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Liouvillian and analytic integrability in a modified Philips model
Chaos, Solitons & Fractals 45 (2012) 1573–1578
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Chaos, Solitons & Fractals Nonlinear Science, and Nonequilibrium and Complex Phenomena journal homepage: www.elsevier.com/locate/chaos
Liouvillian and analytic integrability in a modified Philips model Claudia Valls Departamento de Matemática, Instituto Superior Técnico, Universidade Técnica de Lisboa, Av. Rovisco Pais, 1049-001 Lisboa, Portugal
a r t i c l e
i n f o
a b s t r a c t
Article history: Received 27 March 2012 Accepted 8 October 2012 Available online 15 November 2012
We characterize the Liouvillian and analytic first integrals for a polynomial modified Phillips model. Ó 2012 Elsevier Ltd. All rights reserved.
1. Introduction and statement of the main results Nonlinear chaotic systems have attracted much attention in economics since it has been apparent that they are good models to understand the highly complex dynamics of real financial and economics systems, particularly due to the fact that they may exhibit a very rich dynamics together with a high sensitivity to the initial conditions. Thus, it is not surprising that in recent studies there has been a growing interest in the applications of nonlinear dynamics to economic modeling. We consider in this paper a modification of the wellknown Philips continuous time multiplier-accelerator model in [19] (see also [14] for a related discussion and the references therein). We consider that consumption C depends on the income as C(t) = c Y(t) with 0 < c 6 1, where Y is net income. The capital stock Kd depends linearly on the income as Kd(t) = v Y(t) with v > 0. It is assumed that firms change their capital stocks as soon as the actual stock differs from the desired one
K_ ¼ IðtÞ ¼ bðK d ðtÞ KðtÞÞ ¼ bðv YðtÞ KðtÞÞ
ð1Þ
with I as net investment. The coefficient b is an adjustment parameter and expresses the reaction speed of investment in response to a discrepancy between actual and desired stock. We assume that income changes according to the excess demand, C(t) + I(t) Y(t) in the goods market.
Y_ ¼ aðCðtÞ þ IðtÞ YðtÞÞ;
a>0
ð2Þ
with the coefficient a being an adjustment parameter and such that 1 av > 0. Differentiating (1) with respect to the time t we get
_ I_ ¼ bðv YðtÞ IðtÞÞ and then from (2) we obtain the linear second order differential equation
€ þ ðað1 cÞ þ b abv ÞYðtÞ _ YðtÞ þ abð1 cÞYðtÞ ¼ 0:
Let y = Y Y⁄ with Y⁄ as the fixed-point value of the income Eq. (3) becomes
€ þ ðað1 cÞ þ bð1 av ÞÞy_ þ abð1 cÞy ¼ 0: y
0960-0779/$ - see front matter Ó 2012 Elsevier Ltd. All rights reserved. http://dx.doi.org/10.1016/j.chaos.2012.10.001
ð4Þ
We consider in this paper a modification of (4) in which b = h(y) is a function depending on the income, that is, investment responds linearly to gaps between the desired and the actual capital stock. For definiteness we will assume that h is a polynomial of degree greater or equal than one and with h(0) > 0. We stress that the main advantage of nonlinear cycle models, like the one presented in this paper, over linear models like the original Philips model, is that the linear ones do not rely on precise parameter constellations in order to generate persistent fluctuations, and thus provide worse models for real financial and economic systems. A first step to nonlinearize the Philips model consists precisely in considering b as a polynomial of degree greater or equal than one (and not only a constant). Then (4) becomes
y_ ¼ x; E-mail address: [email protected]
ð3Þ
x_ ¼ ðað1 cÞ þ ð1 av ÞhðyÞÞx að1 cÞhðyÞy ð5Þ
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C. Valls / Chaos, Solitons & Fractals 45 (2012) 1573–1578
that we call a polynomial modification of the Philips model. Here the dot denotes derivative with respect to t. We note that system (5) is a generalized Liénard equation. We want to understand its complex dynamics by characterizing the existence or not of first integrals. Note that the existence of one first integral for system (5) solves completely the problem in the sense that the phase portratis are completely determined. In general the problem of the characterization of a first integral is a very difficult problem. The vector field of system (5) is
X ¼x
@ @ ððað1 cÞ þ ð1 av ÞhðyÞÞx þ að1 cÞhðyÞyÞ : @y @x
Let U R2 be an open and dense set in R2 . We say that the non-constant function H : R2 ! R is a first integral of the polynomial vector field X on U, if H(x(t), y(t)) = constant for all values of t for which the solution (x(t), y(t)) of X is defined on U. Clearly H is a first integral of X on U if and only if X H ¼ 0 on U. An analytic first integral is a first integral which is an analytic function. As a particular case, a first integral which is a polynomial is called a polynomial first integral. A Liouvillian first integral is a first integral H which is a Liouvillian function, that is, roughly speaking which can be obtained ‘‘by quadratures’’ of elementary functions. For a precise definition see [20]. The study of the Liouvillian first integrals is a classical problem of the integrability theory of the differential equations which goes back to Liouville, see for details again [20]. As far as we know the Liouvillian and analytic first integrals of some multi-parameter family of planar polynomial differential systems has only been classified for few differential systems, see for instance [3,7–13,15–18]. Theorem 1. System (5) with c = 1 has a polynomial first integral of the form
x þ ð1 av Þ
Z
hðyÞ dy:
ð6Þ
It is easy to verify that the function in (6) is a polynomial first integral of system (5) with c = 1.
say that S = 0 is an invariant algebraic curve of the vector field X associated to system (5) if it satisfies
x
@S @S ððað1 cÞ þ ð1 av ÞhðyÞÞx þ að1 cÞhðyÞyÞ ¼ KS; @y @x
the polynomial K ¼ Kðx; yÞ 2 C½x; y is called the cofactor of S = 0 and has degree at most n, where n is the degree of h, i.e., n = deg h. By assumptions n P 1. We also say that S is a Darboux polynomial of system (5). Note that a polynomial first integral is a Darboux polynomial with zero cofactor. The invariant algebraic curves are important because a sufficient number of them forces the existence of a first integral. This result is the basis of the Darboux theory of integrability, see for instance [3,5,7]. An exponential factor E of system (5) is a function of the form E ¼ expðg 0 =g 1 Þ R C with g 0 ; g 1 2 C½x; y coprime satisfying that
x
@E @E ððað1 cÞ þ ð1 av ÞhðyÞÞx þ að1 cÞhðyÞyÞ ¼ LE; @y @x
for some polynomial L = L(x,y) of degree at most n = deg h, called the cofactor of E. We will use the following known result whose proof and geometrical meaning is given in [5]. Proposition 3. The following statements hold. (a) If E = exp(g0/g1) is an exponential factor for the polynomial system (5) and g1 is not a constant polynomial, then g1 = 0 is an invariant algebraic curve. (b) Eventually eg0 can be an exponential factor, coming from the multiplicity of the infinite invariant straight line. We note that Eq. (5) is a generalized Liénard system of the form
y_ ¼ x;
x_ ¼ f ðyÞx gðyÞ:
ð7Þ
The following two results on generalized Liénard systems characterize the existence of analytic first integrals and of algebraic curves. For a proof of the first one see [2] and for a proof of the second one see [6]. We set
Z
Z
From now on we will restrict to the case c – 1. Note that v, a > 0 and 1 av > 0.
