STRONG MAXIMUM PRINCIPLE FOR
GENERAL QUASILINEAR PROBLEMS
IN NON DIVERGENCE-FORM
Rabah TAHRAOUI
CEREMADE - Universit´e Paris 9 - Dauphine, UMR 7534
Place du Mar´echal de Lattre de Tassigny
75775 Paris Cedex 16 - France
and
I.U.F.M. Rouen
2, rue du Tronquet - 76131 Mont-St-Aignan - France
I - Introduction
In this paper we deal with the study of the following question : does a positive non null solution u of the problem
−X ij aij(x, u, ∇u) ∂2u ∂xi∂xj ≥ F (x, u, ∇u) in IRN u(x) ≥ 0 ∀x, (1)
vanish at some point x0 ∈ IRN ? The answer depends on the behaviour of
F (x, η, p) with respect to the behaviour of λ1(x, η, p) and λN(x, η, p),
respec-tively the lowest and highest eigenvalue of the matrix a = (aij(x, η, p))ij. The
answer depends also crucially on the behaviour of λN(x, η, p) λ1(x, η, p)
. This allows us to consider some general interesting operators. Let us recall that in [PU1] and [PU2] the authors consider some particular operators in divergence form
depending on x and u : −X i ∂ ∂xi (aij(x, u).A(|∇u|) ∂u ∂xj ). Contrary to the previous works the functiton η → F (x, η, p) is not necessarily monotone and the operator coefficients aij depend on x and u. All the previous papers [VA],
[PU], [FE], [MO], [DO] are devoted to quasilinear problems in divergence-form like, for instance,
−div(ϕ(|∇u|)∇u) + h(u) = 0 in IRN, (2)
where the function η → h(η) is monotone in some neighbourhood of 0 and is such that h(0) = 0. The first result of this kind seems to come from [VA] and [RE]. Roughly speaking, in [VA], [PU], [MO], [DO] the result is linked to the behaviour of some integral like, for instance,
Z δ 0
ds
Φ−1(|H(s)|) for some real δ > 0
(I) where H(t) = Z t 0 h(s)ds, Φ(t) = t 2ϕ(t) −Z t 0 sϕ(s)ds.
Let us recall briefly the main result in the case (2). If
Z δ
0
ds
Φ−1(|H(s)|) = +∞ the strong maximum principle holds for (2). And,
if
Z δ 0
ds
Φ−1(|H(s)|) < +∞, any non negative solution of (2), which approaches
0 at infinity, must vanish for all sufficiently large |x|. The integral like (I) was first introduced by [VA]. For equation involving a gradient term like, for instance
−div(|∇u|m.∇u) + G(|∇u|) + h(u) = 0, (3)
the previous integral condition does not seem to be appropriate, as it is showed in [FE], where systems are also studied - (for systems cf also [BU]). For (3), instead of the integral condition, the authors use the multiplicity or the uniqueness of solution of an ordinary differential equation associated to (3). In [MO] the previous problem is studied in Orlicz-Sobolev space, using the theory of monotone operators. A precise study in the radial case is given in [DO]. As remarked in [DO], for radial solution u with compact support in IRN, it is not necessary to assume that u is positive : in this work the authors consider nodal radial solutions for some quasilinear elliptic
equations involving p-Laplacian. This remark is still valid in non radial cases. For an exhaustive review and some improvements about this aera we can see [PU2]. Finally let us point out that our framework is completely different from the ones of the previous papers, and in the regular case - (u ∈ c2) - we
rediscover the results for operators of divergence form.
Our paper is structured as follows. In a first part we prove theorem 1 giving the strong maximum principle result in the case where u is of class c2. As the notion of viscosity solution [CR] is well adapted to the existence
theory of operators like −X
ij
aij(x, u, ∇u)
∂2u
∂xi∂xj
, we show that our result is also valid if u is a viscosity solution of class c1. The second part is devoted
to the study of compact support principle. in a first time we prove theorem 2 for u of class c2. And next we show that this result remain valid if u is a
viscosity solution of class c1. Also we give, for some class of right hand side
F (., ., .), a necessary and sufficient condition ensuring the strong maximum principle. This is the same kind of condition as the one used in divergence form operators. And finally in our frame work, for simplicity, we choose Ω = IRN. But actually our results are valid for Ω ⊆ IRN\B(x0, R) where
B(x0, R) is some ball. In this case we use the same idea as in the previous
works, for instance [PU2].
II - Setting of the problem, Notations and Hypothesis.
We consider inequality (1) in the following sense : we assume f and a = (aij)ij as regular as necessary and in a first time, we suppose that u is c
2.
In a second time we assume that u is a viscosity solution of class c1. Our goal
is to give some sufficient conditions ensuring the strong maximum principle or the so called compact support principle [PU1] for u ; that is to say if u is a non negative solution of (1), then u(x) > 0 for any x ∈ IRN. Also we will study the case when the solution u has a compact support in IRN.
A - The strong maximum principle. Let us consider a function f : IRNx × IRη × IRs× IRNp → IR sufficiently regular, satisfying f (x, η, o, p) =
0 ∀(x, η, p) and such that
(
for any (x, η, p) ∈ IRN × IR × IRN the map s → f (x, η, s, p) is non increasing.
The function giving the right hand side of (1) is defined from f as follows : F (x, s, p) = f (x, s, s, p) ∀(x, s, p) ∈ IRN × IR × IRN.
(H2)
It is clear that for any (x, p) the map s → F (x, s, p) is not necessarily mono-tone. About f we assume the following hypothesis.
H1 : −f
−(x, η, s, p)
λ1(x, η, p)
≥ g(s) ∀s ∈ [0, b]
for any (x, η, p) belonging to IRN× IR × IRN and where λ1(x, η, p) is the first
eigenvalue of the symmetric matrix a = (aij)ij ; f
−denotes the negative part
of f . The matrix a is in L∞(IRN × IR × IR × IRN) and is assumed to be
symmetric non negative.
H’1 : The function g(.) is continuous, decreasing on [0, b] and such that g(0) = 0.
H2 : T r a(x, η, p) λ1(x, η, p)
− 1 ≤ β(|p|),
for any (x, η, p) ∈ IRN × IR × IRN, the positive continuous function β having the following behaviour :
H3 :
(
β(t) ≤ At + B ∀t ≥ t0,
A > 0, and B > 0 for simplicity. H4 : β(t) ≤ cβ Log 1 t ∀t ∈]0 t0], β is non increasing in ]0, t0] Remarks : i) we have β(|p|) ≥ N − 1, ∀p ∈ IRN
ii) we can replace H2 by the simpler hypothesis H’2 : N λN(x, η, p)
λ1(x, η, p)
− 1 ≤ β(|p|).
