PORTUGALIAE MATHEMATICA
Vol. 56 Fasc. 1 – 1999
EXISTENCE RESULTS FOR SOME QUASILINEAR ELLIPTIC
PROBLEMS WITH RIGHT HANDSIDE IN L1
N. Grenon and O.A. Isselkou
Abstract: We study the existence of unbounded renormalized solutions, for quasilinear elliptic equations in a bounded domain. In a first part, we introduce the symmetrized
problem, and we get an existence result assuming the existence of a renormalized supersolution of the symmetrized problem. Afterwards, we get a sub-super solution theorem
for an equation with a more general right handside.
1 – Introduction
Let Ω be an open bounded set of RN with N ≥ 1. We consider the following
problem:
(
− div A(x, u, Du) = F (x, u) in Ω,
(1.1)
u = 0 on ∂Ω .
We assume that:
(1.2)
A(x, s, ξ) is a Caratheodory function: Ω × RN +1 → RN ,
(1.3)
D
(1.4)
α |ξ|p ≤ hA(x, s, ξ), ξi
E
A(x, s, ξ)−A(x, s, ξ 0 ), ξ −ξ 0 > 0
a.e. x ∈ Ω, ∀ s ∈ R, ∀ ξ, ξ 0 ∈ RN , ξ 6= ξ 0 ,
a.e. x ∈ Ω, ∀ s ∈ R, ∀ ξ ∈ RN ,
|A(x, s, ξ)| ≤ β(|s|) (|ξ|p−1 + b(x))
(1.5)
a.e. x ∈ Ω, ∀ s ∈ R, ∀ ξ ∈ RN
where β is a function: [0, +∞[ → [0, +∞[ defined
everywhere and bounded on the bounded intervalls
0
and where b is a positive function of Lp (Ω) ,
Received : October 14, 1996; Revised : April 7, 1998.
AMS Subject Classification: 35D05, 35J60.
40
N. GRENON and O.A. ISSELKOU
F (x, s) is a Caratheodory function: Ω × R → R+ ,
(1.6)
m
X
fi (x) × gi (s)
i=0
∈ L1 (Ω), fi (x) ≥
0 ≤ F (x, s) ≤
(1.7)
where m ∈ N and fi (x)
0, 0 ≤ i ≤ m and,
for 0 ≤ i ≤ m, gi : R → ]0, +∞[, continous, nondecreasing .
We shall denote by f ? (s) the unidimensional decreasing rearrangement of f ,
that is to say, the unique decreasing function such that |f > t| = |f ? > t| for
every t. We shall denote by f˜(x) the spherical decreasing rearrangement of f ,
that is to say f˜(x) = f ? (ωN |x|N ) for every x in Ω̃, where Ω̃ is the ball of RN
centered at the origin, such that |Ω̃| = |Ω|, and where ωN is the measure of the
unit ball in RN . For all the definitions and properties concerning symetrization
see [5].
Let us consider the symmetrized problem:




(1.8)
m
X
−α ∆p u =
f˜i (x) gi (u)
in Ω̃,
i=0


u = 0
on ∂ Ω̃ ,
where ∆p u = Div(|Du|p−2 Du). We shall use the following notations and definitions:
We note:

