R0 algebras and pseudo-R0 algebras are obtained, and the mutually independence of axioms is proved. We introduce the notions of filters and normal filters in pseudo-weak-R0 algebras. The structures and properties of the generated filters and generated normal filters in pseudo-weak-R0 algebras are obtained. These can be seen as noncommutative generalizations of the corresponding ones in weak-R0 algebras.The most simplified axiom systems of pseudo-weak- Flondor et al.  = 0.97∨(0.97⇝0.9) ≤ 0.97∨0.96 = 0.97 < 1.(v)A is a bounded lattice given by Suppose that Then −,  ~, →, and ⇝ on A are defined by the following:The operations  A satisfies (P1)–(P3) and (P5), but not (P4): (a → b)∨(a → c) = (a−∨b)∨(a−∨c) = (d∨b)∨(d∨c) = b∨a = e, but a → b∨c = a → e = 1.Then R0 algebras and investigate the structures and properties of the generated filters and generated normal filters in pseudo-weak-R0 algebras.We introduce the notions of filters and normal filters in pseudo-weak-F of a pseudo-weak-R0 algebra A is said to be a filter of A if it satisfies(F1)x, y ∈ F⇒x⊙y ∈ F,(F2)x ∈ F, x ≤ y⇒y ∈ F.A nonempty subset F of a pseudo-weak-R0 algebra A, the following are equivalent:F is a filter,F and x, x⇝y ∈ F⇒y ∈ F,1 ∈ F and x, x → y ∈ F⇒y ∈ F.1 ∈ For a subset F. By (F1), x, x⇝y ∈ F⇒x⊙(x⇝y) ∈ F. By (38) and (F2), x∧y ∈ F, and so y ∈ F.(i)⇒(ii). By (F2), we have 1 ∈ x, x → y ∈ F, by (19), x ≤ (x → y)⇝y. By (4), x⇝((x → y)⇝y) = 1 ∈ F. By (ii), y ∈ F.(ii)⇒(iii). If x ∈ F, x ≤ y, then x → y = 1 ∈ F, so y ∈ F; that is, (F2) holds; if x, y ∈ F, by (41), x → (y → (x⊙y)) = (x⊙y)→(x⊙y) = 1 ∈ F, and so x⊙y ∈ F, which means (F1) holds.(iii)⇒(i). If A are both filters of a pseudo-weak-R0 algebra A.Clearly, {1} and For a subset F of a pseudo-weak-R0 algebra A, the following are equivalent:F is a filter,x, y ∈ F, y ≤ x → z⇒z ∈ F,x, y ∈ F, y ≤ x⇝z⇒z ∈ F.x, y ∈ F, y ≤ x → z, by (F2) and z ∈ F. Conversely, if x ∈ F, by x ≤ x → 1, we have 1 ∈ F; suppose that x, x → y ∈ F, by x → y ≤ x → y, we have y ∈ F. By F is a filter.(i)⇔(ii). If (i)⇔(iii). Similarly.A is also a filter. If S⊆A, the least filter containing S; that is, the intersection of all filters of A containing S is called the filter generated by S and denoted by [S). If S = {a}, [{a}) is written [a). ClearlyNext, we consider filter generated by a set. It is easy to verify that the intersection of filters of A be a pseudo-weak-R0 algebra and let S be a nonempty subset of A. ThenLet B denote the right side of the first equality. If x, y ∈ B, then there are a1, a2,…, an, b1, b2,…, bm ∈ S such that a1⊙⋯⊙an ≤ x and b1⊙⋯⊙bm ≤ y. By (37), a1⊙⋯⊙an⊙b1⊙⋯⊙bm ≤ x⊙y, so x⊙y ∈ B. If x ∈ B and x ≤ y, we have a1⊙⋯⊙an ≤ x ≤ y, so y ∈ B. Hence B is a filter. If C is a filter and S⊆C, for any x ∈ B, there are a1, a2,…, an ∈ S such that a1⊙⋯⊙an ≤ x. By (F2), x ∈ C, hence B⊆C.Only prove the first equality. Using (34) to the first equality, we can get the rest of the two equalities. Let a0 : = 1; For convenience, we shall write A is a pseudo-weak-R0 algebra and a ∈ A, thenIf Let F be a filter of a pseudo-weak-R0 algebra A and a ∈ A; thenF be a filter of a pseudo-weak-R0 algebra A and a, b ∈ A; thenLet x ∈ [F ∪ {a})∩[F ∪ {b}), by n1,…, nm, l1,…, lk ≥ 0, s1,…, sm, t1,…, tk ∈ F such thatp = s1⊙⋯⊙sm⊙t1⊙⋯⊙tk and q = max{n1,…, nm, l1,…, lk}, and thenx ≥ (p⊙aq)m∨(p⊙bq)k ≥ ((p⊙aq)m∨(p⊙bq))k ≥ ((p⊙aq)∨(p⊙bq))mk = (p⊙(aq∨bq))mk ≥ (p⊙(a∨b)q2)mk. x ∈ [F ∪ {a∨b}). Hence [F ∪ {a})∩[F ∪ {b})⊆[F ∪ {a∨b}). Inverse contains is obvious.Assume that F be a filter of a pseudo-weak-R0 algebra A and a, b ∈ A. If a∨b ∈ F, thenLet A be a pseudo-weak-R0 algebra and a, b ∈ A; then [a)∩[b) = [a∨b).Let F = {1} in Taking R0 algebra.Next we introduce the notion of normal filters in a pseudo-weak-F of a pseudo-weak-R0 algebra A is called normal if x, y ∈ A, x → y ∈ F if and only if x⇝y ∈ F.A filter F be a normal filter of a pseudo-weak-R0 algebra A. Then there is s ∈ F such that b⊙s ≤ c if and only if there is t ∈ F such that t⊙b ≤ c.Let s ∈ F such that b⊙s ≤ c, by (34), s ≤ b⇝c. By s ∈ F, we have b⇝c ∈ F, and so b → c ∈ F. Put b → c = t ∈ F, and then t⊙b ≤ c. Converse is similar.If there is If F is a normal filter of a pseudo-weak-R0 algebra A and a ∈ A, thent2 ∈ F such thatt2,…, tm ∈ F such thats = tm + ⋯+t2 + s1 ∈ F and n = n1 + ⋯+nm, we have s⊙an ≤ x. That is that the first equality holds.We show the first equality. By By the first equality and If F is a normal filter of a pseudo-weak-R0 algebra A and a ∈ A, thenBy s ∈ F such that s⊙an ≤ x, if and only if there is s ∈ F such that s ∈ F such that Since there is s ∈ F such that an⊙s ≤ x, if and only if there is s ∈ F such that s ∈ F such that Similarly, byIf F is a normal filter of a pseudo-weak-R0 algebra A and a ∈ A, thens ∈ F such that s⊙an ≤ x, if and only if there is s ∈ F such that s ≤ an → x, if and only if an → x ∈ F. There is s ∈ F such that an⊙s ≤ x, if and only if there is s ∈ F such that s ≤ an⇝x, if and only if an⇝x ∈ F. By There is R0 algebras and pseudo-R0 algebras and proved the mutually independence of axioms. We introduced the notions of filters and normal filters in pseudo-weak-R0 algebras and gave the structures and properties of the generated filters and generated normal filters in pseudo-weak-R0 algebras. These will be conducive to further study pseudo-weak-R0 algebras  and pseudo-R0 algebras . In the future, we will investigate relations between various kinds of filters of pseudo-logic algebras. We may also study fuzzy type of filters of pseudo-weak-R0 algebras and pseudo-R0 algebras.We obtained the most simplified axiom systems of pseudo-weak-


In this paper, generalized synchronization (GS) is extended from real space to complex space, resulting in a new synchronization scheme, complex generalized synchronization (CGS). Based on Lyapunov stability theory, an adaptive controller and parameter update laws are designed to realize CGS and parameter identification of two nonidentical chaotic (hyperchaotic) complex systems with respect to a given complex map vector. This scheme is applied to synchronize a memristor-based hyperchaotic complex Lü system and a memristor-based chaotic complex Lorenz system, a chaotic complex Chen system and a memristor-based chaotic complex Lorenz system, as well as a memristor-based hyperchaotic complex Lü system and a chaotic complex Lü system with fully unknown parameters. The corresponding numerical simulations illustrate the feasibility and effectiveness of the proposed scheme. By taking the derivative of J(ϕ)∈Cm×n is the Jacobian matrix of ϕ(x), and J(ϕ) = Jr(ϕ) + jJi(ϕ). By substituting Eqs Define the complex CGS error vector asere e = e, e2, ⋯emTherefore, the problem of CGS for two nonidentical complex systems  and 2)  2) is trTheorem 1 For a given complex map vector ϕ(x), CGS and parameter identification of the response system , Kθ = diag and Kδ = diag are error feedback control strength whose elements are all positive constants, which can adjust converging velocity.e system  and the Proof We introduce a positive Lyapunov function asV(t) along the trajectories of the error dynamical system  and Based on Lyapunov stability theory, since x1, x2∈C, x3, x4∈R, x1, x2. a1, a2, a3 and a4 are unknown real parameters, α1 and β1 are considered as known positive constants. When α1 = 4, β1 = 0.01, a1 = 36, a2 = 20, a3 = 3.2, a4 = 3 and x(0) = T, the Lyapunov exponents of system  = T, the system +u4where y1θ = T, δ = T, u = T, andThe drive system  and respThe complex map vector is given byThe Jacobian matrix of the complex map vector is calculated ase1,r = y1,r − x1,r+x2,i, e1,i = y1,i − x1,i − x2,r, e2,r = y2,r − 2x2,r − x2,i, e2,i = y2,i − 2x2,i+x2,r, e3 = y3 − x3 − x4, According to Eqs α1 = 4, β1 = 0.01, α2 = 0.67 × 10−3, β2 = 0.02 × 10−3, the true values of unknown parameters are θ = T, δ = T. The initial conditions of system (x(0) = T, y(0) = T. The initial values of all unknown parameters are randomly chosen as zero, and the control strength is set as K = diag, Kθ = diag, Kδ = diag. The corresponding simulation results are shown in Figs In order to verify the validity and effectiveness of CGS between system  and 13)13) with f system  and (13)= 8, 11, 0, 8/3T. x1, x2∈C, x3∈R, c1, c2 and c3 are unknown real parameters. When c1 = 27, c2 = 23, c3 = 1, and x(0) = T, the complex Chen system −c3x3where x1n system , is alsoθ = T, δ = T, u = T, andThe drive system  and respThe complex map vector is given byThe Jacobian matrix of the complex map vector is calculated ase1,r = y1,r − x1,i, e1,i = y1,i+x1,r, e2,r = y2,r − x2,i, e2,i = y2,i+x2,r, e3 = y3+x3, e3 = y4 − x3.According to Eqs α2 = 0.67 × 10−3, β2 = 0.02 × 10−3, the true values of unknown parameters θ = T, the initial conditions of system (x(o) = T, y(0) = T, the initial values of unknown parameters K = diag, Kθ = diag, Kδ = diag. The corresponding simulation results are shown in Figs y1,r, y2,r, y4 are synchronized with x1,i, x2,i, x3, and y1,i, y2,i, y3 are antisynchronized with x1,r, x2,r, x3. The synchronization errors, as shown in Numerical simulations are presented to verify the validity and effectiveness of CGS between systems  and 13)13) with ters θ = ,23,1T, δf system  and (13)y1, y2∈C, y3∈R, d1, d2 and d3 are unknown real parameters, u1, u2 and u3 are controllers. When d1 = 29, d2 = 21, d3 = 2, and y(0) = T, the complex Lü system −d3y3+u3where y1θ = T, δ = T, u = T, andThe drive system  and respThe complex map vector is given byThe Jacobian matrix of the complex map vector is calculated asAccording to Eqs α1 = 4, β1 = 0.01, θ = T, δ = T, y(0) = ,K = diag, Kθ = diag, Kδ = diag. The CGS process is plotted in y1,r, y2,r are antisynchronized with x2,i, x1,i, y1,i, y2,i are synchronized with x2,r, x1,r, and y3 is synchronized with In order to verify the validity and effectiveness of CGS between Systems  and 25)25) with 3]T, δ = ,21,2T, xThis paper investigates a novel synchronization scheme named complex generalized synchronization, and its application to synchronization and parameter identification of two nonidentical complex nonlinear systems with fully unknown parameters. An adaptive controller and a parameter estimator are proposed and proved theoretically based on Lyapunov stability theory. Three illustrative examples are presented to verify the correctness and effectiveness of the proposed scheme, namely, CGS of a memristor-based hyperchaotic complex Lü system and a memristor-based chaotic complex Lorenz system, CGS of a chaotic complex Chen system and a memristor-based chaotic complex Lorenz system, as well as CGS of a memristor-based hyperchaotic complex Lü system and a chaotic complex Lü system. The proposed CGS scheme has some advantages, for instance, it can be applied to synchronize complex systems with different orders , can be transformed to other types of synchronization with different given complex map vectors (feasibility), can be achieved in a short time with the appropriate control strength (timelines), and can be almost impossibly predicted with the complex map vector (security). So, CGS has extensively potential applications to secure communication, digital cryptography, and so on, which will be involved in our future works.


