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Darboux integrability of polynomial differential systems in [R.sup.3].

1 Introduction and statement of the main results

Recently the Hopf bifurcations have been studied intensively in two dimensional differential systems with one slow and one fast variable, see for instance [1, 5, 6, 10, 8]. Less analysis has been done of the Hopf bifurcations in slow-fast systems in [R.sup.3] with two slow variables and one first variable, see [7, 9,13,14]. Guuckenheimer in [9] reduces the study of this Hopf bifurcation to study the zero Hopf bifurcation of the differential system

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (1)

where (x, y, z) [member of] [R.sup.3], a, b, c, d [member of] R and the dot denotes derivative with respect to the independent variable t.

The vector field associated to (1) is

[chi] = (y - [chi square])[[partial derivative]/[partial derivative]x] + (z - x) [[partial derivative]/[partial derivative]y] - (d + ax + by + cz)[[partial derivative]/[partial derivative]z].

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Let U be an open subset in [R.sup.3] such that [R.sup.3] \ U has zero Lebesgue measure. We say that a real function H = H(x, y, z): U [subset] [R.sup.3] [right arrow] R non-constant in any open subset of U is a first integral if H (x(t), y(t), z(t)) is constant on all solutions (x(t), y(t),z(t)) of [chi] contained in U, i.e. [chi]H|[sub.U] = 0. The existence of a first integral for a differential system in [R.sup.3] allows to reduce its study in one dimension. This is the main reason to look for first integrals.

One of the more classical problems in the qualitative theory of differential equations depending on parameters is to characterize the existence or not of first integrals in function of these parameters. This is a very difficult problem and not many results can be found in the literature. One of the best tools to look for first integrals is the Darboux theory of integrability.

Our objective is to study the Darboux integrability of system (1). Probably the more interesting novelty of this paper is not the characterization of the Darboux integrability of system (1), but the method for reaching this result, which can be applied to other polynomial differential systems in [R.sup.3].

Now we shall introduce the basic notions of the Darboux theory of integrability restricted to system (1). Let C [x, y, z] be the ring of polynomials in the variables x, y, z with coefficients in C. We say that f [member of] C[x, y, z] is a Darboux polynomial of the vector field X if there exists a polynomial K [member of] C [x, y, z] such that [chi]f = Kf, i.e.

(y - [chi square])[[partial derivative]h/[partial derivative]x] + (z - x) [[partial derivative]h/[partial derivative]y] - (d + ax + by + cz)[[partial derivative]h/[partial derivative]z] = Kh. (2)

The polynomial K = K(x, y, z) is called the cofactor of f. It is easy to show that the cofactor of a Darboux polynomial of the vector field [chi] has degree at most one, i.e. K = [k.sub.0] + [k.sub.1] x + [k.sub.2]y + [k.sub.3]z with [k.sub.i] [member of] C for i = 0, ..., 3. Note that we look for complex Darboux polynomials in real differential systems. The reason is that frequently the complex structure forces the existence of real first integrals, and sometimes if we only work with the real Darboux polynomials, we cannot detect all the real first integrals.

If f [member of] C [x, y, z] is a Darboux polynomial of the vector field [chi] then f (x, y, z) = 0 is an invariant algebraic surface for the differential system (1), i.e. if an orbit has a point in this surface then the whole orbit is contained in it. Note that a Darboux polynomial with zero cofactor is a polynomial first integral.

The Darboux polynomials for system (1) with non-zero cofactor are characterized in the next result.

Theorem 1. System (1) has an irreducible Darboux polynomial with non-zero cofactor if and only if a = b = 0 and c [not equal to] 0. In this case the irreducible Darboux polynomial is cz + d with cofactor -c.

Theorem 1 is proved in section 3.

Let I(x,y,z) be the maximal interval of definition of the solution of system (1) such that at time zero pass through the point (x, y, z). We say that a real function I = I(x, y, z, t): U x [I.sub.(x,y,z)] [right arrow] R non-constant in any open subset of U is an invariant if it is constant on every solution (x(t), y(t), z(t)) contained in U, i.e. if

[dI/dt]|[sub.U] = [chi]I + [[partial derivative]I/[partial derivative]t]|[sub.U] = 0.

Corollary 2. System (1) with a = b = 0 and c [not equal to] 0 has the invariant I = (cz + d)[e.sup.ct].

The proof of Corollary 2 is immediate from Theorem 1 and the definition of invariant.

To know and invariant of a differential system is important because it allows to compute either the [alpha] -limits, or the [omega]-limits of the orbits of the system. More precisely, if c > 0 then when t [right arrow][bar.[alpha]]the orbit of system (1) having maximal interval of definition ([bar.[alpha]], [bar.[omega]]), under the assumption of Corollary 2, tends to the invariant plane cz + d = 0, and studying the dynamics on this invariant plane we can determine the [alpha]-limit sets. A similar study can be done if c < 0 for the [omega]-limit sets. For a definition of [alpha]- or [omega]-limit sets see [4].

