We prove the Hyers-Ulam stability for this quartic functional equation by the directed method and the fixed point method on real Banach spaces. We also investigate the Hyers-Ulam stability for the mentioned quartic functional equation in non-Archimedean spaces.We obtain the general solution of the generalized quartic functional equation  ℱ is stable if any function f satisfying the equation ℱ  approximately is near to exact solution of ℱ. Moreover, a functional equation ℱ is hyperstable if any function f satisfying the equation ℱ  approximately is a true solution of ℱ.We say a functional equation The study of stability problems for functional equations is related to a question of Ulam  concernif(2x) = 16f(x), we getThe oldest quartic functional equation was introduced by Rassias in  and thenf : ℝ → ℝ is a solution of (f(x) = Q where the function Q : ℝ4 → ℝ is symmetric and additive in each variable. The fact that every solution of +fdirect method,” initiated by Hyers in 1941, had been applied by Cădariu and Radu for the first time in [Bodaghi et al.  applied  time in . In othe time in  and of t time in  (for mor time in –20).m. In case m = 2, then , an, an3), af(x+my)+ff(x+my)+ff(x+my)+fTo achieve our aim in this section, we need the following lemma.𝒳 and 𝒴 be real vector spaces. If a function f : 𝒳 → 𝒴 satisfies the functional equation  = m4f(x) for all integers m ≥ 2.Let equation  for all x = y = 0 in (f(0) = 0. Once more, by putting y = 0 in  = 34f(x). Now, assume that, for every k ≤ m − 1, we have f(kx) = k4f(x). If m = 2n, then f(mx) = f(2nx) = 24f(nx). Since n ≤ m − 1, we have f(nx) = n4f(x), and thus f(mx) = m4f(x). Let m = 2n + 1. Then, by substituting x,  y by mx,  x in  and m = 2n. Replacing these equalities in (f(mx) = m4f(x). This completes the proof.Letting y = 0 in , we get y = 0 in , we obtax,  x in , respect = 0 in (f(2x)=24fx,  x in , respect = 0 in (f(2x)=24ff : 𝒳 → 𝒴 satisfies the functional equation (f satisfiesIt is shown in [equation  if and o) =2m2m2−f(y)−2(m2atisfiesf(x+my)+fequation .𝒳 and 𝒴 be real vector spaces. Then a mapping f : 𝒳 → 𝒴 satisfies the functional equation  = 0. Let y = 0 in (f(2x) = 16f(x) for all x ∈ 𝒳. Setting x = 0 in (f(y) = 16f(y), we have f(−y) = f(y). Letting y = x in (f(3x) = 81f(x) for all x ∈ 𝒳. By induction, we obtain f(kx) = k4f(x) for all positive integers k. Replacing x by x + y and x − y in . Hence for each x, y ∈ 𝒳, we havef((m − 1)y) = (m − 1)4f(y) and sof satisfies the functional equation  and sof such thatx, y ∈ 𝒳, where m is an integer with m ≥ 2. Then there exists a unique quartic mapping 𝒯 : 𝒳 → 𝒴 such thatx ∈ 𝒳.Let y = 0 in /2n4} is Cauchy by  = 0 for all positive integers m ≥ 2 and all x, y ∈ 𝒳. Hence, by 𝒯 : 𝒳 → 𝒴 is a quartic mapping. Now, let 𝒯′ : 𝒳 → 𝒴 be another quartic mapping satisfying y = 0 in y 2x in (||f(2nx)2 so thatlimn→∞f = β||x||𝒳r + γ||y||𝒳s in Setting ows from  that such thatx, y ∈ 𝒳, where m is an integer with m ≥ 2. Then there exists a unique quartic mapping 𝒯 : 𝒳 → 𝒴 such thatx ∈ 𝒳.Suppose that ϕ = 0. Thus from (f(0) = 0. Putting y = 0 in 2, we havex ∈ 𝒳. If we show that the sequence {2n4f(x/2n)} is Cauchy, then it will be convergent by the completeness of 𝒴. For this, if we replace x by x/2l in −24β, γ, r, and s be nonnegative real numbers such that r, s > 4. Suppose that f : 𝒳 → 𝒴 is a mapping fulfillingx, y ∈ 𝒳, where m is an integer with m ≥ 2. Then there exists a unique quartic mapping 𝒯 : 𝒳 → 𝒴 such thatx ∈ 𝒳.Let x = y = 0 in (f(0) = 0. Taking ϕ = β||x||𝒳r + γ||y||𝒳s in First, we note that if we put y = 0 in , we haveWe are going to investigate the hyperstability of the given quartic functional equation  by usingd) be a complete generalized metric space and let 𝒥 : Δ → Δ be a mapping with Lipschitz constant L < 1. Then, for each element α ∈ Δ, either d = ∞ for all n ≥ 0 or there exists a natural number n0 such thatd < ∞ for all n ≥ n0;𝒥nα} is convergent to a fixed point β* of 𝒥;the sequence {β* is the unique fixed point of 𝒥 in the set Δ1 = {β ∈ Δ : d < ∞};d≤(1/(1 − L))d for all β ∈ Δ1.Let  = 0 and let φ : 𝒳 × 𝒳 → [0, ∞) be a function such thatx, y ∈ 𝒳, where m is an integer with m ≥ 2. If there exists a constant M ∈ , such thatx, y ∈ 𝒳, then there exists a unique quartic mapping 𝒯 : 𝒳 → 𝒴 such thatx ∈ 𝒳.Let 𝔇 on Δ × Δ as follows:C, and 𝔇 = ∞, otherwise. In a similar way to the proof of [𝔇 is a generalized metric on Δ and the metric space  is complete. Here, we define the mapping 𝒥 : Δ → Δ byg, h ∈ Δ such that 𝔇 < C, by definitions of 𝔇 and 𝒥, we havex ∈ 𝒳. Using  ≤ M𝔇 for all g, h ∈ Δ. Hence, 𝒥 is a strictly contractive mapping on Δ with a Lipschitz constant M. We now show that 𝔇 < ∞. Putting y = 0 in  < ∞ for all n ≥ 0, and thus in this theorem we have n0 = 0. Consequently, the parts (iii) and (iv) of 𝒯 : 𝒳 → 𝒴 such that 𝒯 is a fixed point of 𝒥 and that 𝒥nf → 𝒯 as n → ∞. Thusx ∈ 𝒳, and sox ∈ 𝒳. Now, it follows from  = 0 for all integers  m ≥ 2  and all  x, y ∈ 𝒳. It follows from 𝒯 : 𝒳 → 𝒴 is a quartic mapping which is unique.We wish to make the conditions of 𝒳. Using , we gety = 0 in , we obtahow that  is true ows from  that = 0. If we put φ = α||x||𝒳r + β||y||𝒳r in Note that inequality  implies In the next result, we prove the hyperstability of quartic functional equations under some conditions.r, s, and α be nonnegative real numbers with 0 < r + s ≠ 4 and let f : 𝒳 → 𝒴 be a mapping such thatx, y ∈ 𝒳. Then f is a quartic mapping on 𝒳.Let x = y = 0 in (f(0) = 0. Again, if we put y = 0 in (f(2x) = 16f(x) for all x ∈ 𝒳. It is easy to check that f(2nx) = 2n4f(x), and so f(x) = f(2nx)/2n4 for all x ∈ 𝒳 and n ∈ ℕ. Now, it follows from f is a quartic mapping when φ = α||x||𝒳r||y||𝒳s.Putting y = 0 in , we get y = 0 in , then we𝕂 equipped with a function  |·| from 𝕂 into [0, ∞) such that |r| = 0 if and only if r = 0, |rs| = |r||s|, and |r + s|≤ max{|r|, |s|} for all r, s ∈ 𝕂. Clearly |1| = |−1| = 1 and |n| ≤ 1 for all n ∈ ℕ.We recall some basic facts concerning non-Archimedean spaces and some preliminary results. By a non-Archimedean field we mean a field 𝒳 be a vector space over a scalar field 𝕂 with a non-Archimedean nontrivial valuation |·|. A function ||·|| : 𝒳 → ℝ is a non-Archimedean norm  if it satisfies the following conditions:(i)x|| = 0 if and only if x = 0;||(ii)rx|| = |r|||x||, ;||(iii)the strong triangle inequality (ultrametric); namely,Then  is called a non-Archimedean space. Due to the fact thatxn} is Cauchy if and only if {xn+1 − xn} converges to zero in a non-Archimedean normed space 𝒳. By a complete non-Archimedean normed space we mean one in which every Cauchy sequence is convergent.Let p-adic numbers as a number theoretical analogue of power series in complex analysis. The most interesting example of non-Archimedean spaces is p-adic numbers. A key property of p-adic numbers is that they do not satisfy the Archimedean axiom: for all x, y > 0, there exists an integer n such that x < ny.In , Hensel p be a prime number. For any nonzero rational number x = pr(m/n) in which m and n are coprime to the prime number p. Consider the p-adic absolute value |x|p = pr on ℚ. It is easy to check that |·| is a non-Archimedean norm on ℚ. The completion of ℚ with respect to |·| which is denoted by ℚp is said to be the p-adic number field. One should remember that if p > 3, then |2n| = 1 for all integers n. In [Let rs n. In , the stars n. In ).𝒳 is a normed space and 𝒴 is a complete non-Archimedean space unless otherwise stated explicitly. In the upcoming theorem, we prove the stability of the functional equation  such thatx, y ∈ 𝒳. Suppose that f : 𝒳 → 𝒴 is a mapping satisfying the equalityx, y ∈ 𝒳, where m is an integer with m ≥ 2. Then there exists a unique quartic mapping Q : 𝒳 → 𝒴 such thatx ∈ 𝒳 where Let y = 0 in /16n} is Cauchy by −1auchy by  and 69)y = 0 in  we have||f(x)−f, a, ay = 0  so thatlimn→∞f→[0, ∞) be a function satisfying Γ(|r|s) ≤ Γ(|r|)Γ(s) for all r, s ∈ [0, ∞) for which Γ(|2|) < |16|. Suppose that f : 𝒳 → 𝒴 is a mapping satisfying the inequalityx, y ∈ 𝒳, where m is an integer with m ≥ 2. Then there exists a unique quartic mapping Q : 𝒳 → 𝒴 such thatx ∈ 𝒳.Let ϕ : 𝒳 × 𝒳 → [0, ∞) by ϕ = α(Γ(||x||) + Γ(||y||)), we havex, y ∈ 𝒳. We also havex ∈ 𝒳. Now, Defining We have the following result which is analogous to ϕ : 𝒳 × 𝒳 → [0, ∞) such thatx, y ∈ 𝒳. Suppose that f : 𝒳 → 𝒴 is a mapping satisfying the inequalityx, y ∈ 𝒳, where m is an integer with m ≥ 2. Then there exists a unique quartic mapping Q : 𝒳 → 𝒴 such thatx ∈ 𝒳 where Let x ∈ 𝒳. If we replace x by x/2n+1 in the above inequality and multiply both sides of } is Cauchy. Since the non-Archimedean space 𝒴 is complete, this sequence converges in 𝒴 to the mapping Q. Indeed,x ∈ 𝒳 and non-negative integers n. Since the right hand side of inequality −1ude from  and 82)81)||f(2||f(2(81)ion and (||f(x)−16 Indeed,Q(x)=limapplying . Now, inα > 0, 𝒳 be a non-Archimedean space and let Γ : [0, ∞)→[0, ∞) be a function satisfying Γ(|r|s) ≤ Γ(|r|)Γ(s) for all r, s ∈ [0, ∞) for which Γ(|2|−1) < |16|−1. Suppose that f : 𝒳 → 𝒴 is a mapping satisfying the inequalityx, y ∈ 𝒳, where m is an integer with m ≥ 2. Then there exists a unique quartic mapping Q : 𝒳 → 𝒴 such thatx ∈ 𝒳.Let The proof is a direct consequence of


