On the minimal solution for some variational inequalities

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On the minimal solution for some variational inequalities Michel Chipot a , Senoussi Guesmia b,c,∗ , Soumia Harkat d a Institute of Mathematics, University of Zurich, Winterthurerstrasse 190, CH-8057, Zurich, Switzerland b College of Sciences, Mathematics Department, Qassim University, Saudi Arabia c Mathematics Department, University of the Bahamas, Nassau, Bahamas d Department of Mathematics and Computer Science, University of Oum El Bouaghi, Algeria

Received 19 August 2017; revised 27 May 2018

Abstract Applying an asymptotic method, the existence of the minimal solution to some variational elliptic inequalities defined on bounded or unbounded domains is established. The minimal solution is obtained as limit of solutions to some classical variational inequalities defined on domains becoming unbounded when some parameter tends to infinity. The considered quasilinear operators are only monotone (not strictly) and noncoercive. Some related comparison principles are also investigated. © 2018 Published by Elsevier Inc. MSC: 35J87; 47J20; 47H05; 35J62; 35J70 Keywords: Variational inequalities; Minimal solutions; Monotone operators; Elliptic problems on domains becoming unbounded; Asymptotic behaviour; Anisotropic

1. Introduction and preliminary example In the present work, we are concerned with the existence of minimal solution to some variational elliptic inequalities, in particular noncoercive ones, when the strict monotonicity that * Corresponding author.

E-mail addresses: [email protected] (M. Chipot), [email protected] (S. Guesmia), [email protected] (S. Harkat). https://doi.org/10.1016/j.jde.2018.07.052 0022-0396/© 2018 Published by Elsevier Inc.

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guarantees the uniqueness of the solution is violated. We will investigate variational inequalities defined on bounded domains where the existence of solutions is already ensured and we will deal with the minimal solution. Variational inequalities defined on unbounded domains, for which both the existence of solutions and the minimal solution have to be dealt with, will be also investigated. Since the operators are assumed to be only monotone or even noncoercive, the comparison between the solutions of such inequalities is not always possible and the weak maximum principle for example might fail. In order to illustrate this kind of situation that we want to treat in general context, let us first consider the following example for which we will determine the minimal solution. Let u be a weak solution of the following one dimensional Dirichlet boundary value problem      − a u = 2χ 1  in (−1, 1) ,  − 2 ,1   u (±1) = 0,

(1.1)

where χB denotes the characteristic function of a set B and a : R → R is a single-valued function which graph is depicted in Fig. 1.

Fig. 1. The function a.

Clearly |a(ξ )| ≤ |ξ | and ⎧ 2 ⎪ ⎨ξ if ξ2 ≤ 1, a(ξ ) · ξ = ξ ≥ ξ2 if 1 ≤ ξ ≤ 2, ⎪ 2 ⎩ (ξ − 1)ξ ≥ ξ2 if 2 ≤ ξ, i.e. a(ξ ) · ξ ≥

ξ2 2 .

Let us now find all solutions to (1.1) in the weak sense, i.e. such that

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  u ∈ H 1 (−1, 1) , 0   1 1   a u  v  dx = 2vdx ∀v ∈ H 1 (−1, 1) .  0  −1 −1 2

If u be a solution to (1.1), one has    − a u = 0 in





    1 1 −1, − = 2 in − , 1 . , − a u 2 2

This implies   a u = c1 in





  1 1 −1, − , a u = −2x + c0 in − , 1 , 2 2

  where c0 and c1 are constants. In order to guarantee (1.1) in the distributional sense, a u has to be continuous i.e. 1 + c0 = c1 . We claim that c0 = 0. Indeed, let us suppose c0 = 0 and consider the three cases below.   i. Suppose first that c0 < 0. It follows that a u < 1 a.e. on (−1, 1). Thus we have   a u = u = 1 + c0 in





  1 1  −1, − , a u = u = −2x + c0 in − , 1 . 2 2

Therefore, due to the zero boundary condition at the point −1,

1 u (x) = (1 + c0 ) (x + 1) in −1, − , 2 and for some constant c2 u (x) = −x + c0 x + c2 in 2

1 − ,1 . 2

Due to the zero boundary condition at 1 we have c2 = 1 − c0 and

1 u (x) = −x + c0 x + 1 − c0 in − , 1 . 2 2

The continuity of u at − 12 implies that c0 = 18 which contradicts the fact that c0 < 0. ii. Suppose now that 0 < c0 < 3. Then we have

    −1 c0 − 1 c0 − 1 c0 − 1 < < 1, a u > 1 for x ∈ −1, , a u < 1 for x ∈ ,1 . 2 2 2 2 This implies

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1 −1, − , 2

  1 c0 − 1  a u = u − 1 = −2x + c0 in − , , 2 2

  c0 − 1 a u = u = −2x + c0 in ,1 2   a u = u − 1 = 1 + c0 in

and thus ⎧   (2 + c0 ) x + c2 in −1, − 12 , ⎪ ⎪ ⎪ ⎪   ⎨ 1 c0 −1 2 u (x) = −x + (1 + c0 ) x + c3 in − 2 , 2 , ⎪ ⎪   ⎪ ⎪ ⎩ −x 2 + c0 x + c4 in c0 −1 , 1 . 2 Since u (±1) = 0 and u is continuous at the point − 12 , it follows that c2 = 2 + c0 , c3 = c0 + 74 1 and c4 = 1 − c0 . We use now the continuity of u at the point c02−1 to obtain c0 = − 10 , which is impossible.   iii. Assume that c0 ≥ 3. Then one has a u > 1 a.e. on (−1, 1) and hence we have since    finally a u = u − 1 u = (2 + c0 ) χ

−1,− 12



+ (−2x + c0 + 1) χ

. − 12 ,1

This implies that u (x) = (2 + c0 ) (x + 1) χ

−1,− 12



  + −x 2 + (c0 + 1) x − c0 χ

. − 12 ,1

Since u is continuous, it follows that c0 = − 78 which is also impossible here. Thus we have necessarily c0 = 0 and   a u = 1 in





  1 1 −1, − , a u = −2x in − , 1 . 2 2

    Since a u < 1 a.e. on − 12 , 1 and u (1) = 0 we derive u (x) = 1 − x on 2



1 1  − , 1 , u ∈ [1, 2] on −1, − . 2 2

Thus, the family of solutions to (1.1) is given by ⎛ u (x) = ⎝

x −1

⎞ w (s) ds ⎠ χ

−1,− 12



  + 1 − x 2 χ

, − 12 ,1

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  where w ∈ L2 −1, − 12 is such that w ∈ [1, 2] a.e. on

1

− 2 1 3 −1, − w (s) ds = . , 2 4 −1

It is here clear the weak maximum principle does Then in particular since u ∈  that   not1hold. 1 3 [1, 2], u − 2 = 4 and u (−1) = 0, we have for x ∈ −1, − 2 x x + 1 ≤ u (x) =

1

u (s) ds ≤ 2 (x + 1) ,

−1

3 7 + 2x ≤ u (x) = − 4 4

− 2

u (s) ds ≤

5 + x. 4

x

This gives for minimal solution (Fig. 2) um (x) = (x

 + 1) χ −1,− 34



  7 + + 2x χ 3 1  + 1 − x 2 χ 1  − 4 ,− 2 − 2 ,1 4

and as maximal solution uM (x) = 2 (x

 + 1) χ −1,− 34



  5 + + x χ 3 1  + 1 − x 2 χ 1  . − 4 ,− 2 − 2 ,1 4

Fig. 2. Solutions of (1.1).

