Choose an orthonormal frames field {e1,…, en, en+1} along M such that {e1,…, en} are tangent to M and {en+1} is normal to M. Their dual frames are {θ1,…, θn} and {θn+1}, respectively; obviously, θn+1 = 0 when it is restricted over M. Let h denote the second fundamental form of the immersion φ : M → Sn+1(1). WriteH, S, Pk to be mean curvature, square of the length, and kth power polynomial of second fundamental form, respectively. It is well known that hypersurface φ : Mn → Sn+1(1) is minimal when H ≡ 0 and is totally geodesic when S ≡ 0. For invariant ρ, it is obvious that ρ(Q) ≥ 0 for all Q ∈ M, ρ(Q) = 0 if and only if Q is an umbilical point of M, and 0 ≤ ρ(Q) ≤ C for all Q ∈ M and some positive constant C since M is compact.Let Wn,n/2)(φ) ≥ 4π2 holds for all immersed tori φ : M → S3; recently, it has been completely solved by Marques and Neves in article [n ≥ 3, the functional Wn,n/2)((φ) was studied extensively in [In differential geometry, there is a famous classical Willmore functional of hypersurface which is defined as2dv.See –4 for th article  using Miively in –11.Due to the importance of Willmore conjecture, many geometric experts generalized the classic Willmore functional to a wide range and some interesting results have been obtained.Wp):k1 and k2 denote the principal curvatures of M and k1 ≥ k2. Under some proper conditions, Cai obtained some interesting inequalities.In , Cai stuWn,1):In , Guo andenon. In , Xu and Wn,r):In , Wu reseWn,F)((φ) is called Wn,F) be an n-dimensional hypersurface in an (n + 1)-dimensional unit sphere Sn+1(1); then M is a Wn,F) hypersurface, then M is a Wn,F)(-Willmore hypersurface if and only ifLet F(ρ) = ρn/2, ρ, ρr, When M is an n-dimensional compact minimal hypersurface in (n + 1)-dimensional unit sphere Sn+1(1), thenIt is well known that Simons' integral inequality plays an important role in the study of minimal hypersurface. It says that if M is an n-dimensional compact classical Willmore hypersurface in unit sphere Sn+1(1), thenWn,1), , 15 obtaM be an n-dimensional compact Wn,F)(-Willmore hypersurface in sphere Sn+1(1); then we have Simons' type equality and can give a discussion according to the sign of F, F′, and F′′:ρ ≡ n, there holdsρ ≡ 0, there holdsF′ ≡ 0, F ≡ c ≠ 0, there holdsnF − 2uF′ ≥ 0, F′ ≥ 0, F′′ ≥ 0, there holdsnF − 2uF′ ≤ 0, F′ ≥ 0, F′′ ≥ 0, there holdsLet F(ρ) = ρn/2, ρ, ρr, F, F′, and F′′.When Using the integral equalities in M be an n-dimensional closed Wn,F)(-Willmore hypersurface in unit sphere Sn+1(1), we have the following.F′ ≡ 0, F ≡ c ≠ 0 over , if 0 ≤ ρ ≤ n, then ρ = 0 or ρ = n.When ρ = 0, then H = 0, ρ = 0, S = 0, and M is totally geodesic;For ρ = n, then H = 0, ρ = n, S = n, and m with 1 ≤ m ≤ n − 1.For nF − 2uF′ > 0,  F′ > 0,  F′′ ≥ 0 over , if 0 ≤ ρ ≤ n, then ρ = 0 or ρ = n.When ρ = 0, then M is totally umbilical;For ρ = n, then n ≡ 0(mod  2), H = 0, ρ = n, S = n, and For nF − 2uF′ = 0, F′ > 0, F′′ ≥ 0 over , if 0 ≤ ρ ≤ n, then F(u) = cun/2, c > 0, ρ = 0, or ρ = n.When ρ = 0, M is totally umbilical;For ρ = n, m with 1 ≤ m ≤ n − 1.For nF − 2uF′ < 0, F′ > 0, F′′ ≥ 0 over , if H = 0, 0 ≤ ρ ≤ n, then H = 0, ρ = 0 or H = 0, ρ = n.When H = 0, ρ = 0, then M is totally geodesic;For H = 0, ρ = n, then n ≡ 0(mod  2), H = 0, ρ = n, S = n, For Let F(ρ) = ρn/2, ρ, ρr, F, F′, and F′′. The examples appeared in When φ : Mn → Sn+1(1) be an n-dimensional closed hypersurface in an (n + 1)-dimensional unit sphere and let Φ : M ×  → Sn+1(1) be a variation of φ; it means thatφ0 ≡ φ.Let s =  and σ =  be the orthonormal local frames of TSn+1(1) and T*Sn+1(1), respectively; then e =  = φt−1s is the orthonormal local frames of the pullback vector bundle φt−1TSn+1(1) over M × {t}, such that {e1,…, en} are tangent to M and {en+1} is normal to M due toLet ω to denote the connection form over TSn+1(1); by the pullback operation, we have the following decomposition:θi} are the orthonormal frames of T*M and θn+1 ≡ 0 when it is restricted over M, {Vi, Vn+1 = :f} are the variation vector fields of Φ, {ϕij} is the connection form of TM, and {ϕin+1 = hijθj} is the second fundamental form. In particular, we haveUse Ω and Ω⊤ to denote curvature forms of TSn+1(1) and TM, respectively and write their components:Use ϕij}, the covariant derivatives of hij, H, ρ, Vi, f, and bi can be defined asUsing connection form {dM,  dM× = dM + dt(∂/∂t) and d to denote the differential operators on M, M × , and Sn+1(1), respectively. By calculating directly and comparing both sides of the following equations:We use  article . In fact=dσ−σ∧ω,Φ∗(Ω)=dM×=dσ−σ∧ω,Φ∗(Ω)=dM×article [Φ∗(ΩAB)=Φside of )∧Φ∗(ω),0=dM× be a hypersurface; one has structure equations:Let φ : Mn → Sn+1(1) be a hypersurface; one has Ricci identity:Let φ : Mn → Sn+1(1) be a hypersurface and let V = V⊤ + V⊥ = Viei + fen+1 be a variation vector field; one hasLet With the same notations as above, one hasWn,F) be a hypersurface and let V = V⊤ + V⊥ = Viei + fen+1 be a variation vector field. Suppose that dv = θ1∧⋯∧θn denotes the volume element; one hasLet t)(θi) = θj. Hence,By φ : Mn → Sn+1(1) be a hypersurface and let V = V⊤ + V⊥ = Viei + fen+1 be a variation vector field; one hasLet ρ,  S and H, one hasBy the definition of By Lemmas Sn+1(1), in particular, the isoparametric hypersurfaces. Since all principal curvatures k1,…, kn are constant, then ρ, H, S are constant and H = (1/n)P1, S = P2, ρ = P2 − (1/n)(P1)2. Thus, Wn,F)(-Willmore if and only if F(0) = 0.Obviously, totally geodesic hypersurfaces are n ≡ 0(mod2),P1,  P2,  P3,  ρ, respectively, P1 = 0,  P2 = n,  P3 = 0,  ρ = n. Obviously Cn/2,n/2 always is a Wn,F)(-Willmore hypersurface of Sn+1(1) for any function F.For a particular hypersurface with λ,  μ,  0 < λ, μ < 1, λ2 + μ2 = 1. P1, P2, P3, ρ are, respectively,μ/λ = x > 0, Wn,F), m = n/2.Consider thatSome lemmas are needed for the establishment of Simons' type integral equalities and the discussion of gap phenomenon.ρ, one hash|2 = ∑ijkhij,k2.For ρ and Laplacian operator, we haveT1 = ∑ijkhijhij,kk; by Lemmas T2 = ∑ijkhij; by By the definition of ,kj); by T2=∑ijkhiy Lemmas T1=∑ijkhiituting , one hasFor We know thatM, we haveIntegrating the equality in hij = 0, for all  i ≠ j,  hi = hii.In order to prove x : Mn → Sn+1(1) be a compact hypersurface with ∇h ≡ 0; then there are two cases.h1 = ⋯ = hn = λ = constant, and M is a totally umbilic (λ > 0) or totally geodesic (λ = 0);h1 = ⋯hm = λ = constant > 0, hm+1 = ⋯ = hn = −(1/λ), 1 ≤ m ≤ n − 1, and M is a Riemannian product of M1 × M2, where Let Cm,n−m, 1 ≤ m ≤ n − 1 are the only compact minimal (H = 0) hypersurfaces of dimension n in unit sphere Sn+1(1) satisfying S = n.Clifford torus Obviously, case (1) of nF − 2uF′ > 0,  F′ > 0,  F′′ ≥ 0 over , by the case (4) of ρ ≤ n, the right hand side of identity , H = 0, ρ = n, S = n, and M = Cn/2,n/2.(2) When e (4) of ∫M2ρ(ρ−n)e (4) of ∫M2ρ(ρ−n)nF − 2uF′ = 0, F′ > 0, F′′ ≥ 0  , we know that F(u) = cun/2, c = const > 0. By the case (4) of ρ ≤ n, then ρ = 0 or ρ = n. For ρ = 0, M is totally umbilical; for ρ = n, substituting it into the above equality together with h = 0; by M is a torus; then, by M is a classic Willmore torus Wm,n−m for some m.(3) When


We also introduce common (EA) property for two hybrid pairs F, G : X → 2X and f, g : X → X. We establish some common coupled fixed point theorems for two hybrid pairs of mappings under φ-ψ contraction on noncomplete metric spaces. An example is also given to validate our results. We improve, extend and generalize several known results. The results of this paper generalize the common fixed point theorems for hybrid pairs of mappings and essentially contain fixed point theorems for hybrid pair of mappings.We introduce the concept of (EA) property and occasional  X, d) be a metric space and let CB(X) be the set of all nonempty closed bounded subsets of X. Let D denote the distance from x to A ⊂ X and let H denote the Hausdorff metric induced by d; that is,Let , and let g be a self-mapping on X. An element  ∈ X × X is called(1)F, g} if g(x) ∈ F and g(y) ∈ F,a coupled coincidence point of hybrid pair {(2)F, g} if x = g(x) ∈ F and y = g(y) ∈ F.a common coupled fixed point of hybrid pair {Let F and g by C{F,g}. Note that if  ∈ C{F, g}, then  is also in C{F, g}.We denote the set of coupled coincidence points of mappings F : X × X → 2X be a multivalued mapping and let g be a self-mapping on X. The hybrid pair {F, g} is called w-compatible if g)⊆F whenever  ∈ C{F, g}.Let F : X × X → 2X be a multivalued mapping and let g be a self-mapping on X. The mapping g is called F-weakly commuting at some point  ∈ X × X if g2x ∈ F and g2y ∈ F.Let f : X → X and T : X → 2X. Liu et al. [f : X → X and T : X → 2X.Aamri and El Moutawakil  defined u et al.  introducu et al.  extendedw-compatibility for hybrid pair F : X × X → 2X and f : X → X. We also introduce common (EA) property for two hybrid pairs F, G : X × X → 2X and f, g : X → X. We establish some common coupled fixed point theorems for two hybrid pairs of mappings under φ-ψ contraction on noncomplete metric spaces. The φ-ψ contraction is weaker contraction than the contraction defined in Gnana Bhaskar and Lakshmikantham [In this paper, we introduce the concept of (EA) property and occasional ikantham  and Luonikantham . We imprikantham , Gnana Bikantham , Jain etikantham , Lakshmiikantham , Liu et ikantham , and Luoikantham . The resWe first define the following.f : X → X and F : X × X → CB(X) are said to satisfy the (EA) property if there exist sequences {xn}, {yn} in X, some u, v in X, and A,  B in CB(X) such thatMappings f, g : X → X and F, G : X × X → CB(X). The pairs {F, f} and {G, g} are said to satisfy the common (EA) property if there exist sequences {xn}, {yn}, {un}, and {vn} in X, some u, v in X, and A, B, C, D in CB(X) such thatLet X = [1, +∞) with the usual metric. Define f, g : X → X and F, G : X × X → CB(X) byF, f} and {G, g} are said to satisfy the common (EA) property.Let F : X × X → 2X and f : X → X are said to be occasionally w-compatible if and only if there exists some point  ∈ X × X such that fx ∈ F, fy ∈ F, and fF⊆F.Mappings X = [0, +∞) with usual metric. Define f : X → X, F : X × X → CB(X) byf and F, but fF⊆F and f and F are not w-compatible. However, the pair {F, f} is occasionally w-compatible.Let φ : [0, +∞)→[0, +∞) satisfying the following: φ)  φ is continuous and strictly increasing,(i φ)  φ(t) < t for all t > 0,(ii φ)  φ(t + s) ≤ φ(t) + φ(s) for all t, s > 0.→[0, +∞) which satisfies ψ)  limt→rψ(t) > 0 for all r > 0 and limt→0+ψ(t) = 0,(i ψ)  ψ(t) > 0 for all t > 0 and ψ(0) = 0. and (iiφ), we have that φ(t) = 0 if and only if t = 0. For example, functions φ1(t) = kt where k > 0, φ2(t) = t/(t + 1), φ3(t) = ln(t + 1), and φ4(t) = min{t, 1} are in Φ,  ψ1(t) = kt where k > 0,  ψ2(t) = (ln(2t + 1))/2, and Note that, by  be a metric space. Assume F, G : X × X → CB(X) and f, g : X → X to be mappings satisfying the following.(1)F, f} and {G, g} satisfy the common (EA) property.{(2)x, y, u, v ∈ X, there exist some φ ∈ Φ and some ψ ∈ Ψ such thatFor all (3)f(X) and g(X) are closed subsets of X. Then(a)F and f have a coupled coincidence point,(b)G and g have a coupled coincidence point,(c)F and f have a common coupled fixed point, if f is F-weakly commuting at  and f2x = fx and f2y = fy for  ∈ C{F, f},(d)G and g have a common coupled fixed point, if g is G-weakly commuting at (e)F, G, f,  and g have common coupled fixed point provided that both (c) and (d) are true.Let  property, there exist sequences {xn}, {yn}, {un}, and {vn} in X, some u, v in X, and A, B, C, D in CB(X) such thatf(X) and g(X) are closed subsets of X, then there exist n → ∞ in the above inequality, by using (φ),  (iiφ), and (iψ), we obtainφ) and (iiφ), impliesfx ∈ B and fy ∈ D, it follows thatx, y) is a coupled coincidence point of F and f. This proves (a). Again, by using condition (2) of n → ∞ in the above inequality, by using (φ),  (iiφ), and (iψ), we obtainφ) and (iiφ), impliesG and g. This proves (b).Since {uch thatlimn→∞Fxx~,y~∈X,u=fx=gx~,uch thatlimn→∞Fxx~,y~∈X,u=fx=gx~,f which is F-weakly commuting at ; that is, f2x ∈ F, f2y ∈ F and f2x = fx, f2y = fy. Thus, fx = f2x ∈ F and fy = f2y ∈ F; that is, u = fu ∈ F and v = fv ∈ F. This proves (c). A similar argument proves (d). Then (e) holds immediately.Furthermore, from condition (c), we have f = g in Put X, d) be a metric space. Assume F, G : X × X → CB(X) and g : X → X to be mappings such that(1)F, g} and {G, g} satisfy the common (EA) property,{(2)x,y,u,v ∈ X, there exist some φ ∈ Φ and some ψ ∈ Ψ such thatfor all (3)g(X) is a closed subset of X. Then(a)F and g have a coupled coincidence point,(b)G and g have a coupled coincidence point,(c)F and g have a common coupled fixed point, if g is F-weakly commuting at  and g2x = gx and g2y = gy for  ∈ C{F, g},(d)G and g have a common coupled fixed point, if g is G-weakly commuting at (e)F, G,  and g have common coupled fixed point provided that both (c) and (d) are true.Let  be a metric space. Assume F : X × X → CB(X) and g : X → X to be mappings such that(1)F, g} satisfies the (EA) property,{(2)x, y, u, v ∈ X, there exist some φ ∈ Φ and some ψ ∈ Ψ such thatfor all  If (3) of (a)F and g have a coupled coincidence point,(b)F and g have a common coupled fixed point, if g is F-weakly commuting at  and g2x = gx and g2y = gy for  ∈ C{F, g}.Let  be a metric space. Assume F, G : X × X → CB(X) and f, g : X → X to be mappings satisfying (1) of (1)x,y,u,v ∈ X, there exists some ψ ∈ Ψ such thatfor all  If (3) of (a)F and f have a coupled coincidence point,(b)G and g have a coupled coincidence point,(c)F and f have a common coupled fixed point, if f is F-weakly commuting at  and f2x = fx and f2y = fy for  ∈ C{F, f},(d)G and g have a common coupled fixed point, if g is G-weakly commuting at (e)F, G, f, and  g have common coupled fixed point provided that both (c) and (d) are true.Let  of φ(t) = (1/2)t, t ∈ [0, +∞), and then the above condition reduces to condition (2) of ψ1 = (1/2)ψ and hence by If f = g in Put X, d) be a metric space. Assume F, G : X × X → CB(X) and g : X → X to be mappings satisfying (1) of (1)x, y, u, v ∈ X, there exists some ψ ∈ Ψ such thatfor all  If (3) of (a)F and g have a coupled coincidence point,(b)G and g have a coupled coincidence point,(c)F and g have a common coupled fixed point, if g is F-weakly commuting at  and g2x = gx and g2y = gy for  ∈ C{F, g},(d)G and g have a common coupled fixed point, if g is G-weakly commuting at (e)F, G, and  g have common coupled fixed point provided that both (c) and (d) are true.Let  be a metric space. Assume F : X × X → CB(X) and g : X → X to be mappings satisfying (1) of (1)x, y, u, v ∈ X, there exists some ψ ∈ Ψ such thatfor all  If (3) of (a)F and g have a coupled coincidence point,(b)F and g have a common coupled fixed point, if g is F-weakly commuting at  and g2x = gx and g2y = gy for  ∈ C{F, g}.Let  be a metric space. Assume F, G : X × X → CB(X) and f, g : X → X to be mappings satisfying (1) of (1)F, f} and {G, g} are w-compatible.