A kth-order slant weighted Toeplitz operator on L2(β) is given by Uϕ = WkMϕ, where Mϕ is the multiplication on L2(β) and Wk is an operator on L2(β) given by Wkenk(z) = (βn/βnk)en(z), {en(z) = zk/βk}k∈ℤ being the orthonormal basis for L2(β). In this paper, we define a kth-order slant weighted Toeplitz matrix and characterise Uϕ in terms of this matrix. We further prove some properties of Uϕ using this characterisation.Let  H2 and L2 spaces. During the past few decades, different generalisations of these spaces, like the weighted function spaces HwP and LwP and the weighted sequence spaces H2(β) and L2(β) have developed. Shields [L2(β). Lauric [H2(β). Motivated by these studies, we introduced and studied the notion of a slant weighted Toeplitz operator [L2(β). In this paper, we study a kth-order slant weighted Toeplitz operator on the space L2(β). We begin with the following preliminaries.Toeplitz operators and slant Toeplitz operators  have bee Shields  has made. Lauric  has discoperator  on L2 ≤ 1 when n ≥ 0 and 0 < (βn/βn−1) ≤ 1 when n ≤ 0.Let k ≥ 2, βkn/βn ≤ M < ∞. Consider the following spaces [Throughout the paper, we assume that for a fixed integer g spaces , 6:(1)LL2(β), ||·||β) is a Hilbert space [ek(z) = zk/βk}k∈ℤ and with an inner product defined byH2(β) is a subspace of L2(β). Further, the spaceP : L2(β) → H2(β) is the orthogonal projection of L2(β) onto H2(β). For a given ϕ ∈ L∞(β), the induced weighted multiplication operator [Mϕ and is given by Mϕ : L2(β) → L2(β) such thatThen, (rt space  with an operator  is denotϕ1(z) = z, then Mϕ1 = Mz is the operator defined as Mzek(z) = wkek+1(z), where wk = βk+1/βk for all k ∈ ℤ, and it is known as a weighted shift [If we put ed shift .Aϕ is an operator on L2(β) defined as Aϕ : L2(β) → L2(β) such that Aϕek(z) = (1/βk)∑n=−∞∞an−k2βnen(z).The slant weighted Toeplitz operator  Aϕ is aW : L2(β) → L2(β) is such thatAϕ can be alternately defined byIf Wk : L2(β) → L2(β) is such thatWk isWk is given bySuppose that the operator k ≥ 2, the kth order slant weighted Toeplitz operator Uϕ : L2(β) → L2(β) is such that Uϕ(f) = WkMϕ(f) for all f ∈ L2(β). Thus, Uϕej(z) = (1/βj)∑n=−∞∞ank−jβnen(z).For an integer i, j)th entry of the matrix of Uϕ is given byUϕ with respect to this basis is as follows:The en(z) is not divisible by k. But since this is true for all n ∈ ℤ, the second possibility is ruled out. Hence, ϕ − ψ = 0 or ϕ = ψ.Hence,MzWk = WkMzk; (ii) MzmWk = WkMzkm, m ∈ ℤ. Consider the following: (i) (i) It is sufficient to prove thatn is not a multiple of k. Then,Suppose that n = pk (multiple of k),Now, when On the other hand,n ∈ ℤ,Hence from  and 18)18), we gm. (ii) We prove the result by induction on m = 1, the result is true by part (i). For m = l. Then,Suppose that the result is true for Nowm ∈ Z+.Hence by induction, the result is true for all m = −1 and 1 ≤ j ≤ k − 1,m = −1 and j = nk (a multiple of k),m = −1 also. Therefore, by using induction, we can prove it for all negative integers m. The case when m = 0 is trivially true.For ≤ k − 1,MzmWkenk+e of k),MzmWkej = ∑n=−∞∞anzn ∈ L∞(β). Then,Let Mϕ(z)Uψ(z) = Uϕ(zk)ψ(z).Consider the following: The proof is as follows:wn = βn+1/βn for all n ∈ ℤ. Then, the kth order slant weighted Toeplitz matrix corresponding to the weight sequence 〈wn〉 is a bilaterally infinite matrix 〈λij〉 such thatLet U is a kth order slant weighted Toeplitz operator if and only if MzU = UMzk.We have proved earlier  that U ikth order slant weighted Toeplitz operator in terms of the matrix previously defined.Next, we prove a characterisation of the U on L2(β) is a kthorder slant weighted Toeplitz operator is that its matrix with respect to the orthonormal basis {ek(z) = zk/βk}k∈ℤ is a kthorder slant weighted Toeplitz matrix.A necessary and sufficient condition that an operator U is a kth order slant weighted Toeplitz operator. Then, the corresponding matrix 〈λij〉 is given bywn = βn+1/βn for all n ∈ ℤ. Thus, the matrix of  U is a kth order slant weighted Toeplitz matrix. Conversely, we assume that U is an operator on L2(β) whose matrix is a kth order slant weighted Toeplitz matrix. This means that for all i, j ∈ ℤ, we haveMzU = UMzk. Therefore, we conclude that U is a kth order slant weighted Toeplitz operator.Suppose that S : L2(β) → L2(β) given by Sej = (1/wj)ej+1. Then, S*ej = (1/wj−1)ej−1.Next, consider the operator S is bounded as 〈wn〉 is always positive and bounded.Now, S* = Mz−1.Consider the following: The proof is as follows:U on L2(β) is a kthorder slant weighted Toeplitz operator if and only if  U = Mz−1UMzk, where Mz and Mzk, are the multiplication operators induced by z and zk, respectively.A bounded operator U be a kth order slant weighted Toeplitz operator on L2(β). Then, from  satisfying U = Mz−1UMzk for some fixed integer k ≥ 2. Then, for all i, j ∈ ℤ,U is a kth order slant weighted Toeplitz matrix. Hence, by U is a kth order slant weighted Toeplitz operator.For the converse part, we assume that k ≥ 2, the set of all kthorder slant weighted Toeplitz operators is weakly closed and hence strongly closed.For a fixed integer n, Un is a kth order slant weighted Toeplitz operator and Un → U weakly. Then, for f, g ∈ L2(β), we get that 〈Unf, g〉→〈Uf, g〉.We assume that for each positive integer n, Un = Mz−1UnMzk.From the previous theorem, this implies thatU = Mz−1UMzk. Therefore, U is a kth order slant weighted Toeplitz operator.Hence, kth order slant weighted Toeplitz operator on H2(β) is denoted by Vϕ and defined as Vϕej(z) = (1/βj)∑n=0∞ank−jβnen(z).  j = 0,1, 2,….The compression of a Tϕ : H2(β) → H2(β) is the Toeplitz operator on H2(β) induced by ϕ ∈ L∞(β). The matrix of Vϕ is unilaterally infinite and has the formϕ = ∑n=−∞∞anzn.Alternatively,ϕ → Vϕ is linear and one to one. The mapping TϕWk = PMϕWk = PWkMϕ(zk) = PUϕ(zk) = Vϕ(zk).Further, we observe the following.


This paper studies impulsive control systems with finite andinfinite delays. Several stability criteria are established by employingthe largest and smallest eigenvalue of matrix. Our sufficient conditionsare less restrictive than the ones in the earlier literature. Moreover,it is shown that by using impulsive control, the delay systems can bestabilized even if it contains no stable matrix. Finally, some numericalexamples are discussed to illustrate the theoretical results. As is known, a well-developed theory of impulsive control systems has come into existence x(t − u)du, where s ∈  is a Volterra type functional. In this case, when α = −∞, the interval  is understood to be replaced by  ∈ C, and ∫0r|h(u)|du < ∞.Consider the delay control systemtk, U)} (refer the reader to  → ℝn is continuous.Let ϕ :  → ℝn is continuous and ∫0∞+|h(u)|du < ∞.Furthermore, we can investigate the following impulsive control system with infinite delays:A ∈ ℝn×n, λmax(A) denotes the largest eigenvalue of A and similarly λmin(A) denotes the smallest eigenvalue of A. The norm of matrix is A ∈ ℝn×n, A, which is defined to be the unique positive definite matrix satisfying For any matrix A=A (see ).Let us define the following class of functions for later use:In order to prove our main results, we need the following definitions and lemmas.A ∈ ℝn×n is stable (or Hurwitz) if the eigenvalues of the matrix A all have negative real parts and there is a unique positive definite matrix P ∈ ℝn×n that solves the Lyapunov equationQ is any n × n positive definite matrix.A matrix A ∈ ℝn×n is unstable if the eigenvalues of the matrix A all have positive real parts and there is a unique positive definite matrix P ∈ ℝn×n that solves the Lyapunov equationQ is any n × n positive definite matrix.A matrix x(t) = x is the solution of . Then the trivial solution of (H1) > 0 such that ||ϕ|| < δ implies ||x|| < ɛ, t ≥ t0;stable, if for any H2)(δ in (H1) is independent of t0;uniformly stable, if the H3)(H2) holds and there exists some η > 0 such that for any ɛ > 0 there exists some T = T > 0 such that t0 ≥ 0 and ||ϕ|| < η together imply ||x|| < ɛ, t ≥ t0 + T.uniformly asymptotically stable, if (Assume that ution of  and 4)  x(t) = xution of  and (4) a, b, c ∈ K1, p ∈ PC, and s > 0. Suppose further that V : ;tk, ψ) ∈ ℝ+ × PC) for which ψ(0−) = ψ(0);τ = infk∈N{tk − tk−1} > 0,  M2 = supq>0∫qg(q)(ds/c(s)) < ∞, and M1 = inft≥0∫tt+τp(s)ds > M2.Then the trivial solution of , and g ∈ K2 and V : ;V + I) ≤ g)) for all  ∈ ℝ+ × PC) for which ψ(0−) = ψ(0);τ = supk∈N{tk − tk−1} < ∞, M1 = supt≥0∫tt+τp(s)ds < ∞, and M2 = infq>0∫g(q)q(ds/c(s)) > M1.Then the trivial solution of , and g ∈ K2 and V : ;0, ϕ) of , wheneveV) ≤ g) for all k ∈ N and all x ∈ S(ρ1);M1 = supk∈N∫tktk+1p(s)ds < ∞ and M2 = infu>0∫g(u)u(ds/c(s)) > M1.Then the trivial solution of ..4). Our τ = infk∈N{tk − tk−1} > 0 and assume all the eigenvalues of A have negative real parts. Suppose that there exist a constant γ ≥ 1 and two symmetrical positive definite matrices P, Q ∈ ℝn×n such that𝕄 = ∫0γ|h(u)|du. Then the trivial solution of  = β2s and g(V(ψ(0))) ≥ V(ψ(s)), s ∈ , so we have β2V(ψ(0)) ≥ V(ψ(−r)) and β2V(ψ(0)) ≥ V(ψ(−u)). Calculating the derivative of V along solutions of ,1lnβτ−λmimax(P),lnβτ−λmiax(P),1lnβτ−λmiA is unstable.Our next theorem establishes conditions for uniform asymptotic stability of trivial solution of  when matτ = supk∈N{tk − tk−1} < ∞ and assume all the eigenvalues of A have positive real parts. Suppose that there exist a constant 0 < γ < 1 and two symmetrical positive definite matrices P, Q ∈ ℝn×n such that𝕄 = ∫0γ|h(u)|du. Then the trivial solution of  = γ2s. Then if V(ψ(0)) ≥ g(V(ψ(s))), for s ∈ , we have V(ψ(−r)) ≤ V(ψ(0))/γ2 and V(ψ(−u)) ≤ V(ψ(0))/γ2. Calculating the derivative of V along solutions of |du. Then the trivial solution of  = γ2s. Then if V(ψ(0)) ≥ g(V(ψ(s))), for s ∈ , we have V(ψ(−r)) ≤ V(ψ(0))/γ2 and V(ψ(−u)) ≤ V(ψ(0))/γ2. Calculating the derivative of V along solutions of =xTPxequality  on γ. Thtisfies2λmin(P)|In this paper, a new technique is offered to establish the stability criteria for finite and infinite delay impulsive control systems. In the above theories, rather than employing Razumikhin techniques and Lyapunov functions, we adopt the largest and smallest eigenvalue so that not only can they be easier constructed, but also the conditions ensuring the required stability are less restrictive. Furthermore, in Consider the following impulsive control systems:h(u) = eu−,  u > 0 and with the following parameter matrices:γ ∈  such thatThe trivial solution of the system  is unifoP as follows:Q = ATP + PA. So by  = 2eu−, u > 0 and with the following parameter matrices:Consider the impulsive control systems  where h such thatThe trivial solution of the system  is unifoP as follows:Q = ATP + PA. So by (ρ = 0.009. Also, a simple check shows that both of these conditions are satisfied by choosing γ = 0.6. Therefore, robust exponential stability of the trivial solution of (Obviously,ution of  can be o


𝒮(ℝn) and 𝒮′(ℝn).Parseval's formula and inversion formula for the S-transform are given. A relation between the S-transform and pseudodifferential operators is obtained. The S-transform is studied on the spaces  The space of all rapidly decreasing functions on ℝn is denoted by 𝒮(ℝn) or simply 𝒮. Elements in the dual space 𝒮′ of 𝒮 are called tempered distribution.A function ω appears as a superposition of time-frequency shifts as follows:Some properties of S-transform can be found in –8 and ceω = m(ξ), that is, independent of t, thenω is a multiplication operator. In particular, if ω = 1, then (Sωϕ) = (ℱϕ)(ξ).If ω = m(t), thenIf ω1 and ω2 be the window functions such thatϕ1, ϕ2 ∈ L2(ℝn) and let (Sω1ϕ1) and (Sω2ϕ2) be the S-transforms of ϕ1 and ϕ2, respectively. ThenLet This immediately implies the Plancherel formulaConsiderϕ ∈ L2(ℝn) and window function ω satisfy the condition (If ondition  of the pBy the previous theorem we can writeHenceω be a window function and Sω is the S-transform. Then the transform Sω* defined byω. If ϕ ∈ L2(ℝn) and ψ ∈ L2(ℝn × ℝn), then , thenLet ondition . If ψ1,Considerω satisfies the condition  function or a tempered distribution on ℝn. Then the operatorLet The pseudodifferential operator plays an important role in the theory of partial differential equations. The pseudodifferential operator has been studied on function and distribution spaces by many authors. Details of the concept can be found in , 10.ϕ with respect to a window function ω is given byσ = ω.Here we give a direct relation between S-transform and pseudodifferential operator which will may be very useful in the study of S-transform of distribution spaces. The continuous S-transform of a function In this section we will investigate the S-transform of tempered distribution by means of the Fourier transform.ω ∈ 𝒮(ℝn2), then Sω maps 𝒮(ℝn) into 𝒮(ℝn2).If ωϕ) ∈ 𝒮(ℝn2), since the Fourier transform is continuous isomorphism from 𝒮(ℝn) to 𝒮(ℝn), and its inverse is also a continuous isomorphism from 𝒮(ℝn) to 𝒮(ℝn) (see [By  we haveℝn) , then Sω maps 𝒮′(ℝn) into 𝒮′(ℝn2).If f ∈ 𝒮′(ℝn) and ψ ∈ 𝒮(ℝn2), we haveωf) ∈ 𝒮′(ℝn2).For any ω ∈ 𝒮′(ℝn2), then Sω maps 𝒮(ℝn) into 𝒮′(ℝn2).If ϕ ∈ 𝒮(ℝn) and ψ ∈ 𝒮(ℝn2), thenω ∈ 𝒮′(ℝn2), we haveℱ1(Sωϕ) ∈ 𝒮′(ℝn2) and hence (Sωϕ) ∈ 𝒮′(ℝn2).If