The hypothesis H’2 shows that β(.) controls the behaviour of λN(x, η, p) λ1(x, η, p)
as |p| goes to zero and to infinity.
iii) we shall see later -(cf. remark 3.1) - that (H1) can be replaced by the following :
H”1 : −f −(x, η, s, p) λ1(x, η, p) ≥ g(s) γ(|p|)∀s ∈ [0, b], ∀x, ∀η, ∀p ∈ IR N, with γ(|p|) ≥ γ0 > 0 ∀p ∈ IRN.
H5 : there exists δ > 0 such that lim
λ→0+
g(λ) λ1+δ = 0
For technical reasons let us introduce the following function
e g(t) = g(t) ∀t ∈ [0, b], g(b) ∀t ≥ b, g(−t) ∀t ∈ [−b, 0], g(−b) ∀t, t ≤ −b.
For any λ, 0 < λ < b let us introduce the two following constants k0(λ, a, T ) = k0(λ) = min
h λ
T − a, −g(λ) · (T − a)
i
= −g(λ) · (T − a), by H5 for λ small enough.
k1(λ, a, T ) = k1(λ) = eQ(a,T,β,λ) (T − a) · [λ + (T − a) 2(−g(λ))eQ(a,T,β,λ)] where 0 < a < T , Q(a, T, β, λ) = β a(k0(λ)) · (T − a) + A a + B a(T − a). We will see later - (cf Remark 2) - the meaning of k0(λ) and k1(λ). Let us
give some examples that we can handle in our framework. Example 1 : In the following particular case
aij(x, s, p) = δij
f (x, η, s, p) = g(s)
the hypothesis (H5) is very close to the integral condition
Z 1 0
ds
q
since q−G(s) = h−
Z s 0 g(t)dt
i1/2
≈ cs1+θ − (θ > 0)− in a neighbourhood
of 0. This allows us to rediscover the previous results in the divergence form case. Example 2 : Au = (1 + |∇u|θ+ |∇u|−θ)∂2u ∂x2 + 2(1 + |∇u| θ+ |∇u|−θ)∂2u ∂y2
+(1 + |∇u|θ) + |∇u|−θ) max[2, Log 1
|∇u|] ∂2u ∂z2 f (x, η, s, p) = f (x, s, p) ∀(x, η, s, p) f (x, s, p) = g(s). θ > 0, g(0) = 0, g(s) ≤ 0 ∀s ∈ [0, 1] λ1(p) + λ2(p) + λ3(p) λ1(p) − 1 = 2 + max[2, Log 1 |p|] = β(|p|) f (x, s, p) λ1(p) = g(s) 1 + |p|θ+ |p|−θ ≥ g(s) ∀s ∈ [0, 1] since g(s) ≤ 0 ∀s ∈ [0, 1]. 2 Let us point out that if we have for instance
−Au = g(u) in IR3 , u(x) ≥ 0 ∀x ∈ IR3, lim |x|→+∞u(x) = 0
and if there exists x0 such that u(x0) = 0, then ∇u(x0) = 0. In this case
λ1(|∇u|) + λ2(|∇u|) + λ3(|∇u|)
λ1(|∇u|)
− 1 = β(|∇u|)
is not bounded in a neighbourhood of x0. This point procures difficulties to
state our main result. Example 3.
Let us consider a regular increasing function ϕ : IR+ → IR+
verifying the following assumption : ∀λ > 0 we have sup
t≥0
ϕ(λt)
ϕ(t) < +∞. This hypothesis is satisfied if we have for instance :
lim t→0+ ϕ(t) tθ1 = δ1 ≥ 0 , θ1 > 0 and lim t→+∞ ϕ(t) tθ2 = δ2 with δ2 > 0 if θ2 = 0 and δ2 ≥ 0 if θ2 > 0.
Let us consider a function a : IR3×IR → IR+such that 0 < a0 ≤ a(x, s) ≤
1 ∀(x, s), where a0 is a positive real number, and a function h : IR → IR,
decreasing on [0, 1] and satisfying h(0) = 0. Let b(·) a bounded function defined on IR3 to IR+ : |b(x)| ≤ ||b||∞ ∀x. The operator that we consider is
of the form : −a(x, u) · ϕ(|∇u|)∂ 2u ∂x2 − ϕ(2|∇u|) ∂2u ∂y2
−ϕ(3|∇u|) · [max(1, c0Log
1
|∇u|) + |∇u|] ∂2u
∂z2
= b(x) · h(u) · ϕ(4|∇u|) in IR3, with h(s) ≤ 0 ∀s ∈ [0, 1], h(0) = 0. We have b(x)h(s)ϕ(4|p|) a(x, s)ϕ(|p|) ≥ ||b||∞ a0 Φ4,∞·h(s) = g(s), ∀s ∈ [0, 1] with g(s) ≤ 0, ∀s ∈ [0, 1] 3 · ϕ(3|p|) · [max(1, c0Log 1 |p|) + |p|] a(x, s)ϕ(|p|) − 1 ≤ 3Φ3,∞ a0 · [max(1, c0Log 1 |p|) + |p|] − 1 = β(|p|) where, λ > 0 and Φλ,∞ = sup
t≥0
ϕ(λt)
ϕ(t) which is finite from the assumptions made on ϕ.
Example 4.
We can treat some operators of divergence form in the case u ∈ c2. g : IR → IR such that g(s) ≤ 0 ∀s ∈ [0, 1]
g(0) = 0, g decreasing in [0, 1]. ϕ : IR+→ IR+ such that t → ϕ(t)
t2 is bounded in [0, +∞[.
Let u be a function belonging to c2(IR2) which is a solution of
−div(|∇u|2· ∇u) = ϕ(|∇u|) · g(u) in IR2
u(x) ≥ 0 ∀x ∈ IR2, lim
|x|→+∞u(x) = 0.
This equation can be written − 2 X ij=1 aij(|∇u|) ∂2u ∂xi∂xj
= ϕ(|∇u|) · g(u) = f (u, ∇u) where a11(|p|) = |p|2+ 2p21, a22(|p|) = |p|2+ 2p22 a12(p) = a21(p) = 2p1p2 λ1(|p|) = |p|2, λ2 = 3|p|2, 2 λ2(|p|) λ1(|p|) − 1 = β(|p|) = 5 with p = (p1, p2) and f (s, p) λ1(|p|) ≥ c0 g(s) with c0 = sup t≥0 ϕ(t) t2 .