{(2)Suppose that either(a)g(X) is a closed subset of X and G(X × X)⊆f(X) or(b)f(X) is a closed subset of X and F(X × X)⊆g(X).Then F, G, f, and  g have a common coupled fixed point.Let  property, there exist sequences {xn}, {yn}, {un}, and {vn} in X, some u, v in X, and A, B, C, D in CB(X) satisfying (g(X) is a closed subset of X, and then there exist G and g. Hence, w-compatibility of {G, g}, we have gu ∈ G and gv ∈ G. Now, we shall show that u = gu and v = gv. Suppose, not. Then, by condition (2) of n → ∞ in the above inequality, by using (φ), we obtainu ∈ A, v ∈ C, gu ∈ G, and gv ∈ G, therefore, by (iψ), we getu = gu and v = gv. Hence, we haveG(X × X)⊆f(X), then there exist x, y ∈ X such that fx = u = gu ∈ G and fy = v = gv ∈ G. Now, by condition (2) of φ),  (iiφ), and (iiψ), we getφ) and (iiφ), impliesx, y) is a coupled coincidence point of F and f. Hence,  ∈ C{F, f}. From w-compatibility of {F, f}, we have fF⊆F; hence f2x ∈ F and f2y ∈ F; that is, fu ∈ F and fv ∈ F. Now, we shall show that fu = u and fv = v. Suppose, not. Then, by condition (2) of ψ), we getfu = u and fv = v. Hence, we haveu, v) is a common coupled fixed point of the pairs {F, f} and {G, g}. The proof is similar when (b) holds.Since {tisfying . Supposeby using  and (iφ)f = g in If we put X, d) be a metric space. Assume F, G : X × X → CB(X) and g : X → X to be mappings satisfying (1) of (1)F, g} and {G, g} are w-compatible;{(2)suppose that either(a)g(X) is a closed subset of X and G(X × X)⊆g(X) or(b)g(X) is a closed subset of X and F(X × X)⊆g(X).Then F, G, and  g have a common coupled fixed point.Let  be a metric space. Assume F : X × X → CB(X) and g : X → X to be mappings satisfying (1) of (1)F, g} is w-compatible;{(2)g(X) is a closed subset of X and F(X × X)⊆g(X).Then F and g have a common coupled fixed point.Let  be a metric space. Assume F, G : X × X → CB(X) and f, g : X → X to be mappings satisfying (1) of F, G, f,  and g have a common coupled fixed point.Let  of φ(t) = (1/2)t, t ∈ [0, +∞), then it reduces to condition (2) of ψ1 = (1/2)ψ and hence by If f = g in If we put X, d) be a metric space. Assume F, G : X × X → CB(X) and g : X → X to be mappings satisfying (1) of F, G, and g have a common coupled fixed point.Let  be a metric space. Assume F : X × X → CB(X) and g : X → X to be mappings satisfying (1) of F and g have a common coupled fixed point.Let  be a metric space. Assume F, G : X × X → CB(X) and f, g : X → X to be mappings satisfying (2) of (1)F, f} and {G, g} are occasionally w-compatible.{Then F, G, f,  and g have a common coupled fixed point.Let ,ψ), we haveu = fu = gu and v = fv = gv. Suppose, not. Then, by condition (2) of ψ), we haveu, v) is a common coupled fixed point of F, G, f,  and g.Since the pairs {ows thatf2x∈Ffx,fhus, by  be a metric space. Assume F, G : X × X → CB(X) and g : X → X to be mappings satisfying (2) of (1)F, g} and {G, g} are occasionally w-compatible.{Then F, G,  and g have a common coupled fixed point.Let  be a metric space. Assume F : X × X → CB(X) and g : X → X to be mappings satisfying (2) of (1)F, g} is occasionally w-compatible.{Then F and g have a common coupled fixed point.Let  be a metric space. Assume F, G : X × X → CB(X) and f, g : X → X to be mappings satisfying (1) of F, G, f, and  g have a common coupled fixed point.Let  of φ(t) = (1/2)t, t ∈ [0, +∞), then it reduces to condition (2) of ψ1 = (1/2)ψ and hence by If f = g in Put X, d) be a metric space. Assume F, G : X × X → CB(X) and g : X → X to be mappings satisfying (1) of F, G, and g have a common coupled fixed point.Let  be a metric space. Assume F : X × X → CB(X) and g : X → X to be mappings satisfying (1) of F and g have a common coupled fixed point.Let  defined as d = max{x, y} and d = 0 for all x, y ∈ X. Let F, G : X × X → CB(X) be defined asf, g : X → X be defined asφ : [0, +∞)→[0, +∞) byψ : [0, +∞)→[0, +∞) byx, y, u, v ∈ X with x, y, u, v ∈ [0,1), we haveSuppose that Case (a). If (x2 + y2)/4 = (u + v)/8, thenCase (b). If (x2 + y2)/4 ≠ (u + v)/8 with (x2 + y2)/4 < (u + v)/8, thenu + v)/8<(x2 + y2)/4. Thus, the contractive condition (2) of x, y, u, v ∈ X with x, y, u, v ∈ [0,1). Again, for all x, y, u, v ∈ X with x, y ∈ [0,1) and u, v = 1, we havex, y, u, v ∈ X with x, y ∈ [0,1) and u, v = 1. Similarly, we can see that the contractive condition (2) of x, y, u, v ∈ X with x, y, u, v = 1. Hence, the hybrid pairs {F, f} and {G, g} satisfy condition (2) of x, y, u, v ∈ X. In addition, all the other conditions of z =  is a common coupled fixed point of F, G, f, and  g.


U, let k such that U can be decomposed into a product of k rotations about either U. Here, a rotation means an element D of the special orthogonal group SO(3) or an element of the special unitary group SU(2) that corresponds to D. Decompositions of U attaining the minimum number For any pair of three-dimensional real unit vectors  However, we follow the custom, in quantum physics, to call not only an element of SO(3) but also that of the special unitary group SU(2) a rotation. This is justified by the well-known homomorphism from SU(2) onto SO(3)  are presented explicitly.Then, a natural question arises: What is the least value, r angles ,4. MoreoIn this work, not only explicit constructions but also simple inequalities on geometric quantities, which directly show lower bounds on the number of constituent rotations, will be presented. Remarkably, the proposed explicit constructions meet the obtained lower bounds, which shows both the optimality of the constructions and the tightness of the bounds.D∈SO(3). That interesting result . The Hermitian conjugate of a matrix U is denoted by U†.The notation to be used includes the following: I denotes the 2×2 identity matrix; X, Y and Z denote the following Pauli matrices:θ , where For F from SU(2) onto SO(3) to be defined in §D∈SO(3) andUsing the homomorphism 3.2This work's results lead to an elementary self-contained proof of the following known theorem (appendix F).For anywith3.3The following lemma presents a well-known parametrization of SU(2) elements.For any elementU∈SU(2), there exist someandβ∈ such thatα,β and γ in this lemma are often called Euler angles.2 The lemma can be rephrased as follows: any matrix in SU(2) can be written asa and b such that |a|2+|b|2=1  such thatF is onto SO(3), there exists an element U∈SU(2) such that 4 With this element U, some β∈, write U†VU=Rz(α)Ry(β)Rz(γ) in terms of the parametrization (3.6). Then, since As We also have the following lemma, which is easy but worth recognizing.Let arbitraryandU∈SU(2) be given. PutandThen, for anyif and only if (iff)This readily follows from 4.Here, we present the result establishing U uniquely, For Functions For anywithand β∈, ifthenwhereandα,β and γ vary, in general, if Note that there is no loss of generality in assuming We give two constructions or decompositions, which will turn out to attain the minimum number Given arbitrarywithandβ∈, putandThen, for anyandβ1,…,βk∈ holds for some β1,…,βk∈ ⌉.5 Hence, this proposition gives a decomposition of an arbitrary element β/(2δ)⌉+1 rotations.6The least value of β/2≤δ≤π/2, δ≠0, and αj, γj and θj for which (4.5) holds is given by T=σtj, wherej=1,…,k. (These make (4.6) hold.)For Given anywithputandFor an arbitraryU∈SU(2), choose parametersandβ′∈ such thatThen,Furthermore, for anyandβ′1,…,β′k′∈ holds for some β′1,…,β′k′∈ /(2δ)⌉=⌈β′/(2δ)+1/2⌉. Moreover, if β′≥δ and k′=⌈β′/(2δ)+1/2⌉, the parameter α′1 can be chosen so that it satisfies α′1=0 as well as (4.11) and (4.12). Hence, when β′≥δ, this proposition and the fact just mentioned give a decomposition of an arbitrary element β′<δ, a decomposition of U into the product of four rotations.The least value of α′j,γ′j and θ′j, j=1,…,k′, for which (4.11) and (4.12) hold is given by T= σtj, where j=1,…,k′.An explicit instance of the set of parameters 5.S2, which is denoted by d. Specifically,S2. We have the following lemma.  and (5.4) will be used in the following forms:D and D′∈SO(3) equal the product of 2k−1 rotations and that of 2k rotations, respectively, in lemma k is an integer). It will turn out that these bounds are tight.This can be shown easily by induction on 6.6.1Here, the structure of the whole proof of the results in this work is described. Theorem 6.2The following lemma is fundamental to the results in this work.For anyand for anysuch thatthe following two conditions are equivalent.There exist somesuch thatI. II. U∈SU(2) such thatvx,vy and vz such that(1) Take an element (2) A direct calculation shows(3) We shall prove I ⇒ II. On each side of (6.7) and (6.8), squaring and summing the resultant pair, we have(4) Next, we shall prove II ⇒ I.α,β) into , where the two pairs are related byTransforming (η satisfying (6.12) and (6.15). From II, however, we have (6.10), and hence, η.Now suppose ζ satisfying (6.13) and (6.14). But (6.17) follows again from II or (6.10) since ζ similarly. ▪In a similar way, if θj such that βj∈Rz(α′)Ry(−δ)=Rv(α′), where α,β,γ) replaced by , it readily follows that there exist some α′j,γ′j and j=1,…,k′, that satisfy the following: j=1,…,k′, andNote α′1 in remark β′1=2δ and t1=π/2) or, more directly, from an equation Ry(2δ)=Rv(π)Rz(−π), where Remarks 6.3U∈SU(2);D∈SO(3). The following lemma largely solves the issue of determining the optimal number Let Letandδbe as in theorem Then, for anyandβ∈,andwhereis as defined in theorem Letandδbe as in theorem Then, for anyandβ∈,β=0, since β>0.In the case where To establish (6.19), we shall show the first and third inequalities inj=2k−1 with ν is odd and Note first that remark β≤2(k−1)δ by (5.2) of lemma β/(2δ)⌉≤k−1, and therefore,We shall evaluate f≥δ. Recalling that To establish (6.20), we shall first treat the major case where β′ in proposition β′=f when β′=f, rewrite (4.8), using lemma β′=f in view of (3.6).Note that remark j=2k with ν is odd and β′≤(2k−1)δ by (5.4) of lemma β′+δ)/(2δ)⌉≤k, and, therefore,f≥δ. The proof of (6.20) in the other case is given in appendix C. This completes the proof of the lemma. The proved lemma immediately implies the corollary. ▪To prove the first inequality in (6.25), assume (6.23) holds for some U∈SU(2),U in terms of three parametric expressions:b. (This is because writing U in (4.1) as Note that for any U∈SU,Nm^,n^ showsU∈SU(2), where we assume U† in place of U [note 7 See appendix G for a detailed description of the above construction method.From the viewpoint of construction, we summarize the (most directly) suggested way to obtain an optimal construction of a given element 7.k such that U can be decomposed into the product of k rotations about either U in SU(2), or in SO(3), where U attaining the minimum number This work has established the least value 8.U is an arbitrary fixed rotation and A⊂S2 with |A|=2  is considered. In the series of Brezov et al. [Naturally, the present author could not find any (explicit or implicit) indication that Brezov et al. –12 suggev et al. –12 may bv et al. –12, theyet al. [Despite such differences in essence and background, note in the proof of this paper's formula 6.20) for the minimum even number of factors in lemma 0 for theet al. –12.