The purpose of this paper is to investigate the problem of finding the approximate element of the common set of solutions of a split equilibrium problem and a hierarchical fixed point problem in a real Hilbert space. We establish the strong convergence of the proposed method under some mild conditions. Several special cases are also discussed. Our main result extends and improves some well-known results in the literature. H be a real Hilbert space, whose inner product and norm are denoted by 〈·, ·〉 and ||·||. Let C be a nonempty closed convex subset of H. We introduce the following definitions which are useful in the following analysis.Let T:  C → H is said to be (a)monotone, if(b)α > 0 such thatstrongly monotone, if there exists (c)α-inverse strongly monotone, if there exists α > 0 such that(d)nonexpansive, if(e)k-Lipschitz continuous, if there exists a constant k > 0 such that(f)C, if there exists a constant 0 ≤ k < 1 such thatcontraction on It is easy to observe that every α-inverse strongly monotone T is monotone and Lipschitz continuous. It is well known that every nonexpansive operator T:  H → H satisfies, for all  ∈ H × H, the inequalityx, y) ∈ H × Fix(T),The mapping tone, if〈Tx−Ty,x−uch that〈Tx−Ty,x−T is to find x ∈ C such thatF(T) the set of solutions of (F(T) is closed and convex and PF(T) is well defined (see [The fixed point problem for the mapping uch thatTx=x.We ned (see ).x ∈ C such thatEP(F). Numerous problems in physics, optimization, and economics reduce to finding a solution of (EP(F) is nonempty. In 2007, Plubtieng and Punpaeng [F(T)∩EP(F).The equilibrium problem denoted by EP is to find uch thatF≥0,uch thatF≥0,Punpaeng  introducH1 and H2 be two real Hilbert spaces. Given operators f:  H1 → H1 and g:  H2 → H2, a bounded linear operator A:  H1 → H2, and nonempty, closed, and convex subsets C⊆H1 and Q⊆H2, the SVIP is formulated as follows: find a point x* ∈ C such thatBi: Hi → 2Hi is a set-valued mapping for i = 1,2. Later Byrne et al. [Recently, Censor et al.  introduc∀y∈Q.In , Moudafi∀y∈Q.In  for the uch that〈f(x∗),x−e et al.  and Moude et al. .F1:  C × C → R and F2: Q × Q → R be nonlinear bifunctions and let A: H1 → H2 be a bounded linear operator; then, the split equilibrium problem (SEP) is to find x* ∈ C such thatp ∈ EP(F1) : Ap ∈ EP(F2)}.Very recently, Kazmi and Rizvi  studied uch thatF1≥uch thaty∗=Ax∗∈Q S: C → H be a nonexpansive mapping. The following problem is called a hierarchical fixed point problem: find x ∈ F(T) such thatf:  C → H is a contraction mapping and {αn} and {βn} are two sequences in . Under some certain restrictions on parameters, Yao et al. proved that the sequence {xn} generated by (z ∈ F(T), which is the unique solution of the following variational inequality:U is a Lipschitzian mapping and F is a Lipschitzian and strongly monotone mapping. They proved that under some approximate assumptions on the operators and parameters, the sequence {xn} generated by  and andS: C uch that〈x−Sx,y−xuch that〈x−Sx,y−xquality:〈(I−f)z,yated by (〈ρU(z)−μFC.In this section, we recall some basic definitions and properties, which will be frequently used in our later analysis. Some useful results proved already in the literature are also summarized. The first lemma provides some basic properties of projection onto PC denote the projection of H onto C. Then, one has the following inequalities:Let F:  C × C → ℝ be a bifunction satisfying the following assumptions: (i)F = 0, for all x ∈ C;(ii)F is monotone; that is, F + F ≤ 0, for all x, y ∈ C;(iii)x, y, z ∈ C, limt→0F(tz + (1 − t)x, y) ≤ F;for each (iv)x ∈ C, y → F is convex and lower semicontinuous;for each (v)r > 0 and z ∈ C, there exists a bounded subset K of H1 and x ∈ C∩K such thatfor fixed Let F1: C × C → ℝ satisfies r > 0 and for all x ∈ H1, define a mapping TrF1: H1 → C as follows:(i)TrF1 is nonempty and single-valued;(ii)TrF1 is firmly nonexpansive; that is,(iii)F(TrF1) = EP(F1);(iv)EP(F1) is closed and convex.Assume that F2:  Q × Q → ℝ satisfies s > 0 and for all u ∈ H2, define a mapping TsF2:  H2 → Q as follows:TsF2 satisfies conditions (i)–(iv) of F(TsF2) = EP, where EP is the solution set of the following equilibrium problem:Assume that F1:  C × C → ℝ satisfies TrF1 be defined as in x, y ∈ H1 and r1, r2 > 0. ThenAssume that C be a nonempty closed convex subset of a real Hilbert space H. If  T:  C → C is a nonexpansive mapping with Fix(T) ≠ ∅, then the mapping I − T is demiclosed at 0; that is, if {xn} is a sequence in C weakly converging to x and if {(I − T)xn} converges strongly to 0, then (I − T)x = 0.Let U:  C → H be τ-Lipschitzian mapping and let F:  C → H be a k-Lipschitzian and η-strongly monotone mapping; then for 0 ≤ ρτ < μη, μF − ρU is μη − ρτ-strongly monotone; that is,Let λ ∈  and μ > 0. Let F:  C → H be an k-Lipschitzian and η-strongly monotone operator. In association with nonexpansive mapping T:  C → C, define the mapping Tλ:  C → H byTλ is a contraction provided that μ < (2η/k2); that is,Suppose that an} is a sequence of nonnegative real numbers such thatγn} is a sequence in  and δn is a sequence such thatn=1∞γn = ∞;∑n→∞δn/γn ≤ 0 or ∑n=1∞|δn| < ∞.limsupThen limn→∞an = 0.Assume that {C be a closed convex subset of H. Let {xn} be a bounded sequence in H. Assume that w-limit set ww(xn) ⊂ C where ww(xn) = {x:  xni⇀x};the weak z ∈ C, limn→∞||xn − z|| exists.for each  Then {xn} is weakly convergent to a point in C.Let In this section, we suggest and analyze our method and we prove a strong convergence theorem for finding the common solutions of the split equilibrium problem -16) and and16) aH1 and H2 be two real Hilbert spaces and let C⊆H1 and Q⊆H2 be nonempty closed convex subsets of Hilbert spaces H1 and H2, respectively. Let A:  H1 → H2 be a bounded linear operator. Assume that F1: C × C → ℝ and F2: Q × Q → ℝ are the bifunctions satisfying F2 is upper semicontinuous in first argument. Let S, T:  C → C be a nonexpansive mapping such that Λ∩F(T) ≠ ∅. Let F:  C → C be an k-Lipschitzian mapping and η-strongly monotone and let U:  C → C be τ-Lipschitzian mapping. Now we introduce the proposed method as follows.Let x0 ∈ C arbitrarily, let the iterative sequences {un}, {xn}, and {yn} be generated byrn}⊂ and γ ∈ , L is the spectral radius of the operator A*A, and A* is the adjoint of A. Suppose that the parameters satisfy 0 < μ < (2η/k2), 0 ≤ ρτ < ν, where αn} and {βn} are sequences in  satisfying the following conditions: n→∞αn = 0 and ∑n=1∞αn = ∞;limn→∞(βn/αn) = 0;limn=1∞ | αn−1 − αn | <∞ and ∑n=1∞|βn−1 − βn| < ∞;∑n→∞rn < limsupn→∞rn < 2ς and ∑n=1∞|rn−1 − rn| < ∞.liminfFor a given The proposed method is an extension and improvement of the method of Wang and Xu  for findU = f, F = I, ρ = μ = 1, we obtain an extension and improvement of the method of Yao et al. [If the Lipschitzian mapping o et al.  for findf with a coefficient α ∈ [0,1) in other papers .The contractive mapping ers see , 22, 27), 27f wit This shows that Our method can be viewed as extension and improvement for some well-known results as follows x* ∈ Λ∩F(T). Then {xn},  {un}, and {yn} are bounded.Let x* ∈ Λ∩F(T); we have x* = TrnF1(x*) and Ax* = TrnF2(Ax*). ThenL, it follows thatγ, we getVn = αnρU(xn)+(I − αnμF)(T(yn)). Next, we prove that the sequence {xn} is bounded; without loss of generality we can assume that βn ≤ αn for all n ≥ 1. From (Let ows from  that(36ows thatγ2〈(TrnF2*). Then||un−x∗|| 1. From , we haven, we obtain ||xn − x*|| ≤ max{||x0 − x*||, (1/(1 − ρ))(||(ρU − μF)x*|| + ||Sx* − x*||)}, for n ≥ 0 and x0 ∈ C. Hence {xn} is bounded and, consequently, we deduce that {un}, {yn}, {S(xn)}, {T(xn)}, {F(T(yn))}, and {U(xn)} are bounded.By induction on x* ∈ Λ∩F(T) and {xn} the sequence generated by the n→∞||xn+1 − xn|| = 0;limw-limit set ww(xn) ⊂ F(T), (ww(xn) = {x : xni⇀x}).the weak Let un = TrnF1(xn + γA*(TrnF2 − I)Axn) and un−1 = Trn−1F1(xn−1 + γA*(Trn−1F2 − I)Axn−1)  it follows from σn : = ||TrnF2Axn − Axn|| and χn : = ||TrnF1(xn + γA*(TrnF2 − I)Axn) − (xn + γA*(TrnF2 − I)Axn)||. Without loss of generality, let us assume that there exists a real number μ such that rn > μ > 0, for all positive integers n. Then we getn→∞||un − xn|| = 0. Since x* ∈ Λ∩F(T) by using (γ(1 − Lγ) > 0, limn→∞||xn+1 − xn|| = 0, αn → 0, and βn → 0, we obtainTrnF1 is firmly nonexpansive, we haven→∞||xn+1 − xn|| = 0, αn → 0, βn → 0, and limn→∞||(TrnF2 − I)Axn|| = 0, we obtainz ∈ Λ∩F(T); since T(xn) ∈ C, we haven→∞||xn+1 − xn|| = 0,  αn → 0,  βn → 0,  ||ρU(xn) − μF(T(yn))||, and ||Sxn − xn|| are bounded and limn→∞||xn − un|| = 0, we obtainxn} is bounded, without loss of generality, we can assume that xn⇀x* ∈ C. It follows from x* ∈ F(T). Therefore ww(xn) ⊂ F(T).Since n).From  and the ).From (||yn−yn−1estimate||xn+1−xnby using  and 37)un = Trows from  and (37)y using . Next, we show that w ∈ EP(F1). Since un = TrnF1(xn + γA*(TrnF2 − I)Axn), we haveF1 thatn→∞||un − xn|| = 0, limn→∞||(TrnF2 − I)Axn|| = 0, and xn⇀w, it easy to observe that unk → w. It follows by F1 ≤ 0, for all y ∈ C.Since {t ≤ 1 and y ∈ C, let yt = ty + (1 − t)w; we have yt ∈ C. Then, from Assumptions F1 ≥ 0. From F1 ≥ 0, which implies that w ∈ EP(F1).For any 0 < Aw ∈ EP(F2). Since {xn} is bounded and xn⇀w, there exists a subsequence {xnk} of  {xn} such that xnk → w and since A is a bounded linear operator, Axnk → Aw. Now set vnk = Axnk − TrnkF2Axnk. It follows from  and hence w ∈ Λ.Next, we show that ows from  that limρτ < ν andμF − ρU is μη − ρτ strongly monotone, and we get the uniqueness of the solution of the variational inequality (z ∈ Λ  ∩  F(T).Thus we haveequality  and denon→∞〈ρU(z) − μF(z), xn − z〉≤0. Since {xn} is bounded, there exists a subsequence {xnk} of {xn} such thatxn → z. Considerγn = αn(ν − ρτ) and δn = (2αn(ν − ρτ)/(1 + αn(ν − ρτ))){(1/(ν − ρτ))〈ρU(z) − μF(z), xn+1 − z〉 + ((1 − αnν)βn/αn(ν − ρτ))||Sz − z||||xn+1 − z||}.Next, we claim that limsupxn → z. This completes the proof.SinceS = I − (ρU − μF), then we can get the variational inequality , x − z〉 ≥ 0, for all x ∈ Λ∩F(T), which is just the variational inequality studied by Suzuki [In hierarchical fixed point problem , if S = equality . In 58)S = I −  and and16) a


The aim of this paper is to give several characterizations for the property of weak exponential expansiveness for evolution families in Banach spaces. Variants for weak exponential expansiveness of some well-known results in stability theory (Datko (1973), Rolewicz (1986), Ichikawa (1984), and Megan et al. (2003)) are obtained. T(t)}t≥0 is uniformly exponentially stable if and only if for each vector x from the Banach space X the function t → ||T(t)x|| lies in L2(ℝ+). Later, Pazy generalizes the result in )In the present paper, we introduce the concept of weak exponential expansiveness for evolution families which is an extension of classical concept of exponential expansiveness. Our main objective is to give some characterizations for weak exponential expansiveness properties of evolution families in Banach spaces, and variants for weak exponential expansiveness of some well-known results in stability theory  of all bounded linear operators on X will be denoted by ||·||.Let 𝒰 = {U}t≥s≥0 of bounded linear operators is called an evolution family if the following conditions are satisfied:U = I, the identity operator on X, for all t ≥ 0;UU = U for all t ≥ r ≥ s ≥ 0;M ≥ 1 and ω > 0 such that ||Ux0|| ≤ Meω(t−s)||x0|| for all t ≥ s ≥ 0 and x0 ∈ X;there exist x0 ∈ X and every t0 ≥ 0, the mapping r ↦ ||Ux0|| is continuous on  and τ ↦ ||U|| is continuous on  for all n ∈ ℕ, so there exist Ln > 0 such thatL = infnLn. We suppose that F ∈ ℱ1, and from supλ>0((1 − eλ−)/λ) < ∞.Sufficiency. Let δ > 0 be such that eαδ > 1 + 3αδK, where K and α are given by (𝒰 is not weakly exponentially expansive. Then, by c = 3, there exists x0 ∈ X such that for all t0 ≥ 0 and all u ∈ (0, δ] there is r ≥ t0 witht = r + δ, we havegiven by . We suppequality , the pro


