The discrete Markus–Yamabe problem

Nonlinear Analysis 35 (1999) 343–354

The discrete Markus–Yamabe problem1

Anna Cima a, Armengol Gasull b, Francesc Manosas b;∗

aDepartament de Matem�atica Aplicada II, E. T. S. d’Enginyers Industrials de Terrassa,Universitat Polit�ecnica de Catalunya, Colom, 11 08222 Terrassa, Barcelona, SpainbDepartament de Matem�atiques, Edi�ci C, Universitat Aut�onoma de Barcelona,

08193 Bellaterra, Barcelona, Spain

Received 21 January 1997; accepted 24 June 1997

Keywords: Asymptotic stability; Markus–Yamabe conjecture; Jacobian conjecture

1. Statement of the problem

Let F :Rn→Rn be a C1 map and consider the di�erential system

x=F(x): (1)

Assume that p is a critical point of Eq. (1), i.e., F(p)= 0. We say that p is a globalattractor of the continuous dynamical system induced by Eq. (1) if �(t; x) tends to pas t tends to in�nity for each x∈Rn, where �(t; x) is the solution of Eq. (1) with�(0; x)= x.The next conjecture was explicitly stated by Markus and Yamabe (see [15]) in 1960.

MYC (n) (Markus–Yamabe Conjecture). Let F be a C1 map from Rn to itself suchthat for any x∈Rn; the Jacobian of F at x has all its eigenvalues with negative realpart. If F(p)= 0; then p is a global attractor of x=F(x).

This conjecture was proved for planar polynomial maps in 1988 [16] and for planarC1 maps in 1993 [9, 11] and in 1994 see [10]. In [1, 3] there are examples of smoothvector �elds of Rn; n≥ 4 satisfying the hypothesis of the Conjecture which have aperiodic orbit and in [4] there is an example of a polynomial vector �elds of Rn; n≥ 3satisfying the same hypothesis which has some orbits that scape at in�nity. Therefore

∗ Corresponding author.1 Partially supported by the DGICYT grant number PB96-1153.

0362-546X/98/$19.00 ? 1998 Elsevier Science Ltd. All rights reserved.PII: S0362 -546X(97)00715 -3

344 A. Cima et al. / Nonlinear Analysis 35 (1999) 343–354

the conjecture is only true for n=2. On the other hand in [6] van den Essen showsthat a conjecture weaker than MYC implies the classical Jacobian Conjecture.Our aim is to discuss a natural translation of MYC to the dynamics of the iterations

of maps. Let F :Rn→Rn be a C1 map and consider the sequence:

x(m+1) =F(x(m)); x(0) ∈Rn: (2)

Let p be a �xed point of F , i.e., F(p)=p. We say that p is a global attractor of thediscrete dynamical system (2) if the sequence x(m) tends to p as m tends to in�nityfor any x(0) ∈Rn. The question is the following:DMYQ(n) (Discrete Markus–Yamabe Question). Let F be a C1 map from Rn intoitself such that F(0)= 0 and for any x∈Rn, JF(x) has all its eigenvalues with modulusless than one. Is it true that 0 is a global attractor for the discrete dynamical systemgenerated by F?

The answer of DMYQ(1) is trivially a�rmative. Szlenk worked on this problemfor n=2 and he explained us his result: an example of a rational map F :R2→R2

which gives a negative answer to DMYQ(2). Unfortunately Szlenk suddenly died inJuly 1995. We reproduce his example in Section 5 (see Theorem D) and we prove itsproperties in the Appendix.Szlenk’s example is neither a polynomial nor a di�eomorphism. However, it can

be modi�ed slightly in order to obtain a di�eomorphism which also gives a negativeanswer to DMYQ(2) (see Theorem E in Section 5).If we restrict our attention to the family of polynomial maps, we prove that the

answer to the DMYQ(2) is a�rmative (see Theorem B in Section 2). A negativeanswer to DMYQ(n) for n ≥ 4 is given in [7]. Later in [4] there are examples whichgive a negative answer for n ≥ 3. So the problem is completely solved.The paper is organized as follows. In the next section we prove that the answer to the