if u ≥ k,

 k
u
Tk u =


L0 (Ω),
−k
if − k < u < k,
if u ≤ −k ,
and
the space of measurable functions wich are finite a.e. in Ω. Let us
recall the definition of [7]:
Definition 1.1.
that:
We call renormalized solution of (1.1) a function u such
u ∈ L0 (Ω) ,
Tk u ∈ W01,p (Ω),
1
k
Z
Z
|Du|p dx → 0
∀ k ∈ R+ ,
when k → +∞ ,
k≤|u|≤2k
A(x, u, Du) Du h0 (u) w dx +
Ω
Z
A(x, u, Du) Dw h(u) dx =
Ω
Z
F (x, u) h(u) w dx ,
Ω
∀ w ∈ W01,p (Ω) ∩ L∞ (Ω) and ∀ h ∈ C 1 (R) or piecewise affine
and with compact support.
QUASILINEAR ELLIPTIC PROBLEMS WITH RIGHT HANDSIDE IN L1
41
In the same way, we define a renormalized supersolution:
Definition 1.2.
such that:
We call renormalized supersolution of (1.1) a function ψ
ψ ∈ L0 (Ω) ,
Tk ψ ∈ W 1,p (Ω),
∀ k ∈ R+ ,
∃ Cψ ∈ R+ such that, ∀ k ∈ R+ , 0 ≤ ψ ≤ Cψ on ∂Ω ,
1
k
Z
Z
|Dψ|p dx → 0
when k → +∞ ,
k≤ψ≤2k
Z
A(x, ψ, Dψ) Dψ h0 (ψ) w dx + A(x, ψ, Dψ) Dw h(ψ) dx ≥
Ω
Z
F (x, ψ) h(ψ) w dx ,
Ω
Ω
∀ w ∈ W01,p (Ω) ∩ L∞ (Ω) and ∀ h ∈ C 1 (R) or piecewise affine
and with compact support.
The definition of a renormalized subsolution is obtained exchanging ≥ by ≤.
Let us remark that if a renormalized solution u is in W01,p (Ω) ∩ L∞ (Ω), then u is
an ordinary weak solution, that is to say u verifies:
Z
A(x, u, Du) Dϕ =
Ω
Z
F (x, u) ϕ
Ω
∀ ϕ ∈ W01,p (Ω) .
This is also true for sub and supersolutions. The main result of this work is the
following:
Theorem 1.1. We suppose that A satisfies (1.2), (1.3), (1.4), (1.5), and that
F verifies (1.6), (1.7). If there exists a supersolution ψ ≥ 0 for the problem (1.8),
then there exists a renormalized nonnegative solution u for problem (1.1) such
that |u > t| ≤ |ψ > t|.
Theorem 1.1 is a generalization of ([6]). In this paper the functions fi are
supposed to be in Lq (Ω) with q ≥ max(p0 , N/p) and ψ in L∞ (Ω) and of course
u is also in L∞ (Ω), moreover in [6], A is roughly independent of u. Notice that
q ≥ max(p0 , N/p) insure that the problem:
(1.9)
(
−α ∆p u = f (x)
u=0
in Ω,
on ∂Ω ,
has a solution in W01,p (Ω) ∩ L∞ (Ω) if f ∈ Lq (Ω). Here, f is in L1 (Ω), and then the
solution of (1.9) is no more in L∞ (Ω). Such problems with right handside in L1
have been studied in [1] and in [7] in which renormalized solutions are introduced.
42
N. GRENON and O.A. ISSELKOU
To prove this theorem, we shall first get a comparison result with the symmetrized
problem, and in a second time we shall prove a sub-super solution theorem.
2 – Comparison with the symmetrized problem
