Growing Sandpile Problem with Local and Non-Local Boundaries Conditions

Abstract

In this paper, we study a class of Prigozhin equation for growing sandpile problem subject to local and a non-local boundary condition. The problem is a generalized model for a growing sandpile problem with Neumann boundary condition (see [1]). By the semi-group theory, we prove the existence and uniqueness of the solution for the model and thanks to a duality method we do the numerical analysis of the problem. We finish our work by doing numerical simulations to validate our theoretical results.

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Traoré, U. (2021) Growing Sandpile Problem with Local and Non-Local Boundaries Conditions. Journal of Applied Mathematics and Physics, 9, 2414-2429. doi: 10.4236/jamp.2021.910153.

1. Introduction

The mathematical modeling of the dynamics of granular materials is a fascinating subject as evidenced by the numerous studies which treat this subject in the literature. In this work, we are interested in the formation of a pile of sand under the effect of a vertical source. Regarding this phenomenon, there are mainly two models based on partial differential equations. The first model is that of Hadeler and Kuttler (see [2] [3] [4] ) where the authors model the pile of sand using two layers: a fixed layer above which a mobile layer moves. The second approach to which this work relates is the variational model of Prigozhin (see [5] [6] ) where the surface flow of sand is supposed to exist only at critical slope, that is the maximal admissible slope α for any stationary configuration of sand (here for simplicity it is assumed α = 1 ).

In the particular case of the Dirichlet boundary condition; for that model there are a sufficiently developed theory and efficient numerical schemes which use duality arguments (see [7] [8] [9] ). However, very few results exist in the literature on the model of Prighozin in the case where one considers other types of boundary conditions other than the Dirichlet boundary condition, since it raises many difficult questions: how to define the notion of the solution for the model and how to approach the problem numerically. We have recently partially provided solutions to these questions where we study the Prigozhin model with Neumann and Robin boundaries conditions (see [10] ). These types of conditions are interpreted by the presence of an obstacle for example a wall at the boundary and which prevents the sand escaping. In this work, we look at the model with a non-local boundary condition which generalizes the Neumann and Robin boundary condition:

u t ( m u ) = f in ( 0, T ) × Ω | u | 1, m 0, m ( | u | 1 ) = 0 in ( 0, T ) × Ω u = 0 on [ 0, T ] × Γ 0

m u ν = g on ( 0, T ) × Γ 1 Γ 2 m u ν = a ( t ) ( given function of t ) on [ 0, T ] × Γ 2 u = c ( t ) ( unknown constant of t ) on ( 0, T ) × Γ 2 u ( 0 ) = u 0 , (1)

with Ω N a bounded open domain with a boundary Γ = Γ 0 Γ 1 Γ 2 such that Γ 0 Γ 1 Γ 2 = . The solution u is the height of the surface andf is the source, g L 2 ( Γ 1 ) and a is a function defined on .

In this work, we consider a mixed boundary condition: Dirichlet, Neumann and a non-local boundaries condition:

· on Γ 0 , we apply homogeneous Dirichlet boundary condition. In other words, we assume that at this boundary, the sand falls down, that is

u = 0 on ( 0, T ) × Γ 0 ;

· on Γ 1 , there is a wall that prevents the sand escaping:

m u ν = g on ( 0, T ) × Γ 1 ;

· we control the outgoing flow through the boundary Γ 2 , which results in the following non-local condition:

Γ 2 m u ν d s = A ( t ) on [ 0, T ] × Γ 2 ;

Beside the mathematical interest of non-local conditions, it seems that this type of boundary condition is also encountered in other physical applications. For example, in petroleum engineering model for well modelling in a 3D stratified petroleum reservoir with arbitrary geometry; this kind of boundary condition also arises in petroleum engineering, in the simulation of wells performance, since a nonlinear relation exists between the performance pressure tangential gradient and the fluid velocity along the well (see [11] [12] for details). Another application of this type of the boundary condition is in the study of the heat conduction within linear thermoelasticity (see [13] [14] [15] ), and for the reaction-diffusion equation (see [16] [17] ). One could find other applications in other fields of physics in the papers [18] [19] [20].

