On the symmetry and periodicity of solutions of differential systems

On the symmetry and periodicity of solutions of differential systems

Nonlinear Analysis: Real World Applications 17 (2014) 64–70 Contents lists available at ScienceDirect Nonlinear Analysis: Real World Applications jo...

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Nonlinear Analysis: Real World Applications 17 (2014) 64–70

Contents lists available at ScienceDirect

Nonlinear Analysis: Real World Applications journal homepage: www.elsevier.com/locate/nonrwa

On the symmetry and periodicity of solutions of differential systems✩ Zhengxin Zhou ∗ School of Mathematical Sciences, Yangzhou University, Yangzhou 225002, China

article

info

Article history: Received 15 April 2013 Accepted 21 October 2013

abstract This article deals with the structure of the reflective function of the higher degree polynomial differential systems. The obtained results are applied to discussion of the symmetry and periodicity of the solutions of these systems. © 2013 Elsevier Ltd. All rights reserved.

1. Introduction Studying the property of the solutions of the differential system x′ = X (t , x)

(1)

is very important not only for the theory of ordinary differential equations but also for practical reasons. If X (t + 2ω, x) = X (t , x) (ω is a positive constant), to study the solutions behavior of (1), we could use, as introduced in [1], the Poincaré mapping. But it is very difficult to find the Poincaré mapping for many systems which cannot be integrated in quadratures. The Russian mathematician Mironenko [1,2] first established the theory of reflective functions. Since then a new method to establish the Poincaré mapping of (1) has been found. In the present section, we introduce the concept of the reflective function, which will be used throughout the rest of this article. Consider differential system (1) which has a continuous differentiable right-hand side and general solution ϕ(t ; t0 , x0 ). For each such system, the reflecting function (RF) is defined as F (t , x) := ϕ(−t , t , x) [1,2]. Therefore, for any solution x(t ) of (1), we have F (t , x(t )) = x(−t ). If system (1) is 2ω-periodic with respect to t, then T (x) := F (−ω, x) = ϕ(ω; −ω, x) is the Poincaré mapping of (1) over the period [−ω, ω]. Thus, the solution x = ϕ(t ; −ω, x0 ) of (1) defined on [−ω, ω] is 2ω-periodic if and only if x0 is a fixed point of T (x). The stability of this periodic solution is equivalent to the stability of the fixed point x0 . A differentiable function F (t , x) is a reflecting function of system (1) if and only if it is a solution of the Cauchy problem Ft′ + Fx′ X (t , x) + X (−t , F ) = 0,

F (0, x) = x.

(2)

This implies that sometimes for non-integrable periodic systems we can find out its Poincaré mapping. If, for example, X (t , x) + X (−t , x) = 0, then T (x) = x. If F (t , x) is the RF of (1), then it is also the RF of the system x′ = X (t , x) + Fx−1 R(t , x) − R(−t , F (t , x))

✩ The work supported by the NSF of Jiangsu of China under grant BK2012682 and the NSF of China under grant 11271026.



Tel.: +86 51487975401. E-mail address: [email protected].

1468-1218/$ – see front matter © 2013 Elsevier Ltd. All rights reserved. http://dx.doi.org/10.1016/j.nonrwa.2013.10.006

Z. Zhou / Nonlinear Analysis: Real World Applications 17 (2014) 64–70

65

where R(t , x) is an arbitrary vector function such that the solutions of the above systems are uniquely determined by their initial conditions. Therefore, all these 2ω-periodic systems have a common Poincaré mapping over the period [−ω, ω], and the behavior of the periodic solutions of these systems are the same. So, to find out the reflective function is very important for studying the qualitative behavior of solutions of differential systems. There are many papers which are also devoted to investigations of qualitative behavior of solutions of differential systems by help of reflective functions. Mironenko [1–5] combined the theory of RF with the integral manifolds theory to discuss the symmetry and other geometric properties of solutions of (1), and obtained a lot of excellent new conclusions. Alisevich [6] has discussed when a linear system has triangular reflective function. Musafirov [7] has studied when a linear system has reflective function which can be expressed as a product of three exponential matrices. Veresovich [8] has researched when the nonautonomous two-dimensional quadric systems are equivalent to a linear system. Maiorovskaya [9] has established the sufficient conditions under which the quadratic systems have linear reflecting function. Zhou [10,11] has discussed the structure of reflective function of quadratic systems, and applied the obtained conclusions to study the qualitative behavior of solutions of differential systems. Now, we consider a higher degree polynomial differential system