It is easy to see that we are in the setting of our work. Example 5. A degenerate problem in divergence form.
||p||δ =p21+ p22
δ 2
, p = (p1, p2), δ > 0
−div||∇u||δ.∇u= f (x, u, ∇u) = g(u) · h(x, ∇u) .
We suppose u ∈ c2(IR2). After some elementary computations we see that u
satisfies −a11(∇u) ∂2u ∂x2 − 2a12(∇u) ∂2u ∂x∂y − a22(∇u) ∂2u ∂y2 = f (x, u, ∇u), with a11(p) = δp21· ||p||δ−2+ ||p||δ,
a22(p) = δp22||p||δ−2+ ||p||δ, a12 = a21= 2δp1p2||p||δ−2, and we have λ2(p) = (1 + δ)||p||δ, λ1(p) = ||p||δ β(|p|) = 2 λ2(p) λ1(p) − 1 = 2δ + 1 f (x, s, p) λ1(p) = f (s, s, p) ||p||δ = g(s) · h(x, p)
with h(x, p) ≥ 0, g(s) ≤ 0 on [0, a] and max
(x,p)
h(x, p)
||p||δ ≤ k, some constant. For
instance we can take the particular function h(x, p) = Log(1 + ||p||δ· a(x))
with a(x) ≥ a0 > 0.
Example 6. Let γ be a continuous function IR+ → IR+ such that γ(0) = 0, lim
t→+∞γ(t) = 0. Let β be a function IR
+ → IR+, continuous such that
β(t) ≥ 2. Let us set a11(p) = γ(|p|) · β(|p|) − α, a22(p) = γ(|p|) + α with
α = α(x, s, p) = γ(|p|)
1 + γ(|p|) + a(x, s, p) and a(x, s, p) ≥ 0 ∀(x, s, p), a12 = a21 = {α · [γ(β − 1) − α]}1/2. After some computations we obtain that
λ2(x, s, p) = γ(|p|) · β(|p|), λ1(x, s, p) = γ(|p|). Let us suppose that we have
f (x, s, p) = g(s) · h(x, p), with sup
(x,p)
h(x, p)
γ(|p|) = k > 0 g(s) ≤ 0 ∀s ∈ [0, 1], h(x, p) ≥ 0 ∀(x, p).
Our assumptions appear to be natural. indeed in the case of divergence form equation - (see examples 1, 4, 5) - we easely check that our hypothesis become equivalent to the standard ones ([VA], [PU1] for instance).
To prove the strong maximum principle for (1), the main idea, used in all the previous works, is to construct local sub or super solution of (1). Roughly speaking this means that we have to compare, in some sense, the operator A = X
ij
aij
∂2
∂xi∂xj
with a radial one, for instance. This idea, already used in [TA1, TA2, TA3, TA4], reduces the search of a subsolution of (1) to the existence of a solution for an ordinary differential equation. This method is systematic.
III - Preliminary results : an ordinary differential equation asso-ciated to (1). To establish a “positive local sub solution” of (1), we will see that it is enough to prove the existence of a solution for the following ordinary differential equation :
−ω00 − β a(|ω 0|) · ω0 = ˜g(ω), in ]a, T [ , ω(a) = λ, ω(T ) = 0, ω0(t) < 0, ω00(t) ≥ 0, ω ∈ c2([a, T ]), (4)
where the parameters λ, a, T are choosen in a way somewhat arbitrary. This choice is very crucial to prove the existence result for (4). Let us point out that many of the previous papers use an ordinary differential equation which is somewhat similar to (4). And one of the main differences between our framework and the frameworks of [PU], [FE], [MO] is the presence of the function β(·) and its behaviour given by hypothesis H4. We can compare, for instance, with [PU1] Lemma 2, p. 774, [FE] equation (1.10) p. 27, [MO] lemma 8 p. 441.
Remark 2 : Let us consider (4). k0(λ) is an -(a priori) - estimate from below
to the derivative −ω0(t) and k
1(λ) is its - (a priori) - estimate from above.
We shall see this result in remark 3. We need these estimates to establish the existence of a solution of (4).
For any λ, 0 < λ < λ0 ≤ b, let us consider the functional set
Wλ = {v ∈ c1([a, T ]) /0 ≤ v(t) ≤ λ, v convex decreasing
such that v(a) = λ, k0(λ) ≤ −v0(t)}
Proposition 1. Let a, T be two positive real numbers such that 0 < a < T , T ≤ ah1 + 1 cβ 1 − 1 1 + δ i . (4.1)
Assume hypothesis H’1 and H6. Then we have the following estimate : For any λ, 0 < λ ≤ min(b, λ0) ≤ 1 and for any v ∈ Wλ
I(λ) ≤ K(a, T, β) ·h− g(λ)i1−(T −a)
cβ a
with I(λ) = Z T a e −Rasβa(−v0(θ))dθ ds · Z T a e RT a β a(−v0(θ))dθ(−g(λ))ds
and where K(a, T, β) is a constant independent of λ. Proof. We setβ(t) =e β(t)
a . For any v ∈ Wλ, let us denote by [δ1 ≤ −v
0 ≤ δ 2]
the subset {θ ∈ [a, T ] /δ1 ≤ −v0(θ) ≤ δ2} of [a, T ]. It is clear that
Z T a e −Rasβ(−v˜ 0(θ))dθ ds ≤ (T − a) (6) We have Z T a ˜ β(−v0(θ))dθ = Z [k0(λ)≤−v0<t0] ˜ β(−v0)dθ + Z [−v0≥t0] ˜ β(−v0)dθ because −v0(θ) ≥ k
0(λ) ∀θ ∈ [a, T ]. And since β(·) is decreasing in ]0 t0]
and from (H3) it follows
Z T a ˜ β(−v0)dθ ≤ ˜β(k0(λ)) · (T − a) + Z T a hA a(−v 0) + B a i dθ i.e. Z T a ˜ β(−v0)dθ ≤ ˜β(k0(λ)) · (T − a) + A a + B a(T − a) (7)
since λ0 < 1. From (6) and (7) we obtain
I(λ ≤ Z T a e −Rs aβ(−v˜ 0)dθ ds · Z T a e RT a β(−v˜ 0)dθ · (−g(λ))ds ≤ −g(λ) · (T − a)2· e(T −a) ˜β(k0(λ))· eAa+Ba(T −a). From (H4) we obtain I(λ) ≤ −g(λ) · (T − a)2eAa+ B a(T −a)· 1 [k0(λ)](T −a) cβ a
which can be written
with K(a, T, β) = (T − a) 2eAa+Ba(T −a) (T − a)(T −a) cβ a . 2 Proposition 2. Let λ, a, T be three real numbers verifying (4.1). Then for any v ∈ Wλ there exists a unique solution ω = ωv of
−ω00−β a(|v 0|)ω0 = ˜g(v) in ]a, T [, ω(a) = λ, ω(T ) = 0, −ω0(t) > k 0(λ), ω00(t) ≥ 0. In addition ω ∈ c2([a, T ]). Proof
i) uniqueness. It sufficies to establish that z = 0 is the unique solution
of −z00−β a(|v 0|)z0 = 0, z(a) = z(T ) = 0. (8)
From (8) we have z0(t)eRatβa(−v0)dθ = c
0, some constant. Thus
0 = z(T ) − z(a) =
Z T a z
0(s)ds
which entails c0 = 0. And then z = 0.
ii) existence. We proceed in two steps and use successively the shooting method by introducing the parameter ω0
α(a) = −α, α > 0 -(first step) - and
a continuity argument of the map α → ωα(T ) - (second step) - .
