Darboux integrability of a nonlinear financial system

Applied Mathematics and Computation 218 (2011) 3297–3302

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Applied Mathematics and Computation

journal homepage: www.elsevier .com/ locate/amc

Darboux integrability of a nonlinear financial system

Claudia VallsDepartamento 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

Keywords:Darboux integrabilityNonlinear finance systemInvariant algebraic surfacesDarboux first integralsExponential factors

0096-3003/$ - see front matter � 2011 Elsevier Incdoi:10.1016/j.amc.2011.08.069

E-mail address: [email protected]

We study the complexity of a dynamical financial model recently considered in the litera-ture, namely

. All righ

_x ¼ zþ ðy� aÞx; _y ¼ 1� by� x2; _z ¼ �x� cz:

More precisely, we study the integrability of the system. In particular, we show that it isnot Darboux integrable for any value of the positive real parameters a; b and c.

� 2011 Elsevier Inc. All rights reserved.

1. Introduction

Nonlinear chaotic systems have attracted much attention in economics since it has been apparent that they are goodmodels to understand the highly complex dynamics of real financial and economic systems, particularly due to the fact thatthey may exhibit a very rich dynamics together with a high sensitivity to the initial conditions. Thus, it is not surprising thatin recent studies there has been a growing interest in the applications of nonlinear dynamics to economic modeling.Examples are the forced van der Pol model (see for instance [2–4] and the references therein), the IS-LM model (see for in-stance [1,6,12,13] and the references therein), the Goodwin’s accelerator model [11], among many others. There is also somestudies of applications of geometrical and stochastic techniques to study finance systems (see, for instance [7,14,15]).

In this paper we study a nonlinear system considered by Junhai and Chen in [8,9]. More precisely, it is a nonlinear financesystem (a nonlinear autonomous ordinary differential equation in R2) describing the time variations of three state variables:the interest rate x, the investment demand y, and the price index z. By choosing an appropriate coordinate system andappropriate dimensions for each state variables, is given by the model

_x ¼ zþ ðy� aÞx; _y ¼ 1� by� x2; _z ¼ �x� cz; ð1Þ

where a; b and c are positive real parameters standing respectively for the amount of savings, the unit investment cost andthe demand flexibility. We refer to [8,9] for further details (in particular concerning the topological structure and bifurca-tions of the system).

This paper is a further contribution to the understanding of the complexity, or more precisely of the integrability ofsystem (1) . For a three-dimensional system of differential equations the existence of one first integral reduces the complex-ity of the dynamics and the existence of two first integrals that are functionally independent solves the problem completely(at least theoretically), in the sense that the phase portraits are completely determined. In general, for a given differentialsystem it is a difficult problem to determine the existence or nonexistence of first integrals.

Making the change of variables

X ¼ x; Y ¼ yþ 1b

; Z ¼ z;

ts reserved.

http://dx.doi.org/10.1016/j.amc.2011.08.069

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http://dx.doi.org/10.1016/j.amc.2011.08.069

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3298 C. Valls / Applied Mathematics and Computation 218 (2011) 3297–3302

system (1) becomes

_X ¼ 1b� a

� �X þ Z þ 1

bXY; _Y ¼ �bY � bX2

; _Z ¼ �X � cZ: ð2Þ

A first integral G of system (2) is said to be of Darboux type if it has the form

G ¼ f k11 � � � f

kpp El1

1 � � � Elqq ; ð3Þ

where f1; . . . ; fp are Darboux polynomials (see Section 2 for their definition), E1; . . . ; Eq are exponential factors (see also Sec-tion 2 for their definition), and kj;lk 2 C for j ¼ 1; . . . ; p and k ¼ 1; . . . ; q.

The following is the main result of this paper.

Theorem 1. System (2) has the following properties:

(a) The unique exponential factor is eZ with cofactor �X � cZ.(b) It admits no Darboux first integral.

To prove Theorem 1 we shall use the information about invariant algebraic surfaces of this system. This is the basis of theso called Darboux theory of integrability (see Section 2 for more details).

