We investigate the existence of the periodic solutions of a nonlinear integro-differential system with piecewise alternately advanced and retarded argument of generalized type, in short DEPCAG; that is, the argument is a general step function. We consider the critical case, when associated linear homogeneous system admits nontrivial periodic solutions. Criteria of existence of periodic solutions of such equations are obtained. In the process we use Green's function for periodic solutions and convert the given DEPCAG into an equivalent integral equation. Then we construct appropriate mappings and employ Krasnoselskii's fixed point theorem to show the existence of a periodic solution of this type of nonlinear differential equations. We also use the contraction mapping principle to show the existence of a unique periodic solution. Appropriate examples are given to show the feasibility of our results. Among the functional differential equations, Myshkis  proposediated in , in Wieniated in , the firiated in –21. Appliated in , 22–25. iated in , 26–34.ℤ, ℕ, ℝ, and ℂ be the set of all integers, natural, real, and complex numbers, respectively. Denote by |·| a norm in ℝn, n ∈ ℕ. Fix two real sequences ti, γi, i ∈ ℤ, such that ti < ti+1,  and ti ≤ γi ≤ ti+1 for all i ∈ ℤ, and  ti → ±∞ as i → ±∞.Let γ : ℝ → ℝ be a step function given by γ(t) = γi for t ∈ Ii =  corresponds to γi = ti = i ∈ ℤ, and γ(t) = 2[(t + 1)/2] corresponds to ti = 2i − 1, γi = 2i, i ∈ ℤ. The particular case of DEPCAG, when γi = ti, i ∈ ℤ, an only delayed situation, is considered by first time in Akhmet  the advanced intervals and Ii− =  × D × D1.Butris  investigA : ℝ → ℝn×n, f : ℝ × ℝn × ℝn → ℝn, and g : ℝ × ℝn × ℝn → ℝn are continuous in their respective arguments. In the analysis we use the idea of Green's function for periodic solutions and convert the nonlinear integro-differential systems with DEPCAG =A(tz′(t)=A(tγ(t) is discontinuous. Thus, in general, the right-hand side of the DEPCAG system , i ∈ ℤ. In other words, by a solution z(t) of the DEPCAG system (z′(t) exists at each point t ∈ ℝ, with the possible exception of the points ti ∈ ℝ, i ∈ ℤ, where a one-sided derivative exists, and the nonlinear integro-differential systems with DEPCAG (z(t) on each interval , i ∈ ℤ as well.In our paper we assume that the solutions of the nonlinear integro-differential systems with DEPCAG  are contG system  has discG system  we mean h DEPCAG  are satiω-periodicity of Green's function. The rest of the paper is organized as follows. In In this section we state and define Green's function for periodic solutions of the nonlinear integro-differential system with piecewise alternately advanced and retarded argument .I be the n × n identity matrix. Denote by Φ, Φ = I, t, s ∈ ℝ, the fundamental matrix of solutions of the homogeneous system  ∈ ℤ be the unique integer such that t ∈ Ii = , Green's function for (t) is a fundamental solution of (Suppose that the condition (tion for  is givenution of  and(19)Nω) implies the existence of the matrix D.We note that the condition , we first give the following lemma.To prove double Nω) holds. Let the matrix D be defined by  to study after the existence of periodic solutions.For Nω) holds. Then Green's function G is double ω-periodic; that is, G = G.Suppose that the condition z′ a function such that |y1 − y2| ≤ δ impliesLet where Lipschitz ConditionsLC) such that t, s ∈ ℝ and for any y1, y2 ∈ ℝn we haveThere exists a continuous function Lf) such thatFor Moreover, Lg) such thatFor  Moreover, Moreover, Invariance ConditionsMC) functions and positive constants For every Mf) and positive constants ρ1,  C1,  η2 for whichFor every where Mg) and positive constants ρ2,  C2 for whichFor every where ∫ττ+ω[κ1(s) + κ2(s)]ds ≤ C2, ω ∈ ℝ+, τ ∈ ℝ.where Periodic ConditionsP)(ω > 0 such thatThere exists (1)A(t), f, and g are periodic functions in t with a period ω for all t ≥ τ;(2)C = C for all t ≥ τ;(3)p ∈ ℤ+, for which the sequences {ti}i∈ℤ, {γi}i∈ℤ satisfy the  condition; that is,there exists ω, p) condition is a discrete relation, which moves the interval Ii into Ii+p. Then we have the following consequences.(i)τ ∈ ℝ, the interval  can be decomposed as follows:For any (ii)t ∈  × 𝒞ℛ × 𝒞ℛ and by the periodicity in t, the function g is uniformly continuous on ℝ × 𝒞ℛ × 𝒞ℛ. Thus, for any ϵ′ = (ϵ/cGω) > 0, there exists δ = δ(ϵ) > 0 such that z1, z2 ∈ 𝕊, ||z1 − z2|| ≤ δ implies |g, z1(γ(t))) − g, z2(γ(t)))| ≤ ϵ′ for t ∈ . Then ||ℬz1 − ℬz2|| ≤ ϵ. In fact, by the continuity of g,ℬ is proved.The function Step  2. We show that the image of ℬ is contained in a compact set.x, y ∈ 𝒞ℛ and s ∈ ; for the continuity of the function g, there exists M > 0 such that |g| ≤ M. Let φn ∈ 𝕊 where n is a positive integer; then we haveℬφn(t))′ shows thatA(t) is bounded on  and ℬφn(t), g, φn(γ(t)) are bounded on  × 𝕊 × 𝕊. Thus, the above expression yields ||(ℬφn)′|| ≤ L, for some positive constant L. Hence the sequence (ℬφn) is uniformly bounded and equicontinuous. Ascoli-Arzela's theorem implies that a subsequence (ℬφnk) of (ℬφn) converges uniformly to a continuous ω-periodic function. Thus ℬ is continuous and ℬ(𝕊) is a compact set.Let 𝒜 we obtain the following.In a similar way, for Nω) and (C) hold, 𝒜 is defined by , (P), (Lf), (LC), (Mg) hold. Let R be a positive constant satisfying the inequalityω-periodic solution in 𝕊.Suppose the hypotheses , (P), (Lg), (Mf), (MC), (C) hold. Let R be a positive constant satisfying the inequalityω-periodic solution in 𝕊.Suppose the hypothesis , (P), (Lf), (LC), (Lg) hold. Ifω-periodic solution.Suppose the hypotheses , (P), (Mf), (MC), (Mg), (C) hold. Let R be a positive constant satisfying the inequalityω-periodic solution in 𝕊.Suppose the hypotheses |/(|x| + |y|)) = 0, then the system (Nω), (P3), (Mg), A(t) and g are periodic functions in t with a period ω for all t ≥ τ and |g| ≤ c1(|x| + |y|) + ρ2, with 2c1ω < C2 and ρ2 constants for |x| + |y| ≤ R, R > 0.Considering the DEPCAG system 56)x′(tkii (see ) proved x′(tkii (Nω), (P1), (P3), (Lg), (Mf), and cG(C1 + L2)R + cG(β + ρ1)ω ≤ R with Considering the nonlinear system of differential equations with a general piecewise alternately advanced and retarded argument,x′(t)=A(tP1) is satisfied by ω = ω1, (P2) by ω = ω2, and (P3) by ω = ω3, if ωi/ωj is a rational number for all i, j = 1,2, 3; then (P1), (P2), and (P3) are simultaneously satisfied by ω = l.c.m.{ω1, ω2, ω3}, where l.c.m.{ω1, ω2, ω3} denotes the least common multiple between ω1, ω2 and ω3. In the general case it is possible that there exist five possible periods: ω1 for A, ω2 for f, ω3 for g, ω4 for C, and the sequences {ti}i∈ℤ, {γi}i∈ℤ satisfy the  condition. If ωi/ωj is a rational number for all i, j = 1,2, 3,4, 5, so, in this situation our results insure the existence of ω-periodic solution with ω = l.c.m.{ω1, ω2, ω3, ω4, ω5}. Therefore the above results insure the existence of ω-periodic solutions of the DEPCAG system (Suppose that (G system . These sPω), f, and g are periodic functions in t with a period ω1, ω2, and ω3, respectively, for all t ≥ τ;C = C, for all t ≥ τ;p ∈ ℤ+, for which the sequences {ti}i∈ℤ, {γi}i∈ℤ satisfy the  condition.There exists To determine criteria for the existence and uniqueness of subharmonic solutions of the DEPCAG system , from noAs immediate corollaries of Theorems Nω), (Pω), (Lf), (LC), (Mg) and (𝕊.Suppose the hypotheses (Mg) and  hold. ThMg) and  has at lNω), (Pω), (Lg), (Mf), (MC), (C) and (𝕊.Suppose the hypotheses ( (C) and  hold. Th (C) and  has at lNω), (Pω), (Lf), (LC), (Lg) and (Suppose the hypotheses (Lg) and  hold. ThLg) and  has a unNω), (Pω), (Mf), (MC), (Mg), (C) and (𝕊.Suppose the hypotheses ( (C) and  hold. Th (C) and  has at lWe will introduce appropriate examples in this section. These examples will show the feasibility of our theory.Mathematical modelling of real-life problems usually results in functional equations, like ordinary or partial differential equations, integral and integro-differential equations, and stochastic equations. Many mathematical formulations of physical phenomena contain integro-differential equations; these equations arise in many fields like fluid dynamics, biological models, and chemical kinetics. So, we first consider nonlinear integro-differential equations with a general piecewise constant argument mentioned in the introduction and obtain some new sufficient conditions for the existence of the periodic solutions of these systems.λ : ℝ2 →  satisfies (MC): |C| ≤ c2(R)λ|y| + h for |y| ≤ R, where g : ℝ × ℝn × ℝn → ℝn is continuous and C satisfies (C).The conditions of ɛ > 0, there exists δ > 0 such that |y1 − y2| ≤ δ impliesR such thata* = supt∈ℝ  |a(t)| and a∗ = inft∈ℝ|a(t)|.Indeed, for any π-periodic solution.Then, by Thus many examples can be constructed where our results can be applied.n×n and μ : ℝ → ℝn be two functions satisfyingti}i∈ℤ and {γi}i∈ℤ satisfy the  condition, A, B, Λ, κ, and μ are ω-periodic continuous functions, and z′(t) = A(t)z(t) satisfies (Nω);B is ω-periodic matrix function: |B(t)| ≤ b, g : ℝn × ℝn → ℝn is a continuous function, and |g| ≤ c1(R)|x| + c2(R)|y| + ϑ1 for |x|, |y| ≤ R, where (c1(R) + c1(R))ω ≤ C2;κ(x) − κ(y)| ≤ c3(R)|x − y|, where |Let Λ : ℝ → ℝR such thatω-periodic solution of the DEPCAG system (The hypotheses of G system .Lg) and (MC). On the other hand, the periodic situation of the DEPCAG system (Note that similar results can be obtained under (G system  and 57)Lg) and Let us consider another example for second-order differential equations with a general piecewise constant argument. In this case, we can show the existence and uniqueness of periodic solutions of the following nonlinear DEPCAG system.κ1, κ2 ∈ ℝ, γ(t) = γi if ti ≤ γi < ti+1, i ∈ ℤ, and the sequences {ti}i∈ℤ and {γi}i∈ℤ satisfy the  condition.Consider the following nonlinear DEPCAG system:z′(t) = Az(t) does not admit any nontrivial ω-periodic solution; that is, the condition (Nω) is satisfied. Let φ(t) = (φ1(t), φ2(t)), ψ(t) = (ψ1(t), ψ2(t)) and define 𝕊 = {z ∈ ℙω, ||z|| ≤ R}, where R ∈ ℝ+ satisfies the conditionφ, ψ ∈ 𝕊 we havef we haveπ/ω-periodic solution in 𝕊.We write the DEPCAG system  in the s


As an application, we prove the existence of some new singular directions for a meromorphic function  For the module distribution of a meromorphic function, there are three main theorems, that is, the Picard theorem, the Borel theorem, and the Nevanlinna second fundamental theorem. The fundamental concept in the angular distribution is singular direction. Singular direction is a concept of localizing value distribution in ℂ onto a sector S containing a single ray J : argz = θ emanating from the origin say. A Julia direction and a Borel direction are refinements of the Picard theorem and the Borel theorem, respectively. Corresponding to the Nevanlinna second fundamental theorem, a new singular direction, called T direction, was recently introduced in Zheng  values of a, for any positive number ɛ < π, we havex] implies the maximum integer number which does not exceed x and l is a positive integer.Theorem C was conjectured by Zheng . In 11]11], the f(z) be a meromorphic function and satisfy (f(z) concerning multiple values.Let  satisfy . Then thl + 2)/l] → 2 as l → ∞. Since Zheng [Note that the T direction of meromorphic function concerning multiple values is a refinement of the ordinary T direction since [ be a meromorphic function. A direction argz = θ is called a pseudo-T direction of f(z) if, for any number ɛ  (0 < ɛ < π/2), any system aj   of distinct values, and any system kj   such that kj is a positive integer or +∞ and thatj  (1 ≤ j ≤ q) such thatLet f(z) be a meromorphic function and satisfy (f(z).Let  satisfy . Then thq = 3, kj = ∞  , so Theorem C is a special case of (i) In Theorem C, kj = 1  , then q = 5; if kj = 2  , then q = 4; if kj = l ≥ 3  , then q = 3. So Theorem D is a special case of (ii) If In order to prove r) is a nonnegative increasing function in  and satisfiesE ⊂  such that ∫E(1/rlogr)dr < 1/3, one hasSuppose that Ψ(m  (m ≥ 4) be a fixed positive integer, θ0 = 0, θ1 = 2π/m,…, θm−1 = (m − 1)2π/m, θm = θ0. We put Δ(θi) = {z : |argz − θi| < 2π/m}, Δo(θi) = {z : |argz − θi| < π/m},  i = 0,1,…, m − 1; Δ(θm) = Δ(θ0), Δo(θm) = Δo(θ0). Then among these m angular domains {Δ(θi)}, there is at least an angular domain Δ(θi) such that for any system aj   of distinct values and any system kj   such that kj is a positive integer or +∞ and thatj  (1 ≤ j ≤ q) such thatθi)  (1 ≤ i ≤ m), there is a system aij   of distinct values and a system kij   such that kij is a positive integer or +∞ and thatj  (1 ≤ j ≤ q) we haveo(θi+1), Δ(θi+1), we haveT = ∑i=0m−1T, f) and adding two sides of the above expression from i = 0 to m − 1, we can obtaini, there exists a ri, the inequality T, f) > em3 would bold for r > ri, while the inequality (Eo(θi+1)Δ is the set of r which consists of a series of intervals and satisfiesr0 = max{ri, i = 1,2,…, m}; we have for any i, T, f) > em3; thenE = ∪i=0m−1Eo(θi+1)Δ. Therefore, there exists a sequence rn′ ∈  − E, such thatm, there is at least an angular domain Δ(θi) such that for any system aj   of distinct values and any system kj   such that kj is a positive integer or +∞ and thatj  (1 ≤ j ≤ q) such thatθm}, still denote it {θm}, we assume that θm → θ0. Put L : argz = θ0; then L is a pseudo-T direction that is stated in Firstly, we prove the following statement. Let equality  does notows from , 68), a, am  (m uch thatlimn→∞T(ɛ  (0 < ɛ < π/2), when m is sufficiently large, we have Δ(θm) ⊂ Ω. By . By , we have