The present work is organized as follows. In the next section, we use the nonlinear semi-group theory (see [21] ) to get the existence and uniqueness of a variational solution of (1) and the convergence of the approximate Euler discretization in time solutions to problem (1). In Section 3, we show how to compute the solution of Euler implicit time discretization of (1) using duality argument and in the last section, some results of numerical results are given.

2. Global Existence

Given ε > 0 , we say that ( t i , f i ) i = 1 , , n is an ε -discretization for problem (1) if 0 = t 0 < t 1 < < t n 1 < T = t n with t i t i 1 ε , f 1 , , f n L 2 ( Ω ) , g 1 , , g n L 2 ( Γ 1 ) such that

i = 1 n t i 1 t i f ( t ) f i L 2 ( Ω ) ε ; i = 1 n t i 1 t i g ( t ) g i L 2 ( Γ 1 ) ε

Definition 1. For any ε > 0 , ( u ε ) is an ε -approximate solution of (1), if there exists ( t i , f i , g i ) i = 1 , , n an ε -discretization of problem (1) such that

u ε = ( u 0 for t ( 0, t 1 ] u i for t ( t i 1 , t i ] , i = 1, , n (2)

and u i solves the following Euler implicit time discretization of (1):

u i ε ( m i u i ) = ε f i + u i 1 in Ω | u i | 1, m i 0, m i ( | u i | 1 ) = 0 in Ω u i = 0 on Γ 0 m i u i ν = g i on Γ 1 Γ 2 m u i ν = a i on Γ 2 u i = c i ( unknown constant ) on Γ 2 (3)

The generic problem associated with (3) is given by

v ( m v ) = F in Ω | v | 1, m 0, m ( | v | 1 ) = 0 in Ω v = 0 on Γ 0 m u ν = G Γ 2 m v ν = A ( given constant ) on Γ 2 v = C ( unknown constant ) on Γ 2 (4)

with F L 2 ( Ω ) , G L 2 ( Γ 1 ) .

For convenience, we note the L 2 ( Ω ) scalar product by ( , ) and the euclidean norm on N by . We introduce the convex set K :

K = { z W 1, ( Ω ) H D 1 ( Ω ) : | z ( x ) | 1 a .e x Ω }

where

H D 1 ( Ω ) = { z H 1 ( Ω ) : z = 0 on Γ 0 z constant on Γ 2 } .

We also introduce the function Π K defined on L 2 ( Ω )

Π K ( z ) = ( Γ 1 G z d s A z | Γ 2 if z K , + otherwise ,

The sub-differential of Π K is defined in L 2 ( Ω ) by

Π K ( v ) = { w K : z K , Π K ( z ) Π K ( v ) + ( w , z v ) } .

Remark. In order to define our notions of variational solutions for the above problems, we make the following observations:

(1) Assuming that v K is a solution of (4) in the following sense: there exists a measurable function m satisfying the properties m v ( L 1 ( Ω ) ) d , m ( | v | 1 ) = 0 a.e. in Ω and

Ω v z d x + Ω m v z d x = Ω F z d x + Γ 1 G z d s + A z | Γ 2 z H D 1 ( Ω ) ; (5)

then, v is also a solution of the following optimization problem

max z K { Ω ( f v ) z d x + Γ 1 G z d s + A z | Γ 2 } . (6)

Indeed, taking z K as test function in (5), we get

Ω ( F v ) z d x + Γ 1 g z d s + A z | Γ 2 = Ω m v z d x . (7)

Thus, using the fact that | z | 1 a.e. x Ω , we get

Ω ( F v ) z d x + Γ 1 G z d s + A z | Γ 2 Ω m | v | d x (8)

and taking z = v in (7), we obtain

Ω ( F v ) v d x Γ 1 G z d s A z | Γ 2 = Ω m | v | 2 d x . (9)

So, adding (8) and (9) and using the fact that m ( | v | 1 ) = 0 a.e. in Ω , we obtain

Ω ( F v ) z d x + Γ 1 G z d s + A z | Γ 2 ( Ω ( F v ) v d x + Γ 1 G v d s + A v | Γ 2 ) 0. (10)

(2) It is not clear that there exists ( m , v ) such that (5) is fulfilled, however we can find v K , solution of the problem (6), which is a consequence of the following result.