In fact, thanks to the following lemma um is also the minimal solution of the variational inequality     u ∈ K := u ∈ H 1 (−1, 1) : u ≥ 0 for a.e. x ∈ (−1, 1) , 0   1 1   a u  (v − u) dx ≥ 2 (v − u) dx ∀v ∈ K.   −1 −1 2

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Lemma 1. Let  be an open bounded interval, f ∈ Lq () and a is a monotone continuous function on R such that for some α0 , β0 > 0, a (ξ ) ξ ≥ α0 |ξ |2 , |a (ξ )| ≤ β0 (1 + |ξ |) ∀ξ ∈ R. Assume that the following Dirichlet problem   u ∈ H 1 () , 0       a u v  dx = f vdx ∀v ∈ H01 () ,   

(1.2)



has a positive minimal solution u, then u is also a minimal solution of the variational inequality ¯ ¯   u ∈ K,      (1.3)  a u (v − u) dx ≥ f (v − u) dx ∀v ∈ K,      where K := u ∈ H01 () : u ≥ 0 for a.e. x ∈  . Proof. Since u is also solution to (1.3), let us show that u ≤ u for any u solution to (1.3). Let u ¯ that be a solution to¯ (1.3). Taking v = 2u, v = 0 in (1.3) one sees 

  a u u dx =



 f udx 

and thus u satisfies 

  a u v  dx ≥





 f vdx =



  a u v  dx ∀v ∈ K. ¯



Taking v = (u − u)+ with v + = max(0, v), one derives ¯       a u − a u (u − u) dx ≤ 0, ¯ ¯ {u>u} ¯

that is to say – due to the monotonicity of a        a u − a u u − u = 0 on {u > u} . ¯ ¯ ¯            If u = u one has a u = a u andalso of course when u = u . Thus it holds a u = a u ¯ on {u >¯u}. It follows¯that a (u ∧ u) = a u in  and thus ¯ ¯ ¯ ¯         − a (u ∧ u)  = − a u  = f,  ¯  u ∧ u ∈ H 1 ¯() , 0 ¯

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where u ∧ v := min (u, v). Since u is the minimal solution to (1.3) one has ¯ u ≥ u ∧ u ≥ u. ¯ ¯ This achieves the proof of Lemma. 2 To deal with the general case we adapt the ideas of [9–17]. That is to say we perturb the given inequalities to construct a family of elliptic nonlinear problems defined on large cylindrical domains with (strictly) monotone operators. When the size of the cylinders becomes unbounded the minimal solution is obtained as the limit of the solution to the perturbed problems. Similarly, the existence of solutions of some variational inequalities with linear operator in unbounded domains is investigated in [14]. Likewise the asymptotic behaviour of variational inequalities with nonhomogeneous Dirichlet boundary conditions and pointwise constraints, as the domain becomes unbounded, is established in [15,16] where different rates of convergence are established. In the case of unbounded domains we apply the same arguments to deduce the existence of nonnegative solutions and a minimal solution. In [2,5], the sub-supersolution method has been used to prove the existence of solutions and extremal solutions, confined between their sub-and supersolutions, for a class of noncoercive variational inequalities involving monotone operators. Different methods, as topological fixed point approaches, bifurcation techniques, recession arguments and variational approaches are adapted to deal with the solvability of noncoercive variational inequalities (see [3,4,21,22,24] and the references therein). In [6,7] the existence of maximal and minimal solutions for some quasi-linear elliptic equations with pseudomonotone operators are proved, by using a different method, under fairly general conditions. Also, comparison results for maximal and minimal solutions are proved in the same papers. More information and details about this subject can be found in [5] and the references therein. Our approach is totally different from the above. Here we also take into account bounded and unbounded domains and sometimes the hypotheses are related to the construction. The manuscript is structured as follows. In the next section, we start with the coercive monotone case. Some comparison results between different solutions are established as tools to pass to the limit in the perturbed problems and show the existence of the minimal solution defined on bounded domains. The second section prepares the way to deal with more general problems involving noncoercive operators in the third section. The last section is devoted to adapt the tools and the arguments used in the bounded case to show the existence of nonnegative solutions and their minimal solutions for more complicated variational inequalities involving unbounded domains. In both cases, coercive and noncoercive operators are handled. Comparison results between minimal solutions are also proved under only a monotonicity assumption in all sections. Throughout the paper |.|p,O (resp. |.|1,p,O ) denotes the usual norm in Lp (O) (resp. W 1,p (O)) and ., . O denotes the duality bracket between W0 (O) and W −1,q (O). Also, we denote by C a generic positive constant, not necessarily the same at each occurrence. 1,p

2. Variational inequalities in bounded domains Let us start by describing and defining the setting of the problem. Let  be a bounded open set of Rn , p be a real number with 1 < p < ∞ and q its conjugate. We denote by K a closed 1,p convex subset of W0 () containing 0 and satisfying

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max (u, v) , min (u, v) ∈ K ∀u, v ∈ K.

(2.1)

Note that we can also write max (u, v) = u + (v − u)+ and min (u, v) = v − (v − u)+ with u+ = max(u, 0). This type of lattice convex sets usually occurs in applications. For example: 1,p

- equations; K =W0

(),  1,p - obstacle problems; K = u ∈ W0 () : u (x) ≥ ψ (x) , for a.e. x ∈ 0 , 0 is a subset of  and ψ is a given function on 0 such  that 0 ≥ ψ on 0 and ∂,  1,p - elasto-plastic torsion problem; K = u ∈ W0 () : |∇u (x)| ≤ c, for a.e. x ∈ 0 , c ≥ 0. (See [1,5,8,23].)

Now let a(x, ξ ) = (ai (x, ξ ))1≤i≤n and a0 (x, ξ ) be a family of Carathéodory’s functions defined suitable coerciveness, monotonicity and growth conditions, i.e. for on  × Rn+1 and satisfying   all ξ = (ξi )i , ξ  = ξi i ∈ Rn+1 and for a.e. x in , there exist nonnegative constants α, β such that 

ai (x, ξ )ξi ≥ α

0≤i≤n

 

0≤i≤n



|ξi |p ,

(2.2)

1≤i≤n

  ai (x, ξ ) − ai (x, ξ  ) ξi − ξi ≥ 0,

(2.3)

(x, ξ ) −→ ai (x, ξ ) is measurable on  × Rn+1 , i = 0, . . . , n,

(2.4)

ξ −→ ai (x, ξ ) is continuous on Rn+1 , i = 0, . . . , n,  |ai (x, ξ0 , ξ1 , . . . , ξn )| ≤ ϑ (x) + β |ξi |p−1 with ϑ ∈ Lq ().

(2.5) (2.6)

0≤i≤n

Then for f in Lq (), we consider u solution of the following nonlinear variational inequality   u ∈ K,     Au, v − u  ≥ f (v − u) dx ∀v ∈ K,    1,p

where A is the nonlinear operator defined from W0

(2.7)

() into its dual by

Au (x) := − div a(x, u, ∇u) + a0 (x, u, ∇u).

(2.8)

Here, for simplicity we set ., .  := ., . , |.|1,p, := |.|1,p and note that the above duality is equivalent to  Au, v =

 a(x, u, ∇u) · ∇vdx +



1,p

a0 (x, u, ∇u)vdx ∀v ∈ W0 

() .

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So, the existence of a solution for this problem is a classical result due to [8,18] while the uniqueness cannot be guaranteed in general, as it is shown for instance in the example above. Let > 0 be a real number. We denote by  the cylinder defined as  = (− , ) × . The points in Rn+1 are denoted by (y, x) with x = (x1 , ..., xn ) ∈ Rn and the gradient operator defined over Rn+1 is also denoted by     ∇  = ∂y , ∇ with ∇ = ∂x1 , ∂x2 , ..., ∂xn . We set   1,p K = v ∈ W0 ( ) | v(y, .) ∈ K a.e. in (− , ) . 1,p

This is a closed convex subset of W0 following variational inequality

( ). Then, for f ∈ Lq () let u be the solution of the

  u ∈ K ,   

     ∂y u p−2 ∂y u ∂y (v − u ) dxdy + Au , v − u dy ≥ f (x) (v − u ) dxdy ∀v ∈ K .   −





(2.9) It is clear that all theforegoing hypotheses assumed on the monotone operator A can be adapted   p−2 to the operator −∂y ∂y v  ∂y v + Av in addition to the strict monotonicity. Therefore, there exists a unique solution u of (2.9). Formally, if we pass to the limit when → ∞, the function limit u˜ is a solution to (2.7). In Theorem 1, the main result of this section, we will prove that when f is nonnegative this limit is also the minimal solution to Problem (2.7). The following lemmas will be used in the proof of our main result. Lemma 2. Suppose that f ∈ Lq () is nonnegative and the assumptions (2.1)–(2.6) are satisfied. Then (i) (u ) >0 is a nondecreasing sequence of nonnegative functions bounded above by any solution of Problem (2.7), (ii) for all 0 > 0, there exists a constant C ( 0 ) independent of such that |u |1,p, ≤ C ( 0 ) . 0

Proof. (i) Taking v = u+

∈ K , the nonnegative part of u , in (2.9) we get  − 

 − p ∂y u  dxdy −





−

  −  A −u−

, −u dy ≥

 

f (x)u−

dxdy ≥ 0,

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+ where u−

= u − u . Using the coerciveness condition (2.2), we easily obtain



 − p ∂y u  dxdy +



 p  α u−

1,p dy ≤ 0.