The paper is organized as follows. In Section 2 we introduce some basic definitions and results related to the Darbouxtheory of integrability that are needed to prove our main result. In Section 3 we prove Theorem 1.

2. Preliminary results

In recent years the interest in the study of integrability of differential equations has attracted much attention. In par-ticular, the Darboux theory of integrability plays a central role in the integrability of polynomial differential models, by giv-ing sufficiently conditions for the integrability in the family of Darboux functions. More precisely, the significance of themethod is that one can compute Darboux first integrals by knowing a sufficient number of algebraic invariant surfaces(the so-called Darboux polynomials) and of the so-called exponential factors. We emphasize that the method applies bothto real and complex polynomial ordinary differential equations. The study of complex invariant algebraic curves is neces-sary for obtaining all real first integrals of a real polynomial differential equation. For more details we refer the reader to[10].

We associate to system (1) the vector field

x ¼ 1b� a

� �X þ Z þ 1

bXY

� �@

@X� bðY þ X2Þ @

@Y� ðX þ cZÞ @

@Z: ð4Þ

Now let U � R3 be an open subset. We say that a nonconstant function H : U ! R is a first integral of the vector field (4), orof system (2), if HðXðtÞ;YðtÞ; ZðtÞÞ ¼ constant for all values of t for which some solution ðXðtÞ;YðtÞ; ZðtÞÞ of Eq. (2) belongs to U.Clearly, H is a first integral of x on U if and only if xH ¼ 0 on U. When H is a polynomial, we say that H is a polynomial firstintegral.

Let h ¼ hðX;Y ; ZÞ 2 C½X;Y; Z� be a nonconstant polynomial. We say that h ¼ 0 is an invariant algebraic surface of the vectorfield x in (4) if xh ¼ Kh for some polynomial K ¼ KðX; Y; ZÞ 2 C½X;Y ; Z�. Then h is called a Darboux polynomial and K is calledthe cofactor of h. We note that K has degree at most 1. In particular, a Darboux polynomial with a zero cofactor is a polyno-mial first integral.

Now let g;h 2 C½X;Y; Z� be coprime polynomials. We say that the nonconstant function E ¼ eh=g is an exponential factor ofthe vector field x in (4) if xE ¼ LE, for some polynomial L ¼ LðX;Y; ZÞ 2 C½X;Y ; Z� of degree at most 1, called the cofactor of E.We note that this is equivalent to

1b� a

� �X þ Z þ 1

bXY

� �@ðh=gÞ@X

� bðY þ X2Þ @ðh=gÞ@Y

� ðX þ cZÞ @ðh=gÞ@Z

¼ L:

We refer the reader to [5] for the geometric and algebraic meaning of exponential factors, and for the proof of the followingresult.

Proposition 2. The following properties hold:

(a) If E ¼ eh=g is an exponential factor of the polynomial system (1) and h is not a constant polynomial, then h ¼ 0 is an invariantalgebraic curve;

(b) Eventually eh can be an exponential factor, coming from the multiplicity of the infinite invariant straight line.

In [16] the author proved the following result on the existence of Darboux polynomials.

C. Valls / Applied Mathematics and Computation 218 (2011) 3297–3302 3299

Proposition 3. System (2) has a Darboux polynomial if and only if c ¼ b ¼ ða�ffiffiffiffiffiffiffiffiffiffiffiffiffiffia2 � 4p

Þ=2, in which case the Darbouxpolynomial is f ¼ b2X2 þ b2Z2 þ Y2 and the cofactor is �2c.

3. Proof of Theorem 1

It follows from Propositions 3 and 2 that if E is an exponential factor of system (2), then it must be of the form

E ¼ expðhÞ; h 2 C½X;Y; Z�

or when b ¼ c ¼ a�ffiffiffiffiffiffiffiffia2�4p

2 then it can also be of the form

E ¼ expðh=gÞ; h 2 C½X;Y ; Z�; g ¼ ðb2X2 þ Y2 þ b2Z2Þ‘; ‘ 2 N n f0g; ðg;hÞ ¼ 1:

We separate the study of exponential factors of system (2) into two propositions, in which we consider separately thecases ‘ ¼ 0 and ‘ P 1.