Apart from establishing coefficient bounds for these classes, we establish inclusion relationships involving ( 𝒜(n) denote the class of functions of the formLet f(z) ∈ 𝒜(n) is star-like of complex order b, denoted as f(z) ∈ S*(b) if and only if it satisfiesA function f(z) ∈ 𝒜(n) is convex of complex order b, denoted as f(z) ∈ C(b) if and only if it satisfiesS*(b), and C(b) are introduced and studied by Nasr and Aouf [A function and Aouf .fj given byf1∗f2)(z), is given byFor the two functions f(z) of the form  ase(z) = z,Given the form  and δ ≥ Nn,δ of a function f is introduced and studied by Ruscheweyh [The concept of Neighborhood scheweyh  and extescheweyh .α1, α2,…, αq and β1, β2,…, βs , we define the generalized hypergeometric function qFs asℕ denotes the set of all positive integers and (x)k is the Pochhammer symbol defined in terms of gamma functions asgq,s defined by𝒟λ,μmf(z) : 𝒜 → 𝒜 is defined byμ ≤ λ ≤ 1 and m ∈ ℕ0. By the above definition, it is easy to note thatαi′s, βj′s, q, s, m, λ, and μ, we can deduce several operators such as Sălăgean derivative operator [For complex numbers ently in , an operoperator , Ruschewoperator , fractiooperator , Carlsons; z) asFsqk=Γ consisting of functions of the form f(z) are as given in (For 0 ≤ given in . α ≤ 1 we let B be the subclass of 𝒜(n), consisting of functions of the form f(z) are as given in  denote the Sălăgean derivative of order m given byBy specializing the parameters involved in the above definitions, we could arrive at several known as well as new classes. For example, by taking q = 2, s = 1, α1 = η − 1, (η > −1), α2 = 1, β1 = 1, one getsDmf(z) is the operator introduced and studied by Abubaker and Darus [Similarly, on taking nd Darus  given bym = 0 in the definition of the classes A and B, we could arrive at Sn and Rn which were introduced and studied by Murugusundaramoorthy et al. [Further, by taking y et al. .A and B and several inclusion relationships involving n-δ neighborhoods of analytic univalent functions with negative and missing coefficients belonging to these classes.In this paper, we establish the coefficient inequalities for the classes f ∈ 𝒜(n) as given in  satisfies (Let the functions of form  belong t view of  and 16)A. Then, ssertion . Conversatisfies , then, iSimilarly, we prove the following.f ∈ 𝒜(n) be as defined in  as given in (f(z) ∈ A1 if and only ifLet the function given in . Then, ff ∈ 𝒜(n) be as defined in  as given in  ∈ 𝒜(n) be as defined in .Iff ∈ A. Then, in view of .IfA1 ⊂ Nn,δ(e).IfB1 ⊂ Nn,δ(e).IfA2 ⊂ Nn,δ(e). IfB2 ⊂ Nn,δ(e). IfAσ and Bσ. Here, the classes Aσ consist of functions f ∈ 𝒜(n) for which there exists a function g(z) ∈ A such thatBσ, consisting of functions f(z) ∈ 𝒜(n) for which there exists another function g(z) ∈ B such thatIn this section, we determine the neighborhood properties of g ∈ A andNn,δ(g) ⊂ Aσ.If f ∈ Nn,δ(g), thenSuppose g ∈ A, we havef ∈ Aσ for σ given by  ⊂ Bσ. If g ∈ A1 andNn,δ(g) ⊂ A1σ. If g ∈ B1 andNn,δ(g) ⊂ B1σ. If g ∈ A2 andNn,δ(g) ⊂ A2σ. If g ∈ B2 andNn,δ(g) ⊂ B2σ.If


H space of Chen's expanded mixed element method. We study the new expanded mixed element method for convection-dominated Sobolev equation, prove the existence and uniqueness for finite element solution, and introduce a new expanded mixed projection. We derive the optimal a priori error estimates in L2-norm for the scalar unknown u and a priori error estimates in (L2)2-norm for its gradient λ and its flux σ. Moreover, we obtain the optimal a priori error estimates in H1-norm for the scalar unknown u. Finally, we obtained some numerical results to illustrate efficiency of the new method.We propose and analyze a new expanded mixed element method, whose gradient belongs to the simple square integrable space instead of the classical  Ω is a bounded convex polygonal domain in R2 with Lipschitz continuous boundary ∂Ω and J =  and f are given functions, coefficients a = a, b = b are smooth and bounded functions, coefficient c(x) = (c1(x), c2(x)) is a bounded vector, anda0, a1, b0, b1, and d1.We consider the following Sobolev equation with convection term:Sobolev equations are a class of important evolution partial differential equations and have a lot of applications in many physical problems, such as the porous theories concerned with percolation into rocks with cracks, the heat conduction problems in different mediums, and the transport problems of humidity in soil. In , the finIn 1994, Chen , 9 develH space. In this paper, we will study the new expanded mixed element method for convection-dominated Sobolev equation. We will give the proof for the existence and uniqueness of the solution for semidiscrete scheme and a new expanded mixed projection and the proof of its uniqueness. We will prove the optimal a priori error estimates in L2-norm for the scalar unknown u and a priori error estimates in (L2)2-norm for its gradient λ and its flux σ. In particular, we obtained the optimal a priori error estimates in H1-norm for the scalar unknown u. Finally, we obtained some numerical results to confirm our theoretical analysis.In 2011, we developed and analyzed a new expanded mixed finite element method  for elliC will denote a generic positive constant which is free of the space-time parameters h and Δt. At the same time, we denote the natural inner product in L2(Ω) or (L2(Ω))2 by  with the corresponding norm ||·||. The other notations and definitions of Sobolev spaces as in  ↦ H01 × (L2(Ω))2 × (L2(Ω))2 such thatuh, λh, σh}: ↦ Vh × Wh × Wh such thatVh, Wh) is chosen as the finite element pair P1 − P02 as follows:Vh, Wh) satisfies the so-called discrete Ladyzhenskaya-Babuska-Brezzi condition.In this paper, the new expanded mixed weak formulation of  is to fiuch that(a) )2 instead of the classical H space. Obviously, the regularity requirements on the solution λ = ∇u reduced.Compared to Chen's expanded mixed weak formulation , the graThere exists a unique discrete solution to semidiscrete scheme .ψi(x)}i=1n1 and {φj(x)}j=1n2 be bases of Vh and Wh, respectively. Letvh = ψm, wh = φl, and zh = φs, the problems  Au~′(Substitute (b) into I is a unit matrix.Substituting (b) and  into 1111(a), weThus, by the theory of differential equations , 13) ha ha13) hat); then  and 11)u~(t); thvalently  has a unIn order to analyze the convergence of the method, we first introduce the new expanded mixed elliptic projection associated with our equations.uh, λh, σh): → Vh × Wh × Wh be given by the following mixed relations:Let , wh=σh=σ~h in (b), and h=λ~h in (c) to obs to get||a1/2λ~h=∇u~h in  and usin∇u~h in (||u~h||≤Co obtain2a0∫0t||λstitute (∫0t||λ~h|∇u~h in (||u~h||≤C0.From (λ~h=0.Coombining (c), we g0.From )2 → Wh such thatThere exists a linear operator Πh of For the linear operator ΠPh : H01(Ω) → Vh such thatThere exists a linear operator From , 21, we h and Ph, we rewrite η, δ, and ρ asηm, δm, and ρm are known, it is enough to estimate ηe, δe, and ρe. Using Lemmas Using the definition of Πg Lemmas (a), wh = ρe in (b), and = δe in (c) to obt to obtainwh = ρe, we obtainIntegrate  with resntiating (b) and tvh = (ηe)t in (zh = (δe)t in (Choose ηe)t in (a) and zδe)t in (c) to obCombining  and 39)39), we hSubstitute  into 4040 to obtzh = ρe in (Choose = ρe in (c) and u= ρe in  to obtaiwh = ∇ηe in (Choose  ∇ηe in (b) and u ∇ηe in  and CaucCombining , 42), , , and LemC independent of h such thatThere is a constant η||, ||ηt||, and ||η||1, we consider the following auxiliary elliptic problem:To estimate terms ||problem:−∇·(a∇χ)=problem:5−∇·(a∇χ)From , we obtaη||L2, we can obtainUsing similar method to ||Combining  and 47)47), we oUsing –49), we, we49), For a priori error estimates, we decompose the errors asUsing -6) and  and 6) aWe will prove the error estimates for semidiscrete scheme.u||H2 + ||ut||H2 + ||λ||H1(Ω))2( + ||λt||H1(Ω))2(.Assume that vh = ς in (wh = ξ in (zh = θ in (Choose h = ς in (a), wh h = ξ in (b), and h = θ in (c) to obt to obtainwh = ∇ς in  to hantiating (b) and t= ςt in (a) and z= θt in (c) to obe obtain  − θt in ((a) ||ςt| θt in ((a) ||ςt|e obtain||ς||2+∫0h = ξ in (c) and ue obtain||ς||2+∫0 we have||ςt||2+| ∇ς in −e obtain||ς||2+∫0= ∇ς in  and In this section, we get the error estimates of fully discrete schemes. For the backward Euler procedure, let 0 = uhn, λhn, σhn) ∈ Vh × Wh × Wh,  such thatEquation  has the We will prove the theorem for the fully discrete error estimates.C independent of h and Δt such thatAssume that t = tn, we get the error equationsUsing , 63), a, a63), avh = ςn in , (b0 − CΔt) > 0. Then we use Cronwall's lemma to obtainwh = ∇ςn in (wh = ξn in (vh = ∂tςn in (zh = ∂tθn in (Choose = ςn in (a), wh = ξn in (b), and = θn in (c) to obe obtain12Δt(||ςnote that||R1n||2≤o obtain||ςJ||2+| ∇ςn in (b), we h||.From (b), we g|.From tςn in (a), and tθn in (c) to obAdding the three equations, we obtainUsing  and 74)74), we gzh = ξn in (Taking = ξn in (c), we gSubstitute  into 8080 to getCombining -24), , Ω = ×, J =  = 1 + 2x12 + x22, b = 1 + x12 + 2x22, and c(x) = T, and f is chosen so that the exact solution for the scalar unknown function isIn order to illustrate the efficiency of the new expanded mixed element method, we consider the following initial-boundary value problem of 2D Sobolev equation with the convection term:Ω into the triangulations of mesh size h uniformly, consider the piecewise linear space Vh with index k = 1 for the scalar unknown function u and the piecewise constant space Wh with index k = 0 for the gradient λ and the flux σ, use the backward Euler procedure with uniform time step length Δt = 1/M, and obtain some convergence results for ||u−uh||L∞(L2(Ω)), ||u−uh||L∞(H1(Ω)), ||λ−λh||L∞((L2(Ω))2), and ||σ−σh||L∞((L2(Ω))2) with u, λ, andσ in Figures uh, λh, and σh in Figures t = 1, and We divide the domain u in L2-norm, H1-norm, and the error estimates for λ and σ in (L2)2-norm, which confirm the theoretical results in this paper, in It is easy to see that we obtained the optimal error estimates for L2 for the scalar unknown u and the a priori error estimates in (L2)2-norm for its gradient λ and its flux σ are proved. Especially, the optimal a priori error estimates in H1-norm for the scalar unknown u are derived. Finally, some numerical results are provided to confirm our theoretical analysis.In this paper, a new expanded mixed finite element method is proposed and studied for Sobolev equation with convection-term. The proof for the existence and uniqueness of the solution for semidiscrete scheme, the new expanded mixed projection, and the proof of its uniqueness are given. The optimal a priori error estimates in u, λ, σ}: ↦ H01 × (L2(Ω))2 × (L2(Ω))2 such thatIn the near future, the new expanded mixed method will be applied to other evolution equations such as evolution integrodifferential equations, hyperbolic wave equations, and nonlinear evolution equations. And the new characteristic expanded mixed finite element method for Sobolev equation will be studied. The new expanded characteristic-mixed weak formulation is to find {In another article, we will give the error estimates for the new characteristic expanded mixed finite element method.


