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Advances in Nonlinear Analysis

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Lewy-Stampacchia’s inequality for a pseudomonotone parabolic problem

Olivier Guibé
  • Laboratoire de Mathématiques Raphaël Salem, UMR 6085 CNRS, Av. de ľUniversité, BP.12, 76801, Saint-Étienne-du-Rouvray, France
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/ A. Mokrane
  • Laboratoire ďéquations aux dérivées partielles non linéaires et histoire des mathématiques, École Normale Supérieure, B.P. 92, Vieux Kouba, 16050, Alger, Algérie
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/ Y. Tahraoui
  • Laboratoire ďéquations aux dérivées partielles non linéaires et histoire des mathématiques, École Normale Supérieure, B.P. 92, Vieux Kouba, 16050, Alger, Algérie
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/ G. Vallet
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  • Laboratoire de Mathématiques et Applications de Pau, UMR CNRS 5142, BP1155, 64013, Pau cedex, France
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Published Online: 2019-06-23 | DOI: https://doi.org/10.1515/anona-2020-0015


The main aim of this paper is to extend to the case of a pseudomonotone operator Lewy-Stampacchia’s inequality proposed by F. Donati [7] in the framework of monotone operators. For that, an ad hoc type of perturbation of the operator is proposed.

Keywords: Variational inequalities; penalization; pseudomonotone operator; Lewy-Stampacchia’s inequality

MSC 2010: 35K86; 35R35

1 Introduction

The aim of this paper is to prove the existence of a solution to the parabolic variational inequality


and, especially, to give the associated inequality of Lewy-Stampacchia


where u ↦ −div[a(t, x, u, ∇ u)] is a pseudomonotone operator under the constraint uψ.

After the first results of H. Lewy and G. Stampacchia [14] concerning inequalities in the context of superharmonic problems, many authors have been interested in the so-called Lewy-Stampacchia’s inequality associated with obstacle problems. Without exhaustiveness, let us cite the monograph of J.F. Rodrigues [21] and the papers of A. Mokrane and F. Murat [18] for pseudomonotone elliptic problems, A. Mokrane and G. Vallet [19] in the context of Sobolev spaces with variable exponents, A. Pinamonti and E. Valdinoci [20] in the framework of Heisenberg group, R. Servadei and E. Valdinoci [24] for nonlocal operators or N. Gigli and S. Mosconi [11] concerning an abstract presentation.

The literature on Lewy-Stampacchia’s inequality is mainly aimed at elliptic problems, or close to elliptic problems and fewer papers are concerned with other type of problems. Let us cite J. F. Rodrigues [22] for hyperbolic problems, F. Donati [7] for parabolic problems with a monotone operator or L. Mastroeni and M. Matzeu [17] in the case of a double obstacle.

There is a large literature on parabolic problems with constraints. To cite some recent ones, consider [6, 13] where the main operator is monotone, associated with a nonlinear, possible graph, reaction term.

Concerning Lewy-Stampacchia’s inequality, to the best of the author’s knowledge, F. Donati’s work [7] has not been extended to pseudomonotone parabolic problems with a Leray Lions operator. In this paper, we propose such a result, with very general assumptions on the Carathéodory function a, by using a method of penalization of the constraint associated with a suitable perturbation of the operator. As proposed e.g. by [12, p.102] and [4] for sub/super solutions to obstacle quasilinear elliptic problems, this perturbation is one of the main new point of the proof. Indeed, without it, one is usually only concerned by Lewy-Stampacchia’s inequality in the elliptic case, and one needs to assume, as in [18], some additional, now useless, Hölder-continuity assumptions for a with respect to u and ∇u. Thus, this perturbation allows us on the one hand to prove Lewy-Stampacchia’s inequality in the pseudomonotone parabolic case, and on the other hand to reduce significantly the list of assumptions. Let us mention also that, with this method, one is to revisit Lewy-Stampacchia’s inequality proposed in [18, 19] by assuming only basic assumptions. The second essential result is an extension of the formula of time-integration by parts of Mignot-Bamberger[2] & Alt-Luckhaus[1] to non-classical situations. Some information are given too about the time-continuity of an element u when u and t u are not in spaces in duality relation.

The paper is organized in the following way: after giving the hypotheses and the main result (Theorem 2.2) in Section 2, Section 3 is devoted to the proof of this result. A first step is devoted to the existence of a solution to the penalized/perturbed problem associated with a parameter ε; then, some a priori estimates and passage to the limit with respect to ε are considered when g is a regular non-negative element. A first proof of Lewy-Stampacchia’s inequality is given when g is still regular; finally, the proof of Lewy-Stampacchia’s inequality is extended to the general case. A last part, Section 4, presents an annex containing technical results used in the proofs, in particular the time-integration by part and the time-continuity mentioned above.

