In this paper, the structural characteristics of I. Schur type inequalities are exploited to generalize the corresponding inequalities by variable parameter techniques. Some novel upper and lower bounds for the I. Schur inequality have also been obtained and the upper bounds may be obtained with the help of Maple and automated proving package (Bottema). Numerical examples are employed to demonstrate the reliability of the approximation of these new upper and lower bounds, which improve some known results in the recent literature.Recently, extensive researches on estimating the value of  In fact, extensive researches for the estimated value of e have been studied )bn = (1 + (α/n))n+β and Fα,β(x) = (1 + (α/x))x+β; see xn, λ ∈ ℝ; then we have the following.In fact, the previous theorems can be viewed as the different improvements of original I. Schur inequality. The above improved inequalities follow the motivation of researching mean value sequences of the upper and lower bounds for original inequalities; the authors study the monotonicity of the sequences constructed by the mean value of upper and lower bounds for I. Schur inequality and its relationship with n ∈ N*, the sequence Wn is monotone decreasing, and the inequality xn[1 + λ/(λ + 1)n] < e holds.For d(x) = ln(x + λx + λ) − ln(x + λx) + xln(x + 1) − xlnx  (x > 0); thenIx(λ) = (3x2 + 3x + x3 + 1)λ2 + (2x + 2x2)λ − x3. This is a quadratic function of λ, and its discriminant isIx(λ) areIx(λ) ≤ 0 always holds for x ≥ 1 and x→∞d′(x) = 0, thus d′(x) ≥ 0. It shows that sequence Wn = [1 + λ/(λ + 1)n]xn is monotone increasing under this condition, and the inequality in this theorem can be proved by using limn→∞Wn = e.We consider −lnx−1,d′′(x)=−)n and jn = (1 + (λ/n))n+1, where λ > 0, and it is easy to obtain limn→∞in = limn→∞jn = eλ. Next, we consider their arithmetic mean value sequence, geometric mean value sequence, and harmonic mean value sequence, respectively; considerHere we study the monotonicity of two new sequences and their relationships with λ ≤ 1 and n ∈ N*, the sequences A1 = (1 + (λ/2n))(1 + (λ/n))n are monotone decreasing, and the inequality eλ < (1 + (λ/2n))(1 + (λ/n))n holds.For 0 < M(x) = xln(x + λ) − xlnx + ln(2x + λ) − ln(2x), x > 0; thenM′′(x) in (Y(x): = 4x3 − 4x3λ + 9x2λ − 4x2λ2 + 6xλ2 − xλ3 + λ3. Take ∀ p ≥ 0; then λ : = 1/(p + 1) ∈  ≥ 0. Furthermore, it implies thatM′(x) is monotone increasing, and limx→∞M′(x) = 0 has been verified, such that M′(x) < 0, and A1 = (1 + (λ/2n))(1 + (λ/n))n are monotone decreasing; hence eλ < (1 + (λ/2n))(1 + (λ/n))n can be proved because of limn→∞A1 = eλ.We consider  0; thenM′(x)=lnλ ≤ 1 and n ∈ N*, the sequences G1 = (1 + (λ/n))1/2(1 + (λ/n))n are monotone decreasing, and the inequality (1 + (λ/n))1/2(1 + (λ/n))n > eλ holds.For 0 < f1 = xln(1 + (λ/x)) + (1/2)ln(1 + (λ/x)), x ≥ 1; thenδ(λ) = (−2x + 1)λ2 + 2xλ, which is a quadratic function of λ. And its discriminant is Δ1 = 4x2 > 0. Then two roots of δ(λ) arex/(2x − 1)}x=1∞ is strict monotone decreasing and 1 < (2x/(2x − 1)) ≤ 2, x ≥ 1. According to the characteristic of parabola curve, δ(λ) = (−2x + 1)λ2 + 2xλ > 0 always holds for 0 < λ ≤ 1 and x ≥ 1; then from limx→∞f1′(x) = 0, f1′(x) ≤ 0. It shows that the sequences G1 are monotone decreasing.We consider +(λ/x)),f1′′(x)=−λ ≥ 1, then H1 = (2injn)/(in + jn) = (1 + λ/(2n + λ))(1 + (λ/n))n are monotone increasing, and the following inequalityn ∈ N*.If φ(x) = ln(2x + 2λ) − ln(2x + λ) + xln(x + λ) − xlnx, x > 0; thenZ(x)∶ = −4x2 − 3λx + 4λx2 + 4xλ2 + λ3. Since τ + 1∶ = λ ≥ 1, that is, τ ≥ 0, we haveγ(x) = −λZ(x) ≤ 0 always holds for x > 0, λ ≥ 1. Then φ′(x) is monotone decreasing, and according to limx→∞φ′(x) = 0, we have φ′(x) > 0. Hence, H1 = (1 + λ/(2n + λ))(1 + (λ/n))n are monotone increasing sequences, and we prove thatWe consider −lnx−1,φ′′(x)=−4λ < 1, are the H1 monotone increasing or monotone decreasing? With the help of the Bottema   ≥ 0 holds for 0 < λ ≤ (4/5), x ≥ 1. Moreover, it can be also shown in A remaining issue of mme; see  and refeH1, n = 1,2,…, is monotone decreasing under the previous discussion in the proof of the Then we can conclude that L0–L4 here mean the corresponding real continuous function for L0 = xn(1 + (1/2n)), L1 = xn(n + 1)ln(1 + (1/n)), L2 = (xn/(nln(1 + (1/n)))), L3 = xn(1 + ((4n + 1)/(8n2 + 4n))), and L4 = xn(1 + ((4n + 3)/(8n2 + 8n + 1))), respectively.In this section, we will display some new upper and lower bounds of the I. Schur inequality by using the Matlab 2011b in a personal computer. We give some figures to show their variation trend; the symbols L0 of the I. Schur inequality is not better than the other novel upper bounds L1–L4; the L1 is the best upper bound among five upper bounds. Moreover, we note that the variation curves of L3, L4 are greatly similar; they are both the other alternatives. This figure also shows that the improvement of the upper bound of the I. Schur inequality is beneficial.Form T1 = xn(1 + (1/(2n + 1))),   T2 = xn(1 + (1/n))1/8, T3 = xn(1 + (1/n))1/4,   T4 = xn(1 + ((1/100)/(1 + (1/100))n)), and T5 = xn(1 + (−1/3)/(1 − (1/3))n), respectively. From T1 of the I. Schur inequality is better than the other novel lower bounds T2–T5; the T3 is an alternative among other novel lower bounds. This figure shows that we improve the lower bound of the I. Schur inequality and it is not very successful. But we give some generalized upper and lower bounds for the eλ, λ ∈ ; this work is meaningful in terms of the generalized type of the I. Schur inequality.Firstly, we define the corresponding real continuous function for H1 can be viewed as a good lower bound of eλ, λ ≥ 1 in terms of approximating precision, especially for the case with λ = 1. It is favorable to note that the reliability of the approximation of eλ based on H1 depends on the closeness between λ and 1.From xn and yn, such as the arithmetic mean value, geometric mean value, logarithmic mean value, and harmonic mean value. Besides, we have given some extensions and remarks for known results in the recent literature. At the same time, we extend some ideas and conclusions in [In the previous sections, we have studied the monotonicity of the sequences and their variants based on various mean values of sions in , 30–32.


In , Toader In recent years, there have been plenty of literature, such as –6, dedicp ∈ ℝ and a, b > 0, the centroidal mean pth power mean Mp are, respectively, defined byFor a, b > 0 with a ≠ b. This conjecture was verified by Qiu and Shen .For positive numbers λ and the least value μ, such that the double inequality α ∈  and a, b > 0 with a ≠ b. As applications, we also present new bounds for the complete elliptic integral of the second kind.The main purpose of the paper is to find the greatest value  In order to establish our main result, we need several formulas and Lemmas below.r < 1 and For 0 < fined in , 13 by/r2 is strictly increasing from  to , and the function 2ℰ − r′2𝒦 is increasing from  to .The function   andfu,α > 0, for all r ∈  if and only if u ≥ 3(1 − α)(4/π − 1), and fu,α < 0, for all r ∈  if and only if u ≤ 3(1 − α)/4.  Let g(r) = (1/π)((ℰ − r′2𝒦)/r2). From , one hasWe divide the proof into four cases.Case 1 (u ≥ 3(1 − α)/π).  From (g(r), we clearly see that fu,α(r) is strictly increasing on . Therefore, fu,α(r) > 0, for all r ∈ .).  From  and LemmCase 2 (u ≤ 3(1 − α)/4). From (g(r), we obtain that fu,α(r) is strictly decreasing on . Therefore, fu,α(r) < 0, for all r ∈ .4). From  and LemmCase 3 (3(1 − α)/4 < u ≤ 3(1 − α)(4/π − 1)). From (g(r), we see that there exists λ ∈ , such that fu,α(r) is strictly increasing in  andfu,α(r) leads to the conclusion that there exists 0 < λ < η < 1, such that fu,α(r) > 0 for r ∈  and fu,α(r) < 0 for r ∈ .)). From  and 14)Case 3  andfu,α(1−)≤Case 4 (3(1 − α)(4/π − 1) ≤ u < 3(1 − α)/π). Equation (Equation  leads tog(r), we clearly see that there exists λ ∈ , such that fu,α(r) is strictly increasing in . Therefore, fu,α(r) > 0 for r ∈  follows from . From  and 14)14) togetows from  and 16)g(r), we Now, we are in a position to state and prove our main results. α ∈  and λ, μ ∈ , then the double inequalitya, b > 0 with a ≠ b if and only if If A, T, and a > b. Let p ∈ , t = b/a ∈ , and r = (1 − t)/(1 + t). Then, Since ). Then,C¯ of the second kind in terms of elementary functions as follows.Let r ∈  and  For r ∈ . In the recent past, the complete elliptic integrals have attracted the attention of numerous mathematicians. In , it was r ∈ .Guo and Qi  proved tr ∈ .Yin and Qi  presenteℰ(r) are better than the bounds in .It was pointed out in  that theounds in  for someℰ(r) is better than the lower bound in . The lower bound in  for ℰ(r)bound in . Indeed,x ∈  show that the lower bound in (ℰ(r) is better than the lower bound in bound in :(25)5+6


We investigate the positive solutions of the semilinear parabolicsystem with coupled nonlinear nonlocal sources subject to weighted nonlocal Dirichlet boundaryconditions. The blow-up and global existence criteria are obtained. Ω is a bounded domain in ℝN, N ≥ 1, with smooth boundary ∂Ω. The exponents pi > 0, qi ≥ 0. The weighted functions φi in the boundary conditions are continuous, nonnegative on Ωφidy > 0 on ∂Ω. The initial data ν < 1, ui0(x) ≥ 0, ≢0, and satisfy the compatibility conditions.In this paper, we consider the positive solutions of the semilinear parabolic system with coupled nonlinear nonlocal sources subject to weighted nonlocal Dirichlet boundary conditions:u0(x) ≥ 0, g(t) > 0 or g(t) = (k/|Ω|)∫Ωutdx with k > 0. Chadam et al. [f(u) = 0 and g(t) = ∫Ωψ)dx and proved that the blow-up set is the whole region (including the homogeneous Neumann boundary conditions). Souplet [g(t). Pao [Many physical phenomena were formulated into nonlocal mathematical models and studied by many authors –13. For m et al.  studied ut−Δu=f. Pao  discusseThe problems with both nonlocal sources and nonlocal boundary conditions have been studied as well. To motivate our study, we give a short review of examples of such parabolic equations or systems studied in the literature. For example, Lin and Liu  studied p, l > 0. And some criteria for the existence of global solution as well as for the solution to blow up in finite time were obtained.Gladkov and Kim  considerk = 2:(i)m, q < 1 and np ≤ (1 − m)(1 − q) hold; then the solution of (Assume that ution of  exists g(ii)If one of the following conditions holds:ution of  blows up(iii)Ωφdy ≥ 1 and ∫Ωψdy ≥ 1 for all x ∈ ∂Ω and one of , and  ST = ∂Ω ×  with 0 < T < ∞ for convenience.In the following, we set pi, qi ≥ 1, i = 1,2,…, k.It is known by the standard theory , 23 thatT < ∞, thenProblem  has a popi, qi, i = 1,2,…, k satisfyu1, u2,…, uk) of dy < 1, i = 1,2,…, k, for all x ∈ ∂Ω, then the solution of  satisfiesIf the initial data H) holds. Then the solution of  ≥ 0 on Ωgidy > 0 on ∂Ω, i = 1,2,…, k, j = 1,2,…, N. If, for every i = 1,2,…, k, wk+1 = w1, then wi > 0, i = 1,2, on Suppose that zi = eKt−wi with zi > 0, i = 1,2,…, there exists δ > 0 such that zi > 0 for zi ≥ 0 on zi vanishes at z1 ≡ 0 in Ωg1dy > 0 on ∂Ω. This proves that z1 > 0 and consequently w1 > 0. We complete the proof.Set k}. Thenzit−Δzi+, bi are continuous, nonnegative functions in gi ≥ 0 on Ωgidy < 1 on ∂Ω, and there exist positive constants Ci such that ∫Ω + bi)dx ≤ Ci. Then wi ≥ 0, i = 1,2, on Suppose that, for every wi > 0. Now we consider the general case. Setɛ is any fixed positive constant, and K = 1 + max{∫Ω + bi) dx, i = 1,2,…, k}. By  ≥ 0 on QT. Letting ɛ → 0+, we get the desired result.Suppose that the strict inequalities of  hold; by, k}. By , we get,Ωgidy < 1 is not necessarily valid, we have the following result. The argument of its proof can be referred to [If the boundary condition ∫fi ≥ 0, ci, di, are nonnegative and bounded in QT, gi ≥ 0 on Ωgidy > 0 on ∂Ω, i = 1,2,…, k, j = 1,2,…, N. If, for every i = 1,2,…, k,  wk+1 = w1, then wi ≥ 0, i = 1,2, on Suppose that By r ∈ {1,2,…, k},pr < 1,qr < 1,Let QT.then Before proving w0(x) and φ be continuous, nonnegative functions on θij satisfy 0 < θi1 + θi2 ≤ 1. Then the solutions of the nonlocal problemLet θ > 0 be large enough such thatz = eθt2ψ(x) for  ∈ Ω × , one readily checks thatThe augment is similar to the proof of , Lemma 6ution of . Clearlyai ∈ , i = 1,2,…, k, such thatα = ∑i=1k1/ai. Let Φ ≥ max{φi, i = 1,2,…, k} be a continuous function defined for z solvesz is global. Moreover, z > 1 in i = 1,2,…, k. By  ≤ C0 + M in C0 to be large enough such thatx, t) ∈ ST, it follows thatx, t) ∈ QT. Set Li = (1 + C0 + M)pi+qi | Ω| for convenience. A simple computation yieldsDefineqr > 1, no matter qr+1 > 1 or qr+1 ≤ 1, we can choose br to be small enough such that brqr1− ≥ br+1prLr. For fixed br, there exist bi, i = 1,2,…, r − 1, r + 1,…, k, satisfying biqi1− ≥ bi+1piLi, i = 1,2,…, k. It follows that(a) If qi ≤ 1, i = 1,2,…, k and p1p2 ⋯ pk > (1 − q1)(1 − q2)⋯(1 − qk), we can choose b1 to be small enough such thatbi > 0, i = 2,3,…, k, bk+1 = b1 satisfying biqi1− ≥ bi+1piLi, i = 1,2,…, k. Hence ((b) If k. Hence  holds toui,0(x) ≤ bi(1 + w(x)), i = 1,2,…, k for x ∈ Ω.By  and 37)37), in au, v) is a positive solution of  the first eigenvalue and the corresponding eigenfunction of the linear elliptic problem:ϕ1(x) satisfiesγ = min{αi(qi − 1) + αi+1pi + 1, i = 1,2,…, k}.We denote by qr ≥ 1, we claim that there exist positive constants αi > 1, i = 1,2,…, k, such that the inequalityi = r, /pr+1}. That is,  such that ((a) If equalityαi(qi−1)+equalityαi(qi−1)+equalityαi(qi−1)+equalityαi(qi−1)+equalityαi(qi−1)+qi < 1, i = 1,2,…, k, and p1p2 ⋯ pk > (1 − q1)(1 − q2) ⋯ (1 − qk), we can choose αi > 1 such that(b) If 1.Hence  holds toγ > 1. Now let s(t) be the unique solution of the ODE probleml = min{(1/αi)∫Ωϕ1αiqi+αi+1pi, i = 1,2,…, k}. Then s(t) blows up in finite time T(s0) with s0 being large enough.Hence, for the case (a) or (b), we all have ui,0(x) ≥ sαi(0)ϕ1αi(x), i = 1,2,…, k for x ∈ Ω. We complete the proof.Setui,0 > 0 in Ω, ∫Ωφrdy > 0 on ∂Ω, andui,0 > 0 on ∂Ω. Denote by η the positive constant such that ui,0 > η on H) implies that (ui)t > 0 by the comparison principle, and in turn ui > η, i = 1,2,…, k on ur satisfieszr(t) be the solution of the following Cauchy problem:zr(t) blows up under the conditionΩφrdy ≥ 1, by ur ≥ zr as long as both ur and zr exist, and thus ur blows up for any positive initial data. The proof now is completed.Since