DMYQ is yes for triangular maps de�ned on Rn (see Theorem A in Section 2) and alsofor polynomial maps de�ned on R2. In Sections 3 and 4 we relate the problem with theJacobian Conjecture and with the Markus–Yamabe Conjecture, respectively. Section 5is devoted to obtaining a di�eomorphism of R2 which gives a negative answer toDMYQ(2). Lastly in the Appendix we present the Slenzk’s example mentioned above.After this paper was written G. Meisters told us that a question similar to DMYQ wasalready stated in La Salle’s book [17] p. 20.

2. The triangular case

We say that F is a triangular map if it takes the form:

F(x)= (F1(x1); F2(x1; x2); : : : ; Fn(x1; x2; : : : ; xn)):

We say that F is linearly triangularizable if there exists a linear change such that itmakes F triangular.

A. Cima et al. / Nonlinear Analysis 35 (1999) 343–354 345

The �rst result which we obtain is the following:

Theorem A. Let F :Rn→Rn be a C1 triangular map with all the eigenvalues ofJF(x) with modulus less than one at each x∈Rn. Assume that F(0)= 0. Then 0 is aglobal attractor for the discrete dynamical system generated by F .

Proof. We prove the result by induction. The hypothesis of induction is the following:Consider F :Rm→Rm a triangular C1 map such that∣∣∣∣@F1@x1

∣∣∣∣¡1;∣∣∣∣@F2@x2

∣∣∣∣¡1; : : : ;∣∣∣∣@Fm@xm

∣∣∣∣¡1 and F(0)= 0:

Fix x(0) ∈Rm and set x(k+1) =F(x(k)). Then there exist k0 ∈N big enough, M ∈R+and K ∈R; 0¡K¡1, such that

|x(k+k0)i | ≤ MKk for all i=1; 2; : : : ; m:

Obviously if we prove the above induction hypothesis the theorem follows.The proof for the case m=1 is trivial. Assume that it is true for m= s− 1 and we

prove it for m= s.Fix x(0) ∈R5. By considering F(x1; x2; : : : ; xs−1)≡ (F1(x1); F2(x1; x2); : : : ; Fs−1(x1;

x2; : : : ; xs−1)) and applying the induction hypothesis we have that there exist k0, Mand K such that

|x(k+k0)i | ≤ MKk for i=1; 2; : : : ; s− 1 and all K ∈ NIt is not restrictive to change x(0) by x(k0) and assume that

|x(k)i | ≤ MKk for i=1; 2; : : : ; s− 1: (3)

Consider

x(k)s = Fs(x(k−1))=Fs(x(k−1)1 ; x(k−1)2 ; : : : ; x(k−1)s )

=∫ x(k−1)

s

0

@Fs@xs(x(k−1)1 ; x(k−2)2 ; : : : ; x(k−1)s−1 ; t) dt

+∫ x(k−1)

s−1

0

@Fs@xs−1

(x(k−1)1 ; x(k−2)2 ; : : : ; x(k−1)s−2 ; t; 0) dt + · · ·

+∫ x(k−1)

1

0

@Fs@x1

(t; 0; : : : ; 0) dt:

Then

|x(k)s | ≤ |x(k−1)s |+ |x(k−1)s−1 |+ · · ·+ |x(k−1)1 |; (4)

where we have used that for any i=1; 2; : : : ; s− 1

max{∣∣∣∣@Fs@xi (y1; : : : ; yi; 0; : : : ; 0)

∣∣∣∣ : (y1; y2; : : : ; yi−1)∈Ri−1 and −x(0)i ≤yi≤ x(0)i}

¡1:


Now, from Eqs. (3) and (4) we have:

|x(k)s | ≤ |x(k−1)s |+ (s− 1)MKk−1: (5)

Denoting (k − 1)M by C we can write

|x(k)s | ≤ |x(0)s |+ C(Ks−1 + Ks−2 + · · ·+ 1) ≤ |x(0)s |+ C1− K :

