Implicit equations involving the $p$-Laplacian operator

In this work we study the existence of solutions $u \in W^{1,p}_0(\Omega)$ to the implicit elliptic problem $ f(x, u, \nabla u, \Delta_p u)= 0$ in $ \Omega $, where $ \Omega $ is a bounded domain in $ \mathbb R^N $, $ N \ge 2 $, with smooth boundary $ \partial \Omega $, $ 1<p<+\infty $, and $ f\colon \Omega \times \mathbb R \times \mathbb R^N \times \R \to \R $. We choose the particular case when the function $ f $ can be expressed in the form $ f(x, z, w, y)= \varphi(x, z, w)- \psi(y) $, where the function $ \psi $ depends only on the $p$-Laplacian $ \Delta_p u $. We also show some applications of our results.

We focus on the particular case when the function f can be expressed in the form f (x, z, w, y) = ϕ(x, z, w)−ψ(y), where ϕ is a real-valued function defined on Ω×R×R N , and ψ is a real-valued function defined on Y , where Y is a nonempty interval of R (which will be specified later). We require that ψ depends only on the p-Laplacian ∆ p u. We further distinguish among the case where ϕ is a Carathéodory function depends on x, u, and ∇u, and the case where ϕ is allowed to be highly discontinuous in each variable, for which the dependance on the gradient is not permitted.
In both cases we first reduce problem (1.1) to an elliptic differential inclusion, but methods used are different, depending on the regularity of the function ϕ and on the structure of the problem.
More precisely, in the first case we make use of a result in [14] to obtain the inclusion where F is a lower semicontinuous selection of the multifunction (x, z, w) → {y ∈ Y : ϕ(x, z, w) − ψ(y) = 0}.
We start with the general case when Y coincides with the whole space R, and after we deduce, as a particular case, the existence result when Y is a closed interval of R.
The main tool to obtain existence of solutions to (1.2) is the result below, [12, Theorem 3.1], which deals with the existence of solutions for elliptic differential inclusions with lower semicontinuous right-hand side and is based on a selection theorem for decomposable-valued multifunctions (see [1] and [8]). |y| < a(x) + b|z| p−1 + c|w| p−1 in Ω × R × R N .
Then, (1.2) has a solution u ∈ W 1,p 0 (Ω). Here, λ 1,p is the first eigenvalue of the p-Laplacian in the space W 1,p 0 (Ω). The following is our main result, which extends [8,Theorem 3.2] to the case p = 2. Theorem 1.2. Let ϕ : Ω × R × R N → R be a Carathéodory function and let ψ : R → R be continuous. Suppose that: Then, there exists u ∈ W 1,p 0 (Ω) such that When ϕ is discontinuous we essentially follow [11,Theorem 3.1] to construct an appropriate upper semicontinuous multifunction F related with ψ −1 and ϕ, and then we solve the elliptic differential inclusion −∆ p u ∈ F (x, u) using the following [9, Theorem 2.2] Theorem 1.3. Let U be a nonempty set, Φ : U → W 1,p 0 (Ω), Ψ : U → L p ′ (Ω) two operators and F : Ω × R → 2 R a convex closed-valued multifunction. Suppose that the following conditions hold true: has a closed graph for almost every x ∈ Ω; (i 4 ) There exists r > 0 such that the function belongs to L p ′ (Ω) and ρ p ′ ≤ r. Then, the problem Ψ(u) ∈ F (x, Φ(u)) has at least one solution u ∈ U satisfying |Ψ(u)(x)| ≤ ρ(x) for almost every x ∈ Ω.
1.1. Structure of the paper. In Section 2 we will introduce the functional analytic setting we will use throughout the work. Section 3 is devoted to the case ϕ(x, ·, ·) continuous. Here we will distinguish some cases, which depend on the growth conditions on ϕ or on the choice of the set Y . We also will give some examples where these situations apply. In Section 4 we will consider the discontinuous framework.