2 Notation, hypotheses and main result

Let us denote by ω ⊂ ℝd a Lipschitz bounded domain, for any T > 0, by Q = (0, T) × ω and by p ∈ (1, +∞). As usual, p′ denotes the conjugate exponent of p, V = W01,p(ω) if p ≥ 2 and V = W01,p(ω) ∩ L2(ω) with the graph-norm else. Then, the corresponding dual spaces are V′ = W−1,p(ω) if p ≥ 2 and V′ = W−1,p(ω)+L2(ω) else (cf. e.g. [10, p.24]).

In this situation, the Lions-Gelfand triple [23, §. 7.2]


holds and one denotes, as usually, by


Assume in the sequel the following:

  • H1

    : A is a Leray-Lions pseudomonotone operator of the form


    which acts from W1,p(ω) into W−1,p(ω) where

  • H1,1

    a : (t, x, u, ξ⃗) ∈ Q × ℝ × ℝda(t, x, u, ξ⃗) ∈ ℝd is a Carathéodory function on Q × ℝd+1,

  • H1,2

    a is strictly monotone with respect to its last argument:

    (t,x)Q a.e.,uR,ξ,ηRd,ξη[a(t,x,u,ξ)a(t,x,u,η)](ξη)>0.

  • H1,3

    a is coercive and bounded: there exist constants > 0, β̄ > 0 and ȳ ≥ 0, a function in L1(Q) and a function in Lp(Q) and two exponents q, r < p such that, for a.e. (t, x) ∈ Q, for all u ∈ ℝ and for all ξ⃗ ∈ ℝd,



  • H2

    : assume that the obstacle ψ belongs to Lp(0, T; W1,p(ω)) ∩ Lp(0, T; L2(ω)); that t ψ belongs to Lp(0, T; V′) and ψ ≤ 0 on ω (See Section 4.4 for some comments on the time regularity of such elements).

  • H3

    : the right hand side f, which is assumed to be such that


    belongs to the order dual


    where (Lp(0, T; V′))+ denotes the non-negative elements of Lp(0, T; V′).

  • H4

    : u0L2(ω) satisfies the constraint, i.e. u0ψ(0).

As usual concerning obstacle problems one denotes by


Remark 2.1

𝓚(ψ) is a not empty convex set.


Indeed, if one denotes by v*, the solution in W(0, T) to


−(v*ψ)Lp(0, T; V) is an admissible test-function and one has that


Then, Corollary 4.5 with β = 1 and α = 1 yields for any t ∈ (0, T)


since (v*ψ)(0) = 0. As a consequence, v*ψ and v* ∈ 𝓚(ψ).□

Our aim is to prove the following result.

Theorem 2.2

Under the above assumptions (H1)-(H4), there exists at least u in 𝓚(ψ) with u(t = 0) = u0 and such that, for any vLp(0, T; V), vψ implies that


Moreover, the following Lewy-Stampacchia’s inequality holds


3 Proof of Theorem 2.2

Theorem 2.2 will be proved in four steps.

In a first part, we establish the existence of a solution to a problem where the constraint uψ is penalized. Moreover, the crucial point in the method developed in the present paper is to replace a(⋅, ⋅, u, ξ⃗) by a(⋅, ⋅, max(u, ψ), ξ⃗). The aim of this additional perturbation is to ensure, formally, a monotone behavior of the operator when u violates the constraint. This is the aim of Theorem 3.2.

For technical reasons, some a priori estimates and the passage to the limit will be obtained firstly by assuming that g is regular. This is the object of Lemmas 3.3, 3.4, 3.5 and Theorem 3.7. Then a proof of Lewy-Stampacchia’s inequality, still with a regular g, will be presented in Lemma 3.9.

Finally, one will be able to prove Lewy-Stampacchia’s inequality in the general case.

3.1 Penalization

Denote by = min(p, 2) and let us define the function Θ


and the perturbed operator


Remark 3.1

We wish to draw the reader’s attention to the fact that with the proposed perturbation: ã (t, x, u, ξ⃗) = a(t, x, max(u, ψ), ξ⃗), the idea is to make formally the operator monotone and not pseudomonotone any more on the free-set where the constraint is violated.

We define 𝓐:Lp(0, T; V) → Lp(0, T; V′) such that [𝓐(u)](t) := Ã(u(t)) = −div[ã(t, x, u, ∇ u)]. Note that, the above assumption H1 still holds. Indeed,



Since ∣max(u, ψ)∣q ≤ ∣uq + ∣ ψq, ∣max(u, ψ)∣r/p ≤ ∣ur/p + ∣ ψr/p, (1) and (2) are satisfied by replacing by + ȳψq and by + ∣ψr/p.

