Monotone iterations of two obstacle problems with different operators

In this paper we analyze iterations of the obstacle problem for two different operators. We solve iteratively the obstacle problem from above or below for two different differential operators with obstacles given by the previous functions in the iterative process. When we start the iterations with a super or a subsolution of one of the operators this procedure generates two monotone sequences of functions that we show that converge to a solution to the two membranes problem for the two different operators. We perform our analysis in both the variational and the viscosity settings


Introduction.
The main goal in this paper is to find solutions to the two membranes problem as limits of sequences obtained by iterating the obstacle problem.Let us first describe the obstacle problem (from above or below) and the then two membranes problem (as described in [8]).
1.1.The obstacle problem.The obstacle problem is one of the main problems in the mathematical study of variational inequalities and free boundary problems.The problem is to find the equilibrium position of an elastic membrane whose boundary is fixed, and which is constrained to lie above a given obstacle.In mathematical terms, given an operator L (notice that here we can consider fully nonlinear problems of the form Lu = F (D 2 u, Du, u, x)) that describes the elastic configuration of the membrane, a bounded Lipschitz domain Ω and a boundary datum f , the obstacle problem from below (here solutions are assumed to be above the obstacle) reads as (1.1) or equivalently min{Lu, u − φ} = 0 in Ω, u = f on ∂Ω.
We will denote by u = O(L, φ, f ) the solution to (1.1).
When the operator L is in divergence form and is associated to an energy functional E(u) the problem becomes a variational problem and a solution can be obtained minimizing E in the set of functions in the appropriate Sobolev space that are above the obstacle and take the boundary datum.We refer to Section 2 for details.
The obstacle problem can be also stated as follows: we look for the smallest supersolution of L (with boundary datum f ) that is above the obstacle.This formulation is quite convenient when dealing with fully nonlinear problems using viscosity solutions.We will assume here that the problem (1.1) has a unique continuous viscosity solution (for general theory of viscosity solutions we refer to [9] and [12]).This is guaranteed if L has a comparison principle and one can construct barriers close to the boundary so that the boundary datum f is taken continuously.See Section 3.
We can also consider the obstacle problem from above (here solutions are assumed to be below the obstacle) or equivalently max{Lv, v − ϕ} = 0 in Ω, v = g on ∂Ω.
In this case, the obstacle problem can be viewed as follows: we look for the largest subsolution of L (with boundary datum g) that is below the obstacle.We will denote by v = O(L, ϕ, g) the solution to (1.2).
For general references on the obstacle problem (including regularity of solutions, that in some cases are proved to be optimally C 1,1 ) we just mention [5], [6], [13], [24], the survey [21] and references therein.
1.2.The two membranes problem.Closely related to the obstacle problem, one of the systems that attracted the attention of the PDE community is the two membranes problem.This problem models the behaviour of two elastic membranes that are clamped at the boundary of a prescribed domain, they are assumed to be ordered, one membrane is above the other, and they are subject to different external forces (the membrane that is on top is pushed down and the one that is below is pushed up).The main assumption here is that the two membranes do not penetrate each other (they are assumed to be ordered in the whole domain).This situation can be modeled by two obstacle problems; the lower membrane acts as an obstacle from below for the free elastic equation that describes the location of the upper membrane, while, conversely, the upper membrane is an obstacle from above for the equation for the lower membrane.When the equations that obey the two membranes have a variational structure this problem can be tackled using calculus of variations (one aims to minimize the sum of the two energies subject to the constraint that the functions that describe the position of the membranes are always ordered inside the domain, one is bigger or equal than the other), see [27].On the other hand, when the involved equations are not variational the analysis relies on monotonicity arguments (using the maximum principle).Once existence of a solution (in an appropriate sense) is obtained a lot of interesting questions arise, like uniqueness, regularity of the involved functions, a description of the contact set, the regularity of the contact set, etc. See [7,8,26], the dissertation [28] and references therein.More concretely, given two differential operators L 1 (D 2 u, Du, u, x) and L 2 (D 2 v, Dv, v, x) the mathematical formulation the two membranes problem (with Dirichlet boundary conditions) is the following: With our previous notation, this system can be written as Remark that, in general, the two membranes problem as stated in Definition 1.1 does not have uniqueness.To see this, just take a solution to L 1 (D 2 u) = 0 with u| ∂Ω = f and v the solution to the obstacle problem for L 2 (D 2 v) with u as obstacle from above, that is, v = O(L 2 , u, g).One can easily check that the pair (u, v) is a solution to the general formulation of the two membranes problem stated Definition 1.1.Analogously, one can take a solution to L 2 (D 2 v) = 0 with v| ∂Ω = g and u the solution to the corresponding obstacle problem, u = O(L 1 , v, f ), to obtain another solution to the two membranes problem.In general, these two pairs do not coincide.For example, consider Ω as the interval (0, 1), f ≡ 1, g ≡ 0 on ∂Ω and the operators L 1 (u ′′ ) = u ′′ − 10 and ) and (ũ, ṽ) are solutions to the two membranes problem but they are different.Since u is a solution and ũ is a supersolution to L 1 with the same boundary datum, by the comparison principle, we get u ≤ ũ.On the other hand, since ũ is the solution to the obstacle problem for L 1 (u ′′ ) with ṽ as obstacle from below, ũ ≥ ṽ.Since u and ṽ are not ordered, we obtain that u ≡ ũ.
