BOUNDEDNESS OF SOLUTIONS AND EXISTENCE OF LIMIT CYCLES FOR A NONLINEAR SYSTEM OF DIFFERENTIAL EQUATIONS

1998, 18( 1): 113-120

BOUNDEDNESS OF SOLUTIONS AND EXISTENCE OF LIMIT CYCLES FOR A NONLINEAR SYSTEM OF DIFFERENTIAL EQUATIONS 1 Huang Lihong (-Jt ±% ) Department of Applied Mathematics, Hunan University, changsha 410082, China

Chen Mingpo ( $1lJl if- ) Institute of Mathematics, Academia Sinica, Nankang, Taipei,11529 Taiwan, China

Abstract In this paper, author obtain sufficient conditions for the boundedness of solutions and the existence of limit cycles of the nonlinear differential system

dx/dt

= p(y),dy/dt = -q(y)h(x,y) -

without the traditional assumptions "h(x,y)

g(x)

2: o for Ixl sufficiently large" and "Jo±oo g(x)dx

=+00". Key words Nonlinear differential system, boundcdness, existence, limit cycle.

1 Introduction The autonomous planar system of the form

dx/dt

= y,dy/dt = -yh(x, y) -

g(x)

(1.1)

was first studied by Levinson and Smith in the classical paper [2] as the equivalent system of the equation of Lienard type ( 1.2)

and later on some authors have contributed to the theory of this system with respect to qualitative behavior of solutions[1-9]. In the present paper we study a system of a slightly more general type, namely,

dx/dt = p(y),

dy/dt = -q(y)h(x, y) - g(x),

( 1.3)

where the functions p(y), q(y), g(x) and h( x, y) are continuous for all value of their arguments, and are subject to the conditions which ensure that the existence of unique solution to the initial value problem. 1 Received

Mar.25,1996. This work was supported by the NNSF of China

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Our aim in this paper is to obtain some sufficient conditions for the boundedness of solutions and the existence of limit cycles of (1.3). Our results are different from those of [1-9] in some sence even for the special case (1.1). In particular, as in [1] and [6], all our results also allow to avoid the classical assumptions:

h( x, y) > O( or

~

0) for Ix I sufficiently large,

and

r±1XJ

G( ±oo):= Jo

(1.4)

g(x )dx = +00,

(1.5 )

(see, e.g., [3,4,5,7,8] etc.). To compare our theorems with previous results, two simple examples are given in section 4. Here we mention some assumptions which will be used in the sequel.

(C1) yp(y) > 0 and yq(y) > 0 for all y

Jr+1XJ

( C2)'

rl1U.(Y) d1J q Y'

= +00, r-r

lXJ

,• -

P((Y)l (II; q Y •

=I=-

0, and ~::;j~ Iq(y)1 > O.

= -00, and the limits lim

f/

P((,Y)j dlJ k->O+.I1,; q Y '

and lim ('-,r ,:->0+ .

-h'

rl1U. dy exist. q(y)

(C3) xg(x) > 0 for all x

(x, y) E R

=I=-

O.

there exists a continuous function f(x) : R

(C4) Z

•

R such that h(x, y)

--->

> f(x)

for all

,

(C5) liminf F(x) > -00, lim sup F(x) < +00, where F(x) x--oo

X~+OO

:= J~r

f(z)dz.

(C6) limsup(lF(x)1 + G(x)) = +00, where G(x) := J~" g(z)dz. x-+±oo

(C7) h(O,O) < O.

2 Lemmas In this section, we establish several lemmas which will play an important role in the proof" of our main results in the sequel. First of all, we introduce the following notations:

Dr := {(x, y) : x D 3 :=

~

{(x,y):~;:::;

0, y > O}, D z := {(x, y) : x < 0, Y?: O}, O,y < 0},D4

:={(x,y): x> O,y:::; O}.

Furthermore, for simplicity, we denote throughout this paper by L+(A) and L-(A), respectively, the positive and negative semitrajectory of (1.3) passing through an arbitary point A at time to, and let [to, T A ) denote the right-maximal interval of L+(A), where T A :::; +00. Lemma 1 Suppose that the conditions (C1)-(C6) hold, A = (xo, Yo) E D I . Then L+(A) must intersect the positive x-axis. Proof Suppose not. Since dxjdt

> 0 in Dr,

it follows that L+(A) will stay in the region

Dr for all t E [to, T A ) . Let (x(t), y(t)) be the coordinates of L+(A) at time t. Then y(t) > 0 = p(y(t)) > 0 for all t E [to, T.4.), and so the function z = x(t) is increasing for t E [to,TA ) . It follows that the inverse function t = :e-I(z) of z = ~:(t) exists for t E [to,T.4.)' Again since there is no critical point of (1.3) under the conditions of Lemma 1, according to and dx(t)jdt

the theory of limit-sets, it is certain that L + (A) is unbounded.