Hence, {x(k)s } is a bounded sequence. Since the same is true for {x(k)i } for i=1;2; : : : ; s−1 there exists L¿0 such that |x(k)i | ≤ L for all i=1; 2; : : : ; s and for all k ∈N.So, we can de�ne

D= max{∣∣∣∣@Fs@xs (y1; y2 : : : ; ys)

∣∣∣∣: |yi| ≤ L; i=1; 2; : : : ; s}

and assert that D is strictly less than one. It implies that inequality (4) can now besubstituted by

|x(k)s | ≤ D|x(k−1)s |+ |x(k−1)s−1 |+ · · ·+ |x(k−1)1 |and inequality (5) by

|x(k)s | ≤ D|x(k−1)s |+ Kk−1C:Hence,

|x(k)s | ≤D(D|x(k−2)s |+ Kk−2C) + Kk−1C...

≤Dk |x(0)s |+ C(Kk−1 + DKk−2 + · · ·+ Dk−1)≤Dk |x(0)s |+ kC(max(K;D))k−1

≤ (max(K;D))k−1(D|x(0)s |+ kC)= (max(K;D))kM = M Kk

for some M , where max(K;D)¡K¡1, and the induction is �nished.

The next goal is to show that the answer to DMYQ(2) for polynomial maps is a�r-mative. First of all notice that in the polynomial case the assumption of the existenceof a �xed point can be removed in the hypothesis of DMYQ(n) (this is due to thefact that injective polynomial maps of Rn are also exhaustive).Notice that for n=2, the results of [9,10] or [11] imply that any map (not necessarily

polynomial) satisfying the assumptions of DMYQ(2) has only one �xed point. To seethis if F is such a map it su�ces to consider the map F − I and conclude that it isinjective.

Lemma 1.1. Let F :Rn→Rn (resp. F :Cn→Cn) be a polynomial map such thatJF(x) has all its eigenvalues with modulus less than one at each x∈Rn (resp. x∈Cn).Then the characteristic polynomial of JF(x) is independent on x.


Proof. Let Px(�) be the characteristic polynomial of JF(x) and let �1; �2; : : : ; �n be theroots of Px(�). Then

Px(�)= �n − t1�n−1 + · · ·+ (−1)ntnwhere

tj =n∑

i1¡i2¡···¡iji1 ; i1 ; :::; ij=1

�i1�i2 · · · �ij :

Since |�i|¡1 for all i=1; 2; : : : ; n we have that |tj|¡kj for all j=1; 2; : : : ; n. On theother hand since the components of F are polynomials and each tj can be described asthe sum of all the minors of order j which have its diagonal on the principal diagonalof JF(x), we conclude that tj is a polynomial in x for all j=1; 2; : : : ; n. Since the onlybounded polynomials are the constants, the result follows.

Theorem B. Let F :R2→R2 be a polynomial map with all the eigenvalues of JF(x)with modulus less than one at each x∈R2. Then there exists a unique �xed point ofF which is a global attractor for the discrete dynamical system generated by F .

Proof. Set F =(P;Q). From Lemma 1.1 we know that

t1 =@P@x+@Q@y

and t2 =@P@x@Q@y

− @P@y@Q@x

are constants. Consider G=F − (t1=2)I and set G=( �P; �Q). Then,

�t1 =@ �P@x+@ �Q@y=0 and �t2 =

@ �P@x@ �Q@y

− @ �P@y@ �Q@x= t2 − t21

4:

Since �t1 = 0, there exists a polynomial H (x; y) such that

�P(x; y)=−@H@y

and �Q(x; y)=@H@x:

Then the hessian of H is �t2. If �t2 6=0, by applying the result of [5] we know that upa complex a�ne transformation,

H (u; v)=√

− �t2uv+ h(u)where h is a polynomial in one variable u.Assume �t2¡0. Then, H (u; v)∈R and from the proof of [5] we see that the a�ne

transformation can be taken with real coe�cients. Through this transformation we havethat F(x; y)= (P(x; y); Q(x; y)) can be written as (P(u; v); Q(u; v)) where

P(u; v)= k1u+ k2 and Q(u; v)= k3v+ p(u) + k4

with p a polynomial in one variable. With these coordinates (P; Q) has an unique�xed point ( �u; �v). By doing the translation which sends ( �u; �v) to (0; 0) and applyingTheorem 1, the result follows.


