On Hopf's Lemma and the Strong Maximum Principle

0.29 MB·English

by  S. Bertone

ON HOPF’S LEMMA AND THE STRONG MAXIMUM
PRINCIPLE
5
0
0
S.BERTONE,A.CELLINA,ANDE.M.MARCHINI
2

n
a Abstract. InthispaperweconsiderHopf’sLemmaandtheStrongMaximum
J Principleforsupersolutionsto
N
8
2 gi(u2xi)uxixi =0
i=1
X
] undersuitablehypotheses thatallowgi toassumevaluezeroatzero.
P
A
.
h
t 1. Introduction
a
m LetΩ RN beaconnected,openandboundedset;wecallΩregularifforevery
⊂
[ z ∂Ω, there exists a tangent plane, continously depending on z. We say that Ω
∈
satisfies the interior ball condition at z if there exists an open ball B Ω with
1 ⊂
z ∂B. On Ω, consider the operator
v ∈
3 N
0 (1) F(u)= g (u2 )u ,
5 i xi xixi
i=1
1 X
0 where gi :[0,+ ) [0,+ ) are continuous functions.
∞ → ∞
5 When F is elliptic, two classical results hold.
0
/
h Hopf’s Lemma:
t LetΩ beregular, letu besuchthat F(u) 0on Ω. Supposethat thereexistsz ∂Ω
a ≤ ∈
m such that
u(z)<u(x), for all x in Ω.
:
v
If, in addition, Ω satisfies the interior ball condition at z, we have
i
X
∂u
r (z)<0,
a ∂ν
where ν is the outer unit normal to B at z.
The Strong Maximum Principle:
LetubesuchthatF(u) 0onΩ,thenifuitattainsminimuminΩ,itisaconstant.
≤
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,
to extend the validity of the previous results, as in [1, 5, 6, 8, 9, 10, 12, 13, 14, 15,
16, 18].
Key words and phrases. Strongmaximumprinciple.
1
2 S.BERTONE,A.CELLINA,ANDE.M.MARCHINI
In the case in equation (1) we have g 1, for every i, then F(u)=∆u, and we
i
≡
find the classical problem of the Laplacian, see [7, 9].
On the other hand, when there exists i 1,...,N such that g 0 on an
i
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
such that N g (u2 )u = 0. For instance, let g (t) = 0, for every t [0,2].
i=1 i xi xixi N ∈
The function
P
(x2 1)4 if 1 x 1
u(x1,...,xN)= −0 N − oth−erw≤ise N ≤ ,
(cid:26)
satisfies (1) in RN.
We are interested in the case when 0 g (t) 1 and it does not exist i such
i
≤ ≤
that g 0 on an interval. Since g could assume value zero, the equation (1) is
i i
≡
non elliptic.
The results known so far, for the validity of Hopf’s Lemma and of the Strong
Maximum Principle, suggest that, for possibly non elliptic equations, but arising
from a functional having rotational symmetry, this validity shall depend only on
the behaviour of the functions g near zero, see [4].
i
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 g (t) that goes
i
fastesttozero,astgoestozero. Hopf’slemmaandtheStrongMaximumPrinciple
areessentiallythesameresultaslongaswecanbuildsubsolutionswhoselevellines
canhavearbitrarilylargecurvature. Thisneednotbealwayspossibleforproblems
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 g , for i = 1,...,N 1, are 1 and only g is allowed to go
i N
−
to zero. In this simpler class of equations we are able to show that the condition
(g (t))3/2
N
lim >0
t→0+ tgN′ (t)
is at once necessaryfor the validity ofHopf’s Lemma and sufficientfor 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 g (t) g (t) 1;
N i
≤ ≤ ≤
ii) g is continuous on [0,t]; positive and differentiable on (0,t];
N
iii) on (0,t], the function t g (t)+g (t)t is non decreasing.