We construct some results on the regularity of solutions and the approximate controllability for neutral functional differential equations with unbounded principal operators in Hilbert spaces. In order to establish the controllability of the neutral equations, we first consider the existence and regularity of solutions of the neutral control system by using fractional power of operators and the local Lipschitz continuity of nonlinear term. Our purpose is to obtain the existence of solutions and the approximate controllability for neutral functional differential control systems without using many of the strong restrictions considered in the previous literature. Finally we give a simple example to which our main result can be applied. H and V be real Hilbert spaces such that V is a dense subspace in H. Let U be a Banach space of control variables. In this paper, we are concerned with the global existence of solution and the approximate controllability for the following abstract neutral functional differential system in a Hilbert space H:A is an operator associated with a sesquilinear form on V × V satisfying Gårding's inequality, f is a nonlinear mapping of  × V into H satisfying the local Lipschitz continuity, B : L2 → L2 and C : L2 → L2 are appropriate bounded linear mapping.Let This kind of equations arises in population dynamics, in heat conduction in material with memory and in control systems with hereditary feedback control governed by an integrodifferential law.Recently, the existence of solutions for mild solutions for neutral differential equations with state-dependence delay has been studied in the literature in , 2. As fL2-regularity remain valid under the above formulation of the neutral differential equation ∩W1,2↪C for some T > 0 by using fractional power of operators and Sadvoskii's fixed point theorem. Thereafter, by showing some variations of constant formula of solutions, we will obtain the global existence of solutions of  ↦ x is Lipschitz continuous, which is applicable for control problems and the optimal control problem of systems governed by nonlinear properties.In this paper, we propose a different approach from the earlier works  is the set of all m-times continuously differential functions on Ω.  C0m(Ω) will denote the subspace of Cm(Ω) consisting of these functions which have compact support in Ω.  Wm,p(Ω) is the set of all functions f = f(x) whose derivative Dαf up to degree m in distribution sense belong to Lp(Ω). As usual, the norm is then given byD0f = f. In particular, Wp0,(Ω) = Lp(Ω) with the norm ||·||p,Ω.  W0m,p(Ω) is the closure of C0∞(Ω) in Wm,p(Ω). p = 2 we denote Wm,2(Ω) = Hm(Ω) and W0p2,(Ω) = H0m(Ω).For  p′ = p/(p − 1),  1 < p < ∞. Wp−1,(Ω) stands for the dual space W0p′1,(Ω)* of W0p′1,(Ω) whose norm is denoted by ||·||p,∞−1,.Let Let X is a Banach space and 1 < p < ∞, Lp is the collection of all strongly measurable functions from  to X, the pth powers of norms are integrable, Cm will denote the set of all m-times continuously differentiable functions from  to X. X and Y are two Banach spaces, B is the collection of all bounded linear operators from X to Y, and B is simply written as B(X).If  X0 and X1, θ,p and θ denote the real and complex interpolation spaces between X0 and X1, respectively.For an interpolation couple of Banach spaces If A be a closed linear operator in a Banach space. Then D(A) denotes the domain of (A) and R(A) the range of A; ρ(A) denotes the resolvent set of A, σ(A) the spectrum of A, and σp(A) the point spectrum of A; x ∈ D(A) : Ax = 0} of A is denoted by Ker(A).the kernel or null space {Let H is identified with its dual space we may write V ⊂ H ⊂ V* densely and the corresponding injections are continuous. The norm on V, H, and V* will be denoted by ||·||, |·| and ||·||∗, respectively. The duality pairing between the element v1 of V* and the element v2 of V is denoted by , which is the ordinary inner product in H if v1, v2 ∈ H.If l ∈ V* we denote  by the value l(v) of l at v ∈ V. The norm of l as element of V* is given byV has a stronger topology than H and, for brevity, we may considerFor a be a bounded sesquilinear form defined in V × V and satisfying Gårding's inequality:A be the operator associated with this sesquilinear form:A is a bounded linear operator from V to V* by the Lax-Milgram theorem. The realization of A in H which is the restriction of A toA. From the following inequalitiesD(A), it follows that there exists a constant C0 > 0 such thatLet V,V*)1/2,2 denotes the real interpolation space between V and V*, o, o12), o1.3.3 of ).A generates an analytic semigroup S(t) in both H and V*. The following lemma is from Lemma 3.6.2 of  → ℝ with b(0) = 0 such thatLet f is a nonlinear mapping of  × V into H satisfying the following.(i)L1 : ℝ+ → ℝ such thatx||≤r and ||y||≤r.There exists a function (ii)t ∈  and x ∈ V.The inequalityLet us rewrite (Fx)(t) = f) for each x ∈ L2. Then there is a constant, denoted again by L1(r), such thatx ∈ L2 and x1,x2 ∈ Br(T) = {x ∈ L2:||x||L2 ≤ r}.From now on, we establish the following results on the solvability of .x0 ∈ H, k ∈ L2 for T > 0. Then, there exists a solution x of (Σ) ⊂ Σ, → L2 asJ has a fixed point in the space L2. By Assumptions B and F, it is easily seen that J is continuous from C in itself. LetL2. From . From  it folloy0|.By , 25), a, a(42)irtue of  in Lemma.Since  and Assuch thatT13β/2, let xt*(x) = (K1x)(t) for every x ∈ L2. Then xt* ∈ L2 and we have limn→∞xt*(xn) = xt*(x∞) since w − limn→∞xn = x∞. Hence,n→∞||K1xn||L2 = ||K1x∞||L2. Since L2 is a Hilbert space, the relation , a, aK1 + ow thatlimn→∞milar to  and 52)K1 + K2ondition  the contondition  exists uK2 is contractive. Thus, 𝒲1(T1).So by virtue of condition , K2 is x be a solution of (x0 ∈ H. Then we have that from (C3 such thatT].From now on we establish a variation of constant formula  of solutution of  and x0 hat from -52) it  it x be  account  there exequality . Since tequality  and 44)x be a soequality (63)|xM0b(T1) < 1, then the uniqueness of the solution of (𝒲1(T) is obtained.If ution of  in 𝒲1M0L < 1ch thatM0b(T1)+ondition  exists ux0,k) ∈ D(A) × L2. Then by the argument of the proof of x of  ∈ H × H × L2. Then the solution x of (x ∈ 𝒲1(T) ≡ L2∩W1,2 and the mappingLet Assumptions B and F be satisfied and  ∈ H × L2, then x belongs to 𝒲1(T). Let  ∈ H × H × L2 and let xi ∈ 𝒲1(T) be the solution of  in place of  for i = 1,2. Let xi   ∈ Σ. Then as seen in From ution of  with , let xk be the solution of  let xk be the solution of  to L2. Moreover, if one defines the operator ℱ byℱ is also a compact mapping from L2 to L2.Let one assume that the embedding ution of . Then thx0,k) ∈ H × L2, then in view of xk ∈ L2, we have f ∈ L2. Consequently, by (xk ∈ 𝒲1(T) ⊂ C. With aid of (a) of xk||L2 ≤ ||xk||𝒲1(T), we havek is bounded in L2, then so is xk in 𝒲1(T) ≡ L2∩W1,2. Since V is compactly embedded in H by assumption, the embeddingk ↦ xk is compact from L2. Moreover, we have that ℱ is a compact mapping ofL2 to L2.If  to L2, which is called a controller. For x ∈ L2 we setN :  if for every zT ∈ H and ϵ > 0 there exists a control function u ∈ L2 such that the solution x of  − zT | <ϵ; that is, R(T) in H.The system  is said ,f,u) of  satisfieL2 to H byp ∈ L2.We define the linear operator We need the following hypothesis.ɛ > 0 and p ∈ L2, there exists a u ∈ L2 such thatq1 is a constant independent of p.(i) For any f is a nonlinear mapping of  × H into H satisfying the following.(ii) L1 : ℝ+ → ℝ such thatx | ≤r and |y | ≤r.There exists a function AS(t − s)Bx = S(t − s)ABx for each x ∈ V. Therefore, the system  such thatFx)(t) = f) for t ≥ 0. Throughout this section, invoking (r1 such thatBy virtue of condition (i) of Assumption S we note that e system  is approinvoking , we can u1 and u2 be in L2. Then under the Assumption S, one has that, for 0 ≤ t ≤ T,M2 = eM(MNT+L1(r1)).Let x(t) = x and x2(t) = x. Then for 0 ≤ t ≤ T, we haveLet we havex1(t)−x2T].Under Assumption S, the system  is approɛ > 0 and zT ∈ D(A), there exists u ∈ L2 such thatzT ∈ D(A) there exists p ∈ L2 such thatp(s) = {(zT + (Bx)(T)) − sA(zT + (Bx)(T))} − S(s)(x0 + y0)/T. Let u1 ∈ L2 be arbitrary fixed. Since by Assumption S there exists u2 ∈ L2 such thatw2 ∈ L2 by Assumption S such thatt ≤ T. Therefore, in view of u3 = u2 − w2. We determine w3 such thatt ≤ T. Hence, we haveu* ∈ L2 such thatwn ∈ L2 by Assumption B, such thatun+1 = un − wn, we haveɛ > 0 there exists integer N such thatT] as N tends to infinity.We will show that ws that|zT+(Bx)ch that|S^(G(·xe system  is approA are λn = −n2 and ϕn(y) = (2/π)1/2sinny, respectively. Moreover, (a)ϕn : n ∈ N} is an orthogonal basis of H,{(b)S(t)x = ∑n=1∞en2tϕn,  ∀x ∈ H, t > 0,(c)α < 1; then the fractional power Aα : D(Aα) ⊂ H → H of A is given bylet 0 < LetA−1/2x = ∑n=1∞(1/n)ϕn and ||A−1/2|| = 1.In particular, g is a real valued function belonging to C2) which satisfies the following conditions:(i)g(0) = 0, g(r) ≥ 0 for r > 0,(ii)g′(r) ≤ c(r + 1) and |g′′(r)|≤c for r ≥ 0 and c > 0. If we presentf is a mapping from the whole V to H by Sobolev's imbedding theorem  = μ2r + η2r2/2 (μ and η are constants). In addition, we need to impose the following conditions ons see , 25)..g′(r) ≤ (iii)b is measurable andThe function (iv)2/∂z2)b is measurable, b = b, andThe function (∂(v)C : L2 → L2 is a bounded linear operator.We define B : L2 → L2 byB is bounded linear andB1 is bounded linear with B ∈ D(A1/2) and ||A1/2Bx|| = ||B1x||. Therefore from x of (T].Consider the following neutral differential control system:∂∂t[x(t,∂∂t[x(t,


F : Ω → ℝn is a continuous quasi-bounded functional which satisfies a local Lipschitzcondition with respect to the second argument and Ω is a subset in ℝ × Crn, Crn : = C,yt∈Crn, and yt(θ) = y(t + θ), θ ∈ . A monotone iterative method is proposed to provethe existence of a solution defined for t → ∞ with the graph coordinates lying betweengraph coordinates of two (lower and upper) auxiliary vector functions. This result is applied to scalar advanced linear differential equations. Criteria of existence of positive solutions are given and their asymptotic behavior is discussed.We study asymptotic behavior of solutions of general advanced differential systems  C, where a, b ∈ ℝ, a < b, be the Banach space of the continuous mappings from the interval  to ℝn equipped with the supremum normn. In the case of a = 0 and b = r > 0, we will denote this space as Crn; that is,Let σ ∈ ℝ, A ≥ 0, and y ∈ C, then, for each t ∈ , we define yt ∈ Crn by yt(θ) = y(t + θ), θ ∈ .If F : Ω → ℝn is a continuous quasi-bounded functional which satisfies a local Lipschitz condition with respect to the second argument and Ω is a subset in ℝ × Crn. We recall that the functional F is quasi-bounded if F is bounded on every set of the form  × CrLn ⊂ Ω, where t1 < t2, CrLn : = C, and L is a closed bounded subset of ℝn  with A > 0 if y ∈ C,  ∈ Ω for t ∈ ⊆𝕃 → 𝕃 is a compact monotone increasing operator on a real Banach space 𝕃 with normal order cone 𝕂. If u0 is a subsolution of  we denote the set of all component-wise nonnegative (positive) vectors v in ℝn; that is, v =  with vi ≥ 0 (vi > 0) for i = 1,…, n. For u, v ∈ ℝn, we say that u ≤ v if v − u ∈ ℝ≥0n, u ≪ v if v − u ∈ ℝ>0n, u < v if u ≤ v, and u ≠ v.By ℝt* ∈ ℝ we put Ω : =  × Crn. In the following, let t0 ∈ ℝ and t0 > t*.For a fixed k ∈ ℝ>0n, we consider two systems of the integrofunctional inequalitiest0, ∞), t0 ∈ ℝn for λ1, λ2 ∈ C, ℝn), λ1(t) < λ2(t), wherei = 1,…, n, t ∈  and any function λ ∈ C with ||λ||C ≤ M, where the norm is defined by (a = t0 and b = ϑ + r. For any (ii)k ∈ ℝ>0n and functions λj ∈ C, ℝn), j = 1,2 satisfying λ1(t) ≤ λ2(t) on  into ℝn equipped with the maximum norm and the normal conet0, ϑ] into ℝ≥0n. By the cone 𝕂ϑ, a partial ordering ≤ in 𝕃ϑ is given. For λ, μ ∈ 𝕃ϑ we say that λ ≤ μ if and only if μ − λ ∈ 𝕂ϑ. We introduce the operator Tϑ : 𝕃ϑ → 𝕃ϑ defined byWe prove that  has a coλ, μ ∈ 𝕃ϑ with λ ≤ μ. Condition (iii) implies that Tϑλ ≤ Tϑμ if λ ≤ μ. LetTϑ0μϑ,i = μϑ,i and Tϑ1μϑ,i = Tμϑ,i, i = 1,2) to the fixed points of operator Tϑ, we need to prove that Tϑ is continuous and compact. The first property is obvious (due to continuity of F and I). Let us prove compactness. To this end, let ℒ be a bounded subset of 𝕃ϑ. We have to show that Tϑℒ is a relatively compact subset of 𝕃ϑ. Due to the theorem of Arzelà and Ascoli, it suffices to show that Tϑℒ is bounded and equicontinuous. That Tϑℒ is bounded follows from condition , we haveϑ, j = 1,2. Thus the functions j = 1,2, t ∈ [t0, ∞), satisfyt0, ∞) andj = 1,2, t ∈ [t0, ∞). Now choosing, for example, y = I (see substitution (I is strictly increasing with respect to the second argument.Now we are in a position to apply titution ), the prF  are not explicitly mentioned. Nevertheless these assumptions can be viewed as almost necessary for validity of inequalities F  → ℝ is locally Lipschitz continuous, |d(t)| < ɛ < c, and (c + ɛ) < 1/(re).Let us consider linear advanced equation :(39)y˙(x(t) = exp(λt) or z(t) = exp(λt) we get corresponding transcendental equations:c + ɛ) < 1/(re) is here substantial).To apply λx1, λx2, λx1 < λx2, and (t) defined by formula  → ℝ, j = 1,2, λ1*(t) ≤ λ2*(t) for t ∈ [t0, ∞), satisfying the inequalitiest0, ∞). Then there exists a solution y = y(t) of  such thatLet there be continuous functions ualities5λ1∗(t)≤, where Ld is a suitable constant. The boundedness of operator T  obvious:Firstly we verify conditions  and 23)23) of Thondition ) is on [ξ lying betweenNow, condition  holds. LT .y = y(t) on [t0, ∞) satisfying inequalitiesλx1, λz1 are defined in Equation  has a pot0 is large enough for the asymptotic relations and inequalities to be valid. Now we employ λ1*(t), λ2*(t) satisfy inequalities (y(t) inequalities . W. Wt0 isequality  is fulfiualities  hold.γ : [t0, ∞) → ℝ+ is bounded, locally Lipschitz continuous and r is a positive constant. This equation can be viewed in a sense as a generalization of (c = 0 and d(t) = γ(t)). In the following criterion of existence of positive solutions of (Let us consider linear advanced equation :(61)y˙(ation of  (with c tions of  it is imtions of  and 46)(61)y˙(t)y = y(t) of , the existence of a continuous function λ : [t0, ∞) → ℝ is necessary and sufficient, satisfying inequalityt0, ∞). The mentioned positive solution y = y(t) satisfies inequalitiesFor existence of a positive solution  y(t) of  on [t0,Sufficiency. The proof uses t0, ∞). We set λ1*(t) ≡ 0 and λ2*(t) ≡ λ(t). Then inequalities (oof uses λ1∗(t)≤γexp+r0 ds),λ(t)≥γ(t)equality . Then inequality  is a conequality . Since tequality  must be equality  is obvioNecessity. If a positive solution y = y(t) of  exists, we set y(t) of  on [t0,i = 1,2Then both inequalities  hold simλ(t) in a suitable way we can get sufficient conditions of existence of positive solutions of . The following corollary illustrates such possibility.Taking tions of  on [t0,y = y(t) of  inequalityt0, ∞) sufficient.For existence of a positive solution  y(t) of  on [t0,λ(t) = eγ(t).The corollary is a consequence of  Note that criterion  of existResults obtained in the paper are sharp in a meaning. Indeed, in accordance with [equality , guarantγ(t) = 1/er. It means that inequality . Nevertheless, as it was demonstrated by advanced equation (In connection with the statement of quations  and 42)λx1, λualities  and 46)λx1, λequation . This isy = y2(t) on [t0, ∞) satisfying inequalitiesλx2, λz2 are defined in Equation  has a po


