Hopf maximum principle violation for elliptic equations with non-Lipschitz nonlinearity

0.22 MB·English

by  Yavdat Il'yasov

Hopf maximum prin
iple violation for ellipti
equations with non-Lips
hitz nonlinearity
9
0 ∗
0 Y. Il'yasov Y. Egorov
2
Institute of Mathemati
s RAS, Université Paul Sabatier,
n
a Ufa, Russia Toulouse, Fran
e
J

7
2

]
P
A Abstra
t
.
h We
onsider ellipti
equations with non-Lips
hitz nonlinearity
at −∆u= λ|u|β−1u−|u|α−1u
m
Ω ⊂ Rn n ≥ 3
[ in a smooth bounded domain , , with Diri
hlet bound-
0 < α < β < 1

1 ary
onditions; here . We prove the existen
e of a
v
weak nonnegative solution whi
h does not satisfy the Hopf bound-
1 λ n >
ary maximum prin
iple, provided that is large enough and
9 2(1+α)(1+β)/(1−α)(1−β)
1 .
4
.
1
0
1 Introdu
tion
9
0 Ω Rn n ≥ 3
v: L∂Ωet be a bounded domain in , with a smooth boRunndary
, whi
h is stri
tly star-shaped with respe
t to the origin in . We
i
X

onsider the following problem:
r −∆u= λ|u|β−1u−|u|α−1u Ω,
a
in

 (1.1)
u= 0 ∂Ω.
on

λ 0 < α < β < 1
Here isareal parameter and , sothatthe nonlinearity
f(λ,u) := λ|u|β−1u−|u|α−1u
on the right-hand side of (1.1) is non-
Lips
hitzean at zero.
∗
The(cid:28)rstauthorwaspartlysupportedbygrantsRFBR08-01-00441-a,08-01-97020-p-a
1
Our interest to this problem has been indu
ed by investigations
n = 1
of J.I. Díaz, J. Hernández in paper [1℄. In
ase of dimension
Ω = (−1,1)
when , among other results, they showed that for
ertain
λ > 0 u(x) x ∈ (−1,1)
values equation (1.1) possesses solutions , with
a spe
ial feature
u(−1) = u(1) = 0, u′(−1) = u′(1) = 0.
(1.2)
x= −1
This meansa Hopfboundary maximum prin
iple violation on ,
x = 1
and a loss of the uniqueness for initial value problem to (1.1)
u(−1) = u′(−1) = 0 u ≡ 0
with , sin
e satis(cid:28)es also to (1.1). Further-
more, it
an easily be shown that the existen
e of su
h a solution with
λ > 0
0
yields the existen
e of a set
ontinuum nonnegative solutions
λ> λ
0
of this boundary value problem for any . Observe that property
u
(1.2) implieRs that a fun
tion is also a weaklyfso(lλu,tuio)n of (1.1) on the
whole line . Note that when the nonlinearity is a lo
ally Lip-
s
hitzfun
tion su
haphenomenon isimpossibledueto theuniqueness
solution of initial value problem and/or a Hopf boundary maximum
prin
iple.
Thisriseaquestionastowhether thesimilarphenomenamaybeo
-
n> 1

