L 1 Existence and Uniqueness Results for Quasi-linear Elliptic Equations with Nonlinear Boundary Conditions

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1 L Existence Uniqueness Results for Quasi-linear Elliptic Equations with Nonlinear Boundary Conditions Résultats d Éxistence et d Unicité dans L pour des Equations Elliptiques Quasi-linéaires avec des Conditions au Bord non Linéaires F. Andreu a, N. Igbida b, J. M. Mazón a, J. Toledo a Article à paraître dans Annales de l IHP, C : Analyse non linéaire a b Departamento de Análisis Matemático, Universitat de València, Dr. Moliner 50, 4600 Burjassot (Valencia), Spain. Fuensanta.Andreu@uv.es, mazon@uv.es, toledojj@uv.es LAMFA, CNRS-UMR 640, Université de Picardie Jules Verne, 33 rue Saint Leu, Amiens, France. noureddine.igbida@u-picardie.fr Abstract In this paper we study the questions of existence uniqueness of weak entropy solutions for equations of type div a(x, Du) + γ(u) φ, posed in an open bounded subset of R N, with nonlinear boundary conditions of the form a(x, Du) η + β(u) ψ. The nonlinear elliptic operator div a(x, Du) modeled on the p-laplacian operator p(u) = div ( Du p 2 Du), with p >, γ β maximal monotone graphs in R 2 such that 0 γ(0) 0 β(0), the data φ L () ψ L ( ). Résumé Dans ce papier nous étudions les questions d existence et d unicité de solution faibles et entropiques pour des équations elliptiques de la forme div a(x, Du) + γ(u) φ, dans un domaine borné R N, avec des conditions au bord générale de la forme a(x, Du) η +β(u) ψ. L opérateur div a(x, Du) généralise l operateur p-laplacien p(u) = div ( Du p 2 Du), avec p >, γ et β sont des graphes maximaux monotonnes dans R 2 tels que 0 γ(0) β(0), et les données φ et ψ sont des fonctions L. Mathematics Subject Classification (2000): 35J60, 35D02 Introduction Let be a bounded domain in R N with smooth boundary < p < N, let a : R N R N be a Carathéodory function satisfying (H ) there exists λ > 0 such that a(x, ξ) ξ λ ξ p for a.e. x for all ξ R N,

2 (H 2 ) there exists σ > 0 θ L p () such that a(x, ξ) σ(θ(x) + ξ p ) for a.e. x for all ξ R N, where p = p p, (H 3 ) (a(x, ξ ) a(x, ξ 2 )) (ξ ξ 2 ) > 0 for a.e. x for all ξ, ξ 2 R N, ξ ξ 2. The hypotheses (H H 3 ) are classical in the study of nonlinear operators in divergent form (cf. [23] or [5]). The model example of function a satisfying these hypotheses is a(x, ξ) = ξ p 2 ξ. The corresponding operator is the p-laplacian operator p (u) = div( Du p 2 Du). We are interested in the study of existence uniqueness of weak entropy solutions for the elliptic problem div a(x, Du) + γ(u) φ in (S γ,β ) a(x, Du) η + β(u) ψ on, where η is the unit outward normal on, ψ L ( ) φ L (). The nonlinearities γ β are maximal monotone graphs in R 2 (see, e.g. [2]) such that 0 γ(0) 0 β(0). In particular, they may be multivalued this allows to include the Dirichlet condition (taking β to be the monotone graph D defined by D(0) = R) the Neumann condition (taking β to be the monotone graph N defined by N(r) = 0 for all r R) as well as many other nonlinear fluxes on the boundary that occur in some problems in Mechanic Physics (see, e.g., [6] or []). Note also that, since γ may be multivalued, problems of type (S γ,β ) appears in various phenomena with changes of state like multiphase Stefan problem (cf [4]) in the weak formulation of the mathematical model of the so called Hele Shaw problem (cf. [5] [7]). Particular instances of problem (S γ,β ) have been studied in [9], [5], [3] []. Let us describe their results in some detail. The work of Bénilan, Crall Sacks [9] was pioneer in this kind of problems. They study problem (S γ,β φ,0 ) for any γ β maximal monotone graphs in R2 such that 0 γ(0) 0 β(0), for the Laplacian operator, i.e., for a(x, ξ) = ξ, prove, between other results, that, for any φ L () satisfying the range condition inf{ran(γ)}meas() + inf{ran(β)}meas( ) < φ < sup{ran(γ)}meas() + sup{ran(β)}meas( ), there exists a unique, up to a constant for u, named weak solution, [u, z, w] W, () L () L ( ), z(x) γ(u(x)) a.e. in, w(x) β(u(x)) a.e. in, such that Du Dv + zv + wv = φv, for all v W, (). For nonhomogeneous boundary condition, i.e. ψ 0, one can see [8] for ψ Ran(β), [9, 20] for some particular situations of β γ. Another important work in the L -Theory for p-laplacian type equations is [5], where problem div a(x, Du) + γ(u) φ in (D γ φ ) u = 0 on is studied for any γ maximal monotone graph in R 2 such that 0 γ(0). It is proved that, for any φ L (), there exists a unique, named entropy solution, [u, z] T,p 0 () L (), z(x) γ(u(x)) a.e. in, such that a(., Du) DT k (u v) + zt k (u v) φt k (u v) k > 0, () 2

3 for all v L () W,p 0 () (see Section 2 for the definition of T,p 0 ()). Following [5], problems (S id,β φ,0 ) (S id,β ), where id(r) = r for all r R, are studied in [3] [] respectively, for any β maximal monotone graph in R 2 with closed domain such that 0 β(0). It is proved that, for any φ L () ψ L ( ), there exists a unique u T,p tr (), there exists w L ( ), w(x) β(u(x)) a.e. in, such that a(., Du) DT k (u v) + ut k (u v) + wt k (u v) ψt k (u v) + φt k (u v) k > 0, for all v L () W,p (), v(x) β(u(x)) a.e. in. Our aim is to prove existence uniqueness of weak entropy solutions for the general elliptic problem (S γ,β ). The main interest in our work is that we are dealing with general nonlinear operators div a(x, Du) with nonhomogeneous boundary conditions general nonlinearities β γ. As in [9], a range condition relating the average of φ ψ to the range of β γ is necessary for existence of weak solution entropy solution (see Remark 3.3). However, in contrast to the smooth homogeneous case, a smooth ψ = 0, for the nonhomogeneous case this range condition is not sufficient for the existence of weak solution. Indeed, in general, the intersection of the domains of β γ seems to create some obstruction phenomena for the existence of these solutions. In general, even if D(β) = R, it does not exist weak solution, as the following example shows. Let γ be such that D(γ) = [0, ], β = R {0}, let φ L (), φ 0 a.e. in, ψ L ( ), ψ 0 a.e. in. If there exists [u, z, w] weak solution of problem (S γ,β ) (see Definition 3.), then z γ(u), therefore 0 u a.e. in, w = 0, it holds that for any v W,p () L (), a(x, Du)Dv + zv = ψv + φv. Taking v = u, as u 0, we get 0 a(x, Du)Du + zu = ψu + φu 0. Therefore, we obtain that Du p = 0, so u is constant zv = ψv + φv, for any v W,p () L (), in particular, for any v W,p 0 () L (). Consequently, φ = z a.e. in, ψ must be 0 a.e. in. The main applications we have in mind is the study of doubly nonlinear evolution problems of elliptic-parabolic type degenerate parabolic problems of Stefan or Hele-Shaw type, with nonhomogeneous boundary conditions /or dynamical boundary conditions (see [2]). Notice that in all these applications one has D(γ) = R, which is sufficiently covered in this paper. The results we obtain have an interpretation in terms of accretive operators. Indeed, we can define the (possibly multivalued) operator B γ,β in X := L () L ( ) as { B γ,β := ((v, w), (ˆv, ŵ)) X X : u T,p tr (), with [u, v, w] an entropy solution of (S γ,β v+ˆv,w+ŵ }. ) 3