For any positive ε, a cosmetic modification of [23, Section 8.4] (see also [15, Chap. 3]) yields the following result.

Theorem 3.2

There exists uW(0, T) such that u(t = 0) = u0 and


3.2 The regular case: gL(Q) ↪ Lp(Q)

Following Assumption H3 let us recall that ftψAψ = g = g+g belongs to the order dual Lp(0, T; V)*. In this subsection we impose an additional regularity on g, namely 0 ≤ gL(Q) ↪ Lp(Q).

3.2.1 A priori estimates with respect to ε

Let us test the penalized problem (6) with uεv*,


Thus, by using (1), for any positive δ1, there exists Cδ1 depending on δ1 and ω such that


For the third term, Θ ≤ 0 and v*ψ yield


By using (2), for any positive δ2, there exists Cδ2 depending on δ2 and ω such that


Finally, for any positive δ3, there exists Cδ3 depending on δ3 and ω such that


In conclusion we have


Then, using Young’s inequality and a convenient choice of the parameters δ1,δ2, δ3 yield that for any positive δ there exists C depending on the listed parameters such that


Lemma 3.3

There exists a constant C1 depending onv*W(0,T), ∥ψLp(0,T;V), ∥Lp(ω), ∥L1(Q) andfLp(0,T;V′) such that, for any ε > 0,



If p ≥ 2, W01,p(Ω) = V so that Lemma 3.3 is a straightforward consequence of (7).

If p < 2, it is enough to remark that


It is worth noting that Lemma 3.3 gives that 1εQ((uεψ))q¯dxdt is bounded (with respect to ε) so that we cannot expect to have a bound of the penalized term 1εΘ(uεψ) in Lp(Q) nor in Lp(0, T; V′).

Using the additional regularity gL(Q) we prove in the following lemma more precise estimates on (uεψ).

Lemma 3.4

There exists a constant C2 depending on C1 of Lemma 3.3, such that for any ε > 0,





With the admissible test-function (uεψ), one gets that


Then, since (uεψ)Lp(0, T; V) with (uεψ)(0) = 0, Corollary 4.5 yields: for any t ∈ (0, T),


In view of the definition of ã we have ã(t, x, uε, ∇ uε) = a(t, x, ψ, ∇ uε) in the set {uε < ψ}. Therefore using assumption H1,2 we obtain


We recall that Lemma 3.3 yielded (uεψ)Lq~(Q)q~C1ε so that


and Lemma 3.4 holds.□

Gathering Lemmas 3.3 and 3.4 we prove the following estimates

Lemma 3.5

There exists a constant C3 depending on C1, C2 andgLp(Q) such that for any ε > 0



The growth condition (5) on ã and Lemma 3.3 imply that


and then ã(t, x, uε, ∇ uε) is bounded in Lp(Q)d. The boundedness of ∥Ã(uε)∥Lp(0,T;V′) is a direct consequence of the above inequality.

Recalling that t uε = fÃ(uε) − 1ε Θ(uεψ) it remains to estimate 1ε Θ(uεψ) in Lp(0, T;V′). We distinguish the two cases p ≥ 2 and 1 < p < 2.

If p ≥ 2 then = 2. From Lemma 3.4 we have 1ε ∥(uεψ)L2(Q)C and since


it follows that 1ε Θ(uεψ) is bounded in Lp(0, T; V′).

If 1 < p < 2 then = p. From Lemma 3.4 we have 1ε(uεψ)Lp(Q)p1C and we have


which concludes the proof of Lemma 3.5.□

3.2.2 At the limit when ε → 0

The sequence (uε) is bounded in W(0, T), therefore, up to a subsequence denoted the same, there exists uW(0, T) such that uε converges weakly to u in W(0, T). In particular, one gets that u(t = 0) = u0.

Then, by classical compactness arguments of type Aubin-Lions-Simon [26], the convergence is strong in Lp(Q), and a.e. in Q.

Therefore, (uεψ) → (uψ) in Lp(Q) and thanks to Lemma 3.4, one gets that (uψ) = 0 i.e. u ∈ 𝓚(ψ).

Moreover from Lemma 3.5 there exists ξ⃗Lp(Q)d such that

a~(,,uε,uε) converges weakly to ξ in Lp(Q)d.(12)

By (2), the following estimate holds for any vLp(0, T; V),


so that, since u ∈ ℝ ↦ a(t, x, u, ∇ v) is a continuous function, the theory of Nemytskii operators gives that

a~(t,x,uε,u)a~(t,x,u,u) in Lp(Q)d(13)



Testing the penalized equation (6) introduced in Theorem 3.2 by uεu yields


Since 0tf,uεuds0, the following decomposition


leads to

lim supε[0ttuε,uεuds+0tΩa~(t,x,uε,uε)(uεu)dxds]0.