−



Hence, we derive that u is nonnegative. The same result for u follows by testing (2.7) with u+ and following the same argument as above. Of course this is one of the comparison principles that still takes place when the strict monotonicity is missing, since it only requires the coerciveness condition (2.2). Now, we prove that the sequence (u ) >0 is nondecreasing in and bounded. Let <  . Extending u by 0 on   and since u  is nonnegative we take v = u − (u − u  )+ ∈ K in (2.9) and v = u  + (u − u  )+ ∈ K  in (2.9) written for u  and add the two inequalities, it comes        ∂y u p−2 ∂y u − ∂y u  p−2 ∂y u  ∂y (u − u  )+ dxdy 



+

Au − Au  , (u − u  )+ dy ≤ 0.

−

Thanks to the monotonicity condition (2.3) we deduce        ∂y u p−2 ∂y u − ∂y u  p−2 ∂y u  ∂y (u − u  )+ dxdy ≤ 0. 

This implies – see [10] – 

      ∂y u  + ∂y u   p−2 ∂y (u − u  )+ 2 dxdy ≤ 0



and this leads to ∂y (u − u  )+ = 0 in  . Applying the Poincaré inequality we get (u − u  )+ = 0 in  , which shows that u is a nondecreasing sequence in . On the other hand, taking v = u + (u (y, .) − u)+ ∈ K as a test function in (2.7) we derive, for almost all y, Au, (u (y, .) − u)+ ≥

 

f (u (y, .) − u)+ dx.

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Integrating in y we obtain 



+

Au, (u − u) dy ≥ −

f (u − u)+ dxdy.



One has also u − (u − u)+ ∈ K and from (2.9) we derive  −

  ∂y u p−2 ∂y u ∂y (u − u)+ dxdy −



Au , (u − u)+ dy ≥ −

−





f (u − u)+ dxdy.



Adding the last two inequalities and using the fact that u is independent of y and (2.3) we arrive to        ∂y u p−2 ∂y u − ∂y up−2 ∂y u ∂y (u − u)+ dxdy ≤ 0. 

One concludes as above that (u − u)+ = 0, which means that u ≤ u a.e. on  and for all u solution to (2.7). (ii) Let now 0 > 0 and ρ ∈ D (−2 0 , 2 0 ) such that 0 ≤ ρ ≤ 1 and ρ = 1 on (− 0 , 0 ) .

(2.10)

Then u − ρ p (u − u) ∈ K and from (2.9), we derive 

    ∂y u p−2 ∂y u ∂y −ρ p (u − u) dxdy +





≥



Au , −ρ p (u − u) dy

−

  f −ρ p (u − u) dxdy ≥ 0.



Using the growth condition (2.6) and the fact that u is independent of y it follows that  ρ



p ∂y u  dxdy +

p



ρ p Au , u dy

−



 ≤ −p

  ∂y u p−2 ∂y u ρ p−1 ∂y ρ (u − u) dxdy +

 

ρ p Au , u dy

−



≤C



  ∂y u p−1 ρ p−1 udxdy + C



−

  p−1 ρ p |u |1,p + 1 |u|1,p dy,

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for some constant C independent of (recall u ≥ u ). Applying the Young inequality we easily get from (2.2) that 

 p ρ p ∂y u  dxdy + α



⎛ ⎜ p ρ p |u |1,p dy ≤ ε ⎝

−





 p ρ p ∂y u  dxdy +

⎜ + Cε ⎝

⎞ ⎟ p ρ p |u |1,p dy ⎠

−



⎛







|u|p dxdy +





⎞

⎟ p ρ p |u|1,p + 1 dy ⎠ .

−

2 0

Since ρ = 1 on (− 0 , 0 ), choosing ε small enough we get 

p  |u |p + ∇  u  dxdy ≤ C ( 0 ) .

 0

This achieves the proof of Lemma 2.

2

Now, using the above lemma we can prove the following: Lemma 3. Under the assumptions of Lemma 2, the solution u of (2.9) converges to u, ˜ as goes to ∞, a solution of (2.7). Proof. We start by applying Lemma 2, it follows that u is converging towards some function u. ˜ Next, we show that u˜ is independent of y. We use the idea introduced, for example, in [13]. Let h > 0. The functions T±h u (y, x) and (T±h u (y, x) − u +h (y, x))+ are supported in the closure of ±h

:= (− ∓ h, ∓ h) × , where T±h u denotes the translated of u by ±h in the y-direction i.e. T±h u (y, x) = u (y ± h, x) . Then from (2.9), we have 

  ∂y T±h u p−2 ∂y T±h u ∂y (v − T±h u ) dxdy +

 ∓h AT±h u , v − T±h u dy

− ∓h

±h

 ≥

f (x) (v − T±h u ) dxdy ∀v ∈ K ,±h ,

(2.11)

±h

where     1,p K ,±h := {T±h v | v ∈ K } = v ∈ W0 ±h | v .) ∈ K a.e. in ∓ h,

∓ h) . (y, (−



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Choosing v = T±h u − (T±h u − u +h )+ ∈ K ,±h in (2.11) and v = u +h + (T±h u − u +h )+ ∈ K +h in (2.9) written for u +h and then adding the two inequalities, we obtain        ∂y T±h u p−2 ∂y T±h u − ∂y u +h p−2 ∂y u +h ∂y (T±h u − u +h )+ dxdy ±h

 ∓h +

AT±h u − Au +h , (T±h u − u +h )+ dy ≤ 0.

− ∓h

Using the monotonicity condition (2.3) we deduce        ∂y T±h u p−2 ∂y T±h u − ∂y u +h p−2 ∂y u +h ∂y (T±h u − u +h )+ dxdy ≤ 0. ±h

By the same arguments as in Lemma 2, one can show that u (y ± h, x) ≤ u +h (y, x) . Passing to the limit as → ∞, we get u˜ (y + h, x) ≤ u˜ (y, x) , where h can be now chosen arbitrary. This of course implies that u˜ (y, x) = u˜ (x) . Finally, we use the Minty–Browder technique (see [19,20]) to show that the limit u˜ is a solu p−2 2 that ∂y u  ∂y u and tion to (2.7). Let 0 ∈ R, for large enough, it follows from Lemma  {ai (x, u , ∇u )}i=0,··· ,n are bounded in the Banach space Lq  0 . Therefore   u → u, ˜ ∇u ∇ u˜ in Lp  0 ,     ∂y u p−2 ∂y u d, ai (x, u , ∇u ) di in Lq  . 0

(2.12)

The two first convergences hold for the whole sequence since (u ) >0 is nondecreasing, which guarantees the uniqueness of the limit and the last two convergences hold up to a subsequence. Once the limit is uniquely identified, the previous convergences will also take place for the whole sequence. Now, let φ be a nonnegative function in D (− 0 , 0 ), then under the above convergences we claim, up to a subsequence  0 lim

→∞ − 0

 φAu , u dy =  0

φ

 0in

di ∂xi udxdy ˜

(∂x0 u˜ = u), ˜

(2.13)

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14

 lim

→∞  0

p  φ ∂y u  dxdy = 0.

(2.14)

The last limit means that d = 0. Indeed, using the monotonicity condition (2.3) we get Au , u ≥ Au , u ˜ + Au, ˜ u − u . ˜ Thus one easily derives  0 lim inf

→∞

 0 φAu , u dy ≥ lim inf

→∞

− 0

On the other hand, since u − 

φ |φ|∞

 φAu , u dy ˜ =

− 0



φ

di ∂xi udxdy. ˜

(2.15)

0in

 0

˜ ∈ K (note that u˜ ∈ K), one has from (2.9) (u − u)

  ∂y u p−2 ∂y u ∂y {φ (u − u)} ˜ dxdy +

 0

 φAu , u − u dy ˜ ≤

− 0

 0

φf (u − u) ˜ dxdy ≤ 0

 0

and thus 

 p φ ∂y u  dxdy +

 0

 φAu , u dy ≤

− 0

 0

  ∂y u p−2 ∂y u ∂y φ (u˜ − u ) dxdy

 0

 0 +

φAu , u dy. ˜

− 0

Passing to the lim sup as → ∞, we get ⎡ ⎢ lim sup ⎣

→∞

  0

 p φ ∂y u  dxdy +

 0

⎤ ⎥ φAu , u dy ⎦ ≤

− 0



 0

φ



di ∂xi udxdy. ˜

0in

Combining this with (2.15) we end up with (2.13) and (2.14).   Now noting that for ψ ∈ K and nonnegative function φ ∈ D − 20 , 20 (φ ≡ 0), u + φ |φ|∞

  0 2

(ψ − u ) ∈ K and thus we can take it as a test function in (2.9) to get

0

  ∂y u p−2 ∂y u ∂y {φ (ψ − u )} dxdy +

2

−

 φAu , ψ − u dy ≥

0 2

f φ (ψ − u ) dxdy.

