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Maximal and minimal weak solutions for elliptic problems with nonlinearity on the boundary

  • This paper deals with the existence of weak solutions for semilinear elliptic equation with nonlinearity on the boundary. We establish the existence of a maximal and a minimal weak solution between an ordered pair of sub- and supersolution for both monotone and nonmonotone nonlinearities. We use iteration argument when the nonlinearity is monotone. For the nonmonotone case, we utilize the surjectivity of a pseudomonotone and coercive operator, Zorn's lemma and a version of Kato's inequality.

    Citation: S. Bandyopadhyay, M. Chhetri, B. B. Delgado, N. Mavinga, R. Pardo. Maximal and minimal weak solutions for elliptic problems with nonlinearity on the boundary[J]. Electronic Research Archive, 2022, 30(6): 2121-2137. doi: 10.3934/era.2022107

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  • This paper deals with the existence of weak solutions for semilinear elliptic equation with nonlinearity on the boundary. We establish the existence of a maximal and a minimal weak solution between an ordered pair of sub- and supersolution for both monotone and nonmonotone nonlinearities. We use iteration argument when the nonlinearity is monotone. For the nonmonotone case, we utilize the surjectivity of a pseudomonotone and coercive operator, Zorn's lemma and a version of Kato's inequality.



    We consider an elliptic equation with nonlinear boundary condition of the form

    {Δu+u=0inΩ;uη=f(x,u)onΩ, (1.1)

    where ΩRN(N2) is a bounded domain with C2,α (0<α<1) boundary Ω, and /η:=η(x) denotes the outer normal derivative on the boundary Ω. Here f:Ω×RR is a Carathéodory function, that is, f(,u) is measurable for each u and f(x,) is continuous for a.e. xΩ.

    In this paper, we investigate the existence of maximal and minimal weak solutions (to be clarified later) between an ordered pair of sub- and supersolution of (1.1) for both monotone and nonmonotone nonlinearities. We use monotone iteration procedure when the nonlinearity is monotone. The nonmonotone case required a careful use of the surjectivity of a bounded, pseudomonotone and coercive operator, Zorn's lemma and a version of Kato's inequality up to the boundary. This proof, for the nonmonotone case, is motivated by the works in [1] and [2].

    Elliptic equations with nonlinear boundary conditions have attracted a lot of attention over the last decades, see for instance [3,4,5,6,7,8,9] and references therein. Motivation to study equations with nonlinear boundary conditions stems from the fact that, when the reaction near the boundary depends on the density itself, linear boundary conditions (Dirichlet, Neumann, or Robin) are often inadequate to study chemical, biological, or ecological processes, see [10,11,12,13] and references therein, for specific applications.

    The existence of a solution between an ordered pair of sub- and supersolution of elliptic boundary value problems has been studied extensively. For the linear boundary conditions, the sub–supersolution method for classical solutions were developed in [14,15,16] to study the solvability of quasi-linear and semi-linear equations using monotone iteration method. This method also yields the existence of a maximal and a minimal solution. These iterative methods can be thought as a generalization of the Perron arguments on sub- and superharmonic functions for existence of solutions of the boundary value problem. For relatively recent results on the existence of maximal and minimal solutions, for the linear boundary conditions, we refer readers to [2,17,18] for the Laplacian case, and [1,19] for the p-Laplacian case.

    For the nonlinear boundary case, see [20] and [13,Ch. 4] where the existence of maximal and minimal classical solutions was established for the monotone case. To the best of our knowledge, our results concerning the existence of maximal and minimal weak solutions are new for both monotone and nonmonotone cases.

    We begin with the definitions of weak solution and weak sub- and supersolution. For this, we make use of the real Lebesgue space Lr(Ω) and the Sobolev space H1(Ω).

    Definition 1.1. We say that a function uH1(Ω) is a weak solution to (1.1) whenever:

    (i) f(.,u(.))L2(N1)N(Ω) if N>2, and f(.,u(.))Lr(Ω) for r>1 if N=2, and

    (ii) Ω(uψ+uψ)=Ωf(x,u)ψforall ψH1(Ω).