We will establish some sufficient conditions for nonoscillatory solutions with the property limt→∞xi(t) = 0,  i = 1, 2,…, n.We will discuss nonoscillatory solutions to the  Using the idea and method of 𝕋, which is called a close interval in 𝕋. Open intervals and half-open intervals and so forth are defined accordingly.A time scale z(t) is defined for x1(t) asA function x =  is a solution to the system (t1 ∈ 𝕋 such that functions xi(t), i = 1,2,…, n, are continuously differentiable and x satisfies 𝕋. Let W be the set of all solutions x =  to the system (i=1n|xi(t)| : t ∈  > 0 on  such that [lnx(t)]Δ = xΔ(t)/X(ξt). Because δ(t) > σ(t) ≥ ξt ≥ t for all t ∈ 𝕋κ, x(δ(t)) ≥ x(σ(t)) ≥ X(ξt) ≥ x(t) > 0 for all t ∈ 𝕋κ. Therefore,σ(t) to δ(t) on 𝕋, we get, for t ∈ 𝕋κ,t1 ∈ 𝕋 and a constant c such that ∫σ(t)δ(t)q(s)Δs ≥ c > 1/e holds for t ∈ 𝕋, we get thatl − 1)th equations of 𝕋, thatlth,…, nth equation of , t ∈ [t1, ∞)𝕋, thatn − l + 1 inequalities above and 𝕋,x1(h(t)) ≥ z(h(t)) hold. So we have that, for t ≥ s and s, t ∈ [t1, ∞)𝕋,s = h−1(t)∈[t1, ∞)𝕋, using the monotonicity of z(h(t)), then we have, for some sufficiently large t2 ∈ [t1, ∞)𝕋, thatt ∈ [t2, ∞)𝕋,N4+ ∪ N6+ ∪ ⋯∪Nn−1+ = ∅.get thatz(t)≥∫stp)𝕋, thatz(t)≥∫stpote that  and the tradicts  and henc(III) x ∈ Nn+ on [t1, ∞)𝕋. In this case, we have that, for t ∈ [t1, ∞)𝕋,s = h−1(t), we can obtain thatNn+ = ∅.ion with , and so  (IV) x ∈ N1− on [t1, ∞)𝕋. In this case, we have thatt→∞x1(t) = 0. From (t→∞ | z(t)| = L1 < +∞. Then it follows, by t→∞xi(t) = 0, i = 2,3,…, n. The proof is complete.ave thatx1(t)>0, Similarly to n is even, ι = −1 in the system (x ∈ W to (t→∞xi(t) = 0, i = 1,2,…, n.Assume that e system –80) hol holn is q = 2. Let a(t) = 8, g(t) = 16t, h(t) = 512t, p1(t) = (1/4)t, p2(t) = (7/8)t, p3(t) = (31/32)t, p4(t) = (127/128)t, p5(t) = 511/t9, f(y) = y, K = 1, α(t) = 2t, ι = −1, and n = 5; that is to say, the system has the following form:We consider a system on the time scale x ∈ W to (t→∞xi(t) = 0, i = 1,2,…, 5. In fact, functions x1(t) = 1/t, x2(t) = −1/t3, x3(t) = 1/t5, x4(t) = −1/t7, and x5(t) = 1/t9 are particular components of such a kind of solutions.It follows, from x ∈ W to , limt→∞q = 2. Let a(t) = 2, g(t) = 4t, h(t) = 128t, p1(t) = (1/8)t, p2(t) = (1/4)t, p3(t) = (7/32)t, p4(t) = 31/t7, f(y) = 127y, K = 127, α(t) = 16t, ι = 1, and n = 4; that is to say, the system has the following form:We also consider a system on the time scale x ∈ W to (t→∞xi(t) = 0, i = 1,2,…, 4. In fact, functions x1(t) = 1/t, x2(t) = −2/t3, x3(t) = 7/t5, and x4(t) = −31/t7 are particular components of such a kind of solutions.It follows, from x ∈ W to , limt→∞


A, the smallest nonnegative integer k such that rank (Ak) =rank (Ak+1) is called the Drazin index of A. In this paper, we give some results on the Drazin indices of sum and product of square matrices.For a square matrix  A ∈ ℂn×n, the smallest nonnegative integer k such that rank(Ak) = rank(Ak+1) is called the Drazin index of A, denoted by ind(A). The Drazin inverse of A is the unique matrix AD satisfying Ak+1AD = Ak,  ADAAD = AD, and AAD = ADA, where k = ind(A) (see [A) ⩽ 1, AD is called the group inverse of A, denoted by A#. If ind(A) = 0, then AD = A# = A−1. The theory of generalized inverses is an active research field in computational mathematics, and the Drazin index plays an important role in the study of the Drazin (group) inverse. Some results on the Drazin indices of matrices (operators) can be found in [For rankAk+ is calle(A) (see ). When ifound in –9.A ∈ ℂn×n, it is known [P, Δ and nilpotent matrix N such that A) is the smallest nonnegative integer k such that Nk = 0; that is, ind(A) = ind(N). Hence, ind(A) is the smallest nonnegative integer k such that the group inverse of Ak exists.For any is known  that theIn this paper, we give some results on the Drazin indices of sum and product of square matrices.In order to prove our main results, we give some lemmas as follows.A ∈ ℂn×n and nonnegative integer p, the limit  limε→0+εp(A + εI)−1 exists if and only if p⩾ind(A).For any A is square. Then,Let A is square. Then,Let A is square. Then, M# exists if and only if A# exists and rank(M) = rank(A).Let A is square. Then, M# exists if and only if A#, C# exist and rank(M) = rank(A) + rank(C).Let A ∈ ℂm×n,  B ∈ ℂn×m, |ind(AB) − ind(BA)|⩽1.For any A, let Aπ = I − AAD. We first give an upper bound for the Drazin index of the sum of two square matrices.For a square matrix P, Q be square complex matrices such that PDQ = 0 and PQPπ = 0. Then,Let U, Δ and nilpotent matrix N such that P) = ind(N). Suppose that Q1 is a square matrix with the same order as Δ. By PDQ = 0 we get Q1 = 0, Q2 = 0. Then, PQPπ = 0 we get NQ4 = 0.There exist nonsingular matrices P + Q) is the smallest nonnegative integer k such that the group inverse of (P + Q)k exists. Since P + Q) = ind(N + Q4). Since NQ4 = 0, we have ε(N + Q4 + εI) = (N + εI)(Q4 + εI). Let m = ind(Q4) + ind(N) − 1. By P + Q) = ind(N + Q4) ⩽ m = ind(Q4) + ind(N) − 1. By Q4) ⩽ ind(Q). Since ind(P) = ind(N), we have ind(P + Q) ⩽ ind(P) + ind(Q) − 1.Note that ind(Now we give an example to show that the upper bound in P) = ind(Q) = 2 and PDQ = 0 and PQPπ = 0. The Drazin index of P + Q) = ind(P) + ind(Q) − 1.Let We can obtain the following result from P, Q be square complex matrices such that PQ = 0. Then,Let CA = 0, CB = 0 or BC = 0, DC = 0 (see [It is known that = 0 (see ). A bettA is square. Then, ind(M) ⩽ ind(A) + ind(D) + 1 if one of the following holds:BC = 0, DC = 0;BC = 0, BD = 0;CA = 0, CB = 0;AB = 0, CB = 0.Let We only prove part (1). Parts (2)–(4) can be obtained in the same way.M = P + Q. By BC = 0, DC = 0, we get PQ = 0. By Let A, B ∈ ℂn×n, if s⩾max{ind(AB), ind(BA)}, then (AB)s and (BA)s are similar.For AB)0 and (BA)0 are similar. So we only consider the case s⩾1. There exist nonsingular matrices P, Q such that Ir is the identity matrix of order r = rank(A). Suppose that B1 ∈ ℂr×r. Then,s⩾max{ind(AB), ind(BA)}, the group inverses of (AB)s and (BA)s both exist. Lemmas X, Y such that B1sX = B1s−1B2 and YB1s = B3B1s−1. Then,AB)s and (BA)s are similar.Clearly  = s, then (AB)l and (BA)l are similar for any l⩾s + 1.For AB)l and (BA)l are similar for any l⩾s + 1.By The following corollary is a special case of A, B ∈ ℂn×n, if (AB)# and (BA)# both exist, then AB and BA are similar.For


The functions Φ and w are connected through the distributional identity tn4(hμ′Φ)(t) = 1/w(t), where hμ′ denotes the generalized Hankel transform of order μ. In this paper, we use the projection operators associated with an appropriate direct sum decomposition of the Zemanian space ℋμ in order to derive explicit representations of the derivatives SμmΦ and their Hankel transforms, the former ones being valid when m ∈ ℤ+ is restricted to a suitable interval for which SμmΦ is continuous. Here, Sμm denotes the mth iterate of the Bessel differential operator Sμ if m ∈ ℕ, while Sμ0 is the identity operator. These formulas, which can be regarded as inverses of generalizations of the equation (hμ′Φ)(t) = 1/tn4w(t), will allow us to get some polynomial bounds for such derivatives. Corresponding results are obtained for the members of the interpolation space Yn.For  The d-dimensional Fourier transform of a radial function is also radial and reduces to a 1-dimensional Hankel transform of order d/2 − 1 0, ∞. Then:(i)x, t ∈ I, one hasfor all (ii)x ∈ I and 0 < t ≤ a, one hasfor Suppose that the function  Qn cf.  also satx, t ∈ I. Equation . Hencet ∈ I, whilet ≤ a for some a ∈ I.Assume now 0 < n m ≥ n,  yields ≤ 1  (x ∈ I) and suppρ ⊂ , so that Throughout this section we will assume that the function First we prove a regularity result, along with a Hankel inversion formula and a polynomial estimate, in a general setting.f ∈ ℋμ′ be such that the distribution hμ′f is regular on ℋμ,n. Then, for all m ∈ ℤ+,r ∈ ℤ+ and a function G integrable on I for whichC is independent of t, then Sμmf ∈ 𝒞, withLet m ∈ ℤ+ and let ψ ∈ ℋμ. We havef and since Qnψ ∈ ℋμ,n, we may writeFix  We have〈hμ′(Sμmfay write〈(hμ′f)(ξation of .hμ′f)Sμ,xm ∈ 𝒞 as a function of x ∈ I ensure that𝒞 as a function of x ∈ I 〈hμEquations , 93), a, a93), aAt this point, let us formalize the definition of a basis distribution.ℋμ′ a basis distribution if tn4(hμ′Φ)(t) = 1/w(t) for some weight w  = O(tγ−) as t → ∞.We call Φ ∈ ht w cf.  such thaπμ,2n−1. Although The existence of basis distributions is guaranteed by the next F ∈ Lμ,l1 and there exists γ ∈ ℝ such that F(x) = O(xγ−) as x → ∞. Let r ∈ ℤ+. On ℋμ,r we define the linear functional Fr by(i)Fr ∈ ℋμ,r′.(ii) Any extension Fre ∈ ℋμ′ of Fr to ℋμ satisfies(iii) If Fr2e ∈ ℋμ′ is an extension of Fr2 ∈ ℋμ,2r′ to ℋμ, then(iv) If Fr1 and Fr2 are two extensions of Fr to ℋμ, then hμ′(Fr1) − hμ′(Fr2) ∈ πμ,r−1.Assume hμ′Sμr(hμ′Fre)(t) = (−t2)rFre(t) = F(t). Therefore, part (iv) is a consequence of Note that (ii) gives a, C > 0 be such that F(x) ≤ Cxγ−  (x > a). To prove (i), take ψ ∈ ℋμ,r and writeF ∈ Lμ,l1, for the first integral we obtaink ∈ ℤ+ is chosen so that k > −2r − γ + μ + 3/2. A combination of .Let nd write〈Fr,ψ〉=∫0e obtain|∫0aF(t)(r = 0 in (i) we obtain F ∈ ℋμ′. Next, let Fre ∈ ℋμ′ be an extension of Fr to ℋμ, and let φ ∈ ℋμ. Since (−t2)rφ(t) ∈ ℋμ,r, we may writeφ ∈ ℋμ gives (ii).Now we establish (ii). First of all, we note that specializing Fr2 ∈ ℋμ,2r′ by . Define μ,2r′ by , and letWhen applied to a basis distribution Φ, w with the properties that 1/w ∈ Lμ,l1 and 1/w(t) = O(tγ−) as t → ∞ for some γ ∈ ℝ. Let Φ ∈ ℋμ′ satisfy tn4(hμ′Φ)(t) = 1/w(t), so that hμ′Φ ∈ ℋμ,2n′ andm ∈ ℤ+ one hasSμmΦ ∈ 𝒞,r ∈ ℤ+.Pick a weight function φ∈ℋμ,2n). Then, ff.In order to derive this result from C > 0,  a > 1 be such that 1/w(t) ≤ Ctγ−  (t > a), with γ ∈ ℝ. Use r ∈ ℤ+ satisfyingLet t ≤ a, whilet > a. SetI as long as 1/w ∈ Lμ,l1 and , , t ≤ a, a, while|Sμ,xm[Q2w as in SμmΦ ∈ 𝒞 whenever m ∈ ℤ+ satisfies 2m < 4n + γ − μ − 3/2, a condition in fact weaker than  = O(tγ−) as t → ∞. Let f ∈ Yn, so that (−t2)n(hμ′f)(t) ∈ Lμ,l1. Then hμ′f ∈ ℋμ,n′, and for all m ∈ ℤ+ one hasSμmf ∈ 𝒞, withr ∈ ℤ+.Assume 1/hμ′(Sμnf) ∈ Lμ,w2, we haveC > 0, a > 1 such that 1/w(t) ≤ Ctγ−  (t > a), with γ ∈ ℝ. For any ψ ∈ ℋμ,n, the Cauchy-Schwarz inequality givesk ∈ ℤ+ is chosen so that 2k > −4n − γ + μ + 3/2. This proves that hμ′f ∈ ℋμ,n′. Now  = |F(t)|H(t)  (t ∈ I), from (t)This ends the proof.w as in f ∈ Yn has the property that Sμmf ∈ 𝒞 whenever m ∈ ℤ+ satisfies 4m < 4n + γ − μ − 3/2. Although this condition is actually weaker than (Sμ-derivatives.In [ker than , Theorem


