ON HOPF’S LEMMA AND THE STRONG MAXIMUM
PRINCIPLE
S. BERTONE, A. CELLINA, AND E. M. MARCHINI
Abstract. In this paper we consider Hopf’s Lemma and the Strong Maximum
Principle for supersolutions to
N
X
gi (u2xi )uxi xi = 0
i=1
under suitable hypotheses that allow gi to assume value zero at zero.
1. Introduction
Let Ω ⊂ R
operator
(1)
N
be a connected, open and bounded set and, on Ω, consider the
F (u) =
N
X
gi (u2xi )uxi xi ,
i=1
where gi : [0, +∞) → [0, +∞) are continuous functions.
Ω is called regular if, for every z ∈ ∂Ω, there exists a tangent plane, continuously
depending on z, and Ω satisfies the interior ball condition at z if there exists an
open ball B ⊂ Ω with z ∈ ∂B. When F is elliptic, two classical results hold.
Hopf ’s Lemma:
Let Ω be regular, let u be such that F (u) ≤ 0 on Ω. Suppose that there exists z ∈ ∂Ω
such that
u(z) < u(x),
for all x in Ω.
If, in addition, Ω satisfies the interior ball condition at z, we have
∂u
(z) < 0,
∂ν
where ν is the outer unit normal to B at z.
The Strong Maximum Principle:
Let u be such that F (u) ≤ 0 on Ω, then if u it attains minimum in Ω, it is a constant.
In 1927 Hopf proved the Strong Maximum Principle in the case of second order
elliptic partial differential equations, by applying a comparison technique, see [11].
For the class of quasilinear elliptic problems, many contributions have been given,
1991 Mathematics Subject Classification. 35B05.
Key words and phrases. Strong maximum principle.
The present version of the paper owes much to the precise and kind remarks of two anonymous
referees.
1
2
S. BERTONE, A. CELLINA, AND E. M. MARCHINI
to extend the validity of the previous results, as in [1, 5, 6, 8, 9, 10, 12, 13, 14, 15,
16, 18].
In the case in equation (1) we have gi ≡ 1, for every i, then F (u) = ∆u, and we
find the classical problem of the Laplacian, see [7, 9].
On the other hand, when there exists i ∈ {1, . . . , N } such that gi ≡ 0 on an
interval I = [0, T ] ⊂ R, the Strong Maximum Principle does not hold. Indeed, in
this case, it is always possible to define a function u assuming minimum in Ω and
PN
such that i=1 gi (u2xi )uxi xi = 0. For instance, let gN (t) = 0, for every t ∈ [0, 2].
The function
−(x2N − 1)4 if − 1 ≤ xN ≤ 1
u(x1 , . . . , xN ) =
,
0
otherwise
satisfies (1) in RN .
We are interested in the case when 0 ≤ gi (t) ≤ 1 and it does not exist i such
that gi ≡ 0 on an interval. Since gi could assume value zero, the equation (1) is
degenerate elliptic.
Our interest in the problem comes, partially, from the calculus of variations. We
can consider the equation
N
X
gi (u2xi )uxi xi = 0
i=1
as the Euler-Lagrange equation associated to the functional
!
Z
Z
N
1 X
2
2
J(u) =
L(∇u(x))dx =
fi (uxi )uxi dx,
Ω
Ω 2
i=1
where L(∇u) is strictly convex in {(ux1 , . . . , uxN ) : u2xi ≤ t, for every i = 1, . . . , N }.
Indeed, fix i. Let fi be a solution to the differential equation
gi (t) = fi (t) + 5tfi0 (t) + 2t2 fi00 (t),
(2)
for t ∈ [0, t ]. Since
∂2L 2
(u ) = fi (u2xi ) + 5u2xi fi0 (u2xi ) + 2u4xi fi00 (u2xi ) = gi (u2xi ),
∂u2xi xi
we have that
div∇∇u L(∇u) =
N
N
X
X
∂2L
=
gi (u2xi )uxi xi .
u
x
x
i i
2
∂u
x
i
i=1
i=1
Since J is convex, solutions to the Euler-Lagrange equation are minima (among
those functions satisfying the same boundary conditions.) From our assumptions,
it follows that v ≡ 0 is a solution. The Strong Maximum Principle can be seen as
a strong uniqueness result among solutions to a variational problem: two solutions
u and v are given, not crossing each other: if the two solutions meet at one point,
they are identically equal. For a functional J possessing rotational symmetry, simple
necessary and sufficient condition for the validity of the Strong Maximum Principle
are presented in [4]. The purpose of the present paper is to investigate the same
problem in the class of functional that are convex, but non possessing rotational
symmetry.
ON HOPF’S LEMMA AND THE STRONG MAXIMUM PRINCIPLE
3
The results of [4] for the validity of the Strong Maximum Principle, for equations
possibly non elliptic but arising from a functional having rotational symmetry, show
that this validity depends only on the behaviour of the functions gi near zero.
In this paper, we prove, in section 3, a sufficient condition for the validity of
the Hopf’s Lemma and of the Strong Maximum Principle; a remarkable feature of
this condition is that it concerns only the behaviour of the function gi (t) that goes
fastest to zero, as t goes to zero. Hopf’s lemma and the Strong Maximum Principle
are essentially the same result as long as we can build subsolutions whose level lines
can have arbitrarily large curvature. This need not be always possible for problems
not possessing rotational symmetry. This difficulty will be evident in sections 4
and 5. In these sections, a more restricted class of equations is considered, namely
when all the functions gi , for i = 1, . . . , N − 1, are 1 and only gN is allowed to go
to zero. In this simpler class of equations we are able to show that the condition
3/2
lim+
t→0
(gN (t))
0 (t)
tgN
>0
is at once necessary for the validity of Hopf’s Lemma and sufficient for the validity
of the Strong Maximum Principle.
2. Preliminary results
We impose the following local assumptions.
Assumptions (L):
There exists t > 0 such that:
i) on [0, t ], for every i = 1, . . . , N − 1,
0 ≤ gN (t) ≤ gi (t) ≤ 1;
ii) gN is continuous on [0, t ]; positive and differentiable on (0, t ];
0
iii) on (0, t ], the function t → gN (t) + gN
(t)t is non decreasing.
Notice that, in case ii) above is violated, the Strong Maximum Principle does
not hold; and that condition iii) above includes the case of the Laplacian, gi (t) ≡ 1;
and, finally, that under these assumptions, gi could assume value zero at most for
t = 0.
Moreover, the strict convexity of L(∇u) in {(ux1 , . . . , uxN ) : u2xi ≤ t, for every i =
1, . . . , N } follows by the fact that gi is positive in (0, t ].
Since we will need general comparison theorems that depend on the global properties of the solutions, i.e. on their belonging to a Sobolev space, we will need
also a growth assumption on gi (assumption (G)) to insure these properties of the
solutions.
Assumption (G):
Each function fi as defined in (2), is bounded away from zero and fi (u2xi )u2xi is
strictly convex.
4
S. BERTONE, A. CELLINA, AND E. M. MARCHINI
Any function gi satisfying assumptions (L) on [0, t ] can be modified so as to
satisfy assumption (G) on [0, +∞). In fact, taking fi such that fi (t) = gi ( t ), for
t > t, it is enough to modify gi to ( t, +∞) by setting gi (t) = gi ( t ), for t > t.
Definition 1. Let Ω be open, and let u ∈ W 1,2 (Ω). The map u is a weak solution
to the equation F (u) = 0 if, for every η ∈ C0∞ (Ω),
Z
h∇L(∇u(x)), ∇η(x)idx = 0.
Ω
u is a weak subsolution (F (u) ≥ 0) if, for every η ∈ C0∞ (Ω), η ≥ 0,
Z
h∇L(∇u(x)), ∇η(x)idx ≤ 0.
Ω
u is a weak supersolution (F (u) ≤ 0) if, for every η ∈ C0∞ (Ω), η ≥ 0,
Z
h∇L(∇u(x)), ∇η(x)idx ≥ 0.
Ω
We say that a function w ∈ W 1,2 (Ω) is such that w|∂Ω ≤ 0 if w+ ∈ W01,2 (Ω).
The growth assumption (G) assures that, if u ∈ W 1,2 (Ω), then ∇L(∇u(x)) ∈
L (Ω). The strict convexity of L implies the following comparison lemma.
2
Lemma 1. Let Ω be an open and bounded set, let v ∈ W 1,2 (Ω) be a subsolution
and let u ∈ W 1,2 (Ω) be a supersolution to the equation F (u) = 0. If v|∂Ω ≤ u|∂Ω ,
then v ≤ u a.e. in Ω.
The following technical lemmas will be used later.
Lemma 2. Let n = 2, . . . , N and set
t(1 − a)
(1 − a) + gN (ta)a.
hn (a) = gN
n−1
For every 0 < t ≤ t ( t defined in assumptions (L)), hn (a) ≥ hn (1/n), for every a
in [0, 1].
0
Proof. Since, on (0, t ], the function t → gN (t) + gN
(t)t is non decreasing, we have
that
t(1 − a)
t(1 − a) t(1 − a)
0
0
− gN
h0n (a) = −gN
+ gN (ta) + gN
(ta)ta ≥ 0
n−1
n−1
n−1
if and only if a ≥ 1/n, so that hn (a) ≥ hn (1/n), for every a ∈ [0, 1].
Lemma 3. For every 0 < t ≤ t ( t defined in assumptions (L)), we have that
2 ! 2
N
X
xi
t
xi
≥ gN
.
gN t
ρ
ρ
N
i=1
Proof. We prove the claim by induction on N .
ON HOPF’S LEMMA AND THE STRONG MAXIMUM PRINCIPLE
5
Let N = 2. Set a = sin2 θ1 . Applying Lemma 2 we obtain that
2 ! 2
2 ! 2
x1
x2
x1
x2
gN t
+ gN t
=
ρ
ρ
ρ
ρ
N (t(1 − a))(1 − a) + gN (ta)a ≥ gN
t
.
2
Suppose that the claim is true for N − 1, i.e.
2 ! 2
N
−1
X
xi
xi
t
gN t
.
≥ gN
ρ
ρ
N −1
i=1
Let us prove it for N . Set

