The bifurcations ofheteroclinic loop with one nonhyperbolic equilibrium and onehyperbolic saddle are considered, where the nonhyperbolicequilibrium is supposed to undergo a transcritical bifurcation;moreover, the heteroclinic loop has an orbit flip and an inclinationflip. When the nonhyperbolic equilibrium does not undergo atranscritical bifurcation, we establish the coexistence andnoncoexistence of the periodic orbits and homoclinic orbits. Whilethe nonhyperbolic equilibrium undergoes the transcriticalbifurcation, we obtain the noncoexistence of the periodic orbits andhomoclinic orbits and the existence of two or three heteroclinicorbits. In recent years, a great deal of mathematical efforts has been devoted to the bifurcation problems of homoclinic and heteroclinic orbits with high codimension, for example, the bifurcations of homoclinic or heteroclinic loop with orbit flip, the bifurcations of homoclinic or heteroclinic loop with inclination flip, and so forth; see –5 and thCr (r ≥ 5) systemz ∈ ℝ4, the vector field  g  depends on the parameters , λ ∈ ℝ, μ ∈ ℝl, l ≥ 2, 0 ≤ λ, |μ | ≪1, g = f(z), g = 0, and g = 0. Moreover, the parameter  λ  governs bifurcation of the nonhyperbolic equilibrium, while  μ  controls bifurcations of the heteroclinic orbits.Consider the following p1, p2, where Γ = Γ1⋃Γ2, Γi = {z = ri(t) : t ∈ ℝ}, ri(+∞) = ri+1(−∞) = pi+1, i = 1,2, r3(t) = r1(t), and p3 = p1. Furthermore, the linearization  Df(p1)  has real eigenvalues  0, λ11, −ρ11, and − ρ12  satisfying − ρ12 < −ρ11 < 0 < λ11; Df(p2) has simple real eigenvalues λ21, λ22, −ρ21, and −ρ22 fulfilling −ρ22 < −ρ21 < 0 < λ21 < λ22.Assuming system  has a heH1) is the unstable  manifold of pi, and Wiuu  is the strong unstable  manifold of pi, i = 1,2. Moreover,H2) manifold W2s  of  p2  are fulfilling the strong inclination property. And the fourth equation implies that the stable manifold W1s is of inclination flip as t → −∞.The following conditions hold in the whole paper:m ≥ 1 and n ≥ 1, if we assume dim(W1u) = dim(W2uu) = m and dim(W1ss) = dim(W2ss) = n, then all the results achieved in this paper are still valid.It is worthy of noting that, for any integers λ ∈ ℝ be a parameter to control the transcritical bifurcation of system  be the vector field defined on the center manifold; then by [H3) = 0, (∂θ/∂x) = 0, (∂2θ/∂x2) > 0, (∂2θ/∂x∂λ) < 0, (∂2θ/∂x∂μ) = 0, where xp1 is the x component of p1.Let f system , let the then by , we may H3) is true, then system ; there are two hyperbolic saddles p10 and p11 bifurcated from p1. Denote by p10 = p1 = * and p11 = p1 + *, where λp = θ0λ + O(λ2) + O(λμ) and θ0 = −(∂2θ/∂x∂λ)/(∂2θ/∂x2). Moreover, dim(Wp10s) = 3, dim(Wp10u) = 1, and dim(Wp11u) = dim(Wp11s) = 2.If x and λ → −θxλ−1λ, system , λp = λ + O(λ2) + O(λμ), λ11(0) = λ11, ρij(0) = ρij, j = 1,2,  i = 1,2, λ2j(0) = λ2j,  j = 1,2.Let the neighborhood H1), we may select −Ti and Ti such thatδ > 0 is small enough such that {:|x | , |y | , |u | , |v | <2δ} ⊂ Ui and |δu| = o(δ), |δv | = o(δ).From the normal form , 7), an, an7), ai = 1,2, where (Df(ri(t)))* is the transposed matrix of Df(ri(t)).Consider the linear variational systemZi(t) = (zi1(t), zi2(t), zi3(t), zi4(t)) is a fundamental solution matrix of  –(H3) are satisfied, thenIf conditions (1 satisfying(1) there exists a fundamental solution matrix of (9)2 has a fundamental solution matrix fulfillingwi21 < 0, w112wi33w214w242 ≠ 0, |(wi33)−1wij3 | ≪1, j = 1,2, 4, i = 1,2.(2)  (9)zi1(t), zi2(t), zi3(t), zi4(t)) be a new local active coordinate system along Γi. Given Φi(t) = (ϕi1(t), ϕi2(t), ϕi3(t), ϕi4(t)) = (Zi−1(t))*, then Φi(t) is the fundamental solution matrix of  + Zi(t)Ni(t)≜hi(t), where Ni(t) = *, i = 1,2. Defining the cross sectionsi at t = −Ti  and t = Ti, respectively, i = 1,2.Let qi0 ∈ Si0 and qi1 ∈ Si1, thenZi(−Ti) and Zi(Ti), we get their new coordinates of qi0* and qi1*; that is,Now that if Next, we divide our establishment of the Poincaré  map in the new coordinate system in three steps.Fi1 : Si0 ↦ Si1. Put z = hi(t) into  = Zi−1(t), thenFi1 : Si0 ↦ Si1, ↦.First, consider the map her with  and ((21Fi0 : Si−11 ↦ Si0 (where  S01 = S21). Let τi,  i = 1,2 be the flying time from qi−11* to qi0*; set s1 = eρ11(α)τ1− and s2 = eρ21(α)τ2−. By virtue of the approximate solution of system ,,Fi0 : F20 : S11 ↦ S21 issi, ui0, vi−11), i = 1,2 are called Shilnikov coordinates, andp1 undergoes a transcritical bifurcation based on the structure of orbits in U1, we may see that the equation x01 ≈ x10/h(s1) holds only for x01 ≥ λp. While for x01 ∈ [−β, λp)  (0 < β ≪ 1), the map F10 is well defined only if s1 = 0  asThe final step is to compose the maps Gi = Fi(qi−11) − qi1, i = 1,2. Combing  means that system ,,Gi = F Combing , 27), a, aGi =  Combing , we deritions of = 0 has a unique solution . And putting it into  = 0, then we obtain the following bifurcation equations:Based on the expressions of the successor functions and the implicit function theorem, we know that the equation −1δlns1) = 0, it then follows thatFirstly, we consider the case ppen. By  and 25)λ = 0, whppen. By  turns toFrom the above bifurcation equations, we obtain the following results immediately.H1)–(H3) be true and Miμ1 ≠ 0, i = 1,2. Then, for λ = 0 and 0 < |μ| ≪ 1, one hasLet the conditions  = 2, there exists a codimension 2 surfaceμ ∈ L12, where the surface L12 has a normal plane span{Mμ11, Mμ21} at μ = 0.(i) for rank(l − 1)-dimensional surfacep1  near Γ if and only if μ ∈ L12 .(ii) there exists an l − -dimensios1 = s2 = 0 into (The result (i) will be proved by putting = 0 into .s1 = 0 and s2 > 0 in -dimensional surface L12 given by  > 0 as  μ ∈ L12 and 0 < |μ | ≪1. This implies system  it has a normal vector Mμ11  at μ = 0.There is no difficulty to see that H1)–(H3) hold and Miμ1 ≠ 0, i = 1,2. Then for λ = 0, μ ∈ L12, and 0 < |μ | ≪1, the periodic orbit and homoclinic loop with p1 of system (Assume the conditions H–(H3) hoμ ∈ L12 and 0 < |μ | ≪1, then system −1s1 + δ−1Mμ11μ + h.o.t.>0, and μ ∈ L12, then (V1(0) = N1(0) and2, then  is reducw112w212 < 0, then V1′(s1)N1′(s1) < 0; it is obvious that V1(s1) = N1(s1) has no sufficiently small positive solutions.If ρ22 > λ21, then |V1′(s1)|≪1 and |N1′(s1)|≫1 hold for 0 < s1 ≪ 1, which shows that V1(s1) = N1(s1) has no sufficiently small positive solution.While ρ22 ≤ λ21 and w112w212 > 0. As μ ∈ L12, we have Mμ11μ > 0, and then, for wi12 > 0, i = 1,2 we see thatρ22 < λ21 yields that lims1→0+s1ρ22/λ21−1 = +∞, lims1→0+N1′(s1) = +∞, and lims1→0s1ρ22/λ21−1/N1′(s1) = 0, which shows V1(s1) = N1(s1) has no sufficiently small positive solutions. Obviously, the conclusion is correct as  ρ22 = λ21.Next, we only consider the case ρ22 < λ21, wi12 < 0, i = 1,2, there does not exist a small positive solution for V1(s1) = N1(s1).Similarly, for The proof is then completed.H1)–(H3) hold and Miμ1 ≠ 0, i = 1,2. Let λ = 0, μ ∈ L21, and 0 < |μ | ≪1; then p2 of system  = N2(0) as μ ∈ L21. Moreover,(i) Let  s1  in , we deriρ22 ≥ λ21 and N2′(s2) ≪ 1 = V2′(s2), this means V2(s2) = N2(s2) has no sufficiently small positive solutions.For w112w212 < 0, since we are interested in sufficiently small positive solutions of (V2(s2) = N2(s2) satisfying w112(s2λ21/ρ22 − δ−1Mμ11μ) > 0, which implies that s2λ21/ρ22 − δ−1Mμ11μ < 0  for w112 < 0 . It is easy to see that V2(s2) = N2(s2) has no sufficiently small positive solutions as w112w212 < 0.Now we turn to the case tions of , it suffρ22 < λ21, we have V2′(0) = 1 > 0 = N2′(0), which implies that there exists an V2(s2) > N2(s2) for (ii) For w112 > 0, sow212 > 0. As a result, N2(s2) = V2(s2) has at least one solution Choosing s2 must fulfill 0 < s2 < (δ−1Mμ11μ)ρ22/λ21 as w112 < 0; with similar arguments in proof of (ii), we can prove that there exists a 0 < s2* ≪ 1 such that V2(s2*) = N2(s2*) for 0 < s2* < (δ−1Mμ11μ)ρ22/λ21 ≪ 1. It is easy to compute that N2′′(s2) > 0 for w212 < 0, 0 < s2 < (δ−1Mμ11μ)ρ22/λ21, and μ ∈ L21. Combining with the fact V2(0) = N2(0), N2′(s2) > 0, and V2′(s2) = 1, we immediately know that s2* is unique.(iii) This completes the proof.λ > 0, when p1 undergoes a transcritical bifurcation. From λ > 0, after the creation of the equilibria p10 and p11, there always exists a straight segment orbit heteroclinic to p11 and p10, its length is λp, and we denote this heteroclinic orbit by Γ*. Moreover, x01 = λp is a critical position.Now, we turn to discussing the bifurcations of the heteroclinic loop for x01 ≥ λp. In this case, (s = s1λp/ρ11 (s = 0 means s1 = 0 and vice versa); by virtue of Taylor's development for δλp/(δ − (δ − λp)s1λp/ρ11), we haveFirstly, we take into account the case is case,  becomesλ = 0, we may easily obtain the following results.With similar arguments to H1)–(H3) hold, 0 < λ ≪ 1; thenSuppose the conditions  = 2, there exists an  (l − 2)-dimensional surfaceμ ∈ L12λ and 0 < |μ | ≪1;(i) if rank(l − 1)-dimensional surfacep11  if and only if μ ∈ Lλ12 and 0 < |μ | ≪1.(ii) there exists an l − -dimensioH1)–(H3) hold, Miμ1 ≠ 0, i = 1,2, 0 < λ, |μ | ≪1, and w112w212 < 0. Then, except the homoclinic loop connecting p11 , system .Suppose hypotheses H–(H3) hop10 and heteroclinic loop joining p10, p2 cannot be bifurcated from Γ, which is exactly determined by the generic condition (H1).It is easy to see that homoclinic loop connecting β ≤ x01 < λp. Due to Finally, we consider the case −H1)–(H3) are true, rank > 0 and rank > 0. Then, (i)p11, p2, p10 as μ ∈ Σ1;there exists a surface5Σ1={(ii)λ, μ) spacep11 and p10 for  ∈ Δ.there exists a region in the :−β≤w22−1δ such that , then system (p11  and  p2 and the other is heteroclinic to p2 and  p10.(i) If 2 = 0 in , then ∈ Δ, system (p11 and p10.(ii) If 2 > 0 in , one attg s2 in , we achip10 will go into p10 in different ways according to different fields of x01; see All the heteroclinic orbits joining