In order to solve the large scale linear systems, backward and Jacobi iteration algorithms are employed. The convergence is the most important issue. In this paper, a unified backward iterative matrix is proposed. It shows that some well-known iterative algorithms can be deduced with it. The most important result is that the convergence results have been proved. Firstly, the spectral radius of the Jacobi iterative matrix is positive and the one of backward iterative matrix is strongly positive (lager than a positive constant). Secondly, the mentioned two iterations have the same convergence results (convergence or divergence simultaneously). Finally, some numerical experiments show that the proposed algorithms are correct and have the merit of backward methods. A is a given n × n complex or real matrix.The primal goal of this paper is to study the iterative methods of the linear systems:A in  and ordinary differential equations (ODE) –3 it is A in . In the A in . In the A in . TherefoA in  is very The methods to solve linear systems can be roughly divided into two categories: direct methods and iterative methods 1. Iterative methods are more suitable than direct methods for large linear systems , 12. TheIn this paper, we do some research with the iterative algorithm. To this end, the paper is organized as follows. In D = diag(A) is a diagonal matrix obtained from A and nonsingular and CL and CU are strictly lower and upper triangular matrices obtained from A, (D−1A = (I − L − U).The basic idea to solve  is matriThe Jacobi iterative matrix isudied in –17.τ > 0 is a real constant, obviously . Matrix A, vector b, τ > 0, ω1, ω2, algorithm stop cutoff ϵ. Step  1 . Compute D = diag(A), CL, CU, L = D−1CL, U = D−1CU, x(0) = 0, and set i≔0. Step  2. Compute matrix rding to . Step  3. Compute xi+1)( with (+1) with . Step  4. If ||xi+1)(−xi)||2ω1,   ω2, and τ, we have the following.ω1 = 0, ω2 = 0, and τ = 1, we obtain the Jacobi iterative method.When ω1 = 0, ω2 = 0, and τ = ω, we obtain the backward JOR iterative method.When ω1 = 1, ω2 = 0, and τ = 1, we obtain the backward G-S iterative method.When ω1 = ω, ω2 = 0, and τ = ω, we obtain the backward SOR iterative method.When ω1 = ω, ω2 = 0, and τ = α, we obtain the backward AOR iterative method.When ω1 = ω, ω2 = ω, and τ = ω(2 − ω), we obtain the backward SSOR iterative method.When ω1 = ω, ω2 = ω, and τ = ω, we obtain the backward EMA iterative method.When ω1 = ω, ω2 = ω, and τ = α, we obtain the backward PSD iterative method.When ω1 = ω, ω2 = ω, and τ = 1, we obtain the backward PJ iterative method.When With special values of The convergence relationship between the Gauss-Seidel iterative matrix and the Jacobi iterative matrix is studied in , and theIn order to obtain the convergence results, we give some well-known results which will be used in the proof of A = M − N with A and M nonsingular is called a regular splitting if M−1 ≥ 0 and N ≥ 0. It is called a weak regular splitting if M−1 ≥ 0 and M−1N ≥ 0.The splitting It is obvious that a regular splitting is a weak regular splitting.T ∈ Rn×n is convergent; that is, ρ(T) < 1 if and only if (I − T)−1 exists and (I − T)−1 = ∑k=1∞Tk ≥ 0.The nonnegative matrix A = M − N be a weak regular splitting of A, H = M−1N. Then the following statements are equivalent.(1) A−1 ≥ 0; that is, A is inverse-positive.(2) A−1N ≥ 0.(3) ρ(H) = ρ(A−1N)/(1 + ρ(A−1N)) so that ρ(H) < 1.Let A ≥ 0 be an irreducible n × n matrix. Then(1) A has a positive real eigenvalue equal to its spectral radius;(2) ρ(A), there corresponds an eigenvector x > 0,to (3) ρ(A) increases when any entry of A increases,(4) ρ(A) is a simple eigenvalue of A.Let A = (aij) ≥ 0 be an irreducible n × n matrix. Then for any x > 0, eitherLet By the lemmas above, we give the convergence theorem in the following.A of  ρ(B) > 0,  (2)One and only one of the following mutually exclusive relations is valid.(i)(ii)(iii)Let the coefficient matrix A of  be irredThus, The Jacobi iterative method and the backward MPSD iterative method are either both convergent or both divergent.ρ(ω2L) = ρ(ω1U) = 0 with I − ω2L)−1 ≥ 0, (I − ω1U)−1 ≥ 0, andCombining aii ≠ 0 and A is irreducible, I − L − U = D−1A and B = L + U are irreducible. By 0 ≤ ωk < τ ≤ 1, k = 1,2, we have (1 − τ)I + (τ − ω1)U + (τ − ω2)L ≥ 0 and irreducible. Thus, by T > 0, such that ηk = τ − ωk + λωk, k = 1,2; by calculation,Since Thus, by , S~τ,ω1,B ≥ 0 is irreducible, by ρ(B) > 0. If ωk < τ ≤ 1, k = 1,2, λ − (1 − τ) ≥ 0 because the left side of  Since  side of  is nonne − τ. By ,16)η1Uη1UB ≥ 0 ηk > 0, k = 1,2, B = (bij) ≥ 0 and x > 0, we obtain that B = 0. Thus, ρ(B) = 0. This contradicts ρ(B) > 0. So, Since (2) For mutually exclusive relations, consider the following.ρ(B) < 1, let(i) If 0 < M−1 = (I − ω2L)−1(I − ω1U)−1 ≥ 0 and N ≥ 0, T = M − N is a regular splitting:Since B ≥ 0, 0 < ρ(B) < 1, and 0 < τ ≤ 1, we know that T−1 = (1/τ)(I − B)−1 ≥ 0. By By ηk > 0, k = 1,2, bii = 0, ∀ i,If 2)<1, by , we haveλ < 1 and 1 − ωk > 0  , there isBy ρ(B) < 1.Combining  with 2929, we haBx = x. Since x > 0, we haveρ(B) = 1.(ii) If 2)=1, by , we haveηk > 0, k = 1,2, bii = 0, ∀ i,(iii) If 2)>1, by , we haveλ > 1 and 1 − ωk > 0  , there isBy ρ(B) > 1.Combining  with 3333, we haρ(B) = 1 and ρ(B) < 1 or ρ(B) > 1. This contradicts ρ(B) = 1. So, If ρ(B) > 1 and ρ(B) ≤ 1. This contradicts ρ(B) > 1. So, If ω1, ω2, and τ, we have the following corollaries.With special values of A of  ρ(B) > 0, (2) One and only one of the following mutually exclusive relations is valid. (i)(ii)(iii)Let the coefficient matrix A of  be irredThus, The Jacobi iterative method and the backward JOR iterative method are either both convergent or both divergent.A of  ρ(B) > 0, (2) One and only one of the following mutually exclusive relations is valid.Thus, The Jacobi iterative method and the backward Gauss-Seidel iterative method are either both convergent or both divergent.A of  ρ(B) > 0, (2) One and only one of the following mutually exclusive relations is valid.  Thus, The Jacobi iterative method and the backward SOR iterative method are either both convergent or both divergent.A of  ρ(B) > 0, (2) One and only one of the following mutually exclusive relations is valid.(i)(ii)(iii)Let the coefficient matrix A of  be irredThus, The Jacobi iterative method and the backward AOR iterative method are either both convergent or both divergent.A of  ρ(B) > 0, (2) One and only one of the following mutually exclusive relations is valid.  Thus, The Jacobi iterative method and the backward SSOR iterative method are either both convergent or both divergent.A of  ρ(B) > 0, (2) One and only one of the following mutually exclusive relations is valid.  (i)(ii)(iii)Let the coefficient matrix A of  be irredThus, The Jacobi iterative method and the backward EMA iterative method are either both convergent or both divergent.A of  ρ(B) > 0, (2) One and only one of the following mutually exclusive relations is valid. (i)(ii)(iii)Let the coefficient matrix A of  be irredThus, The Jacobi iterative method and the backward PSD iterative method are either both convergent or both divergent.A of  ρ(B) > 0, (2) One and only one of the following mutually exclusive relations is valid.    (i)(ii)(iii)Let the coefficient matrix A of  be irredThus, The Jacobi iterative method and the backward PJ iterative method are either both convergent or both divergent.The convergence results between the backward MPSD and Jacobi iterative matrix are proposed, and The convergence results between some special cases of backward MPSD  and Jacobi iterative matrix are obtained. These results involve some special iterative methods which are proposed in the references.In this section, we show five examples. The first three examples are used to show the convergence of the proposed iterative methods. A and the vector b of (Let the coefficient matrix tor b of  be(35)AThe Jacobi iterative matrix isρ(B) = 1/2 < 1.(1)τ = α = 1/2, ω1 = ω2 = ω = 1/4. We obtain the backward PSD iterative matrix Let (2)τ = 1, ω1 = ω2 = ω = 1/2. We obtain the backward PJ iterative matrix Let (3)τ = ω = 1/2, ω1 = ω2 = 0. We obtain the backward JOR iterative matrix Let (4)τ = ω1 = ω2 = ω = 1/2. We obtain the backward EMA iterative matrix Let  With these iterative methods and the presented algorithm, the solution is x = T.By caculation, we obtain 0 < A and the vector b of ∂4∂2u∂ρ(B) = 1159/1601 < 1.(1)τ = α = 1/2, ω1 = ω2 = ω = 1/4. We obtain the backward PSD iterative matrix Let (2)τ = 1, ω1 = ω2 = ω = 1/2. We obtain the backward PJ iterative matrix Let (3)τ = ω = 1/2, ω1 = ω2 = 0. We obtain the backward JOR iterative matrix Let (4)τ = ω1 = ω2 = ω = 1/2. We obtain the backward EMA iterative matrix Let The Jacobi iterative matrix isω1, ω2, and τ in this example.From Figures A of , and s = −p/(n + 2) [n = 100 and p = 1. By caculation, we obtain 0 < ρ(B) = 199/203 < 1.(1)τ = α = 1/2, ω1 = ω2 = ω = 1/4. We obtain the backward PSD iterative matrix Let (2)τ = 1, ω1 = ω2 = ω = 1/2. We obtain the backward PJ iterative matrix Let (3)τ = ω = 1/2, ω1 = ω2 = 0. We obtain the backward JOR iterative matrix Let (4)τ = ω1 = ω2 = ω = 1/2. We obtain the backward EMA iterative matrix Let Let the coefficient matrix A of  be(48)A/(n + 2) . Here, wA and the vector b of (ρ(B) = 987/305 > 1.(1)τ = α = 1/2, ω1 = ω2 = ω = 1/4. We obtain the backward PSD iterative matrix Let (2)τ = 1, ω1 = ω2 = ω = 1/2. We obtain the backward PJ iterative matrix Let (3)τ = ω = 1/2, ω1 = ω2 = 0. We obtain the backward JOR iterative matrix Let (4)τ = ω1 = ω2 = ω = 1/2. We obtain the backward EMA iterative matrix Let  It shows that the backward MPSD iteration is invalid for this example.Let the coefficient matrix tor b of  be(53)AA of (Let the coefficient matrix A of  be(59)A=A in [ρ(B) = 811/822 < 1.We can see the analogous matrix A in . By cacuτ = 1, ω1 = 1, ω2 = 0. We obtain the backward Gauss-Seidel iterative matrix Let From The Jacobi iteration is the basic iteration for linear systems and easier to the analysis of the convergence than other iterations. In the paper, we proposed the backward MPSD iteration and obtained the convergence result between backward MPSD iteration  and Jacobi iteration. We pointed out that the backward MPSD iteration and the Jacobi iteration are either both convergent or both divergent under the assumptions in