Let us consider the following problem:

m
X

 − div A(x, u, Du) =
fi (x) gi (u)
i=0


(2.1)
u=0
in Ω,
on ∂Ω .
Theorem 2.1. We suppose that A satisfies (1.2), (1.3), (1.4), (1.5), and
that the functions fi and gi satisfy (1.7). If problem (1.8) has a renormalized
supersolution ψ ≥ 0, then problem (2.1) has a nonnegative renormalized solution
u such that |u > t| ≤ |ψ > t|, for all t ≥ 0.
Proof: Let n ∈ N , we set, for 0 ≤ i ≤ m, fi,n (x) = inf(f (x), n).
Let v ∈ L∞ (Ω), we consider the problem:
(2.2)

m
X

 −Div A(x, u, Du) =
fi,n (x) gi (v)
i=0


u=0
in Ω,
on ∂Ω .
Recall that a weak subsolution of (2.2), is a function v ∈ W 1,p (Ω) ∩ L∞ (Ω) which
verifies:
(2.3)
Z
Z X
m


fi,n (x) gi (v) dx ϕ
A(x, v, Dv) Dϕ dx ≤
Ω
Ω i=0


v≤0
∀ ϕ ∈ W01,p (Ω),
on ∂Ω .
We prove the following lemma:
Lemma 2.1. We suppose that A satisfies (1.2), (1.3), (1.4), (1.5), and that
the functions fi and gi satisfy (1.7). Moreover we suppose that v ≥ 0 verifies
(2.3), then there exists a nonnegative weak solution u ∈ W01,p (Ω) ∩ L∞ (Ω) of
(2.2) such that u ≥ v.
43
QUASILINEAR ELLIPTIC PROBLEMS WITH RIGHT HANDSIDE IN L1
Let M > 0 such that: 0 ≤ v(x) ≤ M . We set:
³
´
ĀM x, u(x), Du(x) =
´
 ³

A
x,
M,
Du(x)



 ³
´
if u(x) ≥ M,
A x, u(x), Du(x)
if v(x) ≤ u(x) ≤ M,


³
´


 A x, v(x), Du(x)
then the problem:
if u(x) ≤ v(x) ;

m
X

 − div Ā(x, u, Du) =
fi,n (x) gi (v)
i=0


(2.4)
u=0
in Ω,
on ∂Ω ,
has at least one nonnegative weak solution u ∈ W01,p (Ω) ∩ L∞ , such that:
kuk∞ ≤
Z
|Ω|
0
0
0
−p0 /N −p0 +p0 /N
α−p/p N −p ωN
s
µZ s ³X
m
0
= Cn .
i=0
´∗
fi,n gi (v) (σ) dσ
¶p0 /p
ds
The existence comes from the theorem of [4, p. 180], moreover u is nonnegative
because the right handside is nonnegative, and L∞ estimate can be proved by
symmetrization techniques (see for instance [5] and the demonstrations below).
Remark that Cn is independent of M , and then we can choose M such that:
(2.5)
M > Cn .
We are now going to prove that u ≥ v. We take (v − u)+ as test function in
(2.3) and (2.2), then,
Z ³
Ω
´
A(x, v, Dv) − ĀM (x, u, Du) D(v − u)+ ≤ 0
but on {x ∈ Ω, v ≥ u} we have ĀM (x, u, Du) = A(x, v, Du), then from (1.3), we
obtain:
(v − u)+ = 0
and so,
(2.6)
u≥v .
From (2.5) and (2.6), we can deduce that ĀM (x, u, Du) = A(x, u, Du) and so u
is in fact solution of (2.2). This proves Lemma 2.1.
44
N. GRENON and O.A. ISSELKOU
We are now going to construct a sequence (un ) in the following way:
we set
u0 = 0 ;
suppose that the sequence is defined until un−1 then un is a solution of:
(2.7)

