Uniqueness for Neumann problems for nonlinear elliptic equations

In the present paper we prove uniqueness results for solutions to a class of Neumann boundary value problems whose prototype is --div((1 + |$\nabla$u| 2) (p--2)/2 $\nabla$u) -- div(c(x)|u| p--2 u) = f in $\Omega$, (1 + |$\nabla$u| 2) (p--2)/2 $\nabla$u + c(x)|u| p--2 u $\times$ n = 0 on $\partial$$\Omega$,

Finally the coefficient c(x) belongs to an appropriate Lebesgue space which will be specified later.
The main difficulties in studying existence or uniqueness for this type of problems are due to the presence of a lower order term, the lower summability of the datum f and the boundary Neumann conditions. 1 The existence for Neumann boundary value problems with L 1 -data when c = 0 has been treated in various contests. In [3], [14], [19], [20] and [29] the existence of a distributional solution which belongs to a suitable Sobolev space and which has null mean value is proved. Nevertheless when p is close to 1, i.e. p ≤ 2 − 1/N, the distributional solution to problem (1.1) does not belong to a Sobolev space and in general is not a summable function; this implies that its mean value has not meaning and any existence result for distributional solution with null mean value cannot hold. This difficulty is overcome in [18] by considering solutions u which are not in L 1 (Ω), but for which Φ(u) is in L 1 (Ω), where Φ(t) = t 0 ds (1+|s|) α with appropriate α > 1. In [1] the case where both the datum f and the domain Ω are not smooth enough is studied and the existence and continuity with respect to the data of solutions whose median is equal to zero is proved with a natural process of approximations and symmetrization techniques. We recall that the median of u is defined by (1.2) med(u) = sup{t ∈ R : meas{u > t} > meas(Ω)/2} .
The existence for solutions having null median to problem (1.1) when c = 0 are proved in [8].
We explicitly remark that when the datum f has a lower summability, i.e. it is just an L 1 -function, one has to give a meaning to the notion of solution; such a question has been faced already in the case where Dirichlet boundary conditions are prescribed, by introducing different notion of solutions (cf. [7], [17], [27], [28] ). Such notion turn out to be equivalent, at least when the datum is an L 1 -function. In the present paper, when f ∈ L 1 (Ω), we refer to the so-called renormalized solutions (see [16], [27], [28]) whose precise definition is recalled in Section 2.
The main novelty of this article is to prove uniqueness (up to additive constants) results for renormalized solutions to problem (1.1) having null median and whose existence has been proved in [8].
To our knowledge uniqueness results for problem (1.1) are new even in the variational case, i.e. when f belongs to L (p * ) ′ (Ω) and the usual notion of weak solution is considered. When c(x) = 0 and f is an element of the dual space of the Sobolev space W 1,p (Ω), the existence and uniqueness (up to additive constants) of weak solutions to problem (1.1) is consequence of the classical theory of pseudo monotone operators (cfr. [25], [26]), while existence results for weak solutions to problem (1.1) when the lower order term appears have been proved in [8].
As pointed out we will prove different results according to the summability of f , i.e. f ∈ L (p * ) ′ (Ω) or f ∈ L 1 (Ω) and to the value of p, i.e. p ≤ 2 and p ≥ 2. As far as p is concerned such a difference is due to the principal part of the operator, which we consider. Actually we assume that the principal part −div(a(x, Du)) is not degenerate when p > 2, i.e. in the model case −div(a(x, ∇u)) = −div((1 + |∇u| 2 ) (p−2)/2 ∇u). But such an assumption is not required when p ≤ 2, that is for such values of p we prove uniqueness results for operators whose prototype is the so-called p-Laplace operator, −∆ p u = −div(|∇u| p−2 ∇u).
