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Existence and multiplicity of positive solutions for one-dimensional p-Laplacian problem with sign-changing weight


  • In this paper, we show the positive solutions set for one-dimensional p-Laplacian problem with sign-changing weight contains a reversed S-shaped continuum. By figuring the shape of unbounded continuum of positive solutions, we identify the interval of bifurcation parameter in which the p-Laplacian problem has one or two or three positive solutions according to the asymptotic behavior of nonlinear term at 0 and . The proof of the main result is based upon bifurcation technique.

    Citation: Liangying Miao, Man Xu, Zhiqian He. Existence and multiplicity of positive solutions for one-dimensional p-Laplacian problem with sign-changing weight[J]. Electronic Research Archive, 2023, 31(6): 3086-3096. doi: 10.3934/era.2023156

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  • In this paper, we show the positive solutions set for one-dimensional p-Laplacian problem with sign-changing weight contains a reversed S-shaped continuum. By figuring the shape of unbounded continuum of positive solutions, we identify the interval of bifurcation parameter in which the p-Laplacian problem has one or two or three positive solutions according to the asymptotic behavior of nonlinear term at 0 and . The proof of the main result is based upon bifurcation technique.



    Let W be a set and H:WW be a mapping. A point wW is called a fixed point of H if w=Hw. Fixed point theory plays a fundamental role in functional analysis (see [15]). Shoaib [17] introduced the concept of α-dominated mapping and obtained some fixed point results (see also [1,2]). George et al. [11] introduced a new space and called it rectangular b-metric space (r.b.m. space). The triangle inequality in the b-metric space was replaced by rectangle inequality. Useful results on r.b.m. spaces can be seen in ([5,6,8,9,10]). Ćirić introduced new types of contraction and proved some metrical fixed point results (see [4]). In this article, we introduce Ćirić type rational contractions for α -dominated mappings in r.b.m. spaces and proved some metrical fixed point results. New interesting results in metric spaces, rectangular metric spaces and b-metric spaces can be obtained as applications of our results.

    Definition 1.1. [11] Let U be a nonempty set. A function dlb:U×U[0,) is said to be a rectangular b-metric if there exists b1 such that

    (ⅰ) dlb(θ,ν)=dlb(ν,θ);

    (ⅱ) dlb(θ,ν)=0 if and only if θ=ν;

    (ⅲ) dlb(θ,ν)b[dlb(θ,q)+dlb(q,l)+dlb(l,ν)] for all θ,νU and all distinct points q,lU{θ,ν}.

    The pair (U,dlb) is said a rectangular b-metric space (in short, r.b.m. space) with coefficient b.

    Definition 1.2. [11] Let (U,dlb) be an r.b.m. space with coefficient b.

    (ⅰ) A sequence {θn} in (U,dlb) is said to be Cauchy sequence if for each ε>0, there corresponds n0N such that for all n,mn0 we have dlb(θm,θn)<ε or limn,m+dlb(θn,θm)=0.

    (ⅱ) A sequence {θn} is rectangular b-convergent (for short, (dlb)-converges) to θ if limn+dlb(θn,θ)=0. In this case θ is called a (dlb)-limit of {θn}.

    (ⅲ) (U,dlb) is complete if every Cauchy sequence in Udlb-converges to a point θU.

    Let ϖb, where b1, denote the family of all nondecreasing functions δb:[0,+)[0,+) such that +k=1bkδkb(t)<+ and bδb(t)<t for all t>0, where δkb is the kth iterate of δb. Also bn+1δn+1b(t)=bnbδb(δnb(t))<bnδnb(t).

    Example 1.3. [11] Let U=N. Define dlb:U×UR+{0} such that dlb(u,v)=dlb(v,u) for all u,vU and α>0

    dlb(u,v)={0, if u=v;10α, if u=1, v=2;α, if u{1,2} and v{3};2α, if u{1,2,3} and v{4};3α, if u or v{1,2,3,4} and uv.