An exponential factor F(x, y, z) of the vector field [chi] is an exponential function of the form exp(g/h) with g and h coprime polynomials in C[x, y, z] and satisfying [chi]F = LF for some L [member of] C[x,y,z] with degree one. The exponential factors appear when some Darboux polynomial has multiplicity larger than one, or when the multiplicity of the plane at infinity is larger than one, for more details see [2, 12].

A first integral of system (1) is called of Darboux type if it is a first integral of the form

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

where [f.sub.1], ..., [f.sub.p] are Darboux polynomials and [F.sub.1], ..., [F.sub.q] are exponential factors.

The next theorem is the main result of the paper, and it characterizes the first integrals of Darboux type for system (1).

Theorem 3. System (1) has Darboux first integrals if and only if b = d = 0 and a + c = 0. Moreover, when system (1) has Darboux first integrals these are functions of Darboux type in the variable ay - z.

Theorem 3 is proved in section 4. In section 2 we introduce some auxiliary results that will be used all through the paper.

2 Auxiliary results

In the rest of this paper we will use the following well known result of the Darboux theory of integrability, see for instance Chapter 8 of [4].

Theorem 4 (Darboux theory of integrability). Suppose that a polynomial vector field [chi] defined in [R.sup.n] of degree m admits p Darboux polynomials [f.sub.i] with cofactors [K.sub.i] for i = 1,p and q exponential factors [F.sub.j] = exp([g.sub.j]/[h.sub.j]) with cofactors [L.sub.j] for j = 1, ..., q. If there exist [[lambda].sub.i], [[mu].sub.j] [member of] C not all zero such that

[p.summation over (i=1)] [[lambda].sub.i][K.sub.i]+ [q.summation over (j=1)] [[mu].sub.j][L.sub.j] = 0, (3)

then the following real (multivalued) function of Darboux type

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

substituting [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], is a first integral of the vector field [chi].

For a proof of the next result see [11, 12].

Proposition 5. The following statements hold.

(a) If [e.sup.g/h] is an exponential factor for the polynomial differential system (1) and h is not a constant polynomial, then h = 0 is an invariant algebraic surface.

(b) Eventually [e.sup.g] can be an exponential factor, coming from the multiplicity of the infinite invariant plane.

The proof of the next result can be found in Chapter 8 of [4].

Lemma 6. Let f be a polynomial and f = [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] its decomposition into irreducible factors in C [x, y, z]. Then f is a Darboux polynomial of system (1) if and only if all the [f.sub.j] are Darboux polynomials of system (1). Moreover if K and [K.sub.j] are the cofactors of f and [f.sub.j,] then K = [[summation].sup.s.sub.j=1] [[alpha].sub.j][K.sub.j] .

We note that in view of Lemma 6 to study the Darboux polynomials of system (1) it is enough to study the irreducible ones.

To prove Theorem 3 we also need one auxiliary result proved in [3]. We recall that a generalized rational function is a function which is the quotient of two analytic functions. In particular rational first integrals and analytic first integrals are particular cases of generalized rational first integrals. Clearly a Darboux type function is a generalized rational function.

Theorem 7. Assume that the differential system (1) has p as a singular point and let [[lambda].sub.1], [[lambda].sub.2], [[lambda].sub.3] be the eigenvalues of the linear part of system (1) at p. Then the number of functionally independent generalized rational integrals of system (1) is at most the dimension of the minimal vector subspace of[R.sup.3] containing the set

{([k.sub.1], [k.sub.2], [k.sub.3]) [member of] [Z.sup.3] : [k.sub.1][[lambda].sub.1] + [k.sub.2][[lambda].sub.2] + [k.sub.3][[lambda].sub.3] = 0, ([k.sub.1],[k.sub.2],[k.sub.3]) [not equal to](0,0,0)}.

3 Proof of Theorem 1

To prove Theorem 1 we state and prove some auxiliary results. We denote by N the set of positive integers.

Lemma 8. If h is a Darboux polynomial of system (1) with non-zero cofactor K then K = [k.sub.0] - mx for some [k.sub.0] [member of] C and m [member of] N [union] {0}.

Proof. Let h be a Darboux polynomial of system (1) with non-zero cofactor K then K = [k.sub.0] + [k.sub.1]x + [k.sub.2]y + [k.sub.3]z for some [k.sub.0], [k.sub.1], [k.sub.2], [k.sub.3] [member of] C.

Let n be the degree of h. We write h in sum of its homogeneous parts as h = [[summation].sup.n.sub.i=1] [h.sub.i] where each [h.sub.i] is a homogenous polynomial of degree i. Without loss of generality we can assume that [h.sub.n] [not equal to] 0 and n [greater than or equal to] 1.

Computing the terms of degree n + 1 in (2) we get that

-[chi square][[partial derivative][h.sub.n]/[partial derivative]x] = ([k.sub.1]x + [k.sub.2]y + [k.sub.3]z)[h.sub.n].

