# The Integrability and Existence of Periodic Solutions on a First-Order Nonlinear Differential Equation with a Polynomial Nonlinear Term

Authors: Ni Hua; Tian Li-Xin
Journal: Mathematical Problems in Engineering
 (2012)

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## Abstract

This paper deals with a first-order differential equation with a polynomial nonlinear term. The integrability and existence of periodic solutions of the equation are obtained, and the stability of periodic solutions of the equation is derived.


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## Body

## 1. Introduction

Consider the following first-order nonlinear differential equation:(1.1)dxdt=∑k=0nak(t)xn-k(n∈N,n≥2),
when n=2, (1.1) becomes Ricatti’s equation, when n=3, (1.1) becomes the following nonlinear Abel type differential equation:(1.2)dxdt=a(t)x3+b(t)x2+c(t)x+d(t).
The nonlinear Abel type differential equation plays an important role in many physical and technical applications [1–9]. The mathematical properties of (1.2) have been intensively investigated in the mathematical and physical literature. Matsuno [10] analyzed a two-dimensional dynamical system associated with Abel nonlinear equation. Strobel and Reid [11], Reid and Strobel [12] have obtained superposition rules (prescriptions for combining a finite number of known particular solutions in such a way to obtain the general solution to a (system of) differential equation(s) without operation of integration) for the Abel type equation, involving four or two particular solutions. Mak et al. [13], Mak and Harko [14] have presented a solution-generating technique for Abel type ordinary differential equation, both suppose that y=y1(x) is a particular solution of (1.2), by means of the transformations methods, and present an alternative method of generating the general solution of (1.2) from a particular one.Zheltukhin and Trzetrzelewski [15] developed the geometric approach to study the dynamics of U(1)-invariant membranes. The approach reveals an important role of the Abel nonlinear differential equation of the first type with variable coefficients depending on time and one of the membrane extendedness parameters. The general solution of the Abel equation was constructed.However, little work was done about the integrability and periodicity of (1.1). In this paper, we discuss the integrability and the periodic solutions of (1.1); the sufficient conditions which guarantee the integrability and the existence of the periodic solutions for (1.1) are obtained, and the stability of the periodic solutions of (1.1) is discussed. To the best of authors’ knowledge, this is the first paper considering the three periodic solutions of (1.1), some new results are obtained.The present paper is organized as follows. In Section2, we give three lemmas to be used later. In Section 3, the integrability of (1.1) is derived. In Section 4, the existence and stability of the periodic solutions of (1.1) are obtained. In Section 5, we conclude our results.


## 2. Preliminary Lemmas

For the sake of convenience, suppose thatf is a continuous ω-periodic function defined on R, we denote fM=supt∈[0,ω]f(t) and fL=inft∈[0,ω]f(t); m(f(t))=(1/ω)∫0ωf(t)dt.Consider the following:(2.1)dxdt=x[a(t)+b(t)xn-1],n∈N, n≥2, a(t), b(t) are continuous functions defined on R.Lemma 2.1.
The domainR+={x∣x>0} is positive invariant with respect to (2.1).Proof.
By (2.1), it follows that
(2.2)x(t)=x(t0)exp⁡{∫t0t[a(s)+b(s)xn-1(s)]ds},
the assertion is valid for all x(t0)>0,t≥t0. The proof is completed.Lemma 2.2.
The domainR-={x∣x<0} is negative invariant with respect to (2.1).Lemma 2.3.
Consider the following:(2.3)dxdt=a(t)x+b(t),a(t), b(t) are continuous ω-periodic functions, if m(a(t))≠0, then (2.3) exists a unique ω-periodic solution η(t), and η(t) can be written as follows:
(2.4)η(t)=∫tt+ωb(s)e∫s-ωta(τ)dτds1-e∫0ωa(τ)dτ,m(a(t))≠0
or
(2.5)η(t)={∫-∞te∫sta(τ)dτb(s)ds,m(a(t))<0,-∫t+∞e∫sta(τ)dτb(s)ds,m(a(t))>0.Proof.
The proof of Lemma2.3 is a few, such as that of the papers [16, 17] and others, But in [16], the author only proved the case of the almost periodic equation; in [17], the authors just proved that Lotka-volterra equation has a unique globally attractive periodic solution, but they did not give the very two expressions of the periodic solution as above. In order to make the following proof clear, here we give our proof.From (2.3), it is easy for us to get the unique solution x(t) with the initial value x(t0)=x0 which can be written as follows:(2.6)x(t)=e∫t0ta(τ)dτx0+∫t0tb(s)e∫sta(τ)dτds,
thus we have(2.7)x(t+ω)=e∫t0t+ωa(τ)dτx0+∫t0t+ωb(s)e∫st+ωa(τ)dτds.
Let t=t0, by (2.6) and (2.7), if x(t) is an ω-periodic function, then it follows that x(t0)=x(t0+ω), thus we can get(2.8)x0=∫t0t0+ωb(s)e∫st0+ωa(τ)dτds1-e∫0ωa(τ)dτ.
Substitute (2.8) into (2.6), it follows that(2.9)x(t)=e∫t0ta(τ)dτ∫t0t0+ωb(s)e∫st0+ωa(τ)dτds1-e∫0ωa(τ)dτ+∫t0tb(s)e∫sta(τ)dτds=∫t0t0+wb(s)e∫s-ωta(τ)dτds+∫t0tb(s)e∫sta(τ)dτds-e∫0ωa(τ)dτ∫t0tb(s)e∫sta(τ)dτds1-e∫0ωa(τ)dτ=∫t0t0+wb(s)e∫s-ωta(τ)dτds+∫t0+ωt+ωb(s)e∫s-ωta(τ)dτds-∫t0tb(s)e∫s-ωta(τ)dτds1-e∫0ωa(τ)dτ=∫tt+ωb(s)e∫s-ωta(τ)dτds1-e∫0ωa(τ)dτ.Whenm(a(t))<0, it follows that(2.10)x(t)=∫tt+ωb(s)e∫s-ωta(τ)dτds1-e∫0ωa(τ)dτ=∫-∞t+ωb(s)e∫s-ωta(τ)dτds-∫-∞tb(s)e∫s-ωta(τ)dτds1-e∫0ωa(τ)dτ=∫-∞tb(s)e∫sta(τ)dτds-∫-∞tb(s)e∫s-ωta(τ)dτds1-e∫0ωa(τ)dτ=∫-∞tb(s)e∫sta(τ)dτds-e∫0ωa(τ)dτ∫-∞tb(s)e∫sta(τ)dτds1-e∫0ωa(τ)dτ=∫-∞tb(s)e∫sta(τ)dτds,thus we have(2.11)x(t+ω)=∫-∞t+ωb(s)e∫st+ωa(τ)dτds=∫-∞tb(s)e∫sta(τ)dτds=x(t).Hence,x(t) is a unique ω-periodic solution of (2.3); we rewrite it as follows:(2.12)η(t)=∫-∞tb(s)e∫sta(τ)dτds.Similarly, we can prove that whenm(a(t))>0, (2.3) has a unique ω-periodic solution as follows:(2.13)η(t)=-∫t+∞b(s)e∫sta(τ)dτds.Remark 2.4.
Under the conditions of Lemma2.3, all the infinite integrals above are convergent.


