L-fuzzy lattice is presented by means of an L-fuzzy partially ordered set. An L-fuzzy partially ordered set A is an L-fuzzy lattice if and only if one of Aa][, and Aa)( is a lattice. The concept of  L-fuzzy lattice based on a fuzzy set of a crisp lattice, so the scope is very limited, and they gave its characterizations by only one kind of its cut sets. To overcome this shortcoming, in this paper we try to present a new definition of an L-fuzzy lattice in an L-fuzzy subset of a general set in terms of an L-poset. Furthermore, its many characterizations are given using the theory of Lβ-sets and Lα-sets proposed by Shi Fu-Gui in 1995. In this way, we obtain more generalized conclusion about L-fuzzy lattice in order to make their applications becoming comprehensive.Many concepts of fuzzy algebra were presented, since the concept of fuzzy subgroups was introduced by Rosenfeld. The concept of fuzzy lattice was also introduced by Tepavajkovski . But theajkovski  defined L denotes a completely distributive lattice, and M(L) denotes the set of all nonzero ∨-irreducible elements in L. P(L) denotes the set of all nonunit prime elements in L. X denotes a nonempty usual sets. LX is the set of all L-fuzzy sets on X. We will not differ a crisp set from its character function. For empty set ∅ ⊂ L, we define ⋀∅ = 1 and ⋁∅ = 0. According to [ ⊂ Aa)( ⊂ Aa][ and a ∈ α(b) implies Aa][ ⊂ Ab)( = Aa) = ⋁a∈M(L)(a∧Aa][) = ⋀a∈P(L)(a∨Aa][ is a mapping from Aa][ into Ba][(Ca][(Ca][)⊆f(C)a][)−1(Da][)−1(Da][)⊆(f−1(D))a], where x, y, and  z are different and  is an interval. We define the order in L as follow.Let e ∈ ,  e ≤ c = 1/2,  c < a,  c < b,  a≰b,  b≰a,  a < 1,  and  b < 1. The order in  is as usual. Then L is a completely distributive lattice. Take R ∈ LX×X such thatR is an L-fuzzy partial order on X. But it is easy to check thatR(1/2) is not a partial order on X.For all A ∈ LX, x, y, s ∈ X, R be an L-fuzzy partial order on A. s is called an L-fuzzy supremum of x, y if the following conditions are true:(S1)A(s) ≥ A(x)∧A(y)(S2)R ≥ R;  R ≥ R,(S3)R ≥ R∧R.Let A ∈ LX, x, y, t ∈ X, R be an L-fuzzy partial order on A. t is called an L-fuzzy infimum of x, y if the following conditions are true:(T1)A(t) ≥ A(x)∧A(y),(T2)R ≥ R;  R ≥ R,(T3)R ≥ R∧R. Let L-fuzzy partially ordered set  is called an L-fuzzy lattice on X if for any x, y ∈ A(0), both L-fuzzy supremum and L-fuzzy infimum of x, y exist.An A, R) be an L-fuzzy partially ordered set. Then the following conditions are equivalent.A, R) is an L-fuzzy lattice on X. is a lattice.For any a ∈ M(L),  is a lattice.For any a ∈ α(0),  is a lattice.For any a ∈ α*(0),  is a lattice.For any a ∈ P(L), (Aa) is a lattice.For any Let ⇒(2)⇒(3)⇒(6)⇒(1). Since above-mentioned sets have been posets, we only need to prove that supremum and infimum of a ∈ L∖{0}, let x, y ∈ A(0) and x, y ∈ Aa][. Analogously we can prove that  ∈ Ra][ in Aa][ in Aa][ in Aa][ and  ∈ Ra][ = Ab]∪{a, b, 1X}, where a≰b, b≰a.  A ∈ LX  is a sublattice of X; that is, A is a fuzzy lattice of X. While for 1/2 ∈ β*(1), A(1/2) = {e, f, g, 1X} is not a sublattice of X. This implies that (1)⇏(6).Then The following condition makes (1)⇒(6) be true.b, c ∈ L, β(b∧c) = β(b) ⋂ β(c), then (1)⇒(6) in the previous theorem is true.If for all a ∈ L, x, y ∈ Aa)), a ∈ β(A(y)). Therefore, a ∈ β(A(x)∧A(y)) = β(A(x))∩β(A(y)). From (1) we know that A(x∨y) ≥ A(x)∧A(y) and A(x∧y) ≥ A(x)∧A(y). So we have that a ∈ β(A(x∨y)) and a ∈ β(A(x∧y)). This shows that (Aa) is a sublattice of .(1)⇒(6). For each X be a nonempty set, let A, B be fuzzy lattices of X, and if B ≤ A, we call B as a fuzzy sublattice of A.Let X, Y be nonempty sets, and let A, B be fuzzy sublattices of X, Y, respectively. L-fuzzy mapping f : A → B is named a fuzzy lattice homomorphism if it satisfies that for all a ∈ P(L), fa)( : Aa)( → Ba) is a fuzzy sublattice of B;if D is a fuzzy sublattice of B, then f−1(D) is a fuzzy sublattice of A.if Let X, Y be nonempty sets, and let A, B be fuzzy sublattices of X, Y, respectively. f : A → B is a fuzzy mapping, then the following conditions are true:f : A → B is a fuzzy lattice homomorphism;a ∈ M(L), fa][ is a lattice homomorphism from Aa][ to Ba][. Thereforefa][(x∧y) = fa][(x)∧fa][(y). This implies that fa][ : Aa][ → Ba][⊆Aa)(. From (2) we know that fb][ : Ab][ → Bb][⊆fa) ∈ fb][⊆fa)(. Hencefa][(x∧y) = fa][(x)∧fa][ is a lattice homomorphism from Aa][ to Ba][ is a lattice homomorphism from Aa][ to Ba][.for all Let A = X, and then the conditions are true obviously. So we can get another special case of bi-fuzzy lattice-fuzzy lattice. Now we give its definition and corresponding theorems.In Definitions L be a completely distributive lattice, X ≠ ∅, and let R be an L-fuzzy relation on X, if R satisfies for any x, y, z ∈ X, there exist s, t ∈ X such that(S1)R ≥ R, R ≥ R,(S2)R ≥ R∧R,(T1)R ≥ R, R ≥ R,(T2)R ≥ R∧R. then we call s, t as supremum and infimum of x, y with respect to R, respectively.Let L be a completely distributive lattice, X ≠ ∅, and let R be an L-fuzzy relation on X, if for any x, y ∈ X, both supremum and infimum of x, y with respect to R exist, and then we call X as a fuzzy lattice with respect to R.Let Same to the corresponding theorems in last section,we have the following theorem.L be a completely distributive lattice, X ≠ ∅, and let R be an L-fuzzy relation; then (1), (2), (3), (6), (7), and (8) of the following conditions are equivalent, and (4)⇒(5)⇒(1) is true.X is a fuzzy lattice with respect to R.a ∈ L∖{0},  is a lattice.For each a ∈ M(L),  is a lattice.For each a ∈ β(1),  is a lattice.For each a ∈ β*(1),  is a lattice.For each a ∈ α(0),  is a lattice.For each a ∈ α*(0),  is a lattice.For each a ∈ P(L),  is a lattice.For each Let