Solving it we get

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

where [C.sub.n] is any function in the variables y, z. Since [h.sub.n] must be a homogeneous polynomial we must have [k.sub.1] = -m with m [member of] N [union]{0}, [k.sub.2] = [k.sub.3] = 0 and [C.sub.n] (y, z) [member of] C[y,z] \ {0}. This concludes the proof of the lemma.

Proposition 9. System (1) with either [a.sup.2] + [b.sup.2] [not equal to] 0, or a = b = c = 0 has no Darboux polynomials with non-zero cofactor.

Proof. For simplifying the computations we introduce the weight change of variables

x = [[mu].sup.-1] X, y = [[mu].sup.-2]Y, z = [[mu].sup.-1] Z, t = [mu]T,

with [mu] [member of] R\ {0}. Then system (1) becomes

X' = Y - [X.sup.2],

Y' = [[mu].sup.2] (Z - X),

Z' = -bY - [mu](aX + cZ) - [[mu].sup.2]d,

where the prime denotes derivative with respect to the variable T.

A polynomial H(X, Y, Z) is said to be weight-homogeneous of degree r [member of] N with respect to the weight exponent s = ([s.sub.1], [s.sub.2], [s.sub.3]) for all [mu] [member of] R \ {0} if we have [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Let h(x, y, z) be a Darboux polynomial of system (1) with cofactor k(x, y, z). Without loss of generality, we can assume that k(x, y, z) = [k.sub.0] + [k.sub.1]x + [k.sub.2]y + [k.sub.3]z. In view of Lemma 8 we get [k.sub.2] = [k.sub.3] = 0 and [k.sub.1] = -m with m [member of] N [union] {0}. Set H(X, Y, Z) = [[mu].sup.n]h([[mu].sup.-1] X, [[mu].sup.-2]Y, [[mu].sup.-1] Z) = [[summation].sup.n.sub.i=0] [[mu].sup.i][H.sub.i] (X, Y, Z) where [H.sub.i] is the weight-homogeneous part with weight degree n - i of H and n is the weight degree of H with weight exponent s = (-1, -2, -1). We also set K(X, Y, Z) = [mu]k([[mu].sup.-1]X, [[mu].sup.-2]Y, [[mu].sup.-1]Z) = [mu]([k.sub.0] - m[[mu].sup.-1] X) = [mu][k.sub.0] - mX. From the definition of a Darboux polynomial, we have

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (4)

where we still use x, y, z instead of X,Y, Z.

Equating in (4) the terms with [[mu].sup.i] for i = 0,1, ..., n + 2 we get

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (5)

for j = 3, ..., n + 2, where [H.sub.j] = 0 for j > n, and L is the linear partial differential operator of the form

L = (y - [chi square]) [[partial derivative]/[partial derivative]x] - by [[partial derivative]/[partial derivative]z].

Consider the first equation of (5), that is L[[H.sub.0]] = -mx[H.sub.0], where [H.sub.0] is a weight-homogeneous polynomial of degree n.

By direct computation we get

[H.sub.0] = [([chi square] - y).sup.m/2] [f.sub.0] (y,v),

where [f.sub.0](y,v) is a differentiable function in y and v = z + b [square root of y] arctanh(x/[square root of y]). We consider different cases.

Case 1: b [not equal to] 0. In this case since [H.sub.0] is a weight-homogeneous polynomial we must have [f.sub.0] = [f.sub.0](y) being [f.sub.0] a homogeneous polynomial, i.e. [f.sub.0](y) = [C.sub.0][y.sup.l] for some [C.sub.0] [member of] C \{0}. Thus,

[H.sub.0] = [C.sub.0] [([chi square] - y).sup.m/2][y.sup.l], [C.sub.0] [member of] C \{0},

and the weight degree of [H.sub.0] is n = m + 2l.

From the second equation of (5) we get

L[[H.sub.1]] = -mx[H.sub.1] + [k.sub.0] [H.sub.0].

Solving it we obtain

[H.sub.1] = [C.sub.0][k.sub.0][([chi square]-y).sup.m/2][y.sup.l-1/2] arctanh[([x/[square root of y]]) + [[chi square] - y).sup.m/2][f.sub.1] (y,v),

where [f.sub.1] (y, v) is a differentiable function. Again since [H.sub.1] is a weight-homogenous polynomial we must have [C.sub.0][k.sub.0] = 0 (which yields [k.sub.0] = 0 because [C.sub.0][not equal to] 0), and [f.sub.1] must be a function of y but since the weight-degree of [H.sub.1] is n - 1 = m + 2l - 1 we get that [f.sub.1] = 0 and thus [H.sub.1] = 0. In short, [k.sub.0] = 0 and [H.sub.1] = 0.

The third equation in (5) is

L[[H.sub.2]] = - mx[H.sub.2] - (z - x) [[partial derivative][H.sub.0]/[partial derivative]y].

