Second order conformally invariant elliptic equations Yanyan Li - PowerPoint PPT Presentation

Second order conformally invariant elliptic equations Yanyan Li Rutgers University May 22, 2017. ICTP, Trieste, Italy

• Yamabe problem 4 ( M n , g ), n ≥ 3, Riemannian, compact, ∃ ? ˜ n − 2 g ) such g ∼ g (˜ g = u that g ≡ Constant . R ˜ • PDE: n − 2 n +2 4( n − 1) R g u = ¯ n − 2 , − ∆ g u + Ru u > 0 , on M , ¯ R = − 1 , 0 , 1.

• Einstein-Hilbert functional: � 2 − n E ( g ) = Vol ( g ) R g dv g . n M • Euler-Lagrange equation: Ric g = λ g .

• Restricting to a conformal class of metrics : 4 n − 2 g � � u ∈ C ∞ ( M ) , u > 0 � � [ g ] := g = u ˜ � 4( n − 1) R g u 2 � |∇ g u | 2 + n − 2 � dv g M Y g ( u ) ≡ E (˜ g ) = . � n − 2 �� 2 n n n − 2 dv g M u • Euler-Lagrange equation for E | [ g ] : 4 n − 2 g ) . g = λ (˜ g = u R ˜ • A critical point u leads to a solution of the PDE in u : n − 2 n +2 4( n − 1) R g u = ¯ n − 2 , − ∆ g u + u > 0 , on M , Ru ¯ R = − 1 , 0 , 1.

• Three mutually exclusive cases: The conformal Laplacian: n − 2 − L g := − ∆ g + 4( n − 1) R g . n − 2 n +2 n − 2 , − ∆ g u + 4( n − 1) R g u = λ 1 ( − L g ) u u > 0 , on M . λ 1 ( − L g ) —- the sign of λ 1 ( − L g ). • Solution of the Yamabe problem: Yamabe(1960), Trudinger (1968), Aubin (1976), Schoen (1984).

• Solution space. • If λ 1 ( − L g ) = 0, n − 2 − ∆ g u + 4( n − 1) R g u = 0 , u > 0 , on M . • Eigenspace for the first eigenvalue is 1 − dimensional. • Existence is also clear.

• If λ 1 ( − L g ) < 0, n − 2 n +2 n − 2 , − ∆ g u + 4( n − 1) R g u = − u u > 0 , on M . • Uniqueness of solution — maximum principle. u = 1 • Existence is also clear: u = ǫ > 0 subsolution, ¯ ǫ supersolution.

• If λ 1 ( − L g ) > 0, n − 2 n +2 n − 2 , − ∆ g u + 4( n − 1) R g u = u u > 0 , on M . • If ( M n , g ) ∼ ( S n , g 0 ), all solutions are known, and unique modulo conformal diffeomorphism of ( S n , g 0 ). • If ( M n , g ) not ( S n , g 0 ), solution space more complicated. • Question. { u ∈ C ∞ ( M ) | u solution } is bounded in L ∞ ( M )?

Much work has been done: Yes, if ( M n , g ) is locally conformally flat. • Schoen (1991) • L. and Zhu (1999) Yes, if n = 3. • Independent works by three groups: • Druet (2004) Yes, if n ≤ 4. • Marques (2005) Yes, if n ≤ 7. • L. and Zhang (2005) Yes, if n ≤ 9. After the above: • L. and Zhang (2006) Yes, if n ≤ 11. • Khuri-Marques-Schoen (2009) Yes, if n ≤ 24. • Brendle (2009) No, if n ≥ 52. • Brendle-Marques (2009) No, if n ≥ 25. • For 8 ≤ n ≤ 24, “Yes” provided the Positive Mass Theorem.

A fully nonlinear Yamabe problem • Schouten tensor: A g = ( n − 2) − 1 ( Ric g − [2( n − 1)] − 1 R g g ) , Let λ ( A g ) = ( λ 1 , · · · , λ n ) = eigenvalues of A g . Then λ 1 ( A g ) + · · · + λ n ( A g ) = R g .

• Yamabe problem on ( M , g ) when λ 1 ( − L g ) > 0: Assume λ 1 ( A g ) + · · · + λ n ( A g ) > 0, ∃ ? ˜ g ∼ g such that λ 1 ( A ˜ g ) + · · · + λ n ( A ˜ g ) = 1. • A more general question on ( M , g ): Assume f ( λ ( A g )) > 0, ∃ ? g ∼ g such that f ( λ ( A ˜ ˜ g )) = 1. • A second order fully nonlinear elliptic PDE: 2 2 n n − 2 u − 1 ∇ 2 A ˜ g = − g u + ( n − 2) 2 ∇ g u ⊗ ∇ g u ( n − 2) 2 u − 2 |∇ g u | 2 2 − g g + A g .

• More precisely: Let Γ ⊂ R n open,convex,symmetric cone,vertex at origin Γ n ⊂ Γ ⊂ Γ 1 n Γ n := { λ ∈ R n | λ i > 0 ∀ i } , Γ 1 := { λ ∈ R n | � λ i > 0 } i =1 f ∈ C 1 (Γ) ∩ C 0 (Γ) symmetric function f λ i > 0 in Γ ∀ i , f > 0 in Γ , f = 0 on Γ

• Illuminating examples: 1 ( f , Γ) = ( σ k , Γ k ), 1 ≤ k ≤ n k � σ k ( λ ) := λ i 1 · · · λ i k , λ i 1 < ··· <λ ik the k -th elementary symmetric function Γ k : the connected component of { λ ∈ R n | σ k ( λ ) > 0 } containing the positive cone Γ n = { λ ∈ R n | λ i > 0 ∀ i }

• A fully nonlinear Yamabe problem Assume λ ( A g ) ∈ Γ on M , 4 n − 2 g such that does there exist ˜ g = u f ( λ ( A ˜ g )) = 1 , λ ( A ˜ g ) ∈ Γ , on M ? • If ( f , Γ) = ( σ 1 , Γ 1 ), the Yamabe problem.

