Table Of ContentAN EXISTENCE RESULT FOR THE SANDPILE PROBLEM
ON FLAT TABLES WITH WALLS
GRAZIANO CRASTA AND STEFANO FINZI VITA
Abstract. We derive an existence result for solutions of a differential system which
characterizestheequilibriaofaparticularmodelingranularmattertheory,theso-called
partially open table problem for growing sandpiles. Such result generalizes a recent
theorem of [6] established for the totally open table problem. Here, due to the presence
ofwallsattheboundary,thesurfaceflowdensityattheequilibriummayresultnomore
8
continuous nor bounded, and its explicit mathematical characterization is obtained by
0
domaindecompositiontechniques. Atthesametimeweshowhowthesesolutionscanbe
0
2 numericallycomputedasstationarysolutionsofadynamicaltwo-layermodelforgrowing
sandpiles and we present the results of some simulations.
n
a
J
4
2 1. Introduction
] In the last years an increasing attention has been devoted towards the study of differen-
P
tial models in granular matter theory (see, e.g., [2] for an overview of different theoretical
A
approaches and models). This field of research, which is of course of strong relevance in
.
h the applications, has also been the source of many new and challenging problems in the
t
a theory of partial differential equations (see, e.g., [3, 6, 11, 16]).
m In this paper we deal with the rather simple phenomenon of the evolution of a sandpile
[ created by pouring dry matter on a flat bounded table. In such a model the table is
represented by a bounded domain Ω ⊂ R2, and the time-independent vertical matter
1
v source by a nonnegative function f ∈ L1(Ω). Recently, Hadeler and Kuttler [15] proposed
9 a new model, extending the ones studied in [4] and [5], where the description of the heap
0
evolution is based on the observation that granular matter forms heaps and slopes (the
8
3 so-called standing layer), while small amounts of matter move down along the slopes,
. forming the so-called rolling layer. We also mention [16], where Prigozhin has studied,
1
0 both from the theoretical and numerical points of view, a degenerate parabolic problem
8 anditsequivalentformulationasavariationalinequality,and[3],whereasimilarapproach
0
hasbeenusedforgrowingsandpilesonthewholeplane. Itisworthtoremarkthatthetwo
:
v different dynamical models of [15] and [16] (see for example [17] for a comparison between
i
X them) have theoretically the same set of admissible equilibria.
r Letusdenotebyu(t,x)andv(t,x),t ≥ 0,x ∈ Ω,respectivelytheheightsofthestanding
a
and rolling layers. Neglecting wind effects, the local dynamics depends on the local slope
|Du|andonthelocaldensityofrollingmatterv (thatweshallcalltransportdensity). For
stability reasons, at any equilibrium configuration the local slope |Du| cannot exceed a
fixedconstant(thecriticalslope),thatwenormalizetothevalue1,anditmustbemaximal
where transport occurs (that is, where v > 0). Moreover, we assume that the matter falls
down the table when the base of the heap touches a portion Γ of the boundary of Ω. From
Date: February 3, 2008.
2000 Mathematics Subject Classification. Primary 35C15, 49J10; Secondary 35Q99, 49J30.
Key words and phrases. Distance function, granular matter, flat tray problem, Hamilton-Jacobi equa-
tions, mass transport.
1
2 G.CRASTAANDS.FINZIVITA
the model point of view we are thus assuming that on ∂Ω\Γ we have a (arbitrarily high)
vertical wall, while on Γ the table is “open”. In the following, we shall refer to the open
table problem in the case Γ = ∂Ω, whereas in the partially open table problem, Γ will be
a nonempty closed subset of ∂Ω.
The dynamical model proposed in [15] deals with the open table problem, but it can be
extended to the partially open table problem in the following way:
∂ v = div(vDu)−(1−|Du|)v+f in [0,∞)×Ω
∂ttu = (1−|Du|)v in [0,∞)×Ω
(1) u(0,·) = v(0,·) = 0 in Ω
u = 0 on Γ , v∂u = 0 on ∂Ω\Γ,
∂ν
where ∂u denotes the normal derivative of u. The nonlinear term which appears in the
∂ν
previousequationswithoppositesignsrepresentstheexchangetermbetweenthetwolayers
during the growth process. Before the equilibrium has been reached, a partial surface flow
is allowed also at sub-critical slopes. As far as we know, a rigorous theory for this model is
not known, and its equilibrium configurations have not been characterized in the general
case. For the open table case, a finite difference scheme has been proposed in [13], which
offers at the same time a tool for the numerical description of stationary solutions.