The functions f1(x), f2(x), g(x), h1(x), h2(x), l1(x), andl2(x) satisfy some suitable conditions. We will prove that the problem has at least two nontrivial solutions by using Mountain Pass Theorem and Ekeland's variational principle.We consider the multiplicity of nontrivial solutions of the following quasilinear elliptic system −div(| Problems of this type are motivated by mathematical physics, since certain stationary waves in nonlinear Klein-Gordon or Schrödinger equations can be reduced to this form . Probleorm  = f2(x) = 0. By applying Nehari manifold method, the author proved the problem (λp/(p−q) + μp/(p−q) < λ1 and has at least two positive solutions when 0 < λp/(p−q) + μp/(p−q) < λ2.In , Hsu stu−div(|∇u−div(|∇uThis paper is motivated by the results of –12. ReplH1)(hi ∈ L∞(ℝN), hi(x)|x|bγ ∈ Lθ(ℝN), θ = p*/(p* − γ), i = 1,2.H2)(g(x)|x|b(α+β) ∈ Lδ(ℝN)∩L∞(ℝN), δ = p*/(p* − α − β).H3)(l1(x)|x|b, l2(x)|x|b ∈ Lσ(Ω), σ = p*/(p* − 1), l1(x) > 0, l2(x) > 0.Throughout this paper, we make the following hypotheses:Lbp(ℝN) the completion of the space C0∞(ℝN) with respect to the norm ||u||Lbp = (∫Nℝ|x|bp−|u|pdx)p1/, and let Wap1,(ℝN) the completion of C0∞(ℝN) with respect to the norm ||u||Wap1, = (∫Nℝ | x|ap− | ∇u|pdx)p1/.In this paper, we let θ = p*/(p* − γ), σ = p*/(p* − 1).We defineS > 0 such that∞ < a < (N − p)/p, a ≤ b < a + 1, d = a + 1 − b, and p* = pN/(N − pd).The following inequality is called Caffarelli-Kohn-Nirenberg inequality . There iX = Wap1,(ℝN) × Wap1,(ℝN) denote the usual Sobolev space with the normX is a reflexive and separable Banach space endowed with the norm ||||. Let u, v) ∈ X is a weak solution of problem  ∈ X, there holdsWe say that  and for all  ∈ X, there holdsJλ,μ correspond to the weak solutions of problem  admits a sequence of critical points in X. Our main result is the following.It is clear that problem  has a vare holds〈Jλ,μ′–(H3) hold. There exist c0, c1 > 0 such that if λ, μ > 0 satisfy 0 < λhθ1 + μhθ2 < c0 and 0 ≤ lσ1p/(p−1) + lσ2p/(p−1) ≤ c1, then the problem (Assume that H–(H3) ho In this section, we first introduce a compact embedding theory which is an extension of the classical Rellich-Kondrachov compactness theorem; see Xuan .Ω ∈ ℝN is an open bounded domain with C1 boundary and 0 ∈ Ω, N ≥ 3, −∞ < a < pN/(N − p). Then the embedding Wap1,(Ω)↪Lαγ(Ω) is continuous if 1 < γ ≤ Np/(N − p) and 0 ≤ α ≤ (1 + a)γ + N(1 − γ/p) and is compact if 1 ≤ γ < Np/(N − p) and 0 ≤ α < (1 + a)γ + N(1 − γ/p).Assume that  The following lemma is called the Mountain Pass Theorem, see   with J = 0. Suppose that J satisfies (PS) condition andA1) ≥ α0 when ||||X = ρ; there are A2) ∈ X, ||||X > ρ such that J < 0. there is .DefineH1)–(H3) hold. There exist c0, c1 > 0 such that if the parameters λ, μ satisfy 0 < λhθ1 + μhθ2 < c0, and 0 ≤ lσ1p/(p−1) + lσ2p/(p−1) ≤ c1, then Jλ,μ satisfies the assumptions (A1)–(A2) in Assume that  ∈ X. By (H2) and the Hölder inequality, we havegδ = (∫Nℝ(g(x) | x|b(α+β))δdx)δ1/, δ = p*/(p* − α − β).Let Jλ,h1(x) | x|bγ ∈ Lθ(ℝN)∩L∞(ℝN), we obtain from the Hölder inequality and (16)∫ℝNg in g(z1) < 1/2p for some z1 = |||| > 0.Letg(z)→+∞ where z → 0+ or z → +∞. Then g(z) has a minimum at z1 > 0. In order to find z1, we haveg′(z1) = 0 andg(z1) < 1/2p implies thatc3 = (2gδ/(α + β))Sα+βc2α+β−p + (1/γ)Sγc2γ−p.Note that c0, c1, α0 > 0 such that Jλ,μ ≥ α0 with 0 < λhθ1 + μhθ2 < c0 and 0 ≤ lσ1p/(p−1) + lσ2p/(p−1) ≤ c1. Then Jλ,μ satisfies the assumption (A1) in Thus, it follows from  and 27)27) that A2) in φ1, φ2) ∈ C0∞(ℝN)  ×  C0∞(ℝN) such thatJλ,μ→−∞ (t → +∞), since α + β > p > γ. Therefore, there exists t large enough, such that Jλ,μ < 0. Then we take e =  ∈ X and Jλ,μ(e) < 0. The condition (A2) of We now verify (H1)-(H2) hold.Let , then there exist subsequences, still denoted {un} and {vn}, and u ∈ Wap1,(ℝN), v ∈ Wap1,(ℝN) such that as n → ∞If {Uk = {x ∈ ℝN | |x | <k}, k = 1,2,….Let un} is bounded in Wap1,(ℝN), {un} is bounded in Wap1,(Uk) for for all k ≥ 1. By un} has a subsequence {un(1)} which converges Lγ(U1). Likewise, the subsequence {un(1)} is bounded in Wap1,(U2). So that it has a subsequence {un(2)} which converges Lγ(U2). Since {un(2)} is a subsequence of {un(1)}, 1, whereumm).Since {h1(x) | x|bγ ∈ Lθ(ℝN) gives that limk→∞(∫Ukc(|h1 | |x|bγ)θdx)θ1/ = 0.The fact Then,  implies is gives .un} is bounded in Wap1,(ℝN), we can assume (up to a subsequence) that un⇀u weakly in Wap1,(ℝN) and ||un||Wap1, ≤ C0 for some constant C0 > 0 and all n ≥ 1.In the following, we show thatε > 0, there exists kε > 0 so large thatWap1,(Ukε)↪Lγ(Ukε) is compact  as n → ∞.By , we knowpact see , we haveBy Brezis-Lieb's Lemma in , we obtaSimilarly, we can proveBy Brezis-Lieb's Lemma, it follows from  that(45H1)–(H3) hold. Then Jλ,μ defined by (PS) condition on X.Assume that } be a (PS)c sequence of Jλ,μ in X; that is, Jλ,μ → c, Jλ,μ′ → 0, in X*. We first claim that {} is bounded in X. In fact, for large n, we obtainp > γ > 1, we conclude that {} is bounded. Then, the {un} and {vn} are bounded in Wap1,(ℝN), respectively. Thus, there exist {} ∈ X and a subsequence of {}, still denoted by {}, such thatJλ,μ′ → 0 in X* implies Pn → 0 as n → ∞.Let {, we see thatTn → 0, as n → ∞.Since that ows from , 49) to toun⇀u, we have∫ℝN|h2||vN given byun − u, vn − v)|| → 0, as n → ∞.SoTn−Qn=∫ℝNJλ,μ satisfies (PS)c condition on X.Thus, Jλ,μ satisfies all assumptions in u1, v1) ∈ X such that  is a solution of problem  ≥ α0 > 0.By Lemmas u2, v2) of problem  ∈ C0∞(ℝN) × C0∞(ℝN) such that ∫Nℝl1(x)φ1dx + ∫Nℝl2(x)φ2dx > 0, and thent > 0 and thus for any open ball Bτ ⊂ X such thatρ > 0, such thatεn ↓ 0 such thatun, vn) ∈ Bρ. We now consider the functional F < F, u ≠ un, v ≠ vn, and thus  is a strict local minimum of F. Moreover,t → 0+; then,φ1, φ2) in , we getJλ,μ′|| ≤ εn.We now seek a solution <cρ+εn,Jλ,μ} ⊂ Bρ such that Jλ,μ → cρ < 0, and Jλ,μ′ → 0 in X* as n → ∞. By un, vn)} has a convergent subsequence in X, still denoted by {}, such that → in X. Thus  is a solution of  < 0. Then the proof of Therefore, there is a sequence {(ution of  with Jλ


