1. Introduction As we know, it is very important to study the properties of the solution of differential system (1)x′=Xt,xfor both the theory and application of an ordinary differential equation.

If X(t+2ω,x)=X(t,x) (ω is a positive constant), we can use the Poincaré mapping introduced in [1] to study the behavior of the solutions of (1). But it is very difficult to find the Poincaré mapping for many systems which cannot be integrated in quadratures. In the 1980s, the Russian mathematician Mironenko first established the theory of reflective functions (RF) in [2, 3]. Since then a quite new method to study (1) has been found. In recent years, more and more experts and scholars have achieved many good results; refer to [4–9] in this direction.

The aim of this article is to use the theory of the generalized reflective function to study the behavior of the solutions of differential systems. The obtained results will provide a new theoretical basis and criterion to further explain the laws of the movement of objects.

In the present section, we first recall some basic notions and results of the generalized reflective function (GRF), which will be used throughout the rest of this article.

Now consider system (1) with a continuous differentiable right-hand side and with a general solution φ(t;t0,x0). For each such system, the GRF of system (1) is defined as F(t,x)∶=φ(α(t);t,x), (t,x)∈D⊂R×Rn, where α(t) is a continuous differentiable function such that α(α(t))=t, α(0)=0. Then for any solution x(t) of (1), we have F(t,x(t))=x(α(t)) and F(α(t),F(t,x))≡F(0,x)≡x. By the definition in [3], a continuous differentiable vector function F(t,x) on R×Rn is called GRF if and only if it is a solution of the Cauchy problem(2)Ftt,x+Fxt,xXt,x=α′tXαt,Ft,x,F0,x=x.Relation (2) is called a basic relation (BR).

Besides this, suppose that system (1) is 2ω-periodic with respect to the variable t, and F is its GRF; if there exists a number η on R such that α(η)=2ω+η, then T(x)=F(η,x)=φ(α(η);η,x) is the Poincaré mapping of (1) over the period η,η+2ω. So, for any solution x(t) of (1) defined on η,η+2ω, it will be 2ω-periodic if and only if F(η,x)=x which is called a basic lemma (BL).

Now, we consider the higher dimensional polynomial differential system(3)x′=a1+a2y+a3z≜At,x,y,zy′=b1+b2y+b3z+b4y2+b5yz+b6z2≜Bt,x,y,zz′=c1+c2y+c3z+c4y2+c5yz+c6z2≜Ct,x,y,z,where ai=ai(t,x), bj=bj(t,x), and cj=cj(t,x) (i=1,2,3; j=1,2,...,6) are continuously differentiable functions in R2, a22+a32≠0 (in some deleted neighborhood of t=0 and t being small enough but different from zero), and there exists a unique solution for the initial value problem of (3). And suppose that(4)Ft,x,y,z=F1t,x,y,z,F2t,x,y,z,F3t,x,y,zTis the GRF of (3).

In this article, we discuss the structure of Fi(t,x,y,z) (i=2,3) when F1(t,x,y,z)=f(t,x) and obtain the good results Fi(t,x,y,z)=fi1(t,x)+fi2(t,x)y+fi3(t,x)z (i=2,3) which are useful for the research of the existence of periodic solutions and establishing the sufficient conditions of system (3) with the form of GRF.

In the following, we denote(5)a-i=aiαt,x;b-j=bjαt,x;c-j=cjαt,x;α=αt;Fi=Fit,x,y,z(i=1,2,3; j=1,2,...,6). The notation “ai(t,x)≠0” means that, in some deleted neighborhood of t=0, ai(t,x) is small enough but different from zero.

2. Main Results Without loss of generality, we suppose that f(t,x)=x. Otherwise, we can take the transformation ξ=f(t,x), η=y, ζ=z.

Now, we consider system (3).

Lemma 1. For system (3), let F1=x, α′(0)=-1; then(6)ai0,x=0, i=1,2,3.

Proof. Using relation (2), we can get(7)At,x,y,z-α′tAα,x,F2,F3=0;

i.e.,(8)a1-α′a-1+a2y+a3z-α′a-2F2-α′a-3F3=0.

Putting t=0, we have a1(0,x)+a2(0,x)y+a3(0,x)z≡0, ∀x,y,z, which implies that relation (6) is valid.

In the following discussion, we always assume (6) holds without further mentioning.