FðyÞ ¼
Theorem 2. The following holds for system (5) with c – 1.
Theorem 4 (Center theorem for analytic Liénard systems). Let f and g in system (7) be analytic functions defined in a neighborhood of zero with g(0) = 0 and g0 (0) > 0. Then the generalized Liénard system (7) has an analytic first integral if and only if F(y) = U(G(y)), for some analytic function U with U(0) = 0.
(a) It has no analytic first integrals. (b) It has no Liouvillian first integrals. We recall that on Theorem 2 only conditions on h(0) are required while other than this, h can be any polynomial of degree greater than or equal than one. The proof of Theorem 2 is given in Section 3 and in Section 2 we introduce some preliminary results.
f ðyÞ dy and GðyÞ ¼
Theorem 5. System (7) with deg f + 1 = deg g has an invariant algebraic curve if and only if there exists an invariant curve x P(y) satisfying
gðyÞ ¼ ½f ðyÞ þ P0 ðyÞPðyÞ; 2. Preliminary results Let Sðx; yÞ 2 C½x; y n C. As usual C½x; y denotes the ring of all complex polynomials in the variables x and y. We
gðyÞ dy:
where
PðyÞ ¼ a0 þ a1 y; or
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C. Valls / Chaos, Solitons & Fractals 45 (2012) 1573–1578
PðyÞ ¼ FðyÞ þ a0 þ a1 y ¼
Z
with cofactor K = h0 (0)(1 av)y if
f ðyÞ dy þ a0 þ a1 y;
a2 ð1 cÞv 0 þ h ð0Þy: ð1 av Þ2 að1 cÞðn þ av Þ ð1 av Þ nþ1 y f ¼ xþ yþC nð1 av Þ nþ1 hðyÞ ¼
for some a0 ; a1 2 C. 2. 3. Proof of Theorem 2
ð9Þ
Note that system (5) can be written as system (7) with
gðyÞ ¼ að1 cÞhðyÞy and f ðyÞ ¼ að1 cÞ þ ð1 av ÞhðyÞ:
with cofactor K = a(1 c)(n + 1)/(1 av) if
hðyÞ ¼
ðn þ 1Það1 cÞðn þ av Þ nð1 av Þ2
3.1. Proof of Statement (a) of Theorem 2
þ Cyn
ð10Þ
for some constant C > 0 and some integer n P 1. Since h(0) > 0, we are under the assumptions of Theorem 4. Hence, system (5) has a center at the origin if and only if there exists an analytic function U with U(0) = 0 such that
FðyÞ ¼ UðGðyÞÞ:
ð8Þ
We will see that this is not possible. We proceed by contradiction. Assume that there exists, indeed, an analytic function U = U(u) with U(0) = 0 such that (8) holds. Then if we write it as a Taylor series as
UðuÞ ¼
X aj uj ;
gðyÞ ¼ ðf ðyÞ þ a1 Þða0 þ a1 yÞ; that is,
að1 cÞhðyÞy ¼ ðað1 cÞ þ a1 þ ð1 av ÞhðyÞÞða0 þ a1 yÞ; ð11Þ or
jP1
gðyÞ ¼ ðf ðyÞ þ PðyÞ0 ÞPðyÞ ¼ a1 ðFðyÞ þ a0 þ a1 yÞ;
taking into account that
GðyÞ ¼ að1 cÞ
Proof. From Theorem 5 system (5) has a Darboux polynomial if and only if either
Z
that is
2
hðyÞy dy ¼ að1 cÞhð0Þy =2 þ h:o:tðyÞ
Z
(where h.o.t(y) denotes the higher order terms in the variable y), we have that
UðGðyÞÞ ¼ a1 að1 cÞhð0Þ However,
FðyÞ ¼ að1 cÞy þ ð1 av Þ
ð12Þ We consider the two cases separately. First case. In this case, the terms of degree n + 1 in (11) yield
y2 þ h:o:tðyÞ: 2 Z
að1 cÞ ¼ a1 ð1 av Þ;
that is a1 ¼
hðyÞ dy
að1 cÞ : 1 av
ð13Þ
Now evaluating (11) on y = 0 we obtain
¼ að1 cÞy þ ð1 av Þhð0Þy þ h:o:tðyÞ;
a0 ðað1 cÞ þ a1 þ ð1 av Þhð0ÞÞ ¼ 0;
and in order that (8) holds we must have, in particular, that the terms of degree one are zero, that is,
að1 cÞ þ ð1 av Þhð0Þ ¼ 0; i:e: hð0Þ ¼
að1 cÞhðyÞy ¼ a1 a0 a1 y þ að1 cÞy þ ð1 av Þ hðyÞdy :
að1 cÞ ; 1 av
which is not possible due to the fact that a(1 c) (1 av) > 0 and h(0) > 0. This concludes the proof of statement (a) of Theorem 4. 3.2. Proof of Statement (b) of Theorem 2
that is, either a0 = 0 or a1 = a(1 c) (1 av)h(0). If a0 = 0, computing the terms of degree one in (11) we get
að1 cÞhð0Þ ¼ a1 ðað1 cÞ þ a1 þ ð1 av Þhð0ÞÞ: Using the value of a1 in (13) and simplifying by a(1 c), we obtain
hð0Þ ¼
1 að1 cÞ að1 cÞ þ ð1 av Þhð0Þ ; 1 av 1 av
First we study the Darboux polynomials of system (5). We have the following proposition.
that is
Proposition 6. The unique Darboux polynomials f of system (5) with c – 1 are:
which is not possible due to the fact that a > 0, c – 1 and av > 0. If a0 – 0, then a1 = a(1 c) (1 av)h(0). Moreover, using again (13) we can rewrite Eq. (11) as
1.
