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Estimate for the number of limit cycles of the equation dy/dx=Q(x,y)/P(x,y), where Q(x,y) and P(x,y) are polynomials

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Selçuk J. Appl. Math. Selçuk Journal of Vol. 5. No. 2. pp. 3-14, 2004 Applied Mathematics

Estimate for the Number of Limit Cycles of the Equation dy dx =

Q(x; y) P (x; y) whereQ(x; y) and P (x; y) are Polynomials

Ovezmamed Huday-Verenov

The Turkmen Polytechnic Institute, Ashgabat, Turkmenistan, e-mail:hudayver@ hotm ail.com

Received: May 14, 2004

Summary. The article gives the estimate for the number of limit cycles of the equation:

dy dx =

P (x; y) Q(x; y)

where Q(x; y) and P (x; y) are polynomials. In addition, it is shown that limit cycles lie on separatrix solutions, and the element of the separatrix solution is being presented by concrete series in a small neighbourhood of a singular point. Key words: Di¤erential Equations, Number of Limit Cycles

1. Introduction

The purpose of our study is to estimate the number of limit cycles lying on integral curves of the equation

(1) dy

dx =

P (x; y) Q(x; y)

considered in the complex space C2, where x and y are variables, and P and Q

are polynomials of the degree n with complex coe¢ cients. The integral curve of the equation (1) can be considered as a two-dimensional surface in the four-dimensional space with its own homology group. Alongside with (1), we will study its parametric representations:

where N is an n-degree polynomial with complex coe¢ cients. We shall de…ne P; Q and N as polynomials in a general form if coe¢ cients of these polynomials allow the ful…llment of the following conditions:

(2)

(2) dx dt = P (x; y) N (x; y); dy dt = Q(x; y) N (x; y) (3) dx dz = P (x; y); dy dz = Q(x; y)

1. Every decision of (1) is extended to the whole plane x, except a countable number of the singular points.

2. Every decision of (1) is dense everywhere in C2.

3. In the neighbourhood of a singular point of the equation (1), be it …nite or in…nite, there is the Poincaré ’s expansion.

4. Every decision of (2) is extended to the whole plane t, except a countable number of the algebraic singular points.

Without going into details, we may say that the set of polynomials P; Q and N , which do not satisfy the afore-mentioned conditions, comprise a set of a dimension lower than that of the set of all P; Q and N . (please see [1], [2]).

Let us consider the concrete solution y = y(x) of (1) and the corresponding solution x = x(t); y = y(t) of the system (2). Let a complex, non-homotopic to zero cycle lie on y = y(x). As shown in [2,3], the components of the solution x = x(t); y = y(t) will be pseudoperiodic functions with the period one. Should Z1; Z2; : : : ; Zsbe non-homotopic to zero and to each other cycles on y(x), then

x(t) and y(t) will be pseudoperiodic functions with the periods lA1; : : : ; lAs, i.e.

each cycle has a corresponding period. As concluded from the system (2), lAi=

Z

Z

N Pdx

where Z coincides with Zi or with a free, homotopic to Zi cycle. It may happen that as a result of parameterization of the solutions using the system (2) or (3), a projection of the cycle will be a closed curve on the plane t or z. Obviously, it may happen only for exclusive P and Q. We a shall assume that non-homotopic to zero cycles on the studied solution y = y(x) of (1) have non-closed projections on t and z. In other words, the solution of (2) or (3), which is corresponding to y = y(x), has non-equal to zero periods.