D+0ν1y1(t) = λ1a1(t)f(y1(t), y2(t)), − D+0ν2y2(t) = λ2a2(t)g(y1(t), y2(t)), where D+0ν is the standard Riemann-Liouville fractional derivative, ν1, ν2 ∈ ((0) = 0 = y2i)((0), for 0 ≤ i ≤ n − 2, and [D+0αy1(t)]t=1 = 0 = [D+0αy2(t)]t=1, for 1 ≤ α ≤ n − 2, or y1i)((0) = 0 = y2i)((0), for 0 ≤ i ≤ n − 2, and [D+0αy1(t)]t=1 = ϕ1(y1), [D+0αy2(t)]t=1 = ϕ2(y2), for 1 ≤ α ≤ n − 2, ϕ1, ϕ2 ∈ C. Our results are new and complement previously known results. As an application, we also give an example to demonstrate our result.By using Krasnoselskii's fixed point theorem, we study the existence of at least one or two positive solutions to a system of fractional boundary value problems given by − Assume that Ω1,  Ω2 are open bounded subsets of E with 0 ∈ Ω1, Tu|| ≤ ||u||, ∀u ∈ P⋂ ∂Ω1, and ||Tu|| ≥ ||u||, ∀u ∈ P⋂ ∂Ω2; or||Tu|| ≥ ||u||, ∀u ∈ P⋂ ∂Ω1, and ||Tu|| ≤ ||u||, ∀u ∈ P⋂ ∂Ω2.||Let T has a fixed point in Then In this section, we apply E represent the Banach space of C when equipped with the usual supremum norm, ||·||. Then put X : = E × E, where X is equipped with the norm |||| : = ||y1|| + ||y2|| for  ∈ X. Observe that X is also a Banach space  is the Green function of ν replaced by ν1 and, likewise, G2 is the Green function of ν replaced by ν2. Now, we define an operator S : X → X byy1, y2) ∈ X is a fixed point of the operator defined in (y1(t) and y2(t) solve problems . In addX → X bySy1,y2). For. ForS, bFor the sake of convenience, we setF1);F2);F3), f and a1(t).The operator Ω⊆K be bounded; that is, there exists a positive constant M > 0 such that |||| ≤ M, for all  ∈ Ω. Let L = maxt≤1, 0≤||||≤M0≤ | a1(t)f(y1(t), y2(t))|+1; then, for  ∈ Ω, we haveT1(Ω) is bounded.Let ε > 0, setting δ = min{(1/2)(Γ(ν1)ε/LΦ2)ν1−1)1//(ν1 − 1)LΦ2}, then, for each  ∈ Ω, t1, t2 ∈ , t1 < t2, and t2 − t1 < δ, one has |T1(t2) − T1(t1)|<ε. That is to say, T(Ω) is equicontinuity. In fact,On the other hand, given In the following, we divide the proof into two cases.Case 1. If δ ≤ t1 < t2 < 1, then we havetξ ∈ .Case 2. If 0 ≤ t1 < δ, t2 < 2δ, then we haveT1 is completely continuous. Similarly, T2 is completely continuous. Consequently, S : K → K is a completely continuous operator. This completes the proof.In , GoodricF1)–(F3) are satisfied. Then problem (Suppose that F–(F3) ar problem  has at lf0 = f∞ = g0 = g∞ = 0  or  f0 = f∞ = g0 = g∞ = ∞? In the rest of this paper, we give some answers to this problem.From H1)  , (P1) are satisfied. Then problem , Suppose that  ∞ and H, (P1) a problem  has at lS is a completely continuous operator. At first, in view of f0 = g0 = ∞, we have f ≥ M(y1 + y2), for 0 < |||| < r1* < ρ1; g ≥ M(y1 + y2), for 0 < |||| < r2* < ρ1, where M satisfies M ≥ 1. Set ρ0 : = min{r1*, r2*}. So we define Ωρ0 by Ωρ0 : = { ∈ X : |||| < ρ0}. Then for each  ∈ K⋂∂Ωρ0, we find thatT1|| ≥ (1/2)|||| for  ∈ K⋂∂Ωρ0.From T2|| ≥ (1/2)|||| for  ∈ K⋂∂Ωρ0. Consequently,y1, y2) ∈ K⋂∂Ωρ0. Thus, S is cone expansion on K⋂∂Ωρ0.Similarly, we find that ||f∞ = g∞ = ∞, we have f ≥ M1(y1 + y2) for y1 + y2 ≥ r1** > ρ1;   g ≥ M1(y1 + y2) for y1 + y2 ≥ r2** > ρ1, where M1 satisfies M1 ≥ 1. Set ρ10 : = max{r1**, r2**}. Let Ωρ0* : = { ∈ X : |||| < ρ0*}. Then  ∈ K⋂∂Ωρ0* impliesT1|| ≥ (1/2)|||| for  ∈ K⋂∂Ωρ0*.Next, since T2|| ≥ (1/2)|||| for  ∈ K⋂∂Ωρ0*.Similarly, we find that ||S2|| ≥ ||||, whenever  ∈ K⋂∂Ωρ0*. Thus, S is cone expansion on K⋂∂Ωρ0*.Consequently, ||Ωρ1 : = { ∈ X : |||| < ρ1}. For  ∈ K⋂∂Ωρ1, from (H1), (P1), we haveT2|| ≤ (1/2)|||| for  ∈ K⋂∂Ωρ1.Finally, let S|| ≤ ||||, whenever  ∈ K⋂∂Ωρ1. Thus, S is cone compression on K⋂∂Ωρ1.Consequently, ||S has a fixed point So, from d point y0,y20∈K⋂(f0 = f∞ = g0 = g∞ = 0 and (H2), (P2) are satisfied. Then problem , Suppose that  problem -2) has  has f0 f0 = g0 = 0, we have f < ε(y1 + y2), g < ε(y1 + y2), for 0 < |||| ≤ ρ < ρ2, where ε satisfies ε ≤ 1. Let Ωρ : = { ∈ X : |||| < ρ}.At first, in view of y1, y2) ∈ K⋂∂Ωρ, we find thatS|| ≤ |||| for  ∈ K⋂∂Ωρ.Then for each  < ε1(y1 + y2), g < ε1(y1 + y2), for y1 + y2 ≥ ρ′ > ρ2, where ε1 satisfies ε1 ≤ 1. We consider two cases.Next, in view of Case 1. Suppose that f is unbounded; there exists ρ* > ρ′ such thatρ* > ρ′, one has f ≤ f < ε1(y1* + y2*) for 0 ≤ |||| ≤ ρ*. Then, for  ∈ K and |||| = ρ*, we obtainCase 2. Suppose that f is bounded; there exists L1 such that f ≤ L1 for all  ∈ K. Taking ρ* ≥ max{2ρ2, L1}, for  ∈ K and |||| = ρ*, we obtainΩρ* : = { ∈ X : |||| < ρ*} such that ||T1|| ≤ (1/2)|||| for  ∈ K⋂∂Ωρ*. Like S|| ≤ ||||, for  ∈ K⋂∂Ωρ*.Ωρ2 : = { ∈ X : |||| < ρ2}. Then  ∈ K⋂∂Ωρ2 impliesT1|| ≥ (1/2)|||| for  ∈ K⋂∂Ωρ2. Like S|| ≥ |||| for  ∈ K⋂∂Ωρ2.Finally, set S has a fixed point So, from d point y0,y20∈K⋂(ρ1 ≠ ρ2 such thatH3), g ≤ B1−1ρ1, for 0 ≤ |||| ≤ ρ1;H4), g ≥ B2−1ρ2, for In the following, for the sake of convenience, setH3) and (H4) are satisfied. Then problem  such that |||| between ρ1 and ρ2.Suppose that ( problem -2), in , in H3)ρ1 < ρ2.With loss of generality, we may assume that Ωρ1 : = { ∈ X : |||| < ρ1}. For  ∈ K⋂ ∂Ωρ1, one hasS|| ≤ |||| for  ∈ K⋂∂Ωρ1.Let Ωρ2 : = { ∈ X : |||| < ρ2}. Then for  ∈ K⋂∂Ωρ2, one hasS|| ≥ |||| for  ∈ K⋂∂Ωρ2. Hence, from Now, set λ1 = λ2 = 1 is not considered.In , problemConsider the following.y1, y2) ∈ X is a solution of  is a fixed point of the operator U : X → X defined byβ1, β2 : → are defined byA pair of functions y, y2 ∈ Xution of  if and oβ1(t) and β2(t) is strictly increasing in t and satisfies β1(0) = β2(0) = 0 and β1(1), β2(1)∈. Moreover, there exist constants Mβ1 and Mβ2 satisfying Mβ1, Mβ2 ∈  such that mint∈β1(t) ≥ Mβ1||β1|| and mint∈β2(t) ≥ Mβ2||β2||.Each of K1 byγ0 ∈ .Let one define a new cone U : K1 → K1 is a completely continuous operator.D1)(ϕ1(y1) ≤ ||y1||/4, ϕ2(y2) ≤ ||y2||/4 for each  ∈ K1;P3), (D1), (P3) are satisfied. Then problem ,Suppose that  ∞ and H, (D1), (D1), P are satif0 = g0 = ∞, we have f ≥ M(y1 + y2), g ≥ M(y1 + y2), for 0 < |||| ≤ ρ0 < ρ1, where M satisfies M ≥ 1.At first, in view of Ωρ0 : = { ∈ X : |||| < ρ0}. Then for each  ∈ K1⋂∂Ωρ0, we find thatU1|| ≥ (1/2)|||| for  ∈ K1⋂∂Ωρ0.Let U2|| ≥ (1/2)|||| for  ∈ K1⋂∂Ωρ0. Consequently,y1, y2) ∈ K1⋂∂Ωρ0. Thus, U is cone expansion on K1⋂∂Ωρ0.Similarly, we find that ||f∞ = g∞ = ∞, we get f ≥ M1(y1 + y2), g ≥ M1(y1 + y2), for y1 + y2 ≥ ρ10 > ρ1, where M1 satisfies M1 ≥ 1. Let ρ0* = max{2ρ1, (ρ10/γ0)} and Ωρ0* : = { ∈ X : |||| < ρ0*}; then,  ∈ K⋂∂Ωρ0* implies y1, y2) ∈ K1⋂∂Ωρ0*, we obtainU1|| ≥ (1/2)|||| for  ∈ K1⋂∂Ωρ0*.Next, since U2|| ≥ (1/2)|||| for  ∈ K1⋂∂Ωρ0*. Consequently, ||U|| ≥ ||||, whenever  ∈ K1⋂∂Ωρ0*. Thus, U is cone expansion on K1⋂∂Ωρ0*.Similarly, we find that ||Ωρ1 : = { ∈ X : |||| < ρ1}. For  ∈ K⋂∂Ωρ1, from (H1), (D1), and (P3), we haveU2|| ≤ (1/2)|||| for  ∈ K1⋂∂Ωρ1. Consequently, ||U|| ≤ ||||, whenever  ∈ K1⋂∂Ωρ1. Thus, U is cone compression on K1⋂∂Ωρ1.Finally, let U has a fixed point So, from d point y0,y20∈K1⋂s of BVP  with(51f0 = f∞ = g0 = g∞ = 0 and (H2), (D1), (P4) are satisfied. Then problem , Suppose that (H2), D, (P4) a problem  has at lf0 = g0 = 0, we have f < ε(y1 + y2), g < ε(y1 + y2) for |||| ≤ ρ < ρ2, where ε satisfies ε ≤ 1. Let Ωρ : = { ∈ X : |||| < ρ}. Then for each  ∈ K1⋂ ∂Ωρ, we find thatU|| ≤ |||| for  ∈ K1⋂ ∂Ωρ.At first, in view of f∞ = g∞ = 0, we have f < ε1(y1 + y2), g < ε1(y1 + y2), for y1 + y2 ≥ ρ′ > ρ2, where ε1 satisfies ε1 ≤ 1. We consider two cases.Next, in view of Case 1. Suppose that f is unbounded and there exists ρ* > ρ′ such thatρ* > ρ′, one has f ≤ f < ε1(y1* + y2*) for 0 ≤ |||| ≤ ρ*.y1, y2) ∈ K1 and |||| = ρ*, we obtainThen, for  ≤ L1 for all  ∈ K1. Taking ρ* ≥ max{2ρ2, L1}, for  ∈ K1 and |||| = ρ*, we obtainΩρ* : = { ∈ X : |||| < ρ*} such that ||U1|| ≤ (1/2)|||| for  ∈ K1⋂ ∂Ωρ*.U|| ≤ ||||, for  ∈ K1⋂ ∂Ωρ*.Like Ωρ2 : = { ∈ X : |||| < ρ2}. Then  ∈ K1⋂ ∂Ωρ2 impliesU1|| ≥ (1/2)|||| for  ∈ K1⋂ ∂Ωρ2.Finally, set U|| ≥ |||| for  ∈ K1⋂ ∂Ωρ2. So, from U has a fixed point Like Uy, y2|| ≥s of BVP  with 0<|λ1 = λ2 = 1 having at least one positive solution. In the following, we also establish the existence of one positive solution to problem , g ≤ B3−1ρ1 for 0 ≤ |||| ≤ ρ1;H6), g ≥ B2−1ρ2 for γ0ρ2 ≤ |||| ≤ ρ2.For the sake of convenience, setH5),  (H6), and (D1) are satisfied. Then problem  such that |||| between ρ1 and ρ2.Suppose that , and D are sati problem , in the ρ1 < ρ2. Let Ωρ1 : = { ∈ X : |||| < ρ1}. For  ∈ K1⋂ ∂Ωρ1, from (H7), (D1), one hasU|| ≤ |||| for  ∈ K1⋂ ∂Ωρ1.With loss of generality, we may assume that Ωρ2 : = { ∈ X : |||| < ρ2}. For  ∈ K1⋂ ∂Ωρ2, one hasH8), we getU|| ≥ |||| for  ∈ K1⋂ ∂Ωρ2. Hence, from Now, set To illustrate how our main results can be used in practice, we present one example.t ∈ :f, g : , we getH1) holds.Now, observe that λ1, λ2, as given by condition (P1), observe by numerical approximation, we find thatλ1, λ2 satisfying 163530 < λ1, λ2 < 164547.25, condition (P1) will be satisfied.On the other hand, to calculate the admissible range of the eigenvalues Consequently, by


The split equality problem (SEP) has extraordinary utility and broad applicability in many areas of applied mathematics. Recently, Byrne and Moudafi (2013)proposed a CQ algorithm for solving it. In this paper, we propose a modification for the CQ algorithm, which computes the stepsize adaptively and performs an additional projection step onto two half-spaces in each iteration. We further propose a relaxation scheme for the self-adaptive projection algorithm by using projections onto half-spaces instead of those onto the original convex sets, which is much more practical. Weak convergence results for both algorithms are analyzed. H1, H2, and H3 be real Hilbert spaces; let C ⊂ H1 and Q ⊂ H2 be two nonempty closed convex sets; let A : H1 → H3 and B : H2 → H3 be two bounded linear operators. The SEP can mathematically be formulated as the problem of finding x and y with the propertyx and y. If H2 = H3 and B = I, then the split equality problem  was introduced by Moudafi  and its CQ algorithm:γk ∈  − ɛ) and λA and λB are the spectral radii of A*A and B*B, respectively. By studying the projected Landweber algorithm of the SEP ) − ɛ). It is easy to see that the alternating CQ algorithm  norms ||A|| and ||B|| . This means that, in order to implement the alternating CQ algorithm ,,3), the lgorithm , one hasg and He  proposedCQ algorithm  in H1 × H2, ωw stands for the set of cluster points in the weak topology. “xk → x”  means the strong  convergence of (xk) to x.Let xk) is said to be asymptotically regular ifA sequence (yn ∈ T(xn) with yn strongly converging to y and xn weakly converging to x; then y ∈ T(x).The graph of an operator is called to be weakly-strongly closed if The next lemma is well known see , 12) and and12]) H be a Hilbert space and let T : H⇉H be a maximal monotone mapping. If (xk) is a sequence in H bounded in norm and converging weakly to some x and (wk) is a sequence in H converging strongly to some w and wk ∈ T(xk) for all k, then w ∈ T(x).Let Ω be a closed convex subset of real Hilbert space H. Recall that the (nearest point or metric) projection from H onto Ω, denoted by  PΩ, is defined in such a way that, for each x ∈ H, PΩx is the unique point in Ω such thatThe projection is an important tool for our work in this paper. Let The following two lemmas are useful characterizations of projections.x ∈ H and z ∈ Ω, then z = PΩx if and only ifGiven x, y ∈ H and z ∈ Ω, it holds PΩ(x)−PΩ(y)||2 ≤ 〈PΩ(x) − PΩ(y), x − y〉;||PΩ(x)−z||2 ≤ ||x−z||2 − ||PΩ(x)−x||2.||For any ιC(x) is an indicator function of the set C defined byγ > 0 and β > 0,I + γNC)−1 = PC and (I + βNQ)−1 = PQ, we haveThroughout this paper, assume that the split equality problem  is consi problem  can be wquations  are exacx* ∈ H1 and y* ∈ H2, then  solves the SEP  solves the fixed point equations ,  θ ∈  and ρ ∈ , let x0 ∈ H1 and y0 ∈ H2 be arbitrary. For k = 0,1, 2,…, computeγk is chosen to be the largest γ ∈ {σk, σkβ, σkβ2,…} satisfyingXk and Yk, the bounding hyperplanes of which support C and Q at uk and vk, respectively,σk = σ0; otherwise, set σk = γk.Given constants Xk  rather than onto the set C  and it is obvious that projections on X  are very simple. It is easy to show C ⊂ Xk and Q ⊂ Yk. The last step is used to reduce the inner iterations for searching the stepsize γk.In this algorithm,  involvesThe search rule  is well γk ≤ σ0. If γk = σ0, then this lemma is proved; otherwise, if γk < σ0, by the search rule  be the sequence generated by X and Y be nonempty closed convex sets in H1 and H2 with simple structures, respectively. If (X × Y)∩Γ is nonempty, then  converges weakly to a solution of the SEP  ∈ Γ; that is, x* ∈ C, y* ∈ Q, and Ax* = By*. Define sk = uk − γk − F); then we havePXk. Similarly, defining tk = vk − γk − G), we getsk = uk − γk − F) and tk = vk − γk − G), the second inequality follows from ∶ = ||xk−x*||2 + ||yk−y*||2 is decreasing and lower bounded by 0 and thus converges to some finite limit, say, l. Moreover, (xk) and (yk) are bounded. This implies thatLet  and (ykl) of (xk) and (yk) which converge weakly to Axkl − Bykl) to Let  the SEP . The weaNC and NQ, are weakly-strongly closed and by passing to the limit in the last inclusions, we obtain, from . By passing to the limit in the relationxk, yk) weakly converges to a solution of the SEP  and (Qk) be two sequences of closed convex sets defined byξk ∈ ∂c(xk) andηk ∈ ∂q(yk).In the Ck⊃C and Qk⊃Q for every k ≥ 0.It is easy to see that σ0 > 0,  β ∈ ,  θ ∈ ,  and ρ ∈ , let x0 ∈ H1 and y0 ∈ H2 be arbitrary. For k = 0,1, 2,…, computeγk is chosen to be the largest γ ∈ {σk, σkβ, σkβ2,…} satisfyingXk and Yk the bounding hyperplanes of which support Ck and Qk at uk and vk, respectively,σk = σ0; otherwise, set σk = γk.Given constants Following the proof of The search rule  is well xk, yk) be the sequence generated by X and Y be nonempty closed convex sets in H1 and H2 with simple structures, respectively. If (X × Y)∩Γ is nonempty, then  converges weakly to a solution of the SEP  ∈ Γ; that is, x* ∈ C, y* ∈ Q, and Ax* = By*. Following the similar proof of k∶ = ||xk−x*||2 + ||yk−y*||2. Then the sequence Γk is decreasing and lower bounded by 0 for that μ ∈  and thus converges to some finite limit, say, l. Moreover, (xk) and (yk) are bounded. This implies thatLet  generated by xkl) and (ykl) of (xk) and (yk) which converge weakly to Axkl − Bykl) to Next we show that the sequence  weakly converges to a solution of the SEP (Following the same argument of  the SEP , which c


The following errors occurred in μg mL−1 h−2 instead of 37.46 μg mL−1 h−2.AUMC value in healthy ducks is 73.46 MAT value in healthy ducks is 0.74 h instead of 0.31 h and MAT value in renal damaged ducks is 0.70 h instead of 2.08 h, with no significance in MAT between healthy and renal damaged ducks.In Page 4, last line in left column, low MAT (mean absorption time) value is 0.74 h instead of 0.31 h.