urred in
ase of the higher dimensions . More pre
isely whether
n > 1
the Hopf boundary maximum prin
iple holds for (1.1) when
f(λ,u)
and the nonlinearity is non-Lips
hitz. To (cid:28)nd an answer to
this question is a main goal in the present work.
u ∈
Let us state our main result. We
onsider a weak solution
H1 := H1(Ω) H1(Ω) C∞(Ω)
0 0 0 0
, where denotes the
losure in standard
H1(Ω) ||·||
1
Sobolevspa
e withthenorm . Wesaythataweaksolution
u∈ H1 u∈ C1(Ω) ∂u =0 ∂Ω
0 of (1.1) is a non-regular, if and ∂ν on .
Our main result is the following
Ω Rn n ≥ 3
Theorem 1.1 Let be a bounded domain in , with smooth
boundary, whi
h is stri
tly star-shaped with respe
t to the origin. As-
0 < α < β < 1 n > 2(1+α)(1 +β)/(1 −α)(1−β)
sume that and .
λ∗ > 0 λ ≥ λ∗
Then there exists su
h that for all problem (1.1) has a
u Ω
λ
non-regular solution solution , whi
h isnonnegative in . Moreover,
λ> λ∗
the number of su
h solutions for is in(cid:28)nite.
λ > λ∗
Furthermore, for any problem (1.1) has a weak solution
w ∈ C1(Ω) Ω
λ
∂wλ 6= 0, whi
h is nonnegaUtiv⊆e i∂nΩ but is not n(onn−-re1g)ular solution,
i.e. ∂ν on some subset of positive -dimensional
Lebesgue measure.
2
The proof of thetheorem relieson the variational arguments. Further-
more,basi
ingredientsintheproof
onsistinusingPohozaev'sidentity
[4℄
orresponding to (1.1) and in applying the spe
tral analysis with
respe
t to the (cid:28)bering pro
edure introdu
ed in [3℄.
Remark 1.1. If one
onsiders the radial symmetri
solutions of
B
R
(1.1) in the ball , then the se
ond part of Theorem 1.1 implies that
w B
λ R
the weak radial solution of (1.1) is positive in th∂ueλb(Ral)l < 0 and
satis(cid:28)es the Hoph boundary maximum prin
iple, i.e. ∂ν .
Remark 1.2. In the theory of integrable systems, the non-regular
type solutions are known as the
ompa
tons: solitary waves with
om-
pa
t support [5℄.
The paper is organized as follows. In Se
tion 2, we apply the
spe
tral analysis related with the (cid:28)bering pro
edure [3℄ to introdu
e
Λ ,Λ
0 1
two spe
tral points whi
h play basi
role in the proof of the
mainresult. InSe
tion3,wederivesomeimportant
onsequen
esfrom
Pohozaev's identity. In Se
tion 4, we prove existen
e of the solution to
anauxiliary
onstrained minimization problem. InSe
tion 5, we prove
Theorem 1.1. Se
tion 6 is an appendix where some te
hni
al result is
proved.
2 Spe
tral analysis with respe
t to the
(cid:28)bering pro
edure
In this se
tion we apply the spe
tral analysis with respe
t to the (cid:28)ber-
ing pro
edure [3℄ to introdu
e two spe
tral points whi
h will play im-
portant roles in the proof of the main result.
Observe that problem (1.1) is the Euler-Lagrange equation of the
fun
tional
1 1 1
E (u) = T(u)−λ B(u)+ A(u),
λ
2 β+1 α+1 (2.1)
where we use the notations
T(u) = |∇u|2dx, B(u) = uβ+1dx, A(u) = uα+1dx.
Z Z Z
Ω Ω Ω
u ∈ H1 e (t) := E (tu) t ∈
0 λ λ
R+Let . Consider the fuQn
(tuio)n= e′ (t)| , L (due)(cid:28)=nede′′f(otr)|
. Introdu
e the fun
tionals λ λ t=1 λ λ t=1
u∈ H1
for 0. Then
Q (u) = T(u)−λB(u)+A(u), L (u) := T(u)−λβB(u)+αA(u).
λ λ
3
u ∈ H1\{0}
Let 0 . Following the spe
tral analysis [3℄, we solve the
system
Q (tu) = t2T(u)−λt1+βB(u)+t1+αA(u) = 0
λ

 E (tu) = t2T(u)−λt1+βB(u)+ t1+αA(u) = 0 (2.2)
λ 2 1+β 1+α


and (cid:28)nd the
orresponding solution
1
2(β −α) A(u) 1−α
t (u) = ,
0
(cid:18)(1+α)(1−β)T(u)(cid:19) (2.3)
α,β
λ (u) = c λ(u),
0 0
where β−α
α,β (1−α)(1+β) (1+α)(1−β) 1−α
c =
0 (1−β)(1+α) (cid:18) 2(β −α) (cid:19)
and 1−β β−α
A(u)1−αT(u)1−α
λ(u) = .
B(u) (2.4)
Thus with respe
t to the (cid:28)bering pro
edure, we have the following
spe
tral point
Λ = inf λ (u).
0 0
H1\{0} (2.5)
0
Λ u ∈ H1 \{0}
Introdu
e the se
ond point 1. Let 0 . Consider now
the following system
Q (tu) = t2T(u)−λt1+βB(u)+t1+αA(u) = 0
λ