Let us explain the main ideas of our results. To this aim let us consider the simpler case of weak solutions and p = 2. When a Dirichlet boundary value problem is considered, following an idea of Artola [4] (see also [13,15]), denoted by u and v two solutions, one can use the test function T k (u − v) and obtain from which one can deduce that u = v a.e. in Ω. In contrast when we consider Neumann boundary conditions and two solutions u, v ∈ H 1 (Ω) having null median, by using T k (u − v) we can prove equality (1.3), but Poincaré-Wirtinger inequality does not allow to get Ω |sign(u − v)| 2 dx = 0 and therefore that u = v a.e. in Ω. However (1.3) and Poincaré-Wirtinger inequality, allow to deduce that u = v a.e. in Ω either u > v a.e. in Ω either u < v a.e. in Ω. Then we prove that u > v a.e. in Ω or u < v a.e. in Ω leads to a contradiction: it is done through a new test function where T k denote the truncate function at height k and Neumann problems have been studied by a different point of view in [21], [22], while existence or uniqueness results for Dirichlet boundary value problems for nonlinear elliptic equations with L 1 -data are treated in [11], [12] and was continued in various contributions, including [2], [6], [7], [9], [10], [16], [5] [17], [23], [24]; mixed boundary value problems have been also studied, for example, in [6], [19].
The paper is organized as follows. In Section 2 we detail the assumptions and we give the definition of a renormalized solution to (1.1). Section 3 is devoted to prove two uniqueness results for weak solutions when the datum is in the Lebesgue space L (p * ) ′ (Ω). In Section 4 we state our main results, Theorem 4.1, Theorem 4.2, where we prove the uniqueness of a renormalized solution to (1.1) when datum is a L 1 function.

Assumptions and definitions
Let us consider the following nonlinear elliptic Neumann problem where Ω is a bounded domain of R N , N ≥ 2, having finite Lebesgue measure and Lipschitz boundary, n is the outer unit normal to ∂Ω. We assume that p is a real number such that 1 < p < N. The function a : Ω × R N → R N is a Carathéodory function such that for almost every x ∈ Ω and for every ξ ∈ R N . Moreover a is strongly monotone, that is a constant β > 0 exists such that for almost every x ∈ Ω and for every ξ, η ∈ R N , ξ = η. We assume that Φ : Ω × R → R N is a Carathéodory function which satisfies the following "growth condition" for a.e. x ∈ Ω and for every s ∈ R. Moreover we assume that such function is locally Lipschitz continuous with respect to the second variable, that is for almost every x ∈ Ω, for every s, z ∈ R.
Finally we assume that the datum f is a measurable function in a Lebesgue space L r (Ω), 1 ≤ r ≤ +∞, which belongs to the dual space of the classical Sobolev space W 1,p (Ω) or is just an L 1 − function. Moreover it satisfies the compatibility condition As explained in the Introduction we deal with solutions whose median is equal to zero. Let us recall that if u is a measurable function, we denote the median of u by (2.9) med(u) = sup t ∈ R : meas{x ∈ Ω : u(x) > t} > meas(Ω) 2 .
Let us explicitely observe that if med(u) = 0 then In this case a Poincaré-Wirtinger inequality holds (see e.g. [30]): where C is a constant depending on p, N, Ω.
When the datum f is not an element of the dual space of the classical Sobolev space W 1,p (Ω), the classical notion of weak solution does not fit. We will refer to the notion of renormalized solution to (2.1) (see [16,28] for elliptic equations with Dirichlet boundary conditions) which we give below.
In the whole paper, T k , k ≥ 0, denotes the truncation at height k that is T k (s) = min(k, max(s, −k)), ∀s ∈ R. and if for every function h belonging to W 1,∞ (R) with compact support and for every ϕ ∈ L ∞ (Ω) ∩ W 1,p (Ω), we have

Remark 2.3.