Preliminaries
Let X be a topological space and let V ⊆ X. We denote by int(V ) the interior of V and byV the closure of V . The symbol B(X) is used to denote the Borel σ-algebra of X.
If (X, d) is a metric space, for every x ∈ X, r ≥ 0 and every nonempty set V ⊆ X, we define Let X and Z be two nonempty sets. A multifunction Φ from X into Z (symbolically Φ : X → 2 Z ) is a function from X into the family of all subsets of Z. A function When (X, A) is a measurable space, Z is a topological space and for every open set W ⊆ Z we have Φ − (W ) ∈ A, we say that the multifunction Φ is measurable. If X and Z are two topological spaces and, for every open (resp. closed) set W ⊆ Z, the set Φ − (W ) is open (resp. closed) in X, we say that Φ is lower semicontinuous (resp. upper semicontinuous). When (Z, δ) is a metric space, the multifunction Φ is lower semicontinuous if and only if, for every z ∈ Z, the real-valued function x → δ(z, Φ(x)), x ∈ X, is upper semicontinuous (see [15,Theorem 1.1]). If, moreover, X is first countable, then the multifunction Φ is lower semicontinuous if and only if, for every x ∈ X, every sequence {x k } in X converging to x and every z ∈ Φ(x), there exists a sequence {z k } in Z converging to z and such that z k ∈ Φ(x k ), for all k ∈ N (see [ Then, the multifunction Q is lower semicontinuous, with nonempty closed values. From now on, Ω is a bounded domain in R N , N ≥ 2, with a smooth boundary ∂Ω, the symbol L(Ω) (respectively, m(Ω)) denotes the Lebesgue σ-algebra (respectively, measure) of Ω, while W 1,p 0 (Ω) stands for the closure of C ∞ 0 (Ω) in W 1,p (Ω). On W 1,p 0 (Ω) we introduce the norm Let p * be the critical exponent for the Sobolev embedding W 1,p 0 (Ω) ⊆ L r (Ω). Recall that Let A p : W 1,p 0 (Ω) → W −1,p ′ (Ω) be the nonlinear operator stemming from the negative p-Laplacian, i.e., and let λ 1,p be its first eigenvalue in W 1,p 0 (Ω). The following facts are well known (see, e.g., [13], Appendix A): (p 1 ) A p is bijective and uniformly continuous on bounded sets;

The case when ϕ is a Carathéodory function
This section deals with the existence of solutions to the equation We first consider the case Y = R. Throughout the section, p ∈ ]1, +∞[ and the following assumptions will be posited: (i) for every (x, z, w) ∈ Ω × R × R N , the set {y ∈ R : ϕ(x, z, w) − ψ(y) = 0} has empty interior; Since, by (i), the set R \ U has empty interior, it follows that U is dense in R, as desired.
Let us next analyze the set turns out to be open, because ϕ(x, ·, ·) is continuous. Otherwise it is empty. So, the set (3.2) is the whole space R × R, and (c) follows.
Moreover, for all y ′ , y ′′ ∈ R we have This actually implies that Finally, fix any y ∈ F (x, z, w). In other words, y ∈ ψ −1 (ϕ(x, z, w)), therefore hy- So all the hypotheses of Theorem 1.1 are fulfilled, and we get a solution u ∈ W 1,p 0 (Ω) to equation (1.2). Taking into account the definition of F , we have ψ(−∆ p u) = ϕ(x, u, ∇u), that is the thesis.