    Definition 1.2. We say that a function ¯uH1(Ω) is a weak supersolution to (1.1) whenever:

    (i) f(.,¯u(.))L2(N1)N(Ω) if N>2, and f(.,¯u(.))Lr(Ω) for r>1 if N=2, and

    (ii) Ω(¯uψ+¯uψ)Ωf(x,¯u)ψ for all 0ψH1(Ω).

    A weak subsolution u_ is defined by reversing the inequality in (ii) above.

    Remark 1.3. Let Γ:H1(Ω)Lr(Ω) be the trace operator given by Γu=u|Ω. It is known that, see e.g. [21], [22,Thm 2.79], and [23,Chapter 6], Γ is continuous (compact) if

    {1r2(N1)N2(1r<2(N1)N2) if N>2r1(r1) if N=2. (1.2)

    Therefore, the integrals on the right hand side of (ii) of Definition 1.1 and Definition 1.2 make sense since (i) holds, and 2(N1)N is the conjugate of 2(N1)N2 when N>2.

    We state and prove our results for the case N>2, since the case N=2 follows clearly using (1.2). We state our first result concerning maximal and minimal solutions for the monotone case.

    Theorem 1.4. Suppose there exists a pair of weak sub- and supersolution u_ and ¯u, respectively, satisfying u_¯u in ¯Ω. Assume that

    (H1) there exists  k0 such that the map sf(x,s)+ks is nondecreasing for all u_s¯u, and for all xΩ.

    Then, there exist a minimal weak solution u and a maximal weak solution u to (1.1), in the sense that if u is any weak solution to (1.1) such that u_u¯u, then uuu.

    Next, we note that if f is locally Lipschitz with respect to the second variable u, and the interval [u_,¯u] is bounded, then f satisfies the hypothesis (H1). For functions f that do not satisfy the monotonicity condition given in (H1), we have the following existence result.

    Theorem 1.5. Suppose there exists a pair of weak sub- and supersolution u_ and ¯u, respectively, satisfying u_¯u in ¯Ω. Assume that

    (H2) there exists a KLr(Ω), r>2(N1)N, such that |f(x,s)|K(x) a.e. xΩ, for all s satisfying u_(x)s¯u(x).

    Then (1.1) has at least one weak solution u such that u_u¯u.

    Finally, we state a result that guarantees the existence of a maximal and a minimal weak solution without assuming monotonicity condition (H1) on the nonlinearity f.

    Theorem 1.6. Assume hypotheses of Theorem 1.5 hold. Then, there exist a minimal weak solution u and a maximal weak solution u to (1.1), in the sense that if u is any weak solution to (1.1) such that u_u¯u, then uuu.

    In [24], Hess proved the existence of a solution, assuming that

    Ω supu_(x)s¯u(x)|f(x,s)|q<, (1.3)

    for q=2. Our result, Theorem 1.4, is sharper, needing only that condition (1.3) hold for q=2(N1)N=22N<2.

    In Section 2, we collect some known results that will be helpful in the sequel. We also state and prove a version of Kato's inequality for our setting, see Theorem 2.4 and Corollary 2.5. In Section 3, we prove Theorem 1.4 using monotone iteration method. In Section 4, we prove Theorem 1.5 by showing that an appropriately defined operator is surjective. We also prove Theorem 1.6 in Section 4 by utilizing Theorem 1.5, Zorn's Lemma and Theorem 2.4. In Section 5, we discuss applications of our results.

    Here we collect some results that we use in the sequel. First, we recall an existence and uniqueness result for a linear problem.

    Proposition 2.1. ([4,8]) Let hLq(Ω) for q1. Then, the linear problem

    {Δv+v=0inΩ;vη=honΩ

    has a unique solution vW1,m(Ω) and

    vW1,m(Ω)ChLq(Ω),where1mNq/(N1).

    In particular, if q=2(N1)N, then uH1(Ω).

    Next, let X be a reflexive Banach space and A:XX. We say that the operator A is coercive if

    A(ψ),ψψX as ψX.

    We say that A is pseudomonotone, whenever

    vnvinXandlim supnA(vn),vnv0imply
    lim infnA(vn),vnψA(v),vψfor anyψX. (2.1)

    We will utilize the following surjectivity result in the proof of Theorem 1.5.