Now assume that �t2¿0. From the classical result of J�orgens (see [14]), we knowthat H (x; y) is quadratic. It implies that F =(P;Q) is a polynomial map of degree one.For this type of maps the result can be easily proved.If �t2 = 0 from the proof of Dillen it follows that, up to an a�ne transformation,

H (u; v)= kv+ h(u); k ∈R, and the result follows as above.

3. The Jacobian Conjecture and the �xed point conjecture

In this section we restrict our attention to polynomial maps. Let F :Rn→Rn be apolynomial map such that JF(x) has all its eigenvalues with modulus less than one. InSection 1 we ask for the existence of a �xed point which is a global attractor. Nowwe formulate a weaker problem as follows:

FPC (Fixed Point Conjecture). Let F :Rn→Rn be a polynomial map such that JF(x)has all its eigenvalues with modulus less than one at each x∈Rn. Then F has a unique�xed point.

Considering the real and the imaginary part of the components of a complex mapF :Cn→Cn and using standard arguments of linear algebra it is easy to see that thisconjecture can be formulated in the following equivalent form:

FPC (Fixed Point Conjecture). Let F :Cn→Cn be a polynomial map such that JF(x)has all its eigenvalues with modulus less than one at each x∈Cn. Then F has aunique �xed point.

Theorem C shows that this problem is equivalent to the celebrated JacobianConjecture, which can be established as follows.

JC (Jacobian Conjecture). Let F :Cn→Cn be a polynomial map with det JF(x)∈C∗

=C\{0} at each x∈Cn. Then F is invertible.

Theorem C.

JC is equivalent to FPC:

Proof. Assume that the JC holds and let F satisfy the hypothesis of FPC for some n.Consider G(x)=F(x)−x. Then the eigenvalues of JG(x) are the eigenvalues of JF(x)minus one. Hence by using Lemma 1.1 we have that det JG(x) is constant; from thehypothesis on F we know that this constant cannot be zero. So, G is invertible andthere exists a unique x0 zero of G, which is the unique �xed point of F(x).Now assume that JC fails for some m. From the Reduction Theorem (see [2]) it

means that there exists n∈N and G :Cn→Cn noninvertible such that

G(x)= x + H (x)

with JH (x) a nilpotent matrix at each x∈Cn. Now set g(x)= 12G(x) and let y; z ∈Cn;

y 6= z with g(y)= g(z)=p. Denoting by h(x) the expression x + p − g(x), we have


that h(y)=y and h(z)= z. On the other hand, since JH (x) is a nilpotent matrix ateach x∈Cn, all its eigenvalues are zero. From the de�nition of h(x) we obtain

Jh(x)= I − Jg(x)= I − ( 12 I + 12JH (x))=

12 I − 1

2JH (x)

which implies that all the eigenvalues of Jh(x) are 12 at each point x∈Cn. Hence, h(x)

satis�es the hypothesis of FPC and it has two di�erent �xed points.

Remember that the results of [9, 10] or [11] imply that FPC for R2 is true. Fromthis fact, and the proof of Theorem C it can be deduced that the JC is true for somespecial subcases and it cannot be deduced that it is true for n=2 as can be thoughtat a �rst look.To end this section we give a conjecture which implies the JC. Firstly we explain

the way we arrive at the conjecture. Notice that the DMYQ has a�rmative answerfor triangularizable maps. So we thought that the next step could be to consider it formaps that take the form F(x)= (f(x); g(xn)) with g :R→R and f :Rn→Rn−1. Wehad some problems to solve DMYQ for this case and we decided to consider the easiestcase in which F is polynomial and all the eigenvalues of JF(x) are zero. That is, wetried to solve DMYQ for polynomial maps of the form F(x)= (f(x); 0) where JF(x)is nilpotent. At this point we wonder if all polynomial maps with nilpotent Jacobiancan be reduced to that case:NC (Nilpotent Conjecture). Let F :Cn→Cn be a polynomial map such that JF isnilpotent. Then there exists a linear change of coordinates A such that AFA−1(x)=(f(x); 0) where f is a polynomial map of Cn into Cn−1.We have not been able to solve the above conjecture. van den Essen has indepen-

dently arrived at the next result.