R be a ring and U be a Lie ideal of R. Suppose that σ, τ are endomorphisms of R. A family D = {dn}n ∈ Nof additive mappings dn:R → R is said to be a - higher derivation of U into R if d0 = IR, the identity map on R and a, b ∈ U and for each n ∈ N. A family F = {fn}n ∈ Nof additive mappings fn:R → R is said to be a generalized - higher derivation -higher derivation) of U into R if there exists a - higher derivation D = {dn}n ∈ Nof U into R such that, f0 = IR and a, b ∈ U and for each n ∈ N. It can be easily observed that every generalized -higher derivation of U into R is a generalized Jordan -higher derivation of U into R but not conversely. In the present paper we shall obtain the conditions under which every generalized Jordan - higher derivation of U into R is a generalized -higher derivation of U into R.Let  For any x,y ∈ R denote the commutator xy − yx by . Recall that a ring R is prime if aRb = {0} implies that a = 0 or b = 0. An additive subgroup U of R is said to be a Lie ideal of R if  ⊆ U. A Lie ideal U of R is said to be a square closed Lie ideal of R if u2∈U for all u∈U. An additive mapping d:R→R is said to be a derivation (resp. Jordan derivation) of R if d(xy) = d(x)y + xd(y)(resp. d(x2) = d(x)x + xd(x)) holds for all x,y ∈ R. Now let D = {dn}n ∈ N be a family of additive mappings dn:R. Following Hasse and Schimdt (→ RD is said to be a higher derivation (resp. Jordan higher derivation) on R if d0 = IR(the identity map on R) and a, b ∈ R and for each n ∈ N. Several interesting results on higher derivation can be seen in Haetinger -derivation -derivation) of R if d(xy) = σ(x)d(y) + d(x)τ(y) (resp. d(x2) = σ(x)d(x) + d(x)τ(x)) holds for all x,y ∈ R. For a fixed a ∈ R, the map da:R→Rgiven by da(x) = aτ (x) − σ(x)a for all x ∈ R is a - derivation which is said to be a -inner derivation determined by a.Throughout - derivation the authors together with Haetinger - higher derivation as follows: A family D = {dn}n ∈ N of additive mappings dn:R → R is said to be a - higher derivation of R if d0 = IR and a, b ∈ R and for each n ∈ N - higher derivation of U into R if the corresponding conditions are satisfied for all a, b ∈ U).Inspired by the notion of  = F(x)y + xd(y) holds for all x, y ∈ R. Motivated by the definition of generalized derivation the notion of generalized higher derivation was introduced by Cortes and Haetinger -inner derivation if F(x) = σ(x)b + aτ(x) holds for some fixed a, b ∈ R and for all x ∈ R. If F is a generalized -inner derivation, a simple computation yields that F(xy) = σ(x)db−(y) + F(x)τ(y), where db− is a -inner derivation. With this point of view an additive mapping F:R→R is said to be a generalized -derivation on R if there exists a -derivation d:R→R such that F(xy) = σ(x)d(y) + F(x)τ(y) holds for all x,y ∈ R. For such an example let S be any ring and F:R→R such that σ,τ:R→R such that F is a generalized -derivation on R with associated - derivation d:R→R such that An additive mapping σ,τ)-higher derivation in a more general setting.In view of the above definition we introduce the analogue of -higher derivation -higher derivation) of R if there exists a -higher derivation D = {dn}n ∈ N of R such that f0 = IR and a, b ∈ R and for each n ∈ N.Let U be a Lie ideal of R. Then F is said to be a generalized -higher derivation -higher derivation) of U into R if the above corresponding conditions are satisfied for all a, b ∈ U.Let Example 1.1. Let R be an algebra over the field of rationals and σ,τ be the endomorphisms of R. Define n ∈ N, where f is a generalized - derivation on R with associated -derivation δsuch that fσ = σf and δτ = τδ-derivation ensures the existence of such f). Consider the sequence {Fn}n ∈ N, this defines a generalized - higher derivation with associated -higher derivation f to be a generalized Jordan −derivation on R which is not a generalized −derivation on R then one can easily construct an example of a generalized Jordan − higher derivation on R which is not a generalized − higher derivation on R.If we choose the underlying R is a Jordan derivation but the converse need not be true in general. A well-known result due to Herstein -higher derivation of R. The main objective of the present paper is to find the conditions on R under which every generalized Jordan -higher derivation of R is a generalized -higher derivation of R. In fact our results generalize, extend and compliment several results obtained earlier in this direction.It is easy to see that every derivation on Herstein  states tHerstein  generaliHerstein  further R is a prime ring of characteristic different from 2 and U a square closed Lie ideal such that U⫅̸Z(R). Then every Jordan higher derivation of U into R is a higher derivation of U into R. The following theorem shows that the above result still holds for arbitrary square closed Lie ideal of R that is, U may be central.Recently, Haetinger  proved tTheorem 2.1. Let R be a prime ring such that char(R) ≠ 2 and U be a square closed Lie ideal of R. Suppose that σ,τ are endomorphisms of R such that στ = τσ and τ is one-one & onto. Then every generalized Jordan -higher derivation of U into R is a generalized -higher derivation of U into R.In order to develop the proof of the theorem, we begin with the following known lemma:Lemma 2.1. . Let G1,G2,⋯,Gnbe additive groups, S:G1×G2×⋯×Gn→R and T:G1×G2×⋯×Gn→ R be the mappings which are additive in each argument. If SxT = 0 for every x ∈ U, ai∈Gi, i = 1,2,⋯,n then S = 0 for every ai∈Gi, i = 1,2,⋯,nor T = 0 for every bi∈Gi, i = 1,2,⋯,n.Lemma 2.2. Let R be a ring and σ, τ be endomorphisms of R such that στ = τσ and F = {fn}n ∈ Nbe a generalized Jordan -higher derivation of U into R with associated -higher derivation D = {dn}n ∈ N of U into R. Then for all u, v, w ∈ U and each fixed n ∈ N we have for all u, v, w ∈ U.(i) R is a 2-torsion free ring then,If (ii) (iii) Proof. (i) For u, v ∈ U, n ∈ N we have, u we obtain u, v ∈ U. Again;  for all u, v ∈ U.Comparing the above expressions and reordering the indices we obtain the required result. for all uv + vu = (u + v)2−u2−v2∈U, using (i) and replacing v by uv + vu we find that, (ii) Since On the other hand, R is 2-torsion free we get the required result.Comparing the equations (2.1) and (2.2) and reordering the indices and using the fact that (iii) Linearizing the above result, we have Again, R is 2-torsion free we get the required result.Comparing (2.3) & (2.4) and using the fact that n ∈ N and for each u,v∈U we denote by Φn the element of R such that Φn = 0, then F = {fn}n ∈ N is a generalized -higher derivation of U into R. Trivially, by using Lemma 2.2(i) we get Φn = −Φn, for all u,v∈U.For every fixed Φnis additive in both the arguments.It can also be seen that the function Lemma 2.3. Let R be a 2-torsion free ring and σ,τbe endomorphisms of R such that στ = τσ. Let F = {fn}n ∈ Nbe a generalized Jordan -higher derivation of U into R with associated -higher derivation D = {dn}n ∈ Nof U into R. If Φm = 0, for each m<n and for all u,v∈U, then Φnτn = 0, for all u,v∈U(i) Φnτn(w)τn = 0, for all u,v,w∈U.(ii) Proof. u,v∈U, uv + vu∈U and uv−vu∈U, we find that 2uv∈U. Suppose β = 4(uv(uv) + (uv)vu)∈U. Using Lemma 2.2(iii) we have, (i) Since for any Φm = 0 for all m<n we have, Using the fact that On the other hand, fn(β)we get Φn ×  = 0 for all u, v ∈ U.Comparing (2.5) with (2.6) for χ = 4(uvwvu + vuwuv) for u, v, w ∈ U. Then by Lemma 2.2(ii) we obtain (ii) Let Again consider, iii), we have Applying Lemma 2.2(R is 2 torsion free we find that Equating (2.7) & (2.8) and using the fact that Initially calculating the first term we have Φm = 0 for all m < n. Using the hypothesis that Similarly the second term reduces to στ = τσ we obtain Now, subtracting the equation (2.11) from (2.10) and using the hypothesis that Similarly, the difference of the last two terms of equation (2.9) yields Φnτn(w)τn = 0 for all u, v ∈ U.Thus, (2.9) becomes We are now well equipped to prove our main theorem.Proof of Theorem 2.1. Suppose that U is commutative. If U is commutative then U is also central -higher derivation reduces to generalized Jordan -derivation and hence using Lemma 2.2(iii) of . We’ll U is commutative, in view of Lemma 2.2(i) of  = d(vu) = σ(v)d(u) + d(v)τ(u), the above equation can be rewritten as Since, Comparing the equations (2.12) and (2.13) we obtain w is central and since τ is one-one and onto hence τ(w) is central but the center of a prime ring is free from zero divisors, the above equation implies that Φ1 = 0 for all u, v ∈ U. Let Φm = 0 for all u, v ∈ U and each m < n then from Lemma 2.2(iii), we have As u, v, w ∈ U.for all i) and commutativity of U, we get Again by using Lemma 2.2(U is commutative we find that dj(τn−j(uv)) = dj(τn−j(vu)) and as dj is a -higher derivation the above equation can be written as Since Comparing equations (2.15) and (2.14) we find that Φm = 0 for all u, v ∈ U, m < n, the above equation reduces to Since w is central and as τ is one-one and onto, τn(w) is also central. But the center of a prime ring is free from the zero divisors, equation (2.16) implies that Φn = 0 for all u, v ∈ U and each n ∈ N.Since U is non-commutative hence U⫅̸Z(R). Using Lemma 2.3(ii) we have Φnτn(w)τn = 0 for all u,v,w∈U. Since τ is one-one and onto, the relation yields that τn−)w = 0 for all u,v,w∈U. Hence by Lemma 2.1, τn−) = 0 for all u,v∈U or  = 0 for all u,v∈U. But since U is non-commutative, we find that Φn = 0, for all u,v∈U and each n ∈ N.This completes the proof of our theorem.Now consider the possibility that As a consequence of the above result we find the following corollaries. Corollary 2.1 settle the conjecture given in  derivation on U then F is a generalized  derivation on U.Corollary 2.2. Let R be a prime ring such that char(R) ≠ 2 and U a square closed Lie ideal of R. Then every Jordan higher derivation of U into R is a higher derivation of U into R.In the above theorem if the underlying ring is arbitrary, then we can prove the following:Theorem 2.2. Let R be a 2- torsion free ring and U be a square closed Lie ideal of R. Suppose that σ,τ are endomorphisms of R such that στ = τσ and τ is one-one & onto. If U has a commutator which is not a right zero divisor, then every generalized Jordan -higher derivation of U into R is a generalized -higher derivation of U into R.Proof. Let x,y∈U be the fixed elements such that c=0⇒c = 0 for every c∈R. We’ll prove the result by induction on n.n = 0, Φ0 = 0. Hence proceeding by induction we can assume that Φm = 0 for all m<n. Using Lemma 2.3(i) we have We know that for τn−) = 0 for all u, v ∈ U. Hence in particular, τn−) = 0. This implies that, The above equation implies that u by u + x in (2.17) we get Replacing v by y in (2.19) and using (2.18) we get Φn = 0, for every u ∈ U,i.e., τn−) = 0. This yields that Again replacing v by v + y in (2.19), to get Replace u by x in (2.21) and using (2.18) we obtain Φn = 0 for all v ∈U. This yields that Replacing Φn = 0 i.e., τn−) = 0. Hence, we conclude that Φn = 0 for all u,v∈U.This completes the proof of our theorem. □Combining (2.20), (2.21) and (2.22) we find that Some special cases of the above theorem are already known and are of great interest.Corollary 2.3.  derivation of U into R then F is a generalized  derivation of U into R.Corollary 2.4. ((Cortes and Haetinger R be a 2 torsion free ring such that R has a commutator which is not a right divisor and U a square closed Lie ideal of R. Then every generalized Jordan higher derivation of U into R is a generalized higher derivation of U into R.


A class related to this function will be introduced and the properties will be discussed.By using a linear operator, we obtain some new results for a normalized analytic function  A simpler definition states that a meromorphic function the formf(z)=g(z)An equivalent definition of a meromorphic function is a complex analytic map to the Riemann sphere. For example, the Gamma function is meromorphic in the whole complex plane; see , 2.Latf(z) will be discussed.In the present paper, we will derive some properties of univalent functions defined by means of the Hadamard product of Hurwitz Zeta function; a class related to this function will be introduced and the properties of the liner operator f(z) normalized byU*. For 0 ≤ β, we denote by S*(β) and k(β) the subclasses of Σ consisting of all meromorphic functions which are, respectively, starlike of order β and convex of order β in U*.Let Σ denote the class of meromorphic functions fj(z)   defined byf1(z) and f2(z) byFor functions β ≠ 0, −1, −2,…, and α ∈ ℂ/{0}, where (λ)n = λ(λ + 1)n+1 is the Pochhammer symbol. We note thatLet us define the function a ∈ ℂ/ℤ0−, ℤ0− = {0, −1, −2,…}; t ∈ ℂ when z ∈ U = U* ⊂ {0}; ℜ(t) > 1 when z ∈ ∂U).We recall here a general Hurwitz-Lerch-Zeta function, which is defined in , 4 by thz, t, a) include, for example, the Riemann zeta function ζ(t) = Φ, the Hurwitz zeta function ζ = Φ, the Lerch zeta function lt(ζ) = Φ,  > 1), and the polylogarithm Lti(z) = zΦ. Recent results on Φ can be found in the expositions [Gt,a(z) and using the Hadamard product for f(z) ∈ Σ, we define a new linear operator Lt,a on Σ by the following series:Important special cases of the function Φ = 1 + qnzn + qn+1zn+1 + ⋯ be analytic functions in U = U* ∪ {0} with q(z) ≠ 0 for z ∈ U. IfLet a > 0, and2,thenℜ{1q(z)}>We begin with the following theorem.α + 1 > 0, Latf(z)/Latf(z) ≠ 0 for z ∈ U* and suppose thatLet 2.Thenℜ{Lat(α+1q(z) byq(z) = 1 + qnzn + qn+1zn+1 + ⋯ analytic function in U* with q(z) ≠ 0 for z ∈ U*. It follows from (a = 1/(1 + α), we get the required result.Define the function q(z) byq(z)=Lat(ows from  that(19identity  in (19),α = β = 1 in Letting Gt,a(z)/z)′ ≠ 0 for z ∈ U* and suppose thatLet 2.Thenℜ{z/z)′ ≠ 0 and t = 0 for z ∈ U* and suppose thatf(z) is starlike in U*.Let δ(α + 1) > 0, zLatf(z) ≠ 0 for z ∈ U* and suppose thatLet 2.Thenℜ(zLat(α+q(z) byq(z) = 1 + qnzn + qn+1zn+1 + ⋯ analytic function in U* with q(z) ≠ 0 for z ∈ U*. It follows from (a = 1/(1 + α), we get the required result.Define the function q(z) byq(z)=(zLaidentity  in 30),,q(z) byα = β = 1 in Letting δ > 0, Gt,a(z) ≠ 0 for z ∈ U* and suppose thatLet The bound in  is the bδ = 1, M = 1 + n/8log2, and t = 0 in Letting f′(z) ≠ 0 for z ∈ U* and suppose thatLet ξ > 0, zf(z))′/Latf(z) ≠ 0 for z ∈ U* and suppose thatLet 2.Thenℜ byq(z) = 1 + qnzn + qn+1zn+1 + ⋯ analytic function in U* with q(z) ≠ 0 for z ∈ U*. Also by a simple computation and by making use of the familiar identity  byq(z)=)′/Gt,a(z) ≠ 0 for z ∈ U* and suppose thatLet 2.Thenℜξ = 1, M = 1 + n/2log2, and t = 0 in Letting zf′(z)/f(z) ≠ 0 for z ∈ U* and suppose thatLet