Lemma 2. Π K is a maximal monotone graph in L 2 ( Ω ) .

Proof. The proof is the same as in our previous paper.

We are now in a position to define our notion of solution to problems (4) and (1).

Definition 3 (see [21] ). For given F L 2 ( Ω ) , G L 2 ( Γ 1 ) , we say that v is a variational solution of (4) if v K and

Ω ( F v ) ( z v ) d x + Γ 1 g ( z v ) d s + A ( z | Γ 2 v | Γ 2 ) 0, for all  z K . (11)

Definition 4 (see [21] ). Given f L ( 0, T ; L 2 ( Ω ) ) , and u 0 K , a variational solution (resp. ε -approximate solution) of (1) is a function u W 1,1 ( 0, T ; L 2 ( Ω ) ) satisfying for any t ( 0, T ) , u ( t ) K and

Ω ( f u t ( t ) ) ( z u ( t ) ) d x + Γ 1 g ( z u ( t ) ) d s + A ( z | Γ 2 u ( t ) | Γ 2 ) 0,

for any z K (resp. u ε given by (2) with u i a variational solution of (3)).

Since Π K is a maximal monotone graph in L 2 ( Ω ) , then, thanks to [21] for any F L 2 ( Ω ) there exists a unique solution v of

v + Π K ( v ) F , (12)

which is equivalent to saying that (4) admits a unique variational solution. Moreover, if v i is the solution corresponding to g i for i = 1 , 2 , then

v 1 v 2 2 F 1 F 2 2 . (13)

As in the works of Igbida and al, we get the following result

Theorem 5. Let u 0 K , T > 0 , f L 2 ( 0, T ; L 2 ( Ω ) ) . Then,

1. for any ε > 0 and any ε -discretization of (1), there exists a unique ε -approximate variational solution of (1).

2. There exists u C ( [ 0, T ) ; L 2 ( Ω ) ) such that u ( 0 ) = u 0 and as ε 0 ,

u ε u in C ( [ 0, T ) ; L 2 ( Ω ) ) .

3. The function u given by (2) is the unique variational solution of (1). Moreover, if for i = 1,2 , u i is a solution corresponding to f i , then

d d t Ω ( u 1 u 2 ) + Ω ( f 1 f 2 ) + in D ( 0, T ) .

In particular, if f 0 , then u 0 a.e. in Ω .

3. Numerical Analysis

Following the ideas developed in our previous work we show in this section how to approximate the solution of problem (1). In our previous works, we are introduced a numerical method based on Fenchel duality theory (see [22] ) for numerical approximation of problem (4), in the special case where the homogeneous Dirichlet boundary condition is applied. In order to use the same approach, we introduce the following result.

Lemma 6. The problem (4) is equivalent to: find v K solution of the following of

arg min z K { 1 2 Ω z g L 2 ( Ω ) 2 + Π K ( z ) } . (14)

Proof. Let v K be a solution of (4). Then, we have

Ω ( F v ) ( z v ) d x + Γ 1 G ( z v ) d s + A ( z | Γ 2 v | Γ 2 ) 0, z K . (15)

We use the well know inequality

v F L 2 ( Ω ) 2 z F L 2 ( Ω ) 2 = 2 ( F v , z v ) v z L 2 ( Ω ) 2 ,

to obtain

1 2 Ω | v F | 2 d x 1 2 Ω | z F | 2 d x + Γ 1 G ( z v ) d s + A ( z | Γ 2 v | Γ 2 ) = Ω ( F v ) ( z v ) d x + Γ 1 G ( z v ) d s + A ( z | Γ 2 v | Γ 2 ) 1 2 Ω | v z | 2 d x .

From (15), we deduce that

1 2 Ω | v F | 2 1 2 Ω | z F | 2 + Γ 1 G ( z v ) d s + A ( z | Γ 2 v | Γ 2 ) 0 ; (16)

i.e.

1 2 v F L 2 ( Ω ) 2 + Π K ( v ) 1 2 z F L 2 ( Ω ) 2 + Π K ( z ) for any z K .