Fromthephysicalconsiderationsabove, anequilibriumconfiguration(u,v)for(1), with
u,v nonnegative functions in Ω, must satisfy
−div(vDu) = f in Ω,
|Du| ≤ 1 a.e. in Ω,
(2)
|Du| = 1 in {v > 0},
u = 0 on Γ, v ∂u = 0 on ∂Ω\Γ.
∂ν
In the case of the open table problem (i.e., Γ = ∂Ω), solutions of (2) have been completely
characterized by Cannarsa and Cardaliaguet [6] (see also [7] for an extension to higher
dimensions). More precisely, denoting by d : Ω → R the distance function from the
∂Ω
boundary of Ω, they proved that there exists a nonnegative continuous function v (with
f
an explicit integral representation) such that (d ,v ) is a solution of (2), and any other
∂Ω f
solution (u,v) satisfies v = v in Ω and u = d in {v > 0}.
f ∂Ω f
Aim of this paper is to extend these results to the case of the partially open table
problem. Asweshallseeinthesequel,thepresenceofverticalwallshasarelevantinfluence
on the regularity of stationary solutions. Namely, we cannot expect the transport density
v to be a continuous function as in the case of the open table problem. This fact has
several consequences.
First of all, we cannot give a pointwise meaning to the boundary conditions in (2). Our
choice here it to set v in L1(Ω), and to consider a weak formulation of the problem (see
(9) below). Another possible choice, which we do not pursue here, could be to set v in
the class of functions with bounded variation, so that the trace of v on ∂Ω is well defined.
Our main result (see Theorem 2.2 below) states that there exists a nonnegative function
v ∈ L1(Ω), of which we give an explicit integral representation, such that (d ,v ) is a
f Γ f
solution to the weak formulation of the problem, where d denotes the distance function
Γ
from Γ (see (5) for its precise definition).
The second major consequence of the lack of continuity of v concerns the uniqueness of
solutions. Namely, the uniqueness of the transport density v for the open table problem
was proved in [6] using a blow-up argument, first introduced in [11] for the analysis of
SANDPILES ON FLAT TABLES WITH WALLS 3
mass transport problems in the framework of the Monge-Kantorovich theory, that relies
onthecontinuityofv. Inouropinionthisargumentcannotbeadaptedtothecasev ∈ L1.
Nevertheless, it could be possible to prove a uniqueness result in a restricted class of more
regular functions F ⊂ L1(Ω), provided that one can show that v ∈ F.
f
A precise formulation of the existence result mentioned above requires some notation.
The table Ω ⊂ R2 will be a Lipschitz domain, i.e. an open bounded connected set with
Lipschitz boundary, and the open boundary Γ ⊂ ∂Ω will be a nonempty closed subset of
∂Ω. Let d: Ω×Ω → [0,∞) denote the path distance in Ω, defined by
(3) d(x,y) = inf{length(γ); γ: [0,1] → Ω Lipschitz path joining x to y},
and let
(4) Lip1(Ω) = {u: Ω → R; u Lipschitz, u(x)−u(y) ≤ d(x,y) ∀x,y ∈ Ω}
be the set of 1-Lipschitz functions in Ω with respect to the path-metric d. Let us denote
by d : Ω → [0,∞) the path distance function from Γ, defined by
Γ
(5) d (x) = inf d(x,y), x ∈ Ω.
Γ
y∈Γ
It is easily seen that, if Γ = ∂Ω, then d is the Euclidean distance from the boundary of
Γ
Ω. Moreover, d belongs to the space of functions
Γ
(6) Lip1(Ω) = {u ∈ Lip1(Ω); u = 0 on Γ}.
Γ
It is also known that d is the maximal function among all functions in Lip1(Ω). Thus d
Γ Γ Γ
is the maximal size of the standing layer.
Givenanonnegativefunctionf ∈ L1(Ω),letu : Ω → [0,∞)denotethefunctiondefined
f
by
(7) u (x) = max{G (x); z ∈ supp(f)} x ∈ Ω,
f z
where, for every x,z ∈ Ω,
(cid:40)
d (z)−d(z,x), if d(z,x) ≤ d (z),
Γ Γ
(8) G (x) =
z
0, otherwise,
and supp(f) denotes the (essential) support of f, that is, the complement in Ω of the
union of all relatively open subsets A ⊆ Ω such that f = 0 a.e. in A.