f ¼ xþ
að1 cÞhð0Þ að1 cÞ y þ 0 ð1 av Þh ð0Þ 1 av
að1 cÞ 1
1 1 av
¼ 0;
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að1 cÞhðyÞy ¼ ð1 av ÞðhðyÞ hð0ÞÞða0 þ a1 yÞ að1 cÞ ¼ ð1 av ÞðhðyÞ hð0ÞÞ a0 y 1 av
f ¼ xþ
f ðyÞ dy a0 a1 y
¼ x þ að1 cÞy þ ð1 av Þ
¼ ð1 av ÞðhðyÞ hð0ÞÞa0 að1 cÞyhðyÞ
¼ xþ
þ að1 cÞhð0Þy; and hence
Z
Z
hðyÞ dy
að1 cÞðn þ 1Þ y 1 av
að1 cÞðn þ av Þ ð1 av Þ nþ1 y ; yþC nð1 av Þ nþ1
ð14Þ
with cofactor K = a(1 c)(n + 1)/(1 av). This completes the proof of the proposition. h
If deg h P 2, then the coefficient of degree n in (14) yields a0(1 av) = 0 which is not possible. If deg h = 1, that is, h(y) = h(0) + h0 (0)y with h0 (0) – 0, then (14) becomes, after simplifying by y,
We first recall that a non-constant complex function R : C2 ! C is an integrating factor of the polynomial vector field X on U, if one of the following three equivalent conditions holds
0 ¼ ð1 av ÞðhðyÞ hð0ÞÞa0 þ að1 cÞhð0Þy:
0
0 ¼ ð1 av Þh ð0Þa0 þ að1 cÞhð0Þ
@ðRPÞ @ðRQÞ ¼ ; @x @y
which yields
a0 ¼
að1 cÞhð0Þ : 0 ð1 av Þh ð0Þ
div ðP; Q Þ ¼
að1 cÞhð0Þ að1 cÞ f ¼ x a0 a1 y ¼ x þ y þ 0 ð1 av Þh ð0Þ 1 av with cofactor K = h0 (0)(1 av)y. Second case. In the second case, evaluating (12) on y = 0 we get that a0a1 = 0. Since a1 – 0 we must have a0 = 0. Then (12) becomes
Z
hðyÞdy : ð15Þ
Taking derivatives in (15) we get
að1 cÞh0 ðyÞy þ að1 cÞhðyÞ ¼ a1 ða1 þ að1 cÞ þ ð1 av ÞhðyÞÞ; that is
hðyÞ a1 ð1 av Þ a1 a1 ¼ : 1 h ðyÞ þ 1 y að1 cÞ y að1 cÞ 0
Solving this linear equation we get
hðyÞ ¼
a1 a1 1 að1cÞ
av Þ 1 a1að1 ð1cÞ
þ Cy
a1 ð1av Þ að1cÞ 1
;
for some constant C > 0. Since h is a polynomial of degree greater or equal one, we must have
a1 ð1 av Þ ¼ 1 þ n; að1 cÞ for some integer n P 1. Then
hðyÞ ¼
ðn þ 1Það1 cÞðn þ av Þ nð1 av Þ2
þ Cyn :
ðn þ 1Það1 cÞðn þ av Þ nð1 av Þ2
The Darboux polynomial is
@P @Q þ ¼ að1 cÞ ð1 av ÞhðyÞ: @x @y
From the Darboux theory of integrability we have the next result proved, for instance, in [4]. Theorem 7. Suppose that the polynomial vector field X of degree m defined in C2 admits p invariant algebraic curves fi = 0 with cofactors Ki for i = 1, . . . ,p, and q exponential factors Ej = exp(gj/hj) with (gj,hj) = 1 and cofactors Lj for j = 1, . . . , q. Then there exist ki ; lj 2 C not all zero such that p q X X ki K i þ lj Lj ¼ div ðP; Q Þ; i¼1
j¼1
if and only if the function of Darboux type k
l
l
f1k1 fp p E1 1 Eq q
ð16Þ
is an integrating factor of the vector field X . To prove the results related with Liouvillian first integrals we use the following result proved in [1,20]. Theorem 8. The polynomial differential system (5) has a Liouvillian first integral if and only if it has an integrating factor of Darboux type (see (16)). In order that system (5) has a Liouvillian first integral, by Theorem 8, system (5) must have an integrating factor of Darboux type. From Proposition 6, if h is neither of the form (9) nor of the form (10), then system (5) has no Darboux polynomials and thus using Proposition 3 if it has an exponential factor it must be of the form E = exp(lg0) with g 0 2 C½x; y n C, l 2 C and cofactor lL of degree n. Then system (5) has an integrating factor of Darboux type if and only if
lL ¼ að1 cÞ þ ð1 av ÞhðyÞ; l 2 C: Thus, using that E = expG(0) = exp(lg0) is an exponential factor, it must satisfy
Note that in this case
hð0Þ ¼
XR ¼ Rdiv ðP; Q Þ;
on U with P = (a(1 c) + (1 av)h(y))x a(1 c)h(y)y and Q = x. As usual the divergence of the vector field X is given by
In this case we get the Darboux polynomial
að1 cÞhðyÞy ¼ a1 a1 y þ að1 cÞy þ ð1 av Þ
div ðRP; RQÞ ¼ 0;
> 0:
x
@G0 @G0 ððað1 cÞ þ ð1 av ÞhðyÞÞx þ að1 cÞhðyÞyÞ ¼ að1 cÞ þ ð1 av ÞhðyÞ: @y @x
ð17Þ
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kh ð0Þð1 av Þy þ lL ¼ að1 cÞ þ ð1 av ÞhðyÞ; k; l 2 C:
Evaluating (17) on x = 0 we get
@G0 að1 cÞhðyÞy ¼ að1 cÞ þ ð1 av ÞhðyÞ: @x x¼0
ð18Þ
Since the left-hand side of (18) has degree at least n + 1 while the right-hand side has degree at most n, we must have
@G0 ¼ 0 and að1 cÞ þ ð1 av ÞhðyÞ ¼ 0; @x x¼0
¼X
Y0
að1 cÞhð0Þ að1 cÞ Y: 0 ð1 av Þh ð0Þ 1 av
Now if we write G0 = G 0(X,Y) = g0(x,y), then G0 is coprime with X. Moreover, we can write E = exp(G0/Xm) with m P 1, and after simplifying by Xm, G0 satisfies
að1 cÞhð0Þ að1 cÞ @G0 0 X Y h ð0Þð1 0 @Y ð1 av Þh ð0Þ 1 av @G0 0 av ÞYX þ mh ð0Þð1 av ÞYG0 @X ¼ LðX; YÞX;
0 ¼ að1 cÞ þ ð1 av Þhð0Þ;
ð19Þ
0
ð20Þ
2
f :¼ F 1 ¼ x þ
0
mh ð0Þð1 av Þ2 Y að1 cÞ
!
