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

Second order conformally invariant elliptic equations Yanyan Li Rutgers University May 24, 2017. ICTP, Trieste, Italy

• Theorem 3-1 (Luc Nguyen, L. ) (Blow up analysis) Assume { u k } ∈ C 2 ( B 2 ), f ( λ ( A u k )) = 1 , u k > 0 , in B 2 , sup u k → ∞ . B 1 Then ∀ ǫ > 0, after passing to a subsequence, ∃ { x 1 k , · · · , x m k } ⊂ B 2 (0), 1 ≤ m ≤ ¯ m , k | ≥ K − 1 > 0 , k − x j | x i ∀ k , i � = j , u k ( x i k ) = sup u k . B δ ( x i k ) K − 1 ≤ u k ( x i k ) ≤ K , ∀ i , j , k , u k ( x j k ) � u k ( x ) − U x i k , u k ( x i k ) ( x ) | ≤ ǫ U x i k , u k ( x i k ) ( x ) , ∀ x ∈ B δ ( x i � k ) . 1 K 2 (0) \∪ m i =1 B δ ( x i k ) ≤ u k ( x ) ≤ k ) , in B 3 k ) , ∀ k . K δ n − 2 u k ( x 1 δ n − 2 u k ( x 1 2 —– U ¯ x ,µ ( x ) = µ U ( µ n − 2 ( x − ¯ x )), 2 − n —– U ( x ) = (1 + | x | 2 ) satisfies f ( λ ( A U )) = 1, 2 —– ¯ m , K depend only on ( f , Γ), δ depends on ( f , Γ) and ǫ .

Proposition 3-1. (Strengthened Liouville type theorem) Assume 0 < v ∈ C 0 ( R n ), 0 < v k ∈ C 2 ( B R k ), R k → ∞ , v k → v in C 0 f ( λ ( A v k )) = 1 in B R k , loc ( R n ) . a � n − 2 2 , x ∈ R n . � v ( x ) = a > 0 , ¯ Then 1 + a 2 | x − ¯ x | 2 • v satisfies f ( λ ( A v )) = 1, in R n in viscosity sense. Open Problem. Let 0 < v ∈ C 0 loc ( R n ) satisfy f ( λ ( A v )) = 1 , in R n in viscosity sense . Is it true that a � n − 2 2 , x ∈ R n ? � v ( x ) = a > 0 , ¯ 1 + a 2 | x − ¯ x | 2

Proof of Proposition 3-1. 0 < v superharmonic, so | y | n − 2 v ( y ) ≥ 2 c 0 > 0 , ∀ | y | ≥ 1 . Passing to subsequence, shrinking R k , shrinking c 0 , may assume | v k ( y ) − v ( y ) | ≤ ( R k ) − n , v k ( y ) ≥ c 0 ( R k ) 2 − n , ∀ | y | ≤ R k .

• Define for x ∈ R n , | x | + 1 ≤ R k / 4, λ k ( x ) = sup { 0 < µ ≤ R k ¯ 4 | ( v k ) x ,λ ≤ v k in B R k (0) \ B λ ( x ) , ∀ 0 < λ < µ } , | y − x | ) n − 2 v k ( x + λ 2 ( y − x ) λ where ( v k ) x ,λ ( y ) := ( | y − x | 2 ), the Kelvin transformation. • ¯ λ k ( x ) well defined and ∃ C ( x ) > 0 such that 1 λ k ( x ) ≤ R k C ( x ) ≤ ¯ 0 < 4 , ∀ k . —– Proof based on : • Local gradient estimates: u ∈ C 2 ( B 2 ) , f ( λ ( A u )) = 1 , 0 < u ≤ b , in B 2 implies |∇ log u | ≤ C in B 1 —– C depends only on ( f , Γ) and b .

• Set ¯ ¯ λ ( x ) =lim inf λ k ( x ) ∈ (0 , ∞ ]. k →∞ • Can prove (maximum principle, Hopf Lemma): either ¯ λ ( x ) ≡ ∞ ∀ x or ¯ λ ( x ) < ∞ ∀ x . ¯ λ ( x ) ≡ ∞ leads to: v ≡ Constant, which can be ruled out. ¯ λ ( x ) < ∞ ∀ x leads to: | y |→∞ | y | n − 2 v x , ¯ | y |→∞ | y | n − 2 v ( y ) < ∞ , ∀ x . lim λ ( x ) ( y ) = α := lim inf

• We have arrived at : 0 < v ∈ C 0 , 1 loc ( R n ), ∆ v ≤ 0 in R n , for every x ∈ R n , there exists 0 < ¯ λ ( x ) < ∞ such that λ ( x ) ( y ) ≤ v ( y ) , ∀ | y − x | ≥ ¯ λ ( x ) , v x , ¯ | y |→∞ | y | n − 2 v x , ¯ | y |→∞ | y | n − 2 v ( y ) , lim λ ( x ) ( y ) = α := lim inf ∀ x . • Claim. We can deduce from the above that a � n − 2 2 , � x ∈ R n . v ( x ) = b a , b > 0 , ¯ 1 + a 2 | x − ¯ x | 2 Since v k → v in C 0 loc ( R n ) and f ( λ ( A v k )) = 1, can prove that b = 1.

