∂x} u^{4} − 2 p u^{4} + 4 r u^{3} + 2 r u^{2}) e^{−2 x− \frac{2}{u}} \\ & − \frac{∂r}{∂u} u^{4} + 2 \frac{∂p}{∂x} u^{4} − \frac{∂r}{∂u} u^{3} + 4 r u^{3} + 2 \frac{∂p}{∂x} u^{3} + 3 r u^{2} + \frac{∂^{2}r}{∂x^{2}} = 0. \end{aligned} (104)

From equation e1, we get

p = \frac{p_{2}}{u^{2}} + \frac{p_{1}}{2}  (105)

where p_{1}, p_{2} are arbitrary functions on x. By substituting this value into equation e2 we obtain:

\frac{∂^{2}r}{∂u^{2}} u^{3} − 3 \frac{∂r}{∂u} u^{2} + 3 r u + 4 \frac{dp_{2}}{dx} = 0.  (106)

Therefore

r = r_{1} u^{3} + r_{2} u − \frac{1}{2 u} \frac{dp_{2}}{dx} ,  (107)

where r_{1}, r_{2} are functions depending on x. Since equation e3 becomes

−12 p_{2} u^{2}(ue^{−2 x− \frac{2}{u}} + u + 1) − 8 \frac{dr_{2}}{dx} u^{2} − \frac{d^{2}p_{1}}{dx^{2}} u^{2} + 6 \frac{d^{2}p_{2}}{dx^{2}} = 0,  (108)

it follows that p_{2} = 0. Therefore, equation e4 can be written as

\begin{matrix}(r_{1} u^{6} + 2 r_{1} u^{5} + 3 r_{2} u^{4} + \frac{dp_{1}}{dx} u^{4} − p_{1} u^{4} + 2 r_{2} u^{3}) e^{−2 x− \frac{2}{u}} \\ +r_{1} u^{6} + 3 r_{2} u^{4} + \frac{dp_{1}}{dx} u^{4} + 2 r_{2} u^{3} + \frac{d^{2}r_{1}}{dx^{2}} u^{3} + \frac{dp_{1}}{dx} u^{3} + \frac{d^{2}r_{2}}{dx^{2}} u = 0. \end{matrix}  (109)

We finally deduce that r_{1} = 0, r_{2} = 0 and p_{1} = 0. By (107), r = 0 and by (105), p = 0. Therefore, equation (79) has no Lie point symmetries.

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C. Muriel and J. L. Romero Dpto. Matemáticas Facultad de Ciencias Universidad de Cádiz Apto. 40, 11510 Puerto Real Cádiz, Spain concepcion.muriel@uca.es

Received October 17, 2001 and in final form Mai 2, 2002