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Research article

Multistep hybrid viscosity method for split monotone variational inclusion and fixed point problems in Hilbert spaces

  • Received: 27 April 2020 Accepted: 08 July 2020 Published: 22 July 2020
  • MSC : 47H05, 47H10, 47J22, 49J40

  • In this paper, we present a multi-step hybrid iterative method. It is proven that under appropriate assumptions, the proposed iterative method converges strongly to a common element of fixed point of a finite family of nonexpansive mappings, the solution set of split monotone variational inclusion problem and the solution set of triple hierarchical variational inequality problem (THVI) in real Hilbert spaces. In addition, we give a numerical example of a triple hierarchical system derived from our generalization.

    Citation: Jamilu Abubakar, Poom Kumam, Jitsupa Deepho. Multistep hybrid viscosity method for split monotone variational inclusion and fixed point problems in Hilbert spaces[J]. AIMS Mathematics, 2020, 5(6): 5969-5992. doi: 10.3934/math.2020382

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  • In this paper, we present a multi-step hybrid iterative method. It is proven that under appropriate assumptions, the proposed iterative method converges strongly to a common element of fixed point of a finite family of nonexpansive mappings, the solution set of split monotone variational inclusion problem and the solution set of triple hierarchical variational inequality problem (THVI) in real Hilbert spaces. In addition, we give a numerical example of a triple hierarchical system derived from our generalization.


    In this paper, we are concerned with the uniqueness and multiplicity of positive solutions for the quasilinear two-point boundary value problem

    {(u1u2)=λf(u),u(x)>0,    1<x<1,u(1)=u(1)=0,       (1.1)

    where λ>0 is a parameter, f:R[0,). This is the one-dimensional version of the Dirichlet problem associated with the prescribed mean curvature equation in Minkowski space

    {div(u1|u|2)=λf(x,u)   in Ω,u=0                                       on Ω. (1.2)

    The study of spacelike submanifolds of codimension one in the flat Minkowski space LN+1 with prescribed mean extrinsic curvature can lead to the type of problems (1.2), where

    LN+1:={(x,t):xRN, tR}

    endowed with the Lorentzian metric

    Ni=1(dxi)2(dt)2,

    and (x,t) is the canonical coordinate in RN+1 (see [1]). Problem (1.1) is also related to classical relativity [2], the Born-Infeld model in the theory of nonlinear electrodynamics [3,4] and cosmology [5,6].

    In 1934, in order to combine Maxwell's equation with the quantum theory, Born [7] proposed a modified energy function

    L(a)=1+a2(H2E2)1a2,

    where a is a constant of the dimensions r20/e with e= charge, r0= radius of the electron, H is the magnetic field and E is the electric field. Then, it is easy to verify that lima0L(a)=L=12(H2E2). Here L is the Lagrangian of Maxwell's equation. When H=0 and a=1, the Poisson equation of L(a) is

    div(u1|u|2)=ρ,

    where ρ denotes the charge density and u=E. This is closely related to the first equation of problem (1.2). See [7,8] for more details.

    The existence/ multiplicity of positive solutions for this kind of problems have been extensively studied by using the method of lower and upper solutions, variational methods, fixed point theorem in cones as well as the bifurcation theory, see [9,10,11,12,13,14] and the references therein.

    Recently, the authors of [9] obtained some existence results for positive radial solutions of problem (1.2) without parameter λ. In another paper, they [10] established some nonexistence, existence and multiplicity results for positive radial solutions of problem (1.2) with λf(|x|,s)=λμ(|x|)sq, where q>1,μ:[0,)R is continuous, strictly positive on (0,). In [11], under some assumptions on f, the authors proved that problem (1.2) has infinitely many radial solutions. In [12], Coelho et al. discussed the existence of either one, or two, or three, or infinitely many positive solutions of the quasilinear two-point boundary value problem (1.1).