 0 2

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15

By the monotonicity condition (2.3), this leads to 

0

  ∂y u p−2 ∂y u ∂y {φ (ψ − u )} dxdy +

 0

2

−

2

 φAψ, ψ − u dy ≥

f φ (ψ − u ) dxdy.

 0

0 2

2

  From (2.14) it is clear ∂y u → 0 in Lp  0 , thus passing to the limit in the above inequality 2

as → ∞ yields

0

0

2

2 φAψ, ψ − u dy ˜ ≥

−

0 2

−

 f (ψ − u) ˜ dxdy.

φ 

0 2

This implies  Aψ, ψ − u ˜ ≥

f (ψ − u) ˜ dx ∀ψ ∈ K. 

Choosing ψ = u˜ + t (v − u), ˜ where 0 < t < 1 and v ∈ K we deduce  A (u˜ + t (v − u)) ˜ , v − u ˜ ≥

f (v − u) ˜ dx ∀v ∈ K. 

Passing to the limit as t → 0, taking into account the Carathéodory conditions (2.4)–(2.5), we get  Au, ˜ v − u ˜ ≥

f (v − u) ˜ dx ∀v ∈ K. 

Then Lemma 3 is proved.

2

Remark 1. Through this Lemma we are also practically answering the following question. What about the limit of the solution to (2.9) when its limit problem has more than one solution? We are now ready to prove the main result of this section. Theorem 1. Suppose that f ∈ Lq () is nonnegative and the assumptions (2.1)–(2.6) are satisfied. Then, there exists a minimal solution of (2.7) i.e. u˜ (x) = min {u (x) , u solution to (2.7)} , u˜ ∈ K is solution to (2.7). Moreover, if u1 and u2 are the minimal solutions of (2.7) obtained by replacing f with f1 and f2 respectively, then if f1 ≤ f2 we have u1 ≤ u2 .

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16

Proof. Let u be an arbitrary solution of the problem (2.7) and u be the solution to (2.9). Then from Lemma 2 we have u (y, x) ≤ u (x) for a.e. (y, x) ∈  .

(2.16)

Passing to the limit as → ∞, we derive from Lemma 3 that u (y, .) converges towards some u˜ ∈ K solution to (2.7). Combining this with (2.16) yields u˜ ≤ u a.e. in . This means that u˜ is the minimal solution of the problem (2.7). Furthermore, let u ,1 and u ,2 be the solutions of (2.9) obtained if we replace f by f1 and f2 respectively. Taking v = u ,1 − + +   u ,1 − u ,2 and v = u ,2 + u ,1 − u ,2 in (2.9) for f1 and f2 respectively, we get          ∂y u ,1 p−2 ∂y u ,1 − ∂y u ,2 p−2 ∂y u ,2 ∂y u ,1 − u ,2 + dxdy 



+

+  Au ,1 − Au ,2 , u ,1 − u ,2 dy

−



≤

+  (f1 − f2 ) u ,1 − u ,2 dxdy ≤ 0.



Using the monotonicity condition (2.3) we obtain          ∂y u ,1 p−2 ∂y u ,1 − ∂y u ,2 p−2 ∂y u ,2 ∂y u ,1 − u ,2 + dxdy ≤ 0. 

This implies u ,1 ≤ u ,2 in  . Passing to the limit as → ∞, using the above argument we get u1 ≤ u2 in . This completes the proof of Theorem 1.

2

Remark 2. The results of Theorem 1 remain true for a nonnegative distribution f in W −1,q ().

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17

3. Noncoercive variational inequalities Employing the results of the previous section, we aim to extend the study to more general variational inequalities. We keep here the notation and assumptions of section 2 and consider the following extension of the nonlinear problem (2.7)   u ∈ K,     Au, v − u ≥ F (x, u) (v − u) dx ∀v ∈ K,   

(3.1)

where F :  × R → R is a nonnegative function satisfying   F (x, ·) : R → R is continuous and nondecreasing for a.e. x ∈ ,   F (·, r) :  → R is measurable ∀r ∈ R, ∗

F (x, u) ∈ Lq () ∀u ∈ Lp () ,

1 1 1 = − . ∗ p p n

(3.2) (3.3)

The existence of a solution u ∈ K to the problem (3.1) requires more assumptions and it is investigated in different situations (see [5]). Here we will give a general condition related to our technique of construction but the existence of the minimal solution remains our main goal. Let us first define the sequence of functions un as follows   u0 = 0,   un ∈ K,     Aun , v − un ≥ F (x, un−1 ) (v − un ) dx ∀v ∈ K,   

(3.4)

where un is the minimal solution of the variational inequality in the last line of (3.4). Its existence is guaranteed by Theorem 1 since F (x, un−1 ) ∈ Lq (). It is clear that the sequence of functions (un )n∈N is nonnegative and in particular we have, F (x, u0 ) ≤ F (x, u1 ) a.e. x ∈ , since F is nondecreasing in the second variable. This implies that u1 ≤ u2 a.e. x ∈ , thanks to the comparison principle of the minimal solutions shown in Theorem 1. By induction we can easily show that (un )n∈N is nondecreasing sequence i.e. un−1 ≤ un a.e. x ∈ , ∀n ≥ 1. Let us then denote by u∞ the pointwise nonnegative limit of (un )n∈N which is not necessarily in Lp () and may equal ∞. We also denote F∞ := lim F (., un ), n→∞

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18

which may also be infinite on some subset. Assume that F∞ ∈ Lq () .

(3.5)

Note that the above assumption is satisfied, for example, if supr≥0 F (., r) ∈ Lq (). Then the following lemma gives a characteristic property about the existence of a solution to (3.1) related to the above scheme. Lemma 4. Let F be a nonnegative function satisfying the hypotheses (3.2), (3.3) and suppose that the assumptions (2.2)–(2.6) are fulfilled. If (3.5) is satisfied then u∞ , the limit of un , belongs to K and is a solution to (3.1). Proof. Let u˜ ∞ ∈ K be the minimal solution of   u˜ ∞ ∈ K,     Au˜ ∞ , v − u˜ ∞ ≥ F∞ (v − u˜ ∞ ) dx ∀v ∈ K.   

(3.6)

The existence of u˜ ∞ is insured by Theorem 1. Then since F∞ ≥ F (., un−1 ) a.e. on , ∀n ∈ N∗ and thanks to the comparison principle of the minimal solutions of Theorem 1, we deduce that un ≤ u˜ ∞ a.e. on , ∀n ∈ N.

(3.7)

We can now affirm that u∞ is finite almost everywhere and u∞ ≤ u˜ ∞ , F∞ = F (·, u∞ ) a.e. on , which, in particular, proves that u∞ ∈ Lp (). All this is not enough to pass to the limit in (3.4). We need the boundedness of the gradient. Then taking v = u˜ ∞ in (3.4), using the coerciveness, the growth condition (2.2) and (2.6), (3.7) with the monotonicity of F we obtain 

p

α |un |1,p ≤ Aun , un ≤ ≤C



F (x, un−1 ) (un − u˜ ∞ ) dx + Aun , u˜ ∞ 

p−1 |un |1,p

 + 1 |u˜ ∞ |1,p ,

for some constant C independent of n. Applying the Young inequality we end up with |un |1,p ≤ C, for some constant C independent of n. Combining now all the above results, it follows, as n → ∞, that

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un → u∞ in , un → u∞ in Lp () , un u∞ in W 1,p () .

19

(3.8)

1,p

Since K is a closed convex subset of W0 () it is also weakly closed and we have in particular u∞ ∈ K. Now in order to pass to the limit in (3.4), we use the Minty–Browder technique and exploit the above convergences. From the monotonicity condition (2.3) and (3.4), it follows, for w ∈ K, that  F (x, un−1 ) (un − w) dx + Aw, w − un ≥ Aw − Aun , w − un ≥ 0. 

Passing to the limit as n → ∞, we get from (3.8) that  Aw, w − u∞ ≥

F (x, u∞ ) (w − u∞ ) dx. 