Two types of distributed time delays are incorporated into the model to describe the time needed for infection of target cell and virus replication. Using the method of Lyapunov functional, we have established that the global stability of the model is determined by two threshold numbers, the basic reproduction number R0and the immune response reproduction number R0∗. We have proven that, if R0 ≤ 1, then the uninfected steady state is globally asymptotically stable (GAS), if R0* ≤ 1 < R0, then the infected steady state without CTL immune response is GAS, and, if R0* > 1, then the infected steady state with CTL immune response is GAS.We study the global stability of a human immunodeficiency virus (HIV) infection model with Cytotoxic T Lymphocytes (CTL) immune response. The model describesthe interaction of the HIV with two classes of target cells, CD4 Mathematical modeling and model analysis of HIV dynamics are important to discover the dynamical behaviors of the viral infection process and estimating key parameter values which leads to development of efficient antiviral drug therapies. Several mathematical models have been proposed to describe the HIV dynamics with CD4+ T cells , and  hi  is limit superior to this delay. The factor  emiτ−  accounts for the loss of target cells during this delay period, where  mi  is constant. On the other hand, to consider the delay between viral RNA transcription and viral release and maturation, we let  τ  be the random variable; that is, the time between these two events with a probability distribution  g(τ)  over the interval  , and  h3  is limit superior to this delay   into  R+6. By the fundamental theory of functional differential equations   for some constant  ρ  and let  ti* ∈   be such that  xi(ti*) = 0.  From (xi(t) > 0  for some  t ∈ , where  ϵ > 0  is sufficiently small. This leads to a contradiction and hence  xi(t) > 0, for all  t ≥ 0.  Further, from (yi(t) ≥ 0,  i = 1,2, and  v(t) ≥ 0  for all  t ∈ . By a recursive argument, we obtain  yi(t) ≥ 0,  i = 1,2, and  v(t) ≥ 0  for all  t ≥ 0.  Now from (z(t) ≥ 0, for all  t ≥ 0.First, we prove that  0.  From  we have er, from  and 7)  xi(t) >Now from  we havei = 1,2. This implies  limsupt→∞xi(t) ≤ λi/di,  i = 1,2.Next we show the boundedness of the solutions of system –8). Fro. Fro8). X(t) = ∑i=12[Fixi(t) + yi(t + τ)] + (p/c)z(t + τ). Thenσ = min{d1, d2, a, b}. Hence  limsupt→∞X(t) ≤ L, where  L = ∑i=12Li = ∑i=12(Fiλi/σ). Since  xi(t) > 0,  yi(t) ≥ 0  and  z(t) ≥ 0 then  limsupt→∞∑i=12yi(t) ≤ L,  limsupt→∞yi(t) ≤ L,  i = 1,2  and  limsupt→∞z(t) ≤ cL/p. On the other hand,t→∞v(t) ≤ kGL/r. Therefore,  x1(t), y1(t), x2(t), y2(t), v(t), and  z(t)  are ultimately bounded.Let  R0  and immune response reproduction number  R0*  of system  as.(i) If  R0 > 1, then there exist  E0  and an infected steady state without CTL immune response(ii) If  R0* > 1, then there exist  E0,  E1, and an infected steady state with CTL immune response  E2 = .(iii) If  z = 0  orz = 0, then from (xi  and  yi  asv = 0  or  ∑i=12(kGFiβiλi/a(di + βiv)) − r = 0.The steady states of –8) sati sati8) syi−rv=0,∑i=12cyiz0.From , where  xi0 = λi/di,  i = 1,2. If  v ≠ 0, then we haveδi = βi/di, i = 1,2.  The solution of .If  ng it in  leads to we have∑i=12kGFin write (R011+δ1v+n system –8) has  has v = hen from  we haveit into (v∗=kGbrc.rc.From  and 20)v = 0, then from   for any  u ∈ {xi, yi, v, z, i = 1,2}. We also define a function  H : →[0, ∞)  as  H(u) = u − 1 − lnu. It is clear that  H(u) ≥ 0  for any  u > 0  and  H  has the global minimum  H(1) = 0.In this section, we establish the global stability of the three steady states of system –8) empl empl8) eR0 ≤ 1, then  E0  is GAS.If  W0  as follows:γi = kGFi/a,  i = 1,2.Define a Lyapunov functional  W0  along the trajectories of  = 0. Since  yi ≥ 0  for  i = 1,2, then  y1 = y2 = 0. Hence  dW0/dt = 0  if and only if  xi = xi0,  yi = 0,  i = 1,2,  v = 0, and  z = 0. From LaSalle's Invariance Principle,  E0  is GAS.The time derivative of  ories of –8) sati satiW0 ries of  we  we (39)Wz.From , we, we56)We obtaindW2dt=∑i=e obtaindW2dt=∑i=.3.1  in , the sol6We obtai obtain5dW2dt=∑i=hen from  we have that is,yi∗xi(t−τer, from  we have+ T cells and macrophages, taking into account the CTL immune response. Two types of distributed time delays have been incorporated into the model to describe the time needed for infection of target cell and virus replication. The global stability of the three steady states of the model has been established by constructing suitable Lyapunov functionals and using LaSalle's Invariant Principle. We have proven thatR0 ≤ 1, then the uninfected steady state  E0  is GAS;if  R0 > 1 ≥ R0*, then the infected steady state without CTL immune response  E1  is GAS;if  R0 > R0* > 1, then the infected steady state with CTL immune response  E2  is GAS.if  In this paper, we have proposed an HIV infection model describing the interaction of the HIV with two classes of target cells, CD4


The purpose of this paper is to introduce the concept of partial rectangular metric spaces as a generalization of rectangular metric and partial metric spaces. Some properties of partial rectangular metric spaces and some fixed point results for quasitype contraction in partial rectangular metric spaces are proved. Some examples are given to illustrate the observed results. In 1906, the famous French mathematician Fréchet  introducIn this paper, we generalize the concept of rectangular metric space and extend the concept of partial metric space by introducing the partial rectangular metric space. A fixed point theorem for quasitype contraction is also proved in the partial rectangular metric space which generalizes several known results in metric, partial metric, and rectangular metric spaces. Results are illustrated by some examples.First, we recall some definitions from partial metric and rectangular metric spaces see , 3)..3]).X is a mapping p : X × X → ℝ such that, for all x, y, z ∈ X,(P1)p ≥ 0;(P2)x = y if and only if p = p = p;(P3)p ≤ p;(P4)p = p;(P5)p ≤ p + p − p.A partial metric space is a pair  such that X is a nonempty set and p is a partial metric on X.A partial metric on a nonempty set X be a nonempty set and let d : X × X → ℝ be a mapping such that(R1)d, for all x, y ∈ X and d = 0, if and only if x = y;0 ≤ (R2)d = d, for all x, y ∈ X;(R3)d ≤ d + d + d, for all x, y ∈ X and for all distinct points w, z ∈ X − {x, y} (rectangular property).Then, d is called a rectangular metric on X, and  is called a rectangular metric space. A sequence {xn} in X is called convergent and converges to x ∈ X, if, for every ɛ > 0, there exists n0 ∈ ℕ such that d < ɛ for all n > n0. Sequence {xn} is called a Cauchy sequence if, for every ɛ > 0, there exists n0 ∈ ℕ such that d < ɛ, for all n, m > n0. A rectangular metric space  is called complete, if every Cauchy sequence in X converges in X.Let In this section, we define partial rectangular metric spaces and prove some properties of partial rectangular metric spaces.X be a nonempty set and let ρ : X × X → ℝ be a mapping such thatρ1) ≥ 0, for all x, y ∈ X;ρ2) = ρ = ρ, for all x, y ∈ X;ρ3) ≤ ρ, for all x, y ∈ X;ρ4) = ρ, for all x, y ∈ X;ρ5) ≤ ρ + ρ + ρ − ρ − ρ, for all x, y ∈ X and for all distinct points w, z ∈ X∖{x, y}.Then, ρ is called a partial rectangular metric on X and the pair  is called a partial rectangular metric space.Let X, ρ), if x, y ∈ X and ρ = 0, then x = y but the converse may not be true.In a partial rectangular metric space  is a partial rectangular metric space, but it is not a rectangular metric space, because ρ ≠ 0, for all x > 0. Also,  is not a partial metric space because it lacks the property (P5). Indeed,Let Next proposition shows that every partial rectangular metric space induces a rectangular metric space.X, ρ), the pair  is a rectangular metric space, whereFor each partial rectangular metric space  and (R2). For (R3), let x, y, z, w ∈ X, where w and z are distinct and w, z ∈ X∖{x, y}; then we haveX, ρr) is a rectangular metric space.By definition of X, ρr) is called the induced rectangular metric space, and ρr is the induced rectangular metric. In further discussion until specified,  will represent induced rectangular metric space.Here,  be a rectangular metric space and there exists x0 ∈ X such that d ≤ d, for all distinct x, y ∈ X. Then, the pair  is a partial rectangular metric space, whereρ is the rectangular metric d.Let (ρ1) and (ρ4) follow by the definition of ρ. For (ρ2), note that ρ = d, for all x ∈ X; therefore, if x, y ∈ X and ρ = ρ = ρ, we have (1/2)[d + d + d] = d = d, which implies that d = 0; that is, x = y.The properties  follows immediately. For (ρ5), let x, y, w, z ∈ X,  x ≠ y,  w,  z ∈ X∖{x, y}. Then, by definition of ρ, we haveX, ρ) is a partial rectangular metric space. Now, by the definition, it is easy to verify that d is the rectangular metric induced by ρ.By the choice of X = {1/n : n ∈ ℕ} ∪ {0} and define d : X × X → ℝ byX, d) is a rectangular metric space. Note that d ≤ d for all distinct x, y ∈ X and so ρ : X × X → ℝ defined byX,   is a partial rectangular metric space, and d is the rectangular metric induced by ρ.Let The proof of following proposition is straightforward and, by using it, one can obtain some more examples of partial rectangular metric space.X, d) and constant α ≥ 0, the pair  is a partial rectangular metric space, whereFor any rectangular metric space  be a partial rectangular metric space, {xn} a sequence in X, and x ∈ X. Then,(i)xn} is said to be convergent and converges to x ∈ X, if limn→∞ρ = ρ;the sequence {(ii)xn} is said to be Cauchy in , if limn,m→∞ρ exists and is finite;the sequence {(iii)X, ρ) is said to be a complete partial rectangular metric space, if, for every Cauchy sequence {xn} in X, there exists x ∈ X such that is a partial rectangular metric space. Consider the sequence {xn} in X, where xn = 1/n. Then, limn→∞ρ = limn→∞[1/n + α] = α = ρ and limn→∞ρ = limn→∞[1/n + α] = α = ρ. Therefore, {xn} has two limits, namely, 0 and 2.Let X, ρ) be a partial rectangular metric space and let {xn} be a sequence in X. Then, the sequence {xn} converges in  and converges to x ∈ X; that is, limn→∞ρr = 0, if and only if limn→∞ρ = limn→∞ρ = ρ.Let  be a partial rectangular metric space and let {xn} be a sequence in X. Then, the sequence {xn} is a Cauchy sequence in  if and only if it is a Cauchy sequence in .Let ; that is,ρ − ρ|≤ρr, for all n, m ∈ ℕ; therefore, the sequence {ρ} is a Cauchy sequence in ℝ+ and so we have limn→∞ρ ∈ ℝ+. Therefore, it follows from  ∈ ℝ+.Let {that is,limn,m→∞xn} is a Cauchy sequence in , that is, limn,m→∞ρ ∈ ℝ+, then again limn→∞ρ ∈ ℝ+ and by definition limn,m→∞ρr = 0.Conversely, if {The proof of the following lemma follows from Lemmas A partial rectangular metric space is complete, if and only if its induced rectangular metric space is complete.In this section, some fixed point theorems in partial rectangular metric spaces are proved.X, ρ) be a partial rectangular metric space and let T : X → X be a mapping. For A ⊂ X, we denote the diameter of A by δ[A] andA = {x} ⊂ X, then δ[A] need not be zero, but δ[A] = ρ.Let  be a complete partial rectangular metric space and let T : X → X be a mapping. Then, T is called a quasicontraction on X with constant λ, if it satisfies the following property:x, y ∈ X, where λ ∈ . Furthermore there exists a positive integer k such that k ∈ {1,2,…, n} and δ[O] = ρ.Let  and so, by , we have ρ < δ[O] and so, by the definition of orbit of T and the diameter δ of a set, we must have δ[O] = ρ, where k ∈ {1,2,…, n}.Suppose d so, by , we haveThe next lemma shows that the orbit of a quasicontraction is necessarily bounded.x ∈ X.Suppose that all the conditions of x ∈ X be arbitrary. Note that {δ[O]} is a nondecreasing sequence of real numbers, son = 1. If n = 2, then from (ρ3) and , we must have ρ ≤ max{ρ, ρ}. Also, since ρ ≤ ρ, for all y, z ∈ X, therefore ](ρ3) and , we havehow thatδ[O]n ≥ 3, then, by