4 Then, under certain assumptions, B γ,β is an m-t-accretive operator in X. Therefore, by the theory of Evolution Equations Governed by Accretive Operators (see, [4], [8] or [3]), for any (v 0, w 0 ) D(B γ,β ) X any (f, g) L (0, T ; L ()) L (0, T ; L ( )), there exists a unique mild-solution of the problem V + B γ,β (V ) (f, g), V (0) = (v 0, w 0 ), which rewrites, as an abstract Cauchy problem in X, the following degenerate elliptic-parabolic problem with nonlinear dynamical boundary conditions v t div a(x, Du) = f, v γ(u), in (0, T ) DP (γ, β) w t + a(x, Du) η = g, w β(u), on (0, T ) v(0) = v 0 in, w(0) = w 0 in. In principle, it is not clear how these mild solutions have to be interpreted respect to the problem DP (γ, β). In a next paper ([2]) we characterize these mild solutions. Let us briefly summarize the contents of the paper. In Section 2 we fix the notation give some preliminaries. In Section 3 we give the definitions of the different concepts of solution we use state the main results. The next section is devoted to prove the uniqueness results. In the last section we prove the existence results. First, we study the existence of solutions of approximated problems, next we prove the existence of weak solutions for data in L p finally the existence of entropy solutions for data in L. 2 Preliminaires For p < +, L p () W,p () denote respectively the stard Lebesgue space Sobolev space, W,p 0 () is the closure of D() in W,p (). For u W,p (), we denote by u or τ(u) the trace of u on in the usual sense by W p,p ( ) the set τ(w,p ()). Recall that Ker(τ) = W,p 0 (). In [5], the authors introduce the set T,p () = {u : R measurable such that T k (u) W,p () k > 0}, where T k (s) = sup( k, inf(s, k)). They also prove that given u T,p (), there exists a unique measurable function v : R N such that DT k (u) = vχ { v <k} k > 0. This function v will be denoted by Du. It is clear that if u W,p (), then v L p () v = Du in the usual sense. As in [3], T,p tr () denotes the set of functions u in T,p () satisfying the following conditions, there exists a sequence u n in W,p () such that (a) u n converges to u a.e. in, (b) DT k (u n ) converges to DT k (u) in L () for all k > 0, (c) there exists a measurable function v on, such that u n converges to v a.e. in. The function v is the trace of u in the generalized sense introduced in [3]. In the sequel, the trace of u T,p tr () on will be denoted by tr(u) or u. Let us recall that in the case where u W,p (), tr(u) coincides with the trace of u, τ(u), in the usual sense, the space T,p 0 (), introduced in [5] to study 4

5 (D γ φ,p ), is equal to Ker(tr). Moreover, for every u Ttr () every k > 0, τ(t k (u)) = T k (tr(u)),, if φ W,p () L (), then u φ T,p tr () tr(u φ) = tr(u) τ(φ). We denote V,p () := V,p ( ) := { φ L () : M > 0 such that { ψ L ( ) : M > 0 such that } φv M v W,p () v W,p () } ψv M v W,p () v W,p (). V,p () is a Banach space endowed with the norm φ V,p () := inf{m > 0 : φv M v W,p () v W,p ()}, V,p ( ) is a Banach space endowed with the norm ψ V,p ( ) := inf{m > 0 : ψv M v W,p () v W,p ()}. Observe that, Sobolev embeddings Trace theorems imply, for p < N, Also, L p () L (Np/(N p)) () V,p () L p ( ) L ((N )p/(n p)) ( ) V,p ( ). V,p () = L () V,p ( ) = L ( ) when p > N, L q () V,N () L q ( ) V,N ( ) for any q >. We say that a is smooth (see [3]) when, for any φ L () such that there exists a bounded weak solution u of the homogeneous Dirichlet problem { div a(x, Du) = φ in (D) u = 0 on, there exists ψ L ( ) such that u is also a weak solution of the Neumann problem (N) { div a(x, Du) = φ in a(x, Du) η = ψ on. Functions a corresponding to linear operators with smooth coefficients p-laplacian type operators are smooth (see [] [22]). The smoothness of the Laplacian operator is even stronger than this, in fact, there is a bounded linear mapping T : L () L ( ), such that the weak solution of (D) for φ L () is also a weak solution of (N) for ψ = T (φ) (see [9]). For a maximal monotone graph γ in R R r N we denote by γ r the Yosida approximation of γ, given by γ r = r(i (I + r γ) ). The function γ r is maximal monotone Lipschitz. We recall the definition of the main section γ 0 of γ the element of minimal absolute value of γ(s) if γ(s) γ 0 (s) := + if [s, + ) D(γ) = if (, s] D(γ) =. 5

6 If s D(γ), γ r (s) γ 0 (s) γ r (s) γ 0 (s) as r +, if s / D(γ), γ r (s) + as r +. We will denote by P 0 the following set of functions, P 0 = {q C (R) : 0 q, supp(q ) is compact, 0 / supp(q)}. In [7] the following relation for u, v L () is defined, (u k) + the following facts are proved. Proposition 2. Let be a bounded domain in R N. u v if (v k) + (u + k) (v + k) for any k > 0, (i) For any u, v L (), if uq(u) vq(u) for all q P 0, then u v. (ii) If u, v L () u v, then u p v p for any p [, + ]. (iii) If v L (), then {u L () : u v} is a weakly compact subset of L (). 3 The main results In this section we give the different concepts of solutions we use state the main results. Definition 3. Let φ L () ψ L ( ). A triple of functions [u, z, w] W,p () L () L ( ) is a weak solution of problem (S γ,β ) if z(x) γ(u(x)) a.e. in, w(x) β(u(x)) a.e. in, a(x, Du) Dv + zv + wv = ψv + φv, () for all v L () W,p (). In general, as it is remarked in [5], for < p 2 N, there exists f L () such that the problem u W, loc (), u p(u) = f in D (), has no solution. In [5], to overcome this difficulty to get uniqueness, it was introduced a new concept of solution, named entropy solution. As in [3] or [], following these ideas, we introduce the following concept of solution. Definition 3.2 Let φ L () ψ L ( ). A triple of functions [u, z, w] T,p tr () L () L ( ) is an entropy solution of problem (S γ,β ) if z(x) γ(u(x)) a.e. in, w(x) β(u(x)) a.e. in a(x, Du) DT k (u v) + zt k (u v) + wt k (u v) for all v L () W,p (). ψt k (u v) + φt k (u v) k > 0, (2) 6

7 Obviously, every weak solution is an entropy solution an entropy solution with u W,p () is a weak solution. Remark 3.3 If we take v = T h (u) ± as test functions in (2) let h go to +, we get that z + w = ψ + φ. Then necessarily φ ψ must satisfy where R + γ,β R γ,β R γ,β ψ + φ R + γ,β, := sup{ran(γ)}meas() + sup{ran(β)}meas( ) := inf{ran(γ)}meas() + inf{ran(β)}meas( ). We will write R γ,β :=]R γ,β, R+ γ,β [ when R γ,β < R+ γ,β. Remark 3.4 Let φ V,p () ψ V,p ( ). Then, if [u, z, w] is a weak solution of problem (S γ,β ), it is easy to see that a(x, Du) Du + zu + wu = ψu + φu. Moreover, if D(γ) {0} D(β) {0}, it follows that z V,p (), w V,p ( ) a(x, Du) Dv + zv + wv = ψv + φv, for any v W,p (). In fact, let v W,p () take T k ( v ) r T r(u) as test function in (). Then, letting r go to 0, there exists M > 0 such that z T k ( v ) + w T k ( v ) M v W,p (). {x :u(x) 0} {x :u(x) 0} Letting now k go to +, applying Fatou s Lemma, we get z v + w v M v W,p (). {x :u(x) 0} {x :u(x) 0} If β(0) is bounded, there exists M 2 > 0 such that w v M 2 v W,p (). {x :u(x)=0} In the case β(0) is unbounded from above (a similar argument can be done in the case of being unbounded from below) let us take T k ( v )S r (u) as test function in (), where S r (s) := s+r r χ [ r,0](s) + χ [0,+ [ (s), then, letting r go to 0, there exists M 2 > 0 such that wt k ( v ) M 2 v W,p (), {x :u(x)=0} 7