að1 cÞhð0Þ að1 cÞ Y þ 0 ð1 av Þh ð0Þ 1 av
að1 cÞðn þ av Þ ð1 av Þ nþ1 y ; yþC nð1 av Þ nþ1
with cofactor K = a(1 c)(n + 1)/(1 av). Then in view of Proposition 3 it can have an exponential factor of the form: either E = exp(g0) or E ¼ expðg 0 =F m 1 Þ with m P 1 and such that F1 does not divide g 0 2 C½x; y. We first prove that system (5) has no exponential factors of the second kind. Assume that system (5) has an exponential factor of the form E ¼ expðg 0 =F m 1 Þ with m P 1 and that F1 does not divide g 0 2 C½x; y. We introduce the change of variables X = F1 and Y = y. Then system (5) becomes
að1 cÞðn þ 1Þ X; Y 0 1 av að1 cÞðn þ av Þ ð1 av Þ nþ1 Y : ¼X Y C nð1 av Þ nþ1
X0 ¼
að1 cÞðn þ av Þ ð1 av Þ nþ1 @G0 Y X Y C nð1 av Þ nþ1 @Y
av Þ2 mhð0Þð1 aðc1Þ
;
að1 cÞðn þ 1Þ @G0 mað1 cÞðn þ 1Þ X G0 þ 1 av 1 av @X
¼ L1 ðX; YÞX;
ð22Þ
where L1 has degree n. Let now G0 ¼ G0 ðYÞ, the restriction of G0 to X = 0. Then G0 – 0, since otherwise G0 would be divisible by X which is not possible. Now, evaluating (22) on X = 0 we get
Solving this linear equation we get G0 ðYÞ ¼ C 1 exp
að1 cÞ : 1 av
Now if we write G0 = G0(X,Y) = g0(x,y), then G0 is coprime with X. Moreover we can write E = exp(G0/Xm) with m P 1, and after simplifying by Xm, G0 satisfies
að1 cÞhð0Þ að1 cÞ Y G00 ðYÞ þ 0 ð1 av Þh ð0Þ 1 av þ mh ð0Þð1 av ÞYG0 ðYÞ ¼ 0:
that is hð0Þ ¼
ð1cÞv Since hð0Þ ¼ að1 , we get a contradiction. Hence, this av Þ2 case does not provide any Liouvillian first integral. If h is of the form in (10), then in view of Proposition 6, system (5) has a Darboux polynomial
where L has degree n. Let now G0 ¼ G0 ðYÞ, the restriction of G0 to X = 0. Then G0 – 0, since otherwise G0 would be divisible by X which is not possible. Now, evaluating (19) on X = 0 we get
ð21Þ
Evaluating (21) on x = y = 0 we get
with cofactor K = h0 (0)(1 av)y. Then in view of Proposition 3 it can have an exponential factor of the form: either E = exp(g0) or E = exp(g0/Fm) with m P 1 and such that F does not divide g 0 2 C½x; y. We first prove that system (5) has no exponential factors of the second kind. Assume that system (5) has an exponential factor of the form E = exp(g0/Fm) with m P 1 such that F does not divide g 0 2 C½x; y. We introduce the change of variables X = F and Y = y. Then, in this case, system (5) becomes 0
@G0 @G0 ððað1 cÞ þ ð1 av ÞhðyÞÞx þ að1 cÞhðyÞyÞ @y @x 0
að1 cÞhð0Þ að1 cÞ y þ 0 ð1 av Þh ð0Þ 1 av
X 0 ¼ h ð0Þð1 av ÞYX;
x
¼ að1 cÞ þ ð1 av ÞhðyÞ þ kh ð0Þð1 av Þy:
which is not possible since h has degree greater or equal than one and av – 1. This completes the proof of the theorem in this case. If h is of the form in (9), then in view of Proposition 6, system (5) has a Darboux polynomial
f :¼ F ¼ x þ
Therefore, in view of Proposition 3 if there exists an exponential factor it must be of the form E = exp(G0) = exp(lg0) with g 0 2 C½x; y n C, l 2 C and
að1 cÞðn þ av Þ ð1 av Þ nþ1 0 Y G0 ðYÞ Y þC nð1 av Þ nþ1 mað1 cÞðn þ 1Þ þ G0 ðYÞ ¼ 0: 1 av
ð23Þ
Solving this linear equation we get for some constant C1. Since G0 must be a polynomial and m h0 (0)(1 av) – 0, we must have C1 = 0, but then G0 ¼ 0, which is not possible. In summary, if h is of the form in (9) then system (5) has an integrating factor of Darboux type if and only if
G0 ðYÞ ¼ C 1 Y
mðnþ1Þn nþav
að1 cÞðn þ av Þ 1 av n þC Y nð1 av Þ nþ1
mðnþ1Þ nþav ;
for some constant C1. Since G0 must be a polynomial and
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C. Valls / Chaos, Solitons & Fractals 45 (2012) 1573–1578
mðn þ 1Þn mðn þ 1Þ m2 ðn þ 1Þ2 n ¼ < 0; n þ av n þ av ðn þ av Þ2
References
we must have C1 = 0, but then G0 ¼ 0, which is not possible. In summary, if h is of the form in (10) then system (5) has an integrating factor of Darboux type if and only if
kað1 cÞðn þ 1Þ þ lL ¼ að1 cÞ 1 av þ ð1 av ÞhðyÞ; k; l 2 C:
Therefore, in view of Proposition 3 if there exists an exponential factor it must be of the form E = exp(G0) = exp(lg0) with g 0 2 C½x; y n C, l 2 C and
x
@G0 @G0 ððað1 cÞ þ ð1 av ÞhðyÞÞx þ að1 cÞhðyÞyÞ @y @x kað1 cÞðn þ 1Þ ¼ að1 cÞ þ ð1 av ÞhðyÞ þ : ð24Þ 1 av
Evaluating (24) on x = 0 we get
að1 cÞhðyÞyÞ
@G0 ¼ að1 cÞ þ ð1 av ÞhðyÞ @x x¼0 þ
kað1 cÞðn þ 1Þ : 1 av
ð25Þ
Since the left-hand side of (25) has degree at least n + 1 while the right-hand side has degree n we must have
@G0 ¼ 0 and að1 cÞ þ ð1 av ÞhðyÞ @x x¼0 kað1 cÞðn þ 1Þ ¼ 0; þ 1 av which is not possible since h has degree at least one and av – 1. This completes the proof of statement (b) of Theorem 2. Acknowledgments Supported by the FCT Grant PTDC/MAT/117106/2010 and again by FCT through CAMGDS, Lisbon.
[1] Christopher C. Liouvillian first integral of second order polynomial differential systems. Electron J Differ Equat 1999;1999:1–7. [2] Christopher C. An algebraic approach to the classification of centers in polynomial Liénard systems. J Math Anal Appl 1999; 229:319–29. [3] Cairó L, Giacomini H, Llibre J. Liouvillian first integrals for the planar Lotka–Volterra system. Rend Circ Mat Palermo 2003;52: 389–418. [4] Christopher C, Llibre J. Integrability via invariant algebraic curves for planar polynomial differential systems. Ann Differ Equat 2000;16: 5–19. [5] Christopher C, Llibre J, Pereira JV. Multiplicity of invariant algebraic curves and Darboux integrability. Pacific J Math 2007;229:63–117. [6] Makoto H. On Polynomial Liénard systems which have invariant algebraic curves. Funkcialaj Ekvacioj 1996;39:403–8. [7] Labrunie S. On the polynomial first integrals of the (a, b, c) Lotka– Volterra system. J Math Phys 1996;37:5539–50. [8] Llibre J, Valls C. Global analytic first integrals for the real planar Lotka–Volterra system. J Math Phys 2007;48(3):033507. 13 pp. [9] Llibre J, Valls C. Global analytic first integrals for the simplified multistrain/two-stream model for tuberculosis and dengue fever. J Nonlinear Math Phys 2009;16:505–16. [10] Llibre J, Valls C. Liouvillian first integrals for Liénard polynomial differential systems. Proc Am Math Soc 2010;138(9):3229–39. [11] Llibre J, Valls C. Liouvillian integrability of the FitzHugh–Nagumo systems. J Geom Phys 2010;60(12):1974–83. [12] Llibre J, Valls C. On the local analytic integrability at the singular point of a class of Liénard analytic differential systems. Proc Am Math Soc 2010;138(1):253–61. [13] Llibre J, Valls C. Liouvillian first integrals of quadratic-linear polynomial differential systems. J Math Anal Appl 2011;379(1): 188–99. [14] Lorenz HW. On the uniqueness of limit cycles in business cycle theory. Metroeconomica 1987;38:281–93. [15] Moulin Ollagnier J. Polynomial first integrals of the Lotka–Volterra system. Bull Sci Math 1997;121:463–76. [16] Moulin Ollagnier J. Rational integration of the Lotka–Volterra system. Bull Sci Math 1999;123:437–66. [17] Moulin Ollagnier J. Liouvillian integration of the Lotka–Volterra system. Qual Theory Dynam Syst 2001;2:307–58. [18] Moulin Ollagnier J. Corrections and complements to: Liouvillian integration of the Lotka–Volterra system. Qual Theory Dynam Syst 2001;2:307–58; Moulin Ollagnier J. Corrections and complements to: Liouvillian integration of the Lotka–Volterra system. Qual Theory Dynam Syst 2004;5:275–84. [19] Phillips AW. Stabilisation policy in a closed economy. Econ J 1954;64:290–323. [20] Singer MF. Liouvillian first integrals of differential systems. Trans Am Math Soc 1992;333:673–88.