A renormalized solution is not in general an L 1 loc (Ω)function and therefore it has not a distributional gradient. Condition (2.12) allows to define a generalized gradient of u according to Lemma 2.1 of [7], which asserts the existence of a unique measurable function v defined in Ω such that ∇T k (u) = χ {|u|<k} v a.e. in Ω, ∀k > 0. This function v is the generalized gradient of u and it is denoted by ∇u.
Equality (2.14) is formally obtained by using in (2.1) the test function ϕh(u) and by taking into account Neumann boundary conditions. Actually in a standard way one can check that every term in (2.14) is well-defined under the structural assumptions on the elliptic operator.
Let us recall Theorem 4.1 of [8]; under assumptions (2.2)-(2.8) there exists at least one renormalized solution u having null median of problem (2.1). Moreover any renormalized solution to (2.1) verifies the following proposition n Ω |Φ(x, u)| |∇T n (u)|dx = 0, where C is a positive constant depending only on m, f , Ω, α and Φ Sketch of the proof. For the proof of (2.15) see Remark 2.4 of [8].
The estimate (2.16) is related to the Boccardo-Gallouët estimates [11], and it is obtained through a usual process. Indeed since m > 0, we can use the renormalized formulation (2.14) with h = h n and ϕ = T 2n (u) 0 ds (1 + |s|) 1+m . In view of (2.13) and (2.15), the growth condition (2.5) on Φ allows one to pass to the limit as n → +∞ and to obtain (2.16). As far as (2.17) Using again that (2.17) and Hölder inequality allow one to deduce (2.18).

Uniqueness results for weak solution
In this section we assume that the right-hand side f is an element of the dual space L (p * ) ′ (Ω). In [8] an existence result for weak solution to problem (2.1) having null median has been proved. Such a weak solution u is a function such that In this section we assume a suitable growth condition on Φ, that is a bound on τ in (2.7) is assumed and the following assumption on the datum is made Now we prove two uniqueness results depending on the values of p: 2 , we have uniqueness results under the assumption that c belongs to L Proof of Theorem 3.1. Since for every fixed k > 0, T k (u−v) ∈ W 1,p (Ω), it can be used as test function in the equation satisfied by u and in the equation satisfied by v. Then by subtracting the two equations, we get We proceed by dividing the proof by steps.
Step 1. We prove that By the assumptions on the strong monotonicity on the operator (2.4) and the local Lipschitz condition on Φ (2.7) with τ which satisfies (3.2), we get The assumption on τ assures that the right-hand side of the previous inequality is finite. Moreover by Hölder inequality and assumption on τ , we obtain Hölder inequality assures that the integral in the right-hand side is finite. Since χ {0<|u−v|<k} tends to 0 a.e. in Ω as k goes to 0, this implies Moreover by Hölder inequality we get which implies (3.6) by (3.10).
Step 2. We prove that either a.e. in Ω.
Therefore, by Step 1, we deduce that and, up to a subsequence, by (3.13) for a suitable constant γ ∈ R, |γ| ≤ 1. On the other hand, we have Therefore, up to subsequence, by (3.13) we get This means that either a.e. in Ω.
Step 3. We prove that u < v, a.e. in Ω or u > v , a.e. in Ω can not occur.
We assume that (3.14) u > v , a.e. in Ω and we prove that this yields a contradiction. The same arguments prove that u < v a.e. in Ω can not be verified.
On the other hand, we have we conclude that This means that "u and v have the same sign". Now let us consider the test function in Ω, one can verify that this means med (w k,δ ) = 0 .
Therefore by Poincaré-Wirtinger inequality we deduce We now evaluate the gradient of w k,δ , in Ω.
Since u and v "have the same sign", then, for every fixed k > 0, it results then for fixed k > 0, we have in Ω .

Moreover we have also
and since |∇u|, |∇v| ∈ L p (Ω), we can apply Lebesgue dominated convergence Theorem, i.e. Since Step 1, we conclude that Now we can pass to the limit in (3.17) as δ → 0 first and then as k → 0 and we get We deduce that χ { u>0} = χ { v<0} a.e. in Ω; this yields a contradiction since we have proved that u and v have the same sign.