M(∫Ω | x|ap− | ∇u|p)div(|x|ap− | ∇u|p−2∇u) = λh(x) | u|r−2u, x ∈ Ω, M(∫Ω | x|ap− | ∇u|p) | x|ap− | ∇u|p−2 (∂u/∂ν) = g(x) | u|q−2u, on ∂Ω, where 1 < (N + 1)/2 < p < N, a < (N − p)/p. By the variational method on the Nehari manifold, we prove that the problem has at least two positive solutions when some conditions are satisfied.The paper considers the existence of multiple solutions of the singular nonlocal elliptic problem − N + 1)/2 < p < N, a < (N − p)/p, λ > 0, Ω is an exterior domain of ℝN: that is, and Ω = ℝN∖D, where D is a bounded domain in ℝN with the smooth boundary ∂D(  = ∂Ω), and 0 ∈ Ω. g(x) and h(x) are continuous functions, M(s) = αs + β with the parameters α, β > 0. In this paper, we consider the existence of multiple solutions for the singular elliptic problem:ρ, P0, h, E, and L are all positive constants. This equation extends the classical d'Alembert's wave equation by considering the effects of changes in the length of the strings during the vibrations. Problem  in (f1)  f is continuous function on h(x) and g(x) are permitted to change sign. Thus, the methods in  and decreases for t ∈  and increases for t ∈  and increases for t ∈  and decreases for t ∈ , we obtain that 1 < t0+ < tα,max(u0+) and φk(t) < 0 for t ∈ . Then, by  on Nλ+. Since Jλ(u0+) = Jλ(|u0+|) and |u0+ | ∈Nλ+, we may assume by u0+ is a positive solution of problem =αt2A3) and 0 < λ < (p/r)λ1. Then, the functional Jλ(u) has a minimizer u0− ∈ Nλ− and Jλ(u0−) = δλ−,u0− is a positive weak solution of problem ( problem . Assume (Jλ(u) is bounded on Nλ−, there exists a minimizing sequence {uk} ∈ Nλ− such thatuk⇀u0− weakly in E. For uk ∈ Nλ−, we deduce by Ω∂g(x) | uk|qdσ > 0; furthermore, ∫Ω∂g(x) | u0−|qdσ > 0. We also want to prove that uk → u0− strongly in E. In fact, if not, we havet0− > 0 such that Since uch thatlimk→∞ Jows thatJλ(uk)≥Jλλ* = min{λ1, λ0}. When 0 < λ < λ*, by Lemmas u0+ ∈ Nλ+, u0− ∈ Nλ−. Nλ+∩Nλ− = ∅; then u0+ and u0− are distinct. Furthermore, since Jλ(u0±) = Jλ(|u0±|) and |u0± | ∈Nλ±, we can assume that the solutions u0+ and u0− are positive. This completes the proof. We set  The object of this paper is to prove the existence of multiple solutions for the nonlinear Kirchhoff-type problem . By the


Our main tool is based on a recent three-critical-point theorem obtained by Ricceri. We also give some examples to illustrate the obtained results.Existence results of three weak solutions for a Dirichlet double eigenvalue quasilinear elliptic system involving the ( Here, F, G : Ω × ℝn → ℝ are measurable functions with respect to x ∈ Ω for every  ∈ ℝn and are C1 with respect to  ∈ ℝn for a.e. x ∈ Ω, and Fui and Gui denote the partial derivatives of F and G with respect to ui, respectively.The aim of this paper is to investigate the existence of at least three weak solutions for the following Dirichlet double eigenvalue quasilinear elliptic system:F and G satisfy the following additional assumptions:1) = 0 for a.e. x ∈ Ω; (G)M > 0 and every 1 ≤ i ≤ n, for every Moreover, X be the Cartesian product of the n Sobolev spaces W0pi1,(Ω) for 1 ≤ i ≤ n; that is, X = W0p11,(Ω) × W0p21,(Ω) × ⋯×W0pn1,(Ω) equipped with the normi ≤ n,pi > N for 1 ≤ i ≤ n, X is compactly embedded in c < +∞. In addition, it is known  is the Lebesgue measure of the set Ω, and equality occurs when Ω is a ball.Here and in the following, we let x0 ∈ Ω such that B⊆Ω, where B stands for the open ball in ℝN of radius r centered at x.Moreover, leti ≤ n.Putu =  ∈ X such thatv =  ∈ X.By a (weak) solution of system , we meanIn the literature many papers (a2)  the setγ.In the present section we discuss the existence of multiple solutions for system . For anyWe formulate our main result as follows.θ and δ with ∑i=1n((δκi)pi/pi) > (θ/∏i=1npi) such that 1) ≥ 0 for a.e. x ∈ Ω∖B and all ti ∈  for 1 ≤ i ≤ n; 2)(b3) ∈ X, we let the functionals Φ, Ψ : X → ℝ be defined byX and it is known that Φ and Ψ are well-defined and continuously Gâteaux differentiable functionals whose derivatives at the point u =  ∈ X are the functionals Φ′(u) and Ψ′(u) given byv =  ∈ X, and Φ is sequentially weakly lower semicontinuous  of  for 1 ≤ i ≤ n; 5)(b6)→/∑i=1n(|ti|pi/pi) ≤ 0.  limsup Then, settinga, b]⊆Λ, there exists ρ > 0 with the following property: for every λ ∈ , there exists δ > 0 such that, for each μ ∈ , the systemi ≤ n, admits at least three weak solutions in X whose norms are less than ρ. Let F = F for all x ∈ Ω and ti ∈ ℝ for 1 ≤ i ≤ n. Since ∫BFdx = (πN/2/Γ(1 + N/2))(D/2)NF, Set κ = κ1 and p = p1. Then we have the following existence result.Let f : ℝ → ℝ be a continuous function and let g : Ω × ℝ → ℝ be an L1-Carathéodory function. Put F(t) = ∫0tf(ξ)dξ for each t ∈ ℝ and assume that there exist two positive constants θ and δ with (δκ)p > θ such that 7)(bF(t) ≥ 0 for all t ∈ ;  8)(b9)(bt|→+∞|F(t)/|t|p ≤ 0.  limsup Then, settinga, b]⊆Λ, there exists ρ > 0 with the following property: for every λ ∈ , there exists δ > 0 such that, for each μ ∈ , the problemW0p1,(Ω) whose norms are less than ρ. Let N = 1 and p = 2. For simplicity, we fix Ω =  and note that in this situation, c = (β − α)/4 and κ = ((β − α)/D)1/2.Now, we want to point out a simple consequence of f : ℝ → ℝ be a continuous function and g :  × ℝ → ℝ an L1-Carathéodory function. Put F(t) = ∫0tf(ξ)dξ for each t ∈ ℝ and assume that there exist two positive constants θ and δ with (δ2(β − α)/D) > θ such that assumption (b7) in 10)(b11)(bt|→+∞|(F(t)/t2) ≤ 0.  limsup Then, settinga, b]⊆Λ, there exists ρ > 0 with the following property: for every λ ∈ , there exists δ > 0 such that, for each μ ∈ , the problemC2 whose norms are less than ρ. Let  First, we present an example of the application of Ω = { ∈ ℝ2 : x2 + y2 < 4}. Consider the systemG : Ω × ℝ2 → ℝ is an arbitrary function which is measurable with respect to  ∈ Ω for every  ∈ ℝ2 and is C1 with respect to  ∈ ℝ2 for a.e.  ∈ Ω, satisfyingM > 0 and i = 1,2. Taking into account c = 4/π, picking x0 =  andt1, t2) ∈ ℝ2, so that D = 2 and δ = θ = 3, we have F ≥ 0 for all ∈× andp1 = p2 = 3, are satisfied. So, settinga, b]⊆Λ, there exists ρ > 0 with the following property: for every λ ∈ , there exists δ > 0 such that, for each μ ∈ , system  × W01,3(Ω) whose norms are less than ρ. Let e system−Δ3u1=λe1The next example follows directly by f(t) = (1/10)t9et−(10 − t) for all t ∈ ℝ. Then one has F(t) = (1/10)t10et− for all t ∈ ℝ. Now, consider the following two-point boundary value problemg :  × ℝ → ℝ is an arbitrary L1-Carathéodory function. Note that, in this case, x0 = 0, D = 1, c = 1/2, and θ = 1 and δ = 2, we have F(t) ≥ 0 for all t ∈ . Also10) is fulfilled. Furthermore, we have limsupt|→+∞|(F(t)/t2) = 0. Thus, all hypotheses of e2/27), (5/e)⊆Λ, there exists ρ > 0 with the following property: for every λ ∈ , there exists δ > 0 such that, for each μ ∈ , problem  whose norms are less than ρ.Suppose  problem−u′′=λf whose norms are less than ρ.In particular, there exist two positive constants


Our results in this paper are generalized and unify several results in the literature as the result of Geraghty (1973) and the Banach contraction principle.We study the generalized Ulam-Hyers stability, the well-posedness, and the limit shadowing of the fixed point problem for new type of generalized contraction mapping, the so-called  The stability problem of functional equations, first initial from a question of Ulam  in 1940,On the other hand, the notion of well-posedness and limit shadowing property of a fixed point problem have evoked much interest to many researchers, for example, De Blassi and Myjak , Reich aRecently, Samet et al.  introducX be a nonempty set and α : X × X → [0, ∞) be a mapping. A mapping f : X → X is said to be α-admissible if it satisfies the following condition:Let X = . Define f : X → X and α : X × X → [0, ∞) by f(x) = 2x for all x ∈ X andf is α-admissible.Let X = [1, ∞). Define f : X → X and α : X × X → [0, ∞) byf is α-admissible.Let f is α-admissible.Every nondecreasing self-mapping α-admissible mapping. Also, they applied these results to derive fixed point theorems in partially ordered metric spaces. As application, they studied the ordinary differential equations via the main results. Several researchers studied and improved contraction mappings via the concept of α-admissible mapping in metric spaces and other spaces →[0,1) which satisfies the following condition.Let tn} of nonnegative real numbers, we haveFor any sequence {This class is first introduced by Geraghty  in 1973.Υ.(i)β1(t) = k for all t ∈ [0, ∞), where k ∈ [0,1).Consider (ii)ConsiderThe following are examples of some functions in First we give the following definition as a generalization of Banach contraction mappings.X, d) be a metric space and f : X → X a given mapping. One says that f is an α-β-contraction mapping if there exist two functions α : X × X → [0, ∞) and β ∈ Υ such thatx, y ∈ X, where 1 < δ ≤ δ∗.Let  = 1 for all x, y ∈ X.It is easy to check that an Next, we introduce the transitive mapping which is useful for our main result.X be a nonempty set. A mapping α : X × X → [0, ∞) is called transitive if it satisfies the following condition:Let Our first main result is the following.X, d) be a complete metric space and f : X → X an α-β-contraction mapping satisfying the following conditions: f is α-admissible;α is transitive;x0 ∈ X such that α) ≥ 1;there exists f is continuous. Then the fixed point problem of f has a solution; that is, there exists x* ∈ X such that x* = f(x*).Let ) ≥ 1 (such a point exists from condition (iii)). Define the sequence {xn} in X byxn = xn−1 for some n ∈ ℕ, then xn = f(xn); that is, xn is a fixed point of f and thus the proof ends. Therefore, we may assume thatf is α-admissible and α = α) ≥ 1, we get α = α(f(x0), f(x1)) ≥ 1. By induction, we getn ∈ ℕ, we haven ∈ ℕ. Therefore, the sequence {d} is strictly decreasing and so d → d as n → ∞ for some d ≥ 0. Next, we claim that d = 0. Assume on the contrary that d > 0. On taking limit as n → ∞ in  = 0, which is a contradiction. Therefore, d = 0 and thusLet ies thatd = 0 which contradicts with xn} is af by adding some condition.In the next theorem, we omit the continuity hypothesis of X, d) be a complete metric space and f : X → X an α-β-contraction mapping satisfying the following conditions:f is α-admissible;α is transitive;x0 ∈ X such that α) ≥ 1;there exists xn} is a sequence in X such that α ≥ 1 for all n ∈ ℕ and xn → x ∈ X as n → ∞, then α ≥ 1 for all n ∈ ℕ.if { Then the fixed point problem of f has a solution; that is, there exists x* ∈ X such that x* = f(x*).Let ) = 0; that is, x* = f(x*). Therefore, the fixed point problem of f has a solution. This completes the proof.On the other hand, from  and hypoquality, , and (25We obtain that Theorems 0) ≥ 1 for all a, b ∈ X, where a, b are fixed points of fAdding condition 1) ≥ 1 and α ≥ 1for all to the hypotheses of f.or x* and y* are two fixed points of f. If condition (H0) holds, then we get the uniqueness of the fixed point of f from (1) holds. From condition (H1), there exists z ∈ X such thatx*, y* are fixed points of f and f is α-admissible, from ≥1e, from ) = 0. Assume on the contrary that,n → ∞ in }, we have x* = y*. This completes the proof.Next, we claim that limn → ∞ in , we gety, using  and 6),,n→∞d be a metric space and f : X → X a mapping. The fixed point problemψ : [0, ∞)→[0, ∞) which is increasing, continuous at 0 and ψ(0) = 0 such that for each ε > 0 and for each w* ∈ X which is an ε-solution of the fixed point equation is problemx=f(x)isψ is defined by ψ(t) = ct for all t ≥ 0, where c > 0, then the fixed point equation  be a metric space and f : X → X a mapping. The fixed point problem of f is said to be well posed if it satisfies the following conditions:f has a unique fixed point x* in X;xn} in X such that limn→∞d) = 0, one has limn→∞d = 0.for any sequence {Let  be a metric space and f : X → X a mapping. We say that the fixed point problem of f has the limit shadowing property in X if, for any sequence {xn} in X satisfying limn→∞d) = 0, it follows that there exists z ∈ X such that limn→∞d(fn(z), xn) = 0.Let  be a complete metric space. Suppose that all the hypotheses of β(0) = 0 and the function ξ : [0, ∞)→[0, ∞) is defined by ξ(t): = t − β(t) which is strictly increasing and onto. Thenα ≥ 1 for all a, b which are an ε-solution of the fixed point equation  ≥ 1 for all n ∈ ℕ such that {xn} is sequence in X in which limn→∞d) = 0 and x* is a fixed point of f, then the fixed point problem of f is well posed;if α ≥ 1 for all n ∈ ℕ such that {xn} is sequence in X in which limn→∞d) = 0 and x* is a fixed point of f, then the fixed point problem of f has the limit shadowing property in X.if Let  ≥ 1. Now we haveξ−1 is increasing, continuous at 0 and ξ−1(0) = 0. Consequently, the fixed point problem of f is generalized Ulam-Hyers stability.From the hypothesis in (a), we claim that the fixed point problem of ution of ; that isequality . From hyf is well posed under the assumption in (b). Let {xn} be sequence in X such that limn→∞d) = 0. From assumption, we get α ≥ 1 for all n ∈ ℕ. Now, we obtain thatn ∈ ℕ. This implies thatn ∈ ℕ. Now we claim that limn→∞d = 0. Assume on the contrary thatn→∞β) = 1. Since β ∈ Υ, we obtain that limn→∞d = 0 which contradicts with  = 0 and so the fixed point problem of f is well posed.Next, we prove that the fixed point problem of ies thatd≤ary that0<limn→∞f has a limit shadowing under assumption (c). Let {xn} be sequence in X such that limn→∞d) = 0. Similar to case (b), we get limn→∞d = 0. Since x* is a fixed point of f, we havef has the limit shadowing property.Finally, we prove that Some Open Problems0) and (H1) by other conditions or more general conditions?In In In b-metric space, and circular metric space?Can we extend the result in this paper to other spaces as cone metric space, complex valued metric space, partial metric space,


