Download Fundamental Solutions and Liouville type Theorems

Fundamental Solutions
and Liouville type Theorems
Alexander Quaas
Departamento Matemática, UTFSM
La Laguna 11 de julio 2013
Alexander Quaas
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Liouville type Theorems
Liouville type Theorems
Laplacian case
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Liouville type Theorems
For the equation
∆u + u p = 0
en RN
non-existence result are very well-know ( Gidas and Spruck )
(Chen-LI 91): Let
N +2
,
p∗ =
N −2
1) If p < p ∗ then no positive solution and
2) if p ≥ p ∗ then positive solutions exist.
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Extremal Pucci Operator
Problem
2
p
N
M+
Λ (D u) + u = 0 in R ,
0 < 1 ≤ Λ y N ≥ 3, p > 1
If u(r ) is a radial function increasing and convex then
2
00
M+
Λ (D u) = Λu (r ) + (N − 1)
Where
N+ =
1
(N − 1) + 1
Λ
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u 0 (r )
r
Extremal Pucci Operator
Conjecture
1<p<
N+ + 2
.
N+ − 2
the equation
2
p
N
M+
Λ (D u) + u = 0 in R
has no positive solution.
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Extremal Pucci Operator
Teorema (Felmer y Quaas 2003)
∗ such that
Assume N+ > 2. Then there exists u p+
max{
N+
N +2
N+ + 2
∗
,
} < p+
<
,
N+ − 2 N − 2
N+ − 2
∗ then non radial solution y
1) If 1 < p < p+
∗ radial solutions exist.
2) Si p ≥ p+
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Open problem
2
p
M+
Λ (D u) + u = 0,
u>0
Without radial symmetry.
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en RN ,
The second critical exponent
pS∗ =
N
,
N −2
For 1 < p ≤ pS∗ , the equation
∆u + u p ≤ 0 en RN .
has no positive solution.(The results is optimal) Gidas
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The second critical exponent
Cutri and Leoni (2000) proof Gidas result for
2
p
N
M+
Λ (D u) + u ≤ 0 in R ,
where
∗
pS,+
=
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N+
N+ − 2
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(0.1)
More general Liouville type Theorems
Felmer , Quaas, 2009.
Armstrong , Sirakov 2011
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Laplacian Case: ∆u + u p ≤ 0
Ideas of Proof.
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Laplacian Case: ∆u + u p ≤ 0
Ideas of Proof.
Define
m(r ) = inf u(x).
|x|≤r
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Laplacian Case: ∆u + u p ≤ 0
Ideas of Proof.
Define
m(r ) = inf u(x).
|x|≤r
( Hadamard Property) Since −∆u ≥ 0 in IR N , and
Φ(|x|) = |x|2−N is a fundamental solution of the Laplace
equation, by comparison we find that m(r )r N−2 is an
increasing function .
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Assume that there exists a solution u > 0 .
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Assume that there exists a solution u > 0 .
Let η : [0, ∞) → R such that 0 ≤ η(r ) ≤ 1, η ∈ C ∞ , η
non-increasing, η(r ) = 1 if 0 ≤ r ≤ 1/2 and η(r ) = 0 if r ≥ 1.
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Assume that there exists a solution u > 0 .
Let η : [0, ∞) → R such that 0 ≤ η(r ) ≤ 1, η ∈ C ∞ , η
non-increasing, η(r ) = 1 if 0 ≤ r ≤ 1/2 and η(r ) = 0 if r ≥ 1.
It is obvious that there exists C > 0 such that