    However, as far as we know, there are few works on the uniqueness of positive solutions of problem (1.1). Very recently, Zhang and Feng [13] discussed the exactness of positive solutions (1.1) under some special cases of f. For example, f(u)=up(p>0), up+uq(0<p<q1).

    Let us briefly recall their main results.

    Theorem A. Assume that f(u)=up with p<1, then for any λ>0 (1.1) has exactly one positive solution.

    Theorem B. Assume that f(u)=up+uq with 0<p<q1, then for any λ>0 (1.1) has exactly one positive solution.

    Motivated on these studies, we are interested in investigating a general class of f, and we point out f(u)=up with 0<p<1, f(u)=up+uq with 0<p<q1 as the special case. In addition, we consider the case where f has n zeros and obtain the existences of arbitrarily many positive solutions of the problem (1.1).

    Throughout this paper, we assume that

    (H1): f:R[0,) is continuous function, and f is not identical to zero.

    If Ω is an open bounded subset in a Banach space E and T:ˉΩE is compact, with 0(IT)(Ω), then i(T,Ω,E) is the fixed point index of T on Ω with respect to E. For the definition and properties of the fixed point index we refer the reader to e.g., [15].

    The rest of the paper is organized as follows. In section 2, we give some preliminary results. In section 3, under some assumptions on f, we prove the uniqueness of positive solutions for problem (1.1). In section 4, we prove the existence of arbitrarily many positive solutions for problem (1.1).

    Below, let X be the Banach space C[0,1] endowed with the norm ||v||=supt[0,1]|v(t)|. Let P={vX:v(t) is decreasing in t,v(1)=0,and v(t)||v||(1t) for all t[0,1]}, then P is a cone in X, and the corresponding open ball of center 0 and radius r>0 will be denoted by Br.

    By an argument similar to that of Lemma 2.4 of [14], we have the following

    Lemma 2.1. Let (λ,u) be a positive solution of the problem (1.1) with ||u||=ρ1 and λ>0. Let x0(1,1) be such that u(x0)=0. Then

    (ⅰ) x0=0;

    (ⅱ) x0 is the unique point on which u attains its maximum;

    (ⅲ) u(t)>0,t(1,0) and u(t)<0,t(0,1).

    By a solution of problem (1.1), we understand that it is a function which belongs to C1[0,1] with ||u||<1, such that u/1u2 is differentiable and problem (1.1) is satisfied.

    By Lemma 2.1, (λ,u) is a positive solution of (1.1) if and only if (λ,u) is a positive solution of the mixed boundary value problem

    {(u1u2)=λf(u),u(0)=0,  u(1)=0. (2.1)

    Now we treat positive classical solutions of (2.1). The following well-known result of the fixed point index is crucial in our arguments.

    Lemma 2.2. [15,16,17] Let E be a Banach space and K a cone in E. For r>0, define Kr=KBr. Assume that T:ˉKrK is completely continuous such that Txx for xKr={xK:x=r}.

    (ⅰ) If Txx for xKr, then i(T,Kr,K) = 0.

    (ⅱ) If Txx for xKr, then i(T,Kr,K) = 1.

    Now we define a nonlinear operator Tλ on PB1 as follows:

    (Tλu)(t)=1tϕ1(s0λf(u(τ))dτ)ds,    uPB1,

    where ϕ(s)=s/1s2.

    We point out that u is a positive solution of problem (2.1) if uPB1 is a fixed point of the nonlinear operator Tλ.

    Next, we give an important Lemma which will be used later.

    Lemma 2.3. Let vX with v(t)0 for t[0,1]. If v concave on [0,1], then

    v(t)min{t,1t}v,   t[0,1].

    In particular, for any pair 0<α<β<1 we have

    minαtβ v(t)min{α,1β}||v||. (2.2)

    Furthermore, if v(0)=v, then we have

    v(t)v(1t) for all t[0,1]. (2.3)

    Proof. It is an immediate consequence of the fact that u is concave down in [0,1].