2. Uniformization of the (1) solution and the corresponding (2) solu-tion

Let consider the solution of (1) y = y(x) and the corresponding solution of the system (2) x = x(t); y = y(t). Let uniformize y = y(x) with the following

(3)

functions x = '( ); y = ( ), where ' and single-valued meromorphic func-tions in the domain K. K is either a circle with an unitary radius, or a …nite plane, or an extended plane. As known [4], ' and are automorphic functions and are invariant to the group G of the linear (fractional) transformation. G is isomorphic to the fundamental group 1( ) a Riemann surface of y = y(x):

If is unicoherent, then G is a trivial group consisting of an identitical trans-formation. Thus, there is a non-trivial element of the group G corresponding to each non-homotopic to zero cycle on : Let T be an element of the group G. As it stands 8 2 K

'(T ) = '( ); (T ) = ( )

If we make substitutions in the system (2) x = '( ); y = ( ), then

(4) d' d = dt d P ('; ) N ('; ); d d = dt d N ('; ) Q('; ) From it, we deduce

(5) dt d = d' d N ('; ) P ('; ); dt d = d d N ('; ) Q('; ) Consequently, dt d is a meromorphic in K function.

Let us consider the case when a branch y = y(x) is regular in the neighbor-hood of the …nite or in…nite singular point M0of the equation (1). M0is viewed

as an ordinary point of y = y(x), to which there is a corresponding meaning of 02 K; where '( ) and ( ) are the meromorphic functions. Furthermore,

since M0 is the singular point of the equation (1) t( ) and t0( ) ! 1 when

! 0. The contrary is true also. If t0( ) ! 1 when ! 0, then certainly

some branch y = y(x) is regular in the neighborhood of the singular point of the equation (1). Really, let ' and be regular in the neighbourhood of 0

addition, from (5) it follows that if P (x0; y0) 6= 0 or Q(x0; y0) 6= 0, then

dt d is regular in the neighborhood of 0, which contradicts our assumption. The

case when ' and are non-limited in the neighbourhood of '0is being studied similarly. Thus, we come to

Lemma 1 If 9 0 2 K that t0( ) ! 1 when ! 0, then the solution y =

y(x) is certainly regular in the neighborhood of one of the singular points, i.e. y = y(x) is the separatrix solution of (1).

We will clarify the feature of t( ) concerning the group G. We specify that t( ) is being determined from (5). For 8A 2 G t(A )

(4)

[t(A )] = t(A )(A ). From (5) it follows [t(A )]== '0(A )N ('(A ); (A )) P ('(A ); (A )) (A ) 0 ='0(A )N ('( ); ( )) P ('( ); ( )) (A ) 0 Since '(A ) = '( ); (A ) = ( ) we have

['(A )]0 = '`(A )(A )0 = '`(t)

[ (A )]0 = `(A )(A )0 = `(t) Consequently, [t(A )]0 =' 0( )N ('; ) P ('; ) = t 0( ) or [t(A )]0 t0( ) = 0; [t(A ) t( )]0 = 0; t(A ) t( ) = 1A Thus, we prove

Lemma 2 Let y(x) be the solution of (1), and x(t); y(t) be the corresponding solution of (2). Then the parameter t is being determined as a function from the system (5) and satis…es the following ratio:

(5)

We shall de…ne the constant lA more precisely. Let Z be a cycle on y(x),

and the corresponding element be from G. It means that when the invariable runs through the segment [ 0; A 0] beginning with 02 K; x; y will run through

the closed curve Z , which freely homotopic to the cycle Z on y(x). Hence, Z [ 0;A o] t0( )d = Z Z N (x; y) P (x; y)dx; t(A 0) t( 0) = Z Z N (x; y) P (x; y)dx = lA

Therefore, the number lA does not depend on the way of uniformization

of the solution y = y(x). Let denote a set of separatrix solutions of (1) and solutions of (2) coming to x = const; y = const by S.

Let y(x) =2 S. Then it is obvious that t( ) determined from (5) is a mero-morphic function, and the following lemma is true

Lemma 3 If y(x) =2 S, then three functions x = '( ); y = ( ) and t = t( ) uniformize simultaneously the functions y = y(x); x = x(t) and y = y(t). We shall note that in the case y(x) =2 S the function t( ) has, generally speaking, the logarithmic singularity.

As for the sepratrix solution, we believe that a singular point, in which neighbourhood a branch of this solution is regular, is lying on y(x), and cycles, which are divided by the point, are being considered as non-homotopic.

3. Features of the function $( ) = t0( )=N('; ).

Theorem 4 Let y(x) =2 S, then $( ) 6= 0; $( ) 6= 1 8 2 K.