## 3. A Sufficient Condition of Integrability of the First-Order Nonlinear Differential Equation

In this section, we discuss the integrability of (1.1)Theorem 3.1.
Consider (1.1), where a0(t), an(t) are continuous functions defined on certain interval I, if there is a constant γ such that the following conditions hold: (H1)
an-1(t)=(-1)n-1(n-1)a0(t)(∫γtan(s)ds)n-1,(H2)
ak(t)=(-1)kCnka0(t)(∫γtan(s)ds)k, k=1,2,…,n-2,
then
(i)
the general solution of (1.1) can be written as follows:(3.1)x(t)=±(e∫an-1(t)dt(C-(n-1)∫a0(t)e-∫an-1(t)dtdt))-1/(n-1)+∫γtan(s)ds,
where C is an integration constant which makes the number of root greater than zero if n is an odd number.(ii)
Equation (1.1) has a particular solution(3.2)η(t)=∫γtan(s)ds.Proof.
(i) Since the conditions (H1), (H2) hold, (1.1) can be written as follows:
(3.3)d(x-∫γtan(s)ds)dt=(x-∫γtan(s)ds)×[(-1)na0(t)(∫rtan(s)ds)n-1+a0(t)(x-∫γtan(s)ds)n-1].
Lety(t)=x-∫γtan(s)ds, (3.3) can be changed as follows:
(3.4)dydt=y[(-1)na0(t)(∫rtan(s)ds)n-1+a0(t)yn-1].
This is the Bernoulli-type equation, set u(t)=y1-n(t), the Bernoulli-type equation (3.4) becomes
(3.5)dudt=(-1)n-1(n-1)a0(t)(∫rtan(s)ds)n-1u-(n-1)a0(t)=an-1(t)u-(n-1)a0(t),
according to the formula of general solution of linear differential equation, we can get that the general solution of (3.5) is given by
(3.6)u(t)=e∫an-1(t)dt(C-(n-1)∫a0(t)e-∫an-1(t)dtdt).
By the variable transformation of u(t)=y1-n(t), it follows that
(3.7)y(t)=±(e∫an-1(t)dt(C-(n-1)∫a0(t)e-∫an-1(t)dtdt))-1/(n-1),
since y(t)=x-∫γtan(s)ds, we can get
(3.8)x(t)=±(e∫an-1(t)dt(C-(n-1)∫a0(t)e-∫an-1(t)dtdt))-1/(n-1)+∫γtan(s)ds.
Remark 3.2.
Ifn is an even number, take “+” in the front of above-corresponding formulas, if n is an odd number, take “±”.
(ii) By (1.1), we set f(t,x)=∑k=0nak(t)xn-k, thus f(t,x) is continuous on I×R, fx(t,x) is also continuous on I×R, hence (1.1) satisfies the principle of existence and uniqueness of differential equation, therefore, (1.1) has no singular solution.
From the variable transformation ofu(t)=y1-n(t), it requires y≠0, but from (3.4), we know y=0 is indeed a solution of (3.4). Since y=x-∫γtan(s)ds, substitute η(t)=∫γtan(s)ds into (1.1), it just satisfies (1.1), so it is a particular solution of (1.1).
This is the end of the proof of Theorem3.1.Corollary 3.3.
Consider Abel type differential equation (1.2), a(t), d(t) are continuous functions defined on certain interval I,  if there is a constant γ such that the following conditions hold:(H1)
b(t)=-3a(t)∫γtd(s)ds,(H2)
c(t)=2a(t)(∫γtd(s)ds)2,then the general solution of (1.2) can be written as follows:
(3.9)x(t)=±(e∫c(t)dt(C-2∫a(t)e-∫c(t)dtdt))-1/2+∫γtd(s)ds,
and ∫γtd(s)ds is a particular solution of (1.2).Corollary 3.4.
Consider the following Ricatti’s equation:(3.10)dxdt=a(t)x2+b(t)x+c(t),a(t), c(t) are continuous functions defined on certain interval I, if there is a constant γ such that the following condition holds:(H1)
b(t)=-a(t)∫γtc(s)ds,then the general solution of (3.10) can be written as follows:
(3.11)x(t)=(e∫b(t)dt(C-∫a(t)e-∫b(t)dtdt))-1+∫γtc(s)ds,
and η(t)=∫γtc(s)ds is a particular solution of (3.10).