The authors prove that the exact difference Δf =(z+1) - f(z)has infinitely many fixed points, if a ∈ ℂ and ∞ are Borel exceptional values  of f. These results extend the related results obtained by Chen and Shon.Let  We use the notations σ(f) to denote the order of f(z), λ, and λ(1/f), respectively, to denote the exponent of convergence of zeros of f(z) − a and poles of f(z). Especially, if a = 0, we denote λ = λ(f). A point z ∈ ℂ is called as a fixed point of f(z) if f(z) = z. There is a considerable number of results on the fixed points for meromorphic functions in the plane; we refer the reader to Chuang and Yang )and Yang . It folland Yang ; we use f be a transcendental meromorphic function in the complex plane ℂ. The exact differences Δf are defined by Δf = f(z + 1) − f(z).Let f, Chen and Shon have proved the following.Recently, there are a number of papers (including –16) focuf be a transcendental entire function of order of growth σ(f) = 1 and have infinitely many zeros with the exponent of convergence of zeros λ(f) < 1. Then Δf has infinitely many zeros and infinitely many fixed points.Let f is less than 1, Chen and Shon have proved the following.When the order of f be a transcendental meromorphic function of order of growth σ(f) ≤ 1. Suppose that f satisfies λ(1/f) < λ(f) < 1 or has infinitely many zeros (with λ(f) = 0) and finitely many poles. Then Δf has infinitely many fixed points and satisfies the exponent of convergence of fixed points τ(Δf) = σ(f).Let f be a transcendental meromorphic function of order of growth σ(f) < 1, is there a similar result as that in Theorem B if λ(1/f) ≥ λ(f) or f has infinitely many zeros (with λ(f) = 0) and infinitely many poles?A natural question is, letting In this paper, we will prove the following theorem to answer the question.f be a transcendental meromorphic function of order of growth σ(f) < 1 and a ∈ ℂ. Suppose that f satisfies λ(1/f) < σ(f) and λ < σ(f). Then Δf has infinitely many fixed points and satisfies the exponent of convergence of fixed points τ(Δf) = σ(f).Let From f be a transcendental meromorphic function of order of growth σ(f) < 1. Suppose that f satisfies λ(f) ≤ λ(1/f) < σ(f). Then Δf has infinitely many fixed points and satisfies the exponent of convergence of fixed points τ(Δf) = σ(f).Let f satisfies λ(1/f) < σ(f) and λ < σ(f). That is to say ∞ and a are Borel exceptional values of f. If we suppose that ∞ and a are Nevanlinna deficiency values of f, is there a similar result as that in Theorem B? In the following, we give In f(z) be a meromorphic function in the complex plane ℂ and a ∈ ℂ∞ = ℂ ∪ {∞}. Nevanlinna's deficiency of f with respect to a is defined byLet a = ∞, then one should replace N) in the above formula by N. If δ > 0, then a is called a Nevanlinna deficiency value of f.If f be a transcendental meromorphic function of order of growth σ(f) < 1 and a ∈ ℂ. Suppose that f satisfies δ = 1 and a is a Nevanlinna deficiency value of f. Then Δf has infinitely many fixed points.Let f be a transcendental entire function of order of growth σ(f) < 1 and a ∈ ℂ. Suppose that δ > 0. Then Δf has infinitely many fixed points.Let f(z) be a meromorphic function. If the function f(z) has finite order, thenk.Let f(z) be a meromorphic function with the exponent of convergence of poles λ(1/f) = λ < +∞ and let c be a nonzero complex number. Then for each ɛ > 0, we haveLet f be a transcendental meromorphic function of order of growth σ(f) < 1 and let c be a nonzero complex number. ThenLet σ(f)≔σ < 1, then λ(1/f) = λ ≤ σ < 1. Therefore, for any 0 < ɛ < 1 − σ, it follows from Since the order f be a function transcendental and meromorphic in the plane which satisfiesf is transcendental.Let f be a transcendental meromorphic function of order of growth σ(f) = σ < 1. Then Δf is transcendental.Let σ(f)≔σ < 1, then, for any positive ɛ(0 < ɛ < 1 − σ), there exists R > 0 such that for any r > R we haveSince the order Therefore,f(z) be a meromorphic function of finite order, then σ(Δf) ≤ σ(f).Let f be a transcendental meromorphic function of order of growth σ(f) < 1. Then for any ɛ > 0 and any positive integer k, there exists a set E ⊂  that depends on f and has finite logarithmic measure, such that for all z satisfying |z | = r ∉ E ∪  we haveLet It is easy to derive the following lemma from f be a transcendental meromorphic function of order of growth σ(f) < 1. Then for any positive integer k there exists a set E ⊂  that depends on f and has finite logarithmic measure, such thatLet SincethenApplying the first fundamental theorem, we getCombining -15) we  we 15) wSincethen, we can getTherefore,Thus from f − z is a transcendental meromorphic function of order of growth σ(Δf − z) ≤ σ(f) < 1. It follows from E ⊂  that has finite logarithmic measure, such that for any r ∉ E we haveBy Lemmas From  and 21)21)-22),,22), we g ≡ f − a by (Denoting f − a by  we derivτ(Δf) ≤ σ(f). If τ(Δf) < σ(f), by λ(1/f) < σ(f) and λ < σ(f), there exists a number η < σ(f), such that for any sufficient  r we haveBy τ(Δf) = σ(f).Combining  and 25)25), we cδ = 1, then N = o). By (Since  f)). By , we can δ > 0, then there is a positive number θ < 1 such thatSince f has only a finite number of fixed points, then from (If Δhen from  and 27)f has onlf being transcendental. Therefore, Δf has infinitely many fixed points.This contradicts


A mathematical model for the relationship between the populations of giant pandas and two kinds of bamboo is established. We use the impulsive perturbations to take into account the effect of a sudden collapse of bamboo as a food source. We show that this system is uniformly bounded. Using the Floquet theory and comparison techniques of impulsive equations, we find conditions for the local and global stabilities of the giant panda-free periodic solution. Moreover, we obtain sufficient conditions for the system to be permanent. The results provide a theoretical basis for giant panda habitat protection. The giant panda is a highly specialized Ursid, approximately its 99% of their diet is bamboo . Many ofAiluropoda melanoleuca) and bamboo by adding a correction term which takes into account the effect of a sudden collapse of bamboo as a food source  × ℝ+3 ∪ ((n + l − 1)T, nT] × ℝ+3 and limt,y)→ = V exists, where t0 = (n + l − 1)T+ and nT+.V is locally Lipschitzian in x.Let ℝV ∈ V0, for ∈((n − 1)T, (n + l − 1)T] × ℝ+3, and the upper right derivative of V with respect to the impulsive differential system (Let  x)∈ is continuous on ((n − 1)T, (n + l − 1)T]∪((n + l − 1)T, nT] and X(t0+) = limt→t0X(t) exists, where t0 = (n + l − 1)T+ and nT+. The smoothness properties of f guarantee the global existence and uniqueness of solution of system (X(t) of  with a > 0, we say that a solution Y(t) of (X(t) if Y(t) is defined on  × ℝ+ ∪ ((n + l − 1)T, nT] × ℝ+ and the limit limt,y)→ = g exists, where t0 = (n + l − 1)T+ and nT+, and is finite for x ∈ ℝ+ and n ∈ ℕ, and ψ1, ψ2 : ℝ+ → ℝ+ are nondecreasing for all n ∈ ℕ. Let r(t) be the maximal solution for the impulsive Cauchy problem:(H):  ∞). Then V ≤ u0 implies that V) ≤ r(t), t ≥ 0, where x(t) is any solution of T, (n + l − 1)T] × ℝ+ ∪ ((n + l − 1)T, nT] × ℝ+ etc. see Remark 2.3 and Theorem 2.3 of  → ℝ3 a saturated solution of system . By t < T0,x] represents the largest integer not exceeding x. That is, (x1(t), x2(t), x3(t)) remains strictly positive on .Let us consider (x1(t), x2(t), x3(t)) of  ∈ ℝ+3 are bounded.All solutions (x1(t), x2(t), x3(t)) be a solution of  and lett ≠ nT, t ≠ (n + l − 1)T, and t > 0, we obtain thatD, it follows thatLet (By t → ∞ isV(t) is bounded in its domain. Consequently, (x1(t), x2(t), x3(t)) are bounded by a constant “D/a30” for sufficiently lager t.which yieldsV(t)≤V is eradicated, it is easy to see that the equations in (When the giant panda tions in  decoupleα) + a10T > 0, then the system (x1*(t) with this notation, and the following properties are satisfied:x1(t) of (x10.Suppose that ln(1 − e system  has a pe1(t) of  startingSimilarly, we have the following β) + a20T > 0, then the system (x2*(t) with this notation, and the following properties are satisfied:x2(t) of (x20.Suppose that ln(1 − e system  has a pe2(t) of  startingx1*(t), x2*(t), 0).It follows from Lemmas x1*(t), x2*(t), 0) by means of the Floquent theory. α) + a10T > 0 and ln(1 − β) + a20T > 0 andx1*(t), x2*(t), 0) is locally stable.Suppose that ln(1 − x1*(t), x2*(t), 0) may be determined by considering the behavior of small-amplitude perturbations of the solution. DefineThe local stability of the periodic solution ( Definex1(t)=u(t) satisfieswhere Φ = 0t(−a30 + a31x1*(s) + a32x2*(s))ds).For the upper triangular matrix, there is no need to calculate the exact forms of ∗ and ∗∗ as it is not required in the analysis that follows. And Δ = exp + a10T > 0, ln(1 − β) + a20T > 0, and condition , x2*(t), 0) is stable.areλ3 > 1 and (x1*(t), x2*(t), 0) is unstable.If the reverse of  is satisx1*(t), x2*(t), 0) is globally asymptotically stable.If the conditions of ɛ1 > 0, ɛ2 > 0 small enough that if condition (x1*(t) is the periodic solution of the systemu(t) → x1*(t), t → ∞, andx2*(t) is the periodic solution of the systemv(t) → x2*(t), t → ∞, andn + l − 1)T, (n + l)T) yieldsρk → 0 as k → ∞. This implies that x3(t) → 0 as t → ∞.Choose ondition  holds,refore,dx3dt≤x3x1(t) → x1*(t) and x2(t) → x2*(t) as t → ∞ if limt→∞x3(t) = 0. For ɛ3 > 0 there exists a T1 > 0 such that 0 < x3(t) < ɛ3, t > T1. Without loss of generality, we may assume that 0 < x3(t) < ɛ3 for all t ≥ 0. Then we havet → ∞, andɛ1 > 0, we havet large enough. Let ɛ3 → 0, and we get x1*(t) − ɛ1 ≤ x1(t) ≤ x1*(t) + ɛ1 for large t, which implies x1(t) → x1*(t) as t → ∞.Next, we prove that x2(t) → x2*(t) as t → ∞. This completes the proof.Similarly, we can get that T* = ((a31/a11)ln(1 − α) + (a32/a22)ln(1 − β))/(a30 − (a31/a11)a10 − (a32/a22)a20), and we find that when T < T*, the giant panda-free periodic solution is globally asymptotically stable. That is to say, in this case, the giant panda will be extinct. In biology, when the period of bamboo flowing is smaller than the threshold T*, the bamboo cannot be revived to support giant panda again, so giant panda will die by starvation.Condition  can be rWe make mention of the definition of permanence before starting the permanence of system .m and M such that every positive solution (x1(t), x2(t), x3(t)) of system  ≤ M, m ≤ x2(t) ≤ M, and m ≤ x3(t) ≤ M for sufficiently large t.System  is said α)+(a10 − a13M)T > 0, ln(1 − β)+(a20 − a23)T > 0, andM is upper bound of the solution of system (Suppose that ln1 −  − α+(a1x1(t), x2(t), x3(t)) be a solution of (M > 0 such that x1(t) ≤ M, x2(t) ≤ M, x3(t) ≤ M for each solution X = (x1(t), x2(t), x3(t)) of (t. The first equation of (u*(t) is the periodic solution of system:u(t) → u*(t), t → ∞, andβ) + (a20 − a23M)T > 0 holds, we can get x2(t) > v*(t) − ɛ ≡ m2, where v*(t) is the periodic solution of system:v(t) → v*(t), t → ∞, andLet (m3 > 0 such that x3(t) ≥ m3 for sufficiently large t. This can be done in the following two steps.Therefore, it is necessary only to find an Step 1. Choose m0 > 0, ɛ4 > 0, ɛ5 > 0 small enough that if condition  ≥ m0 for some t1 > 0. Assuming the contrary, x3(t) < 0 for all t, from system (x1(t) ≥ u(t), x2(t) > v(t), and x3(t) ≥ w(t). By n + l − 1)T, (n + l)T) yieldsThis step will show that .Thusdwdt≥ → ∞ as t → ∞, which contradicts the boundedness of x3(t).Since Step  2. If x3(t) ≥ m0 for all t ≥ t1, then the proof is complete. If not, let t* = inft≥t1{x3(t) < m0}, and then x3(t) ≥ m0 for t ∈ , (n ∈ ℕ), such that x3(t) ≥ m0, t ∈ . Let T′ = sup|tn+1 − tn|, n ∈ ℕ. If T′ = ∞, there must exist a subsequence tni such that tni+1 − tni → ∞ as ni → ∞. From Step 1, this can lead to a contradiction with the boundedness of x3(t); therefore, T′ < ∞. Thenm = min{m1, m2, m3}, and thenNext, we consider the following two subsystems:Imitating the proof of Subsystem  is permaSubsystem  is permaIn this paper, our purpose is that of considering the survival of the giant panda. From Theorems T*. When T < T*, the giant panda-free periodic solution is globally asymptotically stable. That is to say, the giant panda will be extinct if the period of bamboo flowering is smaller than the threshold T*, because the bamboo cannot be revived to support giant panda again. Comparing In this paper, we consider an impulsive differential system of the population ecology on the three populations of the giant panda and two kinds of bamboo. The local and global stability of the giant panda-free periodic solution are obtained and we find the threshold value