Proposition 3.1.NC implies JC.

Proof. It is known that it su�ces to prove JC for maps F such that JF = I + N withN nilpotent and homogeneous of degree 3. Therefore, assuming that NC holds, wewill prove by induction over n next statement: Any map F from Cn into Cn such thatJF = I + N with N nilpotent is injective.For n=1 the above statement is trivial. Suppose that it is true in dimension n and let

F :Cn+1→Cn+1 be such that JF = I + N with N nilpotent. Assuming that NC holds:the map F , by means a linear change of coordinates, can be written as

F(x1; : : : ; xn+1)= (x1 + P1(x1; : : : ; xn+1); x2 + P2(x1; : : : ; xn+1); : : : ; xn+1):

To show that this map is injective it su�ces to show that for any a∈C the mapFa(x1; : : : ; xn) = (x1 + P1(x1; : : : ; xn; a); x2

+P2(x1; : : : ; xn; a); : : : ; xn + Pn(x1; : : : ; xn; a))

is injective. But clearly Fa satis�es the induction hypothesis and we are done.


After the �rst version of this paper was �nished van den Essen and Hubbers havegiven a counterexample of NC, for all n≥ 2 and F of degree greater or equal than 4;see [8]. Observe that following the proof of Proposition 3.1 we can prove next result,which shows that NC is yet an open approach to JC.

Proposition 3.2. NC for polynomial maps of degree less or equal than 3 implies JC.

4. Relation with the Markus–Yamabe conjecture

In this section we relate the MYC(n) with DMYQ(n). Consider the di�erential sys-tem

x=F(x) (6)

and let �(t; x) be the solution of Eq. (6) with �(0; x)= x.Assume that �(T; x) is de�ned for some T¿0 and for some x∈Rn. Then we can

consider the discrete dynamical system given by the ow at time T :

�(T; ·) :V →Rn

where V is a neighbourhood of x.

Lemma 4.1. Let F be a C1 map from Rn to itself and let �(t; x) be the solution ofx=F(x) with �(0; x)= x. Then the following hold:(i) If for all x∈Rn JF(x) has all its eigenvalues with negative real part; then given

U a bounded open subset of Rn there is T¿0 such that for all t ∈ (0; T ) theJacobian of �(t; ) :U→Rn has all its eigenvalues with modulus less than one atany point of U .

(ii) Assume that (d=dx)�(t; x) has all its eigenvalues with modulus less than one fort ∈ (0; tx]. Then JF(x) has all its eigenvalues with nonpositive real parts.

Proof. Since �(t; x) is the solution of x=F(x) with �(0; x)= x we have that

ddt�(t; x)=F(�(t; x)):

Taking the derivatives with respect to x and evaluating in t=0 we can write

ddt

(ddx(�(t; x)

)∣∣∣∣t=0

= JF(x):

This last equality can be rewritten as

limt→0

ddx (�(t; x))− I

t= JF(x); (7)

where I means the scalar identity matrix. From the above relation it is clear that foreach y∈Cl(U ) (Cl(U ) means the topological closure of U ) there exists a neighbour-hood of y, Uy and a positive real number Ty such that for each z ∈Uy the matrix


(d=dx)�(t; x)(z) has all its eigenvalues with real part less than one (resp. greater thanone), for each t ∈ (0; Ty) (resp. t ∈ (−Ty; 0)). Since � is continuous and �(0; x)= x,there exist Wy neighbourhood of y and a positive real number y with y¡Ty such that�(t; z)∈Uy for all z ∈Wy and for all t ∈ (− y; y).Let z ∈Wy and t ∈(0; y). Since �(−t; �(t; x))= x, by the chain ruleddx�(−t; x)(�(t; x)) d

dx�(t; x)(x)= I:

Let �= a + bi be an eigenvalue of (d=dx)�(t; x)(z). Then �−1 = (a− bi)=(a2 + b2) isan eigenvalue of (d=dx)�(−t; x)(�(t; z)). Since z; �(t; z)∈Uy and t ∈ (0; Ty) we obtainRe �¡1 and Re �−1¿1. This clearly implies that |�|= a2 + b2¡1.We have seen that for each y∈Cl(U ) there exists Wy neighbourhood of y and a

positive number y such that (d=dx)�(t; x)(z) has its eigenvalues with modulus lessthan one for each z ∈Wy and for all t ∈ (0; y). Let Wy1 ; : : : ; Wyk be a �nite cover ofCl(U ) and T =min{ y1 ; : : : ; yk}. Clearly T¿0 and (i) follows.To prove (ii) it is enough to consider equality (7).

The set of the values of t for which part (1) of the above lemma holds cannot beextended to R+, because of the known following lemma and the counterexamples withperiodic orbits; see [3].

Lemma 4.2 (Hartman [13]). Assume that system (1) has a periodic orbit of period T .Then the ow at time T has an eigenvalue equal to one.

If the answer to the DMYQ for ows (which are special cases of di�eomorphisms)was a�rmative, then we could conclude that system (6) has a global attractor underthe hypotheses that for any x∈Rn; �(t; x) is de�ned for some t¿0 and �(t; ·) haseigenvalues with modulus less than one. Observe that these last hypotheses are strongerthan the Markus–Yamabe ones.

5. The answer to DMYQ(n) for di�eomorphisms

In this section we obtain a rational di�eomorphism of Rn (n≥ 2) which satis�esthe hypothesis of DMYQ and has a periodic orbit. This example is a modi�cation ofSzlenk’s example which is described in the following theorem that will be proved inthe Appendix.

Theorem D (Szlenk). Let F :R2→R2 be de�ned by

F(x; y)=(− ky3

1 + x2 + y2;

k x3

1 + x2 + y2

)where k ∈

(1;2√3

):

The map F satis�es the following properties:(1) Set p=(x; y)∈R2 and let � be an eigenvalue of JF(p). If xy=0 then �=0.

Otherwise � =∈R and |�|¡√3k=2.


(2) F4(1=√k − 1; 0)= (1=√k − 1; 0).

(3) F is injective.

Let Ga :Rn→Rn be de�ned by

Ga(x1; x2; x3; : : : ; xn)=(− kx321 + x21 + x

22− ax1; k x31

1 + x21 + x22− ax2; cx3; : : : ; cxn

)

where |c|¡1 and k ∈ (1; 2=√3).

Theorem E. For a small enough; the map Ga is a global di�eomorphism from Rn

into itself which satis�es the following properties:(1) For all x∈Rn and for all � eigenvalue of (DGa)(x); |�|¡1.(2) Ga(0)= 0 and there exists p∈Rn; p 6=0 which satis�es G4a(p)=p.

Proof. Clearly we can assume that n=2. Note that Ga=F − aI where F is as inTheorem D.Since JF has at each point eigenvalues with modulus less than (

√3=2)k, it follows

that for a small enough JGa has at each point eigenvalues with modulus less thanone. Note that since JF has all eigenvalues di�erent from a, JGa has at any pointeigenvalues di�erent from zero. So Ga is a local di�eomorphism at every point of R2.In order to see that Ga is a global di�eomorphism we will apply Hadamard’s Theorem[12] which asserts that a smooth map from Rn to itself is a global di�eomorphism ifand only if it is a local di�eomorphism at each point of Rn and it is proper (i.e. thepreimage of a compact set is also a compact set). To see that Ga is proper it su�cesto show that

lim|x; y|→∞

|Ga(x; y)|=∞:

Easy computations show that

|Ga(x; y)|2 = k2r6(cos6 �+ sin6 �) + a2r2(1 + r2)2 − kar4 sin 4�(1 + r2)=2

(1 + r2)2

where r and � denote the usual polar coordinates associated to (x; y). Using thatcos6 �+ sin6 �≥ 1=4; sin 4�≤ 1 and 1 + r2¡2r2 for r big enough we get

|Ga(x; y)|2 ≥ k2r6

4(1 + r2)2+ a2r2 − kar4

2(1 + r2)

≥ k2r6

16r4+ a2r2 − kar2

2

= (k=4− a)2r2

and hence limr→∞Ga(x; y)=∞. Therefore Ga is a global di�eomorphism for all a 6=0.


Set p=(1=√k − 1; 0). Easy computations show that

JF4(p)=((3− (2=k))4 0

0 0

):

Therefore the eigenvalues of JF4(p) are of modulus di�erent from one and so theperiodic orbit is hyperbolic. Hence the periodic orbit remains by small perturbationsof the map. Hence for a small enough Ga has also a four periodic orbit. This ends theproof of the theorem.

Acknowledgements

We want to thank Arno van den Essen for interesting discussions. His help has beenvery important in several parts of this paper, specially in the proof of Theorems Band E.

Appendix (W. Szlenk)

This appendix is devoted to proving Theorem D which is due to W. Szlenk. Thistheorem gives an example of a global smooth homeomorphism (not di�eomorphism)which gives a negative answer to DMYQ(2).

Proof of Theorem D. We only prove (1). The proof of (2) and (3) are simple com-probations. By easy computations we get

det JF(x; y)=3k2x2y2(x2 + y2 + 3)(1 + x2 + y2)3

and trJF(x; y)=−2kxy(x2 − y2)(1 + x2 + y2)2

;

where tr(·) denotes the trace operator. Using polar coordinates we obtain

det JF(r; �) =3k2 sin2 � cos2 �r4(r2 + 3)

(1 + r2)3

=3k2 sin2 2�(r6 + 3r4)4(r6 + 3r4 + 3r2 + 1)

¡3k2

4≤ 1:

On the other hand we obtain that �(x; y), the discriminant of the characteristicpolynomial of JF(x; y), satis�es

�(x; y) = tr2 JF(x; y)− 4det JF(x; y)=−4k2x2y2 2x

4 + 8x2y2 + 2y4 + 12x2 + 12y2 + 9(1 + x2 + y2)4

≤ 0:


If x·y=0 then det JF(x; y)= 0 and tr JF(x; y)= 0 and hence �=0. Otherwise �(x; y)¡0 and hence the eigenvalues of JF(x; y) are not real. Therefore, in this case

|�|=√det JF(x; y)¡

√3k2

4=k√32:

Thus (1) follows.

References

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Markus-Yamabe Conjecture, Adv. Math. 131 (1997) 453–457.[5] F. Dillen, Polynomials with constant hessian determinant, J. Pure Appl. Algebra 71 (1991) 13–18.[6] A. van den Essen, Conjectures and problems surrounding the Jacobian conjecture, in: M. Sabatini (Ed.),

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[9] R. Fessler, A proof of the two-dimensional Markus–Yamabe stability conjecture, Ann. Polon. Math. 62(1995) 45–75.

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[11] C. Gutierrez, A solution to the bidimensional Global Asymptotic conjecture, Ann. Inst. Poincar�e Anal.Non Lin�eaire 12 (1995) 627–671.

[12] P. Hadamard, Sur les transformations punctuelles, Bull. Soc. Math. France 34 (1906) 349–363.[13] P. Hartman, Ordinary Di�erential Equations, Wiley, New York, 1964.[14] K. J�orgens, �Uber die L�osungen der Di�erentialgleichung rt − s2 = 1, Math. Ann. 127 (1954) 130–134.[15] L. Markus, H. Yamabe, Global stability criteria for di�erential systems, Osaka Math. J. 12 (2) (1960)

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The discrete Markus–Yamabe problem