## 4. The Existence and Stability of Periodic Solutions of the First-Order Nonlinear Differential Equation

Define(4.1)Ω1={x∣x>∫γt0an(s)ds},Ω2={x∣x>inf⁡t∈[0,ω]∫γtan(s)ds},Ω3={x∣x<∫γt0an(s)ds},Ω4={x∣x<sup⁡t∈[0,ω]∫γtan(s)ds},wherean(t), γ, ω are the same function and numbers as the following Theorem 4.1, and t0 is any given initial time of (1.1).Theorem 4.1.
Consider (1.1), n is an odd number, and n>2, a0(t), an(t) are continuous ω-periodic functions defined on R, if there is a constant γ such that the following conditions hold:(H1)
an-1(t)=(n-1)a0(t)(∫γtan(s)ds)n-1,(H2)
ak(t)=(-1)kCnka0(t)(∫γtan(s)ds)k, k=1,2,…,n-2,(H3)
m(an(t))=0, an(t)≢0,(H4)
a0(t)<0,
then
(i)
equation (1.1) has three ω-periodic solutions η(t), ϕ(t), χ(t), and they can be written as follows:(4.2)η(t)=∫γtan(s)ds,ϕ(t)=(-(n-1)∫-∞ta0(s)e∫stan-1(τ)dτds)-1/(n-1)+∫γtan(s)ds,χ(t)=-(-(n-1)∫-∞ta0(s)e∫stan-1(τ)dτds)-1/(n-1)+∫γtan(s)ds;(ii)
if given any initial valuex(t0)∈Ω1, then the periodic solution ϕ(t) of (1.1) is globally attractive in Ω2;(iii)
if given any initial valuex(t0)∈Ω3, then the periodic solution χ(t) of (1.1) is globally attractive in Ω4.Proof.
(i) Since the conditions (H1), (H2) hold, (1.1) can be written as follows:
(4.3)d(x-∫γtan(s)ds)dt=(x-∫γtan(s)ds)×[-a0(t)(∫rtan(s)ds)n-1+a0(t)(x-∫γtan(s)ds)n-1].
Let y(t)=x-∫γtan(s)ds, (4.3) can be changed as follows:
(4.4)dydt=y[-a0(t)(∫rtan(s)ds)n-1+a0(t)yn-1].
This is the Bernoulli-type equation, set u(t)=y1-n(t), the Bernoulli-type equation (4.4) becomes
(4.5)dudt=(n-1)a0(t)(∫rtan(s)ds)n-1u-(n-1)a0(t),dudt=an-1(t)u-(n-1)a0(t).
By (H3), (H4), it follows that
(4.6)∫γt+ωan(s)ds=∫γtan(s)ds+∫tt+ωan(s)ds=∫γtan(s)ds+∫0ωan(s)ds=∫γtan(s)ds,
thus ∫γtan(s)ds is an ω-periodic function, therefore, (n-1)a0(t)(∫γtan(s)ds)n-1 is also an ω-periodic function and
(4.7)an-1(t)=(n-1)a0(t)(∫γtan(s)ds)n-1≤0,
since a0(t)<0, an(t)≠0 (not identically equal to zero), thus we have
(4.8)m(an-1(t))<0,
according to Lemma 2.3, it follows that (4.5) has a unique ω-periodic solution as follows
(4.9)φ(t)=-(n-1)∫-∞ta0(s)e∫stan-1(τ)dτds,
from (4.9), it is easy to know that φ(t)>0, since φ(t) is periodic on R, it is bounded on R, so it is positive and bounded, since u(t)=y1-n(t), and by Lemmas 2.1 and 2.2, it follows that (4.4) has two ω-periodic solutions as follows:
(4.10)ψ(t)=(-(n-1)∫-∞ta0(s)e∫stan-1(τ)dτds)-1/(n-1)(ψ(t0)>0),ψ*(t)=-(-(n-1)∫-∞ta0(s)e∫stan-1(τ)dτds)-1/(n-1)(ψ*(t0)<0),
thus ψ(t) is also positive and bounded, and ψ*(t) is negative and bounded, by virtue of y(t)=x-∫γtan(s)ds, (1.1) has two ω-periodic solutions as follows:
(4.11)ϕ(t)=(-(n-1)∫-∞ta0(s)e∫stan-1(τ)dτds)-1/(n-1)+∫γtan(s)ds,χ(t)=-(-(n-1)∫-∞ta0(s)e∫stan-1(τ)dτds)-1/(n-1)+∫γtan(s)ds,
and ϕ(t0)∈Ω1, ϕ(t)∈Ω2, χ(t0)∈Ω3, χ(t)∈Ω4.
By (1.1), we set f(t,x)=∑k=0nak(t)xn-k, thus f(t,x) is continuous on R×R, and fx(t,x) is also continuous on R×R, hence (1.1) satisfies the principle of existence and uniqueness of differential equation, therefore, (1.1) has no singular solution.
From the variable transformation ofu(t)=y1-n(t), it requires y≠0, but from (4.4), we know y=0 is indeed a solution of (4.4), since y(t)=x-∫γtan(s)ds, hence η(t)=∫γtan(s)ds is a periodic solution of (1.1). Substitute η(t)=∫γtan(s)ds into (1.1), it just satisfies (1.1), so η(t)=∫γtan(s)ds is indeed a periodic particular solution of (1.1), therefore, we have proved that (1.1) has three ω-periodic solutions η(t), ϕ(t), and χ(t).
(ii) According to the theories of linear differential equation, we know that the unique solutionu(t) of (4.5) with positive initial value u(t0)=u0 is given by
