Semilinear elliptic problems involving Leray-Hardy potential and - PowerPoint PPT Presentation

Backgrounds Isolated singular solutions semilinear Hardy problem Singular point on the boundary Semilinear elliptic problems involving Leray-Hardy potential and measure data Huyuan Chen Jiangxi Normal Univeristy Workshop: Singular Problems associated to Quasilinear Equations In celebration of Marie Francoise Bidaut-V´ eron and Laurent V´ eron’s 70th birthday

Backgrounds Isolated singular solutions semilinear Hardy problem Singular point on the boundary Dear Prof. Bidaut-V´ eron and Prof. V´ eron, it is a great pleasure for me to participate in this wonderful meeting to celebrate such an important birthday. I would like to take this opportunity to express my gratitude to you for your guidance and lots of assistance. I was most fortunate to be your and Prof. Felmer’s PhD student.

Backgrounds Isolated singular solutions semilinear Hardy problem Singular point on the boundary We will talk about • elliptic equation with absorption nonlinearity and measure data, and elliptic equations with Hardy operators • Isolated singular solutions of nonhomogeneous Hardy problem µ L µ u := − ∆ u + | x | 2 u = f in Ω \ { 0 } , u = 0 on ∂ Ω • semilinear Hardy equation involving measures L µ u + g ( u ) = ν in Ω \ { 0 } , u = 0 on ∂ Ω • solutions of nonhomogeneous Hardy problem with the origin on the boundary

Backgrounds Isolated singular solutions semilinear Hardy problem Singular point on the boundary Outline Backgrounds 1 Laplacian operator Hardy operator Isolated singular solutions 2 Fundamental solution Nonhomogeneous problem Idea of proofs semilinear Hardy problem 3 Main results The ideas of the proofs Singular point on the boundary 4

Backgrounds Isolated singular solutions semilinear Hardy problem Singular point on the boundary Laplacian operator • Benilan-Brezis-Crandall, Ann Sc Norm Sup Pisa (1975); Brezis, Appl Math Opim (1984) For p > 1 , f ∈ L 1 loc ( R N ) , the problem − ∆ u + | u | p − 1 u = f R N in (1.1) has a unique solution u . Moreover, u ≥ 0 if f ≥ 0 . • Lieb-Simon, Adv. Math (1977) The Thomas-Fermi equation, Thomas-Fermi theory of atoms, molecules l 3 � R 3 , − ∆ u + ( u − λ ) + = 2 m i δ a i in (1.2) i =1 where λ ≥ 0 , m i > 0 and δ a i is the Dirac mass at a i ∈ R 3 . The distributional solution of (1.2 ) is a classical solution of 3 R 3 \ { a 1 , · · · , a l } . − ∆ u + ( u − λ ) + = 0 2 in (1.3)

Backgrounds Isolated singular solutions semilinear Hardy problem Singular point on the boundary Laplacian operator A nature question is what difference between Dirac mass source and L 1 source. • Benilan-Brezis, J. Evol. Eq. (2004) (finished 1975) answered this N question, when N ≥ 3 , p ≥ N − 2 , k > 0 , the problem − ∆ u + | u | p − 1 u = kδ 0 in Ω , u = 0 on ∂ Ω (1.4) has no solution. • Brezis-V´ eron, ARMA (1980): when N ≥ 3 , p ≥ N/ ( N − 2) , the basic model − ∆ u + | u | p − 1 u = 0 in Ω \ { 0 } , u = 0 on ∂ Ω (1.5) admits only the zero nonnegative solution.

Backgrounds Isolated singular solutions semilinear Hardy problem Singular point on the boundary Laplacian operator A nature question is what difference between Dirac mass source and L 1 source. • Benilan-Brezis, J. Evol. Eq. (2004) (finished 1975) answered this N question, when N ≥ 3 , p ≥ N − 2 , k > 0 , the problem − ∆ u + | u | p − 1 u = kδ 0 in Ω , u = 0 on ∂ Ω (1.4) has no solution. • Brezis-V´ eron, ARMA (1980): when N ≥ 3 , p ≥ N/ ( N − 2) , the basic model − ∆ u + | u | p − 1 u = 0 in Ω \ { 0 } , u = 0 on ∂ Ω (1.5) admits only the zero nonnegative solution.

Backgrounds Isolated singular solutions semilinear Hardy problem Singular point on the boundary Laplacian operator • Veron, NA (1981) For singularities of positive solutions of (1.5 ) for 1 < p < N/ ( N − 2) (1 < p < ∞ if N = 2 ) , (when ( N + 1) / ( N − 1) ≤ p < N/ ( N − 2) the assumption of positivity is unnecessary) and that two types of singular behaviour occur: ◦ either u ( x ) ∼ c N k | x | 2 − N if N ≥ 3 u ( x ) ∼ ( − c N k ln | x | ) if N = 2 as | x | → 0 and k can take any positive value; u is said to have a weak singularity at 0 , and actually u = u k , u k is a distributional solution of (1.4 ); 2 ◦ or u ( x ) ∼ c N,p | x | − p − 1 as x → 0 ; u is said to have a strong singularity at 0 , and u = u ∞ := lim k →∞ u k .