We established some theorems under the aim of deriving variants of the Banach contraction principle, using the classes of inner contractions and outer contractions, on the structure of fuzzy modular spaces. X, ρ) is called a probabilistic modular space if X is a real vector space and ρ is a mapping from X into the set of all distribution functions  is denoted by ρx, and ρx(t) is the value ρx at t ∈ ℝ) satisfying the following conditions:  ρx(0) = 0; (PM1)  ρx(t) = 1 for all t > 0 if and only if x = θ; (PM2)  ρx−(t) = ρx(t); (PM3)  ραx+βy(s + t) ≥ ρx(s)∧ρy(t) for all x, y ∈ X and α, β, s, t ∈ ℝ0+, α + β = 1.(PM4) For every x ∈ X, t > 0 and α ∈ ℝ∖{0}, ifβ ∈  is β-homogeneous. The concept of a modular space was introduced by Nakano . Soon afRecently, further studies have been made on the probabilistic modular spaces. Nourouzi  extendedfuzzy set A in X is a function with domain X and value in . A triangular norm is a function ∗ : ×→ satisfying, for each a, b, c, d ∈ , the following conditions: a∗1 = a; a∗b ≤ c∗d whenever a ≤ c, b ≤ d; a∗b = b∗a and (a∗b)∗c = a∗(b∗c).A V be a real or complex vector space with a zero θ, ∗ a continuous triangular norm, and μ a fuzzy set on the product V × ℝ+. Suppose that the following properties hold for x, y ∈ V and s, t > 0:  μ > 0; (FM1)  μ = 1 for all t > 0 if and only if x = θ; (FM2)  μ = μ; (FM3)  μ ≥ μ∗μ whenever z is the convex combination between x and y; (FM4)  t ↦ μ is continuous at each fixed x ∈ V. (FM5) the mapping Then, we write  to represent the space with the pre-defined properties. In particular, we call μ a fuzzy modular and the triple  a fuzzy modular space. Let t > 0.It is worth noting that every fuzzy modular is non-decreasing with respect to X be a real or complex vector space and ρ be a modular on X. Take the t-norm a∗b = min{a, b}. For every t ∈ , define μ = t/(t + ρ(x)) for all x ∈ X. Then  is a F-modular space.Let t-norm is replaced by a∗b = a · b and a∗b = max{a + b − 1,0}, respectively.Note that the above conclusion still holds even if the x ∈ V and a non-zero real λ, the equalityβ ∈  is the set of the formr ∈  and t > 0. Now, suppose that μ is β-homogeneous for some β ∈  in V μ-converges   to its μ-limit x ∈ V if and only if μ → 1 as n → ∞ for all t > 0. Note here that the μ-limit is unique if it does exists after all. It is then natural to say that (xn) is μ-Cauchy if for any given ɛ ∈  and t > 0, there exists N ∈ ℕ with μ > 1 − ɛ whenever m, n > N.The and Chen , the fama∗b) = min{a, b}. This triangular norm has a very special property that if ∗′ is an arbitrary triangular norm, then (a∗′b)≤(a∗b) for all a, b ∈ . With this property, it is suitable to call this ∗ a strongest triangular norm. As is claimed by Shen and Chen [V is a real vector space equipped with a β-homogeneous fuzzy modular μ and a strongest triangular norm ∗, then a μ-convergent sequence is μ-Cauchy. The authors also mentioned that if ∗ is not the strongest one, such implementation is not always true.At this point, let us turn to a typical example of a triangular norm which is defined by  is said to be μ-complete if μ-Cauchy sequences actually μ-converges. We note that it makes more sense if we require a complete fuzzy modular space to be equipped with the strongest triangular norm.The space  is μ-complete. Suppose that at each x ∈ V, μ → 1 as t → ∞. If f : V → V is an inner contraction with constant q ∈ , then f has a unique fixed point.Let x0 ∈ V, we suppose that fnx0 ≠ fn+1x0 for all n ∈ ℕ. Let t > 0, observe thatn → ∞, we have fnx0 − fn+1x0 → θ for every t > 0. That is, for any given t > 0 and ɛ ∈ , there exists N ∈ ℕ such thatGiven a point μ > 1 − ɛ for all p ∈ ℕ. Let us assume first that μ > 1 − ɛ holds at some j ∈ ℕ. Observe thatWe now claim to show by induction that fnx0) is Cauchy. Let t > 0 and ɛ ∈  be arbitrary, and we choose N ∈ ℕ according to the claim given above. For n > m > N, we may write m = N + s and n = N + s + t, for some s, t ∈ ℕ. Now, consider thatfnx0) is Cauchy, and so the μ-completeness yields that fnx0 → x∗ for some x∗ ∈ V. It follows thatfx∗ = x∗, since 𝒯μ is Hausdorff. To show that the fixed point of f is unique, assume that y∗ ∈ V is a fixed point of f as well. Finally, we obtain thatx∗ = y∗, and so the conclusion is fulfilled. Next, we shall show that  is called a μ-G-Cauchy sequence if for each fixed p ∈ ℕ and t > 0, we have limn→∞ = 1. If every μ-G-Cauchy sequence μ-converges, V is said to be μ-G-complete. It is to be noted that the notion of μ-G-completeness is slightly stronger than the ordinary completeness. It is enough to see that every μ-Cauchy sequence is also a μ-G-Cauchy sequence. For our result, it is still a question whether or not the μ-G-completeness assumption can be weakened.For this part, we consider a weaker form of a V be a real vector space equipped with a β-homogeneous fuzzy modular μ and the strongest triangular norm ∗ such that  is μ-G-complete. If f : V → V is an outer contraction with constant q ∈ , then f has a unique fixed point.Let x0 ∈ V, we suppose that fnx0 ≠ fn+1x0 for all n ∈ ℕ. By the definition of an outer contraction, we can rewrite this notion in the following:x, y ∈ V and t > 0.Given a point t > 0, observe thatn → ∞, we havefnx0) is Cauchy. Let t > 0 and ɛ ∈  be arbitrary, and we choose N ∈ ℕ. For n > N, n ∈ ℕ and for each p > 0. Now, consider thatfnx0) is Cauchy, and so the μ-completeness yields that fnx0 → x∗ for some x∗ ∈ V. It follows thatn → ∞, we havefx∗ = x∗, since 𝒯μ is Hausdorff. To show that the fixed point of f is unique, assume that y∗ ∈ V is a fixed point of f as well. Finally, we obtain thatμ(1 − q)≥(1 − q) which implies that μ = 1. Therefore, it must be the case that x∗ = y∗, and so the conclusion is fulfilled. Let V is μ-complete as in Is


N-fractional calculus operator Nν method, we derive the fractional solutions of the equation.This paper deals with the design fractional solution of Bessel equation. We obtain explicit solutions of the equation with the help of fractional calculus techniques. Using the  Fractional calculus has an important place in the field of math. Firstly, L'Hospital and Leibniz were interested in the topic in 1695, , NishimoN-Fractional calculus is a very interesting method because this method is applied to singular equation. Note that fractional solutions can be obtained for kinds of singular equation via this method [s method –12, 17. λ and p are real numbers. By means of the substitution Now, consider the following the Bessel equation:xd2zdx2+dBessel equation for having the analogous singularity is given in .The differintegration operators and their generalizations , 17, 18 Rzυ  (υ ∈ ℂ) and the Weyl operator Wzυ  (υ ∈ ℂ), which are defined by [Two of the most commonly encountered tools in the theory and applications of fractional calculus are provided by the Riemann-Liouville operator fined by , 19.Considerℕ being the set of positive integers.provided that the defining integrals in  and 4)  4) existC− is a curve along the cut joining two points z and −∞ + i Im(z),  C+ is a curve along the cut joining two points z and ∞ + i Im(z),  D− is a domain surrounded by C−, and D+ is a domain surrounded by C+. (Here D contains the points over the curve C.)Letf = f(z) be a regular function in D  (z ∈ D),t ≠ z,fν(z)  (ν > 0) is said to be the fractional derivative of f(z) of order ν and fν(z)  (ν < 0) is said to be the fractional integral of f(z) of order −ν, provided (in each case) that |fν(z)| < ∞  (ν ∈ ℝ).Moreover, let Nν be defined by   by cf. –10)(8)N(8)NNν We find it to be worthwhile to recall here the following useful lemmas and properties associated with the fractional differintegration which was defined earlier .f(z) and g(z) are single-valued and analytic in some domain Ω⊆ℂ, thenh1 and h2.If the functions f(z) is single-valued and analytic in some domain Ω⊆ℂ, thenIf the function f(z) and g(z) are single-valued and analytic in some domain Ω⊆ℂ, thengn(z) is the ordinary derivative of g(z) of order n  (n ∈ ℕ0 : = ℕ ∪ {0}), being tacitly assumed (for simplicity) that g(z) is the polynomial part (if any) of the product f(z)g(z).If the functions λ,For a constant λ,For a constant λ,For a constant N-fractional method to nonhomogeneous Bessel equation.Now, let apply y ∈ {y : 0 ≠ |yν| < ∞; ν ∈ ℝ} and f ∈ {f : 0 ≠ |fν| < ∞; ν ∈ ℝ}. We consider the nonhomogeneous Bessel equation:y2 = d2y/dx2,  y = y (z)  (z ∈ ℂ),  f = f(z) (an arbitrary given function), and p, λ are given constants.Let p = 0 of (The cases p = 0 of  and 20)p = 0 of p = 0 of  and 18)p = 0 of Sety=xηψ, ψ=x).Thusy1=ηxη−1ψPutting , we obtaWith some rearrangement of the terms in , we haveη such thatHere, we choose η = p + (1/2). From ((I) Let 2). From  and 24)η = p + eμxϕ twing from , we can ing from  as(32)ϕμ such thatChoose (I) (i): For instance, taking −μxfrom  and 32)μ=-λi, weNν to both members of , we, we12),  Making use of the relations , rewriteν such thatChoose 1/2from .Next, writingity from :(43)ω1+yı, we finally obtain the solution , a, a(44)ω=solution . Inversely,  satisfieatisfies . Therefoatisfies  satisfieatisfies  because atisfies , 35), ϕ=atisfies , and (45ϕ=atisfie(I) (ii): In the case when iλxfrom  and 32)μ=λi, we Nν to both members of ,,50), 4646, and .η = −p + (1/2).(II) Let p by −p in (I) (i) and (I) (ii), we have other solutions (p ≠ 0.With the help of the similar method in (I), replacing olutions  and 20)p by −p iolutions  and 18)p by −p ip ≠ 0, where k is an arbitrary constant.If p = 0, p = 0, p = 0, (5f = 0 in When stead of  and 51)f = 0 in Therefore, we get  for 58)58) and  for 59)59).η = −p + (1/2), replacing p by −p in η = −p + by −p in  and 57)η = −p + Let It is clear by Theorems Application 1. If we substitute p = 0,  λ = 1/4, and f = ix−1/2ei/2)x−( in (i/2)x in , then weBy performing the necessary operations in , we getNow, let us show that the last equality is the solution of :(67)y2=Obviously, if  and 67)67) are pApplication   2. If we substitute p = −1 and  λ = 0 in y2=dt=πx24,y=kx3/2π4y2=dt=πx2usly, if , it is susly, if  is givenp by ip (−λ instead of λ) in  in ; that isp by ip in ((i) Therefore, the solutions for  are giveby ip in , 18),  In the same way, for the solutions for , substitν as follows:Choose ν = pi + (1/2). From (Let 2). From  and 77)ν = pi + Next, set ; then 8 is rewrielations  and 31)82)(eμxψ(eμxψ(82)μ as follows:Choose (ii. 1) In the case when eλxfrom  and 83)μ=-λ, we Nν to both members of , a, a10), aν such thatChoose 1/2from .Next, writingx−1from . This is−1from (u=[(fx(1/ψ1/2−pi=u/2.from  and 80)(92)ψ1/2−(ii. 2) Similarly, in the case when ν = −ip + (1/2). In the same way as in the procedure in (ii), replacing ip by –ip  (ii. 1) and (ii. 2), we can obtain yııı)( and yıv), (, (f = 0,olutions , and (57N-fractional calculus operator Nν-method is applied to the nonhomogeneous and homogeneous Bessel equation. Explicit fractional solutions of Bessel equations are obtained. Furthermore, similar solutions were obtained for the modified same equation by using the method.The