y1 = ρ cos θN −2 . . . cos θ2 cos θ1



y2 = ρ cos θN −2 . . . cos θ2 sin θ1
...



yN −1 = ρ sin θN −2
and set a = sin2 θN −1 . Applying Lemma 2 we obtain that
2 ! 2
N
X
xi
xi
gN t
=
ρ
ρ
i=1
! 2
2
yi
yi
gN t
(1 − a)
(1 − a) + gN (ta)a ≥
ρ
ρ
i=1
t
t(1 − a)
(1 − a) + gN (ta)a ≥ gN
,
gN
N −1
N
N
−1
X
and the claim is proved.
3. A sufficient condition for the validity of Hopf’s Lemma and of
the Strong Maximum Principle
Consider the improper Riemann integral
Z ξ
Z ξ
gN (ζ 2 /N )
gN (ζ 2 /N )
dζ = lim
dζ
b
ζ
ζ
ξ→0
0
ξb
as an extended valued function G,
G(ξ) =
Z
ξ
0
gN (ζ 2 /N )
dζ,
ζ
where we mean that G(ξ) ≡ +∞ whenever the integral diverges.
Given a continuous function u, a direction ν and a point z, we denote
(3)
∂+u
u(z + hν) − u(z)
(z) = lim sup
.
∂ν
h
h→0−
We wish to prove the following lemma.
6
S. BERTONE, A. CELLINA, AND E. M. MARCHINI
Lemma 4 (Hopf’s Lemma).
Let Ω ⊂ RN be a connected, open and bounded set.
1,2
Let u ∈ W (Ω) ∩ C Ω be a weak solution to
N
X
i=1
gi (u2xi )uxi xi ≤ 0.
In addition to the assumptions (L) and (G) on gi , assume that G(ξ) ≡ +∞. Suppose that there exists z ∈ ∂Ω such that
u(z) < u(x),
for all x in Ω
and that Ω satisfies the interior ball condition at z. Then
∂+u
(z) < 0,
∂ν
where ν is the outer unit normal to B at z.
Note that the interior ball condition automatically holds if ∂U is C 2 .
As an example of an equation satisfying the assumptions of the theorem above,
consider the Laplace equation ∆u = 0. The functions gi ≡ 1 satisfy the assumptions
(L) and (G), and
Z ξ
1
G(ξ) =
dζ = +∞.
ζ
0
Another example is obtained setting
gN (t) =
1
| ln(t)|
for 0 ≤ t ≤ 1/e. The assumptions (L) and (G) are satisfied; moreover, for 0 ≤
ξ 2 /N ≤ 1/e,
Z ξ
dζ
= +∞.
G(ξ) =
2
0 ζ| ln(ζ /N )|
Remark 1. In this paper we will consider the operator
F (v) =
N
X
gi (vx2i )vxi xi
i=1
in polar coordinates. Set

x1 = ρ cos θN −1 . . . cos θ2 cos θ1



x2 = ρ cos θN −1 . . . cos θ2 sin θ1
...



xN = ρ sin θN −1
so that
v xi
xi
= vρ
ρ
and
vxi xi = vρρ
xi
ρ
2
"
2 #
vρ
xi
+
1−
.
ρ
ρ
ON HOPF’S LEMMA AND THE STRONG MAXIMUM PRINCIPLE
When v is a radial function, F reduces to
2 ! "
2
N
X
xi
xi
vρ
2
+
F (v) =
gi vρ
vρρ
ρ
ρ
ρ
i=1
vρρ
N
X
gi
i=1
vρ2
xi
ρ
2 ! xi
ρ
2
1−
N
vρ X
gi
+
ρ i=1
xi
ρ
vρ2
2 !#
xi
ρ
7
=
2 !
1−
xi
ρ
2 !
.
In general, we do not expect that the equation F (v) = 0 admits radial solutions.
However we will use the expression of F valid for radial functions in order to reach
our results.
Proof of Lemma 4. a) We assume that u(z) = 0 and that B = B(O, r). Let
n
o
= min u(x) : x ∈ B(O, r/2) > 0 and ω = B(O, r) \ B(O, r/2).
b) We seek a radial function v ∈ W 1,2 (ω) ∩ C(ω)

v is a weak solution to F (v) ≥ 0




 v>0
v=0
(4)


 v≤


vρ (z) < 0.
satisfying
in
in
in
in
ω
ω
∂B(O, r)
∂B(O, r/2)
Consider the Cauchy problem

N −1
ζ
0


 ζ = − ρ gN (ζ 2 /N )
(5)