    Then (U,dlb) is an r.b.m. space with b=2>1. Note that

    d(1,4)+d(4,3)+d(3,2)=5α<10α=d(1,2).

    Thus dlb is not a rectangular metric.

    Definition 1.4. [17] Let (U,dlb) be an r.b.m. space with coefficient b. Let S:UU be a mapping and α:U×U[0,+). If AU, we say that the S is α-dominated on A, whenever α(i,Si)1 for all iA. If A=U, we say that S is α-dominated.

    For θ,νU, a>0, we define Dlb(θ,ν) as

    Dlb(θ,ν)=max{dlb(θ,ν),dlb(θ,Sθ).dlb(ν,Sν)a+dlb(θ,ν),dlb(θ,Sθ),dlb(ν,Sν)}.

    Now, we present our main result.

    Theorem 2.1. Let (U,dlb) be a complete r.b.m. space with coefficient b, α:U×U[0,),S:UU, {θn} be a Picard sequence and S be a α-dominated mapping on {θn}. Suppose that, for some δbϖb, we have

    dlb(Sθ,Sν)δb(Dlb(θ,ν)), (2.1)

    for all θ,ν{θn} with α(θ,ν)1. Then {θn} converges to θU. Also, if (2.1) holds for θ and α(θn,θ)1 for all nN{0}, then S has a fixed point θ in U.

    Proof. Let θ0U be arbitrary. Define the sequence {θn} by θn+1=Sθn for all nN{0}. We shall show that {θn} is a Cauchy sequence. If θn=θn+1, for some nN, then θn is a fixed point of S. So, suppose that any two consecutive terms of the sequence are not equal. Since S:UU be an α-dominated mapping on {θn}, α(θn,Sθn)1 for all nN{0} and then α(θn,θn+1)1 for all nN{0}. Now by using inequality (2.1), we obtain

    dlb(θn+1,θn+2)=dlb(Sθn,Sθn+1)δb(Dlb(θn,θn+1))δb(max{dlb(θn,θn+1),dlb(θn,θn+1).dlb(θn+1,θn+2)a+dlb(θn,θn+1),dlb(θn,θn+1),dlb(θn+1,θn+2)})δb(max{dlb(θn,θn+1),dlb(θn+1,θn+2)}).

    If max{dlb(θn,θn+1),dlb(θn+1,θn+2)}=dlb(θn+1,θn+2), then

    dlb(θn+1,θn+2)δb(dlb(θn+1,θn+2))bδb(dlb(θn+1,θn+2)).

    This is the contradiction to the fact that bδb(t)<t for all t>0. So

    max{dlb(θn,θn+1),dlb(θn+1,θn+2)}=dlb(θn,θn+1).

    Hence, we obtain

    dlb(θn+1,θn+2)δb(dlb(θn,θn+1))δ2b(dlb(θn1,θn))

    Continuing in this way, we obtain

    dlb(θn+1,θn+2)δn+1b(dlb(θ0,θ1)). (2.2)

    Suppose for some n,mN with m>n, we have θn=θm. Then by (2.2)

    dlb(θn,θn+1)=dlb(θn,Sθn)=dlb(θm,Sθm)=dlb(θm,θm+1)δmnb(dlb(θn,θn+1))<bδb(dlb(θn,θn+1))

    As dlb(θn,θn+1)>0, so this is not true, because bδb(t)<t for all t>0. Therefore, θnθm for any n,mN. Since +k=1bkδkb(t)<+, for some νN, the series +k=1bkδkb(δν1b(dlb(θ0,θ1))) converges. As bδb(t)<t, so

    bn+1δn+1b(δν1b(dlb(θ0,θ1)))<bnδnb(δν1b(dlb(θ0,θ1))), for all nN.