It is readily seen that the graph of the positive part of G is, in the path metric, the
z
maximal cone of unitary slope, with apex in z, whose base is contained in Ω and touches
Γ at some point. It is clear that 0 ≤ u ≤ d , and that u ∈ Lip1(Ω). Moreover, since u
f Γ f f
is the sup-envelope of all the cones G with z ∈ supp(f), it is plain that u represents the
z f
minimal standing layer for an equilibrium configuration with respect to the given support
of the source.
We can now state the weak formulation of problem (2): Find nonnegative functions
u,v: Ω → [0,+∞) such that
u(cid:90) ∈ Lip1Γ(Ω), v(cid:90)∈ L1(Ω), u,v ≥ 0,
(9) vDu·Dφ = fφ ∀φ ∈ C∞(R2\Γ),
c
|DΩu| = 1 Ω a.e. in {v > 0},
where C∞(R2 \Γ) denotes the space of C∞ functions φ: R2 → R with compact support
c
in R2\Γ. In this weak formulation the boundary conditions on u and v are embedded in
4 G.CRASTAANDS.FINZIVITA
the choice of the test functions space C∞(R2\Γ) and in the condition u ∈ Lip1(Ω). It is
c Γ
readily seen that, if u and v are smooth enough, then problem (9) is equivalent to (2).
The plan of the paper is the following. In Section 2 we state the main existence result,
constructing explicitly a transport density v (see formula (16) below) and showing that
f
a pair (u,v ) is a solution to problem (9) for every u ∈ Lip1(Ω) satisfying u ≤ u ≤ d .
f Γ f Γ
Moreover, we show that no other function u ∈ Lip1(Ω) can be the standing layer for the
Γ
problem. Section3containstheproofsoftheseresults, thataremainlybasedonaChange
of Variables formula of some independent interest (see Theorem 3.3 below). In Section 4
we compute the explicit solution in a simple case, which will be compared in Section 5 to
the numerical equilibrium solution of the dynamical model (1) obtained via some finite
difference schemes. Indeed, in this last section the specific difficulties of such a numerical
approximation are discussed in details.
2. Existence of a solution
Throughout this paper, Ω ⊂ R2 and Γ ⊂ ∂Ω satisfy the following assumptions.
(H1) Ω ⊂ R2 is a Lipschitz domain, i.e. a nonempty open bounded connected set with
Lipschitz boundary.
(H2) Γ = (cid:83)N Γ is a nonempty closed subset of ∂Ω, with Γ ,...,Γ connected arcs
i=1 i 1 N
of ∂Ω, pairwise disjoint (up to the endpoints) and of class C2. We denote by A ,
i
B the endpoints of the arc Γ , i = 1,...,N, and by Γe the collection of all these
i i
endpoints.
(H3) For every x ∈ Ω there exists y ∈ Γ such that d (x) = |x−y|.
Γ
Remark 2.1. Observe that (H2) does not prevent the intersection of two arcs at the end-
points. This is the case, for example, if Ω is a square and Γ , i = 1,...,4, its sides.
i
Nevertheless, in order to simplify the proofs, in the following we assume that the arcs Γ
i
do not intersect at the endpoints. The general case can be treated with minor modifica-
tions.
Condition (H3) says that, for every x ∈ Ω, there is a point y ∈ Γ such that the closed
segment [[x,y]] with endpoints x and y is a path of minimal length joining x to Γ.
There are three relevant cases where these conditions are satisfied:
(a) Γ = ∂Ω, and Ω is a domain of class C2.
(b) Γ = ∂Ω, and Ω is a domain with piecewise C2 boundary.
(c) Ω is a non-empty open bounded convex set, and Γ satisfies condition (H2) above.
Case (a) and (b) refer to the open table problem. They were considered respectively in
[6](seealso[7]foranextensiontoΩ ⊂ Rn)and[14]inthecaseofpiecewiseC2,1 boundary
with outer (i.e. “convex”) corners.
Let Γ∗ = Γ\Γe. For every y ∈ Γ∗ let ν(y) denote the inward unit normal vector of ∂Ω,
and let κ(y) denote the curvature of ∂Ω at y.