Step 1 : for α > k0(λ), let us consider ω = ωα the unique solution of
−ω00−β a(|v 0|)ω0 = ˜g(v) in ]a, T [ ω(a) = λ, ω0(a) = −α (9)
which has the following representation formula ωα(t) = λ − α Z t a e −Rs aβ(θ)dθ˜ ds (10) + Z t a e −Rasβ(θ)dθ˜ ·h−Z s a e Rσ a β(θ)dθ˜ g(v(σ))dσ i ds
where we denote by ˜β(θ) the term ˜β(−v0(θ)) = β a(−v 0(θ)). Also we have ωα0(t) = −αe− Rt aβ(θ)dθ˜ + e− Rt aβ(θ)dθ˜ · h − Z t a e Rσ a β(θ)dθ˜ g(v(σ))dσ i . (11) Taking α > −g(λ) · Z T a e RT a β(θ)dθ˜ ds, (12) we see that ω0
α(t) < 0 ∀t ∈]a, T [. From (10) the map α ∈ IR+,∗→ ωα(T ) ∈
IR is continuous and for α1 large enough we have ωα1(T ) < 0 and (12). To
prove that there exists ¯α satisfying (12) and ωα¯(T ) = 0, it sufficies to establish
that there exists some α2 verifying (12) and ωα2(T ) > 0. This means that
such a real number α2 verifies simultaneously (12) and
α · Z T a e −Rs aβ(θ)dθ˜ ds < λ + Z T a e −Rs aβ(θ)dθ˜ · h − Z s a e Rσ a β(θ)dθ˜ g(v(σ))dσ i ds. (13) From (5) it follows I(λ) λ ≤ K(a, T, β) h−g(λ) λν i1−(T −a)cβa with ν = 1/(1 − (T − a)cβ α) or again I(λ) λ ≤ K(a, T, β) h−g(λ λ1+δ i1−(T −a)cβa (14)
because a and T are such that ν < 1 + δ since from (4.1) we have T a − 1 = 1 a(T − a) ≤ 1 cβ (1 − 1 1 + δ) < 1 cβ and λ0 < 1.
Consequently, reducing λ0 if necessary, we obtain from (14) and H6 :
I(λ) λ ≤ K(a, T, β) h−g(λ λ1+δ i1−(T −a)cβ a < 1, ∀λ, 0 < λ < λ0. (15)
The relations (14) and (15) entail I(λ) λ = −g(λ) λ Z T a e RT a β(θ)dθ˜ ds · Z T a e −Rs aβ(θ)dθ˜ ds
< 1 + 1 λ Z T a e −Rs aβ(θ)dθ˜ h − Z s a e Rσ a β(θ)dθ˜ g(v(σ))dσ i ds;
From this there exists some real number α2 such that (12) and (13) hold
simultaneously ; i.e. α2 > −g(λ) Z T a e RT a β(θ)dθ˜ ds and ωα 2(T ) > 0.
Conse-quently there exists ¯α ∈]α2 α1[ such that ωα¯(T ) = 0. And since ωα¯(T ) = 0,
we have from (10) ¯ α · (T − a) ≥ ¯α · Z T a e −Rs aβ(θ)dθ˜ ds ≥ λ i.e. ¯ α ≥ λ T − a ≥ k0(λ). Since v ∈ c1([a, T ]), from (11) it follows that ω
¯
α ∈ c1([a, T ]). It remains to
prove that ωα¯ is convex, decreasing. It is clear from (11) that the function
t → π(t) = eR t aβ(θ)dθ˜ · ω0 ¯ α(t) = − ¯α + Z t a −e Rσ a β(θ)dθ˜ g(v(σ))dσ is increasing.
From the previous results there exist α1 > 0 and α2 > 0 such :
α1 > α2 > −g(λ) Z T a e RT a β(θ)dθ˜ ds. Thus ¯α > −g(λ) Z T a e RT
a β(θ)dθ˜ ds, which implies that π(T ) < 0. This means
that ω0 ¯
α(t) < 0 ∀t. The decrease of ωα¯ and the equation
ω00α¯(t) = −˜g(v(t)) − β a(|v
0(t)|) · ω0 ¯ α(t)
prove that we have ω00 ¯
α(t) ≥ 0 ∀t i.e. ωα¯ is decreasing convex.
2 Comments about hypothesis H5
In the proof of the proposition 2 we have used H5 to establish that there exists some admissible real ¯α such that ωα¯(T ) = 0, where ωα¯(·) is given by
(10). Let us consider (11) and (13). Throughout this proof we see easily that the optimal condition to prove the existence of ¯α is the following : for any
λ, 0 < λ < λ0 and any v ∈ Wλ we have − Z T a e Rσ a β a (−v0(θ))dθ g(v(σ))dσ · Z T a e −Rs a β a (−v0 (0))dθ ds (15.1) < λ + Z T a e Rs a β a (−v0(θ))dθh − Z s a e Rσ a β a (−v0(θ))dθ g(v(σ))dσids .
For instance, from (15.1) we see that we can replace (H5) by the more satis-factory hypothesis : lim λ→0+ Λa,T(λ) λ = 0 with Λa,T(λ) = sup v∈Wλ Z T a e Rσ a β a (−v0 (θ))dθ − g(v(σ))dσ. 2 Now let us consider the convex subset of c1([a, T ]) :
f
Wλ =
n
v ∈ Wλ /v(T ) = 0, −v0(t) ≤ k1(λ)
o
and the map
Γ : ˜Wλ −→ ˜Wλ , v −→ ωv = Γ(v) the unique solution of
−ω00−β a(|v 0|) · ω0 = ˜g(v) in ]a T [ , ω(a) = λ, ω(T ) = 0, −ω0(t) > k 0(λ), ω00(t) ≥ 0, (16) for any λ, 0 < λ < λ0.