There are errors in the published paper. “Section 2.4.1” should be labelled as subsection “Disturbance patterns” and “item 2.2” should be labelled as subsection “Forest inventory.”


Under some hypotheses, it is proved that the boundary value problem has a unique solution.We mainly deal with the boundary value problem for triharmonic function with value in a universal Clifford algebra: Δ It iA and Aj are invertible constants; we denote the inverse elements as A−1 and Aj−1.  u(x), (Dju)(x), g(x), fj(x) ∈ Hβ), j = 1,…, 5, 0 < β ≤ 1. The explicit solutions for (In  and 2),,2), A anVn,s (0 ≤ s ≤ n) be an n-dimensional (n ≥ 1) real linear space with basis {e1, e2,…, en}, and Cl the universal Clifford algebra over Vn,s. For more information on Cl (0 ≤ s ≤ n), we refer to  = 0 in Rn; then, for any x ∈ Rn,ωn denotes the area of the unit sphere in Rn.Suppose Δ3[u] = 0 in Rn and |u(x)| = O(|x|) (|x | →∞); then Δ[u] = 0 and Δ2[u] = 0 in Rn.Suppose Δ3[u] = 0 in Rn and u(x) is bounded in Rn; then u(x) ≡ C.Suppose Δ3[u] = 0 in Rn∖∂Ω, and for x ∈ ∂Ω, u ∈ C6), [u]+(x) = [u]−(x) ∈ C5), and, moreover, Dj[u]+(x) = Dj[u]−(x) ∈ Hαj), 0 < αj ≤ 1, where j = 0,1,…, 5. Then Δ3[u] = 0 in Rn.Suppose Δ3[u] = 0 in Ω−, Dj[u](x) ∈ Hαj), 0 < αj ≤ 1 , and |u(x)| = O(1) (|x| → ∞); then for x ∈ Ω−Hj(y − x) is as in  = −Ψj(y), . For x ∈ Rn∖∂Ω, denoting3[u*] = 0 in Rn∖∂Ω. By using Plemelj formula, combining with weak singularity of Hj(x) (j ≥ 2), we obtain the following:αj′ ≤ 1 . By using 3[u*] = 0 in Rn. It is clear that we have |u*(x)| = O(1) (|x | →∞). In view of For 3[u] = 0 in Ω−, Dj[u](x) ∈ Hαj), 0 < αj ≤ 1 , and |u(x)| = O(1) (|x | →∞), thenLet u(x)| = O(1) (|x | →∞) in u(∞) = 0, then the results in When the condition |In this section, we will consider the Riemann boundary value problem ; the expThe Riemann boundary value problem  is solvaG(x) ∈ Hα)  0 < α < 1 and G(x) satisfies the following condition:M is the positive constant mentioned in Suppose w(x) = D5[u](x) then w+(x) = w−(x)A5 + f5(x), x ∈ ∂Ω. Moreover, by D5[u](∞) = 0. we get the following:1(x) being as in (2u(x) − Δ2Ψ1(x)∶ = φ1(x), x ∈ Rn∖∂Ω, and using (Δ2u)+(x) = (Δ2u)−(x)A4 + f4(x), x ∈ ∂Ω, we conclude thatφ1(∞) = 0, and we then get the following representation formula:2(x) from (2u(x) − Δ2Ψ1(x) − Δ2Ψ2(x) = 0, x ∈ Rn∖∂Ω. DenoteD3u)+(x) = (D3u)−(x)A3 + f3(x), x ∈ ∂Ω, we obtain thatφ2(∞) = 0, and we then get the following representation formula:Denoting ng as in , it is eng as in . By Coro(x) from  that Δ2ng as in . By Coro3(x) being as in (D3u(x) − D3Ψ1(x) − D3Ψ2(x) − D3Ψ3(x) = 0, x ∈ Rn∖∂Ω. We denoteu)+(x) = (Δu)−(x)A2 + f2(x), x ∈ ∂Ω. We conclude thatφ3(∞) = 0; then4(x) being as in (u(x) − ΔΨ1(x) − ΔΨ2(x) − ΔΨ3(x) − ΔΨ4(x) = 0, x ∈ Rn∖∂Ω. DenotingDu)+(x) = (Du)−(x)A1 + f1(x), x ∈ ∂Ω, we getφ4(∞) = 0, and we then get the following representation formula:Here Ψng as in , then weng as in . Corollang as in , that Δuken from . By Coro5(x) as defined in (Du(x) − ∑i=14DΨi(x) = 0, x ∈ Rn∖∂Ω. Defineu+(x) = u−(x)G(x) + g(x) we arrive atφ5(∞) = 0. We obtain thatFinally, we use Ψfined in  and get ken from . It is cφ6(y) is a Hölder continuous function to be determined on ∂Ω. Then, by using Plemelj formula, ), we haveT is a contraction operator mapping the Banach space Hα) into itself, which has a unique fixed point for the operator T. Thus, there exists a unique solution to (Letting  side of , we getondition , the intution to . The pro


Res. Natl. Inst. Stand. Technol. Volume 100, Number 3, May–June 1995, p. Page 211, above Eq. (3).Replace “the magnetic field within the balance chamber is vertical with the form…” by “the vertical magnetic field strength within the balance chamber has the form…”.Page 217, Eq. (8).Ia” by “Ia”.Replace “μmHmax” by “μ0Hmax”.Page 218, two paragraphs above Sec. 5.1. Replace “Page 223, first paragraph of Sec. 11.1.Replace “: it must be less than that based on a sum…; it must be greater than that based on a sum…” by “: it must be greater than that based on a sum…; it must be less than that based on a sum…”.Page 224, Table 5, column 4, row 2.Replace “0.1116” by “−0.1116”.Page 224, Eq. (10) should be:χ” by “This again underestimates χ”.Page 225, two paragraphs above the Acknowledgment. Replace “This again overestimates


G = (V(G), E(G)) be a simple connected graph with vertex set V(G) = {v1, v2,…, vn} and edge set E(G). Its adjacency matrix A(G) = (aij) is defined as n × n matrix (aij), where aij = 1 if vi is adjacent to vj and aij = 0, otherwise. Denote by d(vi) or dG(vi) the degree of the vertex vi. It is well known that A(G) is a real symmetric matrix. Hence, the eigenvalues of A(G) can be ordered asA(G) is called the spectral radius of G, denoted by ρ(G). It is easy to see that if G is connected, then A(G) is nonnegative irreducible matrix. By the Perron-Frobenius theory, ρ(G) has multiplicity one and exists a unique positive unit eigenvector corresponding to ρ(G). We refer to such an eigenvector corresponding to ρ(G) as the Perron vector of G.Let Pn and Cn the path and the cycle on n vertices, respectively. The characteristic polynomial of A(G) is det(xI − A(G)), which is denoted by Φ(G) or Φ. Let X be an eigenvector of G corresponding to ρ(G). It will be convenient to associate with X a labelling of G in which vertex vi is labelled xi (or xvi). Such labellings are sometimes called “valuation” [Denote by luation” .ω are given. And, in [ω the minimum value of the spectral radius is attained for a kite graph PKn−ω,ω, where PKn−ω,ω is a graph on n vertices obtained from the path Pn−ω and the complete graph Kω by adding an edge between an end vertex of Pn−ω and a vertex of Kω  = ω − 1. If G ∈ Iω+1,ω, then, from the Perron-Frobenius theorem, the first ω − 1 smallest values of the spectral radius of Iω+1,ω are PKω;i1, (0 ≤ i ≤ ω − 2), respectively, where PKω;i1, is the graph obtained from PKω1, by adding i (0 ≤ i ≤ ω − 2) edges. So in the following, we consider that n ≥ ω + 2.Let In order to complete the proof of our main result, we need the following lemmas.v be a vertex of the graph G. Then the inequalitiesG is connected, then λ1(G) > λ1(G − v).Let For the spectral radius of a graph, by the well-known Perron-Frobenius theory, we have the following.G be a connected graph and H a proper subgraph of G. Then ρ(H) < ρ(G).Let G be a graph on n vertices, thenG is a regular graph.Let v be a vertex of a graph G and suppose that two new paths P = v(vk+1)vk ⋯ v2v1 and Q = v(ul+1)ul ⋯ u2u1 of lengths k and l (k ≥ l ≥ 1) are attached to G at v( = vk+1 = ul+1), respectively, to form a new graph Gk,l  be the graphs as defined above. Then ρ > ρ.Let v be a vertex of the graph G and N(v) the set of vertices adjacent to v.Let G be a connected graph, and let u, v be two vertices of G. Suppose that v1, v2,…, vs ∈ N(v)∖(N(u)⋃{u}) (1 ≤ s ≤ d(v)) and x =  is the Perron vector of G, where xi corresponds to the vertex vi (1 ≤ i ≤ n). Let G* be the graph obtained from G by deleting the edges vvi and adding the edges uvi (1 ≤ i ≤ s). If xu ≥ xv, then ρ(G) < ρ(G*).Let v be a vertex of G, let φ(v) be the collection of circuits containing v, and let V(Z) denote the set of vertices in the circuit Z. Then the characteristic polynomial Φ(G) satisfiesw adjacent to v, and the second summation extends over all Z ∈ φ(v).Let G is a sequence of vertices v1, v2,…, vk with k ≥ 2 such thatv1 = vk);the vertices in the sequence are distinct ;d(vi) satisfy d(v1) ≥ 3, d(v2) = ⋯ = d(vk−1) = 2 (unless k = 2) and d(vk) ≥ 3.the vertex degrees An internal path of a graph Wn be the tree on n vertices obtained from Pn−4 by attaching two new pendant edges to each end vertex of Pn−4, respectively.Let G ≠ Wn is a connected graph and uv is an edge on an internal path of G. Let Guv be the graph obtained from G by subdivision of the edge uv. Then ρ(Guv) < ρ(G).Suppose that H1 be the graph obtained from Kω and a path P4 : v1v2v3v4 by joining a vertex of Kω and a nonpendant vertex, say, v2, of P4 by a path with length 2 and let H2 be the graph obtained from Kω by attaching two pendant edges at two different vertices of Kω  > ρ(H1).Let bove see . If ω ≥ ω ≥ 3, by direct computations, we have ρ(H2) > ρ(H1). In the following, we suppose that ω ≥ 6. From For 5 ≥ ω > ρ(H1) ≥ ρ(Kω) = ω − 1 ≥ λ2(H1) and ω > ρ(H2) ≥ ρ(Kω) = ω − 1. Then from (ρ(H2) > ω − 1 + (2/ω2) > ρ(H1).By direct calculation, we haveg1  see . It is eLet -1vn see .PKn−ω,ωi be the graphs defined as above  ≥ ρ ≈ 2.00659 > ρ(Cn).Clearly, G1 = PKn−3,3n−3 − vn−1vn−2 + vn−3vn−1, let G2 = PKn−3,3n−3 + vn−1vn, and let Cn−1,1 be the graph obtained from Cn−1 and an isolated vertex by adding an edge between some vertex of Cn−1 and the isolated vertex  < ρ < ρ(Wn) = ρ(Cn) < ρ. Thus, we only need to prove that ρ(G) > ρ if G ≠ Pn, PKn−2,2n−2, Wn, Cn, PKn−2,2n−3. If G is a tree, note that G ≠ Pn, PKn−2,2n−2, Wn, PKn−2,2n−3, then, from ρ(G) > ρ. If G contains some cycle as a subgraph, then, from Lemmas ρ(G) ≥ ρ > ρ.Let PKn−ω,ωi, G1 and G2 be the graphs defined as above  < ρ(G1). If n ≥ 12, from Lemmas ρ < ρ < ρ < 2.23808. From x < 2.23808, we haveρ < ρ and 2.23601 < ρ < 2.23808. Then, we haveρ < ρ(G1). By similar method, we have for n ≥ 8For 8 ≤ H3 be the graph obtained from Kω by attaching two pendant edges at some vertex of Kω; let H4 be the graph obtained from Kω and P2 by adding two edges between two vertices of Kω and two end vertices of P2  > ρ if G ≠ PKn−3,3, PKn−3,3n−2, PKn−3,3n−3, PKn−3,3n−4.Let We distinguish the following three cases.Case  1. If there exist at least two vertices outside of K3 that are adjacent to some vertices of K3, then we have that G contains either H2 (ω = 3) or H3 (ω = 3) as a proper subgraph. If G contains H2 (ω = 3) as a proper subgraph, from Lemmas G contains H3 (ω = 3) as a proper subgraph, from Lemmas  Lemmas 1ρ(G)>ρ is a tree. If there exist two vertices u, r ∈ V(G − V(K3)) such that d(u) ≥ 3 and d(r) ≥ 3, then, from Lemmas ρ(G) > ρ. If there exists only one vertex u ∈ V(G − V(K3)) such that d(u) ≥ 4, then, from Lemmas ρ(G) ≥ ρ(G1) > ρ. If there exists exactly one vertex u ∈ V(G − V(K3)) such that d(u) = 3, note that G ≠ PKn−3,3n−2, PKn−3,3n−3, PKn−3,3n−4, then from Lemmas ρ(G) > ρ.Subcase 2. Suppose that G − V(K3) contains cycle Cg as a subgraph. If g = 3,4, then, from Lemmas ρ(G) ≥ ρ(G2) > ρ. If g ≥ 5, then, from Fg from G by deleting vertices such that ρ(G) ≥ ρ(Fg), where Fg is the graph obtained from K3 and a cycle Cg by joining a vertex of K3 and a vertex of Cg with a path and |V(Fg)|≤n , (1 ≤ i ≤ g − 1), v1vg ∈ E(Cg), and d(v1) = 3. Then, from Lemmas ρ(Fg − v2v3) > ρ. Thus, we have ρ(G) > ρ.)