In this work, nonparametric nonsymmetric measure of divergence, a particular part of Vajda generalized divergence at m = 4, is taken and characterized. Its bounds are studied in terms of some well-known symmetric and nonsymmetric divergence measures of Csiszar's class by using well-known information inequalities. Comparison of this divergence with others is done. Numerical illustrations (verification) regarding bounds of this divergence are presented as well.Vajda (1972) studied a generalized divergence measure of Csiszar's class, so called “Chi- If we take pi ≥ 0 for some i = 1, 2, 3,…, n, then we have to suppose that 0f(0) = 0f(0/0) = 0.Let Γf-divergence measure, which is given byf :  → R  is a real, continuous, and convex function and P = , Q =  ∈ Γn, where pi and qi are probability mass functions. Many known divergences can be obtained from this generalized measure by suitably defining the convex function f. Some of those are as follows.Csiszar  introducP, Q ∈ Γn. These measures are as follows (see , Δ = 2[1 − W], h = 1 − B, and I = (1/2)[F + F], where W = 2∑i=1n(piqi/(pi + qi)) is the harmonic mean divergence, F = ∑i=1npilog(2pi/(pi + qi)) is the relative JS divergence [Nonsymmetric measures are those that are not symmetric with respect to probability distributions ows see –10):(7):(7)P, Qvergence . EquatioIn this section, we obtain nonsymmetric divergence measure for convex function and further define the properties of function and divergence. Firstly, f is convex and normalized, that is, f(1) = 0, then Cf and its adjoint Cf are both nonnegative and convex in the pair of probability distribution  ∈ Γn × Γn.If the function f :  → R be a function defined as(a)f1(1) = 0, f1(t) is a normalized function.Since (b)f1′′(t) ≥ 0 for all t ∈ ⇒f1(t) is a convex function as well.Since (c)f1′(t) < 0 at  and f1′(t) > 0 at ⇒f1(t) is monotonically decreasing in  and monotonically increasing in  and f1′(1) = 0, f1′′(1) = 0.Since Now, let fined asft=f1t=t−f1(t) in  is called adjoint of χ4 = ∑i=1n((pi − qi)4/qi3)  in  = ∑i=1n(|pi − qi|m/qim−1), m ≥ 1.Now put 1(t) in χ4 is convex and nonnegative in the pair of probability distribution  ∈ Γn × Γn.In view of (b)χ4 = 0 if P = Q or pi = qi .(c)χ4 ≠ χ4⇒χ4 is a nonsymmetric divergence measure.Since Properties of the divergence measure defined in  are as fCf, which is given by f-divergence measures.In this section, we are taking well-known information inequalities on f1, f2 : I ⊂ R+ → R be two convex and normalized functions, that is, f1(1) = f2(1) = 0, and suppose the following assumptions.(a)f1 and f2 are twice differentiable on  where 0 < α ≤ 1 ≤ β < ∞, α ≠ β.(b)m, M such that m < M andf2′′(t) > 0 for all t ∈ .There exist the real constants Let P, Q ∈ Γn and satisfying the assumption 0 < α ≤ pi/qi ≤ β < ∞, then one has the following inequalities:Cf is given by  and χ4 be defined as in α < 1, thenIf 0 < (b)α = 1, thenIf Let Δ(ed as in  and 11)P, Q) andf2′′(t) > 0 for all t > 0 and f2(1) = 0, f2(t) is a convex and normalized function, respectively. Now put f2(t) in (g(t) = f1′′(t)/f2′′(t) = 3(t − 1)2(t + 1)3/2t5, where f1′′(t) and f2′′(t) are given by (g′(t) = 0⇒t = 1, t = −1, and t = 5.Let us consider6f2t=t−12given by  and 16)(16)f2t=tg(t) is decreasing in  and [5, ∞) but increasing in [1,5).It is clear that g(t) has a minimum and maximum value at t = 1 and t = 5, respectively, because g′′(1) = 24 > 0 and g′′(5) = −216/15625 < 0, so(a)α < 1, thenif 0 < (b)α = 1, thenif Results (P and Q.Also Results  and 15)g(t) has Results , 17),  h < 0, som=inft∈0 1, thenM=supt∈αh and χ4 be defined as in α < 1, thenIf 0 < (b)α = 1, thenIf Let ed as in  and 11)h af2′′(t) > 0 for all t > 0 and f2(1) = 0, f2(t) is a convex and normalized function, respectively. Now put f2(t) in (g(t) = f1′′(t)/f2′′(t) = 48(t − 1)2/t7/2, where f1′′(t) and f2′′(t) are given by (g′(t) = 0⇒t = 1 and t = 7/3.Let us considergiven by  and 24)(24)f2t=1g(t) is decreasing in  and [7/3, ∞) but increasing in [1,7/3).It is clear that g(t) has a minimum and maximum value at t = 1 and t = 7/3, respectively, because g′′(1) = 96 > 0 and g′′(2.33) = −1951/913 < 0, so(a)α < 1, thenif 0 < (b)α = 1, thenif Results (P and Q.Also Results  and 23)g(t) has Results , 25),  h < 0, som=inft∈0 1, thenM=supt∈α 1, thenM=supt∈1I and χ4 be defined as in α < 1, thenIf 0 < (b)α = 1, thenIf Let ed as in , respectf2′′(t) > 0 for all t > 0 and f2(1) = 0, f2(t) is a convex and normalized function, respectively. Now put f2(t) in (g(t) = f1′′(t)/f2′′(t) = 24(t − 1)2(t + 1)/t4, where f1′′(t) and f2′′(t) are given by (g′(t) = 0  ⇒  t = 1, Let us considergiven by  and 32)(32)f2t=tg(t) is decreasing in  and It is clear that g(t) has a minimum and maximum value at t = 1 and g′′(1) = 96 > 0 and (a)α < 1, thenif 0 < (b)α = 1, thenif Results (P and Q.Also Results  and 31)g(t) has Results , 33),  h65<0, som=inft∈0 1, thenM=supt∈α 1, thenM=supt∈1Results , after iP* and χ4 be defined as in (Let ed as in  and 11)P*P, Q ∈ Γn, one hasFor f2′′(t) ≥ 0 for all t > 0 and f2(1) = 0, f2(t) is a convex and normalized function, respectively. Now put f2(t) in (g(t) = f1′′(t)/f2′′(t) = 2/(2t5 + t4 + t3 + t2 + t + 2), where f1′′(t) and f2′′(t) are given by (g(t) is always decreasing in , soP and Q.Let us considergiven by  and 39)(39)f2t=t.Result  is obtai.Result , 40), a, a(39)f2, ∞), som=inft∈αE* and χ4 be defined as in E*f2′′(t) ≥ 0 for all t > 0 and f2(1) = 0, f2(t) is a convex and normalized function, respectively. Now put f2(t) in (g(t) = f1′′(t)/f2′′(t) = 16/t3/2(5t2 + 6t + 5), where f1′′(t) and f2′′(t) are given by (g(t) is always decreasing in , soP and Q.Let us considergiven by  and 44)(44)f2t=t.Result  is obtai.Result , 45), a, a(44)f2, ∞), som=inft∈αNow in this section, we obtain bounds of divergence measure  in termsJR and χ4 be defined as in α < 1, thenIf 0 < (b)α = 1, thenIf Let ed as in  and 11)JRf2′′(t) > 0 for all t > 0 and f2(1) = 0, f2(t) is a convex and normalized function, respectively. Now put f2(t) in (g(t) = f1′′(t)/f2′′(t) = 12(t2 − 1)2/(t + 3)t5, where f1′′(t) and f2′′(t) are given by (g′(t) = 0⇒t = 1,  t = −1, and t ≈ 116/59.Let us considergiven by  and 50)(50)f2t=tg(t) is decreasing in  and [116/59, ∞) but increasing in [1,116/59).It is clear that g(t) has a minimum and maximum value at t = 1 and t = 116/59, respectively, because g′′(1) = 24 > 0 and g′′(116/59) ≈ −7/10 < 0, so(a)α < 1, thenif 0 < (b)α = 1, thenif Results (P and Q.Also Results  and 49)g(t) has Results , 53),  h 1, thenM=supt∈α 1, thenM=supt∈1Results , after iK and χ4 be defined as in α < 1, thenIf 0 < (b)α = 1, thenIf Let ed as in  and 11)K af2′′(t) > 0 for all t > 0 and f2(1) = 0, f2(t) is a convex and normalized function, respectively. Now put f2(t) in (g(t) = f1′′(t)/f2′′(t) = 12(t − 1)2/t4, where f1′′(t) and f2′′(t) are given by (g′(t) = 0⇒t = 1 and t = 2.Let us considergiven by  and 58)(58)f2t=tg(t) is decreasing in  and [2, ∞) but increasing in [1,2).It is clear that g(t) has a minimum and maximum value at t = 1 and t = 2, respectively, because g′′(1) = 24 > 0 and g′′(2) = −3/4 < 0, so(a)α < 1, thenif 0 < (b)α = 1, thenif Results (P and Q.Also Results  and 57)g(t) has Results , 59),  h < 0, som=inft∈0 1, thenM=supt∈α 1, thenM=supt∈1Results , after iG and χ4 be defined as in α < 1, thenIf 0 < (b)α = 1, thenIf Let ed as in  and 11)G af2′′(t) > 0 for all t > 0 and f2(1) = 0, f2(t) is a convex and normalized function, respectively. Now put f2(t) in (g(t) = f1′′(t)/f2′′(t) = 24(t − 1)2(t + 1)/t3, where f1′′(t) and f2′′(t) are given by (g′(t) = 0⇒t = 1 and t = −3.Let us considergiven by  and 66)(66)f2t=tg(t) is decreasing in  and increasing in [1, ∞).It is clear that g(t) has a minimum value at t = 1, because g′′(1) = 120 > 0, so(a)α < 1, thenif 0 < (b)α = 1, thenif Results (P and Q.Also Results  and 65)g(t) has Results , 67),  h > 0, som=inft∈0 1, thenM=supt∈α 1, thenM=supt∈1Results , after iP, Q), h, JR, K, and χ4 and verify inequalities .In this section, we give two examples for calculating the divergences Δ, ,ualities , and 56P, Q, h and Q its approximated Poisson probability distribution with parameter (λ = np = 5), for the random variable X; then we have Let By using p = 1/2.Put the approximated numerical values from  in 14),,14), 2222, 48),,48), andP be the binomial probability distribution with parameters  and Q its approximated Poisson probability distribution with parameter (λ = np = 7), for the random variable X; then we have Let p = 7/10.By using α=5772≤piues from , and ) ) , 38), ) , and (64∞). Function is decreasing in  and increasing in . χ4, E*, K, Δ, and G. We have considered pi = , qi = , where a ∈ . It is clear from χ4 has a steeper slope than E*, K, Δ, and G.Many research papers have been studied by Taneja, Kumar, Dragomir, Jain, and others, who gave the idea of divergence measures, their properties, their bounds, and relations with other measures. Taneja especially did a lot of quality work in this field: for instance, in  he derivDivergence measures have been demonstrated to be very useful in a variety of disciplines such as anthropology, genetics, finance, economics and political science, biology, analysis of contingency tables, approximation of probability distributions, signal processing, pattern recognition, sensor networks, testing of the order in a Markov chain, risk for binary experiments, region segmentation, and estimation.This paper also defines the properties and bounds of Vajda's divergence and derives new relations with other symmetric and nonsymmetric well-known divergence measures.