The Factorization Method is a noniterative method to detect the shape and position of conductivity anomalies inside an object. Themethod was introduced by Kirsch for inverse scattering problems and extended to electrical impedance tomography (EIT) by Brühl and Hanke. Since these pioneering works, substantial progress has been made on the theoretical foundations of the method. The necessary assumptions have been weakened, and the proofs have been considerably simplified. In this work, we aim to summarize this progress and present a state-of-the-art formulation of the Factorization Method for EIT with continuous data. In particular, we formulate the method for general piecewise analytic conductivities and give short and self-contained proofs. Ω⊆ℝn from current-voltage measurements on a part of its surface Σ⊆∂Ω. Mathematically, this leads to the problem of recovering the coefficient σ(x) in the elliptic partial differential equationuσg is the solution of  aims to reconstruct the spatial conductivity distribution inside an imaging subject equation∇·σ∇uσg=0σ) is compared to a reference NtD Λ(σ0) in order to determine, if and where σ differs from a known background conductivity σ0. This problem also appears in time-difference EIT, where measurements at different times are compared to monitor temporal conductivity changes. These applications lead to the shape reconstruction problem of determining the support of σ − σ0 from Λ(σ) and Λ(σ0).In several applications, EIT is used to determine the position of conductivity changes. This includes anomaly detection problems, where Λ denotes the subspace of L∞(Ω)-functions with positive essential infima. H⋄1(Ω) and L⋄2(Σ) denote the spaces of H1- and L2-functions with vanishing integral mean on ∂Ω .We start by making the mathematical setting precise. Let σ ∈ L+∞(Ω) and g ∈ L⋄2(Σ), there exists a unique solution uσg ∈ H⋄1(Ω) of the elliptic partial differential equationuσg ∈ H⋄1(Ω) solves (σ) is a self-adjoint, compact linear operator.For equation∇·σ∇uσg=0σ0 ∈ L+∞(Ω) be piecewise analytic. For each point z ∈ Ω that has a neighborhood in which σ0 is analytic, and each unit vector d ∈ ℝn, ||d|| = 1, let Φz,d be the solution ofz,d is called a dipole function.Let monotony lemma since it shows that a larger conductivity leads to a smaller NtD. More precisely, it shows a relation between the difference of two NtDs and the difference of the corresponding conductivities and the interior energy of an electric potential. The second lemma shows that this energy term is the image of the adjoint of an auxiliary virtual measurement operator that is defined on a subregion of Ω. The third lemma is a functional analytic relation between the norm of an image of an operator and the range of its adjoint. Together with the first two lemmas, it implies that the range of the auxiliary virtual measurement operator can be calculated from the NtDs. Finally, using the previous dipole functions, the last lemma shows that the range of the auxiliary virtual measurement operator determines the region on which they are defined.Our presentation of the Factorization Method in the next section relies on the following four lemmas. The first lemma is frequently called a We start with the monotony lemma.σ1, σ0 ∈ L+∞(Ω). Then, for all g ∈ L⋄2(∂Ω), j : = Λ(σj), j = 0,1, and u0 : = uσ0g. Let g ∈ L⋄2(∂Ω), we have thatThe lemma seems to go back to Ikehata, Kang, Seo, and Sheen , 23, cf.σ1 and σ0, we conclude thatBy interchanging σ0 ∈ L+∞(Ω) and a measurable subset D⊆Ω, we define the virtual measurement operator LD byv ∈ H⋄1(Ω) solvesu0|2 in Given a reference conductivity LD is given byu0 ∈ H⋄1(Ω) solvesThe adjoint operator of g ∈ L⋄2(Σ) and F ∈ L2(D)n, we have thatFor all The following functional analytic lemma uses bounds on the image of an operator to characterize the range of its dual operator.X and Y be real Hilbert spaces with inner products X and Y, respectively. Let A ∈ ℒ and x′ ∈ X. Then, Let X, Y1, and Y2 are three real Hilbert spaces, Ai ∈ ℒ, i = 1,2, and if there exists C > 0 withℛ(A1)⊆ℛ(A2).The assertion can be generalized to Banach spaces, and, in that context, it is called the “14th important property of Banach spaces” in Bourbaki . For thex′ ∈ ℛ(A*), then there exists y′ ∈ Y such that x′ = A*y′. Hence,C = ||y′||.If x′ ∈ X be such that there exists C > 0 with |X | ≤C||Ax|| for all x ∈ X. We definef is a well-defined, continuous linear functional on ℛ(A). By setting it to zero on ℛ(A)⊥, we can extend f to a continuous linear functional on Y. Using the Riesz theorem, it follows that there exists y′ ∈ Y withx ∈ X, we havex′ = A*y′ ∈ ℛ(A*). Now let LD determines the region D on which it is defined. We state the lemma for a simple special case, a generalized version of the lemma will be formulated in The last lemma shows that the range of the virtual measurement operator σ0 = 1, D⊆Ω be open, and Let d ∈ ℝn, ||d|| = 1, and every point z ∈ Ω∖∂D, it holds thatThen, for all unit vectors The proof is similar to the one of [z ∈ D and ϵ > 0 be such that F ∈ L2(D)n be the zero continuation of ∇(f1 − f2) to D.First, let v ∈ H⋄1(Ω), and, for all w ∈ H⋄1(Ω),z,d|Σ = v|Σ = LD(F) ∈ ℛ(LD).Then, the functionz,d|Σ ∈ ℛ(LD). Let v ∈ H⋄1(Ω) be the function from the definition of LD. Then,v = Φz,d in the connected set Now let Φv = Φz,d in If σ differs from a reference conductivity σ0 by a range criterion. Before we turn to a new general formulation of the method, we first state it for a special case that is similar to the one that was treated in the original works of Brühl and Hanke [Now we will formulate the Factorization Method and characterize a region where a conductivity nd Hanke , 4.σ0 = 1 and σ = 1 + χD, where D⊆Ω is an open set so that z ∈ Ω, z ∉ ∂D, and all dipole directions d ∈ ℝn, ||d|| = 1,Let g ∈ L⋄2(∂Ω),σ) − Λ(1)| = Λ(1) − Λ(σ), and, using D and ℛ(LD) in The monotony σ(x) = 1 + κ(x)χD(x), when there exists a conductivity jump ϵ > 0 so that eitherObviously, the same arguments can be used to treat the case σ0 is a piecewise analytic function and that either σ − σ0 ≥ 0 or σ − σ0 ≤ 0. Roughly speaking, under this general assumption, the Factorization Method then characterizes the support of σ − σ0 up to holes in the support that have no connections to Σ. For a precise formulation, we use the concept of the inner and outer support from [Now we drop the assumptions that the background is constant, that there is a clear conductivity jump, and that the complement of the inclusions is connected. We will merely assume that the reference conductivity ort from  that hasort from ; see alsort from , 30.connected to Σ if U∩Ω is connected and U∩Σ ≠ ∅.A relatively open set κ : Ω → ℝ, we define support supp(κ) as the complement |>0,  the outer support outΣ suppκ as the complement (in κ|U ≡ 0.  the The interior of a set M⊆Ω is denoted by intM and its closure (with respect to ℝn) by M is measurable, we also define(d)Σ M = outΣsuppχM. out It is easily checked that outΣ(suppκ) = outΣsuppκ. For a measurable function With this concept, we can extend the range characterization in σ0 ∈ L+∞(Ω) be piecewise analytic. Let D⊆Ω be measurable. Let d ∈ ℝn, ||d|| = 1, and every point z ∈ Ω that has a neighborhood in which σ0 is analytic,Then, for all unit vectors z ∈ intD, then there exists a small ball If z,d|Σ ∈ ℛ(LD), and let v ∈ H⋄1(Ω) be the function from the definition of LD, so that , and let σ0 ∈ L+∞(Ω) be a piecewise analytic function. Let eitherz ∈ Ω that have a neighborhood in which σ0 is analytic, as well as all unit vectors d ∈ ℝn, ||d|| = 1, Let z ∈ Ω have a neighborhood in which σ0 is analytic, and let d ∈ ℝn be a unit vector with ||d|| = 1. We only prove the assertions for σ ≥ σ0. The other case is completely analogous.Let z ∈ inn supp  (σ − σ0). Then there exists a small ball Bϵ(z) and δ > 0 so that σ − σ0 ≥ δ on Bϵ(z). Using the monotony g ∈ L⋄2(∂Ω),D : = supp(σ − σ0), the monotony g ∈ L⋄2(∂Ω),First, let D with d ∈ ℝn, ||d|| = 1,A result of Hyvönen and the author [definiteness assumption that σ ≥ σ0 on Ω or σ ≤ σ0 on Ω. However, Grinberg, Kirsch, and Schmitt [exclude a region E⊆Ω from Ω, in such a way that the Factorization Method only requires the definiteness assumption on Ω∖E. In this subsection, we show how their idea can be incorporated into our formulation of the method.It is a long standing open theoretical problem whether the range criterion of the Factorization Method holds true without the  Schmitt , 31 showσ = 1 + χD+ − (1/2)χD−, it is not known whetherD+ and D−. More precisely, if we know a subset E that contains D− without intersecting D+, then we can use the Factorization Method to find D+ ∪ E (and thus D+).To point out the main idea, we first formulate the result for a simple special case. Let us stress that, for σ0 = 1 and σ = 1 + χD+ − (1/2)χD−, where D+, D−⊆Ω are open. Let E⊆Ω be an open set.(a)D+⊆E and z ∈ Ω, z ∉ ∂(D− ∪ E), and all dipole directions d ∈ ℝn, ||d|| = 1, If (b)D−⊆E and z ∈ Ω, z ∉ ∂(D+ ∪ E), and all dipole directions d ∈ ℝn, ||d|| = 1, If Let g ∈ L⋄2(∂Ω),The monotony ince cf. (54)∫Σg(In case (b), we obtain thatWe can also extend these ideas to the general setting of σ ∈ L+∞(Ω) and let σ0 ∈ L+∞(Ω) be a piecewise analytic function. Let E⊆Ω be a measurable set.Let α, β ∈ ℝ such that(a)σ ≤ σ0 on Ω∖E, then for all z ∈ Ω that have a neighborhood in which σ0 is analytic, as well as all unit vectors d ∈ ℝn, ||d|| = 1, If  and(b)σ ≥ σ0 on Ω∖E, then for all z ∈ Ω that have a neighborhood in which σ0 is analytic, as well as all unit vectors d ∈ ℝn, ||d|| = 1, If  andChoose z ∈ inn supp(σ − σ0) with z ∉ E, there exists a small ball Bϵ(z) and δ > 0 so that σ0 − σ ≥ δ on Bϵ(z). Using the monotony g ∈ L⋄2(∂Ω),For every 2(∂Ω),6∫Σg(Λ(σ)−The monotony  The Factorization Method can be used to detect regions in which a conductivity differs from a known reference conductivity. In this work, we summarized the progress on the method's theoretical foundation. We formulated the method for general piecewise analytic conductivities and gave comparatively simple and self-contained proofs. We also showed how the idea of excluding a part of the imaging region can be incorporated into this formulation.σ0 and the existence of the dipole functions.The regularity assumptions can be weakened even further. Our proofs only require unique continuation arguments for the reference conductivity Two major open theoretical questions still exist in the context of the Factorization Method. The theoretical justification of the method requires a definiteness condition (on the whole domain or after excluding an a priori known part of the domain). It is unknown whether the method's range criterion holds without such a definiteness condition. The second open question concerns the numerical stability of the method's range criterion. So far, there are no rigorous convergence results for numerical implementations of this range criterion (see, however, Lechleiter  for a fi