    Proposition 2.2. ([25,Thm. II. 2.8], [22,Thm. 2.99]) Let X be a reflexive Banach space. If A:XX is a bounded, pseudomonotone and coercive operator, then for each bX, Au=b has a solution.

    Finally, we say that a subset Y of a partially ordered set (X,) is a chain if xy or yx for every x,yY. Then, to prove Theorem 1.6, we use the following version of Zorn's lemma (see [22]):

    Proposition 2.3 (Zorn's lemma). If in a partially ordered set (X,), every chain Y has an upper bound, then X possesses a maximal element.

    In [26], authors established Kato's inequality up to the boundary for a function uW1,1(Ω). Here, we state and prove a version of Kato's inequality up to the boundary, that is necessary in the proof of Theorem 1.6. This result can be rephrased as the maximum of two weak subsolutions is also a weak subsolution. In particular, the maximum of two weak solutions is a weak subsolution.

    Theorem 2.4. Let u1 and u2 be functions in H1(Ω) such that there exist f1 and f2 in Lr(Ω), for r2(N1)N, satisfying

    Ω(uiψ+uiψ)Ωfiψforall 0ψH1(Ω), (2.2)

    for i=1,2. Then, u:=max{u1,u2} satisfies

    Ω(uψ+uψ)Ωfψforall 0ψH1(Ω),

    where f(x):={f1(x)if u1(x)>u2(x)f2(x)if u1(x)u2(x), a.e. xΩ.

    Proof. Define

    Ω1:={xΩ:u1(x)>u2(x)}and Ω2:=ΩΩ1

    and

    Γ1:={xΩ:u1(x)>u2(x)}andΓ2:=ΩΓ1.

    Fix 0ψH1(Ω). Then,

    I=Ωuψ+Ωuψ=Ω1(u1ψ+u1ψ)I1+Ω2(u2ψ+u2ψ)I2.

    Consider a sequence ξnC1(R) such that

    ξn(t):={1if t1/n0if t0,

    and ξn>0 on (0,1/n). Then, define the sequence of functions

    rn(x):=ξn((u1u2)(x))forx¯Ω.

    Observe that rnH1(Ω) and rn converges pointwise to χΩ1Γ1, where the characteristic function is defined as χΩ1Γ1(x):={1if xΩ1Γ10if otherwise. Moreover, ||rn||L(Ω)L(Ω)1 and supp(rn)¯Dn, where Dn:={xΩ:0<u1(x)u2(x)<1n}. Then, using Lebesgue Dominated Convergence Theorem, we have that

    I1=limn[Ωrnu1ψ+Ωrnu1ψ].

    Since rnH1(Ω)L(Ω)L(Ω), it follows that rnψH1(Ω) for any test function ψH1(Ω)L(Ω). Recalling that rn=0 on ΩDn, and that u1 satisfies (2.2), we can write

    Ωrnu1ψ+rnu1ψ=Ωu1(rnψ)+u1(rnψ)Dnψu1rnΩf1rnψDnψu1rn. (2.3)

    Taking the limit as n in the first term of the right-hand side of (2.3), using the Lebesgue Dominated Convergence Theorem, we get

    limnΩf1rnψ=Γ1f1ψ.

    Likewise, for I2 we have

    I2=limn[Ω(1rn)u2ψ+Ω(1rn)u2ψ],

    and

    Ω(1rn)u2ψ+Ω(1rn)u2ψ=Ωu2[(1rn)ψ]+u2(1rn)ψ+Dnψu2rnΩf2(1rn)ψ+Dnψu2rn. (2.4)

    Taking the limit as n in the first term of the right-hand side of (2.4) and using the Lebesgue Dominated Convergence Theorem, we get

    limnΩf2(1rn)ψ=Γ2f2ψ.

    Using the fact that rn=ξn(u1u2)(u1u2), the sum of the second terms of the right-hand side of (2.3) and (2.4) yields

    Dnψu1rn+Dnψu2rn=Dnψ(u1u2)rn=Dnψξn(u1u2)|(u1u2)|20, (2.5)

    since ψ0andξn0. Adding (2.3) and (2.4), taking the limit, and using (2.5), we get

    I=I1+I2Γ1f1ψ+Γ2f2ψ=Ωfψ.