Using (14) we obtain

lim supε[0tt(uεu),uεuds+0tΩ[a~(t,x,uε,uε)a~(t,x,uε,u)](uεu)dxds]0.

The monotone character of the operator ã(x, t, u, ξ⃗) with respect to ξ⃗ (see Assumption H1,2 and (3)) implies

12lim supε(uεu)(t)L2(Ω)2=lim supε0tt(uεu),uεuds0



It follows that

uε(t)u(t) in L2(Ω)foranyt(16)

and in view of (14)


Set v⃗Lp(Q)d. Since


using (15) and information similar to (14) allow one to pass to the limit and to conclude that


By the classical Minty’s trick, considering v⃗ = ∇ u + λ w⃗, w⃗Lp(Q)d and λ ∈ ℝ, we have necessarily


Thus, a classical property of radial continuity coming from the assumptions on a yields, for any w⃗Lp(Q)d,


i.e. ξ = ã(t, x, u, ∇ u) = a(t, x, u, ∇ u), since uψ.

Remark 3.6

Note that, following [3, Proof of Lemma 1], (15) yields the convergence in measure, then the a.e. convergence ofuε tou (up to a subsequence if needed), so that this is also a way to identify ξ⃗ has being a(t, x, u, ∇ u).

We are now in a position to pass to the limit in the penalized problem and to conclude the existence of a solution to the obstacle problem under the additional regularity on g.

Let us consider vLp(0, T; V), vψ as a test function in the penalized problem (6),


In view of (16) we have


From (17) and the identification ξ⃗ = ã(t, x, u, ∇ u) = a(t, x, u, ∇ u) it follows that


The weak convergence of uε to u in Lp(0, T; V) yields that


At last splitting the penalized term in the following way

1εQΘ(uεψ)(vuε)dxdt=1εQ[(uεψ)]q~1(vψ)dxdt01ε(uεψ)Lq~(Q)q~0 thanks to (11)

allows one to pass to the limit in (18). One concludes that a solution exists, i.e.

Theorem 3.7

Assume H1H4, ft ψA ψ = g = g+gLp(0, T; V)* where gLp (Q) ∩ L2(Q). There exists at least u ∈ 𝓚(ψ) with u(t = 0) = u0 such that, for any vLp(0, T; V) with vψ,


Remark 3.8

Note that the pseudomonotone assumption of the operator doesn’t ensure the uniqueness of the solution. Observe that under additional assumptions on the operator a, namely a local Lipschitz continuity with respect to the third variable, standard arguments allow one to prove the uniqueness of the solution obtained in Theorem 3.7.

3.3 Lewy-Stampacchia’s inequality for a regular g

Note that με : = t uε − div[ã(⋅, ⋅, uε, ∇ uε)] − f = 1ε [(uεψ)]−1 ≥ 0, so that the limit μ := t u − div[ã(⋅, ⋅, u, ∇ u)] − f is a non-negative Radon measure which is by Lemma 3.5 an element of Lp (0, T; V′).

Using an idea from A. Mokrane and F. Murat [18], denote by zε:=g1ε[(uεψ)]q~1, we have


Observing that


as in [18] in the elliptic case and under more restrictive assumptions on the operator a, proving that zε converges to 0 in an appropriate space leads to the Lewy-Stampacchia’s inequality. Due to the time variable and the weak assumption on a we have to face to additional difficulties. For technical reasons, we will assume in this section only that, on top of gLp(Q) ∩ Lp(0, T; V), g ≥ 0, that t gL(Q). Roughly speaking it allows one to use a test function depending on g and together with Lemma 4.3 to perform an integration by part formula and then the convergence analysis of zε.

The general case will be obtained in the next section by a regularization argument based on Lemma 4.1 of the Annex.

Our aim is now to show the convergence of zε to 0 in L2(Q) to prove the following lemma.

Lemma 3.9

Under the assumptions of Theorem 3.7 and assuming moreover that gLp(Q) ∩ Lp(0, T; V), g ≥ 0 with t gL(Q), the solution u satisfies


A priori, following Lemma’s 4.3 notations, one should denote by


For that, we need Ψ(t, x, u) to be a test-function.

Since x ↦ [x]−1 is not a priori a Lipschitz-continuous function (e.g. if p < 2 ), therefore, for any positive k, we will denote by ηk(x) = ( − 1) 0x+ min(k, s−2)ds, Ψk(t, x, λ) = (g1εηk(λ)) and Λk(t, x, λ) = 0λΨk(t,x,σ)dσ.