 ζ(r/2) = − .
r
There exists a unique local solution ζ of (5), such that
Z ρ
Z ζ(ρ)
N −1
2ρ
gN (ζ 2 /N )
dζ =
−
ds = −(N − 1) ln
.
ζ
s
r
r/2
−r
We claim that ζ is defined in [r/2, +∞). Indeed, suppose that ζ is defined in
[r/2, τ ), with τ < +∞. Since ζ 0 > 0, ζ is an increasing function, so that τ < +∞
if and only if limρ→τ ζ(ρ) = 0. But
Z ζ(ρ)
2ρ
gN (ζ 2 /N )
dζ = lim −(N − 1) ln
−∞ = lim
,
ρ→τ
ρ→τ − ζ
r
r
a contradiction. Hence, the solution ζ of (5) is defined in [r/2, +∞).
Setting vρ = ζ, since, for every ρ ∈ (r/2, r),
− < vρ (ρ) < 0,
r
we have that the function
Z ρ
v(ρ) =
vρ (s)ds
r
solves the problem
vρρ gN
vρ2
N
!
+
vρ
(N − 1) = 0,
ρ
8
S. BERTONE, A. CELLINA, AND E. M. MARCHINI
in particular, for every ρ ∈ (r/2, r), v(ρ) > 0 and vρ (ρ) ≤ −η < 0; moreover
v(r) = 0 and v(r/2) ≤ .
√
Since vρρ ≥ 0 and − t ≤ vρ ≤ 0, for every ρ ∈ (r/2, r), by the hypotheses on gi
and by Lemma 3, we have that
2 ! 2
2 !
2 !
N
N
X
X
x
x
x
v
xi
i
i
i
ρ
gi vρ2
F (v) = vρρ
+
gi vρ2
1−
≥
ρ
ρ
ρ
ρ
ρ
i=1
i=1
vρρ
N
X
gN
vρ2
xi
ρ
2 ! xi
ρ
2
gN
vρ2
xi
ρ
2 ! 2
N
vρ X
xi
1−
+
ρ i=1
ρ
xi
ρ
2
+
i=1
vρρ
N
X
i=1
vρρ gN
vρ2
N
!
+
!
=
vρ
(N − 1) ≥
ρ
vρ
(N − 1).
ρ
The function v solves (4), indeed, v is in C 2 (ω) and it is such that F (v) ≥ 0 and
v > 0 in ω, v(r) = 0, v(r/2) ≤ and vρ (z) < 0.
c) Since u, v ∈ W 1,2 (ω) ∩ C(ω), v is a weak subsolution and u is a weak solution
to F (u) = 0, and v|∂ω ≤ u|∂ω , applying Lemma 1, we obtain that v ≤ u in ω.
We have
∂v
v(z) = u(z) and vρ (z) =
(z) ≤ −η.
∂ν
If
∂ +u
(z) > −η,
∂ν
+
where ∂∂νu has been defined in (3), then there exists h < 0 such that v(z + hν) >
u(z + hν), in contradiction with the fact that v ≤ u in ω.
From Hopf’s Lemma we derive:
Theorem 1 (Strong Maximum Principle).
Let Ω ⊂ RN be a connected, open and
1,2
bounded set. Let u ∈ W (Ω) ∩ C Ω be a weak supersolution to
N
X
gi (u2xi )uxi xi = 0.
i=1
In addition to the assumptions (L) and (G) on gi , assume that G(ξ) ≡ +∞. Then,
if u attains its minimum in Ω, it is a constant.
Proof. a) Assume minΩ u = 0 and set C = {x ∈ Ω : u(x) = 0}. By contradiction,
suppose that the open set Ω \ C 6= ∅.
b) Since Ω is a connected set, there exist s ∈ C and R > 0 such that B(s, R) ⊂ Ω
and B(s, R) ∩ (Ω \ C) 6= ∅. Let p ∈ B(s, R) ∩ (Ω \ C). Consider the line ps. Moving
p along this line, we can assume that B(p, d(p, C)) ⊂ (Ω \ C) and that there exists
a point z ∈ C such that d(p, C) = d(p, z). Set r = d(p, C). W.l.o.g. suppose that
ON HOPF’S LEMMA AND THE STRONG MAXIMUM PRINCIPLE
9
p = O.
c) The set Ω \ C satisfies the interior ball condition at z, hence Hopf’s Lemma
implies
∂+u
(z) < 0.
∂ν
But this is a contradiction: since u attains minimum at z ∈ Ω, we have that
Du(z) = 0.
4. A necessary condition for the validity of Hopf’s Lemma
In this and the following section we consider the operator
(6)
F (u) =
N
−1
X
uxi xi + g(u2xN )uxN xN ,
i=1
We wish to provide a necessary condition for the validity of Hopf’s Lemma in a
class of degenerate elliptic equations.
Consider the case
Z ξ
g(ζ 2 /N )
G(ξ) =
dζ < +∞.
ζ
0
Theorem 2. Let g, satisfying assumptions (L) and (G), be such that, on (0, t ], ( t
defined in assumptions (L)),
g 0 (t) > 0,
and
lim+
t→0
g(t) + g 0 (t)t ≤ 1,
(g(t))3/2
= 0.
tg 0 (t)
Then there exist: an open regular region Ω ⊂ RN ; a radial function u ∈ C 2 (Ω),
such that F (u) ≤ 0 in Ω, and a point z ∈ ∂Ω such that u(z) = 0, u(z) < u(x) for
every x ∈ Ω and
∂u
(z) = 0
∂ν
where ν is the outer unit normal to Ω at z.
If, in addition, we assume that
(7)
g(t)
tg 0 (t)
is bounded in
(0, t ]
then Ω satisfies the interior ball condition at z.
Remark 2. When limt→0+ g 0 (t)t exists, it follows that
3/2
lim
t→0+
(g(t))
tg 0 (t)
exists, and that
g(t) + g 0 (t)t ≤ 1 , on (0, t ].
10
S. BERTONE, A. CELLINA, AND E. M. MARCHINI
Indeed, we have that
lim (g(t) + g 0 (t)t) = 0.
t→0+
From
Z
Z ξ
g(t)
g(ζ 2 /N )
dt = 2
dζ < +∞,
t
ζ
0
0
it follows that limt→0+ g(t) = 0. If limt→0+ g 0 (t)t > 0, there exists K > 0 such that,
when 0 < t ≤ t, g 0 (t)t ≥ K, so that
Z t
g(t)t =
(g(s) + g 0 (s)s) ds ≥ Kt,
ξ2
0
and g(t) ≥ K, a contradiction.
The map
1
,
| ln(t)|k
with k > 2, for 0 ≤ t ≤ 1/e, satisfies the assumption
g(t) =
3/2
lim
t→0+
(g(t))
tg 0 (t)
= 0.
The following lemma is instrumental to the proofs of the main results.
Lemma 5. Let g satisfies assumptions (L) and (G). Suppose that for every 0 <
t ≤ t,
g 0 (t) ≥ 0 and g(t) + g 0 (t)t ≤ 1.
Set
k1 (a) = (1 − a) + ag(ta)
and
k2 (a) = −a − (1 − a)g(ta)
For every 0 < t ≤ t ( t defined in assumptions (L)), k1 and k2 are non increasing
in [0, 1].
Proof. Since, for every 0 < t ≤ t,
g 0 (t) ≥ 0
we have that, for every 0 ≤ a ≤ 1,
and
g(t) + g 0 (t)t ≤ 1,
k10 (a) = −1 + g(ta) + g 0 (ta)ta ≤ 0
and
k20 (a) = −1 + g(ta) − (1 − a)g 0 (ta)t = −1 + g(ta) + g 0 (ta)ta − g 0 (ta)t ≤ 0.
Proof of Theorem 2. a) Let v be a radial function. Setting a = sin2 θN −1 , (6)
reduces to
vρ
N − 2 + a + (1 − a)g(vρ2 a) .
F (v) = vρρ 1 − a + ag(vρ2 a) +
ρ
Let a = 1, we seek a solution to
(8)
vρρ g(vρ2 ) + (N − 1)
vρ
=0
ρ
ON HOPF’S LEMMA AND THE STRONG MAXIMUM PRINCIPLE
11
and a radius R(1) > 0 such that vρ (R(1) + 1) = 0 and vρ (ρ) < 0, for every
ρ ∈ [2, R(1) + 1). Consider the Cauchy problem

N −1 ζ


 ζ0 = −
ρ g(ζ 2 )
(9)