The two membranes problem for the Laplacian with a right hand side, that is, for was considered in [27] using variational arguments.Latter, in [7] the authors solve the two membranes problem for two different fractional Laplacians of different order (two linear non-local operators defined by two different kernels).Notice that in this case the problem is still variational.In these cases an extra condition appears, namely, the sum of the two operators vanishes, L 1 (D 2 u) + L 2 (D 2 v) = 0, inside Ω.Moreover, this extra condition together with the variational structure is used to prove a C 1,γ regularity result for the solution.
The two membranes problem for a fully nonlinear operator was studied in [7,8,26].In particular, in [8] the authors consider a version of the two membranes problem for two different fully nonlinear operators, L 1 (D 2 u) and L 2 (D 2 v).Assuming that L 1 is convex and that L 2 (X) = −L 1 (−X), they prove that solutions are C 1,1 smooth.
We also mention that a more general version of the two membranes problem involving more than two membranes was considered by several authors (see for example [1,10,11]).
1.3.Description of the main results.As we mentioned at the beginning of the introduction, our main goal is to obtain solutions to the two membranes problems as limits of iterations of the obstacle problem.For iterations of the obstacle problem in a different context we refer to [4].
Let us consider two different operators L 1 and L 2 (with boundary data f and g respectively, we assume that f > g) and generate two sequences iterating the obstacle problems from above and below.Given an initial function v 0 , take u 0 as the solution to the obstacle problem from below for L 1 with boundary datum f and obstacle v 0 , that is, Now, we use this u 0 as the obstacle from above for L 2 with datum g and obtain We can iterate this procedure (solving the obstacle problem for L 1 or L 2 with boundary data f or g and obstacle the previous u or v) to obtain two sequences {u n } n , {v n } n , given by Our main goal here is to show that, when the initial function v 0 is a subsolution for L 2 with boundary datum g, then both sequences of functions {u n } n , {v n } n are nondecreasing sequences that converge to a limit pair that gives a solution of the two membranes problem, i.e., there exists a pair of functions (u ∞ , v ∞ ) such that and the limit functions satisfy An analogous result can be obtained when we start the iteration with u 0 a supersolution to L 1 with boundary datum f and consider {u n } n , {v n } n , given by In this case the sequences are non-increasing sequences and also converge to a solution to the two membranes problem.
We will study this iterative scheme using two very different frameworks for the solutions.First, we deal with variational methods and understand solutions in the weak sense (this imposes that L 1 and L 2 must be in divergence form) and next we deal with viscosity solutions (allowing L 1 and L 2 to be general elliptic operators).
Variational operators.To simplify the notation and the arguments, when we deal with the problem using variational methods, we will concentrate in the particular choice of L 1 and L 2 as two different p−Laplacians; that is, we let where ∆ p w = div(|∇w| p−2 ∇w), ∆ q w = div(|∇w| q−2 ∇w) are two different p−Laplacians and h p , h q are two given functions.These operators are associated to the energies These energies are naturally well-posed in the Sobolev spaces W 1,p (Ω) and W 1,q (Ω), respectively.
In this variational framework the obstacle problem can be solved minimizing the energy in the appropriate set of functions.In fact, the solution to the obstacle problem from below, u = O(L p , ϕ, f ), can be obtained as the minimizer of the energy E p among functions are constrained to be above the obstacle, that is, e Ω .The minimizer in (1.3) exists and is unique provided the set Λ p f,ϕ is not empty.We refer to Section 2 for details and precise definitions.
Due to the fact that the problem satisfied by the limit of the sequences constructed iterating the obstacle problem from above and below does not have uniqueness (see the comments after Definition 1.1) we need to rely on monotonicity of the sequences to obtain convergence (rather than use compactness arguments that provide only convergence along subsequences).
We notice at this point that, in this variational setting, one can solve the two membranes problem just minimizing the total energy The fact that there is a unique pair that minimizes E(u, v) follows from the strict convexity of the functional using the direct method of calculus of variations.One can check that the minimizing pair is in fact a solution to the two membranes problem given in Definition 1.1.However, in general, there are other solutions to the two membranes problem in the sense of Definition 1.1 that are not minimizers of E(u, v).Since solutions according to Definition 1.1 are, in general, not unique, we observe that the limit that we prove to exist for the sequences that we construct iterating the obstacle problem from above and below does not necessarily converge to the unique minimizer to E(u, v).
Fully nonlinear operators.On the other hand, when we deal with viscosity solutions to the obstacle problems we use the general framework described in [12] and we understand sub and supersolutions applying the operator to a smooth test function that touches the graph of the sub/supersolution from above or below at some point in the domain.Remark that here we can consider fully nonlinear elliptic operators that need not be in divergence form.
When we develop the viscosity theory of the iterations in this general viscosity setting it is more difficult to obtain estimates in order to pass to the limit in the sequences {u n } n , {v n } n and hence we have to rely again on monotonicity (that is obtained using comparison arguments).
In the viscosity framework there is no gain in considering two particular operators.Therefore, we will state and prove the results for two general second order elliptic operators that satisfy the comparison principle.However, if one looks for a model problem one can think on two normalized p−Laplacians, that is, let Here ∆w = trace(D 2 w) (the usual Laplacian), ∆ ∞ w = D 2 w ∇w |∇w| , ∇w |∇w| (the normalized infinity Laplacian), α i , β i are nonnegative coefficients and h p , h q are continuous functions.This operator, the normalized p−Laplacian, appears naturally when one considers game theory to solve nonlinear PDEs, we refer to [19,22,23] and the recent books [3,18].