 n  dx    = pi (t , x)yi = P (t , x, y),   dt i=0

(3)

m   dy   qj (t , x)yj = Q (t , x, y),   dt = j =0

∂ P (t ,x,y)

where pi (t , x); qj (t , x) (i = 0, 1, 2, . . . , n, j = 0, 1, 2, . . . , m) are continuously differentiable functions in R2 and ∂ y ̸= 0, and there exists a unique solution for the initial value problem of (3). n and m are positive integers. In the following, suppose that F (t , x, y) = F1 (t , x, y), F2 (t , x, y)T is the RF of (3). In this paper, we will discuss the structure of F2 (t , x, y) when F1 (t , x, y) = f (t , x). At the same time, we obtain the good result that F2 (t , x, y) = f0 (t , x) + f1 (t , x)y, which extends and improves the conclusion of literatures [10,11]. The obtained results are used for research of problems of the existence of a periodic solution of system (3) and establish the sufficient conditions under which the first component of the solution of (3) is an even function with respect to time t. The obtained conclusions extend and improve the known results. 2. Main results Without loss of generality, we assume that f (t , x) = x. Otherwise, we take the transformation u = f (t , x), v = y. Let F1 (t , x, y) = x. Then by the relation (2) we get P (t , x, y) + P (−t , x, F2 (t , x, y)) = 0, i.e., n 

Ai F2i = 0,

(4)

i =0

where A0 =

1 pn (−t , x)

[p0 (−t , x) + P (t , x, y)];

Ai =

pi (−t , x) pn (−t , x)

,

i = 1, 2, . . . , n − 1.

Here suppose that in some deleted neighborhood of t = 0 and |t | being small enough pn (t , x) is different from zero. As F2 (0, x, y) = y, from (4) implies the following lemmas directly. Lemma 1. For the system (3). If F1 = x, then pi (0, x) = 0,

i = 0, 1, 2, . . . , n.

Lemma 2. Let for the system (3) F1 = x and limt →0 lim

t →0

pi (t , x) + pi (−t , x) pn (t , x)

(5) pi (t ,x) pn (t ,x)

(i = 1, 2, . . . , n − 1) exist. Then

= 0 (i = 0, 1, 2, . . . , n − 1),

lim

t →0

pn (t , x) pn (−t , x)

= −1.

(6)

In the following discussion, we always assume (5) holds without further mention. Theorem 1. Suppose that F = (x, F2 )T is the RF of system (3) and F2 satisfies F22 + A1 F2 + A0 = 0,

(7)

66

Z. Zhou / Nonlinear Analysis: Real World Applications 17 (2014) 64–70

where A1 = a10 (t , x) + a11 (t , x)y, A0 = a00 (t , x) + a01 (t , x)y + a02 (t , x)y2 and lim (a10 (t , x) + a01 (t , x)) = 0,

lim (a11 (t , x) + a02 (t , x)) = −1,

t →0

aij (t , x) and

a10 (t ,x)a11 (t ,x)−2a01 (t ,x) a211 (t ,x)−4a02 (t ,x)

Proof. Taking G = F2 +

A1 , 2

t →0

lim a00 (t , x) = 0.

t →0

are continuously differentiable functions in R2 . Then F2 (t , x, y) = f0 (t , x) + f1 (t , x)y.

the identity (7) becomes to

G2 = δ,

(8)

where δ = 14 A21 − A0 . Differentiating relation (8) respect to t implies

 2G

1 2

DA1 +

m 

 di (t , x)G

i

= Dδ,

(9)

i=0

where

∂ A1 ∂ A1 + P (t , x, y); ∂t ∂x ∂δ ∂δ ∂δ Dδ = + P (t , x, y) + Q (t , x, y); ∂t ∂x ∂y  ∂ k Q (−t , x, y)  dk (t , x) = − (k = 1, 2, . . . , m);  A k! ∂ yk y=− 1 DA1 =

d0 (t , x) = −Q

2



−t , x , −

A1



2

.