 ζ(2) = −1.
We are interested in a negative solution ζ. Define R(1) to be the unique positive
real solution to
R(1) + 1
G(−1) − (N − 1) ln
= 0,
2
i.e.
G(−1)
R(1) = 2e N −1 − 1.
The solution ζ of (9), satisfies
Z ρ
Z ζ(ρ)
ρ
N −1
g(t2 )
−
dt =
ds = −(N − 1) ln
.
G(ζ(ρ)) − G(−1) =
t
s
2
2
ζ(2)
Then, for every ρ ∈ (2, R(1) + 1), G(ζ(ρ)) > 0 and ζ(ρ) < 0, while ζ(R(1) + 1) = 0.
Setting vρ (ρ) = ζ(ρ) and
Z ρ
vρ (s)ds,
v(ρ) =
R(1)+1
we obtain that v solves (8) and, for every ρ ∈ (2, R(1) + 1),
vρ (ρ) < vρ (R(1) + 1) = 0
and
v(ρ) > v(R(1) + 1) = 0.
b) Set, for ρ ∈ (1, R(1)], u(ρ) = v(ρ + 1). Since, for the function v, we have
vρρ (ρ)g(vρ2 (ρ)) + (N − 1)
vρ (ρ)
= 0,
ρ
uρρ (ρ)g(u2ρ (ρ)) + (N − 1)
uρ (ρ)
= 0.
ρ+1
at ρ + 1 we obtain
(10)
This equality yields, for ρ ∈ (1, R(1)),
uρρ g(u2ρ )
uρ
+ (N − 1)
= −(N − 1)uρ
ρ
1
1
−
ρ+1 ρ
< 0.
c) Let 1/2 < a < 1. We wish to find R(a) ≤ R(1) such that u is a solution to
uρ
F (u) = uρρ 1 − a + ag(u2ρ a) +
N − 2 + a + (1 − a)g(u2ρ a) ≤ 0,
ρ
for ρ ∈ (1, R(a)). Since
F (u) =
setting
−uρ
ρ(N − 1) 1 − a + ag((uρ )2 a) −
ρ(ρ + 1)g((uρ )2 )
(ρ + 1)g((uρ )2 ) N − 2 + a + (1 − a)g((uρ )2 a)
,
k(ρ) = ρ(N −1) 1 − a + ag((uρ )2 a) −(ρ+1)g((uρ )2 ) N − 2 + a + (1 − a)g((uρ )2 a) ,
12
S. BERTONE, A. CELLINA, AND E. M. MARCHINI
we obtain that F (u) ≤ 0 if and only if k(ρ) ≤ 0. We have
k 0 (ρ) = (N − 1) 1 − a + ag((uρ )2 a) − g((uρ )2 ) N − 2 + a + (1 − a)g((uρ )2 a) +
ρ(N − 1)a
d
d
g((uρ )2 a) − (ρ + 1) g((uρ )2 ) N − 2 + a + (1 − a)g((uρ )2 a) −
dρ
dρ
d
g((uρ )2 a) =
dρ
(N − 1) 1 − a + ag((uρ )2 a) − g((uρ )2 ) N − 2 + a + (1 − a)g((uρ )2 a) −
(ρ + 1)g((uρ )2 )(1 − a)
d
d
g((uρ )2 ) N − 2 + a + (1 − a)g((uρ )2 a) − g((uρ )2 )(1 − a) g((uρ )2 a)+
dρ
dρ
d
d
ρ (N − 1)a g((uρ )2 a) − g((uρ )2 ) N − 2 + a + (1 − a)g((uρ )2 a) −
dρ
dρ
g((uρ )2 )(1 − a)
d
g((uρ )2 a) .
dρ
Assume N ≥ 2. Applying Lemma 5, we obtain
(N − 1) 1 − a + ag((uρ )2 a) − g((uρ )2 ) N − 2 + a + (1 − a)g((uρ )2 a) ≥
(N − 1)g((uρ )2 ) − g((uρ )2 )(N − 1) = 0;
from
d
d
g((uρ )2 ) ≤
g((uρ )2 a) ≤ 0,
dρ
dρ
(11)
we have
d
d
− g((uρ )2 ) N − 2 + a + (1 − a)g((uρ )2 a) − g((uρ )2 )(1 − a) g((uρ )2 a) ≥ 0;
dρ
dρ
since N − 2 + a ≥ (N − 1)a, from (11), we deduce
d
d
ρ (N − 1)a g((uρ )2 a) − g((uρ )2 )(N − 2 + a + (1 − a)g((uρ )2 a))−
dρ
dρ
g((uρ )2 )(1 − a)
d
g((uρ )2 a) ≥
dρ
d
d
ρ (N − 1)a g((uρ )2 a) − g((uρ )2 )(N − 2 + a) ≥
dρ
dρ
ρ(N − 1)a
d
d
g((uρ )2 a) − g((uρ )2 )
dρ
dρ
≥ 0.
The previous inequalities imply that k 0 (ρ) ≥ 0. It follows that F (u) ≤ 0, for every
ρ ∈ (1, R(a)), if and only if
k(R(a)) ≤ 0.
ON HOPF’S LEMMA AND THE STRONG MAXIMUM PRINCIPLE
13
We have that
k(R(a)) = R(a)(N − 1) 1 − a + ag((uρ (R(a)))2 a) −
(R(a) + 1)g((uρ (R(a)))2 ) N − 2 + a + (1 − a)g((uρ (R(a)))2 a) ≤
(N − 1) R(a)(1 − a) − g((uρ (R(a)))2 )a ≤
c(N − 1) R(1)(1 − a) − g((uρ (R(a)))2 )a ≤
g((uρ (R(a)))2 )
.
(N − 1) R(1)(1 − a) −
2
We define R(a) to be a solution to
g((uρ (R(a)))2 )
= 0.
2
d) In order to solve (12) for the unknown R(a), recalling that 1−a = cos2 θN −1 =
2
c , let
s
(12)
R(1)(1 − a) −
g((uρ (r))2 )
.
2R(1)
The function h is decreasing, differentiable and with inverse differentiable. We have
that |c| = h(R(1 − c2 )), so that R(1 − c2 ) = h−1 (|c|), R(1 − c2 ) is increasing in |c|
and
lim R(1 − c2 ) = R(1).
h(r) =
c→0
Let 0 < |c̄| < 1/2 be such that R(1 − c̄2 ) ≥ 1, so that, for c2 ≤ c̄2 , we have
R(1 − c2 ) ≥ R(1 − c̄2 ) ≥ 1. We have obtained that, for every 1 ≥ a ≥ 1 − |c|2 , there
exists R(a) such that (12) holds. It follows that
g((uρ (R(a)))2 )
k(R(a)) ≤ R(1)(1 − a) −
= 0,
2
so that the function u solves F (u) ≤ 0 for every ρ ∈ (1, R(a)).
N
e) Set Ω = {(x1 , . . . , xN ) ∈ RN : ρ ∈ (1, R(1 − c2 )) and |c|
< |c̄|}. Ω ⊂ R is a
1,2
connected, open and bounded set and u ∈ W (Ω) ∩ C Ω is a weak solution to
F (u) ≤ 0. The point z = (R, 0, . . . , 0) ∈ ∂Ω is such that u(z) < u(x), for all x ∈ Ω.
We wish to show that Ω is regular in a neighborhood of z = (R, 0, . . . , 0). Since
d
2
dc R(1 − c ) exists, in (0, |c̄|), to prove our claim it is sufficient to show that
lim
c→0
d
R(1 − c2 ) = 0.
dc
Recalling (10), we have that
0
1
d
R(1 − c2 ) = h−1 (c) = 0
=
dc
h (R(1 − c2 ))
(13)
3/2
√
g((uρ (R(1 − c2 )))2 )
2R(1)
2
−
R(1 − c ) + 1
.
N −1
(uρ (R(1 − c2 )))2 g 0 ((uρ (R(1 − c2 )))2 )
Since
lim uρ (R(1 − c2 )) = 0
c→0
14
S. BERTONE, A. CELLINA, AND E. M. MARCHINI
and
3/2
lim
t→0+
(g(t))
tg 0 (t)
= 0,
it follows that
d
R(1 − c2 ) = 0.
c→0 dc
lim
d
cos θN −1 |θN −1 = π = 1, this shows that
Since dθ
2
regular.
d
dθ R(1
− cos2 θN −1 ) = 0, and Ω is
f) To prove the validity of the interior ball condition at z = (R, 0, . . . , 0), it is
enough to show that the second derivative of R(1 − c2 ) is bounded at c = 0, i.e.
that
1 d
2 c dc R(1 − c )
is bounded. Set
t(c) = (uρ (R(1 − c2 )))2 ,
from (13) we obtain
and from (10)
√
(g(t(c)))3/2
2R(1)
d
R(1 − c2 ) + 1
R(1 − c2 ) = −
,
dc
N −1
t(c)g 0 (t(c))
dt(c)
d
= 2uρ (R(1 − c2 ))uρρ (R(1 − c2 )) (R(1 − c2 )) =
dc
dc
√
(g(t(c)))1/2
2 2R(1)
g 0 (t(c))
and
g 0 (t(c)) dt(c) √
d
(g(t(c)))1/2 =
= 2R(1).
dc
2(g(t(c)))1/2 dc
From g(t(0)) = 0, we obtain that
(g(t(c)))1/2 =
√
2R(1)c
and
√
(g(t(c)))3/2
g(t(c))
= 2R(1)
.
0
ct(c)g (t(c))
t(c)g 0 (t(c))
From condition (7) we obtain
1 d
2 R(1
−
c
)
c dc
≤ M.
ON HOPF’S LEMMA AND THE STRONG MAXIMUM PRINCIPLE
15
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1
• O
Figure 1. Ω in the case N = 2.
5. A sufficient condition for the validity of the Strong Maximum
Principle
Consider the case
Z
ξ
g(ζ 2 /N )
dζ < +∞.
ζ
0
We wish to prove the following theorem.
G(ξ) =
Theorem 3 (The Strong Maximum Principle).
Let Ω ⊂ RN be a connected, open
1,2
and bounded set. Let u ∈ W (Ω) ∩ C Ω be a weak supersolution to
N
−1
X
uxi xi + g(u2xN )uxN xN = 0,
i=1
where g satisfies assumptions (L) and (G) and, on (0, t ] ( t defined in assumptions
(L)),
g 0 (t) ≥ 0 and g(t) + g 0 (t)t ≤ 1.
Moreover, suppose that there exists K > 0 such that, for every 0 < ξ 2 /N ≤ t, we
have
p
(14)
g(ξ 2 /N ) ≤ K eG(ξ) − 1 .
16
S. BERTONE, A. CELLINA, AND E. M. MARCHINI
Then, if u attains its minimum in Ω, it is a constant.
Remark 3. When the function g satisfies the condition
g 0 (t)t ≤ 2K (g(t))
3/2
,
for every 0 < t ≤ t, then it satisfies
g(t) + g 0 (t)t ≤ 1
and
for every 0 < ξ 2 /N ≤ t.
p
g(ξ 2 /N ) ≤ K eG(ξ) − 1 ,
Indeed, since G(ξ) < +∞, we have that limt→0 g(t) = 0. Hence, we can suppose
that g(t) ≤ 1/(2K + 1), for 0 < t ≤ t, so that
g(t) + g 0 (t)t ≤ g(t) + 2K (g(t))3/2 ≤ 1.
Moreover, since g(0) = 0, G(0) = 0 and
p
0
0
g(ξ 2 /N )
g(ξ 2 /N ) ≤ 2K
≤ K eG(ξ) − 1 ,
ξ
we obtain that
p
g(ξ 2 /N ) ≤ K eG(ξ) − 1 .
Remark 4. Among the functions g such that
lim+
t→0
(g(t))3/2
tg 0 (t)
exists, there exists K > 0 such that, for every 0 < ξ 2 /N ≤ t, we have
p
g(ξ 2 /N ) ≤ K eG(ξ) − 1
if and only if
lim
t→0+
(g(t))3/2
> 0.
tg 0 (t)
An example of a map satisfying the assumptions of the theorem above, is given
by
1
,
(ln(t))2
for 0 ≤ t ≤ 1/e4. For 0 ≤ ξ 2 /N ≤ 1/e4 , we have
Z ξ
1
1
G(ξ) =
dζ = −
,
2 /N ))2
2 /N )
ζ(ln(ζ
2
ln(ξ
0
g(t) =
and
Set
p
g(ξ 2 /N ) =
1
1
2| ln(ξ2 /N )| − 1
= eG(ξ) − 1.
≤
2
e
| ln(ξ 2 /N )|
R(λ, λN ) = {(x1 , . . . , xN ) : |xi | ≤ λ, for i = 1, . . . , N − 1, |xN | ≤ λN }.
Differently from the proof of Lemma 4 and Theorem 1, we will build a subsolution
that is not radially symmetric. This construction is provided by next theorem.
ON HOPF’S LEMMA AND THE STRONG MAXIMUM PRINCIPLE
17
Theorem 4. Under the same assumptions on g as on Theorem 3, for every r > 0
and every , there exist: l, lN ; an open convex region A ⊂ R(l, lN ); a function
v ∈ W 1,2 (ω) ∩ C 1 (ω) ∩ C(ω), where ω = B(A, r) \ A, such that
i) 0 ≤ l ≤ 2Kr, and 0 ≤ lN ≤ r/4;
ii)