    Thus, u:=max{u1,u2} satisfies

    Ω(uψ+uψ)Ωfψfor all  0ψH1(Ω),

    completing the proof of Theorem 2.4.

    Likewise, we have a result for the minimum of two supersolutions.

    Corollary 2.5. Let u1 and u2 be functions in H1(Ω) such that there exist f1 and f2 in Lr(Ω), for r2(N1)N, satisfying

    Ω(uiψ+uiψ)Ωfiψfor all 0ψH1(Ω),

    for i=1,2. Then, u:=min{u1,u2} satisfies

    Ω(uψ+uψ)Ωfψ,forall 0ψH1(Ω),

    where

    f(x):={f1(x)ifu1(x)<u2(x)f2(x)ifu1(x)u2(x),a.e.xΩ.

    Proof. Using the fact that min{u1,u2}=max{u1,u2}, the proof follows from Theorem 2.4.

    We will construct a monotone operator, and show that the iterative scheme starting with a weak subsolution (supersolution) will converge to a minimal (maximal) weak solution.

    Let J:={uH1(Ω):u_u¯u}. Define the linear map T:JH1(Ω) by T(u)=v, where v satisfies

    {Δv+v=0inΩ;vη+kv=f(x,u)+kuonΩ.

    Step 1. T is well-defined and maps J into itself.

    For every uJ, we have u_u¯u. Then using (H1) and the fact that u_ and ¯u are sub and supersolutions, we get

    f(x,u_)+ku_f(x,u)+kuf(x,¯u)+k¯u,

    and

    0|u|max{|u_|,|¯u|}|u_|+|¯u|.

    Taking into account the definitions of u_ and ¯u, we have that f(.,u_(.)),f(.,¯u(.)) are in L2(N1)N(Ω). Since u_,¯uH1(Ω), then by the continuity of the trace operator (1.2) and the embedding of L2(N1)N2(Ω) into L2(N1)N(Ω), for every uJ, we have

    f(x,u)+kuL2(N1)N(Ω)f(x,u_)+ku_L2(N1)N(Ω)+f(x,¯u)+k¯uL2(N1)N(Ω)C. (3.1)

    Therefore, f(.,u(.))+ku(.)L2(N1)N(Ω). Then, Proposition 2.1 implies that v=T(u)H1(Ω) is unique. Thus, the map T is well-defined.

    Further, if u,wJ with uw, then by the weak maximum principle and the fact that f satisfies (H1), T(u)T(w), that is, the map T is nondecreasing. Moreover, repeating the argument and using Definition 1.2(ii), it follows that

    u_T(u_)T(¯u)¯u. (3.2)

    Hence, T maps J to J.

    Step 2. There exist weakly convergent monotone sequences in H1(Ω).

    Let's construct monotone sequences {un} and {wn} successively from the (linear) iteration process

    un=T(un1) with u0=u_ and wn=T(wn1) with w0=¯u.

    Using (3.2) and the monotonicity of T, we get

    u_=u0unwnw0=¯u. (3.3)

    We show that {un} is convergent. The proof for {wn} is analogous. We see that un=T(un1) satisfies

    Ω(unψ+unψ)+kΩunψ=Ω(f(x,un1)+kun1)ψ,

    for all ψH1(Ω). Letting un=T(un1) as a test function, we get

    Ω(|un|2+u2n)+kΩu2n=Ω(f(x,un1)+kun1)un. (3.4)

    Since un1,unJ, using Hölder's inequality in (3.4), and the bound (3.1), we have

    un2H1(Ω)un2H1(Ω)+kun2L2(Ω)f(x,un1)+kun1L2(N1)N(Ω)unL2(N1)N2(Ω)C(¯uL2(N1)N2(Ω)+u_L2(N1)N2(Ω)).