Substituting (8) into (9), we obtain C0 + C1 G = 0, when m = 2k, C0 = −Dδ + 2(d1 δ + d3 δ 2 + · · · + d2k−1 δ k ),

C1 = DA1 + 2(d0 + d2 δ + · · · + d2k δ k ),

when m = 2k + 1, C0 = −Dδ + 2(d1 δ + d3 δ 2 + · · · + d2k+1 δ k+1 ),

C1 = DA1 + 2(d0 + d2 δ + · · · + d2k δ k ).

Now we only consider m is an odd number. When m is an even number, a similar discussion can come to the same conclusion. C Case 1. If C1 ̸= 0. It is easy to see that C0 and C1 are the polynomials with respect to variable y. Substituting G = − C0 1

into (8) we have C02 = C12 δ , which implies that C0 is divisible by C1 , thus, G(t , x, y) = g0 (t , x) + g1 (t , x)y. Therefore, F2 = f0 (t , x) + f1 (t , x)y. Case 2. If C1 ≡ 0, then C0 ≡ 0, i.e., Dδ = 2(d1 + d3 δ + · · · + d2k+1 δ k )δ.

(10)

As

δ=

A21 4

 − A0 =

where ϵ(t , x) =

a210 4

1 4

a211

− a02

y+

(a10 a11 −2a01 )2 . 4(a211 −4a02 ) a10 a11 −2a01 = − a2 −4a 02 11

− a00 −

In the Eq. (10), taking y



a10 a11 − 2a01 a211 − 4a02

2

+ ϵ(t , x),

By the conditions of the present theorem, we have limt →0 ϵ(t , x) = 0.

=: y∗ , we get

k  ∂ϵ(t , x) ∂ϵ(t , x) + P (t , x, y∗ ) = 2ϵ(t , x) d2i+1 ϵ i (t , x). ∂t ∂x i=0

Z. Zhou / Nonlinear Analysis: Real World Applications 17 (2014) 64–70

67

By the uniqueness of solution of the initial problem of the above partial differential equation, we have ϵ(t , x) ≡ 0. Thus A21

δ=

4

 − A0 =

1 4

a211 − a02

 y+

a10 a11 − 2a01

2

a211 − 4a02

,

and

 F 2 ( t , x) =

1 4

a211

 − a02 y +

a10 a11 − 2a01



a211 − 4a02

Summarizing, the above the proof is completed.

= f0 (t , x) + f1 (t , x)y. 

Similarly, we obtain the following conclusion. Theorem 2. Suppose that F = (x, F2 )T is the RF of system (3) and F2 satisfies F22 + A1 F2 + A0 = 0,

(11)

where A1 = A1 (t , x), A0 = a00 (t , x) + a01 (t , x)y + a02 (t , x)y2 and lim (A1 (t , x) + a01 (t , x)) = 0,

t →0

A1 (t , x), aij (t , x) and

a01 (t ,x) a02 (t ,x)

lim a02 (t , x) = −1,

lim a00 (t , x) = 0.

t →0

t →0

are continuously differentiable functions in R2 . Then F2 (t , x, y) = f0 (t , x) + f1 (t , x)y.

Remark 1. Clearly, Theorem 1 of papers [10,11] can be obtained by taking n = 2 (n = m = 2) in the present Theorem 2. And the proof of the present theorem improve the proof of the theorem in [10,11]. Here we use a simpler method to solve completely the problem on the structure of F2 (t , x, y) when F1 = x and n = 2. By the algebra fundamental theorem, we know that any polynomial one can write down as a product of linear and quadratic polynomials. Using Theorems 1 and 2, we get: Theorem 3. If F = (x, F2 (t , x, y)) is the RF of (3) and F2 (t , x, y) satisfies n 

Ai F2i =

i =0



(F22 + A1i (t , x, y)F2 + A0i (t , x, y))