    Lemma 2.4. Tλ(PB1)P and the map Tλ:PBρP for all ρ(0,1) is completely continuous.

    Proof. From the condition (H1) one can easily see that TλvX with Tλv(1)=0, and for vP, a simple computation shows that

    (Tλv)(t)=λϕ1(s0f(u(τ))dτ)ds.

    So Tλv(t) is decreasing on [0, 1], which implies that

    Tλv(t)Tλv(1)=0, for t[0,1].

    Moreover, it is easy to see that (Tλv) decreasing in [0,1], then it follows from Lemma 2.3 that

    v(t)v(1t) for all t[0,1].

    Therefore, we obtain that Tλ(PB1)P. Moreover, the operator Tλ is compact on PˉBρ is an immediate consequence of [9,Lemma 2].

    Simple computations leads lead to the following lemma.

    Lemma 2.5. Let ϕ(s):=s/1s2. Then ϕ1(s)=s/1+s2 and

    ϕ1(s1)ϕ1(s2)ϕ1(s1s2)s1s2,  s1,s2[0,).

    In particular, for 0<s11 we have

    ϕ1(s1s2)s1ϕ1(s2).

    In this section, we are going to study the positive solution of problem (1.1). Results to be proved in this section are true for any positive parameter λ. So, we may assume λ=1 for simplicity and therefore consider

    {(u1u2)=f(u),u(0)=0,  u(1)=0. (3.1)

    and the corresponding operator

    (T1u)(t)=1tϕ1(s0f(u(τ))dτ)ds,    t[0,1], (3.2)

    defined on PB1.

    Now, we present the main results of this section as follows.

    Theorem 3.1. Assume (H1), and f is increasing, such that for any u>0 and t(0,1) there always exists some ξ>1 such that

    f(tu)(1+ξ)tf(u). (3.3)

    Then problem (3.1) can have at most one positive solution.

    Remark 3.1. Since f is increasing, condition (3.3) implies that f(u)>0 when u>0, and can be replaced by the following simpler condition:

    f(tu)>tf(u).

    Remark 3.2. To compare with the Theorem 3.1 and Theorem 3.4 in [13], ours in this section cover a very broad class of functions of f. For example, f(u)=i=mi=1upi with 0<pi<1 or f(u)=1+u.

    Theorem 3.1 is proved via a sequence of Lemmas. But we need to have a definition first.

    Definition 3.1. [16,17] Let K be a cone in real Banach space Y and be the partial ordering defined by K. Let A:KK and u0>θ, where θ denotes the zero element Y.

    (a) for any x>θ, there exist α, β>0 such that

    αu0A(x)βu0;

    (b) for any αu0xβu0 and t(0,1), there exists some η>0 such that

    A(tx)(1+η)tAx.

    Then A is called u0-sublinear

    Lemma 3.2. Suppose that f satisfies the conditions of Theorem 3.1. Then the operator T1, defined by 3.2, is u0-sublinear with u0=1t.

    Proof. We divide the proof into two steps.

    Step 1. We check condition (a) of Definition 3.1. First, we show that for any u>0 from PB1, there exist α,β>0 such that

    αu0T1(u)βu0. (3.4)

    Let M=maxt[0,1]{f(u(t))}. It follows from Lemma 2.5, we have

    (T1u)(t)1tϕ1(s0Mdτ)ds12M(1t2)M(1t).

    So, we may take β=M.

    Notice that (T1u)(t) is strictly decreasing in t and vanishes at t=1. Choose any c(0,1) and set

    m=c0f(u(t))dt.

    Then for all t[c,1], we have

    (T1u)(t)1tϕ1(m)ds=m1+m2(1t).

    Since (T1u)(t)(T1u)(c)=m1+m2(1c) for all t[0,c], we have

    (T1u)(t)m1+m2(1c)(1t)

    for all t[0,1]. So, we may choose α=m1+m2(1c) and (3.4) is proved.