|≤n see . SupposePKn−ω,ωi and ω ≥ 4).Let X = T be the Perron vector of xi corresponds to vi. It is easy to prove that xn = xn−1. From ω ≥ 4 we have Let Mω2 (ω ≥ 4) be the graph as shown in Let n vertices with maximum clique size ω ≥ 4 and n ≥ ω + 5, the first four smallest spectral radii are exactly obtained for PKn−ω,ω, PKn−ω,ωn−2, PKn−ω,ωn−3, respectively.Among all connected graphs on G be a connected graph with maximum clique size ω ≥ 4 and n ≥ ω + 5 vertices. Suppose that Kω is a maximum clique of G. From Lemmas ρ(G) > ρ if G ≠ PKn−ω,ω, PKn−ω,ωn−2, PKn−ω,ωn−3. We distinguish the following three cases.Let Case  1. If there exist at least two vertices outside of Kω that are adjacent to some vertices of Kω, then G contains either H2 or H3 as a proper subgraph. If G contains H2 as a proper subgraph, from Lemmas G contains H3 as a proper subgraph, from Lemmas Case  2. Suppose that there exists a vertex, say, u, which does not belong to Kω, such that u is adjacent to at least two vertices of Kω. From Lemmas Case  3. Suppose that there uniquely exists a vertex u which does not belong to Kω such that u is adjacent to a vertex of Kω. If G − V(Kω) is a tree, note that G ≠ PKn−ω,ω, PKn−ω,ωn−2, PKn−ω,ωn−3, then, from Lemmas ρ(G) > ρ. Suppose that G − V(Kω) contains cycle Cg as a subgraph. If g = 3, note that ρ(G) > ρ(G*) > ρ, where G* = PKn−ω,ωn−3 + vn−1vn. If g ≥ 4, then by the similar reasoning as that of Subcase 2 of Case 3 of ρ(G) > ρ.H3 and H4 be the graphs defined as above T be the Perron vector of H3, where xi corresponds to vi. From AX = ρ(H3)X, we havex > ω − 1 and ω ≥ 3, we haveρ(H3) > ρ(Kω) = ω − 1. From (ρ(H3) which is the largest root of equation r1(x) = 0. Similarly, we have ρ(H4) which is the largest root of equationx > ω − 1,ρ(H3) < ρ(H4).Let −4.Thenr1(ω−1)=− we haver1′(x)=3xG be a graph on n vertices with maximum clique size ω ≥ 3 and n = ω + 2. Let PKω2,, H2, H3, and H4 be the graphs defined as above  > ρ(H4). We distinguish the following two cases.From Lemmas Case  1. Suppose that there exists exactly one vertex outside of Kω that is adjacent to at least two vertices of Kω. Then G contains Mω2 (see ρ(Mω2) > ρ(H4). ω2 see  as a subCase  2. Suppose that the two vertices outside of Kω that are all adjacent to some vertices of Kω. Note that G ≠ H2, H3, H4. Then G contains one of graphs Mω2 as a subgraph, where H3 by adding an edge between two pendant vertices. From ρ(G) > ρ(Mω2) > ρ(H4).H5 be the graph obtained from H2 and an isolated vertex by adding an edge between a pendant vertex of H2 and the isolated vertex; let H6 be the graphs as shown in Let H5 be the graphs defined as above T be the Perron vector of xi corresponds to vi. It is easy to see that x1 = x5. From ω ≥ 4, we haveLet H5 and H6 be the graphs defined as above  > ρ(H5). In the following, we suppose that ω ≥ 5. Then, from Lemmas ω > ρ(H5) > ρ(Kω) = ω − 1 ≥ 4. Let X = T be the Perron vector of H5, where xi corresponds to vi. From AX = ρ(H5)X, we haveω > ρ(H5) > ω − 1 ≥ 4,ρ(H6) = ρ(H5 − v1v2 + v1v6) > ρ(H5).For H7 be the graph obtained from H3 and an isolated vertex by adding an edge between vω and the isolated vertex; let H8 be the graph obtained from H3 and an isolated vertex by adding an edge between v2 and the isolated vertex; let H9 be the graph obtained from H3 and an isolated vertex by adding an edge between one pendant vertex and the isolated vertex; and let H10 be the graph obtained from PKω3,ω+1 and an isolated vertex by adding an edge between vω+1 and the isolated vertex , the first four smallest spectral radii are obtained for PKω3,, PKω3,ω+1, H5, respectively.Let ρ(G) > ρ(H5) if G ≠ PKω3,, PKω3,ω+1, H5. We distinguish the following four cases.From Lemmas Case  1. There exists exactly one vertex outside of Kω that is adjacent to only one vertex of Kω. Then G must be one of graphs PKω3,, PKω3,ω+1, and Case  2. There exists one vertex outside of Kω that is adjacent to at least two vertices of Kω. Then G contains Mω2 (see ρ(G) > ρ(Mω2) > ρ(H5).ω2 see  as a proCase  3. If there exactly exist two vertices outside of Kω that are adjacent to some vertices of Kω, then G contains H5 or H9  ≥ ρ(H9) > ρ(H5). If G contains H5 as a subgraph, note that G ≠ H5, then, from ρ(G) > ρ(H5).Case  4. If there exist three vertices outside of Kω that are adjacent to some vertices of Kω, then G contains one of graphs H6, H7, and H8  > ρ(H7) > ρ(H6) > ρ(H5). Then, from ρ(G) > ρ(H5).PKn−ω,ωi and Let x > ω − 1 (ω ≥ 4), we haveρ > ρ(Kω) = ω − 1. Thus, for x > ω − 1 (ω ≥ 4), we have ω ≥ 4).From PKn−ω,ωi and H2 be the graphs defined as above  > ρ. In the following, we suppose that ω ≥ 6. From ω ≥ 6, we haveω > ρ ≥ ρ(Kω) = ω − 1 ≥ λ2. Then from  > ω − 1 + 1/ω2. From the proof of ρ(H2) > ω − 1 + 2/ω2 (ω ≥ 6). The result follows.For PK4,ωω+. In the  we haveg3(ω−1+1ωn vertices with maximum clique size ω and n = ω + 4 (ω ≥ 4), the first four smallest spectral radii are obtained for PKω4,, PKω4,ω+2, PKω4,ω+1  > ρ if G ≠ PKω4,, PKω4,ω+2, PKω4,ω+1. We distinguish the following three cases.Let Case  1. There exists exactly one vertex outside of Kω that is adjacent to one vertex of Kω.Subcase 1. Suppose that G − V(Kω) is a tree. If G contains exactly one pendant vertex, then G = PKω4,. If G contains exactly two pendant vertices, then G = PKω4,ω+1 or G = PKω4,ω+2. If G contains three pendant vertices, then G = H10 (see ρ(H10) > ρ.H10 see . From LeSubcase 2. Suppose that G − V(Kω) contains a cycle. If G − V(Kω) contains C4, then G contains H11 as a subgraph, where H11 is obtained from PKω4,ω+1 by adding an edge between two pendant vertices. From ρ(H11) > ρ. If G − V(Kω) does not contain C4, then G contains ρ(G) > ρ.Case  2. There exists at least one vertex outside of Kω that is adjacent to at least two vertices of Kω. Then G contains Mω2 (see ρ(G) > ρ(Mω2) > ρ(H2) > ρ.ω2 see  as a subCase  3. There exist at least two vertices outside of Kω that are adjacent to some vertices of Kω. Then G contains H2 or H3 as a subgraph  > ρ(H2) > ρ. Thus, from ρ(G) > ρ.n with maximum clique size ω ≥ 2, are determined.In this paper, the first four graphs, which have the smallest values of the spectral radius among all connected graphs of order


Particularly, two coverings with the same reductions are proved to generate the same f-lower and f-upper approximations. Finally, we discuss the relationships between the new model and some other variable precision rough set models.Classical rough set theory is a technique of granular computing for handling the uncertainty, vagueness, and granularity in information systems. Covering-based rough sets are proposed to generalize this theory for dealing with covering data. By introducing a concept of misclassification rate functions, an extended variable precision covering-based rough set model is proposed in this paper. In addition, we define the  In the era of big data, it is difficult to obtain useful information in huge data. Many researchers have proposed lots of efficient means of dealing with the difficulty. As one of these efficient means, classical rough set theory based on equivalence relations is proposed by Pawlak , 2 in thIn variable precision rough sets, the misclassification rate of equivalence classes of all elements in a universe is identical. Similarly, in variable precision covering-based rough sets, the misclassification rate of neighborhoods of all elements in a universe is identical too. However, in practical applications, since there are different understanding or demands about equivalence classes or neighborhoods of different elements, the misclassification rate usually varies. Hence, it is necessary to propose misclassification rate functions.f-lower and f-upper approximations in terms of neighborhoods and investigate their properties.To address the above issue, we propose a variable precision covering-based rough set model based on functions by introducing misclassification rate functions in this paper. We present the concepts of the f-lower and f-upper approximations. Moreover, the relationships between this model and some other variable precision rough set models are exhibited. The rest of this paper is arranged as follows. U is a non-empty finite set.In this section, we present some fundamental concepts and existing results of classical rough sets, covering-based rough sets, and variable precision covering-based rough sets. Throughout this paper, the universe U be a universe and R an equivalence relation on U. R will generate a partition U/R. The elements in U/R are called equivalence classes. Let [x]R denote an equivalence class which includes x ∈ U. For any X⊆U, we can describe X by the two sets,Let X with respect to R, respectively.They are called the lower and upper approximations of Covering-based rough sets are presented as the extension of classical rough sets by extending partitions to coverings on a universe.U be a universe of discourse and C a family of subsets of U. If none subsets in C is empty and ⋃C = U, then C is called a covering of U. The pair  is called a covering approximation space.Let Neighborhoods are important concepts in covering-based rough sets.C be a covering of U and x ∈ U. NC(x) = ⋂{K ∈ C∣x ∈ K} is called the neighborhood of x with respect to C. When there is no confusion, we omit the subscript C.Let C be a covering of U and x ∈ U. For any X⊆U, the lower and upper approximations of X with respect to  are defined as follows, respectively:  Let C is a partition of  U, then covering-based rough sets degenerate into classical rough sets.Clearly, if C be a covering of U and K ∈ C. If K is a union of some sets in C − {K}, we say K is a reducible element of C; otherwise K is an irreducible element of C. The family of all irreducible elements of C is called the reduct of C, denoted as Reduct  (C).Let β  (0 ≤ β < 0.5) in classical rough sets, namely, some degrees of misclassification are allowed, Ziarko proposed variable precision rough sets.By introducing the parameter X and Y be two non-empty subsets of  U. We say that X is included in Y, if every element of X is an element of Y. In this case, the inclusion relation is certain. That is to say, there does not exist the slightest misclassification between X and Y. However, slight misclassification is allowed in practical applications. Therefore, we present the majority inclusion relation which is a generalized definition of inclusion relation. Before the majority inclusion relation is presented, we introduce the measure of the relative degree of misclassification of one set with respect to others.Let X and Y be two subsets of a universe U. The measure c of the relative degree of misclassification of the set X with respect to set Y is defined asX| denotes the cardinality