→ N N′
ON HOPF’S LEMMA AND THE STRONG MAXIMUM PRINCIPLE 3
Notice that, in case ii) above is violated, the Strong Maximum Principle does
nothold;andthatconditioniii)aboveincludesthecaseoftheLaplacian,g (t) 1;
i
≡
and, finally, that under these assumptions, g could assume value zero at most for
i
t=0.
Moreover,we can consider the equation
N
g (u2 )u =0
i xi xixi
i=1
X
as the Euler-Lagrangeequation associated to the functional
N
1
J(u)= L( u)dΩ= f (u2 )u2 dΩ,
ZΩ ∇ ZΩ 2 i=1 i xi xi!
X
whereL( u)isstrictlyconvexin (u ,...,u ):u2 t, for every i=1,...,N .
∇ { x1 xN xi ≤ }
Indeed, fix i. Let f be a solution to the differential equation
i
(2) g (t)=f (t)+5tf (t)+2t2f (t),
i i i′ i′′
for t [0,t]. Since
∈
∂2L
(u2 )=f (u2 )+5u2 f (u2 )+2u4 f (u2 )=g (u2 ),
∂u2 xi i xi xi i′ xi xi i′′ xi i xi
xi
we have that
N ∂2L N
div L( u)= u = g (u2 )u .
∇∇u ∇ ∂u2 xixi i xi xixi
i=1 xi i=1
X X
Moreover,the strict convexity of L( u) in (u ,...,u ):u2 t, for every i=
∇ { x1 xN xi ≤
1,...,N follows by the fact that g is positive in (0,t].
i
}
Sincewewillneedgeneralcomparisontheoremsthatdependontheglobalprop-
erties of the solutions, i.e. on their belonging to a Sobolev space, we will need
also a growth assumption on g (assumption (G)) to insure these properties of the
i
solutions.
Assumption (G):
Each function f as defined in (2), is bounded and f (u2 )u2 is strictly convex.
i i xi xi
Any function g satisfying assumptions (L) on [0,t] can be extended so as to
i
satisfy assumption (G) on [0,+ ). In fact, it is enough to extend g to (t,+ )
i
∞ ∞
by setting g (t)=f (t), for t>t.
i i
Definition 1. Let Ω be open, and let u W1,2(Ω). The map u is a weak solution
∈
to the equation F(u)=0 if, for every η C (Ω),
∈ 0∞
L( u(x)), η(x) dx =0.
h∇ ∇ ∇ i
ZΩ
u is a weak subsolution (F(u) 0) if, for every η C (Ω), η 0,
≥ ∈ 0∞ ≥
L( u(x)), η(x) dx 0.
h∇ ∇ ∇ i ≤
ZΩ
4 S.BERTONE,A.CELLINA,ANDE.M.MARCHINI
u is a weak supersolution (F(u) 0) if, for every η C (Ω), η 0,
≤ ∈ 0∞ ≥
L( u(x)), η(x) dx 0.
h∇ ∇ ∇ i ≥
ZΩ
We say that a function w W1,2(Ω) is such that w 0 if w+ W1,2(Ω).
∈ |∂Ω ≤ ∈ 0
The growth assumption (G) assures that, if u W1,2(Ω), then L( u(x))
∈ ∇ ∇ ∈
L2(Ω). The strict convexity of L implies the following comparison lemma.
Lemma 1. Let Ω be a open and bounded set, let v W1,2(Ω) be a subsolution and
∈
let u W1,2(Ω) be a supersolution to the equation F(u) = 0. If v u , then
∂Ω ∂Ω
v u∈a.e. in Ω. | ≤ |
≤
We wish to express the operator
N
F(v)= g (v2 )v
i xi xixi
i=1
X
in polar coordinates. Set
x =ρcosθ ...cosθ cosθ
1 N 1 2 1
−
x =ρcosθ ...cosθ sinθ
2 N 1 2 1
 −
...
 x =ρsinθ
N N 1
−
so that 
2 2
x x v x
i i ρ i
v =v and v =v + 1 .
xi ρ ρ xixi ρρ(cid:18) ρ (cid:19) ρ " −(cid:18) ρ (cid:19) #
When v is a radial function, F reduces to
N 2 2 2
x x v x
F(v)= g v2 i v i + ρ 1 i =
i=1 i ρ(cid:18) ρ (cid:19) !" ρρ(cid:18) ρ (cid:19) ρ −(cid:18) ρ (cid:19) !#
X
N 2 2 N 2 2
x x v x x
v g v2 i i + ρ g v2 i 1 i .