For every x ∈ Ω we denote by
Π (x) = {y ∈ Γ; d (x) = d(x,y)}
Γ Γ
the set of all projections of x on Γ. By (H3) it is clear that
Π (x) = {y ∈ Γ; d (x) = |x−y|}.
Γ Γ
SANDPILES ON FLAT TABLES WITH WALLS 5
R Γ R
R
R
R
Γ
Γ Γ
Figure 1. OpenboundaryΓ(dotted), extendedridgeR(bold)andtrans-
port rays
For every x ∈ Ω and every y ∈ Π (x), let us define
Γ
(10) l(x) = d (x)·sup{t > 0; y+t(x−y) ∈ Ω and y ∈ Π (y+t(x−y))},
Γ Γ
x−y
(11) m(x) = y+l(x) ,
|x−y|
(12) R = ]]y,m(x)[[,
x
where]]x,y[[denotesthesegmentjoiningxtoy withouttheendpoints. ThesetR iscalled
x
a distance ray (or transport ray) through x and, in general, it depends on the projection
point y. On the other hand, it is not difficult to prove that m(x) and l(x) do not depend
on the choice of y ∈ Π (x). It is readily seen that l(x) is the length of the transport ray
Γ
R . The set
x
(13) R = {x ∈ Ω; d (x) = l(x)}
Γ
will be called the extended ridge of Ω (see Figure 1). It coincides with the usual definition
of ridge when Γ = ∂Ω (see [12]). Finally let τ: Ω → [0,∞) denote the normal distance to
the ridge, defined by
(14) τ(x) = l(x)−d (x), x ∈ Ω.
Γ
This function is clearly bounded from above by max{d (x); x ∈ Ω}.
Γ
Let Ω∗ ⊂ Ω be the set of regular points of d , that is the set of all x ∈ Ω such that the
Γ
projection Π (x) is a singleton. It is well-known that L2(Ω\Ω∗) = 0.
Γ
Let x ∈ Ω∗, let Π (x) = {y}, and define for every t ∈ [0,τ(x)]
Γ
d (x)+t
Γd (x) , if y ∈ Γe,
(15) M (t) = Γ
x 1−(d (x)+t)κ(y)
Γ , if y ∈ Γ∗.
1−d (x)κ(y)
Γ
Let us define the function
(cid:90) τ(x)
f(x+tDd (x))M (t)dt, if x ∈ Ω∗,
(16) v (x) = Γ x
f 0
0, if x ∈ Ω\Ω∗.
We remark that, if ]]y,z[[ is a transport ray, then z ∈ R, and
lim v (x) = 0.
f
x→z, x∈]]y,z[[
In other words, the transport density v vanishes at the end of each transport ray.
f
The main theoretical contribution of this paper is the following existence result.
6 G.CRASTAANDS.FINZIVITA
B
1
B
Γ 2
1 Ω∗ ΩΒ1 ΩΒ
1 2
A
1
ΩΑ Ω∗
1 ΩΑ 2
2
Γ
2
A
2
Figure 2. An example of domain decomposition
Theorem 2.2 (Existence). Let Ω and Γ satisfy (H1)-(H3). Then the pair (u,v ) is a
f
solution of (9) for every u ∈ Lip1(Ω) satisfying u ≤ u ≤ d . Furthermore, if (u,v) is
Γ f Γ
any solution of (9), then u ≤ u ≤ d .
f Γ
It is apparent that the result above characterizes all the possible standing layers as the
functions u ∈ Lip1(Ω) satisfying u ≤ u ≤ d . On the other hand, the uniqueness of the
Γ f Γ
transport density v remains an open problem.
f
3. Proof of Theorem 2.2
In what follows we shall always use the following decomposition of the set Ω∗ of regular
points of d . For every i = 1,...,N, let us define the sets
Γ
Ω∗ = {x ∈ Ω∗; Π (x) ∈ Γ∗},
i Γ i
(17)
ΩA = {x ∈ Ω∗; Π (x) = {A }}, ΩB = {x ∈ Ω∗; Π (x) = {B }}.
i Γ i i Γ i
Then (see Fig. 2 for an example)
N
Ω∗ = (cid:91) (cid:16)Ω∗∪ΩA∪ΩB(cid:17) .