    Hence, there exists a uniform constant C>0, depending on Ω, f, k, u_ and ¯u, such that

    unH1(Ω)C. (3.5)

    By the reflexivity of H1(Ω), (3.5), there is a subsequence (relabeled) un which converges weakly to u in H1(Ω).

    Step 3. f(x,un)+kun converges weakly to f(x,u)+ku in L2(N1)N(Ω).

    Since the sequence un in Step 2 is nondecreasing and bounded (see (3.3)), it converges pointwise to u, that is,

    u(x)=limnun(x)J. (3.6)

    Using the fact that f is continuous in the second variable u for a.e xΩ and (3.6), we have that

    f(x,u(x))+ku=limnf(x,un(x))+kun(x).

    By (3.1), f(x,un)+kun is bounded in L2(N1)N(Ω). Then, Lebesgue Dominated Convergence Theorem yields

    (f(x,un)+kun)(f(x,u)+ku)L2(N1)N(Ω)0 as n.

    Therefore, f(x,un)+kun converges strongly (hence weakly) to f(x,u)+ku in L2(N1)N(Ω). Thus, for all ψH1(Ω)L2(N1)N2(Ω), we have

    limnΩ(f(x,un)+kun)ψ=Ω(f(x,u)+ku)ψ. (3.7)

    Step 4. u is a weak solution to (1.1).

    First, since uH1(Ω), the continuity of the trace (1.2) and L2(N1)N2(Ω)L2(N1)N(Ω), imply that uL2(N1)N(Ω). Therefore, for some positive constant C, we have

    f(x,u)L2(N1)N(Ω),=f(x,u)+kukuL2(N1)N(Ω),f(x,u)+kuL2(N1)N(Ω),+kuL2(N1)N(Ω),C.

    Second, from the monotone iteration, we know that un=T(un1) satisfies

    Ω(unψ+unψ)+kΩunψ=Ω(f(x,un1)+kun1)ψ.

    Observe that un converges weakly to u in H1(Ω), strongly in L2(Ω) (see Step 2) and f(x,un)+kunconverges weakly to f(x,u)+kuin L2(N1)N(Ω) (see Step 3). Then taking the limit as n and using (3.7), we get for any ψH1(Ω)

    Ω(uψ+uψ)+Ωkuψ=limn(Ω(unψ+unψ)+Ωkunψ)=limn(Ω(f(x,un1)+kun1)ψ)=Ω(f(x,u)+ku)ψ.

    Hence,

    Ω(uψ+uψ)=Ωf(x,u)ψfor allψH1(Ω).

    Moreover, we also have f(x,u)L2(N1)N(Ω). Thus u is a weak solution to (1.1).

    Step 5. u is the minimal weak solution in the interval [u_,¯u].

    Let v be a weak solution to (1.1) with u_v¯u. Then v is a weak supersolution, and u_v. Repeating the above iteration procedure with u0=u_, we get u_uv. Thus u is a weak minimal solution.

    Similarly, we can construct the maximal weak solution u from the sequence {wn} with w0=¯u. This completes the proof of Theorem 1.4.

    We prove Theorem 1.5 by applying Proposition 2.2 to an appropriate operator related to our problem (1.1). Then, Theorem 1.6 is proved by using Zorn's lemma and Theorem 2.4. Theorem 1.5 guarantees that the set defined for the Zorn's lemma in the proof of Theorem 1.6 is nonempty.

    Let us consider a modified problem

    {Δu+u=0inΩ;uη=g(x,u)onΩ, (4.1)

    where

    g(x,s):={f(x,u_(x)),s<u_(x),f(x,s),u_(x)s¯u(x),f(x,¯u(x)),s>¯u(x) (4.2)

    is the truncated function. We observe that g is a Carathéodory function, since f is a Carathéodory function. We note that a weak solution u of (4.1) is a weak solution of (1.1) whenever u_u¯u.

    Our plan is to establish the existence of a weak solution u of (4.1), and verify that u_u¯u. For the existence part, we use Proposition 2.2. For this, we define the map B:H1(Ω)(H1(Ω)) given by

    B(v),ψ:=Ω(vψ+vψ)Ωg(x,v)ψ, (4.3)

    for all ψH1(Ω).