δ[O] = ρ, where k ∈ {1,2,…, n}. If k = 1 or k = 2, we have finished. Suppose k ≥ 3. If Tx = x or T2x = x or T2x = Tx, then we have Tkx ∈ {x, Tx} and  ≤ max{ρ, ρ}, therefore, from (ρ5), ] = ρ ≤ λδ[O] ≤ λδ[O] = λρ; therefore, it follows from the above inequality thatSuppose  Tx} and  holds. Sx, then  holds. Tom (ρ5), , 21), a, an ≥ 3,In the next theorem, existence and uniqueness of fixed point of quasicontraction in complete partial rectangular metric spaces are proved.X, ρ) be a complete partial rectangular metric space and let T : X → X be a quasicontraction on X with constant λ. Then, T has a unique fixed point u ∈ X and ρ = 0.Let  > 0. Then, from  = 0; that is, u = v. Thus, if fixed point of T exists, then it is unique. Further, if u ∈ X is a fixed point of T and ρ > 0, then, from  = 0. So, if u is a fixed point of T, then ρ = 0.Let us first show that, if fixed point of en, from , we haveen, from , we haveT. For arbitrary x ∈ X, we will show that the iterative sequence {Tnx} is a Cauchy sequence. Let m, n ∈ ℕ with m > n. Then, using k ∈ {1,2,…, m − n + 1} such that δ[O] = ρ; therefore, using λ ∈ [0,1), it follows from , there exists u ∈ X such thatu is the fixed point of T. Suppose ρ > 0. Without loss of generality, we can assume that Tnx ≠ Tn+1x, for all n ∈ ℕ, also, there exists n0 ∈ ℕ such that Tnx ∉ {u, Tu}, for all n > n0. Therefore, it follows from (ρ5) thatn, from  ≤ ρ, then, from  = 0. If ρ ≤ ρ, then, for sufficiently large n, we haveρ = 0. Therefore, we must have Tu = u. Thus, u is the unique fixed point of T.Now, we prove existence of fixed point of her with ρ thatρ≤ρn, from  thatρ≤ρn, from  is a complete partial rectangular metric space. Since, for all x ∈ X, x > 0,  ρ = x > 0, therefore  is not a rectangular metric space. Also,  is not a partial metric space because it lacks the property (P5). Indeed,T : X → X byT is a quasicontraction with constant λ ∈ [11/14, 1). All the conditions of T has a unique fixed point u = 0. Note that the rectangular metric induced by ρ is given byT is not a quasicontraction with respect to ρr. Indeed, for x = 4, y = 5, we have ρr = 5 and ρr = 5, ρr = 5, ρr = 5, ρr = 0, and  ρr = 5. Therefore, there exists no λ ∈ [0,1) such thatx, y ∈ X. Thus, T is not a quasicontraction in the induced rectangular metric space.Let T is a Banach contraction in the partial rectangular metric space ; that is, ρ ≤ λρ, for all x, y ∈ X with λ ∈ [9/10, 1), while it is not even a quasicontraction in the induced rectangular metric space . Also  is not a partial metric space; therefore, the results of Branciari [Note that, in the above example, the mapping ranciari  and Mattranciari  are not The following corollaries generalize the Banach, Kannan, Reich, Chatterjea, and Hardy-Rogers fixed point results  in partX, ρ) be a complete partial rectangular metric space and let T : X → X be a mapping. Suppose that the following condition is satisfied:λ ∈ [0,1). Then, T has a unique fixed point u ∈ X and ρ = 0.Let  be a complete partial rectangular metric space and let T : X → X be a mapping. Suppose that the following condition is satisfied:x, y ∈ X, where α ∈ [0, 1/2). Then, T has a unique fixed point u ∈ X and ρ = 0.Let  be a complete partial rectangular metric space and let T : X → X be a mapping. Suppose that the following condition is satisfied:x, y ∈ X, where α, β, and γ are nonnegative constants such that α + β + γ < 1. Then, T has a unique fixed point u ∈ X and ρ = 0.Let  be a complete partial rectangular metric space and let T : X → X be a mapping. Suppose that the following condition is satisfied:x, y ∈ X, where α ∈ [0, 1/2). Then, T has a unique fixed point u ∈ X and ρ = 0.Let  be a complete partial rectangular metric space and let T : X → X be a mapping. Suppose that the following condition is satisfied:x, y ∈ X, where α, β, γ, μ, and λ are nonnegative constants such that α + β + γ + μ + λ < 1. Then, T has a unique fixed point u ∈ X and ρ = 0.Let  be a complete partial rectangular metric space and let T : X → X be a mapping. Suppose that for some positive integer n, the following condition is satisfied:x, y ∈ X, where λ ∈ [0,1). Then, T has a unique fixed point u ∈ X and ρ = 0.Let  = 0. Now, TnTu = TTnu = Tu; therefore, Tu is another fixed point of Tn and, by uniqueness, we have Tu = u. Thus, u is a fixed point of T. Since every fixed point of T is also a fixed point of Tn, the fixed point of  T is unique.We note that ondition  of Theor


We build a Taylor's expansion for composite functions. Some applications are introduced, where the proposed technique allows the authors to obtain an asymptotic expansion of high order in many small parameters of solutions. So, first in However, for given in , it is vΩ be an open subset of ℝn  and 0 ∈ Ω. Let  g =  ∈ CN+1  and  f ∈ CN+1.Seek the representation formula for fog, such that for ||x|| being small enough,dα, |α| ≤ N are calculated from the values of the given functions f; g1,…, gp  and of their derivatives at a suitable point.Let Next in α =  ∈ ℤ+n and x =  ∈ ℝn, we putWe use the following notations. For a multi-index The following lemma is useful to solve the Problems m, N ∈ ℕ and aα ∈ ℝ, α ∈ ℤ+n, 1 ≤ |α| ≤ N. ThenTNm), α ∈ ℤ+n, 1 ≤ |α| ≤ N are defined by the recurrent formulasLet The proof of f  around the point g =  ∈ ℝp, we obtain thatNow, using Taylor's expansion of the function gi as follows:Similarly, we use Maclaurin's expansion of Substituting  into 1212, we gea = σi)( = (σβi), σβi)( = (1/β!)Dβgi(0), 1 ≤ |β| ≤ N, it implies thatApplying Hence,Consequently,Clearly, the f = f ∈ CN+1, for each  fixed  ∈  × ℝ+; gi ∈ CN+1, i = 1,2, 3, for each  fixed  ∈  × ℝ+ : gi ∈ CN+1, i = 4,5, 6, for each  fixed t ∈ ℝ+, we now investigate the  As an application of the method used in x, t) ∈  × ℝ+, using Taylor's expansion of the function f around the point g =  ∈ ℝ3 × ℝ+3 up to order N, we obtain thatH =  ∈ ℝ6, ||H|| is small enough, Diαif = ∂f/∂gi, i = 1,2,…, 6.For each  fixed  ∈  × ℝ+: we have the following.For each fixed (gi(ε) = gi, i = 1,2, 3.(i) The representation formula for g1(ε) as followsWe rewrite g2(ε), g3(ε); we writeIt is similar to t ∈ ℝ+: we have the following.For each  fixed gi(ε) = gi, i = 4,5, 6.(ii) The representation formula for WriteSimilarly,Then  , 24), , , 28), a, a28), aWe also need the following lemma.α =  ∈ ℤ+n, |α | ≥ 1, thenFor all We haveRj = O(||ε||N+1), j = 4,5, 6, soNote that Combining  and 36)36) yieldSubstituting  into 2020, we obuα ∈ W = {v ∈ L∞) : vt ∈ L∞)}, α ∈ ℤ+n, |α| ≤ N, hence uα, uαx, uαt ∈ L∞) ⊂ L∞(QT). Then from uα ∈ W =n = 1.(i) This result improves the one in , for n =F = μ, where μ ∈ CN+1, n = p, this result is also obtained in [(ii) In case of ained in .


A high-order finite difference scheme is proposed for solving time fractional heat equations. The time fractional derivative is described in the Riemann-Liouville sense. In the proposed scheme a new second-order discretization, which is based on Crank-Nicholson method, is applied for the time fractional part and fourth-order accuracy compact approximation is applied for the second-order space derivative. The spectral stability and the Fourier stability analysis of the difference scheme are shown. Finally a detailed numerical analysis, including tables, figures, and error comparison, is given to demonstrate the theoretical results and high accuracy of the proposed scheme. In the last decades, more and more attention has been placed on the development and research of fractional differential equations, because they can describe many phenomena, physical and chemical processes more accurately than classical integer order differential equations –4. And tO(τμ + h4), where 1 ≤ μ < 2. Here, we propose a method for the time fractional differential heat equations with the accuracy of order O(τ2 + h4).There are many different discretizations in time variable equipped with the compact difference scheme in spatial variable. The approximations given in –9 are ofMαu/∂tα denotes α-order modifying Riemann-Liouville fractional derivative , thenSuppose xi, and then we obtain the following truncation error for any i, where 1 ≤ i ≤ M − 1:We use the Taylor expansion of each term about κ to both side of (O(τ2 + h4);φj = T, φj0 = r(xj), φjk = f, 1 ≤ k ≤ N, 1 ≤ j ≤ M, and Uj = T.If we apply the operator  side of ,(8)κ∂MαAN+1)×(N+1)( and BN+1)×(N+1)×(N+1)( and β1 = ON+1)×1αj+αj (1 ≤ j ≤ M) of the scheme  denote the spectral radius of a matrix A, that is, the maximum of the absolute value of the eigenvalues of the matrix A.Let ρ(αj) < 1, (1 ≤ j ≤ M), by inductionWe will prove that α1 is a zero matrix ρ(α1) = 0 < 1.since α2 = −B−1A, and α2 is a lower triangular matrix of the following form:Moreover, ρ(α2) = ρ(−B−1A) = |(1/2h2 − w0/12)/(1/h2 + 10w0/12)| < |(1/2h2 − w0/12)/(1/2h2 + 9w0/12 + 1/2h2 + w0/12)|<|(1/2h2 − w0/12)/(1/2h2 + w0/12)|.Therefore h2 > 0, we can write ρ(α2) < 1.Sinceρ(αj) < 1. After some calculations we find thatAi,i = w0/12 − 1/2h2, Bi,i = 10w0/12 + 1/h2 and αji,i = λj, nonzero eigenvalue of αj for 2 ≤ i ≤ N + 1. h2 − w0/12 ≥ 0, then we have two subcases. (a)λj < 1,If 0 < (b)λj < 0,If −1 < If 1/2h2 − w0/12 < 0, then we have two subcases.(a)λj < 1,If 0 < (b)λj < 0,If −1 < Now, assume ρ(αj) < 1 then it follows that ρ(αj+1) < 1. So ρ(αj) < 1 for any j, where 1 ≤ j ≤ M.So, we have proved that whenever A ∈ ℝN×N, Am → 0 as m → ∞ if and only if ρ(A) ≤ 1. We note that if A is normal, then ||A|| = ρ(A) but when the matrix A is not normal the spectral radius gives no indication of the magnitude of the roundoff error for finite M. In this case a condition of the form ρ(A) ≤ 1 guarantees eventual decay of the errors, but does not control the intermediate growth of the errors. Then, it is easy to understand that ρ(A) ≤ 1 is a necessary condition for stability but not always sufficient.It is well known that for any Ujk be the approximate solution and define ρjk = ujk − Ujk, k = 0,1,…, N − 1, j = 1,…, M − 1. Then, we write ρjk = dkeijhβ and obtain the following roundoff error equation for 112ρk(x) can be expanded in a Fourier series as follows:dk(l) = 1/L∫0Lρk(x)ei2πlx/L− and we introduce the following norm:0L|ρk(x)|2dx = ∑l=−∞∞|dk(l)|2, we obtainWe now define the grid functions:ρjk = dkeijhβ, where β = 2πl/L and L = 1. Substituting the above expression into  = (t4 + 1)x5(1 − x). The solution by the proposed scheme is given in Exact solution of this problem is The errors in It can be concluded from In this work, the compact difference scheme was successfully applied to solve the time fractional heat equations. The second order approximation for the Riemann-Liouville fractional derivative is equipped with the higher order compact difference schemes. The Fourier analysis and the spectral stability method are used to show that the proposed scheme is unconditionally stable. Numerical results are in good agreement with the theoretical results.