v is a weak solution to F (v) ≥ 0 in ω



v>0
in ω
(15)
v
=
0
in ∂B(A, r)



v≤
in ∂A.
To give a sketch of the proof we consider the construction in R2 . Using polar
coordinates (ρ, θ) and setting a = sin2 θ, the operator F , when computed on a
radial function w, reduces to
wρ
F (w) = wρρ 1 − a + ag(wρ2 a) +
a + (1 − a)g(wρ2 a) .
ρ
We will use this expression of F , valid only for radial functions, to build a non-radial
solution of F (u) ≥ 0.
We set a = 1 and w a solution to F (w) = 0 on [R(1), R(1) + r), where R(1) is
such that w(R(1)) ≤ and w(R(1) + r) = 0. The radial function w would solve
F (w) ≥ 0, for 0 ≤ a ≤ 1, but the domain we would obtain is too large for our
purposes.
Fix a1 < 1 and find the smallest R(a1 ) > 0 such that, setting
wa1 (ρ) = w(ρ − R(a1 ) + R(1)),
the function wa1 is a solution to F (w a1 ) ≥ 0, for every a < a1 . We also set
√
D0 = {(x1 , x2 ) : R(1) < ρ < R(1) + r, a1 ≤ sin θ ≤ 1}
and on D0 we define v0 (x1 , x2 ) = w(ρ(x1 , x2 )). This function v0 is a restriction of
the function v we seek.
Fix a2 < a1 and set
√
√
D1 = {(x1 , x2 ) : R(a1 ) < ρ1 < R(a1 ) + r, a2 ≤ sin θ ≤ a2 },
√
√ where ρ1 = d((x1 , x2 ), O1 ) and O1 = (R(1) − R(a1 )) 1 − a, a , see figure 2, and
on D1 , set v1 (x1 , x2 ) = wa1 (ρ1 (x1 , x2 )). The formula
v0 (x1 , x2 ) on D0
v̄(x1 , x2 ) =
v1 (x1 , x2 ) on D1
defines a function v̄ that it is a weak solution to F (v̄) ≥ 0 on int(D0 ∪ D1 ).
Iterating this procedure, we construct a function v on a domain ω, as in figure
2, solving (15).
Condition (14) implies that the domain ω is sufficiently small.
Proof of Theorem 4. Fix r; we can assume that is such that 0 < 2 /r2 ≤ t and
that
s s 2 1
1
2
2 2 g
+
g
.
ln
g
≤
2
2
2
r
2
r
r
4K
18
S. BERTONE, A. CELLINA, AND E. M. MARCHINI
Fix the origin O0 = (0, . . . , 0), and set polar coordinates as

x1 = ρ cos θN −1 . . . cos θ2 cos θ1



x2 = ρ cos θN −1 . . . cos θ2 sin θ1
...



xN = ρ sin θN −1 .
1) When w is a radial function, setting a = sin2 θ1 , F reduces to
wρ
N − 2 + a + (1 − a)g(wρ2 a) .
F (w) = wρρ 1 − a + ag(wρ2 a) +
ρ
For a = 1, we seek a solution to
wρ
+ wρρ g(wρ2 ) = 0.
(16)
(N − 1)
ρ
such that wρ (R(1)) = −/r and wρ (ρ) < 0, for every ρ ∈ [R(1), R(1) + r). Consider
the Cauchy problem

N −1 ζ
0


 ζ = − ρ g(ζ 2 )
(17)