0≤i≤r



(F2 + bj0 (t , x) + bj1 (t , x)y) = 0,

0≤j≤s

where A1i (t , x, y), A0i (t , x, y) are the same as A1 (t , x, y), A0 (t , x, y) satisfy the conditions of Theorem 1 or Theorem 2. Then F2 = f0 (t , x) + f1 (t , x)y. Now let us discuss the system (3). Suppose that in some deleted neighborhood of t = 0 and |t | being small enough, pn (t , x) is different from zero. Theorem 4. Suppose that F = (x, F2 )T is the RF of system (3) and Proof. Putting G = F2 +

An−1 , n

∂ An−1 ∂x

̸= 0. Then F2 = f0 (t , x) + f1 (t , x)y.

the relation (4) becomes

G n + δn−2 G n−2 + δn−3 G n−3 + · · · + δ1 G + δ0 = 0,

(12)

where

  An−1 δ0 = A0 + Φ t , x, − ; n

Φ (t , x, y) =

n 

δk =

 ∂ Φ (t , x, y)  k! ∂ yk y=− An−1

(k = 1, 2, . . . , n);

n

Ai yi .

i =1

Differentiating relation (12) respect to t implies m+n−3

 j =0

Bj G j = 0,

(13)

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Z. Zhou / Nonlinear Analysis: Real World Applications 17 (2014) 64–70

here Bj = Dδj + (j + 1)dˆ 0 δj+1 −

j  (n − i)δi dj+1−i (j = 0, 1, 2, . . . , n + m − 3), i=1

in which

∂δ0 ∂δ0 ∂δ0 + P (t , x, y) + Q (t , x, y); ∂t ∂x ∂y ∂δj ∂δj Dδj = + P (t , x, y) (j = 1, 2, .., n − 2); ∂t ∂x    An−1 ∂ k Q (−t , x, y)  d0 = −Q −t , x, − ; dk = −  A n k! ∂ yk y=− n−1 Dδ0 =

(k = 1, 2, . . . , m);

n

1

dˆ 0 = d0 +

n

DAn−1 ,

dj = 0, when j > m; δn−1 Computing (12), we get Gn = −

n −2 

δj G j =

j=0

G n +k =

n −1 

∂ An−1 ∂ An−1 + P (t , x, y); ∂t ∂x = 0, δn = 1; δj = 0, when j > n. DAn−1 =

n −1 

g0j G j ,

j=0

gk j G j ,

k = 1, 2, . . . , m − 3,

j=0

where gj+1 0 = gj n−1 g00 , g0 j = −δj ,

gj+1 i = gj n−1 g0 i + gj i−1 ,

j = 0, 1, 2, . . . , n − 2,

i = 1, 2, . . . , n − 1;

g0 n−1 = 0.

Substituting the above relations into (13) yields C0 + C1 G + · · · + Cn−1 G n−1 = 0,

(14)

where m−3

Cj = Bj +



Bn+i gij ,

j = 0, 1, 2, . . . , n − 1.

(15)

i =0

By the assumptions of the present theorem implies Cn−1 ̸= 0, and G n−1 = −

1 Cn−1

(C0 + C1 G + · · · + Cn−2 G n−2 ).

Substituting it into (12), we get D0 + D1 G + · · · + Dn−2 G n−2 = 0,

(16)

where Dj = δj Cn2−1 + Cn−2 Cj − Cn−1 Cj−1 ,

j = 0, 1, 2, . . . , n − 2, C−1 = 0.

If Dj ≡ 0 (j = 0, 1, 2, . . . , n − 2), then the polynomial G n + G=

Cn−2 . Cn−1

n−2 j =0

δj G j is divisible by polynomial

n−1 j=0

Cj G j , it yields that

Substituting it into (12) implies the polynomial Cn−2 is divisible by polynomial Cn−1 , thus, G is a polynomial with

respect to variable y, substituting it into (12) again, we obtain G = g0 (t , x) + g1 (t , x)y. Therefor, F2 = f0 (t , x) + f1 (t , x)y. Hence, in this case the present theorem is true. If there is a Dk ̸= 0 and Dj ≡ 0, j > k, 1 ≤ k ≤ n − 2. Similar as above we can substitute (16) into (14) implies a k − 1-degrees equation with respect to variable G. And so on, finally, we will come to E0 + E1 G + E2 G 2 + · · · + Er G r = 0,