Let us finish the introduction with a short paragraph concerning the equivalence between weak and viscosity solutions for divergence form operators.For the Dirichlet problem ∆u = 0 the notions of weak and viscosity solutions coincide (and in fact the Dirichlet problem has a unique classical solution), see [16] and [25].Moreover, the equivalence between weak and viscosity solutions include quasi-linear equations, [17,20], and some non-local equations, [2,14].
The paper is organized in two sections; in Section 2 we deal with that variational setting of the problem and in Section 3 we analyze the problem using viscosity theory.

Variational solutions
In this section we will obtain two sequences of solutions to variational obstacle problems whose limits constitute a pair of functions that is a variational solution to the two membranes problem.
Let us start by introducing some notations and definitions.We follow [15].For short, we write for the operators and their associated energies.In this variational context, we now introduce the definition of weak super and subsolutions.
for all non-negative w ∈ W 1,p 0 (Ω) and in this case we writte L p u ≥ 0 (resp.L p u ≤ 0) weakly in Ω.We say that u ∈ W 1,p (Ω) is a weak solution for L p if When Λ p f,ϕ = ∅, we say u ∈ Λ p f,ϕ is the solution of the p -Laplacian upper obstacle problem in a varational sense with obstacle ϕ and boundary datum f if i.e, u minimizes the energy associated with the operator L p := −∆ p + h p in the set of functions that are above the obstacle and take the boundary datum, Λ p f,ϕ .In this case, we denote u = O(L p , ϕ, f ).In Proposition 1 we show that this solution exists and is unique provided the set Λ p f,ϕ is not empty.Analogously, for g When Λ q g,ϕ = ∅, we say that v ∈ Λ q g,ϕ is the solution of the q -Laplacian lower obstacle problem in a varational sense with obstacle ϕ and boundary g data if For short, we denote v as v = O(L q , ϕ, g).
Proof.Let us prove existence and uniqueness of the solution to the obstacle problem from below.The proof is contained in Theorem 3.21 in [15] but we reproduce some details here for completeness.Suppose that Λ p f,ϕ = ∅.Since E p is a coercive and strictly convex functional in W 1,p (Ω), we obtain that all minimizing sequences u j ∈ Λ p f,ϕ converge weakly to the same limit u ∈ W 1,p (Ω) (that we want to show that is the unique minimizer of E p in Λ p f,ϕ ).Let us show that u ∈ Λ p f,ϕ .Rellich-Kondrachov's Theorem implies that, taking a subsequence, u j → u strongly in L p (Ω).Then, taking a subsequence, we have that u j → u a.e. in Ω.Thus, since u j ≥ ϕ a.e.Ω and u j − f ∈ W 1,p 0 (Ω) for all j ≥ 1, we get u ≥ ϕ a.e.Ω and u − f ∈ W 1,p 0 (Ω).Let us prove that u verifies the properties stated in the proposition.Let us start with the fact that for all v ∈ Λ p f,ϕ and t ∈ [0, 1].Thanks to the fact that E p reaches its minimum in the set Λ p f,ϕ at u, we have that Taking t = 0, we get If we consider w ∈ W 1,p 0 (Ω) w ≥ 0 a.e.Ω, and v = u + tw ∈ Λ p f,ϕ for t ≥ 0 small enough, if we come back to (2.4), we obtain Finally, we show that L p u = 0 weakly in {u(x) > ϕ(x)} when ϕ is continuous.From Section 3.26 in [15] we have that the solution to the obstacle problem with continuous obstacle becomes also continuous (after a redefinition in a set of measure zero).Therefore, for a continuous obstacle the set {x : u(x) > ϕ(x)} is an open set.Take a ball and consider a nonnegative w ∈ W 1,p 0 (B r ) ∩ C 0 (B r ).Using that ϕ < u in ∂B r , we get v = u + tw ∈ Λ p f,ϕ for |t| small enough.Then, let us come back to (2.4) and obtain t ˆBr In this case t can be positive or negative.Thus, we get weakly in B r and therefore in {u > ϕ}.
The other case (an obstacle from above) is analogous.
This minimizer of the energy is in fact the infimum of weak supersolutions are above/below the obstacle.
Proof.Let us consider u = inf w ∈ Λ p f,ϕ : L p w ≥ 0 weakly in Ω .We will prove that u = u in Ω.
Let us start proving that u ≥ u.By Proposition 1, we have u ∈ w ∈ Λ p f,ϕ : L p w ≥ 0 weakly in Ω , and then we get and then, using (2.4), we get and, using (2.7), we obtain Substracting these inequalities we conclude that The other case is analogous.
Remark 1.Notice that Proposition 2 implies that the solution to the obstacle problem is monotone with respect to the obstacle in the sense that, for ϕ 1 ≥ ϕ 2 , we have Now, we are ready to introduce the definition of a weak solution to the two membranes problem.Definition 2.3.Let f ∈ W 1,p (Ω) and g ∈ W 1,q (Ω) be two functions such that f ≥ g in ∂Ω in the sense of traces.Take h p ∈ W −1,p (Ω) and h q ∈ W −1,q (Ω).We say that the pair (u, v) Remark 2. In general, given a pair (f, g), the solution to the two membranes problem is not unique.
2.1.Iterative method.The main result of this section reads as follows.
Theorem 2.4.For p ≥ q let f ∈ W 1,p (Ω) and g ∈ W 1,q (Ω) be two functions such that f ≥ g in ∂Ω in the sense of traces and take h p ∈ W −1,p (Ω) and h q ∈ W −1,q (Ω).