 ζ(R(1)) = − .
r
We are interested in a negative solution ζ. Define R(1) to be the unique positive
real solution to
R(1) + r
= 0,
G(−/r) − (N − 1) ln
R(1)
i.e.
r
R(1) = G(−/r)
.
e N −1 − 1
Consider the unique solution ζ of (17), such that ζ(R(1)) = −/r, i.e., such that
Z ρ
Z ζ(ρ)
N −1
ρ
g(t2 )
dt =
−
ds = −(N − 1) ln
G(ζ(ρ)) − G(−/r) =
.
t
s
R(1)
R(1)
ζ(R(1))
Then, for every ρ ∈ [R(1), R(1) + r), G(ζ(ρ)) > 0 and ζ(ρ) < 0, while ζ(R(1) + r) =
0. Setting wρ (ρ) = ζ(ρ) and
Z ρ
w(ρ) =
wρ (s)ds,
R(1)+r
we obtain that w solves (16) and, for every ρ ∈ (R(1), R(1) + r),
and
−/r = wρ (R(1)) < wρ (ρ) < wρ (R(1) + r) = 0
0 = w(R(1) + r) < w(ρ) < w(R(1)) ≤ .
2) Applying Lemma 5, we infer that the function w defined in 1) is actually a
solution to
wρ
wρ
(N − 2 + a) + g(wρ2 a) wρρ a +
(1 − a) ≥ 0,
F (w) = wρρ (1 − a) +
ρ
ρ
for every 0 ≤ a ≤ 1 and every ρ ∈ (R(1), R(1) + r).
ON HOPF’S LEMMA AND THE STRONG MAXIMUM PRINCIPLE
19
3) Let ā < 1. We wish to find the smallest R(ā) > 0 such that, setting
wā (ρ) = w(ρ − R(ā) + R(1)),
the function wā is a solution to
wρā
wρā
ā
ā
ā 2
ā
(18) F (w ) = wρρ (1 − ā) +
(N − 2 + ā) + g((wρ ) ā) wρρ ā +
(1 − ā) ≥ 0,
ρ
ρ
for every ρ ∈ (R(ā), R(ā) + r).
Since, for the function w, we have
(N − 1)
wρ (ρ)
+ g(wρ2 (ρ))wρρ (ρ) = 0,
ρ
at ρ − R(ā) + R(1) we obtain
(N − 1)
This equality yields
(N − 2 + ā)
wρā (ρ)
ā
+ g((wρā (ρ))2 )wρρ
(ρ) = 0.
ρ − R(ā) + R(1)
wρā
ā
+ āg((wρā )2 ā)wρρ
=
ρ
wρā
ρ(ρ − R(ā) + R(1))
and
(1 − ā)
ā
wρρ
g((wρā )2 ā)
(N + 2 − ā)(ρ − R(ā) + R(1)) − āρ(N − 1)
g((wρā )2 )
wρā
+
g((wρā )2 ā)
ρ
(1 − ā)wρā
ρ(ρ − R(ā) + R(1))
Since
F (wā ) = (N − 2 + ā)
=
(ρ − R(ā) + R(1))g((wρā )2 ā) −
(N − 1)ρ
g((wρā )2 )
(ρ − R(ā) + R(1)) N − 2 + ā + (1 − ā)g((wρā )2 ā) −
Since
.
wρā
wρā
ā
ā
+ āg((wρā )2 ā)wρρ
+ (1 − ā) wρρ
+
g((wρā )2 ) ,
ρ
ρ
we obtain that F (w ā ) ≥ 0 if and only if
Set
!
(N − 1)ρ
1 − ā + āg((wρā )2 ā) ≤ 0.
ā
2
g((wρ ) )
k(ρ) = (ρ − R(ā) + R(1)) N − 2 + ā + (1 − ā)g((wρā )2 ā) −
(N − 1)ρ
1 − ā + āg((wρā )2 ā)
ā
2
g((wρ ) )
d
d
g((wρā )2 ) ≤
g((wρā )2 ā) ≤ 0,
dρ
dρ
20
S. BERTONE, A. CELLINA, AND E. M. MARCHINI
applying Lemma 5, we have that
d
k 0 (ρ) = N − 2 + ā + (1 − ā)g((wρā )2 ā) + (ρ − R(ā) + R(1))(1 − ā) g((wρā )2 ā)−
dρ
1
ρ
d
ā 2
(N − 1)
1 − ā + āg((wρā )2 ā) −
−
g((w
)
)
ρ
ā
2
ā
2
2
g((wρ ) ) (g((wρ ) )) dρ
(N − 1)
ρ
d
g((wρā (ρ))2 ā)ā ≤
g((wρā )2 ) dρ
(N − 1)ρ d
(N − 1)ρ d
g((wρā )2 ) 1 − ā + āg((wρā )2 ā) −
g((wρā (ρ))2 ā)ā ≤
ā
2
2
(g((wρ ) )) dρ
g((wρā )2 ) dρ
(N − 1)ρ d
g((wρā )2 ) 1 − ā + āg((wρā )2 ā) − āg((wρā )2 ) ≤
(g((wρā )2 ))2 dρ
(N − 1)ρ
d
(1 − ā)g((wρā )2 ) g((wρā )2 ) ≤ 0.
ā
2
2
(g((wρ ) ))
dρ
Since the function k(ρ) is non increasing, it follows that F (w ā ) ≥ 0, for every
ρ ∈ (R(ā), R(ā) + r), if and only if
R(1) N − 2 + ā + (1 − ā)g((wρā (R(ā)))2 ā) −
(N − 1)R(ā)
1 − ā + āg((wρā (R(ā)))2 ā) =
ā
2
g((wρ (R(ā)) )
R(1) N − 2 + ā + (1 − ā)g
2
ā
r2
2 R(1) N − 2 + ā + g r2 ā (1 − ā)
.
2 N −1
1 − ā + g r 2 ā ā
2 N
−
2
+
ā
+
g
r 2 ā (1 − ā)
R(1)
.
2 N −1
1 − ā + g r 2 ā ā
i.e. if and only if
R(ā) ≥ g
2
r2
Hence, we define
R(ā) = g
2
r2
(N − 1)R(ā)
−
2
g r2
1 − ā + āg
2
ā
r2
≤ 0,
4) The function wā defined in point 3) is a solution to
ā
F (w ) =
ā
wρρ
(1
wρā
wρā
ā 2
ā
− a) +
(N − 2 + a) + g((wρ ) a) wρρ a +
(1 − a) ≥ 0,
ρ
ρ
ON HOPF’S LEMMA AND THE STRONG MAXIMUM PRINCIPLE
for every a < ā.
(R(ā), R(ā) + r),
21
Indeed, applying Lemma 5 we obtain that, for every ρ ∈
(ρ − R(ā) + R(1)) N − 2 + a + (1 − a)g((wρā )2 a) −
(N − 1)ρ
1 − a + ag((wρā )2 a) ≤
ā
2
g((wρ ) )
(ρ − R(ā) + R(1)) N − 2 + ā + (1 − ā)g((wρā )2 ā) −
(N − 1)ρ
1 − ā + āg((wρā )2 ā) ≤ 0.
g((wρā )2 )
5) Assume we have a partition α of [0, π/2], α = {0 = αn < · · · < α1 <
α0 = π/2}. This partition defines two partitions of [0, 1], given by ci = cos αi and
si = sin αi .
Consider the sums
S1 (α) =
n
X
i=1
and
n−1
X
R(1 − c2i ) (ci+1 − ci )
R(1 − c2i−1 ) − R(1 − c2i ) ci = R(1)c1 +
i=1
n
X
S2 (α) =
R(s2i−1 )(si−1
i=1
− si ) = R(1)(1 − s1 ) +
n−1
X
i=1
R(s2i )(si − si+1 ),
where, in the previous equalities, we have taken into account that R(1 − c2n ) =
R(0) = 0. Our purpose is to provide a partition α and corresponding estimates for
S1 (α) and S2 (α) that are independent of .
The sums
n−1
X
i=1
R(1 − c2i )(ci+1 − ci )
and
n−1
X
i=1
are Riemann sums for the integrals
Z 1
R(1 − c2 )dc
and
c1
Z
s1
R(s2i )(si − si+1 )
R(s2 )ds.
0
Consider the first integral. From
N − 1 − c2 + g
c2 + g
2
r 2 (1
we obtain that
Z
1
c1
2
r 2 (1
− c2 ) (1 − c2 )
r2
R(1)
N −1
R(1)g
2
r2
R(1 − c2 )dc = g
2
− c 2 ) c2
Z
Z
1
c1
1
≤
N −1
c2
N − 1 − c2 + g
c1
dc
.
c2
c2 + g
2
r 2 (1
2
r 2 (1
− c 2 ) c2
dc ≤
− c2 ) (1 − c2 )
22
S. BERTONE, A. CELLINA, AND E. M. MARCHINI
Set
Sx∗ (c) = R(1)c + R(1)g
R(1)g
2
r2
2
r2
Z
1
c
2
db
1
=
R(1)
c
+
g
−
1
=
b2
r2
c
!
1
−1 .
2 +
c
g r2
c
Evaluating the last term at the minimum point c =
Sx∗
q
2
r2
g
, we obtain


s !
2
2
2

q
g
= R(1)g
−1 =
r2
r2
2
g
r2
2r
q
2
r2
N − 2 + s2 + g
2
r2
g
− R(1)g
eG(/r) − 1
q
q
2
2
We fix c1 = g r 2 , so that α1 = arccos g r 2 .
Consider the second integral
Z s1
R(s2 )ds.
.
0
From
N − 2 + s2 + g
2 2
r2 s
2
r2
1 − s2 + g
we obtain that
Z
s1
2
R(s )ds = g
0
g
(1 − s2 )
s2
r2
R(1)
N −1
Z
2
r2
R(1)
N −1
Z
2
≤
s1
1 − s2
N − 2 + s2 + g
2 2
r2 s
N − 2 + s2 + g
2
r2
1 − s2
0
2 2
r2 s
1 − s2 + g
0
s1
2 2
r2 s
(1 − s2 )
(1 − s2 )
s2
(1 − s2 )
ds ≤
ds.
Set
Sy∗ (s) = R(1)(1 − s) + g
R(1)(1 − s) + g
2
r2
2
r2
Since
1−
s
R(1)
N −1
R(1)
N −1
1−g
Z
s
N − 2 + b2 + g
2
r2
1 − b2
0
(1 − b2 )
db =
2 N −1
1+s
g
−1 s+
.
ln
r2
2
1−s
2
r2
≤g
2
r2
,
ON HOPF’S LEMMA AND THE STRONG MAXIMUM PRINCIPLE
q
evaluating the last term at the point s1 = sin α1 = 1 − g
s
s
2 !
2 !
Sx∗
= R(1) 1 − 1 − g
+
1−g
r2
r2
g
2
r2
2
r2
23
, we obtain
q