Contents lists available at SciVerse ScienceDirect
Chaos, Solitons & Fractals Nonlinear Science, and Nonequilibrium and Complex Phenomena journal homepage: www.elsevier.com/locate/chaos
Liouvillian and analytic integrability in a modified Philips model Claudia Valls Departamento de Matemática, Instituto Superior Técnico, Universidade Técnica de Lisboa, Av. Rovisco Pais, 1049-001 Lisboa, Portugal
a r t i c l e
i n f o
a b s t r a c t
Article history: Received 27 March 2012 Accepted 8 October 2012 Available online 15 November 2012
We characterize the Liouvillian and analytic first integrals for a polynomial modified Phillips model. Ó 2012 Elsevier Ltd. All rights reserved.
1. Introduction and statement of the main results Nonlinear chaotic systems have attracted much attention in economics since it has been apparent that they are good models to understand the highly complex dynamics of real financial and economics systems, particularly due to the fact that they may exhibit a very rich dynamics together with a high sensitivity to the initial conditions. Thus, it is not surprising that in recent studies there has been a growing interest in the applications of nonlinear dynamics to economic modeling. We consider in this paper a modification of the wellknown Philips continuous time multiplier-accelerator model in [19] (see also [14] for a related discussion and the references therein). We consider that consumption C depends on the income as C(t) = c Y(t) with 0 < c 6 1, where Y is net income. The capital stock Kd depends linearly on the income as Kd(t) = v Y(t) with v > 0. It is assumed that firms change their capital stocks as soon as the actual stock differs from the desired one
K_ ¼ IðtÞ ¼ bðK d ðtÞ KðtÞÞ ¼ bðv YðtÞ KðtÞÞ
ð1Þ
with I as net investment. The coefficient b is an adjustment parameter and expresses the reaction speed of investment in response to a discrepancy between actual and desired stock. We assume that income changes according to the excess demand, C(t) + I(t) Y(t) in the goods market.
Y_ ¼ aðCðtÞ þ IðtÞ YðtÞÞ;
a>0
ð2Þ
with the coefficient a being an adjustment parameter and such that 1 av > 0. Differentiating (1) with respect to the time t we get
_ I_ ¼ bðv YðtÞ IðtÞÞ and then from (2) we obtain the linear second order differential equation
€ þ ðað1 cÞ þ b abv ÞYðtÞ _ YðtÞ þ abð1 cÞYðtÞ ¼ 0:
Let y = Y Y⁄ with Y⁄ as the fixed-point value of the income Eq. (3) becomes
€ þ ðað1 cÞ þ bð1 av ÞÞy_ þ abð1 cÞy ¼ 0: y
0960-0779/$ - see front matter Ó 2012 Elsevier Ltd. All rights reserved. http://dx.doi.org/10.1016/j.chaos.2012.10.001
ð4Þ
We consider in this paper a modification of (4) in which b = h(y) is a function depending on the income, that is, investment responds linearly to gaps between the desired and the actual capital stock. For definiteness we will assume that h is a polynomial of degree greater or equal than one and with h(0) > 0. We stress that the main advantage of nonlinear cycle models, like the one presented in this paper, over linear models like the original Philips model, is that the linear ones do not rely on precise parameter constellations in order to generate persistent fluctuations, and thus provide worse models for real financial and economic systems. A first step to nonlinearize the Philips model consists precisely in considering b as a polynomial of degree greater or equal than one (and not only a constant). Then (4) becomes
y_ ¼ x; E-mail address: [email protected]
ð3Þ
x_ ¼ ðað1 cÞ þ ð1 av ÞhðyÞÞx að1 cÞhðyÞy ð5Þ
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C. Valls / Chaos, Solitons & Fractals 45 (2012) 1573–1578
that we call a polynomial modification of the Philips model. Here the dot denotes derivative with respect to t. We note that system (5) is a generalized Liénard equation. We want to understand its complex dynamics by characterizing the existence or not of first integrals. Note that the existence of one first integral for system (5) solves completely the problem in the sense that the phase portratis are completely determined. In general the problem of the characterization of a first integral is a very difficult problem. The vector field of system (5) is
X ¼x
@ @ ððað1 cÞ þ ð1 av ÞhðyÞÞx þ að1 cÞhðyÞyÞ : @y @x
Let U R2 be an open and dense set in R2 . We say that the non-constant function H : R2 ! R is a first integral of the polynomial vector field X on U, if H(x(t), y(t)) = constant for all values of t for which the solution (x(t), y(t)) of X is defined on U. Clearly H is a first integral of X on U if and only if X H ¼ 0 on U. An analytic first integral is a first integral which is an analytic function. As a particular case, a first integral which is a polynomial is called a polynomial first integral. A Liouvillian first integral is a first integral H which is a Liouvillian function, that is, roughly speaking which can be obtained ‘‘by quadratures’’ of elementary functions. For a precise definition see [20]. The study of the Liouvillian first integrals is a classical problem of the integrability theory of the differential equations which goes back to Liouville, see for details again [20]. As far as we know the Liouvillian and analytic first integrals of some multi-parameter family of planar polynomial differential systems has only been classified for few differential systems, see for instance [3,7–13,15–18]. Theorem 1. System (5) with c = 1 has a polynomial first integral of the form
x þ ð1 av Þ
Z
hðyÞ dy:
ð6Þ
It is easy to verify that the function in (6) is a polynomial first integral of system (5) with c = 1.