(4.12)u(t)=u0e∫t0tan-1(τ)dτ-(n-1)∫t0ta0(s)e∫stan-1(τ)dτds.
Since
(4.13)an-1(t)=(n-1)a0(t)(∫γtan(τ)dτ)n-1,
thus an-1(t) is an ω-periodic function on R, and by (4.7), we know an-1(t)≤0, thus (an-1)L<0.
Following we prove that there is a positive numberT>t0, such that
(4.14)(n-1)a0M(an-1)L(1-ε)≤u(t)≤u0-2ωa0L1-∫0ωan-1(τ)dτ,
as t>T, here ε is any small positive number and ε<1.
From (4.12), given any positive number u0>0, by the condition a0(t)<0 of Theorem 4.1, it follows that u(t)>0; following we first prove that u(t) has upper bound on [t0,+∞)(4.15)u(t)=u0e∫t0tan-1(τ)dτ-(n-1)∫t0ta0(s)e∫stan-1(τ)dτds≤u0-(n-1)∫-∞ta0(s)e∫stan-1(τ)dτds.
According to Lemma 2.3, it follows that
(4.16)u(t)≤u0-(n-1)∫-∞ta0(s)e∫stan-1(τ)dτds=u0-(n-1)∫tt+ωa0(s)e∫s-ωtan-1(τ)dτds1-e∫0ωan-1(τ)dτ=u0-ω(n-1)a0(ξ)e∫ξ-ωtan-1(τ)dτ1-e∫0ωan-1(τ)dτ,(t<ξ<t+ω)=u0-ω(n-1)a0(ξ)e∫ξtan-1(τ)dτe∫0ωan-1(τ)dτ1-e∫0ωan-1(τ)dτ≤u0-ω(n-1)a0(ξ)e∫0ω|an-1(τ)|dτe∫0ωan-1(τ)dτ1-e∫0ωan-1(τ)dτ=u0-ω(n-1)a0(ξ)e∫0ω(|an-1(τ)|+an-1(τ))dτ1-e∫0ωan-1(τ)dτ=u0-ω(n-1)a0(ξ)1-e∫0ωan-1(τ)dτ≤u0-ω(n-1)a0L1-e∫0ωan-1(τ)dτ.
So u(t) has upper bound on [t0,+∞).
Secondly, we prove thatu(t)≥(n-1)(a0M/(an-1)L)(1-ε) as t>T(4.17)u(t)=u0e∫t0tan-1(τ)dτ-(n-1)∫t0ta0(s)e∫stan-1(τ)dτds≥-(n-1)∫t0ta0(s)e∫stan-1(τ)dτds≥-(n-1)∫t0ta0Me(an-1)L(t-s)ds=-(n-1)(-a0M(an-1)L)(1-e(an-1)L(t-t0))=(n-1)a0M(an-1)L(1-e(an-1)L(t-t0)).
Since limt→+∞e(an-1)L(t-t0)=0, for all ε>0(ε<1), there must be T>t0, when t>T, we have e(an-1)L(t-t0)<ε, hence when t>T, we can get
(4.18)u(t)≥(n-1)a0M(an-1)L(1-ε),
therefore, we have proved that there is a positive number T>t0, as t>T, it follows that
(4.19)(n-1)a0M(an-1)L(1-ε)≤u(t)≤u0-ω(n-1)a0L1-e∫0ωan-1(τ)dτ,(t>T),
under the transformation of u(t)=y1-n(t), y(t)=x(t)-∫γtan(s)ds, we can get the following.
If given the initial value(4.20)x(t0)=(u0)-1/(n-1)+∫γt0an(s)ds∈Ω1,
then the unique solution x(t) of (1.1) is given by
(4.21)x(t)=(u0e∫t0tan-1(τ)dτ-(n-1)∫t0ta0(s)e∫stan-1(τ)dτds)-1/(n-1)+∫γtan(s)ds,
if given the initial value
(4.22)x(t0)=-(u0)-1/(n-1)+∫γt0an(s)ds∈Ω3,
then the unique solution x*(t) of (1.1) is given by
(4.23)x*(t)=-(u0e∫t0tan-1(τ)dτ-(n-1)∫t0ta0(s)e∫stan-1(τ)dτds)-1/(n-1)+∫γtan(s)ds,
and it is easy to know x(t)∈Ω2 and x*(t)∈Ω4, respectively.
Define a Lyapunov function as follows:(4.24)V(t,u,φ)=(u-φ)2,
where u(t) is the unique solution with the positive initial value u(t0)=u0 of (4.5), and φ is the unique positive ω-periodic solution of (4.5), differentiating both sides of (4.24) along the solution of (4.5), we get
(4.25)dV(t,u,φ)dt=2(u-φ)(dudt-dφdt)=2an-1(t)(u-φ)2,
by (4.24), we have
(4.26)dV(t,u,φ)V(t,u,φ)=2an-1(t)dt,
thus
(4.27)V(t,u,φ)=V(t0,u(t0),φ(t0))e∫t0t2an-1(s)ds,
by (4.24), we can get
(4.28)(u-φ)2=(u(t0)-φ(t0))2e∫t0t2an-1(s)ds,
by (4.28), that it follows
(4.29)|u-φ|=|u(t0)-φ(t0)|e∫t0tan-1(s)ds,
since
(4.30)x(t)=u-1/(n-1)(t)+∫γtan(s)ds,ϕ(t)=φ-1/(n-1)(t)+∫γtan(s)ds,
it follows that
(4.31)u(t)=1(x(t)-∫γtan(s)ds)n-1,φ(t)=1(ϕ(t)-∫γtan(s)ds)n-1,
substitute (4.31) into (4.29), it follows that
(4.32)|1(x(t)-∫γtan(s)ds)n-1-1(ϕ(t)-∫γtan(s)ds)n-1|=|1(x(t0)-∫γt0an(s)ds)n-1-1(ϕ(t0)-∫γt0an(s)ds)n-1|e∫t0tan-1(s)ds.
According to mean value theorem, we can get
(4.33)|-(n-1)ξ-n(x(t)-ϕ(t))|=|-(n-1)ζ-n(x(t0)-ϕ(t0))|e∫t0tan-1(s)ds,
where
(4.34)x(t)-∫γtan(s)ds<ξ<ϕ(t)-∫γtan(s)ds
or
(4.35)ϕ(t)-∫γtan(s)ds<ξ<x(t)-∫γtan(s)ds,x(t0)-∫γt0an(s)ds<ζ<ϕ(t0)-∫γt0an(s)ds
or
(4.36)ϕ(t0)-∫γt0an(s)ds<ζ<x(t0)-∫γt0an(s)ds.