8 consequently, since β(0) must be bounded from below (because D(β) {0}), there exists M 3 > 0 such that w T k ( v ) M 3 v W,p (). {x :u(x)=0} Letting now k go to +, applying Fatou s Lemma, we get w v M 4 v W,p (). {x :u(x)=0} Similarly, there exists M 5 > 0 such that {x :u(x)=0} z v M 5 v W,p (). We shall state now the uniqueness result of entropy solutions. Since every weak solution is an entropy solution of problem (S γ,β ), the same result is true for weak solutions. Theorem 3.5 Let φ L () ψ L ( ), let [u, z, w ] [u 2, z 2, w 2 ] be entropy solutions of problem (S γ,β ). Then, there exists a constant c R such that u u 2 = c a.e. in, z z 2 = 0 a.e. in. w w 2 = 0 a.e. in. Moreover, if c 0, there exists a constant k R such that z = z 2 = k. Respect to the existence of weak solutions we obtain the following results. Theorem 3.6 Assume D(γ) = R R γ,β < R+ γ,β. Let D(β) = R or a smooth. (i) For any φ V,p () ψ V,p ( ) with φ + there exists a weak solution [u, z, w] of problem (S γ,β ). ψ R γ,β, (3) (ii) For any [u, z, w ] weak solution of problem (S γ,β φ,ψ ), φ V,p () ψ V,p ( ) satisfying (3), any [u 2, z 2, w 2 ] weak solution of problem (S γ,β φ 2,ψ 2 ), φ 2 V,p () ψ 2 V,p ( ) satisfying (3), we have that (z z 2 ) + + (w w 2 ) + (ψ ψ 2 ) + + (φ φ 2 ) +. In the case R γ,β = R+ γ,β, that is, when γ(r) = β(r) = 0 for any r R, existence uniqueness of weak solutions are also obtained. 8

9 Theorem 3.7 For any φ V,p () ψ V,p ( ) with φ + ψ = 0, (4) there exists a unique (up to a constant) weak solution u W,p () of the problem div a(x, Du) = φ in a(x, Du) η = ψ on in the sense that for all v W,p (). a(x, Du) Dv = ψv + φv, In the line of Proposition C (iv) of [9] given for the Laplacian operator, as a consequence of Theorem 3.6 we have the following result. Corollary 3.8 a is smooth if only if for any φ V,p () there exists T (φ) V,p ( ) such that the weak solution u of div a(x, Du) = φ in is a weak solution of u = 0 div a(x, Du) = φ a(x, Du) η = T (φ) on, in on. Moreover, the map T : V,p () V.p ( ) satisfies (T (φ ) T (φ 2 )) + (φ φ 2 ) +, for all φ, φ 2 V,p (). In the case ψ = 0 we have the following result without imposing any condition on γ, in the same line to the one obtained by Bénilan, Crall Sack in [9] for the Laplacian operator L -data. Theorem 3.9 Assume D(β) = R or a smooth. Let R γ,β < R+ γ,β. (i) For any φ V,p () such that φ R γ,β, there exists a weak solution [u, z, w] of problem (S γ,β φ,0 ), with z << φ. (ii) For any [u, z, w ] weak solution of problem (S γ,β φ ), φ,0 V,p (), φ R γ,β, any [u 2, z 2, w 2 ] weak solution of problem (S γ,β φ ), φ 2,0 2 V,p (), φ 2 R γ,β, we have that (z z 2 ) + + (w w 2 ) + (φ φ 2 ) +. For Dirichlet boundary condition we have the following result. 9

10 Theorem 3.0 Assume D(β) = {0}. For any φ V,p (), there exists a unique [u, z] = [u, z ] W,p 0 () V,p (), z γ(u) a.e. in, such that a(x, Du) Dv + zv = φv, for all v W,p 0 (). Moreover, if φ, φ 2 V,p (), then (z z φ2,ψ 2 ) + Let us now state the existence results of entropy solutions for data in L. (φ φ 2 ) +. (5) Theorem 3. Assume D(γ) = R, D(β) = R or a smooth. Let also assume that, if [0, + [ D(β), lim k + γ0 (k) = + lim k + β0 (k) = +, (6) if ], 0] D(β), Then, lim k γ0 (k) = lim k β0 (k) =. (7) (i) for any φ L () ψ L ( ), there exists an entropy solution [u, z, w] of problem (S γ,β ). (ii) For any [u, z, w ] entropy solution of problem (S γ,β φ,ψ ), φ L (), ψ L ( ), any [u 2, z 2, w 2 ] entropy solution of problem (S γ,β φ 2,ψ 2 ), φ 2 L (), ψ 2 L ( ), we have that (z z 2 ) + + (w w 2 ) + (ψ ψ 2 ) + + (φ φ 2 ) +. Taking into account Theorem 3. Corollary 3.8 we have the following result. Corollary 3.2 a is smooth if only if for any φ L () there exists T (φ) L ( ) such that the entropy solution u of div a(x, Du) = φ in is an entropy solution of u = 0 div a(x, Du) = φ a(x, Du) η = T (φ) on, in on. Moreover, the map T : L () L ( ) satisfies (T (φ ) T (φ 2 )) + (φ φ 2 ) +, for all φ, φ 2 L (), T ( V,p () ) V,p ( ). In the homogeneous case without any condition on γ we also obtain the following result. 0

11 Theorem 3.3 Assume D(β) = R or a is smooth. Let also assume that, if [0, + [ D(γ) D(β), the assumption (6) holds,, if ], 0] D(γ) D(β) the assumption (7) holds. Then, (i) for any φ L (), there exists an entropy solution [u, z, w] of problem (S γ,β φ,0 ), with z << φ. (ii) For any [u, z, w ] entropy solution of problem (S γ,β φ ), φ,0 L (), any [u 2, z 2, w 2 ] entropy solution of problem (S γ,β φ ), φ 2,0 2 L (), we have that (z z 2 ) + + (w w 2 ) + (φ φ 2 ) +. As we mention in Remark 5.9, different conditions to (6) (7) can be used in order to get Theorems We also obtain the following result given by Bénilan et al. in [5] for Dirichlet boundary condition. Theorem 3.4 Assume D(β) = {0}. For any φ L (), there exists a unique entropy solution [u, z] of div a(x, Du) + γ(u) φ in u = 0 on, in the sense given by Bénilan et al. in [5] (see () in the Introduction). 4 Proof of the uniqueness result This section deals with the proof of the uniqueness result Theorem 3.5. We firstly need the following lemma. Lemma 4. Let [u, z, w] be an entropy solution of problem (S γ,β ). Then, for all h > 0, λ Du p k ψ + k φ. {h< u <h+k} { u h} { u h} Proof. Taking T h (u) as test function in (2), we have a(x, Du) DT k (u T h (u)) + zt k (u T h (u)) + ψt k (u T h (u)) + φt k (u T h (u)). wt k (u T h (u)) Now, using (H ) the positivity of the second third terms, it follows that λ Du p k ψ + k φ. {h< u <h+k} { u h} { u h}