of X.Let X and Y be two subsets of a universe U, and 0 ≤ β < 0.5. The majority inclusion relation is defined asLet X⊆Y if and only if c = 0.By the definition, it is clear that β1 < β2 < 0.5, then If 0 ≤ U be a universe of discourse, R an equivalence relation, and 0 ≤ β < 0.5. For any X⊆U, the β-lower and β-upper approximations of X with respect to R are defined as follows, respectively:  Let β = 0, it is easy to see that variable precision rough sets become classical rough sets. By extending partitions to coverings, variable precision rough sets are generalized to variable precision covering-based rough sets. We present the definition of variable precision covering-based rough sets as follows.When C be a covering of U and 0 ≤ β < 0.5. For any X⊆U, the β-lower and β-upper approximations of X with respect to  are defined as follows, respectively:  Let β = 0, then variable precision covering-based rough sets are covering-based rough sets; if C is a partition of U, then variable precision covering-based rough sets are variable precision rough sets; if C is a partition of U and β = 0, then variable precision covering-based rough sets become classical rough sets.Clearly, if In variable precision rough sets, the misclassification rate of equivalence classes of all elements in a universe is identical. Similarly, in variable precision covering-based rough sets, the misclassification rate of neighborhoods of all elements in a universe is identical too. However, in practical applications, we have different understanding or demands about equivalence classes or neighborhoods of different elements. That means the misclassification rate usually varies. Therefore, we present variable precision covering-based rough sets based on functions by introducing a concept of misclassification rate functions.C be a covering of U and NC(x) the neighborhood of x. It is obvious that ∪x∈UNC(x) = U. That is, the collection of neighborhoods of all elements in U is still a covering of U. It is called neighborhood covering of U with respect to C and denoted by Cov(C) [Let y Cov(C) .C be a covering of U. A function f defined from Cov(C) to [0,0.5) is called misclassification rate function.Let C be a covering of U and f a misclassification rate function. For any X⊆U, the f-lower and f-upper approximations of X with respect to  are defined as follows, respectively:  Let U = {x1, x2, x3, x4, x5, x6, x7, x8} and C = {K1, K2, K3}, where K1 = {x1, x2, x3, x4, x5}, K2 = {x4, x5, x6, x7} and K3 = {x6, x8}. Suppose that f(N(x)) = |N(x)|/2|U| and X = {x2, x3, x4, x5, x6}. Through the definition of neighborhoods, we have N(x1) = N(x2) = N(x3) = {x1, x2, x3, x4, x5}, N(x4) = N(x5) = {x4, x5}, N(x6) = {x6}, N(x7) = {x4, x5, x6, x7} and N(x8) = {x6, x8}. Therefore, f(N(x1)) = f(N(x2)) = f(N(x3)) = 5/16, f(N(x4)) = f(N(x5)) = f(N(x8)) = 2/16, f(N(x6)) = 1/16 and f(N(x7)) = 4/16. By c(N(xi), X) = 1/5  , c(N(xj), X) = 0  , c(N(x7), X) = 1/4 and c(N(x8), X) = 1/2. Hence f ≡ 0, then f ≡ β = 1/5, then Let In practical applications, according to various needs, the different misclassification rate functions can be given by workers or researchers.f(N(xi)) = 0.1   and f(N(x7)) = f(N(x8)) = 0.2. By Suppose that f-positive region Pos cf(X), f-boundary region Bn cf(X) and f-negative region Neg cf(X) of X with respect to  by the following definition.Similar to classical rough sets or variable precision rough sets, we present the concepts of U be a universe of discourse and f a misclassification rate function. For any X⊆U, the f-positive region Pos cf(X), f-boundary region Bn cf(X) and f-negative region Neg cf(X) of X with respect to  are defined as follows, respectively:Let U be a universe of discourse. For any X⊆U, the following conclusions are true,cf(X) = {x ∈ U∣c(N(x), X) ≥ 1 − f(N(x))},Neg cf(X) = {x ∈ U∣f(N(x)) < c(N(x), X) < 1 − f(N(x))},Bn cf(−X) =Negcf(X).Pos Let x ∈ U∣c(N(x), X) ≥ 1 − f(N(x))}.(1) x ∈ U∣f(N(x)) < c(N(x), X)<1 − f(N(x))}.(2) x ∈ U, c(N(x), −X) = 1 − c(N(x), X). x ∈ U∣1 − c(N(x), X) ≤ f(N(x))}  =  {x ∈ U∣c(N(x), X) ≥ 1 − f(N(x))} = Neg cf(X).(3) For any f-accuracy and f-roughness, which are important numerical characteristics of this type of rough sets.In the following definition, we present the C be a covering of U. For any X⊆U, the f-accuracy and f-roughness are defined as follows, respectively:  ρf(X) = 1 − λf(X).Let In this subsection, we present the properties and some significant results concerning the new model.C be a covering of  U and f a misclassification rate function. For all X, Y⊆U, the following conclusions are true,X⊆Y, then If Let x ∈ U, f(N(x)) < 0.5 < 1 − f(N(x)). By (1) For any x ∈ U, since c(N(x), ∅) = 1, it follows that c(N(x), U) = 0, it follows that (2) For any X⊆Y, then c(N(x), Y) ≤ c(N(x), X). Hence, (3) If X, Y⊆U, since c(N(x), X ∪ Y) ≤ c(N(x), X) and c(N(x), X ∪ Y) ≤ c(N(x), Y), it follows that (4) For all (5) Similar to the proof of (4).X, Y⊆U, since c(N(x), X) ≤ c(N(x), X∩Y) and c(N(x), Y) ≤ c(N(x), X∩Y), it follows that (6) For all (7) Similar to the proof of (6).x ∈ U, c(N(x), −X) = 1 − c(N(x), X). Hence, (8) For any C be a covering of U and f, g two misclassification rate functions. Suppose that f(N(x)) ≤ g(N(x)) for any x ∈ U. For every X⊆U, the following relationships are true,Let According to C be a covering of U and f a misclassification rate function. Denote α1 = min{f(N(x))∣N(x) ∈Cov(C)} and α2 = max{f(N(x))∣N(x) ∈Cov(C)}. For any X⊆U, thenLet C, D be two different coverings of U and f a misclassification rate function. If Reduct  (C) = Reduct  (D), then X⊆U.Let C) = Cov(D). Suppose that K1 is a reducible element of C. For any x ∈ U, x ∈ K1 or x ∉ K1. If x ∉ K1, it is straightforward that NC(x) = NC−{K1}(x). Otherwise, there exists K′ ⊂ K1 such that x ∈ K′ and K′ is an irreducible element of C. Thus, NC(x) = ⋂x∈K∈CK = ⋂K′⋂K1 = ⋂K′ = NC−{K1}(x). Hence NC(x) = NC−{K}(x) for any reducible element K ∈ C. According to this method, it is easy to see that NC(x) = NC)Reduct  ((x). From the arbitrariness of x, Cov(C) = Cov(Reduct  (C)). Since Reduct  (C) = Reduct  (D), it follows that Cov(C) = Cov(D). This completes the proof.From U = {x1, x2, x3, x4, x5, x6, x7, x8}. C = {K1, K2, K3, K4}, where K1 = {x1, x2, x3}, K2 = {x3, x4, x5}, K3 = {x4, x5, x6, x7} and K4 = {x6, x8}. D = {K1, K2, K3, K4}, where K1 = {x1, x2, x3, x4, x5}, K2 = {x4, x5, x6, x7}, K3 = {x6, x8} and K4 = {x4, x5, x6, x7, x8}. Clearly, Reduct  (C) = Reduct  (D). Suppose X = {x2, x3, x4, x5, x6}. Hence Let However, the converse proposition of C be the covering in D = {K1, K2, K3, K4}, where K1 = {x1, x2, x3, x4, x5}, K2 = {x4, x5, x6, x7}, K3 = {x6, x8} and K4 = {x1, x2, x3, x4, x5, x6}. Suppose that X = {x2, x3, x4, x5, x6}. It is easy to know that C) ≠ Reduct  (D).Let f be a misclassification rate function and β a constant, where 0 ≤ β < 0.5.“1”:C) is a partition of U;Cov is a partition of U and f ≡ 0;Cov(“4”:C) is a partition of U and f ≡ β;Cov(“5”:C) is a partition of U;Cov(“6”:β = 0;“7”:f ≡ 0;“8”:f ≡ β.In this subsection, we will use a figure  to explaf-lower and f-upper approximations. Moreover, we exhibited the relationships between this model and some other variable precision rough set models. In future work, we will seek more specific misclassification rate functions for dealing with more types of data.In this paper, we proposed the variable precision covering-based rough set model based on functions as a generalization of a variable precision covering-based rough set model and studied its properties. Through the concept of reductions, we obtained that two coverings with the same reductions generate the same


The main result is illustrated with the aid of an example.By employing a nonlinear alternative for contractive maps, we investigate the existence of solutions for a boundary value problem of fractional  Here, we emphasize that the nonlocal conditions are regarded as more plausible than the standard initial conditions for the description of some physical phenomena. In (g(x) may be understood as g(x) = ∑j=1pαjx(tj), where αj, j = 1,…, p, are given constants and 0 < t1 < ⋯<tp ≤ 1. For more details, we refer to the work by Byszewski and Lakshmikantham q is defined byFor a real parameter q-analogue of the Pochhammer symbol  is defined asq-analogue of the exponent (x − y)k isq-gamma function Γq(y) is defined asy ∈ R∖{0, −1, −2,…}. Observe that Γq(y + 1) = [y]qΓq(y).The f be a function defined on . The fractional q-integral of the Riemann-Liouville type of order β ≥ 0 is (Iq0f)(t) = f(t) andLet β = 1 in q-integral:q-integrals and fractional q-integrals, see Sections 1.3 and 4.2, respectively, in  × R → P(R) is said to be a Carathéodory function if t ↦ F is measurable for each x ∈ R;x ↦ F is upper semicontinuous for almost all t ∈ .Further, a Carathéodory function F is called L1-Carathéodory if (iii)α > 0, there exists φα ∈ L1 such thatfor each  x‖ ≤ α and for a.e. t ∈ .for all ‖A multivalued map G to be the set Gr(G) = { ∈ X × Y,  y ∈ G(x)} and recall two results for closed graphs and upper-semicontinuity.We define the graph of G : X → Pcl(Y) is u.s.c., then Gr(G) is a closed subset of X × Y, that is, for every sequence {xn}n∈N ⊂ X and {yn}n∈N ⊂ Y, if n → ∞, xn → x∗, yn → y∗, and yn ∈ G(xn), then y∗ ∈ G(x∗). Conversely, if G is completely continuous and has a closed graph, then it is upper semicontinuous.If X be a separable Banach space. Let F :  × X → Pcp,c(X) be an L1-Carathéodory function. Then, the operatorC × C.Let x ∈ AC1 is called a solution of problem  with f(t) ∈ F), a.e. on  such that Dqυx(t) = f(t), a.e. on  and x(0) = g(x) and x(ω) = b∫δ1x(s)dqs.A function To prove our main result in this section we will use the following form of the nonlinear alternative for contractive maps , we haveStep 2. F2 is compact, convex valued, and completely continuous. This will be established in several claims.Claim 1. F2 maps bounded sets into bounded sets in C. For that, let Bρ = {x ∈ C : ‖x‖ ≤ ρ} be a bounded set in C. Then, for each h ∈ F2(x), x ∈ Bρ, we haveh ∈ F2(Bρ), we haveClaim 2. F2 maps bounded sets into equicontinuous sets. As before, let Bρ be a bounded set and let h ∈ F2(x) for x ∈ Bρ. Let t1, t2 ∈  with t1 < t2 and x ∈ Bρ. Then, for each h ∈ F2(x), we obtainx and tends to zero as t2 − t1 → 0. Therefore, it follows by the Arzelá-Ascoli theorem that F2 : C → P is completely continuous.Claim 3. F2 has a closed graph. Let xn → x∗, hn ∈ F2(xn), and hn → h∗. Then, we need to show that h∗ ∈ F2(x∗). Associated with hn ∈ F2(xn), there exists fn ∈ SF,xn such that, for each t ∈ ,f∗ ∈ SF,x∗ such that, for each t ∈ ,L1 → C defined byLet us consider the continuous linear operator Θ : n → ∞. Thus, it follows by SF is a closed graph operator. Further, we have hn(t) ∈ Θ. Since xn → x∗, therefore, we havef∗ ∈ SF,x∗. Hence, F2 has a closed graph . In consequence, the operator F2 is compact valued.Observe thatF1 and F2 satisfy hypotheses of x ∈ λF1(x) + λF2(x) for λ ∈ , then there exists f ∈ SF,x such that x = λF(x), that is,λ ∈  and x ∈ ∂BM with x = λF(x), where BM = {x ∈ C : ‖x‖ ≤ M}. Then, x is a solution of x = λF(x) with ‖x‖ = M. Now, by the last inequality, we getF has a fixed point on  by Thus, the operators tradicts . Hence, q-fractional boundary value problem:Consider the following υ = 3/2, q = 1/2, b = 1/5, ω = 1/4, δ = 3/4, l = 1/15, andϑ = 0.191667, μ0 ≈ 1.3990, k0 = 3.9564, ||p|| = 43/48, ψ(||x||) = 1, and M > 1.7026 by (H3). Thus, all the conditions of Here,