Lipα and Lip  classes are the particular cases ofLip(ξ(t), p) class. The main result of this paper generalizes some well-known results in this direction.Approximation theory is a very important field which has various applications in pure and applied mathematics. The present study deals with a new theorem on the approximation of functions of Lipschitz class by using Euler's mean of conjugate series of Fourier series. In this paper, the degree of approximation by using Euler's means of conjugate of functions belonging toLip( The Fourier series associated with  f  at the point  x  is given bysn. The conjugate series of .Let  given byf(x)~12a0f ∈Lipα  ifA function  f ∈Lip,  p > 1  consider that if(Definition  5.38  of Chandra ).ξ(t),  f ∈Lip(ξ(t), p),M  is a positive number independent of  x  and  t.Given a positive increasing function  ξ(t) = tα, thenLip(ξ(t), p)  coincides withLip. If  p → ∞  inLip, then it coincides withLipα.In case  L∞-norm of a function  f : R → R  is defined byLp-norm is defined byf : R → R  by a trigonometric polynomial  tn  of order  n  under sup norm || ||∞ is defined by (, then the degree of approximation of  f  by the Nörlund means of the Fourier series for f is given byTn  are the    means of Fourier series of  f.If  δ  and consider a function  f ∈Lipα,  0 < α ≤ 1. Working in same direction we prove the following theorem.Hölland et al.  have shof : R → R  is a  2π  periodic, Lebesgue integrable and belonging toLip(ξ(t), p)  for > 1  and ifx, then degree of approximation off ∈Lip{ξ(t), p}, by Euler  meanIf    and if{∫01/n(t|(ξ(1n)),{∫1/nπ < 1  when  0 < x < 1. Therefore,We havekth  partial sum of the conjugate series of the Fourier series  means, we getThe  r series  is givenf ∈Lip(ξ(t), p)⇒ψ ∈Lip(ξ(t), p).Clearly,ψ(t) ∈Lip(ξ(t), p), condition (t ≥ (2t/π), lemma, and second mean value theorem for integrals, we haveψ(t) ∈Lip(ξ(t), p), and condition  = tα, then the degree of approximation of a functionf ∈Lip,  1/p < α < 1, by Euler's means    of the conjugate series of the Fourier series (If  r series  is givenp → ∞  in α < 1,If


We extend the Exp-function method to fractional partial differential equations in the sense of modified Riemann-Liouville derivative based on nonlinear fractional complex transformation. For illustrating the validity of this method, we apply it to the space-time fractional Fokas equation and the nonlinear fractional Sharma-Tasso-Olver (STO) equation. As a result, some new exact solutions for them are successfully established. F-expansion method [G′/G)-expansion method [Nonlinear differential equations of integer order (NLDEs) can be used to describe many nonlinear phenomena such as fluid mechanics, plasma physics, optical fibers, biology, solid state physics, chemical kinematics, and chemical physics. In the research of the theory of NLDEs, searching for more explicit exact solutions to NLDEs is one of the most fundamental and significant studies in recent years. With the help of computerized symbolic computation, much work has focused on the various extensions and applications of the known algebraic methods to construct the solutions to NLDEs. There have been a variety of powerful methods. For example, these methods include the generalized Riccati subequation method , 2, the n method , the Expn method –9, and tn method , 11.Fractional differential equations are generalizations of classical differential equations of integer order and have recently proved to be valuable tools to the modeling of many physical phenomena and have been the focus of many studies due to their frequent appearance in various applications in physics, biology, engineering, signal processing, systems identification, control theory, finance, and fractional dynamics. In order to obtain exact solutions for fractional differential equations, many powerful and efficient methods have been proposed so far . Butξ = ξ, a given fractional partial differential equation expressed in independent variables t, x1, x2,…, xn can be turned into another ordinary differential equation of integer order, whose solutions are supposed to have one of the following forms:m1, m2, n1, and n2 are unknown positive integers that will be determined by using the balance method. Determine the highest order nonlinear term and the linear term of highest order in the ordinary differential equation and express them in terms of Dtαf follows –24, 26):In the following, we will apply the Exp-function method to find exact solutions for the space-time fractional Fokas equation and the nonlinear fractional Sharma-Tasso-Olver (STO) equation.q = U(ξ), where ξ = (k1x1α/Γ(1 + α)) + (k2x2α/Γ(1 + α)) + (l1y1α/Γ(1 + α)) + (l2y2α/Γ(1 + α)) + (ctα/Γ(1 + α)) + ξ0, k1, k2, l1, l2, c, ξ0 are all constants with k1, k2, l1, l2, c ≠ 0. Then by the use of 4∂2αq<α≤1.In , the aut4∂2αq<α≤1e use of , (6) canm2, n2, we need to balance the linear term of the highest order with the highest order nonlinear term in  and UU′′, we have 15n2 + m2 = 14n2 + 2m2, which implies m2 = n2. Similarly, balancing the lowest order Exp-function in U(4) and UU′′, we have 15n1 + m1 = 14n1 + 2m1, which implies m1 = n1. For simplicity, we will proceed in two selected cases.First we suppose that the solution of  can be e term in . By simpm1 = n1 = 1, m2 = n2 = 1. Thenepξ together, equating each coefficient to zero, yield a set of algebraic equations. Solving these equations with the aid of the mathematical software Maple yields the following family of values of ai, bi, i = −1,0, 1. Let  1. ThenU(ξ)=a−1etituting , eliminaFamily  1. Considerb1, b0 are arbitrary constants.ξ = (k1x1α/Γ(1 + α)) + (k2x2α/Γ(1 + α)) + (l1y1α/Γ(1 + α)) + (l2y2α/Γ(1 + α)) + (ctα/Γ(1 + α)) + ξ0. If we especially take b0 = 2b1, then we obtain the following hyperbolic function solitary wave solution:Substituting the result above into , we can m1 = n1 = m, m2 = n2 = m, and for simplicity, we takeepnξ together, equating each coefficient to zero, yield a set of algebraic equations. Solving these equations yields the following result. Let  we takeU(ξ)=a−metituting , eliminaFamily  2. Considerbm, b0 are arbitrary constants.ξ = (k1x1α/Γ(1 + α)) + (k2x2α/Γ(1 + α)) + (l1y1α/Γ(1 + α)) + (l2y2α/Γ(1 + α)) + (ctα/Γ(1 + α)) + ξ0. If we especially take b0 = 2bm, then we obtain the following solitary wave solution:Substituting the result above into , we can m1 = n1 and m2 = n2. Similar to the above, we only consider the following two cases. Now we suppose that the solution of  can be e1U(ξ)=∑p=−m1 = n1 = 1, m2 = n2 = 1. Theneipξ together, equating each coefficient to zero, yield a set of algebraic equations. Solving these equations yields the following. Let  1. ThenU(ξ)=a−1etituting , eliminaFamily  3. Considerb1, b0 are arbitrary constants.ξ = (k1x1α/Γ(1 + α)) + (k2x2α/Γ(1 + α)) + (l1y1α/Γ(1 + α)) + (l2y2α/Γ(1 + α)) + (ctα/Γ(1 + α)) + ξ0. If we especially take b0 = 2b1, then we obtain the following trigonometric function solution:Substituting the result above into , we can eipnξ together, equating each coefficient to zero, yield a set of algebraic equations. Solving these equations, we obtain another family of values of ai, bi, i = −m, 0, m. LetU(ξ)=a−metituting , eliminaFamily  4. Considerξ = (k1x1α/Γ(1 + α)) + (k2x2α/Γ(1 + α)) + (l1y1α/Γ(1 + α)) + (l2y2α/Γ(1 + α)) + (ctα/Γ(1 + α)) + ξ0. If we especially take b0 = 2bm, then we obtain the following solitary wave solution:ove into :(25)U7 = U(ξ), where ξ = kx + (ctα/Γ(1 + α)) + ξ0, k, c, ξ0 are all constants with k, c ≠ 0. Then by use of  equation with time-fractional derivative , 27: caU′′′ and U2U′, we have m2 = n2, while we obtain m1 = n1 by balancing the lowest order Exp-function in U′′′ and U2U′. Similar to the solving process for the space-time fractional Fokas equation, we will also proceed the computation under two selected cases. First we suppose that the solution of  can be em1 = n1 = 1, m2 = n2 = 1. Thenepξ together, equating each coefficient to zero, yield a set of algebraic equations. Solving these equations yields two families of values for ai, bi, i = −1,0, 1.Let  1. ThenU(ξ)=a−1etituting , eliminaFamily  1. Considerb1, b−1 are arbitrary constants. Family  2. Considerb1, b0, b−1 are arbitrary constants.ξ = kx + (ctα/Γ(1 + α)) + ξ0. If we especially take c = −ak3, b1 = b−1 or c = −ak3, b1 = −b−1 in ==a−metituting , eliminaFamily  3. Considerbm, bm− are arbitrary constants. Family  4. Considerbm, b0, bm− are arbitrary constants.ξ = kx + (ctα/Γ(1 + α)) + ξ0. If we especially take c = −am2k3, bm = bm− or c = −am2k3, bm = −bm− in  ==a−1etituting , eliminaFamily  5. Considerb1, b−1 are arbitrary constants. Family  6. Considerb1, b0, b−1 are arbitrary constants.Substituting the results above into , we can c = −ak3, b1 = b−1 or c = −ak3, b1 = −b−1 in =a−metituting , eliminaFamily  7. Considerbm, bm− are arbitrary constants. Family  8. Considerbm, b0, bm− are arbitrary constants.c = −am2k3, bm = bm− or c = −am2k3, bm = −bm− in −1,U14(ξ) = re rem = 2,tions by , (54) artions by , 34), , ,  ar arm = 2, We have extended the Exp-function method to solve fractional partial differential equations successfully. As applications, some generalized and new exact solutions for the space-time fractional Fokas equation and the nonlinear fractional Sharma-Tasso-Olver (STO) equation have been successfully found. As one can see, this method is based on the homogeneous balancing principle. So it can also be applied to other fractional partial differential equations where the homogeneous balancing principle is satisfied.


AbstractPseudaspidapion botanicum sp. n. from China is described and figured. Its host plant is Grewia biloba G. Don var. parviflora (Bunge) Hand.-Mazz . The genus Harpapion Voss, 1966 is recorded as new for China and Vietnam and two comb. n. are proposed: Harpapion vietnamense  (from Aspidapion) and Harpapion coelebs  (from Pseudaspidapion). A key to the known species of the genus Pseudaspidapion from China is presented. Pseudaspidapion relating it to the genus Aspidapion Schilsky, 1901. Aspidapiini PageBreakAlonso-Zarazaga, 1990 of the subfamily Apioninae Schoenherr, 1823 . This genus is quite similar to Aspidapion, but it can be distinguished from the latter by relatively equal width of elytral striae and interstriae, the presence of fenestrae in the tegminal plate, and the absence of protibial mucrones in males, among other features , the Museo Nacional de Ciencias Naturales, Madrid (MNCN), the Zoological Institute of Russian Academy of Sciences, Moscow (ZIN), the Museum of Natural History, University of Wrocław (MNHW) and the Beijing Botanical Garden, Beijing (BBG).Type specimens were obtained from ZIN on loan or belong to IZCAS and their data are summarised in Descriptions were made and photographs were taken with a CCD Qimagine MicroPublisher 5.0 RTV mounted on a Zeiss SteREO Discovery V.12. Extended focus images were generated with Auto-Montage Pro 5.03.0061 and edited with Adobe Photoshop CS 5.0 if required. Microscopic slides were studied under a Leica DM 2500 microscope and photos were taken with a Nikon CoolPix 5400. The map was made with the software ArcGIS 9.3. Drawings were made from the original photographs by using the software Adobe Illustrator CS5.0, or directly by using a drawing tube linked to the microscope.Nomenclature of the rostral parts follows The dissecting method used follows After description, the genitalia and other parts of each specimen were placed in DMHF on a plastic card for long term conservation .PageBreakLabels are described as they are (in Chinese), with pinyin romanization or comments in square brackets; labels are separated by semicolons and lines by slashes.Alonso-Zarazaga & Wang sp. n.urn:lsid:zoobank.org:act:88656E61-918F-4F1B-A3BD-6A3FA714A237http://species-id.net/wiki/Pseudaspidapion_botanicumPseudaspidapion yunnanicum, but it can be distinguished from the latter by the characters in the  This new species resembles Measurements (in mm): Standard length: 1.82. Rostrum: length: 0.71, maximum width: 0.17. Pronotum: median length: 0.49, maximum width: 0.59. Elytra: median length: 1.38, maximum width: 1.08. . Pseudaspidapion as described in With the general characters of genus Integument. Generally piceous black  semicircular, 0.69× as long as wide, apical edge of arc with a flat and glabrous side, punctate medially with interspace microreticulate . Ninth sternite  Y-shaped and slightly winged near arm base, manubrium ca. 2.00× as long as arms  (n=10): Standard length: 1.38–2.05 (mean= 1.735). Rostrum: length: 0.52–0.82 (mean= 0.722), maximum width: 0.13–0.19 (mean= 0.164). Pronotum: median length: 0.36–0.53 (mean= 0.479), maximum width: 0.44–0.68 (mean= 0.594). Elytra: median length: 1.06–1.54 (mean= 1.406), maximum width: 0.78–1.18 (mean= 1.038). Rostrum 1.44–1.58× as long as pronotum, 4.00–4.63× as long as wide. Ventrite 5 sometimes slightly subtriangular. Otherwise as in holotype.Female paratypes , Haidian district, Beijing (P. R. of China). Paratypes (132♂137♀): 4♂7♀: : 北京三堡 [Běijīng Sānpù]; : 1964.VII.20 / leg. 