Coupled nonlinear dynamical systems have been widely studied recently. However, the dynamical properties of these systems are difficult to deal with. The local activity of cellular neural network (CNN) has provided a powerful tool for studying the emergence of complex patterns in a homogeneous lattice, which is composed of coupled cells. In this paper, the analytical criteria for the local activity in reaction-diffusion CNN with five state variables and one port are presented, which consists of four theorems, including a serial of inequalities involving CNN parameters. These theorems can be used for calculating the bifurcation diagram to determine or analyze the emergence of complex dynamic patterns, such as chaos. As a case study, a reaction-diffusion CNN of hepatitis B Virus (HBV) mutation-selection model is analyzed and simulated, the bifurcation diagram is calculated. Using the diagram, numerical simulations of this CNN model provide reasonable explanations of complex mutant phenomena during therapy. Therefore, it is demonstrated that the local activity of CNN provides a practical tool for the complex dynamics study of some coupled nonlinear systems. Coupled nonlinear dynamical systems have been widely studied in recent years. However, the dynamical properties of these systems are difficult to deal with. Although the research on emergence and complexity has gained much attention during the past decades, the determination, prediction, and control of the complex patterns generated from high-dimensional coupled nonlinear systems are still far from perfect. Nature abounds with complex patterns and structures emerging from homogeneous media, and the local activity is the origin of these complexities , 2. The Although Chua presents the main theorem of local activity at a cell equilibrium point , 2, it iThe remaining of this paper is organized as follows. The local activity of CNN is introduced in M × N array of cells. Each cell is denoted by C, where i = 1,2,…, M, j = 1,2,…, N. The dynamics of each cell is given by the equation:The CNN architecture is composed of a two-dimensional xij, yij, uij are the state, output, and input variables of the cell, respectively. ak,l, bk,l, zij are the elements of the A-template, the B-template, and threshold, respectively. r is the radius of influence sphere. The output yij is the piece-wise linear function given byA vast majority of active homogeneous media that are known to exhibit complexity are modeled by a reaction-diffusion partial differential equation (PDE):X =  is state variables,  is spatial coordinates, fi is a coupled nonlinear vector function called the kinetic term, and D1, D2,…, Dn are constants called diffusion coefficients. Replacing the Laplace in above formulation by its discrete version yieldsD = diag, From Chua and his collaborators' point, PDEs are merely mathematical abstractions of nature, and the concept of a continuum is in fact an idealization of reality. Even the collection of all electrons in a solid does not form a continuum, because much volume separating the electrons from the nucleus represents a vast empty space . Reaction state variables, but only m (m ≤ n) state variables couple directly to their nearest neighbors via “reaction-diffusion”. Consequently, each cell has the following state equations:Generally speaking, in a reaction-diffusion CNN, every cell has Qi = (∈Rn) of equation (The cell equilibrium point equation  can be dQi has the following form:The Jacobian matrix at the equilibrium point kl(Qi) are called cell parameters andQi are defined viaQi.Qi if and only if, its admittance matrix at Qi satisfies at least one of the following four conditions [YQ(s) has a pole in Re[s] > 0.ω = ω0, where ω0 is any real number.YQ(s) has a simple pole s = iωp on the imaginary axis, where its associated residue matrix:YQ(s) has a multiple pole on the imaginary axis.A reaction-diffusion CNN cell is called locally active at the equilibrium point nditions .YQ(s) Qi is called stable if and only if, all the real parts of eigenvalue λi of Jacobian matrix at the equilibrium point Qi are negative  > 0), and any one of the following conditions holds.f(s) ≠ 0.f(s) = 0, and m > n, where s is m and n orders zero point of g(s) and f(s), respectively, where f(s) = T1s3 + K1s2 + L1s + Δ1, g(s) = s4 + Ts3 + Ks2 + Ls + Δ.A necessary and sufficient condition for Obviously proved.DenoteYQH(iw) < 0 for some w = w0 ∈ R if any one of the following conditions holds.a11 > 0. a11 = 0, TT1 − K1 > 0. a11 = 0, TT1 − K1 = 0, Q > 0. a11 = 0, TT1 − K1 = 0, Q < 0, ΔΔ1 > 0. a11 = 0, TT1 − K1 = 0, Q < 0, P ≥ 0, ΔΔ1 − P2/Q/4 > 0, ΔΔ1 ≤ 0,a11 = 0, TT1 − K1 = 0, Q = 0, P > 0. a11 = 0, TT1 − K1 = 0, Q = 0, P ≤ 0, ΔΔ1 > 0. a11 = 0, TT1 − K1 < 0, ΔΔ1 > 0. a11 = 0, TT1 − K1 < 0, ΔΔ1 ≤ 0, and λj* ≥ 0, h(λj*) < 0, for j = 1 or 2.a11 < 0, D > 0, Ω1 > 0, g(Ω1) < 0. a11 < 0, D < 0, and Ωj ≥ 0, g(Ωj) < 0, for j = 1, 2 or 3.a11 < 0, D = 0, p = q = 0, g(−F/4E) < 0. a11 < 0, D = 0, q2/4 = −p3/27 ≠ 0, and Ωj ≥ 0, g(Ωj) < 0, for j = 1 or 2.Let the following parameters be defined as in YQ(iω) to satisfy condition (2) in Re[YQ(iω)] < 0,a11 > 0, then Re[YQ(iω)] < 0 when ω is large enough (See (1) of If a11 = 0, thenIf Let f(λ) = −Qλ2 − Pλ − ΔΔ1,TT1 − K1 > 0, then Re[YQ(iω)] < 0 when ω is large enough (See (2) of  If TT1 − K1 = 0, thenIf Q > 0, then Re[YQ(iω)] < 0 when ω is large enough (See (3) of If Q < 0,If 1 > 0, ∃ ω0 ∈ R, such that Re[YQ(iω0)] < 0 (See (4) of If ΔΔ1 ≤ 0, solve f′(λ*) = 0, we can get λ* = −0.5P/Q, f(λ*) = 0.25P2/Q − ΔΔ1.If ΔΔP ≥ 0, ΔΔ1 − 0.25P2/Q > 0, ∃ ω0 > λ*, such that Re[YQ(iω0)] < 0 (See (5) of Then when Q = 0, then Re[YQ(iω)] = −Pω2 − ΔΔ1.If P > 0, then Re[YQ(iω)] < 0 when ω is large enough (See (6) of If P ≤ 0, ΔΔ1 > 0, then ∃ω0, such that Re[YQ(iω0)] = −ΔΔ1 < 0 (See (7) of If TT1 − K1 < 0, let h(λ) = −(TT1 − K1)λ3 − Qλ2 − Pλ − ΔΔ1,If 1 > 0, then ∃ω0, such that Re[YQ(iω0)] < 0 (See (8) of If ΔΔ1 ≤ 0, solve h(λ*) = 0, we can geti = 1,2, if λi* ≥ 0, h(λi*) < 0, then ∃ω0, such that Re[YQ(iω0)] < 0 (See (9) of If ΔΔa11 < 0, let g(Q) = EQ4 + FQ3 + GQ2 + HQ + I, then g′(Q) = 4EQ3 + 3FQ2 + 2GQ + H. Let x = Ω + (F/4E), then the above becomes g′(Q) = 4E(x3 + px + q) = 4Ef(x), then xi, i = 1,2, 3 are the roots of f(x) = 0, Ωi are the roots of g′(Ω) = 0. If any one of the (10)–(13) of Re[YQ(iω0)] < 0.If So, if any one of conditions (1)–(13) holds, Re[YQ(iω0)] < 0. YQ(s) Satisfies condition (2) in j = 1, or 2, letYQ(s) satisfies condition (3) of K1w12 − Δ1 ≠ 0. K1w12 − Δ1 = 0, (L1 − T1w12)(w22 − w12) > 0. K1w22 − Δ1 ≠ 0. K1w22 − Δ1 = 0, (L1 − T1w22)(w12 − w22) > 0. K > 0, Δ1 ≠ 0, Δ = 0, L = KT ≠ 0, and any one of the following conditions holds.TΔ1 < 0. T(K − T1K) − Δ1 + KK1 ≠ 0.   T(K − T1K) − Δ1 + KK1 = 0, T(Δ1 − KK1) + K(L1 − T1K) < 0.   1L > 0.   Δ = 0, ΔK > 0, K2 − 4Δ > 0, and j = 1, or 2.Δ < 0, or  AjBj1 − Aj1Bj ≠ 0. AjBj1 − Aj1Bj = 0, AjAj1 − BjBj1 > 0.   For f(s) = T1s3 + K1s2 + L1s + Δ1, g(s) = s4 + Ts3 + Ks2 + Ls + Δ, obviously, ∞ is not a single pole of YQ(s) on the imaginary axis.Let YQ(s) has a simple pole s = iω on the imaginary axis, where its associated residuek1 ≠ 0, so f(s) ≠ 0, which implies that iω is not a zero point of f(s) = 0, iω  is not a removed pole of YQ(s).(I)YQ(s) has four poles s = ±iω1, ±iω2  (ω1 ≠ ω2 ≠ 0) on the imaginary axis. In this case,  g(s) = (s2 + ω12)(s2 + ω22) = s4 + (ω12 + ω22)s2 + ω12ω22. Hence we obtain T = L = 0, K = ω12 + ω22 > 0, Δ = ω12ω22 > 0. Then, we can get k1 is a complex number or a negative real number. YQ(s) satisfies condition (3) in  If (II)YQ(s) has a simple pole s = 0 and two conjugate poles ±iω  (ω ≠ 0) on the imaginary axis, and another pole is a ≠ 0.If 1 ≠ 0, and g(s) has the form:T = −a, K = ω2 > 0, L = −aω2 = KT, Δ = 0, Δ1 = 0, Therefore,In this case, it follows that Δ = 0, Δ(1)YQ(s) at s = 0 isK > 0, Δ1 ≠ 0, Δ = 0, L = KT ≠ 0, TΔ1 < 0, k1 is a negative real number. YQ(s) satisfies condition (3) in The residue of (2)YQ(s) at The residue of k1 is either an imaginary number or a negative real number. YQ(s) satisfies condition (3) in (III)YQ(s) has a simple pole s = 0 on the imaginary axis, and the other poles are ai, Re[ai] ≠ 0, i = 1,2, 3, it follows that Δ1 ≠ 0, Δ = 0, and g(s) has the formT = −(a1 + a2 + a3), K = a1a2 + a1a3 + a2a3, L = −a1a2a3 ≠ 0, hence the reside of YQ(s) at s = 0 is1L > 0, k1 is a negative real number. YQ(s) satisfies condition (3) in If (IV)YQ(s) has two conjugate poles ±iω(ω > 0) on the imaginary axis, and the other poles are Re[a] ≠ 0, Re[b] ≠ 0. In this case, g(s) has the formT = −(a + b), K = ab + ω2, L = −(a + b)ω2, Δ = abω2 ≠ 0. Then, ab = K − ω2, Δ = (K − ω2)ω2⇔ω4 − Kω2 + Δ = 0. Solving it, we haveK > 0, K2 − 4Δ ≥ 0, and T = −(a + b) = L/ω2. Then, the residue of YQ(s) at s = ±ωj* isk1 < 0 or it is an imaginary number. YQ(s) satisfies condition (3) in If If YQ(s) satisfies condition (3) in Therefore, when any one of conditions (I)–(IV) holds, YQ(s) has a multiple pole on the imaginary axis if any one of the following conditions holds.L = 0, Δ1 ≠ 0. Δ = L = K = 0, and  Δ1 ≠ 0  or  L1 ≠ 0.   Δ = L = K = T = 0, and Δ1 ≠ 0 or L1 ≠ 0 or K1 ≠ 0.  Δ = T = L = 0, K > 0, Δ = (K/2)2, and any one of the following conditions holds.1 ≠ KK1. 2ΔL1 ≠ KT1.   2f(s) = T1s3 + K1s2 + L1s + Δ1, g(s) = s4 + Ts3 + Ks2 + Ls + Δ. Obviously, when the conditions (I)–(III) hold, 0 is the multiply poles of YQ(s).Let YQ(s) has two multiply nonzero poles ±iω  (ω > 0), then g(s) has the form:T = L = 0, K = 2ω2 > 0, Δ = ω4 = (K/2)2 > 0.If iω are the multiply poles of YQ(s), thenf(iω) ≠ 0 whereIf ±f(iω) ≠ 0, any one of ±iω is a multiply pole of YQ(s).Obviously, when any one of (1)-(2) of (IV) in YQ(s) satisfies condition (4) in So, when any one of condition (I)–(IV) holds, YQ(s) satisfies When any one of Theorems These theorems can be implemented by a computer program for calculating the bifurcation diagram of the general corresponding CNN to determine emergence of complex dynamic patterns of the corresponding CNN.Life systems consist of locally coupled homogeneous media. Mostly, dynamics of life systems are suitable to be described via locally connected reaction-diffusion CNNs. It may be expected that reaction-diffusion CNN will become a promising candidate for modeling life phenomena.x, y, v, yn, vn represent the numbers of uninfected cells, infected cells infected by normal virus, normal virus, infected cells infected by mutated virus, and mutant viruses, respectively. λ is the rate of reproduction of uninfected cells. Uninfected cells die at rate dx and become infected at rate bxv by normal virus and infected at rate bnxvn by mutated virus. Infected cells infected by normal and mutated virus are removed at rate ay and ayn, respectively. Normal virus is produced at rate ky and removed at rate uv, mutated virus is produced at rate kyn and removed at rate uvn. e is the rate constant describing the probability of mutation of virus , a, b, bn, d, e, k, kn, u, λ are positive constants. The model was briefly analyzed in Nowak's book.In Chapter 11 “Timing the emergence of resistance” (Page 110) of the book “Virus dynamic: mathematical principles of immunology and virology” (Oxford university press), Nowak et al. proposed a mathematical model which describes the mutation selection of HBV infection during the therapy :(37)dxdThe reaction-diffusion CNN of HBV mutation selection of model has the form:2xij = xi−1j + xi+1j + xij−1 + xij+1 − 4xij.D1 = 0) and solve it, we can get the two equilibrium points:Let equation  be zerosx0 = au/((1 − e)bk) and Q1, Q2 stand for the patient's complete recovery and HBV persistent infection, respectively.Qi   isConsequently, the Jacobian matrix at the equilibrium point k, u as variables, and λ = 10, a = 0.5, b = 0.01, bn = 0.005, e = 0.0001, kn = 10, and d = 0.01, using Theorems Q1 and Q2 at k ∈ , u ∈ , see Figures Taking Q1 does not exist at the edge of chaos domain.In Figures λ = 10, k = 0.01, a = 0.5, b = 0.01, bn = 0.005, kn = 10,  e = 0.0001, and  k = 1.0,3.0,4.9,5.1,10,24,39, u = 2,5, 9, we model the dynamic trajectories of equation , when these parameters are located in the green domain ;u, the dynamic pattern of equation  but less than a threshold value .u, the dynamic pattern of equation ,When these parameters are located in the red domain (edge of chaos);u value, the dynamic pattern of equation (regardless of the equation  is conveThe No. 2 and No. 3 variables in equation  increaseThe local activity of CNN has provided a powerful tool for studying the emergence of complex patterns in a homogeneous lattice formed by coupled cells. Based on the local activity principle, the analytic criteria for the local activity in reaction-diffusion CNN with five state variables and one port are set up. The analytical criteria include four theorems, which provide the inequalities involving the parameters of the CNN. The inequalities can be used for calculating the bifurcation diagram to determine emergence of complex dynamic patterns of the reaction-diffusion CNN. As an application example, a reaction-diffusion CNN of HBV mutation-selection model is analyzed and simulated, and the bifurcation diagrams are calculated. Numerical simulations show this CNN model may explain certain complex mutant conditions during the therapy. We conclude that the local activity theory provides a practical tool for the study of the complex dynamics of certain coupled nonlinear systems.