Let us consider v 0 a weak subsolution of L q v = 0 in Ω such that v 0 − g ∈ W 1,q 0 (Ω) and then define inductively the sequences Ω) converge strongly in W 1,p (Ω) and W 1,q (Ω), respectively.Moreover, the limits of the sequences are a solution of the two membranes problem.That is, there exists a pair of functions and, in addition, the limit pair Proof.Notice that from the definition of the sequences as solutions to the obstacle problem we have v n ≤ u n a.e.Ω for each n ∈ N. Since we assumed that the boundary data are ordered, f ≥ g, and functions in W 1,p (Ω) and in W 1,q (Ω) have a trace on the boundary, we can also say that u n ≥ v n a.e. on the boundary (with respect to the (N − 1)dimensional measure).From the construction of the sequences we obtain that the sets Λ p f,vn and Λ q g,un are not empty, in fact we have that u n−1 ∈ Λ p f,vn (since v n solves an obstacle problem from above with obstacle u n−1 , we have v n ≤ u n−1 ) and v n ∈ Λ q g,un (since u n solves an obstacle problem from below with obstacle v n , we have v n ≤ u n ).Hence the sequences {u n } and {v n } are well defined.
We will divide our arguments in several steps.
First step: First, we prove a monotonicity result.The two sequences are nondecreasing.
Let us see that v 0 ≤ v 1 a.e.Ω.This is due to the fact that v 0 is a weak subsolution of L q v = 0 and v 0 ∈ Λ q g,u0 and by (2.6) we have Then, we conclude that v 0 ≤ v 1 .
Also we can see that u 0 ≤ u 1 a.e.Ω.In fact, from (2.5), we have u 0 = inf w ∈ Λ p f,v0 : L p w ≥ 0 weakly in Ω and then, using that v 1 ≥ v 0 a.e.Ω, we obtain that u 1 ∈ Λ q f,v0 and u 1 is a weak supersolution for L p u = 0.Then, u 0 ≤ u 1 a.e. in Ω.Now, to obtain the general case, we just use an inductive argument.Suppose v n−1 ≤ v n and u n−1 ≤ u n a.e.Ω.Since v n ≤ u n , we have v n ∈ Λ q g,un .From (2.6) we have v n+1 = sup w ∈ Λ q g,un : L q w ≤ 0 weakly in Ω and L q v n ≤ 0 weakly in Ω which implies v n ≤ v n+1 a.e.Ω.On the other hand, the last inequality and the fact that v n+1 ≤ u n+1 give us u n+1 ∈ Λ p f,vn .Hence, (2.5) implies that and we conclude that u n ≤ u n+1 a.e.Ω.
Second step: Our next step is to show that the sequences {u n } ∞ n=1 and {v n } ∞ n=1 are bounded in W 1,p (Ω) and W 1,q (Ω), respectively.To prove this fact let us take w ∈ W 1,q (Ω) the solution to L q w = 0 with boundary datum g.That is, w satisfies L q w = 0 weakly in Ω and w = g in ∂Ω in the sense of traces.
By (2.6), we have that L q v n ≤ 0 weakly in Ω, and thanks to v n − w ∈ W 1,q 0 (Ω) and the comparison principle for L q , we obtain that v n ≤ w a.e.Ω.Let us consider z = O(L p , w, f ).Since v n ≤ w we have z ∈ Λ p f,vn for all n ∈ N. Now, since u n is the solution to the obstacle problem, u n minimizes the energy among functions in the set Λ p f,vn .Then, as z is in the former set, we get for all n ∈ N, that is, we have We can rewrite this inequality to obtain Using that u n − f ∈ W 1,p 0 (Ω), from Poincare's inequality, we obtain Then, by the triangle inequality we get If we come back to (2.8), we obtain Thus, we get the desired bound for u n , there exists a constant C independent of n such that u n W 1,p (Ω) ≤ C for all n ∈ N.

Now, we prove that
Now, we just proceed as before to obtain Third step: We show strong convergence to a solution of the two membranes problem.
Since {u n } and {v n } are bounded in W 1,p (Ω) and W 1,q (Ω) respectively, there exist subsequences {u nj } ⊂ {u n } and {v nj } ⊂ {v n } such that By the Rellich-Kondrachov theorem, there exists a subsequence {u nj } that converges strongly in L p (Ω) and a subsequence {v nj } that converges strongly in L q (Ω).Extracting again a subsequence if needed, we get as n j → ∞.Using that {u n } and {v n } are increasing we obtain that the entire sequences converge pointwise (and weakly) to the unique limits (that are given by the supremums of the sequences).This implies that u ∞ ≥ v ∞ a.e.Ω, u ∞ = f and v ∞ = g a.e.∂Ω in the sense of traces.This gives us u ∞ ∈ Λ p f,v∞ and v ∞ ∈ Λ q g,u∞ .
Let us define u = O(L p , v ∞ , f ) and v = O(L q , u ∞ , g).Our next step is to prove that u = u ∞ and v = v ∞ a.e.Ω and to get that the convergence is strong in the corresponding Sobolev spaces.