say that S = 0 is an invariant algebraic curve of the vector field X associated to system (5) if it satisfies
x
@S @S ððað1 cÞ þ ð1 av ÞhðyÞÞx þ að1 cÞhðyÞyÞ ¼ KS; @y @x
the polynomial K ¼ Kðx; yÞ 2 C½x; y is called the cofactor of S = 0 and has degree at most n, where n is the degree of h, i.e., n = deg h. By assumptions n P 1. We also say that S is a Darboux polynomial of system (5). Note that a polynomial first integral is a Darboux polynomial with zero cofactor. The invariant algebraic curves are important because a sufficient number of them forces the existence of a first integral. This result is the basis of the Darboux theory of integrability, see for instance [3,5,7]. An exponential factor E of system (5) is a function of the form E ¼ expðg 0 =g 1 Þ R C with g 0 ; g 1 2 C½x; y coprime satisfying that
x
@E @E ððað1 cÞ þ ð1 av ÞhðyÞÞx þ að1 cÞhðyÞyÞ ¼ LE; @y @x
for some polynomial L = L(x,y) of degree at most n = deg h, called the cofactor of E. We will use the following known result whose proof and geometrical meaning is given in [5]. Proposition 3. The following statements hold. (a) If E = exp(g0/g1) is an exponential factor for the polynomial system (5) and g1 is not a constant polynomial, then g1 = 0 is an invariant algebraic curve. (b) Eventually eg0 can be an exponential factor, coming from the multiplicity of the infinite invariant straight line. We note that Eq. (5) is a generalized Liénard system of the form
y_ ¼ x;
x_ ¼ f ðyÞx gðyÞ:
ð7Þ
The following two results on generalized Liénard systems characterize the existence of analytic first integrals and of algebraic curves. For a proof of the first one see [2] and for a proof of the second one see [6]. We set
Z
Z
From now on we will restrict to the case c – 1. Note that v, a > 0 and 1 av > 0.
FðyÞ ¼
Theorem 2. The following holds for system (5) with c – 1.
Theorem 4 (Center theorem for analytic Liénard systems). Let f and g in system (7) be analytic functions defined in a neighborhood of zero with g(0) = 0 and g0 (0) > 0. Then the generalized Liénard system (7) has an analytic first integral if and only if F(y) = U(G(y)), for some analytic function U with U(0) = 0.
(a) It has no analytic first integrals. (b) It has no Liouvillian first integrals. We recall that on Theorem 2 only conditions on h(0) are required while other than this, h can be any polynomial of degree greater than or equal than one. The proof of Theorem 2 is given in Section 3 and in Section 2 we introduce some preliminary results.
f ðyÞ dy and GðyÞ ¼
Theorem 5. System (7) with deg f + 1 = deg g has an invariant algebraic curve if and only if there exists an invariant curve x P(y) satisfying
gðyÞ ¼ ½f ðyÞ þ P0 ðyÞPðyÞ; 2. Preliminary results Let Sðx; yÞ 2 C½x; y n C. As usual C½x; y denotes the ring of all complex polynomials in the variables x and y. We
gðyÞ dy:
where
PðyÞ ¼ a0 þ a1 y; or
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C. Valls / Chaos, Solitons & Fractals 45 (2012) 1573–1578
PðyÞ ¼ FðyÞ þ a0 þ a1 y ¼
Z
with cofactor K = h0 (0)(1 av)y if
f ðyÞ dy þ a0 þ a1 y;
a2 ð1 cÞv 0 þ h ð0Þy: ð1 av Þ2 að1 cÞðn þ av Þ ð1 av Þ nþ1 y f ¼ xþ yþC nð1 av Þ nþ1 hðyÞ ¼
for some a0 ; a1 2 C. 2. 3. Proof of Theorem 2
ð9Þ
Note that system (5) can be written as system (7) with
gðyÞ ¼ að1 cÞhðyÞy and f ðyÞ ¼ að1 cÞ þ ð1 av ÞhðyÞ:
with cofactor K = a(1 c)(n + 1)/(1 av) if
hðyÞ ¼
ðn þ 1Það1 cÞðn þ av Þ nð1 av Þ2
3.1. Proof of Statement (a) of Theorem 2
þ Cyn
ð10Þ
for some constant C > 0 and some integer n P 1. Since h(0) > 0, we are under the assumptions of Theorem 4. Hence, system (5) has a center at the origin if and only if there exists an analytic function U with U(0) = 0 such that
FðyÞ ¼ UðGðyÞÞ:
ð8Þ
We will see that this is not possible. We proceed by contradiction. Assume that there exists, indeed, an analytic function U = U(u) with U(0) = 0 such that (8) holds. Then if we write it as a Taylor series as
UðuÞ ¼
X aj uj ;
gðyÞ ¼ ðf ðyÞ þ a1 Þða0 þ a1 yÞ; that is,
að1 cÞhðyÞy ¼ ðað1 cÞ þ a1 þ ð1 av ÞhðyÞÞða0 þ a1 yÞ; ð11Þ or
jP1
gðyÞ ¼ ðf ðyÞ þ PðyÞ0 ÞPðyÞ ¼ a1 ðFðyÞ þ a0 þ a1 yÞ;
taking into account that
GðyÞ ¼ að1 cÞ
Proof. From Theorem 5 system (5) has a Darboux polynomial if and only if either
Z
that is
2
hðyÞy dy ¼ að1 cÞhð0Þy =2 þ h:o:tðyÞ
Z
(where h.o.t(y) denotes the higher order terms in the variable y), we have that
UðGðyÞÞ ¼ a1 að1 cÞhð0Þ However,
FðyÞ ¼ að1 cÞy þ ð1 av Þ
ð12Þ We consider the two cases separately. First case. In this case, the terms of degree n + 1 in (11) yield
y2 þ h:o:tðyÞ: 2 Z
að1 cÞ ¼ a1 ð1 av Þ;
that is a1 ¼
hðyÞ dy
að1 cÞ : 1 av
ð13Þ
Now evaluating (11) on y = 0 we obtain
¼ að1 cÞy þ ð1 av Þhð0Þy þ h:o:tðyÞ;
a0 ðað1 cÞ þ a1 þ ð1 av Þhð0ÞÞ ¼ 0;
and in order that (8) holds we must have, in particular, that the terms of degree one are zero, that is,
að1 cÞ þ ð1 av Þhð0Þ ¼ 0; i:e: hð0Þ ¼
að1 cÞhðyÞy ¼ a1 a0 a1 y þ að1 cÞy þ ð1 av Þ hðyÞdy :
að1 cÞ ; 1 av
which is not possible due to the fact that a(1 c) (1 av) > 0 and h(0) > 0. This concludes the proof of statement (a) of Theorem 4. 3.2. Proof of Statement (b) of Theorem 2
that is, either a0 = 0 or a1 = a(1 c) (1 av)h(0). If a0 = 0, computing the terms of degree one in (11) we get
að1 cÞhð0Þ ¼ a1 ðað1 cÞ þ a1 þ ð1 av Þhð0ÞÞ: Using the value of a1 in (13) and simplifying by a(1 c), we obtain
hð0Þ ¼
1 að1 cÞ að1 cÞ þ ð1 av Þhð0Þ ; 1 av 1 av
First we study the Darboux polynomials of system (5). We have the following proposition.