Remark 3.2. A very simple situation when hypothesis (iii) occurs is the following.
Suppose As an application of the previous result, we consider the following example.
Since lim y→±∞ (y − λ sin y) = ±∞, the function y → ϕ(x, z, w) − ψ(y) surely changes sign. Moreover, since it vanishes only at points of R and not in intervals, the set has empty interior in R. Hence, hypotheses (i) and (ii) are fulfilled.
Fix now y ∈ F (x, z, w). Since, in other words, y ∈ ψ −1 (ϕ(x, z, w)), hypothesis (iii) ′ implies that inf Taking into account (iv), we see that all the hypotheses of [12, Theorem 3.4] are fulfilled. Therefore, there exists u ∈ W 1,p 0 (Ω) such that −∆ p u ∈ F (x, u, ∇u). Exploiting the definition of F , this means that u is a solution to equation (3.1).
Moreover, in order to verify hypothesis (iv), we have to check if there exists R > 0 such that If 0 < R << 1, choosingh in such a way that h L q (Ω) < R 2 , we have that (3.9) is immediately satisfied, since the terms containing R 2 and R 3 are negligible with respect to R. So all the hypotheses of Theorem 3.5 are fulfilled, and we get the thesis.
The next result provides solutions to equation (3.1) when the function ψ is of the form y → y − h(y), with h continuous and bounded. Note that here we have to require a specific growth condition on ϕ. Theorem 3.7. Let ϕ : Ω × R × R N → R be a Carathéodory function and let h ∈ L ∞ (R) be continuous. Suppose that (i)-(ii) hold true and, moreover, Then, there exists a solution u ∈ W 1,p 0 (Ω) to the equation Proof. Fix x ∈ Ω and define, for all (z, w) ∈ R × R N , Reasoning as in the above proofs ensures that F is lower semicontinuous w.r.t. (z, w), with nonempty closed values, and L(Ω) ⊗ B(R × R N )-measurable. Fix If we choose ε in such a way that hypothesis (h3) of Theorem 1.1 is fulfilled, with a := 2f + C ε ∈ L p ′ (Ω, R + 0 ) and b := c := 2 γ−1 µε.
In both cases, there exists u ∈ W 1,p 0 (Ω) such that −∆ p u ∈ F (x, u, ∇u). Through a familiar argument, this entails that u is a solution to equation (3.10).
We conclude this section considering the case when Y is a closed interval of R. Observe that here no growth conditions on ϕ are required. (1) for every (x, z, w) ∈ Ω × R × R N , the set {y ∈ [α, β] : ϕ(x, z, w) − ψ(y) = 0} has empty interior; Then, there exists a solution u ∈ W 1,p 0 (Ω) to equation (3.1). Proof. As before, fix x ∈ Ω, and for all (z, w) ∈ R × R N define A familiar argument ensures that F takes nonempty closed values, is lower semicontinuous w.r.t. (z, w) and L(Ω) ⊗ B(R × R N )-measurable.
As application of the previous theorem, we consider two examples, which differ by the behavior of the function ψ. In both cases, the condition which permits to get a solution is the boundedness of ϕ. Example 3.9. Let f ∈ L ∞ (Ω), k ∈ N, k even and such that kπ > f ∞ , and let ψ : [−kπ, kπ] → R be defined by ψ(y) = y cos y. Then, there exists a solution u ∈ W 1,p 0 (Ω) to the equation Proof. Observe that assumption (1) is clearly satisfied. Moreover, for every x ∈ Ω, we have Therefore, hypothesis (2) is also satisfied. Thanks to Theorem 3.8, there exists at least a solution u ∈ W 1,p 0 (Ω) to equation (3.11).
Note that the interval [α, β] could be unbounded, as the following example shows.
Reasoning as in [10, Theorem 3.1], we see that ρ p ′ ≤ r once the same property holds true for the function x → j(x) := sup |z|≤g(r) |f (x, z)|.
Hypothesis (iv) and the assumption 0 / ∈ (α, β) are essential to obtain the existence of a solution for equations as in the previous theorem. Below we list two examples, apparently very similar, and such that one admits a solution while the other one doesn't.  Note that equation (4.4) doesn't have any solution u ∈ W 1,p 0 (Ω). Suppose on the contrary that u is such a solution. Since ϕ(u) ≥ 0, then from equation (4.4) we have −∆ p u ≥ 0, and so the Strong Maximum Principle implies that u ≡ 0 or u > 0. Suppose that u ≡ 0, then this would imply that −∆ p u ≡ 0, which is in contrast with (4.4). Suppose now that u > 0. Then, taking into account the definition of ϕ and equation (4.4), we have −∆ p u = 0. This fact, jointly with the boundary condition u| ∂Ω = 0, implies u ≡ 0 which is again impossible.
It is also evident from the definition of ϕ that hypothesis (iv) and 0 / ∈ (α, β) cannot be verified simultaneously.