马文珍 [Mă Wénzhēn]; : IOZ(E)1638668-1638678; 3♀: PageBreak: 北京三堡 [Běijīng Sānpù] / 600m; : 1964.VIII.21 / leg. 李铁生 [Lí Tiéshēn]; : IOZ(E)1638679-1638681; 1♂1♀: : 北京三堡  / 1979.VIII.8; : leg. 廖素柏 [Liào Sùbăi]; : IOZ(E)1638682-1638683; 5♂5♀: : 北京上方山 [Běijīng Shàngfāngshān]; : 1980.VIII.2 / leg. 廖素柏 [Liào Sùbăi];  IOZ(E)1638684-1638693; 2♂: : 北京上方山 [Běijīng Shàngfāngshān]; : 1979.VII.25 / leg. 陈元清 [Chén Yuánqīng]; : IOZ(E)1638694-1638695; 2♂2♀: : 北京居庸关 [Běijīng Jūyōngguān] / 500m; : 1964.VIII.20 / leg. 李铁生 [Lí Tiéshēn]; : IOZ(E)1638696-1638699; 1♂: : 北京八达岭 [Běijīng Bādálīng] / 700m; : 1963.VII.25 / leg. 李铁生 [Lí Tiéshēn]; : IOZ(E)1638700; 8♂8♀: : 北京香山 [Běijīng Xiāngshān]; : 1963.V.30 / leg. 李铁生 [Lí Tiéshēn]; : IOZ(E)1638701-1638716; 1♀: : 北京香山 [Běijīng Xiāngshān]; : 1964.V.5 / leg. 马文珍 [Mă Wénzhēn]; : IOZ(E)1638717; 1♀: : 北京香山 [Běijīng Xiāngshān] / 1957.VIII.16; : IOZ(E)1638718; 1♂: : 北京卧佛寺 [Běijīng Wòfósì] / 50m; : 1962.VI.25 / leg. 王春光 [Wáng Chūnguāng]; : IOZ(E)1638719; 1♀: : 北京卧佛寺 [Běijīng Wòfósì]; : 1963.IX.3 / leg. 姜胜巧 [Jiāng Shèngqiáo]; : IOZ(E)1638720; 1♀: : 北京卧佛寺 [Běijīng Wòfósì] / 50m; : 1962.VIII.31 / leg. 谢汝忠 [Xieruzhong]; : IOZ(E)1638721; 1♂: : 北京潭柘寺 [Běijīng Tánzhèsì] / 1975.VII.24; : leg. 王书永 [Wang Shuyong]; : IOZ(E)1638722; 1♀: : 1964.VII.20 / leg. 马文珍 PageBreakPageBreak[Mă Wénzhēn]; : IOZ(E)1638723; 2♂3♀: : 北京圆明园 [Běijīng Yuánmíngyuán] / 1980. VII.7 / leg. 姜胜巧 [Jiāng Shèngqiáo]; : IOZ(E)1638724-1638728; 2♂10♀: : 北京门头沟军庄西杨坨村 [Běijīng Méntóugōu Jūnzhuāng Xīyángtún] / 2008.VII.27 / leg. 王志良 [Wáng Zhìliáng]; : 寄主: 小花扁担木 [Jìzhū: Xiăohuābiăndànmù] / Grewia biloba G. Don var. parviflora (Bunge) Hand.-Mazz; : IOZ(E)1638729-1638740; 5♂8♀: : 北京怀柔九渡河镇怀九河 [Běijīng Huáiróu Jiūdùhézhèn Huáijiūhé] / 2008.VI.15 / leg. 王志良 [Wáng Zhìliáng]; : IOZ(E)1638741-1638753; 1♀: : 北京怀柔三渡河 [Běijīng Huáiróu Sāndùhé] / 2008.V.24 / leg. 王志良 [Wáng Zhìliáng];  IOZ(E)1638754; 46♂24♀: : 北京香山植物园樱桃沟 [Běijīng Xiāngshān Zhíwùyuán Yīngtáogōu] / 2008.VIII.19 / leg. 王志良 [Wáng Zhìliáng]; : IOZ(E)1638755-1638770, IOZ(E)1638784-1638790, IOZ(E)1638804-1638850; 28♂41♀: : 北京植物园樱桃沟 [Běijīng zhíwùyuán Yīngtáogōu] / 寄主：小花扁担木 [Jìzhū: Xiăohuābiăndànmù] / 2008.VI.1 / Leg. 王志良 [Wáng Zhìliáng]; : IOZ(E)1638851-1638900, IOZ(E)1638911, IOZ(E)1638928-1638945; 7♂7♀: : 北京海淀百望山 [Běijīng Hăidiàn Băiwàngshān] / 2009.VI.21 / leg. 杨干燕 [Yáng Gànyàn]; : IOZ(E)1639912-1638918, IOZ(E)16398920-1638926; 2♂1♀: : 陕西华阴县孟塬 [Shaănxi Huáyīn Mèngyuán] / 450m / 1972.VIII.9; : IOZ(E)1639562-1639563, IOZ(E)1639540, deposited in IZCAS.Holotype: ♂: : 北京植物园樱桃沟 [Běijīng zhíwùyuán Yīngtáogōu] / 2008.VI.1 / Leg. 王志良 [Wáng Zhìliáng]; : IOZ(E)1638667, deposited in IZCAS. This is the Beijing Botanical Garden at Yingtaogou : 北京香山植物园樱桃沟 [Běijīng Xiāngshān Zhíwùyuán Yīngtáogōu] / 2008.VIII.19 / leg. 王志良 [Wáng Zhìliáng]; : IOZ(E)1638781, IOZ(E)1638801, IOZ(E)1638771-1638779, IOZ(E)1638791-1638799, to be deposited in MNCN.1♂1♀: : 北京香山植物园樱桃沟 [Běijīng Xiāngshān Zhíwùyuán Yīngtáogōu] / 2008.VIII.19 / leg. 王志良 [Wáng Zhìliáng]; : IOZ(E)1638780, IOZ(E)1638800, to be deposited in BBG.1♂1♀: : 北京香山植物园樱桃沟 [Běijīng Xiāngshān Zhíwùyuán Yīngtáogōu] / 2008.VIII.19 / leg. 王志良 [Wáng Zhìliáng]; : IOZ(E)1638783, IOZ(E)1638803, to be deposited in ZINM.1♂1♀: : 北京香山植物园樱桃沟 [Běijīng Xiāngshān Zhíwùyuán Yīngtáogōu] / 2008.VIII.19 / leg. 王志良 [Wáng Zhìliáng]; : IOZ(E)1638782, IOZ(E)1638802, to be deposited in MNHW.The male holotype has not been dissected to avoid any damage, since there are many specimens collected with it. The new species is named after the first locality where it was found during a collecting visit: the Beijing Botanical Garden. It is a Latin adjective. The species is known for the moment only from the municipality of Beijing and the province of Shaanxi.Pseudaspidapion botanicum sp. n. was collected from Grewia biloba G. Don var. parviflora (Bunge) Hand.-Mazz   , which i to July , this bePseudaspidapion and Aspidapion in China in the recent Palaearctic catalogue  of Aspidapion vietnamense, which has been previously dissected and whose pygidium and tegmen were not conserved, shows several differences from Aspidapion, namely, the metatibial mucros are evidently elongate and knicked at their apices (subdentate at the outer margin), the rostrum is clearly dilated at the antennal insertion and distinctly constricted apicad, the setae on the front margin of pronotum are parallel to it, and the apex of the penis is distinctly curved in lateral view, dorsally dentate near the apex. All these characters suggest it is a Harpapion Voss, 1966. After a study of the type species of this genus, Harpapion considerandum , we consider that, even in the absence of these diagnostic parts, it can be unmistakeably considered a member of this genus and consequently is here transferred: Harpapion vietnamense (Korotyaev), comb. n. The same can be said about Pseudaspidapion coelebs , which show also the characters of the latter genus, and is here formally transferred as well: Harpapion coelebs , comb. n. This genus will be the subject of a forthcoming paper.During this research, we were able to study types of several of the species placed in the genera atalogue  , from Altay prefecture in northern Xinjiang PageBreak(China). Apart from the species Aspidapion panfilovi, whose placement may need a revision as mentioned above, these specimens represent the easternmost distribution record of the otherwise Western Palaearctic genus Aspidapion. Therefore, the distribution pattern of Pseudaspidapion is mainly Palaeotropical but has expanded to Philippines . The two genera can still be distinctly separated from each other. Also, we suppose that the blank distributional area of Pseudaspidapion between Shaanxi and South China represents a lack of collection records rather than an actual distributional range limit, because the vegetation between these two points includes most of the range of Grewia biloba  and many other congeneric species of the host genus.Previously, the genus c region . Thus thlippines  and to NPseudaspidapion rufopiceum , a related species, which is included below from the data in its original description.PageBreakSome of these species are described only from one sex, which makes it difficult to key all of the Chinese species. However, we have tried to provide a key to the known Chinese species (see below). Consequently, more field work is needed to gather specimens of both sexes, and complete our knowledge. We have not studied specimens of Aspidapion panfilovi Korotyaev, 1985 , which seems to be related to these species. Body length excludes rostrum.This key includes also


This paper deals with the study of a generalized function of Mittag-Leffler type. Various properties including usual differentiation and integration, Euler(Beta) transforms, Laplace transforms, Whittaker transforms, generalized hypergeometric series form with their several special cases are obtained and relationship with Wright hypergeometric function and Laguerre polynomials is also established.33C45, 47G20, 26A33. In 1903, the Swedish mathematician Gosta Mittag-Leffler  was studied by Wiman n is the Pochhammer symbol (Rainville (where (ainville ) α) > 0,Re(β) > 0,Re(γ) > 0 and In 2007, Shukla and Prajapati , Re(β), Re(γ), Re(δ)) > 0where In this paper a new definition of generalized Mittag-Leffler function is investigated and defined as where Further the generalization of definition (1.7) is investigated and defined as follows where, and The definition (1.9) is a generalization of all above functions defined by (1.1)-(1.7).μ = ν, ρ = σ, it reduces to Eq.  will be the result.Setting s to Eq.  defined s to Eq. , moreoveThroughout this investigation, we need the following well-known facts to study the various properties and relation formulas of the function f(z) is defined as Beta(Euler) transforms (Sneddon ) of the f(z) is defined as Laplace transforms (Sneddon ) of the f(z) is defined as Mellin- transforms of the function and the inverse Mellin-transform is given by Whittaker transform (Whittaker and Watson ) 1.16Wλ,μ(t) is the Whittaker confluent hypergeometric function.where The generalized hypergeometric function (Rainville ) is defiWright generalized hypergeometric function (Srivastava and Manocha ) is defiFox’s H-function (Saigo and Kilbas ) is giveGeneralized Laguerre polynomials (Rainville ). These γ and is defined by Incomplete Gamma function (Rainville ). This iAs a consequence of definitions (1.1)-(1.9) the following results hold:α) > 0, Re(β) > 0, Re(γ) > 0, Re(δ) > 0, Re(μ) > 0, Re(ν) > 0, Re(ρ) > 0, Re(σ) > 0 and p, q > 0 andq ≤ Re(α) + p, thenIf In particular, which is (2.1.1).The proof of (2.1.2) can easily be followed from the definition (1.9). Now which proves (2.1.3). □μ = ν, ρ = σand p = 1 in (2.1.1) immediately leads to (2.1.4).Substituting μ = ν, ρ = σand p = 1 in (2.1.2) immediately leads to (2.1.5).Substituting μ = ν,ρ = σand p = 1 in (2.1.3) immediately leads to (2.1.6).Putting α) > 0, Re(β) > 0, Re(γ) > 0, Re(δ) > 0, Re(μ) > 0, Re(ν) > 0, Re(ρ) > 0, Re(σ) > 0, Re(w) > 0; andq ≤ Re(α) + p then for If In particular, From (1.9), which is the proof of (2.2.1).Again using (1.9) and term by term differentiation under the sign summation(which is possible in accordance with the uniform convergence of the series (1.9) in any compact set which is the proof of (2.2.2). □μ = ρ,ν = σ, in (2.2.1), we get (2.2.3).Setting μ = ρ, ν = σ, in (2.2.2), we get (2.2.4).Setting α + p), thenIf which proves (2.3.1). □μ = ν, ρ = σ, δ = p = 1, (2.3.1) reduces to the known result of Shukla and Prajapati Shukla and Prajapati (For rajapati  (2.3.1).μ = ν, ρ = σand p = 1 in (2.3.1), we get (2.3.2).Setting Special Properties: Setting putting μ = ν, ρ = σand p = q = δ = 1 in (2.3.1), we haveβ = γ = δ = q = 1 in (2.3.2), we have For If In particular, which proves (2.4.1).s to Now change the variable from which proves (2.4.2).Now which proves (2.4.3).q = δ = 1 and γ = q = δ = 1 in (2.4.1) and (2.4.3) yields (2.4.4) and (2.4.5) respectively. □Putting Using (1.9) with Δ is a l-tupple Δ is a q-tupple Δ is a k-tupple where q+l+1Fk+p+m: (i) If q + l + 1 ≤ k + p + m, the function q+l+1Fk+p+m converges for all finite z. (ii) If q + l + 1 = k + p + m + 1, the function q+l+1Fk+p+m converges for |z| < 1 and diverges for |z| > 1(iii) If q + l + 1 > k + p + m + 1, the function q+1+1Fk+p+m+1 is divergent for |z| ≠ 0(iv) If q + l + 1 = k + p + m + 1, the function q+l+1Fk+p+m+1 is absolutely convergent on the circle for |z| = 1, if Convergence criterion of generalized Mittag-leffler function In this section we discuss some useful integral transforms like Euler transform, laplace transform and Whittaker transform of Mellin-Barnes integral representation of q < Re(α) + p. Then the function Let (1.9) and (1.10) be satified and z)| < 1; the contour of integration beginning at −i∞ and ending at +i∞, and indented to separate the poles of the integrand at where | arg as the sum of the residues at the poles which completes the proof. □μ = ρ, ν = σand p = 1, we get the Melin Barne’s integral of the function Setting (Mellin transform) ofFrom Theorem 4.1, we have where is in the form of inverse Mellin-Transform (1.15). So applying the Mellin-transform (1.14) yields directly the required result. □(Euler(Beta)transform) offrom which the result follows. □Special properties: (i) For q = 1, (4.3.2) reduces to Tariq OSalim . 4.(ii) For δ = q = 1 in (4.3.2), we have a = β, α = σ, then (4.3.2) reduces to If α = β = γ = δ = q = 1 in (4.3.2), we have Putting (Laplace transform)from which the result follows. □q = 1, (4.4.2) reduces to Tariq O Salim .(Whittaker transform)ϕt = v in L.H.S. of Theorem 4.5, we have Substituting from which the result follows. □Special properties :(i) Putting q = δ = 1 in (4.5.2), we have (ii) For q = γ = δ = 1 in (4.5.2), we have (iii) Now putting q = β = α = γ = δ = 1 in (4.5.2), we have If the condition (1.10) be satisfied, then (1.9) can be written as Using (4.1.1), we have from α = k, β = μ + 1, γ = − m, q ∈ N with q|m and replacing z by zk in (1.6), we get Putting et alwhere zk.Note that Further for which is the required relationship.