Since φ(t) is positive and bounded, and from (4.19), (4.31), we know that
(4.37)x(t)-∫γtan(s)ds,ϕ(t)-∫γtan(s)ds,x(t0)-∫γt0an(s)ds,ϕ(t0)-∫γt0an(s)ds
are positive and bounded as t>T, hence ξ, ζ are positive and bounded as t>T, so we have
(4.38)|x(t)-ϕ(t)|=(ξζ)n|x(t0)-ϕ(t0)|e∫t0tan-1(s)ds,
by (4.8), it follows that
(4.39)∫t0tan-1(s)ds⟶-∞,
as t→+∞, and by (4.38), we can get that the ω-periodic solution ϕ(t) of (1.1) is globally attractive in Ω2.
(iii) The proof of the periodic solutionχ(t) of (1.1) being globally attractive in Ω4 is similar to that of the periodic solution ϕ(t) of (1.1) being globally attractive in Ω2, so we omit it here.
This is the end of the proof of Theorem4.1.Remark 4.2.
In Theorem4.1, since the transformation of u=y1-n, it follows u>0, thus we suppose the initial value u(t0)>0.Theorem 4.3.
Consider (1.1), n is an even number, and n≥2, a0(t), an(t) are continuous ω-periodic functions defined on R, if there is a constant γ such that the following conditions hold:(H1)
an-1(t)=(1-n)a0(t)(∫γtan(s)ds)n-1,(H2)
ak(t)=(-1)kCnka0(t)(∫γtan(s)ds)k, k=1,2,…,n-2,(H3)
m(an(t))=0, an(t)≢0,(H4)
a0(t)<0, ∫γtan(s)ds≤0,
then
(i)
equation (1.1) has two ω-periodic solutions η(t), ϕ(t), and they can be written as follows:
(4.40)η(t)=∫γtan(s)ds,ϕ(t)=(-(n-1)∫-∞ta0(s)e∫stan-1(τ)dτds)-1/(n-1)+∫γtan(s)ds.(ii)
If given any initial valuex(t0)∈Ω1={x∣x>∫γt0an(s)ds}, then the periodic solution ϕ(t) of (1.1) is globally attractive in Ω2={x∣x>inft∈[0,ω]∫γtan(s)ds}.Proof.
The proof of Theorem4.3 is similar to that of Theorem 4.1, so we omit it here.Theorem 4.4.
Consider (1.1), n is an even number, and n≥2, a0(t), an(t) are continuous ω-periodic functions defined on R, if there is a constant γ such that the following conditions hold:(H1)
an-1(t)=(1-n)a0(t)(∫γtan(s)ds)n-1,(H2)
ak(t)=(-1)kCnka0(t)(∫γtan(s)ds)k, k=1,2,…,n-2,(H3)
m(an(t))=0, an(t)≢0,(H4)
a0(t)>0, ∫γtan(s)ds≥0,
then
(i)
equation (1.1) has two ω-periodic solutions η(t), ϕ(t), and they can be written as follows:(4.41)η(t)=∫γtan(s)ds,ϕ(t)=(-(n-1)∫-∞ta0(s)e∫stan-1(τ)dτds)-1/(n-1)+∫γtan(s)ds.(ii)
If given any initial valuex(t0)∈Ω1={x∣x<∫γt0an(s)ds}, then the periodic solution ϕ(t) of (1.1) is globally attractive in Ω2={x∣x<supt∈[0,ω]∫γtan(s)ds}.Proof.
The proof of Theorem4.4 is similar to that of Theorem 4.1, so we omit it here.
From Theorem4.1, if n=3, it is easy for us to draw the following corollary.
Define(4.42)Ω1={x∣x>∫γt0d(s)ds},Ω2={x∣x>inf⁡t∈[0,ω]∫γtd(s)ds},Ω3={x∣x<∫γt0d(s)ds},Ω4={x∣x<sup⁡t∈[0,ω]∫γtd(s)ds},
where d(t), γ, ω are the same function and numbers as the following Corollary 4.5, and t0 is any given initial time of (1.2).Corollary 4.5.
Consider Abel type differential equation (1.2), a(t), d(t) are ω-periodic continuous functions defined on R, if there is a constant γ such that the following conditions hold:(H1)
b(t)=-3a(t)∫γtd(s)ds;(H2)
c(t)=2a(t)(∫γtd(s)ds)2;(H3)
m(d(t))=0, d(t)≢0;(H4)
a(t)<0;
then
(i)
equation (1.2) has three ω-periodic solutions η(t), ϕ(t), χ(t), and they can be written as follows:
(4.43)η(t)=∫γtd(s)ds,ϕ(t)=(-2∫-∞ta(s)e∫stc(τ)dτds)-1/2+∫γtd(s)ds,χ(t)=-(-2∫-∞ta(s)e∫stc(τ)dτds)-1/2+∫γtd(s)ds.(ii)
If given any initial valuex(t0)∈Ω1, then the periodic solution ϕ(t) of (1.2) is globally attractive in Ω2.(iii)
If given any initial valuex(t0)∈Ω3, then the periodic solution χ(t) of (1.2) is globally attractive in Ω4.From Theorems4.3 and 4.4, if n=2, it is easy for us to draw the following corollaries.Corollary 4.6.