We study the norm induced by the supremum metric on the space of fuzzy numbers. And then we propose a method for constructing a norm on the quotient space of fuzzy numbers. This norm is very natural and works well with the induced metric on the quotient space. Since the concept of fuzzy numbers was firstly introduced in the 1970s, it has been studied extensively from many different aspects of the theory and applications such as fuzzy topology, fuzzy analysis, fuzzy logic, and fuzzy decision making . The opX is a normed space with norm ||·||, it is readily checked that the formula d = ||x − y||, for x, y ∈ X, defines a metric d on X. Thus a normed space is naturally a metric space and all metric space concepts are meaningful. However, we will show that such proposition does not hold true for the well known supremum metric on the space of fuzzy numbers. To overcome this weakness, we will consider the quotient space of fuzzy numbers up to an equivalence relation which is introduced by Mareš . For each such fuzzy set μ, we denote by [μ]a = {x ∈ ℝ : μ(x) ≥ a} for any a ∈ , μR(a)] for all a ∈ . We denote the class of fuzzy numbers by ℱ. Notice that the real numbers ℝ can be imbedded in ℱ by defining a fuzzy number as followsa ∈ ℝ.A fuzzy set of ℝ is a function μ, ν ∈ ℱ and a ∈ ℝ, owing to Zadeh's extension principle  → ℝ by assigning the midpoint of each a-level set to μM(a) for all a ∈ , that is,For a fuzzy number μ, ν ∈ ℱ, μ ~ ν if and only if μM = νM.For any ℱ/𝒮, +, ×) is a real vector space.d∞, μR(a0)]≠{0}. Thus we have ||μ|| > 0.(i)  It is obvious that ||μ ∈ ℱ and b ∈ ℝ, we have(ii)  For all μ, ν ∈ ℱ, we have thatℱ.(iii)  For all ℱ, the function d : ℱ × ℱ → ℝ induced by ||·|| as d = ||μ − ν|| is not a metric on ℱ. Consequently, we get that d∞ ≠ ||μ − ν||.Although ||·|| is a norm on d has the following properties:d ≥ 0, for any μ, ν ∈ ℱ;d = d, for any μ, ν ∈ ℱ;d ≤ d + d, for any μ, ν, ω ∈ ℱ;d ≠ 0, for any fuzzy number μ ∉ ℝ.The function d, it is obvious that d ≥ 0, for any μ, ν ∈ ℱ.(i)  By the definition of μ, ν ∈ ℱ, we have that(ii)  For all a ∈ , and for any μ, ν, ω ∈ ℱ, we only proof the following six cases. Similarly, the others can be proved.(iii)  In order to prove the triangle inequality, for any fixed Case  1 ( μL(a) ≤ νL(a) ≤ μR(a) ≤ νR(a) and ωL(a) ≤ μL(a)). In this case we have thatCase  2 ( μL(a) ≤ νL(a) ≤ μR(a) ≤ νR(a) and μL(a) ≤ ωL(a) ≤ ωR(a) ≤ νR(a)). In this case we have thatCase  3 ( μL(a) ≤ νL(a) ≤ νR(a) ≤ μR(a) and ωL(a) ≤ μL(a)). In this case we have thatCase  4 ( μL(a) ≤ ωL(a) ≤ ωR(a) ≤ νL(a) ≤ νR(a) ≤ μR(a)). If max{|μL(a) − νR(a)|, |μR(a) − νL(a)|} = |μL(a) − νR(a)|, then we have thatCase  5 ( μL(a) ≤ νL(a) ≤ ωL(a) ≤ ωR(a) ≤ νR(a) ≤ μR(a)). In this case we have thatCase  6 ( μL(a) ≤ ωL(a) ≤ νL(a) ≤ νR(a) ≤ ωR(a) ≤ μR(a)). In this case we have thatμ ∉ ℝ, there exists a0 ∈  such that μR(a0) − μL(a0) > 0. Thus we have that(iv)  Since ℱ/𝒮 → ℝ asμ〉∈ℱ/𝒮.In order to induce a metric which is compatible with the norm, we consider the quotient space of fuzzy numbers. It is very natural to define a function ||·|| : ℱ/𝒮.The function ||·|| in  is a norμ〉|| ≥ 0 for all μ ∈ ℱ and ||〈μ〉|| = 0 if 〈μ〉 = 〈0〉. Conversely, if 〈μ〉≠〈0〉, from μM ≠ 0M. Thus we have(i)  It is obvious that ||〈μ〉∈ℱ/𝒮 and b ∈ ℝ, we have(ii)  For all 〈μ〉, 〈ν〉∈ℱ/𝒮, we have thatℱ/𝒮.(iii)  For all 〈ρ : ℱ/𝒮 × ℱ/𝒮 → ℝ induced by ||·|| as ρ = ||〈μ − ν〉|| is exactly a metric on ℱ/𝒮.Now we show that the function ρ is a metric on ℱ/𝒮.The function ρ, it is obvious that ρ ≥ 0, for any 〈μ〉, 〈ν〉∈ℱ/𝒮. If ρ = 0, then ||〈μ − ν〉|| = 0. Thus by μ〉 = 〈ν〉. In addition, by ρ = ||〈μ − μ〉|| = 0.(i)  By the definition of μ〉, 〈ν〉∈ℱ/𝒮, we have that(ii)  For all 〈μ〉, 〈ν〉, 〈ω〉∈ℱ/𝒮, we haveρ is a metric on ℱ/𝒮.(iii)  For any 〈d∞ on the space of fuzzy numbers. And then we proposed a method for constructing a norm on the quotient space of fuzzy numbers. This norm is very natural and works well with the induced metric on ℱ/𝒮. The works in this paper enable us to study the fuzzy numbers in the new environment. We hope that our results in this paper may lead to significant, new, and innovative results in other related fields.In this present investigation, we studied the norm induced by the supremum metric


For the continued fraction expansion arising from τm, we solve a Gauss-Kuzmin-type problem.We consider a family { 𝒪(qn) as n → ∞ with 0 < q < 1.Chan considered some continued fraction expansions related to random Fibonacci-type sequences , 2. In , he studm ≥ 2. In this section, we show our motivation and main theorems.The purpose of this paper is to prove a Gauss-Kuzmin-type problem for the continued fraction expansions of real numbers in  whose digits are differences of consecutive nonpositive integer powers of an integer x < 1 can be written as the infinite regular continued fractionan ∈ ℕ+≔{1,2, 3,…}.One of the first and still one of the most important results in the metrical theory of continued fractions is the so-called Gauss-Kuzmin theorem. Any irrational 0 < an)n∈ℕ+. It started on October 25, 1800, with a note by Gauss in his mathematical diary (entry 113)  byλ denotes the Lebesgue measure on I and τn is the nth iterate of τ.Roughly speaking, the metrical theory  of continued fraction expansions is about properties of the sequence  0 < q < 1. This has been called the Gauss-Kuzmin theorem or the Kuzmin theorem.Twelve years later, in a letter dated January 30, 1812, Gauss wrote to Laplace that he did not succeed in solving satisfactorily “a curious problem” and that his efforts “were unfruitful.” In modern notation, this problem is to estimate the erroren(x)≔λ can be written as the forman's are nonnegative integers. Such an's are also called incomplete quotients (or continued fraction digits) of x with respect to the expansion in I, ℬI, τm) is defined as follows: I≔, where ℬI denotes the σ-algebra of all Borel subsets of I and τm is the transformationIi≔{x ∈ I : mi+1)−I, ℬI, γm, τm) as  with the following probability measure γm on :In addition to (i), one writes (Fix an integer quantized index map  ηm : I → ℕ byηm(mα−) = ⌊α⌋. By using τm and ηm, the sequence (an)n∈ℕ+ in (τm0(x) = x. In this way, τm gives the algorithm of Chan's continued fraction expansion (Define the )n∈ℕ+ in  is obtaixpansion .I, ℬI, γm, τm) be as in I, ℬI, γm, τm) is ergodic. = γm(τm−1(A)) for any A ∈ ℬI.The measure Let  is a “dynamical system” in the sense of m ∈ . As it can be seen, χm is a constant independent of the value of x.Continued fraction expansions of almost all irrational numbers are not periodic. Nevertheless, we readily reproduce another famous probabilistic result. It is the asymptotic value given in . This isμ-measure preserving operator T : → is given by the beautiful formulaτm in ≔∫ablgiven in , 26), a, aμ-measgiven in , respectm ≥ 2. Let km be as in  be as in x has the expansion in −( < τmn < mi−). We will show that the event τmn ≤ x has the following asymptotic probability:τm: if a random variable X in the unit interval has the density ρm, and then so does τm. The reason for this invariance is that, for 0 ≤ x < x + h ≤ 1, τm lies between x and x + h if and only if there exists i ≥ 1, so that X lies between 1/(x + i + h) and 1/(x + i). Thush → ∞ gives that, for an arbitrary probability density function f for X, the corresponding density Gf for τm is given a.e. in I by the equationG : L1 → L1 admits the density function ρm as an eigenfunction corresponding to the eigenvalue 1; that is, Gρm = ρm. Here L1 denotes the Banach space of all complex functions f : I → ℂ for which ∫I|f|dλ < ∞.We show our main theorems in this subsection. Fix an integer be as in  and let nsion in  and τm is as in , then thG is 1 and this eigenvalue is simple.The only eigenvalue of modulus 1 of τm is an ergodic operator on the unit interval [ρm is the density of the invariant measure, and G is called transfer operator for τm [G has the same analytical expression as the Perron-Frobenius operator of τm under the Lebesgue measure [From another perspective, the operator interval , ρm is  for τm . The tra measure .Our main result is the following theorem.τm and ωm be as in  is given, define functions Fm,n (n ≥ 0) on I byx ∈ I. Then there exists a constant 0 < qm < 1 such that Fm,n is written asLet be as in  and 31)τm and ωγm is the measure defined in ((i) From , we see fined in . In factan)n∈ℕ in (ψ-mixing under γm (and under many other probability measures including λ) [A1 ∈ σ , A2 ∈ σ, and k ∈ ℕ+, with suitable positive constants q < 1 and C.(ii) The solution of this problem implies that (n)n∈ℕ in  is exponuding λ) , 12; thaψ-mixing implies lots of limit theorems in both classical and functional versions. To form an idea of the results to be expected it is sufficient to look at the corresponding results for the regular continued fraction expansions [In turn, pansions .τm [τm is ergodic  In  we emphaτm , 6. Thes  of  f” .τm for m ≥ 3. For example, study the optimality of the convergence rate. Use the same strategy as in [(i) Solve the Gauss-Kuzmin-Lévy problem of gy as in .τ in ((ii) It is known that the Riemann zeta function is written by using a kind of Mellin transformation of the Gauss transformation τ in  as folloτ. Then, by replacing τ with τm in  ≤ x}, and Im,n,i = {x ∈ Im,n :   αi/(1 + (m − 1)x) < τmn(x) < αi}.Let Im,n,i and  = μ andFrom  and 12)12), we sτmn(x)=m−i).Then  holds bep ∈ ℕ, the derivative Fp′ exists everywhere in I and is bounded. Then it is easy to see by induction that Fm,p+n′ exists and is bounded for all n ∈ ℕ+. This allows us to differentiate  is given in  byFor fm,n} in |. ThenFor {m,n} in , define We haveNow from  and by cNow  impliesδi = αi − αi2 and 0 ≤ x ≤ 1.First, we note thatx ∈ I and i ∈ ℕ. Hence, hm(x) ≤ hm(0). This leads toNext, observe that the functionelations , 59), a, a(60)hmleads toαiPmi((m−elations  and 54)(60)hm(x)Rm,n(x) such thatIntroduce a function Fm,n(0) = 0 and Fm,n(1) = 1, we have Rm,n(0) = Rm,n(1) = 0. To prove qm < 1 such thatBecause fm,n(x) = km + 𝒪(qmn), then its integration will show  has this desired form, it suffices to prove the following lemma.To demonstrate that x ∈ I and n ∈ ℕ, there exists a constant 0 < qm < 1 such thatFor any qm be as in (qm < 1. To this end, for i ≥ 2, observe thatm ∈ ℕ, m ≥ 2.Let be as in . Using Lbe as in  it is en