    Fix ε>0. Then ε2=ε>0. For ε, there exists ν(ε)N such that

    bδb(δν(ε)1b(dlb(θ0,θ1)))+b2δ2b(δν(ε)1b(dlb(θ0,θ1)))+<ε (2.3)

    Now, we suppose that any two terms of the sequence {θn} are not equal. Let n,mN with m>n>ν(ε). Now, if m>n+2,

    dlb(θn,θm)b[dlb(θn,θn+1)+dlb(θn+1,θn+2)+dlb(θn+2,θm)]b[dlb(θn,θn+1)+dlb(θn+1,θn+2)]+b2[dlb(θn+2,θn+3)+dlb(θn+3,θn+4)+dlb(θn+4,θm)]b[δnb(dlb(θ0,θ1))+δn+1b(dlb(θ0,θ1))]+b2[δn+2b(dlb(θ0,θ1))+δn+3b(dlb(θ0,θ1))]+b3[δn+4b(dlb(θ0,θ1))+δn+5b(dlb(θ0,θ1))]+bδnb(dlb(θ0,θ1))+b2δn+1b(dlb(θ0,θ1))+b3δn+2b(dlb(θ0,θ1))+=bδb(δn1b(dlb(θ0,θ1)))+b2δ2b(δn1b(dlb(θ0,θ1)))+.

    By using (2.3), we have

    dlb(θn,θm)<bδb(δν(ε)1b(dlb(θ0,θ1)))+b2δ2b(δν(ε)1b(dlb(θ0,θ1)))+<ε<ε.

    Now, if m=n+2, then we obtain

    dlb(θn,θn+2)b[dlb(θn,θn+1)+dlb(θn+1,θn+3)+dlb(θn+3,θn+2)]b[dlb(θn,θn+1)+b[dlb(θn+1,θn+2)+dlb(θn+2,θn+4)+dlb(θn+4,θn+3)]+dlb(θn+3,θn+2)]bdlb(θn,θn+1)+b2dlb(θn+1,θn+2)+bdlb(θn+2,θn+3)+b2dlb(θn+3,θn+4)+b3[dlb(θn+2,θn+3)+dlb(θn+3,θn+5)+dlb(θn+5,θn+4)]bdlb(θn,θn+1)+b2dlb(θn+1,θn+2)+(b+b3)dlb(θn+2,θn+3)+b2dlb(θn+3,θn+4)+b3dlb(θn+5,θn+4)+b4[dlb(θn+3,θn+4)+dlb(θn+4,θn+6)+dlb(θn+6,θn+5)]bdlb(θn,θn+1)+b2dlb(θn+1,θn+2)+(b+b3)dlb(θn+2,θn+3)+(b2+b4)dlb(θn+3,θn+4)+b3dlb(θn+5,θn+4)+b4dlb(θn+6,θn+5)+b5[dlb(θn+4,θn+5)+dlb(θn+5,θn+7)+dlb(θn+7,θn+6)]bdlb(θn,θn+1)+b2dlb(θn+1,θn+2)+(b+b3)dlb(θn+2,θn+3)+(b2+b4)dlb(θn+3,θn+4)+(b3+b5)dlb(θn+4,θn+5)+<2[bdlb(θn,θn+1)+b2dlb(θn+1,θn+2)+b3dlb(θn+2,θn+3)+b4dlb(θn+3,θn+4)+b5dlb(θn+4,θn+5)+]2[bδnb(dlb(θ0,θ1))+b2δn+1b(dlb(θ0,θ1))+b3δn+2b(dlb(θ0,θ1))+]<2[bδb(δν(ε)1b(dlb(θ0,θ1)))+b2δ2b(δν(ε)1b(dlb(θ0,θ1)))+]<2ε=ε.

    It follows that

    limn,m+dlb(θn,θm)=0. (2.4)

    Thus {θn} is a Cauchy sequence in (U,dlb). As (U,dlb) is complete, so there exists θ in U such that {θn} converges to θ, that is,

    limn+dlb(θn,θ)=0. (2.5)

    Now, suppose that dlb(θ,Sθ)>0. Then

    dlb(θ,Sθ)b[dlb(θ,θn)+dlb(θn,θn+1)+dlb(θn+1,Sθ)b[dlb(θ,θn+1)+dlb(θn,θn+1)+dlb(Sθn,Sθ).