Consider Ricatti’s equation (3.10), a(t), c(t) are ω-periodic continuous functions defined on R, if there is a constant γ such that the following conditions hold:(H1)
b(t)=-a(t)∫γtc(s)ds;(H2)
m(c(t))=0, c(t)≢0;(H3)
a(t)<0, ∫γtc(s)ds≤0;
then
(i)
equation (3.10) has two ω-periodic solutions η(t), ϕ(t), and they can be written as follows:(4.44)η(t)=∫γtc(s)ds,ϕ(t)=(-∫-∞ta(s)e∫stb(τ)dτds)-1+∫γtc(s)ds.(ii)
If given any initial valuex(t0)∈Ω1={x∣x>∫γt0c(s)ds}, then the periodic solution ϕ(t) of (3.10) is globally attractive in Ω2={x∣x>inft∈[0,ω]∫γtc(s)ds}.Corollary 4.7.
Consider Ricatti’s equation (3.10) a(t), c(t) are ω-periodic continuous functions defined on R, if there is a constant γ such that the following conditions hold:(H1)
b(t)=-a(t)∫γtc(s)ds;(H2)
m(c(t))=0, c(t)≢0;(H3)
a(t)>0, ∫γtc(s)ds≥0;
then,
(i)
equation (3.10) has two ω-periodic solutions η(t), ϕ(t), and they can be written as follows:(4.45)η(t)=∫γtc(s)ds,ϕ(t)=(-∫-∞ta(s)e∫stb(τ)dτds)-1+∫γtc(s)ds.(ii)
If given any initial valuex(t0)∈Ω1={x∣x<∫γt0c(s)ds}, then the periodic solution ϕ(t) of (3.10) is globally attractive in Ω2={x∣x<supt∈[0,ω]∫γtc(s)ds}.Theorem 4.8.
Consider (1.1), n is an odd number, and n>2, a0(t), an(t) are continuous ω-periodic functions defined on R, if there is a constant γ such that the following conditions hold:(H1)
an-1(t)=(n-1)a0(t)(∫γtan(s)ds)n-1;(H2)
ak(t)=(-1)kCnka0(t)(∫γtan(s)ds)k, k=1,2,…,n-2;(H3)
m(an(t))=0, an(t)≢0;(H4)
a0(t)>0;
then,
(i)
equation (1.1) has three ω-periodic solutions η(t), ϕ(t), χ(t), and they can be written as follows:(4.46)η(t)=∫γtan(s)ds,ϕ(t)=((n-1)∫t+∞a0(s)e∫stan-1(τ)dτds)-1/(n-1)+∫γtan(s)ds,χ(t)=-((n-1)∫t+∞a0(s)e∫stan-1(τ)dτds)-1/(n-1)+∫γtan(s)ds.(ii)
if given any initial valuex(t0)∈Ω1∪Ω2, then the periodic solution η(t) of (1.1) is globally attractive in Ω3∪Ω4, where(4.47)Ω1={x∣∫γt0an(s)ds≤x<(φ(t0))-1/n-1+∫γt0an(s)ds},Ω2={x∣∫γt0an(s)ds≥x>-(φ(t0))-1/n-1+∫γt0an(s)ds},Ω3={x∣inf⁡t∈[0,ω]∫γtan(s)ds≤x<sup⁡t∈[0,ω]((φ(t))-1/(n-1)+∫γtan(s)ds)},Ω4={x∣sup⁡t∈[0,ω]∫γtan(s)ds≥x>inf⁡t∈[0,ω](-(φ(t))-1/(n-1)+∫γtan(s)ds)},φ(t)=(n-1)∫t+∞a0(s)e∫stan-1(τ)dτds.Remark 4.9.
Whereφ(t) above is the same function as that of the following formula (4.54), and t0 is any given initial time of (1.1).Proof.
(i) Since the conditions (H1), (H2) hold, (1.1) can be written as follows:
(4.48)d(x-∫γtan(s)ds)dt=(x-∫γtan(s)ds)×[-a0(t)(∫rtan(s)ds)n-1+a0(t)(x-∫γtan(s)ds)n-1].
Lety(t)=x-∫γtan(s)ds, (4.48) can be changed as follows:
(4.49)dydt=y[-a0(t)(∫rtan(s)ds)n-1+a0(t)yn-1].
This is the Bernoulli-type equation, set u(t)=y1-n(t), the Bernoulli-type equation (4.49) becomes
(4.50)dudt=(n-1)a0(t)(∫rtan(s)ds)n-1u-(n-1)a0(t),dudt=an-1(t)u-(n-1)a0(t).
By (H3), (H4), it follows that
(4.51)∫γt+ωan(s)ds=∫γtan(s)ds+∫tt+ωan(s)ds=∫γtan(s)ds+∫0ωan(s)ds=∫γtan(s)ds,
thus ∫γtan(s)ds is an ω-periodic function, therefore, (n-1)a0(t)(∫γtan(s)ds)n-1 is also an ω-periodic function and
(4.52)an-1(t)=(n-1)a0(t)(∫γtan(s)ds)n-1≥0,
since a0(t)>0, an(t)≠0 (not identically equal to zero), thus we have
(4.53)m(an-1(t))>0,
according to Lemma 2.3, it follows that (4.50) has a unique ω-periodic solution as follows:
(4.54)φ(t)=(n-1)∫t+∞a0(s)e∫stan-1(τ)dτds,
from (4.54), it is easy to know that φ(t)>0, since φ(t) is periodic on R, it is bounded on R, so it is positive and bounded, since u(t)=y1-n(t), and by Lemmas 2.1 and 2.2, it follows that (4.49) has two ω-periodic solutions as follows:
(4.55)ψ(t)=((n-1)∫t+∞a0(s)e∫stan-1(τ)dτds)-1/(n-1),(ψ(t0)>0),ψ*(t)=-((n-1)∫t+∞a0(s)e∫stan-1(τ)dτds)-1/(n-1),(ψ*(t0)<0),