Solving it we get

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

where [f.sub.2] (y, v) is a differentiable function. Since [H.sub.2] is a weight-homogenous polynomial of degree n - 2 = m + 2l - 2, we must have [f.sub.2] = [f.sub.2] (y) and 4l = m + 4l = 0. In short, l = m = 0 and [k.sub.0] = 0, which yields K = 0, in contradiction with the fact that h is a Darboux polynomial with non-zero cofactor. Hence if b [not equal to] 0 there are no Darboux polynomials with non-zero cofactor.

Case 2: b = 0. In this case we consider two subcases.

Subcase 2.1: a [not equal to] 0. The first equation in (5), i.e. L[[H.sub.0]] = -mx[H.sub.0], together with the fact that [H.sub.0] is a homogeneous polynomial with weight-degree n yields

[H.sub.0] = [([chi square] - y).sup.m/2][f.sub.0](y, z),

where [g.sub.0] [member of] C[y, z] and has weight-degree n - m. The second equation in (5) yields

L[[H.sub.1]] = -mx[H.sub.1] + [k.sub.0][H.sub.0] + (ax + cz)[[partial derivative][H.sub.0]/[partial derivative]z].

Solving it we get

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

where [f.sub.1] is a smooth function in the variables y, z. Taking into account that [H.sub.1] is a weight-homogeneous polynomial with weight-degree n- 1 and that a [not equal to] 0 we get [partial deivative][f.sub.0] /[partial derivative]z = 0 and [k.sub.0] = 0. Hence [H.sub.0] = [C.sub.0] [([chi square] - y).sup.m/2][y.sup.l] with [C.sub.0] [member of] C \ {0}, n = m + 2l and [H.sub.1] = [([x.sup.2] - y).sup.m/2][f.sub.1] (y,z) with [f.sub.1] [member of] C[y,z] of weight-degree n - m - 1.

The third equation in (5) yields

L[[H.sub.2]] = -mx[H.sub.2] + (ax + cz) [[partial derivative][H.sub.1]/[partial derivative]z] - (z - x)[[partial derivative][H.sub.0]/[partial derivative]y].

Solving it we get

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

where [f.sub.2] is a smooth function in the variables y, z. Since [H.sub.2] must be a weight-homogeneous polynomial, we must have [ly.sup.l] + ay[partial derivative][f.sub.1]/[partial derivative]z = 0 and (m + 4l)[y.sup.l] -4cy[partial derivative][f.sub.1] /[partial derivative]z = 0. Using that a [not equal to] 0 we obtain [f.sub.1] = -[ly.sup.l-1]z/a + [g.sub.1] (y) and am + 4(a + c)l = 0. Since [f.sub.1] has weight degree n - m - 1 = 2l - 1 and y has weight degree 2, we must have [g.sub.1] = 0. Furthermore, if a + c = 0, then m = 0 which is not possible because then K = 0 in contradiction with the fact that the cofactor is nonzero. Hence, a + c [not equal to] 0 and l = -am/(4(a + c)). In short, [f.sub.1] = -[ly.sup.l-1] z/a, a + c [not equal to] 0, l = -am/(4(a + c)) and [H.sub.2] = m[([x.sup.2] - y).sup.m/2-1] [y.sup.l-1] (y - xz)/4 + [f.sub.2](y,z) where [f.sub.2] [member of] C [y, z] is a weight-homogeneous polynomial with weight-degree n - 2.

The fourth equation in (5) yields

L[[H.sub.3]] = -mx[H.sub.3] + (ax + cz) [[partial derivative][H.sub.2]/[partial derivative]z] - (z - x)[[partial derivative][H.sub.1]/[partial derivative]y] + d[[partial derivative][H.sub.1]/[partial derivative]z].

Solving it we get

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

where [f.sub.3] is a function in the variables y, z. Since [H.sub.3] must be a weight-homogeneous polynomial, we have

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (6)

?From the first identity in (6) we get

[f.sub.2](y,z) = [m(4c + a(m + 4))/8a[(a + c).sup.2]] [y.sup.l-2][z.sup.2]+[g.sub.2](y).

Substituting it in the second identity in (6) we obtain

[2([a + c])m/a] (a(a(a+c) + 2d)y + 2(a + c)[z.sup.2]) = 0.

Since a(a + c) [not equal to] 0 we must have m = 0. In short, [k.sub.0] = m = 0, which yields K = 0, in contradiction with the fact that h is a Darboux polynomial with nonzero cofactor. Hence if b = 0 and a [not equal to] 0 there are no Darboux polynomials with non-zero cofactor.

Subcase 2.2: a = 0. By hypothesis we also have c = 0. Proceeding as in Subcase 2.1 we get that [H.sub.0] = [([x.sup.2] - y).sup.m/2][f.sub.0] (y,z) where [f.sub.0] [member of] C[y,z] \ {0} is a weighthomogeneous polynomial of weight degree n m. The second equation in (5) becomes

L[[H.sub.1]] = -mx[H.sub.1] + [k.sub.0][H.sub.0].