The point is, the delta shock wave is not the one of transport equations, which is obviously different from cases of some other systems such as Euler equations or relativistic Euler equations. For the generalized Chaplygin gas, unlike the polytropic or isothermal gas, there exists a certain critical value ϵ2 depending only on the Riemann initial data, such that when ϵ drops to ϵ2, the delta shock wave appears as u− > u+, which is actually a delta solution of the same system in one critical case. Then as ϵ becomes smaller and goes to zero at last, the delta shock wave solution is the exact one of transport equations. Furthermore, the vacuum states and contact discontinuities can be obtained as the limit of Riemann solutions when u− < u+ and u− = u+, respectively.The limit of Riemann solutions to the nonsymmetric system of Keyfitz-Kranzer type with a scaled pressure is considered for both polytropic gas and generalized Chaplygin gas. In the former case, the delta shock wave can be obtained as the limit of shock wave and contact discontinuity when  The nonsymmetric system of Keyfitz-Kranzer type can be written asρt+ = u − p, and p = p(ρ), system  = ργ, γ > 0 is smooth and strictly increasing withp into ϵp and taking p(ρ) = (1/ρ − 1/ρ*), ρ ≤ ρ*, where ρ* is the maximal density which corresponds to a total traffic jam and is assumed to be a fixed constant although it should depend on the velocity in practice. Then, Shen and Sun  = ρ+ − ρ−. By simple calculation, we obtainρ− ≠ ρ+, andρ− = ρ+.With these definitions, one can construct a We can also justify that the delta shock wave satisfies the entropy condition:In this section, we analyze some basic properties and solve the Riemann problem for .System  and 10)10) have So the 1-characteristic field is genuinely nonlinear, and the 2-characteristic field is always linearly degenerate.t → βt and x → βx, β > 0, the solution is only connected with ξ = x/t. Thus we should seek the self-similar solutionu, ρ)(±∞) = .Since -10) and and10) a problem -(10) and problem  can be rFor smooth solutions, system  can be rξ = σϵ, the Rankine-Hugoniot conditionρ] = ρ − ρ− and σϵ is the velocity of the discontinuity. From  coincid in the phase plane; that is,  bel belξ = σu−, ρ−), we draw the curve u = u− for ρ > 0 in the phase plane, which is parallel to the ρ-axis. We denote it by R when ρ < ρ− and S when ρ > ρ−. Through the point , we draw the curve , denoted by J. Then the phase plane is divided into four regions  ∈ I, that is, u+ > u−  and  u+ < u− + ϵ(ργ − ρ−γ), the solution is S + J;when  ∈ II, that is, u+ > u−  and  u+ > u− + ϵ(ργ − ρ−γ), the solution is R + J;when  ∈ III, that is, u+ < u−  and  u+ > u− + ϵ(ργ − ρ−γ), the solution is R + J;when  ∈ IV, that is, u+ < u−  and  u+ < u− + ϵ(ργ − ρ−γ), the solution is S + J.when (Through the point (he curve  which inions see . Thus weions see -10) as  as u−, α < 1, while both the two characteristic fields are fully linearly degenerate as α = 1.Systems  and 11)11) have α < 1, we get rarefaction wave and shock wave which can be expressed byWhen 0 < α < 1, through the point , we draw the curve u = u− for ρ > 0 in the phase plane, denoted by R when ρ < ρ− and S when ρ > ρ−. Through the point , we draw the curve , we draw the curve , the Riemann solution is showed as follows:u+, ρ+) ∈ I, that is, u+ > u− and u+ < u− + ϵ(ρ−α− − ρα−), the solution is S + J;when  ∈ II, that is, u+ > u− and u+ > u− + ϵ(ρ−α− − ρα−), the solution is R + J;when  ∈ III, that is, u+ < u− and u+ > u− + ϵ(ρ−α− − ρα−), the solution is R + J;when  ∈ IV, that is, u+ < u− and u+ < u− + ϵ(ρ−α− − ρα−), the solution is S + J.when  is given byFor any given  ∈ V, we introduce a definition of δ-measure solution, in which we introduce a definition of a generalized solution  = ρ+ − ρ−, 0 < α < 1.Suppose that f system  and 11)Ω ⊂ ℝ × ℝρ− ≠ ρ+, andρ− = ρ+.From , we obtaWe also can justify that the delta shock wave satisfies the entropy condition:α = 1, the detailed study can be found in H(ξ − σ) and H is the Heaviside function.The first integral in  can be d{∫−∞σϵ+∫σerms in .The second term in  can be cen, from 1, 84), , (85)∫σϵe obtainlimϵ→ϵ2∫With the same reason as above, we haveρϵ and ρϵuϵ as ϵ → ϵ2, by tracing the time-dependence of weights of the δ-measure.Finally, we study the limits of φ ∈ C0∞×[0, ∞)) and set Let e obtainlimϵ→ϵ2∫er hand,limϵ→ϵ2∫With the same reason as before, we obtainu− > u+ and 0 < ϵ < ϵ2,  ∈ V. So the Riemann solution of . We want to observe the behavior of strength and propagation speed of the delta shock wave when ϵ decreases and finally tends to zero.When ution of  and 11)u− > u+ρ+ ≠ ρ−, letting ϵ → 0 in 11) are iu+, ρ+) ∈ IV, we conclude that the shock wave and a contact discontinuity coincide as a delta shock wave when ϵ → ϵ2. As ϵ continues to drop and goes to zero eventually, the delta shock solution is nothing but the Riemann solution to transport equations (Combining the results of the above, when (quations .u− > u+, u− < u+, and u− = u+, respectively. For the polytropic gas, different from cases of some other systems such as Euler equations or relativistic Euler equations, the delta shock wave is not the one of transport equations as parameter ϵ tends to zero. For the generalized Chaplygin gas, the delta shock wave appears as parameter ϵ tends to ϵ2, depending only on the Riemann initial data. Then as ϵ becomes smaller and goes to zero at last, the delta shock wave solution is the exact one of transport equations.So far, the discussion for limit of Riemann solutions to the nonsymmetric system of Keyfitz-Kranzer type with both the polytropic gas and generalized Chaplygin gas has been completed. From the above analysis, as the pressure vanishes, there appear delta shock wave, vacuum state, and contact discontinuity when


X, d) and for the family 𝒜 of subsets of X established a new cone metric H : 𝒜 × 𝒜 → E and obtained fixed point of set-valued contraction of Nadler type. Further, it is noticed in the work of Janković et al., 2011 that the fixed-point problem in the setting of cone metric spaces is appropriate only in the case when the underlying cone is nonnormal. In the present paper we improve Wardowski's result by proving the same without the assumption of normality on cones.Wardowski (2011) in this paper for a normal cone metric space ( Huang and Zhang  generaliE be a Banach Space and P a subset of E. Then, P is called a cone wheneverP  is closed, nonempty, and P ≠ {θ},ax + by ∈ P for all x, y ∈ P and nonnegative real numbers a, b,P∩(−P) = {θ}.Let P induces a partial ordering ≼ on E by x≼y if and only if y − x ∈ P. So x < y will stand for x≼y and x ≠ y, while x ≪ y will stand for y − x ∈ intP, where intP denotes the interior of P. The cone P is called normal if there is a number K > 0 such that, for all x, y ∈ E,K  satisfying  for all x, y ∈ X and d = θ if and only if x = y,2) = d for all x, y ∈ X,3)≼d + d for all x, y, z ∈ X. Then, d is called a cone metric on X, and  is called a cone metric space.  Let X, d) be a cone metric space, x ∈ X and {xn}n≥1 a sequence in X. Then, {xn}n≥1 converges to x whenever for every c ∈ E with θ ≪ c there is a natural number N such that d ≪ c for all n ≥ N. We denote this by limn→∞xn = x or xn → x. {xn}n≥1 is a Cauchy sequence whenever for every c ∈ E with θ ≪ c there is a natural number N such that d ≪ c for all n, m ≥ N.   is called a complete cone metric space if every Cauchy sequence in X is convergent.Let  is called closed if, for any sequence {xn}⊂A  convergent to x, we have x∈A. Denote by N(X) the collection of all nonempty subsets of X and by C(X) a collection of all nonempty closed subsets of X. Denote by FixT a set of all fixed points of a mapping T. In the present paper, we assume that E is a real Banach space, P is a cone in E with nonempty interior , and ≼ is a partial ordering with respect to P. In accordance with d ∈ V, for all n ≥ N1. Then, (λn/2/(1 − λ1/2))d ≪ c, for all n ≥ N1. Thus,m > n. Therefore, {xn}n≥1 is a Cauchy sequence. Since X is complete, there exists u ∈ X such that xn → u. Sincen, xn+1 ∈ Txn, we have yn ∈ Tu such that N2  such thatn ≥ N2,yn → u, and it implies that u ∈ Tu.Let X = , E = CR2  with the norm ||x|| = ||x||∞ + ||x/||∞, P = {x ∈ E : x ≥ 0}, x(t) = t, and y(t) = tK2. Then,  0 ≤ x ≤ y,  ||x|| = 2, and ||y|| = 1 + 2K. For all K ≥ 1, since K||x|| < ||y||. Therefore, P is non-normal. Define d : X × X → E as follows:𝒜 be a family of subsets of X of the form 𝒜 = { : x ∈ X} ∪ {{x} : x ∈ X}, and define H : 𝒜 × 𝒜 → E as follows:H  satisfies (H1)–(H4) of T : X → 𝒜  asT satisfies the conditions of λ = 1/2 and 0∈FixT. Suppose


We investigate the moduli spaces of stable sheaves on a smooth quadric surface with linear Hilbert bipolynomial in some specialcases and describe their geometry in terms of the locally free resolution of the sheaves. ℂ, the field of complex numbers.Throughout the paper, our base field is X with a fixed Hilbert polynomial, which is itself a projective variety, and the moduli space has been studied quite intensively in the last decade for the case with linear Hilbert polynomial over projective spaces  of degree 1 such thatu, v) ∈ ℤ⊕2.For a purely 1-dimensional sheaf mt + c is the Hilbert polynomial of ℱ with respect to the ample line bundle 𝒪Q. Let us take any D ∈ |𝒪Q|, T ∈ |𝒪Q|, and a smooth conic C ∈ |𝒪Q| such that neither D, T, nor C is contained in the 1-dimensional reduced curve Supp(ℱ).Let us assume that D, T, and C induce maps jD : ℱ → ℱ,  jT : ℱ → ℱ, and jC : ℱ → ℱ. Since neither D nor T is contained in the 1-dimensional reduced curve Supp(ℱ), we have jD ≠ 0 and jT ≠ 0. Since ℱ is pure, we obtain that jD, jT, and jC are injective. Thus, there are exact sequencesThe curves a : = h0 ⊗ 𝒪D) and b : = h0 ⊗ 𝒪T). The sheaves ℱ ⊗ 𝒪D, ℱ ⊗ 𝒪T, and ℱ ⊗ 𝒪C have finite supports and thus the dimensions of their cohomology H0 do not change even if we twist them by a line bundle on Q. From  ⊗ 𝒪C) = m.Let us set  Q. From , we get χ) = av + bu + c for all  ∈ ℤ⊕2. If u = v, then the claim is true. Now assume that u ≠ v, say u > v. We use u − v exact sequences like  instead of ℱ with 0 ≤ c < u − v to get χ) = χ+(u − v)b.We claim that ces like  with ℱ ∈ ℚ of ℱ to be a linear bipolynomial such thatℱ with respect to 𝒪Q is defined to be χℱ(t) = χℱ.One defines the Hilbert bipolynomial χℱ is a linear function, that is, χℱ = mx + ny + t for some  ∈ ℤ⊕3.We are mainly interested in the case when ℱ be a pure sheaf of dimension 1 on Q with χℱ = mx + ny + t. The p-slope of ℱ is defined to be p(ℱ) = t/(m + n). ℱ is called semistable (stable) with respect to the ample line bundle 𝒪Q if (1)ℱ does not have any 0-dimensional torsion,(2)ℱ′, one hasfor any proper subsheaf  χℱ′ = m′x + n′y + t′.where Let ℱ with χℱ = mx + ny + t, let us define Cℱ : = Supp(ℱ) to be its scheme-theoretic support and then we have Cℱ ∈ |𝒪Q|. We often use slope stability and slope semistability instead of Gieseker stability or Gieseker semistability just to simplify the notation; they should be the same because the support is 1-dimensional, and from mt + χ and m′t + χ′, the inequality for Hilbert and slopes χ/m the same.For every semistable 1-dimensional sheaf M be the moduli space of semistable sheaves on Q with linear Hilbert bipolynomial χ = mx + ny + t.Let M in a different way as a subvariety of MQ,ℙ3, the moduli space of semistable sheaves on ℙ3 with linear Hilbert polynomial χ(x) = mx + t, which are 𝒪Q-sheaves. To be precise, if ℱ is 𝒪Q-sheaf, then all of its 𝒪ℙ3-subsheaves are also 𝒪Q-sheaves. It implies that the notions of p-stability and μ-stability of ℱ are the same and thus MQ,ℙ3 may be defined without using ℙ3. Moreover, the sheaf with linear Hilbert bipolynomial χ = ax + by + c has Hilbert polynomial χ(x) = (a + b)x + c with respect to 𝒪Q and thus we have a natural decompositionM is a subvariety of MQ,ℙ3.We can define ℱ be any purely 1-dimensional