We have discussed some important problems about the spaces  For example, quite recently, Kayaduman and Şengönül introduced the spaces x = (xk) such that (∑j=0k(xj/(j + 1))) ∈ f, f0 and gave some important results on those spaces in , and also the inequalities qσ(Ax) ≤ L(x) (σ-core of Ax⊆K-core of x), qσ(Ax) ≤ qσ(x) (σ-core of Ax⊆σ-core of x), for all x ∈ ℓ∞, have been studied. Here, the Knopp core, in short K-core of x, is the interval  (see , (see , and also determined necessary and sufficient conditions for a matrix A to yield K-core(Ax)⊆st-core(x) for all x ∈ ℓ∞.Let = 0 (see ). In thixn+i.In , the σ-cx)] (see ). When σ)], (see ). The co authors , 15–19 a authors  have intIn this paper, we define the spaces of Cesàro sigma convergent and Cesàro null sequences and give some interesting theorems.x = (xk) is said to be Cesàro sigma convergent to the number α if and only if limn→∞∑k=0n∑j=0k(xσj(p)/(n + 1)(k + 1)) = α uniformly in p, and the set of all such sequences is denoted with α = 0, then we write α = σC-limx, respectively.A bounded sequence x = (xk) is said to be Cesàro sigma bounded if and only if supn|∑k=0n∑j=0k(xσj(p)/(n + 1)(k + 1))| < ∞, and the set of all such sequences is denoted with A bounded sequence C denotes the Cesàro matrix of one order. Define the sequence y = (yn), which will be frequently used as the C-transform of a sequence x = (xk); that is,σ(p) = p + 1, then the spaces With the notation of , we can Now, we begin with the following theorem.Vσ and Vσ0, respectively; that is, The sequence spaces Vσ. In order to prove the fact Vσ. Consider the transformation of C defined with the notation of  whenever Cx = θ and hence C is injective. Let y ∈ Vσ and define the sequence x byC is surjective. Hence, C is linear bijection which therefore shows that the spaces Vσ are linearly isomorphic, as desired. This completes the proof. The fact that the spaces Vσ0 are linearly isomorphic can also be proved by the similar way, so we omit it.We consider only the spaces ation of  from V~σVσ and Vσ0 are BK-spaces with the norm ||·||ℓ∞.The spaces The sets Vσ and Vσ0 are BK-spaces with the norm ||·||∞,  = {) : x ∈ λ} is subspace of λ × μ and the set G(T) is normed space with the normT is a linear transformation from λ to μ.It is known that, if G(C) of the transformation C is closed subspace in The graph Vσ are Banach spaces ) is convergent to  in G(C) for y ∈ Vσ. With this supposition, and from the equalityxn → x and C(xn) → y as n → ∞. Also, since C is continuous and from the definition of the sequential continuous, we obtain that Cx = y and this completes the proof.We know that the spaces ces see,  and the Following Başar , we starA = (ank) and B = (bnk) map the sequences x = (xk) and y = (yk) which are connected by the relation (z = (zn) and t = (tn), respectively; that is,B is applied to the C-transform of the sequence x = (xk) while the method A is directly applied to the entries of the sequence x = (xk). So, the methods A and B are essentially different. Let us assume that the matrix product BC exists which is a much weaker assumption than the conditions on the matrix B belonging to any matrix class, in general. The methods A and B in  under the application of formal summation by parts. This leads us to the fact that BC exists and is equal to A and (BC)x = B(Cx) formally holds, if one side exists. This statement is equivalent to the following relation between the entries of the matrices A = (ank) and B = (bnk):n, k ∈ ℕ.Let us suppose that the infinite matrices relation  to the sthat is,zn=(Ax)n=A ∈ , if and only ifA ∈ , if and only if ( only if  and 17)A ∈ , if and only if (I is identity matrix. only if  and 17)A ∈  and E = (enk) are connected with the relationμ be any given sequence space. Then, E ∈ (μ : Vσ).Suppose that the entries of the infinite matrices x = (xk) ∈ μ and consider the following equality with ∑j=σ(p) = p + 1, then If we take Now, right here, we have stated two theorem which are natural consequences of the A = (ank) be an infinite matrix real or complex numbers. Then, n∈ℕ∑k|∑j=k∞(anj/(j + 1))| < ∞,  supσ-limn→∞∑k∑j=k∞(anj/(j + 1)) = α,  Let I is the identity matrix.where A = (ank) be an infinite matrix real or complex numbers. Then, (4)n∈ℕ∑k|∑j=k∞(anj/(j + 1))| < ∞,sup(5)σ-limn→∞∑j=k∞(anj/(j + 1)) = αk, exists for each fixed k ∈ ℕ,(6)m→∞∑k|∑i=0m∑j=k∞(aσi(n),j/(m + 1)(j + 1)) − αk| = 0, uniformly in n.limLet The proof is clear from A = (ank) be an infinite matrix real or complex numbers. Then, (7)n∈ℕ∑k|∑j=k∞(anj/(j + 1))| < ∞,sup(8)σ-limn→∞∑j=k∞(anj/(j + 1)) = αk exists for each fixed k ∈ ℕ,(9)σ-limn→∞∑k∑j=k∞(anj/(j + 1)) = α. Let The proof is clear from Furthermore, from the Lemma 3.2 of , we haveA = (ank) and B = (bnk) are connected with the relationμ be any given sequence space. Then, if (10)αk = limn∑j=k∞(bnj/(j + 1)) exists for every k ∈ ℕ, (11)n(∑k | ∑j=k∞(bnj/(j + 1)) − αk|) < ∞, sup(12)α = ∑k | αk | <∞.Suppose that the entries of the infinite matrices In this section, we use matrices classes x ∈ ℓ∞. Then, σC-core of x is defined by the closed interval σC-core of x is α if and only if σC-limx = α.Let We need the following lemma due to Das  for the C|| = ||cni(p)|| < ∞ and let limn→∞supp∈ℕ | cni(p) | = 0. Then, there is a y = (yi) ∈ ℓ∞ such that ||y|| ≤ 1 andLet ||P and Q be sublinear functionals on a linear space X. Then, {X, P}⊂{X, Q} if and only if P(x) ≤ Q(x) for all x ∈ X.Let x ∈ ℓ∞ if and only if n, i, p ∈ ℕ.Consider the following.Sufficiency. Since x ∈ ℓ∞, it is known that for any given ϵ > 0, there exists a positive integer i0 such that xi ≤ L(x) + ϵ whenever i ≥ i0. Now, let us writen → ∞, uniformly in p. Therefore, we obtain by applying the operator limsupn→∞supp∈ℕ to the equality .Since  holds, wNecessity. We observe by inserting −x = (−xi) in place of x = (xi) in the inequality x ∈ c, then ℓ(x) = L(x) = limx and so σC-limAx = lim x. Hence, p. On the other hand, if we choose the sequence of matrices n, i, p ∈ ℕ, then clearly 𝒜 satisfies the conditions of y = (yi) ∈ ℓ∞ with ||y|| ≤ 1 andy in the hypothesis thatear thatliminfn→σ(p) = p + 1, we also have the following.In the special case BC-core(Ax)⊆K-core(x) for all x ∈ ℓ∞ if and only if x ∈ ℓ∞ if and only if )reg and  holds.Consider the following.Necessity. By the similar way used in the proof of necessity of qσ(x) ≤ L(x) for any sequence x, condition  = qσ(x) [)reg and  hold. Sieg, from u(x)=inf we haveu(x)≥inf= qσ(x) , the proσ(p) = p + 1, we also have the following theorem.In the special case BC-core(Ax)⊆B-core(x) for all x ∈ ℓ∞ if and only if One can see that )reg and  holds.E⊆ℕ with natural density zero.c ⊂ st∩ℓ∞. Now, define a sequence t = (ti) via x ∈ ℓ∞ asE is any subset of ℕ with δ(E) = 0. Then, st-limti = 0 and t ∈ st0, so we have At)n = ∑i∈Eaniti, the matrix B = (bni), defined byn, must belong to the class x ∈ st∩ℓ∞ and let st-lim x = ℓ. Write E = {i : |xi − ℓ | ≥ε} for any given ε > 0, so that δ(E) = 0. Since σC-limn→∞∑iani = 1, we haveσC-lim(Ax) = st-lim x; that is, Let ssity of  follows )reg and  holds. Londition  implies σ(p) = p + 1, we also have the following theorem.In the special case Consider that E⊆ℕ with natural density zero.x ∈ ℓ∞ if and only if Consider that )reg and  holds.Consider the following.Necessity. Firstly, assume that x ∈ ℓ∞ where β(x) = st-lim sup x. Hence, since β(x) = st-lim sup x ≤ L(x) for all x ∈ ℓ∞ , we havSufficiency. Let x ∈ ℓ∞, then β(x) is finite. Let E be a subset of ℕ defined by E = {i : xi > β(x) + ε} for a given ε > 0. Then, it is obvious that δ(E) = 0 and xi ≤ β(x) + ε if i ∉ E. For any real number λ, we write λ+ = max{λ, 0} and λ− = max{−λ, 0} whence |λ | = λ+ + λ−, λ = λ+ − λ− and |λ | −λ = 2λ−. Now, we can writen→∞supp∈ℕ, we obtained from hypothesis that ε is arbitrary. and let  holds. Iσ(p) = p + 1, we also have the following theorem.In the special case BC-core(Ax)⊆st-core(x) for all x ∈ ℓ∞ if and only if Consider that )reg and  holds.


𝒜ε(t) of equation x ∈ Ω, converge to the global attractor 𝒜 of the above-mentioned equation with ε = 0 for any t ∈ ℝ.We will study the upper semicontinuity of pullback attractors for the 3D nonautonomouss Benjamin-Bona-Mahony equations with external force perturbation terms. Under some regular assumptions, we can prove the pullback attractors  Ω ⊂ ℝ3 is a bounded domain with sufficiently smooth boundary ∂Ω; u = , u2, u3) is the velocity vector field; ν > 0 is the kinematic viscosity; ɛ ≥ 0 is a small nonnegative parameter; the external force g is locally square integrable in time for  ∈ Ω × ℝ, that is, for any t ∈ ℝ, g ∈ Lloc2, where H = (L2(Ω))3, V = (H01(Ω))3, and  and ||·|| are the inner product and norm of H, respectively.In this paper, we will consider the upper semicontinuity of pullback attractors for the following 3D Benjamin-Bona-Mahony equation:The Benjamin-Bona-Mahony (BBM) equation is a well-known model in physical applications which incorporates dispersive effects for long waves in shallow water that was introduced by Benjamin et al.  as an imFor the well-posedness of global solutions for BBM equation, we can refer to –7. For t2, t ∈ ℝ. Here b ≠ 0, a ∈ ℝ2, and p ≥ 3 is an integer. The author proved the supremum norms of the solutions with small initial data decay to zero like t−2/3 as t tends to infinity.Biler  investigHper2(Ω) and Hper1(Ω), respectively. Moreover, Wang et al. [By energy equation and weak continuous method, Wang  and Wangg et al.  got the By the decomposition of the semigroup, Wang  studied on, Wang  also obtH01 under certain assumptions, here w is the two-sided real-valued Wiener process on a probability space. He also proved the random attractor is invariant and attracts every pulled-back tempered random set under the forward flow. The asymptotic compactness of the random dynamical system is established by a tail-estimates method, which shows that the solutions are uniformly asymptotically small when space and time variables approach infinity.Wang  considerH1(ℝ3) by showing the solutions are point dissipative and asymptotic compactg ∈ L2(ℝ3) and f(u) = u + (1/2)u2. Stanislavova [Stanislavova et al.  first prislavova  investigcompactut−Δut−νBy the method of orthogonal decomposition, Zhu , 18 obtau and Mu  deduced H01(Ω). Qin et al. [H02(Ω) by weak continuous method. Zhao et al. [J. Park and S. Park  studied n et al.  derived ut−Δut−νo et al.  investigMoreover, Çelebi et al.  deduced For the upper semicontinuity of corresponding attractors between autonomous and perturb nonautonomous systems, we can refer to Bao , Hale anTo our knowledge, there are less results on the upper semicontinuity of pullback attractors for the 3D nonautonomous BBM equations with the nonautonomous perturbation; we will pay attention to this issue in the sequel.This paper is organized as following. In 𝒜ɛ = {Aɛ(t)}t∈ℝ for the perturbed nonautonomous system with ɛ > 0 and global attractor 𝒜 for the unperturbed autonomous system with ɛ = 0 of the following equation:ɛ = 0.In this section, we will consider the relationship between pullback attractors X be a Banach space with norm ||·||X. The Hausdorff semidistance distX in X between B1⊆X and B2⊆X is defined bydX denotes the distance between two points x and y.Let S(t) : X → X (t ∈ ℝ) is a C0-semigroup defined on X. If the global attractor 𝒜 for S(t) exists, then it has the following properties: (1) 𝒜 is an invariant, compact set; (2) 𝒜 attracts every bounded sets in X, that is, limt→+∞dist(S(t)B, 𝒜) = 0 for all bounded subsets B ⊂ X.For an autonomous system, U}t≥τ is said to be a process in X ifU is continuous in X.For a nonautonomous system, the two-parameter mapping class {Now we will recall some definitions and framework on the existence theory of pullback attractors.𝒜 = {A(t)}t∈ℝ is said to be a pullback attractor for the continuous process {U} if it satisfies the following:𝒜 is invariant for all t ≥ τ.𝒜 is pullback attracting, that is, limτ→+∞distB, A(t)) = 0 for all bounded subsets B ⊂ X.A family of compact sets ℬ = {B(t)}t∈ℝ is said to be pullback absorbing for the process U, if for every t ∈ ℝ and all bounded subsets B ⊂ X, there exists a time T > 0, such thatThe family of subsets ℬ = {B(t)}t∈ℝ be a family of subsets in X. A process U is said to be pullback ℬ-asymptotically compact in X if for all t ∈ ℝ, any sequences τn → ∞ and xn ∈ B(t − τn); the sequence {Uxn} is precompact in X.Let ℬ = {B(t)}t∈ℝ be pullback absorbing set for the process U and U is pullback ℬ-asymptotically compact in X. Then, the family 𝒜 = {A(t)}t∈ℝ that is defined by A(t) = Λ is a pullback attractor for U in X for the process {U}, whereLet the family of sets ℬ-asymptotic compactness in terms of the noncompact measure.In the following, we will characterize the pullback B ⊂ X, ℬ = {B(t)}t∈ℝ be a family of sets in X. A process U is said to be pullback ℬ-κ contracting, if for any t ∈ ℝ, ɛ > 0, there exists a time Tℬ > 0, such thatκ(B) is the Kuratowski noncompact measure defined asLet ℬ = {B(t)}t∈ℝ, X and satisfy that for any t ∈ ℝ, there exists a time U is pullback ℬ-asymptotically compact, if it is pullback κ contracting.Let See, for example, Wang and Qin .U is pullback κ contracting and the family of sets ℬ = {B(t)}t∈ℝ is pullback absorbing for U, then the process U possesses a pullback attractor.Assume that the assumptions in See, for example, Wang  and  Qin .ℬ = {B(t)}t∈ℝ be a family of sets in X. Suppose U = U1 + U2 : ℝ × ℝ × X → X satisfies(i)t ∈ ℝ,for any  + satisfies limτ→+∞Φ = 0 for each t ∈ ℝ;where Φ : ℝ × ℝ → ℝ(ii)t ∈ ℝ and T ≥ 0, ⋃τ≤T0≤U2B(t − τ) is bounded and U2B(t − τ) is precompact in X for any τ > 0.for any Then the process U is pullback ℬ-κ contracting in X.Let See, for example, Wang and Qin .ɛ ∈ .We now perturb the nonautonomous term with a small parameter t ∈ ℝ,  τ ∈ ℝ, and x ∈ X, we haveX.For each H1) holds, and for any ɛ ∈ }t∈ℝ for all ɛ > 0. If there exists a compact set K ⊂ X, such that𝒜ɛ and 𝒜 have the upper semicontinuity, that is,Assume that  for the process generated by the nonautonomous dissipative system.In order to apply ℬ = {B(t)}t∈ℝ is pullback absorbing for U, and for each ɛ ∈ }t∈ℝ is a family of compact sets in X. Suppose Uɛ = Uɛ1, + Uɛ2, : ℝ × ℝ × X → X satisfies(i)t ∈ ℝ and any ɛ ∈  = 0 for each t ∈ ℝ;where Φ : ℝ × ℝ → ℝ(ii)t ∈ ℝ and any T ≥ 0, ∪τ≤T0≤Uɛ2,B(t − τ) is bounded, and for any t ∈ ℝ, there exists a time Tℬ(t) > 0, which is independent of ɛ, such thatfor any  K ⊂ X, such thatand there exists a compact set  ɛ ∈ }t∈ℝ and (H2) holds.Then for each Assume that the family See, for example, Wang and Qin .In this section, firstly, we recall some notations about the functional spaces which will be used later to discuss the regularity of pullback attracting set.A is denoted by A = −Δ with domain D(A) = H2(Ω)⋂H01(Ω) and λ is the first eigenvalue of A; we consider the family of Hilbert spacesD(As/2)↪D(Ar/2) for any s > r and the continuous embeddings ∈ [0, 3/2), ℋ2 = H2(Ω)⋂H01(Ω).The operator 𝒫 to the systems  = 𝒫g.Then, applying the Helmholtz-Leray projector  systems –3), we , we 𝒫 to systems –(3)(29uτ ∈ H01(Ω), the external force g ∈ Lloc2. Also we assume that there exist constants β > 0, 0 ≤ α < σ/2, and σ = 2ν/((2/λ) + 2), such thatf is locally square integrable in time; that is, f ∈ Lloc2 and satisfiesη ≤ min{λν, ν, ν/((2/λ) + 2)} and any t ∈ ℝ.Assume that es that∫−∞teσs|Fi  are smooth functions satisfyings ∈ ℝ, where C1, C2, and σ are positive constants.For the nonlinear vector function At last, we will state the main result and the proof of this paper as the following.uτ ∈ V, then the pullback attractors 𝒜ɛ = {𝒜ɛ(t)}t∈ℝ for  is defined on the Banach space V.Assume that –36) hol hol36) ht∈ℝ for  (which it∈ℝ for ) with ɛ or 𝒜 for  with ɛ =ɛ = 0 and pullback attractors for nonautonomous system  (t ∈ ℝ) generated by problem  hol hol36) h problem  (or prob problem –(3)) witUsing similar technique as in , 17, 18,uτ ∈ V, then problem  (ɛ ≥ 0) satisfyingUɛ} generated by the global solutions possess pullback attractors 𝒜ɛ for all ɛ ≥ 0 in V.Assume that –36) hol hol36) h problem  possesseSee, for example, .uɛ(t) = Uɛuτ of  > 0, such thatRɛ(t) = Cɛeηt−∫∞−teηs||f(s)||H2ds, and C is a positive constant independent of B, t, τ.Suppose that –36) hol hol36) hσ = 2ν/((2/λ) + 2), Rσ = {r : R →  | limt→−∞eσtr2(t) = 0} and denote by 𝒟σ the class of families V centered at zero with radius We choose t ∈ ℝ,  τ ∈ ℝ, and uτ ∈ V be fixed, and denoteu ∈ C; V), then for all u ∈ V, we derive thatτ ∈ ℝ.Let σ, we easily getuτ ∈ D(τ),  t ≥ τ.Let Rɛ(t) the nonnegative number given for each t ∈ ℝ byV defined by𝒟σ-pullback absorbing for the process {U}.Setting ℬɛ = {Bɛ(t)}t∈ℝ is pullback absorbing in V easily. Moreover,SettingRɛ(t), Bɛ(t) be given as above. For any t ∈ ℝ, the solution v(t) = Uɛ1,u(t − τ) of (τ ≥ 0 and ut−τ ∈ Dɛ(t − τ).Let  − τ) of  satisfiev and integrating over Ω, we deriveB(·) and ℱi(0) = 0 asMultiplying equation in  with v aη ≤ min{λ1ν, ν}.Using Poincaré's inequality, it followst − τ to t, we gett ≥ τ, which completes our proof.Integrating  from t −ℬɛ(t) = {Bɛ(t)}t∈ℝ be given by  > 0 and a function Iɛ(t) > 0, such that the solution Uɛ2,uτ = w(t) of  and any ut−τ ∈ Bɛ(t − τ).Let given by  and 52)ℬɛ(t)  w(t) of  satisfieAσw(t) in H, we deriveC depends on ||ut−τ||V2, σ, and the first eigenvalue λ of the operator A.Taking the inner product of equation in  with A derive12ddt,Arding to , we havet − τ to t, we conclude thatt > τ. This completes the proof of desiring lemma.Integrating  from t −t ∈ ℝ, any τ > 0, if u0 varies in bounded sets, then the solution uɛ(t) = Uɛu0 of problem (u(t) = S(t)u0 of the unperturbed problem  satisfiesyɛ(t), using  + 2)}, we knowf ∈ Lloc2, using Lemmas ɛ → 0+, which implies (Denotetisfiesytϵ+Aytɛ), using –36) and and(64),yɛ|∂Ω=0,.Using –(36) andhat is,ddt(||yɛ implies .D(As/2)↪(Ls)6/(3−2(Ω))3 is compact, combining Lemmas Since the embedding