Which end the first part of the proof. Now assume v is the solution of the minimization problem (14). Let z 0 K , since K is convex, then for any t [ 0,1 ) , z = ( 1 t ) v + t z 0 K . Then, we have

1 2 Ω | v F | 2 d x Γ 1 G v d s A v | Γ 2 1 2 Ω | F ( ( 1 t ) v + t z 0 ) | 2 d x Γ 1 G ( ( 1 t ) v + t z 0 ) d s A ( ( 1 t ) v + t z 0 ) | Γ 2 1 2 F v t ( z 0 v ) L 2 ( Ω ) 2 Γ 1 G v d s A v | Γ 2 t Γ 1 G ( z 0 v ) d s t A ( z 0 | Γ 2 v | Γ 2 )

Which implies that

1 2 Ω | v F | 2 d x Γ 1 G v d s A v | Γ 2 1 2 Ω | v F | 2 d x t ( F v , z 0 v ) + t 2 2 Ω | z 0 v | 2 d x Γ 1 G v d s A v | Γ 2 t Γ 1 G ( z 0 v ) d s t A ( z 0 | Γ 2 v | Γ 2 ) .

Therefore,

( F v , z 0 v ) + Γ 1 G ( z 0 v ) d s + A ( z 0 | Γ 2 v | Γ 2 ) t 2 Ω | z 0 v | 2 d x . (17)

Hence, by letting t 0 in (17), we obtain

Ω ( F v ) ( z 0 v ) d x + Γ 1 G ( z 0 v ) d s + A ( z 0 | Γ 2 v | Γ 2 ) 0,

for any z 0 K , so v is a solution of (4).

Inspired by the works of Igbida and al, we consider as a dual problem associated with problem (14), the following optimization problem

sup { G ( w ) : w H d i v , K ( Ω ) } , (18)

where

G ( w ) = 1 2 Ω ( div ( w ) ) 2 d x + Ω g div ( w ) d x + Ω | w | d x (19)

and

H div , K ( Ω ) = { w H div ( Ω ) : Ω [ div ( w ) ] ξ d x = Ω w ξ d x Γ 1 G ξ d s A ξ | Γ 2 , ξ H D 1 ( Ω ) } ,

H div ( Ω ) : = { w ( L 2 ( Ω ) ) N ; div ( w ) L 2 ( Ω ) } .

In the following, we prove the connection between problems (18) and (14). We also present a method for numerical approximation of the solution of problem (4) by computing sup { G ( w ) : w H div , K ( Ω ) } . We first show the following result.

Lemma 7. Let F L 2 ( Ω ) , G L 2 ( Γ 1 ) , w H div , K ( Ω ) and z K , we have

G ( w ) J ( z ) ,

where J ( z ) = 1 2 Ω | z F | 2 d x + Π K ( z ) .

Proof. Taking w H div , K ( Ω ) , z K and using the fact that 1 2 ( div ( w ) ( z g ) ) 2 0 , we get

1 2 Ω ( div ( w ) ) 2 d x Ω div ( w ) g d x 1 2 Ω ( z g ) 2 d x Ω div ( w ) z d x .

Since w H div , K ( Ω ) , we have

1 2 Ω ( div ( w ) ) 2 d x Ω div ( w ) g d x 1 2 Ω ( z g ) 2 d x + Ω w z d x Γ 1 G z d s A z | Γ 2 .

Thus,

G ( w ) 1 2 Ω ( z g ) 2 d x Γ 1 G z d s A z | Γ 2 + Ω w z d x Ω | w | d x . (20)

As z belongs to K , it follows that

Ω w z d x Ω | w | d x 0. (21)

Then,

G ( w ) 1 2 Ω ( z g ) 2 d x Γ 1 G z d s A z | Γ 2

The connection between the problem (14) and (18) is given by the following results.

Theorem 8. Let F L 2 ( Ω ) , G L 2 ( Γ 1 ) and v the variational solution of (14). Then, there exists a sequence ( w ε ) ε > 0 in H div , K ( Ω ) , ω ( M ( Ω ) ) s such that, as ε 0 ,

Ω | ω ε | d x Ω v ( g v ) d x + Γ 1 G v d s + A v | Γ 2 , (22)

div ( ω ε ) v F L 2 ( Ω ) (23)

ω ε ω ( M ( Ω ) ) s weak . (24)

Moreover, we have

lim ε 0 G ( ω ε ) = inf ω H div , K ( Ω ) G ( ω ) = min z K J ( z ) = [ 1 2 Ω | v F | 2 d x Γ 1 G v d s A v | Γ 2 ] . (25)

Let recall that, M ( Ω ) denote the space of all bounded Radon measures in Ω .