12 Proof of Theorem 3.5. Let [u, z, w ] [u 2, z 2, w 2 ] be entropy solutions of problem (S γ,β ). For every h > 0, we have that a(x, Du ) DT k (u T h (u 2 )) + z T k (u T h (u 2 )) + w T k (u T h (u 2 )) ψt k (u T h (u 2 )) + φt k (u T h (u 2 )) a(x, Du 2 ) DT k (u 2 T h (u )) + z 2 T k (u 2 T h (u )) + w 2 T k (u 2 T h (u )) ψt k (u 2 T h (u )) + φt k (u 2 T h (u )) Adding both inequalities taking limits when h goes to, on account of the monotonicity of γ β, if I h,k := a(x, Du ) DT k (u T h (u 2 )) + a(x, Du 2 ) DT k (u 2 T h (u )), we get lim sup I h,k (z z 2 )T k (u u 2 ) (w w 2 )T k (u u 2 ) 0. (8) h Let us see that To prove this, we split where Ih,k 2 := Ih,k 3 := Ih,k := { u <h, u 2 h} { u h, u 2 <h} { u <h, u 2 <h} lim inf h I h,k 0 for any k. (9) I h,k = I h,k + I 2 h,k + I 3 h,k + I 4 h,k, (a(x, Du ) a(x, Du 2 )) DT k (u u 2 ), a(x, Du ) DT k (u h sign(u 2 )) + { u <h, u 2 h} a(x, Du ) DT k (u u 2 ) + Ih,k 4 := Combining the above estimates we get { u h, u 2 <h} { u h, u 2 h} { u <h, u 2 h} a(x, Du 2 ) DT k (u 2 u ), { u h, u 2 <h} a(x, Du ) DT k (u u 2 ) a(x, Du ) DT k (u h sign(u 2 )) + a(x, Du 2 ) DT k (u 2 h sign(u )) 0. { u h, u 2 h} a(x, Du 2 ) DT k (u 2 u ) a(x, Du 2 ) DT k (u 2 h sign(u )) I h,k I h,k + L h,k + L 2 h,k, (0) 2

13 where L h,k := L 2 h,k := { u <h, u 2 h} { u h, u 2 <h} a(x, Du 2 ) DT k (u 2 u ), a(x, Du ) DT k (u u 2 ) Ih,k is non negative non decreasing in h. Now, if we set we have C(h, k) := {h < u < k + h} {h k < u 2 < h}, L 2 h,k C(h,k) Then, by Hölder s inequality, we get { u u 2 <k, u h, u 2 <h} a(x, Du ) Du + a(x, Du ) (Du Du 2 ) C(h,k) a(x, Du ) Du 2. ( ) /p (( ) /p ( ) /p L 2 h,k a(x, Du ) p Du p + Du 2 ). p C(h,k) C(h,k) C(h,k) Now, by (H 2 ), ( ) /p ( ( ) p a(x, Du ) p σ p θ(x) + Du p C(h,k) C(h,k) σ2 p ( θ p p + {h< u <k+h} ) /p Du p. ) /p On the other h, by Lemma 4., we obtain ( Du p k ψ + λ { u h} {h< u <k+h} {h k< u 2 <h} ( Du 2 p k ψ + λ { u 2 h k} Then, since φ L (), ψ L ( ) having in mind that since u i T,p tr (), we obtain that lim meas{x : u i(x) r} = 0 r + lim meas{x : u i(x) r} = 0, r + lim h L2 h,k = 0. { u h} φ { u 2 h k} Similarly, lim h L h,k = 0. Therefore by (0), (9) holds. Now, from (9), (8) (0), we have that lim (a(x, Du ) a(x, Du 2 )) DT k (u u 2 ) = 0. h + { u <h, u 2 <h} 3 ) φ ).

14 Therefore, for any h > 0, DT h (u ) = DT h (u 2 ) a.e. in. Consequently, there exists a constant c such that u u 2 = c a.e. in. Moreover, by (8) (9), we have (z z 2 )T k (u u 2 ) + from where it follows that (w w 2 )T k (u u 2 ) = 0 k > 0, () (w w 2 )χ {u u 2 0} = 0 a.e. in, (z z 2 )χ {u u 2 0} = 0 a.e. in. Then, if c 0 it follows that w = w 2, z = z 2. In order to see that z = z 2 in the case c = 0, we take T h (u ) ϕ T h (u ) + ϕ, ϕ D(), as test functions in (2) for the solution [u, z, w ] [u, z 2, w 2 ], respectively, adding these inequalities letting h go to +, if k > ϕ, we get where J h,k = = { u >h} lim J h,k + (z z 2 )ϕ 0, h a(x, Du ) [DT k (u T h (u ) + ϕ) + DT k (u T h (u ) ϕ)] a(x, Du ) [DT k (u T h (u ) + ϕ) + DT k (u T h (u ) ϕ)]. Then, using Hölder s inequality Lemma 4., we obtain that Hence Similarly, Therefore z = z 2. lim J h,k = 0. h z ϕ z 2 ϕ If c 0, following the arguments of Lemma 3.5 of [6], we have that z = z 2 is constant. In fact, let j(r) = r 0 γ0 (s)ds, therefore, γ = j, the subdifferential of j. Now, z (x) γ(u (x)) γ(u (x) + c) a.e. x, consequently, j(u (x) + c) j(u (x)) = cz (x) a.e. in. Moreover, if γ(r) is bounded, j is Lipschitz continuous, j(t k (u ) + c), j(t k (u )) W,p () (j(t k (u ) + c) j(t k (u ))) = 0 a.e. in. The above identity is obvious when u k, in the case u < k, we have (j(u + c) j(u )) = 0. Therefore j(t k (u ) + c) j(t k (u )) is constant (this constant, in fact, does not depend on k) consequently cz is constant. As c 0, z is constant. In the case γ is not bounded, we work, again as in Lemma 3.5 of [6], truncating γ. Finally, in order to see that w = w 2, we use the fact that we can take as test function in (2), for the corresponding (S γ,β ), v = T h(u i ) ± ϕ, for any ϕ W,p () L (). Then, since u = u 2 + c z = z 2, we get w ϕ = w 2 ϕ. Therefore w = w 2. z 2 ϕ. z ϕ. 4

15 5 Proofs of the existence results In this section we give the proofs of the existence results. In order to get the existence of week solutions, the main idea is to consider the approximated problem div a(x, Du) + γ m,n (u) φ m,n in (S γm,n,βm,n φ m,n,ψ m,n ) a(x, Du) η + β m,n (u) ψ m,n on, where γ m,n β m,n are approximations of γ β given by γ m,n (r) = γ(r) + m r+ n r β m,n (r) = β(r) + m r+ n r respectively, m, n N, we are approximating φ ψ by φ m,n = sup{inf{m, φ}, n} ψ m,n = sup{inf{m, ψ}, n} respectively, m, n N. For these approximated problems we obtain existence of weak solutions with appropriated estimates monotone properties, which allow us to pass to the limit. 5. Approximated problems Proposition 5. Assume D(γ) = D(β) = R. Let m, n N, m n. Then, the following hold. (i) For φ L () ψ L ( ), there exist u = u,m,n W,p () L (), z = z,m,n L (), z(x) γ(u(x)) a.e. in, w = w,m,n L ( ), w(x) β(u(x)) a.e. in, such that [u, z, w] is a weak solution of (S γm,n,βm,n ). Moreover, if M := φ + ψ, nm u nm, there exists c(, N, p) > 0 such that γ 0 ( nm) z γ 0 (nm), Du p c(, N, p) ( ) L p () φ V,p () + ψ V,p ( ). λ (ii) If m m 2 n 2 n, φ, φ 2 L (), ψ, ψ 2 L ( ) then (z,m,n z φ2,ψ 2,m 2,n 2 ) + + (w,m,n w φ2,ψ 2,m 2,n 2 ) + (ψ ψ 2 ) + + (φ φ 2 ) +. 5