r ≤ k − 1,

where Ej (j = 0, 1, 2, . . . , r ) are the polynomials with respect to variable y. Either Ej (j = 0, 1, 2, .., r ) are equal to zero, or at least exists one of Ej (1 ≤ j ≤ r ) is not equal to zero. In the first case, similar discussions as above, we can get the conclusion of present theorem. In the second case, we can get S0 + S1 G = 0 and S0 , S1 are the polynomial with respect to variable y and S1 ̸= 0. By using (12) yields G = g0 (t , x) + g1 (t , x)y. Thus, F2 (t , x, y) = f0 (t , x) + f1 (t , x)y. Summarizing the above the proof of the present is finished 

Z. Zhou / Nonlinear Analysis: Real World Applications 17 (2014) 64–70

69

In the same way we can get: Theorem 5. Suppose that F = (x, F2 )T is the RF of system (3) and there exists a Ck ̸= 0, 1 ≤ k ≤ n − 1. Then F2 = f0 (t , x) + f1 (t , x)y. Theorem 6. If n is an odd number and the conditions of Lemma 2 are satisfied, Cj ≡ 0 (j = 0, 1, 2, . . . , n − 1) and limt →0

∂ k Φ (t ,x,y) ∂ yk

= 0 (k = 1, 2, 3, . . . , n − 2). Then F2 = f0 (t , x) + f1 (t , x)y.

Proof. Since n is an odd number, then there is a real function y = y∗ (t , x) such that δ0 (t , x, y∗ ) = 0. In the identifies Cj = 0 (j = 1, 2, . . . , n − 2) taking y = y∗ we get Dδj = −(j + 1)dˆ 0 δj+1 +

j 

m−3

(n − i)δi dj+1−i −



i=1

Bn+i gji ,

j = 1, 2, . . . , n − 2.

i=1

In view of the hypothesis, we see limt →0 δj = 0, j = 1, 2, . . . , n − 2. By the uniqueness of solution of the initial problem of the above partial differential equation, we get δj ≡ 0 (j = 1, 2, . . . , n − 2). Thus,

δ0 =

pn (t , x)



pn (−t , x)

y+

An−1 (−t , x)

n

n

+ εˆ ,

(17)

where

εˆ =

p0 (t , x) + p0 (−t , x)

+

pn (−t , x)

pn (t , x) pn (−t , x)

    An−1 (−t , x) An−1 (t , x) Φ −t , x , − + Φ −t , x, − , n

n

where Φ (t , x, y) = j =1 A j y . By Lemma 2, we have limt →0 εˆ = 0. A (−t , x) In the identity C0 = 0 taking y = − n−1 n =: yˆ , we obtain

n

j

∂ εˆ ∂ εˆ + P (t , x, yˆ ) = εˆ ∂t ∂x



m−3

nd1 +

 j =0

   Bn+j gj−1 n−1  

. y=ˆy

By the uniqueness of solution of the initial problem  of the above partial differential equation, we get εˆ = 0. Thus,

δ0 =

pn (t ,x) pn (−t ,x)

(y +

An−1 (−t ,x) n . n

) From (12), we get G =

The proof is finished.

n

An−1 (−t ,x) n

( t ,x ) − ppnn(− (y + t ,x )

), and F = G −

An−1 (t ,x) n

= f0 (t , x) + f1 (t , x)y.



Theorem 7. Suppose that lim

t →0

pn (t , x) pn (−t , x)

pk (t , x) +

n 

= −1,

lim

pn−1 (t , x) + pn−1 (−t , x) pn (−t , x)

t →0 j−k k f1

pj (−t , x)Cjk f0

= 0,

= 0,

k = 0, 1, 2, . . . , n − 2;

j=k

′ f0t′ + f0x p0 (t , x) + f1 q0 (t , x) +

m 

j

qj (−t , x)f0 = 0;

j=0

′ ′ f1t′ + f0x p1 (t , x) + f1x p0 (t , x) + f1 q1 (t , x) +

m 

j −1

qj (−t , x)Cj1 f0

f1 = 0;

j =1

′ ′ f0x pk (t , x) + f1x pk−1 (t , x) + f1 qk (t , x) +

m 

j −k k f1

qj (−t , x)Cjk f0

= 0,

k = 2, 3, . . . , m,

j =k

pj (t , x) = 0, when j > n. where

 f1 =

n



pn (t , x) pn (−t , x)