Backgrounds Isolated singular solutions semilinear Hardy problem Singular point on the boundary Laplacian operator • Chen-Matano-Veron, JFA (1989): Anisotropic singularities When 1 < p < ( N + 1) / ( N − 1) , u is a solution of (1.5 ), then 2 p − 1 u ( r, θ ) ∼ ω ( θ ) , where ω is a solution of ◦ either r − ∆ S N − 1 ω + | ω | p − 1 ω = l p ω S N − 1 ; in 2 ◦ or there exists an integer k < p − 1 and θ 0 ∈ [0 , 2 π ) such that u ( r, θ ) ∼ c N,q kr k sin( kθ + θ 0 ) as r = | x | → 0 ; ◦ or u ( x ) ∼ − c N k ln | x | as | x | → 0 . • Veron, Handb. Differ. Eq., North-Holland 2004: For N ≥ 3 , the problem − ∆ u + g ( u ) = ν in Ω , u = 0 on ∂ Ω (1.6) has a unique distributional solution u ν if ν is a bounded Radon measure, g is nondecreasing locally Lipchitz continuous, g (0) = 0 and � ∞ N ( g ( s ) − g ( − s )) s − 1 − N − 2 ds < + ∞ . 1

Backgrounds Isolated singular solutions semilinear Hardy problem Singular point on the boundary Laplacian operator • V` azquez, Proc. Royal Soc. Edinburgh. A (1983) When N = 2 , introduced the exponential orders of growth of g defined by � ∞ � � | g ( ± t ) | e − bt dt < ∞ β ± ( g ) = ± inf b > 0 : (1.7) 1 if ν is any bounded measure in Ω with Lebesgue decomposition � ν = ν r + α j δ a j , j ∈ N where ν r is part of ν with no atom, a j ∈ Ω and α j ∈ R satisfy 4 π 4 π β − ( g ) ≤ α j ≤ β + ( g ) , (1.8) then − ∆ u + g ( u ) = ν in Ω , u = 0 on ∂ Ω (1.9) admits a unique weak solution. • Baras and Pierre , Ann Inst Fourier Grenoble (1984) When g ( u ) = | u | p − 1 u for p > 1 and they discovered that if p ≥ N N − 2 the problem is well posed if and only if ν is absolutely continuous with respect to the Bessel capacity c 2 ,p ′ with p ′ = p p − 1 .

Backgrounds Isolated singular solutions semilinear Hardy problem Singular point on the boundary Hardy operator Hardy inqualities The Hardy inequalities ( N − 2) 2 ξ 2 � � |∇ ξ | 2 dx, ∀ ξ ∈ H 1 | x | 2 dx ≤ 0 (Ω); 4 Ω Ω Improved Hardy inequality ( N − 2) 2 ξ 2 � � � ξ 2 dx ≤ |∇ ξ | 2 dx, ∀ ξ ∈ H 1 | x | 2 dx + c 0 (Ω); 4 Ω Ω Ω Denote µ 0 = − ( N − 2) 2 . 4 Note that µ 0 < 0 if N ≥ 3 and µ 0 = 0 if N = 2 . Let Hardy operator be defined by µ L µ = − ∆ + | x | 2 . (1.10)

Backgrounds Isolated singular solutions semilinear Hardy problem Singular point on the boundary Hardy operator Singular radial solutions of L µ When µ ≥ µ 0 R N \ { 0 } L µ u = 0 in (1.11) has two branches of radial solutions with the explicit formulas that � | x | τ − ( µ ) if µ < µ 0 Γ µ ( x ) = | x | τ + ( µ ) , Φ µ ( x ) = and − | x | τ − ( µ ) ln | x | if µ = µ 0 (1.12) where − √ µ − µ 0 + √ µ − µ 0 . τ − ( µ ) = − N − 2 τ + ( µ ) = − N − 2 and 2 2 Here the τ − ( µ ) and τ + ( µ ) are the zero points of τ ( τ + N − 2) − µ = 0 . In the following, we use the notations τ − = τ − ( µ ) and τ + = τ + ( µ ) .

Backgrounds Isolated singular solutions semilinear Hardy problem Singular point on the boundary Hardy operator semilinear Hardy problem • Dupaigne, JAM (2002) the strong, H 1 0 and distributional solutions of L µ u = u p + tf, u > 0 in Ω , u = 0 on ∂ Ω . (1.13) ◦ a classical solution u is a C 2 (¯ Ω \ { 0 } ) function verifies the equation pointwise in Ω \ { 0 } and u ( x ) ≤ c Γ µ for some c > 0 ; ◦ a H 1 solution u is a H 1 0 (Ω) function verifies the identity � µ � ( u p + tf ) ξ, ∀ ξ ∈ H 1 ( ∇ u ∇ ξ − | x | 2 uξ ) = 0 (Ω); Ω Ω u ◦ a distributional solution u , if u ∈ L 1 (Ω) , | x | 2 ∈ L 1 (Ω , ρdx ) and u verifies that � � ( u p + tf ) ξ, ∀ ξ ∈ C 2 (¯ u L µ ξ = Ω) ∩ C 0 (Ω) , Ω Ω where ρ ( x ) = dist( x, ∂ Ω) .

Backgrounds Isolated singular solutions semilinear Hardy problem Singular point on the boundary Hardy operator Dupaigne’s main results Theorem Assume that N ≥ 3 , µ ∈ [ µ 0 , 0) , f is a smooth, bounded and nonnegative function and 2 q ∗ µ = 1 + − τ + ( µ ) For 1 < p < q ∗ µ , there exists t 0 such that ( i ) if 0 < t < t 0 , problem (1.13 ) has a minimal classical solution; ( ii ) if t = t 0 , problem (1.13 ) has a minimal distributional solution; ( iii ) if t > t 0 , problem (1.13 ) has no distributional solution. • Brezis-Dupaigne-Tesei Sel Math (2005) When t = 0 , (1.13 ) has a nontrivial nonnegative solution of for p < q ∗ µ and does not have nonnegative distributional solutions for p ≥ q ∗ µ .