First, we begin with the case of the upper membranes u n .Since u = O(L p , v ∞ , f ) and u ∞ ∈ Λ p f,v∞ we have that E p (u) ≤ E p (u ∞ ).On the other hand, since {v n } is an increasing sequence we have u ∞ ≥ v ∞ ≥ v n for each n ∈ N wich implies u ∈ Λ p f,vn .Then, we have that E p (u n ) ≤ E p (u).On the other hand, by the semicontinuity of the norm, u n = u ∞ in ∂Ω and by weak convergence, we obtain Then, we have (2.9) Thus, E p (u ∞ ) = E p (u).And since the energy minimizer is unique, u ∞ = u a.e.Ω.Moreover, by (2.9) and the weak convergence, we have obtained that {u n } is a minimizing sequence of the energy and the gradient of u n converges strongly to the gradient of u ∞ as n goes to infinity.Then, we conclude that Once we have seen that Note that we cannot repeat the previous argument step by sept because we do not know if v is in Λ q g,un−1 .Though, it is enough to change a little our strategy (we use at this point that p ≥ q).We construct a subsequence {ṽ n } such that ṽn ∈ Λ q g,un−1 and E p (ṽ n ) converges to E p (v) as n goes to infinity.We define {ṽ n } as ṽn = v − u ∞ + u n .Since p ≥ q we have that ṽn ∈ W 1,q (Ω), ṽn − g ∈ W 1,q 0 (Ω) and E q (ṽ n ) goes to E q (v) as n → ∞ due to the Rellick-Kondrachov Compactness Theorem and the fact that u n converges to u ∞ in W 1,p (Ω).Also we have the inequality ṽn ≤ u n because v ≤ u ∞ .Then, ṽn ∈ Λ q g,un which implies E q (v n+1 ) ≤ E q (ṽ n ).Besides, by the semicontinuity of the norm, the fact that v n = v ∞ on ∂Ω and the weak convergence, we obtain Then, E q (v ∞ ) = E q (v).By the uniqueness of the minimizer, v ∞ = v a.e.Ω.Furthermore, from the same reasons as in the case of the upper membranes, we have that the entire sequence converges strongly, This ends the proof.
If p < q we can also construct a pair of sequences that converges in W 1,p (Ω) and W 1,q (Ω) to a solution of the two membranes problem.In this case we just have to start the iterations of the obstacle problems with u 0 a weak supersolution to L p u = 0. Theorem 2.5.For p < q let f ∈ W 1,p (Ω) and g ∈ W 1,q (Ω) be two functions such that f ≥ g in ∂Ω in the sense of traces and take h p ∈ W −1,p (Ω) and h q ∈ W −1,q (Ω).Take u 0 a weak supersolution to Ω) converge strongly in W 1,p (Ω) and W 1,q (Ω), respectively.Moreover, the limits of the sequences are a solution of the two membranes problem.That is, there exists a pair of functions u ∞ ∈ Λ p f,v∞ and v ∞ ∈ Λ q g,u∞ such that u n → u ∞ strongly in W 1,p (Ω) and v n → v ∞ strongly in W 1,q (Ω), and, in addition, the limit pair Proof.The difference between these sequences and the sequences defined in proposition (2.4) is that, here, {u n } and {v n } are decreasing sequences.
Due to the monotonicity of the sequences, we can prove that u n converges weakly to some u ∞ in W 1,p (Ω) and v n converges to some v ∞ weakly in W 1,q (Ω) as n goes to infinity.
The strong convergence of v n to v ∞ is given in the same way that we got strong convergence for u n to u ∞ in proposition (2.4) thanks to {v n } is a decreasing sequence.On the other hand, for {u n } we can reproduce the proof for {v n } in proposition (2.4) because W 1,q (Ω) ֒→ W 1,p (Ω) continuously.Remark 3.An alternative idea to deal with the case p < q runs as follows: It can be easily proved that Then, if p < q we consider For the problem with the new set of parameters, p ′ , q ′ , h p ′ , h q ′ , f ′ and g ′ , we can apply the iterative method described in Theorem (2.4) and get a pair and Λ q g,u∞ respectively and i.e, (u ∞ , v ∞ ) is a solution for the original two membranes problem.
Remark 4. We remark that the limit depends strongly on the initial function from where we start the iterations.We would like to highlight that {u n } and {v n } converge to u ∞ and v ∞ strongly in W 1,p (Ω) and W 1,q (Ω) respectively due to the fact that both sequences are monotone.The monotonicity, in turn, depends on the initial function for the iteration.Specifically, when p ≥ q, the monotonicity arises because the initial datum v 0 is a weak subsolution of L q v = 0. On the other hand, when p ≤ q it stems from the fact that u 0 is a weak supersolution of L p u = 0.
If one considers the pair (u, v) obtained as follows: let u be the solution to L p u = 0 in Ω with u = f on ∂Ω and v the solution to the corresponding obstacle problem v = O(L q , u, g), then, as we have mentioned in the introduction this pair (u, v) is a solution to the two membranes problem.Now, if one starts the iteration procedure with u 0 = u we obtain that the sequences converge after only one step.In fact we get v 1 = O(L q , u, g) and next the iteration gives again, that is, we have u = O(L p , v 1 , f ).This is due to the fact that u is a solution to L p u = 0 in the whole Ω with u ≥ v 1 and therefore it is the solution to the obstacle problem (with v 1 as obstacle from below).Analogously, if one starts with v the solution to L q v = 0 in Ω with v = g on ∂Ω we obtain u as the solution to the corresponding obstacle problem u = O(L p , v, f ).This pair (u, v) is also a solution to the two membranes problem, but, in general the two pairs are different.

Extension to the two membranes problem for nonlocal operators. If we consider
L p := (−∆ p ) s + h p and L q := (−∆ q the fractional p−Laplacian.Notice that L p and L q are associated with the energies Our iterative method of the Theorem (2.4) gives us a pair (u ∞ , v ∞ ) that is solution of the "new" two membranes problem whenever W s,p (Ω) ֒→ W t,q (Ω) continuosly.