Case I (<inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M63"><mml:msub><mml:mrow><mml:mi>a</mml:mi></mml:mrow><mml:mrow><mml:mn mathvariant="normal">3</mml:mn></mml:mrow></mml:msub><mml:mo>≠</mml:mo><mml:mn mathvariant="normal">0</mml:mn></mml:math></inline-formula>). From relation (8), we get(9)F3=l1+l2F2,where (10)l1=l11+l12y+l13z,l11=a1-α′a-1α′a-3,l12=a2α′a-3,l13=a3α′a-3,l2=-a-2a-3.Differentiating relation (9) with respect to t implies(11)∑i=02PiF2i=0,where(12)P0=p01+p02y+p03z+p04y2+p05yz+p06z2,p01=α′c-1-l2b-1+l11c-3-l2b-3+l112c-6-l2b-6-l11t′+l11x′a1+l12b1+l13c1,p02=α′l12c-3-l2b-3+2l11l12c-6-l2b-6-l12t′+l11x′a2+l12x′a1+l12b2+l13c2,p03=α′l13c-3-l2b-3+2l11l13c-6-l2b-6-l13t′+l11x′a3+l13x′a1+l12b3+l13c3,p04=α′l122c-6-l2b-6-l12x′a2+l12b4+l13c4,p05=2α′l12l13c-6-l2b-6-l12x′a3+l13x′a2+l12b5+l13c5,p06=α′l132c-6-l2b-6-l13x′a3+l12b6+l13c6,P1=p11+p12y+p13z,p11=α′c-2-l2b-2+l2c-3-l3b-3+l11c-5-l2b-5+2l2l11c-6-l2b-6-l2t′+l2x′a1,p12=α′l12c-5-l2b-5+2l2l12c-6-l2b-6-l2x′a2,p13=α′l13c-5-l2b-5+2l2l13c-6-l2b-6-l2x′a3,P2=α′c-4-l2b-4+l2c-5-l2b-5+l22c-6-l2b-6.

Lemma 2. For system (3), suppose P2≠0, F1=x and the limit limt→0p0i/P2 (i=1,2,...,6) exist; then(13)Limt→0p0jP2=0 j=1,3,6,limt→0p02+p11P2=0,limt→0p05+p13P2=0,limt→0p04+p12P2=-1.

Proof. Using relation (11), we have limt→0F22+limt→0P1/P2F2+limt→0P0/P2=0, i.e.(14)limt→0F22+limt→0p11+p12y+p13zP2F2+limt→0p01+p02y+p03z+p04y2+p05yz+p06z2P2=0,by F20,x,y,z=y, which implies that the results of Lemma 2 are true.

Theorem 3. For system (3), suppose that limt→0p12/P2+2>0 and all the conditions of Lemmas 1 and 2 are satisfied; then(15)Fit,x,y,z=fi0t,x+fi1t,xy+fi2t,xz i=2,3.

Proof. As P2≠0, by relation (11), we have(16)F22=-P1P2F2-P0P2.F23=-P1P2F22-P0P2F2=P12P22-P0P2F2+P0P1P22.Differentiating relation (11) with respect to t, we can obtain(17)DP0+DP1F2+α′P1Bα,x,F2,F3+DP2F22+2α′P2Bα,x,F2,F3F2=0;Substituting (9) into the above, we get(18)∑i=03QiF2i=0,where(19)Q0=DP0+α′μ1P1,Q1=DP1+α′μ2P1+2α′μ1P2,Q2=DP2+α′μ3P1+2α′μ2P2,Q3=2α′μ3P2;μ1=b-1+l1b-3+l12b-6,μ2=b-2+l2b-3+l1b-5+2l1l2b-6,μ3=b-4+l2b-5+l22b-6;DPi=∂Pi∂t+∂Pi∂xAt,x,y,z+∂Pi∂yBt,x,y,z+∂Pi∂zCt,x,y,z i=0,1,2.Substituting (16) into (18), we have (20)R0+R1F2=0,in which(21)R0=Q0-Q2P0P2+Q3P0P1P22,R1=Q1-Q2P1P2+Q3P12P22-P0P2.