16 Proof. Observe that m s+ n s = m s + ( m ) n s = ( m n) s + + n s. Let us take c r > sup{nm, γ r (nm), γ r ( nm), β r (nm), β r ( nm)}, where γ r β r are the Yosida approximations of γ β, respectively. For r N, it is easy to see that the operator B r : W,p () (W,p ()) defined by B r u, v = a(x, Du) Dv + T cr (γ r (u))v + u p 2 uv r + T cr (u + )v T cr (u )v + T cr (β r (u))v m n + T cr (u + )v T cr (u )v ψv φv, m n is bounded, coercive, monotone hemicontinuous. Then, by a classical result of Browder ([2]), there exists u r = u,m,n,r W,p (), such that a(x, Du r ) Dv + T cr (γ r (u r ))v + u r p 2 u r v r + T cr ((u r ) + )v T cr ((u r ) )v m n + T cr (β r (u r ))v + T cr ((u r ) + )v (2) T cr ((u r ) )v m n = ψv + φv, for all v W,p (). Taking v = T k ((u r mm) + ) in (2), misleading non negative terms, dividing by k, taking limits as k goes to 0, we get T cr (u r )sign + (u r mm) + T cr (u r )sign + (u r mm) m m ψ sign + (u r mm) + φ sign + (u r mm). Consequently (T cr (u r ) mm)sign + (u r mm) + (T cr (u r ) mm)sign + (u r mm) (mψ mm)sign + (u r mm) + (mφ mm)sign + (u r mm) 0. Therefore, since m n, u r (x) nm a.e. in. Similarly, taking v = T k ((u r + nm) ) in (2), we get u r (x) nm a.e. in. 6

17 Consequently, (2) yields for all v W,p (). a(x, Du r ) Dv + γ r (u r )v + u r p 2 u r v + r m + β r (u r )v + u + r v u r v = m n u r nm, (3) u + r v n ψv + φv, Taking v = T k ((u r ) + ) in (4), disregarding some positive terms, dividing by k letting k go to we get that, similarly, taking T k ((u r ) ) we get u r v (4) u + r + γ r (u r ) + + β r (u r ) + φ + + ψ +, (5) m u r + γ r (u r ) + β r (u r ) φ + ψ. (6) n Taking v = u r meas( ) u r as test function in (4) having in mind that = ( γ r (u r ) u r (β r (u r ) β r ( ( ) β r (u r ) u r u r meas( ) meas( ) u r )) (u r ) ( u r = (γ r (u r ) γ r meas( ) ( γ r (u r ) u r meas( ) ( γ r (u r ) u r meas( ) working similarly with the other terms, we get ( λ Du r p φ u r meas( ) m + n ( γ r (u r ) meas() u + r u r ( meas() ( meas() meas() meas( ) meas() meas() u r ) 0; u r )) (u r u r ) u r ) ) ( ) u r + ψ u r u r meas( ) ) u r u r meas( ) u r u r 7 meas( ) u r ) ) u r. meas( ) ) u r meas()

18 Now, by Poincaré s inequality the trace Theorem, there exists c = c (, N, p) > 0 such that ( ) φ u r u r c φ V meas( ),p () Du r L p (), ( ) ψ u r u r c ψ V,p ( ) Du r Lp (). meas( ) On the other h, by (5) (6), ( γ r (u r ) u r meas() ( u + r u r m meas() meas( ) + ( u r u r n meas() meas( ) ( ) 2 ψ + φ u r meas() ) u r meas( ) ) u r meas( ) Moreover, applying again the generalized Poincaré inequality, there exists c 2 = c 2 (, N, p) > 0 such that u r u r meas() meas( ) ( ) meas() u r u r + p meas() Lp u r u r () meas( ) Lp () c 2 Du r L p (). Therefore, there exists c 3 = c 3 (, N, p) > 0, such that u r ) u r. Du r p L p () c 3 ( ) φ V,p () + ψ V,p ( ). (7) λ As a consequence of (3) (7) we can suppose that there exists a subsequence, still denoted u r, such that converges weakly in W,p () to u W,p (), u r u r converges in L q () a.e. in to u, for any q, with u r converges in L p ( ) a.e. to u, nm u nm. (8) Taking into account (8), we get that γ r (u r ) is uniformly bounded. Consequently, we can assume that γ r (u r ) z L () weakly, moreover γ 0 ( nm) z γ 0 (nm). Since u r u in L (), applying [9, Lemma G], it follows that z(x) γ(u(x)) a.e. on. 8

19 On the other h, since β r (u r ) is also uniformly bounded, we can assume that β r (u r ) w L ( ) weakly. Again, applying [9, Lemma G], it follows that w(x) β(u(x)) a.e. in. Let us see now that {Du r } converges in measure to Du. We follow the technique used in [0] (see also [3]). Since Du r converges to Du weakly in L p (), it is enough to show that {Du r } is a Cauchy sequence in measure. Let t ɛ > 0. For some A >, we set C(x, A, t) := inf{(a(x, ξ) a(x, η)) (ξ η) : ξ A, η A, ξ η t }. Having in mind that the function ψ a(x, ψ) is continuous for almost all x the set {(ξ, η) : ξ A, η A, ξ η t } is compact, the infimum in the definition of C(x, A, t) is a minimum. Hence, by (H 3 ), it follows that Now, for r, s N any k > 0, the following inclusion holds where C(x, A, t) > 0 for almost all x. (9) { Du r Du s > t} { Du r A} { Du s A} { u r u s k 2 } {C(x, A, t) k} G, G = { u r u s k 2, C(x, A, t) k, Du r A, Du s A, Du r Du s > t}. Since the sequence Du r is bounded in L p () we can choose A large enough in order to have (20) meas({ Du r A} { Du s A}) ɛ 4 for all r, s N. (2) By (9), we can choose k small enough in order to have meas({c(x, A, t) k}) ɛ 4. (22) On the other h, if we use T k (u r u s ) T k (u r u s ) as test functions in (4) for u r respectively, we obtain a(x, Du r ) DT k (u r u s ) + γ r (u r )T k (u r u s ) + u r p 2 u r T k (u r u s ) + u + r T k (u r u s ) u r T k (u r u s ) r m n + β r (u r )T k (u r u s ) + u + r T k (u r u s ) u r T k (u r u s ) m n = ψt k (u r u s ) + φt k (u r u s ), a(x, Du s ) DT k (u r u s ) γ s (u s )T k (u r u s ) u s p 2 u s T k (u r u s ) u + s T k (u r u s ) + u s T k (u r u s ) s m n β s (u s )T k (u r u s ) u + s T k (u r u s ) + u s T k (u r u s ) m n = ψt k (u r u s ) φt k (u r u s ). u s (23) (24) 9