To be able to define a variational solution of the obstacle problem for these new L p and L q is nedeed that Λ p f,ϕ and Λ q g,ϕ are closed sets under the weak topology.This is given by Mazur's Theorem if W s,p (Ω) and W t,q (Ω) are reflexive Banach space and Λ p f,ϕ and Λ q g,ϕ are closed convex sets (with the norm) in W s,p (Ω) and W t,q (Ω) respectively.

Viscosity solutions
In this section we aim to prove similar convergence results for iterations of the obstacle problem to a pair that solves the two membranes problems when the involved operators are not variational.Here we will consider two nonlinear elliptic operators and we understand solutions in the viscosity sense.To define the sequences as solutions to the corresponding obstacle problems we need to assume that the involved operators verify the following set of conditions: Hypothesis: The operators L 1 and L 2 verify To be precise, working in the viscosity sense, we need to introduce the definition of semicontinuous functions.Definition 3.1.f : Ω −→ R is a lower semicontinuous function, l.s.c, at x ∈ Ω if for each ε > 0 there exists a δ > 0 such that If a function f is l.s.c. at every x ∈ Ω, we say that f is l.s.c. in Ω.The lower semicontinuous envelope of f is If a functions g is u.s.c. at every x ∈ Ω, we say that it is u.s.c. in Ω.The upper semicontinuous envelope of g is Now, we give the precise definition of sub and supersolutions in the viscosity sense.
Definition 3.2.The function u : Ω −→ R is called a L-viscosity supersolution if its lower semicontinuous envelope function, u * , satisfies the following: for every φ ∈ C 2 (Ω) such that φ touches u * at x ∈ Ω strictly from below, that is, u * − φ has a strict minimum at x with u * (x) = φ(x), we have Lφ(x) ≥ 0.
In this case, we write Lu ≥ 0 in the viscosity sense.
Conversely, u is called a L-viscosity subsolution if its upper semicontinuous envelope function, u * , satisfies that for every φ ∈ C 2 (Ω) such that φ touches u * at x ∈ Ω strictly from above, that is, φ − u * has a strict minimum at x with u * (x) = φ(x), we have Lφ(x) ≤ 0 and we write Lu ≤ 0 in the viscosity sense.
Finally, u is a L-viscosity solution if it is both a L-viscosity supersolution and a L-viscosity subsolution and we denote Lu = 0 in the viscosity sense.Now, let us introduce three results concerning the limit of non-decreasing sequences of viscosity sub or supersolutions.Although these results are not difficult to prove, they will be crucial for the proof of our main Theorem (3.9) in this section.
On the other hand, by the definition of pointwise value of the lower semicontinuous envelope, Then, given ε > 0 there exists a sequence In short, for each ε > 0 there exists a y ∈ B r/2 (x 0 ) and n 0 ∈ N such that ) for all n ≥ n 0 thanks to the above inequality and (3.10).Note that n 0 depends on r, so we write n 0 = n 0 (r).We have n 0 (r) → ∞ and x r n → x 0 as r → 0 + .Due to the fact that v n is a Lviscosity subsolution, we have Since L is a continuous operator we have Therefore v ∞ is a L-viscosity subsolution.
Lemma 3.3.Let {u n } ⊂ C(Ω) a non-decreasing sequence of continuous up to the boundary functions.If u n converge pointwise to some function u ∞ , then u ∞ is a lower semicontinuous function in Ω.
Proof.Let be x 0 ∈ Ω.Since u n −→ u ∞ pointwise, we have that for each ε > 0 there exists a n 0 ∈ N such that u ∞ (x 0 ) ≤ u n0 (x 0 ) + ε.On the other hand, due to the fact that u n0 is continuous there exist a δ such that u n0 (x 0 ) ≤ u n0 (y) + ε for all y ∈ B δ (x 0 ).Moreover, thanks to the fact that the sequence {u n } is non-decreasing we get that, for every ε > 0 there exists δ > 0 such that The proof is finished.
Proposition 4. Let {u n } ∞ n=0 ⊂ C(Ω) be a non-decreasing sequence of continuous up to the boundary functions such that u n is a L-viscosity supersolution for each n ∈ N and u n −→ u ∞ pointwise.Then, u ∞ is a L-viscosity supersolution.
Proof.By Lemma 3.3 we know that u ∞ is a lower semicontinuous function.To prove that u ∞ is a L-viscosity supersolution let us consider φ ∈ C 2 (Ω) such that u ∞ − φ has a strict minimum at x 0 ∈ Ω such that (u ∞ − φ)(x 0 ) = 0. We want to obtain that (Lφ)(x 0 ) ≥ 0.
Let r > 0 be a fixed small radius.Since u ∞ − φ is lower semicontinuous, these functions reach a positive minimum in Ω \ B r (x 0 ), say in as n to infinity.Let us continue the proof assuming this fact and prove it after finishing our argument.
The claim implies that there exists n 0 = n 0 (r) such that (3.11) 0 for all x ∈ Ω \ B r (x 0 ) and n ≥ n 0 .