For the proof of this result, we analyze the following problem

v ε w ε = F in Ω w ε = ϕ ε ( v ε ) in Ω v ε = 0 on Γ 0 w ε η = G on Γ 1 , Γ 2 w ε η d s = A on Γ 2 , v ε constanton Γ 2 (26)

where for any ε > 0 , ϕ ε : N s is given by

ϕ ε ( r ) = 1 ε ( | r | 1 ) + r | r | forall r N

and satisfies the following properties.

(i) for any r 1 , r 2 N , ( ϕ ε ( r 1 ) ϕ ε ( r 2 ) ) ( r 1 r 2 ) 0 ;

(ii) there exists ε 0 > 0 and ρ > d such that ϕ ε ( r ) r | r | 2 for any | r | > ρ and ε < ε 0 ;

(iii) for any ε > 0 and r N , | ϕ ε ( r ) | ϕ ε ( r ) r .

Lemma 9. There exists a unique weak solution ( v ε ) 0 < ε < ε 0 to problem (26) in the sense that v ε H D 1 ( Ω ) , w ε = ϕ ε ( v ε ) ( L 2 ( Ω ) ) N and for all z H D 1 ( Ω ) ,

Ω v ε z d x + Ω ϕ ε ( v ε ) z d x = Ω F z d x + Γ 1 G z d s + A z | Γ 2 . (27)

Moreover,

(1) ( v ε ) 0 < ε < ε 0 is bounded in H D 1 ( Ω ) .

(2) ( w ε ) 0 < ε < ε 0 is bounded in ( L 1 ( Ω ) ) N .

(3) For any Borel set B Ω ,

lim inf ε 0 B | v ε | d x d | B | . (28)

Proof. (1) We define the operator A ε : H D 1 ( Ω ) ( H D 1 ( Ω ) ) by

A ε v , ξ = Ω v ξ d x + Ω ϕ ε ( v ) ξ d x .

It’s not difficult to see that the operator A ε is monotone, coercive, hemicontinous and bounded.

Defining C : H D 1 ( Ω ) defined by

C ( z ) = Ω G ξ d x + Γ 1 G ξ d s + A ξ | Γ 2 .

then thanks to [23], there exists v ε H D 1 ( Ω ) such that

A v ε , ξ = C , ξ , for all ξ H D 1 ( Ω )

i.e.

Ω v ε ξ d x + Ω ϕ ε ( v ε ) ξ d x = Ω F ξ d x + Γ 1 G ξ d s + A ξ | Γ 2 .

Which end the proof of the existence. Now, if (26) admits two solutions u and v, then we subtract the two equations obtained by replacing respectively v ε by u and v in (27) to get

Ω ( u v ) ξ d x + Ω ( ϕ ε ( u ) ϕ ε ( v ) ) ξ d x = 0, ξ H D 1 ( Ω ) . (29)

We take u v as a test function in (29) to obtain

Ω ( u v ) 2 d x + Ω ( ϕ ε ( u ) ϕ ε ( v ) ) ( u v ) d x = 0. (30)

Thus, using the property (i) of ϕ ε , we deduce that u = v . a.e. Ω

(1) Taking v ε as test function in (27), we have

Ω v ε 2 d x + 1 ε Ω ( | v ε | 1 ) + | v ε | d x = Ω F v ε d x + Γ 1 G v ε d s + A v ε | Γ 2 .

Since v ε is constant on Γ 1 , there exists C such that

Ω v ε 2 d x Ω F v ε d x + Γ 1 G v ε d s + A v ε | Γ 2 F L 2 ( Ω ) v ε L 2 ( Ω ) + G L 2 ( Γ N ) v ε L 2 ( Γ N ) + | A C | F L 2 ( Ω ) v ε H 1 ( Ω ) + G L 2 ( Γ N ) v ε L 2 ( Γ N ) + | A C | .