Some new results on existence and uniqueness of a solution are established by using fixed point theorems. Some illustrative examples are also presented. We extend previous results even in the integer case α = 1.We study a boundary value problem for the system of nonlinear impulsive fractional differential equations of order  For the last decades, fractional calculus has received a great attention because fractional derivatives provide an excellent tool for the description of memory and hereditary properties of various processes of science and engineering. Indeed, we can find numerous applications in viscoelasticity –3, dynamOn the other hand, the study of dynamical systems with impulsive effects has been an object of intensive investigations in physics, biology, engineering, and so forth. The interest in the study of them is that the impulsive differential systems can be used to model processes which are subject to abrupt changes and which cannot be described by the classical differential problems  and f : J × R → R is a given continuous function.Fečkan et al.  investigcD0+α is the Caputo fractional derivative of order α ∈  with the lower limit zero, f : J × R → R is jointly continuous, tk satisfy 0 = t < t1 < ⋯<tp < tp+1 = T, x(tk+) = limε→0+x(tk + ε) and x(tk−) = limε→0+x(tk − ε) represent the right and left limits of x(t) at t = tk, Ik ∈ C, and a, b, c are real constants with a + b ≠ 0.In , Guo andn-dimensional analogues of the problem ≠0. Here f, g :  × Rn → Rn and Ii : Rn → Rn are given functions.Motivated by the papers above, in this paper, we study impulsive fractional differential equations with the two-point and integral boundary conditions in the following form:The rest of the paper is organized as follows. In C we denote the Banach space of all continuous functions from J to Rn with the normRn. We also introduce the Banach spaceA ∈ Rn×n, then ||A|| is the norm of A.In this section, we introduce notations, definitions, and preliminary facts that will be used in the remainder of this paper. By Let us recall the following known definitions and results. For more details see , 16.g ∈ C and α > 0, then the Riemann-Liouville fractional integral is defined byz asIf α > 0 of a continuous function g :  → R is defined byn = [α] + 1 (the notation [α] stands for the largest integer not greater than α).The Caputo fractional derivative of order g(t), the Caputo fractional derivative becomes the conventional integer order derivative of the function g(t) as α → n.Under natural conditions on α, β > 0 and n = [α] + 1; then the following relations hold:Let α > 0, g(t) ∈ C⋂L1, the homogeneous fractional differential equation,ci ∈ R, i = 0,1,…, n − 1, and n = [α] + 1.For g(t) ∈ C⋂L1, with derivative of order n that belongs to C⋂L1; thenci ∈ R, i = 0,1,…, n − 1, and n = [α] + 1.Assume that p, q ≥ 0, f ∈ L1. ThenT]. Moreover, if f ∈ C, then , then cD0+αI0+αf(t) = f(t) for all t ∈ .If We define a solution problem  as follox ∈ PC is said to be a solution of problem (cD0+αx(t) = f), for t ∈ , t ≠ ti, i = 1,2,…, p, and for each i = 1,2,…, p, x(ti+) − x(ti) = Ii(x(ti)), 0 = t0 < t1 < t2 < ⋯<tp < tp+1 = T, and the boundary conditions Ax(0) + Bx(T) = ∫0Tg)ds are satisfied.A function  problem  if cD0We have the following result which is useful in what follows.f, g ∈ C. Then the function x is a solution of the boundary value problem for impulsive differential equationLet x is a solution of the boundary value problem  is still an arbitrary constant vector. For determining x(0) we use the boundary value condition Ax(0) + Bx(T) = ∫0Tg(s)ds:t ∈  = f(t), t0 < t ≤ t1. If t ∈  = f(t), tk < t ≤ tk+1, and x(tk+) − x(tk) = Ik(x(tk)). The lemma is proved.Conversely, assume that atisfies . If t ∈ X be a Banach space and W ⊂ PC. If the following conditions are satisfied,W is uniformly bounded subset of PC,W is equicontinuous in , k = 0,1, 2,…, p, where t0 = 0, tp+1 = T,W(t) = {u(t) : u ∈ W, t ∈ J′}, W(tk+) = {u(tk+) : u ∈ W}, and W(tk−) = {u(tk−) : u ∈ W} are relatively compact subsets of X,then W is a relatively compact subset of PC.Let (H1)f, g : J × Rn → Rn are continuous functions.(H2)Lf > 0 and Lg > 0 such thatThere are constants  t ∈  and all x, y ∈ Rn.for each (H3)li > 0, i = 1,2,…, p such thatThere exist constants  x, y ∈ Rn.for all For brevity, letOur first result is based on Banach fixed point theorem. Before stating and proving the main results, we introduce the following hypotheses.J.Assume that (H1)–H3) hold. If hold. IfThe proof is based on the classical Banach fixed theorem for contractions. Let us setQ be the following operator:Q maps Br into Br. It is clear that Q is well defined on PC. Moreover for x ∈ Br and t ∈  into itself.Let Q is a contraction. Let x, y ∈ PC. Then, for each t ∈  due to condition  which is a unique solution to (Next we will show that ondition  and the  we have6|(Qx)(t)(H4)Nf > 0, Ng > 0 such that |f| ≤ Nf, |g| ≤ Ng for each t ∈ J and all x ∈ Rn.There exist constants (H5)Ik ∈ C. The second result is based on the Schaefer fixed point theorem. We introduce the following assumptions.J.Assume that (H1), (H4), and (H5) hold. Then the boundary value problem  has at lQ defined by . Then, for each k = 0,1, 2,…, p and for all t ∈ .η > 0, there exists a positive constant l such that, for each x ∈ Bη = {x ∈ PC : ||x||PC ≤ η}, we have ||Q(x)||PC ≤ l. By (H4), (H5) we have, for each k = 1,2,…, p and for all t ∈ .τ1, τ2 ∈  as in Step  2, and let x ∈ Bη. Thenτ1 → τ2, the right-hand side of the above inequality tends to zero.Let X = Rn), we can conclude that the operator Q : PC → PC is completely continuous.As a consequence of Steps  1 to 3 together with the Arzela-Ascoli theorem (Step  4. One has a priori bounds.Now it remains to show that the setx = λQ(x) for some 0 < λ < 1. Thus, for each t ∈ |=|λIn this section, we give some examples to illustrate our main results.f1 = cos((1/10)x2(t)),  f2 = (et−/(9 + et))·(|x1|/(1 + |x1|)), and  T = 1,  p = 1,  Consider problem  with f1α = 0,2. Indeed,Evidently,ondition  is satisThen, by Considerα ≤ 1,  f1 = et−/(1 + x22(t)),  f2 = sinx1(t),  T = 1,  p = 1,  Nf = 1, Ng = 0), and consequently boundary value problem (Here 0 <  problem  has at l


RN. For time t ≥ 0, we can define a functional H(t) associated with the solution of the equations and some testing function f. When the pressure function P of the governing equations is of the form P = Kργ, where ρ is the density function, K is a constant, and γ > 1, we can show that the nontrivial C1 solutions with nonslip boundary condition will blow up in finite time if H(0) satisfies some initial functional conditions defined by the integrals of f. Examples of the testing functions include rN−1ln(r + 1), rN−1er, rN−1(r3 − 3r2 + 3r + ε), rN−1sin((π/2)(r/R)), and rN−1sinh r. The corresponding blowup result for the 1-dimensional nonradial symmetric case is also given.We study, in the radial symmetric case, the finite time life span of the compressible Euler or Euler-Poisson equations in  As usual, ρ = ρ ≥ 0 and u = u ∈ RN are the density and the velocity, respectively. P = P(ρ) is the pressure function. The γ-law for the pressure term P(ρ) can be expressed asγ ≥ 1. If K > 0, it is a system with pressure. If K = 0, it is a pressureless system.The compressible isentropic Euler 1/2.The solutions in radial symmetry are expressed by3 becomesThe Poisson equation 3 becomesThe equations in radial symmetry can be expressed in the following form:δ = 0), Makino et al.  testing function f(r) satisfying the following properties:r→0(f(r)/rN−1) exists,limf(r)/r is increasing,then the solutions blow up in finite time.(a) For the attractive forces (δ = 0 or 1), if H0 satisfies the following initial functional condition:T = (2R∫0Rf(r)dr)/H0.(b) For the nonattractive forces (The key ideas in obtaining the above results are (i) to design the right form of generalized functional and find the right class of testing functions and (ii) to transform the nonlinear partial differential equations into the Riccati inequality.ρ) conserves its nonnegative nature.The density function 1ρ0) ≥ 0.The mass equation 1(11)DρDρ0(r) ≠ 0, we havef(r) on both sides.)For the nontrivial density initial condition in radial symmetry, r from 0 to R for γ > 1 or K = 0:δ = −1, we havedF(r) = f(r)dr and γ > 1 or K = 0.Subsequently, we take integration with respect to o deduce∫0Rf(r)Vtwe have1∫0Rf(r)Vtf(0) = 0 by property 1 and f is increasing by property 2.Note that R > 0,f′(r)≥(1/r)f(r) by property 2.Now, we define the assistant functional:2f(r)dr,H2(t)2R∫0 view of , we getT.It is well known that, with the initial conditionequality  will bloδ = 0 or 1, by a similar analysis, one can show thatT = (2R∫0Rf(r)dr)/H0.(b) For The proof is completed.H(t), readers may refer to Sideris' paper  and vanish at the boundaries:t ≥ 0. By considering u and ρ instead, one may suppose a ≥ 0. Let f(x) be a nonnegative and nonzero C1 testing function, such that f(x)/x is increasing for x > a and the functional is given by(a)δ = 1 or −1, if the initial functional H0 satisfies For  then the solutions blow up in finite time.(b)δ = 0, if H0 > 0, then the solutions blow up on or before the finite time T = (2b∫abf(x)dx)/H0. For Suppose 2 becomesγ ≠ 1, one hasf(x) on both sides, taking integration with respect to x from a to b and using integration by parts, to yieldu = u = ρ = ρ = 0, for all t, we getf(x) and the Cauchy-Schwarz inequality  For δ = 0,(b) For