Proposition 4. The unique exponential factor of system (2) of the form E ¼ eh with h 2 C½X;Y ; Z� is eZ, with cofactor �X � cZ.

Proof. Let E ¼ expðhÞ with h 2 C½X;Y ; Z� be an exponential factor of system (2). We write h in the form

h ¼Xn

j¼1

hjðX;Y; ZÞ; ð5Þ

where each hj ¼ hjðX;Y; ZÞ is a homogeneous polynomial of degree j. Without loss of generality one can assume that hn – 0,and that the cofactor of E is of the form

LðX;Y; ZÞ ¼ a0 þ a1X þ a2Y þ a3Z

with ai 2 C, for i ¼ 0;1;2;3. Then h satisfies

1b� a

� �X þ Z þ 1

bXY

� �@h@X� bðY þ X2Þ @h

@Y� ðX þ cZÞ @h

@Z¼ a0 þ a1X þ a2Y þ a3Z: ð6Þ

Evaluating (6) at X ¼ Y ¼ Z ¼ 0 we obtain a0 ¼ 0.Computing the terms of degree nþ 1 in (6) yields

1b

YX@hn

@X� bX2 @hn

@Y¼ 0: ð7Þ

Now we use the method of characteristic curves for solving partial differential equations. The characteristic associated to (7) is

dXdY¼ � Y

b2X;

whose general solution is b2X2 þ Y2 ¼ d, where d is an arbitrary nonnegative constant. We consider the change of variables

u ¼ b2X2 þ Y2 and v ¼ X: ð8Þ

The inverse transformation is

Y ¼ �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiu� b2v2

qand X ¼ v ;

and one can rewrite (7) in the form

vb�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiu� b2v2

q� �@�hn

@v ¼ 0; that is; �hn ¼ �hnðu; ZÞ:

Since �hnðu;v ; ZÞ ¼ hnðX;Y; ZÞ and hn must be a polynomial of degree n, we conclude that

hnðX;Y ; ZÞ ¼X½n=2�

i¼0

an�2iZn�2iðb2X2 þ Y2Þi; an�2i 2 R; i ¼ 0; . . . ; ½n=2�: ð9Þ

Substituting h into Eq. (6) with a0 ¼ 0, we obtain

1b XY @hn

@X � bX2 @hn@Y ¼ 0;

1b XY @hj

@X � bX2 @hj

@Y þ 1b � a� �

X þ Z� � @hjþ1

@X � bY @hjþ1@Y � ðX þ cZÞ @hjþ1

@Z ¼ 0;1b � a� �

X þ Z� �

@h1@X � bY @h1

@Y � ðX þ cZÞ @h1@Z ¼ a1X þ a2Y þ a3Z;

ð10Þ


where the second equation holds for j ¼ ½n=2�; . . . ;1.Introducing hn into Eq. (10), after some calculations, we have

1b

XY@hn�1

@X� bX2 @hn�1

@Y¼ X

X½n=2��1

i¼0

½ðn�2iÞan�2i �2ðiþ 1Þb2an�2i�2�Zn�2i�1ðb2X2 þ Y2Þi þ kXa1ðb2X2 þ Y2Þ½n=2�

þ Y2X½n=2�

i¼1

b� aþ 1b

� �2ian�2iZ

n�2iðb2X2 þ Y2Þi�1 þX½n=2�

i¼0

2i a�1b� c

� �þ nc

� �an�2iZ

n�2iðb2X2 þ Y2Þi;

where

k ¼0; if n iseven;1; if n is odd:

Using (8) and setting �hnðu;v ; ZÞ ¼ hnðX;Y; ZÞ, by the last equation we obtain the ordinary differential equation