Using the following trace inequality (see [24] )

Γ | v ε | 2 d s C 1 ( Ω | v ε | 2 d x + Ω | v ε | 2 d x ) ,

where C 1 is a positive constant. One obtains

v ε L 2 ( Ω ) 2 ( F L 2 ( Ω ) + C 1 G L 2 ( Γ N ) ) v ε H 1 ( Ω ) + | A C | . (31)

On the other hand

1 ε Ω ( | v ε | 1 ) + | v ε | d x ( F L 2 ( Ω ) + C 1 G L 2 ( Γ N ) ) v ε H 1 ( Ω ) + | A C | . (32)

Combining (32) and property (ii) of ϕ ε , for any 0 < ε < ε 0 , we get

Ω | v ε | 2 d x [ | v ε | ρ ] | v ε | 2 d x + [ | v ε | > ρ ] | v ε | 2 d x [ | v ε | ρ ] | v ε | 2 d x + 1 ε Ω ( | v ε | 1 ) + | v ε | d x | ρ | 2 | Ω | + ( F L 2 ( Ω ) + C 1 G L 2 ( Γ N ) ) v ε H 1 ( Ω ) + | A C | . (33)

Thus, adding (31) and (33), we obtain

v ε H 1 ( Ω ) 2 | ρ | 2 | Ω | + 2 ( F L 2 ( Ω ) + C 1 G L 2 ( Γ N ) ) v ε H 1 ( Ω ) + 2 | A C | .

Using the Young’s inequality, we get

v ε H 1 ( Ω ) 2 | ρ | 2 | Ω | + 2 | A C | + 1 2 [ 2 ( F L 2 ( Ω ) + C 1 G L 2 ( Γ N ) ) ] 2 + 1 2 v ε H 1 ( Ω ) 2 ,

Consequently

v ε H 1 ( Ω ) 2 2 | α | 2 | Ω | + 4 | A C | + 4 ( F L 2 ( Ω ) + C 1 G L 2 ( Γ N ) ) 2 . (34)

Then ( v ε ) is bounded in H D 1 ( Ω ) .

(2) Using (32) and the property (iii) of ( ϕ ε ) , we have

Ω | w ε | d x = Ω | ϕ ε ( v ε ) | d x Ω ϕ ε ( v ε ) v ε d x 1 ε Ω ( | v ε | 1 ) + | v ε | d x ( F L 2 ( Ω ) + C 1 G L 2 ( Γ N ) ) v ε H 1 ( Ω ) (35)

and since ( v ε ) is bounded in H D 1 ( Ω ) , it follows that ( w ε ) is bounded in ( L 1 ( Ω ) ) N .

(3) Now, let B Ω be a fixed Borel set. We have

v ε L 1 ( B ) ( | v ε | 1 ) + + 1 L 1 ( B ) ( | v ε | 1 ) + L 1 ( B ) + 1 L 1 ( B ) B | ( | v ε | 1 ) + | d x + | B | B ( | v ε | 1 ) + | v ε | d x + | B | ε ( F L 2 ( Ω ) + C 1 G L 2 ( Γ N ) ) v ε H 1 ( Ω ) + | B | .

Letting ε 0 and using the fact that v ε is bounded in H 1 ( Ω ) , we obtain

lim inf ε 0 Ω | v ε | d x | B | .

Proof of the Theorem 8. Thanks to Lemma 9, there exists v ˜ in H D 1 ( Ω ) , ω ( M b ( Ω ) ) N and a subsequence that we denote again by ε , such that

w ε ω in ( M b ( Ω ) ) N weak ,

v ε v ˜ in H D 1 ( Ω ) weakandin L 2 ( Ω ) .

Hence, by problem (26), we deduce that

div ( w ε ) v ˜ F in L 2 ( Ω ) .