Remark 3 : As mentioned previously in remark 2, k0(λ) and k1(λ) are a
priori estimates of the derivative of the solution ωv = Γ(v). It is clear for
k0(λ). For k1(λ) we proceed from (10) and as in proposition 1 :
¯ α · Z T a e −Rs aβ(θ)dθ˜ ds = λ + Z T a e −Rs aβ(θ)dθ˜ × h − Z s a e Rσ a β(θ)dθ˜ g(v(σ))dσ i ds, or ¯ α · Z T a e −RT a β(θ)dθ˜ ds ≤ λ + Z T a h Z T a e RT a β(θ)dθ˜ · (−g(λ))dσ i ds,
or again ¯ α · Z T a e −RT a β(θ)dθ˜ ds ≤ λ + (T − a)2· (−g(λ))e RT a β(θ)dθ˜ .
But from (7) we have
Z T a β a(−v 0(θ))dθ =Z T a ˜ β(θ)dθ ≤ β a(k0(λ)) · (T − a) + A a + B a(T − a) = Q(A, T, β, λ). Thus ¯
α · e−Q(a,T,β,λ) ≤ λ + (T − a)2· (−g(λ))eQ(a,T,β,λ) i.e. ¯ α ≤ e Q(a,T,β,λ) T − a h λ + (T − a)2· (−g(λ)) · eQ(a,T,β,λ)i= k1(λ) 2 Proposition 3. Let λ0 be a positive real number, small enough and such
that λ0 < b. Then for any λ, 0 < λ ≤ λ0, for any (a, T ) ∈ (IR+)2 such that
0 < a < T , T a − 1 ≤ 1 cβ (1 − 1 1 + δ), there exists a c 2-function w belonging to f Wλ, which is a solution of −ω00− β a(|ω 0|) · ω0 = g(ω) in ]a, T [ ω(a) = λ, ω(T ) = 0, −ω0(t) > k 0(λ) (E) Proof.
We use the fixed point theorem of Schauder in Wfλ equiped with the
natural topology of c1(([a, T )]. It is clear that for any v ∈ Wf
λ, the solution Γ(v) = ωv of (16) satisfies suph|ω00(t)| /t ∈ [a, T ]i≤ β∞ a k1(λ) + |g(λ)| (17) with β∞ = sup[β(t) /k0(λ ≤ t ≤ k1(λ)]
i) Γ is a compact operator. Let vn be a sequence belonging to Wfλ.
The sequence ωn= Γ(vn) verifies :
kω0 nk∞ ≤ k1(λ), kω00 nk∞ ≤ β∞ a k1(λ) + |g(λ)|, from (17), kωnk∞ ≤ λ. (18)
From ArzelaAscoli Theorem there exists a subsequence (labelled again) -ωnwhich converges to some ω in c1([a, T ]). And it is easy to see that ω ∈Wfλ;
i.e. Γ is compact.
ii) Γ is continuous on λ. Let ve n be a sequence in Wfλ converging to
v ∈ Wfλ. Let ω be the solution corresponding to v : ω = Γ(v). Let us set
ωn = Γ(vn). Thanks to (18) there exists a subsequence of {ωn}n converging
to ω : it is easy to pass to the limit in the equation (16). As ω is the unique solution of (16) corresponding to v, it follows that any subsequence of ωn
which converges, tends to ω as n goes to intinity. This means that the whole sequence ωn tends to ω ; i.e. Γ is continuous onWfλ equiped with the natural
topology of c1([a, T ]).
The Schauder Theorem can be applied : there exists ω ∈ Wfλ which is a
solution of −ω00− β a(|ω 0|)ω0 = ˜g(ω) in [a, T ] , ω(a) = λ, ω(T ) = 0, −ω0(t) ≥ k 0(λ)
and thus ω is a solution of (E) since 0 ≤ ω(t) ≤ λ ≤ b and ˜g(s) = g(s) ∀s ∈ [0, b]. The c2-regularity of ω follows from (E).
2 IV - Strong Maximum Principle
A - The regular case
Theorem 1. Let us assume that the matrix a(x, η, p) is continuously diffe-rentiable with respect to the variable p in IRN × IR × IRN. Assume the hy-pothesis H1 to H5, H1 and suppose that
−X ij aij(x, u(x)∇u(x)) ∂2u ∂xixj ≥ F (x, u, ∇u) in IRN , u(x) ≥ 0 ∀x ∈ IRN, (19)
has a c2 solution. Then we have either u(x) = 0 ∀x ∈ IRN or u(x) >
0 ∀x ∈ IRN.
Proof. We argue by contradiction. Let us suppose there exists x0 ∈ IRN such
that u(x0) = 0 and u 6≡ 0. Then there is some x1 ∈ IRN such that u(x1) > 0
and there exists T0 > 0 such that we have : u(x) > 0 ∀x ∈ B(x1, T0), where
B(x1, T0) is the open ball centred at x1, with radius T0. We set
T1 = T0 h 1 + 1 cβ (1 − 1 1 + δ) i .
