Ammari H., Benkirane A., Touzani A. (editors) Recent Developments in Nonlinear Analysis
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Let E be any measurable set in IR
N
, such that 0 < λ(E) < ∞. Then there
exists a positive constant C such that
Z
E
|u −u
E
| dλ ≤ C |||∇u|||
A,E
,
for all u ∈ W
1
A,loc
(IR
N
).
3) Let A be an N-function such that A and A
∗
satisfy the ∆
2
condition.
Suppose that IR
N
is A-hyperbolic. Let E be any non empty bounded domain
in IR
N
. Then there exists a positive constant C such that for all u ∈ E
0
A
(E)
Z
E
|u| dλ ≤ C |||∇u|||
A
.
Recall that for all f ∈ L
A
such that |||f|||
A
> 1, we have
R
A ◦ f dλ >
|||f|||
A
. We set s(A) = inf
n
log
R
A◦f dλ
log|||f|||
A
− 1, f ∈ L
A
, |||f|||
A
> 1
o
. Hence
s(A) ≥ 0. One can consult [37] for some examples of N-functions satisfying
s(A) > 0.
Now we are ready to solve the A-Laplace equation.
Theorem 9.3. Let L
A
be a reflexive Orlicz space such that s(A) > 0. Let
h ∈ L
∞
(IR
N
) have compact support. Then the equation ∆
A
u + h = 0 has a
weak solution u ∈ L
1
A
(IR
N
) if IR
N
is A-hyperbolic.
Remark 9.2. 1) We have in fact solved the equation in the space E
0
A
(D) ⊂
L
1
A
(IR
N
).
2) When A(t) = p
−1
t
p
, p > 1, L
A
= L
p
is the usual Lebesgue space,
we have s(A) = p − 1 > 0. Thus we recover the result in [94], when the
manifold M is IR
N
.
Corollary 9.2. Let L
A
(IR
N
) be a reflexive Orlicz space such that s(A) > 0
and α < N. Suppose that h ∈ L
∞
(IR
N
) has compact support. Then the
equation ∆
A
u + h = 0 has a weak solution u ∈ L
1
A
(IR
N
).
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266
Back on stochastic model for sandpile
Noureddine Igbida
LAMFA, UMR 6140, Universit´e de Picardie Jules Verne
33 rue Saint Leu, 80038 Amiens, France
E-mail: noureddine.igbida@u-picardie.fr
Our aim in this note is to give a simplified proof of the convergence of Evans-
Rezakhanlou Stochastic Model to the evolution surfaces model of sandpile.
Keywords: Evans-Rezakhanlou Stochastic Model; Convergence.
1. Introduction
The stochastic model for sandpile was introduced by Evans and Rezakhan-
lou in
6
as a variant of physical models for sandpile. It corresponds to a
Markov process
X(t)
t≥0
defined by an infinitesimal generator describ-
ing the evolution of stack of unit cubes resting on the plane when new
cubes are being added to the pile, by being placed either upon a heretofore
unoccupied square in the plane or else upon the top of a current column.
At each time t > 0, the configuration X(t) needs to be stable which means
that the heights of any two adjacent columns of cubes can differ by at most
December 28, 2009 12:15 WSPC - Proceedings Trim Size: 9in x 6in recent
267
one. So adding new cubes on existing pile, we ordain two possibilities to
each cube:
• if the configuration is stable, then the cube remains in place
• otherwise, the cube has several downhill ”staircases” along which it can
move, and the cube will randomly select among the allowable downhill
paths.
If we consider the case where the cubes can be also taken away, then a third
possibility may be ordain
• the cube has several ”upward staircases” along which it can move, and
the cube will randomly select among the allowable upward paths.
Assuming that we continuously add cubes at random locations on a starting
empty stack, Evans and Rezakhanlou studies the limit when one rascals
in both space and time, so as to consider growing piles of more and more
smaller and smaller cubes ? They prove that the macroscopic limit is rather
simple and very connected Prigozhin model for sandpile (cf.
8
and
2
).
Since the work of Prigohzin (cf.
8
see also
2
), this has been well known
(see also,
36
and the references therein) that the evolution of the surface of
the sandpile when the angle of stability is equal to π/4 can be described by
the following evolution problem
∂
t
u + ∂II
K
(u) 3 f
ˆu(0) = 0,
(1)
where K =
n
z ∈ W
1,∞
(IR
2
) ∩ L
2
(IR
2
) ; |∇z| ≤ 1
o
, ∂II
K
denotes the sub-
differential operator of the indicator function (cf.
4
) of K and f describes a
source term . Solution u is the height of the surface that grows up (resp.
grows down) under sand addition (resp. sand removal) by a source called f .
This is a critical slope model obtained by using the continuity equation and
the gradient constraint |∇u| ≤ 1 a.e. in IR
2
(see
8
and
2
for more details).