(1) If R1≠0, we get F2=-R0/R1 by relation (20). Through the expression of Pi, Qj (i=0,1,2; j=0,1,2,3), we know that R1 is a quadratic polynomial with respect to y, z and R0 is a cubic polynomial with respect to y, z. Substituting F2=-R0/R1 into (11), we obtain P2R02=-R1P0R1-P1R0 which implies that R1∣R0 or R1∣P2 and F2=∑i+j=03f2ij(t,x)yizj. Substituting the relations into (11) and equating the coefficients of the same powers of y and z, we have f2ij(t,x)=0, i+j>1. Finally, we arrive at(22)Fit,x,y,z=fi0t,x+fi1t,xy+fi2t,xz i=2,3

(2) If R1=0, it follows R0=0 from (20). By simple computation, we obtain(23)DP0P2=-α′μ1P1P2+2α′μ2P0P2-α′μ3P0P1P22,(24)DP1P2=-2α′μ1+α′μ2P1P2-α′μ3P12P22-2P0P2

Let Δ=P12/P22-4P0/P2; we have(25)DΔ=DP12P22-4P0P2=2P1P2DP1P2-4DP0P2=2α′μ2-μ3P1P2Δ

by (23) and (24).

Since(26)Δ=P12P22-4P0P2=1P22p11+p12y+p13z2-4P2p01+p02y+p03z+p04y2+p05yz+p06z2=s1+s2y+s3z+s4y2+s5yz+s6z2=s4y+s52s4z+s22s42+wt,x,z,where(27)s1=1P22p112-4P2p01,s2=1P222p11p12-4P2p02,s3=1P222p11p13-4P2p03,s4=1P22p122-4P2p04,s5=1P222p12p13-4P2p05,s6=1P22p132-4P2p06;wt,x,z=w0+w1z+w2z2,w0=14s44s1s4-s22,w1=12s42s3s4-s2s5,w2=14s44s4s6-s52,

we can obtain limt→04s4s6-s52=0, limt→02s3s4-s2s5=0, and limt→04s1s4-s22=0 by taking advantage of Lemma 2. Thus limt→0w1t,x,z=0. It is easy to verify that(28)DΔt,x,y,zy=-s5/2s4z-s2/2s4=Δt+ΔxA+ΔyB+ΔzCy=-s5/2s4z-s2/2s4=wt+wxAt,x,-s52s4z-s22s4,z+wzCt,x,-s52s4z-s22s4,z.

Using identity (25), we get (29)wt+wxAt,x,-s52s4z-s22s4,z+wzCt,x,-s52s4z-s22s4,z=2α′wμ2-μ3P1P2y=-s5/2s4z-s2/2s4.

By the uniqueness of solutions of the initial problem of linear partial differential equations, we have w=0. Therefore(30)Δ=s4y+s52s4z+s22s42.

So, using relation (11), we get (31)F2=s24s4s4-P12P2+12s4y+s54s4s4z=f20t,x+f21t,xy+f22t,xz.Finally, we get F3=l1+l2F2=f30(t,x)+f31(t,x)y+f32(t,x)z by relation (9).

Summarizing the above, the proof is finished.

Theorem 4. For system (3), let F1(t,x,y,z)=x, P1≠0, P2=0, and ai(0,x)=0 (i=1,2,3); then(32)F2t,x,y,z=-P0P1=-p01+p02y+p03z+p04y2+p05yz+p06z2p11+p12y+p13z,F3=l1+l2F2.

Case II (<inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M132"><mml:msub><mml:mrow><mml:mi>a</mml:mi></mml:mrow><mml:mrow><mml:mn mathvariant="normal">3</mml:mn></mml:mrow></mml:msub><mml:mo>≡</mml:mo><mml:mn mathvariant="normal">0</mml:mn></mml:math></inline-formula>, <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M133"><mml:msub><mml:mrow><mml:mi>a</mml:mi></mml:mrow><mml:mrow><mml:mn mathvariant="normal">2</mml:mn></mml:mrow></mml:msub><mml:mo>≠</mml:mo><mml:mn mathvariant="normal">0</mml:mn></mml:math></inline-formula>). From relation (8), we have(33)F2=δ1+δ2y,where (34)δ1=a1-α′a-1α′a-2,δ2=a2α′a-2.Differentiating this identity with respect to t gives(35)∑i=02ViF3i=0,where(36)V0=v01+v02y+v03z+v04y2+v05yz+v06z2,v01=α′b-1+δ1b-2+δ12b-4-δ1t′+δ1x′a1+δ2b1v02=α′δ2b-2+2δ1δ2b-4-δ2t′+δ2x′a1+δ1x′a2+δ2b2,v03=-δ1x′a3-δ2b3,v04=α′δ22b-4-δ2x′a2-δ2b4,v05=-δ2x′a3-δ2b5,v06=-δ2b6.V1=v11+v12y,v11=α′b-3+δ1b-5,v12=α′δ2b-5,V2=α′b-6.