i i i
i=1
Next, we need an approximation of this decomposition in terms of sets that can be
easily parameterized (see Theorem 3.2 below). For every (cid:15) > 0 let us define the sets
(cid:26) (cid:27)
(18) G(cid:15) = x ∈ R2; min|x−y| < (cid:15) , Γ(cid:15) = ∂G(cid:15), i = 1,...,N.
i i i
y∈Γi
From Remark 2.1 and (H3), there exists r > 0 with the following property: for every (cid:15) ∈
(0,r], the sets G(cid:15) are pairwise disjoint and of class C1,1. Given (cid:15) ∈ (0,r] and i = 1,...,N,
i
for every y ∈ Γ(cid:15) let ν(cid:15)(y) denote the outer (with respect to G(cid:15)) unit normal vector to Γ(cid:15)
i i i
at y. In the following, (cid:15) will always denote a number in (0,r].
Let us define the maps
(19) Φ(cid:15): Γ(cid:15)×R → R2, Φ(cid:15)(y,t) = y+(t−(cid:15))ν(cid:15)(y), y ∈ Γ(cid:15), t ∈ R, i = 1,...,N.
i i i i
The following lemma states that Ω can be parameterized using the maps Φ(cid:15). Moreover,
i
this parametrization is independent of the choice of (cid:15). Let us define
N N
Γ(cid:101)(cid:15) = Γ(cid:15)∩Ω, i = 1,...,N, Γ(cid:15) = (cid:91) Γ(cid:15), Γ(cid:101)(cid:15) = (cid:91) Γ(cid:101)(cid:15),
i i i i
i=1 i=1
and denote by Π(cid:15) = Π the projection operator on Γ(cid:15).
Γ(cid:15)
SANDPILES ON FLAT TABLES WITH WALLS 7
ε
Ω
ε
Γ
Γε
Figure 3. The set Ω
(cid:15)
Lemma 3.1. Let x ∈ Ω and let y ∈ Π (x). If y ∈ Γ , then for every (cid:15) ∈ (0,r) there exists
Γ i
a unique y ∈ Γ(cid:15) such that y ∈ Π(cid:15)(x) and x = Φ(cid:15)(y ,d (x)). Moreover, x ∈ ]]y,y [[, if
(cid:15) i (cid:15) i (cid:15) Γ (cid:15)
dΓ(x) < (cid:15), and y(cid:15) ∈ ]]y,x[[, y(cid:15) ∈ Γ(cid:101)(cid:15)i, if dΓ(x) > (cid:15).
Proof. Let y be the intersection of Γ(cid:15) with the ray starting from y and passing through
(cid:15) i
x. From the definition of Γ(cid:15) it is straightforward to check that y satisfies all the stated
i (cid:15)
properties. (cid:3)
Let y(cid:15) ∈ Γ(cid:101)(cid:15), and let y ∈ Γ be the unique projection of y(cid:15) on Γ. Let us define
(cid:40)
−1/(cid:15), if y ∈ Γe,
(20) κ(cid:15)(y ) =
(cid:15) κ(y)/(1−(cid:15)κ(y)), if y ∈ Γ∗.
It can be checked that κ(cid:15)(y ) is the curvature of Γ(cid:15) at every point y where the curvature
(cid:15) (cid:15)
is defined. We remark that, for every i = 1,...,N, κ(cid:15) is defined at all but four points of
Γ(cid:101)(cid:15)i. Let x ∈ Ω∗, let Π(cid:15)(x) = {y(cid:15)}, and define
1−(d (x)+t−(cid:15))κ(cid:15)(y )
(21) M(cid:15)(t) = Γ (cid:15) , t ∈ [0,τ(x)].
x 1−(d (x)−(cid:15))κ(cid:15)(y )
Γ (cid:15)
It is easy to check that, if Π (x) = {y}, then
Γ
(22) M(cid:15)(t) = M (t), ∀t ∈ [0,τ(x)].
x x
The main tool needed for the proof of Theorem 2.2 is a Change of Variables formula
(see Theorem 3.3 below). As a first step, we need the following approximation result (see
Figure 3).
Theorem 3.2. For every (cid:15) ∈ (0,r] and i = 1,...,N, let
N
Ω(cid:15) = {Φ(cid:15)(y,t); y ∈ Γ(cid:101)(cid:15), 0 < t < l(y)}, Ω(cid:15) = (cid:91) Ω(cid:15).
i i i
i=1
Then Ω(cid:15) ⊂ Ω, and lim L2(Ω\Ω(cid:15)) = 0.