We establish the stabilities and blowup results for the nonisentropic Euler-Poisson equations by the energy method. By analysing the second inertia, we show that the classical solutions of the system with attractive forces blow up in finite time in some special dimensions when the energy is negative. Moreover, we obtain the stabilities results for the system in the cases of attractive and repulsive forces. As usual, ρ = ρ ≥ 0, u = u ∈ ℝN, and S are the density, the velocity, and the entropy, respectively. P is the pressure function, for which the constants K ≥ 0 and γ ≥ 1.The compressible nonisentropic Euler  are C1 solutions with compact support Ω = Ω(t) for each fixed time t. We also denote the total mass by M, where ρ0 = ρ0(x): = ρ.Lastly, we will denote γ > 1; namely, In this section, we establish some lemmas for the proof of the main results. The following lemma will be used to derive the energy functional for ρ, u, S) of system  of system     =−δρ∇Appendix:∫Ω12u2∇· of system γ = 1. Coρ, u, S) of system  of system  with respect to t.For the classical solution (f system  with γ >E(t) is a decreasing function and is conserved if the system is not damped.Thus, By ρ, u, S) of system  with respect to t.For the classical solution (f system  with γ =E(t) is a decreasing function and is conserved if the system is not damped.Thus, By ρ, u, S) is a classical solution of system =∫Ωρxx is the gradient operator with respect to the spatial variable x.Firstly,N = 2, N = 2 is established.Note thatN ≥ 3, N ≥ 3 are also established.For Now, we are ready to present the stability results.Considering the classical solutions of system , we haveCase 1. For δ = 1, β = 0, N = 3 or 4, γ ≥ 2(N − 1)/N, and E(0) ≥ 0, we have Case 2. For δ = −1, β = 0, N ≥ 4, and γ ≥ (N + 2)/N, we haveCase 3. For δ = −1, β = 0, N = 2, and γ > 1, we have First of all, by definitions , 6), an, an6), aCase 1. By Propositions equality , we haveCase 2. By Propositions N ≥ 3 by /N, and E(0) < 0, then the classical solutions of (If    Case 1 (β = 0). As before, we have, from Propositions t, we see that H(t) is negative as the leading coefficient of the right hand side of (H(t) is nonnegative by definition (ows that H(t)≤(N−2finition . This isCase 2 (β ≠ 0). Now (eβt on both sides and taking integration that A1 and A2. Note that this implies that H(t) is negative for sufficient large t as β > 0 and (N − 2)E(0) < 0. Therefore, the solutions blow up in finite time. 0). Now  becomes(


By using the concept of exceptional family, we proposea sufficient condition of a solution to general order complementarity problems (denoted by GOCP) in Banach space, which is weaker than that in Németh, 2010 (Theorem 3.1). Thenwe study some sufficient conditions for the nonexistence of exceptional family forGOCP in Hilbert space. Moreover, we prove that without exceptional family isa sufficient and necessary condition for the solvability of pseudomonotone generalorder complementarity problems. There are several types of order complementarity problems in real world applications. Among them, the linear order complementarity problem was systematically studied (see ). The prRn to general Hilbert space  weakly proper to CCP . Smi. Smi15])ace (see ). Using ace (see ) and Kalace (see ), some eace see , 21). In. InRn tace (see ). In 201CCP (see ). In 201CCP (see ).Motivated and inspired by the works mentioned above, in this paper, by using the concept of exceptional family in , we propThe remainder of this paper is organized as follows. The preliminary results which will be used in this paper are stated in In this section, we recall some background materials and preliminary results used in the subsequent sections. Firstly, we give some concepts from .X be a Banach space whose norm is denoted by ||·||. Let K ⊂ H be a closed set. K is called a wedge, if for any λ ≥ 0 and x, y ∈ K, λx ∈ K and x + y ∈ K. A wedge K is called a cone if K∩(−K) = {0}.Let X is called an order if it meetsx⪯x for all x ∈ X;reflexivity; that is, x⪯y and y⪯x, then x = y;antisymmetry; that is, if x⪯y and y⪯z, then x⪯z.transitivity; that is, if A relation ⪯ on X is induced by a cone K ⊂ X; that is, x⪯y if and only if y − x ∈ K. Hence K = {x ∈ X : 0⪯x} by using the relation ⪯ on X. Then we denote an ordered Banach space by .We say a relation ⪯ on X is induced by a cone K ⊂ X if and only if it isx⪯y, then x + z⪯y + z for all z ∈ X;translation invariant; that is, if x⪯y, then λx⪯λy for any λ > 0;scale invariant; that is, if xn}n>0 and {yn}n>0 in X with xn⪯yn for all n > 0, then x*⪯y*, where x* and y* are the limits of {xn}n>0 and {yn}n>0, respectively.continuous; that is, if for any two convergent sequences {A relation ⪯ on X, ||·||, K) is called a vector lattice if for every x, y ∈ X there exists x∧y : = inf{x, y} with respect to the order induced by K. In this case we say that the cone K is latticial. By the above concepts, we give the following property from  × Ω → X be a completely continuous mapping. If y ∉ ht(∂Ω), then deg is a constant for 0 ≤ t ≤ 1, where ht(x) = x − H.Let Ω⊆X be an open bounded subset, I : X → X an identity mapping, and f  : I − f, Ω, y) ≠ 0, then equation x − f(x) = y has at least one solution in Ω.Let First we recall the definition of general order complementarity problems see , 5, 6) a, 6 a5, 6X, ||·||) be a Banach space ordered by the latticial cone K ⊂ X and D ⊂ X a nonempty closed convex set. Consider m mappings f1, f2,…, fm : X → X. The general order complementarity problem defined by the family of mappings {fi}i=1m and the set D isLet  be a Banach space ordered by the latticial cone K ⊂ X and D ⊂ X a nonempty closed convex set. Consider m mappings f1, f2,…, fm : X → X. A sequence {xr}r>0⊆D is said to be an exceptional family for GOCP if the following conditions are satisfied:xr|| → +∞ as r → +∞,||r > 0, there exists a real number μr > 0 such that ur1∧⋯∧urm = 0, with uri = μrxr + fi(xr), for i = 1,2,…, m.for every real number Let (The following lemma comes from the property proved in . If there exists an r > 0 such thatht(x), Ur, 0) is constant for t ∈ . This together with x − ϕ(x), Ur, 0) = deg = 1. Therefore, we know that problem ϕ(x) = x is solvable from From r > 0, there exist a vector xr ∈ Br and a scalar tr ∈  such that htr(xr) = 0; that is, xr − (1 − tr)ϕ(xr) = 0.On the other hand, for every tr = 0, then xr − ϕ(xr) = 0, which again implies solvability of the problem GOCP. If tr = 1, then xr = 0, which contradict with the fact xr ∈ Br. Hence tr ≠ 1. If 0 < tr < 1, then from the definition of ϕ(x) we gettr, we obtainμr = tr/(1 − tr). Let uri = μrxr + fi(xr), i = 1,2,…, m; thenxr}r>0 is an exceptional family for GOCP. The proof is complete.If fi and Si = I − fi being completely continuous operators for all i = 1,2,…, m. Moreover, [Sm(D) + K ⊂ D holds, which does not need this condition in Notice that, in , they usH whose inner product and norm are denoted by 〈·, ·〉 and ||·||, respectively. We propose some sufficient conditions and prove that they guarantee existence of solutions to the general order complementarity problem. Firstly, we give the condition as follows.In this section, we consider the general order complementarity problems in Hilbert space H, 〈·, ·〉) be a Hilbert space ordered by the latticial cone K ⊂ H and D ⊂ H a nonempty set. f1, f2,…, fm : H → H satisfy the following condition: there exists ρ > 0 such that for all x ∈ D with ||x|| > ρ, there exists y ∈ H with ||y|| < ||x|| such thatLet  be a Hilbert space ordered by the latticial cone K ⊂ H, D ⊂ H an unbounded closed convex set and fi, Si = I − fi are completely continuous for all i = 1,2,…, m. If fi}i=1m, D) and hence, GOCP is solvable.Let  has an exceptional family {xr}r>0 ⊂ D. By Suppose that GOCP∧⋯∧fm(xr)〉≥0. We havefi}i=1m, D). Then the problem is solvable.Take H, 〈·, ·〉) be a Hilbert space ordered by the latticial cone K ⊂ H and D ⊂ H a nonempty set. f1, f2,…, fm : H → H satisfy the following condition: there exists a nonempty bounded subset C ⊂ D such that for every x ∈ D∖C, there exists y ∈ C such thatLet  be a Hilbert space ordered by the latticial cone K ⊂ H and D ⊂ H an unbounded closed convex set and fi, Si = I − fi are completely continuous for all i = 1,2,…, m. If fi}i=1m, D) and hence, GOCP is solvable.Let  such that 〈x − y, f1(x)∧⋯∧fm(x)〉≥0. Hence Let p-order coercivity condition  to  to p-ordH, 〈·, ·〉) be a Hilbert space ordered by the latticial cone K ⊂ H and D ⊂ H an unbounded set. Consider m mappings f1, f2,…, fm : H → H. f1(x)∧⋯∧fm(x) is said to be p-order coercive with respect to D, if there exists p ∈  be a Hilbert space ordered by the latticial cone K ⊂ H and D ⊂ H an unbounded closed convex set and fi, Si = I − fi are completely continuous for all i = 1,2,…, m. If there exists some p ∈ ∧⋯∧fm(x) is p-order coercive with respect to D, then there exists no exceptional family for GOCP and hence, GOCP is solvable.Let (f1(x)∧⋯∧fm(x) is p-order coercive with respect to D for some p ∈  weakly proper to GOCP.H, 〈·, ·〉) be a Hilbert space ordered by the latticial cone K ⊂ H and D ⊂ H a nonempty set. Consider m mappings f1, f2,…, fm : H → H. f1(x)∧⋯∧fm(x) is said to be(a)D if, for every x, y ∈ D, x ≠ y, one haspseudomonotone on (b)D if, for every x, y ∈ D, x ≠ y, one hasquasiomonotone on Let  be a Hilbert space ordered by the latticial cone K ⊂ H and D ⊂ H an unbounded set. Consider m mappings f1, f2,…, fm : H → H. f1(x)∧⋯∧fm(x) is said to be(a)D, if for every sequence {xr} ⊂ D with limr→+∞||xr|| = +∞, there exists a y ∈ H and some r such thatweakly proper on (b)D, if for every sequence {xr} ⊂ D with limr→+∞||xr|| = +∞, there exists a y ∈ H and some r such thatstrictly weakly proper on Let  be a Hilbert space ordered by the latticial cone K ⊂ H and D ⊂ H an unbounded closed convex set and fi, Si = I − fi are completely continuous for all i = 1,2,…, m. If f1(x)∧⋯∧fm(x) is pseudomonotone on D, then the following conditions are equivalent: fi}i=1m, D) has no exceptional family;GOCP has at least a solution;GOCP({f1(x)∧⋯∧fm(x) is weakly proper on D.Let ((1)⇒(2) follows from fi}i=1m, D) has at least a solution, there exists x* ∈ D such that f1(x*)∧⋯∧fm(x*) = 0. Then for every sequence {xr} ⊂ D with limr→+∞||xr|| = +∞, we havef1(x)∧⋯∧fm(x) is weakly proper on D.(2)⇒(3). Since GOCP has an exceptional family. Then there exists {xr}r>0⊆D, ||xr|| → +∞  (r → +∞), and μr > 0 such thatf1(x)∧⋯∧fm(x) is weakly proper on D, then there exists a y ∈ H and some r such thatf1(x)∧⋯∧fm(x) is pseudomonotone on D yieldsfi}i=1m, D) has no exceptional family. From the above, we complete the proof.(3)⇒(1). Suppose that GOCP∧⋯∧f1(x)∧⋯∧fm(x) is pseudomonotone on D, GOCP has no exceptional family ⇔ GOCP which has at least a solution.The above theorem shows that if H, 〈·, ·〉) be a Hilbert space ordered by the latticial cone K ⊂ H and D ⊂ H an unbounded closed convex set and fi, Si = I − fi are completely continuous for all i = 1,2,…, m. If f1(x)∧⋯∧fm(x) is quasimonotone on D, then there exists no exceptional family for GOCP and hence, GOCP is solvable.Let  has an exceptional family. Then from the proof of f1(x)∧⋯∧fm(x) is strictly weakly proper on D, then there exists a y ∈ H and some r such thatf1(x)∧⋯∧fm(x) is quasimonotone on D yieldsfi}i=1m, D) has no exceptional family. From the above, we complete the proof.Suppose that GOCP({)〉≥0.By  we get(In this paper, by using the concept of exceptional family in , we prop