Consider λ such that 0 < λ < minnu(x) / x ∈ ∂B(x1, T0)
o
. We choose also 0 < λ < b. Considering, if necessary, the change of function ˜u(x) = u(x1+ x) in (19), we can suppose that x1 ≡ 0 i.e. u(0) > 0 : there is no
loss of generality. From Proposition 3, consider ω a solution of
−ω00− β T0 (|ω0|) · ω0 = g(ω) in ]T0T1[ ω(T0) = λ, ω(T1) = 0, −ω0(t) < 0, ω ∈ c2([T0, T1]), ω00(t) ≥ 0. (20)
Setting ω(x) = ω(|x|), we obtain from (20) −ω00(|x|) − β(|ω 0(|x|)) |x| · ω 0(|x|) ≤ g(ω(|x|)) ∀x, T0 < |x| < T1 i.e. ∀x ∈ B(0, T1)\B(0, T0). By H2 and H1 we obtain −ω00(|x|) − h 1 T r(a(x, u(x), ∇ω(x)) |x|2 X ij aij(x, u(x), ∇ω(x))xixj − 1i ω 0(|x|) |x| ≤ g(ω(|x|)) ≤ −f −(x, u(x), ω(x), ∇ω(x)) λ1(x, u(x), ∇ω(x)) ≤ − f −(x, u(x), ω(x), ∇ω(x)) 1 |x|2 X ij aij(x, u(x), ∇ω(x))xixj ≤ 1 f (x, u(x), ω(x), ∇ω(x)) |x|2 X ij aij(x, u(x), ∇ω(x))xixj ∀x ∈ B(0, T1)\B(0, T0)
i.e. −ω 00(|x|) |x|2 · X ij aij(x, u(x), ∇ω(x)) xixj −hT r(a, (x, u(x), ∇ω(x)) −X ij aij(x, u(x), ∇ω(x)) |x|2 i ω0(|x|) |x| ≤ f (x, u(x), ω(x), ∇ω(x))
and since ω ∈ c2, we obtain
−X ij aij(x, u(x), ∇ω(x)) ∂2ω ∂xi∂xj ≤ f (x, u(x), ω(x), ∇ω(x)) ω/∂B(0,T0)= λ, ω/∂B(0,T1) = 0, ∂ω ∂ν < 0 on ∂B(0, T1), (20.1)
where ν stands for the external normal unit vector to ∂B(0, T1). Since
k0(λ) ≤ |∇ω(x)| ≤ k1(λ) the matrix a(x, u(x), ∇ω(x)) is uniformly
ellip-tic. By the comparison principle given in [GT] p. 207-208, we obtain ω(x) ≤ u(x) ∀x ∈ B(0, T1)\B(0, T0)
We claim that u(x) > 0 ∀x ∈ B(0, T1). For this purpose suppose that
there exists x2 ∈ ∂B(0, T1) such that u(x2) = 0. We have ∇u(x2) = 0 since
u(x) ≥ 0, ∀x. For t > 0 small enough we have ω(x2− tν) − ω(x2)
t ≤
u(x2− tν) − u(x2)
t
where ν stands for the exterior normal unit vector to the boundary ∂B(0, T1)
at x2. Passing to the limit as t goes to 0, we obtain
0 < −ω0(T1) = −
∂ω
∂ν(x2) ≤ −
∂u(x2)
∂ν = 0 since ∇u(x2) = 0
It is a contradiction. Thus u(x) > 0 ∀x ∈ B(0, T1). Let us consider T2 such
that T2 = T1 h 1 + 1 cβ (1 − 1 1 + δ) i .
As previously we can establish that u(x) > 0 ∀x ∈ B(0, T2). So there is a
sequence of positive real numbers Tn such that
Tn = Tn−1 h 1 + 1 cβ (1 − 1 1 + δ) i = T0 h 1 + 1 cβ (1 − 1 1 + δ) in
with u(x) > 0 ∀x ∈ B(0, Tn). For n large enough we have x0 ∈ B(0, Tn).
This entail that u(x0) > 0. Thus it is a contradiction.
2 Remark 3.1. For convenience we have simplified the hypothesis H1. Instead of H1 we can suppose the more general assumption
H001 : − f
−(x, η, s, p)
λ1(x, η, p)
≥ g(s)
γ(|p|) ∀s ∈ [0, b] ,
∀(x, η, p) ∈ IRN × IR × IRN, γ(|p|) ≥ γ0 > 0 ∀p ∈ IRN. In fact to see this,
from the proposition 3 there exists a c2 function ω such that
−ω00− δ 0· β(|ω0|)γ(|ω0|) a ω 0 = δ 0· g(ω) in [a, T ] ω(a) = λ , ω(T ) = 0 , ω0(t) < 0 , ω00(t) ≥ 0
where δ0 > 0 is a some real number such that δ0· γ(|p|) ≥ 1 ∀p. To prove
that ω satisfy (20.1) it is sufficient to see that we have −δ0 · γ(|ω0|)ω00− δ0·
β(|ω0|)γ(|ω0|)ω0
a ≤ δ0· g(ω) since −ω00(t) ≤ 0 and δ0· γ(|ω0|) ≥ 1, i.e.
−ω00(|x|) − β(ω 0(|x|)) a · ω 0(|x|) ≤ g(ω) γ(|ω0|) ≤ − f−(x, u(x), ω(x), ∇ω(x)) λ1(x, u(x), ∇ω(x)) . The sequel remain to be inchanged.
B - The case of viscosity solution. Let us recall the following
Definition. Assume u of class c1. We say that u is a viscosity super solution of (1) if, for any c2 test function ϕ such that u − ϕ has a local minimizer x
0,
we have
−Σaij(x0, u(x0), ∇u(x0))
∂2ϕ
∂xi∂xj
(x0) ≥ F (x0, u(x0), ∇u(x0))
(∗)
Corollary 1. Assume that the inequality (H1) is strict and u is a c1 viscosity
supersolution of (1). Then under the assumptions (H1) to (H5) we have either u(x) = 0 ∀x ∈ IRN or u(x) > 0 ∀x ∈ IRN.
Proof. We proceed as previously. We argue by contradiction. Suppose that u(x0) = 0 and u 6≡ 0. Then there exists a ball B(x1, T0) such that
u(x) > 0 ∀x ∈ B(x1, T0). We set T1 = T0 h 1 + 1 cβ (1 − 1 1 + δ) i . (∗∗)
Consider λ > 0 such that 0 < λ < minhb, min(u(x)/x ∈ ∂B(x1, T0))
i
. Let us consider ϕ = ω and let x2 be such that
u(x2) − ω(x2) = min
h
u(x) − ω(x)/x ∈ B(x1, T1)\B(x1, T0)
i
. i) if x2 ∈ ∂B(x0, T0) we have u(x) > 0 ∀x ∈ B(x1, T1) since we have
u(x) ≥ ω(x) + u(x2) − ω(x2) > ω(x) > 0 ∀x ∈ B(x1, T1)\B(x1, T0)
ii) if x2 ∈ ∂B(x1, T1), we proced as previously and we prove that there is
a contradiction.
iii) if x2 ∈ B(, T1)\B(x1, T0), then u − ω has a strict minimizer x2 and
from the very definition of viscosity supersolution we obtain (∗) at x2 which
is a contradiction since ω is a strict subsolution of (1).
Finally we have proved that u(x) > 0 ∀x ∈ B(x1, T1). Now from T1 we
define T2 as in (∗∗) ; and so on. The last part of the proof remain to be
V - Solutions with compact support. A - The regular case.
In the case the maximum principle does not hold, a natural question is: under what conditions any non null solution of (1) has a compact support? In the sequel our purpose is to give a satisfactory answer to the above question. The inequality we are going to treat is of the form
−Pijaij(x, u(x), ∇u) ∂2u ∂xi∂xj ≤ F (x, u, ∇u) in IRN u(x) ≥ 0 ∀x ∈ IRN. (22)
As in section II-A, F is defined from some function f : IRNx×IRη×IRs×IRNp →
IR as follows F (x, s, p) = f (x, s, s, p) ∀x, ∀s, ∀p. About f we assume the following hypothesies :
˜
H1 : f : IRN×IR×IR×IRN → IR, continuous ; there exist a, b, −b < 0 < a such that ∀x ∈ IRN, ∀η, ∀p ∈ IRN the function s → f (x, η, s, p) is decreasing and f (x, η, 0, p) = 0 ∀x, ∀p, ∀η.