G1 be a group and let G2 be a metric group with the metric d. Given ε > 0, does there exist a δ > 0 such that if a function h : G1 → G2 satisfies the inequality d(h(xy), h(x)h(y)) < δ, for all x, y ∈ G1, then there exists a homomorphism H : G1 → G2 with d(h(x), H(x)) < ε, for all x ∈ G1?  Let In 1940, Ulam [G1 and G2 are Banach spaces. Indeed, he proved that each solution of the inequality ||f(x + y) − f(x) − f(y)|| ≤ ε, for all x and y, can be approximated by an exact solution, say an additive function. In this case, the Cauchy additive functional equation, f(x + y) = f(x) + f(y), is said to have the Hyers-Ulam stability.The case of approximately additive functions was solved by Hyers  under thRassias  attemptestigated –9.The terminologies, the generalized Hyers-Ulam stability, and the Hyers-Ulam stability can also be applied to the case of other functional equations, differential equations, and various integral equations. c > 0, the partial differential equationutt and uxx denote the second time derivative and the second space derivative of u, respectively.Given a real number φ : ℝ × ℝ → [0, ∞) be a function. If, for each twice continuously differentiable function u : ℝ × ℝ → ℂ satisfyingu0 : ℝ × ℝ → ℂ of the  wave equation (∞) such thatx, t) is independent of u and u0, then we say that the wave equation , we prove the generalized Hyers-Ulam stability of the  wave equation .φ : ℝ × ℝ → [0, ∞) be given such that the double integrala, b ∈ ℝ. If a twice continuously differentiable function u : ℝ × ℝ → ℂ satisfies the inequalityx, t ∈ ℝ, then there exists a solution u0 : ℝ × ℝ → ℂ of the wave equation  = v andx, t ∈ ℝ. Hence, we havex, t ∈ ℝ. Thus, it follows from inequality  is a solution of the wave equation  be given asu : ℝ × ℝ → ℂ satisfies inequality =αa, b ∈ ℝ, in view of Since


Due to a formatting error, the “permille” symbol (‰) was omitted whenever it was used in the text. Below are the locations of the errors:1. On Page 3, left column, last paragraph, “lowest 1” should be “lowest 1‰”:1a. “For each target, the averages over the lowest 1‰ and 1% CARMSD to native are reported.”1b. “We also report the average CARMSD of the lowest 1‰ and 1% Rosetta energies.”2. On Page 3, left column, last paragraph, “average 1” should be “average 1‰”:2a. “Even if Rosetta’s coarse-grained energy is noisy, EdaFold still takes advantage of it and achieves a better average 1‰ and 1% CARMSD”3. On Page 3, right column, last paragraph, “best 1” should be “best 1‰”:3a. “EdaFold outperforms Rosetta on a majority of targets either in terms of best 1‰ and 1% average CARMSD to native”4. On Page 4, table 1, “top 1” should be “top 1‰” in the row that starts with “Target”.5. On Page 6, legend of figure 4, “best 1” should be “best 1‰”.There was also an error in the last sentence of the Abstract. The correct link is:http://www.riken.jp/zhangiru/software.html [^]


Many known results appear as special consequences of our work. Some applications of our work to the generalized integral operator are also given.We obtain certain simple sufficiency criteria for a class of  Ap(n) denote the class of functions f(z) of the form𝕌 = {z ∈ ℂ : |z| < 1}. We note that A1(n) = A(n) and A1(1) = A. Also let Sp*, Cp and Mp, , denote the subclasses of Ap(n) consisting of all functions f(z), of the form  ≡ S*, M1 ≡ C.Let the formf(z)=zp+∑Mp of multivalent alpha convex functions of order β. We also consider some special cases of our results which lead to various interesting corollaries, and relevances of some of these with other known results are also mentioned.Sufficient condition was studied by various authors for different subclasses of analytic and multivalent functions; for some of the related work, see –9. The on ∈ ℕ, 0 ≤ α ≤ 1, 0 ≤ β < p, and p ∈ ℕ.We will assume throughout our discussion, unless otherwise stated, that To obtain our main results, we need the following lemmas due to Mocanu .q(z) ∈ A(n) satisfies the conditionIf q(z) ∈ A(n) satisfies the conditionδn is the unique root of the equationIf f(z) ∈ Ap(n) satisfiesf(z) ∈ Mp.If q(z) byf(z) ∈ Ap(n). Then clearly (q(z) ∈ A(n).Let us set a function q(z) byq(z)=z(f(q(z) ∈ S*.Differentiating  logarithus using , we haveq(z) ∈ S*, it implies that Re(zq′(z)/q(z)) > 0. Therefore, we getf(z) ∈ Mp.From , we can α = 0 and α = 1 in By taking f(z) ∈ Ap(n) satisfiesβ < p, then f(z) ∈ Sp*.If f(z) ∈ Ap(n) satisfiesβ < p, then f(z) ∈ Cp.If n = 1 and p = 1 in Corollaries Further if we take f(z) ∈ A satisfiesβ < 1, then f(z) ∈ S*(β).If f(z) ∈ A satisfiesβ < 1, then f(z) ∈ C(β).If β = 0 and p = 1 in Corollaries If we put f(z) ∈ Ap(n) satisfiesδn is the unique root of (f(z) ∈ Mp.If  root of , then f(f(z) ∈ Ap(n). Then clearly (q(z) ∈ A(n).Let us set clearly  shows thδn is the root of (q(z) ∈ S*. Now by the same arguments as in the proof of Differentiating , we haveus using , we have root of . Hence, α = 0 in Making f(z) ∈ Ap(n) satisfiesf(z) ∈ Sp*.If n = 1 = p in Further if we take f(z) ∈ A satisfiesδ1 is the unique root of the equationf(z) belongs to the class of starlike functions of order β.If α = 1 in Taking f(z) ∈ Ap(n) satisfiesf(z) ∈ Cp.If n = 1 and p = 1 in β = 0, we get the result proved by Mocanu [If we take y Mocanu .f(z) ∈ Ap(n), we considerG(z) ∈ A(n), and when p = 1, γ = 1, and δ = 1, then (For considerG(z)={δ∫0γ ≥ δ/p, δ > 0, and f(z) ∈ Ap(n) satisfyf(z) ∈ Sp*.If p = 1, β = 0, and α = 1/δ, we have  G(z) ∈ M1.From , we getGδ−1(z)G′ave used . By usinf(z) ∈ Sp*, where γ ≥ δ/p.From , we can γ ≥ δ/p, δ > 0, and f(z) ∈ Ap(n) satisfyδn is the unique root of (f(z) ∈ Sp*.If satisfy3|arg(f(z)p = 1, β = 0, and α = 1/δ.The result follows on similar lines as in the last theorem using


Fixed point results for a self-map satisfying locally contractive conditions on a closed ball in an ordered 0-complete quasi-partial metric space havebeen established. Instead of monotone mapping, the notion of dominated mappings is applied. We have used weaker metric, weaker contractive conditions, andweaker restrictions to obtain unique fixed points. An example is given which shows that how this result can be used when the corresponding results cannot. Our results generalize, extend, and improve several well-known conventional results. Fixed points results of mappings satisfying certain contractive conditions on the entire domain have been at the centre of vigorous research activity, see –3) and i and i3])Recently, many results appeared related to fixed point theorem in complete metric spaces endowed with a partial ordering in the literature. Ran and Reurings  proved aOn the other hand notion of a partial metric space was introduced by Matthews in . In partT is a contraction not on the entire space X but merely on a subset Y of X. However, if Y is closed and a sequence {xn} in X converges to some x in X, then by imposing a subtle restriction on the choice of x0, one may force the sequence to stay eventually in Y. In this case, one can establish the existence of a fixed point of T. Arshad et al. ,  x, y ∈ X is a partial metric on X. The function dpq : X × X → R+ defined by dpq = q + q − q − q is a  metric on X. The ball x ∈ X and ε > 0.Note that if X, q) be a quasi-partial metric. Then, we have the following; xn} in  converges to a point x ∈ X if and only if limn→∞  q = limn→∞  q = q.A sequence {xn} in  is called a Cauchy sequence if the limits limn,m→∞  q and limn,m→∞  q exist (and are finite).A sequence {X, q) is said to be complete if every Cauchy sequence {xn} in  converges to a point x ∈ X such that  = limn,m→∞  q = limn,m→∞  q.The space  be a quasi-partial metric space, let  be the corresponding partial metric space, and let  be the corresponding metric space. These statements are equivalent. (i) The sequence {xn} is Cauchy in . (ii) The sequence {xn} is Cauchy in . (iii) The sequence {xn} is Cauchy in . These statements are also equivalent. (i)  is complete. (ii)  is complete. (iii)  is complete.Let  be a partial ordered set. Then x, y ∈ X are called comparable if x⪯y or y⪯x holds.Let  be a partially ordered set. A self-mapping f on X is called dominated if fx⪯x for each x in X.Let  be a complete metric space, S : X → X a mapping, r > 0, and x0 an arbitrary point in X. Suppose there exists k ∈ [0,1) withd<(1 − k)r. Then there exists a unique point x* in x* = Sx*.Let  be a quasi-partial metric space.xn} in  is called 0-Cauchy if limn,m→∞  q = 0 or limn,m→∞  q = 0.A sequence {X, q) is called 0-complete if every 0-Cauchy sequence in X converges to a point x ∈ X such that q = 0.The space  is Cauchy in  and if  is complete, then it is 0-complete but the converse assertions do not hold. For example, the space X = [0, +∞)∩Q with q = |x − y | +|x| is a 0-complete quasi-partial metric space but it is not complete  = 2 | x − y| and  is not complete).It is easy to see that every 0-Cauchy sequence in  is complete quasimetric space which is also quasi-partial, then it is 0-complete, quasi-partial metric space.Moreover, if  is called an ordered quasi-partial metric space if (i) q is a quasi-partial metric on X and (ii) ⪯ is a partial order on X.Let X, ⪯, q) be a 0-complete ordered quasi-partial metric space, S : X → X a dominated map, and x0 an arbitrary point in X. Suppose that there exists k ∈ [0,1) such thatxn} in xn} → u implies that u⪯xn, then there exists a point x* in x* = Sx* and q = 0. Moreover, x* is unique, if for every pair of elements x, y in X  there exists a point z ∈ X such that z⪯x and z⪯y.Let ¯ withq=Uniqueness. Let y be another point in y = Sy. If x* and y are comparable, thenx* = y. Now if x* and y are not comparable then there exists a point x* and y, that is, z⪯x* and z⪯y. Moreover by assumption z⪯x*⪯xn ⋯ ⪯x0. Now we will prove that n ∈ N. Now as S is dominated, it follows that Sn−1z⪯Sn−2z⪯⋯⪯z⪯x* and Sn−1z⪯y for all n ∈ N, which further implies Sn−1z⪯Snx* and Sn−1z⪯Sny for all n ∈ N as Snx* = x* and Sny = y for all n ∈ N:x* = y.X = [0, +∞)∩Q be endowed with order x⪯y if q ≤ q and let q : X × X → R+ be the 0-complete ordered quasi-partial metric on X defined by q = max{y − x, 0} + x. DefineS is dominated mapping. Take x0 = 1/2,  r = 1/2; then q = 1/2,  k = 4/9 withx, y ∈ ∩X, then,X in each case.Let S and q = 0.Now ifx, y in X.In X, ⪯, q) be a 0-complete ordered quasi-partial metric space, S : X → X a dominated map, and x0 an arbitrary point in X. Suppose there exists k ∈ [0,1) withxn} in X,  {xn} → u implies that u⪯xn, then there exists a point x* in X such that x* = Sx* and q = 0. Moreover, x* is unique, if for every pair of elements x, y in X, there exists a point z ∈ X such that z⪯x and z⪯y.Let  be a 0-complete partial metric space, S : X → X a map, and x0 an arbitrary point in X. Suppose there exists k ∈ [0,1) withx* in x* = Sx*. Further q = 0.Let  be a complete ordered metric space, S : X → X a dominated map, and x0 an arbitrary point in X. Suppose that there exists k ∈ [0,1) such thatxn} in xn} → u implies that u⪯xn, then there exists a point x* in x* = Sx* and q = 0. Moreover, x* is unique, if for every pair of elements x, y in X there exists a point z ∈ X such that z⪯x and z⪯y.Let  be a 0-complete ordered quasi-partial metric space, S : X → X a dominated map, and x0 an arbitrary point in X. Suppose that there exists b ∈ [0, 1/2) such thatx, y in Let (k = b/(1 − b). If, for a nonincreasing sequence {xn} in xn} → u implies that u⪯xn, then there exists a point x* in x* = Sx* and q = 0. Moreover, x* is unique, if for any two points x, y in z⪯x and z⪯y, andAndxn+1 = Sxn with initial guess x0, as xn+1 = Sxn⪯xn for all n ∈ {0} ∪ N. We will prove that n ∈ N by mathematical induction. By using inequality  ≤ q, we have q = 0. Similarly, q = 0. Now if x* and y are not comparable to then there exists a point x* and y. Moreover by assumption Sz⪯z⪯x*⪯xn ⋯ ⪯x0. Now by using inequality  ≤ q, we have q = 0. Similarly, q = 0. Hence x* = y.equality , we haveequality . Let S2zies thatq be a complete dislocated quasimetric space, S : X → X a map, and x0 an arbitrary point in X. Suppose there exists k ∈ [0, 1/2) withx, y in θ = k/(1 − k). Then there exists a unique point x* in x* = Sx* and dq = 0.Let  be a 0-complete ordered quasi-partial metric space and let S : X → X be a dominated map. Suppose that there exists b ∈ [0, 1/2) such thatx, y in X. If, for a nonincreasing sequence {xn} in X,  {xn} → u implies that u⪯xn, then there exists a point x* in X such that x* = Sx* and q = 0. Moreover, x* is unique, if for any two points x, y in z⪯x and z⪯y.Let  be a 0-complete ordered quasi-partial metric space, S : X → X a dominated map, and x0 an arbitrary point in X. Suppose that there exists c ∈ [0, 1/2) such thatx, y in Let (k = c/(1 − c). If, for a nonincreasing sequence {xn} in xn} → u implies that u⪯xn, then there exists a point x* in x* = Sx*. Further q = 0.Andxn+1 = Sxn with initial guess x0. As xn+1 = Sxn⪯xn for all n ∈ {0} ∪ N. We will prove that n ∈ N by mathematical induction. By using inequality (j ∈ N. As xn+1⪯xn, so using inequality (n ∈ N. Also xn+1⪯xn for all n ∈ N. It implies thatxn} is a 0-Cauchy sequence in n → ∞ and using the fact that x*⪯xn⪯xn−1, when xn → x*, we havex* = Sx*.Consider a Picard sequence equality , we haveequality , we obtaWe can obtain the partial metric, quasi-metric, and metric version of all theorems which are still not present in the literature.