2
2
s
1
−
1−g
R(1) 
1
q
g
1−g
−1
− ln 
2
2
N −1
r
r
2
1+ 1−g

2
r2
2
r2
s s s 2 !
1
2
2
2 R(1) g
+
.
g
g
ln g
2
2
2
r
r
2
r
r2 
 ≤
To define the other points of the required partition α, consider the integrals
Z 1
Z s1
R(1 − c2 )dc and
R(s2 )ds.
0
c1
Set
σ = R(1)g
2
r2
.
By the basic theorem of Riemann integration, taking a partition α with mesh size
small enough, the value of the Riemann sums
n−1
X
i=1
differs from
R(1 −
Z
c2i )(ci+1
1
c1
− ci )
and
R(1 − c2 )dc
and
n−1
X
i=1
R(1)c1 +
Z
Z
R(s2 )ds
s1
0
R(1 − c2i ) (ci+1 − ci ) ≤
1
c1
R(s2i )(si − si+1 )
i=1
by less than σ. In particular we obtain
S1 (α) = R(1)c1 +
n−1
X
2
R(1 − c )dc + σ ≤
2r
e
R(1)(1 − s1 ) +
r
e
q
g
G(/r)
N −1
2
r2
−1
n−1
X
i=1
Z
2
r2
−1
≤ 2Kr
R(s2i )(si − si+1 ) ≤
s1
0
g
G(/r)
N −1
and
S2 (α) = R(1)(1 − s1 ) +
q
R(s2 )ds + σ ≤
s s 2 !
2
2 r
1
2 g
g
ln g
≤ .
+
2
2
2
r
2
r
r
4
24
S. BERTONE, A. CELLINA, AND E. M. MARCHINI
6) With respect to the coordinates fixed at the beginning of the proof, consider
xi ≥ 0. Set
√
√
D0 = {(x1 , . . . , xN ) : R(1) < ρ < R(1) + r, a1 ≤ sin θN −1 ≤ a0 = 1}
and on D0 define the function
v0 (x1 , . . . , xN ) = w(ρ(x1 , . . . , xN )).
By point 1), the function v0 is of class C 2 (int(D0 )) and satisfies, pointwise, the
inequality F (v0 ) ≥ 0. Having defined v0 , define v1 as follows. Set:
for N = 2,
√
√ O1 = (Ox1 1 , Ox1 2 ) = (R(1) − R(a1 )) 1 − a1 , a1 ,
for N > 2,
O1 (θ1 , . . . , θN −2 ) = (Ox1 1 , . . . , Ox1 N ) =
(R(1) − R(a1 ))
for N ≥ 2,
√
√
√ 1 − a1 cos θN −2 . . . cos θ1 , 1 − a1 cos θN −2 . . . sin θ1 , . . . , a1 ,
ρ1 (x1 , . . . , xN ) =
and
q
(x1 − Ox1 1 )2 + · · · + (xN − Ox1 N )2
1
sin θN
−1 (x1 , . . . , xN ) =
xN − Ox1 N
.
ρ1 (x1 , . . . , xN )
Recalling the definition of w a1 in 3), consider
D1 = {(x1 , . . . , xN ) : R(a1 ) < ρ1 (x1 , . . . , xN ) < R(a1 ) + r,
√
and, on D1 , set
1
a2 ≤ sin θN
−1 (x1 , . . . , xN ) ≤
√ a1
v1 (x1 , . . . , xN ) = wa1 (ρ1 (x1 , . . . , xN )).
The function v1 is of class C 2 (int(D1 )). We claim that v1 still satisfies F (v1 ) ≥ 0.
Remark that the set of the points O 1 is equal to
√
O1 = {(x1 , . . . , xN ) : ρ = R(1) − R(a1 ), sin θN −1 = a1 }
and that for every point p ∈ D1 , the corresponding point O 1 (p) is the projection of
p on O1 , while ρ1 (p) = d(p, O 1 ). Then we obtain
∂ρ1
=0
∂θi
for every i = 1, . . . , N − 2,
∂Ox1 i
≥0
∂xi
for every i = 1, . . . , N − 1,
∂Ox1 N
= 0.
∂xN
ON HOPF’S LEMMA AND THE STRONG MAXIMUM PRINCIPLE
25
and
wρa1 (ρ1 )
ρ1
∇v1 =
x1 − Ox1 1 , . . . , xN − Ox1 N =
1
1
1
wρa1 (ρ1 ) cos θN
−1 . . . cos θ1 , cos θN −1 . . . sin θ1 , . . . , sin θN −1 ,
(v1 )xi xi =
a1
wρρ
(ρ1 )
xi − Ox1 i
ρ1
2
wρa1 (ρ1 )
+
ρ1
1−
xi − Ox1 i
ρ1
2
∂Ox1 i
−
∂xi
!
=
a1
wρρ
(ρ1 ) cos2 θN −1 . . . sin2 θi−1 +
wρa1 (ρ1 )
ρ1
∂Ox1 i
1 − cos θN −1 . . . sin θi−1 −
∂xi
2
2
.
Then
F (v1 ) =
N
−1
X
(v1 )xi xi + (v1 )xN xN g((v1 )2xN ) =
i=1
2 1
2 1
a1
1
a1
2
wρρ
(ρ1 ) cos2 θN
−1 + sin θN −1 g (wρ (ρ1 )) sin θN −1
+
wρa1 (ρ1 )
2 1
1
2 1
a1
2
N − 2 + sin2 θN
−
−1 + cos θN −1 g (wρ (ρ1 )) sin θN −1
ρ1
−1
wρa1 (ρ1 ) NX
∂Ox1 i
≥0
ρ1
∂xi
i=1
since wa1 (ρ1 ) verifies equation (18).
1
The sets D0 and D1 intersect on sin θN −1 (x1 , . . . , xN ) = sin θN
−1 (x1 , . . . , xN ) =
√
a1 . For a point (x1 , . . . , xN ) in this intersection we have
ρ1 (x1 , . . . , xN ) = ρ(x1 , . . . , xN ) − (R(1) − R(a1 )).
Hence, on D0 ∩ D1
R(a1 ) ≤ ρ1 (x1 , . . . , xN ) ≤ R(a1 ) + r
if and only if
R(1) ≤ ρ(x1 , . . . , xN ) ≤ R(1) + r,
and the functions v0 and v1 coincide:
v1 (x1 , . . . , xN ) = wa1 (ρ1 (x1 , . . . , xN )) = w(ρ1 (x1 , . . . , xN ) + R(1) − R(a1 )) =
w(ρ(x1 , . . . , xN )) = v0 (x1 , . . . , xN ).
The formula
v̄(x1 , . . . , xN ) =
v0 (x1 , . . . , xN )
v1 (x1 , . . . , xN )
on D0
on D1
defines a function v̄ in C 0 (int(D0 ∪D1 )). We claim that it is also in C 1 (int(D0 ∪D1 )).
26
S. BERTONE, A. CELLINA, AND E. M. MARCHINI
In fact, we have
∇v0 (x1 , . . . , xN ) =
wρ (ρ(x1 , . . . , xN ))
(x1 , . . . , xN ),
ρ(x1 , . . . , xN )
∇v1 (x1 , . . . , xN ) =
wρa1 (ρ1 (x1 , . . . , xN ))
x1 − Ox1 1 , . . . , xN − Ox1 N
ρ1 (x1 , . . . , xN )
and, on D0 ∩ D1 ,
1
1
(x1 , . . . , xN ).
x1 − Ox1 1 , . . . , xN − Ox1 N =
ρ1 (x1 , . . . , xN )
ρ(x1 , . . . , xN )
On int(D0 ) and int(D1 ), the function v̄ is of class C 2 and satisfies, pointwise, the
inequality F (v̄) ≥ 0. We claim that v̄ is also in W 1,2 (int(D0 ∪ D1 )) and that it is a
weak solution to F (v̄) ≥ 0 on int(D0 ∪D1 ). In fact, for every η ∈ C0∞ (int(D0 ∪D1 )),
applying the divergence theorem separately to int(D0 ) and to int(D1 ), we obtain
Z
[div∇∇v L(∇v̄(x))η(x) + h∇L(∇v̄(x)), ∇η(x)i] dx =
Z
Z
Z
int(D0 ∪D1 )
int(D0 )∪int(D1 )
[div∇∇v L(∇v̄(x))η(x) + h∇L(∇v̄(x)), ∇η(x)i] dx =
η(x)h∇L(∇v̄(x)), n0 (x)idl +
∂(int(D0 ))
√
D0 ∩{sin θN −1 = a1 }
Z
√
1
D1 ∩{sin θN
−1 = a1 }
Z
η(x)h∇L(∇v̄(x)), n1 (x)idl =
∂(int(D1 ))
η(x)h∇L(∇v̄(x)), n0 (x)idl+
η(x)h∇L(∇v̄(x)), n1 (x)idl,
where n0 , n1 are the outer unit normal respectively to ∂(int(D0 )), ∂(int(D1 )). The
last term equals zero, since v̄ ∈ C 1 (int(D0 ∪D1 )) and n0 = −n1 on D0 ∩{sin θN −1 =
√
√
1
a1 } = D1 ∩ {sin θN
a1 }. Hence, when η ≥ 0, we have that
−1 =
Z
int(D0 ∪D1 )
h∇L(∇v̄(x)), ∇η(x)idx ≤ 0,
as we wanted to show.
Assuming defined On−2 (θ1 , . . . , θN −2 ) and a function v ∈ C 1 (int(D0 ∪ · · · ∪
Dn−2 )), consider
)=
, . . . , Oxn−1
On−1 (θ1 , . . . , θN −2 ) = (Oxn−1
N
1
On−2 (θ1 , . . . , θN −2 ) + (R(an−2 ) − R(an−1 ))
p
p
√
1 − an−1 cos θN −2 . . . cos θ1 , 1 − an−1 cos θN −2 . . . sin θ1 , . . . , an−1 .
ON HOPF’S LEMMA AND THE STRONG MAXIMUM PRINCIPLE
Set
ρn−1 (x1 , . . . , xN ) =
q
27
2
(x1 − Oxn−1
)2 + · · · + (xN − Oxn−1
1
N ) ,
n−1
sin θN
−1 (x1 , . . . , xN ) =
xN − Oxn−1
N
ρn−1 (x1 , . . . , xN )
Dn−1 = {(x1 , . . . , xN ) : R(an ) < ρn−1 (x1 , . . . , xN ) < R(an−1 ) + r,
0=
√
n−1
an ≤ sin θN
−1 (x1 , . . . , xN ) ≤
√
an−1
and define on Dn−1 the function
vn−1 (x1 , . . . , xN ) = wan−1 (ρn−1 (x1 , . . . , xN )).
Set D = int(D0 ∪ · · · ∪ Dn−1 ), the same considerations as before imply that the
function