The continuity and the monotonicity of F in the second variable, (3.7) and (3.8) are used to ensure the strong convergence of F (·, un ) in Lq (). Taking w = u∞ + t (v − u∞ ) with 0 < t < 1 and v ∈ K, then passing to the limit as t → 0 yields  Au∞ , v − u∞ ≥

F (x, u∞ ) (v − u∞ ) dx ∀v ∈ K. 

This completes the proof of Lemma 4.

2

We will see in the following theorem that (3.5) is more than just a simple condition and u∞ is more than just a simple solution of (3.1). Theorem 2. Under the assumptions (2.1)–(2.6), (3.2), (3.3), we have the equivalence between the following assertions. i) (3.1) has at least one solution, ii) (3.1) has a minimal solution, iii) the hypothesis (3.5) holds. Moreover if the hypothesis (3.5) holds, then u∞ , the limit of un , belongs to K and is the minimal solution to (3.1) i.e. u∞ (x) = min {u (x) , u solution to (3.1)} a.e. on . Proof. From the above lemma we can see that iii) ⇒ i) ⇐ ii). In the following we will show

(3.9)

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iii) ⇐ i) ⇒ ii), which achieves the equivalence of the above assertions and (3.9) simultaneously. Suppose that (3.1) has a solution u¯ ∈ K. Let u ∈ K be the minimal solution of  Au, v − u ≥ F (x, u) ¯ (v − u) dx ∀v ∈ K. 

Here u¯ is considered as a data while u is the unknown solution. The existence of u is insured by Theorem 1. Since u¯ ≥ 0 a.e. on  (which also implies that F (., u) ¯ ≥ F (., 0) a.e. on ) and thanks to the comparison principle of the minimal solutions of Theorem 1, we can show, by induction, that un ≤ u ≤ u¯ a.e. on , ∀n ∈ N.

(3.10)

Then it follows, by the monotonicity of F in the second variable, that F (., un ) ≤ F (., u) ¯ a.e. on , ∀n ∈ N, which shows (3.5). Then, thanks to the preceding lemma and (3.10) we derive that u∞ , the limit of un , belongs to K and is a solution to (3.1) and u∞ ≤ u¯ a.e. on , ∀n ∈ N. Since u¯ is an arbitrary solution of (3.1) we deduce that u∞ is the minimal solution of (3.1) and the proof is achieved. 2 Remark 3. It is possible to assume that F (x, 0) ≥ 0 and F (x, ·) is nondecreasing only on R+ for a.e. x ∈ , or to assume that F is defined on  × R+ , to keep all the above results. Let us now give some immediate variant of the above result. Assume that F¯ :  × R+ → R is a Carathéodory function satisfying the following Lipschitz condition:   F¯ (x, r) − F¯ (x, s) ≤ k |r − s| a.e. x ∈ , ∀r, s ∈ R+ ,

(3.11)

for some constant k > 0 and in addition F¯ (x, 0) ∈ Lq () is a nonnegative function.

(3.12)

Then, as a consequence of Theorem 2 (recall Remark 3), we have Corollary 1. Assume that the assumptions (2.1)–(2.6), (3.11) and (3.12) are satisfied and for p ≥ 2n+2 n+2 there exists a solution to   u ∈ K,     Au, v − u ≥ F¯ (x, u) (v − u) dx ∀v ∈ K.   

(3.13)

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21

Then (3.13) has a minimal solution, i.e. u˜ (x) = min {u (x) , u solution to (3.13)} , u˜ ∈ K

(3.14)

is a solution to (3.13). Proof. We first check that F (x, r) = F¯ (x, r) + kr satisfies almost all the hypotheses assumed on F above. Since we have from (3.11) 0 ≤ F (x, r) − F (x, s) ≤ 2k (r − s) a.e. x ∈ , r > s ≥ 0,

(3.15)

it is obvious to see that F (x, .) is continuous and nondecreasing on R+ for a.e. x ∈ . We also deduce from the above inequality and (3.12) that F (x, r) ≥ F (x, 0) ≥ 0 a.e. x ∈ , ∀r ≥ 0. ∗

Now if u ∈ Lp () we have (3.3) i.e. F (x, u) = F (x, u) − F (x, 0) + F (x, 0) ∈ Lq (). This will follow from (3.12) and (3.15), since u ∈ Lq (). Indeed, by the Sobolev embedding 2n ⇔ p1 − n1 ≤ 1 − p1 ⇔ p ∗ ≥ q, this is the case. theorem when p ≥ n+1 Consider the following variational problem   u ∈ K,      Au, v − u + ku (v − u) dx ≥ F (x, u) (v − u) dx ∀v ∈ K.    

(3.16)

In fact this is nothing else than our problem (3.13). On the other hand, we can easily check that the operator Aˆ = A + kI d (I d is the identity mapping) satisfies the assumptions (2.2)–(2.5). The hypothesis (2.6) is assumed to ensure the boundedness of A. Aˆ is also bounded thanks to the 1,p 2n i.e. for any u, v ∈ W0 (), assumption p ≥ n+1  Au, v + k 

  uvdx ≤ C |u|1,p + 1 |v|1,p + k |u|q, |v|p,   ≤ C |u|1,p + 1 |v|1,p ,

since W 1,p () ⊂ Lq (). This completes the proof of the corollary. 2 ∗

p q Remark 4. It is assumed that p ≥ 2n+2 n+2 to have L ( ) ⊂ L ( ) which ensures that the arguments of the previous section can be adapted.

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22

4. Variational inequalities in unbounded domains This section is devoted to study the existence of nonnegative solutions and their minimal solutions for some quasilinear variational inequalities in unbounded domains. We first investigate variational inequalities with coercive operators. 4.1. Variational inequalities with coercive operator Let ω be a bounded open subset of Rn−1 , n ≥ 2, and Kω be a lattice closed convex subset of containing 0 i.e.

1,p W0 (ω)

0, max (u, v) , min (u, v) ∈ Kω ∀u, v ∈ Kω .

(4.1)

For x ∈ R × ω we denote by x1 the first coordinate of x and by X2 the n − 1 last ones, i.e. x = (x1 , X2 ) with X2 = (x2 , ..., xn ) . 1,p

Also, we denote by K the closed convex subset of Wloc (R × ω) defined by   1,p 1,p K := Wloc (R; Kω ) := v ∈ Wloc (R × ω) | v = 0 on R × ∂ω and v(x1 , .) ∈ Kω for a.e. x1 ∈ R . Note that 1,p

u ∈ Wloc (R × ω) ⇐⇒ u ∈ W 1,p ((−a, a) × ω) ∀a > 0. q

Then for a nonnegative f in Lloc (R, Lq (ω)), we consider the following nonlinear variational inequality defined on the infinite cylinder  = R × ω   u ∈ K,        a(x, u, ∇u) · ∇ − u)) dx + a (x, u, ∇u)ϕ − u) dx ≥ f ϕ (v − u) dx, (ϕ (v (v 0  R×ω R×ω R×ω    ∀v ∈ K, ∀ϕ ∈ D (R) , ϕ ≥ 0. (4.2) q

Note that if we have more smoothness on a1 (x, u, ∇u) i.e. if ∂x1 a1 (x, u, ∇u) ∈ Lloc (R, Lq (ω)), the above variational inequality can be written as   ai (x, u, ∇u)∂xi (v − u) (x1 , .) dX2 ω 2in



+



 a0 (x, u, ∇u) − ∂x1 a1 (x, u, ∇u) (v − u) (x1 , .) dX2

ω

 ≥

f (v − u) (x1 , .) dX2 ∀v ∈ K, a.e. x1 in R. ω

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23 1,p

Since the domain is unbounded and f is not necessarily in the dual of W0 (R × ω), the existence of nonnegative solutions to problem (4.2) is not an ordinary issue. Once this is ensured, we can then look for the minimal nonnegative solution. Here, we will use the same approach as in Section 2, to prove these existence results. To this end, in addition to the hypotheses (2.2)–(2.6), assume that ai (x1 , X2 , ξ0 , 0, ξ2 , · · · , ξn ) = ai (X2 , ξ0 , 0, ξ2 , · · · , ξn ) := ai (X2 , ξ0 , ξ2 , · · · , ξn ), ∀ξj ∈ R, j = 0, 2, ..., n, i = 0, ..., n. (4.3) That is to say if ξ1 = 0 then the coefficients ai for i = 0, ..., n are independent of x1 . We also assume that there exists h ∈ Lq (ω) such that f (x1 , X2 ) ≤ h (X2 ) for a.e. (x1 , X2 ) ∈ R × ω.