    Step 2. We check condition (b) of Definition 3.1. We need to show that for any αu0uβu0 and ξ(0,1), there exists some η>0 such that

    T1(ξu)(1+η)ξT1u.

    To this end, due to the conditions satisfied by f, there exists an η>0 such that

    f(ξu)[(1+η)ξ]f(u).

    It follows from Lemma 2.5 that

    (T1(ξu))(t)=1tϕ1(s0f(ξu)dτ)ds1tϕ1[s0(1+η)ξf(u)dτ]ds(1+η)ξ1tϕ1(s0f(u)dτ)ds.

    Thus we have

    T1((ξu))(t)[(1+η)ξ](T1u)(t).

    Therefore, the proof is complete.

    Proof of Theorem 3.1. Let f be from Theorem 3.1. From Lemma 3.2, we know that the operator T1 is increasing and u0-sublinear with u0=1t.

    Next, we show that T1 can have at most one positive fixed point. Assume to the contrary that there exist two positive fixed points u and x. Now, we claim that there exists c>0 such that ucx. By (3.2), we know that

    x(t)=1tϕ1(s0f(x(τ))dτ)ds,    t[0,1],
    u(t)=1tϕ1(s0f(u(τ))dτ)ds,    t[0,1],

    So, x(t) is strictly decreasing in t and vanishes at t=1. There exist σ1,σ2(0,) such that

    σ1(1t)x(t)σ2(1t),   t[0,1].

    Using the same method, there exist δ1,δ2(0,) such that

    δ1(1t)u(t)δ2(1t)   t[0,1].

    Therefore, t[0,1]

    u(t)δ1(1t)=δ1σ2σ2(1t)δ1σ2x(t).

    So, we can choose c=δ1σ2. Set c=sup{t|utx}. We claim that

    c1.

    If not, then there exists some ξ>0 such that T1(cx)(1+ξ)cT1x. This combine the fact that T1 is increasing imply that

    u=T1uT1(cx)(1+ξ)cT1x=(1+ξ)cx,

    which is a contraction since (1+ξ)c>c. So, ux and similarly we can show that xu. Now with the above lemmas in place, Theorem 3.1 follows readily. Therefore, the proof is complete.

    Example 3.1. Let us consider the problem

    {(u1u2)=u,u(0)=0,  u(1)=0. (3.5)

    By using Theorem 3.1, we know that the problem (3.5) has a unique positive solution, this is just the conclusion of [13, Theorem 3.1].

    In this section, we shall investigate the positive solutions of problem (1.1). Furthermore, we are going to examine how the number of zeros of f may have a huge impact on the number of solutions of (1.1). Here is the main result of this section.

    Theorem 4.1. Assume (H1), and there exist two sequences of positive numbers ai and bi, satisfying

    ai<biai+1<bi+1,

    and such that f(ai)=0 and f(bi)=0, and f(u)>0 on (ai,bi), for all i=1,,n (nN={1,2,}). Then there exists λ0 such that for all λλ0 the problem (1.1) has n distinct positive solutions, u1,u2,,un, such that

    ai<uibi

    for each i=1,,n.

    Remark 4.1. Note that f will satisfy the conditions in the theorem if there exist numbers an>an1>>a1>a0=0 such that f(ai)=0 for i=1,,n, and f(u)>0 for ai1<u<ai, i=1,,n.

    Now we define a nonlinear operator Tiλ, i=1,n as follows:

    Tiλu(t)=1tϕ1(s0λfi(u(r))dr)ds,    uPB1. (4.1)

    In order to prove the main result, we define fi by

    fi(u)={f(u),    0ubi,0,       biu.

    For any 0<r<1, define Or by

    Or=PBr.

    Note that Or={uP:u=r}. Similar to T1, it is easy to verify that Tiλ is compact on PBr for all r(0,1).