Demonstration. From Lemma 3, x( ); y( ) and t( ) are holomorphic in K. Hence, $ is the meromorphic in K function. Let $( ) have a pole ($( 0) = 1)

when = 0. It is possible only when N ('( 0); ( 0) = 0. Hence, t0( 0) = 0.

Indeed, if t0( 0) 6= 0, then we can deduce from (3) that for holomorphity of

'( 0); ( 0) in 0 it is necessary that

(6)

However, it is excluded by the condition of the theorem.

Due to the same reason, if N ('; ) has a zero of the second order, then t0( ) also has a zero with an order not less than two. Using a small deformation of N (x; y) we can achieve that N ('; ) 6= 0, i.e. N(x; y) = 0, is not an integral of the equation (1), and we assume this without loss of generality. Therefore, we assert that a pole of $( ) does not exist.

Let not 0 be a zero of $( ). It is possible if t0( 0) = 0 and multiplicity of a

zero of t0( 0) in 0 should be higher than that of N (x( ); y( )) by one order.

By letting the equality t = t( ) in the neighbourhood of 0 and studying the

corresponding solution of (2) x = x(t); y = y(t) in the neighbourhood t = t( 0),

we make certain that the mutual single-valued correspondence between the small neighborhood 0in the circle K and the neighbourhood of the point x( 0); y( 0)

of solution y = y(x) does not exist. However, it contradicts the conditions of uniformization.

Therefore, $( ) 6= 0 8 2 K.

The theorem is proved. We shall introduce the denotation

(6) Z 0 t0( ) N (; )d = Z Since dZ

d = ! 6= 0;then (6) can be solved for in the neighbourhood of every point z 2 K0, where K0 is a domain of values of the integral from (6), when runs through K. It is plane that the function = (z), which was determined as a result of solving (6) for , is a regular function in the small neighbourhood of any point in K0, however, as a whole, this function can be multivalued in K0.

Assuming that (0) = 0, let choose a branch of = (z),

We are considering now the functions '( (z)), ( (z)) in K0. This pair is a

solution of the system:

(3) dx

dz = P (x; y); dy

dz = Q(x; y) We shall introduce the following denotations:

'( (z)) = x1(z); ( (z)) = y1(z);

(7)

The solution fx1(z); y1(z)g is de…ned in K0. However, as a solution of the system

(3), it can be also extended beyond K0.

Lemma 5 Any point z0, to which the solution x(z); y(z) can extend, belongs to

K0.

Demonstration. Due to features of the functions '( ); ( ), there is certainly such 12 K that '( 1) = x1(z0); ( 1) = y1(z0).

We introduce the denotation

1

Z

0

!( )d = z1( 1):

Then (z1) = 1, and we come to:

x1(z1) = '( (z1)) = '( 1) = x1(z0);

y1(z1) = ( (z1)) = ( 1) = x1(z0):

It means that a cycle on the solution y = y(x) corresponds to the route connecting the points Z0 and Z1 on the plane Z. Let A 2 G be an element of

G, which corresponds to a cycle. Then, Z(A 1) = Z( 1) + Z0 Z1 = Z0 t e

Z02 K0.

We proved the lemma.

Assume that (z) is a multivalued in K0 function. Therefore, there is such

a closed route Z 2 K0 that if we run through it starting from the point Z1, we

shall come to 0

2(Z1) , which di¤ers from the starting value 01 = 1(Z1). It

means that there are such points that 0

1; 022 K.

Z( 01) = Z( 02) = Z1

Lemma 6 2 is a holomorphic function of in a small neighbourhood 01:

Demonstration. z( 1) = 1 Z 0 t0( ) N d ; z( 2) = 2 Z 0 t0( ) N d

The equality Z( 1) = Z( 2) is solvable in the neighborhood of the point 0

1; 02 for 2. After di¤erentiating both sides of the equality, we have:

t0( 1)d 1

N ('( 1); ( 1))

= t0( 2)d 2 N ('( 1); '( 2))

(8)

and d 1 d 2 = t0( 2) N ( 2) N ( 1) t0( 1) Hence (7) d 1 d 2 6= 0; d 2 d 1 6= 0:

Due to this reason, (7) is solvable for 2 in a small neighbourhood 01: 2= ( 1)

and ( 1) is a holomorphic function of 1.

