A Sub-Supersolution Approach for Robin Boundary Value Problems with Full Gradient Dependence

: The paper investigates a nonlinear elliptic problem with a Robin boundary condition, which exhibits a convection term with full dependence on the solution and its gradient. A sub-supersolution approach is developed for this type of problems. The main result establishes the existence of a solution enclosed in the ordered interval formed by a sub-supersolution. The result is applied to ﬁnd positive solutions.


Introduction
In this paper we study the following nonlinear elliptic boundary value problem −div(A(x, ∇u)) + α(x)|u| p−2 u = f (x, u, ∇u) in Ω A(x, ∇u) · ν(x) + β(x)|u| p−2 u = 0 on ∂Ω (1) on a bounded domain Ω ⊂ R N with N ≥ 3 and with a boundary ∂Ω of class C 1 . The notation ν(x) stands for the unit exterior normal at any x ∈ ∂Ω and p is a real number with 1 < p < +∞. We note that, in the stated problem, the boundary condition is of Robin type. We describe the data entering our problem. The leading differential part of the equation in (1) is the term div(A(x, ∇u)) in divergence form driven by the map A : Ω × R N → R N which is composed with the (weak) gradient ∇u of the solution u : Ω → R. No homogeneity condition is required for the map A. Precisely, we assume that A : Ω × R N → R N is continuous and fulfills the conditions: (A1) There exist constants c 1 and c 2 with 0 < c 1 ≤ c 2 such that A(x, ξ) · ξ ≥ c 1 |ξ| p and |A(x, ξ)| ≤ c 2 (|ξ| p−1 + 1) for all (x, ξ) ∈ Ω × R N .
Here and subsequently we denote by | · | and · the standard Euclidean norm and scalar product on R N , respectively.

Prerequisites of Sub-Supersolution Method
This section contains preliminaries that will be used in the sequel. First, we fix some notation. For any r ∈ R, we set r + = max{r, 0} (the positive part of r). If r > 1, we also set r = r r−1 (the Hölder conjugate of r). In particular, for p ∈ (1, +∞) we have p = p p−1 . As indicated in Section 1, Ω is a bounded domain in R N with N ≥ 3 whose boundary ∂Ω is of class C 1 . In order to avoid repetitive arguments, we suppose that N > p. The complementary case N ≤ p can be treated along the same lines and actually is easier. By · L r (Ω) we denote the usual norm on the Banach space L r (Ω).
We seek the solutions to problem (1) in the Sobolebv space W 1,p (Ω), which is a Banach space equipped with the norm For our study of problem (1) it is convenient to use the following equivalent norm on W 1,p (Ω) (see, e.g., ([23], Lemma 2.7) or ( [15], Proposition 2.8)) ( The dual space of W 1,p (Ω) is denoted (W 1,p (Ω)) * , while the notation ·, · designates the duality pairing between W 1,p (Ω) and (W 1,p (Ω)) * , we denote by → the strong convergence and by the weak convergence. The Sobolev embedding theorem ensures that the space Corresponding to the map A : Ω × R N → R N describing the principal part of the equation in problem (1), we introduce the operatorÃ : W 1,p (Ω) → (W 1,p (Ω)) * defined by which is well defined thanks to assumption (A1). It turns out from assumption (A2) and the continuity of A that A(x, ξ) is maximal monotone in the variable ξ for all x ∈ Ω. This allows us to invoke ([2], Proposition 10), which yields: Proposition 1. Assume that the continuous map A : Ω × R N → R N satisfies the conditions (A1) and (A2). Then the mapÃ : W 1,p (Ω) → (W 1,p (Ω)) * in (3) has the (S + )-property, that is, any sequence {u n } ⊂ W 1,p (Ω) with u n u in W 1,p (Ω) and lim sup n→+∞ Ã (u n ), u n − u ≤ 0 fulfills u n → u in W 1,p (Ω).
Due to assumption (A1), the integrals in the above definitions exist. We notice that u ∈ W 1,p (Ω) is a solution of (1) if and only if u is simultaneously a subsolution and a supersolution.
We are going to argue with a sub-supersolution for problem (1), that is, an ordered pair of a subsolution u and a supersolution u such that u ≤ u, which means the pointwise inequality u(x) ≤ u(x) for a.e. x ∈ Ω. Then we can associate the ordered interval Our goal is to obtain a solution u ∈ W 1,p (Ω) of problem (1) with the location property u ∈ [u, u], which will be achieved through comparison by means of a truncation operator that we now describe. Corresponding to a subsolution u and a supersolution u satisfying u ≤ u a.e. in Ω, we define the truncation operator T = T(u, u) : for all u ∈ W 1,p (Ω) and a.e. x ∈ Ω. It readily follows that T : W 1,p (Ω) → W 1,p (Ω) is continuous and bounded (in the sense that it maps bounded sets into bounded sets).
Finally, we mention a few things about the pseudomonotone operators. Let X be a Banach space with the norm · and its dual X * . We denote by ·, · the duality pairing between X and X * . A map A : X → X * is called bounded if it maps bounded sets into bounded sets. The map A : X → X * is said to be coercive if The main theorem for pseudomonotone operators reads as follows (see, e.g., ([22], Theorem 2.99)). Theorem 1. Let X be a reflexive Banach space. If A : X → X * is a pseudomonotone, bounded and coercive map, then A is surjective.