The same arguments yield that we can not have u < v a.e. in Ω. The conclusion follows.
Proof of Theorem 3.2. As in the previous proof we arrive to equality (3.5) and we divide the proof by 3 steps.
Step 2. We prove that either a.e. in Ω.
By Poincaré-Wirtinger inequality, we get Therefore, by Step 1. we deduce that and, up to a subsequence, for a suitable constant γ ∈ R, |γ| ≤ 1. On the other hand, we have Therefore, up to subsequence, we get This means that either u = v , a.e. in Ω or u < v , a.e. in Ω or u > v , a.e. in Ω .
Step 3. Arguing as in Step 3 of the previous theorem, we prove that the last two possibilities can not occur. Then conclusion follows.
Remark 3.5. In [8] we estabilished the existence of a weak solution when a(x, ξ) is replaced by a Leray-Lions operator a(x, r, ξ) which depends on x, r and ξ and verifies the standard conditions (see [25]). In the Dirichlet case and 1 < p ≤ 2 it is well known (see [13] [15]) that under suitable assumptions on a(x, r, ξ) the weak solution is unique.
In view of the proofs of Theorem 3.1 and Theorem 3.2 it is possible to obtain the uniqueness of the weak solution having null median of the problem If we assume that a(x, r, ξ) is a Carathéodory function which verifies and moreover a(x, r, ξ) satisfies a Lipschitz condition with respect to r | a(x, s, ξ) − a(x, r, ξ)| ≤ c 2 |s − r|(|ξ| p−1 + |s| p−1 + |r| p−1 + h(x)), for almost every x ∈ Ω, s ∈ R and for every ξ ∈ R N , then Theorem 3.1 and Theorem 3.2 hold true. Indeed the methods developped in [13] allow one to prove Step 1 in Theorem 3.1 namely and Step 1 in Theorem 3.2 namely In both cases the Step 2 and Step 3 remain unchanged.
Remark 3.6. In [8] and in the present paper we have chosen to deal with solutions to (2.1) with null median value instead of null mean value. As explained in Introduction this choice allows one to consider solution to (2.1) for f ∈ L 1 (Ω) even if the solution u does not belong to L 1 (Ω). When f ∈ L (p * ) ′ (Ω) a simply examination of the proof of [8] leads to the existence of solutions to (2.1) such that Ω u dx = 0.
Assuming that (2.2)-(2.7) are in force similar arguments to the one developped in the proof of Theorem 3.1 and Theorem 3.2 yield the uniqueness of solution to (2.1) having a null mean value. Let us explain briefly the case p = 2.
Step 1 remains unchanged so that if u and v are two solutions of (2.1) then we have Poincaré-Wirtinger inequality leads to

As in
Step 2 in Ω, u < v a.e. in Ω, u > v a.e. in Ω. We now show that u < v a.e. in Ω or u > v a.e. in Ω can not occur. The method is similar to Step 3 of the proof of Theorem 3.1: In the case u > v a.e. in Ω, the Lebesque dominated Theorem allows one to conclude that Ω sign(u) − 1 |Ω| Ω sign(u)dy 2 dx = 0 , and then u has a constant sign. Recalling that Ω u dx = Ω v dx = 0 gives a contradiction. Therefore u = v a.e. in Ω.

Uniqueness result for renormalized solution
In this section we prove the uniqueness of the renormalized solution to problem (2.1), when the following assumption on datum is made As in Section 3 we state two uniqueness theorems depending on the values of p:  Proof of Theorem 4.1. Let u and v be two renormalized solutions to (2.1). Let h n defined by (2.19). Since for any k > 0, h n (u)T k (u − v) = h n (v)T k (T 2n (u) − T 2n+k (u)) ∈ L ∞ (Ω) ∩ W 1,p (Ω), we can use h = h n (u) and ϕ = h n (v)T k (u − v) in (2.14) written in u, and we can use h = h n (v) and ϕ = h n (u)T k (u − v) in (2.14) written in v. By substracting the two equations, we get We proceed by dividing the proof into 3 steps.