Likewise, we have the following conclusions.

Lemma 5. Suppose F1=x, a3≡0, a2≠0, b6≠0, and the limits limt→0v0i/b6 (i=1,2,...,5) and limt→0a1/α′a-2 exist; then(37)limt→0v0jα′b-6=0 j=1,2,4,limt→0v03+v11α′b-6=0,Limt→0v05+v12α′b-6=0,limt→0a1-α′a-1α′a-2=0,limt→0a2α′a-2=limt→0b6α′b-6=1.

Theorem 6. Suppose that all the conditions of Lemmas 1 and 5 are satisfied; then(38)F2=δ1t,x+δ2t,xy,F3=f30t,x+f31t,xy+f32t,xz.

Theorem 7. Suppose that F1=x, v112+v122≠0, b3=b6=0. ai(0,x)=0 (i=1,2); then(39)F2=δ1t,x+δ2t,xy,F3t,x,y,z=-V0V1=-v01+v02y+v03z+v04y2+v05yz+v06z2v11+v12y.

Theorem 8. Function F(t,x,y,z)=x,f21y+f22z,g21y+g22zT is the GRF of system (3), if the following conditions are satisfied:(40)a1-α′a-1=0,a2-α′a-2f21-α′a-3g21=0,a3-α′a-3f22-α′a-3g22=0f21f22b1+f22g22c1-α′b-1c-1=0,f21t′f22t′g21t′g22t′+f21x′f22x′g21x′g22x′a1+f21f22g21g22b2b3c2c3-α′b-2b-3c-2c-3f21f22g21g22=0,f21x′a2f22x′a3g21x′a2g22x′a3+f21f22g21g22b4b6c4c6-α′b-4b-6c-4c-6f212f222g212g222-α′f21g21b-5f22g22b-5f21g21c-5f22g22c-5=0,f21x′f22x′g21x′g22x′a3a2+f21f22g21g22b5c5-α′2b-4f21f22+b-5f21g22+f22g21+2b-6g21g222c-4f21f22c-5f21g22+f22g21+2c-6g21g22=0,f21f22g21g22f-21f-22g-21g-22=f210,xf220,xg210,xg220,x=1001.

Proof. By checkout of the BR, it can be proved that the function(41)F=x,f21y+f22z,g21y+g22zTis the GRF of system (3).

Theorem 9. Let the hypotheses of Theorem 8 be satisfied, system (3) is 2ω-periodic with respect to t, and there exists a number η in R such that α(η)=2ω+η; then all the solutions of system (3) defined on the interval η,η+2ω are 2ω-periodic if and only if f212ω+η,x=1, f222ω+η,x=0, g212ω+η,x=0, and g222ω+η,x=1.

Proof. By Theorem 8, the Poincaré mapping of system (3) is T(x,y,z)=F(αη,x,y,z)=x,f212ω+η,xy+f222ω+η,xz,g212ω+η,xy+g222ω+η,xzT.

According to the previous introduction, the solutions of system (3) are 2ω-periodic if and only if T(x,y,z)≡x,y,zT, which implies that the assertions of the present theorem hold. The proof is finished.

Example 10. The differential system(42)x′=1-esint1-x3sinty-x2zsinty′=12ycostx6 sin2 t+x3-1-12x2z coste-sint1+sint+x3 sin2 t+y23x2sint-x5 sin2 t+yze-sintx4 sin2 t-2xsintz′=12x4y costesint1-sint+x3 sin2 tx6 sin2 t+x3-1+12zcostt-x3-x6 sin2 t-y2esintx6 sin2 t-4x3sint-yze-sint3x2sint-x5 sin2 thas GRF F(t,x,y,z)=x,F2,F3T, in which(43)F2t,x,y,z=1+αtx3sintyesint-αtx2zsint,F3=1-αtx3sintze-sint+αtx4ysint, αt=-t.Since this system is a 2π-periodic system, and there exists a number η=-π in R such that α(η)=2π+η, then T(x)=F(-π,x)=x; by Theorem 9, all the considered solutions defined on -π,π are 2π-periodic.