−∆(η|x|) ≤ C .
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Assume that there exists a solution u > 0 .
Let η : [0, ∞) → R such that 0 ≤ η(r ) ≤ 1, η ∈ C ∞ , η
non-increasing, η(r ) = 1 if 0 ≤ r ≤ 1/2 and η(r ) = 0 if r ≥ 1.
It is obvious that there exists C > 0 such that
−∆(η|x|) ≤ C .
Take ξ(x) = m(R/2)η(|x|/R), then by scaling we have
−∆(ξ|x|) ≤
m(R/2)C
,
R2
Moreover,
ξ(x) = 0 ≤ u(x) if |x| > R y
ξ(x) = m(R/2) ≤ u(x) if |x| ≤ R/2.
Thus the function u(x) − ξ(x) has a global minimum at xR with
|xR | < R
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Then,
m(R/2)C
.
R2
m(R)p ≤ u(xR )p ≤ −∆u ≤ −∆(ξ|x|) ≤
Using that m(R/2) ≤ Cm(R), we find
−2
m(R) ≤ C R p−1 .
that is a contradiction with
cR −(N−2) ≤ m(R) ≤ CR
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2
− p−1
,
Fractional elliptic
Operator ( α ∈ (0, 1))
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Class of operator
Let N ≥ 2, α ∈ (0, 1) and Λ ≥ 1, consider
A = {a ∈ L∞ (S N−1 ) / a(ω) ∈ [1, Λ], a.e. in S N−1 }.
For a ∈ A, define the linear operator
Z
a(ŷ ) dy
δ(u, x, y ) N+2α ,
La (u)(x) =
|y |
RN
x ∈ RN ,
where ŷ = y /|y |, for all y ∈ RN \ {0} y
δ(u, x, y ) = u(x + y ) + u(x − y ) − 2u(x),
The kernel is define as
K (y ) =
a(ŷ )
,
|y |N+2α
For the fractional Laplacian we have a ≡ 1.
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x, y ∈ RN .
Class of operator
Take ttwo set of index I and J, consider
K = {ai,j ∈ A / (i, j) ∈ I × J},
Li,j = Lai,j
and assume
1) K is ∗-weakly close in L∞ (S N−1 )
2) K rotationally invariant, that is, if for a rotation matrix R in
RN×N and a ∈ K we define aR (x) = a(Rx) then aR ∈ K for all
a ∈ K.
Define the operator
I(u)(x) = inf sup Li,j u(x),
I
x ∈ R.
J
Properties of I: for each x ∈ RN there exist (i ∗ , j ∗ ) ∈ I × J such
that
Z
ai ∗ ,j ∗ (ŷ ) dy
,
I(u)(x) = Li ∗ ,j ∗ u(x) =
δ(u, x, y )
|y |N+2α
RN
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Class of operator
Example:
α-Laplacian:
α
Z
∆ u(x) =
δ(u, x, y )
RN
dy
|y |N+2α
Caffarelli-Silvestre operator:
Z
M+
S+ (δ(u, x, y ))
(u)(x)
=
A
RN
dy
|y |N+2α
where
S+ (t) = Λt+ + t−
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t ∈ R.
Class of operator
Fractional Pucci Operator:Consider
Pα = {a ∈ A / a(ω) =
1
1
2
− 12
|detA ||A
ω|N+2α
, ω ∈ S N−1 , A ∈ SΛ }.
we define
M+
Pα (u)
Z
δ(u, x, y )
= sup
a∈P
RN
a(ŷ )
dy .
|y |N+2α
We have
+
2
lim (1 − α)M+
Pα (u) = MΛ (D u).
α→1
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Principal Theorem
Gidas type result
Let I as above with α ∈ (0, 1)
Let NI a dimension number for I, That is r −NI +2α fundamental
solution for I (later).
If NI > 2α and p ≤
NI
NI −2α
then:
All viscosity solution u, of
I(u) + u p ≤ 0,
are u ≡ 0. Moreover, if p >
non-trivial solutions.
NI
NI −2α
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u ≥ 0,
Then the equation has
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Principal Theorem
Fractional Laplacian
In this case NI = N let u be a non-negative solution of
(∆)α u + u p ≤ 0,
If
1<p≤
then u ≡ 0
For
p>
N
.
N − 2α
N
,
N − 2α
there are solutions.
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u≥0
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Fundamental Solution
Let −N < σ < 2α and