The adipate and oxalate ions are located on centres of inversion. The Dy3+ cation has a distorted tricapped trigonal–prismatic geometry and is coordinated by nine O atoms, four belonging to three adipate anions, four to two oxalate anions and one from an aqua ligand. The cations are bridged by adipate ligands, generating a two-dimensional network parallel to (010). This network is further extended into three dimensions by coordination of the rigid oxalate ligands and is further consolidated by O—H⋯O hydrogen bonds. A part of the adipate anion is disordered over two positions in a 0.75:0.25 ratio.In the title coordination polymer, [Dy DOI: 10.1107/S1600536814024544/gk2622Isup2.hklStructure factors: contains datablock(s) I. DOI: Click here for additional data file.10.1107/S1600536814024544/gk2622fig1.tifx y z x y z x y - z x y z x y z . DOI: x, 1 − y, 1 − z; (ii) 1 − x, 1 − y, 1 − z; (iii) −x, − y, 1- z; (iv) − x, − y, −z; (v) 1 − x, 1 − y, 2 − z.The fragment of the structure of the title compounds, with the atom-numbering scheme. Displacement ellipsoids are drawn at the 30% probability level and H atoms are shown as small spheres of arbitrary radii. Symmetry code: (i) −1033325CCDC reference:  crystallographic information; 3D view; checkCIF reportAdditional supporting information:


Scientific Reports5: Article number: 1342010.1038/srep13420; published online: 08262015; updated: 10302015In the HTML version of this Article, the legend of Figure 1 contains typographical errors.d) Hex ≈ HC H* ≈ Hsat and e Hex ≥ H* ≈ Hsat HC”“(should read:d) Hex ≈ HC ≪H* ≈ Hsat and (e) Hex ≥ H* ≈ Hsat ≫ HC”“(


The Rosenzweig-MacArthur (1963) criterion is a graphical criterion that has been widely used for elucidating the local stability properties of the Gause (1934) type predator-prey systems. It has not been stated whether a similar criterion holds for models with explicit resource dynamics ), like the chemostat model. In this paper we use the implicit function theorem and implicit derivatives for proving that a similar graphical criterion holds under chemostat conditions, too. The parameters C > 0, D > 0, a > 0, b > 0, m > 0, A > 0, B > 0, and M > 0 stand for concentration, dilution rate, search rate for the prey, handling time for the prey , converℋ1 = 0 is asymptotically invariant for  < 1, 0 ≤ θ < 2π, x3(0) ≤ 0 and positive constant parameters. In this case, consider the functionℋ2 = 0 is asymptotically invariant for (∗-marks in x3 = 0 has some equilibrium on the unit circle as its limit set. But if x3(0) > 0, then the limit set is the whole unit circle. We see, however, that the chain recurrent set [A reduction into two dimensions  is often Waltman  gave the r sin θ:r˙=r1−r,θ r sin θ:r˙=r1−r,θ r sin θ:r˙=r1−r,θ functionH2=x32.Alb ≠ 0, then the growth function of f, ψ, and F are C1),(A-II)f(0) = 0, f(x) > 0 for x > 0,(A-III)x − 1)F(x) < 0 for x ≠ 1,((A-IV)x − λ1)ψ(x) > 0, x ≠ λ1 > 0,(We start this study by relating  to a wid a Gause  type prend Cheng . In geneψ(x) = 0 (the predator isocline) in this case givesλ1 < 1, then solutions of system ) that has the JacobianJ) = λ1f(λ1)ψ′(λ1) > 0 and Tr J) = f(λ1)F′(λ1) with f(λ1) > 0, so the Trace-determinant criterion [x = λ1) intersects the prey isocline (y = F(x)) at point where the prey isocline decreases F′(λ1) < 0 and is unstable when F′(λ1) > 0. In fact, all the topological properties including results of global stability and uniqueness of limit cycles for all parameters of  and finally the chemostat estimateκ = 0 corresponds to the known case b = 0 . We remb/m.Now,  takes thm.Now,  in Kuang [x in f, ψ, ρ, and H are C1),(C-II)f(0) = 0, f(ξ) > 0 for ξ > 0,(C-III)H′(s) < 0, H(1) = 0,(C-IV)ξ − λ2)ψ(ξ) > 0, ξ ≠ λ2 > 0,((C-V)ρ(ξ) > 0, ρ′(ξ) > 0, ξ > 0,(C-VI)f(ξ) + ψ(ξ) < 0, λ2 < ξ ≤ 1.−We rewrite the system on a form allowing isoclines to be analyzable. It is far from clear how this should be done in the chemostat case. We decided to work with the following form:α > μ. Before going further, we prove a basic theorem. We note that system  correspoξ ≥ 0, η ≥ 0, ξ + η ≤ 1. Solutions of  = , and  = . Thus, solutions remain positive. To prove that solutions remain bounded, we assume 1 ≤ ξ + η ≤ γ/κ and consider the series of inequalitiesBy uniqueness of solutions no solutions can intersect the four solutions H decreasing and concave down as far as possible, too. We have one equilibrium at the origin, one at the carrying capacity , and one equilibrium at , where η∗ satisfies the condition η∗ = ρ(λ2)H(λ2 + η∗). We prove first, that the first two equilibria are saddles. The corresponding Jacobians are given byJ also contains the information that the eigenvector corresponding to the positive eigenvalue points into the triangle ξ ≥ 0, η ≥ 0, and ξ + η ≤ 1. This criterion can be formulated asα > μ and f(1) > ψ(1) > 0.We further conclude thatλ2, η∗) we start by doing some estimates concerning its location and define an implicit function ξ ≤ 1. The sign of the derivative of the implicit function λ2, η∗) and get For the interior equilibrium –(C-VI). The interior fixed point of  is localξ > 0, η > 0, ξ + η < 1 is invariant and its boundary is not approached by any of the solutions. We illustrate the graphical conclusions of The last assertion is due to the Poincaré-Bendixson theorem  becauseη(0) = 0 and η(0) = 1, resp., in the above equation and remember the chemostat estimate (βξ2 and −καη2 are of different sign). The level curve consists therefore of either two intersecting lines or a hyperbola. One branch of this hyperbola is the predator isocline curve Finally, we return to the specific expressions that we used as our prototype example  in orderestimate ). We arecause of . We contcause of  can be wen as in , the isoThe prey isocline t system  with (22


Escherichia coli RuvB hexameric ring motor proteins, together with RuvAs, promote branch migration of Holliday junction DNA. Zero mode waveguides (ZMWs) constitute of nanosized holes and enable the visualization of a single fluorescent molecule under micromolar order of the molecules, which is applicable to characterize the formation of RuvA–RuvB–Holliday junction DNA complex. In this study, we used ZMWs and counted the number of RuvBs binding to RuvA–Holliday junction DNA complex. Our data demonstrated that different nucleotide analogs increased the amount of Cy5-RuvBs binding to RuvA–Holliday junction DNA complex in the following order: no nucleotide, ADP, ATPγS, and mixture of ADP and ATPγS. These results suggest that not only ATP binding to RuvB but also ATP hydrolysis by RuvB facilitates a stable RuvA–RuvB–Holliday junction DNA complex formation.The  Escherichia coli, RuvA, RuvB, and RuvC are involved in the processing of Holliday junction DNA into mature recombinant DNA molecules12379Homologous recombination is a crucial biological process not only for the repair of damaged chromosomes, but also for generating genetic diversity. Holliday junction DNA is an important intermediate of the homologous recombination that consists of two homologous duplex DNA molecules linked by a single-stranded crossover. In Thermus themophilus and Thermatoga maritime RuvBs have a crescent–like structure with three domains N, M, and C111311181920Crystallographic studies showed that −15 L), it is very difficult to visualize single fluorescent molecule of interest under submicromolar order concentrations of the fluorescent molecules. To overcome this limitation, zero mode waveguides (ZMWs) have been developed and applied to single molecule real time DNA sequencing2425−19–10−20 L. Thus, the ZMWs is said to enable the visualization of a single fluorescent molecule under micromolar order of the molecules272829The single molecule fluorescence imaging technique using total internal reflection fluorescence (TIRF) microscope is a conventional and powerful method for the characterization of biomolecule interactions in real time22In case of RuvB, approximately submicromolar concentrations of RuvB were required for RuvB binding to a Holliday junction DNA. Thus, using TIRF microscope, the observation of fluorescently labeled RuvB binding to a Holliday junction DNA was very difficult. In this study, to characterize RuvB binding to RuvA–Holliday junction DNA complex, we fabricated ZMWs and labeled RuvB with Cy5. Then, we succeeded in visualizing Cy5-RuvBs binding to a RuvA–Holliday junction DNA complex immobilized on the nanohole. We counted the number of Cy5 photobreaching steps under various nucleotide conditions and determined the most probable numbers of RuvBs binding to the complex. Our data shows that, in the presence of ATPγS and ADP, a more stable RuvA–RuvB–Holliday junction complex was formed, suggesting that ATP synergistically facilitates both RuvB hexameric ring formation and RuvA–RuvB–Holliday junction DNA complex formation, which is crucial for Holliday junction DNA branch migration.E. coli RuvB has no Cys residues.To characterize RuvB binding to Holliday junction DNA using the single molecule fluorescence imaging technique, we constructed and purified a RuvB mutant, RuvB-S39C, and then label the purified RuvB protein with Cy5-maleimide as described in Materials and Methods. We used RuvB-S39C to label RuvB protein by the highly specific conventional reaction between the sulfhydryl group and maleimide group because wild type To determine the effect of Ser-Cys mutation, we measured branch migration activity of the purified RuvB mutant in the presence of RuvA using a stopped-flow system . We usedThe labeling ratio of Cy5-labeled RuvB-S39C was 42%. The Holliday junction DNA branch migration activity of Cy5-labeled RuvB-S39C was comparable with that of unlabeled RuvB-S39C , indicatAs described in the methods, 400 nM of Cy5-labeled RuvB-S39C was used to visualize RuvB binding to the RuvA–Holliday junction DNA complex. Because ZMWs enable us to visualize single fluorescently labeled biomolecules at a high concentration of them, we fabricated ZMWs for single molecule Cy5-RuvB observation. Two methods have primarily been reported for ZMW fabrication31To confirm that RuvA–RuvB complexes is capable of promoting branch migration of Holliday junction DNA in the nanoholes, Cy3-labeled Holliday junction DNA was immobilized on a streptavidin coated glass surface, as described in Materials and Methods. The ratio of fluorescent spots of Holliday junction DNA to nanoholes was about 90% before the addition of RuvB proteins to the nanoholes . After tAs described above, we demonstrated that the RuvA–RuvB complex promoted branch migration of Cy3 labeled Holliday junction DNA immobilized on the nanohole, indicating that the RuvA–RuvB–Holliday junction DNA complex was formed in the nanohole. Next, we performed single molecule characterization of the RuvA–RuvB–Holliday junction DNA complex formation using Cy5-labeled RuvB. In the presence of ATP, the RuvA–RuvB protein complex promotes Holliday junction DNA branch migration, resulting in the disassembly of Holliday junction DNA and the formation of Y-form DNA. Here, we used ATPγS and ADP as nucleotide cofactors. It was impossible for us to visualize Cy5-RuvB binding to the junction in the presence of ATP. We observed bright spots that emitted stable Cy3 and Cy5 fluorescence from nanoholes on which Cy3-Holliday junction DNA was immobilized. Most of the Cy3 and Cy5 fluorescence intensity decreased in a stepwise manner due to photobreaching . The num2+, RuvBs exist as a monomer and/or dimer233We characterized RuvB binding to the complex under conditions without nucleotides, with ADP, ATPγS or both ADP and ATPγS . BecauseIn the case that the RuvB protomer is a dimer, we considered that the distribution of Cy5-RuvB in dimer was random, and as shown in To date, RuvB properties on RuvA–RuvB–Holliday junction complex formation or DNA–binding activity have been largely characterized by an electrophoresis mobility shift assay (EMSA) with glutaraldehyde cross-linking3536Single fluorescence imaging techniques enabled us to characterize the protein–protein or protein–DNA complex in more detail. We could visualize the assembly or disassembly processes of the complex and count the number of molecules constituting the complex in real time. In this study, to characterize the single molecule formation of RuvA–RuvB–Holliday junction DNA complex, we labeled RuvB with Cy5 and fabricated ZMWs. We measured the number of RuvBs binding to a RuvA–Holliday junction DNA complex under various nucleotide conditions . Interes1337381314In the presence of ATPγS and ADP, 97%–98% of Holliday junction DNAs interacted with RuvBs and approximately 