    Since α(θn,θ)1, we obtain

    dlb(θ,Sθ)bdlb(θ,θn+1)+bdlb(θn,θn+1)+bδb(max{dlb(θn,θ),dlb(θ,Sθ).dlb(θn,θn+1)a+dlb(θn,θ), dlb(θn,θn+1) dlb(θ,Sθ)}).

    Letting n+, and using the inequalities (2.4) and (2.5), we obtain dlb(θ,Sθ)bδb(dlb(θ,Sθ)). This is not true, because bδb(t)<t for all t>0 and hence dlb(θ,Sθ)=0 or θ=Sθ. Hence S has a fixed point θ in U.

    Remark 2.2. By taking fourteen different proper subsets of Dlb(θ,ν), we can obtainvnew results as corollaries of our result in a complete r.b.m. space with coefficient b.

    We have the following result without using α-dominated mapping.

    Theorem 2.3. Let (U,dlb) be a complete r.b.m. space with coefficient b,S:UU, {θn} be a Picard sequence. Suppose that, for some δbϖb, we have

    dlb(Sθ,Sν)δb(Dlb(θ,ν)) (2.6)

    for all θ,ν{θn}. Then {θn} converges to θU. Also, if (2.6) holds for θ, then S has a fixed point θ in U.

    We have the following result by taking δb(t)=ct, tR+ with 0<c<1b without using α-dominated mapping.

    Theorem 2.4. Let (U,dlb) be a complete r.b.m. space with coefficient b, S:UU, {θn} be a Picard sequence. Suppose that, for some 0<c<1b, we have

    dlb(Sθ,Sν)c(Dlb(θ,ν)) (2.7)

    for all θ,ν{θn}. Then {θn} converges to θU. Also, if (2.7) holds for θ, then S has a fixed point θ in U.

    Ran and Reurings [16] gave an extension to the results in fixed point theory and obtained results in partially ordered metric spaces. Arshad et al. [3] introduced -dominated mappings and established some results in an ordered complete dislocated metric space. We apply our result to obtain results in ordered complete r.b.m. space.

    Definition 2.5. (U,,dlb) is said to be an ordered complete r.b.m. space with coefficient b if

    (ⅰ) (U,) is a partially ordered set.

    (ⅱ) (U,dlb) is an r.b.m. space.

    Definition 2.6. [3] Let U be a nonempty set, is a partial order on θ. A mapping S:UU is said to be -dominated on A if aSa for each aAθ. If A=U, then S:UU is said to be -dominated.

    We have the following result for -dominated mappings in an ordered complete r.b.m. space with coefficient b.

    Theorem 2.7. Let (U,,dlb) be an ordered complete r.b.m. space with coefficient b, S:UU,{θn} be a Picard sequence and S be a -dominated mapping on {θn}. Suppose that, for some δbϖb, we have

    dlb(Sθ,Sν)δb(Dlb(θ,ν)), (2.8)

    for all θ,ν{θn} with θν. Then {θn} converges to θU. Also, if (2.8) holds for θ and θnθ for all nN{0}. Then S has a fixed point θ in U.

    Proof. Let α:U×U[0,+) be a mapping defined by α(θ,ν)=1 for all θ,νU with θν and α(θ,ν)=411 for all other elements θ,νU. As S is the dominated mappings on {θn}, so θSθ for all θ{θn}. This implies that α(θ,Sθ)=1 for all θ{θn}. So S:UU is the α-dominated mapping on {θn}. Moreover, inequality (2.8) can be written as

    dlb(Sθ,Sν)δb(Dlb(θ,ν))

    for all elements θ,ν in {θn} with α(θ,ν)1. Then, as in Theorem 2.1, {θn} converges to θU. Now, θnθ implies α(θn,θ)1. So all the conditions of Theorem 2.1 are satisfied. Hence, by Theorem 2.1, S has a fixed point θ in U.