We show that these ranks generalize some known rank functions over semirings such as the determinantal rank. We also show that this family of ranks satisfies the rank-sum and Sylvester inequalities. We classify all bijective linear maps which preserve these ranks.We use the  For matrices over semirings, all of these definitions are no longer equivalent and each of these generalizes to a distinct rank function for matrices over semirings. There are many different rank functions for matrices over semirings and their properties and the relationships between them have been much studied . In tt of Tan , 5 to deϵ-determinant of Tan first introduced in  and B = [bij] be m by n matrices over a semiring  and let C = [cij] be an n by p matrix over the same semiring. Then A + B = [aij ⊕ bij] and AC = [⨁k=1naik ⊗ ckj]. The set of n by n matrices over a semiring is itself a semiring. For any k ∈ S, the matrix kA = [k ⊗ aij].Let A = [aij] is an n by n matrix over a commutative ring, then the standard determinant expression of A is Sn is the symmetric group of order n and sgn(σ) = +1 if σ is even permutation and sgn(σ) = −1 if σ is odd permutation. Here sgn(σ)aσ(1)1aσ(2)2 ⋯ anσ(n) is called a term of the determinant.If Since we do not have subtraction in a semiring, we cannot write the determinant of a matrix over a semiring in this form. We split the determinant into two parts, the positive determinant and the negative determinant.A be an n by n matrix over a commutative semiring S. We define the positive and the negative determinant as An is the alternating group of order n, that is, the set of all even permutations of order n and Sn∖An is the set of all odd permutations of order n.Let A over a commutative ring takes the form As such we note that the determinant of a matrix In the semiring case, one cannot subtract the negative determinant from the positive determinant and so the positive determinant and the negative determinant are listed as a pair. This pair is called the bideterminant.A be an n by n matrix over a commutative semiring. The bideterminant of A is (det+(A), det−(A)).Let The definition of the permanent involves no subtractions; hence it carries over to the semiring case unchanged.A = [aij] be an n by n matrix over a semiring; then the permanent of A is Let A) = det+(A) ⊕ det−(A).The permanent of a square matrix is the sum of its positive and negative determinants: per called the difference preorder on semirings.S be a semiring. We define the difference preorder ≥ on S as follows: if x, y ∈ S then x ≥ y if there exists z ∈ S such that x = y ⊕ z.Let The difference preorder may not be a partial order. However for many semirings such as the nonnegative semiring, max-plus semiring, and any Boolean algebra or distributive lattice, the difference semiring corresponds to the natural order on the set.In , 5, Tan S be a semiring. A bijection τ : S → S is called a symmetry if τ(τ(a)) = a for all a ∈ S and τ(a ⊗ b) = a ⊗ τ(b) = τ(a) ⊗ b for all a, b ∈ S.Let ϵ-function in  denote the k by k submatrix of A whose th entry is aαi,βj. We define π(α) = ∑j=1kαj.Tan has shown that the S be a commutative semiring and let ϵ ∈ S satisfy ϵ2 = 1. Let A ∈ Mm,n(S), B ∈ Mn,p(S), 1 ≤ k ≤ min. Let α and β be two subsets of {1,2,…, n} of cardinality k. Then there exists a δ ∈ S such that detϵ((AB)) =  ⊗ detϵ)⊕[δ ⊗ (1⊕ϵ)].Let S be a commutative semiring and let ϵ ∈ S satisfy ϵ2 = 1. If A ∈ Mn(S) and α⊆{1,2,…, n}, then detϵ(A) = ⨁β⊆{1,2,…,n};|β|=|α|(ϵ)π(α)+π(β) ⊗ detϵ(A[α∣β]) ⊗ detϵ(A[αc∣βc]).Let S is a commutative ring and ϵ = −1, the ϵ-determinant reduces to the regular determinant and the two theorems above reduce to the usual Binet-Cauchy theorem and the Laplace expansion.In the case where One corollary of the generalized Binet-Cauchy theorem is the following difference preorder inequality for square matrices.S be a commutative semiring and let ϵ ∈ S satisfy ϵ2 = 1. The inequality detϵ(AB) ≥ detϵ(A) ⊗ detϵ(B) holds for all A, B ∈ Mn(S).Let S is a Boolean algebra (and by necessity ϵ = 1 which means detϵ is the permanent) has appeared in ) ⊕ δ ⊗ [1 ⊕ ϵ]. Since the first summand is 0, so Ir+1ϵ is contained in the ideal generated by 1 ⊕ ϵ. Hence rankdetϵ(A) ≤ r.Let ϵ, I)-rank.We can also prove a version of Sylvester's inequality for the  ≤ min, rankdetϵ,I(B)) holds for all A ∈ Mm,n(S) and B ∈ Mn,p(S).Let r = min, rankdetϵ,I(B)). If r ≥ min we are done so suppose r < min and let α and β be both arbitrary subsets of {1,2,…, n} of cardinality r + 1. Then either Ir+1ϵ(A) or Ir+1ϵ(B) is contained in I. It follows from the Binet-Cauchy theorem that detϵ ∈ I.Let ϵ ∈ I is required for our version of Sylvester's inequality to hold; this is largely our motivation for insisting on this condition.It should be noted that the condition 1 ⊕ We also have the following rank-sum inequality.S be a commutative semiring and let ϵ be an element of S such that ϵ2 = 1 and that 1 ⊕ ϵ is not a unit. Let I be an proper ideal of S which contains 1 ⊕ ϵ. The inequality rankdetϵ,I(A + B) ≤ rankdetϵ,I(A) + rankdetϵ,I(B) holds for all A, B ∈ Mm,n(S).Let m = n and rankdetϵ,I(A) + rankdetϵ,I(B) = n − 1. Hence A, B ∈ Mn(S). Let us suppose that r = rankdetϵ,I(A). This implies that n − r − 1 = rankdetϵ,I(B). We can use ϵ(A + B) ∈ I. Note that every term in the expansion of detϵ(A + B) is of a power of ϵ times detϵ(A[α∣β]) ⊗ detϵ(B[αc∣βc]) where α and β are subsets of {1,2,…, n} satisfying |α | = |β|. Let k = |α | = |β|. If k ≤ r, then n − k ≥ n − r > rankdetϵ,I(B) and since |αc | = |βc | = n − k we must have detϵ(B[αc∣βc]) ∈ I. Similarly, if k > r, then detϵ(A[α∣β]) ∈ I. Therefore every term in the expansion of detϵ(A + B) is in I and hence rankdetϵ,I(A + B) ≤ n − 1 = rankdetϵ,I(A) + rankdetϵ,I(B).We begin by proving the inequality in the special case where r = rankdetϵ,I(A) + rankdetϵ,I(B). If r ≥ min then we are done so suppose r < min. Now let α and β be subsets of {1,2,…, m} and {1,2,…, n}, respectively, both of cardinality r + 1. Then rankdetϵ,I((A + B)[α∣β]) ≤ rankdetϵ,I(A[α∣β]) + rankdetϵ,I(B[α∣β])≤rankdetϵ,I(A) + rankdetϵ,I(B) = r. Hence detϵ((A + B)[α∣β]) ∈ I and since (A + B)[α∣β] is an arbitrary r + 1 by r + 1 submatrix of A + B, we have rankdetϵ,I(A + B) ≤ r = rankdetϵ,I(A) + rankdetϵ,I(B).Now we prove the general case. Let ϵ, I)-rank of matrices over antinegative commutative semiring.In this section, we look at bijective linear operators which preserve (S be a commutative semiring and A be an m by n matrix over S. The term rank of A is the minimum number of lines (rows and columns) needed to include all nonzero entries of A. The term rank of a matrix A is denoted by t(A).Let f(A) ≤ t(A) whenever A is an m by n matrix over S.For any commutative semiring, one has S be a semiring and A, B ∈ Mm,n(S). We write A ≤ B if there exists C ∈ Mm,n(S) such that A ⊕ C = B. We note that the relation (≤) is a reflexive and transitive relation but not antisymmetric in general. Therefore it is a preorder. It is easy to check that any linear operator T : Mm,n(S) → Mm,n(S) preserves this preorder. Further, if S is an antinegative semiring then the term rank is a monotone function; that is, if A ≤ B then t(A) ≤ t(B).Let S be a commutative semiring and A, B ∈ Mm,n(S). The Schur product of A and B, denoted as A∘B, is an m by n matrix whose th entry is aij ⊗ bij.Let S be a commutative semiring. A matrix A ∈ Mm,n(S) is called a submonomial matrix if every line (row or column) of A contains at most one nonzero entry. A matrix A ∈ Mn(S) is called a monomial matrix if every line (row or column) of A contains exactly one nonzero entry.Let P, Q, B) operator is a fundamental concept in the theory of linear preservers over semirings.The concept of  to itself. One says that T is a strong  operator if there exist P ∈ Mm(S), Q ∈ Mn(S), and B ∈ Mm,n(S) such that P and Q are permutation matrices, and all of the entries of B are units and either T(X) = P(X∘B)Q or m = n and T(X) = P(XT∘B)Q.Let We also use a theorem from the same reference. We note though there is an error earlier in  for the S be a commutative antinegative semiring. If rank:Mm,n(S) → ℤ is a function which satisfies 0 ≤ rank(A) ≤ t(A) for all A ∈ Mm,n(S) with equality whenever A is a submonomial matrix, then any bijective linear operator which preserves this rank function must be a strong  operator.Let ϵ, I)-rank satisfies the hypotheses of the above theorem, we now have the following corollary which classifies all bijective linear operators which preserve the -rank.Since the -rank preserver on Mm,n(S) must be a strong  operator.Let ϵ-rank.In this section, we explore connections between the sign pattern matrices and  sign pattern matrix. If A = [aij] is a real matrix, then the sign pattern of A is obtained from A, by replacing each entry by its signs [A is denoted by Sg(A) = [sg(aij)], where A matrix whose entries are from the set {+1, −1,0} is called ats signs , 15. Then by n sign pattern matrices by Qn. Sometimes we may not know the signs of certain entries, so a new symbol, #, has been introduced to denote such entries.Thus in a sign pattern