(cid:15)→0
Proof. It is not difficult to see that the set
N
O(cid:15) := (cid:91){Φ(cid:15)(y,t); y ∈ Γ(cid:101)(cid:15), 0 < t ≤ l(y)}
i
i=1
covers Ω up to a set of Lebesgue measure Nπ(cid:15)2, i.e. L2(Ω\O(cid:15)) < Nπ(cid:15)2. Moreover, from
Lemma 3.1 it is clear that Ω(cid:15) ⊂ O(cid:15) ⊂ Ω. The conclusion now follows observing that
N
O(cid:15)\Ω(cid:15) ⊆ D := (cid:91){Φ(cid:15)(y,l(y)); y ∈ Γ(cid:101)(cid:15)} ⊆ R,
i i
i=1
8 G.CRASTAANDS.FINZIVITA
and that D has vanishing Lebesgue measure (see [9], Corollary 6.8). (cid:3)
Let us define the map Φ(cid:15): Γ(cid:101)(cid:15)×R → R2 by setting Φ(cid:15)(y,t) = Φ(cid:15)(y,t) if y ∈ Γ(cid:101)(cid:15).
i i
Theorem 3.3 (Change of Variables). For every h ∈ L1(Ω) we have
(cid:34) (cid:35)
(cid:90) (cid:90) (cid:90) l(y)
h(x)dx = lim h(Φ(cid:15)(y,t))[1−(t−(cid:15))κ(cid:15)(y)]dt dH1(y)
Ω (cid:15)→0 (cid:101)Γ(cid:15) 0
(23)
(cid:34) (cid:35)
(cid:90) (cid:90) l(y)−(cid:15)
= lim h(y+tν(cid:15)(y))[1−tκ(cid:15)(y)]dt dH1(y).
(cid:15)→0 (cid:101)Γ(cid:15) −(cid:15)
Proof. From Theorem 3.2 we have that
(cid:90) (cid:90) N (cid:90)
(cid:88)
h(x)dx = lim h(x)dx = lim h(x)dx.
Ω (cid:15)→0 Ω(cid:15) (cid:15)→0 Ω(cid:15)
i=1 i
For every i = 1,...,N we have the decomposition
Ω(cid:15) = Ω(cid:15),∗∪Ω(cid:15),A∪Ω(cid:15),B,
i i i i
defined as in (17) replacing Ω∗ with Ω(cid:15).
Using the arguments developed in [9, Sect. 7], it can be checked that
(cid:34) (cid:35)
(cid:90) (cid:90) (cid:90) l(y)
(24) h(x)dx = h(Φ(cid:15)(y,t))[1−(t−(cid:15))κ(cid:15)(y)]dt dH1(y).
Ω(cid:15),∗ (cid:101)Γ(cid:15)∩Ω(cid:15),∗ 0
i i i
We shall now prove that
(cid:34) (cid:35)
(cid:90) (cid:90) (cid:90) l(y)
(25) h(x)dx = h(Φ(cid:15)(y,t))[1−(t−(cid:15))κ(cid:15)(y)]dt dH1(y).
Ω(cid:15),A (cid:101)Γ(cid:15)∩Ω(cid:15),A 0
i i i
Let us define the region
A(cid:15) = {Φ(cid:15)(y,t); y ∈ Γ(cid:101)(cid:15)i ∩Ω(cid:15)i,A, 0 < t < l(y)}.
It is clear that A ⊂ Ω(cid:15),A. Hence, it is enough to prove that
(cid:15) i
(cid:34) (cid:35)
(cid:90) (cid:90) (cid:90) l(y)
(26) h(x)dx = h(Φ(cid:15)(y,t))[1−(t−(cid:15))κ(cid:15)(y)]dt dH1(y).
A(cid:15) (cid:101)Γ(cid:15)i∩Ω(cid:15)i,A 0
Observethat,bytheverydefinitionofκ(cid:15),wehavethatκ(cid:15)(y) = −1/(cid:15)foreveryy ∈ Γ(cid:101)(cid:15)∩ΩA.
i i
Then, the integral in brackets becomes
(cid:90) l(y) (cid:90) l(y) t
h(Φ(cid:15)(y,t))[1−(t−(cid:15))κ(cid:15)(y)]dt = h(Φ(cid:15)(y,t)) dt.