    Now, we present an example of our main result. Note that the results of George et al. [11] and all other results in rectangular b-metric space are not applicable to ensure the existence of the fixed point of the mapping given in the following example.

    Example 2.8. Let U=AB, where A={1n:n{2,3,4,5}} and B=[1,]. Define dl:U×U[0,) such that dl(θ,ν)=dl(ν,θ) for θ,νU and

    {dl(12,13)=dl(14,15)=0.03dl(12,15)=dl(13,14)=0.02dl(12,14)=dl(15,13)=0.6dl(θ,ν)=|θν|2    otherwise

    be a complete r.b.m. space with coefficient b=4>1 but (U,dl) is neither a metric space nor a rectangular metric space. Take δb(t)=t10, then bδb(t)<t. Let S:UU be defined as

    Sθ={15        ifθA13        ifθ=19θ100+85 otherwise.

    Let θ0=1. Then the Picard sequence {θn} is {1,13,15,15,15,}. Define

    α(θ,ν)={85        ifθ,ν{θn}47            otherwise.

    Then S is an α-dominated mapping on {θn}. Now, S satisfies all the conditions of Theorem 2.1. Here 15 is the fixed point in U.

    Jachymski [13] proved the contraction principle for mappings on a metric space with a graph. Let (U,d) be a metric space and represents the diagonal of the cartesian product U×U. Suppose that G be a directed graph having the vertices set V(G) along with U, and the set E(G) denoted the edges of U included all loops, i.e., E(G)⊇△. If G has no parallel edges, then we can unify G with pair (V(G),E(G)). If l and m are the vertices in a graph G, then a path in G from l to m of length N(NN) is a sequence {θi}Ni=o of N+1 vertices such that lo=l,lN=m and (ln1,ln)E(G) where i=1,2,N (see for detail [7,8,12,14,18,19]). Recently, Younis et al. [20] introduced the notion of graphical rectangular b-metric spaces (see also [5,6,21]). Now, we present our result in this direction.

    Definition 3.1. Let θ be a nonempty set and G=(V(G),E(G)) be a graph such that V(G)=U and AU. A mapping S:UU is said to be graph dominated on A if (θ,Sθ)E(G) for all θA.

    Theorem 3.2. Let (U,dlb) be a complete rectangular b -metric space endowed with a graph G, {θn} be a Picard sequence and S:UU be a graph dominated mapping on {θn}. Suppose that the following hold:

    (i) there exists δbϖb such that

    dlb(Sθ,Sν)δb(Dlb(θ,ν)), (3.1)

    for all θ,ν{θn} and (θn,ν)E(G). Then (θn,θn+1)E(G) and {θn} converges to θ. Also, if (3.1) holds for θ and (θn,θ)E(G) for all nN{0}, then S has a fixed point θ in U.

    Proof. Define α:U×U[0,+) by

    α(θ,ν)={1, ifθ,νU, (θ,ν)E(G)14,                  otherwise.

    Since S is a graph dominated on {θn}, for θ{θn},(θ,Sθ)E(G). This implies that α(θ,Sθ)=1 for all θ{θn}. So S:UU is an α-dominated mapping on {θn}. Moreover, inequality (3.1) can be written as

    dlb(Sθ,Sν)δb(Dlb(θ,ν)),

    for all elements θ,ν in {θn} with α(θ,ν)1. Then, by Theorem 2.1, {θn} converges to θU. Now, (θn,θ)E(G) implies that α(θn,θ)1. So all the conditions of Theorem 2.1 are satisfied. Hence, by Theorem 2.1, S has a fixed point θ in U.

    The authors would like to thank the Editor, the Associate Editor and the anonymous referees for sparing their valuable time for reviewing this article. The thoughtful comments of reviewers are very useful to improve and modify this article.

    The authors declare that they have no competing interests.



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