3% of the complexes contained six RuvBs  and 2. T40412+. Thus, in this study, we determined the distribution of the number of RuvBs binding to a RuvA–Holliday junction DNA based on two models. One model is based on the model that the RuvB protomer is a monomer -ΔruvABC100::kan) was used for protein overexpressionThe expression plasmid for RuvB-S39C was constructed by PCR mediated site-directed mutagenesis as described previouslyRuvA and RuvB proteins were purified as previously described3N, N-dimethylformamide (DMF). RuvB-S39C and Cy5-maleimide was mixed at the ratio of 1:5 in 1.0 mL of mixture containing 10 μM RuvB-S39C, 50 μM Cy5-maleimide, 20 mM HEPES-KOH (pH 7.0), and 8% Glycerol. The mixture was incubated for 16 h at 4 °C. After the coupling reaction, purification of Cy5-RuvB was performed using Resource Q (GE) in a purification buffer containing 30 mM Tris-HCl (pH 7.5), 1 mM EDTA and 15% Glycerol. The protein was eluted with a 20 mL linear gradient from 0 M to 1 M NaCl in a purification buffer.Using RuvB-S39C and Cy5-maleimide, Cy5-RuvB was prepared as below. Approximately 1.0 mg of Cy5-maleimide was dissolved in 10 μL of 2, and Holliday junction DNA was constructed as described previously and used for the branch migration assayTwo Holliday junction DNAs were prepared as below. Four oligonucleotides  were mixed in a buffer containing 50 mM Tris-HCl (pH 7.5), 100 mM NaCl, and 10 mM MgCl2 and 1 mM DTT. Solution II contained 400 nM RuvB, 2 mM ATP, 20 mM Tris-HCl (pH 8.0), 10 mM MgCl2 and 1 mM DTT. Branch migration starts by mixing Solution I and Solution II at 25 °C. All data curves represent the average of at least four experiments.Branch migration activity of the RuvA–RuvB complex was carried out using a stopped-flow spectrafluorometer  equipped with a photomultiplier tube. The filter used with the photomultiplier was Semrock FF01-567/15-25. Excitation was at 520 nm for Cy3. Solution I and Solution II were prepared as below. Solution I contained 100 nM RuvA, 10 nM Holliday junction DNA, 20 mM Tris-HCl (pH 8.0), 10 mM MgClFused silica coverslips were immersed in 4% ammonium and 4.3% hydrogen peroxide for 10 min at 75 °C and then washed thoroughly with deionized water. The coverslips were dried with an air blower, and then baked at 200 °C for 10 min. The coverslips were then cleaned by air plasma at 18 W for 10 min. Before coating with Hexamethyldisilazane , the coverslips were immersed in 2-butanone and cleaned by sonication for 5 min. A resist film of ma-N 2403 (Micro resist technology) was then coated on the HMDS coated coverslip with a spin coater. The ESPACER (Showa Denko) was then coated on the ma-N 2403 coated coverslip. Electron beam (EB) lithography (Elionix Inc.) was performed with an accelerating voltage of 80 kV, and a beam current of 100 pA. After EB patterning, the coverslips were immersed in deionized water for 30 s, and the pattern was then developed by immersing it in ma-D 525 (Micro resist technology) for 2 min. After development, the coverslips were washed thoroughly with deionized water for 5 min and dried with an air blower. Aluminum coating was performed in a thermal evaporator using a BN composite boat. The thickness of the coated aluminum was monitored using a thickness monitor (Eiko Engineering Co. Ltd.). After aluminum coating, the remaining photoresist was lifted off by immersing it in 2-butanone with sonication for 5 min.The ZMWs were observed by scanning electron microscopy  and the diameter of each nanohole was measured.Samples were observed at 25 ± 2 °C on an Olympus IX71 inverted microscope with a 100X oil-immersion objective as described previouslyN-hydroxy-succinimidyl (NHS) group  and 0.1 mg NHS-Bio-PEG  dissolved in 50 mM MOPS buffer (pH 7.5). After the coating, the ZMWs were washed with MilliQ water and dried on a clean bench with an air blower.ZMWs were washed with acetone under sonication for 5 min and then washed with 2-propanol under sonication for 5 min. The ZMWs were dried with an air blower and cleaned with air plasma at 18 W for 5 min. The ZMWs were immersed in a preheated 0.6% (vol/vol) aqueous solution of poly(vinylphosphonic acid) (Funakoshi) for 10 min at 90 °C. They were washed briefly with deionized water, dried with an air blower, and annealed on a hot plate at 80 °C for 10 min. The Polyethylene glycol (PEG) coating was then performed as described previously2). PEG-coated ZMWs were incubated for 5 min at room temperature with a drop containing 0.06 mg/ml streptavidin in Buffer A. The ZMWs were then rinsed with Buffer A. A 20 μl mixture containing 400 nM RuvA and 40 nM Holliday junction DNA in Buffer A was added to the ZMWs and incubated for 5 min. The ZMWs were rinced with Buffer A containing 2.5 mM Protocatechuic acid , 250 nM Protocatechuate-3,4-dioxygenase , and 2 mM Trolox (Sigma) to wash out the unbound RuvA and Holliday junction DNA. The ZMW was set on the microscope. A mixture containing 400 nM Cy5-RuvBs and the indicated amount of nucleotides in Buffer A with 2.5 mM PCA, 250 nM PCD, and 2 mM Trolox was added to the ZMW and the fluorescence imaging from Cy3 labeled Holliday junction DNA and Cy5-RuvB were recorded.To observe RuvB binding to RuvA–Hollliday junction DNAs, the DNA was immobilized in the nanoholes. PEG-coated ZMWs were incubated for 5 min at room temperature with a drop containing 1% F127 in Water. The ZMWs were then rinsed with Buffer A (20 mM HEPES-KOH (pH 8.0), 10 mM MgClno-nucleotide, FADP, FATPγS, and FADP and ATPγS, respectively.The calculated data were obtained by a binominal distribution between 42% of Cy5-RuvB and 58% of nonlabeled RuvB as shown in no-nucleotide = (a/100 + 0.58 × b/100 + 0.3364 × c/100 + 0.1551 × d/100 + 0.1132 × e/100 + 0.0656 × f/100 + 0.0381 × g/100 − 0.58)2 + (0.42 × b/100 + 0.4872 × c/100 + 0.4239 × d/100 + 0.3278 × e/100 + 0.2377 × f/100 + 0.1654 × g/100 − 0.35)2 + (0.1764 × c/100 + 0.3069 × d/100 + 0.3560 × e/100 + 0.3442 × f/100 + 0.2994 × g/100 − 0.07)2 + (0.0741 × d/100 + 0.1719 × e/100 + 0.2492 × f/100 + 0.2891 × g/100)2 + (0.0311 × e/100 + 0.0902 × f/100 + 0.1570 × g/100)2 + (0.0131 × f/100 + 0.0455 × g/100)2 + (0.0055 × g/100)2.FADP = (a/100 + 0.58 × b/100 + 0.3364 × c/100 + 0.1551 × d/100 + 0.1132 × e/100 + 0.0656 × f/100 + 0.0381 × g/100 − 0.32)2 + (0.42 × b/100 + 0.4872 × c/100 + 0.4239 × d/100 + 0.3278 × e/100 + 0.2377 × f/100 + 0.1654 × g/100 − 0.37)2 + (0.1764 × c/100 + 0.3069 × d/100 + 0.3560 × e/100 + 0.3442 × f/100 + 0.2994 × g/100 − 0.23)2 + (0.0741 × d/100 + 0.1719 × e/100 + 0.2492 × f/100 + 0.2891 × g/100  − 0.08)2 + (0.0311 × e/100 + 0.0902 × f/100 + 0.1570 × g/100)2 + (0.0131 × f/100 + 0.0455 × g/100)2 + (0.0055 × g/100)2.FATPγS = (a/100 + 0.58 × b/100 + 0.3364 × c/100 + 0.1551 × d/100 + 0.1132 × e/100 + 0.0656 × f/100 + 0.0381 × g/100 − 0.29)2 + (0.42 × b/100 + 0.4872 × c/100 + 0.4239 × d/100 + 0.3278 × e/100 + 0.2377 × f/100 + 0.1654 × g/100 − 0.35)2 + (0.1764 × c/100 + 0.3069 × d/100 + 0.3560 × e/100 + 0.3442 × f/100 + 0.2994 × g/100 − 0.24)2 + (0.0741 × d/100 + 0.1719 × e/100 + 0.2492 × f/100 + 0.2891 × g/100 − 0.10)2 + (0.0311 × e/100 + 0.0902 × f/100 + 0.1570 × g/100 − 0.02)2 + (0.0131 × f/100 + 0.0455 × g/100)2 + (0.0055 × g/100)2.FADP and ATPγS = (a/100 + 0.58 × b/100 + 0.3364 × c/100 + 0.1551 × d/100 + 0.1132 × e/100 + 0.0656 × f/100 + 0.0381 × g/100 − 0.19)2 + (0.42 × b/100 + 0.4872 × c/100 + 0.4239 × d/100 + 0.3278 × e/100 + 0.2377 × f/100 + 0.1654 × g/100 − 0.35)2 + (0.1764 × c/100 + 0.3069 × d/100 + 0.3560 × e/100 + 0.3442 × f/100 + 0.2994 × g/100 − 0.30)2 + (0.0741 × d/100 + 0.1719 × e/100 + 0.2492 × f/100 + 0.2891 × g/100 − 0.13)2 + (0.0311 × e/100 + 0.0902 × f/100 + 0.1570 × g/100 − 0.02)2 + (0.0131 × f/100 + 0.0455 × g/100 − 0.01)2 + (0.0055 × g/100)2.Fa, b, c, d, e, f, and g indicate percentages of 0, 1, 2, 3, 4, 5, and 6 RuvBs binding to a RuvA-Holliday junction DNA. The a, b, c, d, e, f, and g satisfied the following conditions.a, b, c, d, e, f, and g were nonnegative integers.The inequalities as below were not allowed.no-nucleotide, GADP, GATPγS, and GADP and ATPγS, respectively.In case that the RuvB protomer is a dimer containing Cy5-RuvB at random, the calculated data were also obtained by a binominal distribution between 42% of Cy5-RuvB and 58% of nonlabeled RuvB as shown in no-nucleotide = (h/100 + 0.3364 × i/100 + 0.1132 × j/100 + 0.0381 × k/100 − 0.58)2 + (0.4872 × i/100 + 0.3278 × j/100 + 0.1654 × k/100 − 0.35)2 + (0.1764 × i/100 + 0.3560 × j/100 + 0.2994 × k/100 − 0.07)2 + (0.1719 × j/100 + 0.2891 × k/100)2 + (0.0311 × j/100 + 0.1570 × k/100)2 + (0.0455 × k/100)2 + (0.0055 × k/100)2.GADP = (h/100 + 0.3364 × i/100 + 0.1132 × j/100 + 0.0381 × k/100  − 0.32)2 + (0.4872 × i/100 + 0.3278 × j/100 + 0.1654 × k/100 − 0.37)2 + (0.1764 × i/100 + 0.3560 × j/100 + 0.2994 × k/100  − 0.23)2 + (0.1719 × j/100 + 0.2891 × k/100  − 0.08)2 + (0.0311 × j/100 + 0.1570 × k/100)2 + (0.0455 × k/100)2 + (0.0055 × k/100)2.GATPγS = (h/100 + 0.3364 × i/100 + 0.1132 × j/100 + 0.0381 × k/100 − 0.29)2 + (0.4872 × i/100 + 0.3278 × j/100 + 0.1654 × k/100 − 0.35)2 + (0.1764 × i/100 + 0.3560 × j/100 + 0.2994 × k/100 − 0.24)2 + (0.1719 × j/100 + 0.2891 × k/100 − 0.10)2 + (0.0311 × j/100 + 0.1570 × k/100 – 0.02)2 + (0.0455 × k/100)2 + (0.0055 × k/100)2.GADP and ATPγS = (h/100 + 0.3364 × i/100 + 0.1132 × j/100 + 0.0381 × k/100 − 0.19)2 + (0.4872 × i/100 + 0.3278 × j/100 + 0.1654 × k/100 − 0.35)2 + (0.1764 × i/100 + 0.3560 × j/100 + 0.2994 × k/100 − 0.30)2 + (0.1719 × j/100 + 0.2891 × k/100 − 0.13)2 + (0.0311 × j/100 + 0.1570 × k/100 – 0.02)2 + (0.0455 × k/100 – 0.01)2 + (0.0055 × k/100)2.Gh, i, j, and k indicate percentages of 0, 2, 4, and 6 RuvB dimers binding to a RuvA-Holliday junction DNA. The h, i, j, and k satisfied the following conditions.h, i, j, and k were nonnegative integers.The inequalities as below were not allowed.no-nucleotide, HADP, HATPγS, and HADP and ATPγS, respectively.In case that all of Cy5 labeled RuvB dimers contain a Cy5-RuvB and a non-labeled RuvB, the calculated data were obtained by a binominal distribution between 84% of RuvB dimer containing single Cy5-RuvB and 16% of nonlabeled RuvB dimer as shown in no-nucleotide = (l/100 + 0.16 × m/100 + 0.0256 × n/100 + 0.0041 × p/100 + 0.0007 × q/100 + 0.0001 × r/100 − 0.58)2 + (0.84 × m/100 + 0.2688 × n/100 + 0.0645 × p/100 + 0.0138 × q/100 + 0.0028 × r/100 + 0.0005 × s/100 − 0.35)2 + (0.7056 × n/100 + 0.3387 × p/100 + 0.1084 × q/100 + 0.0289 × r/100 + 0.0069 × s/100 − 0.07)2 + (0.5927 × p/100 + 0.3793 × q/100 + 0.1517 × r/100 + 0.0486 × s/100)2 + (0.4978 × q/100 + 0.3983 × r/100 + 0.1912 × s/100)2 + (0.4182 × r/100 + 0.4015 × s/100)2 + (0.3513 × s/100)2.HADP = (l/100 + 0.16 × m/100 + 0.0256 × n/100 + 0.0041 × p/100 + 0.0007 × q/100 + 0.0001 × r/100 − 0.32)2 + (0.84 × m/100 + 0.2688 × n/100 + 0.0645 × p/100 + 0.0138 × q/100 + 0.0028 × r/100 + 0.0005 × s/100 − 0.37)2 + (0.7056 × n/100 + 0.3387 × p/100 + 0.1084 × q/100 + 0.0289 × r/100 + 0.0069 × s/100 − 0.23)2 + (0.5927 × p/100 + 0.3793 × q/100 + 0.1517 × r/100 + 0.0486 × s/100 – 0.08)2 + (0.4978 × q/100 + 0.3983 × r/100 + 0.1912 × s/100)2 + (0.4182 × r/100 + 0.4015 × s/100)2 + (0.3513 × s/100)2.HATPγS = (l/100 + 0.16 × m/100 + 0.0256 × n/100 + 0.0041 × p/100 + 0.0007 × q/100 + 0.0001 × r/100 − 0.29)2 + (0.84 × m/100 + 0.2688 × n/100 + 0.0645 × p/100 + 0.0138 × q/100 + 0.0028 × r/100 + 0.0005 × s/100 − 0.35)2 + (0.7056 × n/100 + 0.3387 × p/100 + 0.1084 × q/100 + 0.0289 × r/100 + 0.0069 × s/100 − 0.24)2 + (0.5927 × p/100 + 0.3793 × q/100 + 0.1517 × r/100 + 0.0486 × s/100 – 0.10)2 + (0.4978 × q/100 + 0.3983 × r/100 + 0.1912 × s/100 – 0.02)2 + (0.4182 × r/100 + 0.4015 × s/100)2 + (0.3513 × s/100)2.HADP and ATPγS = (l/100 + 0.16 × m/100 + 0.0256 × n/100 + 0.0041 × p/100 + 0.0007 × q/100 + 0.0001 × r/100 − 0.19)2 + (0.84 × m/100 + 0.2688 × n/100 + 0.0645 × p/100 + 0.0138 × q/100 + 0.0028 × r/100 + 0.0005 × s/100 − 0.35)2 + (0.7056 × n/100 + 0.3387 × p/100 + 0.1084 × q/100 + 0.0289 × r/100 + 0.0069 × s/100 − 0.30)2 + (0.5927 × p/100 + 0.3793 × q/100 + 0.1517 × r/100 + 0.0486 × s/100 – 0.13)2 + (0.4978 × q/100 + 0.3983 × r/100 + 0.1912 × s/100 – 0.02)2 + (0.4182 × r/100 + 0.4015 × s/100 – 0.01)2 + (0.3513 × s/100)2.Hl, m, n, p, q, r, and s indicate percentages of 0, 1, 2, 3, 4, 5, and 6 RuvB dimers binding to a RuvA-Holliday junction DNA. The l, m, n, p, q, r, and s satisfied the following conditions.l, m, n, p, q, r, and s were nonnegative integers.The inequalities as below were not allowed.How to cite this article: Iwasa, T. et al. Synergistic effect of ATP for RuvA—RuvB—Holliday junction DNA complex formation. Sci. Rep.5, 18177; doi: 10.1038/srep18177 (2015).