Solving it we get

[H.sub.1] = [([x.sup.2] - y).sup.m/2][y.sup.-1/2][k.sub.0][f.sub.0](y,z) arctanh(x/[square root of y]) + [([x.sup.2] - y).sup.m/2][f.sub.1](y,z),

where [f.sub.1] is a function in y, z. Since [H.sub.1] is a weight-homogeneous polynomial of weight degree n - 1, we must have [k.sub.0][f.sub.0] = 0 and [f.sub.1] be a weight-homogeneous polynomial of weight degree n -m - 1. Since [f.sub.0] [not equal to] 0 we must have [k.sub.0] = 0 and [H.sub.1] = [([x.sup.2] - y).sup.m/2] [f.sub.1] (y, v).

From the the third equation in (5) we get

L[[H.sub.2]] = -mx[H.sub.2] - (z - x) [[partial derivative][H.sub.0]/[partial derivative]y] + d[[partial derivative][H.sub.0]/[partial derivative]z].

Solving it we obtain

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

where [f.sub.2] is a function in y, z. Since [H.sub.2] is a weight-homogeneous polynomial of weight degree n - 2, we must have [partial derivative][f.sub.0]/[partial derivative]y = 0 and [mzf.sub.0] 4dy[partial derivative][f.sub.0] /[partial derivative]z = 0. Since [f.sub.0] [not equal to] 0 we must have m = 0. In short, [k.sub.0] = m = 0 which yields K = 0, in contradiction with the fact that h is a Darboux polynomial with non-zero cofactor. Hence if b = a = 0 and c [not equal to] 0 there are no Darboux polynomials with non-zero cofactor.

In view of Proposition 9, to complete the proof of Theorem 1 we should study the case a = b = 0 and c [not equal to] 0 and show that it has the unique irreducible Darboux polynomial cz + d. We introduce the change of variables u = cz + d. Then system (1) becomes

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (7)

We introduce the change of variables For simplifying the computations we introduce the weight change of variables

x = [[mu].sup.-1]X, y = [[mu].sup.-2]Y, u = [[mu].sup.-1]U, t = [mu]T,

with [mu] [member of] R \{0}. Then system (7) becomes

X' = Y - [X.sup.2],

Y' = -[[mu].sup.2] X + [[mu].sup.2] U/c -d/c [[mu].sup.3],

U' = - c[mu]U,

where the prime denotes derivative with respect to the variable T. From the definition of a Darboux polynomial, we have

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII](8)

where we still use x, y, u instead of X, Y, U.

Equating in (8) the terms with [[mu].sup.i] for i = 0,1, ..., n + 2 we get

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (9)

for j = 4, ..., n + 3, where [H.sub.j] = 0 for j > n and L is the linear partial differential operator of the form

L = (y - [x.sup.2]) [[partial derivative]/[partial derivative]x].

Consider the first equation of (9). By direct computation we get

[H.sub.0] = [(y - [x.sup.2]).sup.m/2] [f.sub.0](y,u),

where [f.sub.0] [member of] C [y, u] has weight-degree n - m. The second equation in (5) gives

L[[H.sub.i]] = -mx[H.sub.1] + [k.sub.0][H.sub.0] + cu [[partial derivative][H.sub.0]/[partial derivative]u].

Solving it and using that it has weight-degree n - 1 we get

[H.sub.1] = [([x.sup.2] - y).sup.m/2] [y.sup.-1/2] ([k.sub.0][f.sub.0] (y, u) + cu [[partial derivative][f.sub.0]/[partial derivative]u]arctanh (x/[square root of y]) + [([x.sup.2] - y).sup.m/2] [f.sub.1](y, u),

which yields [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Moreover, [k.sub.0] = -lc, [g.sub.0] has weight-degree n - m - l and [H.sub.1] = [([x.sup.2]-y).sup.m/2] [f.sub.1](y, u) with [f.sub.1] [member of] C[y, u] and has weight-degree n - 1.

The third equation in (5) gives

L[[H.sub.2]] = -mx[H.sub.2] + [k.sub.0][H.sub.1] + cu [[partial derivative][H.sub.1]/[partial derivative]u] -(u/c - x) [[partial derivative][H.sub.0]/[partial derivative]y].