@�hn�1

@v ¼ bX½n=2��1

i¼0

ðn� 2iÞan�2i � 2ðiþ 1Þb2an�2i�2

h iZn�2i�1ui 1

ð�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiu� b2v2

pÞþ bka1u½n=2� 1


pÞ

þ bX½n=2�

i¼1

b� aþ 1b

� �an�2iZ

n�2iui�1 ð�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiu� b2v2

pÞ

v þ bX½n=2�

i¼0

2i a� 1b� c

� �þ nc

� �a2ðm�iÞZ

2ðm�iÞui 1

vð�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiu� b2v2

pÞ:

Solving this equation, we obtain

�hn�1 ¼ bX½n=2��1

i¼0

ðn� 2iÞan�2i � 2ðiþ 1Þb2an�2i�2

h iZn�2i�1ui

Zdv


pÞþ bka1u½n=2�

Zdv


pÞ

þ bX½n=2�

i¼1

b� aþ 1b

� �2ian�2iZ

n�2iui�1Z ð� ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

u� b2v2p

v dv þ bX½n=2�

i¼0

2i a� 1b� c

� �þ nc

� �an�2iZ

n�2iui

�Z

dv


pÞþ �h�n�1ðu; ZÞ; ð11Þ

where �h�n�1 is an arbitrary function of u and Z. Since
Zdvffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
u� b2v2p ¼ 1

barctan

bvffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiu� b2v2

p !

;

Z ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiu� b2v2

pv dv ¼


qþ

ffiffiffiup

loguv

2ðffiffiffiupþ


pÞ

ð12Þ

Zdv

vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiu� b2v2

p ¼ 1ffiffiffiup log

v

2ðffiffiffiupþ


pÞ;

in order that hn�1 is a homogeneous polynomial of degree n� 1, we must have

ðn� 2iÞan�2i � 2ðiþ 1Þb2an�2i�2 ¼ 0, for i ¼ 0; . . . ; ½n=2� � 1, ka1 ¼ 0, 2i b� aþ 1

b

� �an�2i ¼ 0 for i ¼ 1; . . . ; ½n=2�,

2i a� 1b � c

� �þ nc

� �an�2i ¼ 0 for i ¼ 0; . . . ; ½n=2�.

From the third and fourth relations, we get

a P 2; b ¼ c ¼ a�ffiffiffiffiffiffiffiffiffiffiffiffiffiffia2 � 4p

2; nc ¼ 0 or an�2i ¼ 0; i ¼ 0; . . . ; ½n=2�:

Since n P 3 and an�2i – 0 for i ¼ 0;1; . . . ; ½n=2� (otherwise hn ¼ 0), we must have c ¼ 0 and thus also 0 ¼ b ¼ a�ffiffiffiffiffiffiffiffia2�4p

2 , which isimpossible. Therefore, we must have n 6 2, that is, h is a polynomial of degree 2. Introducing h in (6) and using the fact that ithas degree two, we finally obtain that h ¼ Z and L ¼ �X � cZ. This concludes the proof of the proposition. h

Proposition 5. System (2) with b ¼ c ¼ a�ffiffiffiffiffiffiffiffia2�4p

2 has no exponential factors of the form E ¼ eh=ðb2X2þY2þb2Z2Þ‘ with ‘ 2 N and

h 2 C½X;Y; Z� such that h is coprime with b2X2 þ Y2 þ b2Z2.

Proof. We consider system (2) with b ¼ c ¼ a�ffiffiffiffiffiffiffiffia2�4p

2 , that is,

C. Valls / Applied Mathematics and Computation 218 (2011) 3297–3302 3301

_X ¼ bX þ Z þ 1b

XY; _Y ¼ �bY � bX2; _Z ¼ �X � bZ: ð13Þ

We assume that E ¼ expðh=ðb2X2 þ Y2 þ b2Z2Þ‘Þ, with ‘ 2 N and h 2 C½X;Y; Z� such that ðh; gÞ ¼ 1, is a exponential factor ofsystem (13), and we shall reach a contradiction. Without loss of generality we assume that the cofactor L is of the formL ¼ a0 þ a1X þ a2Y þ a3Z, with ai 2 C for i ¼ 0; . . . ;3.