The lemma is proved.

Let x1(z); y1(z) the de…ned solution of (3). Hence,

x1(z( )) = '( ); y1(z( )) = ( ) 8 K:

According to the de…nition of ( ), we have z( ) = z( ( )): At the same time,

x1(z( ( ))) = '( ( )); y1(z( ( ))) = ( ( )):

Therefore,

x1(z( )) = '( ( )); y1(z( )) = ( ( ))

or

'( ) = '( ( )); ( ) = ( ( )):

It means that there is a cycle on the solution y = y(x). If it is homotopic to zero, then certainly ( ) , i.e. = (z) can not be multivalued. If it coincides with one of the non-homotopic to zero cycles lying on y = y(x), then due to z( ) = z( ( )) and the presentation of the solution y = y(x) as x = x1(z); y = y1(z), this cycle is being projected into the closed curve on the

plane z, which is not possible according to our assumption.

Thus, = (z) is a single-valued function in K0. Hence, the pair of functions

x1(z) = '( (z)); y1(z) = ( (z))

(9)

4. Features of the singular points

Let M0(x0; y0) be a singular point of (3). According to our assumptions

regard-ing P and Q, the system (3) can be reduced to the followregard-ing form:

(8) dU

dZ = 1U;

dV

dZ = 2U; where U; V and x; y are interrelated:

(9) x x0= 1 P i+j=1 aijUiVj; y y0= 1 P i+j=1 bijUiVj:

Series on the right sides are convergent when jUj < , jV j < , Where is a positive number. We shall solve (8) and substitute U and V in (9):

(10) x x0= 1 P i+j=1 aij c1e 1z z0 i c2e 2z z0 j ; y y0= 1 P i+j=1 bij c1e 1z z0 i c2e 2z z0 j : Let consider the sector T with its top in z0 on the plane z:

Re 1(z z1) 0; Re 2(z z1) 0

For this sector

e 1(z z0) < 1; e 2(z z0) < 1 and e 1(z z0)

! 0; e 2(z z0)

! 0 when z ! 1 on it.

Therefore, when jc1j < ; jc2j < , functions (10) de…ne the regular in the sector solution (3). In addition, x(z) ! x0; y(z) ! y0 when z ! 1 in this sector.

Let x1(z); y1(z) be single-valued solutions (3) in K0. We shall choose such z0

that the point x1(z0); y1(z0) will get into a small neighbourhood of M0, where

the convergence is taking place. Then this solution, when z ! 1 on the sector, will be converging to M0. If, in addition to this the solution fx1(z); y1(z)g is

periodic, then this sector transforms into semi-plane. The system (3) has n2

singular points. Hence, there is a pair of di¤erent singular points, which sectors intersect, and that is not possible.

The contradiction appeared because of the assumption that the solution y = y(x), on which a non-homotopic to zero cycle is lying, is not a separatrix solution. Thus, we come to

(10)

Theorem 7 Every solution (1), on which a non-homotopic to zero cycle is lying, is a separatrix solution.

We shall study separatrix solutions (1) and try to de…ne a number of non-homotopic to zero and to each other cycles with incommensurable periods lying on them.

5. On a umber of cycles lying on one solutiony = y(x) of the equation (1).

As it was stated at the beginning, the problem is equivalent to the problem of de…nition of a number of independent periods of pseudoperiodic solutions of (2). Let x(t); y(t) be the solution (2) that has 3 independent periods. Thus, as known [3], this solution has an everywhere dense set of periods and is itself dense everywhere in the space (x; y; t).

Let l be one of the periods. We shall take the point t = t0 on the plane, and

let x(t0); y(t0) be one of the values of the solution x(t); y(t). By virtue of the

de…nition 1, there is such a route L on the plane t starting in the point t0 and

…nishing in the point t0+ l that values x(t); y(t), after running through the

route L, will again become x(t0); y(t0). We can say that functions x(t); y(t) are

regular in a certain zone 2" in width and encompassing the line L when running from the point t0 to t0+ l on the interesting for us branch.