coherent sheaf on ℙ3 with Hilbert polynomial mx + ti. Assume that ℱ is not semistable and letℱ  is a projective and irreducible scheme. If mn > 0, then M° is a Zariski dense and open subset of M with dimension 2mn + 1.The moduli Q holds. But it holds, using Castelnuovo-Mumford criterion with the Serre dualityℱ ∈ M and j ≥ −1.The first assertion follows verbatim from the proof of Proposition 2.3 and Theorem 3.1 in , only whℱ′, ℱ′′) to ℱ′ ⊕ ℱ′′, where m = m′ + m′′ and n = n′ + n′′. Then the dimension of the image of this map is at least 2mn − 2m′n′ − 2m′′n′′ − 1 and it is at least 1 if mn > 0. In other words, general sheaf in M is stable.For the second assertion, let us consider a mapℱ on Q with Hilbert bipolynomial χℱ = mx + ny + t, let us defineℱ. Since ℱ is pure, the natural map φℱ : ℱ → ℱDD is injective. Since the support of ℱ is 1-dimensional, φℱ is bijective as in Remark 4 of , where Li is a line in |𝒪Q|.One hast = n and let us choose L ∈ |𝒪Q| and then it fits into𝒪L is stable. For a line L ∈ |𝒪Q|, we haveL, we havef is a zero map. Thus, there exists a nontrivial extension of 𝒪L by 𝒪L and it is 𝒪L2. In particular, 𝒪L⊕2 and 𝒪L2 represent the same point in HilbQ(2y + 2). In general, 𝒪Lk⊕ and 𝒪kL with k ≥ 1 represent the same point in M. Thus, 𝒪L with L ∈ |𝒪Q| is strictly semistable if and only if n ≥ 2. Conversely, let us choose a semistable sheaf ℱ with χℱ = ny + n. In particular, the schematic support L = Supp(ℱ) of ℱ is in |𝒪Q|. Since χ(ℱ) = n > 0, there exists a nontrivial morphism 𝒪Q → ℱ and it induces an injection 𝒪L1 → ℱ, where L1 is a subscheme of L. Here we have L1 ∈ |𝒪Q| for some s ≤ n and so χL1 = sx + s. Thus, the quotient 𝒢 = ℱ/𝒪L1 is a semistable sheaf with χ𝒢 = (n − s)y + (n − s). By induction, we have [𝒢] = [𝒪L2] with L2 ∈ |𝒪Q|. In particular, ℱ is an extension of 𝒪L2 by 𝒪L1 with L1 + L2 ∈ |𝒪Q| and thus ℱ is equivalent to 𝒪L1 ⊕ 𝒪L2.Let us assume that t < n and fix ℱ ∈ M with C : = Cℱ ∈ |𝒪C|. Since χ(ℱ) = t > 0, there is a nonzero map f : 𝒪Q → ℱ. Since ℱ is an OC-sheaf, f induces a nonzero map h : 𝒪C → ℱ. Since 𝒪C has slope 1 > t/n and it is semistable, we get a contradiction. Alternatively, as in Lemma 4.10 of [T⊆C of Im(h) and then use an injective map 𝒪T → ℱ with 𝒪T ∈ |𝒪Q| with 1 ≤ n′ < n, and thus we have μ(𝒪T) = 1.Now, let us assume that 0 <  4.10 of , we may m = 1, that is, χℱ = x + ny + t, it is enough to check the case t = 1 since gcd = 1.For the case of M consists of 𝒪C with C ∈ |𝒪Q|. In particular, one has M≅ℙn+12.χ𝒪C = x + ny + 1 and 𝒪C is semistable by ℱ be a semistable sheaf with χℱ = x + ny + 1 and so C : = Cℱ is a curve in |𝒪Q|. Since we have χ(ℱ) = 1, there exists a nonzero map 𝒪Q → ℱ and it induces a nonzero map h : 𝒪C → ℱ. Note that Im(h) has no 0-dimensional torsion since ℱ is semistable. Since 𝒪C is also semistable, we haveh factors through an injection 𝒪D↪ℱ, where D is a curve contained in C. If D is properly contained in C, we have p(𝒪D) > p(ℱ) contradicting the semistability of ℱ and thus we have D = C; that is, h is an isomorphism from 𝒪C to its image. Since 𝒪C and ℱ have the same Hilbert polynomial, we have ℱ≅𝒪C.From the sequencex + 2y + t, it is enough to investigate the case when t = 1,2. Let us denote the moduli space M by Mt.For the moduli space of semistable sheaves with linear Hilbert bipolynomial 2M1 consists of the unique nontrivial extensions ℱ of 𝒪P by 𝒪C for each curve C ∈ |𝒪Q| and a point P ∈ C, and one also has h0(ℱ) = 1. The moduli space χ(ℱ) = 1, there is a nonzero map 𝒪Q → ℱ, inducing a nonzero map h : 𝒪C → ℱ, where C : = Cℱ ∈ |𝒪Q|. Since χ𝒪C = 2x + 2y, we have p(𝒪C) = 0 < 1/4 = p(ℱ). The map h factors through an injection 𝒪D↪ℱ, where D is a curve contained in C. If D is properly contained in C, we have p(𝒪D) > p(ℱ) contradicting to the semistability of ℱ and thus we have D = C; that is, h is an isomorphism from 𝒪C to its image, that is, we haveχ𝒢 = 1. In particular, we have 𝒢≅𝒪P, the skyscraper sheaf supported on a point P ∈ C. Since ℱ has no 0-dimensional torsion, the sequence does not split. Note that Ext1≅H1(𝒪P)∨ = 0, and thus from the sequence of C we haves is the transpose of Hom, 𝒪P) → Hom which is given by the multiplication by the defining equation of C. Since P is a point on C, the map s is a zero map. In particular, the dimension of Ext1 is 1 and so ℱ corresponds to a unique nontrivial extensionh0(ℱ) ≤ 2 and that h0(ℱ) = 1 if and only if no injective map 𝒪C → ℱ is an isomorphism at P. This is certainly true if ℱ is not locally free of rank 1 at P. Note that ℱ is a line bundle at each point of C∖{P} and thus it is sufficient to prove h0(ℱ) = 1 when ℱ is a line bundle on the curve C. In this case the nonexistence of a section of ℱ that does not vanish at P is equivalent to the nonsplitting of (h0(ℱ) = 0 and so the point P is uniquely determined by ℱ.Since xtension0⟶𝒪C⟶ℱ⟶𝒪Pxtension0⟶𝒪C⟶ℱ⟶𝒪Pℱ is a nontrivial extension of 𝒪P by 𝒪C, where P is a point on C. If ℱ is not semistable, then there exists a subsheaf 𝒦 ⊂ ℱ with p(𝒦) > p(ℱ) = 1/4 and so we have χ𝒦 = m′x + n′y + t′ with ≤ and t′ ≥ 1. If the composite s : 𝒦 → ℱ → 𝒪P is a zero map, then we have an injection 𝒦↪𝒪C, contradicting the semistability of 𝒪C. Thus, the composite is surjective and so we have the following diagram:𝒦′ is the kernel of the map s and ℋ is the quotient ℱ/𝒦. Since χ𝒦′ = m′x + n′y + (t′ − 1) and 𝒪C is semistable, we have t′ = 1 and thus χℋ = (2 − m′)x + (2 − n′)y with no constant term. Since ℋ is the quotient of 𝒪C, it must be 𝒪T for some curve T contained in C. But no such curves have the Hilbert polynomials with no constant term. Hence ℱ is semistable.Conversely, let us assume that M1. Let us assume the existence of a polystable sheaf ℱ = ℱ1 ⊕ ⋯⊕ℱs with s ≥ 2. We have χ(ℱ) = 1 = χ(ℱ1)+⋯+χ(ℱs). If we let χℱi = aix + biy + ci, then we haveci > 0 for all i and thus we have s = 1, a contradiction.There is no strictly semistable sheaf in ℱ is in M1 if and only if it admits the following resolution:𝒜 : = 𝒪Q ⊕ 𝒪Q and f : = h1l2 − h2l1 is a defining equation of Cℱ.A sheaf h0(ℱ) = 1 and so h1(ℱ) = 0. If ℱ admits the sequence . Take any A ∈ |𝒪Q| which is not contained in C and with P ∉ A. The multiplication by an equation of A gives an exact sequenceℱ|A) = deg(A∩C) = 4. Thus we have h1) = 0 and h0) = 5. Together with the exact sequence , we obtain that ℱ is globally generated at P and so we have a surjectionℋ : = ker(ψ) and then ℋ is a torsion-free sheaf of rank 2 on Q with c1 = . By Theorem 19.9 in [ℋ is locally free. Note that χℋ = 2xy + 3x + y + 1. Thus, we have h0) > 0 and so we have an exact sequenceZ is a 0-dimensional subscheme of Q and ∈{, }. If  = , then we have χℐZ = xy + x + 1 and it is absurd since χ𝒪Q = xy + x. Thus, we have  =  and Z = ∅. Since Ext1, 𝒪Q) = 0, we have ℋ≅𝒪Q ⊕ 𝒪Q and the sequence  is given by f : = h1l2 − h2l1 is a defining equation of C = Cℱ.Note that sequence , then ℱ sequence  tensoredjection3ψ:𝒪Q⊕𝒪Q is globally generated and so a surjection φ′ : 𝒪Q ⊕ 𝒪Q → ℱ. In this case, ker(φ′) is no longer a direct sum of two line bundles.Using the proof of Lemma 5.3 in , we can W to beW0 ⊂ W to be the set of φ ∈ W such that h1l2 − h2l1 ≠ 0. Then we have a surjective morphismLet us define a vector space φ1, φ2 ∈ W0 with π(φ1) = π(φ2); that is, we have the following diagram:f is an isomorphism. Since Ext1, 𝒜) = 0, we have a map f1 ∈ End) associated with f. Note that f1 is given by a, b ∈ ℂ× and z ∈ H0). Similarly, we have a map f2 : 𝒜 → 𝒜 which is c1, c2 ∈ ℂ×. In particular, we havec1 = 1. In other words, π(φ1) = π(φ2) if and only if φ1 and φ2 are in the same orbit in W0 under the action byLet us choose π : W0 → M1 is a geometric quotient map by the action of G. In particular, one has M1≅W0/G and so M1 is isomorphic to 𝒫ic1.To get the assertion it suffices to prove that it has local sections as in Lemma 5.1 and Theorem 5.5 in .M1 is stable, M1 has a universal family ℍ1 on M1 × ℙ2  for a unique ℱ ∈ M1, we also have a universal family ℍ5 on M5 × ℙ2 with ℱ as the fibre. Since h0(ℱ) = 1 and h1(ℱ) = 0 for all ℱ ∈ M1, the base change theorem gives that u∗(ℍ1) is a line bundle on M1, where u : M1 × ℙ2 → M1 is the first projection. Since h0) = 5 and h1) = 0 for all ℱ ∈ M1, the base change theorem gives that v∗(ℍ5) is a vector bundle of rank 5 on M1 by identifying M5 with M1, where v : M5 × ℙ2 → M5. For a fixed ℱ ∈ M1 and a matrix φ ∈ π−1(ℱ), let us write f : = h1l2 − h2l1 is a defining equation of Cℱ. Take an open neighborhood U of ℱ in M1 over which u∗(ℍ1) and v∗(ℍ5) are trivial. The matrix φ was constructed starting with a section σ of ℱ which spans ℱ together with the twist σ′ of a nonzero section of ℱ. Since u∗(ℍ1)|U and v∗(ℍ5)|U are trivial, there are maps e1 : 𝒪U → 𝒪U and e2 : 𝒪U → 𝒪U⊕5 with e1(ℱ) = σ′ and e2(ℱ) = σ. Since σ′ and σ span ℱ, there is a neighborhood V of ℱ in U such that the sections e1(𝒢) and e2(𝒢) span every 𝒢 ∈ V. The construction of φ gives that e1 and e2 induce a section of π in a neighborhood of φ whose image by π is V.Since every element of e 180 of ). Since M1 is irreducible and unirational. In fact, we can prove more.As an automatic consequence, we obtain that M1 is rational. Q × Q be the diagonal and denote its ideal sheaf by ℐΔ. Denoting by p1 and p2 the projection from Q × Q to each factor, let us define a sheaf 𝒰 to be p1*𝒪Q)⊠ℐΔ on Q × Q. For each point P ∈ Q, we have 𝒰|Q×{P}≅ℐP. Thus, we have h1(𝒰|Q×{P}) = 0 and so p2∗𝒰 is a vector bundle of rank 8 on Q since h0(𝒰|Q×{P}) = 8. Let us consider the projective bundle𝒵 over a point P ∈ Q is the set of curves of type  on Q passing through P and so there is a natural map from 𝒵 to ℙH0)≅ℙ8. In other words, 𝒵 is the universal curve of type  on Q and it is isomorphic to M1. Since 𝒵 is locally trivial over Q, it is rational.Let Δ ⊂ ℱ ∈ M2 admits one of the following types:𝒪C → ℱ → η → 0, where η is a skyscraper 𝒪C-sheaf with degree 2,0 → 𝒪T1 → ℱ → 𝒪T2 → 0 with T1, T2 ∈ |𝒪Q|,0 → 𝒪T1 → ℱ → 𝒪T2 → 0, where T1 ∈ |𝒪Q| and T2 ∈ |𝒪Q| with  ∈ {, }.0 → Any sheaf χ(ℱ) = 2, we have h0(ℱ) ≥ 2. Thus, there exists a nonzero map 𝒪Q → ℱ and it induces a nonzero map h : 𝒪C → ℱ, where C : = Cℱ ∈ |𝒪Q|. The map h factors through an injection 𝒪T1↪ℱ where T1 is a curve contained in C:T1 = C, that is the map h is an isomorphism from 𝒪C to its image in ℱ, then its cokernel ℋ is the skyscraper sheaf supported on two points, say P1, P2 ∈ C. Thus we have the sequenceSince T1 is properly contained in C and then we obtain that T1 has bidegree , , or  since p(𝒪T1) ≤ p(ℱ) = 1/2 and ℱ is semistable. Let T2 ⊂ Q be the only curve such that T1 + T2 = C. Let ℋ′ be the quotient of ℋ by its torsion τ, that is, ℋ′ : = ℋDD.Let us assume that T1 ∈ |𝒪Q| and so we have χℋ′ = x + y + 1 − deg(τ). Since ℱ is semistable, we get τ = 0. Since every quotient of ℱ has the slope at least 1/2, the same is true for ℋ. Thus, ℋ is semistable and ℋ≅𝒪T2.First, assume T1 ∈ |𝒪Q|, that is, χ𝒪T1 = 2x + y + 1 and so we have χℋ = y + 1. If ℋ has 0-dimensional torsion 𝒯 with length k ≥ 1, then the quotient ℋ/𝒯 is a quotient of ℱ with the p-slope 1 − k ≤ 0, contradicting the semistability of ℱ. Thus ℋ has no 0-dimensional torsion and so we have ℋ≅𝒪T2 for a curve T2 ∈ |𝒪Q| with C = T1 + T2.Now, without loss of generality, let us assume that M2 is globally generated. Every sheaf in ℱ ∈ M2 and then there is no nonzero map ℱ → 𝒪C≅ωC since ℱ is semistable. Thus we have h1(ℱ) = 0 and so h0(ℱ) = 2. It is clear that ℱ of types (B) and (C) is globally generated and so we may assume that ℱ is of type (A), but neither of (B) nor of (C).Let us take ℋ⊆ℱ be the image of the evaluation map H0(ℱ) ⊗ 𝒪Q → ℱ and then ℋ is pure. Assume that ℋ ≠ ℱ. Since ℱ is of type (A), it is globally generated outside at most two points of Cred. In particular, we have χℋ = 2x + 2y + c with c ≤ 1 and deg(ℱ/ℋ) = 2. Since h0(ℋ) = h0(ℱ) = 2, we have h1(ℋ) = 2 − c. Note that every nonzero section of ℋ vanishes at finitely many points since ℱ is neither of types (B) nor (C). Since h0(𝒪C) < h0(ℋ), we have ℋ ≠ 𝒪C and c = 1. A nonzero section of ℋ induces an exact sequence𝒢≅𝒪P for some P ∈ Cred. Since ℋ is pure, this exact sequence does not split. As in the proof of ℋ = ℱ and so ℱ is globally generated.Let ℱ is a sheaf in M2 if and only if it admits a sequencezij ∈ H0) such that f : = z11z22 − z12z21 is a defining equation of Cℱ.