Some new Hermite-Hadamard type inequalities for differentiable convex functions were presented by Xi and Qi. In this paper, we present new generalizations on the Xi-Qi inequalities. Below we recall some Hermite-Hadamard type inequalities.Let f′(x)| is convex on , thenIn 1998, Dragomir and Agarwal  showed tf′(x)|p/(p−1) is convex on  with p > 1, thenand (ii) if |f′(x)|q is convex on  with q ≥ 1, thenIn 2000, Pearce and Pečarić  showed tf′(x)|p/(p−1) is convex on  with p > 1, thenIn 2004, Kirmaci  showed tf′ ∈ L and |f′(x)|q is convex on  with q ≥ 1, thenIn 2010, Sarikaya et al.  showed tλ, μ ∈  and if f′ ∈ L and |f′(x)|q is convex on  with q ≥ 1, thenIn 2012, Xi and Qi  showed tMoreover, for other results involving the Hermite-Hadamard type inequalities, we also refer to –23.In this paper, we generalize the Xi-Qi inequalities.λ,  μ ∈ ℝ and let f : I⊆ℝ → ℝ be differentiable on I° and {a, b}⊆I with a < b. Assume that f′ ∈ L and 0 < ϵ < b − a. ThenLet Integrating by part and changing variable, we haveThus,s > 0 and 0 ≤ ξ ≤ 1. ThenLet λ,  μ ∈  and let f : I⊆ℝ → ℝ be differentiable on I° and {a, b}⊆I with a < b. Assume that f′ ∈ L and 0 < ϵ < b − a. If |f′(x)|q is convex on  with q ≥ 1, thenLet f′(x)|q is convex on  with q ≥ 1. By Suppose that |Case  (q = 1). By the convexity of |f′(x)| and Case  (q > 1). By Hölder's inequality, we havef′(x)|q and By the convexity of |ϵ = (b − a)/2 in It is easy to notice that if we put λ,  μ ∈  and let f : I⊆ℝ → ℝ be differentiable on I° and {a, b}⊆I with a < b. Assume that f′ ∈ L. If |f′(x)|q is convex on  with q ≥ 1, thenLet ϵ = (b − a)/3 in One can easily check that if we put λ,  μ ∈  and let f : I⊆ℝ → ℝ be differentiable on I° and {a, b}⊆I with a < b. Assume that f′ ∈ L. If |f′(x)|q is convex on  with q ≥ 1, thenLet ϵ = 2(b − a)/3 in One can easily check that if we put λ,  μ ∈  and let f : I⊆ℝ → ℝ be differentiable on I° and {a, b}⊆I with a < b. Assume that f′ ∈ L. If |f′(x)|q is convex on  with q ≥ 1, thenLet λ = μ = 0 in It is easy to notice that if we put f : I⊆ℝ → ℝ be differentiable on I° and {a, b}⊆I with a < b. Assume that f′ ∈ L and 0 < ϵ < b − a. If |f′(x)|q is convex on  with q ≥ 1, thenLet λ = μ = 1/2 in It is easy to notice that if we put f : I⊆ℝ → ℝ be differentiable on I° and {a, b}⊆I with a < b. Assume that f′ ∈ L   and 0 < ϵ < b − a. If |f′(x)|q is convex on  with q ≥ 1, thenLet λ = μ = 1 in It is easy to notice that if we put f : I⊆ℝ → ℝ be differentiable on I° and {a, b}⊆I with a < b. Assume that f′ ∈ L and 0 < ϵ < b − a. If |f′(x)|q is convex on  with q ≥ 1, thenLet λ,  μ ∈  and let f : I⊆ℝ → ℝ be differentiable on I° and {a, b}⊆I with a < b. Assume that f′ ∈ L and 0 < ϵ < b − a. If |f′(x)|q is convex on  with q ≥ 1, thenLet f′(x)|q is convex on  with q ≥ 1. If q = 1, then, by Suppose that |q > 1. By Next, we suppose that f′(x)|q and By the convexity of |This proof is completed.ϵ = (b − a)/2 in It is easy to notice that if we put λ,  μ ∈  and let f : I⊆ℝ → ℝ be differentiable on I° and {a, b}⊆I with a < b. Assume that f′ ∈ L. If |f′(x)|q is convex on  with q ≥ 1, thenLet ϵ = (b − a)/3 in One can easily check that if we put λ,  μ ∈  and let f : I⊆ℝ → ℝ be differentiable on I° and {a, b}⊆I with a < b. Assume that f′ ∈ L. If |f′(x)|q is convex on  with q ≥ 1, thenLet ϵ = 2(b − a)/3 in One can easily check that if we put λ,  μ ∈  and let f : I⊆ℝ → ℝ be differentiable on I° and {a, b}⊆I with a < b. Assume that f′ ∈ L  . If |f′(x)|q is convex on  with q ≥ 1, thenLet λ = μ = 0 in It is easy to notice that if we put f : I⊆ℝ → ℝ be differentiable on I° and {a, b}⊆I with a < b. Assume that f′ ∈ L and 0 < ϵ < b − a. If |f′(x)|q is convex on  with q ≥ 1, thenLet λ = μ = 1/2 in It is easy to notice that if we put f : I⊆ℝ → ℝ be differentiable on I° and {a, b}⊆I with a < b. Assume that f′ ∈ L and 0 < ϵ < b − a. If |f′(x)|q is convex on  with q ≥ 1, thenLet λ = μ = 1 in It is easy to notice that if we put f : I⊆ℝ → ℝ be differentiable on I° and {a, b}⊆I with a < b. Assume that f′ ∈ L and 0 < ϵ < b − a. If |f′(x)|q is convex on  with q ≥ 1, thenLet s, q}⊆[1, ∞) and {a, b, wa, wb}⊆ with a < b. Let 0 < ϵ < b − a.In this section, we suppose that {a, b} with weight {wa, wb} is defined byThe weighted arithmetic mean of data {a, b} with weight {wa, wb} is defined byThe weighted geometric mean of data {a, b} is defined byThe generalized logarithmic mean of data {a, b} is defined byThe identric mean of data {f(x) = xs on , we get the following:Applying f(x) = xs on , we get the following:Applying f(x) = lnx on , we get the following:Applying f(x) = lnx on , we get the following:Applying f(x) = xs on , we get the following:Applying f(x) = xs on , we get the following:Applying f(x) = lnx on , we get the following:Applying f(x) = lnx on , we get the following:Applying


ℒg(z) = (1 − z)−1∫z1f(ζ)dζ on a class of “mixed norm” spaces of analytic functions on the unit disk, X = Hα,νp,q, defined by the requirement g ∈ X⇔r ↦ (1 − r)αMp ∈ Lq), where 1 ≤ p ≤ ∞, 0 < q ≤ ∞, α > 0, and ν is a nonnegative integer. This class contains Besov spaces, weighted Bergman spaces, Dirichlet type spaces, Hardy-Sobolev spaces, and so forth. The expression ℒg need not be defined for g analytic in the unit disk, even for g ∈ X. A sufficient, but not necessary, condition is that p, q, α, and ν for which 1°ℒ is well defined on X, 2°ℒ acts from X to X, 3° the implication We consider the action of the operator  H(𝔻) denote the class of all functions holomorphic in the unit disk 𝔻 of the complex plane. In ).ℒ acts as a bounded operator from Ap,β into Ap,β if and only if p > β + 2.Theorem C (see ). ℒ actsIf  for all n, and , then ℒg ∈ 𝒟01 if and only ifTheorem D (see ). If g^ℒg is defined on the space in the sense of ρ ∈ .Here “acts” means, among other things, that g ∈ H1 and therefore for g ∈ Hp ∪ 𝒟01, p ≥ 1, because Hp ⊂ H1 and 𝒟01 ⊂ H1. However, .Let 1/2 ≤ p = ∞.See the proof of X satisfies . Besides H1, we have, for instance, the following.On the other hand, if a space atisfies  for all The space 𝒟01 is contained in H1and satisfies , but ℒ does not map 𝒟01 into H1.Theorem F (see ). The spOne of the aims of the present paper is to extend Theorems A, B, C, Hαp,q , where α > 0 when q < ∞, and α ≥ 0 when q = ∞, the class of those g ∈ H(𝔻) for whichν be a nonnegative integer, we define the spaceHα,νp,q is given byj=0−1 should be interpreted as equal to zero.We denote by It is well known and easy to prove that these spaces are complete.p = q, ν = 0, a more (but not most) natural norm is given byThe norm  is not tgiven by  because Hα,νp,∞ = :Hα,νp is specific in that the set, 𝒫, of all analytic polynomials is not dense in it. The closure of 𝒫 in Hα,νp coincides with the “little oh” spaceThe space From now on, unless specified otherwise, we suppose thatIt is sometimes more convenient to work with the Besov type spaces.β ∈ ℝ and choose any integer ν ≥ 0 such that ν − β > 0. The space 𝔅βp,q is defined asq = ∞, we can assume that ν = β. It is well known that this definition is independent of ν; this follows immediately from Lemma A below. If β > 0, then 𝔅βp,q is “true” Besov spaces; if β < 0, then 𝔅βp,q = Hβ−p,q; if β = 0, then it is called Hardy-Bloch spaces , and, in particular, ℒ maps the ordinary Lipschitz space into itself.This is, maybe, a new result.ℒ does not act as a bounded operator from Aβ1 into Aβ1, for any β > −1.In particular, β = αp − 1; that is, α = (β + 1)/p.This is seen from ℒ maps the Dirichlet space 𝒟βp into itself if and only if p > 1 + β/2. In particular, ℒ maps 𝒟β1 if and only if −1 < β < 0.(a) ℒ maps 𝒟β2 into itself if and only if β < 2.Another case: (b) These facts are, maybe, new.ℒ maps Hαp,q into itself if and only if α < 1 and p > 1/(1 − α).(a) ν = 0. This is related to a result of . In the case q ≥ 1, a direct proof can be found in . Fix n and letσkP are -means of P. It follows thatσkP||1 ≤ ||P||1. Now we use Lagrange's inequality in the formP asLet f ∈ Hβ,νq1,, where β = ν. In this case, by Lemma A, the function f belongs to Hβ,νq1, if and only iff is in Hβ,νq1,, because ɛq > 1.As noted above it is enough to prove that Thus, we have proved the implication (a)⇒(b) of It is clear that (b) implies (a). We have already noted that (c) implies (b). The following assertion shows that (d) implies (c).q ≤ 1,  1 ≤ p ≤ 2,  κp,α,ν = ν + 1 − α − 1/p = 0, and g ∈ Hα,νp,q, thenIf Hα,νq1, ⊂ Hα,ν1,1, it is enough to consider the case q = 1. LetPn∗Vn = Vn, which follows from . We usesee thatLemma  A p ≤ ∞,  0 < q ≤ ∞, and κp,α,ν < 0;1 ≤ p ≤ 2,1 < q ≤ ∞,κp,α,ν = 0;1 ≤ p ≤ ∞,0 < q ≤ ∞,κp,α,ν = 0. 2 < It remains to be proven that (a) implies (d); that is, that (a) does not hold in the following cases:q < ∞.Case (1) is part of q > 1).The following assertion proves the desired result in Cases (2) and .If 1 ≤ Cq is independent of ɛ. This inequality remains true if fɛ,1 is replaced by fɛ,ρ with ρ < 1. The function fɛ,ρ(ρ < 1) belongs to fɛ,ρ : ɛ > 0, ρ < 1} is bounded in Hα,νp,q. On the the other hand,(a) Letq ≤ 1 and p > 2 is more delicate and depends on Khinchin's inequality and a deep result of Khinchin and Kolmogorov  = sign(sin(2nπt)), n ≥ 0, 0 ≤ t ≤ 1. If {cn} is a sequence in ℂ such that ∑n=0∞ | cn|2 = ∞, then the series ∑n=0∞cnRn(t) diverges for almost all t ∈  and moreover the sequence of its partial sums is unbounded a.e. Theorem H.   We also need the following theorem of Khinchin  such that |E | = 1, which follows from Theorem H. (We can assume that E does not contain points where Rn(t) = 0 because the set of such points is denumerable). On the other hand, by gτ ∈ Hα,νp,q for at least one τ ∈ E.Consider first the case 2 < sn|| ≤ C||gτ|| < ∞ , where C is independent of n. On the other hand, as noted before, the sequencep < ∞.To complete the proof we consider the polynomials  ≤ 1 − 3p/2, while this implies p ≤ 2, a contradiction which shows that g ∉ Hαp,∞, for p > 2.If For technical reasons, we introduce the spaceIn the proof of  For any sequence {ck}k=mn there is a polynomial h(z) = ∑k=mnbkzk such that |bk | ≥|ck| andwhere C is an absolute constant.Theorem J.