 L (tu) = t2T(u)−λβt1+βB(u)+αt1+αA(u) = 0 (2.6)
λ

t ∈ R+ λ ∈ R+
for , . Solving this system we (cid:28)nd as above
α,β
λ (u) = c λ(u),
1 1 (2.7)
where β−α
α,β 1−α 1−β 1−α
c = .
1 1−β (cid:18)β−α(cid:19) (2.8)
Then we have
Λ = inf λ (u).
1 1
H1\{0} (2.9)
0
4
0< Λ < Λ < +∞
1 0
Proposition 2.1 .
λ (u) = Cα,βλ (u) u ∈ H1 \{0}
1 0 0
Proof. Observe that for any with
Cα,β = cα,β/cα,β Cα,β < 1
0 1 . It is not hard to show that , therefore
Λ < Λ
1 0
.
Λ < +∞ 0 < Λ
0 1
It is
lear that . Let us show that . Note that
λ(u) H1 \{0}
0
is a zero-homogeneous fun
tion on . Therefore we may
S := {v ∈ H1 : ||v|| = 1}
restri
t the in(cid:28)mum in (2.9) to the set 0 1 .
Set
(1+α)(2∗ −1−β) (1+α) 2∗
γ = , p = , q = ,
(2∗−1−α) γ (1+β −γ) (2.10)
2∗ = 2n/(n − 2) p,q > 1 1/p + 1/q = 1
where . Then , and by the
Hölder inequality we have
B(u)≤ ( u2∗dx)1/q ·A(u)1/p.
Z (2.11)
Ω
By Sobolev's inequality
u2∗dx ≤ C ||u||2∗ = C < +∞.
0 1 0
Z
Ω
u ∈ S C u ∈ H1 u∈ S
for , where 0 does not depend on 0. Hen
e for any
we have λ(u) = A(u)(1−β) ≥c0A(u)−(2(∗2−∗−2)1(−β−α)α),
B(u)(1−α) (2.12)
0 < c < +∞ u ∈ H1 A(u) ≤ C <
0 0 1
where does not depend on . Sin
e
+∞ S Λ > 0
1
on we see from (2.5) that .
3 Pohozaev's identity
We will need the following regularity result
0 < α < β < 1 u ∈ H1
0
Proposition 3.1 Assume that . Suppose that
u∈ C1,κ(Ω) κ ∈ (0,1)
is a weak solution of (1.1). Then for .
u ∈ H1 |f(λ,u)| <
0
CPr(o1o+f. |uL|e)t u ∈ R be a weakCso>lut0ion of (1.1). Sin
e
, with some , then (see Lemma B.3 in [6℄)
u ∈ Lq(Ω) q < ∞ −∆u = f(λ,u) ∈ Lq(Ω)
for any . This implies that
q < ∞
for any . Thus, by the Caldéron-Zygmund inequality (see [2℄)
5
u ∈ H2,q(Ω) u ∈ C1,κ(Ω) κ ∈ (0,1)
, when
e for by the Sobolev em-
bedding theorem.
P
λ
We will denote by the fun
tional
(n−2) 1 1
P (u) := T(u)−λ B(u)+ A(u)
λ
2n β+1 α+1
u∈ H1
de(cid:28)ned for 0.
Ω Rn n≥
L3emma 3.1 Suppose that is a smooth bounded domain in Rn,
, whi
h is stri
tly star-shaped with respe
t to the origin in . Let
u u ∈ H1
0
be a weak solution of (1.1), . Then the following Pohozaev
identity holds 2
1 ∂u
P (u)+ x·νdx = 0.
λ
2n Z (cid:12)∂ν(cid:12) (3.1)
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12) u ∈ H2,2 ∩C1 ∩H1
0
Proof.f(Bλ,yuP)roposition 3.1 we know thatR . Thus,
sin
e isa
ontinuous fun
tion on , we are in position to apply
Lemma 1.4 in [6℄, that
ompletes the proof.
u ∈ H1
Let 0. Based on the ideas of the spe
tral analysis with
respe
t to the (cid:28)bering pro
edure [3℄ we
onsider the following system
of equations
Q (u) := T(u)−λB(u)+A(u) = 0
λ


 L (u) := T(u)−λβB(u)+αA(u) = 0
 λ

 (3.2)
(n−2) 1 1
 Pλ(u) := 2n T(u)−λβ+1B(u)+ α+1A(u) = 0.