Clearly, h satisfies

bX þ Z þ 1b

XY� �

@h@X� bðY þ X2Þ @h

@Y� ðX þ bZÞ @h

@Z¼ �2c‘hþ ða0 þ a1X þ a2Y þ a3ZÞðb2X2 þ Y2 þ b2Z2Þ‘: ð14Þ

We write h as the sum of its homogeneous parts as in (5) and again we assume that hn – 0 and n – 0. We note that if ai ¼ 0for i ¼ 0;1;2;3, then we reach a contradiction. Indeed if ai ¼ 0 for i ¼ 0; . . . ;3, then it follows from (14) that his either 0 or aDarboux polynomial of system (13) with cofactor �2c‘. The first case is impossible, and it follows from Proposition 3 that inthe second case we have h ¼ ðb2X2 þ Y2 þ b2Z2Þ‘, thus contradicting the fact that ðh; gÞ ¼ 1.

We consider five different cases.

Case 1: 2‘ > nþ 1. In this case computing the terms of degree ‘þ 1 in (14), we obtain

ða1X þ a2Y þ a3ZÞðb2X2 þ Y2 þ b2Z2Þ‘ ¼ 0; that is ai ¼ 0; i ¼ 1;2;3; ð15Þ

and computing the terms of degree ‘ in (14) yields

a0ðb2X2 þ Y2 þ b2Z2Þ‘ ¼ 0; that is; a0 ¼ 0:

Therefore, this case is impossible.Case 2: 2‘ ¼ nþ 1. In this case, computing the terms of degree ‘þ 1 in (14), again we obtain (15), and thus ai ¼ 0 for

i ¼ 1;2;3. Computing the terms of degree ‘ in (14) yields

1b

XY@hn

@X� bX2 @hn

@Y¼ a0ðb2X2 þ Y2 þ b2Z2Þ‘:

Evaluating this last equation at X ¼ 0, we obtain a0 ¼ 0. Thus, ai ¼ 0 for i ¼ 0; . . . ;3 and this case is impossible.Case 3: 2‘ ¼ n. Computing the terms of degree nþ 1 in (14) yields

1b

XY@hn

@X� bX2 @hn

@Y¼ ða1X þ a2Y þ a3ZÞðb2X2 þ Y2 þ b2Z2Þ‘:

Using (8), we rewrite it in the form

vbð�


qÞ @

�hn

@v ¼ ða1v þ a2ð�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiu� b2v2

qÞ þ a3ZÞðuþ b2Z2Þ‘:

Solving this differential equation we obtain

�hn ¼ ðuþ b2Z2Þ‘ a1

barctan

bvffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiu� b2v2

p !

þ a2 log v þ a3bZffiffiffiup log

v

uþffiffiffiup ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

u� b2v2p

!þ �Cnðu; ZÞ;

where �Cn is a function in the variables u and Z. Since hnðX;Y ; ZÞ ¼ �hnðu;v; ZÞmust be a polynomial of degree n, we must haveai ¼ 0 for i ¼ 1;2;3.Proceeding as in the proof of Proposition 4, we find that hn ¼ h2‘ has the form in (9) with n ¼ 2‘. Proceeding then as in theproof of Proposition 4 with b ¼ c ¼ a�

ffiffiffiffiffiffiffiffia2�4p

2 and n ¼ 2‘, we find that �hn�1 ¼ �h2‘�1 is equal to (see (11))

X‘�1

i¼0

2ð‘� iÞa2ð‘�iÞ � 2ðiþ 1Þb2a2ð‘�i�1Þ

h iZ2‘�2i�1ui

Zdv


pÞþ a0ðuþ b2Z2Þ‘

Zdv

uð�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiu� b2v2

pÞþ �h�n�1ðu; ZÞ:

Using (12) and the fact that hn�1ðX;Y ; ZÞ ¼ �hn�1ðu;v ; ZÞ is a polynomial of degree n� 1, we find that a0 ¼ 0. Hence ai ¼ 0 fori ¼ 0; . . . ;3, and this case is impossible.