ℱ ∈ M2 be a sheaf of type (A) and then it is globally generated by h0(ℱ) = 2, we have a surjectionℋ : = ker(ψ) and then it is a torsion-free sheaf of rank 2 on Q with c1 = . By Theorem 19.9 in [ℋ is locally free. Note that h0) = 2. From the sequenceH0⊕2) → H0) is an isomorphism and so h1) = 0. Similarly, we have h1) = 0 and h2(ℋ) = h1(ℱ) = 0. By Remark 2.3 in [ℋ is globally generated. Since c1) = 0 or h0) = 2, we have ℋ≅𝒪Q⊕2 and the resolution  and (C), respectively. In particular, we haveLet us define a subscheme ℱ of type (B) are strictly semistable. In particular, they are contained in 𝔅. The sheaves ℱ. Let 𝒦 be a subsheaf of ℱ with p(𝒦) > 1/2 and the quotient sheaf ℋ : = ℱ/𝒦. If the composite map s : 𝒦↪ℱ↠𝒪T2 is a zero map, then 𝒦 is a subsheaf of 𝒪T1, contradicting the semistability of 𝒪T1. The sheaf Im(s) is a subsheaf of 𝒪T2 and so we have p(Im(s)) ≤ 1/2. Similarly, the sheaf ker(s) is a subsheaf of 𝒪T1 and so p(ker(s)) ≤ 1/2. From the exact sequencep(𝒦) ≤ 1/2, a contradiction.It is enough to check the semistability of M2 the closed subscheme of M2, consisting of the strictly semistable sheaves.Let us denote by ∂𝔖2 is the permutation group of order 2. In particular, ∂M2 is a rational variety.One has𝔅 ⊂ ∂M2. Let ℱ be a strictly semistable sheaf and so it has a proper quotient sheaf ℋ with p(ℋ) = 1/2. From the semistability of ℱ, ℋ has no 0-dimensional torsion. From the equality p(ℱ) = p(ℋ), we obtain that ℋ is also semistable. Since p(ℋ) = 1/2, the Hilbert bipolynomial of ℋ is either 2x + 1, 2y + 1 or x + y + 1. The first 2 cases cannot happen due to χℋ = x + y + 1 and so ℋ≅𝒪T2 with T2 ∈ |𝒪Q| and T2 ⊂ Cℱ by 𝒦 is the kernel of the quotient map ℋ → ℋ, then its p-slope is again 1/2 and so 𝒦 is semistable. Similarly as before, we have 𝒦≅𝒪T1 with T1 ∈ |𝒪Q| and Cℱ = T1 + T2. Hence, we have ℱ ∈ 𝔅.Obviously, we have ℱ be a sheaf of type (B), that is, it corresponds to a pair of two curves {T1, T2}. Let us assume that ℱ admits another sequenceT3, T4 ∈ |𝒪Q|. Note that 𝒪Ti is stable for all i. Thus, the composite map s : 𝒪T3 → ℱ → 𝒪T2 is either a zero map or an isomorphism. In the former case, we have 𝒪T3≅𝒪T1 and so 𝒪T4≅𝒪T2. In the latter case, we have 𝒪T3≅𝒪T2 and 𝒪T4≅𝒪T1. Hence, the class of a strictly semistable sheaf ℱ corresponds to a uniquely determined pair of two curves in |𝒪Q| and we have 𝔅≅(ℙ3 × ℙ3)/𝔖2. The second assertion follows from the fact that any symmetric product Sd(ℙN) of any projective space is a rational variety .T1, T2 ∈ |𝒪Q|, one hasFor two curves 𝒪T2, −) to the sequence of T1, we obtain𝒪Q-sheaves, we haveT1 and T2 are reduced, and so the assertion is derived.Note that we haveℱ of type (C), but not of type (B), are stable. In particular, the sheaves of type (C) are semistable.The sheaves 𝒦 of ℱ with p(𝒦) ≥ 1/2 and the quotient sheaf ℋ : = ℱ/𝒦. Since the composite s : 𝒦↪ℱ²↠𝒪T2 is not a zero map, thus we have Im(s)≅𝒪T2(−Z) for a 0-dimensional subscheme Z of T2 with length k. In particular, its Hilbert bipolynomial is y + 1 − k. If we let χ𝒦 = m′x + n′y + t′, then we have p(𝒦) = t′/(m′ + n′) ≥ 1/2. In particular, we have t′ ≥ 1. If we define 𝒦′ to be the kernel of the map s, then it is a subsheaf of 𝒪T1 and thus we have p(𝒦′) = (t′ − 1 + k)/(m′ + n′ − 1) ≤ 1/3. Combining the two inequalities, we have k = 0 and so the map s is surjective. Thus, we have ℋ≅𝒪T1/𝒦′. Note also that t′ can be either 1 or 2. If t′ = 2, then we have m′ = n′ = 2 and so χ𝒦′ = 2x + y + 1 = χ𝒪T1. In particular, we have ℋ = 0 and so 𝒦≅ℱ, a contradiction. Now, assume t′ = 1 and so m′ + n′ ≤ 2. In particular, ℋ is not a 0-dimensional sheaf. Moreover, ℋ is a quotient sheaf of 𝒪T1 with constant term 1 and so we have ℋ≅𝒪T3 with T3 ⊂ T1 and T3 ∈ |𝒪Q|. For example, if T3 = T1, then we have 𝒦′ = 0 and it contradicts the nontriviality of the extension ℱ. Thus, ℱ also admits the sequenceχ𝒦 = x + y + 1. Since 𝒦′ is a subsheaf of 𝒪T1 with χ𝒦′ = x, we have 𝒦′≅𝒪T4, where T4 is a subcurve of T1 such that T1 = T3 + T4. Thus, 𝒦 is an extension of 𝒪T2 by 𝒪T4. It is nontrivial, otherwise we would have 𝒪T2 as a direct factor of ℱ. Since there exists such a unique extension 𝒪T2+T4, ℱ admits an extension of 𝒪T3 by 𝒪T2+T4:ℱ is of type (B).As before let us assume the existence of a proper subsheaf ℱ be a line bundle on a reduced curve C ∈ |𝒪Q| with degree 2. ℱ is semistable if and only if one has:ℱ|T) ≥ 1 for all subcurves T of C in |𝒪Q| with ⪇⪇,deg(ℱ|A) ≥ 0 for each smooth subcurve A of C in |𝒪Q| with ≤⪇.deg(ℱ is stable if and only if deg(ℱ|T) ≥ 1 for all subcurves T of C in |𝒪Q| with ⪇⪇.Let ℱ is not stable  and take a proper subsheaf ℋ of ℱ with p(ℋ) ≥ p(ℱ) (resp. p(ℋ) > p(ℱ)). Taking a saturation of ℋ in ℱ, we may assume that 𝒢 : = ℱ/ℋ is a pure sheaf. Call A the scheme support of ℋ and T the scheme support of 𝒢. The definition of scheme support of a purely 1-dimensional sheaf gives A + T = C as effective divisors. Thus T has one of the types in the assertion. Since C is reduced and ℱ is a line bundle on C, the support of 𝒢 must be a proper subcurve T of C. If T does not have a type of 𝒪Q or 𝒪Q, then we are done. But the case of T having such types is excluded using the argument in the proof of In both parts, the “only if” part is obvious. Assume that 𝔅∩ℭ ≠ ∅.One has B = B′ + T2 with B′ ∈ |𝒪Q| and T2 ∈ |𝒪Q|, and set A ∈ |𝒪Q| to be smooth. For any extension ℱ ∈ 𝔅 of 𝒪B by OA, for example, ℱ = 𝒪A ⊕ 𝒪B, let ℋ be the kernel of the composition ℱ → 𝒪B→𝒪T2 and then ℋ is a pure sheaf with T1 : = A + B′ as its scheme support and has Hilbert bipolynomial χℋ = χ𝒪T1. Note that it has 𝒪A as its subsheaf.Let us set ℋ≅𝒪T1, it is sufficient to prove that ℋ is semistable. Suppose ℋ is not semistable and take a proper saturated stable subsheaf 𝒢 ⊂ ℋ with χ𝒢 = ax + by + c. Its scheme support is contained in T1 and it is of type . Without loss of generality, let us assume that a ≤ b. First, assume  = . In this case, we would have c ≥ 2 because p(𝒢) > p(ℋ) and so we have h0(ℋ) ≥ 2, contradicting the fact that h0(ℱ) = 2 and that ℱ is globally generated. Assume a = b = 1. The map 𝒢 → ℋ on A∖T2 must be just the inclusion 𝒪A → ℋ, because ℋ|A∖T2 is a line bundle. Thus either we have 𝒢 = 𝒪A or 𝒪A is not saturated in ℋ. Hence the saturation 𝒜 of 𝒪A in ℱ has slope greater than 1/2, contradicting the semistability of ℱ. Now assume a = 0 and b = 1, that is, C𝒢 = B′. Since B′ is smooth, 𝒢 is a line bundle on B′. If its degree d is at least 1, then 𝒢 contradicts the semistability of ℱ. If d ≤ 0, then we have p(𝒢) < p(ℋ), a contradiction. Hence ℱ is also contained in ℭ.To prove T1 ∈ |𝒪Q| and T2 ∈ |𝒪Q|, one hasFor 𝒪T2, −) to the sequence of T1, we obtain2 = 0. Note also that Ext1≅H0(𝒪T2) and Ext2)≅H0(𝒪T2)∨. Thus we have the assertion.Applying the functor Hom is the global sheaf of a sheaf with support on P1 and P2 with one copy of ℂ on each point P1, P2,When R be a regular local ring of dimension 2 and take x, y generators of its maximal ideal. All Exti groups are with respect to R. Since R/(y) is Gorenstein, so the duality gives ExtR1(R/(y), R)≅R/(y) and ExtRi(R/(y), R) = 0 for all i ≠ 1. From the exact sequenceu is the multiplication by x, we get that ExtR1(R/(y), R/(x)) is the cokernel of the multiplication by x in R/(y) → R/(y); that is, we have ExtR1(R/(y), R/(x)) = ℂ. The same is true for extensions of 𝒪B by 𝒪A when A and B are transversal.Let ℱ be a sheaf of type (A) with no 0-dimensional torsion. Then ℱ is semistable unless it admits the sequenceT1 ∈ |𝒪Q| and T2 ∈ |𝒪Q| with ∈{, }.Let 𝒦 be a subsheaf ℱ with maximal p-slope p(𝒦) > 1/2 and so the quotient sheaf ℋ : = ℱ/𝒦 has no 0-dimensional torsion. Let us set χ𝒦 = m′x + n′y + t′ with t′ ≥ 1 and ⪇. If the composite s : 𝒦↪ℱ²↠η is a zero map, then 𝒦 is a subsheaf destabilizing 𝒪C, a contradiction. If s is not surjective, for instance, Im(s) = 𝒪P⊊η, then ker(s) is a subsheaf of 𝒪C with Hilbert bipolynomial m′x + n′y + t′ − 1. Thus we have t′ = 1 and the quotient ℋ′ : = 𝒪C/𝒦′ has Hilbert bipolynomial with zero constant term. Since ℋ′ has no 0-dimensional torsion, we have ℋ′≅𝒪D for a curve D contained in C. But the Hilbert bipolynomial of 𝒪D has nonzero constant term, a contradiction. Thus the map s is surjective. Following the same argument before, we obtain that t′ = 1 and m′ + n′ ≤ 1. Without loss of generality, let us assume that  = . Then we have χℋ = 2x + y + 1 and thus we have ℋ≅𝒪T1, where T1 is a curve contained in Cℱ and T1 ∈ |𝒪Q|. Since 𝒦 is a subsheaf of ℱ with χ𝒦 = y + 1, we have 𝒦≅𝒪T2 since 𝒦 has no 0-dimensional torsion. Thus ℱ fits into the sequence  to the sequence of C ∈ |𝒪Q|, we obtainf is the dual of the map Hom, η) → Hom given by the multiplication by the defining equation of C, the map f is a zero map and thus we have Ext1≅H0(η)∨. In particular its dimension is 2.Applying the functor Hom(ℱ be a sheaf of type (B) fitting into an exact sequenceT1, T2 ∈ |𝒪Q|. Then ℱ is of type (A) if and only if T1 and T2 have no common components; that is, Cℱ has no multiple component.Let T1 and T2 have a common component, say T, then ℱ has rank 2 at the general point of T and thus ℱ is not of type (A).If T1∩T2 is finite. Since we have h1(𝒪T1) = 0, the sequence (h0(ℱ) = 2 and ℱ is globally generated. Let σ be a general section of ℱ and then it does not vanish at the general point of any of the components of Cℱ. Since Cℱ is reduced, σ induces an injective map 𝒪Cℱ↪ℱ and thus ℱ has type (A).Conversely, assume that sequence  implies ℱ be a sheaf of type (C). T2 is not a component of T1, then ℱ is of type (A).If T2 is a double component of Cℱ, that is, T2 ⊂ T1, then it is not of type (A).If Let T2 ∈ |𝒪Q|.T2 is not a component of T1, ℱ is a line bundle on C = Cℱ outside finitely many points of C. Moreover, it is not an 𝒪T1-sheaf. Note that ℱ is globally generated since 𝒪T1 and 𝒪T2 are globally generated with h1(𝒪T1) = 0. Thus, a general section of ℱ does not vanish at a general point of T2 and so it does not induce an injection 𝒪T1↪ℱ. Hence, ℱ fits into some sequence |. Let Γ be the projectivisation of Ext1 and in particular we have Γ≅ℙ1 by e ∈ Γ gives a semistable sheaf. Such a sheaf has rank 2 at the points of T2∖(T2∩T3) and, in particular, it is not a line bundle over its support at a general point of T2. Thus, it never fits into an exact sequence 1. Below we give a partial answer to this question in the case of 𝒫ic2.In general, the question whether the variety M2 is unirational with degree 4. C of bidegree  in Q and a point P ∈ C to consider a sheaf 𝒪C(P) ∈ M1. If 𝕋P is the tangent plane of Q at P, then we have 𝕋P∩C = {2P, Q1, Q2} for some points Q1, Q2 on Q since deg(C) = 4. It defines a rational map𝒪C(P) to 𝒪C(Q1 + Q2). Note that 𝒪C(Q1 + Q2) = 𝒪C(−2P). We claim that the map Φ is generically 4 to 1 and so the assertion follows.Let us fix a smooth curve U  be the dense open subset of M1  formed by the sheaves ℱ such that Cℱ is smooth. Each element of U  is uniquely determined by a smooth C ∈ |𝒪Q| and a degree one  line bundle on C. By Riemann-Roch, each degree one line bundle on C is associated with a unique P ∈ C. Then the map Φ sends 𝒪C(P) to ℛ : = 𝒪C(−2P). Fix any degree two line bundle ℳ on C. Since we are in characteristic zero, there are exactly four line bundles 𝒜 on C such that 𝒜⊗2≅𝒪C. Hence, for each ℛ ∈ Pic2(C) there are exactly 4 points P ∈ C such that ℛ≅𝒪C(−2P). Hence, Φ is dominant and the preimage of each element of V has cardinality 4.Let M2 as of now, and we left the rationality question as a conjecture.We did not succeed in getting any smaller degree of unirationality of M2 is rational.