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On the other hand, it follows from (1.3) that

p(y) dy q(y) dt

= -p(y)h(x, y) -

g(x)p(y).

(2.1)

q(y)

Integrating (2.1) from to to t E [to,T A ) along L+(A), we obtain

ruldy .Ify(t) Yo q(y)

= _ .Ito ft

+ g(x(s))p(y(s))] q(y(s))

[h(x(s) ,.y(s))p(y(s))

ds.

::; - It: h(x(s), y(s))p(y(s))ds < _ .Ito ft f(x(s))(dx(s))ds ds =f(z)dz

a:

= -F(x(t))

Define

Q(y)

(2.2)

+ F(xo).

= k:.....O+ lim

l

y

p(z)

ksgny

- () dz. q Z

Then (2.2) implies

Q(y(t)) - Q(yo) ::; -F(x(t))

(2.3)

+ F(xo),

< +00. Therefore, it follows from (C1) and (C2) [to, 7":,d. Thus, it is certain that there exists a constant

which and (C5) imply that Q(y(t)) - Q(yo) that y(t) is bounded from above for t E M

> 0 such that 0 < y(t) ::; M for all t E [to, TA ) .

Notice that L+(A) is unbounded, it is clear that lim x(t) t .....

T;;

= +00.

In the following, we are

going to obtain a contradication. N

Let

=1+

max q(y).

O~y~M

Integrating (2.1) from to to t E [to, T A ) along L+(A), we have

Q(y(t)) - Q(yo)

= - .ftto [h(x(s),y(s)) + :~:i;in p(y(s))ds

::; - it o [f(x(s)) + g(;~S))] p(y(s))ds = - itto [f(x(s)) + g(~S))] (d~~s))ds t

= - jx(t) f(z)dz _ .lN ..rx(t) g(z)dz = -F(x(t)) + F(xo) - kG(x(t)) + kG(xo). Xo

Xo

This implies 1

Q(y(t))::; Q(yo) - N[F(x(t)) In view of (C5),(C6) and lim x(t) limiI~fQ(y(t)) t ..... T A

= -00.

t ..... T;;

+ G(x(t))] -

= +00,

N -1

-pfF(x(t))

1

+ F(xo) + NG(xo).

(2.4)

it is easy to see that the inequality (2.4) implies

But this contradicts the fact that y(t)

> 0 for all t E [to, T A ) , and hence

L+(A) must intersect the positive x-axis. This completes the proof of Lemma 1. Lemma 2 Suppose that the conditions (C1)-(C6) hold, A = (XO,Yo) E D3 . Then L+(A)

must intersect the negative x-axis,

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The proof of Lemma 2 can be given by an analogous argument as in the proof of Lemma 1, and hence it is omitted here. Lemma 3 If the conditions (C1),(C3) and (C7) are satisfied, then the origin is the only critical point of (1.3), and it is locally repulsive. Proof In view of (C1) and (C3), it is obvious that the origin is a unique critical point of

(1.3). Next, we will show that the origin is repulsive. Set V(x, y)

= G(x) +

l

Y

p(z)dz.

From (Cl) and (C3), it is clear that for 0 < c « 1 the curve V(x,y) = c is a closed curve surrounding the origin. By (Cl) and (C7), for 0 < c« 1, along the closed curve V(x,y) = c we have

dV(X,IJ) I . dt.

I (1.3)

= -p(y)q(y)h(x,y) > O,when

y

i- O.

Hence the origin is locally repulsive.

Lemma 4 Assume that the conditions (C1)-(C7) hold and that A = (xo, Yo) E D z . Then L+(A) must intersect the positive y-axis. Proof Let (x(t),y(t)) be the coordinates of L+(A) at time t. Suppose that L+(A) does not intersect the positive y-axis for all t E [to, T A ) . Then we have (x(t), y(t)) E D z for all t E [to, r.d. It follows that :1;(t) < 0, y(t) > 0 and dx(t)jdt = p(y(t)) > 0 for t E [to, T A ) , and so :ro "::: :1;(t) < 0 for all t E [to, T A ) . By Lemma 3 and the theory of limit-sets, it is easy to see that limsupy(t) = +00. Again from the fact that dx(t)jdt > 0 for all t E [to,TA ) , it is obvious

--»:

that the limit lim x(t) exists. Let t~T;;-

= a.

lim :1;(t) »-r: Then

:1;0

< a "::: O. Next, we are going to prove that liminfy(t) > O. t~T;;

Otherwise, then there must be a time sequence {t n that

dy.(t) I dt t=t"

Notice that

=0

and

}

satisfying t n

lim y(t n ) = O.

n~+oo

-+

T A as n

-+

+00

such (2.5)

dy(t) -----;.]t = -q(y(t))h(x(t), y(t)) - g(x(t)),

we have

q(y(tn))h(x(t n), y(t n)) Letting n

-+

+ g(x(t n)) = O.