Step 1. By passing to the limit in (4.5) first as n → +∞, then as k → 0 this step is to devoted to prove that (4.6) lim We first study the behaviour of the last two integrals in (4.5) as n goes to +∞ by showing (4.7) and by symmetry with respect to u and v (4.8) By assumption (2.3) and Hölder inequality we have Using (2.2) and (2.13) we deduce that Therefore recalling that a 0 ∈ L p ′ (Ω) we conclude that To prove that (4.7) holds it remains to control Ω h ′ n (u)h n (v)T k (u − v)Φ(x, v) · ∇u dx. By assumption (2.5) and Hölder inequality we have and Poincaré-Wirtinger inequality leads to where C > 0 is independent of n and k. It follows that Therefore (4.11) leads to and then (4.7) holds. We observe that (4.8) is obtained by analogous argument. Then by (4.5), (4.7), (4.8), using the assumptions on the strong monotonicity on the operator (2.4), the local Lipschitz condition on Φ (2.7) with τ which satisfies (4.2) and Young inequality we get where lim n ω k (n) = 0. We now prove that (4.15) |c(x)| 2 (1 + |u| + |v|) 2τ (|∇u| + |∇v|) 2−p ∈ L 1 (Ω), so that we can pass to the limit in (4.14) as n → +∞. By Hölder inequality we get This choice is possible since (4.2) holds and in view of (2.17) and (2.18) of Proposition 2.4 we have Passing to the limit as n goes to +∞, assumption (2.6) on c and Fatou Lemma yield that |c(x)| 2 (1 + |u| + |v|) 2τ (|∇u| + |∇v|) 2−p ∈ L 1 (Ω).
Then we can pass to the limit as n → +∞ in (4.14), and dividing (4.14) by k 2 and using Fatou Lemma we get Recalling that χ {0<|u−v|<k} converges to 0 a.e. as k goes to zero, Lebesgue dominated Theorem and (4.15) allow one to conclude that (4.6) holds.
Observe that for k < n Then by Poincaré-Wirtinger inequality, we get Let us evaluate the integral at the right-hand side. We show that it goes to zero first as k → 0 and then as n → +∞. Since in Ω, Let us evaluate the second integral in the right hand side of (4.21). By Hölder inequality we obtain , that is, if n is fixed, where C n > 0 is a constant depending on n (and independent of k). Therefore, by Step 1, we deduce that weak- * , we deduce from (4.20) and (4.22) that for fixed n, as k → 0 We now pass to the limit as n → +∞. By the definition of h n we have so that (2.13) and (4.22) lead to Therefore using (4.19), we deduce that It follows that, up to a subsequence for a suitable constant γ ∈ R, |γ| ≤ 1.
On the other hand since u is finite a.e. we have Then, up to subsequence, by (4.23) we get This implies and means that either in Ω or u < v , a.e. in Ω or u > v , a.e. in Ω .
Step 3. We prove that u < v , a.e. in Ω or u > v , a.e. in Ω can not occur. We assume that (4.24) u > v , a.e. in Ω and we prove that this yields a contradiction. The arguments used in Step 3 of Theorem 3.1 allow us to prove that "u and v have the same sign".
We now evaluate the gradient of w n,k,δ : in Ω.
On the other hand since which means med(w n,k,δ ) ≥ 0. We can conclude that med(w n,k,δ ) = 0.