ρρ i=1 i ρ(cid:18) ρ (cid:19) !(cid:18) ρ (cid:19) ρ i=1 i ρ(cid:18) ρ (cid:19) ! −(cid:18) ρ (cid:19) !
X X
In general, we do not expect that the equation F(v) = 0 admits radial solutions.
Howeverwe will use the expressionof F validfor radialfunctions in orderto reach
our results.
The following technical lemmas will be used later.
Lemma 2. Let n=2,...,N and set
t(1 a)
h (a)=g − (1 a)+g (ta)a.
n N N
n 1 −
(cid:18) − (cid:19)
For every 0 < t t (t defined in assumptions (L)), h (a) h (1/n), for every a
n n
≤ ≥
in [0,1].
Proof. Since, on (0,t], the function t g (t)+g (t)t is non decreasing, we have
→ N N′
that
t(1 a) t(1 a) t(1 a)
h′n(a)=−gN n −1 −gN′ n −1 n −1 +gN(ta)+gN′ (ta)ta≥0
(cid:18) − (cid:19) (cid:18) − (cid:19) −
ON HOPF’S LEMMA AND THE STRONG MAXIMUM PRINCIPLE 5
if and only if a 1/n, so that h (a) h (1/n), for every a [0,1].
n n
≥ ≥ ∈ (cid:3)
Lemma 3. For every 0<t t (t defined in assumptions (L)), we have that
≤
N 2 2
x x t
i i
g t g .
N N
i=1 (cid:18) ρ (cid:19) !(cid:18) ρ (cid:19) ≥ (cid:18)N(cid:19)
X
Proof. We prove the claim by induction on N.
Let N =2. Set a=sin2θ . Applying Lemma 2 we obtain that
1
2 2 2 2
x x x x
1 1 2 2
g t +g t =
N N
ρ ρ ρ ρ
(cid:18) (cid:19) !(cid:18) (cid:19) (cid:18) (cid:19) !(cid:18) (cid:19)
t
g (t(1 a))(1 a)+g (ta)a g .
N N N
− − ≥ 2
(cid:18) (cid:19)
Suppose that the claim is true for N 1, i.e.
−
N 1 2 2
− xi xi t
g t g .
N N
i=1 (cid:18) ρ (cid:19) !(cid:18) ρ (cid:19) ≥ (cid:18)N −1(cid:19)
X
Let us prove it for N. Set
y =ρcosθ ...cosθ cosθ
1 N 2 2 1
−
y =ρcosθ ...cosθ sinθ
2 N 2 2 1
 −
...
 y =ρsinθ
N 1 N 2
− −
and set a=sin2θN 1. Applying Lemma 2 we obtain that
−
N 2 2
x x
i i
g t =
N
ρ ρ
i=1 (cid:18) (cid:19) !(cid:18) (cid:19)
X
N 1 2 2
− yi yi
g t (1 a) (1 a)+g (ta)a
N N
i=1 (cid:18)ρ(cid:19) − !(cid:18)ρ(cid:19) − ≥
X
t(1 a) t
g − (1 a)+g (ta)a g ,
N N N
N 1 − ≥ N
(cid:18) − (cid:19) (cid:18) (cid:19)
and the claim is proved.