The Associated Operator Equation
Assume that a subsolution u and a supersolution u for problem (1) with u ≤ u are given and that f : Ω × R × R N → R satisfies the following growth condition adapted to the ordered interval [u, u]: (H) There exist a function σ ∈ L r (Ω) with r ∈ (1, p * ) and constants a > 0 and We introduce the cut-off function π : where r 1 > 0 is the constant postulated in hypothesis (H). From (9) and the fact that u, u ∈ L p * (Ω) we infer that π verifies the growth condition with a constant c > 0 and a function ∈ L p * (p−r 1 ) r 1 (Ω). Now for every λ > 0 we define the nonlinear operator A λ : Hypothesis (H) guarantees that the operator A λ in (11) is well defined.
For every λ > 0, the operator A λ : W 1,p (Ω) → (W 1,p (Ω)) * in (11) has the expression The composition N f • T makes sense because T takes values in the ordered interval [u, u] as seen from (7). The following theorem asserts the solvability of the equation Theorem 2. Assume that the conditions (A1), (A2) and (H) are satisfied. Then Equation (14) possesses at least a solution u ∈ W 1,p (Ω) provided λ > 0 is sufficiently large.
The composed operator N f • T is bounded because T is bounded and N f is completely continuous. Since B, Π and Γ are completely continuous, it follows from (13) that A λ : W 1,p (Ω) → (W 1,p (Ω)) * is bounded.
We claim that A λ : W 1,p (Ω) → (W 1,p (Ω)) * is a pseudomonotone operator. Let a sequence {u n } ⊂ W 1,p (Ω) satisfy u n u in W 1,p (Ω) and The complete continuity of the operators B, Π and Γ yields the strong convergent sequences for all v ∈ W 1,p (Ω). We infer that Inequality (17) enables us to apply Proposition 1 ensuring that the strong convergence u n → u in W 1,p (Ω) holds.

Gathering (19)-(22) leads to
for all u ∈ W 1,p (Ω), with a constant c 0 > 0. Due to the fact that p > 1, it turns out thereby the operator A λ is coercive. Summarizing, we have proved that the operator A λ : W 1,p (Ω) → (W 1,p (Ω)) * is bounded, pseudomonotone and coercive. This allows us to apply Theorem 1 with A = A λ for λ > 0 sufficiently large. The surjectivity of A λ implies the existence of a solution u ∈ W 1,p (Ω) of Equation (14), thus completing the proof.

Remark 1.
As a consequence of (23), we can precisely determine the threshold of λ > 0 in the statement of Theorem 2. (1) Our result regarding the method of sub-supersolution for problem (1) is stated as follows.

Main Abstract Result for Problem
Theorem 3. Assume that the conditions (A1), (A2) and (H) are satisfied. Then problem (P) possesses a solution u ∈ W 1,p (Ω) satisfying u ≤ u ≤ u a.e. in Ω, where u and u are the subsolution and the supersolution that are postulated in assumption (H).
Next we show that u ≤ u a.e in Ω. Setting v = (u − u) + ∈ W 1,p (Ω) in (5) and (24) produces and By subtracting (29) from (28) and taking into account (3) we arrive at Along (9) and proceeding as above, (30) results in which entails that u ≤ u a.e in Ω, thus proving the claim. Therefore the solution u ∈ W 1,p (Ω) of the operator equation (14) verifies the enclosure property u ≤ u ≤ u a.e. in Ω. Then we obtain from (7) and (9) that Tu = u and Π(u) = 0. Hence for our function u the equalities (24) and (4) coincide. We see that u ∈ W 1,p (Ω) is a solution of the original problem (1) fulfilling in addition u ≤ u ≤ u a.e. in Ω. This completes the proof.

An Application
The aim of this section is to apply Theorem 3 to establish the existence of positive solutions of Robin problem (1). The main point is to find appropriate ordered sub-supersolutions. The approach can be used to get other types of solutions.
In order to simplify the presentation, we focus on problem (1) driven by the Robin p-Laplacian, 1 < p < +∞, and when α(x) ≡ 0 and the x-dependence in the convection term f (x, s, ξ) is dropped. We emphasize that α ≡ 0 marks a sharp distinction in regard to the Neumann problem. Specifically, we consider the (purely) Robin problem with β(x) ≥ 0 for a.e. x ∈ ∂Ω, β ≡ 0.
We suppose that f : R × R N → R is a continuous function verifying the following assumption: (H ) There exist constants a 0 > 0, a 1 > 0, b > 0 and r 1 ∈ (0, p (p * ) ) such that and The condition (33) involves the first eigenvalue λ 1 of the (negative) Dirichlet p-Laplacian as given in (8). Let us note that u = b is not a solution to problem (31) because the boundary condition is not verified. We formulate the following result concerning problem (31). Proof. Fix an eigenfunction ϕ 1 of −∆ p on W 1,p 0 (Ω), with ϕ 1 > 0 in Ω, corresponding to the first eigenvalue λ 1 (see (8) and the related comments). Since ϕ 1 ∈ C 1 (Ω), we can choose an ε > 0 such that where a 0 is the positive constant prescribed in hypothesis (H ). We note that u = εϕ 1 is a subsolution in the sense of (5) for the Robin problem (31). Indeed, by (8), (33) and (35) and since the trace of u on ∂Ω vanishes, we infer that Ω |∇u| p−2 ∇u · ∇vdx + ∂Ω β(x)|u| p−2 uvdσ = ε p−1 Ω |∇ϕ 1 (x)| p−2 ∇ϕ 1 (x) · ∇v(x)dx This proves that u = εϕ 1 is a subsolution of problem (31). Now we observe that the constant function u = b is a supersolution of problem (31). Indeed, let us notice from assumption (34) that for all v ∈ W 1,p (Ω) with v ≥ 0, which confirms that u = b is a supersolution of problem (31) in the sense of (6).
Hypothesis (H ) is verified taking a 0 = 1 and b = 2. Theorem 4 can be applied to problem (31) with f (s, ξ) given above.