(4.4)

For > 0, we set  = (− , )2 × ω

− ., . ω dx1 . We denote 1,p of W0 ( ) defined by

and for simplicity we also set ., . ,ω instead of points in  and by K the closed convex subset

by (y, x1 , X2 ) the

  1,p K := v ∈ W0 ( ) | v(y, x1 , .) ∈ Kω , a.e. in (− , )2 . Then consider uω and u respectively solutions of the following variational inequalities   uω ∈ Kω ,      Aω uω , v − uω ω ≥ h (v − uω ) dX2 ∀v ∈ Kω   ω

(4.5)

and   u ∈ K ,    

     ∂y u p−2 ∂y u ∂y (v − u ) dxdy + Au , v − u ,ω dy ≥ f (x1 , X2 ) (v − u ) dxdy,   −





  ∀v ∈ K , (4.6) where Aω u = −



∂xi ai (X2 , u, ∇X2 u) + a0 (X2 , u, ∇X2 u) and A is the nonlinear operator

2in

given by (2.8). Under the above assumptions, the problems (4.5) and (4.6) have respectively a solution uω ∈ Kω and a unique solution u ∈ K . Then, we have

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24

Theorem 3. Suppose that the assumptions (2.2)–(2.6), (4.1), (4.3) and (4.4) are satisfied. Then ˜ as goes to (u ) >0 is a nonnegative nondecreasing sequence in converging towards some u, ∞, a nonnegative solution of (4.2). Proof. The proof will be broken into several steps. Step 1. u is nondecreasing sequence in bounded by any solution uω to (4.5). Since f is nonnegative, arguing as in Lemma 2, it follows easily that u and uω are nonnegative and (u ) is nondecreasing sequence. Now, taking v = uω + (u (y, x1 , .) − uω )+ ∈ Kω as a test function in (4.5), then integrating on (− , )2 and taking into account the assumption (4.3), we derive 

!

Auω , (u − uω )

+



"

,ω

dy ≥

−

h (u − uω )+ dxdy.



At the same time, choosing v = u − (u − uω )+ ∈ K in (4.6). Then, adding the resulting inequality with the above one yields 

  ∂y u p−2 ∂y u ∂y (u − uω )+ dxdy +



!

Au − Auω , (u − uω )+

−



 ≤

"

,ω

dy

(f − h) (u − uω )+ dxdy.



Using the monotonicity condition (2.3) and the assumption (4.4), we get 

  ∂y u p−2 ∂y u ∂y (u − uω )+ dxdy ≤ 0.



Since uω is independent of y, this implies that        ∂y u p−2 ∂y u − ∂y uω p−2 ∂y uω ∂y (u − uω )+ dxdy ≤ 0. 

One derives easily (u − uω )+ = 0 and thus u ≤ uω .

(4.7)

Step 2. u, ˜ the pointwise limit of u , is independent of y. It is now established that 0 ≤ u ≤ u  ≤ uω ∀ ≤  , then it follows that u possesses a pointwise nonnegative limit that we will denote by u˜ i.e.

(4.8)

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u → u˜ in R2 × ω.

25

(4.9)

Following the same arguments as in the proof of Lemma 3 we get u (y + h, x1 , X2 ) ≤ u +h (y, x1 , X2 ) ∀h ∈ R, a.e. in h := (− − h, − h) × (− , ) × ω. Passing to the limit as → ∞, we easily obtain u˜ (y, x1 , X2 ) = u˜ (x1 , X2 ) . Step 3. For all 0 > 0, there exists a constant C 0 independent of such that |u |1,p, ≤ C 0 .

(4.10)

0

  Let  ∈ D (−2 0 , 2 0 )2 such that 0 ≤  ≤ 1 and  = 1 on (− 0 , 0 )2 . Taking v = u − p (u − uω ) ∈ K in (4.6) then following the same arguments as in the proof of the second part of Lemma 2 we end up with (4.10). Step 4. u˜ is a solution to (4.2). The proof of this part is based on the Minty–Browder technique and follows the proof of the last step of Lemma 3 with some modifications. First, for 0 > 0 we have from (4.9) and (4.10)     u → u˜ in Lp  0 and u u˜ in W 1,p  0 .

(4.11)

Note that since     K˜ 0 := v ∈ W 1,p  0 | v(y, x1 , .) ∈ Kω , a.e. in (− 0 , 0 )2 is closed and convex, it is also weakly closed and by consequence u˜ ∈ K˜ 0 i.e. u(x ˜ 1 , .) ∈ Kω a.e. in (− 0 , 0 ). Then by using the above convergence results, we can prove as in (2.13) and (2.14) that    lim φ {a(x, u , ∇u ) · ∇u + a0 (x, u , ∇u )u } dxdy = φ di ∂xi udxdy, ˜ (4.12)

→∞  0

 lim

→∞  0

 0

p  φ ∂y u  dxdy → 0,

0in

  ∀φ ∈ D (− 0 , 0 )2 , φ ≥ 0,

(4.13)

  where di is the weak limit of ai (x, u , ∇u ) in Lq  0 .  2

0 0 Let now φ ≡ 0 be a nonnegative function in D − 2 , 2 , w ∈ K and taking v = u + φ |φ|∞

(w − u ) ∈ K as a test function in (4.6) we get

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26

0



  ∂y u p−1 ∂y u ∂y (φ (w − u )) dxdy +

 0 2

2

Au , φ (w − u ) 0 ,ω dy 2

−



0 2

φf (w − u ) dxdy.

≥  0 2

Passing to the limit as → ∞, taking into account (4.11), (4.12) and (4.13) we obtain   0



 φf (w − u) ˜ dxdy.

di ∂xi (φ (w − u)) ˜ dxdy ≥

0in

(4.14)

 0

2

2

On the other hand, we claim by using the Minty–Browder technique that

0

2

 Au, ˜ φ (w − u) ˜ 0 ,ω dy ≥ 2

−

0 2

 0



di ∂xi (φ (w − u)) ˜ dxdy.

(4.15)

0in

2

Indeed, let t > 0 and ψ be a nonnegative function in D



− 20 , 20

2

, then it follows from the

monotonicity condition (2.3) that  ψ {a (x, u˜ + tφ (w − u) ˜ , ∇ (u˜ + tφ (w − u))) ˜ − a(x, u , ∇u )} · ∇ (u˜ − u ) dxdy  0 2

 ψ {a0 (x, u˜ + tφ (w − u) ˜ , ∇ (u˜ + tφ (w − u))) ˜ − a0 (x, u , ∇u )} (u˜ − u ) dxdy

+  0 2



ψ {a (x, u˜ + tφ (w − u) ˜ , ∇ (u˜ + tφ (w − u))) ˜ − a(x, u , ∇u )} · ∇ (φ (w − u)) ˜ dxdy

+t  0



2

ψ {a0 (x, u˜ + tφ (w − u) ˜ dxdy ≥ 0. ˜ , ∇ (u˜ + tφ (w − u))) ˜ − a0 (x, u , ∇u )} φ (w − u)

+t  0 2

Passing to the limit as → ∞, taking into account (4.11) and (4.12) we get  ψ  0 2

 0in

˜ , ∇ (u˜ + tφ (w − u))) ˜ − di ) ∂xi (φ (w − u)) ˜ dxdy ≥ 0. (ai (x, u˜ + tφ (w − u)

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Letting t → 0, we easily obtain  ψ



# ˜ ∇ u) ˜ − di ) ∂xi (φ (w − u)) ˜ dxdy ≥ 0 ∀ψ ∈ D (ai (x, u,

0in

 0

0 0 − , 2 2

2 $ , ψ ≥ 0.

2

This is nothing else but (4.15). Now, combining (4.14) and (4.15) we end up with

0

2

#

 Au, ˜ φ (w − u) ˜ 0 ,ω dy ≥

φf (w − u) ˜ dxdy ∀w ∈ K, φ ∈ D

2

0 0 − , 2 2

2 $ , φ ≥ 0.

 0

− 20

2

(4.16) Taking φ (y,  x1 ) = ϕ˜ (y) ϕ (x1 ) in (4.16) where ϕ˜ ≡ 0 and ϕ are nonnegative functions in 

0 0 D − 2 , 2 , we derive

0

2

 ϕ˜ (y) Au, ˜ ϕ (w − u) ˜ 0 ,ω dy ≥

ϕ˜ (y) ϕf (w − u) ˜ dxdy.