The
omputation of the
orresponding determinant shows that this
system is solvable if and only if
θ ≡2(1+α)(1+β)−n(1−α)(1−β)= 0.
(3.3)
θ < 0
Note that if and only if
n > 2(1+α)(1+β)/(1−α)(1−β).
e′ (t) = 0 t > 0
Ot1b(use)r:v=e tt1h(aut)t∈heR+equatiotn2(uλ) := t2(u,) ∈ R+ has at mto1s(tut)w≤otr2o(out)s
λ and λ su
h that ,
e′′(t1(u)) ≤ 0 e′′(t2(u)) ≥ 0
λ and λ .
6
θ < 0 u ∈ H1 \{0} t > 0
Proposition 3.2 Assume that . If 0 and are
Q (tu) = 0 P (tu) ≤ 0
λ λ
su
h that and then we have
L (tu) > 0.
λ
u∈ H1\{0} t > 0
Proof. Let 0 and as in the assumption. Then
T(u) = λtβ−1B(u)−tα−1A(u)
(3.4)
(1−β)
L (tu) > 0 ⇔ λtβ−α B(u) > A(u).
λ
(1−α) (3.5)
P (t1(u)u) ≤ 0
λ
Equality (3.4) implies that holds if and only if
[2(1+β)+n(1−β)](1+α)
λtβ−α B(u) ≥ A(u).
[2(1+α)+n(1−α)](1+β) (3.6)
θ < 0
Observe, that the inequality implies
[2(1+β)+n(1−β)](1+α) (1−β)
< .
[2(1+α)+n(1−α)](1+β) (1−α) (3.7)
Thus (3.7) and (3.6) give
(1−β)
λtβ−α B(u) > A(u)
(1−α)
and therefore by (3.5) the proof is
omplete.
u
0
Corollary 3.1 If is a non-regular solution solution of (1.1) then
E (u ) > 0 θ <0
λ 0
. Furthermore, if in addition , then
Q (u ) = 0, P (u )= 0, L (u ) > 0.
λ 0 λ 0 λ 0
u
0
Proof. Observe that if is the non-regular solution solution of (1.1),
P (u ) = 0 E (u) = P (u) +
λ 0 λ λ
then by (3.1) we have . Hen
e using
(1/2n)T(u)
we get 1
E (u ) = T(u )> 0.
λ 0 0
n
Q (u ) = 0 u P (u ) = 0
λ 0 0 λ 0
Note that if is a solution of (1.1), and if
in addition this solution is the non-regular solution solution. Hen
e
θ < 0 L (u ) > 0
λ 0
assumption and Proposition 3.2 imply that .
7
4 Constrained minimization problems
Consider the following
onstrained minimization problem:
E (u) → min
λ