    First, we show that B is well-defined and bounded. The first integral of (4.3) is well-defined since v,ψH1(Ω). By the Hölder's inequality combined with the continuity of trace operator (1.2) and hypothesis (H2), we get

    {u_v¯u}|f(x,v)ψ|KLr(Ω)ψLr(Ω), (4.4)

    where r<2(N1)N2 is the conjugate of r. Then, the definition of g given in (4.2), Definition 1.2(i), and (4.4) yield

    |Ωg(x,v)ψ|{v<u_}|f(x,u_)ψ|+{u_v¯u}|f(x,v)ψ|+{v>¯u}|f(x,¯u)ψ|C2ψH1(Ω), (4.5)

    where the last inequalities of (4.5) follow by (4.4) and (1.2), and the constant C2 depends only on K and Ω.

    Second, we show that B is pseudomonotone, see definition (2.1). For this, we set B=LG, where L,G:H1(Ω)(H1(Ω)) are defined by

    L(v),ψ:=Ω(vψ+vψ) and G(v),ψ:=Ωg(x,v)ψ,

    for all ψH1(Ω). Then we show that B is pseudomonotone in the following steps. Let vnv in H1(Ω).

    Step 1: LvnLv in (H1(Ω)).

    Since vnv, L(vn)L(v),ψ0 as n for all ψH1(Ω). Hence,

    L(vn)L(v)(H1(Ω))=supψH1(Ω)1|L(vn)L(v),ψ|0 as n,

    as desired.

    Step 2: G(vn)G(v) in (H1(Ω)).

    Suppose that vnv in H1(Ω) but G(vn)G(v) in (H1(Ω)). Then there exists ε0>0 and a subsequence {vnj} such that

    G(vnj)G(v)(H1(Ω))ε0. (4.6)

    Using the fact that {vnj} is bounded in H1(Ω) and the compactness of the trace operator (1.2), there exists a subsequence {vnj} such that vnjv in Lr(Ω), where r<2(N1)N2. By [27,Theorem 4.9], there exists a subsequence {vnj} such that

    vnj(x)v(x) a.e. xΩ.

    Since g(x,.) is continuous for a.e. xΩ, then g(x,vnj(x))g(x,v(x)) a.e. xΩ and g(x,vnj(x)) is bounded in Lr(Ω) by (H2). Using the Lebesgue Dominated Convergence Theorem, we get

    g(.,vnj(.))g(.,v(.))Lr(Ω)0 as n.

    By the Hölder's inequality, for all ψH1(Ω), we get

    G(vnj)G(v),ψ0 as j.

    Therefore, G(vnj)G(v)(H1(Ω))=supψH1(Ω)1|G(vnj)G(v),ψ|0 as j. Hence, G(vnj)G(v) in (H1(Ω)) as j, a contradiction to (4.6).

    Step 3: B is pseudomonotone.

    Let vnv in H1(Ω). Using Step 1-Step 2, we get that

    B(vn)B(v)in(H1(Ω)).

    Therefore, B(vn),ψB(v),ψ as n for all ψH1(Ω). Furthermore, by [27,Proposition 3.5 (iv)], B(vn),vnB(v),v as n. Hence,

    B(vn),vnψB(v),vψ as n,

    establishing that B is pseudomonotone.

    Finally, we show that B is coercive, i.e., B(ψ),ψ/ψH1(Ω) as ψH1(Ω). For any ψH1(Ω), using (4.5) in the definition of the operator B, we have

    B(ψ),ψψ2H1(Ω)C2ψH1(Ω)12ψ2H1(Ω)C3.

    Hence B is coercive. Thus B satisfies the hypotheses of Proposition 2.2 with X=H1(Ω). Therefore, for b=0(H1(Ω)), there exists uH1(Ω) such that

    B(u),ψ=0ψH1(Ω).

    Moreover, g(x,.) is bounded in L2(N1)N2(Ω) by (H2), and therefore in L2(N1)N(Ω) by continuous embedding of L2(N1)N2(Ω) into L2(N1)N(Ω). Hence u is a weak solution of (4.1). It remains to prove that u is a weak solution of (1.1). For this, we will show that u_u¯u in ¯Ω, so that g=f in (4.1).