2

 0

− 20

2

Since u˜ is independent of y this implies that  Au, ˜ ϕ (w − u) ˜ 0 ,ω ≥ 2

 



− 20 , 20 ×ω



0 0 ϕf (w − u) ˜ dx, ∀w ∈ K, ∀ϕ ∈ D − , , ϕ≥0 2 2

and since 0 is arbitrary this means that 

 a(x, u, ∇u) · ∇ (ϕ (w − u)) dx +

R×ω

 a0 (x, u, ∇u)ϕ (w − u) dx ≥

R×ω

f ϕ (w − u) dx,

R×ω

∀w ∈ K, ∀ϕ ∈ D (R) , ϕ ≥ 0, which completes the proof of Theorem 3.

2

As we saw above, u˜ is more than a usual solution and we can state the following result Theorem 4. Suppose that the assumptions (2.2)–(2.6), (4.1), (4.3) and (4.4) are satisfied. Then, there exists a minimal nonnegative solution to (4.2) i.e. u˜ = min {u solution to (4.2), u ≥ 0} ∈ K is a solution to (4.2). Moreover, let u1 and u2 be the minimum of nonnegative solutions to (4.2) obtained by replacing f with f1 and f2 respectively. Then if f1 ≤ f2 we have u1 ≤ u2 .

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Proof. Let u be an arbitrary nonnegative solution of (4.2) and u be the solution of the problem (4.6). Since u ≥ 0 choosing v = u + (u − u)+ ∈ K as a test function in (4.2), taking into account the fact that supp (u − u)+ ⊂  , then integrating on (− , ) we get 

!

+

Au, (u − u)

"

,ω



f (u − u)+ dxdy,

dy ≥

−

(4.17)



we took ϕ in (4.2) such that ϕ = 1 on (− , ). At the same time, testing (4.6) by v = u − (u − u)+ ∈ K then summing the produced inequality with (4.17) we obtain 

  ∂y u p−2 ∂y u ∂y (u − u)+ dxdy +



!

Au − Au, (u − u)+

"

,ω

dy ≤ 0.

−



Using the monotonicity condition (2.3) we get        ∂y u p−2 ∂y u − ∂y up−2 ∂y u ∂y (u − u)+ dxdy ≤ 0, 

since u is independent of y. By consequence, it follows that u ≤ u a.e. in  .

(4.18)

Passing to the limit as → ∞, we derive from Theorem 3 that u˜ ≤ u a.e. in R × ω. This means that u˜ is the minimal nonnegative solution of (4.2). For the proof of the maximum principle, we denote by u1 and u2 the converging sequences defined above as solutions to (4.6) for f1 and f2 respectively. Since f1 ≤ f2 we have u1 ≤ u2 . Passing to the limit, we deduce that u1 ≤ u2 and this completes the proof of Theorem 4. 2 Remark 5. The maximum principle showed in the above theorem can be extended as follows. Let u1 be the nonnegative minimal solution to (4.2) for some f = f1 satisfying (4.4) and u2 be an arbitrary nonnegative solution to (4.2) for some f = f2 that does not necessarily satisfy +  (4.4). If we assume that f1 ≤ f2 , then u1 ≤ u2 . Indeed, taking v = u1 − u1 − u2 ∈ K (resp. +  v = u2 + u1 − u2 ∈ K) as a test function in (4.6) for f = f1 (resp. in (4.2) for f = f2 ), choosing in (4.2) ϕ ∈ D (R) such that ϕ = 1 on (− , ) and integrating on (− , ), then summing  + the produced inequalities we derive, as supp u1 − u2 ⊂  ,

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   %    + + &  1 p−2 1 1 1 1 ∂y u ∂y u − u2 dxdy + Au − Au2 , u − u2 ∂y u  

 ≤

dy

,ω

−

 + (f1 (x) − f2 (x)) u1 − u2 dxdy.



This implies, as u2 is independent of y,

    +  p−2  1 p−2 1   ∂y u − ∂y u2 ∂y u2 ∂y u1 − u2 dxdy ≤ 0, ∂y u  

we also used the monotonicity condition (2.3). Thus u1 = lim u1 ≤ u2 .

→∞

Consider now the following nonlinear elliptic problem defined on the infinite cylinder R × ω   u ∈ W 1,p (R × ω) , u = 0 on R × ∂ω,  loc   Au = f in R × ω.

(4.19)

The solution is understood in the following sense 

 a(x, u, ∇u) · ∇φdx +

R×ω

 a0 (x, u, ∇u)φdx =

R×ω 1,p

f φdx ∀φ ∈ D (R × ω) .

(4.20)

R×ω 1,p

We take Kω = W0 (ω), then any solution of problem (4.2) for Kω = W0 (ω) is a solution of (4.19) and vice versa. Thus the existence of nonnegative solutions of problem (4.19) is proved in Theorem 3. Indeed, to see this let u ∈ K be a solution of (4.2). By choosing in (4.2) v = u ± φ with φ ∈ D (R × ω) and ϕ = 1 on the support of φ, we end up with the weak formulation above of problem (4.19). The converse is an immediate consequence of (4.20). Thus, we have the following result as an immediate consequence of Theorem 4. Corollary 2. Under the assumptions of Theorem 4, there exists a minimal nonnegative solution to (4.19) i.e. u˜ = min {u solution to (4.19), u ≥ 0} is a solution to (4.19). Moreover, if u1 and u2 are the minimal nonnegative solutions to (4.19) obtained by replacing f with f1 and f2 respectively, then if f1 ≤ f2 we have u1 ≤ u2 . 4.2. Noncoercive variational inequalities We keep the notation and the assumptions of the subsection 4.1. Then we consider the following nonlinear variational inequality defined on the infinite cylinder R × ω

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  u ∈ K,       a(x, u, ∇u) · ∇ (ϕ (v − u)) dx + a0 (x, u, ∇u)ϕ (v − u) dx  R×ω R×ω      ≥ F (x, u) ϕ (v − u) dx, ∀v ∈ K ∀ϕ ∈ D (R) , ϕ ≥ 0,  

(4.21)

R×ω

where the function F is defined as in Section 3, replacing  by R × ω. In addition, we assume that h (X2 , r) := sup F (x1 , X2 , r) , x1 ∈R

satisfies (3.3), i.e. ∗

h (X2 , u) ∈ Lq (ω) ∀u ∈ Lp (ω) .

(4.22)

Of course we do not have any proof neither on the existence of nonnegative solutions of problem (4.21) nor of their minimal solution.   We claim that we can define sequences uˆ n n∈N and (un )n∈N such that uˆ 0 = u0 = 0 and for n ≥ 1, uˆ n ∈ Kω , un ∈ K are respectively the minimal nonnegative solutions to  Aω uˆ n , v − uˆ n ω ≥

  h(X2 , uˆ n−1 ) v − uˆ n dX2 ∀v ∈ Kω

(4.23)

ω

and 

 a(x, un , ∇un ) · ∇ (ϕ (v − un )) dx +

R×ω

R×ω

 F (x, un−1 ) ϕ (v − un ) dx,

≥

a0 (x, un , ∇un )ϕ (v − un ) dx ∀v ∈ K, ∀ϕ ∈ D (R) , ϕ ≥ 0.

(4.24)

R×ω

The existence of the minimal solution uˆ n ∈ Kω (resp. un ∈ K), to (4.23) (resp. to (4.24)) will be ensured by Theorem 1 (resp. Theorem 4). More precisely we have: Lemma 5. Let F and h be nonnegative functions satisfy   the hypotheses above. Then under the assumptions (2.2)–(2.6) and (4.1), the sequences uˆ n n∈N and (un )n∈N are well defined and nondecreasing satisfying un ≤ uˆ n ,

  F (x, un ) ≤ h X2 , uˆ n , ∀n ∈ N, a.e. on R × ω.

(4.25)

Proof. Of course u0 , uˆ 0 satisfy Suppose that un−1 , uˆ n−1 are defined and satisfy  clearly (4.25).  (4.25). One has, by (4.22), h X2 , uˆ n−1 ∈ Lq (ω) and uˆ n exists by Theorem 1. Moreover by (4.25) written for n − 1 one has

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  F (x, un−1 ) ≤ h X2 , uˆ n−1 ∈ Lq (ω) and by (4.4) the existence of un follows. Arguing as in Theorem 3 (step 1) one deduces easily that un ≤ uˆ n and (4.25) follows. The monotonicity follows by induction from our previous comparison principles. 2 Now setting h∞ = limn→∞ h(·, uˆ n ), F∞ = limn→∞ F (·, un ) and we assume that h∞ ∈ Lq (ω) .