˜
H2 : There exist two continuous functions γ : IR+→ IR+, g : IR → IR such that γ(t) > 0 ∀t > 0, g(0) = 0 g(s) ≤ 0 ∀s ∈ [0 a], f (x, η, s, p) λN(x, η, p) ≤ g(s) γ(|p|) ∀x, ∀η, ∀p, ∀s ∈ [0, a] . ˜ H3 : Z a 0 ds
H−1(−G(s)) < +∞ where H and G are defined as following:
h(t) = t(γ(t) + ), ∀t > 0 with > 0 arbitrary small ; H(t) =
Z t
0 h(s)ds, G(s) =
Z s
0 g(θ)dθ.
To simplify the computations let us set ˜
a(x, p) = a(x, u(x), p) ∀x, ∀p ˜
f (x, s, p) = f (x, u(x), s, p) ∀x, ∀s, ∀p
We can compare the function h of this framework with the ones used in the previous papers : [VA], [PU], [FE], [MO], [DO].
Example :
aij = 0 ∀i 6= j aii = ai(|p|), 0 ≤ a1 ≤ a2 ≤ . . . ≤ aN ,
F (x, s, p) = f (x, s, p) = c(|p|) · g(s), c(|p|) > 0, g(s) ≤ 0 ∀s ∈ [0, 1], γ(|p|) = aN(|p|)/c(|p|). For instance we can take c(|p|) = an(|p|) i.e. γ(|p|) =
1.
Remark 4.
1) Consider a regular function v(x) = ϕ(|x|). Then
X ij ˜ aij ∂2v ∂xi∂xj = ϕ00(|x|) ·X ij ˜ aijxixj |x|2 + ϕ 0(|x|) |x| h T r(˜a) −X ij ˜ aijxixj |x|2 i 2) T r(˜a) −X ij ˜ aijxixj |x|2 ≥ N −1X λi ≥ (N − 1)λ1
where 0 ≤ λ1(˜a) ≤ · · · ≤ λN(˜a), stand for the eigenvalues of the matrix
˜
a = (aij). In a first step we use the same idea as in the previous papers
[VA], [PU].
Proposition 4 : There exists a c2-function ω : [0, A] → IR such that
−γ(|ω0(r)|) · ω00 (r) ≥ g(ω(r)) in ]0 A[ ω(0) = a, ω0(A) = ω(A) = 0, ω00 (A) = 0 ω0(r) < 0 ∀r ∈]0 A[, ω00 (r) ≥ 0 ∀r ∈]0A[ , with A = Z a 0 ds H−1(−G(s)).
Remark 4.1. Contrary to [VA] and [PU] we use in proposition 4 an inequal-ity i.e. it is not necessary to solve an equation.
Proof of Proposition 4.
Let > 0 be a small real number. Consider the problem
−hγ(|ω0(r)|) + iω00 (r) = g(ω(r)) in ]0 A[ ω(0) = a, ω0(A) = ω(A) = 0 ω0(r) < 0 ∀r ∈]0 A[.
By a monotony argument we define ω as follows - (cf [VA], PU], [MO], [DO]) r = Z a ω(r) ds H−1(−G(s)) (23)
which give by derivation
1 = −ω0(r) /H−1(−G(ω(r))) (23.1) i.e. −G(ω(r)) = H(−ω0 (r)) (24) or again −g(ω(r)) · ω0(r) = h(−ω0(r)) · (−ω00(r)) . But from (23.1) ω0r) < 0 ∀r. Thus
− ω00(r)h(−ω 0(r)) −ω0(r) = g(ω(r)) r ∈]0 A[ i.e. −( + γ(|ω0|) · ω00 = g(ω) in ]0 A[ (25)
and ω(0) = a , ω(A) = 0 , ω0(A) = 0 by (22) and (24). And since
g(0) = 0, ω00
(A) = 0 follows from (25) ; since 0 ≤ ω(r) ≤ a, (25) imply ω00(r) ≥ 0, ∀r and then we obtain
−γ(|ω0|) · ω00 ≥ g(ω) in ]0 A[.
2 Proposition 5. The function v(x) = ω(|x| − R) satisfies
−X ij aij(x, u(x), ∇v(x)) ∂2v ∂xi∂xj ≥ ˜f (x, v(x), ∇v(x)), x ∈ Bc(0, R) = {x /|x| > R},
where R > 0 is large enough. Proof.
We denote again by ω the extension by zero of ω to IR+; from proposition 4, this extension ω belongs to c2(IR+). Since u(x) ≥ 0 and lim
|x|→+∞u(x) = 0,
there exists R > 0 such that : ∀x ∈ D(R) = {x / |x| > R} we have 0 ≤ u(x) ≤ a. Let us set v(x) = ω(|x| − R) = v(|x|) ∀x ∈ D(R). It is clear that we have
−γ(|v0|)v00 ≥ g(v), in ]R, R + A[ v(R) = a, v0(A + R) = v(A + R) = v00 (A + R) = 0 v0(|x|) < 0 R < |x| < R + A.
Using the remark 4 and the monotony of v, we obtain −v00(|x|) − v 0(|x|) |x| h T r(˜a(, x, ∇v) −X ij ˜ aij(x, ∇v)xixj |x|2 i × 1 1 |x|2 X ij ˜ aij(x, ∇v)xixj ≥ g(v) γ(|∇v|) .
Since 0 ≤ v(|v|) ≤ a and from (H2) we obtain
−v00 (|x|) − v 0(|x|) |x| h T r(˜a) −X ij ˜ aijxixj |x|2 i 1 1 |x|2 X ˜ aijxixj ≥ g(v) γ(|∇v|) ≥ f (x, u(x), v, ∇v) λN(x, u(x), ∇v) ≥ 1f (x, v, ∇v)˜ |x|2 X ˜ aijxixj (26)
since ˜f (x, v(x), ∇v(x)) ≤ 0 ∀ |x| ≥ R, because 0 ≤ v(x) ≤ a. That is to
say −X ij ˜ aij(x, ∇v) ∂2v ∂xj∂xj ≥ ˜f (x, v(x), ∇v(x)) , ∀x, |x| > R . 2
Remark 5. If γ(0) = 0 then the term g(v(x))
γ(|∇v(x)|) in (26) means
g(v(x)) γ(|∇v(x)|) if R < |x| < R + A and 0 if |x| ≥ R + A since g(0) = 0. Now we are able to prove the main result of this section.