The strong order of the convergence of the numerical method is given, and the convergence of the numerical method is obtained. Some earlier results are generalized and improved.This paper is concerned with the convergence of stochastic  Baker and Buckwar x1, x2, x3 ∈ ℜn.There is a positive constant In fact, the global Lipschitz condition  implies τ0 = 0 and τj, j = 1,2,… are the jump times.We note for later reference that  involvesθ-Euler methods . Here h ∈  is a step size, which satisfies M = T/h for some positive integer M, tk = kh  . yk ≈ x(tk), Δwk = w(tk+1) − w(tk), and ΔNk = N(tk+1) − N(tk) are the Brownian and Poisson increments, respectively.One generalisation of stochastic  methods , 21 to st ∈ , we definey(t) immediately provides a result for yk.For ximation . So a coCi, Di, i = 1,2,… denote generic constants, independent of h. The main theorem of the paper is rather technical. We will present a number of useful lemmas in the section and then complete the proof in Throughout our analysis, h* < 1 such that for all 0 < h ≤ h*,Under αβ ≤ |α|2 + |β|2. Now, using E | Δwk|2 = mh and From , we haveAi(h), i = 1,…, 5 is a constant dependent on h and A3(h) = 4K2h2(θ2 + (1 − θ)θ).For the jump, we convert to the compensated Poisson increment ombining , 14), a, aΔN~k∶=ombining  with  ≥ 1/2 and noting that (E | γ(tj)|2 ≤ B1, we obtainNow choosing h* > 0 such that, for all 0 < h ≤ h*,Under t ∈ ⊆. In this interval we haveC3 = 3mK2(1 + 2C1) + 3λ(1 + λ)K2(1 + 2C1 + B1) + 15K2.Consider irtue of –14) and andt ∈ ⊆. By |2 ≤ B1, on , z1 ≡ yk, z2 ≡ yk+1, C5 = 72K2(1 + 2C1) + 24K2λ(2 + λ)(1 + 3C1). In the following we consider C6 = 4C5; the proof is complete.Consider , T]. By , we have T]. By (3y(μt)−z¯We can now state and prove our main result of this paper.q > 1 and Assumptions h* > 0 and C = C(q) such that, for all 0 < h < h*,Under t1 ≤ T we haveThe analysis uses ideas from , where aM′ is the smallest integer such that M′h ≥ t1.By a, b, ɛ > 0 and 1 < q < ∞.Now the number of nonzero terms in the summation in  is a ranɛ = hq − 1)2/q(4 + λT)((q − 1)λ + 2q2B) + 16K1((C3 + C5)(T + 2 + 4λ + λ2T) + TC4 + TC6).Now, substituting , 32), , , and 4141 into  yields(By the Gronwall inequality, we haveθ-methods give strong convergence rate arbitrarily close to order 1/2 under appropriate moment bounds on the jump magnitude. This problem class is now widely used in mathematical finance.By Under The convergent result can be extended to the case of nonlinear coefficients that are local Lipschitz , 7, 12 bUnder the local Lipschitz condition and


Conditions for the existence of locally expansive C1 solutions for such equations are given.Iterative equations which can be expressed by the following form  C be the set of all continuous self-mappings on a topological space X. For any f ∈ C, let fm denote the mth iterate of f; that is, fm = f∘fm−1, f0 = id,  m = 1,2,…. Equations having iteration as their main operation, that is, including iterates of the unknown mapping, are called iterative equations. It is one of the most interesting classes of functional equations . By the choices of H and ϕ, we have λϕ(0) = 0 and λiϕ(0) = Hi′(O), where i = 0,1,…, n − 1.For f of (f(x) = ϕ(c(ϕ−1(x))) by the Schröder transformation, where c is a constant to be determined, then  is a zero of the following polynomial:If function ution of , then weynomial:P(x)=xn−Hlynomial .Finally, we give a basic lemma. D ⊂ Rm be a convex open set, and let a =  and a + h =  belong to D, then there exists a θ ∈  such thatLet S(n) = {1,2,…, n − 1}. Let H ∈ ℋ. Suppose that there is a neighborhood U of  O ∈ Rn that satisfies A1+) ≥ H0′(X) ≥ 0;  for all A2+), Hi′(O) ≥ Hi′(X) ≥ 0. for all  Suppose that C1 solution near 0. Then,  has a lon is odd and H ∈ ℋ. Suppose that there is a neighborhood  U of O ∈ Rn that satisfies A1−) ≤ H0′(X) ≤ 0; for all A2±) ≥ H0′(X) ≥ 0 for all odd i ∈ S(n) and Hi′(O) ≤ Hi′(X) ≤ 0 for all even i ∈ S(n). for all Suppose that C1 solution near 0. Then,  has a lon is even and H ∈ ℋ. Suppose that there is a neighborhood  U of O ∈ Rn that satisfies A1+) ≥ H0′(X) ≥ 0; for all A2∓) ≤ Hi′(X) ≤ 0 for all odd i ∈ S(n) and Hi′(O) ≥ Hi′(X) ≥ 0 for all even i ∈ S(n). for all Suppose that C1 solution near 0.Then,  has a loc > 1  and σ > 0 such that for arbitrary given τ > 0,  = 0 and ϕ′(0) = τ.Under the conditions of n τ > 0,  has a  Cc is real and (C1 solution ϕ with ϕ(0) = 0 and ϕ′(0) ≠ 0, then by differentiating the equation, we can see that c is a root of characteristic polynomial →+∞ when x → +∞, and this means that P has a root c > 1. In the case of Theorems P has a root c < −1. Since for both of the cases c > 1 and c < −1, 0 < |cn+i− | <1,  i = 0,1,…, n − 1 and c is a zero of (Hi′(O),  i = 0,1,…, n − 1. This also means that 1 − ∑i=0n−1|Hi′(O)||cn+2i−2| > 0. Now, we can choose a constant σ1 > 0 such that the following statements are true;A1σ) holds on , where σ ∈ {+, −}; (A2δ) holds on , where δ ∈ {+, ±, ∓};  holds on , by A2δ, δ ∈ {+, ±, ∓}, we haveϕ, we can get thatK2, we get that𝒢) ⊂ 𝒜.Moreover, for all 𝒢 is continuous. Considering ϕ, φ ∈ 𝒜, by A2δ, δ ∈ {+, ±, ∓}, we have𝒢 is evident. By Schauder's fixed point theorem, there exists a ϕ ∈ 𝒜 such that 𝒢ϕ = ϕ. This means that  of 0 such that ϕ−1 exists and is also C1 on J. Without any loss of generality, we can assume that J = ϕ. Hence, ϕ :  → J is a homeomorphism. Moreover, we can choose a neighborhood I ⊂ J of 0 which is so small that ciϕ−1(x)∈ for all x ∈ I,  i = 1,2,…, n. Let f(x) = ϕ(cϕ−1(x)) for x ∈ I. Clearly f is also C1 and invertible on I. Moreover, all iterates fj,  j = 1,2,…, n, are well defined on I, and fj(x) = ϕ(cjϕ−1(x)),  x ∈ I. Obviously, we have f(0) = 0, f′(0) = c, and f is locally expansive. Finally, for any x ∈ I, we havef is a locally expansive C1 solution of  = 2sin(x0) + sin(x2). It is easy to verify that H satisfy the assumptions of C1 solution in a neighborhood of 0. Consider the following equation:H = −2sin(x0) − sin(x2). It is easy to verify that H satisfy the assumptions of C1 solution in a neighborhood of 0. Consider the following equation:H = 2sin(x0) + sin(x2). It is easy to verify that H satisfy the assumptions of C1 solution in a neighborhood of 0. Consider the following equation:


We introduce the concept of Cayley bipolar fuzzy graphs and investigate some of their properties. We present some interesting properties of bipolar fuzzy graphs in terms of algebraic structures. We also discuss connectedness in Cayley bipolar fuzzy graphs. A digraph is a graph whose edges have direction and are called arcs (edges). Arrows on the arcs are used to encode the directional information: an arc from vertex (node) x to vertex y indicates that one may move from x to y but not from y to x. The Cayley graph was first considered for finite groups by Cayley in 1878. Max Dehn in his unpublished lectures on group theory from 1909 to 1910 reintroduced Cayley graphs under the name Gruppenbild (group diagram), which led to the geometric group theory of today. His most important application was the solution of the word problem for the fundamental group of surfaces with genus, which is equivalent to the topological problem of deciding which closed curves on the surface contract to a point  and μRN : X × Y → .A bipolar fuzzy relation R be a bipolar fuzzy relation on universe X. Then R is called a bipolar fuzzy equivalence relation on X if it satisfies the following conditions: R  is bipolar fuzzy reflexive; that is, R =  for each x ∈ X;R is bipolar fuzzy symmetric; that is, R = R for any x, y ∈ X;R is bipolar fuzzy transitive; that is, R ≥ ⋁y∧R).Let R be a bipolar fuzzy relation on universe X. Then R is called a bipolar fuzzy partial order relation on X if it satisfies the following conditions:R is bipolar fuzzy reflexive; that is, R =  for each x ∈ X;R is bipolar fuzzy antisymmetric; that is, R ≠ R for any x, y ∈ X;R is bipolar fuzzy transitive; that is, R ≥ ⋁y∧R).Let R be a bipolar fuzzy relation on universe X. Then R is called a bipolar fuzzy linear order relation on X if it satisfies the following conditions:R is bipolar fuzzy partial relation;μRP∨μR−1P) > 0, (μRN∧μR−1N) < 0 for all x, y ∈ X. where A =  is a bipolar fuzzy set in V and B =  is a bipolar fuzzy relation on E such thatxy ∈ E. We note that B need not to be symmetric.A D be a bipolar fuzzy digraph. The indegree of a vertex x in D is defined by ind(x) = (indμP(x), indμN(x)), where indμP(x) = ∑y≠xμAP(xy) and indμN(x) = ∑y≠xμAN(xy). Similarly, the outdegree of a vertex x in D is defined by outd(x) = (outdμP(x), outdμN(x)), where outdμP(x) = ∑y≠xμAP(xy) and outdμN(x) = ∑y≠xμAN(xy). A bipolar fuzzy digraph in which each vertex has the same outdegree r is called an outregular digraph with index of outregularity r. In-regular digraphs are defined similarly.Let V, ∗) be a group and let A =  be the bipolar fuzzy subset of V. Then the bipolar fuzzy relation R defined on V byG =  called the Cayley bipolar fuzzy graph induced by the .Let  be a group and let A =  be a bipolar fuzzy subset of V. Then the bipolar fuzzy relation R on V defined byG = , called the Cayley bipolar fuzzy graph induced by the .Let  = μAP(1) = μAP(2) = 0.5, μAN(0) = μAN(1) = μAN(2) = −0.4. Then the Cayley bipolar fuzzy graph G =  induced by  is given by Consider the group R in the above definition describes the strength of each directed edge. Let G denote a bipolar fuzzy graph G =  induced by the triple .We see that Cayley bipolar fuzzy graphs are actually bipolar fuzzy digraphs. Furthermore, the relation G is vertex transitive.The Cayley bipolar fuzzy graph a,  b ∈ V. Define ψ : V → V by ψ(x) = ba−1x for all x ∈ V. Clearly, ψ is a bijective map. For each x, y ∈ V,Let R(ψ(x), ψ(y)) = R. Hence ψ is an automorphism on G. Also ψ(a) = b. Hence G is vertex transitive.Every vertex transitive bipolar fuzzy graph is regular.G =  be any vertex transitive bipolar fuzzy graph. Let u, v ∈ V. Then there is an automorphism f on G such that f(u) = v. Note thatG is regular.Let Cayley bipolar fuzzy graphs are regular.Proof follows from Theorems G =  denote bipolar fuzzy graph. Then bipolar fuzzy relation R is reflexive if and only if μAP(1) = 1 and μAN(1) = −1.Let R is reflexive if and only if R =  for all x ∈ V. NowR is reflexive if and only if μAP(1) = 1 and μAN(1) = −1.G =  denote bipolar fuzzy graph. Then bipolar fuzzy relation R is symmetric if and only if (μAP(x), μAN(x)) = (μAP(x−1), μAN(x−1)) for all x ∈ V.Let R is symmetric. Then for any x ∈ V,μAP(x), μAN(x)) = (μAP(x−1), μAN(x−1)) for all x ∈ V. Then for all x, y ∈ V,R is symmetric.Suppose that R is antisymmetric if and only if {x : (μAP(x), μAN(x)) = (μAP(x−1), μAN(x−1)) = .A bipolar fuzzy relation S, ∗) be a semigroup. Let A =  be a bipolar fuzzy subset of S. Then A is said to be a bipolar fuzzy subsemigroup of S if for all x, y ∈ S, μBP(xy) ≥ μAP(x)∧μAP(y) and μBN(xy) ≤ μAN(x)∨μAN(y).Let  is a bipolar fuzzy subsemigroup of .A bipolar fuzzy relation R is transitive and let x, y ∈ V. Then R2 ≤ R. Now for any x ∈ V, we have R = (μAP(x), μAN(x)). This implies that {R∧R : z ∈ V} = R2 ≤ R. That is ∨{μAP(z)∧μAP(z−1xy) : z ∈ V} ≤ μAP(xy) and ∧{μAN(z)∨μAN(z−1xy) : z ∈ V} ≥ μAN(xy). Hence μAP(xy) ≥ μAP(x)∧μAP(y) and μAN(xy) ≤ μAP(x)∨μAN(y). Hence A =  is a bipolar fuzzy subsemigroup of .Suppose that A =  is a bipolar fuzzy subsemigroup of . That is, for all x, y ∈ VμBP(xy) ≥ μAP(x)∧μAP(y) and μBN(xy) ≤ μAN(x)∨μAN(y). Then for any x, y ∈ V,RμP2 ≤ RμP and RμN2 ≥ RμN. Hence R is transitive.Conversely, suppose that We conclude that.R is a partial order if and only if A =  is a bipolar fuzzy subsemigroup of  satisfyingμAP(1) = 1 and μAN(1) = −1,x : (μAP(x), μAN(x)) = (μAP(x−1), μAN(x−1)) = {1, −1}}.{A bipolar fuzzy relation R is a linear order if and only if  is a bipolar fuzzy subsemigroup of  satisfyingμAP(1) = 1 and μAN(1) = −1,x : (μAP(x), μAN(x)) = (μAP(x−1), μAN(x−1)) = {1, −1}},{x : μAP(x)∨μAP(x−1) > 0, μAN(x)∧μAN(x−1) < 0} = V. {A bipolar fuzzy relation R is a linear order. Then by x ∈ V, (R∨R−1) > 0. This implies that R∨R > 0. Hence {x : μAP(x)∨μAP(x−1) > 0, μAN(x)∧μAN(x−1) < 0}.Suppose R is partial order. Now for any x, y ∈ V, we have (x−1y), (y−1x) ∈ V. Then by condition (iv), {x : μAP(x)∨μAP(x−1) > 0, μAN(x)∧μAN(x−1) < 0}. Therefore R is linear order.Conversely, suppose that conditions (i), (ii), and (iii) hold. Then by R is a equivalence relation if and only if  is a bipolar fuzzy subsemigroup of  satisfyingμAP(1) = 1 and μAN(1) = −1,μAP(x), μAN(x)) = (μAP(x−1), μAN(x−1)) for all x ∈ V. > 0, μAN(xi) < 0, for i = 1,2, 3,…, n, we have, μAP(x1x2 ⋯ xn) = 0 and μAN(x1x2 ⋯ xn) = 0.G is a Hasse diagram and let x1, x2,…, xn be vertices in V with n ≥ 2 and μAP(xi) > 0, μAN(xi) < 0, for i = 1,2, 3,…, n. Then it is obvious that R = (μAP(xi), μAN(xi)), for i = 1,2,…, n, where x0 = 1. Therefore  is a path from 1 to x1x2 ⋯ xn. Since G is a Hasse diagram, we have R = 0. This implies that μAP(x1x2 ⋯ xn) = 0 and μAN(x1x2 ⋯ xn) = 0. Conversely, suppose that for any collection x1, x2,…xn of vertices in V with n ≥ 2 and μAP(xi) > 0, μAN(xi) < 0, for i = 1,2, 3,…, n, we have, μAP(x1x2 ⋯ xn) = 0 and μAN(x1x2 ⋯ xn) = 0. Let  be a path in G from x0 to xn with n ≥ 2. Then R > 0, for i = 1,2,…, n. Therefore μAP(xi−1−1xi) > 0, μAN(xi−1−1xi) < 0, for i = 1,2,…, n. Now consider the elements x0−1x1, x1−1x2,…, xn−1−1xn in V. Then by assumption μAP(x0−1x1x1−1x2 ⋯ xn−1−1xn) = 0 and μAN(x0−1x1x1−1x2 ⋯ xn−1−1xn) = 0. That is, μAP(x0−1xn) = 0 and μAN(x0−1xn) = 0. Hence, R = 0. Thus G is a Hasse diagram.Suppose G =  be any bipolar fuzzy graph; then G is connected  if and only if the induce fuzzy graph  is connected .Let S, ∗) be a semigroup and let A =  be a bipolar fuzzy subset of S. Then the subsemigroup generated by A is the meeting of all bipolar fuzzy subsemigroups of S which contains A. It is denoted by 〈A〉.Let  be a semigroup and A =  be a bipolar fuzzy subset of S. Then bipolar fuzzy subset 〈A〉 is precisely given by 〈μAP〉(x) = ∨{μAP(x1)∧μAP(x2)∧⋯∧μAP(xn) : x = x1x2 ⋯ xn with  μAP(xi) > 0 for i = 1,2,…, n}, 〈μAN〉(x) = ∧{μAN(x1)∨μAN(x2)∨⋯∨μAN(xn) : x = x1x2 ⋯ xn with μAN(xi) < 0 fori = 1,2,…, n} for any x ∈ S.Let (S defined by μAP(xi) > 0 for i = 1,2,…, n}, μAN(xi) < 0 for i = 1,2,…, n}, for any x ∈ S. Let x, y ∈ S. If μAP(x) = 0 or μAP(y) = 0, then μAP(x)∧μAP(y) = 0 and μAN(x) = 0 or μAN(y) = 0, and then μAN(x)∨μAN(y) = 0. Therefore, μAP(x) ≠ 0, μAN(x) ≠ 0, then by definition of S containing . Now let L be any bipolar fuzzy subsemigroup of S containing . Then for any x ∈ S with x = x1x2 ⋯ xn with μAP(xi) > 0, μAN(xi) < 0, for i = 1,2,…, n, we have μLP(xi) ≥ μLP(x1)∧μLP(x2)∧⋯∧μLP(xn) ≥ μAP(x1)∧μAP(x2)∧⋯∧μAP(xn) and μLN(xi) ≥ μLN(x1)∧μLN(x2)∧⋯∧μLN(xn) ≥ μAN(x1)∧μAN(x2)∧⋯∧μAN(xn). Thus μLP(x)≥∨{μAP(x1)∧μAP(x2)∧⋯∧μAP(xn) : x = x1x2 ⋯ xn with μAP(xi) > 0 for i = 1,2,…, n}, μLN(x)≤∧{μAN(x1)∨μAN(x2)∨⋯∨μAN(xn) : x = x1x2 ⋯ xn with μAN(xi) < 0 for i = 1,2,…, n}, for any x ∈ S. Hence x ∈ S. Thus μAP, μAN).Let S, ∗) be a semigroup and A =  be a bipolar fuzzy subset of S. Then for any α ∈ ,  =  and (〈(μ+)αP〉, 〈(μ+)αN〉) = , where  denotes the subsemigroup generated by  and 〈〉 denotes bipolar fuzzy subsemigroup generated by .Let  = . Similarly, we have (〈(μ+)αP〉, 〈(μ+)αN〉) = .S, ∗) be a semigroup and A =  be a bipolar fuzzy subset of S. Then by A) = A+〉 = supp〈A〉.Let  induced by . Then we have the following results.Let A be any subset of V′ and G′ =  be the Cayley graph induced by . Then G′ is connected if and only if 〈A〉⊇V − v1.Let G is connected if and only if  supp〈A〉⊇V − v1.A be any subset of a set V′ and let G′ =  be the Cayley graph induced by the triplet . Then G′ is weakly connected if and only if 〈A ∪ A−1〉⊇V − v1, where A−1 = {x−1 : x ∈ A}.Let S, ∗) be a group and let A be a bipolar fuzzy subset of S. Then we define A−1 as bipolar fuzzy subset of S given by A−1(x) = A(x−1) for all x ∈ S.Let (G is weakly connected if and only if supp(〈A ∪ A−1〉)⊇V − v1.A be any subset of a set V′ and let G′ =  be the Cayley graph induced by the triplet . Then G′ is semiconnected if and only if 〈A〉 ∪ 〈A−1〉⊇V − v1, where A−1 = {x−1 : x ∈ A}.Let G is semi-connected if and only if supp(〈A〉 ∪ 〈A−1〉)⊇V − v1.G′ =  be the Cayley graph induced by the triplet . Then G′ is locally connected if and only if 〈A〉 = 〈A−1〉, where A−1 = (x−1 : x ∈ A).Let G is locally connected if and only if supp(〈A〉) = supp(〈A−1〉).Let G′ =  be the Cayley graph induced by the triplet , where V′ is finite. Then G′ is quasi-connected if and only if it is connected.Let G is quasi-connected if and only if it is connected.A finite Cayley bipolar fuzzy graph μP strength of a path P = v1, v2,…, vn is defined as min) for all i and j and is denoted by SμP. The μN strength of a path P = v1, v2,…, vn is defined as max) for all i and j and is denoted by SμN.The G =  be a bipolar fuzzy graph. Then G is said to beα-connected if for every pair of vertices x, y ∈ G, there is a path P from x to y such that strength (P) ≥ α,weakly α-connected if a bipolar fuzzy graph  is α-connected,semi-α-connected if for every x, y ∈ V, there is a path of strength greater than or equal to α from x to y or from y to x in G,locally α-connected if for every pair of vertices x and y, there is a path P of strength greater than or equal to α from x to y whenever there is a path P′ of strength greater than or equal to α from y to x,quasi-α-connected if for every pair x, y ∈ V, there is some z ∈ V such that there is directed path from z to x of strength greater than or equal to α and there is a directed path from z to y of strength greater than or equal to α.Let G =  be any bipolar fuzzy graph; then G is α-connected  if and only if the induce fuzzy graph  is connected .Let G denote the Cayley bipolar fuzzy graphs G =  induced by . Also for any α ∈ , we have the following results.Let G is α-connected if and only if 〈A〉α⊇V − v1.G is weakly α-connected if and only if 〈A∪A−1〉α⊇V − v1.G is semi-α-connected if and only if (〈A〉α ∪ 〈A−1〉α)⊇V − v1.G be locally α-connected if and only if 〈A〉α = 〈Aα−1〉.Let G is quasi-α-connected if and only if it is α-connected.A finite Cayley bipolar fuzzy graph Fuzzy graph theory is finding an increasing number of applications in modeling real time systems where the level of information inherent in the system varies with different levels of precision. Fuzzy models are becoming useful because of their aim of reducing the differences between the traditional numerical models used in engineering and sciences and the symbolic models used in expert systems. A bipolar fuzzy set is a generalization of the notion of a fuzzy set. We have introduced the notion of Cayley bipolar fuzzy graphs in this paper. The natural extension of this research work is application of bipolar fuzzy digraphs in the area of soft computing including neural networks, decision making, and geographical information systems.


We consider the three-dimensional Boussinesq equations, and obtain an Osgood type regularity criterion in terms of the velocity gradient. Assume that  is the smooth solution to  for 0 ⩽ t < T. Ifu, θ) can be extended after time t = T. Here, Let , we haveThe Osgood type condition  is weakeThe rest of this paper is organized as follows. In 𝒮(ℝ3) be the Schwartz class of rapidly decreasing functions. For f ∈ 𝒮(ℝ3), its Fourier transform φ ∈ 𝒮(ℝ3) such thatj ∈ ℤ, the Littlewood-Paley projection operators Sj and f ∈ L2(ℝ3), thenL2 sense. By telescoping the series, we thus have the following Littlewood-Paley decomposition:f ∈ L2(ℝ3), where the summation is the L2 sense. Notice thatp ⩽ q ⩽ ∞ and C is an absolute constant independent of f and j.Let ∞ < s < ∞, 1 ⩽ p, q ⩽ ∞; the homogeneous Besov space 𝒵′(ℝ3) is the dual space ofLet −refer to  for moreH1 bounds of the solution .This section is devoted to proving 1 with −Δu, , then  is uniformly bounded in L∞). This completes the proof of Taking the inner products of 1 with −Δon as in , we getn as in (∇u=∑l<−qΔimate asJ3⩽∑l>q||see thatJ⩽∑l<−q∫ℝy to getI1⩽12||∇(thering (J⩽[C2−q/2her that12ddt||∇(ituting (12ddt||∇(mate as2J3⩽∑l>q||


F3(·) due to Marichev-Saigo-Maeda, to the product of the generalized Bessel function of the first kind due to Baricz. The results are expressed in terms of the multivariable generalized Lauricella functions. Corresponding assertions in terms of Saigo, Erdélyi-Kober, Riemann-Liouville, and Weyl type of fractional integrals are also presented. Some interesting special cases of our two main results are presented. We also point out that the results presented here, being of general character, are easily reducible to yield many diverse new and known integral formulas involving simpler functions.We apply generalized operators of fractional integration involving Appell's function  The fractional calculus is nowadays one of the most rapidly growing subject of mathematical analysis. It is a field of applied mathematics that deals with derivatives and integrals of arbitrary orders. The fractional integral operator involving various special functions has found significant importance and applications in various subfields of applicable mathematical analysis. Many applications of fractional calculus can be found in turbulence and fluid dynamics, stochastic dynamical system, plasma physics and controlled thermonuclear fusion, nonlinear control theory, image processing, nonlinear biological systems, astrophysics, and in quantum mechanics. Since last four decades, a number of workers like Love , McBridewν(z) has been introduced and studied in [The computation of fractional derivatives  of special functions of one and more variables is important from the point of view of the usefulness of these results in the evaluation of generalized integrals and the solution of differential and integral equations. Motivated by these avenues of applications, a remarkably large number of fractional integral formulas involving a variety of special functions have been developed by many authors . Fraudied in –34. Herex > 0 and α, α′, β, β′, γ ∈ ℂ byF3(·) in the kernel. These operators were rediscovered and studied by Saigo in [Fp;μ . Such fp;μ (see ) were stp;μ (see .F3(·) denotes so-called 3rd Appell function    2) the swν(z), which is defined for z ∈ ℂ∖{0} and b, c, ν ∈ ℂ with ℜ(ν)>−1 by the following series  is the familiar Gamma function   2) with e, e.g., –33, [34,p Ψq in several variables and defined by , βjk , γjk , and δjk , for all k = 1,…, n, are real and positive and (ap) means the array of p-parameters a1,…, ap; with similar interpretations for (bp), (γp11), (αp11), and so forth, and where (a)n is the Pochhammer symbol defined (for a ∈ ℂ) by   2) with Our results in this Section are based on the preliminary assertions giving composition formula of fractional integral  with a pα, α′, β, β′, γ ∈ ℂ and if ℜ(γ) > 0, ℜ(ρ) > max{0, ℜ(α + α′ + β − γ), ℜ(α′ − β′)}, thenLet The Marichev-Saigo-Maeda fractional integration  of produα, α′, β, β′, σ, λ, γ, νj, ρj, b, c ∈ ℂ and satisfying the inequalitiesx > 0, |ax/b | <1, one hasA, B, C, D, E, F, G, and H are given by the followingLet ℐ. Using definition , we obtainFollowing the convergence condition of This, in accordance with , gives tα′ = 0 in wν(z).On setting α, β, β′, σ, λ, γ, νj, ρj, b, c ∈ ℂ and x > 0, |ax/b| < 1A, C, D, F, G, and H are given by , , a, aβ, β′,given by , respectIn the sequel, we use the following results.α, α′, β, β′, γ ∈ ℂ and if ℜ(γ) > 0, ℜ(ρ) < 1 + min{ℜ(−β), ℜ(α + α′ − γ), ℜ(α + β′ − γ)}, thenLet The Marichev-Saigo-Maeda fractional integration  of produα, α′, β, β′, σ, λ, γ, νj, ρj, b, c ∈ ℂ and x > 0, |a/bx | <1 such thatA′, B′, C′, D′, E′, and F′ are given by the following:G, H are given by (Let given by  and 19)α, α′, β,𝒥. Applying definition (b − at)λ− and then changing the order of integration and summation, we findk ∈ ℕ0For convenience, let the left-hand side of the  be denotfinition  and the Now, on making use of This, in accordance with , gives tα′ = 0 in wν(z) as follows.On taking α, β, β′, σ, λ, γ, νj, ρj, b, c ∈ ℂ and x > 0, |a/bx| < 1, andA′, B′, C′, D′, G and H are given by , , ,  are oscillatory and may be regarded as generalizations of trigonometric functions. Indeed, for large argument (z ≥ 1) the function z − πν/2 − π/4). Similarly, modified Bessel functions of the first kind Iν(z), which are Bessel functions of imaginary argument, may be regarded as generalization of exponentials. Exponential functions have the unique and special property that they are particularly easy to multiply and to raise to powers: eazebz = ea+b)z( and (ez)r = erz. Further, it can be easily seen that for c = 1 and b = 1, the Generalized Bessel function of the first kind (Jν(z), and when c = −1 and b = 1 the function wν(z) becomes Iν(z). Similarly, when c = 1 and b = 2, the function wν(z) reduces to c = −1 and b = 2, then wν(z) becomes wν(0) = 0. Therefore, the results presented in this paper are easily converted in terms of the various special Bessel functions after some suitable parametric replacement.In this section, we briefly consider another variation of the results derived in the preceding sections. Bessel functions are important special functions that appear widely in science and engineering. Bessel functions of the first kind rst kind  reduces uel from , we haveν = −b/2, then the generalized Bessel function wν(z) in  in  have fole, e.g., ):ν = −b/2 and c = −c2, wν(z) in  in  in  have relν(z) in  have fole, e.g., ):(47)w1ν = 1 − b/2 and c = −c2 in (wν(z) and hyperbolic sine function as (48)w1−Now, by virtue of the relations  to 48),,48), oneα′ = 0, the results given by Malik et al. [λ = 0 and b = c = 1, then our main results α′ = 0, t results  and (31)Fractional integral formulas involving products of Bessel functions have been developed and play an important role in several physical problems. In fact, Bessel functions are playing the important role in studying solutions of differential equations, integral equations , and the