(4.26)

Then we have Theorem 5. Let F be a nonnegative function satisfying the hypotheses above. Then, under the assumptions (2.2)–(2.6), (4.1) and (4.26), (un )n∈N is a nondecreasing and converging sequence towards u˜ ∈ K, as n goes to ∞, a minimal nonnegative solution to (4.21). Proof. Since F is nondecreasing we have F (·, un−1 ) ≤ F∞ , ∀n ≥ 1, a.e. in R × ω and it follows easily, by using the maximum principle showed in Theorem 4, that un ≤ u˜ ∞ ,

(4.27)

where u˜ ∞ ∈ K is the minimal solution to (4.2) with f = F∞ . Note that F∞ ≤ h∞ a.e. in R × ω and h∞ is independent of x1 which insures (4.4). Using Lemma 5 we establish that un possesses p a pointwise limit that we will denote by u˜ ∈ Lloc (R × ω), i.e. un → u˜ on R × ω. Then, in order to get more estimates, we test (4.24) by v = un − ϕ p (un − u˜ ∞ ) and choose ϕ such that ϕ = 1 on (− 0 , 0 ), we derive, using the growth condition (2.6), 

 ϕ

p+1

a(x, un , ∇un ) · ∇un dx +

R×ω

ϕ p+1 a0 (x, un , ∇un )un dx

R×ω



≤

F (x, un−1 ) ϕ

p+1

R×ω

  a(x, un , ∇un )∇ ϕ p+1 u˜ ∞ dx

(un − u˜ ∞ ) dx + R×ω





a0 (x, un , ∇un )ϕ p+1 u˜ ∞ dx − (p + 1)

+ R×ω

 ≤C



ϕ

 p+1





|un |p + |u˜ ∞ |p + |∇ u˜ ∞ |p + 1 dx + C

R×ω

+

α 2

∂x1 ϕϕ p a1 (x, un , ∇un )un dx

R×ω

 p   ϕ ∂x1 ϕ  |un |p + |u˜ ∞ |p dx

R×ω

 ϕ p+1 |∇un |p dx, R×ω

for some positive constant C independent of n. Finally, thanks to the coerciveness assumption and (4.27), we end up with |un |1,p,(− 0 , 0 )×ω ≤ C ( 0 ) .

(4.28)

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Since 0 is arbitrary we deduce, as n → ∞, that p

1,p

un → u˜ in Lloc (R × ω) , un u˜ in Wloc (R × ω) , u˜ ∈ K.

(4.29)

Now, by following the same arguments as in the proof of Theorem 3, replacing u by un , it is easy to show that u˜ is a nonnegative solution to (4.21). We also claim that u˜ is the minimal solution. Indeed, let u be a nonnegative solution to (4.21). Here we cannot use the maximum principle mentioned in Theorem 4, since for f = F (·, u) the hypothesis (4.4) is not insured. By consequence, we are not sure that for f = F (·, u) the problem (4.2) has a minimal solution. Here Remark 5 will do the trick. Applying iteratively the maximum principle, showed in Remark 5, it follows that un ≤ u. Then letting n → ∞, we end up with our minimal solution. This completes the proof of Theorem 5. 2 Remark 6. Let F1 , F2 : R × ω × R → R be Carathéodory functions and monotone nondecreasing in the last variable. Assume that F1 is nonnegative and satisfy the assumptions (4.22), (4.26) and F2 satisfies the following coerciveness and growth conditions: for all r ∈ R and for a.e. x in R × ω, there exists a nonnegative constant β  such that F2 (x, r) r ≥ 0, |F2 (x, r)| ≤ ϑ (x) + β  |r|p−1 with ϑ ∈ Lloc (R × ω). q

Then Theorem 5 remains true for F = F1 − F2 , since the problem (4.21) is equivalent to   u ∈ K,       a(x, u, ∇u) · ∇ (ϕ (v − u)) dx + (a0 (x, u, ∇u) + F2 (x, u)) ϕ (v − u) dx  R×ω R×ω      F1 (x, u) ϕ (v − u) dx, ∀v ∈ K, ∀ϕ ∈ D (R) , ϕ ≥ 0. ≥   R×ω

Acknowledgments This work was performed, in part, when the first author was visiting the USTC in Hefei and during his part time employment at the S. M. Nikolskii Mathematical Institute of RUDN University, 6 Miklukho-Maklay St, Moscow, 117198. The publication was supported by the Ministry of Education and Science of the Russian Federation. References [1] D.R. Adams, V. Hrynkiv, S. Lenhart, Optimal control of a biharmonic obstacle problem, in: Around the Research of Vladimir Maz’ya, III, in: Int. Math. Ser. (N. Y.), vol. 13, Springer, New York, 2010, pp. 1–24. [2] H. Amann, On the existence of positive solutions of nonlinear elliptic boundary value problems, Indiana Univ. Math. J. 21 (1971) 125–146. [3] D.D. Ang, K. Schmitt, V.K. Le, Noncoercive variational inequalities: some applications, Nonlinear Anal. T.M.A. 15 (1990) 497–512. [4] C. Baiocchi, F. Gastaldi, F. Tomarelli, Some existence results on noncoercive variational inequalities, Ann. Sc. Norm. Super. Pisa Cl. Sci. 13 (1986) 617–659.

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33

[5] S. Carl, V.K. Le, D. Motreanu, Nonsmooth Variational Problems and Their Inequalities: Comparison Principles and Applications, Springer Monogr. Math., Springer, New York, 2007. [6] J. Casado-Diaz, Minimal and maximal solutions for Dirichlet pseudomonotone problems, Nonlinear Anal. T.M.A. 43 (2001) 277–291. [7] J. Casado-Diaz, A. Porretta, Existence and comparison of maximal and minimal solutions for pseudomonotone elliptic problems in L1 , Nonlinear Anal. T.M.A. 53 (2003) 351–373. [8] M. Chipot, Variational Inequalities and Flow in Porous Media, Springer Verlag, New York, 1984. [9] M. Chipot, Goes to Plus Infinity, Birkhäuser, 2002. [10] M. Chipot, Asymptotic Issues for Some Partial Differential Equations, Imperial College Press, World Scientific, 2016. [11] M. Chipot, S. Guesmia, A. Sengouga, Singular perturbation of some nonlinear problems, J. Math. Sci. 176 (6) (2011) 828–843. [12] M. Chipot, A. Mojsic, P. Roy, On some variational problems set on domains tending to infinity, Discrete Contin. Dyn. Syst. Ser. A 36 (7) (2016) 3603–3621. [13] M. Chipot, Y. Xie, Some issues on the p-Laplace equation in cylindrical domains, Proc. Steklov Inst. Math. 261 (2008) 287–294. [14] M. Chipot, K. Yeressian, On some variational inequalities in unbounded domains, Boll. Unione Mat. Ital. (9) V (2012) 243–262. [15] M. Chipot, K. Yeressian, On the asymptotic behavior of variational inequalities set in cylinders, Discrete Contin. Dyn. Syst. Ser. A 33 (11 & 12) (2013) 4875–4890. [16] M. Chipot, K. Yeressian, Asymptotic behaviour of the solution to variational inequalities with joint constraints on its value and its gradient, Contemp. Math. 594 (2013) 137–154. [17] S. Guesmia, Some convergence results for quasilinear parabolic boundary value problems in cylindrical domain of large size, Nonlinear Anal. 70 (9) (2009) 3320–3331. [18] D. Kinderlehrer, G. Stampacchia, An Introduction to Variational Inequalities and Their Applications, Academic Press, New York, 1980. [19] J. Leray, J.L. Lions, Quelques résultats de Visik sur les problèmes elliptiques non linéaires par les méthodes de Minty–Browder, Bull. Soc. Math. France 93 (1965) 97–107. [20] J.L. Lions, Quelques méthodes de résolution des poblèmes aux limites non linéaires, Dunod-Gauthier-Villars, 1969. [21] J.L. Lions, G. Stampacchia, Variational inequalities, Comm. Pure Appl. Math. 20 (1967) 493–519. [22] P. Quittner, Solvability and multiplicity results for variational inequalities, Comment. Math. Univ. Carolin. 30 (1989) 281–302. [23] J.F. Rodrigues, Obstacle Problems in Mathematical Physics, North-Holland, Amsterdam, 1987. [24] A. Szulkin, Positive solutions of variational inequalities: a degree theoretic approach, J. Differential Equations 57 (1985) 90–111.