(cid:3)
3. A sufficient condition for the validity of Hopf’s Lemma and of
the Strong Maximum Principle
Consider the improper Riemann integral
ξ g (ζ2/N) ξ g (ζ2/N)
N N
dζ = lim dζ
ζ ζ
Z0 ξ→0Zξ
as an extended valued function G,
b b
ξ g (ζ2/N)
N
G(ξ)= dζ,
ζ
Z0
where we mean that G(ξ) + whenever the integral diverges.
≡ ∞
6 S.BERTONE,A.CELLINA,ANDE.M.MARCHINI
We wish to prove the following lemma.
Lemma 4 (Hopf’s Lemma). Let Ω RN be a connected, open and bounded set.
⊂
Let u W1,2(Ω) C Ω be a weak solution to
∈ ∩
(cid:0) (cid:1) N
g (u2 )u 0.
i xi xixi ≤
i=1
X
In addition to the assumptions (L) and (G) on g , assume that G(ξ) + . Sup-
i
≡ ∞
pose 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.
As an example of an equation satisfying the assumptions of the theorem above,
considertheLaplaceequation∆u=0. Thefunctionsg 1satisfytheassumptions
i
≡
(L) and (G), and
ξ 1
G(ξ)= dζ =+ .
ζ ∞
Z0
Another example is obtained setting
1
g (t)=
N
ln(t)
| |
for 0 t 1/e. The assumptions (L) and (G) are satisfied; moreover, for 0
≤ ≤ ≤
ξ2/N 1/e,
≤
ξ dζ
G(ξ)= =+ .
ζ ln(ζ2/N) ∞
Z0 | |
Proof of Lemma 4. a) Assume that u(z)=0 and that B =B(O,r). We prove the
claim by contradiction. Suppose that
∂u
(z) 0,
∂ν ≥
where ν is the outer unit normal to B at z. Let ǫ = min u(x):x B(O,r/2) ;
∈
we have that ǫ>0. Set n o
ω =B(O,r) B(O,r/2).
\
b) We seek a radial function v W1,2(ω) C(ω) satisfying
∈ ∩
v is a weak solution to F(v) 0 in ω
≥
v >0 in ω

(3) v =0 in ∂B(O,r)
 v ǫ in ∂B(O,r/2)
≤
v (z)<0.
ρ

ON HOPF’S LEMMA AND THE STRONG MAXIMUM PRINCIPLE 7
Consider the Cauchy problem
N 1 ζ
ζ = −
′ − ρ g (ζ2/N)
N
(4) 
 ζ(r/2)= ǫ.
−r
There exists a unique localsolution ζ of (4), such that
ζ(ρ) g (ζ2/N) ρ N 1 2ρ
N
dζ = − ds= (N 1)ln .
ζ − s − − r
Z−rǫ Zr/2 (cid:18) (cid:19)
We claim that ζ is defined in [r/2,+ ). Indeed, suppose that ζ is defined in
∞
[r/2,τ), with τ < + . Since ζ > 0, ζ is an increasing function, so that τ < +
′
∞ ∞
if and only if lim ζ(ρ)=0. But
ρ τ
→
ζ(ρ) g (ζ2/N) 2ρ
N
= lim dζ = lim (N 1)ln ,
−∞ ρ→τZ−rǫ ζ ρ→τ− − (cid:18) r (cid:19)
a contradiction. Hence, the solution ζ of (4) is defined in [r/2,+ ).
∞
Setting v =ζ, since, for every ρ (r/2,r),
ρ
∈
ǫ
<v (ρ)<0,
ρ
−r
we have that the function
ρ
v(ρ)= v (s)ds
ρ
Zr
solves the problem
v2 v
ρ ρ
v g + (N 1)=0,
ρρ N
N! ρ −
in particular,v(ρ)>0 and v (ρ)<0, for every ρ (r/2,r), v(r)=0 and v(r/2)
ρ
∈ ≤
ǫ.