N governed by a pair of parameters ℓ and γ. We study long-range percolation on the bipartite hierarchical lattice where any edge (running between vertices of unlike bipartition sets) of length k is present with probability pk = 1 − exp(−αβk−), independently of all other edges. The parameter α is the percolation parameter, while β describes the long-range nature of the model. The model exhibits a nontrivial phase transition in the sense that a critical value αc ∈  if and only if ℓ ≥ 1, 1 ≤ γ ≤ N − 1, and β ∈ . Moreover, the infinite component is unique when α > αc.We propose a family of bipartite hierarchical lattice of order  N ≥ 2, the hierarchical lattice of order N is defined byd on ΩN is defined byx, y, z ∈ ΩN. This means that  is an ultrametric space. Roughly speaking, this corresponds to the leaves of an infinite N-ary tree, with metric distance half the graph distance.For an integer ΩN is analyzed in  on a bipartite square lattice with bipartition V1 = {v =  ∈ ℤ2 : vx − vy  is  odd} and V2 = {v =  ∈ ℤ2 : vx − vy  is  even} such that ℤ2 = V1 ∪ V2. In other words, the bond percolation cannot occur on the above bipartite square lattice for occupation probability 2p(1 − p) ≤ 1/2. This is consistent with the classical result which says that bond percolation on ℤ2 does not occur when occupation probability ≤1/2 . Otfound in .P(k) ∝ kγ−. In this model, an edge between vertices with degrees k1 and k2 is occupied with probability proportional to (k1k2)α−. By using generating function method, Hooyberghs et al. [PA(k) ∝ kγA− and PB(k) ∝ kγB−, respectively, has the same critical behaviors with biased percolation on a monopartite scale-free network when γA = γB = γ.Another example is the biased percolation , 19 on iΩN  can be viewed as a spanning subgraph of that from . The uniqueness of the infinite component holds here for the same reason as the uniqueness result for the percolation graph of β = N2 and the graph distance  between 0 and a vertex x. It is also interesting to study the mean field percolation (N → ∞) and compare it with that on ΩN [ΩN (other than the 2-coloring addressed in this paper) are possible.We consider t on ΩN . Directet on ΩN  and otheWe start with some notations. Then we prove x ∈ , define Br(x) as the ball of radius r around x; that is, Br(x) = {y : d ≤ r}. We make the following observations. Firstly, for any vertex x, Br(x) contains Nr vertices. In particular, if r < ℓ, all vertices in the ball have the same type. Secondly, Br(x) = Br(y) if d ≤ r. Finally, for any x, y, and r, we have either Br(x) = Br(y) or Br(x)∩Br(y) = ∅.For a vertex S of vertices, denote by Cn(x) be the component of vertices that are connected to x by a path using only vertices within Bn(x). For disjoint sets S1, S2⊆ΩN, we denote by S1↔S2 the event that at least one edge joins a vertex in S1 to a vertex in S2. S1↮S2 means the event that such an edge does not exist. By definition, if S1, S2⊆ΩNi for i = 1 or 2, then S1↮S2 occurs with probability 1. Let Cnm(x) be the largest component in Bn(x). If there are more than one such components, just take any one of them as Cnm(x). It is clear that |Cnm(x)| = maxy∈Bn(x) | Cn(y)| [For a set  Cn(y)| .Ak be the event that the origin 0 ∈ ΩN1 connects by an edge to at least one vertex at distance k in ΩN2. By construction, for k < ℓ, P(Ak) = 0. For k = ℓ, there are ((N − γ)/(N − 1))(N − 1)Nk−1 vertices in ΩN2 at distance k from 0. Hence,k > ℓ, there are ((N − γ)/N)(N − 1)Nk−1 vertices in ΩN2 at distance k from 0. Similarly, we obtaink > ℓ.Let by using . For k >Ak}k≥1 are independent andβ ≤ N, 1 ≤ γ ≤ N − 1, and α > 0, infinitely many of Ak occur with probability 1 by the second Borel-Cantelli lemma. Thus, θ = 1 for any ℓ > 1, 1 ≤ γ ≤ N − 1, α > 0, and 0 < β ≤ N. The result then follows.Since all the events {αc = ∞ by virtue of the monotonicity. Note that, for j ≥ ℓ, there are (γ/N)Nj vertices in Bj(0) ∩ ΩN1 and ((N − γ)/N)Nj vertices in Bj(0) ∩ ΩN2. Hence, by the comments in the proof of (i) and taking β = N2, for any j ≥ ℓ, we obtainα ≥ 0.We only need to show nℓ = 0 and i such that θ = 0 for all α ≥ 0. This implies αc = ∞.Let  We have{|C(0)|=∞αc is a direct consequence of the proof of Theorem 1(b) in [ΩN1, ΩN2, d) can be viewed as a spanning subgraph of that of , the percolation cluster C(0) is almost surely finite; namely, θ = 0, for α small enough.The positivity of  1(b) in . Since tαc. The main technique to be used is an iteration involving the tail probability of binomial distributions [β < N2, we choose an integer K and a real number δ such thatδ < N. For n ≥ 1, letan is the probability that the largest component of a ball of radius nK contains at least (γ/N)δnK vertices in ΩN1 and at least ((N − γ)/N)δnK vertices in ΩN2. Such a ball is said to be good. We set a0 = b0 = 1 by convention. It is clear that, for α > 0, all an and bn are positive, since nK is a finite number and the connection probability in  for some c > 0.We show that there exists some α > 0 such that liminfn→∞bn > 0. We show that there exists some ℕ the nonnegative integers. We can naturally label the vertices in ΩN via the map f : ΩN → ℕ asnK is said to be very good if it is good and its largest component connects by an edge to the largest component of the first (as per the aforementioned order) good subball in the same ball of radius (n + 1)K. Clearly, the first good subball of radius nK in a ball of radius (n + 1)K is very good. Condition (NK − 1)δnK ≥ δn+1)KK((0) is good if (a) it contains NK − 1 good subballs of radius nK, and (b) all these good subballs are very good.We start with ondition  implies nK in a ball of radius (n + 1)K has a binomial distribution Bin with parameters NK and an. Given the collection of good subballs, the probability that the first such good subball is very good equals to 1. Fix any of the other good subballs, say B; the probability that B is not very good is upper bounded bya and b are the number of vertices in the largest component of the first good subball in ΩN1 and ΩN2, respectively; likewise, c and d are the number of vertices in the largest component of the subball B in ΩN1 and ΩN2, respectively. By definition, we have a, c ≥ (γ/N)δnK, b, d ≥ ((N − γ)/N)δnK, and the distance between two vertices in a ball of radius (n + 1)K is at most (n + 1)K.The number of good subballs of radius B to be very good is at leat 1 − εn. Thus, the number of very good subballs is stochastically larger than a random variable obeying a binomial distribution Bin). From the above comments (a) and (b) and the definition of an, it follows thatξn = 1 − an, we obtainc > 0 large enough so that α large enough so that (c) εn ≤ exp(−c(n + 1)) and (d) ξ1 ≤ exp(−2c) hold. To see (c), note that β < δ2 and thenα→∞ε0 = 0, ξ0 = 0 and by ), thenξn ≤ exp(−c(n + 1)) ≤ exp(−cn) for all n ∈ ℕ. We then finish the proof of According to our above choice of bn, we claim thatCnK(0)∩ΩN1 | ≥(γ/N)δnK and |CnK(0)∩ΩN2 | ≥((N − γ)/N)δnK, then BnK(0) is the first good subball in the derivation above. If this component is connected to at least NK − 2 other large components in Bn+1)K((0) as above, then the component containing the origin in Bn+1)K((0) has (γ/N)δnK(NK − 1)≥(γ/N)δn+1)K( vertices in ΩN1 and ((N − γ)/N)δnK(NK − 1)≥((N − γ)/N)δn+1)K converges to 1 exponentially fast. It then follows from ) > 0 for all n. Since b1 > 0, inequality (A simple coupling givesP(Bin(NK− side of  and the ows from  that(27. Hence,bn+1≥b1∏k


Our results generalize some recent results from cone metric and cone rectangular metric spaces into ordered cone rectangular metric spaces. Examples are provided which illustrate the results.We prove some common fixed-point theorems for the ordered  There are a number of generalizations of metric spaces. One such generalization is obtained by replacing the real valued metric function with a vector valued metric function. In the mid-20th century (see ), the noIn , BranciaOrdered normed spaces and cones have applications in applied mathematics, for instance, in using Newton's approximation method  and in og-weak contraction is introduced by Vetro  in conend Zhang , Azam etnd Zhang , Azam annd Zhang , Malhotrnd Zhang , and thend Zhang  on orderWe need the following definitions and results, consistent with , 20.E be a real Banach space and P a subset of E. The set P is called a cone ifP  is closed, nonempty, and P ≠ {θ}; here θ is the zero vector of E; a, b ∈ ℝ, a, b ≥ 0, x, y ∈ P⇒ax + by ∈ P;x ∈ P and −x ∈ P⇒x = θ.Let P ⊂ E, we define a partial ordering “⪯” with respect to P by x⪯y if and only if y − x ∈ P. We write x≺y to indicate that x⪯y but x ≠ y, while x ≪ y if and only if y − x ∈ P0, where P0 denotes the interior of P.Given a cone P be a cone in a real Banach space E; then P is called normal, if there exists a constant K > 0 such that for all x, y ∈ E,K satisfying the above inequality is called the normal constant of P.Let X be a nonempty set and E a real Banach space. Suppose that the mapping d : X × X → E satisfiesθ⪯d for all x, y ∈ X and d = θ if and only if x = y; d = d for all x, y ∈ X; d⪯d + d for all x, y, z ∈ X. Let d is called a cone metric on X, and  is called a cone metric space. In the following, we always suppose that E is a real Banach space, and P is a solid cone in E; that is, P0 ≠ ϕ and “⪯” is partial ordering with respect to P.Then, E = ℝ and P = }. Then, this cone is not normal .d : X × X → E as follows:X, d) is nonnormal cone rectangular metric space but  is not a cone metric space because it lacks the triangular property. Define f and g be self-mappings of a nonempty set X and C = {x ∈ X : fx = gx}. The pair  is called weakly compatible if fgx = gfx for all x ∈ C. If w = fx = gx for some x in X, then x is called a coincidence point of f and g, and w is called a point of coincidence of f and g.Let X is equipped with a partial order “⊑” and mapping d : X × X → E such that  is a cone rectangular metric space, then  is called an ordered cone rectangular metric space. Let f, g : X → X be two mappings. The mapping f is called nondecreasing with respect to “⊑”, if for each x, y ∈ X, x⊑y implies fx⊑fy. The mapping f is called g-nondecreasing if for each x, y ∈ X, gx⊑gy implies fx⊑fy. A subset 𝒜 of X is called well ordered if for all the elements of 𝒜 are comparable; that is, for all x, y ∈ 𝒜 either x⊑y or y⊑x. 𝒜 is called g-well ordered if all the elements of 𝒜 are g-comparable; that is, for all x, y ∈ 𝒜 either gx⊑gy or gy⊑gx.If a nonempty set g = IX (the identity mapping of X), the g-well orderedness reduces into well orderedness. But, for nontrivial cases, that is, when g ≠ IX the concepts of g-well orderedness and well orderedness are independent.  In the trivial case, that is, for X = {0,1, 2,3, 4}, let “⊑” be a partial order relation on X defined by ⊑ = {, ,, , , , , , }. Let 𝒜 = {0,1, 3}, ℬ = {1,4} and g : X → X be defined by g0 = 1,  g1 = 2,  g2 = 3,  g3 = 3,  g4 = 0. Then it is clear that 𝒜 is not well ordered but it is g-well ordered, while ℬ is not g-well ordered but it is well ordered. Let X, ⊑, d) be an ordered cone rectangular metric space f, g : X → X two mappings. The mapping f is called ordered Reich-type contraction if for all x, y ∈ X with x⊑y,  λ,  μ,  δ ∈  for all x, y ∈ X with gx⊑gy, where λ ∈ }. Define d : X × X → E as follows:X, d) is a complete nonnormal cone rectangular metric space but not cone metric space. Define mappings f,  g : X → X and partial order “⊑” on X as follows:f is an ordered g-weak contraction in  with λ ∈ [1/3,1), μ = δ = 0. Indeed, we have to check the validity of  = , . Then,λ ∈ [1/3,1), μ = δ = 0. Again,λ, δ, and μ such that λ + μ + δ < 1.Let idity of  only forerefore,  holds foerefore,  holds fof and g. Note that f is not an ordered Reich-type contraction. Indeed, for point  =  there are no λ, μ, δ ∈ [0,1) such that condition  be the cone rectangular metric space as in X, d) is a complete nonnormal cone rectangular metric space but not cone metric space. Define mappings f, g : X → X and partial order “⊑” on X as follows:f is an ordered g-weak contraction in  with λ = δ ∈ [1/8,1/2),  μ = 0. Indeed, we have to check the validity of  = , , . Then,λ = δ ∈ [1/8,1/2), μ = 0. Again,λ, δ, and μ such that λ + μ + δ < 1.Let  = , and condition  are satisfied and 3 is a coincidence point of f and g. Note that f3 = g3 = 2; that is,  3 is a coincidence point of f and g but fg3 ≠ gf3; therefore, f and g are not weakly compatible and have no common fixed point. All other conditions of f, “nondecreasing” and “completeness of space,” are replaced by another condition.  In the following theorem, the conditions on X, ⊑, d) be an ordered cone rectangular metric space and f, g : X → X is two mappings such that f(X) ⊂ g(X). Suppose that the following conditions are satisfied: f is an ordered g-weak contraction that satisfies ⪯d for all x ∈ X. there exists Let  = d for all x ∈ X and gz = fu for some z ∈ X  ⊂ g(X)); then F(u)⪯F(x) for all x ∈ X. If F(u) = θ, then gu = fu; that is,  u is a coincidence point of f and g. If θ≺F(u), then by assumption (B) gu⊑fu, so gu⊑gz, and by (A), we obtainF(u) = θ; that is,  gu = fu, and so u is a coincidence point of f and g.Let The existence, necessary and sufficient condition for uniqueness of common fixed point follows from a similar process as used in