K be an arbitrary field and X a square matrix over K. Then X is sum of two square nilpotent matrices over K if and only if, for every algebraic extension L of K and arbitrary nonzero α ∈ L, there exist idempotent matrices P1 and P2 over L such that X = αP1 − αP2.Let  A over a field K is a sum of two nilpotent matrices over K if and only if A is similar to a particular form. In an early paper, Pazzis (see  = {z ∈ Z+ | 1 ≤ z ≤ s} for some s ∈ Z+. Mm,n(K) denotes the space consisting of all m × n matrices over K; Mn(K) = Mn,n(K). r(A) is the rank of A ∈ Mm,n(K). E denotes a vector space over K and dim(E) is the dimension of E. X ∈ Mn(K) is called s2N in Mn(K) if there exist square nilpotent N1 and N2 ∈ Mn(K) such that X = N1 + N2, while X is called an  composite in Mn(K) if there exist idempotent P1 and P2 ∈ Mn(K) such that X = αP1 + βP2, where α, β ∈ K∖{0}  composite in Mn(L) for every algebraic extension L of K and arbitrary nonzero α ∈ L (when car(K) = 2, we still use ±P for the meaning of  composites).If there is no statement, the meanings of notations mentioned in this paragraph hold all over the paper. X ∈ Mn(K), on the one hand, we will prove that X is s2N in Mn(K) implies X is ±P; that is, the form provided by Botha satisfies the condition as in , Ni ∈ Mri(K), ∑i=1sri = n, and both the characteristic polynomial and minimal polynomial of Ni are xri. Furthermore, Ni is similar to C(xri) as follows:C(xri) = E2,1 + E3,2 + ⋯+Eri,ri−1 ∈ Mri(K).For arbitrary field ri is even, C(xri) = ∑j=1ri/2Ej,2j−12 + ∑j=1ri/2−1Ej+1,2j2; when ri is odd, C(xri) = ∑j=1ri−1)/2/2 is s2N, and Ni is s2N follows. Hence N is s2N.When s2N → ±P. Suppose X ∈ Mn(K) is s2N in Mn(K); that is, there exist square nilpotent matrices N1 and N2 ∈ Mn(K) such that X = N1 + N2. It will take two steps to prove X is ±P.X is nonsingular, then X is ±P. If X = N1 + N2 with N12 = N22 = 0, inspect the eigenspaces of N1 and N2. Note that N1 and N2 are square nilpotent matrices, their ranks satisfy the following inequality matrices. r(N1) = r(N2) = n/2.Since X is nonsingular implies 0 is not its eigenvalue. Secondly, if the inequality is strict, then intersection of eigenspaces of N1 and N2 contains nonzero vectors; that is, there exists nonzero x ∈ Mn,1(K) such that N1x = N2x = 0, which implies that 0 is one of eigenvalues of X. This is a contradiction. Hence, r(N1) + r(N2) = n; that is, n is even and N1 and N2 are similar but not equal.At first, N1 is square nilpotent with r(N1) = n/2, we can choose n/2 linear independent vectors from the set of its column vectors which can make up a base of eigenspace of N1 and denote β by the n × (n/2) matrix consisting of these n/2 columns. Correspondingly, we have n × (n/2) matrix γ with all columns from the set of columns of N2. Because 0 is the only vector in the intersection of eigenspaces of N1 and N2, n × n matrix Because N12γ = 0 implies that nonzero column vectors of N1γ are eigenvectors of N1 and r(N1γ) = n/2. Hence; N1γ and β are equal under certain column transformation; that is, there is an invertible matrix T1 such that N1γ = βT1. Correspondingly, there is an invertible matrix T2 such that N2β = γT2.y1 and y2 are (n/2) × n matrices. Naturally, the following equation is true:N1 and N2 as follows:N1γ = βT1 and N2β = γT2, the above three equations imply that N1 is similar to N2 is similar to Let X is similar to L of K and arbitrary nonzero α ∈ L, X is also similar to the following matrix:X is ±P.Hence, X is singular and similar to Y ⊕ N, where Y is nonsingular and N is nilpotent. Then X is ±P. If Y is s2N. Without loss of generality, we assume X = Y ⊕ N in the following proof since s2N holds under similarity transformations.At first, we need to prove that n1 is the same for Y and the order of n4 is the same for N. Then N12 = 0 implies the following equations are true:Let X − N1)2 = N22 = 0, we get the following equations after replacing n1 with Y − n1 and n4 with N − n4 in the previous equations:Since  is ±P. If X is similar to Y ⊕ N, where Y is nonsingular and N is nilpotent, then X is ±P if and only if Y is ±P by Corollaries X is nonsingular. Furthermore, if X is nonsingular and similar to Y1 ⊕ Y2, where all eigenvalues of Y1 are not in K and all eigenvalues of Y2 are in K. Then X is ±P if and only if Y1 is ±P and Y2 is ±P. It will take two steps to prove X is s2N.K) ≠ 2 and all eigenvalues of X are not in K; then for arbitrary nonzero α ∈ K, X is an  composite; that is, there exist idempotent matrices P1 and P2 ∈ Mn(K) such that X = αP1 − αP2. Suppose car(Q1(0) be the eigenspace of P1 with respect to 0, Q1(1) the eigenspace of P1 with respect to 1, let Q2(0) be the eigenspace of P2 with respect to 0, and Q2(1) the eigenspace of P2 with respect to 1. Both α and −α are not eigenvalues of X implies that Q1(0)∩Q2(0) = Q1(1)∩Q2(1) = Q1(0)∩Q2(1) = Q1(1)∩Q2(0) = {0}; then dim(Q1(0)) = dim(Q1(1)) = dim(Q2(0)) = dim(Q2(1)) = n/2 ) ≥ n/2 implies Q1(0)∩Q2(0) ≠ {0} or Q1(0)∩Q2(1) ≠ {0}, etc.); that is, n is even.Let S and T are n × (n/2) matrices with r(S) = r(T) = n/2 satisfying P1S = 0 and P2T = T; then n × n nonsingular matrix. Let P1 and P2 as follows:P1 and P2 are idempotent implies that VP1T and UP2S are idempotent and r(P1) = r(P2) = n/2 implies that VP1T = In/2 and UP2S = 0n/2. Hence, X is similar to the following matrix:X is s2N in Mn(K).Suppose K) = 2, X is  composite for arbitrary nonzero α ∈ K, we can similarly prove that X is s2N in Mn(K) replacing −α with α in the previous proof.When car(K) ≠ 2 and all eigenvalues of X are in K; then by jk = jk for every k ∈ Z+ and arbitrary nonzero α ∈ K. Suppose car(x + αi)]ri = (x2 − αi2)ri with 2∑i=1sri = n and αi ∈ K∖{0} is one of eigenvalues of X for every i ∈ [s]. Without loss of generality, we just need to prove Xi is s2N.Moreover, Xi is similar to C((x2 − αi2)ri) as follows:x2 − αi2)ri = xri2 − ari−22xri−22 − ⋯−a2x2 − a0. We have C((x2 − αi2)ri) = E2,1 + ⋯+Eri,2ri−12 + a0Eri1,2 + a2Eri3,2 + ⋯+ari−22Eri−1,2ri2 = + = N1 + N2. Obviously, both N1 and N2 are square nilpotent matrices; that is, Xi is s2N. Hence, X is s2N in Mn(K).Since K) = 2, all blocks in the Jordan reduction of X with respect to α ∈ K∖{0} have an even size by α are (x + α)si = ((x + α)2)si/2 = (x2 + α2)si/2, where si is even. Similarly, we can also prove that X is s2N in Mn(K).When car(


The monotonicity of the solutions of a class of nonlinear fractional differential equations is studied first, and the existing results were extended. Then we discuss monotonicity, concavity, and convexity of fractional derivative of some functions and derive corresponding criteria. Several examples are provided to illustrate the applications of our results. Fractional calculus is a generalization of the traditional integer order calculus. Recently, fractional differential equations have received increasing attention since behavior of many physical systems can be properly described as fractional differential systems. Most of the present works focused on the existence, uniqueness, and stability of solutions for fractional differential equations, controllability and observability for fractional differential systems, numerical methods for fractional dynamical systems, and so on see the monographs –4 and thIt is well known that the monotonicity, the concavity, and the convexity of a function play an important role in studying the sensitivity analysis for variational inequalities, variational inclusions, and complementarity. Since fractional derivative of a function is usually not an elementary function, its properties are more complicated than those of integer order derivative of the function. The focal point of this paper is to investigate the monotonicity, the concavity, and the convexity of fractional derivative of some functions.Now we recall some definitions and lemmas which will be used later. For more detail, see –4. a, b] of ℝ, the fractional order integral of a function f ∈ L1 of order α ∈ ℝ+ is defined byGiven an interval  can be written asRiemann-Liouville's derivative of order f is defined on the interval  and fn)((t) ∈ L1. The Caputo's fractional derivative of order α with lower limit a for f is defined asn − 1 < α ≤ n.Suppose that a function α ≤ 1, it holds thatParticularly, when 0 < α. Namely,Re(α) denotes the real parts of α.There exists a link between Riemann-Liouville and Caputo's fractional derivative of order α < 1, it holds thatParticularly, for 0 < f :  → ℝ with  ⊂ ℝ is said to be convex if whenever t1 ∈ , t2 ∈ , and θ ∈ , the inequalityA function RLDt0αf(t) and CDt0αf(t). Summarizing this paper forms the content of The rest of this paper is organized as follows. α < 1. The paper  and u(t0) ≥ 0, then the solutions u(t) of  of  are nonng) ≥ 0 and (d/dt)g) ≥ 0 on , then the solution u(t) of  of  is nondeg) ≤ 0 and (d/dt)g) ≤ 0 on , then the solution u(t) of  of  is not iAssume that 0 < tions of  exist.Ifg) ≥ 0, it holds that (1/Γ(α))∫t0t(t − s)α−1g)ds ≥ 0. Noting u(t0) ≥ 0, we have u(t) ≥ 0.The conclusion of (1) is obvious. In fact,  is equivd/dt)g) ≥ 0 on  and g) ≥ 0, thus t0, t1]. Hence u(t) is nondecreasing on , and (2) holds.Now we prove the validity of (2) and (3). First, by the definition of the Caputo's derivative, it holds from  that(10Similar to the proof of (2), we can prove that (3) holds. This completes the proof.g = λu, then (CDt0αu(t) = λu. The solution u(t) of CDt0αu(t) = λu isu(t0) ≥ 0, thenLemma  2.4 in  is a parλu, then  is CDtα < 1. Consider the fractional differential equationt > 0, we have t + sint ≥ 0 and (t + sint)′ = 1 − cost ≥ 0. By u(t) is nondecreasing in t for t ≥ 0. Assume that 0 < α < 1. Consider the fractional differential equationg) = −2, then g) < 0 and t ≥ t0. By f(t) is not increasing. In fact, by computation we get t0, ∞), thus f(t) is decreasing. Assume that 0 < The following fractional comparison principle is an improvement of Lemma  6.1 in  and Theoα < 1 and CDt0αf(t) ≥ CDt0αg(t) on interval . Suppose further that f(t0) ≥ g(t0), then f(t) ≥ g(t) on .Suppose that 0 < CDt0αf(t) − CDt0αg(t) = m(t), t ∈ . ThenIt0α on both sides of (m(t) ≥ 0, thus It0α(m(t)) ≥ 0. Then we havef(t) ≥ g(t) on , and the proof is completed.Set 1]. ThenCDt0α. Theorem  2.6 in  such that f(t0) ≤ 0, t0, t1], then RLDt0αf(t) is nondecreasing on ; f(t0) ≥ 0, t0, t1], then RLDt0αf(t) is not increasing on ; f(t0) > 0, t0, t1] and β ∈  such that RLDt0αf(t) is not increasing on  and is not decreasing on ; f(t0) < 0, η ∈  such that RLDt0αf(t) is nondecreasing on  and RLDt0αf(t) is not increasing on .Assume that 0 < d/dt)(RLDt0αf(t)) ≥ 0 on . Thus RLDt0αf(t) is nondecreasing on . By assumptions in (2), it follows that RLDt0αf(t) is not increasing in t on . Consequently, the conclusions of (1) and (2) are true.Using formula , we havef(t0) > 0 and t → t0. By the fact that δ1 > 0 such that (d/dt)(RLDt0αf(t)) ≤ 0 on . On the other hand, when β ∈  such that (d/dt)(RLDt0αf(t)) ≤ 0 on  and (d/dt)(RLDt0αf(t)) ≥ 0 on . Therefore, the conclusion of (3) is valid.Let us prove (3). Noting formula ,(24)ddtThe proof of (4) is similar to that of (3). This completes the proof. CDt0αf(t). Now we are to investigate the monotonicity of the function α < 1. If there exists an interval  such that t0, t1] and CDt0αf(t) is nondecreasing on . If t0, t1] and CDt0αf(t) is not increasing on . Assume that 0 < t0, t1] and d/dt)(CDt0αf(t)) ≥ 0 in t on . Hence, CDt0αf(t) is nondecreasing on interval . If d/dt)(CDt0αf(t)) ≤ 0. Hence, CDt0αf(t) is not increasing on . The proof is completed. Set The following examples illustrate applications of Theorems α < 1. Consider RLDt0αf(t), where f(t) = et, for all t0 ∈ ℝ. Since β > t0 such that RLDt0α(et) is decreasing on  and is increasing on . Since (sint)′′ ≤ 0 for t ∈  and (sint)′|t=π/2 = 0, by CDπ0.5αsint is decreasing on . By similar argument, CDπ1.5αsint is increasing on t ∈ . Since CDt0αsint = (1/Γ(1 − α))∫t0t(t − τ)α−cosτd, thus (1/Γ(1 − α))∫π0.5t(t − τ)α−cosτdτ is decreasing on  and (1/Γ(1 − α))∫π1.5t(t − τ)α−cosτdτ is increasing on t ∈ .Assume that 0 < α < 1. Consider CDt0αf(t); here t0 = 1 and f(t) = t − t2. Obviously, t ∈ , CDt0αf(t) is not increasing on .Assume that 0 < RLDt0αf(t) and CDt0αf(t). By formula  ≥ 0 on , f(t0) > 0, then RLDt0αf(t) is concave on . If f′′′(t) ≤ 0 on , f(t0) ≤ 0, then RLDt0αf(t) is convex on . Assume that 0 < CDt0αf(t). The next theorem is on the convexity and the concavity of α < 1. If there exists an interval  such that f′′′(t) ≥ 0 on , CDt0αf(t) is concave on . If f′′′(t) ≤ 0 on , CDt0αf(t) is convex on . Assume that 0 < f′′′(t) ≥ 0 on , d2/dt2)(CDt0αf(t)) ≥ 0 on . Hence, CDt0αf(t) is concave in t on . If f′′′(t) ≤ 0, on , d2/dt2)(CDt0αf(t)) ≤ 0 on . Therefore CDt0αf(t) is convex in t on . By formula , we haveα < 1. Consider the fractional differential equation CDt0αf(t); here t0 < 0 and f(t) = t − t2. Obviously, f′′′(t) = 0. For all t0 < 0, it holds that f′′′(t) = 0 on . Then by CDt0α(t − t2) is convex on . Assume that 0 < RLDt0αf(t), where f(t) = et, t0 ∈ ℝ. Obviously, Theorems RLDt0αet. Now we employ the method which is used in the proof of t ∈ . Thus (d2/dt2)(RLDt0αet) > 0 on  and (d2/dt2)(RLDt0αet) ≥ 0 on  and is concave on [β, +∞). Consider the concavity and convexity of the function proof of d2dt2 and CDt0αf(t). Based on the relation between the Riemann-Liouville fractional derivative and the Caputo's derivative, we obtain the criteria on the monotonicity, the concavity, and the convexity of the functions RLDt0αf(t) and CDt0αf(t). In the meantime, five examples are given to illustrate the applications of our criteria.The main part of this paper is to study the monotonicity, the concavity, and the convexity of the functions