Existence, uniqueness and numerical approximation the solution are well
known by now for this model (cf.,
8
,
25
and
7
). More precisely, for any f ∈
BV
loc
(0, T ; L
2
(IR
2
)), we know that that (1) has a unique solution u in the
sense that u ∈ W
1,∞
loc
(0, ∞; L
2
(IR
2
)), u(0) = 0 and, for any t ≥ 0, u(., t) ∈ K
and
Z
IR
2
f(t, x) −∂
t
u(t, x)
u(t, x) − ξ(x)
≥ 0 for any ξ ∈ K. (2)
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268
To give the connection between the two models, let N be a large integer.
Assume the cubes are of side length O(N
−1
) and cubes newly and randomly
added at rate O(N
−1
) are continuously falling downhill. For the description
of the evolving of cubes, the authors of
6
introduce a probabilistic lattice
model. They thus consider a Markov process for the height X(t, i), defined
for times t ≥ 0 and sites i ∈ ZZ
2
. Resacaled source terms f
+
t
N
,
i
N
and
f
−
t
N
,
i
N
control the rate new cubes are added to the pile or removed
from it. Then, they proved that
IE
h
Z
IR
2
1
N
X
N t, [N x]
− u(t, x)
2
i
→ 0 as N → ∞, (3)
where u is the unique solution of (1). As shown in,
6
the key estimates for
the proof (3) are some kind of microscopic version of (2) (see (10)). To
prove these estimates the authors of
6
prove some elementary intermediate
estimates evolving various types of sets. Our aim in this note, is to improve
these key estimates directly by using some simple arguments.
In the next section, we recall the stochastic model of Evans and Reza-
khanlou. In Section 3, we prove the key estimates (10). Then, as in,
6
we
introduce the discrete evolution problem associated with (1). Then, we give
the proof of (3) by using some results of.
1
2. The stochastic model for sandpile problem (cf.
6
)
To describe the stochastic process for sandpile, we consider the lattice ZZ
n
.
We equipped ZZ
n
with an Euclidean norm and we say that i, j ∈ ZZ
n
are
adjacent, written i ∼ j, provided
|i −j| ≤ 1.
Without loose of generality, we restrict ourself to the cases n = 2, and we
write i = (i
1
, i
2
) to denote a typical site in ZZ
2
. Then, we introduce the
Hilbert space
H := l
2
(ZZ
2
) =
n
X : ZZ
2
→ IR ; kXk :=
X
i
X(i)
2
< ∞
o
.
A (stable) configuration is a mapping X : ZZ
2
→ ZZ such that
|X(i) −X(j)| ≤ 1 if i ∼ j
and X has bounded support.
December 28, 2009 12:15 WSPC - Proceedings Trim Size: 9in x 6in recent
269
The state space is
S :=
n
X : ZZ
2
→ ZZ ; X is a configuration
o
,
and the set of stable configuration is
ˆ
K :=
n
X ∈ H ; |X(i) −X(j)| ≤ 1 if i ∼ j
o
.
Let
ˆ
f : (0, ∞) × ZZ
2
→ ZZ be a function, such that f
+
(resp. f
−
) is
controlling the rate new cubes are added (resp. removed) to the pile. The
source of cubes
ˆ
f generates a stochastic process (X(t), t ≥ 0) in the state
space S. It is clear that the probability that X(t) be situated (at time t) in
a given set Γ of E, under the condition that the movement of the system
up to time s (s < t) is completely known, depends only on the state of
the system at time s. In other words (X(t), t ≥ 0) is a Markov process. To
study this process, we need to know its infinitesimal generator A, or more
precisely AF (X(t)) for any F ∈ B(S), the set of bounded functions. To
this aim, let us consider p
+
(i, j, ξ) the probability that a cube placed on
a given configuration ξ ∈ S at the position i will end up at j after it has
fallen downward over the stack with eight ξ. Likewise, let p
−
(i, j, ξ) be the
probability that the removal of a cube from the pile at i will result in a
removal at site j, after the cubes along a staircase each shifts downwards
to fill in the gap created at site i. So, for any i, j ∈ ZZ
2
we have
0 ≤ p
±
(i, j, ξ) ≤ 1 and
X
i∈ZZ
2
p
±
(i, j, ξ) = 1.
Thanks to,
6
we consider c
±
given by
c
±
(j, X, t) =
X
i∈ZZ
2
p
±
(i, j, X(t))
ˆ
f
±
(t, i), for any (t, j) ∈ ZZ
2
× [0, ∞).
(4)
The parameter c
+
(j, X, τ ) (resp. c
−
(j, X, τ )) is highly nonlocal factor and
records the rate, at time τ, new cubes come to rest at the site j after falling
downhill (resp. the rate at which cubes are removed from the site j, at time
τ). In particular, we have
X
j∈ZZ
2
c(j, X, t) =
X
i∈ZZ
2
f(t, i).
Thanks again to,
6
the infinitesimal generator A of the Markov process