(2.6)

+00, we get q(O)h(a, 0) + g(a)

= 0,

and so g(a) = O. It follows from (C3) that a = O. Again by the fact that :1;(t n ) < O,q(y(t,,)) > 0, g(x(tn)) < 0, h(O, 0) < 0, lim y(t n) = 0 and lim x(t n) = a = 0, when n is sufficiently "-+00 "-+00 large we have

q(y(tn))h(x(t,,), y(t,,))

+ g(x(t n ) ) < O.

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Huang & Chen: BOUNDEDNESS OF SOLUTIONS AND EXISTENCE OF LIMIT CYCLES

But this contradicts (2.6). This shows liminfy(t) t--->T;;'

117

> O. Therefore, from the condition (C1) it is

sure that there is a positive constant b such that

where t 1 E (to, T.4 ) , and if Yo f. 0, then we may choose t 1 = to· Now, integrating (2.1) from t1 to t E [t 1, T A ) along L+(A) and using the notation in (2.3), we have

Q(y(t)) - Q(y(td)

=-

:s -

= -

Itt,lh(x(s),y(s))

+ ~i;i:mp(y(s))ds

+ g(xis))]p(y(s))ds J.t [f(x(s)) + g(x(s))]('IX(s))ds . t! b ds Itt! [f(x(s))

-- - Jx(t) f( Z )dZ - 1b •(x(t) ()d x(tll x(til 9 Z z. Notice that Xo :S x(t) < 0 for all t E [to, ~4), from (C1) and (C2) it is easy to see that the above inequality implies that y(t) is bounded from above for t E [h, T A ) . This contradicts limsupy(t)

= +00.

Hence L+(A) must intersect the positive y-axis. The proof of Lemma 4 is

t--->T.:;-

complete. By an argument similar to above we can prove the following Lemma 5. Lemma 5 If the conditions (C1)-(C7) hold and A = (xo, Yo) E D 4 , then L+(A) must intersect the negative y-axis.

3 Main Results In this section, we give our main results about the boundedness of solutions and the existence of limit cycles of (1.3). Theorem 1 Assume that the conditions (C1)-(C7) hold, and that one of the following two conditions is satisfied.

(C+S) There exist a constant 0 :::: 0 and a function r(x) E C 1((- 00, -oj, (0, +ex:.')) such

that

q(1'(x))h(x, r(x))

dr(:r)

+ g(x) + p(1·(x))-.-1r.x

(3.1)

:::: 0,

for all x E (-00, -0]. (C - 8) There exist a constant 0 :::: 0 and a function r(:I.:) E C 1 ([0,

q(-r(x))h(x, 1'(X)) for all

:1.:

E

d-r(x) dx

+ g(x) + p(r(x))-- :S

+X», (- 00, 0))

such that

0,

(3.2)

[o,+X».

Then all solutions of (1.3) are bounded, and (1.3) possesses at least one limit cycle. Proof We only consider the case that (C+S) holds, since the case that (C-S) holds can be treated similarly. According to Lemma 3, the origin is a unique critical point of (1.3) and is repulsive. Therefore, by the theory of limit sets, to prove Theorem 1, we only need to show that all positive sernitrajectories of (1.3) are bounded under the conditions of Theorem 1.

Suppose that there exists some point A = (xo, Yo) E R 2 such that L+(A) is unbounded. Then, from Lemmas 1,2,4 and 5, it is clear that L+(A) is a clockwise spiral surrounding the

118

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ongm. Again by the theory of limit sets, it is certain that one of the following two possible cases must hold. (i) The origin is the only a-limit point of L-(A), and it is a global focus. (ii) There exists a limit cycle CL of (1.3) such that CL is the a -limit set of L-(A), and there is no limit crcle of (1.3) in the exterior of CL. Now choose constants M and N such that

M;::: 1 + max{lxol, <5} and N;::: 1 + r(M). Let B = (-M, N). We next prove that L+(B) is bounded. Otherwise, in view of Lemmas 1,2,4 and 5, it is certain that L + (B) is a clockwise spiral surrounding the origin, and that it must cross the curve y

= r(x)

and intersect the line x

= -M

Yc > N. But this is impossible, since, along the curve y

dy I dx (1.3)

at some point C

= r(x), we have

= (-M,yc)

with

_ d1'(X) = _ q(y)h(x, y) + g(x) dx = -

p(y) q(r(x))h(x, r(x))