that is
Proposition 6. The unique Darboux polynomials f of system (5) with c – 1 are:
which is not possible due to the fact that a > 0, c – 1 and av > 0. If a0 – 0, then a1 = a(1 c) (1 av)h(0). Moreover, using again (13) we can rewrite Eq. (11) as
1.
f ¼ xþ
að1 cÞhð0Þ að1 cÞ y þ 0 ð1 av Þh ð0Þ 1 av
að1 cÞ 1
1 1 av
¼ 0;
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C. Valls / Chaos, Solitons & Fractals 45 (2012) 1573–1578
að1 cÞhðyÞy ¼ ð1 av ÞðhðyÞ hð0ÞÞða0 þ a1 yÞ að1 cÞ ¼ ð1 av ÞðhðyÞ hð0ÞÞ a0 y 1 av
f ¼ xþ
f ðyÞ dy a0 a1 y
¼ x þ að1 cÞy þ ð1 av Þ
¼ ð1 av ÞðhðyÞ hð0ÞÞa0 að1 cÞyhðyÞ
¼ xþ
þ að1 cÞhð0Þy; and hence
Z
Z
hðyÞ dy
að1 cÞðn þ 1Þ y 1 av
að1 cÞðn þ av Þ ð1 av Þ nþ1 y ; yþC nð1 av Þ nþ1
ð14Þ
with cofactor K = a(1 c)(n + 1)/(1 av). This completes the proof of the proposition. h
If deg h P 2, then the coefficient of degree n in (14) yields a0(1 av) = 0 which is not possible. If deg h = 1, that is, h(y) = h(0) + h0 (0)y with h0 (0) – 0, then (14) becomes, after simplifying by y,
We first recall that a non-constant complex function R : C2 ! C is an integrating factor of the polynomial vector field X on U, if one of the following three equivalent conditions holds
0 ¼ ð1 av ÞðhðyÞ hð0ÞÞa0 þ að1 cÞhð0Þy:
0
0 ¼ ð1 av Þh ð0Þa0 þ að1 cÞhð0Þ
@ðRPÞ @ðRQÞ ¼ ; @x @y
which yields
a0 ¼
að1 cÞhð0Þ : 0 ð1 av Þh ð0Þ
div ðP; Q Þ ¼
að1 cÞhð0Þ að1 cÞ f ¼ x a0 a1 y ¼ x þ y þ 0 ð1 av Þh ð0Þ 1 av with cofactor K = h0 (0)(1 av)y. Second case. In the second case, evaluating (12) on y = 0 we get that a0a1 = 0. Since a1 – 0 we must have a0 = 0. Then (12) becomes
Z
hðyÞdy : ð15Þ
Taking derivatives in (15) we get
að1 cÞh0 ðyÞy þ að1 cÞhðyÞ ¼ a1 ða1 þ að1 cÞ þ ð1 av ÞhðyÞÞ; that is
hðyÞ a1 ð1 av Þ a1 a1 ¼ : 1 h ðyÞ þ 1 y að1 cÞ y að1 cÞ 0
Solving this linear equation we get
hðyÞ ¼
a1 a1 1 að1cÞ
av Þ 1 a1að1 ð1cÞ
þ Cy
a1 ð1av Þ að1cÞ 1
;
for some constant C > 0. Since h is a polynomial of degree greater or equal one, we must have
a1 ð1 av Þ ¼ 1 þ n; að1 cÞ for some integer n P 1. Then
hðyÞ ¼
ðn þ 1Það1 cÞðn þ av Þ nð1 av Þ2
þ Cyn :
ðn þ 1Það1 cÞðn þ av Þ nð1 av Þ2
The Darboux polynomial is
@P @Q þ ¼ að1 cÞ ð1 av ÞhðyÞ: @x @y
From the Darboux theory of integrability we have the next result proved, for instance, in [4]. Theorem 7. Suppose that the polynomial vector field X of degree m defined in C2 admits p invariant algebraic curves fi = 0 with cofactors Ki for i = 1, . . . ,p, and q exponential factors Ej = exp(gj/hj) with (gj,hj) = 1 and cofactors Lj for j = 1, . . . , q. Then there exist ki ; lj 2 C not all zero such that p q X X ki K i þ lj Lj ¼ div ðP; Q Þ; i¼1
j¼1
if and only if the function of Darboux type k
l
l
f1k1 fp p E1 1 Eq q
ð16Þ
is an integrating factor of the vector field X . To prove the results related with Liouvillian first integrals we use the following result proved in [1,20]. Theorem 8. The polynomial differential system (5) has a Liouvillian first integral if and only if it has an integrating factor of Darboux type (see (16)). In order that system (5) has a Liouvillian first integral, by Theorem 8, system (5) must have an integrating factor of Darboux type. From Proposition 6, if h is neither of the form (9) nor of the form (10), then system (5) has no Darboux polynomials and thus using Proposition 3 if it has an exponential factor it must be of the form E = exp(lg0) with g 0 2 C½x; y n C, l 2 C and cofactor lL of degree n. Then system (5) has an integrating factor of Darboux type if and only if
lL ¼ að1 cÞ þ ð1 av ÞhðyÞ; l 2 C: Thus, using that E = expG(0) = exp(lg0) is an exponential factor, it must satisfy
Note that in this case
hð0Þ ¼
XR ¼ Rdiv ðP; Q Þ;
on U with P = (a(1 c) + (1 av)h(y))x a(1 c)h(y)y and Q = x. As usual the divergence of the vector field X is given by
In this case we get the Darboux polynomial
að1 cÞhðyÞy ¼ a1 a1 y þ að1 cÞy þ ð1 av Þ
div ðRP; RQÞ ¼ 0;
> 0:
x
@G0 @G0 ððað1 cÞ þ ð1 av ÞhðyÞÞx þ að1 cÞhðyÞyÞ ¼ að1 cÞ þ ð1 av ÞhðyÞ: @y @x
ð17Þ
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C. Valls / Chaos, Solitons & Fractals 45 (2012) 1573–1578 0
kh ð0Þð1 av Þy þ lL ¼ að1 cÞ þ ð1 av ÞhðyÞ; k; l 2 C:
Evaluating (17) on x = 0 we get
@G0 að1 cÞhðyÞy ¼ að1 cÞ þ ð1 av ÞhðyÞ: @x x¼0
ð18Þ
Since the left-hand side of (18) has degree at least n + 1 while the right-hand side has degree at most n, we must have
@G0 ¼ 0 and að1 cÞ þ ð1 av ÞhðyÞ ¼ 0; @x x¼0
¼X
Y0
að1 cÞhð0Þ að1 cÞ Y: 0 ð1 av Þh ð0Þ 1 av
Now if we write G0 = G 0(X,Y) = g0(x,y), then G0 is coprime with X. Moreover, we can write E = exp(G0/Xm) with m P 1, and after simplifying by Xm, G0 satisfies
að1 cÞhð0Þ að1 cÞ @G0 0 X Y h ð0Þð1 0 @Y ð1 av Þh ð0Þ 1 av @G0 0 av ÞYX þ mh ð0Þð1 av ÞYG0 @X ¼ LðX; YÞX;
0 ¼ að1 cÞ þ ð1 av Þhð0Þ;
ð19Þ
0
ð20Þ
2
f :¼ F 1 ¼ x þ
0
mh ð0Þð1 av Þ2 Y að1 cÞ
!
