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Fundamental Solutions and Liouville type Theorems Alexander Quaas Departamento Matemática, UTFSM La Laguna 11 de julio 2013 Alexander Quaas USM Liouville type Theorems Liouville type Theorems Laplacian case Alexander Quaas USM 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. Alexander Quaas USM Alexander Quaas USM 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 Alexander Quaas USM 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. Alexander Quaas USM 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+ Alexander Quaas USM Open problem 2 p M+ Λ (D u) + u = 0, u>0 Without radial symmetry. Alexander Quaas USM 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 Alexander Quaas USM 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,+ = Alexander Quaas N+ N+ − 2 USM (0.1) More general Liouville type Theorems Felmer , Quaas, 2009. Armstrong , Sirakov 2011 Alexander Quaas USM Laplacian Case: ∆u + u p ≤ 0 Ideas of Proof. Alexander Quaas USM Laplacian Case: ∆u + u p ≤ 0 Ideas of Proof. Define m(r ) = inf u(x). |x|≤r Alexander Quaas USM 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 . Alexander Quaas USM Assume that there exists a solution u > 0 . Alexander Quaas USM 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. Alexander Quaas USM 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 . Alexander Quaas USM 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 Alexander Quaas USM 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 Alexander Quaas USM 2 − p−1 , Fractional elliptic Operator ( α ∈ (0, 1)) Alexander Quaas USM 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. Alexander Quaas USM 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 Alexander Quaas USM 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− Alexander Quaas USM 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 Alexander Quaas USM 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 Alexander Quaas u ≥ 0, Then the equation has USM 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. Alexander Quaas u≥0 USM 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). Alexander Quaas USM 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. Alexander Quaas USM Fundamental Solution Alexander Quaas USM 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 Alexander Quaas USM 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 , Alexander Quaas for all USM 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), Alexander Quaas USM 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. Alexander Quaas USM Proof of the Liouville Type Theorem Alexander Quaas USM 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 ≤ Alexander Quaas m(R/2)C . R 2α USM 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 Alexander Quaas USM Case NI ≤ 2α Liouville Property Assume NI ≤ 2α and u is a solution of I(u) ≤ 0, and u ≥ 0, then u is constant. Alexander Quaas USM in RN , Other non-existence results joint with S. Alarcon J. Garcia-Melian Alexander Quaas USM (P) −∆u + |∇u|q ≥ λu p u≥0 where q > 1, p > 0, λ > 0 and N ≥ 3. Alexander Quaas USM en IR N , en IR N , (0.2) Observe that q= 2p q ⇔p= , p+1 2−q is one critical case. Alexander Quaas USM Observe that q= 2p q ⇔p= , p+1 2−q is one critical case. Subcritical case: 0 < p < q 2−q , Alexander Quaas USM 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 . Alexander Quaas USM Dificulties The term |∇u|q , with q > 1: make the operator non-homogenous. Alexander Quaas USM 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 . Alexander Quaas USM 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. Alexander Quaas USM 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) ≤ Alexander Quaas m(R) m(R)q + R2 Rq USM 1 p . Region for non-positive super-solution Alexander Quaas USM Remarks In the critical case λ is very relevant. Alexander Quaas USM Remarks In the critical case λ is very relevant. The existence results are optimal. Alexander Quaas USM 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. Alexander Quaas USM Cone domain and half space Alexander Quaas USM Thanks Alexander Quaas USM