Several new theorems on the existence and multiplicity of positive radial solutions are obtained by means of fixed point index theory. Our conclusions are essential improvements of the results in Lan and Webb (1998), Lee (1997), Mao and Xue (2002), Stańczy (2000), and Han and Wang (2006).A class of elliptic boundary value problem in an exterior domain is considered under some conditions concerning the first eigenvalue of the relevant linear operator, where the variables of nonlinear term  In u(x) = z(|x|) of  = z((1 − t)n)1/(2−) thus reducing  of , where zv(t) = z| for all v ∈ E, P = {v ∈ E : v(t)⩾0  for  t ∈ }, and Q = {v ∈ P : mint∈v(t)⩾min{a, 1 − b}||v||}, where ⊂. It is easy to show that P and Q are cones in E. Let Ωr = {u ∈ E : ||u|| < r} be the open ball of radius r in E. Define a set H byLet h ∈ H, define an operator Th byFor Then we have the following lemma.h ∈ H,(i)Th : E → E is a completely continuous positive linear operator, and the spectral radius r(Th) ≠ 0 and Th has a positive eigenfunction φh1 corresponding to its first eigenvalue λh1 = (r(Th))−1;(ii)Th(P) ⊂ Q;(iii)δ1, δ2 > 0, such thatthere exist (iv)Jh by Jh(v) = ∫01h(t)φh1(t)v(t)dt for v ∈ E; then Jh(Thv) = λh1−1Jh(v) for  v ∈ E;define a functional (v)letthen P0 is a cone in E and Th(P) ⊂ P0, where δ1 is defined by ⩽To prove E is a Banach space, Tn : E → E   are completely continuous operators, T : E → E, andT is a completely continuous operator.Suppose that T : E → E is a completely continuous linear operator and  T(P)⊂(P). If there exist ϕ ∈ E∖(−P) and a constant  c > 0 such that cTϕ⩾ϕ, then the spectral radius  r(T) ≠ 0 and T has a postive eigenfunction corresponding to its first eigenvalue λ1 = (r(T))−1.Suppose that  H that(i) It follows from the definition of Th : E → E. Obviously, Th(P) ⊂ P and Th is a linear operator; namely, Th is a positive linear operator. Next, we will show that Th is completely continuous. For any natural number n  (n⩾2), letHence, by Lebesgue's dominated convergence theorem, it is easy to see that hn : →∪. Therefore, by Th : E → E is a completely continuous operator.It is clear that rding to , 19), a, aThn :t1 ∈  such that kh(t1) > 0. Thus there is ⊂ such that t1 ∈  and kh(s) > 0 for all t, s ∈ . Take ζ ∈ P such that ζ(t1) > 0 and ζ(t) = 0 for all t ∉ . Then for t ∈ ,It is obvious that there exists c > 0 such that c(Thζ)(t)⩾ζ(t) for all t ∈ . From r(Th) ≠ 0 and Th has a positive eigenfunction corresponding to its first eigenvalue λh1 = (r(Th))−1.So there exists a constant Th(P) ⊂ Q, we only need to show(ii) To prove v ∈ P, from k ⩽ k = s(1 − s) for t, s ∈ , we haveIn fact, for every t ∈ ,Notice that, for v ∈ P,It follows from  and 26)26) that,Th maps P into Q.So  holds; tφh1 is a positive eigenfunction of Th, we know from the maximum principle (see ; we have φh1(0) = φh1(1) = 0. This implies that φh1′(0) > 0 and φh1′(1) < 0  Since ple (see ) that φ< 0 (see ). Defineh is continuous on  and Φh(s) > 0 for all s ∈ . So, there exist δ1, δ2 > 0 such that δ1 ⩽ Φh(s) ⩽ δ2 for all s ∈ . Thust, s ∈ .Then it is easy to see that ΦJh(v) = ∫01h(t)φh1(t)v(t)ds ⩽ δ2∫01t(1 − t)h(t)v(t)dt < +∞ for all v ∈ E. So J : E → ℝ1 is well defined. For v ∈ E,(iv) From , Jh(v) P0 is a cone in E. It follows from ((v) It is easy to verify that ows from  and 30)P0 is a Jh(Thv)⩾λh1−1δ1||Thv|| for all v ∈ P. The proof is completed.So Denote1) × ℝ+, ℝ+) and for any M > 0 there exists a function hM ∈ H such that2)(Hh ∈ H such thatthere exists a function 3)(Hh ∈ H such thatthere exists a function 4)/v > M1;liminf5)/v > M1;liminf6)(Hl > 0 such thatthere exists a number 7)(Hh ∈ H such thatthere exists a function 8)(Hh ∈ H with h(t)≢0 for t ∈  and q ∈ C such thatthere exist We list some conditions as follows which will be useful in this section.1) holds. Then A : Q → Q is a completely continuous operator.Assume (H1), A is well defined and for every v ∈ Q, Av is nonnegative and continuous on . Note the property of k; it is easy to see that A(Q) ⊂ Q. (H1) and Lebesgue's dominated convergence theorem ensure the continuity of A. Finally, by using Ascoli-Arzela theorem, we can prove that A is completely continuous.The proof is similar to that of Lemma 3.1 in , so we o1) holds.2) is satisfied, then i = 1 for sufficiently small positive number r;If (H3) is satisfied, then i = 1 for sufficiently large positive number R;if (H4) is satisfied, then i = 0 for sufficiently small positive number r;if (H5) is satisfied, then i = 0 for sufficiently large positive number R;if (H6) is satisfied, then i = 0;if (H7) is satisfied, then i = 0 for sufficiently small positive number r;if (H8) are satisfied, then i = 0 for sufficiently large positive number R.if (HAssume (H2), there exists r > 0 such that(i) By  ⊂ Q and the spectral radial r(Sh) = 1. For every v ∈ Q∩∂Ωr, it follows from . Thus τ1 = μ1−1 > 1 and τ1v1 = Av1 ⩽ Shv1. By induction, we have τ1nv1 ⩽ Shnv1, n = 1,2,…. Then τ1nv1 = Shnv1 ⩽ ||Sh||||v1|| and taking the supremum on  gives τ1n ⩽ ||Shn||. By the spectral radius formula, we havei = 1.If there exist 3), there exist η > 0 and ε0 ∈  such that(ii) By (H1), there is hη ∈ H such that g ⩽ hη(t) for all ∈×. Hence,From  ⊂ Q. Let C1 = ∫01t(1 − t)hη(t)dt < +∞. SetDefine W is bounded. For any v ∈ W, from , the first eigenvalue of Sh, r(Sh)−1 > 1. Therefore, the inverse operator (I − Sh)−1 exists andSince Th(P) ⊂ Q that (I − Sh)−1(P) ⊂ Q. Hence, we have from  = 1.That is, (iii)–(v) have been proved in , so we s7), there exists r > 0 such that(vi) By  = 0.Hence, by  we haveη > 0 and ε0 ∈  such that(vii) From , there eq is bounded on , there is a constant C2 > 0 such thatSince q(v)⩾λh1(1 + ε0)v − C2 for all v ∈ , we have ∫abh(t)φh1(t)dt > 0. Thus from φ1h(t) >i = 0. The proof is completed.This is a contradiction. So, by the property of omitting a direction for fixed point index, we have Now, we are ready to state our main results of this section.1), (H2), (H3), and (H6) hold; then the singular BVP  = 1, and i = 1. From i = 0 and the additivity property of the fix point index, we obtainA has at least two fixed points, one in According to ular BVP  has at l1) and one of the following conditions are satisfied, then the singular BVP (2) and (H5) hold;(H2) and (H6) hold;(H2) and (H8) hold;(H3) and (H4) hold;(H3) and (H6) hold;(H3) and (H7) hold., (H2), and (H3), respectively. And they are somewhat easy to verify. In fact, (H1*) is a key condition in ψ(v); then we have g ⩽ h(t)ψ(v) ⩽ hM(t) for all ∈× and hM ∈ H. Consequently, (H1) holds. From g/h(t)v ⩽ h(t)ψ(v)/h(t)v = ψ(v)/v, it is easy to see that if (H2*) or (H3*) holds, then (H2) or (H3) holds, respectively.If (H1), the key conditions imposed on the nonlinear term g are the following ones:(A2)A ⊂  of positive measures such thatthere exists a set (A3)h ∈ H such thatthere exists a function In , Stańczy(A2′)a, b]⊂ such thatthere exists a set g/v > M1;(HW5) liminf h ∈ H such that(HW2) there exists a function where mh = h(s)ds)−1 for h ∈ H.In a recent paper , Han and6), and h ∈ H such that(HW3) there exists a function Additionally, they obtained a twin-solution theorem for the singular BVP  by usingmh ⩽ λh1, (H2) is an improvement of (HW2). In fact, without loss of generality, suppose φh1, the positive eigenfunction corresponding to λh1, satisfies ||φh1|| = φh1(t0) = 1; thenSince mh ⩽ λh1 and then (HW2) implies (H2). Therefore, Theorems 7) and (H8). These improvements allow us to deal with more singular problems.So α, β, γ, and δ⩾0 and ρ = γβ + αγ + αδ > 0, our results still hold. In fact, the Green functionk is defined in =−v′′(t)=gFor the following singular two-point BVP whose variables of nonlinear term are separated, the hypotheses and results will be more concise:h is a fixed function, Th, λh1, and φh1 are confirmed exclusively. So we skip the subscript h in the following. Corresponding  to (H1)–(H8), we formulate the conditions for singular BVP (1′) and h ∈ H;2′)(Hv→0+g(v)/v < λ1;limsup3′)(Hv→+∞g(v)/v < λ1;limsup4′) > 0 and liminfv→0+g(v)/v > M1)−1;min5′) > 0 and liminfv→+∞g(v)/v > M1)−1;min6′)(Hl > 0 such thatthere exists a number 7′)(Hv→0+g(v)/v > λ1;liminf8′)(Hv→+∞g(v)/v > λ1.liminfSince ular BVP :(H1′)g t1 ∈  with φ1(t1) = mint∈φ1(t). ThenTake φ1(t1), mint∈h(t), and mint∈∫abkds > 0; we haveNotice that 4′) implies (H7′) and (H5′) implies (H8′).So (H8′). In fact, the cone Q will be replaced by the cone P0 which is defined in (8′). See the proof of Observe that the conditionfined in  as we co1′), (H2′), (H3′), and (H6′) hold; then the singular BVP (Assume that (Hular BVP  has at lThis theorem is a direct corollary of 1′)  and one of the following conditions are satisfied, then the singular BVP (2′) and (H5′) hold;(H2′) and (H6′) hold;(H2′) and (H8′) hold;(H3′) and (H4′) hold;(H3′) and (H6′) hold;(H3′) and (H7′) hold.(HIf (Hular BVP  has at l8), (H8′) does not contain the condition that h≢0 for t ∈ .(i), (ii), and (iv)–(vi) are direct corollaries of T(P) ⊂ P0 is completely continuous, we know that A(P) ⊂ P0 is also completely continuous. Similar to item (i) of 2′), it is not difficult to prove i = 1 for sufficiently small positive number r. By (H8′), there exist η > 0 and ε0 ∈  such that(iii) Since  g is bounded on , there is a constant C4 > 0 such that g(v)⩾λ1(1 + ε0)v − C4 for all v ∈ . ThusSince C5 = C4∫01h(t)φ1(t)h(s)ds)dt; then C5 > 0 is a finite constant. TakeLet v1 ∈ P0∩∂ΩR and μ1⩾0 such that v1 = Av1 + μ1φ1; then from 83), we hi = 0. So A has a fixed point in This contradicts with . By the ular BVP  has at lh ∈ L1, h⩾0 a.e. on  and g ∈ C;One of the following conditions holds:(h1)v→0+g(v)/v < α and β < liminfv→+∞g(v)/v;limsup(h2)v→+∞g(v)/v < α and β < liminfv→0+g(v)/v, where α = mh and β = h(s)ds)−1.limsupLan and Webb have studied BVP  in 21]..21]. Theα ⩽ λh1 ⩽ β. Therefore, the items (iii) and (vi) of 2′), (H3′), (H7′), and (H8′) cannot be improved anymore.Under the above condition (i), we still can prove a, b] = .At the end of this section, we present three simple examples to which our theorems can be applied, respectively. We choose [c > 0 is a constant. Obviously, g ⩽ h(t)ψ(v) for all ∈ × ℝ+, where h(t) = t−1(1 − t)−1 andLetλ = 32/3 < 11, if limv→0+ψ(v)/v = c < λh1 and limv→+∞ψ(v)/v = c < λh1, then g satisfies all the conditions of Since c < λh1. Since limv→0+ψ(v)/v = +∞ > λh1 and limv→+∞ψ(v)/v = c < λh1, the item (iv) of Letp, q ∈ , ai > 0 for i = 1,2,…, n, and a1 < λh1. Since limv→0+ψ(v)/v = a1 < λh1 and limv→+∞ψ(v)/v = +∞ > λh1, the item (iii) of LetDefine a setc = (1 − a)n)1/(2− and d = (1 − b)n)1/ × ℝ+, ℝ+) and for any M > 0 there exists a function pM ∈ K such that2)(Cp ∈ K such thatthere exists a function 3)(Cp ∈ K such thatthere exists a function 4)/u > cn2−2(n − 2)2M1;liminf5)/u > cn2−2(n − 2)2M1;liminf6)(Cl > 0 such thatthere exists a number 7)(Cp ∈ K such thatthere exists a function 8)(Cp ∈ K with p(s)≢0 for s ∈  and q ∈ C such thatthere exist According to , we formNow we are ready to state our main results for the elliptic BVP .1), (C2), (C3), and (C6) are satisfied, then the elliptic BVP , (1) and one of the following conditions are satisfied, then the elliptic BVP (2) and (C5) hold;(C2) and (C6) hold;(C2) and (C8) hold;(C3) and (C4) hold;(C3) and (C6) hold;(C3) and (C7) hold., ℝ+) and f ∈ C.For the following BVP whose nonlinear terms are variable-separated, it is easy to give more concise theorems as Theorems One can see that, through variables substitution, it is easy to give some examples like in