On the other hand, since u n converges to u ∞ pointwise and (u ∞ − φ)(x 0 ) = 0, there exists n 1 = n 1 (r) such that Thus, the above and (3.11) imply u n − φ reaches its minimum in Ω in the interior of the ball B r (x 0 ) for all n ≥ n 3 = max{n 0 , n 1 }.Call this point x r n .Then, since u n is a L-viscosity supersolution, we get (Lφ)(x r n ) ≥ 0 for all n ≥ n 3 (r) with x r n ∈ B r (x 0 ).Due to the fact that x r n → x 0 as r → 0 + , we have Therefore, u ∞ is a L-viscosity supersolution in Ω.Now, we will prove the claim.Let us define w ∞ = u ∞ − φ, w n = u n − φ and a n = w n (z n ).Thanks to {u n } is a pointwise increasing sequence, {a n } is also an increasing sequence, Moreover, the sequence is bounded above by w ∞ (z ∞ ) due to the fact that u n converges pointwise to u ∞ .Hence, Then, there exists a number a ∞ ≤ w ∞ (z ∞ ) such that a n converges to a ∞ as n tends to infinity.
Suppose that a ∞ < w ∞ (z ∞ ).Then, because {z n } is a colection of points in a compact set, there exists a subsequence {z nj } ⊂ {z n } and a point z ∈ Ω \ B r (x 0 ) sucht that z nj −→ z as n j → ∞.Since w n is a lower semicontinuous function, for each ε > 0, there exists a δ = δ(ε, z) such that w ∞ ( z) ≤ w ∞ (y) + ε for all y ∈ B δ ( z).Due to the fact that z nj converges to z, we can suppose z nj ∈ B δ ( z) for all n j .Then w ∞ ( z) ≤ w ∞ (z nj ) + ε for all n j .For each n j there exists a m j ∈ N such that Observe that we can construct a sequence {(n j , k j )} where k j ≥ m j for all j ∈ N and (n j , k For that ε 0 the above argument implies that there exists a sequence {(n j , k j )} that goes to (∞, ∞) as j → ∞ such that a ∞ + 2ε 0 < w ∞ (z ∞ ) ≤ w kj (z nj ) + 2ε 0 for all j ∈ N and therefore, a ∞ < w kj (z nj ) for all j ∈ N. On the other hand, since {a n } is a increasing sequence and a kj = w kj (z kj ) we have that w kj (z kj ) ≤ a ∞ < w kj (z nj ) for all j ∈ N.
Thanks to the fact that for each j ∈ N the functions w kj are continuous, Ω \ B r (x 0 ) is connected and z nj and z kj are in Ω \ B r (x 0 ), there exists Thus, there is a subsequence {y ji } ⊂ {y j } and a point y ∈ Ω \ B r (x 0 ) such that y ji → y as i → ∞.Due to the fact that w ∞ is a lower semicontinuous function we have w ∞ (y) ≤ lim yj i →y However, this is a contradiction because y ∈ Ω \ B r (x 0 ) and that implies Therefore, we have the desired equality a ∞ = w ∞ (z ∞ ).Now, we are ready to introduce the definition of a solution to the obstacle problem in the viscosity sense.
in Ω in the viscosity sense, Lu = 0 in {u * > ϕ * } in the viscosity sense.
Whereas, given ψ and g such that f ≤ ψ * on ∂Ω, we say that v is a viscosity solution of the L-upper obstacle problem with obstacle ψ and boundary datum g, and we denote in Ω in the viscosity sense, Lv = 0 in {v * < ψ * } in the viscosity sense.
Remark 5. Note that, when u is a viscosity solution of the L-lower obstacle problem with obstacle ϕ and boundary datum f , then, its lower and upper semicontinuous envelopes, u * and u * , are also viscosity solutions for the same problem.
However, we have uniqueness of solutions to the obstacle problem up to semicontinuous envelopes.Proof.We will only prove item (a), the other item is analogous.
(a) Suppose that there exists two lower semicontinuous functions solutions u 1 and u 2 , let us prove that u 2 ≥ u 1 .
In view of the previous lemma, from now we will suppose that u = O(L, ϕ, f ) is lower semicontinuous and v = O(L, ϕ, f ) is upper semicontinuous.
Next, we show that when the obstacles and the boundary data are ordered then the solutions to the obstacle problems are also ordered.Lemma 3.6.
(a) We will consider two cases: In the set {u In the open set {u 1 > ϕ * 1 } we have that Lu 2 ≥ 0 and Lu 1 = 0 and u 2 ≥ u 1 in ∂{u 1 > ϕ * 1 }.Then, by the Comparison Principle we obtain that In the open set {u > ϕ * } we have that Lw ≥ 0 and Lu = 0 in the viscosity sense and w ≥ u in ∂{u > ϕ * }.Then, by the Comparison Principle we obtain that w ≥ u in {u 1 > ϕ * }.
Then, u ≤ w in Ω. Taking minimum we get u ≤ u.
The other case is analogous.Now, we introduce the definition of a solution to the two membranes problem in the viscosity sense (this is just Definition 1.1 with solutions understood in the viscosity sense).Definition 3.8.Given f, g ∈ C(∂Ω) with f ≥ g, let L 1 and L 2 by two operators that satisfy the hypothesis at the begining of the section.We say a pair of functions (u, v) is a solution of the two membranes problem with boundary data (f, g) if Our main result is the following.Theorem 3.9.Given f, g and L 1 , L 2 , let us consider v 0 ∈ C(Ω) a L 2 -viscosity subsolution such that v 0 ≤ g in ∂Ω and then define inductively the sequences Both sequences of functions {u n } ∞ n=0 ⊂ C(Ω), {v n } ∞ n=0 ⊂ C(Ω) converge to some limit functions u ∞ and v ∞ , respectively.Moreover, theses limits are a solution of the two membranes problem with boundary data (f, g), that is, In addition, the functions u ∞ and v ∞ are lower semicontinuous in Ω and continuous in the interior of the set where u ∞ touches v ∞ , i.e, in the interior of {u ∞ = v ∞ }.