matrix all we know is the sign of each entry. We do not know the exact values of the entries. We denote the set of all The generalized sign pattern matrices are the matrices over the set {+1, −1,0, #}, where # corresponds to entries where the sign is unknown.S = {+1, −1,0, #}, then  is a commutative semiring with identity, where the operations of addition and multiplication are defined as follows: The set {+1, −1,0, #} can be viewed as a semiring. If S. More about the sign pattern semiring can be found in [Clearly all the properties of a semiring are satisfied where 0 is the additive identity and +1 is multiplicative identity. Here +1 and −1 are the units of found in .A be a real matrix. The qualitative class of A is Q(A), the set of all real matrices with the same sign pattern as A.Let A is called sign-nonsingular (SNS) if every matrix in its qualitative class is nonsingular.A sign pattern matrix ϵ-rank. The sign pattern semiring has only two elements whose square is the identity, namely, 1 and −1. The ideal generated by 1 = 1 + 1 is the entire semiring but # = 1 + −1 generates the unique proper ideal {#, 0}. Therefore −1 is the only available choice for ϵ and we have a unique ϵ-rank. Hence detϵ(A) = det+(A)⊕(−1 ⊗ det−(A)). It is easy to show that an n by n sign pattern matrix has ϵ-rank n if and only if it is an SNS matrix. Hence the ϵ-rank of a sign pattern matrix A is the largest integer k for which there exists a k by k SNS submatrix of A.For matrices over the sign pattern semiring, we can give a more concrete interpretation of the detϵ,I(A) ≤ rankdetϵ(A). In other words amongst the family of -ranks, choosing I to be the ideal generated by 1 ⊕ ϵ gives us the largest possible rank function from this family. The minimal rank functions from this family arise from the choice of I to be a maximal ideal which contains 1 ⊕ ϵ. In general, there may be many maximal ideals. In this section, we will look at semirings which have a unique maximal ideal. We will use the term sublocal semiring to denote a semiring which has a unique maximal ideal. Sublocality in semirings is essentially the straightforward generalization of the very useful concept of locality in rings. We use the term sublocal semiring because the term local semiring has been used to define a slightly different concept.It was remarked earlier that rankI of a commutative semiring S is called a k-ideal if for any y ∈ S with x, x + y ∈ I one has y ∈ I.An ideal R to be semiring, the semiring ideals of R are exactly the semiring k-ideals of R which are also exactly the ring ideals of R.We note that if we consider a commutative ring I of a commutative semiring S is called a maximal  ideal if there exists no other proper ideal   J such that I ⊂ J.A proper ideal S be a commutative semiring. One says that S is a local semiring if S has only one k-maximal ideal.Let k-maximal ideals. This is useful as some semirings do not have proper k-ideals. For example, the sign pattern semiring has only one proper ideal {0, #} and this is not a k-ideal.Now we will define sublocal semirings using maximal ideals instead of S be a commutative semiring. One says that S is a sublocal semiring if S has only one maximal ideal.Let I = {0, #}. We note that this maximal ideal is contained in the proper subsemiring P = {0, +1, #}; P however fails to be an ideal in the sign pattern semiring. The set of all natural numbers, ℕ = {0,1, 2,…}, forms a sublocal semiring whose only one maximal ideal I = {ℕ/{1}}. All chain semirings are sublocal semirings with S/{1} as a unique maximal ideal. A semifield is a commutative semiring in which all elements except 0 have a multiplicative inverse. (The Boolean and max-plus semirings are examples of semifields.) All semifields are sublocal semirings as the zero ideal is the unique maximal ideal.We note that both local and sublocal semirings are direct but different semiring generalizations of the concept of a local ring. Local semirings have been useful in semiring theory; see  for examWe begin with the following elementary lemma whose proof is identical to the corresponding result for rings.a of a commutative semiring S is a unit of S if and only if a lies outside all maximal ideals of S.An element a be a unit of S; then the ideal generated by a must be S itself and hence a lies outside all maximal ideals of S. If a is not a unit of S, then 1 ∉ aS and hence there exits a maximal ideal M of S such that a ∈ aS⊆M.Let k-ideal. The analog for sublocal semirings is an easy consequence of One of the key results of  is that S is a sublocal semiring if and only if the set of all nonunits of S forms an ideal.A commutative semiring k-ideal is an ideal, it follows that every local semiring is a sublocal semiring. The converse is false. Consider the nonnegative integers ℕ with the usual addition and multiplication. The unique maximal ideal is ℕ∖{1}; the maximal k-ideals are of the form pℕ for any prime p.Since every Since the set of nonunits in any sublocal semiring is an ideal, the nonunits are closed under addition. Hence we have the following observation which will prove useful later on.S be a sublocal semiring. Let a1, a2 ∈ S. If a1 ⊕ a2 is a unit of S, then either a1 or a2 is a unit of S.Let ϵ satisfying the condition that ϵ2 = 1 and 1 ⊕ ϵ is not a unit. Such an element may not exist in a given semiring; the max-min and max-plus semirings are examples of semiring which lack an ϵ. Fortunately, there is a known construction which allows us to append such an element. This construction is from [The ranks introduced in the previous sections all require an element  is from , in whicS, ⊕, ⊗) is a commutative semiring then S2 = {∣a, b ∈ S} is also a commutative semiring with addition and multiplication defined as follows: for all a, b, c and d ∈ S, S2. Essentially this construction allows us to append an element ϵ =  with the property ϵ2 = 1 to the semiring S in a natural way giving us a way to apply the ϵ-determinant theory to semirings which do not have nontrivial self inversive elements. The ideal in S2 generated by  =  + ϵ is Δ = { : x ∈ S} which we will call the diagonal ideal. The ϵ-determinant in this case is the standard bideterminant and the ϵ-rank is the standard determinantal rank defined as follows.If  ≠ det−(B).Let ϵ, I)-rank of matrices over general semirings.The determinantal rank has been much studied , , , } is also a semiring with the addition and the multiplication defined above for S2. Moreover it is isomorphic to the sign pattern semiring as  = 0,  = +1,  = −1, and  = #.Recall that S2 inherits some important properties from S.We complete our paper by showing that the symmetrized semiring S be a commutative semiring. If S is antinegative and has no zero divisors then S2 is also antinegative and has no zero divisors.Let S is antinegative, only 0 has an additive inverse. Let us suppose that  ∈ S2 has an additive inverse, so there exists  ∈ S2, such that ⊕ = . Consequently a1 ⊕ a2 = 0 and b1 ⊕ b2 = 0. Since S is antinegative so a1 = a2 = b1 = b2 = 0. Hence  = . Thus only the additive identity has an additive inverse in S2 which means S2 is an antinegative semiring. Now suppose that  ∈ S2 is a zero divisor, so there exists a nonzero  ∈ S2, such that ⊗ = . It follows that ((a1 ⊗ a2)⊕(b1 ⊗ b2)) = 0 and ((a1 ⊗ b2)⊕(a2 ⊗ b1)) = 0. Since S is antinegative, a1 ⊗ a2 = 0 and b1 ⊗ b2 = 0. Also S has no zero divisors so either a1 = 0 or a2 = 0 and either b1 = 0 or b2 = 0; combining this with ((a1 ⊗ b2)⊕(a2 ⊗ b1)) = 0, we get either  =  or  = . Hence S2 has no zero divisors.Since S is a semiring, we let U(S) denote the set of units of S. There is a very close relation between the units of S and the units of S2.If U(S2) = { : x ∈ U(S)}⋃{ : x ∈ U(S)}.If S is a commutative antinegative semiring with no zero divisors then x be a unit in S. Then there exists a nonzero element a in S such that x ⊗ a = 1. Consider ⊗ =  =  and ⊗ =  = . Consequently  and  are units of S2. Now we have to prove that these are the only units for S2. Suppose  ∈ S2 is a unit in S2. Then there exists a nonzero element  ∈ S2 such that ⊗ = . Thus ((a ⊗ x)⊕(b ⊗ y), (b ⊗ x)⊕(a ⊗ y)) = . Consequently (a ⊗ x)⊕(b ⊗ y) = 1 and (b ⊗ x)⊕(a ⊗ y) = 0. Since (b ⊗ x)⊕(a ⊗ y) = 0 and S is an antinegative semiring, so b ⊗ x = 0 and a ⊗ y = 0. Also given that S has no zero divisors it follows that either b = 0 or x = 0 (note that both b and x cannot be zero because (a ⊗ x)⊕(b ⊗ y) = 1) and either a = 0 or y = 0 ⊕(b ⊗ y) = 1). Since  and  are nonzero elements of S2 so the units of S2, , have only two choices which are  and . Putting  =  in (a ⊗ x)⊕(b ⊗ y) = 1, we get a ⊗ x = 1, and this means that x is a unit of S. Putting  =  in (a ⊗ x)⊕(b ⊗ y) = 1, we get b ⊗ y = 1, and this means that y is a unit of S. Thus all the units in S2 are of the type  and  where x is a unit in S.Let S is a sublocal antinegative semiring with no zero divisors then so is S2.We can now prove that if S is a sublocal antinegative semiring with no zero divisors then S2 is also a sublocal antinegative semiring with no zero divisors.If S is a sublocal antinegative semiring with no zero divisors; then by S2 is an antinegative semiring with no zero divisors. Hence we only need to prove that S2 is sublocal; we show this by proving that the set of all nonunits in S2 forms an ideal of S2. Let M = { : where    is  not  a  unit  of  S2} be the set of all nonunits of S2. Let  and  ∈ M such that ⊕ = , where x is a unit in S. Hence a1 ⊕ a2 = x and b1 ⊕ b2 = 0. Since S is antinegative so b1 = 0 and b2 = 0, and also a1 ⊕ a2 = x, where x is a unit in S so  = , where a1 is a unit in S, or  = , where a2 is a unit in S. It follows that either  is a unit or  is a unit in S2, which is a contradiction to the fact that both  and  ∈ M. Thus the sum of nonunits in S2 is a nonunit. A similar argument works if ⊕ = , where x is a unit in S. Now suppose that for  ∈ M and  ∈ S2 we have ⊗ = , where x is a unit in S. Then (a ⊗ s1)⊕(b ⊗ s2) = x and (a ⊗ s2)⊕(b ⊗ s1) = 0. Since S is antinegative so a ⊗ s2 = 0 and b ⊗ s1 = 0, and also S has no zero divisors so either a = 0 or s2 = 0 and either b = 0 or s1 = 0. Clearly  or  cannot be  since (a ⊗ s1)⊕(b ⊗ s2) = x, so we have  =  or . Further x is a unit in S and (a ⊗ s1)⊕(b ⊗ s2) = x, so either a ⊗ s1 is a unit in S or b ⊗ s2 is a unit in S. Hence either a is a unit in S or b is a unit in S. We get  =  or  is a unit in S2, which is a contradiction to the fact that  ∈ M. Thus ⊗ ∈ M for all  ∈ M and  ∈ S2. Hence the set of all nonunits in S2 forms an ideal of S2. Consequently S2 is a sublocal semiring. A similar argument works if ⊗ = , where x is a unit in S.Suppose so using  either a