    Clearly, (u¯u)+:=max{0,u¯u}H1(Ω) and (u_u)+:=max{0,u_u}H1(Ω). Then, using the weak formulation of (4.1) with the test function ψ:=(u¯u)+0 in H1(Ω), and the facts that ¯u is a supersolution of (1.1) and (u¯u)+=0 in {u¯u}, we have

    Ω(u(u¯u)++u(u¯u)+)=Ωg(x,u)(u¯u)+={u>¯u}f(x,¯u)(u¯u)+=Ωf(x,¯u)(u¯u)+Ω¯u(u¯u)++Ω¯u(u¯u)+. (4.7)

    Then, (4.7) yields

    0Ω|(u¯u)+|2+Ω|(u¯u)+|2=Ω(u¯u)(u¯u)++Ω(u¯u)(u¯u)+0,

    which implies that (u¯u)+H1(Ω)=0. That is, u¯u a.e. in Ω. Using the the continuity of the trace operator (1.2), we get that (u¯u)+L2(N1)N2(Ω)=0. Hence, u¯u a.e. in ¯Ω.

    Analogously, taking the test function ψ:=(u_u)+0 and using the fact that u_ is a subsolution of (1.1), we obtain that

    0Ω|(u_u)+|2+Ω|(u_u)+|2=Ω(u_u)(u_u)++Ω(u_u)(u_u)+0,

    Therefore, u_u a.e. in ¯Ω, and hence u_u¯u a.e. in ¯Ω. Thus, u is a weak solution of (1.1), completing the proof of Theorem 1.5.

    We will use Zorn's Lemma and Proposition 2.3, to prove our result. Consider the set

    A:={uH1(Ω):u_(x)u(x)¯u(x) a.e. in ¯Ω and u is a weak solution of (1.1)},

    and we note that A is nonempty by Theorem 1.5. Let {ui}iIA be a family of chain. Since ui is a weak solution of (1.1), taking ui as the test function and using (4.4), we get

    uiH1(Ω)=Ω(|ui|2+u2i)=Ωf(x,ui)uiC,

    where C depends on u_,¯u,K,Ω but independent of iI. By the separability and reflexivity of H1(Ω), there exists an increasing sequence un such that

    unu:=supiIui in H1(Ω).

    Clearly, u is an upper bound of the chain {ui}iI. It suffices to show that uA. Since {un} is nondecreasing and u_un¯u, we have that un(x)u(x), and un(x)u(x) for all n, and u_(x)u(x)¯u(x) pointwise a.e. in ¯Ω. Furthermore, since f is Carathéodory, we have that

    f(x,un(x))f(x,u(x)) as n.

    This, in conjunction with (H2), and the Lebesgue Dominated Convergence Theorem yields f(x,un)f(x,u)Lr(Ω)0 as n. Therefore, using Hölder's inequality, we deduce that

    |Ωf(x,un)ψΩf(x,u)ψ|Ω|f(x,un)f(x,u)||ψ|f(x,un)f(x,u)Lr(Ω)ψLr(Ω)0,

    which yields

    limnΩf(x,un)ψ=Ωf(x,u)ψ for all ψH1(Ω).

    Taking the limit as n, we get for any ψH1(Ω)

    Ω(uψ+uψ)=limnΩ(unψ+unψ)=limnΩf(x,un)ψ=Ωf(x,u)ψ.

    Hence, u is a weak solution of (1.1), thus concluding that uA.

    By Zorn's Lemma, there exists a maximal element uA. It remains to show that u is maximal in the sense that if ˆu is any other weak solution of (1.1) between u_ and ¯u, then ˆuu. So, let ˆu be a weak solution of (1.1) between u_ and ¯u, and u is the maximal element of A. By Proposition 2.4, u=max{ˆu,u} is a subsolution of (1.1). Then, by Theorem 1.5, there exists a weak solution u0 of (1.1) satisfying

    u_uu0¯u.

    Thus, u0A. On the other hand, umax{ˆu,u}=uu0. But u is maximal element of A, so necessarily u=u0. Therefore, we readily see that ˆuuu0=u, and hence

    u_ˆuu¯u,

    as desired. The existence of a minimal element u of A is proved analogously. This completes the proof of Theorem 1.6.