N-dimensional Euler equations with spherical symmetry. We first show that there are only trivial solutions when the velocity is of the form c(t)|x|α−1x + b(t)(x/|x|) for any value of α ≠ 1 or any positive integer N ≠ 1. Then, we show that blowup phenomenon occurs when α = N = 1 and γ > 1) and the isothermal case (γ = 1). The blowup phenomenon of solutions is investigated for the initial-boundary value problem (IBVP) of the  N-dimensional Euler equations for compressible fluid:ρ, u, and p represent the density, velocity, and pressure of the fluid, respectively. n is the unit normal vector on the unit sphere. The γ-law for p is given by  and b(t) is a time-dependent drifting function.In , the autf system  in spherρ = ρ and u = u(x/r), system rα + b(t) for any real α and integer N in the isentropic and isothermal cases except the case γ = α = N = 1. For γ = α = N = 1, one has the following two cases.(1)t,c(t) → 0, → as t → ∞.If (2)t,c(t)→−∞ and → as t → ∞.If There are only trivial solutions to the r system  of the fAs a corollary, we also obtain the following.ρ, u) be a solution for (a0≔a(0) > 0, and γ = N = 1. Then a(t) satisfiesλ ∈ ℝ. Furthermore, one has the following five cases.(1)λ < 0, then the solution → as t → T∗ for some finite T∗ > 0.If (2)λ > 0, then ρ is bounded above and the solution → as t → ∞.If (3)λ = 0 and a1 > 0, ρ is bounded above and the solution → as t → ∞.If (4)λ = 0 and a1 = 0, then the solution is trivial.If (5)λ = 0 and a1 < 0, then the solution → as t → −a0/a1.If Let  is set to be positive. Thus, we suppose ρ > 0 in the following to avoid the trivial solutions ρ ≡ 0.It is well-known that γ > 1, one hasFor 1, one hasργ−2. Then, the result follows.From 1, one haγ > 1, one hasFor 2, one hasFrom 2, one haγ = 1.Similarly, we have the following two lemmas for γ = 1, one hasFor γ = 1, one hasu = c(t)r + b(t) and γ > 1.For A1 ≠ A2 or B1 ≠ B2. If A ≠ 0, B ≠ 0, andConsider the following dynamical systemc3+A1cc˙+c3+A1cc˙+A1B2 − A2B1 = 0, then it is clear that c = 0 is the only solution. So we suppose A1B2 − A2B1 ≠ 0. One has from (If has from  that(21)B = 0 and A ≠ 0, then c = 0 is the only solution.If A = 0 and B ≠ 0, then c = 0 is the only solution.If A and B are not zero.So, we suppose both ξ.From , one has1, one hasA(A1B2 − A2B1)/2B2 ≠ 0, then c = 0 is the only solution. So we suppose it is zero; that is, 22), one us, from 1, one hathat is,  holds. T one has  and 19)(23)c¨=−A one has  is satisγ > 1. Then there are only trivial solutions to the N-dimensional Euler system (u = c(t)rα + b(t) with α ≠ 1.Assume r system  of the f4, we haveN > 1, if α = 0, then u = 0 from  is independent of t and r, respectively. Thus, ρ is a constant.First, we set+bt.From 4, we hav, we haveu=ctrα−1. we have2u=ctrα−1.α−1−c˙,2ut+uurt=cα ≠ 0 and −1, after substituting , a, aα ≠ 0 tituting  into (9)tituting , and 7)α ≠ 0 andα ∉ {0,1/3,1/2,2/3,1, 2}, then the powers 3α − 1 and 2α − 1 are different and unique among the powers in (α + (γ − 1)(α + N − 1) and (γ − 1)(2α + 3N − 3) + 4α cannot be both zero for N ≠ 1, we conclude that c = 0. Hence, u = 0 and ρ is a constant.If owers in . In thisα ∈ {1/3,1/2,2/3,2}, we have For α = 1/3 or 2/3, as 2α − 1 is unique among other powers, one hasγ − 1)(2α + 3N − 3) + 4α ≠ 0, we conclude that c = 0. Thus, u = 0 and ρ is a constant.For α = 1/2, as α − 1 and −1 are different and unique among other powers, one hasN ≠ 1. As α + (γ − 1)(N − 1) ≠ 0, we conclude that c = 0. Thus, u = 0 and ρ is a constant.For α = 2, as 3α − 1 is unique among other powers, one hasα + (γ − 1)(α + N − 1) ≠ 0, we conclude that c = 0. Thus, u = 0 and ρ is a constant.For α = −1.Next, we consider the case α = −1, the corresponding equation of (3N − 5) and 2 − (N − 2)(γ − 1) cannot be both zero for N ≠ 1, we conclude that c = 0 and the solutions are trivial.For ation of  is(39)E1N = 1 and α ≠ 1, one can show that there are only trivial solutions with similar procedures. The proof is complete.For γ > 1 in γ = 1.Using similar analysis as that given for the case γ = 1. Then there are only trivial solutions to the N-dimensional Euler system (u = c(t)rα + b(t) with α ≠ 1.Assume r system  of the fα = 1 will be analyzed as follows.Next, the crucial case γ > 1. Then there are only trivial solutions to the N-dimensional Euler system (u = c(t)rα + b(t) with α = 1.Assume r system  of the fγ > 1 and α = 1, one has1 and rα + b(t) with α = 1 and N > 1. For N = α = 1, one has the following two cases.(1)t, c(t) → 0, → as t → ∞.If (2)t, c(t)→−∞ and → as t → ∞.If Assume r system  of the fγ = α = 1, the corresponding system of  in [N in (For ystem of  is46)G1G1γ = α =here47) It is cl1. Then (2c3+4cc˙+n (7) in  when we  It is c2 and  must be zero. More precisely, if T∗ is finite and limt→T∗a(t) > 0, then we can extend the solution by solving (a(T∗)≔limt→T∗a(t) > 0. This contradicts the definition of T∗.Let =a˙/a in 1. One haatisfies . Consideatisfies  with ini solving  with iniλ < 0, and thenμ≔(1/2)a12 − λlna0. It follows thatt2 is negative, we see that a(t) will be negative if t is sufficiently large. This implies that Next, suppose  and then(56)wheru → −∞ and ρ → ∞ as t → T∗.On the other hand, from , one hasλ > 0, one has(i)a ≥ eμ/λ− > 0,(ii)T∗ = ∞,(iii)From (i), (ii), and (iii) above, we see that ρ is bounded above byt→∞a(t) = ∞. This is because if limt→∞a(t) is finite, then a(t) is bounded by some positive number M > 0. But, from 1ρt,r=a0ρλ > 0, → as t → ∞.Next, we show thatlimt→∞a˙λ = 0 can be verified trivially, the proof is complete.As the cases for


Besides giving characterizations of these spaces, we study some of their properties. Also, we investigate the behavior of intuitionistic fuzzy β-compactness, intuitionistic fuzzyβ-almost compactness, and intuitionistic fuzzy β-nearly compactness under several types ofintuitionistic fuzzy continuous mappings.The concept of intuitionistic fuzzy  In this paper some properties of intuitionistic fuzzy β-compactness were investigated. We use the finite intersection property to give characterization of the intuitionistic fuzzy β-compact spaces. Also we introduce intuitionistic fuzzy β-almost compactness and intuitionistic fuzzy β-nearly compactness and established the relationships between these types of compactness.As a generalization of fuzzy sets introduced by Zadeh , the conX be a nonempty fixed set and I the closed interval . An intuitionistic fuzzy set (IFS) A is an object of the following form: μA(x) : X → I and νA(x) : X → I denote the degree of membership, namely, μA(x), and the degree of nonmembership, namely, νA(x), for each element x ∈ X to the set A, respectively, and 0 ≤ μA(x) + νA(x) ≤ 1 for each x ∈ X.Let A and B be IFSs of the forms A = {〈x, μA(x), νA(x)〉; x ∈ X} and B = {〈x, μB(x), νB(x)〉; x ∈ X}. ThenA⊆B if and only if μA(x) ≤ μB(x) and νA(x) ≥ νB(x);A∩B = {〈x, μA(x)∧μB(x), νA(x)∨νB(x)〉; x ∈ X};A ∪ B = {〈x, μA(x)∨μB(x), νA(x)∧νB(x)〉; x ∈ X}.Let A = {〈x, μA, νA〉; x ∈ X} instead of A = {〈x, μA(x), νA(x)〉; x ∈ X}.We will use the notation ~ = {〈x, 0,1〉; x ∈ X} and 1~ = {〈x, 1,0〉; x ∈ X}.Consider 0α, β ∈  such that α + β ≤ 1. An intuitionistic fuzzy point (IFP) pα, β) and  or simply by X and Y, we will denote the intuitionistic fuzzy topological spaces (IFTSs). Each IFS which belongs to τ is called an intuitionistic fuzzy open set (IFOS) in X. The complement X is called an intuitionistic fuzzy closed set (IFCS) in X.An intuitionistic fuzzy topology (IFT) in Coker's sense on a nonempty set X and Y be two nonempty sets and let f : → be a function.Let B = {〈y, μB(y), νB(y)〉; y ∈ Y} is an IFS in Y, then the preimage of B under f is denoted and defined by f−1(B) = {〈x, f−1(μB(x)), f−1(νB(x))〉; x ∈ X}. Since μB(x), νB(x) are fuzzy sets, we explain that f−1(μB(x)) = μB(x)(f(x)) and f−1(νB(x)) = νB(x)(f(x)).If pα,β) of pα,β) be an IFTS and let A = {〈x, μA(x), νA(x)〉; x ∈ X} be an IFS in X. Then the intuitionistic fuzzy closure and intuitionistic fuzzy interior of A are defined byA) = ∩{C : C  is  an  IFCS  in  X  and  C⊇A};cl (A) = ∪{D : D  is  an  IFOS  in  X  and  D⊆A}.int  is an IFCS, int (A) is an IFOS in X, and A is an IFCS in X if and only if cl (A) = A; A is an IFOS in X if and only if int (A) = A.Let  be an IFTS and let A and B be IFSs in X. Then the following properties hold:A)⊆A⊆cl (A).int  if A⊆cl(int(clA)). The complement of an IFβOS A in IFTS X is called an intuitionistic fuzzy β-closed set (IFβCS) in X.An IFS f be a mapping from an IFTS X into an IFTS Y. The mapping f is calledf−1(B) is an IFOS in X, for each IFOS B in Y;intuitionistic fuzzy continuous if and only if β-continuous if and only if f−1(B) is an IFβOS in X, for each IFOS B in Y.intuitionistic fuzzy Let X, τ) be an IFTS and let A = {〈x, μA(x), νA(x)〉; x ∈ X} be an IFS in X. Then the intuitionistic fuzzy β-closure and intuitionistic fuzzy β-interior of A are defined byβcl(A) = ∩{C : C  is  an  IFβCS  in  X  and  C⊇A};βint(A) = ∪{D : D  is  an  IFβOS  in  X  and  D⊆A}.Let → from an intuitionistic fuzzy topological space  to another intuitionistic fuzzy topological space  is said to be intuitionistic fuzzy β-irresolute if f−1(B) is an IFβOS in  for each IFβOS B in .A function X be an IFTS. A family of {〈x, μGi(x), νGi(x)〉; i ∈ J} intuitionistic fuzzy open sets (intuitionistic fuzzy β-open sets) in X satisfies the condition 1~ = ∪{〈x, μGi(x), νGi(x)〉; i ∈ J} which is called an intuitionistic fuzzy open (intuitionistic fuzzy β-open) cover of X. A finite subfamily of an intuitionistic fuzzy open (intuitionistic fuzzy β-open) cover {〈x, μGi(x), νGi(x)〉; i ∈ J} of X which is also an intuitionistic fuzzy open (intuitionistic fuzzy β-open) cover of X is called a finite subcover of {〈x, μGi(x), νGi(x)〉; i ∈ j}.Let X is called intuitionistic fuzzy compact if each fuzzy open cover of X has a finite subcover for X.An IFTS X is called intuitionistic fuzzy almost compact  if, for every IF open cover {Uj : j ∈ J} of X, there exists a finite subfamily J0 ⊂ J such that X = ∪{cl(Uj) : j ∈ J0}.An IFTS X is said to be intuitionistic fuzzy β-compact (IF β-compact) if every IF β-open cover of X has a finite subcover.An IFTS X, τ) and  be two IFTSs. A function f : X → Y is said to be intuitionistic fuzzy weakly continuous (IF weakly continuous) if, for each IFOS V in Y, f−1(V)⊆int (f−1(clV)).Let  is said to be IF β-compact relative to X if, for every collection {Ai; i ∈ I} of IF β-open subsets of X such that B⊆∪{Ai; i ∈ I}, there exists a finite subset I0 of I such that B⊆∪{Ai; i ∈ I0}.An IFTS B of  is said to be IF β-compact if it is IF β-compact as a subspace of X.An IFTS β-closed sets {Ai; i ∈ I} has the finite intersection property (in short FIP) if, for any subset I0 of I, ∩i∈I0Ai ≠ 0~.A family of IF X the following statements are equivalent. X is IF β-compact.