Note that Ψk(t, x, 0) = 0 and t Ψk(t, x, λ) = tg1{g1εηk(λ)<0} so that, since Ψk(t, x, u) is a test-function, by Lemma 4.3, for any t,


We now pass to the limit first as k → ∞ and then as ε → 0. Since g ≥ 0, one has that Ψk(t, x, λ) = 0 if λ ≥ 0 and as uε(0) = u0ψ(0), one gets that


Note that (Ψk(t, x, λ))k is a non-increasing sequence of functions with non-positive values so that by monotone convergence theorem


since the integration holds on the set of negative values of uε(t) − ψ(t).

Due to the definition of zε we have


from which it follows using again the monotone convergence theorem


As far as the first term of (19) is concerned we obtain


For the fourth term of (19) we have the following equality


since in this situation, the integration holds in the set where uεψ. Thus,


We now claim that estimate (10) of Lemma 3.4 which gives

[a~(t,x,ψ,uε)a~(t,x,ψ,ψ)](uεψ)ε0 in L1(Q)

and Assumptions H1,1 to H1,3 imply that, up to a subsequence (still denoted by ε), ∇(uεψ) converges to 0 a.e. in Q.

Indeed, up to a subsequence (still denoted by ε), uε converges to u a.e. in Q with uψ a.e. and |(t, x, ψ, ∇ uε) – (t, x, ψ, ∇ψ)| |∇(uεψ)| → 0 a.e. in Q.

Consider (t, x) such that the above limits hold. Since,




one gets that (∇(uεψ)(t, x))ε is a bounded sequence.

Since ∇(uεψ)(t, x) = –∇(uεψ)(t, x) 1{uε<ψ}(t, x), it converges to 0 if u(t, x) > ψ(t, x).

Else, at the limit, one has that u(t, x) = ψ(t, x). If one assumes that (uεψ)(t, x) is not converging to 0, then there exists a subsequence ε (depending on (t, x)) and a positive δ such that ∥(uεψ)(t, x)∥ ≥ δ > 0. Then, necessarily –(uεψ)(t, x) = (uεψ)(t, x) and, since it is a bounded sequence in ℝd, there exists ξ⃗ ∈ ℝd and a new subsequence still labeled ε such that uε(t, x) converges to ξ⃗, with the additional information: ∥ξ⃗ψ(t, x)∥ ≥ δ > 0. Therefore, since ξ⃗ψ(t, x)


But, this is in contradiction with the convergence of the same sequence to 0 and the result holds.

Note that (t, x) ∈ Q a.e.,


and uε 1{uε<ψ} ψ 1{uε<ψ} converges to 0 with ψ 1{uε<ψ} bounded. Then, the continuity of with respect to its fourth argument can be assumed uniform and [(t, x, ψ, uε) – (t, x, ψ, ∇ψ)] 1{uε<ψ} converges a.e. to 0.

Since it is bounded in Lp(Q), it converges weakly to 0 in Lp(Q) and


As a conclusion, zε converges to 0 in L2(Q). On the one hand we have


On the other hand


Since (⋅, ⋅, u, ∇u) = a(⋅, ⋅, u, ∇u), Lemma 3.9 is proved.

Remark 3.10

Note that, for any φLp(0, T; V),


In such a way, ∥t uε – div[(⋅, ⋅, uε, uε)] – fLp(0, T; V′) ≤ ∥gLp(0, T; V′).

3.4 Proof of the main result

In this section, H3 is assumed and g = ft ψA(ψ) = g+g where g+, g ∈ (Lp(0, T; V))+ are non-negative elements of Lp(0, T; V).

Thanks to Lemma 4.1, there exists positives (gn) ⊂ Lp(Q) such that gng in Lp(0, T; V). Then, by a regularization procedure, one can assume that gnLp(Q) ∩ Lp(0, T; V), gn ≥ 0 with tgnLq~(Q). Then, the corresponding sequence fn converges to f in Lp(0, T; V).

Remark 3.11

In fact, since D(Q)+ is dense in Lp(Q)+, one can consider gn as regular as needed.

Associated with gn, Theorem 3.7 and Lemma 3.9 provide the existence of un ∈ 𝓚(ψ) with un(t = 0) = u0 and such that, for any vLp(0, T; V), vψ implies that


and satisfying the Lewy-Stampacchia’s inequality


Since this solution comes from the above penalization method, and as C1 of Lemma 3.3 can be chosen independent of n, one gets that


Thus, a(⋅, un, un) is bounded in Lp(Q)d and, thanks to the above Lewy-Stampacchia’s inequality, t un is bounded in Lp(0, T; V).

Up to a subsequence denoted similarly, un converges weakly to an element u ∈ 𝓚(ψ) in W(0, T) and strongly in Lp(Q); and a(⋅, un, un) converges to an element ξ⃗ in Lp(Q)d.