Since v 0 and √t v 0, for every ρ (r/2,r), by the hypotheses on g
ρρ ρ i
≥ − ≤ ≤ ∈
and by Lemma 3, we have that
N 2 2 N 2 2
x x v x x
F(v)=v g v2 i i + ρ g v2 i 1 i
ρρi=1 i ρ(cid:18) ρ (cid:19) !(cid:18) ρ (cid:19) ρ i=1 i ρ(cid:18) ρ (cid:19) ! −(cid:18) ρ (cid:19) !≥
X X
N 2 2 N 2
x x v x
v g v2 i i + ρ 1 i =
ρρi=1 N ρ(cid:18) ρ (cid:19) !(cid:18) ρ (cid:19) ρ i=1 −(cid:18) ρ (cid:19) !
X X
N x 2 x 2 v v2 v
v g v2 i i + ρ(N 1) v g ρ + ρ(N 1).
ρρi=1 N ρ(cid:18) ρ (cid:19) !(cid:18) ρ (cid:19) ρ − ≥ ρρ N N! ρ −
X
The function v solves (3), indeed, v is in C2(ω) and it is such that F(v) 0 and
≥
v >0 in ω, v(r)=0, v(r/2) ǫ and v (z)<0.
ρ
≤
c) Since u,v W1,2(ω) C(ω), v is a weaksubsolutionand u is a weak solution
∈ ∩
toF(u)=0,andv u ,applyingLemma1,weobtainthatv u inω. From
∂ω ∂ω
| ≤ | ≤
∂v ∂u
v (z)= (z)< (z),
ρ
∂ν ∂ν
it follows that there exists x0 ω such that v(x0)>u(x0), a contradiction.
∈
8 S.BERTONE,A.CELLINA,ANDE.M.MARCHINI
(cid:3)
From Hopf’s Lemma we derive:
Theorem 1 (Strong Maximum Principle). Let Ω RN be a connected, open and
⊂
bounded set. Let u W1,2(Ω) C Ω be a weak supersolution to
∈ ∩
N (cid:0) (cid:1)
g (u2 )u =0.
i xi xixi
i=1
X
In addition to the assumptions (L) and (G) on g , assume that G(ξ) + . Then,
i
≡ ∞
if u attains its minimum in Ω, it is a constant.
Proof. a) Assume min u = 0 and set = x Ω : u(x) = 0 . By contradiction,
Ω
C { ∈ }
suppose that the open set Ω = .
\C 6 ∅
b)SinceΩisaconnectedset,thereexists andR>0suchthatB(s,R) Ω
∈C ⊂
and B(s,R) (Ω )= . Let p B(s,R) (Ω ). Consider the line ps. Moving
∩ \C 6 ∅ ∈ ∩ \C
p along this line, we can assume that B(p,d(p, )) (Ω ) and that there exists
C ⊂ \C
one point z such that d(p, )= d(p,z). Set r = d(p, ). W.l.o.g. suppose that
∈C C C
p=O.
c) The set Ω satisfies the interior ball condition at z, hence Hopf’s Lemma
\C
implies
∂u
(z)<0.
∂ν
But this is a contradiction: since u attains minimum at z Ω, we have that
∈
Du(z)=0.
(cid:3)
4. A necessary condition for the validity of Hopf’s Lemma
In this and the following section we consider the operator
N 1
−
(5) F(u)= u +g(u2 )u ,
xixi xN xNxN
i=1
X
We wish to provide a necessary condition for the validity of Hopf’s Lemma in a
class of non elliptic equations.
Consider the case
ξ g(ζ2/N)
G(ξ)= dζ <+ .
ζ ∞
Z0
Theorem2. Considertheoperator(5),whereg satisfiesassumptions(L)and(G),
and on (0,t] (t defined in assumptions (L)),
g (t)>0 and g(t)+g (t)t 1.
′ ′
≤
If
(g(t))3/2
lim =0,
t→0+ tg′(t)
ON HOPF’S LEMMA AND THE STRONG MAXIMUM PRINCIPLE 9
then there exist: an open regular region Ω RN; a radial function u C2(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
g(t)
(6) is bounded in (0,t]
tg (t)
′
then Ω satisfies the interior ball condition at z.