 (4.1)
Q (u) = 0.
λ

We denote by
M := {w ∈ H1 : Q (u) = 0}
λ 0 λ
Eˆ := min{E (u) : u ∈ M }
λ λ λ
the admissible set of (4.1), and by
(u )
m
the minimal value in this problem. We say that is a minimizing
sequen
e of (4.1), if
E (u )→ Eˆ m → ∞ u ∈ M , m = 1,2,...
λ m λ m λ
as and (4.2)
λ> Λ M
1 λ
Proposition 4.1 If , then the set is not empty, meanwhile
M λ< Λ
λ 1
the set is empty when .
λ > Λ u ∈ H1 \ {0}
1 0
Proof. Let . Then by (2.5) there exists su
h
Λ < λ(u) < λ L (t(u)u) = 0, Q (t(u)u) = 0
1 λ(u) λ(u)
that and . Hen
e,
Q (t(u)u) < 0 λ(u) < λ t > 0
λ
, sin
e and therefore there exists su
h
Q (tu) = 0 tu ∈ M
λ λ
that , i.e. .
Theproofofthese
ondpartofthePropositionfollowsimmediately
Λ
1
from the de(cid:28)nition (2.5) of .
From here it follows that
Eˆ < +∞ λ> Λ
λ 1
Corollary 4.1 for any .
4.1 Existen
e of the solution of (4.1).
λ > Λ u ∈ H1 \
1 0 0
L{0e}mma E4.1(uFo)r=aEnˆy u ∈prMoblem (4.1) has a solution
λ 0 λ 0 λ
, i.e. and .
λ > Λ M
1 λ
Proof. Let . Then is not empty and there is a minimizing
(u ) t ≥ 0 v ∈ H1 m = 1,2,...,
sequen
e m of (4.1). Set m and m 0, su
h
u = t v ||v || = 1
m m m m 1
that , .
{t }
m
Let us show that is bounded. Observe that
1−λtβ−1B(v )+tα−1A(v )= 0,
m m m m
(4.3)
8
Q (t u ) = 0 m = 1,2,... ||v || = 1
λ m m m 1
sin
e , . Note that sin
e ,
B(v ),A(v )
m m
are bounded.
(t )
m
Suppose that there exists a subsequen
e again denoted su
h
t → ∞ m → +∞
m
that as . Then the left hand side of (4.3) tends
m → +∞ Q (u ) = 0
λ m
to 1 as what
ontradi
ts to the assumption ,
m = 1,2,...
.
(t )
m
Suppose now that there exist subsequen
es again denoted ,
(v ) t → 0 v → 0 H1 m → +∞
m m m 0
su
h that and/or weakly in as .
tα−1A(v ) → C m → ∞ 0 ≤ C < +∞
m m
Assume that as , where .
β−1
λt B(v ) → 1+C m → ∞ B(v ) ≤
m m m
Then as . By (2.11) we have
C ·A(v )1/p 0 < C < +∞ m = 1,2,...
0 m 0
, where does not depend on .
Therefore
β−1+(1−α)
tβ−1B(v ) ≤ C ·tβ−1A(v )1/p = t p (tα−1A(v ))1/p.
m m 0 m m m m m
(4.4)
Let us show that (1−α)
β −1+ > 0.
p (4.5)
(2∗−1−α)
p =
Substituting (2∗−1−β) we get
(1−α) (β −1)(2∗ −1−α)+(1−α)(2∗ −1−β)
β −1+ = .
p (2∗−1−α)
Sin
e
(β −1)(2∗ −1−α)+(1−α)(2∗ −1−β) =
2∗(β −α)−(1+α)(β −1)−(1−α)(1+β) =
2∗(β −α)−2(β −α) = (β −α)(2∗ −2) > 0,
we get the desired
on
lusion. Hen
e the right hand sidein (4.4) tends
β−1
t B(v ) → 0 m → ∞
m m
to zero and therefore as , whi
h
ontradi
ts
our assumption.
tα−1A(v ) → +∞ m → ∞
m m
Assume now that as . Then by (4.3) we
have A(v )
m
→ 1
λtβ−αB(v ) (4.6)
m m
m → ∞
as . Using (2.11) we dedu
e
A(v ) A(v )(p−1)/p (t(α−1)A(v ))(p−1)/p
m m m m
> c = c ,
λtmβ−αB(vm) 0 tmβ−α 0 tm(p−1)(α−1)/p+β−α (4.7)
9
0 < c < +∞ m = 1,2,...
0
where does not depend on . Using (4.5) we
get
(p−1)(α−1) 1−α
+β−α = β −1+ > 0.
p p
+∞
This implies that the right hand side of (4.7) tends to ,
ontrary
to (4.6).
(u ) H1
Thus m is bounded in 0, and hen
e by Sobolev's embedding
(u ) H1
theorem, m has a subsequen
e whi
h
onverges weakly in 0 and
L 1 < p < 2∗ (u )
p m
strongly in , . Denoting this subsequen
e again by
u → u H1 L 1 < p < 2∗
m 0 0 p
we get weakly in and strongly in , for some
u ∈ H1 (t ) (v )
0 0 m m
. By the above,uth6=e 0sequen
esE (u )a≤ndEˆ arQe s(eupa)r≤ate0d
0 λ 0 λ λ 0
from zero and therefore . Thus and .
Q (u ) < 0 Q (t2(u )u ) = 0 t2(u )u ∈ M
Assume λ 0 . Then λ λ 0 0 , i.e. λ 0 0 λ
E (t2(u )u ) < E (u ) ≤ Eˆ
and λ λ 0 0 λ 0 λ. Hen
e we get a
ontradi
tion and
E (u ) = Eˆ Q (u ) = 0
λ 0 λ λ 0
therefore and . This
ompletes the proof of
Lemma 4.1.
Λ
0
From thede(cid:28)nition (2.9) of andusing arguments asin theproof
of Lemma 4.1 it is not hard to derive
λ > Λ Eˆ < 0 Λ < λ < Λ
0 λ 1 0
C0 <orEoˆlla<ry+4∞.2 If λ = Λ, then Eˆ = 0. If , then
λ 0 λ
, and if , then .
4.2 Existen
e of the solution of (1.1).
λ> Λ u ∈ H1\{0}
Let 1 then by Lemma 4.1 there existsa solution 0 0 of
µ µ
1 2
(4.1). This implies that there exist Lagrange multipliers , su
h
that
µ DE (u ) = µ DQ (u ),
1 λ 0 2 λ 0
(4.8)
|µ |+|µ | =6 0
1 2
and .
θ < 0 λ > Λ u ∈ H1
1 0 0
Proposition 4.2 Let , and be a solution
P (u ) ≤ 0 u
λ 0 0
of (4.1). Assume that . Then is a weak nonnegative
solution of (1.1).
L (u ) 6= 0 θ < 0
λ 0
Proof. Note that by Proposition 3.2 we have , sin
e ,
Q (u ) = 0 P (u ) ≤ 0
λ 0 λ 0
and by the assumption . From (4.1) and (4.8)
0 = µ Q (u ) = µ L (u ) L (u ) 6= 0
1 λ 0 2 λ 0 λ 0
we have . But and therefore
µ = 0 DE (u )= 0 E (|u |)= E (u )
2 λ 0 λ 0 λ 0
. Thusby(4.8)wehave . Sin
e ,
Q (|u |) = Q (u ) = 0 u ≥ 0
λ 0 λ 0 0
we may assume that . This
ompletes
10