We define the set A δ by A δ = [ | v ˜ | 1 + δ ] , with δ > 0 . Using the fact that v ε v ˜ in ( L 1 ( Ω ) ) s -weak as ε 0 , it follows that

( 1 + δ ) | A δ | A δ | v ˜ | d x lim inf ε 0 A δ | v ε | d x . (36)

Taking B = A δ in Lemma 7, we obtain

( 1 + δ ) | A δ | | A δ | (37)

which implies that | A δ | = 0 . Since δ > 0 is arbitrary, then, it follows that | v ˜ | 1 a.e. in Ω . Let us now show that v ˜ is also a solution of (4). For all z K , we have

Ω ( F v ˜ ) ( z v ˜ ) d x + Γ 1 G ( z v ˜ ) d s + A ( z | Γ 2 v ˜ | Γ 2 ) = lim ε 0 [ Ω ϕ ε ( v ε ) ( z v ˜ ) d x Γ 1 G ( z v ε ) d s + A ( z | Γ 2 v ε | Γ 2 ) ] + Γ 1 G ( z v ˜ ) d s + A ( z | Γ 2 v ˜ | Γ 2 )

= lim ε 0 Ω ϕ ε ( v ε ) ( z v ˜ ) d x Γ 1 G ( z v ˜ ) d s + A ( z | Γ 2 v ˜ | Γ 2 ) + Γ 1 G ( z v ˜ ) d s + A ( z | Γ 2 v ˜ | Γ 2 ) = lim ε 0 Ω ϕ ε ( v ε ) ( z v ˜ ) d x = lim ε 0 Ω ( ϕ ε ( v ε ) ϕ ε ( z ) ) ( z v ˜ ) d x 0, (38)

i.e. v ˜ is also a solution of (4). Since (4) admits a unique solution then v = v ˜ . Now, we must show that w ε satisfies (22). By (iii), we have

lim sup ε 0 Ω | ω ε | d x = lim sup ε 0 Ω | ϕ ε ( v ε ) | d x lim sup ε 0 Ω ϕ ε ( v ε ) v ε d x lim sup ε 0 ( Ω ( F v ε ) v ε d x + Γ 1 G v ε d s + A v ε | Γ 2 ) Ω ( F v ) v d x + Γ 1 G v d s + A v | Γ 2 . (39)

Moreover taking v as test function in (27) and having in mind that v = v ˜ , we get

Ω ( F v ) v d x + Γ 1 G v d s + A v | Γ 2 (40)

= lim ε 0 Ω ( F v ε ) v d x + Γ 1 G v d s + A v | Γ 2

= lim ε 0 Ω w ε v d x

= lim ε 0 Ω w ε v d x

lim ε 0 Ω | w ε | d x . (41)

Combining (39) and (40), we get

lim ε 0 Ω | w ε | d x = Ω ( F v ) v d x + Γ 1 G v d s + A v | Γ 2 . (42)

Thanks to the Lemma 7 we have G ( w ε ) min z K J ( z ) , hence from (22) and (23), we deduce that, as ε , G ( w ε ) min z K J ( z ) .

Remark. As consequences of Theorem 8, we have another characterization of the solution v of the problem (4):

v div ( ω ) = g in ( H D 1 ( Ω ) ) * | ω | ( Ω ) = Ω | ω ε | d x Ω ( F v ) v d x + Γ 1 G v d s + A v | Γ 2 . (43)

Moreover, we have another characterization of H div , K ( Ω ) which will be useful for the numerical part:

H div , K ( Ω ) = { ϕ + L , ϕ H }

where L is the variational solution to the Laplace problem

Δ L = 0 in Ω L = 0 on Γ 0 L ν = G on Γ 1 , L constanton Γ 2 L ν = A | Γ 2 | on Γ 2 . (44)

and

H = { φ H div ( Ω ) : Ω div ( φ ) ξ = Ω φ ξ , ξ H D 1 ( Ω ) } .

Consequently, the dual problem (18) is equivalent to

inf { G ( ϕ ) = G ( ϕ + L ) : ϕ H } . (45)

For the numerical approximation of the problem (45), we use the finite element method, then we assume that:

· Ω is an open and bounded subset of 2 ;

· T h will be a regular partitioning of Ω ¯ by n disjoint open simplex τ , with Ω ¯ = τ T h τ ¯ .