κp,α,ν > 0. If (1) 1 < q ≤ ∞, 2 < p ≤ ∞, and ν − α ≤ −1/2, or (2) 0 < q ≤ 1, 2 < p ≤ ∞, and ν − α < −1/2, then Hν,αp,q⊈ℓ−11.Let We haveCase (1),  (p < ∞). Let q = ∞ and letck = (k + 1)ξ. Thenξ + 1/2 = α − ν; that is, ξ = α − ν − 1/2 ≥ 0.We have, by εn ∈ {−1,1} such that the functionHα,νp,∞. On the other hand,This implies that there is a sequence q < ∞, we consider the functionIf 1 < Case (1),  (p = ∞). Choose {ck} as above and consider the function h(z) = ∑k=0∞bkzk, where |bk | ≥ck and ||Δnh||∞ ≤ C(∑k∈Inck2)1/2 . Finally we use the inequalityξ < α − ν − 1/2, and repeat the above reasoning to complete the proof.In Case (2) we choose ν − α = −1/2, 0 < q ≤ 1, and 2 < p ≤ ∞, then Hα,νp,q ⊂ ℓ−11.If We havep ≤ ∞, κp,α,ν > 0, 0 < q ≤ ∞, and ν − α > −1/2, then Hα,νp,q ⊂ ℓ−11.If 2 < g ∈ Hα,νp,q. Then g ∈ Hα,νp,∞, and hence g ∈ Hα,ν∞2,. It follows that ||Δng||2 ≤ c2n(α−ν). On the other hand,α − ν < 0. This proves the proposition.Let The following proposition completes the proof of p ≤ 2, κp,α,ν > 0, and 0 < q ≤ ∞, then Hα,νp,q ⊂ ℓ−11.If 1 ≤ g ∈ Hα,νp,q. Choose β so that κp,α,ν = κβ,ν2,; that is, α = β + 1/2 − 1/p. Then, by Hα,νp,q ⊂ Hβ,νq2,. This implies that Hα,νp,q ⊂ Hβ,ν∞2, which means that ||Δng||2 ≤ C2n(β−ν), and soLet kp,α,ν = ν − α + 1 − 1/p > 0; that is, ℒ acts as an operator from Hα,νp,q into Hα,νp,q(α > 0). We want to analyze the proof of p ≤ ∞ and κp,α,ν > 0, so that ℒ maps Hα,νp,q into Hα,νp,q by In this section we suppose that q = ∞ and ν − α ≤ −1/2. Then (i) there exists a function g ∈ Hα,νp,q such that ξ = α − ν − 1/2 ≥ 0. (ii) The exponent ξ is best possible. (iii) If ξ > 0, then there is an η > 0 and a function g ∈ Hα,νp,q such that Let ξ is best possible we use g ∈ Hα,νp,∞ if and only if ||Δng||p ≤ C2n(α−ν), which implies that ||Δng||2 ≤ C2α−ν)n follows from the proof of (iii) This follows from (i).q < ∞ and ν − α ≤ −1/2. Then (i) there exists a function g ∈ Hα,νp,q such that g ∈ Hα,νp,q such that ξ > 0 and 0 < η < ξ, then there exists a function g such that Let 1 < Hα,νp,q ⊂ hα,νp, which implies that ||Δng||2 = o(2n(α−ν)). If o(1), which is impossible.Assertion (i) is part of the proof of In a similar way one proves the following.q ≤ 1 and ξ = α − ν − 1/2 > 0. Then (i) if 0 < η < ξ, then there is a function g ∈ Hα,νp,q such that η cannot be replaced by ξ.Let Combining the above propositions we get the following.ℒ map Hα,νp,q into itself, and p > 2. Then the following statements are equivalent: g ∈ Hα,νp,q such that η > 0;there is a function ν − α < −1/2.  Moreover, if ν − α < −1/2 and η ∈  is arbitrary, then there is g ∈ Hα,νp,q such that q = ∞, we can take η = α − ν − 1/2.Let


Notions of frontier and semifrontier in intuitionistic fuzzy topology have been studied and several of their properties, characterizations, and examples established. Many counter-examples have been presented to point divergences between the IF topology and its classical form. The paper also presents an open problem and one of its weaker forms. IFS is a sufficiently generalized notion to include both fuzzy sets and vague sets. Fuzzy sets are IFSs but the converse is not necessarily true  and α + β ≤ 1. An intuitionistic fuzzy point (IFP for short) xα,β)( ∈ A if α ≤ μA(x) and β ≥ γA(x). Clearly an intuitionistic fuzzy point can be represented by an ordered pair of fuzzy points as follows:X is denoted as IFP(X).Let B = 〈y, μB(y), γB(y)〉 is an IFS in Y, then the preimage of B under f, denoted by f−1(B), is the IFS in X defined byA = 〈x, μA(x), γA(x)〉 is an IFS in X, then the image of A under f, denoted by f(A), is the IFS in Y defined byIf The concept of fuzzy topological space, first introduced by Chang in , was genX is a family of IFSs in X satisfying the following axioms:G1⋂G2 ∈ τ for any G1, G2 ∈ τ,Gi ∈ τ for any arbitrary family {Gi : i ∈ J}⊆τ.⋃An intuitionistic fuzzy topology (IFT for short) on a nonempty set X, τ) is called an intuitionistic fuzzy topological space  and members of τ are called intuitionistic fuzzy open  sets. The complement A is called an intuitionistic fuzzy closed (IFC) set in X. Collection of all IFO  sets in IFTS X is denoted as IFO(X) ).In this case, the pair , then A1⋃A2 ∈ IFC(X),If 𝒜 ⊂ IFC(X), then ⋂𝒜 ∈ IFC(X).Let X, τ) be an IFTS and A = 〈μA, γA〉 an IFS in X. Then the fuzzy interior and fuzzy closure of A are denoted and defined asLet  be an IFTS and A, B be IFSs in X. Then the following properties hold:A⊆A, (A⊆Cl A),Int A = Int A, (Cl Cl A = Cl A),Int Int A⊆B⇒Int A⊆Int B, (A⊆B⇒Cl A⊆Cl B),A⋂B) = Int A⋂Int B(Cl (A⋃B) = Cl A⋃Cl B),Int (A⋃B)⊇Int A⋃Int B, (Cl (A⋂B)⊆Cl A⋂Cl B).Int (Let (X be an IFTS and let A ∈ IFS(X). Then xλ,μ)( ∈ IFP(X) is called an intuitionistic fuzzy frontier point  of A if A is called an IF frontier of A and denoted by Fr A. It is clear that Let A ∈ IFS(X), A⋃Fr A ⊂ Cl A. However, the inclusion cannot be replaced by an equality.For each A in an IFTS X, the following hold:A is IFC then Fr A⊆A,If A is IFO then If For an IFS (1)  A⊆A, if A is IFC set in X.(2) A is IFO implies (3) (4) Converse of (2) and (3) of X, τ) be the IFTS defined by Çoker (Example 3.3 d)⊆Cl f(f−1(B))⊆Cl B = B or f([f−1(B)]d)⊆B gives [f−1(B)]d⊆f−1(f[f−1(B)]d)⊆f−1(B) or [f−1(B)]d⊆f−1(B) implies f−1(B) is IF closed in X. Thus, f is IF continuous.(2)⇒(1) Suppose f : X → Y be an IF continuous mapping. Then Fr f−1(B)⊆f−1(Fr B), for any IFS B in Y.Let f is IF continuous. Let B be an IFS in Y. Thenf−1(B)⊆f−1(Fr B).Suppose that A⊆B and B ∈ IFCS(X). Then Fr A⊆B.Let A⊆B implies Cl A⊆Cl B, we have Since X, τ) and  be two IFTSs and f : X → Y, a function. Then f is said to be fuzzy open if the image of each IFS in τ is in ϕ.Let (f : X → Y be an IFO mapping and B an IFS in Y. Then f−1(Fr B)⊆Fr f−1(B).Let f is IFO and B is an IFS in Y. PutA is IF open and therefore f(A) is IF open in Y. This gives f−1(Fr B)⊆Fr f−1(B).Suppose n Y. PutA=Fr f−1 is called an intuitionistic fuzzy semiopen set (IFSOS) ifA is called an intuitionistic fuzzy semiclosed set if the complement of A is an IFSOS.An IFS A in an IFTS  are denoted and defined asThe semiclosure and semi-interior of an IFS A in IFTS X, the following hold:sCl A = A⋃Int Cl A,sInt A = A⋂Cl Int A.For an IFS A be an IFS in X. From Int (Cl (A⋃Int Cl A)) = Int (Cl A⋃Cl Int Cl A) = Int Cl A⊆A⋃Int Cl A, it follows that A⋃Int Cl A is an IFSC set. Hence, sCl A⊆A⋃Int Cl A. Since sCl A is IFSC, we have Int Cl A⊆Int Cl sCl A⊆sCl A. Thus A⋃Int Cl A⊆sCl A.(1)  Let (2) This can be proved in a similar manner as (1).A be an IFS in IFTS X. Then the intuitionistic fuzzy semifrontier of A is defined as sFr A is an IFSC set.Let In the following theorems, we note that almost all the properties related to intuitionistic fuzzy semi-interior, intuitionistic fuzzy semi-closure and intuitionistic fuzzy semifrontier are analogous to their counterparts in Intuitionistic Fuzzy Topology, and hence proofs of most of them are not given.A and B in an IFTS X, one hassCl  sCl A = sCl A,sInt  sInt A = sInt A,sInt (A⋃B)⊇sInt A⋃sInt B,sInt (A⋂B) = sInt A⋂sInt B,sCl (A⋃B) = sCl A⋃sCl B,sCl (A⋂B)⊆sCl A⋂sCl B.For IFSs sInt A and sInt B are both IFSO sets and A⊆A⋃B, B⊆A⋃B implies sInt A⊆sInt (A⋃B) and sInt B⊆sInt (A ⋃ B). In Combination, sInt A⋃sInt B⊆sInt (A⋃B) or(5) A⋂B⊆A and A⋂B⊆B imply sInt (A⋂B)⊆sInt A⋂sInt B. Conversely, sInt A⊆A and sInt B⊆B imply sInt A⋂sInt B⊆A⋂B and sInt A⋂sInt B is IFSO. But sInt (A⋂B) is the largest IFSO set contained in A⋂B; hence sInt A⋂sInt B⊆sInt (A⋂B). This gives the equality.(6) (7) This follows easily from (2).A⋂B⊆A, A⋂B⊆B(8) Since In the following theorem, (1)–(5) are analogues of A in IFTS X, the following hold:A is IFSC, then sFr A⊆A,if A is IFSO, then if A⊆B and B ∈ IFSC(X) ). Then sFr A⊆B  ) denotes the class of intuitionistic fuzzy semi-closed (resp. intuitionistic fuzzy semiopen) sets in X,let sFr A⊆Fr A,sCl sFr A⊆Fr A.For an IFS sCl A⊆Cl A and (6) Since (7) The converse of (2), (3), (6), and (7) of In Example 3.3 of , we chooThe following is an analogue of A be an IFS in IFTS X. Then one hassFr A = sCl A − sInt A,sFr sInt A⊆sFr A,sFr sCl A⊆sFr A,sInt A⊆A − sFr A. Let To show that (2), (3), and (4) of In general topology, the following hold:In Example 3.3 of , we chooA and B be IFSs in an IFTS X. Then sFr (A⋃B)⊆sFr A⋃sFr B.Let The converse of In Example 3.3 of , we chooHowever, we have the following theorem which is an analogue of A and B in IFTS X, one hasFor IFSs A and B in IFTS X, one hasFor IFSs The analogue of A in IFTS X, one hassFr sFr A⊆sFr A,sFr sFr sFr A⊆sFr sFr A.For an IFS As in the case of In Example 3.3 of , we chooA in IFTS X is called an intuitionistic fuzzy semi-Q-neighborhood of an IFP e if there exists an IFSO set B in X, such that eqB⊆A.An IFS e = xα,β)⊇A. Equivalently, xα,β), whenever e ∈ A. The union of all the semi-accumulation points of A is called the intuitionistic fuzzy semiderived set of A, denoted as Asd. It is evident that Asd⊆sCl A.An IFP A be an IFS in X, then sCl A = A⋃Asd.Let Ω = {e | e  is  a  semi-adherence  point  of  A}. Then from sCl A = ⋃Ω. On the other hand, e ∈ Ω is either e ∈ A or e ∉ A; for the latter case, by e ∈ Asd; hence sCl A = ⋃Ω ⊂ A⋃Asd. The reverse inclusion is obvious.Let A in an IFTS X, A is IFSC if Asd⊆A.For any IFS f : X → Y be a function from an IFTS X to another IFTS Y. Then f is said to be an intuitionistic fuzzy semicontinuous function if f−1(A) is IFSO in X for each IFO set A in Y.Let f : X → Y be a function. Then the following are equivalent:f : X → Y is intuitionistic fuzzy semi-continuous,f(Asd)⊆sCl f(A), for any IFS A in X.Let f is intuitionistic fuzzy semi-continuous. Let A be an IFS in X. Since sCl f(A) is IFC in Y, f−1(sCl f(A)) is IFSC in X. A⊆f−1(sCl f(A)) gives sCl A⊆sCl f−1(sCl f(A)) = f−1(sCl f(A)). Therefore, f(Asd)⊆f(sCl A)⊆ff−1(sCl f(A))⊆sCl f(A). Consequently, f(Asd)⊆sCl f(A).(1)⇒(2) Suppose that f(Asd)⊆sCl f(A). Letting B be any IFC set in Y, we show that f−1(B) is IFSC in X. By our hypothesis, f([f−1(B)]sd)⊆sCl f(f−1(B))⊆sCl B = B or f([f−1(B)]sd)⊆B gives [f−1(B)]sd⊆f−1(f[f−1(B)]sd)⊆f−1(B) or [f−1(B)]sd⊆f−1(B) implies f−1(B) is IFSC in X. Thus, f is intuitionistic fuzzy semi-continuous.(2)⇒(1) Suppose f : X → Y be a intuitionistic fuzzy semi-continuous function. Then one hasB in Y.Let f is intuitionistic fuzzy semi-continuous. Let B be an IFS in Y.Suppose that sFr f−1(B)⊆f−1(sFr B).Then