thus ψ(t) is also positive and bounded, and ψ*(t) is negative and bounded, by virtue of y(t)=x-∫γtan(s)ds, (1.1) has two ω-periodic solutions as follows:
(4.56)ϕ(t)=((n-1)∫t+∞a0(s)e∫stan-1(τ)dτds)-1/(n-1)+∫γtan(s)ds,χ(t)=-((n-1)∫t+∞a0(s)e∫stan-1(τ)dτds)-1/(n-1)+∫γtan(s)ds.
By (1.1), we set f(t,x)=∑k=0nak(t)xn-k, thus f(t,x) is continuous on R×R, and fx(t,x) is also continuous on R×R, hence (1.1) satisfies the principle of existence and uniqueness of differential equation, therefore, (1.1) has no singular solution.
From the variable transformation ofu(t)=y1-n(t), it requires y≠0, but from (4.49), we know y=0 is indeed a solution of (4.49), since y(t)=x-∫γtan(s)ds, hence η(t)=∫γtan(s)ds is a periodic solution of (1.1). Substitute η(t)=∫γtan(s)ds into (1.1), it just satisfies (1.1), so η(t)=∫γtan(s)ds is indeed a periodic particular solution of (1.1), therefore, we have proved that (1.1) has three ω-periodic solutions η(t), ϕ(t) and χ(t).
(ii) According to the theories of linear differential equation, we know that the unique solutionu(t) of (4.50) with the positive initial value u(t0)=u0 is given by
(4.57)u(t)=u0e∫t0tan-1(τ)dτ-(n-1)∫t0ta0(s)e∫stan-1(τ)dτds=e∫t0tan-1(τ)dτ[u0-φ(t0)]+φ(t).
If given the initial valueu0≥φ(t0), then u(t)≥φ(t)>0.
Remark 4.10.
Since the transformation ofu(t)=y1-n(t), it requires u(t)>0, in addition,
(4.58)e∫t0tan-1(τ)dτ=e(n-1)∫t0ta0(τ)(∫γτan(θ)dθ)n-1dτ⟶+∞
as t→+∞, that is to say, it requires u0≥φ(t0).
Under the transformation ofu(t)=y1-n(t), y(t)=x(t)-∫γtan(s)ds, we can get the following.
If given the initial value(4.59)x(t0)=(u0)-1/(n-1)+∫γt0an(s)ds<φ(t0)-1/(n-1)+∫γt0an(s)ds∈Ω1,then the unique solution x(t) of (1.1) is given by
(4.60)x(t)=(u0e∫t0tan-1(τ)dτ-(n-1)∫t0ta0(s)e∫stan-1(τ)dτds)-1/(n-1)+∫γtan(s)ds,
and it is easy for us to know that x(t)∈Ω3.
If given the initial value(4.61)x(t0)=-(u0)-1/(n-1)+∫γt0an(s)ds>-(φ(t0))-1/(n-1)+∫γt0an(s)ds∈Ω2;then the unique solution x(t) of (1.1) is given by
(4.62)x(t)=-(u0e∫t0tan-1(τ)dτ-(n-1)∫t0ta0(s)e∫stan-1(τ)dτds)-1/(n-1)+∫γtan(s)ds
it is easy for us to know that x(t)∈Ω4.
Therefore, we can draw the following conclusion.
If given the initial valuex(t0)∈Ω1∪Ω2, then x(t)∈Ω3∪Ω4.
Next, we prove that the periodic solutionη(t) of (1.1) is globally attractive in Ω3∪Ω4 if given the initial value x(t0)∈Ω1∪Ω2.
(4.63)|x(t)-η(t)|=|±(u0e∫t0tan-1(τ)dτ-(n-1)∫t0ta0(s)e∫stan-1(τ)dτds)-1/(n-1)|=(u0e∫t0tan-1(τ)dτ-(n-1)∫t0ta0(s)e∫stan-1(τ)dτds)-1/(n-1)=(e∫t0tan-1(τ)dτ[u0-φ(t0)]+φ(t))-1/(n-1)=|x(t0)-η(t0)||x(t0)-η(t0)|n-1e∫t0tan-1(τ)dτ[u0-φ(t0)]+φ(t)=|x(t0)-η(t0)|n-1(1/u0)[e∫t0tan-1(τ)dτ[u0-φ(t0)]+φ(t)],
by (4.53), it follows that
(4.64)∫t0tan-1(s)ds⟶+∞,
as t→+∞, from (4.63), we can get that the ω-periodic solution η(t) of (1.1) is globally attractive in Ω3∪Ω4.
This is the end of the proof of Theorem4.8.Theorem 4.11.
Consider (1.1), n is an even number, and n≥2, a0(t), an(t) are continuous ω-periodic functions defined on R, if there is a constant γ such that the following conditions hold:(H1)
an-1(t)=(1-n)a0(t)(∫γtan(s)ds)n-1;(H2)
ak(t)=(-1)kCnka0(t)(∫γtan(s)ds)k, k=1,2,…,n-2;(H3)
m(an(t))=0, an(t)≢0;(H4)
a0(t)<0, ∫γtan(s)ds≥0;
then
(i)
equation (1.1) has two ω-periodic solutions η(t), ϕ(t), and they can be written as follows:(4.65)η(t)=∫γtan(s)ds,ϕ(t)=((n-1)∫t+∞a0(s)e∫stan-1(τ)dτds)-1/(n-1)+∫γtan(s)ds;(ii)
if given any initial valuex(t0)∈Ω1={x∣x>∫γt0an(s)ds}, then the periodic solution η(t) of (1.1) is globally attractive in Ω2={x∣x>inft∈[0,ω]∫γtan(s)ds}.Proof.
The Proof of Theorem4.11 is similar to that of Theorem 4.8, so we omit it here.Theorem 4.12.