(cid:15)
0 0
On the other hand, the integral over A can be computed using polar coordinates (ρ,θ).
(cid:15)
We remark that Γ(cid:101)(cid:15) ∩Ω(cid:15),A is an arc of circumference of radius (cid:15), so that dH1(y) = (cid:15)dθ,
i i
hence formula (26) follows.
It is clear that (25) holds if Ω(cid:15),A is replaced by Ω(cid:15),B. The identity (23) then follows
i i
from (24) and (25). (cid:3)
Lemma 3.4. The function u defined in (7) satisfies u = d in the set {v > 0}.
f f Γ f
Moreover, if u ∈ Lip1(Ω) is a function satisfying u ≤ u ≤ d in Ω, then Du = Dd in
f Γ Γ
{v > 0}.
f
SANDPILES ON FLAT TABLES WITH WALLS 9
Proof. From the very definition of u it is plain that u (x) ≥ d (x) for every x ∈ supp(f).
f f Γ
On the other hand, by the maximality of d among all functions of Lip1(Ω) vanishing on
Γ
Γ, we conclude that
(27) u = d in supp(f).
f Γ
Now let x ∈ {v > 0} and let us prove that u (x) = d (x). From the definition of v
f f Γ f
we have that x ∈ Ω∗, that is, Π (x) is a singleton {y}, with y ∈ Γ. Let [[y,z]] be the
Γ
transport ray through x. Since v (x) > 0, we have that ]]x,z[[∩supp(f) (cid:54)= ∅. Let x be
f 0
a point belonging to this intersection. From (27) we have that u (x ) = d (x ), which in
f 0 Γ 0
turn implies that u = d along the segment [[y,x ]]. Since x ∈ [[y,x ]], we conclude that
f Γ 0 0
u (x) = d (x), and the first part of the lemma is proved.
f Γ
Thesecondpartfollowsfrom[1,Prop.4.2](seealso[8,Lemma7.3]and[10,Lemma4.3])
upon observing that u = d in {v > 0}. (cid:3)
Γ f
Lemma 3.5. Let (u,v) be a solution of (9). Then
(cid:90) (cid:26)(cid:90) (cid:27)
(28) fu = max fw; w ∈ Lip1(Ω) .
Γ
Ω Ω
Proof. By a standard approximation argument we have that
(cid:90) (cid:90)
(29) vDu·Dφ = fφ ∀φ: Ω → R Lipschitz, φ = 0 on Γ.
Ω Ω
Let w ∈ Lip1(Ω). Since v ≥ 0, |Dw| ≤ 1, |Du| ≤ 1, and |Du| = 1 a.e. in {v > 0}, we have
Γ
that
(cid:90)
vDu·(Dw−Du) ≤ 0.
Ω
Hence we infer that
(cid:90) (cid:90) (cid:90)
fu− fw ≥ vDu·(Dw−Du)−f(w−u) = 0,
Ω Ω Ω
where the last equality follows from (29) with φ = w−u. Since this inequality holds for
every w ∈ Lip1(Ω), (28) follows. (cid:3)
Γ
Proof of Theorem 2.2. We divide the proof into three steps. In the first one we shall prove
that (d ,v ) is a solution to (9). Then, in the second step, we shall prove that also (u,v )
Γ f f
is a solution for every u ∈ Lip1(Ω) satisfying u ≤ u ≤ d . Finally, in the last step we
f Γ
shall prove that if u ∈ Lip1(Ω) is a nonnegative function with {u < u } (cid:54)= ∅, then for
Γ f
every nonnegative function v ∈ L1(Ω) the pair (u,v) is not a solution of (9).
Step 1. We give only a sketch of the proof, since it follows the lines of the proof of
Theorem 7.2 in [9].
Given φ ∈ C∞(R2\Γ), we have to prove that (9) holds with u = d . Using the Change
c Γ
of Variables formula (23) we have that
(cid:34) (cid:35)
(cid:90) (cid:90) (cid:90) l(y)
(30) fφ = lim φ(Φ(cid:15)(y,t))f(Φ(cid:15)(y,t))[1−(t−(cid:15))κ(cid:15)(y)]dt dH1(y).