    In order to prove the theorem, we need the following two Lemmas.

    Lemma 4.2. Suppose that f satisfies the conditions of Theorem 4.1. If vPB1 is a positive fixed point of (4.1), i.e. Tiλv=v, then v is a solution of (2.1) such that

    supt[0,1]v(t)bi.

    Proof. Notice that v satisfies

    {(u1u2)=λfi(u),u(0)=0,  u(1)=0.

    If on the contrary that supt[0,1]v(t)=v(0)>bi, then there exists a t0(0,1) such that v(t)>bi for t[0,t0) and v(t0)=bi. It follows that

    (v1v2)=0 for t(0,t0].

    Thus, (v1v2) is constant on [0,t0]. Since v(0)=0, it follows that v(t)=0 for t[0,t0]. Consequently, v(t) is a constant on [0,t0]. This is a contradiction. Therefore, supt[0,1]v(t)bi. On the other hand, since f(v)fi(v) for 0vbi, v is a solution of (2.1).

    Choose any ζ such that

    max{aibi:1in}<ζ<1.

    Lemma 4.3. Let the conditions of Theorem 4.1 be satisfied. For any i{1,,n}, there exists ri such that [ζri,ri](ai,bi). Furthermore, for any vOri we have

    Tiλv12(1ζ)2ϕ1(λωri),

    where ωri=minζrivri{fi(v)}>0.

    Proof. Based on the choice of ζ, the existence of ri with [ζri,ri](ai,bi) is obvious. Notice that for any vP we have that v(t)v(0)(1t) for all t[0,1]. In particular, we have ζv(0)v(t)v(0) for all t[0,1ζ]. Let vOri, then fi(v(t))ωri for t[0,1ζ]. It follows that

    ||Tiλv||1ζ0ϕ1(t0λfi(v(τ))dτ)dt1ζ0ϕ1(t0λwridτ)dt1ζ0tϕ1(λwri)dt=12(1ζ)2ϕ1(λωri).

    Proof of Theorem 4.1. Define λ0 by

    λ0=max{ϕ(2ri(1ζ)2)/wri:i=1,2,,n}.

    For each i=1,,n and λ>λ0, by Lemma 4.3 we infer that

    Tiλv>v for vOri.

    On the other hand, for each fixed λ>λ0, since fi(v) is bounded, this combine the fact that the rage of ϕ1 is (1,1) imply that there is an Ri>ri such that

    Tiλv<v for vORi.

    It follows from Lemma 2.2 that

    i(Tiλ,Ori,P)=0,  i(Tiλ,ORi,P)=1,

    and hence

    i(Tiλ,ORiˉOri,P)=1.

    Thus, Tiλ has a fixed point vi in ORiˉOri. Lemma 4.2 implies that the fixed point vi is a solution of (2.1) such that

    ai<rivibi.

    Consequently, (2.1) has n positive solutions, v1,v2,,vn for each λ>λ0, such that

    ai<supt[0,1]vi(t)bi.

    Therefore, the proof is complete.

    Remark 4.1. It is interesting to compare our main result(Theorem 4.1) to Candito et al. [18]. In [18], the authors showed the existence of three classical solutions of the one dimensional prescribed mean curvature equation in Euclid space by using a variational approach.

    Example 4.1. Let us consider the problem

    {(u1u2)=λg(u),u(0)=0,  u(1)=0, (4.2)

    where g(u)=sin8πu+1.

    Let a1=316, a2=716, a3=1116, a4=1516, we can easily check that g(ai)=0, and g(u)>0 for ai<u<ai+1, i=1,2,3.

    From Theorem 4.1, the problem (4.2) has at least three positive solutions u1,u2,u3 with aj<supr[0,1]uj(r)aj+1, j=1,2,3 provided that λ is large enough.

    The authors are very grateful to the anonymous referees for their valuable suggestions.

    The authors declare that they have no conflicts of interest.



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