G is called a fractional -deleted graph if G − {e} admits a fractional -factor for any e ∈ E(G). A graph G is called a fractional -critical deleted graph if, after deleting any n vertices from G, the resulting graph is still a fractional -deleted graph. The toughness, as the parameter for measuring the vulnerability of communication networks, has received significant attention in computer science. In this paper, we present the relationship between toughness and fractional -critical deleted graphs. It is determined that G is fractional -critical deleted if t(G) ≥ ((b2 − 1 + bn)/a).A graph  G be a graph with the vertex set V(G) and the edge set E(G). For a vertex x ∈ V(G), we use dG(x) and NG(x) to denote the degree and the neighborhood of x in G, respectively. Let δ(G) denote the minimum degree of G. For any S⊆V(G), the subgraph of G induced by S is denoted by G[S].All graphs considered in this paper are finite, are loopless, and are without multiple edges. The notation and terminology used but undefined in this paper can be found in . Let G bThe problem of fractional factor can be considered as a relaxation of the well-known cardinality matching problem. It has wide-ranging applications in areas such as scheduling, network design, and the combinatorial polyhedron. For instance, several large data packets are to be sent to various destinations through several channels in a communication network. The efficiency of this work can be improved if large data packets are to be partitioned into small parcels. The feasible assignment of data packets can be seen as a fractional flow problem and it becomes a fractional factor problem when the destinations and sources of a network are disjoint.g and f are two integer-valued functions on V(G) such that 0 ≤ g(x) ≤ f(x) for all x ∈ V(G). A fractional -factor is a function h that assigns to each edge of a graph G a number in  so that for each vertex x we have g(x) ≤ ∑e∈E(x)h(e) ≤ f(x). If g(x) = a and f(x) = b for all x ∈ V(G), then a fractional -factor is a fractional -factor. Moreover, if g(x) = f(x) = k for all x ∈ V(G), then a fractional -factor is just a fractional k-factor. Throughout this paper, k ≥ 1 is an integer, and we will not reiterate it again.Suppose that G is called a fractional -critical graph if, after deleting any n vertices from G, the resulting graph still has a fractional -factor. A graph G is called a fractional -deleted graph if, after deleting any m edges, the resulting graph still has a -factor. Fractional deleted graph and fractional critical graph, as extensions of the concept of fractional factor, describe the existence of fractional factor in communication networks when certain channels or certain sites are damaged.A graph G is called a fractional -critical deleted graph if, after deleting any n vertices from G, the resulting graph is still a fractional -deleted graph. In particular, the fractional -critical deleted graph is just fractional -critical deleted graph if m = 1.Gao  proposedg, f, n)-critical deleted graph.LetG be a graph and let g, f be two nonnegative integer-valued functions defined on V(G) satisfying g(x) ≤ f(x) for all x ∈ V(G). Let n be a nonnegative integer. Then G is a fractional -critical deleted graph if and only ifS and T of V(G) with |S | ≥n.Let  toughness was first introduced by Chvátal in  such that δ(H) ≥ 1 and 1 ≤ dG(x) ≤ k − 1 for every x ∈ V(H), where T⊆V(G) and k ≥ 2. Let T1,…, Tk−1 be a partition of the vertices of H satisfying dG(x) = j for each x ∈ Tj, where one allows some Tj to be empty. If each component of H has a vertex of degree at most k − 2 in G, then H has a maximal independent set I and a covering set C = V(H) − I such thatcj = |C∩Tj| and ij = |I∩Tj| for every j = 1,…, k − 1.Let The lemma below can be deduced from Lemma 2.2 in .G be a graph and let H = G[T] such that dG(x) = k − 1 for every x ∈ V(H) and no component of H is isomorphic to Kk, where T⊆V(G) and k ≥ 2. Then there exist an independent set I and the covering set C = V(H) − I of H satisfyingIi) = k − i}, 1 ≤ i ≤ k, and ∑i=1k | Ii)-critical deleted graph due to |V(G)|≥n + b + 2. In what follows, we assume that G is not complete.If G satisfies the conditions of g, f, n)-critical graph. By ε ≤ 2, there exist subsets S and T of V(G) such thatS and T such that |T| is minimum. Obviously, we deduce T ≠ ∅ and dG−S(x) ≤ g(x) − 1 ≤ b − 1 for any x ∈ T.Suppose that l be the number of the components of H′ = G[T] which are isomorphic to Kb and let T0 = {x ∈ V(H′)∣dG−S(x) = 0}. Let H be the subgraph obtained from H′ − T0 by deleting those l components isomorphic to Kb.Let V(H)| = 0, then by virtue of (ω(G − S) = |T0 | +l > 1, then t(G) ≤ (|S|/ω(G − S)) ≤ ((b(|T0 | +l) + bn + 1)/a(|T0 | +l)) < ((b + bn + 1)/a), which contradicts t(G) ≥ ((b2 − 1 + bn)/a) and b ≥ 2. If ω(G − S) = |T0 | +l = 1, then |T0 | +l = 1. Since dG−S(x)+|S | ≥dG(x) ≥ δ(G) ≥ 2t(G), we have 2t(G) ≤ b − 1 + |S | ≤b − 1 + ((b(n + 1) + 1)/a), which contradicts t(G) ≥ ((b2 − 1 + bn)/a).If |irtue of  we inferV(H)|>0. Let H = H1 ∪ H2, where H1 is the union of components of H which satisfies that dG−S(x) = b − 1 for every vertex x ∈ V(H1) and H2 = H − H1. In terms of H1 has a maximum independent set I1 and the covering set C1 = V(H1) − I1 such thatIi) = b − i}, 1 ≤ i ≤ b, and ∑i=1b | Ii)( | = |I1|. Let Tj = {x ∈ V(H2)∣dG−S(x) = j} for 1 ≤ j ≤ b − 1. Each component of H2 has a vertex of degree at most b − 2 in G − S by the definitions of H and H2. According to H2 has a maximal independent set I2 and the covering set C2 = V(H2) − I2 such thatcj = |C2∩Tj| and ij = |I2∩Tj| for every j = 1,…, b − 1. Set W = V(G) − S − T and U = S ∪ C1 ∪ (NG(I1)∩W) ∪ C2 ∪ (NG(I2)∩W). We infert0 = |T0|. Let t(G) = t. Then when ω(G − U) > 1, we haveω(G − U) = 1. In terms of (b | T | −dG−S(T) ≥ a | S | −bn − 1, we obtainNow we consider that |We infer|U|≤|S|+| we have|U|≥tω(G−erms of (|S|+|C1| uch that|V(H1)|≤∑uch that∑j=1b−1|+∑rtue of (|V(H1)|+at0 + l.The following proof splits into two cases by the value of at ≥ b2 − 1 + bn, we have (at − b)(t0 + l) − bn − 1 ≥ b2 − b − 2 ≥ 0. Hence,  + j + a ≤ a2 − 1. By t(G) ≥ ((b2 − 1 + bn)/a), we get n = 0 and ∑j=1b−2ij = 0, which contradicts the definition of H2 and the choice of I2 (see j=1b−2ij ≠ 0).There is at least one a < b, then at ≤ b(b − 2)+(a − b + 1) + b = (b2 − 1)+(a − b)+(2 − b) < b2 − 1, which contradicts t(G) ≥ ((b2 − 1 + bn)/a).If t0 + l ≥ 2 or b ≥ 3, then by (bt − a)(t0 + l) − bn − 1 ≥ 1 we haveb ≥ a and h1(2) < 0, if a = 1, we deduceab + b − a − i + 2 − b2 ≤ −b2 + (a + 1)b − a due to i ≥ 2. Letb ≥ a, we inferIf n ≥ 1, we obtainIf n = 0, t0 + l = 1, and  =   = , then h1 < 0 and h2 < 0, a contradiction). Then the result follows from the main result in [G is fractional 2-deleted graph if t(G) ≥ (3/2).In conclusion, we have esult in  which deIn this case,  becomesSubcase 1 (|I1 | = 0). In this subcase,  = , ∑j=2b−1ij = 0, |C2 | ≤|I2|, and |T | ≤2 | I2|. Thus,I2 | = 1, then |S | ≤((2(b − 1) + bn + 1)/a) = 3 + 2n and δ(G)≤|S | +1 ≤ 4 + 2n. This contradicts δ(G) ≥ 2t(G) ≥ 6 + 4n. Hence, |I2 | ≥2.(i) If Z = {x∣x ∈ C2, dG−S(x) = 1} and z = |Z|. Thus, 0 ≤ z ≤ |I2| and NG−S(v) ∈ I2 if v ∈ Z. We obtainZ′ = {x∣x ∈ NG(I2)∩W, dG−S(x) = 1}, we infera, b) =  and |C2 | ≤|I2|, we get 2n(1 − (1/|I2|)) ≤ (1/|I2|) − 1. By |I2 | ≥2, we derive the contradiction.Let a = b, then max{hj} = hb−1 = −bn and the second largest value of hj is hb−2 = −bn − 1. In terms of the analysis of H2, choose a vertex with the smallest degree and add it to I2. Hence, by the definition of H2, we confirm that H2 is connected; each vertex in I2 has degree b − 1 in G − S except that one vertex has degree b − 2 in G − S. This fact impliesI2 | = 1, then |S | ≤1 + n and δ(G)≤|S | +(b − 1) ≤ b + n, which contradicts δ(G) ≥ 2t(G) > b + n. Hence, |I2 | ≥2 andn(1 − (1/|I2|))≤((1/b|I2|) − 1), which contradicts b ≥ 2 and |I2 | ≥2.(ii) If Subcase 2 (|I2 | = 0). In this subcase,  | ≤1, and |I(1) | ≤2. Now, we consider the following three subcases.subcase,  becomesSubcase 2.1 (|I(1) | = 1). In this subcase, we have ∑i=3b | Ii) = k − 1, there exists a vertex y ∈ In such that NHn(x)∩NHn(y) ≠ ∅,” we obtain |I1 | ≥2,bn(1 − (1/|I1|))≤(b − a)(1 − b) + ((1 − a)/|I1|), a contradiction.a 2.2 in : “for eaSubcase 2.2 (|I(1) | = 2). In this subcase, ∑i=3b | Ii)( | = 0. We can get a contradiction via a similar discussion as in Subcase 2.1.Subcase 2.3(|I(1)| = 0). In this subcase, we have ∑i=4b | Ii)( | = 0 and |I(3) | ≤1. If |I1 | = 1, then |S | ≤(((b − 1) + bn + 1)/a). Thus, we inferI1 | ≥2. Let Y = NG(I1)∩W.y ∈ Y such that y only adjacent to one vertex in I1. ResetI1 | ≥2, we obtainbn(1 − (1/|I1|))≤(b − a)(1 − b) + ((1 − a)/|I1|), a contradiction.If there is a vertex Y is adjacent to at least two vertices in I1, we getU = S ∪ C1 ∪ (NG(I1)∩W). Due to |I1 | ≥2, we deducebn(1 − (1/|I1|))≤(b − a)(1 − b)+((1/|I1|) − (a/2)), which contradicts a ≥ 1 and |I1 | ≥2.If each vertex in Subcase 3 (|I1 | ≠0 and |I2 | ≠0). From what we have discussed in Subcase 1, we get ∑j=1b−1(b − 2)(b − j)ij ≤ ∑j=1b−1(at − aj − b + j)ij + bn + 1. Then, we deducei=4b | Ii) | ≤1, |I(1) | ≤2, and n = 0, by what we have discussed in Subcase 2. It is enough to discuss the situation of |I(1) | = 0; other two cases for |I(1) | = 1 and |I(1) | = 2 can be considered in a similar way.I(1) | = 0, we get ∑i=4b | Ii) | ≤1,I1 | +|I2 | ≥2, we getb − a)(b − 1)(|I1 | +|I2|) ≤ 1 − a | I2|, a contradiction.Under the condition of |We complete the proof of the theorem.


The Acknowledgments contain several errors in the “2014 iGEM Interlab Study Contributors” section. In the “ETH_Zurich” subsection, the correct spelling of “Verena Jagger” is “Verena Jäggin”. In the “HUST-China” subsection, the correct spelling of “Yunjun Yang” is “Yunjun Yan”. In the “Leicester” subsection, the name “Ross Campbell” should be listed after “Richard Badge”. In the “Marburg” subsection, the name “Daniel F. Hurtgen” should be listed after “Matthias Franz”.


Scientific Reports5: Article number: 1422910.1038/srep14229; published online: 09182015; updated: 01072016.This Article contains typographical errors in Table 2.−1/2) for samples ‘HPGA-50 (4.9 × 102)’, ‘HPGA-20 (2.1 × 103)’ and ‘GA (4.2 × 103)’ were incorrectly given as ‘HPGA-50 (4.9 × 10−2)’, ‘HPGA-20 (2.1 × 10−3)’ and ‘GA (4.2 × 10−3)’ respectively.Under the column ‘Before Cycling’, the Warburg coefficient (ohm·s−1/2) for samples ‘HPGA-50 (5.4×102)’, ‘HPGA-20 (2.3×103)’ and ‘GA (7.4 × 103)’ were incorrectly given as ‘HPGA-50 (5.4×10−2)‘, ‘HPGA-20 (2.3×10−3)’ and ‘GA (7.4×10−3)’ respectively.In addition, under the column ‘After Cycling’, the Warburg coefficient (ohm·s