20 Adding (23) (24) disregarding some positive terms, we get (a(x, Du r ) a(x, Du s )) DT k (u r u s ) (γ r (u r ) γ s (u s )) T k (u r u s ) ( r u r p 2 u r ) s u s p 2 u s T k (u r u s ) Consequently, there exists a constant ˆM independent of r s such that (a(x, Du r ) a(x, Du s )) DT k (u r u s ) k ˆM. (β r (u r ) β s (u s )) T k (u r u s ). Hence meas(g) for k small enough. meas({ u r u s k 2, (a(x, Du r ) a(x, Du s )) D(u r u s ) k}) = k k { u r u s <k 2 } (a(x, Du r ) a(x, Du s )) D(u r u s ) (a(x, Du r ) a(x, Du s )) DT k 2(u r u s ) k k2 ˆM ɛ 4 Since A k have been already chosen, if r 0 is large enough we have for r, s r 0 the estimate meas({ u r u s k 2 }) ɛ 4. From here, using (20), (2), (22) (25), we can conclude that meas({ Du r Du s t}) ɛ for r, s r 0. From here, up to extraction of a subsequence, we also have a(., Du r ) converges in measure a.e. to a(., Du). Now, by (H 2 ) (7), a(., Du r ) converges weakly in L p () N to a(., Du). Finally, letting r + in (4), we prove (i). In order to prove (ii), we write u,r = u,m,n,r u 2,r = u φ2,ψ 2,m 2,n 2,r. Taking T k ((u,r u 2,r ) + ), with r large enough, as test function in (4) for u,r, m = m n = n, we get a(x, Du,r ) DT k ((u,r u 2,r ) + ) + γ r (u,r )T k ((u,r u 2,r ) + ) + r + u,r p 2 u,r T k ((u,r u 2,r ) + ) + m β r (u,r )T k ((u,r u 2,r ) + ) + = m ψ T k ((u,r u 2,r ) + ) + u +,r T k((u,r u 2,r ) + ) n u +,r T k((u,r u 2,r ) + ) n φ T k ((u,r u 2,r ) + ), taking T k (u,r u 2,r ) + as test function in (4) for u 2,r, m = m 2 n = n 2, we get (25) u,r T k((u,r u 2,r ) + ) u,r T k((u,r u 2,r ) + ) 20

21 r a(x, Du 2,r ) DT k ((u,r u 2,r ) + ) γ r (u 2,r )T k ((u,r u 2,r ) + ) u 2,r p 2 u 2,r T k ((u,r u 2,r ) + ) u + 2,r m T k((u,r u 2,r ) + ) + u 2,r 2 n T k((u,r u 2,r ) + ) 2 β r (u 2,r )T k ((u,r u 2,r ) + ) u + 2,r m T k((u,r u 2,r ) + ) + u 2,r 2 n T k((u,r u 2,r ) + ) 2 = ψ 2 T k ((u,r u 2,r ) + ) φ 2 T k ((u,r u 2,r ) + ). Adding these two inequalities, misleading some non negative terms, dividing by k, letting k 0, we get (γ r (u,r ) γ r (u 2,r )) + + (β r (u,r ) β r (u 2,r )) + (26) (ψ ψ 2 ) + + (φ φ 2 ) +. Therefore, taking into account the above convergence, (ii) is obtained. In the homogeneous case without any condition on γ we obtain the following result. Proposition 5.2 Assume D(β) = R. Let m, n N, m n. Then, the following hold. (i) For φ L (), there exist u = u φ,m,n W,p () L (), z = z φ,m,n L (), z(x) γ(u(x)) a.e. in, w = w φ,m,n L ( ), w(x) β(u(x)) a.e. in, such that [u, z, w] is a weak solution of problem (S γm,n,βm,n φ,0 ), z << φ. (ii) If m m 2 n 2 n, φ, φ 2 L (), then (z φ,m,n z φ2,m 2,n 2 ) + + (w φ,m,n w φ2,m 2,n 2 ) + (φ φ 2 ) +. Proof. Following the proof of Proposition 5. there exists u r = u φ,m,n,r W,p (), such that for all v W,p (). + u r n φ, a(x, Du r ) Dv + u r p 2 u r v r γ r (u r )v + u + r (u r v) u r v m n u + r v u r v = φv, n + β r (u r )v + m We can finish the proof as in Propositions 5. if we prove that γ r (u r ) is weakly convergent in L (). Taking v = q(γ r (u r )), q P 0, as test function in (27) we have that, after misleading non negative terms, γ r (u r )q(γ r (u r )) φq(γ r (u r )), which implies, γ r (u r ) << φ. In particular, see Proposition 2., γ r (u r ) φ γ r (u r ) z L () weakly in L (), with z << φ. (27) 2

22 Remark 5.3 Observe that if D(β) = {0}, for any γ, if we rewrite the proof of Proposition 5.2, using W,p 0 () instead of W,p (), we find u = u φ,m,n W,p 0 () L (), z = z φ,m,n L (), z(x) γ(u(x)) a.e. in, such that a(x, Du) Dv + zv + u + v u v = φv, m n for all v W,p 0 (). Moreover, if m m 2 n 2 n, φ, φ 2 L (), then (z,m,n z φ2,ψ 2,m 2,n 2 ) + (φ φ 2 ) +. Proposition 5.4 Assume D(γ) = R a smooth. Let m, n N, m n. Then, the following hold. (i) For φ L () ψ L ( ), there exist u = u,m,n W,p () L (), z = z,m,n L (), z(x) γ(u(x)) a.e. in, w = w,m,n L ( ), w(x) β(u(x)) a.e. in, such that [u, z, w] is a weak solution of (S γm,n,βm,n ). Moreover, there exists c(, N, p) > 0 such that Du p c(, N, p) ( ) L p () φ V λ,p () + ψ V,p ( ). (ii) If m m 2 n 2 n, φ, φ 2 L (), ψ, ψ 2 L ( ) then (z,m,n z φ2,ψ 2,m 2,n 2 ) + + (w,m,n w φ2,ψ 2,m 2,n 2 ) + (ψ ψ 2 ) + + (φ φ 2 ) +. Proof. Applying Proposition 5. to β r, the Yosida approximation of β, there exists u r = u,m,n,r W,p () L () z r = z,m,n,r L (), z r γ(u r ) a.e. in, such that a(x, Du r ) Dv + z r v + β r (u r )v + m u + r v u r v + u + r v n m n = ψv + φv, for all v W,p (). Moreover, u r is uniformly bounded by nm, M := φ + ψ, γ 0 ( nm) z r γ 0 (nm), z r ± + w r ± ψ ± + φ ±. Let now û L () ẑ γ(û), ẑ L (), be such that û is solution of the Dirichlet problem (see Remark 5.3) div a(x, Dû) + ẑ + mû+ nû = φ in u r v (28) û = 0 on. 22

23 Since a is smooth, there exists ˆψ L ( ) such that a(x, Dû) Dv + ẑv + û + v m n û v = ˆψv + φv, (29) for any v W,p () L (). Taking v = q(β r (u r û)), q P 0, as test function in (28), q(β r (u r û)) as test function in (29), adding both equalities we get, after misleading non negative terms, that β r (u r )q(β r (u r )) (ψ ˆψ)q(β r (u r )), i.e., β r (u r ) << ψ ˆψ, which implies (see Proposition 2.) that β r (u r ) w L ( ) weakly in L ( ). Now, arguing as in the proof of Proposition 5., we obtain (i). To prove (ii), Proposition 5. implies, denoting u i,r = u φi,ψ i,m i,n i,r z i,r = z φi,ψ i,m i,n i,r, i =, 2, (z,r z 2,r ) + + (β r (u,r ) β r (u 2,r )) + (ψ ψ 2 ) + + (φ φ 2 ) +. (30) Taking limits in (30) when r goes to +, (ii) holds. In the case ψ = 0, we have the following result. Proposition 5.5 Assume a smooth. Let m, n N, m n. Then, the following hold. (i) For φ L (), there exist u = u φ,m,n W,p () L (), z = z φ,m,n L (), z(x) γ(u(x)) a.e. in, w = w φ,m,n L ( ), w(x) β(u(x)) a.e. in, such that [u, z, w] is a weak solution of problem (S γm,n,βm,n φ,0 ), with z << φ. (ii) If m m 2 n 2 n, φ, φ 2 L (), then (z φ,m,n z φ2,m 2,n 2 ) + + (w φ,m,n w φ2,m 2,n 2 ) + (φ φ 2 ) Existence of weak solutions Proof of Theorem 3.6. We approximate φ ψ by φ m,n = sup{inf{m, φ}, n} ψ m,n = sup{inf{m, ψ}, n}, respectively. We have, φ m,n L (), ψ m,n L ( ), are non decreasing in m, non increasing in n, φ m,n L p () φ L p () ψ m,n L p ( ) ψ L p ( ). Then, if m n, by Propositions 5. 23