m
X



gi (un−1 ) fi,n (x),

 − div A(x, un , Dun ) =
i=0
un ≥ un−1 ,





1,p
un ∈ W0 (Ω) ∩ L∞ (Ω) .
We have to show that the sequence (un ) is well defined:
For n = 0,

(2.8)
m
X

 − div A(x, 0, 0) = 0 ≤
fi,1 (x) gi (0),
i=0


u0 = 0 ≤ 0
on ∂Ω ,
that is to say, u0 is a subsolution of problem corresponding to u1 , and so from
Lemma 2.1, u1 exists. Suppose that the sequence is defined until un−1 , then:
− div A(x, un−1 , Dun−1 ) =
m
X
gi (un−2 ) fi,n−1 (x)
in Ω
i=0
and
un−1 ≥ un−2 ;
then, as for 0 ≤ i ≤ m, gi is nondecreasing,
− div A(x, un−1 , Dun−1 ) ≤
m
X
gi (un−1 ) fi,n (x)
in Ω
i=0
and then un exists from Lemma 2.1. On another hand we construct a sequence
(vn ), in the following way:
we set
v0 = 0
and vn ∈ W01,p (Ω̃) ∩ L∞ (Ω̃) is a solution of:
−α ∆p vn =
m
X
gi (vn−1 ) f˜i,n (x)
in Ω̃ .
i=0
We are going to prove that the sequence (vn ) has the following property:
vn−1 ≤ vn ≤ ψ
∀n ≥ 1 .
QUASILINEAR ELLIPTIC PROBLEMS WITH RIGHT HANDSIDE IN L1
45
Recall that we suppose that ψ is a renormalized supersolution of problem
(1.8). For n = 0, we have v1 ≥ v0 = 0. In the inequation satisfied by ψ, we take
w = (v1 − ψ)+ which is in W01,p (Ω̃) ∩ L∞ (Ω̃) and for h a function h ∈ C 1 (R) such
that h(s) = 1 if s ≤ kv1 k∞ , and h(s) = 0 if s ≥ kv1 k∞ +1. Then h(ψ)w = w.
In the equation satisfied by v1 we take (v1 − ψ)+ as test fuction. This leads to:
α
Z ³
Ω
´
|Dv1 |p−2 Dv1 − |Dψ|p−2 Dψ D(v1 − ψ)+ ≤
≤
Z µX
m
Ω
gi (v0 ) f˜i,1 (x) −
m
X
¶
gi (ψ) f˜i (x) (v1 − ψ)+ ≤ 0
i=0
i=0
and thus,
v1 ≤ ψ .
Suppose by induction that:
vn−2 ≤ vn−1 ≤ ψ .
Similarly, in the inequation satisfied by ψ, we take w = (vn − ψ)+ which is in
and for h a function h ∈ C 1 (R) such that h(s) = 1 if s ≤ kvn k∞ ,
and h(s) = 0 if s ≥ kvn k∞ +1. In the equation satisfied by vn we take (vn − ψ)+
as test function. As gi is nondecreasing, we obtain:
W01,p (Ω̃)∩L∞ (Ω̃)
Z ³
Ω
´
|Dun |p−2 Dvn − |Dψ|p−2 Dψ D(vn − ψ)+ ≤
≤
Z µX
m
Ω
gi (vn−1 ) f˜i,n (x) −
m
X
¶
gi (ψ) f˜i (x) (vn − ψ)+ ≤ 0 .
i=0
i=0
Now if we take (vn−1 − vn )+ as test function in the equations satisfied by vn−1
and vn , after substraction, we obtain:
Z ³
Ω
´
|Dvn−1 |p−2 Dvn−1 − |Dvn |p−2 Dvn D(vn−1 − vn )+ ≤
≤
Z µX
m
Ω
i=0
gi (vn−2 ) f˜i,n−1 (x) −
m
X
¶
gi (vn−1 ) f˜i,n (x) (vn−1 − vn )+ ,
i=0
gi is nondecreasing, and by induction vn−2 ≤ vn−1 , thus:
Z ³
Ω
´
|Dvn−1 |p−2 Dvn−1 − |Dvn |p−2 Dvn D(vn−1 − vn )+ ≤ 0
and thus,
vn−1 ≤ vn .
46
N. GRENON and O.A. ISSELKOU
For all s in R, we note s− = − inf(s, 0). Let τ be a function of W01,p (Ω̃) such
that 0 ≤ τ ≤ 1, then, (vn − kτ )− ∈ W01,p (Ω̃) ∩ L∞ (Ω̃) and k(vn − kτ )− k∞ ≤ k.
We take −(vn − kτ )− as test function in the equation satisfied by vn , and we note
Ck,τ different constant which depend on k and τ ,
α
Z
{vn ≤kτ }
≤k
Z
{vn ≤kτ }
|Dvn |p−1 |Dτ | dx −
Z
|Dvn |p dx ≤
m
X
{vn ≤kτ } i=0
gi (vn−1 ) f˜i,n (x) (vn − kτ )−
then,
α
Z
{vn ≤kτ }
|Dvn |p dx ≤ Ck,τ
³Z
{vn ≤kτ }
|Dvn |p dx
´ p−1
p
dx + Ck,τ
and thus,
Z
{vn ≤kτ }
|Dvn |p dx ≤ Ck,τ
a fortiori,
Z
(2.9)
{τ ≡1}
|DTk vn |p dx ≤ Ck,τ .
We now specify the choice of τ , we take τ = T1 ((ψ − Cψ − 1)+ ), then w ≡ 1
on {τ < 1}, and w ≡ 0 on {ψ ≥ Cψ +2}. In the equation satisfied by vn , we take
w vn which is in W01,p (Ω) ∩ L∞ (Ω), as test function, and we obtain
α
Z
Ω
|Dvn |p w dx + α
Z
Ω
|Dvn |p−2 Dvn Dw vn dx =
Z X
m
Ω i=0
gi (vn−1 ) f˜i,n (x) w un dx
then,
α
Z
p
{w≡1}
|Dvn | dx + α
≤C
≤C
Z
Z
Z
{w<1}
Ω
|Dvn |p w dx ≤
|Dvn |p−1 |Dw| dx + C
{w≡1}
|Dvn |
p−1
|Dw| dx + C
Z
{w<1}
|Dvn |p−1 |Dw| dx + C
but, {x ∈ Ω, w(x) < 1} ⊂ {x ∈ Ω, τ (x) = 1}, then from (2.9),
α
Z
{w≡1}
p
|Dvn | dx ≤ C
Z
{w≡1}
|Dvn |p−1 |Dw| dx + C
QUASILINEAR ELLIPTIC PROBLEMS WITH RIGHT HANDSIDE IN L1
47
and thus,
α
Z
{w≡1}
|Dvn |p dx ≤ C
but, {x ∈ Ω, τ (x) < 1} ⊂ {x ∈ Ω, w(x) < 1}, then,
α
(2.10)
Z
{w≡1}
|Dvn |p dx ≤ C ;
from (2.9) and (2.10) we deduce that:
TK vn is bounded in W01,p (Ω) .
(2.11)
We are now going to show that ũn ≤ vn a.e. in Ω̃.
For n = 0, ũ0 = 0 = v0 .
We set:

0
if s ≤ t,



1
ϕ(s) =
(s − t)
if t < s ≤ t + h,

h



1
if s > t + h .
We can take ϕ(un ) as test function in the equation satisfied by un , that leads to:
1
h
Z
A(x, un , Dun ) Dun dx =
{t<un ≤t+h}
1
h
+
m
X
Z
{t<un ≤t+h} i=0
Z
m
X
{t+h<un } i=0
From (1.4), and because 0 <
α
h
Z
p
{t<un ≤t+h}
un −t
h
|Dun | ≤
Z
gi (un−1 ) fi,n (x) (un−t) dx
gi (un−1 ) fi,n (x) dx .
≤ 1 on {t < un ≤ h + t}, we get:
m
X
{t<un ≤t+h} i=0
Z
m
X
+
{t+h<un } i=0
gi (un−1 ) fi,n (x) dx
gi (un−1 ) fi,n (x) dx ;
from Hölder,
1
α
h
µ
Z
{t<un ≤t+h}
≤
Z
|Dun |
¶p µ
m
X
{t<un ≤t+h} i=0
1
h
Z
dx
{t<un ≤t+h}
¶− p0
p
gi (un−1 ) fi,n (x) dx +
Z
≤
m
X
{t+h<un } i=0
gi (un−1 ) fi,n (x) dx .
48
N. GRENON and O.A. ISSELKOU
We note ν(t) = |un > t|. Let h tend to zero.
d
α −
dt
µ
Z
{t<un }
|Dun |
¶p ³
0
−ν (t)
´− p0
p
≤
Z
m
X
{t<un } i=0
gi (un−1 ) fi,n (x) dx .
From the definition of the perimeter of De Giorgi, and the isoperimetric
inequality, we have:
−
d
dt
Z
1/N
{t<un }
|Dun | ≥ N ωN ν(t)1−1/N
then,
p/N
α N p ωN ν(t)p−p/N −ν 0 (t)
¡
¢− p0
p
≤
m Z
X
gi (un−1 ) fi,n (x) dx
i=0 t<un
but, from the extension of Hardy–Littlewood theorem, which is proved in [6],
m Z
X
gi (un−1 ) fi,n (x) dx ≤
i=0 t<u
m Z
X
ν(t)
?
(gi (un−1 ))? (σ) fi,n
(σ) dσ .
i=0 0
As gi is nondecreasing, we obtain:
m Z
X
i=0 t<u
gi (un−1 ) fi,n (x) dx ≤
Z
0
m
ν(t) X
?
gi (u?n−1 )(σ) fi,n
(σ) dσ
i=0
thus,
¡
¢p/p0
1
−p/N
1 ≤ N −p ωN ν(t)−p+p/N −ν 0 (t)
α
Z
0
m
ν(t) X
?
gi (u?n−1 )(σ) fi,n
(σ) dσ
m
ν(t) X
?
gi (u?n−1 )(σ) fi,n
(σ) dσ
i=0
and thus,
1 ≤ α
−p0 /p
N
−p0
0
0
−p0 /N
ωN
ν(t)−p +p /N
0
(−ν (t))
µZ
0
i=0
¶p0 /p
then, we integrate between 0 and u?n (s) − ² with ² > 0. We know that:
then,
u?n (s)−²
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯un > u?n (s) − ²¯ = ¯u?n > u?n (s) − ²¯ ≤ ¯u?n > u?n (s)¯ ≤ s
≤ α
−p0 /p
N
−p0
−p0 /N
CN
Z
|Ω
r
s
−p0 +p0 /N
µZ
m
rX
0 i=0
?
gi (u?n−1 )(σ) fi,n
(σ) dσ
¶p0 /p
dr .
49
QUASILINEAR ELLIPTIC PROBLEMS WITH RIGHT HANDSIDE IN L1
As it is true for every ² > 0, we obtain:
u?n (s)
≤α
−p0 /p
N
−p0
−p0 /N
CN
Z
|Ω
r
−p0 +p0 /N
s
µZ
m
rX
0 i=0
?
gi (u?n−1 )(σ) fi,n
(σ) dσ
¶p0 /p
dr .
We suppose by induction that,
?
u?n−1 (σ) ≤ vn−1
(σ)
then,
u?n (s)
≤ α
−p0 /p
N
−p0
−p0 /N
CN
Z
|Ω
r
−p0 +p0 /N
s
µZ
= vn? (s) .
m
rX
0 i=0
?
?
(σ) dσ
)(σ) fi,n
gi (vn−1
¶p0 /p
dr
The last step consists in proving that (un ) converges to a renormalized solution
of (2.1). First we take Tk un as test function in the equation satisfied by un ,
Z
Ω
A(x, un , Dun ) DTk un dx ≤
Z X
m
Ω i=0
gi (un−1 ) fi,n (x) Tk un dx ;
this implies from (1.4) that (we note Ck different constants independent of n, but
which depend on k)
α
Z
Ω
|DTK un |p dx ≤
Z
m
X
un ≤k i=0
+k
Z
gi (un−1 ) fi,n (x) un dx
m
X
un ≥k i=0
gi (un−1 ) fi,n (x) dx .