A new kind of fuzzy module over a fuzzy ring is introduced by generalizing Yuan and Lee's definition of the fuzzy group and Aktaş and Çağman's definition of fuzzy ring. The concepts of fuzzy submodule, and fuzzy module homomorphism are studied and some of their basic properties are presented analogous of ordinary module theory. G is fuzzy and the binary operation on G is nonfuzzy in the classical sense as Rosenfeld's definition is  a  mapping} andLet A = {a} and B = {b} and let R be denoted as a∘b; thenLet Using the notations in , we haveG be a nonempty set and let R be a fuzzy binary operation on G.   is called a fuzzy group if the following conditions are true: (G1)a, b, c, z1, z2 ∈ G, ((a∘b)∘c)(z1) > θ and (a∘(b∘c))(z2) > θ imply z1 = z2;∀(G2)eo ∈ G such that (eo∘a)(a) > θ and (a  ∘  eo)(a) > θ for any a ∈ G ;∃(G3)a ∈ G, ∃b ∈ G such that (a∘b)(eo) > θ and (b∘a)(eo) > θ .∀Let a∘b)∘c)(d) > θ⇔(a∘(b∘c))(d) > θ.Consider (c) > θ  implies  c ∈ H;∀a ∈ H  implies  a−1 ∈ H.H be a fuzzy subgroup of G. LetaH  (Ha) is called a left (right) coset of H.Let H be a fuzzy subgroup of G:H is called a normal fuzzy subgroup of G.Let G, R) be a fuzzy subgroup. IfG, R) is called abelian fuzzy group.Let  and  be two fuzzy groups and let f : G1 → G2 be a mapping. Iff is called a fuzzy (group) homomorphism. If f is 1-1, it is called a fuzzy monomorphism. If f is onto, it is called a fuzzy epimorphism. If f is both 1-1 and onto, it is called a fuzzy isomorphism.Let (G be a fuzzy binary operation on R. Then we have a mappingF(R) = {A | A : R →   is  a  mapping} andA = {a} and B = {b} and let G and H be denoted as a∘b and a∗b, respectively. ThenLet Using the notations of , we haveR be a nonempty set and let G and H be two fuzzy binary operations on R. Then  is called fuzzy ring if the following conditions hold: (R1)R, G) is an abelian fuzzy group;((R2)a, b, c, z1, z2 ∈ R, ((a∗b)∗c)(z1) > θ and (a∗(b∗c))(z2) > θ imply z1 = z2;∀(R3)a, b, c, z1, z2 ∈ R, ((a∘b)∗c)(z1) > θ and ((a∗c)∘(b∗c))(z2) > θ imply z1 = z2; (a∗(b∘c))(z1) > θ and ((a∗b)∘(a∗c))(z2) > θ imply z1 = z2.∀Let R, G, H) be a fuzzy ring. a∗b)(u) > θ⇔(b∗a)(u) > θ, then  is said to be a commutative fuzzy ring.If (e∗ ∈ R such that (a∗e∗)(a) > θ and (e∗∗a)(a) > θ for every a ∈ R, then  is said to be fuzzy ring with identity.If ∃R, G, H) be a fuzzy ring with identity. If (a∗b)(e∗) > θ and (b∗a)(e∗) > θ, ∀a ∈ R, ∃b ∈ R, then b is said to be an inverse element of a and is denoted by a∗−1.Let  be a fuzzy ring and let S be a nonempty subset of R. Then  is a fuzzy subring of R if and only ifa∘b)(c) > θ  implies  c ∈ S  and  (a∗b)(c) > θ  implies  c ∈ S  for all  a, b ∈ S, c ∈ R; is called a fuzzy ideal of R if the following conditions are satisfied. x, y ∈ I, (x∘y)(z) > θ⇒z ∈ I for all z ∈ R;∀x ∈ I, x−1 ∈ I;∀s ∈ I, for all r ∈ R, (r∗s)(x) > θ⇒x ∈ I and (s∗r)(y) > θ⇒y ∈ I, x, y ∈ R.For all A nonempty subset I be a fuzzy ideal of fuzzy ring R and let Ω = {a∘I | a ∈ R}. One defines a relation over Ω:Let R, G, H) be a fuzzy ring and  be an abelian fuzzy group and let p be fuzzy function R × M into M. Then we have a mappingF(R) = {A | A : R →   is  a  mapping} and F(M) = {N | N : M → is  a  mapping}.Let  and J be denoted as r⊙m and a ⊕ b, respectively. ThenLet Using the notations , we haveR, G, H) be a fuzzy ring and let  be an abelian fuzzy group. M is called a (left) fuzzy module over R or (left) R-fmodule together with a fuzzy function p : R × M → M if the following conditions hold. For all r, r1, r2 ∈ R and for all m, m1, m2 ∈ M,(M1)r⊙(m1 ⊕ m2))(x) > θ and ((r⊙m1)⊕(r ⊙ m2))(y) > θ imply x = y;((M2)r1∘r2)⊙m)(x) > θ and ((r1⊙m)⊕(r2⊙m))(y) > θ imply x = y;(((M3)r1∗r2)  ⊙  m)(x) > θ and (r1  ⊙  (r2  ⊙  m))(y) > θ imply x = y. be a fuzzy ring and let  be an R-fmodule; then for all r, r1, r2 ∈ R, m, m1, m2 ∈ M,r ⊙ (m1 ⊕ m2))(x) > θ⇔((r  ⊙  m1)⊕(r  ⊙  m2))(x) > θ;(r1∘r2)⊙m)(x) > θ  ⇔  ((r1⊙m)⊕(r2⊙m))(x) > θ;((r1∗r2)⊙m)(x) > θ  ⇔  (r1⊙(r2⊙m))(x) > θ.((Let (r  ⊙  (m1  ⊕  m2))(x) > θ and let x1, x2, y ∈ M such that p > θ, p > θ, and J > θ. Byx = y from (M1) and so ((r  ⊙  m1) ⊕ (r  ⊙  m2))(x) > θ. Similarly by ((r⊙m1)⊕(r⊙m2))(x) > θ we have (r⊙(m1 ⊕ m2))(x) > θ.(1) Let (It is easy to prove (2) and (3) similar to the proof of (1).R, G, H) be a fuzzy ring with zero element eo and  be a left R-fmodule with identity element eJ. Then for all r ∈ R, m ∈ M,r⊙eJ)(eJ) > θ;(eo⊙m)(eJ) > θ;(r⊙m)(x) > θ⇒(r⊙m−1)(x−1) > θ;(r⊙m)(x) > θ⇒(r−1⊙m)(x−1) > θ.(Let (x ∈ M such that (r⊙eJ)(x) > θ. Then  (1) Let r  ⊙  eJ) ⊕ (r  ⊙  eJ))(x) > θ from It follows that  > θ and x = eJ from Proposition  2.1 in )([x ⊕ B]) > θ and (r⊙[a ⊕ B])([y ⊕ B]) > θ. Then there exist p > θ and p > θ. Since a1 ⊕ B ~ a2 ⊕ B, we have x′ ⊕ B ~ y′ ⊕ B from x ⊕ B] = [y ⊕ B].Let (r⊙([a1 ⊕ B]⊕[a2 ⊕ B]))([x ⊕ B]) > θ and ((r⊙[a1 ⊕ B])⊕(r⊙[a2 ⊕ B]))([y ⊕ B]) > θ and let a ∈ A such that J > θ. Then there exist  x′, x1′, x2′ ∈ A, b1, b2 ∈ B such that(1) Let  > θ. Then by J > θ, J > θ, J > θ, and the proof of Theorem  4.2 in  = [y ⊕ B].It follows that r1∘r2)⊙[a ⊕ B])([x ⊕ B]) > θ and ((r1⊙[a ⊕ B]) ⊕ (r2⊙[a ⊕ B]))([y ⊕ B]) > θ and let r ∈ R such that G > θ. Then we have x2, x3 ∈ A, and b1, b2 ∈ B such that(2) Let ((r1⊙a1) ⊕ (r1⊙b1))(x2) > θ. So there exist u1 ∈ A, v1 ∈ B such thatThen, byJ > θ and so u1 ⊕ B ~ x2 ⊕ B.Thus r2⊙a1)⊕(r2⊙b2))(x3) > θ. So there exist u2 ∈ A, v2 ∈ B such thatSinceJ > θ and so u2 ⊕ B ~ x3 ⊕ B.Similarly, we get r1⊙a1)⊕(r2⊙a1))(x1) > θ. HenceSinceu1 ⊕ B ~ x2 ⊕ B, u2 ⊕ B ~ x3 ⊕ B, J > θ, and J > θ, we obtain x1 ⊕ B ~ y1 ⊕ B and [x ⊕ B] = [y ⊕ B].Therefore, since r1∗r2)  ⊙  [a ⊕ B])([x ⊕ B]) > θ and (r1  ⊙  (r2  ⊙  [a  ⊕ B]))([y  ⊕  B]) > θ and let r ∈ R such that H > θ. Then there exist x2, x3 ∈ A, and b1, b2 ∈ B such thatw ∈ A such that p > θ. Byw = y1 and p > θ.(3) Finally, let  > θ, and p > θ, then we have x1 ⊕ B ~ y1 ⊕ B from x ⊕ B] = [y ⊕ B].Since R, G, H) be fuzzy ring and f : A → B an R-fmodule epimorphism. Then A/K is isomorphic to B where K = Ker  f.Let  > θ.  5.3 in . Then itw ∈ A such that p > θ. Since f is an R-fmodule homomorphism, we get p, f(w)) > θ, so u = f(w).   By a′  ⊕  K ~ a  ⊕  K, p > θ, p > θ, and b′ ⊕ K ~ w ⊕ K and so [b ⊕ K] = [w ⊕ K]. Therefore u = f(b).Let


The authors find some new inequalities of Jordan type for the sine function. These newly established inequalities are of new form and are applied to deduce some known results. The equalities in  in xm for m ≥ 2. A straightforward computation givesLet the function equality . Hence, In the final section of this paper, we will apply In order to prove Theorems f, g :  → ℝ be differentiable on . If g′ ≠ 0 and f′/g′ are decreasing on , then the functionsa, b). Let We are now in a position to prove our theorems.π/2]. A direct calculation givesx > x on  leads toπ/2) for m ≥ 2 and n ≥ 0. As a result, the function f3′(x)/f4′(x) is decreasing on . In virtue of π/2). SinceLetπ/2]. It is easy to see thatπ/2]. Furthermore, we havex > x and the conditions in (f3′(x)/f4′(x)]′ is negative on . This means that the function f3′(x)/f4′(x) is decreasing on . Consequently, making use of π/2). SinceH(π/2) ≤ H(x) ≤ H(0), the inequality (Lettions in , it is nequality  follows.After proving Theorems λ ≤ 1 and A, B > 0 with A + B ≤ π. Then,Let 0 ≤ ce paper , many mace paper  or an evThe first application of k ≥ 2, let Ai > 0 and ∑i=1kAi ≤ π. If 0 ≤ λ ≤ 1, thenn ≥ 0 and m ≥ 2 are integers. For x = λπ/2 in the inequality (i < j ≤ n results in (Substituting equality  reveals e either , , if n ≥ 0 and m ≥ 2 are integers, thenFor This follows from integrating on all sides of the double inequality . n = 0 givesn = 0 and m = 2 yieldsApplying  0 gives41m+1+1≤∫ished in .