Let H h be a finite dimensional subspace of R T 0 ( T h ) H , with a finite dimension equal to N = N ( h ) , where R T 0 ( T h ) is the space of lowest-order Raviart-Thomas finite elements (see [25] ) defined by

H h = { q h ( L 2 ( Ω ) ) 2 : q τ h = a τ + b τ x , a 2 , b , τ T h , and ( q τ h q τ h ) ν τ = 0 on τ τ } ,

where ν τ represents the outward unit normal to τ , the boundary of τ .

By r h we denote the interpolation operator onto H h given in ( [25], Theorem 6.1). Then, thanks to [25], for all w H div ( Ω ) , we have

r h ( w ) w in ( L 2 ( Ω ) ) 2 and div ( r h ( w ) ) div ( ω ) in L 2 ( Ω ) , as h 0. (46)

We have the following convergence properties as h tends to 0.

Theorem 10. Let F L 2 ( Ω ) , G L 2 ( Γ 1 ) , v a solution to the minimization problem (14) and w h a solution of the optimization problem

sup { G ( q h ) : q h H h } . (47)

Then, as h 0 ,

div ( ω h ) v f ˜ in L 2 ( Ω ) (48)

and

G ( ω h ) min z K J ( z ) = 1 2 Ω | v g | 2 d x Γ N ( Ω K ( s , y ) L ( y ) d y L ( s ) ) v ( s ) d s . (49)

The proof of this result is similar to the proof of Theorem 3.8 in the paper of Igbida et al.; therefore, we omit it.

4. Numerical Simulations

We start by approximating the term τ | q h + Z | d x on each element τ of the partitioning T h . We take

τ | q h + L | d x | τ | | q h + L | ( P τ ) ,

where | τ | represents the area of simplex τ and ( P τ ) is one of the vertices of τ . Using this approximation, at each time n × d t , n and d t the time step, the solution of (47) is a minimizer of the non-differentiable functional G : s ,

G ( w h ) : = 1 2 ( A w h , w h ) + ( d t f h n + u n 1 , div ( w h ) ) + τ T h | τ | | w h + L | ( P τ ) .

The minimization of the functional G is done according to the Gauss-Seidel type algorithm.

In all the tests the discretization in time d t = 0.001 and convergence criterion ε = 10 6 , Ω = ( 1,1 ) 2 . For the discretization of the problem, we use the Raviart-Thomas elements of the lowest order [25] on a regular square grid.

After having obtained the minimizer w h of the functional G , the solution u n of the Euler implicit time discretization of (1) is computed by using extremality relation (43) in a weak sense with piecewise finite elements P 0 .

In the first test, the source f ( x , t ) is a constant equal to 1, for all time and for all x Ω . The Neumann boundary conditions q ν = m u ν = 0 are considered on the boundary { ( 1, y ) | y ] 1,1 [ } which simulated the existence of a wall on the boundary. The outgoing flow across the border Γ 1 = { ( 1, y ) | y ] 1,1 [ } at any time t is is Γ 1 m u ν d s = 1 .

Figure 1 shows the configuration of the sandpile at time t = 1.520 when the solution u becomes stationary. On sees also the effect of the non-local boundary condition of the configuration of the sandpile.

In the second test (see Figure 2), one analyzes the impact of the condition of Neumann on the configuration of the sandpile in the absence of the source f. We can also see that the flux is concentrated around Γ 0 = { ( 1, y ) | y ] 1,1 [ } . This example allows us to conclude that the function g behaves likes a source of sand.

In the last simulation, the three data f, g and A act at the same time. Figure 3 and Figure 4 show respectively an intermediate and the final configuration

Figure 1. Final configuration of the sandpile surface at t = 1.520 for f 1 , A 1 and g = 0 .

Figure 2. Sandpile surface and the flux at t = 15 for f 0 , A 0 and g 20 .

Figure 3. Sandpile surface and the flux at t = 0.787 for f 1 , A 1 and g 0.1 .

Figure 4. Sandpile surface and the flux at t = 15 for f 1 , A 1 and g 0.1 .

of the sandpile and the behavior of the flow. We can see here that the final configuration of the sandpile is slightly different from that of the test 1 (see Figure 1). This is due to the sign of the total outflow through Γ 1 . We also remark that the flux has the same direction as the gradient of the surface and almost zero on the axis y = 0 .

Conflicts of Interest

The author declares no conflicts of interest regarding the publication of this paper.

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