Relaxation oscillations of two-dimensional planar singular perturbed systems with a layer equation exhibiting canard cycles are studied. The canard cycles under consideration contain two turning points and two jump points. We suppose that there exist three parameters permitting generic breaking at both the turning points and the connecting fast orbit. The conditions of one  relaxation oscillation near the canard cycles are given by studying a map from the space of phase parameters to the space of breaking parameters. Let H(n) denote the maximum number of limit cycles of a general planar polynomial vector field of degree n. As mentioned in [H(n), but there are many results on the lower bounds of H(n); for example, H(3) ≥ 13, H(4) ≥ 21, and H(5) ≥ 28  and g are two smooth functions with respect to variables  ∈ ℝ2 × ℝk and ɛ is a small real number. As ɛ > 0 and small, we make the time scaling t = τ/ɛ and get the following equivalent standard form of slow-fast system which has the same phase portraits as the one of system ,,ɛ = 0 inn system  and 4)ɛ = 0 iny with the slow variable x acting as a parameter.System  and 4)  4) are, f = 0 defines the critical manifold S of the equilibrium of the layer equation  =  of system , and rescaling of time, system  of system (h1 and h2 are smooth and a(λ0) = 0 and a(λ) is smooth at λ0 in case of a generic turning point.At contact points that are isolated points on the critical manifold where normal hyperbolicity is lost, the blow-up method pioneered by Dumortier and Roussarie  is a powf system  have bee, system  can, locɛ approaches 0 and the limit cycles that are close to slow-fast cycle are called relaxation oscillations. In 2007, Dumortier et al. [f of degree 7 bifurcate from the canard cycle consisting of two jump points by analyzing the zeros of slow divergence integral of canard cycle  is near .Consider the following smooth slow-fast Liénard system Fa(x), g fulfill the following conditions.H1), as a = 0, the function F0(x) has five singular points: two Morse maximum points at p1, p4, two Morse minimum points at p2, p3, and one Morse maximum at p0 with −x0 < p1 < p2 < p0 < p3 < p4 < x0. The parameter a is just the difference a = Fa(p4) − Fa(p1). Let α0 = F0(0), α1 = F0(p2), α2 = F0(p3), α = max{α1, α2}, and β = F0(p4) = F0(p1). Then the values F0(±x0) are assumed to be below the minimum value min{α1, α2}.On the interval (−H2) = 0, g = 0, g = 0, g ≠ 0, and g ≠ 0, but ∂g/∂x ≠ 0, i = 2,3, ∂g/∂b ≠ 0 and ∂g/∂c ≠ 0, g > 0 for p1 < x < p2 or p0 < x < p3, g < 0 for p2 < x < p0 or p3 < x < p4 (see Suppose that  p4 see .H3), there exists a canard cycle containing four horizontal segments: one between the two Morse maxima x = p1, x = p4 denoted by σh, one below the left Morse maximum value and at the height y = u denoted by σl, one below the right Morse maximum value and at the height y = v denoted by σr(v), and one at the height y = w and between p2 and p3 denoted by σm, where u ∈ , v ∈ , w ∈ , and the corresponding canard cycle is denoted by Γuvw.When In the following, we assume that the smooth functions x1, x2 is defined as follows:An essential tool to study the limit cycles bifurcated from the canard cycle is the slow divergence integral see , 13–15);–15;13–15uvw of system (x1(u), x2(u), respectively, denote x-coordinates of intersection points between σl and slow curve y = F0(x), where −x0 < x2(u) < x1(u) < p2; let x1(w), x2(w), respectively, denote x-coordinates of intersection points between σm and y = F0(x), where p2 < x1(w) < x2(w) < p3; let x1(v), x2(v), respectively, denote x-coordinates of intersection points between σr and curve y = F0(x), where p3 < x1(v) < x2(v) < x0. By applying the slow divergence integral formula introduced in , J(v), K(w), L(v), M(u), and N(w), which are the slow divergence integrals of six slow curves contained in Γuvw. In detail, two of these curves are located on the left of p2 and their slow divergence integrals are functions of u, two of them are on the right of p3 and their slow divergence integrals are functions of v, and two of them are between p2 and p3 and their slow divergence integrals are functions of w , g fulfill the conditions H1, H2, H3. Let D = J(v) − I(u) + M(u) − N(w) + K(w) − L(v) denote the total slow divergence integral of Γuvw of system , v0 ∈ , and w0 ∈  such that D ≠ 0, then for ɛ > 0 and small enough  is a regular point of Φɛ, whose explicit expression will be given in uvw.If there exist u0 ∈ , v0 ∈ , and w0 ∈  such that I(u0) − J(v0) > 0, M(u0) − N(w0) = 0, and I(u0) − J(v0) = K(w0) − L(v0), then for ɛ > 0 and small enough there exits (u0(ɛ), v0(ɛ), w0(ɛ)) that is a generic fold singularity of Φɛ. A relaxation oscillation bifurcates from Γuvw and this semistable limit cycle is generically unfolded by the parameter  for ɛ > 0 and small enough, producing a pair of hyperbolic limit cycles of system , v0 ∈ , and w0 ∈  such that J(v0) − I(u0) = 0, M(u0) − N(w0) = 0, L(v0) − K(w0) = 0, and J′(v0)M′(u0)K′(w0) − I′(u0)L′(v0)N′(w0) ≠ 0, then for ɛ > 0 and small enough there exits ɛ. A codimension 2 relaxation oscillation bifurcates from Γuvw and this degenerated limit cycle is generically unfolded by the parameter  for ɛ > 0 and small enough, producing system , with z ∈ ℝp for some p ∈ ℕ, is called ɛ-regularly smooth in z, if f is continuous and all partial derivatives of f with respect to z exist and are continuous in .A function ɛ > 0 small enough, a limit cycle of system , J(v), K(w), L(v), M(u), and N(w) for ɛ = 0.For f system  cuts ∑lCi  , the transitions have the following expressions:l to C1: from ∑l to C2: from ∑m to C2: from ∑m to C3: from ∑r to C1: from ∑r to C3: from ∑where functions fl, gl, fr, gr, hm1, hm2 are ɛ-regularly smooth in ɛ-regularly smooth in From –15, due o(ε) and are ε-regularly smooth in By using the same analysis as , 15, we a, a, ɛ. So one can solve , a, ɛ-flat perturbations of the previous ones. We will continue to call them We can solve this system in an solve  to obtai term of  except tThe proof of ɛ small enough and we view Φɛ as a map from  to ɛ as follows:u, v, and w.We take u, v, w, ɛ) = det/∂). To find the singular points of map Φɛ, by direct computation we get the following formula about the determinant Δ:Let Δ = 0 is equivalent to the equationFor ɛ > 0 and small enough by noting ɛ-regularity of function ɛ is nondegenerated at the point , so from u0v0w0.It follows from  that theSo the conclusion of the first part of In this subsection, we give the proof of the second part of First, we present the following lemma.ɛ > 0 and small enough, if there exists P0(ɛ) = (u0(ɛ), v0(ɛ), w0(ɛ)) that satisfies Δ = 0, then it holds thatgrad Δ =  is the gradient of function Δ and k is nonzero function at (P0(ɛ), ɛ).As E = 0 determines a surface S1 in the neighborhood of point P0 for ɛ > 0 and small. From the fact that Δ = 0 is equivalent to E = 0, we get that the equation Δ = 0 also determines the surface S1. For grad Δ|P0(ɛ), grad E|P0(ɛ) are both the normal vectors of surface S1 at the point P0(ɛ), then we get , ψ, ϕ as follows:ɛ is given in u0, v0, w0) by P0, from assumptions in the second part of E = 0 and (∂E/∂u · ∂E/∂v)|P0,0) = 0. On surface u0(ɛ), v0(ɛ), w0(ɛ)) denoted by P0(ɛ) such that E(P0(ɛ), ɛ) = 0, P0(0) = P0; that is, Δ(P0(ɛ), ɛ) = 0.Denote (φ/∂v)|P0,0) from the first equation of (ϕ/∂w)|P0,0), From  = 0 we compute the second derivative of u asThen, as Δ(D/∂u)|P0 = M′(u0) − I′(u0) < 0, (∂D/∂v)|P0 = J′(v0) − L′(v0) > 0, and (∂D/∂w)|P0 = K′(w0) − N′(w0). So (∂E/∂u)|P0(ɛ),ɛ)( = (∂D/∂u)|P0 + O(ɛ) < 0, (∂E/∂v)|P0(ɛ),ɛ)( = (∂D/∂v)|P0 + O(ɛ) > 0, and (∂E/∂w)|P0(ɛ),ɛ)( = (∂D/∂w)|P0 + O(ɛ).From , we get P0(ɛ),ɛ)( = k((∂D/∂u)|P0 + O(ɛ), (∂D/∂v)|P0 + O(ɛ), (∂D/∂w)|P0 + O(ɛ)).From From , we get I(u0) − J(v0) > 0, M(u0) − N(w0) = 0, we get that ɛ → 0+ and ɛ → 0+.Noticing that ɛ > 0 and small enough we get thatThen for ϕwφv ≠ 0, dΔ/du)|P0(ɛ),ɛ))( ≠ 0. So we get that (u0(ɛ), v0(ɛ), w0(ɛ)) is a fold point of map Φɛ. So from uvw by unfolding the parameters a, b, c.Because ɛ) (see ), we getThe conclusion of the second part of In this subsection, we give the proof of the third part of ɛ > 0 and small enough, if there exists u, v, w, ɛ) = 0, dΔ/du = 0, and As u, v, w, ɛ) = 0 we getFrom the proof of the second part of dΔ/du = 0 is equivalent toFrom dΔ/du = 0 is equivalent to ɛ > 0 and small. That means that both equations dΔ/du = 0 and So rewrite , and we n we get .The proof of Proof of the Third Part of  From Lemmas E = 0, S1 : E = 0 and lɛ which passes through point . Therefore, for ɛ > 0 and small enough, we choose point lɛ. Then It is easy to check that ɛ given in (u at Consider the map Φgiven in . By applFrom From  and 26)26), we gɛ > 0 and small enough we get thatThen for ɛ > 0 and small enough. From a, b, c), for ɛ > 0 and small enough, producing system  = (−(1/6)x6 + (5/4)x4 − 2x2 + ax). Here we use symbols pi, i = 1,2, 3,4 defined in p1 = −2, p4 = 2; the values of F0(x) at these two points are both 4/3 and the values of F0(x) at the points p2 = −1, p3 = 1 are both −11/12 and F0(0) = 0.For this system, V ∈ , the equation F0(x) = V will have three roots on the right side of the y-axis. For the roots, the formula in the above equation contains complex number, so we choose the smallest one of the above three roots and denote it by the variable v, where v ∈ . By noting that relation between u and v is one to one relation, we express the other two roots in the form v ∈ . Using the same way, we take y = U, U ∈ ; then the equation F0(x) = U will have three roots on the left side of the y-axis; we choose the biggest one of the above three roots and denote it by the variable u; we express the other two roots in the form x2(u) < x1(u) < u, u ∈ . Also F0(x) = y is symmetrical about 0, so the straight line y = W, W ∈ , between −1 and 1 will intersect the curve with two points; then we, respectively, denote their x-coordinate by −w, w, where w ∈ . The corresponding canard cycle ΓUVW can be seen in For a given P(x) = (1/8)x8 − (3/2)x6 + 6x4 − 8x2, Q1(x) = −(1/9)x9 + (1/8)x8 + (9/7)x7 − (24/5)x5 − (3/2)x6 + 6x4 + (16/3)x3 − 8x2, Q2(x) = −(1/11)x11 + (3/10)x10 + (2/3)x9 − (13/4)x8 + (3/7)x7 + (21/2)x6 − (56/5)x5 − 6x4 + 16x3 − 8x2; then from the slow divergence integral formula (Let P(x) = 1/x8 − (3/M(u) − N(w) = 0; that is,Firstly, consider c1 = c2 = 0, from the fact that the function P(x) is monotone on the interval  we get c1 and c2 of obtained equation to equal zeros, we get w1(u) = 0, w2(u) = 0. Therefore, we getWhen  it into ; by settI(u) − J(v) = 0; that is,Secondly, consider P(x) is strictly monotone on the interval  and v = −u as c1 = c2 = 0.By noticing that the even function c1 and c2 of obtained equation to equal zeros, we getLet it into ; then byL(v) − K(w); that is,Thirdly, consider function Putting the expressions  and 37)37) into Qi(x), i = 1,2, we get that ζ1(u) · ζ2(u) ≠ 0, u ∈ .From the definition of function Denote θ(u) ≠ 0, u ∈ , then ζ1(u)/ζ2(u) = η1(u)/η2(u).For H(u) = L(v(u)) − K(w(u)), where w(u) and v(u) are, respectively, implicitly determined by M(u) − N(w) = 0 and I(u) − J(v) = 0.Denote By direct computations, we getOn the other hand from expressions , we get H(u) = 0, then we getAs η1(u)/η2(u) and η1′(u)/η2′(u) on the interval  is plotted in The graph of functions u0 ∈ , there exists c2 = −(η2(u0)/η1(u0))c1 + O(c12), 0 < |c1 | ≪1 such that equation H(u) = 0 holds and dH(u0)/du = c1θ(u0)(η1′(u0) + (η1(u0)/η2(u0))η2′(u0)) + O(c12) ≠ 0 for c1 > 0 and small enough.From the graph in c2 = −(η2(u0)/η1(u0))c1 + O(c12), 0 < |c1 | ≪1, it holds that I′(u0)L′(v0)N′(w0) − J′(v0)K′(w0)M′(u0) ≠ 0, where w0 = w(u0), v0 = v(u0). So there exists  which satisfies the conditions in the third part of UVW, where U = F0(u0), V = F0(v(u0)), W = F0(w(u0)), u0 ∈ .In other words, as