Remark 1. When lim g (t)t exists, it follows that
t 0+ ′
→
(g(t))3/2
lim
t 0+ tg′(t)
→
exists, and that
g(t)+g (t)t 1, on (0,t].
′
≤
Indeed, we have that
lim (g(t)+g′(t)t)=0.
t 0+
→
Otherwise, there exists K >0 such that, when 0<t t, g (t)t K, so that
′
≤ ≥
t
g(t)t= (g(s)+g (s)s)ds Kt,
′
≥
Z0
and g(t) K. From
≥
ξ2 g(t)
dt<+ ,
t ∞
Z0
it follows that lim g(t)=0, a contradiction.
t 0+
→
The map
1
g(t)= ,
ln(t)k
| |
with k >2, for 0 t 1/e, satisfies the assumption
≤ ≤
(g(t))3/2
lim =0.
t→0+ tg′(t)
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 (t) 0 and g(t)+g (t)t 1.
′ ′
≥ ≤
Set
k (a)=(1 a)+ag(ta) and k (a)= a (1 a)g(ta)
1 2
− − − −
For every 0 < t t (t defined in assumptions (L)), k and k are non increasing
1 2
≤
in [0,1].
10 S.BERTONE,A.CELLINA,ANDE.M.MARCHINI
Proof. Since, for every 0<t t,
≤
g′(t) 0 and g(t)+g′(t)t 1,
≥ ≤
we have that, for every 0 a 1,
≤ ≤
k1′(a)=−1+g(ta)+g′(ta)ta≤0
and
k2′(a)=−1+g(ta)−(1−a)g′(ta)t=−1+g(ta)+g′(ta)ta−g′(ta)t≤0.
(cid:3)
Proof of Theorem 2. a) Let v be a radial function. Setting a = sin2θ , (5)
N 1
−
reduces to
v
F(v)=v 1 a+ag(v2a) + ρ N 2+a+(1 a)g(v2a) .
ρρ − ρ ρ − − ρ
Let a=1, we seek a(cid:0)solution to (cid:1) (cid:0) (cid:1)
v
(7) v g(v2)+(N 1) ρ =0
ρρ ρ − ρ
suchthat v (R(1)+1)=0 and v (ρ)<0, for every ρ [2,R(1)+1). Consider the
ρ ρ
∈
Cauchy problem
N 1 ζ
ζ = −
′ − ρ g(ζ2)
(8) 
 ζ(2)= 1.
−
We areinterestedinanegativesolutionζ. Define R(1)tobe theunique positive
real solution to
R(1)+1
G( 1) (N 1)ln =0,
− − − 2
(cid:18) (cid:19)
i.e.
G(−1)
R(1)=2e N−1 1.
−
The solution ζ of (8), satisfies
ζ(ρ) g(t2) ρ N 1 ρ
G(ζ(ρ)) G( 1)= dt= − ds= (N 1)ln .
− − t − s − − 2
Zζ(2) Z2 (cid:16) (cid:17)
Then, for everyρ (2,R(1)+1), G(ζ(ρ))>0 and ζ(ρ)<0, while ζ(R(1)+1)=0.
∈
Setting v (ρ)=ζ(ρ) and
ρ
ρ
v(ρ)= v (s)ds,
ρ
ZR(1)+1
we obtain that v solves (7) and, for every ρ (2,R(1)+1),
∈
v (ρ)<v (R(1)+1)=0 and v(ρ)>v(R(1)+1)=0.
ρ ρ
b) Set, for ρ (1,R(1)], u(ρ)=v(ρ+1). Since, for the function v, we have
∈
v (ρ)
v (ρ)g(v2(ρ))+(N 1) ρ =0,
ρρ ρ − ρ
at ρ+1 we obtain
u (ρ)
(9) u (ρ)g(u2(ρ))+(N 1) ρ =0.
ρρ ρ − ρ+1