on D0
 v0 (x1 , . . . , xN )
...
...
v̄(x1 , . . . , xN ) =

vn−1 (x1 , . . . , xN ) on Dn−1
is such that v̄ ∈ W 1,2 (D) ∩ C 1 (D) ∩ C(D) and it is a weak solution to F (v̄) ≥ 0
on D. This completes the construction of v̄ as a weak solution to F (v̄) ≥ 0 on
D0 ∪ · · · ∪ Dn−1 .
Set O∗ = (Ox∗1 , . . . , Ox∗N ) = (0, . . . , 0, R(1) − lN ). We have that
D0 ∪ · · · ∪ Dn−1 ⊂ {(x1 , . . . , xN ) : 0 ≤ xi ≤ l + r for i = 1, . . . , N − 1,
Ox∗N ≤ xN ≤ Ox∗N + lN + r) .
Define the full domain ω and the solution by symmetry with respect to the point
O∗ . Figure 2 shows this construction in dimension N = 2 and for n − 1 = 2. Hence
the solution will be in W 1,2 (ω) ∩ C 1 (ω) ∩ C(ω) and a weak solution of F (v) ≥ 0 on
ω.
7) The previous construction yields a region A centered in O ∗ , a corresponding region ω and a function v that solves (15). The change of coordinates x̂1 =
∗
x1 , . . . , x̂N −1 = xN −1 , x̂N = xN − ON
, centers A at the origin and proves the
theorem.
In order to prove Theorem 3, we need this further lemma.
Lemma 6. Consider the sets A and R(O ∗ , l, lN ), where A, O∗ , l, lN have been
defined in Theorem 3. Then, for every p ∈ ∂R,
d(p, A) < lN .
Proof. Set q = O ∗ + (l, . . . , l, lN ). We prove that
∗
d(q, A) < lN .
Set pi = O + (0, . . . , 0, l, 0, . . . , 0) and let ΠN −1 the hyperplane passing through
p1 , . . . , pN . Since A is convex and pi ∈ A, we obtain that
See Figure 3.
d(q, A) < d(q, ΠN −1 ) < lN .
28
S. BERTONE, A. CELLINA, AND E. M. MARCHINI
...............................................................................
..................
. .......
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..........
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...
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D0
..
.....
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........
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..
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.. ......... D1
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O∗ •
.
...•
..
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..
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....
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....................
.......
.......
........................................................................................
.
........
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.
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.
.
.
...........
............α2
.......
.................
...... ....
1•
..
..............
.
.......................................O
.. ......................
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.
..................
..
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1
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.. .....
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..
α
O•
Figure 2. The sets ω and A in the case N = 2 and n = 3.
R(l, lN )
p1
l
q
..
.............................................................................................................
...
.................
.
........
.
.
...
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.........
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........
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........
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....
........
....
.
........
lN
.
... ...
A
.
........
.
... ....
........
..
.
...
........
....
.
........ ...
.........
...
........
........
..
........ .. p
O∗
........ . 2
...
.....
.
...
...
...
...
...
..
....
...
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.....
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.......
...
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......
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.........
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................................................
Figure 3. A and R(O∗ , l, lN ) in the case N = 2.
ON HOPF’S LEMMA AND THE STRONG MAXIMUM PRINCIPLE
29
Proof of Theorem 3. a) Suppose that u attains its minimum in Ω, and assume
minΩ u = 0 and set C = {x ∈ Ω : u(x) = 0}. By contradiction, suppose that the
open set Ω \ C 6= ∅.
b) Since Ω is a connected set, there exist s ∈ C and R > 0 such that B(s, R) ⊂ Ω
and B(s, R) ∩ (Ω \ C) 6= ∅. Let p ∈ B(s, R) ∩ (Ω \ C). Consider the line ps. Moving
p along this line, we can assume that B(p, d(p, C)) ⊂ (Ω \ C), and that there exists
a point z ∈ C such that d(p, C) = d(p, z).
c) Fix r:
0<r<
Set
d(p, C)
.
32(N − 1)K 2 + 78
r
(r) = min u(z) : z ∈ B p, d(p, C) −
,
4
¯
we have that (r) > 0, and we set = min{(r), r ξ}.
d) For r and as defined in c), consider: l, lN , A and v as defined in Theorem
4. Without loss of generality, since the set A is symmetric with respect to both
coordinate axis, we can suppose that pz belongs to the first quadrant, i.e. that, for
every i = 1, . . . , N , zi ≥ pi , where z = (z1 , . . . , zN ) and p = (p1 , . . . , pN ).
r
Define the point q on the segment pz such that d(q, p) = d(p, z) − . Set
2
q ∗ = q−(l, . . . , l, lN ), R(q ∗ , l, lN ) = q ∗ +R(l, lN ), A∗ = q ∗ +A and v ∗ (x+q ∗ ) = v(x).
We first claim that
r
.
R(q ∗ , l, lN ) ⊂ B p, d(p, C) −
4
Let t ∈ R(q ∗ , l, lN ), then t can be written as (q1 − 2α1 l, . . . , qN −1 − 2αN −1 l, qN −
, we have that
2αN lN ), with 0 ≤ αi ≤ 1, for i = 1, . . . , N . Since r < 32(Nd(p,C)
−1)K 2 + 7
8
d(t, p)2 =
N
X
i=1
d(p, C) −
(qi − 2αi li − pi )2 = d(q, p)2 +
N
X
i=1
4α2i li2 −
N
X
i=1
4αi li (qi − pi ) ≤
N
r 2 X 2 r 2
r2
r 2
+
4li ≤ d(p, C) −
+ 16(N − 1)K 2 r2 +
< d(p, C) −
.
2
2
4
4
i=1
See Figure 4.
Since A∗ ⊂ R(q ∗ , l, lN ), we have obtained that
r
,
A∗ ⊂ B p, d(p, C) −
4
so that u ≥ in ∂A∗ .
By Lemma 6,
r
d(q, A∗ ) < lN ≤ ,
4
we have that
d(z, A∗ ) ≤ d(z, q) + d(q, A∗ ) <
3
r,
4
30
S. BERTONE, A. CELLINA, AND E. M. MARCHINI
B (p, d(p, C))
B p, d(p, C) −
r
4
B p, d(p, C) −
r
2
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l
R
p
l
q
•
• z
•
Figure 4. The set R = R(q ∗ , l, l2 ) in the case N = 2.
so that
z ∈ ω ∗ = B(A∗ , r) \ A∗ .
e) The function v ∗ satisfies
 ∗
v is a weak solution to F (v) ≥ 0


 ∗
v >0
v∗ = 0


 ∗
v ≤
in
in
in
in
ω∗
ω∗
∂B(A∗ , r)
∂A∗ .
Since u, v ∗ ∈ W 1,2 (ω ∗ ) ∩ C(ω ∗ ), v ∗ is a weak subsolution and u is a weak solution
∗
∗
to F (u) = 0, and v|∂ω
in ω ∗ .
∗ ≤ u|∂ω ∗ , applying Lemma 1, we obtain that u ≥ v
But u(z) = 0 < v ∗ (z), a contradiction.
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ON HOPF’S LEMMA AND THE STRONG MAXIMUM PRINCIPLE
31
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Dipartimento di Matematica e Applicazioni, Universitá degli Studi di Milano-Bicocca,
Via R. Cozzi 53, 20125 Milano
E-mail address: [email protected]
Dipartimento di Matematica e Applicazioni, Universitá degli Studi di Milano-Bicocca,
Via R. Cozzi 53, 20125 Milano
E-mail address: [email protected]
Dipartimento di Matematica e Applicazioni, Universitá degli Studi di Milano-Bicocca,
Via R. Cozzi 53, 20125 Milano
E-mail address: [email protected]