Solving it we obtain

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

where [f.sub.2] [member of] C[y, u] and has weight-degree n -2. Since [H.sub.2] is a polynomial and [g.sub.0] [not equal to] 0 we must have [g'.sub.0](y) = 0, i.e, [g.sub.0] = [C.sub.0] [member of] C {0}, m = 0 and [f.sub.1] = [C.sub.1][u.sup.l]. This yields n = l. Moreover, since [H.sub.1] has weight-degree n- 1 = l -1 we must have [C.sub.1] = 0. In short, [H.sub.0] = [C.sub.0][u.sup.n], n = l, [H.sub.1] = 0 and [H.sub.2] = [f.sub.2](y,u) with weight-degree n -2 = l- 2. Proceeding inductively we get that [H.sub.i] = 0 for i = 2, ..., n. Hence the Darboux polynomial is h = [C.sub.0][(d + cz).sup.n] with cofactor - cn. The irreducible Darboux polynomial is d + cz with cofactor - c. This concludes the proof of Theorem 1.

4 Proof of Theorem 3

We will divide the proof of Theorem 3 into different results. The first one characterizes the polynomial first integrals of system (1).

Lemma 10. System (1) has a polynomial first integral if and only if b = d = 0 and a + c = 0. In this case a polynomial first integral is ay - z.

Proof. Let h [member of] C[x,y,z] \ C be a polynomial first integral of system (1). We write h as a polynomial in the variable x as

h = [n.summation over (i=0)] [h.sub.i](y, z) [x.sup.i],

where each [h.sub.i] [member of] C [y, z]. The coefficient of [x.sup.n+1] from (2) with K = 0 is

- [nh.sub.n]- [[partial derivative][h.sub.n]/[partial derivative]y] - a [[partial derivative][h.sub.n]/[partial derivative]z] = 0.

Solving it we get

[h.sub.n] = F(ay - z)[e.sup.-ny],

for some function F. Thus, since [h.sub.n] must be a polynomial we get that n = 0 and F is a polynomial in the variable ay - z. In short h = [h.sub.0] = F(ay - z) being F a polynomial. Then from (2) with K = 0 we get

[dF/d(ay - z)](a(z - x) + (d + ax + by + cz)) = [dF/d(ay - z)](d + by + (c + a)z) = 0.

Since F is a polynomial first integral it is not constant and thus dF/d(ay - z) [not equal to] 0. This implies that d = b = 0 and c = -a.

Theorem 3 follows directly from the following two propositions.

Proposition 11. The unique first integrals of Darboux type for system (1) with b = d = 0 and a + c = 0 are functions of Darboux type in the variable ay - z.

Proof. We consider system (1) with b = d = 0 and a + c = 0, that is,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (10)

It follows from Lemma 10 that this system has the polynomial first integral [H.sub.1] = ay - z, which obviously is a generalized rational first integral. To conclude the proof of the proposition we shall show that system (10) has no other first integrals of Darboux type independent with [H.sub.1] . To prove this we will use Theorem 7. First we note that the singular points of system (10) are of the form (x, [x.sup.2], x) with x [member of] R. We compute the eigenvalues [[lambda].sub.1], [[lambda].sub.2], [[lambda].sub.3] of the Jacobian matrix of this system on these singular points and we get

[[lambda].sub.1] = 0, [[lambda].sub.2,3] = 1/2 (a - 2x [+ or -] [square root of ([(a + 2x).sup.2] - 4))].

Therefore [k.sub.1][[lambda].sub.1] + [k.sub.2][[lambda].sub.2] + [k.sub.3][[lambda].sub.3] = 0 is equivalent to

[k.sub.2] (a -2x+ [square root of ([(a + 2x).sup.2] - 4))] +[k.sub.3] (a- 2x + [square root of ([(a + 2x).sup.2] -4))] = 0,

or in other words,

[k.sub.2]/[k.sub.3] = - [[a - 2x + [square root of ([(a + 2x).sup.2] -4)]]/[a - 2x - [square root of ([(a + 2x).sup.2] -4)]]]. (11)

It is clear that the left-hand side of (11) is a rational number (because [k.sub.2], [k.sub.3] [member of] Z), and that choosing x in a convenient way the right-hand side of (11) is irrational. Therefore (11) cannot hold for this convenient choice of x. Hence for this special singular point (x, [x.sup.2], x), the dimension of the minimal vector space of [R.sup.3] containing the set

{([k.sub.1], [k.sub.2], [k.sub.3]) [member of] [Z.sup.3] : [k.sub.1][[lambda].sub.1] + [k.sub.2][[lambda].sub.2] + [k.sub.3][[lambda].sub.3] = 0, ([k.sub.1], [k.sub.2], [k.sub.3]) [not equal to] (0,0,0)}

is clearly one generated by (1,0,0). Thus it follows from Theorem 7 that system (10) can only have one generalized rational first integral, which is a function of Hi. This completes the proof of the proposition.

In the next result we characterize the exponential factors of system (1) with either b [not equal to] 0, or d [not equal to] 0, or c + a [not equal to] 0.

Lemma 12. The unique exponential factors of system (1) with either b [not equal to] 0, or d [not equal to] 0, or c + a [not equal to] 0 are:

(a) [e.sup.y] and [e.sup.ay-z] with cofactors z - x and d + by + (c + a)z if a + c [not equal to] 0, or a + c = 0 and b [not equal to] 0.