Then from Poincaré-Wirtinger inequality, by using (4.21) and (4.26), we obtain Ω |w n,k,δ | p dx ≤ C Ω |∇w n,k,δ | p dx (4.27) We now prove that (4.28) lim Clearly (4.28) is a consequence of (2.13) in Definition 2.2. As far as (4.29) is concerned, by Hölder inequality we have 1 2 and in view of the definition of h n , if n is fixed, for any k < 1 we have where C n > 0 is a constant depending on n (and independent of k). From (4.6) it follows that for any fixed n > 0 (4.29) holds.
The same arguments yield that we can not have u < v a.e. in Ω. The conclusion follows.
Proof of Theorem 4.2. Arguing as in the previous theorem we obtain (4.5) and we proceed by dividing the proof by steps. The main difference with respect to the proof of Theorem 4.1 is that for p > 2 we have to control quadratic terms in u − v (see (4.32)) while u and v are solutions to a p-growth problem.
Step 1. By passing to the limit in (4.5) first as n → +∞, then as k → 0 this step is to devoted to prove that We pass to the limit in (4.5) first as n → +∞, then as k → 0. Arguing as in Step 1 of the previous theorem we get that Then, using the assumptions on the strong monotonicity on the operator (2.4), the local Lipschitz condition on Φ (2.7) with τ which satisfies (3.3) and Young inequality we get where lim n ω k (n) = 0. We then obtain By Hölder inequality and assumptions on the data we get where ν = 2tτ t−2 . According to the assumption on τ we have We can pass to the limit as n → +∞ in (4.33), then using Fatou Lemma we get Recalling that χ {0<|u−v|<k} converges to 0 a.e. as k goes to zero Lebesgue dominated Theorem and (4.35) allow one to conclude that (4.32) holds.
Let us consider the function h n (u) T k (u−v) k and observe that for k < n h n (u) Since p ≥ 2 the function h n (u) T k (u−v) k belongs to H 1 (Ω) by Poincaré-Wirtinger inequality we get (4.37) Let us evaluate the integral at the right-hand side. We show that it goes to zero first as k → 0 then as n → +∞. Since It is easy to verify that for fixed n, as k → 0 so that (2.13), (4.32) and (4.38) lead to Then, using (4.37), we deduce It follows that, up to a subsequence, by (4.39) for a suitable constant γ ∈ R, |γ| ≤ 1.
On the other hand since u is finite a.e. This means that either u = v , a.e. in Ω or u < v , a.e. in Ω or u > v , a.e. in Ω .

Arguing as in
Step 3 of the previous theorem, we can prove that the last two possibilities can not occur. Then conclusion follows. Remark 4.3. As in the case of weak solutions, the existence of renormalized solutions hold for a class of more general problems (3.25) where f belongs to L 1 (Ω), Φ verifies growth conditions and a(x, r, ξ) is a Leray-Lions operator which depends on x, s and ξ (see [8]). Due to the lack of regularity of u in the L 1 case by using the techniques developped in the present paper it seems not possible to obtain uniqueness result when a verifies In Step 2 and Step 3 the structure of the operator does not play any role. Equation (4.5) in which we pass to the limit first as n → +∞ and then as k → 0 to derive (4.40) becomes Ω h n (u)h n (v)(a(x, u, ∇u) − a(x, v, ∇v)) · ∇T k (u − v) dx (4.41) Since the operator is pseudo-monotone the main obstacle is the control of the first term of (4.41).
It follows that χ {0<|u−v|<k} 1 k 2 |a(x, u, ∇v) − a(x, v, ∇v))| 2 → 0, in L 1 (Ω). Since the other terms in (4.41) can be controlled by similar methods to the one used in Theorem 4.2 we are able to conclude that (4.40) holds and then that u = v a.e. in Ω.
When p ≥ 2 if τ ≤ N(p − 1) N − p with λ > 1 2 , h ≥ 0 and h ∈ L 2 (Ω), then the renormalized solution u with null median of (3.25) is unique. It is worth noting that (4.42) and (4.43) are similar except in the power of |s| and |r| and the regularity of h. The main reason is that for p ≥ 2 we use quadratic method for a p-growth equation.