að1 cÞhð0Þ að1 cÞ Y þ 0 ð1 av Þh ð0Þ 1 av
að1 cÞðn þ av Þ ð1 av Þ nþ1 y ; yþC nð1 av Þ nþ1
with cofactor K = a(1 c)(n + 1)/(1 av). Then in view of Proposition 3 it can have an exponential factor of the form: either E = exp(g0) or E ¼ expðg 0 =F m 1 Þ with m P 1 and such that F1 does not divide g 0 2 C½x; y. We first prove that system (5) has no exponential factors of the second kind. Assume that system (5) has an exponential factor of the form E ¼ expðg 0 =F m 1 Þ with m P 1 and that F1 does not divide g 0 2 C½x; y. We introduce the change of variables X = F1 and Y = y. Then system (5) becomes
að1 cÞðn þ 1Þ X; Y 0 1 av að1 cÞðn þ av Þ ð1 av Þ nþ1 Y : ¼X Y C nð1 av Þ nþ1
X0 ¼
að1 cÞðn þ av Þ ð1 av Þ nþ1 @G0 Y X Y C nð1 av Þ nþ1 @Y
av Þ2 mhð0Þð1 aðc1Þ
;
að1 cÞðn þ 1Þ @G0 mað1 cÞðn þ 1Þ X G0 þ 1 av 1 av @X
¼ L1 ðX; YÞX;
ð22Þ
where L1 has degree n. Let now G0 ¼ G0 ðYÞ, the restriction of G0 to X = 0. Then G0 – 0, since otherwise G0 would be divisible by X which is not possible. Now, evaluating (22) on X = 0 we get
Solving this linear equation we get G0 ðYÞ ¼ C 1 exp
að1 cÞ : 1 av
Now if we write G0 = G0(X,Y) = g0(x,y), then G0 is coprime with X. Moreover we can write E = exp(G0/Xm) with m P 1, and after simplifying by Xm, G0 satisfies
að1 cÞhð0Þ að1 cÞ Y G00 ðYÞ þ 0 ð1 av Þh ð0Þ 1 av þ mh ð0Þð1 av ÞYG0 ðYÞ ¼ 0:
that is hð0Þ ¼
ð1cÞv Since hð0Þ ¼ að1 , we get a contradiction. Hence, this av Þ2 case does not provide any Liouvillian first integral. If h is of the form in (10), then in view of Proposition 6, system (5) has a Darboux polynomial
where L has degree n. Let now G0 ¼ G0 ðYÞ, the restriction of G0 to X = 0. Then G0 – 0, since otherwise G0 would be divisible by X which is not possible. Now, evaluating (19) on X = 0 we get
ð21Þ
Evaluating (21) on x = y = 0 we get
with cofactor K = h0 (0)(1 av)y. Then in view of Proposition 3 it can have an exponential factor of the form: either E = exp(g0) or E = exp(g0/Fm) with m P 1 and such that F does not divide g 0 2 C½x; y. We first prove that system (5) has no exponential factors of the second kind. Assume that system (5) has an exponential factor of the form E = exp(g0/Fm) with m P 1 such that F does not divide g 0 2 C½x; y. We introduce the change of variables X = F and Y = y. Then, in this case, system (5) becomes 0
@G0 @G0 ððað1 cÞ þ ð1 av ÞhðyÞÞx þ að1 cÞhðyÞyÞ @y @x 0
að1 cÞhð0Þ að1 cÞ y þ 0 ð1 av Þh ð0Þ 1 av
X 0 ¼ h ð0Þð1 av ÞYX;
x
¼ að1 cÞ þ ð1 av ÞhðyÞ þ kh ð0Þð1 av Þy:
which is not possible since h has degree greater or equal than one and av – 1. This completes the proof of the theorem in this case. If h is of the form in (9), then in view of Proposition 6, system (5) has a Darboux polynomial
f :¼ F ¼ x þ
Therefore, in view of Proposition 3 if there exists an exponential factor it must be of the form E = exp(G0) = exp(lg0) with g 0 2 C½x; y n C, l 2 C and
að1 cÞðn þ av Þ ð1 av Þ nþ1 0 Y G0 ðYÞ Y þC nð1 av Þ nþ1 mað1 cÞðn þ 1Þ þ G0 ðYÞ ¼ 0: 1 av
ð23Þ
Solving this linear equation we get for some constant C1. Since G0 must be a polynomial and m h0 (0)(1 av) – 0, we must have C1 = 0, but then G0 ¼ 0, which is not possible. In summary, if h is of the form in (9) then system (5) has an integrating factor of Darboux type if and only if
G0 ðYÞ ¼ C 1 Y
mðnþ1Þn nþav
að1 cÞðn þ av Þ 1 av n þC Y nð1 av Þ nþ1
mðnþ1Þ nþav ;
for some constant C1. Since G0 must be a polynomial and
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C. Valls / Chaos, Solitons & Fractals 45 (2012) 1573–1578
mðn þ 1Þn mðn þ 1Þ m2 ðn þ 1Þ2 n ¼ < 0; n þ av n þ av ðn þ av Þ2
References
we must have C1 = 0, but then G0 ¼ 0, which is not possible. In summary, if h is of the form in (10) then system (5) has an integrating factor of Darboux type if and only if
kað1 cÞðn þ 1Þ þ lL ¼ að1 cÞ 1 av þ ð1 av ÞhðyÞ; k; l 2 C:
Therefore, in view of Proposition 3 if there exists an exponential factor it must be of the form E = exp(G0) = exp(lg0) with g 0 2 C½x; y n C, l 2 C and
x
@G0 @G0 ððað1 cÞ þ ð1 av ÞhðyÞÞx þ að1 cÞhðyÞyÞ @y @x kað1 cÞðn þ 1Þ ¼ að1 cÞ þ ð1 av ÞhðyÞ þ : ð24Þ 1 av
Evaluating (24) on x = 0 we get
að1 cÞhðyÞyÞ
@G0 ¼ að1 cÞ þ ð1 av ÞhðyÞ @x x¼0 þ
kað1 cÞðn þ 1Þ : 1 av
ð25Þ
Since the left-hand side of (25) has degree at least n + 1 while the right-hand side has degree n we must have
@G0 ¼ 0 and að1 cÞ þ ð1 av ÞhðyÞ @x x¼0 kað1 cÞðn þ 1Þ ¼ 0; þ 1 av which is not possible since h has degree at least one and av – 1. This completes the proof of statement (b) of Theorem 2. Acknowledgments Supported by the FCT Grant PTDC/MAT/117106/2010 and again by FCT through CAMGDS, Lisbon.
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