Ω (cid:15)→0 (cid:101)Γ(cid:15) 0
For every fixed y ∈ Γ(cid:101)(cid:15), let us integrate by parts the term in brackets. Recalling that
Φ(cid:15)(y,0) = y −(cid:15)ν(cid:15)(y) ∈ Γ, we have that φ(Φ(cid:15)(y,0)) = 0, hence the integration by parts
gives
(cid:90) l(y)
I(y) := φ(Φ(cid:15)(y,t))f(Φ(cid:15)(y,t))[1−(t−(cid:15))κ(cid:15)(y)]dt
0
(cid:90) l(y) (cid:90) l(y)
= Dφ(Φ(cid:15)(y,t))·ν(cid:15)(y) f(Φ(cid:15)(y,s))[1−(s−(cid:15))κ(cid:15)(y)]dsdt.
0 t
10 G.CRASTAANDS.FINZIVITA
From the definition (16) of v and (22) we deduce that
f
(cid:90) l(y) 1−(s−(cid:15))κ(cid:15)(y)
v (Φ(cid:15)(y,t)) = f(Φ(cid:15)(y,s)) ds,
f 1−(t−(cid:15))κ(cid:15)(y)
t
so that
(cid:90) l(y)
I(y) = Dφ(Φ(cid:15)(y,t))·ν(cid:15)(y)v (Φ(cid:15)(y,t))[1−(t−(cid:15))κ(cid:15)(y)]dt.
f
0
Observe now that DdΓ(Φ(cid:15)(y,t)) = ν(cid:15)(y) for every y ∈ Γ(cid:101)(cid:15) and every t ∈ (0,l(y)). Substi-
tuting the last expression for I(y) in (30) and using again the Change of Variables formula
(23) we finally get
(cid:34) (cid:35)
(cid:90) (cid:90) (cid:90) l(y)
fφ = lim (v Dφ·Dd )(Φ(cid:15)(y,t))[1−(t−(cid:15))κ(cid:15)(y)]dt dH1(y)
f Γ
Ω (cid:15)→0 (cid:101)Γ(cid:15) 0
(cid:90)
= v Dφ·Dd .
f Γ
Ω
By the way, choosing φ = d (see (29)) we have that
Γ
(cid:90) (cid:90)
v = fd < +∞,
f Γ
Ω Ω
hence the nonnegative function v belongs to L1(Ω).
f
Step 2. Let u ∈ Lip1(Ω) satisfy u ≤ u ≤ d . From Lemma 3.4 we have that u = d and
f Γ Γ
Du = Dd in the set {v > 0}, hence
Γ f
(cid:90) (cid:90) (cid:90)
v Du·Dφ = v Dd ·Dφ = fφ ∀φ ∈ C∞(R2\Γ),
f f Γ c
Ω Ω Ω
where the last equality follows from Step 1.
Step 3. Let u ∈ Lip1(Ω) be a nonnegative function with {u < u } =(cid:54) ∅. From [10],
Γ f
Proposition 4.4(iii), we deduce that also the set {x ∈ supp(f); u(x) < d (x)} is not
Γ
empty and, in particular, u < d on a set of positive measure where f > 0. Since u ≤ d
Γ Γ
and f ≥ 0, we infer that
(cid:90) (cid:90)
(31) fu < fd .
Γ
Ω Ω
On the other hand (u,v) is a solution to (9), so that (28) holds, in contradiction with
(31). (cid:3)
4. A test example
Inthissectionwedescribeasimpleexamplewhichillustratesverywellhowthepresence
of vertical walls on the boundary can influence the regularity of solutions (u,v) of (9). Let
Ω = (0,1)2 be the unit square of R2, and Γ = {(x,y) ∈ R2 : 0 ≤ x ≤ 0.5 ; y = 0} the only
open part of its boundary. Assume f ≡ 1 in all of Ω. From the picture in Fig. 4 we see
that the sand transport rays behave differently in the two half sides of the table: in the
left-hand side they lay parallel in the direction of Γ, whereas in the right-hand side they
converge all together into the extremal point P = (0.5,0), creating a singularity.
Since f = 1 in Ω, we have that u = d so that the only possible standing layer is
f Γ
u = d . The explicit computation for the solution (d ,v ) of Theorem 2.2 can be done
Γ Γ f
by decomposition of the domain Ω along the segment PQ. Using polar coordinates (r,θ)