24 or 5.4, there exist u m,n W,p () L (), z m,n L (), z m,n (x) γ(u m,n (x)) a.e. in w m,n L ( ), w m,n (x) β(u m,n (x)) a.e. on, such that a(x, Du m,n ) Dv + z m,n v + w m,n v + m for any v W,p (). Moreover, u + m,nv u n m,nv + u + m m,nv n = ψ m,n v + φ m,n v, u m,nv (3) z m,n ± + w m,n ± φ ± + ψ ± (32) Du m,n p c(, N, p) ( ) L p () φ V λ,p () + ψ V,p ( ). (33) Fixed m N, by Propositions 5. or 5.4 (ii), {z m,n } n=m {w m,n } n=m are monotone non increasing. Then, by (32) the Monotone Convergence Theorem, there exists ẑ m L (), ŵ m L ( ) a subsequence n(m), such that z m,n(m) ẑ m m w m,n(m) ŵ m m. Thanks again to Proposition 5. or 5.4 (ii), ẑ m ŵ m are non decreasing in m. Now, by (32), we have that ẑ m ŵ m are bounded. Using again the Monotone Convergence Theorem, there exist z L () w L ( ) such that ẑ m converges a.e. in L () to z Consequently, ŵ m converges a.e. in L ( ) to w. z m := z m,n(m) converges to z a.e. in L () (34) w m := w m,n(m) converges to w a.e. in L ( ). (35) If we set u m := u m,n(m), φ m := φ m,n(m) ψ m := ψ m,n(m), then we have + m a(x, Du m ) Dv + z m v + w m v u + mv n(m) u mv + m = ψ m v + u + mv n(m) φ m v, u mv (36) 24

25 for any v W,p (). As a consequence of (33), { } u m u m meas( ) m is bounded in W,p (). (37) Let us see that { meas( ) u m } : m N is a bounded sequence. (38) If (38) does not hold, then, extracting a subsequence if necessary, we can suppose that u m converges to + (or, respectively). Suppose first that u m converges to +. Hence, by (37) we have converges to + a.e. in, a.e. in. u m Moreover, since for m large enough u m ( u m ) ( u m + meas( ) by (37), we get {, similarly, { u m u m meas( ) } } m N m N u m ) = is bounded is bounded. In the case u m converges to, we similarly obtain that ( ) u m u m, meas( ) u m converges to a.e. in, a.e. in, { u + m } m N { u + m } m N are bounded. Therefore, we have z = sup{ran(γ)} (z = inf{ran(γ)}, respectively) w = sup{ran(β)} (w = inf{ran(β)}, respectively). Now, taking v = as test function in (36), we get u + m u m + u + m u m m n(m) m n(m) = φ m + ψ m z m w m, we get a contradiction with (3). Hence, (38) is true. By (37) (38), we have { u m W,p ()} m is bounded. Therefore, there exists a subsequence, that we denote equal, such that u m u weakly in W,p (), u m u in L p () a.e. in, u m u in L p ( ) a.e. in. 25

26 Moreover, arguing as in Proposition 5., it is not difficult to see that {Du m } is a Cauchy sequence in measure. Then, up to extraction of a subsequence, Du m converges to Du a.e. in. Consequently, we obtain that a(., Du m ) converges weakly in L p () N a.e. in to a(., Du). From these convergences, we finish the proof of existence. The proof of (ii) is a consequence of the existence result, Propositions 5. (ii) or 5.4 (ii), the uniqueness result. Remark 5.6 For positive data φ ψ, it is not necessary the assumption D(γ) = D(β) = R, that is, we can improve the above result in the following way. Assume [0, + [ D(γ) R + γ,β > 0. Let [0, + [ D(β) or a smooth. For any 0 φ V,p ) 0 ψ V,p ( ) with φ + ψ < R+ γ,β, there exists a weak solution of problem (S γ,β ). A similar result holds for non positive data. Proof of Theorem 3.7. Let us approximate φ by φ m = T m (φ) meas() α m ψ by ψ m = T m (ψ), where α m = T m(φ) + T m(ψ). Observe that By Proposition 5., there exist u m W,p () L () such that a(x, Du m ) Dv + u m v + u m v = m m for any v W,p (). lim α m = 0 (39) m + φ m + ψ m = 0. (40) ψ m v + φ m v, (4) Taking v = u m as test function in (4), using (39) the Poincaré inequality, it is easy to see that { } u m u m is bounded in W,p (). (42) meas( ) Let us also see that { meas( ) u m m } : m N is a bounded sequence. (43) If (43) does not hold, then, extracting a subsequence if necessary, we can suppose that u m converges to + (or, respectively). Suppose first that u m converges to +. Hence, as in the proof of Theorem 3.6, we have { } u m m N is bounded. Now, taking v = m in (4) using (40), it follows that u m = +, lim m + which is a contradiction. Similarly, we get a contradiction in the case u m converging to. Hence, (43) is true. By (42) (43), we have { u m W,p ()} m is bounded, we can finish as in the proof of Theorem

27 Remark 5.7 Taking into account the arguments used in Remark 3.4, we get that [u, z, w] in the above results (including also the case β = D) satisfies ( ) zv + wv φv + ψv + σ g L p () + Du p L p () Dv L p () for all v W,p (), for some c(, N, p) > 0. Du p c(, N, p) ( ) L p () φ V λ,p () + ψ V,p ( ), Taking β = D, γ(r) = 0 for all r R, a smooth in Theorem 3.6, by Remark 5.7, it follows Corollary 3.8. The proof of Theorem 3.9 follows in a similar way to the proof of Theorem 3.6 taking into account Propositions Finally, on account of Remark 5.3, it follows Theorem Existence of entropy solutions Proof of Theorem 3.. Observe that, under the assumptions of the theorem, we have R γ,β = R. We divide the proof in several steps. Step. Let us approximate φ by φ m := T m (φ) ψ by ψ m := T m (ψ). Then, by Theorem 3.6, there exist u m W,p (), z m V,p (), z m (x) γ(u m (x)) a.e. in, w m V,p ( ), w m (x) β(u m (x)) a.e. on, such that for any v W,p (). Moreover, Consequently a(x, Du m ) Dv + z m v + w m v = ψ m v + φ m v, (44) z m ± + w m ± ψ m ± + φ ± m (45) z n z m + w n w m ψ n ψ m + φ n φ m. z m z in L () w m w in L ( ). Taking v = T k (u m ) in (44), we obtain λ DT k (u m ) p k ( φ + ψ ), k N. (47) By (47), we have {T k (u m )} is bounded in W,p (). Then, we can suppose that there exists σ k W,p () such that T k (u m ) converges to σ k weakly in W,p (), T k (u m ) converges to σ k in L p () a.e. in (46) 27

28 T k (u m ) converges to σ k in L p ( ) a.e. in. Step 2. Let us see that u m converges almost every where in. If D(β) is bounded from above by r, using the Poincaré inequality (47), meas{x : σ + k (x) = k} C lim inf k p m C k p ( T k ((u m ) + ) + (σ + k )p k p ( ( ( φ + ψ r meas( ) + λ where p = Np N p C is independent of k m. (T k ((u m ) + )) p lim inf m k p ) ) /p p DT k ((u m ) + ) p ) ) /p p k k > 0, If D(β) is unbounded from above, then, we are supposing lim k + γ 0 (k) = +. Therefore, for k > 0 large enough (in order to have γ 0 (k) > 0), by (45) we have meas{x : σ + k (x) = k} = γ 0 + (σk (x)) {x : σ + k (x)=k}} γ 0 (k) γ 0 (k) lim inf γ 0 (T k ((u m ) + )) m γ 0 (k) ( φ + ψ ). Consequently, in any case, there exists g(k) > 0, lim k + g(k) = 0, such that meas{x : σ + k (x) = k} g(k) k > 0. (48) Similarly, if D(β) is bounded from below or assumption (7) holds, we can prove that there exists g(k) as above such that meas{x : σ k (x) = k} g(k) k > 0. (49) Note that we have proved (48) (49) in any case. lim k + g(k) = 0, such that Consequently, there exists g(k) > 0 with meas{x : σ k (x) = k} g(k) k > 0. Therefore, if we define u(x) = σ k (x) on {x : σ k (x) < k}, then u m converges to u a.e. in, (50) we have that T k (u m ) converges weakly in W,p () to T k (u), T k (u m ) converges in L p () a.e. in to T k (u) T k (u m ) converges in L p ( ) a.e. in to T k (u). Consequently, u T,p (). 28