We know that if un (x) ≤ k then un−1 (x) ≤ k, then on {un ≤ k}, we have
gi (un−1 ) ≤ Ck and fi,n (x) ≤ fi (x). Moreover in the second term of the right
handside of the previous inequality, we can use the extension of the Hardy–
Littlewood theorem which is given in [6], and we obtain:
α
Z
Ω
|DTK un |p dx ≤ Ck
We can add
Pm R
Z X
m
Ω i=0
fi (x) dx + k
˜
m Z
X
i=0 {ũn ≥k}
i=0 {ũn <k} gi (ũn−1 ) fi,n (x) ũn (x) dx
gi (ũn−1 ) f˜i,n (x) dx .
which is nonnegative in the
50
N. GRENON and O.A. ISSELKOU
right handside, and so,
α
Z
Ω
|DTK un |p dx ≤ Ck
+
Z X
m
Ω i=0
fi (x) dx + k
m Z
X
i=0 {ũn <k}
= Ck
Z X
m
Ω i=0
m Z
X
i=0 {ũn ≥k}
gi (ũn−1 ) f˜i,n (x) dx
gi (ũn−1 ) f˜i,n (x) ũn (x) dx
fi (x) dx +
m Z
X
i=0 Ω̃
gi (ũn−1 ) f˜i,n (x) Tk ũn (x) dx .
We have seen that, ∀ n ∈ N , ũn ≤ vn ≤ ψ a.e. in Ω̃, then,
α
Z
p
Ω
|DTK un | dx ≤ Ck
= Ck
≤ Ck
Z X
m
Ω i=0
Z X
m
Ω i=0
fi (x) dx +
m Z
X
fi (x) dx +
Z
i=0 Ω̃
Ω̃
gi (vn−1 ) f˜i,n (x) Tk vn dx
|DTk vn |p dx
because of (2.11), so we have proved that,
(2.12)
kDTK un kp ≤ Ck .
As the sequence (un ) is nondecreasing, p.p. x ∈ Ω, un (x) tends to infinity or
converges to a finite limit, we note u(x). Let A = {x ∈ Ω, un (x) → +∞}, and let
Bn,k = {x ∈ Ω, un (x) > k}, then ∀ k ≥ 0,
A⊂
+∞
[
Bn,k
n=0
and
¯ +∞
¯
¯[
¯
Bn,k ¯ = lim |Bn,k |
¯
n→∞
n=0
∀k ≥ 0
because (un ) is nondecreasing. But,
then,
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯{x ∈ Ω, un (x) > k}¯ ≤ ¯{x ∈ Ω̃, vn (x) > k}¯ ≤ ¯{x ∈ Ω̃, ψ(x) > k}¯
and consequently
¯
¯
¯
¯
|A| ≤ ¯{x ∈ Ω̃, ψ(x) > k}¯ ,
¯
¯
∀k ∈ N
¯
¯
|A| ≤ lim ¯{x ∈ Ω̃, ψ(x) > k}¯ = 0
k→+∞
QUASILINEAR ELLIPTIC PROBLEMS WITH RIGHT HANDSIDE IN L1
51
then,
un (x) → u(x)
a.e. x ∈ Ω .
We are now going to show that (un ) converges to a renormalized solution of
(2.1).
From (2.12), we can deduce that DTk un → DTk u in Lp (Ω) weak. We are now
going to show that DTk un → DTk u in Lp (Ω) strong. We take Tk un − Tk u as test
function in the equation satisfied by un , then,
Z
Ω
A(x, un , Dun ) D(Tk un − Tk u) dx =
Z X
m
Ω i=0
gi (un−1 ) fi,n (x) (Tk un − Tk u) dx
thus,
Z ³
Ω
´
A(x, un , Dun ) − A(x, un , DTk u) D(Tk un − Tk u) dx +
+
=
Z X
m
Ω i=0
Z
Ω
A(x, un , DTk u) D(Tk un − Tk u) dx =
gi (un−1 ) fi,n (x) (Tk un − Tk u) dx .
As un (x) ≤ u(x), Tk un − Tk u ≡ 0 on {x ∈ Ω, un (x) ≥ k}. Then the previous
inequality becomes:
Z
{un ≤k}
(2.13)
³
´
A(x, Tk un , DTk un ) − A(x, Tk un , DTk u) D(Tk un − Tk u) dx +
+
=
Z
Z
m
X
{un ≤k} i=0
{un ≤k}
A(x, Tk un , DTk u) D(Tk un − Tk u) dx =
gi (un−1 ) fi,n (x) (Tk un − Tk u) dx .
Let n tend to +∞, by Lebesgue theorem, we can see that:
Z
m
X
{un ≤k} i=0
gi (un−1 ) fi,n (x) (Tk un − Tk u) dx → 0
and
A(x, Tk un , DTk u) ξ{un ≤k} → A(x, Tk u, DTk u) ξ{u≤k}
D(Tk un − Tk u) → 0