G′/G)-expansion method. Abundant traveling wave solutions with arbitrary parameters are successfully obtained by this method and the wave solutions are expressed in terms of the hyperbolic, trigonometric, and rational functions. It is shown that the new approach of generalized (G′/G)-expansion method is a powerful and concise mathematical tool for solving nonlinear partial differential equations in mathematical physics and engineering.Exact solutions of nonlinear evolution equations (NLEEs) play a vital role to reveal the internal mechanism of complex physical phenomena. In this work, the exact traveling wave solutions of the Boussinesq equation is studied by using the new generalized (05.45.Yv, 02.30.Jr, 02.30.Ik G′/G)-expansion method -expansion method -expansion method -expansion method (Naher and Abdullah η))-expansion method  expansion method to construct the exact traveling wave solutions of the Boussinesq equation.The objective of this article is to apply the new generalized (G'/G)-expansion method, we give the description of the new generalized (G'/G) expansion method. In Section Application of the method, we apply this method to the Boussinesq equation, results and discussions and graphical representation of solutions. Conclusions are given in the last section.The outline of this paper is organized as follows: In Section Description of the new generalized  is an unknown function, Ф is a polynomial in v and its derivatives in which highest order derivatives and nonlinear terms are involved and the subscripts stand for the partial derivatives.where Step 1: We combine the real variables x and t by a complex variable ηV is the speed of the traveling wave. The traveling wave transformation (2) converts Eq. (v=v(η):where erts Eq.  into an ψ is a polynomial of v and its derivatives and the superscripts indicate the ordinary derivatives with respect to η.where Step 2: According to possibility, Eq.  and βi  and d are arbitrary constants to be determined and M(η) iswhere either G=G(η) satisfies the following auxiliary nonlinear ordinary differential equation:where η; A, B, C and E are real parameters.where the prime stands for derivative with respect to Step 4: To determine the positive integer N, taking the homogeneous balance between the highest order nonlinear terms and the derivatives of the highest order appearing in Eq. N  and (d+M)-N . Subsequently, we collect each coefficient of the resulted polynomials to zero, yields a set of algebraic equations for αi  and βi , d and V.tute Eq.  and Eq. tute Eq.  includintute Eq.  with theStep 6: Suppose that the value of the constants αi , βi , d and V can be found by solving the algebraic equations obtained in Step 5. Since the general solutions of Eq. , βi , d and V into Eq.  > 0,Family 2: When B ≠ 0, ω=A-C and Ω =B2 + 4E(A-C) < 0,Family 3: When B ≠ 0, ω = A-C and Ω = B2 + 4 E(A-C) = 0,Family 4: When B = 0, ω = A-C and Δ=ωE>0,Family 5: When B = 0, ω = A-C and Δ = ωE < 0,G'/G) expansion method to construct many new and more general traveling wave solutions of the Boussinesq equation. Let us consider the Boussinesq equation,In this section, we will put forth the new generalized N  and (d+M)-N . We collect each coefficient of these resulted polynomials to zero yields a set of simultaneous algebraic equations  for α0, α1, α2, β1, β2d, K and V. Solving these algebraic equations with the help of computer algebra, we obtain following:Substituting Eq.  togetherSet 1:n1 = ( - A2 + V2A2 - 12d2ω2 + 8Eω - 12Bdω - B2), n2 = ( - 2Edω + 3Bd2ω + 2d3ω2 - EB + B2d), n3 = - ( - 2Ed2ω + d4ω2 + 2Bd3ω + E2 + B2d2 - 2BdE), n4 = - ( - 8EB2ω + V4A4 - 2V2A4 - 16E2ω2 + A4 - B4), ω = A - C, V, d, A, B, C, E are free parameters.where Set 2: ω = A-C, V, d, A, B, C, E are free parameters.Set 3: V=V, 1=0,n5 = ((V2 - 1)A2 + 8Eω + 2B2), n6 = - (16E2ω2 + 8EB2ω + B 4), n7 = ((V2 - 1)2A4 - 256E2ω2 - 128B2Eω -16B4), ω = A - C, V, A, B, C, E are free parameters.where C1 = 0 but C2 ≠ 0; C2 = 0 but C1 ≠ 0 respectively:For set 1, substituting Eq.  into Eq.C1 = 0 but C2 ≠ 0; C2 = 0 but C1 ≠ 0 respectively:Substituting Eq.  into Eq.Substituting Eq.  into Eq.C1 = 0 but C2 ≠ 0; C2 = 0 but C1 ≠ 0 respectively:Substituting Eq.  into Eq.C1 = 0 but C2 ≠ 0; C2 = 0 but C1 ≠ 0 respectively:Substituting Eq.  into Eq.x-Vt.where η = C1 = 0 but C2 ≠ 0; C2 = 0 but C1 ≠ 0 respectively:Again for set 2, substituting Eq.  into Eq.C1 = 0 but C2 ≠ 0; C2 = 0 but C1 ≠ 0 respectively:Substituting Eq.  into Eq.Substituting Eq.  into Eq.C1 = 0 but C2 ≠ 0; C2 = 0 but C1 ≠ 0 respectively:Substituting Eq.  into Eq.C1 = 0 but C2 ≠ 0; C2 = 0 but C1 ≠ 0 respectively:Substituting Eq.  into Eq.x-Vt.where η = C1 = 0 but C2 ≠ 0; C2 = 0 but C1 ≠ 0 respectively:Similarly, for set 3, substituting Eq.  into Eq.C1 = 0 but C2 ≠ 0; C2 = 0 but C1 ≠ 0 respectively:Substituting Eq.  into Eq.Substituting Eq.  into Eq.C1 = 0 but C2 ≠ 0; C2 = 0 but C1 ≠ 0 respectively:Substituting Eq.  into Eq.C1 = 0 but C2 ≠ 0; C2 = 0 but C1 ≠ 0 respectively:Substituting Eq.  into Eq.η = x-Vt.where It is worth declaring that some of our obtained solutions are in good agreement with already published results which are presented in the following tables Table .Table 1Beside this table, we obtain further new exact traveling wave solutions The graphical illustrations of the solutions are given below in the figures with the aid of Maple Figures , 4 and 5The solutions corresponding to G'/G) -expansion method. We apply the new approach of generalized (G'/G)-expansion method for the exact solution of this equation and constructed some new solutions which are not found in the previous literature. This study shows that the new generalized (G'/G)-expansion method is quite efficient and practically well suited to be used in finding exact solutions of NLEEs. Also, we observe that the new generalized (G'/G)-expansion method is straightforward and can be applied to many other nonlinear evolution equations.In this paper, we obtain the traveling wave solutions of the Boussinesq equation by using the new approach of generalized (


By making use of basic hypergeometric functions, a class of complex harmonic meromorphic functions with positive coefficients is introduced. We obtain some properties such as coefficient inequality,growth theorems, and extreme points. Back in 1748, Euler considered the infinite product ∞−1 = ∏k=0∞(1 − qk+1)−1 and ever since it became very important to many areas. However, it was stated in the literature that the development of these functions was slower until Heine (1878) converted a simple observation such that limq→1[(1 − qa)/(1 − q)] = a, which then returns the theory of  2ϕ1 basic hypergeometric series to the famous theory of Gauss  2F1 hypergeometric series. Various authors qzk−1 = kzk−1 = h′(z), where h′(z) is the ordinary derivative. For more properties of Dq, see , ℱsrf(z) : MH → MH is defined asℱsr, we introduce the class MH as follows.Using convolution given by  and f(z)given by , a lineaMH denote the family of harmonic meromorphic functions α  (0 ≤ α < 1) and for all z ∈ 𝕌∖{0}.Let r = 1, s = 0, a1 = q, and q → 1, MH is the class of harmonic meromorphic starlike functions of order α and for r = 1, s = 0, a1 = q, α = 0, and q → 1, the class MH gives the class of harmonic meromorphic starlike functions for all z ∈ 𝕌∖{0}.For Denote α < 1, |q | <1), then f is harmonic univalent sense preserving in 𝕌∖{0} and f ∈ MH. If the form  and satiα < 1. Consider the expressionr  (0 ≤ r < 1). Therefore, letting r → 1 in . Note that f is sense preserving in 𝕌∖{0}. This is becauseq→1[|Dqh(z)|≥|Dqg(z)|] = [|h′(z)|≥|g′(z)|]. Hence the theorem.Suppose that  holds trpressionA(z)=|zDqpressionA(z)=|zDqpothesis , it follpothesis  holds, sNow, we prove that condition  is necesLet fined by . Then, fℜ(z)≤|z| for all z, it follows from . Then,| > 1, where i = {2,…, r}, j = {1,…, s}, and 0 < |z | = r < 1, one hasIf fined by  is in thα < 1 show that the bounds given in 𝕌∖{0}.We will only prove the right-hand inequality. The proof for the left-hand inequality is similar and we will omit it:Next, we give the following.λk ≥ 0, γk ≥ 0, and ∑k=0∞(λk + γk) = 1. In particular, the extreme points of hk} and {gk}.Setλk ≤ 1  and 0 ≤ γk ≤ 1 .λ0 ≥ 0. Consequently, we obtainLetq-hypergeometric functions and analytic functions can be found in [Other work related to found in , 17.


We establish new inequalities similar to Hardy-Pachpatte-Copson's type inequalities. These results in special cases yield some of the recent results. The classical Hardy's integral inequality is as follows.p > 1, f(x) ≥ 0 for 0 < x < ∞, and F(x) = (1/x)∫0xf(t)dt, thenf ≡ 0. The constant is the best possible.If Theorem A was first proved by Hardy , in an ap > 1, m ≠ 1,f(x) ≥ 0 for 0 < x < ∞, and F(x) is defined byf ≡ 0. The constant is the best possible.If Inequalities  and 3)  3) whichOur main results are given in the following theorems.a < b < R, c < d < R′, p > 1, q < 1, and α > 0 be constants. Let w be positive and locally absolutely continuous in ×. Let h be a positive continuous function and let H = ∫ax∫cyhds dt, for ∈×. Let f be nonnegative and measurable on ×. Ifx, y) ∈  × , and if F is defined byx, y)∈×, thenLet f, w, h, and r reduce to f(x), w(x), h(x), and r(x), respectively, and with suitable modifications in Let This is just a new inequality established by Pachpatte .Moreover, we note that the inequality established in w(x) = r(x) = 1,  H(R) = R, and α = 1 in  cha chaw(x) This is just a new inequality established by Love .h(x) = 1, a → 0,  b → ∞, and log (R/(x − a)) = 1 in (Let ) = 1 in ; then  be positive and locally absolutely continuous in ×. Let h be a positive continuous function and let H = ∫ax∫cyhds dt, for ∈×. Let f be nonnegative and measurable on ×. Letx, y)∈×. If F is defined byx, y)∈×, thenLet f, w, h, and r reduce to f(x), w(x), h(x), and r(x), respectively, and with suitable modifications in Let This is just a new inequality established by Pachpatte .On the other hand, we note that the inequality established in w(x) = r(x) = 1, H(R) = R, and β = 1 in  ch chw(x) =This is just a new inequality established by Love .u = wFp and in view ofx, y)∈×, thenIf we let H = ∫ax∫cyhds dt, for ∈×, thenLetx, we haveFrom , 20), a, a20), aq < 1, then we observe thatIf p, p/(p − 1) on the right side of  = wFp and in view ofx, y)∈×, thenIf we let H = ∫ax∫cyhds dt, for ∈×, thenLety, we haveFrom , 29), a, a29), aq > 1, then we observe thatIf p, p/(p − 1) on the right side of (By applying Hölder's inequality with indices  side of , we obtapth power, we obtainDividing both sides of  by the s


And this system is of wide interest in different branches of science, such as physics, chemistry, biology, evolutionary game theory, and economics. We refer the reader to the book of Hofbauer and Sigmund [It is well known that  Sigmund  for its  Sigmund –4 and reXi(t) is the number of individuals in the ith population at time t and Xi(t) ≥ 0, βi is the intrinsic growth rate of the ith population, the αij are interaction coefficients measuring the extent to which the jth species affects the growth rate of the ith, βi and αij are parameters, and the values of these parameters are not very small usually.To study the bifurcation of Lotka-Volterra class, we consider three-dimensional generalized Lotka-Volterra systemsOver the last several decades, many researchers have devoted their effort to study the existence and number of isolated periodic solutions for system . There h3 with two small parameters λ1 and λ2 and other bounded parameters. In aijk, bijk, and cijk for i, j, k = 0,1, 2 are functions of the parameters βi and αij in system  = O(|x|2) is C∞ in x ∈ ℝ3, andλ1, λ1, λ2)x we obtainD = diag(A(λ1), λ2), witha1b1c1 ≠ 0 as in [S1, S2, and S3 are 2π periodic in θ, and S1, S2, and S3 = O. By a further scaling of the formπ periodic in θ but may not be well defined at p = 0. Thus, we suppose p ≫ ε > 0 for ×x+x22+ on the half plane p > 0 ifB is The averaging systemAccording to Theorem 4.1.3 in , we can ε ≪ 1. Further, the periodic orbit is stable (resp. unstable) if one (resp. none) of the following conditions holds:Δ = 0 and Δ < 0 and Δ > 0, Suppose that  holds. Twhere Δ is given by .s = x3 + δ1 and θ → ε−1θ and truncating the terms of order ε2, we have from , s(ε)) withThen, by letting ave from 21)dpdθdpdθs = xd hence (dpdθ=ps+ε)) with2p(0)=1,  ained in .0 ≠ 0. Then, for any given ε1 > 0 there exist an ε0 > 0 and a C1 function ϕ0(λ1) = (2d1λ1/a1) + δ0λ12 + O(λ13) and ϕ1(λ1) = (2d1λ1/a1) + δ0′λ12 + O(λ13) such that for 0 < λ12 + λ22 < ε0, (0ϕ1(λ1) − ε1λ12 < Δ0λ2 < Δ0ϕ0(λ1) and has no invariant torus if Δ0λ2 > Δ0ϕ0(λ1). Moreover, the torus, if it exists, is stable (resp. unstable) when Δ0 < 0 (resp. >0).Suppose that  holds an2 < ε0,  has a un+3, where ℝ+ = {x ∈ ℝ : x > 0}. We now look for the conditions for the existence of positive equilibria of system  = (αij(ε))3×3, and we suppose M(ε) is similar toNow, we shall investigate a special form of system  with a sdYidt=T, x = T, and system  = 0, and for simplicity, we write the nonlinear part of (y). By doing the following transformation:P(z) = (P1(z), P2(z), P3(z))T, which is to be determined,  denote the cubic terms in z of .ermined,  becomesrmined, −1=Iy noting(Dh1)−1=Iwherez = w + Q(w) ≡ h2(w), where Q =  is homogeneous cubic polynomial, so that /(v(v + ε)), p12 = −(−v2 + vuε + vε + uε2)/(v(v + ε)), p13 = 2v/(v + ε), p21 = (3v + ε)/(v + ε), p31 = −(6v2 + 6vuε + u2ε2)/2v2, and p32 = −uε(2v + uε)/2v2. Then, in the new variables  system  < u < 0, and then for 0 < ε ≪ ε0,  < u < 0. It can be checked that ε ≪ ε0. Then, by In this example, it is easy to see that /u. From  and 10)δ1 = 1,  satisfy , we needε0) < u < 0 holds, where Δ is given by  and ϕ0(λ1) are defined in λ1 = ε and λ2 = −(24/5)ε. By some easy calculations, we can obtain that for 0 < ε1 < A1 − A2 inequality Δ0ϕ1(λ1) − ε1λ12 < Δ0λ2 < Δ0ϕ0(λ1) holds. Thus, by By  and 29)29), we c