,

f0 = −

1 n−1



pn−1 (−t , x) pn (−t , x)

+

pn−1 (t , x) pn (−t , x)

f11−n



,

Cjm =

j! m! (m − j)!

.

Then F = (x, f0 + f1 y)T is the RF of system (3), and the first component of solution of (3) is an even function with respect to time t.

70

Z. Zhou / Nonlinear Analysis: Real World Applications 17 (2014) 64–70

Proof. It is not difficult to verify that under the above conditions the function F = (x, f0 + f1 y)T is the solution of the Cauchy problem (2). So F = (x, f0 + f1 y)T is the RF of system (3). As F1 = x, then x(t ) = x(−t ), i.e., the first component of solution of (3) is an even function with respect to time t.  Theorem 8. Let all the conditions of Theorem 7 be satisfied and pi (t + 2ω, x) = pi (t , x) (i = 0, 1, 2, . . . , n), qj (t + 2ω, x) = qj (t , x) (j = 0, 1, 2, . . . , m). Then all the solutions of (3) defined on [−ω, ω] are 2ω-periodic. Proof. By Theorem 7, F = (x, f0 + f1 y)T is the RF of system (3). In view of pi (t , x) and qj (t , x) are 2ω-periodic with respect to t, it yields f0 (t , x) and f1 (t , x) are 2ω-periodic in t. On the other hand, as F2 (−t , x, F2 (t , x, y)) = y, it implies f1 (t , x)f1 (−t , x) = 1 and f0 (−t , x) + f0 (t , x)f1 (−t , x) = 0. From these identities follows F (−ω, x, y) ≡ (x, y). Thus, the Poincaré mapping [1,2] of the periodic system (3) can be expressed as T (x, y) = F (−ω, x, y) ≡ (x, y). By the properties of RF [1], the conclusion of the present theorem is true. The proof is complete.  Example. Differential system

 n n   ′ i   x = r ( t , x ) y − r1 i (−t , x)(ye2s + ses (x2 + 1))i , 1 i    i=0 i =0   n   1 2s s 2 −s ′ ( 2cye + e c ( 1 + s )( x + 1 )) − 2xse r1 i (t , x)yi + y =−  1 + e2s  i = 0   m m      e−2s r2 i (t , x)yi − r2 i (−t , x)(ye2s + ses (x2 + 1))i i =0

i=0

has RF F = (x, ye + se (x + 1))T , where s := sin t , c := cos t , ri j (t , x) are arbitrary continuously differentiable functions in R2 . Besides this, if r1 i (t + 2π , x) = r1 i (t , x) (i = 0, 1, 2, . . . , n), r2 j (t + 2π , x) = r2 j (t , x) (j = 0, 1, 2, . . . , m). By Theorem 8, all the solutions of the considered system defined on [−π , π ] are 2π -periodic. 2s

s

2

Remark 2. Consider 3-dimensional polynomial system

 m  dx   aijk xi yj z k ,  = −y + x  dt   i+j+k=0    m  dy  aijk xi yj z k , =x+y dt   i+j+k=0    n   dz    = bijk xi yj z k ,  dt

(18)

i+j+k=1

where aijk , bijk are constants. Taking x = r cos θ , y = r sin θ , system (18) becomes

 n n    i j k i+j  dz = b cos θ sin θ z r = pi (θ , z )r i ,  ijk   dθ i+j+k=1 i=0 m −1 m    dr  i j k i+j   = r a cos θ sin θ z r = qi (θ , z )r i . ijk  dθ i+j+k=0

(19)

i=1

Obviously, system (19) is a 2π -periodic polynomial system. If all the conditions of Theorem 8 are satisfied, then all the solutions of system (18) are 2π -periodic. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11]

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