    In this section, we apply our existence results, Theorem 1.4 and Theorem 1.5, to problems involving sublinear nonlinearities. In particular, in each case we construct an ordered pair of weak sub- and supersolution. We apply Theorem 1.4 to establish Theorem 5.1 and Theorem 1.5 in Remark 5.2 below.

    Theorem 5.1. Consider

    {Δu+u=0inΩ;uη=λf(u)onΩ, (5.1)

    where λ>0 parameter and f:[0,)[0,) is locally Lipschitz continuous function satisfying

    (i) f(0)=0 with f(0)>0, and

    (ii) limsf(s)s=0.

    Then (5.1) has a positive weak solution for λ>μ1f(0), where μ1>0 is the first eigenvalue of the Steklov eigenvalue problem

    {Δφ1+φ1=0inΩ;φ1η=μ1φ1onΩ, (5.2)

    and 0<φ1H1(Ω) is the corresponding eigenfunction.

    Proof. Let λ>μ1f(0) be fixed. Using hypothesis (i), we verify that u_:=ϵφ1 is a subsolution of (5.1) for ϵ0. Indeed, we observe that since λ>μ1f(0) is fixed, ξ(s):=μ1sλf(s) satisfies ξ(0)=0 and ξ(0)<0, then ξ(s)<0 for s0. Therefore, for all 0ψH1(Ω), the following holds for ϵ0

    Ωu_ψ+Ωu_ψ=μ1Ω(ϵφ1)ψλΩf(ϵφ1)ψ=λΩf(u_)ψ.

    Next, using hypothesis (ii), we show that there exists Mλ>0 such that ¯u:=Me is a weak supersolution of (5.1) for all MMλ, where e is the unique positive solution of

    {Δe+e=0in Ω;eη=1on Ω.

    We observe that while f is not assumed to be nondecreasing, ¯f(t):=maxs[0,t]f(s) is nondecreasing, and f(t)¯f(t) for all t0. Moreover, due to hypothesis (ii), ¯f satisfies the sublinear condition at infinity

    limt+¯f(t)t=0.

    Therefore, there exists Mλ>0 such that for all MMλ

    ¯f(MeL(Ω))MeL(Ω)1λeL(Ω) or equivalently λ¯f(MeL(Ω))M.

    Then ¯u=MeH1(Ω) satisfies

    Ω¯uψ+Ω¯uψ=MΩψλΩ¯f(MeL(Ω))ψλΩ¯f(Me)ψλΩf(Me)ψ=λΩf(¯u)ψ

    for all 0ϕH1(Ω). Therefore, ¯u is a weak supersolution of (1.1) for each λ>μ1f(0). Clearly ¯u=Meϵ(λ)φ1=u_ a.e. in ¯Ω. We remark that since f is locally Lipschitz and [u_,¯u] is bounded, f satisfies hypothesis (H1) of Theorem 1.4. Hence, there exists a positive weak solution u of (5.1) such that ϵφ1uMe a.e. in ¯Ω for any λ>μ1f(0). This completes the proof.

    Remark 5.2. On the other hand, if f is continuous (not necessarily Lipschitz), satisfies hypothesis (ii) of Theorem 5.1 and f(s)>0 for s0, the problem (5.1) has a positive weak solution for each λ>0. Indeed, it is easy to see that u_0 is a strict weak subsolution and for each λ>0, there exists Mλ>0 such that ¯u=Me is a weak supersolution for all MMλ, as in the proof of Theorem 5.1. Then, the result follows by Theorem 1.5.

    We would like to thank the referee for carefully reading the manuscript and providing valuable suggestions. We would like to thank Professor Jesús Jaramillo for helpful discussions about Step 2 in the proof of Theorem 1.5. The fifth author is supported by grants PID2019-103860GB-I00, MICINN, Spain, and by UCM-BSCH, Spain, GR58/08, Grupo 920894. All authors acknowledge MSRI for bringing this group together for collaboration.

    The authors declare there is no conflicts of interest.



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