Finally, the embedding of W(0, T) in C([0, T], L2(Ω)) yields the weak convergence of un(t) to u(t) in L2(Ω), for any t.

Since u ∈ 𝓚(ψ),


Therefore, passing to the limit with respect to n in



0Ttu,udt+Qξudxdt+12u0L2(Ω)212u(T)L2(Ω)2+lim supnQa(t,x,un,un)undxdt.

Since 0Ttu,udt=12u(T)L2(Ω)212u0L2(Ω)2, one gets that

lim supnQa(t,x,un,un)undxdtQξudxdt.

Thus, (2) and the continuity property of Nemytskii's operator ensure the following limit argument:



0lim infnQa(t,x,un,un)undxdtQξudxdt.

Then, limnQa(t,x,un,un)undxdt=Qξudxdt and arguments already developed previously based on Minty’s trick for the pseudomonotone operator A yield the identification ξ⃗ = a(t, x, u, ∇u) and one has


From the weak lower semicontinuity of |⋅|L2(Ω), one has

lim supn0Ttun,vundt0Ttu,vudt.

Since 0Tfn,vundt0Tf,vudt, we deduce the existence result of Theorem 2.2 for general f. At last the Lewy-Stampacchia’s inequality is a consequence of passing to the limit in the one satisfied by un.

4 Annex

4.1 Positive cones in the dual

Lemma 4.1

The positive cone of Lp(0, T; V) ∩ L2(Q) is dense in the positive cone of 𝓥′, the dual set of 𝓥 = Lp(0, T; V).

By a truncation argument, the same result holds for the positive cone of Lp(0, T; V) ∩ Lp(Q) when p < 2.


This result is given in [7, Lemma p.593]. We propose here a sketch of a proof following the idea of [18].

Consider fLp(0, T; V) such that f ≥ 0 in the sense: ∀ φLp(0, T; V), φ ≥ 0 ⇒ 〈f, φ〉 ≥ 0.

Let us construct a sequence (fε) ⊂ Lp(0, T; V) with fε ≥ 0 such that fεf in Lp(0, T; V).

Consider the following operator B : Lp(0, T; V) → Lp(0, T; V) defined, for any u, vLp(0, T; V) by


B = DJ where J:u1pQ|u|pdxdt+1p0Tu(t)L2(Ω)pdt, is a G-differentiable, hemi continuous convex function, and, B is a strictly monotone, bounded, continuous and coercive operator from Lp(0, T; V) into Lp(0, T; V). Then ([23, section 2.1]), denote by v the unique solution to Bv = f.

For any ε > 0 and n ∈ ℕ, denote by vεn the solution to Bvεn+1εTn(vεnv)=0inLp(0,T;V) where Tn is the truncation at the height n.

Using vεnv as test function, one has J(vεn)+1εTn(vεnv)L2(Q)2J(v).

Thus, there exists vε, weak limit in Lp(0, T; V) of a subsequence (denoted similarly) of vεn satisfying: J(vε) + 1εvεvL2(Q)2J(v) and, by classical monotony arguments, solution in Lp(0, T; V) to the problem:


Then, up to a subsequence, vεv → 0 in L2(Q), vεv in Lp(0, T; V) and


so that fε = –1ε(vεv) ∈ Lp(0, T; V) ∩ L2(Q) is non-negative.

Finally, as lim supε J(vε) ≤ J(v), an argument of uniform convexity yields the convergence of vε to v in Lp(0, T; V) and –1ε(vεv) = BvεBv = f in Lp(0, T; V).□

4.2 Compactness when p < 2

Concerning the compactness argument in Lp(Q) when p < 2: note that there exists an integer k ≥ 1 such that W0k,p(Ω)denseLp(Ω) so that

W0k,p(Ω)denseVdenseLp(Ω)[Lp(Ω)]Wk,p(Ω) and VWk,p(Ω).

Remark 4.2

Let us justify that the identification Lp(Ω) ≡ [Lp(Ω)]′ is possible if L2(Ω) is already chosen as the pivot-space.

Indeed, one has: Lp(Ω) dense L2(Ω) dense Lp(Ω) with reflexive B-spaces, so that Lp(Ω) dense L2(Ω) dense Lp(Ω).

Consider TLp(Ω) and TnL2(Ω) such that TnT in Lp(Ω). Then, by the pivot-space identification, there exists unL2(Ω) such that Tn = un in the sense of Riesz-identification.

Then, for any vLp(Ω) with norm 1,


By considering v=Sgn(un)|un|p1unLpp1, one has that the sequence (un) is bounded in Lp(Ω) and that, up to a subsequence if needed, it converges weakly to a given u in Lp(Ω).