Theorem 2. Assume u ∈ c2(IRN) and hypothesis ˜H1 - ˜H3. We have 0 ≤
u(x) ≤ v(x) = ω(|x| − R) ∀x, |x| > R. And thus u(x) = 0 ∀x, |x| > R. Proof. We argue by contradiction. Let us suppose that there exists x0,
|x0| > R such that
u(x0) − v(x0) = max
|x|>R(u(x) − v(x)) > 0 .
(27)
As u and v are c2, we obtain
(
∇u(x0) = ∇v(x0)
D2(u(x
0) − v(x0)) ≤ 0.
(28)
There exists a neighbourhood ϑ1(x0) such that
D2(u(x) − v(x)) ≤ 0 ∀x ∈ ϑ1(x0). (29) As a0 = ˜ aij(x0, ∇u(x0)
ij is a non negative matrix it follows
−T ra0· D2(u(x) − v(x)) ≥ 0 ∀x ∈ ϑ1(x0) , i.e. −X ij ˜ aij(x0, ∇u(x0)) ∂2(u(x) − v(x)) ∂xi∂xj ≥ 0, ∀x ∈ ϑ1(x0). (30)
But from (21) and proposition 5 it follows −X ij ˜ aij(x, ∇v(x)) ∂2v ∂xi∂xj +X ij ˜ aij(x, ∇u(x)) ∂2u ∂xi∂xj ≥ ˜f (x, v(x), ∇v(x)) − ˜f (x, u(x), ∇u(x)) = δ(x) where δ(x) = f (x, v(x), ∇v(x)) − ˜˜ f (x, v(x), ∇v(x0))
− hf (x, u(x), ∇u(x)) − ˜˜ f (x, u(x), ∇u(x0))
i
+ hf (x, v(x), ∇v(x˜ 0)) − ˜f (x, u(x), ∇u(x0))
i
For any > 0 ∃ ϑ2(x0) such that
sup
ϑ2(x0)
| ˜f (x, v(x), ∇v(x)) − ˜f (x, v(x), ∇v(x0))| <
(31)
and there exists ϑ3(x0) such that
sup
ϑ3(x0)
| ˜f (x, u(x), ∇u(x)) − ˜f (x, u(x), ∇u(x0)| < .
(32)
But, since u(x) − v(x) ≤ 0 ∀x, |x| = R, we have from (27) u(x0) − v(x0) = max
D(R)
(u(x) − v(x)) > 0;
and as s → ˜f (x, s, p) is decreasing ∀(x, p), there exist ϑ4(x0) and α > 0 such
that u(x) − v(x) > 0 ∀x ∈ ϑ4(x0), and
˜
f (x, v(x), ∇v(x0)) − ˜f (x, u(x), ∇u(x0)) ≥ α ∀x ∈ ϑ4(x0) ,
(33)
because ∇u(x0) = ∇v(x0). Let us set
ϑ(x0) = 4
\
i=1
ϑi(x0).
Then thanks to (31) - (33) the following inequality holds :
δ(x) ≥ α − 2 > 0 ∀x ∈ ϑ(x0) , for small enough .
(34) This gives −X ij ˜ aij(x, ∇v(x)) ∂2v(x) ∂xi∂xj +X ij ˜ aij(x, ∇u(x)) ∂2u ∂xi∂xj > 0 ∀x ∈ ϑ(x0) which entails −X ij ˜ aij(x0, ∇u(x0)) ∂2(v(x 0) − u(x0)) ∂xi∂xj > 0 , since ∇u(x0) = ∇v(x0) . As x → −X ij ˜ a(x0, ∇u(x0)) ∂2 ∂xi∂xj
(v(x) − u(x)) is continuous, ∃W(x0) such
that −X ij ˜ aij(x0, ∇u(x0)) ∂2 ∂xi∂xj (v(x) − u(x)) > 0, ∀x ∈ W(x0). (35)
Finally for any x ∈ ϑ(x0)TW(x0), (30) is in contradiction with (35).
2 B - The viscosity case.
Definition. Assume u of class c1. We say that u is a viscosity subsolution
of (22) if, for any c2 test function ϕ such that u − ϕ has a local maximizer
x0, we have −X ij ˜ aij(x0, ∇u(x0)) ∂2ϕ ∂xi∂xj (x0) ≤ ˜f (x0, u(x0), ∇u(x0)) (36)
Corollary 2. Assume that inequality ( ˜H2) is strict and u is a c1 viscosity
subsolution of (22). Then under the assumptions ( ˜H1) to ( ˜H3), there exists R > 0 such that u(x) = 0 ∀x, |x| > R.
Proof. We proceed as in the Theorem 2, arguing by contradiction. Let us consider Proposition 5 and ϕ = v. We have proved that
˜ f (x, v(x), ∇v(x)) < −X ij ˜ aij(x, ∇v(x)) ∂2v(x) ∂xi∂xj ∀x, |x| > R (37)
since ( ˜H2) is strict. The function v and the real number R are defined in proposition 5. Let x0 be such that
u(x0) − v(x0) = max
|x|>R(u(x) − v(x)) > 0.
Since ∇u(x0) = ∇v(x0), from (36) we obtain
−X ij ˜ aij(x0, ∇v(x0)) ∂2v ∂xi∂xj (x0) ≤ f (x˜ 0, u(x0), ∇v(x0)) (38) ≤ f (x˜ 0), v(x0), ∇v(x0))
since we have ( ˜H1). Thus (38) is in contradiction with (37) at x0, and thus
VI - Necessary and sufficient conditions to obtain the strong maxi-mum principle.
Let us point out that our assumptions (H”1) in section II and ( ˜H2) in section V are quite satisfactory. Indeed let us consider, for instance, the equation −X ij aij(x, ∇u(x)) ∂2u ∂xi∂xj = f (x, u(x), ∇u(x)) u(x) ≥ 0 in IRN. (E)
And assume that we have : there exist g and γ such that g(s) γ(|p|) ≤ −f−(x, s, p) λ1(x, p) , f (x, s, p) λN(x, p) ≤ c. g(s) γ(|p|) (H)
for any s and p, where c is some positive constant. We have : g(s) γ(|p|) ≤ −f−(x, s, p) λ1(x, p) ≤ f (x, z, p) λN(x, p) < c. g(s) γ(|p|).
The other assumptions remain to be inchanged. From theorem 1 and theorem 2, the integral condition
Z +∞
0
ds
H−1(−G(s)) = +∞, - (cf ˜H3) - is necessary
and sufficient to obtain the strong maximum principle for (E).
Acknowledgments : I am grateful to Pierre Cardaliaguet for his valuable remarks especially those concerning viscosity solutions.
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