Case 4: 2‘ ¼ n� 1. Computing the terms of degree nþ 1 in (14) yields (7). Proceeding as in the proof of Proposition 4 withn ¼ 2‘þ 1, we find that hn ¼ h2‘þ1 has the form as in (9). Proceeding again as in the proof of Proposition 4 withn ¼ 2‘þ 1; b ¼ c ¼ a�


2 , we find that �hn�1 ¼ �h2‘ is equal to (see (11))

X‘�1

i¼0

ð2‘þ 1�2iÞa2‘þ1�2i �2ðiþ 1Þb2a2‘�2i�1

h iZ2‘�2iui

Zdv


pÞþ ba1u‘

Zdv


pÞþ bc

X‘i¼0

a2‘þ1�2iZ2‘þ1�2iui

�Z

dv


pÞþ ðu2 þ b2Z2Þ‘ a1

Zdv


pÞþ a2

Zdvv þa3Z

Zdv


pÞ

!þ �h�2‘ðu;ZÞ:


Using (12) and the fact that hn�1ðX; Y; ZÞ ¼ �hn�1ðu;v ; ZÞ must be a polynomial of degree n� 1, we must have

a1 ¼ 0; ð2‘þ 1� iÞa2‘þ1�2i ¼ 2ðiþ 1Þb2a2‘�2i�1:

This yields that a2‘þ1�2i ¼ 0 for i ¼ 0; . . . ; ‘, and thus hn ¼ 0, which is a contradiction. Hence, this case is impossible.Case 5: 2‘ < n� 1. In this case, computing the terms of degree nþ 1 in (14) we obtain (7). Proceeding as in the proof of

Proposition 4, we find that hn has the form as in (9). Proceeding again as in the proof of Proposition 4 withb ¼ c ¼ a�


2 yields (see (11))

�hn�1 ¼X½n=2��1

i¼0

½ðn� 2iÞan�2i � 2ðiþ 1Þb2an�2i�2�Zn�2i�1uiZ

dv


pÞ

þ bka1u½n=2�Z

dv


pÞþ bc

X½n=2�

i¼0

ðn� 2‘Þan�2iZn�2iui

Zdv


pÞþ �h�n�1ðu; ZÞ;

where �h�n�1 is an arbitrary function of u and Z. Using (12) and the fact that n – 2‘ and that hn�1 must be a polynomial of de-gree n� 1, we find that bcðn� 2‘Þan�2i ¼ 0 for i ¼ 0; . . . ; ½n=2�. Since bcðn� 2‘Þ – 0, we find that an�2i ¼ 0 for i ¼ 0; . . . ; ½n=2�,and thus hn ¼ 0, which is a contradiction. This completes the proof of the proposition. h

Proof (Proof of Theorem 1). Statement (a) in the theorem follows readily from Propositions 4 and 5.Now we prove statement (b) by contradiction. Assume that G is a first integral of Darboux type. In view of its definition in

(3) and taking into account Propositions 3–5, it must be of the form

G ¼ ðb2X2 þ b2Z2 þ Y2ÞkelZ ; if b ¼ c ¼ a�ffiffiffiffiffiffiffiffia2�4p

2 ;

elZ ; otherwise;

(

where k;l 2 C. Since G is a first integral, it satisfies —XG ¼ 0, that is,

0 ¼ —XG ¼ �ð2kc þ ðX þ cZÞlÞG; if b ¼ c ¼ a�ffiffiffiffiffiffiffiffia2�4p

2 ;

�ðX þ cZÞlG; otherwise:

(

In both cases we find that l ¼ 0, and in the first case, since c – 0, we also get that k ¼ 0. Therefore G is a constant, contradict-ing the fact that G is a first integral. This concludes the proof of the theorem. h

Acknowledgements

Partially supported by FCT through CAMGDS, Lisbon.

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Darboux integrability of a nonlinear financial system