• Answer is “ Yes ” if: (i) ( f , Γ), f concave, homogeneous of degree 1, ( M , g ) is locally conformally flat, 1 and k ≥ n (ii) ( f , Γ) = ( σ k , Γ k ) , k 2 , 1 (iii) ( f , Γ) = ( σ k , Γ k ) , k = 2 . k • Through work of many: [Alice Chang, Gursky, Paul Yang, 2002], [Pengfei Guan, Guofang Wang, 2003], [Aobing Li, L., 2003, 2005], [Gursky, Viaclovsky, 2004, 2007], [Yuxin Ge, Guofang Wang, 2006], [Weimin Sheng, Trudinger, Xujia Wang, 2007], [Luc Nguyen, L., 2014].

1 k , Γ k ), 3 ≤ k < n • “ Open ” in particular if: ( f , Γ) = ( σ k 2 . • Answer would be “Yes” if can prove a priori estimates: For 4 n − 2 g , u > 0, g = u ˜ f ( λ ( A ˜ g )) = 1 on M implies u ≤ C on M .

• Taking M = R n , or rescaling a blow up sequence of solutions of the geometric equation leads to f ( λ ( A u )) = 1 , in R n , where 2 2 n A u := − n − 2 u − n +2 ( n − 2) 2 u − 2 n n − 2 ∇ 2 u + n − 2 ∇ u ⊗ ∇ u 2 ( n − 2) 2 u − 2 n n − 2 |∇ u | 2 I , −

• Caffarelli, Nirenberg, Spruck, 1985: Introduce ( f , Γ) of such type, pioneering work on existence of smooth solutions for Dirichlet problem: � f ( λ ( ∇ 2 u )) in Ω ⊂ R n , = g ( x ) , u = h ( x ) on ∂ Ω . • Equation f ( λ ( A u )) = 1 resembles the above. • Additional feature: conformal invariance of equation.

In this series of lectures, we • Study estimates and raise open problems

A Liouville type Theorem • Theorem 1. (L., 2006; L., Luc Nguyen, Bo Wang, 2016) 0 < u ∈ C 0 ( R n \ { 0 } ), viscosity solution of λ ( A u ) ∈ ∂ Γ in R n \ { 0 } . Then u is radially symmetric about { 0 } . • Corollary 1. 0 < u ∈ C 0 ( R n ), viscosity solution of λ ( A u ) ∈ ∂ Γ in R n . Then u ≡ u (0). Γ ⊂ R n open,convex,symmetric cone,vertex at origin Γ n ⊂ Γ ⊂ Γ 1 n Γ n := { λ ∈ R n | λ i > 0 ∀ i } , Γ 1 := { λ ∈ R n | � λ i > 0 } i =1 2 2 n A u := − n − 2 u − n +2 ( n − 2) 2 u − 2 n n − 2 ∇ 2 u + n − 2 ∇ u ⊗ ∇ u 2 ( n − 2) 2 u − 2 n n − 2 |∇ u | 2 I , −

We will prove: Proposition 1. Assume Ω ⊂ R n bounded open, { P 1 , · · · , P m } ⊂ Ω. λ ( A u ) ∈ R n \ Γ in Ω , 0 < u ∈ C 2 (Ω) , 0 < v ∈ C 2 (Ω \ { P 1 , · · · , P m } ) , λ ( A v ) ∈ Γ in Ω \ { P 1 , · · · , P m } . v ≥ u on ∂ Ω . Then v ≥ u in Ω \ { P 1 , · · · , P m } .

We first prove: Proposition 1 with no singularity. Assume Ω ⊂ R n bounded open. λ ( A u ) ∈ R n \ Γ , λ ( A v ) ∈ Γ in Ω , 0 < u , v ∈ C 2 (Ω) , v ≥ u on ∂ Ω . Then v ≥ u in Ω .

So λ ( A w + ǫϕ ) ∈ Γ . Take ǫ i = δ i → 0, and let 2 − n − 2 w + ǫ i ϕ i = v . i Then { v i } has the needed approximation property.

A calculation gives w ∇ 2 ϕ + ϕ ∇ 2 w − ∇ w · ∇ ϕ I + ǫ 2 A ϕ . � � A w + ǫϕ = A w + ǫ � � A w + 1 Replacing ∇ 2 w by w − 1 2 |∇ w | 2 I in the above, we have w ∇ 2 ϕ + |∇ w | 2 � � A w + ǫϕ = (1+ ǫ ϕ + ǫ 2 A ϕ . w ) A w + ǫ ϕ I − ∇ w · ∇ ϕ I 2 w Since ∇ 2 ϕ ( y ) = 2 δϕ ( y ) I + 4 δ 2 ϕ ( y ) y ⊗ y . ∇ ϕ ( y ) = 2 δϕ ( y ) y , w ∇ 2 ϕ + |∇ w | 2 ϕ I − ∇ w · ∇ ϕ I ≥ δϕ wI . 2 w