Thus, for any vLp(Ω),


Since this element u is unique in its way, the identification holds.

Then, since the embedding of V is compact in Lp(Ω), by Aubin-Lions-Simon [26] compactness theorems, if a sequence is bounded in W(0, T), it is also bounded in {uLp(0, T; V), t uLp(0, T; Wk,p(Ω))} and relatively compact in Lp(Q).

4.3 On Mignot-Bamberger -Alt-Luckhaus integration by part formula

We propose in next Lemma a time integration by part formula adapted to our situation. Its proof has been inspired by [9].

Lemma 4.3

Consider uLp(0, T; W1,p(Ω)) ∩ Lp(0, T; L2(Ω)) such that t uLp(0, T; V).

Let Ψ : Q × ℝ → ℝ be a function such that (t, x) ↦ Ψ(t, x, λ) is measurable, λΨ(t, x, λ) is non-decreasing (it can be càdlàg§ or càglàd) and denote by Λ : Q × ℝ → ℝ, (t, x, λ) ↦ aλ Ψ(t, x, τ) where a is any arbitrary real number. Assume moreover that t Ψ exists with |Ψ(λ = 0)|+|t Ψ| ≤ hL2(Q). If Ψ(t, x, u) ∈ Lp(0, T; V), then, for any βW1,∞(0, T) and any 0 ≤ s < tT,



Thanks to the assumptions on Ψ, it is a measurable function on Q × ℝ and Λ is a Carathéodory function on Q × ℝ. Moreover,


so that Λ, ΨLloc2(ℝ, L2(Q)) and the Nemytskii operator associated with Λ is continuous from L2(Q) to L1(Q). Moreover, t Λ(t, x, λ) = aλ tΨ(t, x, τ) and

|t Λ(t, x, λ)| ≤ |λa|h(t, x) ≤ |λ|2 + h2(t, x)/4 + |a|h(t, x) so that the Nemytskii operator associated with t Λ is also continuous from L2(Q) to L1(Q).

By assumption, uC([0, T], L2(Ω)) and one extends u to ū in ℝ by ū(t) = u0 if t < 0 and ū(t) = u(T) si t > T. Therefore, if I1 := (–1, T + 1), ūLp(I1, W1,p(Ω)) ∩ C(Ī1, L2(Ω)) such that t ūLp(I1, V) with t ū = 0 when t < 0 or t > T.

Similarly to u, denote by Ψ̄ the extension to I1 of Ψ in the same way and by Λ̄ the corresponding integral as introduced in the Lemma.

For any fixed 0 < h < < 1, let us denote by


Consider β ∈ 𝓓(I1) and h, small enough so that suppβ + [–h, h] ⊂ I1. Then, in L2(Ω),


Thus, vh and wh converge to t ū in 𝓓′[I1, L2(Ω)] and 𝓓′[I1, V], and to t u in 𝓓′[0, T; L2(Ω)] and 𝓓′[0, T; V]. Moreover, by [5, Corollary A.2 p.145], the properties of Bochner integral and since t ū = 0 outside (0, T),


Therefore, vh converges weakly to t ū in Lp[I1, V] and to t u in Lp[0, T; V] (as well as for wh).

For any βD(I1), one has Ψ(⋅, ū)βLp(I1, V), and


Let us recall that a is a given real and Λ̄(t, x, λ) = aλ Ψ̄(t, x, τ) . Since Ψ̄ is a non-decreasing function of its third variable, for any real numbers u and v, one has


Thus, assuming moreover that β is non-negative,


and, for h small enough to have suppβ + [–h, h] ⊂ I1,


so that

lim infhI1×ΩΛ¯(,u¯(t+h))Λ¯(,u¯(t))hβdxdtI1tu¯,Ψ¯(,u¯)βdt=0Ttu,Ψ(,u)βdt,lim suphI1×ΩΛ¯(,u¯(t))Λ¯(,u¯(th))hβdxdtI1tu¯,Ψ¯(,u¯)βdt=0Ttu,Ψ(,u)βdt.

On the other hand,




Then, one gets by passing to the limit when h → 0, and thanks to the time-extension procedure,

lim infhI1×ΩΛ¯(th,x,u¯(t))Λ¯(t,x,u¯(t))hβ(th)dxdt0Ttu,Ψ(,u)βdt+I1×ΩΛ¯(,u¯)βdt=0Ttu,Ψ(,u)βdt+QΛ(,u)βdt+ΩΛ(0,x,u0)β(0)dxΩΛ(T,x,u(T))β(T)dxlim suphI1×ΩΛ¯(t,x,u¯(t))Λ¯(t+h,x,u¯(t))hβ(t+h)dxdt

Note that




is an integrable function, the properties of the point of Lebesgue (Steklov average) yields


Since the same holds for limhI1×ΩΛ¯(t,x,u¯(t))Λ¯(t+h,x,u¯(t))hβ(t+h)dxdt,βD+([0,T])


Since β is involved in linear integral terms, a classical argument of regularization yields the result for any non-negative elements of W1,∞, then for any elements of W1,∞.