(b) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Proof. Let F = exp(g/h) be an exponential factor of system (1) with cofactor L, where g,h [member of] C[x,y,z] with (g,h) = 1. Then from the definition of exponential factor and in view of Proposition 5 we have that either h is a constant the we can take h = 1, or h is a Darboux polynomial of system (1).

Case 1. We first assume that F is of the form F = exp(g) where g = g(x,y,z) [member of] C[x,y,z] \ C, with cofactor L = L(x,y,z) of degree at most one. Without loss of generality we can assume that g has no constant term and that we can write L = [l.sub.0] + [l.sub.1]x + [l.sub.2]y + [l.sub.3]z. So we have

(y - [x.sup.2]) [[partial derivative]g/[partial derivative]x]+ (z - x) [[partial derivative]g/[partial derivative]y] - (d + ax + by + cz) [[partial derivative]g/[partial derivative]z] = [l.sub.0] + [l.sub.1]x + [l.sub.2]y + [l.sub.3]z. (12)

We write g as a polynomial in the variable x as

g = [n.summation over (i=0)] [g.sub.i](y, z) [x.sup.i],

where each [g.sub.i] [member of] C [y, z]. The coefficient of [x.sup.n+1] from (12) if n [greater than or equal to] 1 is

-[ng.sub.n] - [[partial derivative][g.sub.n]/[partial derivative]y] - a [[partial derivative][g.sub.n]/[partial derivative]z] = 0.

Solving it we get

[g.sub.n] = F(ay - z)[e.sup.-ny],

for some function F. Thus [g.sub.n] = 0 if n [greater than or equal to] 1. For n = 0 we have g = [g.sub.0] (y, z) and the coefficients of x from (12) are

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Now solving it we get from the first equation

g = G(ay - z) - [l.sub.1]y;

and from the second, introducing the variable u = ay - z, we obtain

(d + by + (a + c)(ay + u))G' (u) = -[l.sub.0] - [l.sub.2]y - ([l.sub.1] + [l.sub.3])(ay + u). (13)

We consider different subcases.

Subcase 1.1: a + c [not equal to] 0. In this case solving (13) we get

G(u) = - [1/[(a + c).sup.2]] ((a + c)([l.sub.1] + [l.sub.3])u + (-([l.sub.1] + [l.sub.3](d + by)| + (a + c)([l.sub.0] + [l.sub.2]y)) log (d + (a + c)u + (b + a (a + c))y)).

Since G(u) must be a polynomial we get

[l.sub.0] = [d([[l.sub.1] + [l.sub.3]])/[a + c]], [l.sub.2] = [b([[l.sub.1] + [l.sub.3]])/[a + c]] and G(u) = - [[[l.sub.1] + [l.sub.3]]/[a+c]] u.

Then g = -[l.sub.1]y -([l.sub.1] + [l.sub.3])(ay - z)/(a + c).

Subcase 1.2: a + c = 0 and b = 0. In this case, d [not equal to] 0 and solving (13) we get

G(u) = - [u/2d](2[l.sub.0] + ([l.sub.1] + [l.sub.3])u + 2([l.sub.2] + a([l.sub.1] +[l.sub.3]))y).

Since G(u) must be a polynomial independent of y we get

[l.sub.2] = -a{[l.sub.1] + [l.sub.3]) and G(u) = - [u/2d](2[l.sub.0] + ([l.sub.1] + [l.sub.3])u).

Then g = -[l.sub.1]y - (ay - z)(2[l.sub.0] + ([l.sub.1] + [l.sub.3]) (ay - z)) /(2d).

Subcase 1.3: a + c = 0,b [not equal to] 0 and d = 0. In this case solving (13) we get

G(u) = - [u/2by](2[l.sub.0] + ([l.sub.1] + [l.sub.3])u + 2{[l.sub.2] + a([l.sub.1] + [l.sub.3]))y).

Since G(u) must be a polynomial independent of y we get

[l.sub.2] = 0, [l.sub.3] = -[l.sub.1] and G(u) = [[l.sub.2]u/b].

Then g = - [l.sub.1]y - [l.sub.2] (ay -z)/b.

Subcase 1.4: a + c = 0, bd [not equal to] 0. In this case solving (13) we get

G(u) = - [u/2(d + by)](2[l.sub.0] + ([l.sub.1] + [l.sub.3])u + 2([l.sub.2] + a([l.sub.1] + [l.sub.3]))y).

Since G(u) must be a polynomial we get

2[l.sub.0] + ([l.sub.1] + [l.sub.3])u + 2([l.sub.2] + a([l.sub.1] + [l.sub.3]))y = [kappa]2(d + by), [kapp] [member of] C \{0}.

Solving it, we obtain

[kappa] = [[l.sub.0]/d], [l.sub.3] = - [l.sub.1], [l.sub.2] = [b[l.sub.0]/d] and G(u) = -[[l.sub.0]/d] u.