then,
Z
{un ≤k}
0
in Lp (Ω) strong
in Lp (Ω) weak
A(x, Tk un , DTk u) D(Tk un − Tu ) dx → 0 .
52
N. GRENON and O.A. ISSELKOU
We can now use the following lemma which is proved in [2].
Lemma 2.2. Suppose that A verifies (1.2), (1.3), (1.4), (1.5), if (zn ) is a
sequence such that:
• zn is bounded in L∞ (Ω),
• zn → z in W01,p (Ω) weak and a.e. in Ω,
• lim
Z ³
n→0 Ω
´
A(x, zn , Dzn ) − A(x, zn , Dz) D(zn − z) = 0;
then, zn → z in W01,p (Ω) strong.
We can apply this lemma to TK un and deduce that: TK un → TK u in W01,p (Ω)
strong. Let w ∈ W01,p (Ω) ∩ L∞ (Ω) and h ∈ C 1 (R) or piecewise affine, and with
compact support, and let k such that h ≡ 0 on ]− ∞, −k[ ∪ ]k, +∞[
Z
Ω
A(x, un , Dun ) h0 (un ) w dx +
Z
Ω
A(x, un , Dun ) h(un ) Dw dx =
=
Z X
m
ω i=0
gi (un−1 ) fi,n (x) h(un ) w dx
that is to say, from the choice of k,
Z
Z
0
Ω
A(x, Tk un , DTk un ) h (Tk un ) w dx +
(2.14)
Ω
=
A(x, Tk un , DTk un ) h(Tk un ) Dw dx =
Z X
m
Ω i=0
A(x, Tk un , DTK un ) → A(x, Tk u, DTk u)
gi (Tk un−1 ) fi,n (x) h(Tk un ) w dx ,
a.e. in Ω, moreover from (1.5) ,
¯
¯ p0
³
´p 0
0
¯
¯
.
¯A(x, Tk un , DTK un )¯ ≤ β(k)p |DTk un |p−1 + b(x)
0
The right handside converges in L1 (Ω) strong, consequently |A(x, Tk un , DTk un )|p
0
is equiintegrable, and then from Vitali’s lemma |A(x, Tk un , DTk uN )|p →
0
0
|A(x, Tk u, DTk u)|p . Finally, A(x, Tk un , DTk uN ) → A(x, Tk u, DTk u) in Lp (Ω)
strong and we can pass to the limit in (2.14), and we obtain
Z
0
Ω
A(x, Tk u, DTk u) h (Tk u) w dx +
Z
Ω
A(x, Tk u, DTk u) h(Tk u) Dw dx =
=
Z X
m
Ω i=0
gi (Tk u) fi (x) h(Tk u) w dx
QUASILINEAR ELLIPTIC PROBLEMS WITH RIGHT HANDSIDE IN L1
53
that is to say,
Z
0
A(x, u, Du) h (u) w dx +
Ω
Z
A(x, u, Du) h(u) Dw dx =
Ω
=
Z X
m
Ω i=0
gi (u) fi (x) h(u) w dx .
3 – A sub-supersolution theorem
Theorem 3.1. We suppose that A satisfies (1.2), (1.3), (1.4), (1.5), and that
F satisfies (1.6), (1.7), if there exists a nonnegative renormalized supersolution
ψ of (1.1), then there exists a nonnegative renormalized solution u of (1.1), such
that u ≤ ψ a.e. in Ω.
Proof: We can remark that ϕ = 0 is a subsolution of (1.1) (we could remark
that the hypothesis F (x, s) ≥ 0 could be replaced by: there exists a weak subsolution ϕ ∈ W 1,p (Ω) ∩ L∞ (Ω) such that ϕ ≤ ψ).
Let n ≥ 1, we consider the problem:
(
(3.1)
where Fn (x, s) =
− div A(x, u, Du) = Fn (x, u(x))
u=0
in Ω,
on ∂Ω ,
F (x, s)
.
1 + n1 F (x, s)
Lemma 3.1. Under the hypotheses of Theorem 3.1, we suppose that there
exists a weak subsolution v ∈ L∞ (Ω) of problem (3.1), such that 0 ≤ v ≤ ψ,
then there exists a solution u ∈ W01,p (Ω) ∩ L∞ (Ω) of problem (3.1) such that
v ≤ u ≤ ψ.
Proof of the Lemma: Let M such that kvk∞ ≤ M . We set:
³
´
ĀM x, u(x), Du(x) =
and
´
 ³