Note that T being arbitrary, the result holds for any t and s = 0, then for any t and s by subtracting the integral from 0 to s to the one from 0 to t.□

Remark 4.4

As a consequence,

ddt[ΩΛ(t,x,u)dx]=tu,Ψ(t,x,u)+ΩtΛ(t,x,u)dx in D(0,T)

and t ↦ ∫Ω Λ(t, x, u)dx is absolutely-continuous in [0, T].

Corollary 4.5

Consider uLp(0, T; W1,p(Ω)) ∩ L(0, T; L2(Ω)) such that t uLp(0, T; V), αL2(Ω), α ≥ 0 and Ψ : ℝ → ℝ a given non-decreasing function. Assume that Ψ(u) αLp(0, T; V), then, for any βW1,∞(0, T) and any 0 ≤ s < tT,


where a is any arbitrary real number.

Remark 4.6

Note that, by linearity, the same result holds if α = α1α2 with Ψ(u)αiLp(0, T; V) (i = 1, 2) or if Ψ = Ψ1Ψ2 and Ψi (i = 1, 2) satisfies the assumptions.

4.4 Strong continuity in L2(Ω)

We consider the following notations in the sequel: V(Ω) = W1,p(Ω) ∩ L2(Ω), V0(Ω) = W01,p(Ω) ∩ L2(Ω) and V(Ω) = W–1,p(Ω) + L2(Ω).

Let us prove in this section the following result of continuity.

Lemma 4.7

Let Ω ⊂ ℝN be a bounded domain with Lipschitz boundary Ω, then we have


Remark 4.8

This result is not the usual one since u and t u are not in spaces being in duality relation and few words are needed concerning the time-derivative.

Note that both V(Ω) and V0(Ω) are dense subspaces of the chosen pivot space L2(Ω) so that it can be identify to a subspace of V′(Ω) or (V(Ω))′. Therefore, u, as an element of Lp(0, T; V(Ω)) ↪ Lp(0, T, L2(Ω)) ↪ Lp(0, T; V(Ω)), has a time derivative in the sense of distributions in 𝓓′(0, T, V′(Ω)) and it is assumed to belong to Lp(0, T; V(Ω)).


This result is based on a classical method: first in ℝN, then in the half-space R+N and finally in Ω thanks to an atlas of charts.

Obviously, if Ω = ℝN, we have W01,p(ℝN) = W1,p(ℝN), and the result holds by classical arguments (e.g. [25, Prop. 1.2 p. 106]).

If Ω=R+/resp.N = {(x, xN) ∈ ℝN; xN > 0/resp. < 0}, the method is based on a suitable extension of u to ℝN. Following a recommendation of F. Murat, we consider the following extension, proposed in [16, (12.21-22) p.83] and revisited in [8, p.2]:


Note that ũLp(0, T; V(ℝN)) and, thanks to a change of variables, that for any φCc(]0,T[×RN), one gets




By construction, ψ(t, x) = φ(t, x, xN) − 3φ(t, x, −xN) + 2φ(t, x, − xN2) = 0 if xN = 0, as well as tψ, and ψW1,∞(0, T; V0(R+N)) with ψLp(0,T;V0(R+N))CφLp(0,T;V(RN)) for a given constant C.

Therefore, |0TR+Nu~tφdxdt|CtuLp(0,T;V(R+N))φLp(0,T;V(RN)), and tũLp(0, T; V(ℝN)). Then, one concludes that ũC([0, T], L2(ℝN)) i.e. uC([0, T], L2(R+N)).

Finally, the result holds in the general case by considering an atlas of charts as proposed e.g. in [8, p.3]. □


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  • Some arguments are given in Annex 4.2 when p < 2. 

  • = min(2, p) 

  • §

    continu à droite et limite à gauche: right continuous with left limit 

  • continu à gauche et limite à droite : left continuous with right limit 

About the article

Received: 2018-04-23

Accepted: 2018-10-14

Published Online: 2019-06-23

Published in Print: 2019-03-01

Citation Information: Advances in Nonlinear Analysis, Volume 9, Issue 1, Pages 591–612, ISSN (Online) 2191-950X, ISSN (Print) 2191-9496, DOI: https://doi.org/10.1515/anona-2020-0015.

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© 2020 O. Guibé et al., published by De Gruyter. This work is licensed under the Creative Commons Attribution 4.0 Public License. BY 4.0

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