Let’s take in a small neighbourhood of t0the point t0and the point x(t0); y(t0)

belonging to the solution x(t); y(t) and satisfying equalities x(t0) = x(t0); y(t0) =

t(t0); such point exists in virtue of the everywhere density of periods. We

con-tinue the branch passing through the point t = t(t0); x(t0) y(t0) up to the point t0. Consequently, we receive a new point

Let’s consider the route L starting in t and resulted from the parallel translation of L. The end of the route L will be in the point t0+ l. Points t0

and t are so close to each other that the route L lies inside the corresponding to L zone 2" in width. It means that if we run along L from t0 to t0+ l, we

shall start from the point x(t~ 0), ~y(t0) of the solution x(t); y(t) and return back

(11)

A number of points ~x(t0), ~y(t0) with similar features, which correspond to

dif-ferent values t0, is no less than countable. Therefore, because of the analyticity

of the equation, all solutions of the system starting in a small neighborhood of the point (t0; x(t0); y(t0)) shall be pseudoperiodic with the period 1. In other

words, a cycle lying on y = y(x) and corresponding to this period will be an identical cycle. However, the solution y = y(x) of the equation (1) correspond-ing to x(t); y(t) is a separatrix solution in a neighborhood of at least one scorrespond-ingular point of the equation (1). Therefore, a cycle lying on y = y(x) lies simultane-ously in a small neighborhood of the point M0. Obviously, this cycle is not

identical. It contradicts the assumption of existence of three non-homotopic cycles on y = y(x). Thus, we come to

Theorem 8 On condition of the generality of P; Q; N , there are not more than two non-homotopic to zero and to each other cycles on one solution y = y(x), and one of them lies in a small neighborhood of a singular point. From theorem 2 and 3, we can deduce the main theorem

Theorem 9 Almost all equations of the y = P

Q type, where P and Q are poly-nomials of the nthdegree, have following fundamental features:

a) Non-homotopic to zero cycles lies on separatrix solutions; b) There are not more than two cycles on each solution;

c) One cycle on each solution de…nitely lies in a neighbourhood of one of the singular points of the equation (1);

d) A number of separatrix solutions is equal to 2n2+ n + 1.

Theorem 10 A number of non-homotopic to zero and to each other cycles of the equation (1) does not exceed 2(2n2+ n + 1). This result follows from points b) and d) of Theorem 4. As deduced from Lemma 1 from [1], a real limit cycle of the real equation, which lies in a small neighborhood of a singular point of the equation on the separatrix of that singular point, is an algebraic curve. Hence, the following theorem is true.

Theorem 11 A number of real limit cycles of the equation (1) with real coef-…cients does not exceed 2(2n2+ n + 1). At the same time, a number of

(12)

Demonstration. As shown in [1] , a number of real limit cycles does not exceed a number of complex cycles lying on solutions of (1) in the complex domain. At the same time, we can say that cycles bear features listed in Theorem 3. From the above, we obtain our result.

Therefore, in this study we estimate the number of limit cycles of the equa-tion y0 = PQ, where P and Q are polynomials. In addition, and it is of no less importance, the way of determining them is being shown, namely that they lie on separatrix solutions. It is also important to note that the element of the sep-aratrix solution is being presented by concrete series in a small neighbourhood of a singular point.

References

1. Landis, E.M; Petrovsky, I.G.(1955): On the number of limit cycles of the equation dy

dx =

Q(x; y)

P (x; y), Mathematical collection 37(79):2 (in Russian).

2. Landis, E.M; Petrovsky, I.G.(1957): On the number of limit cycles of the equation dy

dx =

Q(x; y)

P (x; y), Mathematical collection 43(85):2 (in Russian).

3. Huday-verenov M.G.(1990): On the functions determined by the di¤erential equa-tions, Ashgabat (in Russian).

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