A
x,
T
ψ(x),
Du(x)

M


 ³
´
A x, u(x), Du(x)


³
´


 A x, v(x), Du(x)
F̄ (x, u(x)) =

F (x, ψ(x))






if u(x) ≥ TM ψ(x),
if v(x) ≤ u(x) ≤ TM ψ(x),
if u(x) ≤ v(x) ,
if u(x) ≥ ψ(x),
F (x, u(x))
if v(x) ≤ u(x) ≤ ψ(x),
F (x, v(x))
if u(x) ≤ v(x) ,
54
N. GRENON and O.A. ISSELKOU
F̄ (x, u(x))
.
1 + n1 F̄ (x, u(x))
F̄n (x, u(x)) =
Then, − div ĀM (x, u(x), Du(x)) − F̄n (x, u(x)) verifies the hypotheses of the theorem of [4, p. 180], and the problem:

³
´
 − div ĀM x, u(x), Du(x) = F̄n (x, u(x)),
 u = 0 on ∂Ω ,
(3.2)
has at least one solution u in W01,p (Ω) ∩ L∞ (Ω) such that:
kuk∞ ≤
0
1
−p0 +
p0
N
+
0
−p0 + pN + pp +1
p0
p
+1
|Ω|
0
0
p0
α−p/p N −p n p = Dn .
In the following we shall suppose that M ≥ Dn . We are going to show that
moreover, u ≤ ψ. We take (u − ψ)+ as test function in (3.2), and in the inequation satisfied by ψ, we take w = (u − ψ)+ ∈ W01,p (Ω) ∩ L∞ (Ω), and a
function h ∈ C 1 (R) such that h(s) = 1 if s ≤ M and h(s) = 0 if s ≥ M + 1.
As in the previous section, that leads to:
Z µ
Ω
³
´
³
ĀM x, u(x), Du(x) − A x, ψ(x), Dψ(x)
≤
Z ³
Ω
´¶
D(u − ψ)+ dx ≤
´
F̄n (x, u(x)) − F (x, ψ(x) (u − ψ)+ dx = 0
thus,
Z µ
Ω
³
´
³
ĀM x, u(x), Du(x) − A x, ψ(x), Dψ(x)
´¶
D(u − ψ)+ dx ≤ 0
and we deduce that (u − ψ)+ = 0 a.e. in Ω.
We take now (v − u)+ as test function in (3.2) and in the inequation satisfied
by v, and we can show like previously that u ≥ v. Finally, we have shown that u
is a solution of (3.1), and proved the lemma.
We construct a sequence (un ), such that:
u0 = 0 ,
un ∈ W01,p (Ω) ∩ L∞ (Ω) ,
− div A(x, un , Dun ) = Fn (x, un (x)) ,
un−1 ≤ un ≤ ψ .
QUASILINEAR ELLIPTIC PROBLEMS WITH RIGHT HANDSIDE IN L1
55
Using the previous lemma, we can show by induction that we can construct this
sequence, if we remark that
− div A(x, un−1 , Dun−1 ) = Fn−1 (x, un−1 (x)) ≤ Fn (x, un−1 (x))
that is to say, un−1 is a subsolution of the equation satisfied by un .
To prove that Tk un is bounded in W01,p (Ω), we use the same method as in
the previous section to prove (2.11). Let τ be a function of W01,p (Ω) such that
0 ≤ τ ≤ 1, then, (un − kτ )− ∈ W01,p (Ω) ∩ L∞ (Ω) and k(un − kτ )− k∞ ≤ k.
We take −(un − kτ )− as test function in the equation satisfied by un , and we
note Ck,τ different constant which depend on k and τ ,
Z
{un ≤kτ }
A(x, un , Dun ) D(un − kτ ) dx = −
Z
Ω
Fn (x, un ) (un − τ k)− dx
then, from (1.4) and (1.5),
α
Z
p
{un ≤kτ }
|Dun | dx ≤ Ck,τ
Z
{un ≤kτ }
³
´
|Dun |p−1 + b(x) Dτ dx + Ck,τ
then,
α
Z
{un ≤kτ }
|Dun |p dx ≤ Ck,τ
and then,
Z
a fortiori,
Z
(3.3)
{un ≤kτ }
{τ ≡1}
³Z
{un ≤kτ }
|Dun |p dx
´ p−1
p
dx + Ck,τ
|Dun |p dx ≤ Ck,τ
|DTk un |p dx ≤ Ck,τ .
We now specify the choice of τ , we take τ = T1 ((ψ − Cψ − 1)+ ), then w ≡ 1
on {τ < 1} and w ≡ 0 on {ψ ≥ Cψ + 2}. In the equation satisfied by un , we take
w un which is in W01,p (Ω) ∩ L∞ (Ω) as test function.
Z
Ω
A(x, un , Dun ) Dun w dx +
Z
Ω
A(x, un , Dun ) Dw un dx =
Z
Ω
Fn (x, un ) w un dx
then,
α
Z
{w≡1}
|Dun |p dx + α
≤ C
Z ³
Ω
≤ C
Z
Z
{w<1}
|Dun |p w dx ≤
´
|Dun |p−1 + b(x) |Dw| dx + C
{w≡1}
|Dun |
p−1
|Dw| dx + C
Z
{w<1}
|Dun |p−1 |Dw| dx + C
56
N. GRENON and O.A. ISSELKOU
but, {x ∈ Ω, w(x) < 1} ⊂ {x ∈ Ω, τ (x) = 1}, then from (3.3),
α
Z
{w≡1}
|Dun |p dx ≤ C
Z
{w≡1}
|Dun |p−1 |Dw| dx + C
and thus,
α
Z
{w≡1}
|Dun |p dx ≤ C
but, {x ∈ Ω, τ (x) < 1} ⊂ {x ∈ Ω, w(x) < 1}, then,
(3.4)
α
Z
{w≡1}
|Dun |p dx ≤ C
from (3.3) and (3.4) we deduce that TK un is bounded in W01,p (Ω). On another
hand, as (un ) is nondecreasing and un ≤ ψ, un converges almost everywhere in
Ω to a function u. This implies that DTK un → DTK u in Lp (Ω) weak. In the
same way, with slight modifications, we can prove as in the previous section that
Tk un → Tk u in W01,p (Ω) strong, and that u is a renormalized solution of (1.1).
This proves Theorem 3.1.
We can now prove Theorem 1.1: suppose that there exists a renormalized
supersolution ψ ≥ 0 for problem (1.8), then problem (2.1) has a renormalized
solution ū such that |ū > t| ≤ |ψ > t|, ∀ t ≥ 0. But, ū is also a renormalized supersolution of (1.1), and then by Theorem 3.1, there exists a nonnegative
renormalized solution u of problem (1.1), such that u ≤ ū a.e. in Ω, and thus
such that |u > t| ≤ |ψ > t|.
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[6] Mossino, J., Moutoussamy, I. and Simon, A. – Un theorème d’existence et de
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N. Grenon,
Université d’Orleans – Bâtiment de Mathématiques,
Rue de Chartres, 45100 Orléans La Source – FRANCE
and
O.A. Isselkou,
Faculté des Sciences de Nouakchott,
BP 5026, Nouakchott – MAURITANIE