Then g = -[l.sub.1]y - [l.sub.0](ay - z)/d.

Case 2. We study the exponential factors of the form exp(g/h) with cofactor L = L(x,y,z) of degree at most one, (g,h) = 1 and h is a Darboux polynomial. Therefore g and h satisfy

(y - [x.sup.2]) [[partial derivative]g/[partial derivative]x] + (z -x) [[partial derivative]g/[partial derivative]y] - (d + ax + by + cz) [[partial derivative]g/[partial derivative]z] = Kg + Lh, (14)

where we have simplified the common factor exp(g/h) and we have used the fact that h is a Darboux polynomial of system (1) with cofactor K. By Theorem 1 and Lemma 6, a = b = 0, c [not equal to] 0, h = [(cz + d).sup.n] with n [greater than or equal to] 1 and K = -nc. Then (14) becomes

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (15)

with n [greater than or equal to] 1. We denote by [bar.g] the restriction of g to z = -d/c. Then [bar.g] [not equal to] 0 (since otherwise (h, g) [not equal to] 1). In this case, [bar.g] satisfies

(y - [x.sup.2]) [[partial derivative][bar.g]/[partial derivative]x] - (d/c + x) [[partial derivative][bar.g]/[partial derivative]y] = -nc[bar.g]. (16)

We write [bar.g] as a polynomial in the variable x that is [bar.g] = [[summation].sup.m.sub.i=0] [[bar.g].sub.i] (y)[x.sup.i] . Then the coefficient of [x.sup.n+1] in (16) satisfy

-m[[bar.g].sub.m] - [d[[bar.g].sub.m]/dy] = 0,

which yields [[bar.g].sub.m] = [C.sub.m] [e.sup.-my]. Since [[bar.g].sub.m] must be a polynomial we have m = 0 and [bar.g] = [[bar.g].sub.0] = [[bar.g].sub.0] (y). Therefore, from (16) we get

-(d\c + x) [d[[bar.g].sub.0]/dy] = -nc[[bar.g].sub.0],

which yields [[bar.g].sub.0] = 0, in contradiction with the fact that [bar.g] [not equal to] 0.

Proposition 13. System (1) with either b [not equal to] 0, or d [not equal to] 0, or c + a [not equal to] 0 has no first integrals of Darboux type.

Proof. It follows from Theorem 4 that system (1) has a first integral of Darboux type if and only if there exists [[lambda].sub.i], [[mu].sub.j] [member of] C not all zero such that equation (3) is satisfied where p, q are the numbers of Darboux polynomials and exponential factors, respectively. Furthermore, [K.sub.j] and [L.sub.j] are the cofactors of Darboux polynomials and exponential factors, respectively. It follows from Theorem 1 that the cofactor of the Darboux polynomials of system (1) when a = b = 0 and c [not equal to] 0 is -c and otherwise there are no Darboux polynomials of system (1). We consider three different cases.

Case 1: a = b = 0 and c [not equal to] 0. In this case a + c [not equal to] 0 and it follows from Theorem 1 and Lemma 12 that equation (3) is equivalent to

-c[[lambda].sub.1] + [[mu].sub.1] (z - x) + [[mu].sub.2] (d + cz) = 0.

Solving it we get [[lambda].sub.1] = [[mu].sub.1] = [[mu].sub.2] = 0. In short there are no first integrals of Darboux type in this case.

Case 2: a + c = 0 and b = 0. In this case d [not equal to] 0 and it follows from Theorem 1 and Lemma 12 that equation (3) is equivalent to

[[mu].sub.1] (z - x) + [[mu].sub.2]d + 2[[mu].sub.3]d(ay - z) = 0.

Solving it we get [[mu].sub.1] = [[mu].sub.2] = [[mu].sub.3] = 0 and there are no first integrals of Darboux type in this case.

Case 3: Remaining cases. It follows from Theorem 1 and Lemma 12 that equation (3) is equivalent to

[[mu].sub.1](z - x) + [[mu].sub.2] (d + by + (c + a)z) = 0.

Solving it we get [[mu].sub.1] = [[mu].sub.2] =0 which concludes the proof of Proposition 13.

Acknowledgements

The first author is supported by a MINECO/FEDER grant number MTM200803437, by an AGAUR grant number 2009SGR 410, by ICREA Academia and by FP7-PEOPLE-2012-IRSES 316338 and 318999

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Departament de Matematiques, Universitat Autoonoma de Barcelona, 08193 Bellaterra, Barcelona, Catalonia, Spain email:jllibre@mat.uab.cat[5mm] Departamento de Matematica, Instituto Superior Tecnico, Universidade Tecnica de Lisboa, Av. Rovisco Pais 1049-001, Lisboa, Portugal email:cvalls@math.ist.utl.pt

Received by the editors in February 2012.

Communicated by J. Mawhin.

2010 Mathematics Subject Classification : Primary 34A05,34A34,34C14.
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