Proof.As before, we divide the proof in several steps.In the open set {v 1 < u 0 } we have that L 2 v 1 = 0 and L 2 v 0 ≤ 0 and v 1 ≥ v 0 in ∂{v 1 > u 0 }.Then, by the Comparison Principle we obtain that v 1 ≥ v 0 in {v 1 > u 0 }.Thus, v 0 ≤ v 1 in Ω.Now, using Lemma 3.6 we have u 0 ≤ u 1 .By induction we can continue and obtain u n ≤ u n+1 and v n ≤ v n+1 for all n ≥ 1.
Second step: Let us prove that {u n } ∞ n=0 and {v n } ∞ n=0 are bounded.Let w ∈ C(Ω) be a L 2 -solution with w = g in ∂Ω.Then, since L 2 w = 0, we get L 2 v n ≤ 0.Moreover, we have w ≥ v n in ∂Ω.Hence, applying the Comparison Principle, we get w ≥ v n in Ω for all n ≥ 0. This implies, together with the fact that v n ≥ v 0 , that {v n } is bounded.Now, if we consider z = O(L 1 , w, f ), using that v n ≤ w by Lemma 3.6 u n ≤ z and z ∈ C(Ω).This implies that {u n } is bounded.for all x ∈ Ω.Using that {u n } are continuous and the sequence is increasing we obtain that u ∞ is lower semincontinuous.We also have u ∞ = f and v ∞ = g in ∂Ω.
This function v is well defined and upper semicontinuous.Our goal is to prove that v = v ∞ .
Using that u n ≤ u ∞ , we get v n ≤ v for all n ≥ 0 by Lemma 3.6.Then, we have that v ∞ ≤ v and v * ∞ ≤ v * = v.This implies that v * ∞ ≤ u ∞ .On the other hand, This also implies that v * ∞ is a continuous function in the interior of {u ∞ = v * ∞ }.Then, v ∞ = v * ∞ there and therefore v ∞ is a continuous function in the interior of the set {u ∞ = v ∞ }.Remark 6.If u ∞ or v ∞ are continuous at ∂{u ∞ = v * ∞ }, then, both functions u ∞ and v ∞ are continuous in Ω and therefore the sequences defined in the previous theorem converge uniformly in the whole Ω.
In fact, without loss of generality, suppose that u ∞ is continuous in ∂{u ∞ = v * ∞ }.We will prove the above using the Comparison Principle for L 1 .Since L 1 u ∞ = 0 in {u ∞ > v * ∞ } in the viscosity sense, we have in particular that u * ∞ is a L 1subsolution and u ∞ * a L 1 -supersolution in {u ∞ > v * ∞ } in the viscosity sense.Moreover, since f > g in ∂Ω, the boundary ∂{u ∞ > v * ∞ } is the disjoint union of ∂Ω and ∂{u ∞ = v * ∞ }.In ∂Ω, we have u * ∞ = u ∞ * by construction of u n .And in ∂{u ∞ = v * ∞ } we have also that u * ∞ = u ∞ * because we have supposed u ∞ continuous across that bounday.Then, Moreover, as we have seen in the proof of the above theorem, u ∞ is also continuous in {u ∞ > v * ∞ }.Then u ∞ is continuous in Ω.As a consequence, using the latest hypothesis concerning the operator L 2 , v ∞ is continuous in the whole Ω, because v ∞ is the solution of the upper obstacle problem with continuous boundary datum and continuous obstacle.
Finally, since the continuous functions u ∞ and v ∞ are the limit of the sequences of continuous functions {u n } and {v n } in a compact set Ω, the convergence is uniform.
Remark 7. As happens in the variational setting, we also have here that the limit depends strongly on the initial function from where we start the iterations.Moreover, the convergence of (u n , v n ) to a solution of the two membranes problem relies on the monotonicity of the sequences {u n } and {v n }.This property comes from the fact that the initial function v 0 is a L 2 -viscosity subsolution.For this discussion in the variational context, we refer to the arguments given in Remark 4. membrane, Atti Accad.Naz.Lincei Rend.Cl.Sci.Fis.Mat.Natur.(8) p−2 ∇w − |∇u| p−2 ∇u) • ∇(u − w)]dx using that (|b| p−2 b − |a| p−2 a)(a − b) ≤ 0 (with a strict inequality for a = b) we obtain |{u > z}| = 0.Then, u = z ≤ w.Thus, we have obtained that u ≤ u.
with F continuous in both coordinates, such that the comparison principle holds.• If f is continuous, there exists a unique w ∈ C(Ω) solution to Lw = 0 in Ω, w = f in ∂Ω.• If ϕ, ψ and f , g are continuous, then the corresponding solutions to the obstacle problems, u = O(L 1 , ϕ, f ) and v = O(L 2 , ψ, g) are continuous.See Definition 3.4 below.

Definition 3 . 4 .
Given ϕ : Ω → R and f ∈ C(∂Ω) such that f ≥ ϕ * in ∂Ω.We say that u : Ω −→ R is a viscosity solution of the L-lower obstacle problem with obstacle ϕ and boundary datum f and we denote u = O(L, ϕ, f ) if u satisfies:

Lemma 3 . 5 .
Given ϕ and f defined as before (a) if f ≥ ϕ * in ∂Ω, there exists at most one lower semicontinuous function u such that u = O(L, ϕ, f ).(b) if f ≤ ϕ * in ∂Ω, there exists at most one upper semicontinuous function u such that u = O(L, ϕ, f ).