σ

r
vσ (r ) = − log r

 σ
−r
if −N < σ < 0
if σ = 0
if 0 < σ < 2α,
we find
Ivσ (x) = c(σ)|x|σ−2α ,
where
Z
c(σ) =
δσ (y )
RN
x ∈ RN \ {0},
aσ (ŷ ) dy
,
|y |N+2α
for some aσ := ai,j , with (i, j) ∈ I × J depending on σ, and

σ
σ

if σ ∈ (−N, 0)
|e1 + y | + |e1 − y | − 2
δσ (y ) = − log |e1 + y | − log |e1 − y | if σ = 0,


−|e1 + y |σ − |e1 − y |σ + 2 if σ ∈ (0, 2α).
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Existence of Fundamental Solution
Theorem (Felmer & Quaas)
For all −N < σ < 2α and for all I in the class, there exists a
unique σ ∈ (−N, min{2α, 1}) such that
I(vσ ) = 0.
That σ is denote by σI dimension number for I is
NI = −σI + 2α.
Labutin, Cutri and Leoni, Capuzzo-Dolcetta y Cutri, Felmer and
Quaas, Armstrong, Sirakov and Smart, Armstrong and Sirakov.
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Fundamental Solution
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Ideas of the Proof.
1) For all −N < σ < 2α, c(σ) is well define
lim c(σ) = ∞ and
σ→−N
lim c(σ) = −∞.
σ→2α
2) The function c has at most one zero in (−N, 2α).
3) If c does not have a zero in(−N, 2α) \ {0}. Then, for each
σ ∈ (−N, 0) we have c(σ) > 0 and then
I(
c(σ)
vσ − 1
) = r σ−2α
≥0
−σ
−σ
and,
c(σ)
vσ + 1
) = r σ−2α
≤ 0.
σ
σ
So uσ− = (vσ − 1)/−σ is a sub-solution for σ ∈ (−N, 0) and
uσ+ = (vσ + 1)/σ is a super-solution for σ ∈ (0, 2α). From here
I(
lim uσ− (r ) = lim+ uσ+ (r ) = − log r ≡ v0 (r ),
σ→0−
σ→0
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Proof of main Theorem
Hadamard Theorem does Work!! (comparison is diferent)
but still we have some properties for
m(r ) = min u(x),
|x|≤r
Lemma 1.
Assume NI > 2α. Then, for all σ ∈ (−N, σI ) and r1 ≥ 1, there
exists c > 0 such that u 6= 0 is a solution of
Iu(x) ≤ 0
en RN
then
m(r ) ≥ cm(r1 )r σ ,
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for all
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r ≥ r1 .
Lemma 2
Lemma 2.
Assume NI > 2α. Then there exists r1 > 0 and a constant c such
that for u solution of
in RN
Iu(x) ≤ 0
we have
m(R/2) ≤ cm(R),
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p R ≥ r1 .
Proof of the Liouville Type Theorem
Proof of the Liouville Type Theorem (the subcritical case).
Let η : [0, ∞) → R such that 0 ≤ η(r ) ≤ 1, η ∈ C ∞ , η
non-increasing, η(r ) = 1 if 0 ≤ r ≤ 1/2 and η(r ) = 0 if r ≥ 1. It is
obvious that there exists C > 0 such that
−M+
λ,Λ (η|x|) ≤ C .
Take ξ(x) = m(R/2)η(|x|/R), then by scaling we have
I(ξ|x|) ≥ −
m(R/2)C
,
R 2α
Moreover,
ξ(x) = 0 ≤ u(x) if |x| > R y
ξ(x) = m(R/2) ≤ u(x) if |x| ≤ R/2.
Thus the function u(x) − ξ(x) has a global minimum at xR with
|xR | < R and the function v (x) := ξ(x) + (u(xR ) − ξ(xR ))
(extended by u in B(0, R)), is a test function.
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Proof of the Liouville Type Theorem
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Therefore
I(v )(xR ) + u(xR )p ≤ 0.
Pero v (x) − ξ(x) ≥ v (xR ) − ξ(xR ) = 0 para todo x ∈ RN , ası́ que
0 ≤ M−
Λ (v − ξ)(xR ) ≤ I(v )(xR ) − I(ξ)(xR )
Then,
m(R)p ≤ u(xR )p ≤
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m(R/2)C
.
R 2α
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Proof of the Liouville Type Theorem
From Lemma 2, we get
−2α
m(R) ≤ C R p−1 .
From Lemma 1, if σ < σI , we find
m(R) ≥ cR σ
and then
σ≤
But p <
NI
NI −2α ,
is σI >
−2α
p−1 ,
−2α
.
p−1
so we find σ < σI such that
σI > σ >
−2α
p−1
which is a contradiction
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Case NI ≤ 2α
Liouville Property
Assume NI ≤ 2α and u is a solution of
I(u) ≤ 0,
and u ≥ 0,
then u is constant.
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in RN ,
Other non-existence results
joint with S. Alarcon J. Garcia-Melian
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(P)
−∆u + |∇u|q ≥ λu p
u≥0
where q > 1, p > 0, λ > 0 and N ≥ 3.
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en IR N ,
en IR N ,
(0.2)
Observe that
q=
2p
q
⇔p=
,
p+1
2−q
is one critical case.
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Observe that
q=
2p
q
⇔p=
,
p+1
2−q
is one critical case.
Subcritical case: 0 < p <
q
2−q ,
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Observe that
q=
2p
q
⇔p=
,
p+1
2−q
is one critical case.
Subcritical case: 0 < p <
Critical case: p =
q
2−q ,
q
2−q .
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Dificulties
The term |∇u|q , with q > 1: make the operator
non-homogenous.
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Dificulties
The term |∇u|q , with q > 1: make the operator
non-homogenous.
The ”fundamental solution” (sub-solution) exists if and
N
only if 1 < q < N−1
.
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Dificulties
The term |∇u|q , with q > 1: make the operator
non-homogenous.
The ”fundamental solution” (sub-solution) exists if and
N
only if 1 < q < N−1
.
Hadamard property not direct.
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Dificulties
The term |∇u|q , with q > 1: make the operator
non-homogenous.
The ”fundamental solution” (sub-solution) exists if and
N
only if 1 < q < N−1
.
Hadamard property not direct.
With the maximum principle we get
m(2R) ≤
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m(R) m(R)q
+
R2
Rq
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1
p
.
Region for non-positive super-solution
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Remarks
In the critical case λ is very relevant.
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Remarks
In the critical case λ is very relevant.
The existence results are optimal.
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Remarks
In the critical case λ is very relevant.
The existence results are optimal.
The results holds if we replace u p by f (u), where
f : [0, +∞) → R is continuos and positive in (0, +∞) and
lim inf
u→0
f (u)
>0
up
for some p > 0.
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Cone domain and half space
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Thanks
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