I-double gradation fuzzy topological spaces and also study these functions in relation to some other types of already known functions.In this paper, we introduce and characterize double fuzzy weakly preopen and double fuzzy weakly preclosed functions between  In the history of science, new theories have always been necessary in order for existing scientific theories to progress and this will continue to be true in the future. Two examples of essentially different mathematical theories that deal with the concept of uncertainty are probability theory and the theory of fuzzy sets. Whereas probability theory has a history of around 360 years, the theory of fuzzy sets is little more than 50 years old. Since the 1960s fuzzy methods have entered the scientific and technological world, good theoretical progress  has been made, and there have been technical advances in various areas .e∞ Theory. Tang . The family of all fuzzy subsets on X denoted by IX. By X. For a fuzzy subset λ∈IX, f(λ) and f−1(λ) define the direct image and the inverse image of f, defined by f−1(ν)(x)=ν(f(x)), for each λ∈IX, ν∈IY, and x∈X, respectively. For fuzzy subsets λ and μ in X, we write λqμ to mean that λ is quasi coincident (q-coincident) with μ, that is, there exists at least one point x∈Xsuch that λ(x) + μ(x)>1. Negation of such a statement is denoted as Throughout this paper, let Samanta and Mondal (Garcia and Rodabaugh (I-double gradation fuzzy topology (ττ∗) on X is a pair of maps τ, τ∗:IX→I, which satisfies the following properties: λ∈IX.(O1) τ(λ1∧λ2)≥τ(λ1)∧τ(λ2) and τ∗(λ1∧λ2)≤τ∗(λ1)∨τ∗(λ2) for each λ1, λ2∈IX.(O2) λi∈IX, i∈Γ.(O3)  Let  λ≤Cτ,τ∗,(C2) Cτ,τ∗∨Cτ,τ∗=Cτ,τ∗,(C3) Cτ,τ∗≤Cτ,τ∗ if r1≤r2and s1≥s2,(C4) Cτ,τ∗,r,s)=Cτ,τ∗.(C5) For  Let (Xττ∗) be an I-dfts. Then for each r∈I0, s∈I1 and λ∈IX, we define an operator Iτ,τ∗:IX×I0×I1→IXas follows: λμ∈IX, rr1r2∈I0and ss1s2∈I1, the operator Iτ,τ∗satisfies the following statements: Iτ,τ∗≤λ,Iτ,τ∗∧Iτ,τ∗=Iτ,τ∗,Iτ,τ∗≥Iτ,τ∗ if r1≤r2and s1≥s2,Iτ,τ∗,r,s)=Iτ,τ∗,Iτ,τ∗,r,s)=λ, then If For X,τ,τ∗) be an I-dfts. For λ∈IX, r∈I0and s∈I1.Let -fuzzy preopen -fpo, for short) if λ≤Iτ,τ∗,r,s). A fuzzy set λ is called -fuzzy preclosed -fpc, for short) iff r,s)-fpo set. The -fuzzy preinterior of λ, denoted by PIτ,τ∗ is defined by (1) r,s)-fuzzy preclosure of λ, denoted by PCτ,τ∗ is defined by The -fuzzy regular open -fro, for short) if λ=Iτ,τ∗,r,s). A fuzzy set λ is called -fuzzy regular closed -frc, for short) iff r,s)-fro set.(2) λ is called -fuzzy α-open -fαo, for short) if λ≤Iτ,τ∗,r,s),r,s). A fuzzy setλis called -fuzzy α-closed -fαc, for short) iff r,s)-fαo set.(3) X,τ,τ∗) be an I-dfts. For λ∈IX, r∈I0and s∈I1. λis -fpo -fpc) iff λ=PIτ,τ∗ ),Iτ,τ∗≤PIτ,τ∗≤λ≤PCτ,τ∗≤Cτ,τ∗,Let  is -fpc set in IYfor each λ∈IX, r∈I0and s∈I1; double fuzzy preclosed if τ2(f(λ))≥τ1(λ) and λ∈IX, r∈I0and s∈I1,double fuzzy open if τ2(f(λ))≥rand r,s)-fro set λ∈IX, r∈I0and s∈I1.double fuzzy almost open if Let I-dfts I-dfts f is called: λ∈IX, r∈I0and s∈I1; τ1(λ)≥rand double fuzzy weakly open if α-open if f(λ) is -fαo in IYfor each λ∈IX, r∈I0and s∈I1; τ1(λ)≥rand double fuzzy Let X,τ,τ∗) be an I-dfts, μ∈IX, xt∈P(X), r∈I0and s∈I1where P(X) is the family of all fuzzy points in X. μis called an -fuzzy open Q-neighborhood of xtif τ(μ)≥r, τ∗(μ)≤sand xtqμ. We denote the set of all -fuzzy open Q-neighborhood of xtby Qτ,τ∗.Let  be an I-dfts, λ∈IX, xt∈P(X), r∈I0and s∈I1. xtis called -fuzzy θ-cluster point of λif for every μ∈Qτ,τ∗, we have Cτ,τ∗qλ. We denote Dτ,τ∗ is called -fuzzy θ-closure of λ.Let  an I-dfts. For λ, μ∈IXand r, s∈I0, we have the following: xtis -fuzzy θ-cluster point of λiff xt∈Dτ,τ∗.Cτ,τ∗≤Dτ,τ∗,τ(λ)≥rand τ∗(λ)≤s, then Cτ,τ∗=Dτ,τ∗,If λis -fpo, then Cτ,τ∗=Dτ,τ∗,If λis -fpo and λ=Cτ,τ∗,r,s), then Dτ,τ∗=λ.If Let -fuzzy θ-closed set is called -fuzzy θ-open and the -fuzzy θ-interior operator denoted by Tτ,τ∗ is defined by The complement of ≤Tτ,τ∗ for any λ∈IX, r∈I0and s∈I1,Tτ,τ∗=Iτ,τ∗ for each λ∈IX, r∈I0and s∈I1; τ(λ)≥rand τ∗(λ)≤s.From Theorem 2.4 It is easy to see that: A function λ∈IX, r∈I0 and s∈I1; τ1(λ)≥r and for each Every double fuzzy weakly open function is double fuzzy preopen and every double fuzzy preopen function is double fuzzy weakly preopen, but the converse need not be true in general.X={a,b,c} and Y={x,y,z}. Fuzzy sets λ1, λ2 and λ3 are defined as: Let τ1and τ2 as follows: Define f(a)=z, f(b)=x and f(c)=y is double fuzzy weakly preopen but not double fuzzy preopen. Where f(λ) is not Then the mapping X={a,b,c} and Y={x,y,z}. Fuzzy sets λ1, λ2 and λ3 are defined as: Let Let f(a)=z, f(b)=xand f(c)=yis double fuzzy weakly preopen but not double fuzzy weakly open. Since Then the mapping f is double fuzzy weakly preopen,λ∈IX, r∈I0and s∈I1,ν∈IY, r∈I0and s∈I1,ν∈IY, r∈I0and s∈I1.For a function λ∈IX and (1)⇒(2) Let f is double fuzzy weakly preopen, Since f(λ),r,s)) and so, and hence μ∈IX; τ1(μ)≥r and (2)⇒(1) Let f is double fuzzy weakly preopen.Hence ν∈IY. By using (2), (2)⇒(3) Let (3)⇒(2) Trivial.ν∈IY. Using (3), we have (3)⇒(4) Let r,s).Therefore, we obtain ν∈IY, r∈I0 and s∈I1, i.e., (4)⇒(3) Similarly we obtain, f is double fuzzy weakly preopen,xt ∈ P(X) and each μ∈IX; τ1(μ)≥rand xt≤μ, there exists -fpo set γsuch that f(xt)≤γand For each For the function xt ∈ P(X) and μ∈IXsuch that τ1(μ)≥r, xt≤μ. Since f is double fuzzy weakly preopen, then μ,r,s)), with f(xt)≤γ.(1)⇒(2) Let μ∈IX; τ1(μ)≥r, ys≤f(μ). It follows from (2) that r,s)-fpo γ∈IYand ys≤γ. Hence we have, f is double fuzzy weakly preopen function. □(2)⇒(1) Let f is double fuzzy weakly preopen;λ∈IX, r∈I0and s∈I1; τ1(λ)≥rand ν∈IX, r∈I0and s∈I1; Let ν∈IX; τ1(ν)≥r and (1)⇒(2) Let and so λ∈IX; τ1(λ)≥r and r,s)-fc set and r,s) by (3) we have (2)⇒(3) Let (3)⇒(2) Trivial.(2)⇒(1) Trivial. □f is double fuzzy weakly preopen;ν∈IX, r∈I0and s∈I1; τ1(ν)≥rand λ∈IX, r∈I0and s∈I1; τ1(λ)≥rand r,s)-fpo set λ∈IX;r,s)-fαo set λ∈IX.For a function ν∈IX, r∈I0 and s∈I1; (1)⇒(2) Let (2)⇒(3) It is clear.λ be -fpo set. Hence by (3), (3)⇒(4) Let (4)⇒(5) and (5)⇒(1) are clear. □λ∈IX, r∈I0and s∈I1.A function f is double fuzzy preopen.If λ∈IX such that τ1(λ)≥r and f is double fuzzy weakly preopen Let f is double fuzzy strongly continuous, then f(λ) is -fpo. □However, since f(λ) is -fpo for each λ∈IX, r∈I0and s∈I1; A function f is double fuzzy weakly preopen function.If λ∈IX; τ1(λ)≥r and Let □The converse of the above theorem need not be true in general as in the following Example.X={a,b,c} and Y={x,y,z}. Define fuzzy sets λ1, λ2 as follows: Let Let f(a)=x, f(b)=yand f(c)=z is double fuzzy weakly preopen but it isn’t double fuzzy contra-preclosed.Then the function I-dfts  is said to be -fuzzy regular space if for each λ∈IX; τ(λ)≥rand τ∗(λ)≤sis a union of -fo sets μi∈IXsuch that Cτ,τ∗≤λfor each i∈J.An X,τ,τ∗) be -regular fuzzy topological space. Then, f is double fuzzy preopen.Let ≥r and xt≤λ, let The sufficiency is clear. For the necessity, let f is double fuzzy preopen. □Thus If λ∈IX; τ1(λ)≥r and τ1∗(λ)≤s. Since f is double fuzzy almost open and r,s)-fro, then Let and hence f is double fuzzy weakly preopen. □This shows that X,τ,τ∗) be an I-dfts, r∈I0and s∈I1. The two fuzzy sets λ, μ∈IXare said to be -fuzzy separated iff r,s)-fuzzy separated sets is said to be -fuzzy connected.Let  an I-dfts. The fuzzy sets λ, μ∈IXsuch that r,s)-pre-separated if r,s)-fpo sets ν, γsuch that λ≤ν, μ≤γ, I-dfts which can not be expressed as a union of two fuzzy -pre-separated sets is said to be fuzzy -pre-connected space.Let -fuzzy pre-connected space r,s)-fuzzy connected.If r,s)-fuzzy connected. Then there exist -fuzzy separated sets β, γ∈IX such that βand γ are -fuzzy separated, there exists λ, μ∈IX; τ1(λ)≥r, τ1(μ)≥r and β≤λ, γ≤μ, f(β)≤f(λ), f(γ)≤f(μ), f is double fuzzy weakly preopen and double fuzzy strongly continuous function, from Theorem 3.10 we have f(λ) and f(μ) are -fpo sets. Therefore, f(β) and f(γ) are -fuzzy pre-separated and Let r,s)-fuzzy pre-connected. Thus r,s)-fuzzy connected. □which is contradiction with A function λ∈IX, r∈I0 and s∈I1; for each Clearly, every double fuzzy preclosed function is double fuzzy weakly preclosed, but the converse need not be true in general, as the next example shows.X={a,b} and Y={x,y}. Fuzzy sets λ1 and λ2 are defined as: Let Let f(a)=x, f(b)=y is double fuzzy weakly preclosed but is not double fuzzy preclosed.Then the function f is double fuzzy weakly preclosed;λ∈IX, r∈I0and s∈I1; τ1(λ)≥rand λ∈IX, r∈I0and s∈I1; r,s)-fpc set λ∈IX, r∈I0and s∈I1;r,s)-fαc λ∈IX, r∈I0and s∈I1.For a function Straightforward. □f is double fuzzy weakly preclosed;r,s)-fro set λ∈IX, r∈I0and s∈I1;ν∈IY, μ∈IX, r∈I0and s∈I1; τ1(μ)≥rand f−1(ν)≤μ, there exists -fpo set γ∈IYwith ν≤γand For each ys ∈ P(Y) and each μ∈IX, r∈I0and s∈I1such that τ1(μ)≥rand f−1(ys)≤μ, there exists -fpo set γ∈IY; ys≤γand For each fuzzy point λ∈IX, r∈I0and s∈I1;λ∈IX, r∈I0and s∈I1;r,s)-fpo set λ∈IX, r∈I0and s∈I1.For a function We will prove (2)⇒(3) and (1)⇒(6).ν∈IY, r∈I0, s∈I1 and let μ∈IX; τ1(μ)≥r and f−1(ν)≤μ. Then r,s)-fro, γ is -fpo with ν≤γ and (2)⇒(3): Let ν∈IY, r∈I0 and s∈I1; r,s)-fpo γ∈IY with ys≤γ and (1)⇒(6): Let ys∈ P(Y) and each r,s)-fpo set γ∈IY; If μ(x) + s>1 and hence there exists t∈ such that μ(x)>t>1−s. Then r,s)-fpo set γ∈IY; yt≤γsuch that γ(y)>tand hence γ(y)>1−s. Thus γis -fpo neighborhood of ys. □Let X,τ,τ∗) be an I-dfts. A fuzzy set λ∈IXis called -fuzzy pre-Q-neighborhood of xtif there exists -fpo set μ∈IXsuch that xtqμ≤λ. We denote the set of all -fuzzy pre-Q-neighborhood of xtby PQτ,τ∗.Let . A fuzzy point xt∈PCτ,τ∗ if and only if for every μ ∈ PQτ,τ∗, μqλis hold.In an Straightforward. □ν∈IX, r∈I0and s∈I1; f is double fuzzy preclosed.If ν∈IX, r∈I0and s∈I1; f is double fuzzy weakly preclosed by using Theorem 3.12, there exists -fuzzy pre-Q-neighborhood γ∈IYwith ys≤γand f(ν) is -fpc and f is double fuzzy preclosed function. □Let τ2(f(λ))≥rand λ∈IX, r∈I0and s∈I1; τ1(λ)≥rand A function f is double fuzzy weakly preclosed.If λ∈IX, r∈I0 and s∈I1 such that Let □ν∈IYand every λ∈IX, r∈I0, s∈I1such that τ1(λ)≥rand f−1(ν)≤λ, there exists -fpc set γ∈IYsuch that ν≤γand If ν∈IYand let λ∈IX, r∈I0and s∈I1such that τ1(λ)≥rand f−1(ν)≤λ. Put γis -fpc set in IYsuch that ν≤γsince f is double fuzzy weakly preclosed, Let ys∈ P(Y) and every λ∈IX, r∈I0and s∈I1such that τ1(λ)≥rand f−1(ys)≤λ, there exists -fpc set γ∈IY; ys≤γsuch that If λ∈IXis called -fuzzy θ-compact if for each family {μi∣i∈J} in x∈X, there exist a finite subset J0of J such that A fuzzy set r,s)-fuzzy θ-closed, then f(λ) is -fpc for each -fuzzy θ-compact λ∈IX, r∈I0and s∈I1.If λ be -fuzzy θ-compact and let xt≤λ there is x∈X and since λ is -fuzzy θ-compact, there is f is double fuzzy weakly preclosed, by using Theorem 3.12 there exists Let ys≤γand f(λ) is -fpc set. □Therefore X,τ,τ∗) be an I-dfts. The fuzzy sets λ, μ∈IXare -fuzzy strongly separated if there exist ν, γ∈IXsuch that τ(ν)≥rand τ∗(ν)≤s, τ(γ)≥rwith λ≤ν, μ≤γand Let  is called -fuzzy pre T2if for each r,s)-fpo sets λ, μ∈IXsuch that An r,s)-fuzzy strongly separated, then r,s)-fuzzy pre-T2.If γ,ν∈IX, r∈I0and s∈I1; τ1(γ)≥r, τ1(ν)≥rand τ1(ν)≤ssuch that r,s)-fpo sets λ, μ∈IYsuch that f is surjective. Thus r,s)-fuzzy pre-T2. □Let I-dfts  is said to be -extremally disconnected if τ)≥rand τ∗)≤sfor each λ∈IX; τ(λ)≥rand τ∗(λ)≤s.an I-dfts  is said to be -fuzzy almost compact if for each -fuzzy open cover {λi∣i∈J} of X, there is a finite subset J0of J such that an λin an I-dfts  is said to be -fuzzy p-compact iff for each family of -fpo sets {μi∣i∈J} satisfies x∈X. There exists finite subfamily J0of J such that x∈X.A fuzzy set r,s)-extremally disconnected I-dfts. Let f−1(ys) is -fuzzy almost compact for each ys ∈ P(Y). If λ∈IYis -fuzzy P-compact. Then f−1(λ) is -fuzzy almost compact.Let νj∣æ∈J} be -fuzzy open cover of f−1(λ). Then for each ys≤λ∧f(X), J(ys) of J. Since r,s)-extremally disconnected each r,s)-fpc set r,s)-fuzzy preclosed cover of λ, K of λ∧f(X). Hence, Let {f−1 (λ) is -fuzzy almost compact. □so