_ d1'(X) dx + g(x) + p(r(x))dr(x)/dx p(r(x)) ::; 0,

for x E (-=, -<5), which shows that L + (B) cannot cross the curve y = r( x) and intersect the line x = -M at some point above B. Therefore , L+(B) is bounded. Again from Lemma 3 and the theory of limit sets, it follows that there exists one limit cycle of (1.3) such that it is the w-limit set of L+(B). If the case (i) occurs, then it is obvious that (1.3) cannot possess limit cycle, and hence the case (i) is impossible. Now suppose that the case (ii) occurs. We choose the above constants M and N are so large that the point B

= (-M, N)

lies in the

exterior of the limit cycle CL. Notice that CL is the a-limit set of L - (A), it is ea.<:,y to see that LC cannot be the w-limit set of L+(B). On the other hand, notice that (1.3) does not exist critical point and limit cycle in the exterior of CL, from the theory of limit sets, it is certain that L + (B) must be unbounded. Thus, we obtain a contradiction, and hence the case (ii) is also impossible. Therefore, (1.3) cannot possess unbounded positive semitrajeetory under the conditions of Theorem 1. This completes the proof of Theorem 1. Theorem 2 Suppose that the conditions (C1)-(C7) hold, and that one of the following

two conditions is satisfied.

(C+9) There exist a constant <5 > 0 and a function r( x) E C 1 (( -=, -<5], (0, +OQ)) such

that

q(r(x)).f(x)

dr(x)

+ g(x) + p(r(x))-l_. ex

;::: 0,

(3.3)

for all x E (-=, -<5].

(C- 9) There exist a constant <5 > 0 and a function r( x) E C 1 ([<5, +=), (-=,0)) such that

q(r(x)).f(x)

dr(x)

+ g(x) + p(1'(X))~

::; 0,

(3.4)

for all x E [<5,+=). Then all solutions of (1.3) are bounded, and (1.3) possesses at least one limit cycle. Obviously, when the conditions (C1) and (C4) hold, the inequalities (3.3) and (3.4) imply (3.1) and (3.2) respectively. Therefore, Theorem 2 is a direct corollary of Theorem 1.

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Theorem 3 Suppose that the conditions (C1)-(C7) are satisfied, and that one of the following two assumptions holds. (C+lO) liminff(x) > 0 and limsup{lg(x)l/f(x)} < sup{q(y)}. x-+-oo

x-+-oo

y>O

x-++oo

x-++oo

y
(C-lO) lim inf j'{z ) > 0 and limsup{lg(x)l/f(x)} < sup{-q(y)}. Then all solutions of (1.3) are bounded, and (1.3) possesses at least one limit cycle. Proof Obviously, (C+10) implies that there exist constants b > 0 and K > 0 such that

f(x) > 0 and q(K) 2: Ig(x)l/ f(x) for all x E (-00, -b]. It follows from (C1) and (C3) that q(K)f(x) + g(x) 2: 0 for x E (-00, -b]. Set r(x) == K. Then (3.4) follows. Thus, (C+lO) implies (C+9) under the assumptions (C1) and (C3). By a similar argument, we can prove that (C-lO) implies (C-9) if (C1) and (C3) hold. Therefore, it follows from Theorem 2 that the conclusions of Theorem 3 are true. Remark 3.1 For the system (1.1), the conditions (C1) and (C2) hold naturally. Therefore, our Theorems 1-3 are available for (1.1). Remark 3.2 In [1], F .Bucci gave a result for the existence of nontrivial periodic solutions of (1.1) under (C3),(C4),(C7),(C+8) and a similar condition to (C6), but she did not consider the boupdedness of solutions of (1.1). As in [1], paper [2-9] except [5] also did not study the boundedness of solutins of (1.1). On the other hand, we also stress the fact that our results are available to more general system (1.3), and, as far as we know, no similar results were given up to now for (1.3).

4 Exaples In order to compare our results with those obtained in [1-9], in this section we give two simple examples. Example 1 Let 0: and (3 be the ratio of two positive odd integers and be a positive integer. Then the system:

0:

2: (3, and let

17),

(4.1) has at least one limit cycle, and all its solutions are bounded. Proof For this system, we have p(y) = yO,q(y) = yf3,h(x,y) = 1- 2cosx + y2m and = xe- x ' . It is obvious that the conditions (C1),(C2),(C3) and (C7) hold. Let f(x) = 1 - 2 cos x. Then (C4) follows. On the other hand, by some simple computations, we have

g(x)

F(x)

=x -

2sinx and G(x)

= ~(1- e- X').

Thus, it is easy to see that-the conditions (C5) and (C6) hold. We next prove that (C+8) also holds. Choose a constant N > 1, and set r( x)

q(r(x))h(x, r(x))

= (N -

=N

- e", Then we have

+ g(x) + p(1'(X)) dr(x)

eX)f3[1 - 2 cos x + (N - eX)2m]

dx

+ xe- x '

-

(N _ eX)Oe x •