β-closed subsets of X satisfying the FIP has a nonempty intersection.Any family of IF For an IFTS X be IF β-compact space and let {Ai; i ∈ I} be a family of IF β-closed subsets of X satisfying the FIP. Suppose ∩i∈IAi = 0~. Then β-open sets cover X, then from IF β-compactness of X there exists a finite subset I0 of I such that i∈I0Ai = 0~ which gives a contradiction and therefore ∩i∈IAi ≠ 0~. Thus (i)⇒(ii).Let Ai; i ∈ I} be a family of IF β-open sets cover X. Suppose that for any finite subset I0 of I we have ∪i∈I0Ai ≠ 1~. Then i∈IAi ≠ 1~ and contradicts that {Ai; i ∈ I} is an IF β-open cover of X. Hence our assumption ∪i∈I0Ai ≠ 1~ is wrong. Thus ∪i∈I0Ai = 1~ which implies that X is IF β-compact. Thus (ii)⇒(i).Let {β-closed subset of an intuitionistic fuzzy β-compact space is intuitionistic fuzzy β-compact relative to X.An intuitionistic fuzzy A be an IF β-closed subset of X. Let {Gi; i ∈ I} be cover of A by IF β-open sets in X. Then the family β-open cover of X. Since X is IF β-compact, there is a finite subfamily {G1, G2,…, Gn} of IF β-open cover, which also covers X. If this cover contains β-open subcover of A. So A is IF β-compact relative to X.Let X, τ) and  be intuitionistic fuzzy topological spaces and let f : → be intuitionistic fuzzy β-irresolute, surjective mapping. If  is IF β-compact space then so is .Let → be intuitionistic fuzzy β-irresolute mapping of an intuitionistic fuzzy β-compact space  onto an IFTS . Let {Ai : i ∈ I} be any intuitionistic fuzzy β-open cover of . Then {f−1(Ai) : i ∈ I} is collection of intuitionistic fuzzy β-open sets which covers X. Since X is intuitionistic fuzzy β-compact, there exists a finite subset I0 of I such that subfamily {f−1(Ai); i ∈ I0} of {f−1(Ai):  i ∈ I} covers X. It follows that {Ai; i ∈ I0} is a finite subfamily of {Ai : i ∈ I} which covers Y. Hence Y is intuitionistic fuzzy β-compact.Let X, τ) and  be intuitionistic fuzzy topological spaces and let f : → be intuitionistic fuzzy β-irresolute mapping. If A is IF β-compact relative to X then f(A) is IF β-compact relative to Y.Let (Ai : i ∈ I} be a family of IF β-open cover of Y such that f(A)⊆∪i∈IAi. Then A⊆f−1(f(A))⊆f−1(∪i∈IAi) = ∪i∈If−1(Ai). Since f is IF β-irresolute, f−1(Ai) is IF β-open cover of X. And A is IF β-compact in ; there exists a finite subset I0 of I such that A⊆∪i∈I0f−1(Ai). Hence f(A)⊆f(∪i∈I0f−1(Ai)) = ∪i∈I0f(f−1(Ai))⊆∪i∈I0(Ai). Thus f(A) is IF β-compact relative to Y.Let {β-continuous image of IF β-compact space is IF compact.An IF f : → be an IF β-continuous from an IF β-compact space X onto IFTS Y. Let {Ai : i ∈ I} be IF open cover of Y. Then {f−1(Ai); i ∈ I} is IF β-open cover of X. Since X is IF β-compact, there exists finite subset I0 of I such that finite family {f−1(Ai); i ∈ I0} covers X. Since f is onto, {Ai : i ∈ I0} is a finite cover of Y. Hence Y is IF compact.Let X, τ) and  be two intuitionistic fuzzy topological spaces. A mapping f : X → Y is said to be intuitionistic fuzzy strongly β-open if f(V) is IFβOS of Y for every IFβOS V of X.Let → be an IF strongly β-open, bijective function and Y is IF β-compact; then X is IF β-compact.Let Ai : i ∈ I} be IF β-open cover of X, and then {f(Ai) : i ∈ I} is IF β-open cover of Y. Since Y is IF β-compact, there is a finite subset I0 of I such that finite family {f(Ai) : i ∈ I0} covers Y. But 1X~ = f−1(1Y~) = f−1f(∪i∈I0(Ai)) = ∪i∈I0Ai and therefore X is IF β-compact.Let {β-compactness, IF β-almost compactness, and IF β-nearly compactness.In this section we investigate the relationships between IF X, τ) is said to be IF β-almost compactness if and only if, for every family of IF β-open cover {Ai : i ∈ I} of X, there exists a finite subset I0 of I such that ∪i∈I0βclAi = 1~.An IFTS  is said to be IF β-nearly compactness if and only if, for every family of IF β-open cover {Ai : i ∈ I} of X, there exists a finite subset I0 of I such that ∪i∈I0βint(βclAi) = 1~.An IFTS  is said to be IF β-regular if, for each IF β-open set A ∈ X, A = ∪{Ai ∈ IX∣Ai  is  IF  β-open, βclAi⊆A}.An IFTS  be IFTS. Then IF β-compactness implies IF β-nearly compactness which implies IF β-almost compactness.Let  be IF β-compact. Then for every IF β-open cover {Ai : i ∈ I} of X, there exists a finite subset ∪i∈I0Ai = 1~. Since Ai is an IFβOS, for each i ∈ I, Ai = βint(Ai) for each i ∈ I. Ai = βintAi⊆βint(β cl Ai) for each i ∈ I. Here 1~ = ∪i∈I0Ai = ∪i∈I0βintAi⊆∪i∈I0βint(β cl Ai). Thus ∪i∈I0βint(βclAi) = 1~ which implies that  is IF β-nearly compactness. Now let  be IF β-nearly compact. Then for every IF β-open cover {Ai : i ∈ I} of X, there exists a finite subset ∪i∈I0βint(βclAi) = 1~. Since βint(βclAi)⊆βclAi for each i ∈ I0, 1~ = ∪i∈I0βint(βclAi)⊆∪i∈I0βclAi. Thus ∪i∈I0βclAi = 1~. Hence  is IF β-almost compact.Let  is IFTS, where τ = {0~, 1~, An}, n ∈ N, where An : X →  is defined by An = {〈x, 1 − 1/n, 1/n〉; x ∈ X}, n ∈ N. The collection {An : n ∈ N} is IF β-open cover of X. But no finite subset of {An : n ∈ N} covers X. Hence X is not IF β-compact. But βclAn = 1~ for n ≥ 3. Thus there exists a finite subfamily {An : n ∈ N0} for N0⊆N such that ∪n∈N0βclAn = 1~. Thus X is IF β-almost compactness. Also βint(βclAn) = βint (1~) = 1~ for n ≥ 3. Thus there exists a finite subfamily {An : n ∈ N0} for N0⊆N such that ∪n∈N0βintβclAn = 1~. Thus X is IF β-nearly compactness.Let X, τ) be IFTS. If  is IF β-almost compact and IF β-regular then  is IF β-compact.Let  is IF β-regular, Ai = ∪{Bi ∈ IX∣Bi  is  IF  β-open, βcl Bi⊆Ai} for each i ∈ I. Since 1~ = ∪i∈I(∪i∈IBi) and  is IF β-almost compact there exists a finite set I0 of I such that ∪i∈I0βclBi = 1~. But βcl(Bi)⊆Ai  (βint(βclBi)⊆βcl(Bi)). We have ∪i∈I0Ai⊇∪i∈I0βclBi = 1~. Thus, ∪i∈I0Ai = 1~. Hence  is IF β-compact.Let {X, τ) be IFTS. If  is IF β-nearly compact and IF β-regular then  is IF β-compact.Let  is IF β-regular, Ai = ∪{Bi ∈ IX∣Bi  is  IF  β-open, β cl Bi⊆Ai} for each i ∈ I. Since 1~ = ∪i∈I(∪i∈IBi) and  is IF β-nearly compact there exists a finite set I0 of I such that ∪i∈I0βint(βclBi) = 1~. But βint (βclBi)⊆βcl(Bi)⊆Ai. We have ∪i∈I0Ai⊇∪i∈I0βclBi⊇∪i∈I0βint(βclBi) = 1~. Thus, ∪i∈I0Ai = 1~. Hence  is IF β-compact.Let {X, τ) is IF β-almost compact, if and only if, for every family {Ai : i ∈ I} of IF β-open sets having the FIP, ∩i∈IβclAi ≠ 0~.An IFTS  is IF β-almost compact, there exists a finite subset I0 of I such that i∈I0βint(β cl Ai) = 0~. But Ai = βintAi⊆βint (βclAi). This implies that ∩i∈I0Ai = 0~ which is in contradiction with FIP of the family. Conversely, let {Ai : i ∈ I} be a family of IF β-open sets such that ∪i∈IAi = 1~. Suppose that there exists no finite subset I0 of I such that ∪i∈I0βclAi = 1~. Since Ai⊆βint (βclAi) for each i ∈ I, ∪i∈IAi ≠ 1~ which is in contradiction with ∪i∈IAi = 1~.Let {X, τ) and  be IFTS and let f : → be intuitionistic fuzzy β-irresolute, surjective mapping. If  is IF β-almost compact space then so is .Let → be intuitionistic fuzzy β-irresolute mapping of an intuitionistic fuzzy β-compact space  onto an IFTS . Let {Ai : i ∈ I} be any intuitionistic fuzzy β-open cover of . Then {f−1(Ai) : i ∈ I} is an intuitionistic fuzzy β-open cover of X. Since X is intuitionistic fuzzy β-almost compact, there exists a finite subset I0 of I such that ∪i∈I0βcl (f−1(Ai)) = 1X~. And f(1X~) = f(∪i∈I0βcl(f−1(Ai))) = ∪i∈I0f(βcl(f−1(Ai))) = 1Y~. But βcl (f−1(Ai))⊆f−1(βcl Ai) and from the surjectivity of f, f(βcl (f−1(Ai)))⊆f(f−1(βclAi)) = βclAi. So ∪i∈I0βclAi⊇∪i∈I0f(βcl(f−1(Ai))) = 1Y~. Thus ∪i∈I0  βclAi = 1Y~. Hence  is IF β-almost compact.Let X, τ) and  be intuitionistic fuzzy topological spaces and let f : → be intuitionistic fuzzy β-continuous, surjective mapping. If  is IF β-almost compact space then  is IF almost compact.Let . Then {f−1(Ai) : i ∈ I} is an intuitionistic fuzzy β-open cover of X. Since X is intuitionistic fuzzy β-almost compact, there exists a finite subset I0 of I such that ∪i∈I0βcl(f−1(Ai)) = 1X~. And from the surjectivity of f, 1Y~ = f(1X~) = f(∪i∈I0βcl(f−1(Ai)))⊆∪i∈I0f(βcl(f−1(Ai)))∪i∈I0βclf(f−1(Ai))⊆∪i∈I0clf(f−1(Ai))⊆∪i∈I0cl Ai which implies that ∪i∈I0clAi = 1Y~. Hence  is IF almost compact.Let {X, τ) and  be two intuitionistic fuzzy topological spaces. A function f : X → Y is said to be intuitionistic fuzzy β-weakly continuous (IF β-weakly continuous) if, for each IFβOS V in Y, f−1(V)⊆βint(f−1(βclV)).Let  to an IFTS  is IF strongly β-open if and only if f(βintV)⊆βintf(V).A mapping f is IF strongly β-open mapping then f(βintV) is an IFβOS in Y for IFβOS V in X. Hence f(βintV) = β intf(βintV) = βintf(V). Thus f(βintV)⊆βintf(V).If V be IFβOS in X and then V = βintV. Then by hypothesis, f(V) = f(βintV)⊆βintf(V). This implies that f(V) is IFβOS in Y.Conversely, let X, τ) and  be IFTS and let f : → be intuitionistic fuzzy β-weakly continuous, surjective mapping. If  is IF β-compact space then  is IF β-almost compact.Let (Ai : i ∈ I} be IF β-open cover of Y such that ∪i∈IAi = 1Y~. Then ∪i∈If−1(Ai) = f−1(∪i∈IAi) = f−1(1Y~) = 1X~.  is IF β-compact, and there exists a finite subset I0 of I such that ∪i∈I0f−1(Ai) = 1X~. Since f is IF β-weakly continuous, f−1(Ai)⊆βint(f−1(βclAi))⊆f−1(βclAi). This implies that ∪i∈I0f−1(βclAi)⊇∪i∈I0f−1(Ai) = 1X~. Thus ∪i∈I0f−1(βclAi) = 1X~. Since f is surjective, 1Y~ = f(1X~) = f(∪i∈I0f−1(βclAi)) = ∪i∈I0f(f−1(βclAi)) = ∪i∈I0βclAi. Hence ∪i∈I0βclAi = 1Y~. Hence  is IF β-almost compact.Let {X, τ) and  be intuitionistic fuzzy topological spaces and let f : → be intuitionistic fuzzy β-irresolute, surjective, and strongly β-open mapping. If  is IF β-nearly compact space then so is .Let . Since f is IF β-irresolute, then {f−1(Ai) : i ∈ I} is an intuitionistic fuzzy β-open cover of X. Since  is IF β-nearly compact, there exists a finite subset I0 of I such that ∪i∈I0βint(βclf−1(Ai)) = 1X~. Since f is surjective, 1Y~ = f(1X~) = f(∪i∈I0βint (βclf−1(Ai))) = ∪i∈I0f(β int (βclf−1(A))). Since f is IF strongly β-open, f(βint(βclf−1(Ai)))⊆βintf(β clf−1(Ai)) for each i ∈ I. Since f is IF β-irresolute, then f(βclf−1(Ai))⊆βclf(f−1(Ai)). Hence we have 1Y~ = ∪i∈I0f(βint(βclf−1(Ai)))  ⊆  ∪i∈I0βintf(βclf−1(Ai))  ⊆  ∪i∈I0βint(βclf(f−1(Ai))) = ∪i∈I0βint(βcl(Ai)). Thus 1Y~ = ∪i∈I0βint(βcl(Ai)). Hence  is IF β-nearly compact.Let {