29 Arguing as in Proposition 5., it is not difficult to see that {Du m } is a Cauchy sequence in measure. Similarly, we can prove that DT k (u m ) converges in measure to DT k (u). Then, up to extraction of a subsequence, Du m converges to Du a.e. in. Consequently, we obtain that a(., DT k (u m )) converges weakly in L p () N a.e. in to a(., DT k (u)). (5) Step 3. Let us see now that u T,p tr (). On the one h we have that u m u a.e. in. On the other h, since DT k (u m ) is bounded in L p () DT k (u m ) DT k (u) in measure, it follows from [5, Lemma 6.] that DT k (u m ) DT k (u) in L (). Next, let us see that u m converges a.e. in. Let suppose first that D(β) is bounded from above by r, then, by (47), there exists a constant C 3 such that meas{x : σ + k (x) = k} σ + k k T k ((u m ) + ) lim inf r meas( ) m k k k > 0. If D(β) is unbounded from above, then, we are supposing lim k + β 0 (k) = +. Therefore, for k > 0 large enough (in order to have β 0 (k) > 0), by (45) we have meas{x : σ + k (x) = k} = β 0 + (σk (x)) {x : σ + k (x)=k}} β 0 (k) β 0 (k) lim inf β 0 (T k ((u m ) + )) m β 0 (k) ( φ + ψ ). We work similarly if D(β) is bounded from below or assumption (7) holds,, in any case, there exists ĝ(k) > 0, lim k + ĝ(k) = 0, such that meas{x : σ k (x) = k} ĝ(k) k > 0. Hence, if we define v(x) = T k (u)(x) on {x : T k (u)(x) < k}, then Consequently, u T,p tr (). u m converges to v a.e. in. (52) Since z m (x) γ(u m (x)) a.e. in w m (x) β(u m (x)) a.e. in, from (46), (50), (52) from the maximal monotonicity of γ β, we deduce that z(x) γ(u(x)) a.e. in w(x) β(u(x)) a.e. in. Step 4. Finally, let us prove that [u, z, w] is an entropy solution relative to D(β) of (S γ,β ). To do that, we introduce the class F of functions S C 2 (R) L (R) satisfying S(0) = 0, 0 S, S (s) = 0 for s large enough, S( s) = S(s), S (s) 0 for s 0. Let v W,p () L (), v(x) D(β) a.e. in, S F. Taking S(u m v) as test function in (44), we get a(x, Du m ) DS(u m v) + z m S(u m v) + w m S(u m v) (53) = ψ m S(u m v) + φ m S(u m v). 29

30 We can write the first term of (53) as a(x, Du m ) Du m S (u m v) a(x, Du m ) DvS (u m v). (54) Since u m u Du m Du a.e., Fatou s Lemma yields a(x, Du) DuS (u v) lim inf a(x, Du m ) Du m S (u m v). m The second term of (54) is estimated as follows. Let r := v + S. By (5) On the other h, a(x, DT r u m ) a(x, DT r u) weakly in L p (). (55) DvS (u m v) Dv L p (). Then, by the Dominated Convergence Theorem, we have DvS (u m v) DvS (u v) in L p () N. (56) Hence, by (55) (56), it follows that a(x, Du m ) DvS (u m v) = lim m a(x, Du) DvS (u v). Therefore, applying again the Dominated Convergence Theorem in the other terms of (53), we obtain a(x, Du) DS(u v) + zs(u v) + ws(u v) ψs(u v) + φs(u v). From here, to conclude, we only need to apply the technique used in the proof of [5, Lemma 3.2]. The proof of (ii) is a consequence of the existence result, Theorem 3.6 (ii), the uniqueness result. Theorem 3.3 Theorem 3.4 follows in a similar way taking into account Theorem 3.9 Theorem 3.0 respectively. Remark 5.8 In Theorem 3., if the data φ ψ are non negative (non positive, respectively), then assumption (7) ((6), respectively) is not necessary. That is, only assuming [0, + [ D(γ), [0, + [ D(β) or a smooth, assumption (6) if [0, + [ D(β), for any 0 φ L () 0 ψ L ( ), there exists an entropy solution of problem (S γ,β ). A similar result holds for non positive data. Remark 5.9 In Theorems , it is not difficult to see that (6) can be substituted by one of the following assumptions, (6 ) 0 < α, r 0 > 0 : γ 0 (r) r α r r 0, (6 ) 0 < α, r 0 > 0 : β 0 (r) r α r r 0 ; (7) can be substituted by one of the following assumptions, (7 ) 0 < α, r 0 > 0 : γ 0 (r) ( r) α r r 0, (7 ) 0 < α, r 0 > 0 : β 0 (r) ( r) α r r 0. 30

31 5.4 Some extensions Following the ideas developed in this work, it is possible to find a larger class of entropy solutions when β is only assumed to have closed domain. Definition 5.0 Let φ L () ψ L ( ). A triple of functions [u, z, w] T,p tr () L () L ( ) is an entropy solution relative to D(β) of problem (S γ,β ) if z(x) γ(u(x)) a.e. in, w(x) β(u(x)) a.e. in a(x, Du) DT k (u v) + zt k (u v) + wt k (u v) (57) ψt k (u v) + φt k (u v) k > 0, for all v L () W,p (), v(x) D(β) a.e. in. For this concept of solution we can prove the following result. Theorem 5. Assume D(β) is closed D(β) D(γ). Let also assume that if [0, + [ D(β) the assumption (6) holds, if ], 0] D(β) the assumption (7) holds. Then, (i) for any φ L () ψ L ( ) there exists an entropy solution [u, z, w] = [u, z, w ] relative to D(β) of problem (S γ,β ). Moreover, β 0 (inf D(β)) w β 0 (sup D(β)) z ± + w ± ψ ± + φ ±. (ii) Given φ, φ 2 L () ψ, ψ 2 L ( ), (z z φ2,ψ 2 ) + + (w 2 w φ2,ψ 2 ) + (ψ ψ 2 ) + + (φ φ 2 ) +. (iii) For any [u, z, w ] entropy solution relative to D(β) of problem (S γ,β φ,ψ ), φ L (), ψ L ( ), any [u 2, z 2, w 2 ] entropy solution relative to D(β) of problem (S γ,β φ 2,ψ 2 ), φ 2 L (), ψ 2 L ( ), we have that (z z 2 ) + (ψ ψ 2 ) + + (φ φ 2 ) +. Remark 5.2 In general, for this concept of solution we do not have uniqueness of w, as the following example shows. Let γ β be such that γ(0) = [0, ] β(0) =], 0] let 0 < φ < ψ 0. Then, for any w such that ψ w 0, [0, φ, w] is an entropy solution relative to D(β). Acknowledgements. This work has been performed during the visits of the first, third fourth authors to the Université de Picardie Jules Verne the visits of the second author to the Universitat de València. They thank these institutions for their support hospitality. The first, third fourth authors have been partially supported by PNPGC project, reference BFM , also, the first third authors by the EC through the RTN Programme Nonlinear Partial Differential Equations Describing Front propagation other Singular Phenomena, HPRN-CT

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