Consider (1.1), n is an even number, and n≥2, a0(t), an(t) are continuous ω-periodic functions defined on R, if there is a constant γ such that the following conditions hold:
(H1)
an-1(t)=(1-n)a0(t)(∫γtan(s)ds)n-1;(H2)
ak(t)=(-1)kCnka0(t)(∫γtan(s)ds)k, k=1,2,…,n-2;(H3)
m(an(t))=0, an(t)≢0;(H4)
a0(t)>0, ∫γtan(s)ds≤0;
then
(i)
equation (1.1) has two ω-periodic solutions η(t),ϕ(t), and they can be written as follows:
(4.66)η(t)=∫γtan(s)ds,ϕ(t)=((n-1)∫t+∞a0(s)e∫stan-1(τ)dτds)-1/(n-1)+∫γtan(s)ds;(ii)
if given any initial valuex(t0)∈Ω1={x∣x<∫γt0an(s)ds}, then the periodic solution η(t) of (1.1) is globally attractive in Ω2={x∣x<supt∈[0,ω]∫γtan(s)ds}.Proof.
The Proof of Theorem4.12 is similar to that of Theorem 4.8, so we omit it here.From Theorem4.8, if n=3, it is easy for us to draw the following corollary.Corollary 4.13.
Consider Abel type differential equation (1.2), a(t), d(t) are ω-periodic continuous functions defined on R, if there is a constant γ such that the following conditions hold:(H1)
b(t)=-3a(t)∫γtd(s)ds;(H2)
c(t)=2a(t)(∫γtd(s)ds)2;(H3)
m(d(t))=0, d(t)≢0;(H4)
a(t)>0;
then
(i)
equation (1.2) has three ω-periodic solutions η(t), ϕ(t), χ(t), and they can be written as follows:
(4.67)η(t)=∫γtd(s)ds,ϕ(t)=(2∫t+∞a(s)e∫stc(τ)dτds)-1/2+∫γtd(s)ds,χ(t)=(2∫t+∞a(s)e∫stc(τ)dτds)-1/2+∫γtd(s)ds.(ii)
If given any initial valuex(t0)∈Ω1∪Ω2, then the periodic solution η(t) of (1.2) is globally attractive in Ω3∪Ω4, where(4.68)Ω1={x∣∫γt0d(s)ds≤x<(φ(t0))-1/2+∫γt0d(s)ds},Ω2={x∣∫γt0d(s)ds≥x>-(φ(t0))-1/2+∫γt0d(s)ds},Ω3={x∣inf⁡t∈[0,ω]∫γtd(s)ds≤x<sup⁡t∈[0,ω]((φ(t))-1/2+∫γtd(s)ds)},Ω4={x∣sup⁡t∈[0,ω]∫γtd(s)ds≥x>inf⁡t∈[0,ω](-(φ(t))-1/2+∫γtd(s)ds)},φ(t)=2∫t+∞a(s)e∫stc(τ)dτds.From Theorem4.11 and 4.6, if n=2, it is easy for us to draw the following corollaries.Corollary 4.14.
Consider Riccati’s equation (3.10), a(t), c(t) are ω-periodic continuous functions defined on R, if there is a constant γ such that the following conditions hold:(H1)
b(t)=-a(t)∫γtc(s)ds;(H2)
m(c(t))=0, c(t)≢0;(H3)
a(t)<0, ∫γtc(s)ds≥0;
then
(i)
equation (3.10) has two ω-periodic solutions η(t),ϕ(t), and they can be written as follows:
(4.69)η(t)=∫γtc(s)ds,ϕ(t)=(∫t+∞a(s)e∫stb(τ)dτds)-1+∫γtc(s)ds.(ii)
If given the initial valuex(t0)∈Ω1={x∣x>∫γt0c(s)ds}, then the periodic solution η(t) of (3.10) is globally attractive in Ω2={x∣x>inft∈[0,ω]∫γtc(s)ds}.Corollary 4.15.
Consider Riccati’s equation (3.10), a(t), c(t) are ω-periodic continuous functions defined on R, if there is a constant γ such that the following conditions hold:(H1)
b(t)=-a(t)∫γtc(s)ds;(H2)
m(c(t))=0, c(t)≢0;(H3)
a(t)>0, ∫γtc(s)ds≤0;
then
(i)
equation (3.10) has two ω-periodic solutions η(t), ϕ(t), and they can be written as follows:(4.70)η(t)=∫γtc(s)ds,ϕ(t)=(∫t+∞a(s)e∫stb(τ)dτds)-1+∫γtc(s)ds.(ii)
If given any initial valuex(t0)∈Ω1={x∣x<∫γt0c(s)ds}, then the periodic solution η(t) of (3.10) is globally attractive in Ω2={x∣x<supt∈[0,ω]∫γtc(s)ds}.


## 5. Concluding Remarks

Generally, just as Ricatti’s equation, (1.1) is nonintegrable, however in the present paper, we have got a sufficient condition which guarantees the integrability and have presented some exact periodic solutions of nonlinear equation if the coefficients of (1.1) satisfy certain conditions; moreover, we get that the periodic solutions are globally attractive in certain conditions. These conclusions will have certain applications.



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*Source: 703474-2012-02-27.xml*