• A Lemma. For n ≥ 2, B 1 ⊂ R n , w 1 , w 2 ∈ C 0 ( B 1 ), w 1 , w 2 differentiable at 0, u ∈ L 1 loc ( B 1 \ { 0 } ), ∆ u ≤ 0 in B 1 \ { 0 } , u ( y ) ≥ max { w 1 ( y ) , w 2 ( y ) } , y ∈ B 1 \ { 0 } , w 1 (0) = w 2 (0) = lim inf y → 0 u ( y ) . Then ∇ w 1 (0) = ∇ w 2 (0) . • Apply the lemma with w 1 = w ( x ) , w 2 = w (˜ x ) , u = v ψ , x , ˜ x ∈ D .

y • Proof of Claim. Let ψ ( y ) = | y | 2 , and w ( x ) := ( v x , ¯ λ ( x ) ) ψ , x ∈ R n . • Then, for some δ ( x ) > 0, v ψ ≥ w ( x ) in B δ ( x ) \ { 0 } , w ( x ) (0) = α = lim inf y → 0 v ψ ( y ) , ∆ v ψ ≤ 0 , in B δ ( x ) \ { 0 } , • Let D = { x ∈ R n | v is differentiable at x } . Since v ∈ C 0 , 1 loc ( R n ), Lebesgue measure | R n \ D | = 0. • By the Lemma , ∇ w ( x ) (0) = ∇ w (˜ x ) (0) , ∀ x , ˜ x ∈ D . Namely, for some V ∈ R n , ∇ w ( x ) (0) = V , ∀ x ∈ D .

• A calculation yields n n ∇ w ( x ) (0) = ( n − 2) α x + α n − 2 v ( x ) n − 2 ∇ v ( x ) . • Thus � n − 2 n − 2 − n − 2 2 α | x | 2 + V · x n n − 2 v ( x ) − � ∇ x α = 0 , ∀ x ∈ D . 2 2 x ∈ R n and d ∈ R , • Consequently, for some ¯ 2 2 x | 2 + d α − 2 v ( x ) − n − 2 ≡ α − n − 2 | x − ¯ n − 2 . • Since v > 0, we must have d > 0, so 2 α n − 2 � n − 2 2 . � v ( x ) ≡ x | 2 d + | x − ¯ Claim proved.

Proposition 3-2. Assume { v k } ∈ C 2 ( B R k ), R k → ∞ , f ( λ ( A v k ))( y ) = 1 , 0 < v k ( y ) ≤ v k (0) = 1 , | y | ≤ R k . (1) 0 ( ǫ ) and δ ′ = δ ′ ( ǫ ) such that ∀ k > k ′ Then ∀ ǫ > 0, ∃ k ′ 0 = k ′ 0 , ∀ | y | ≤ δ ′ R k . | v k ( y ) − U ( y ) | ≤ 2 ǫ U ( y ) , (2) Recall: � n − 2 � 2 1 —– U ( x ) := 1+ | x | 2 —– A U ≡ 2 I , f ( λ ( A U )) ≡ 1

By the local gradient estimates and by Proposition 3-1, after passing to subsequence, in C 0 loc ( R n ) . v k → U , Lemma 1. ∀ ǫ > 0, ∃ k 0 , such that ∀ k ≥ k 0 , | y | = r v k ( y ) ≤ (1 + ǫ ) U ( y ) , min ∀ 0 < r < R k . Proof. • Facts: U 0 ,λ ( y ) < U ( y ) , ∀ 0 < λ < 1 , | y | > λ, U 0 , 1 ≡ U . U 0 ,λ ( y ) > U ( y ) , ∀ λ > 1 , | y | > λ.

• Contradiction argument: If for some ǫ > 0, ∃ r k min v k ( y ) > (1 + ǫ ) U ( y ) . | y | = r k • Then, using the above facts of U , r k → ∞ , and ( v k ) λ ( y ) ≤ v k ( y ) , ∀ 0 < λ < 1 + ǫ 2 , | y | = r k . • Sending k → ∞ . U λ ( y ) ≤ U ( y ) , ∀ 0 < λ < 1 + ǫ 2 , λ < | y | < ∞ . Violating the above facts of U . Lemma 1 proved.

Lemma 2. ∀ ǫ > 0, ∃ small δ 1 > 0, large r 1 > 0, such that for large k , v k ( y ) ≥ (1 − ǫ ) U ( y ) , ∀ | y | ≤ δ 1 R k , � n +2 n − 2 v ≤ ǫ. k r 1 ≤| y |≤ δ 1 R k Proof. Since v k → U in C 0 loc ( R n ), ∃ r 1 such that for large k v k ( y ) ≥ (1 − ǫ 2 ) U ( y ) , ∀ | y | ≤ r 1 , v k ( y ) ≥ (1 − ǫ 2 )( r 1 ) 2 − n , ∀ | y | = r 1 , Superharmonicity of v k , maximum principle, we have | y | 2 − n − ( R k ) 2 − n � v k ( y ) ≥ (1 − ǫ 2 ) � , r 1 ≤ | y | ≤ R k .

2 n − 2 ), Thus, for any δ 1 ∈ (0 , ǫ v k ( y ) ≥ (1 − 2 ǫ 2 ) | y | 2 − n , r 1 ≤ | y | ≤ δ 1 R k . The equation of v k implies that ∃ δ > 0, − ∆ v k ( y ) ≥ n − 2 n +2 δ v k ( y ) in r 1 ≤ | y | ≤ δ 1 R k . n − 2 2 This implies (1 − 2 ǫ 2 ) | y | 2 − n v k ( y ) ≥ + 1 � n − 2 dx , ∀ | y | = δ 1 R k n +2 C | y | 2 − n δ v k ( x ) . 2 2 r 1 ≤| x |≤ δ 1 R k / 8 By Lemma 1, ∀ | y | = δ 1 R k (1 + 2 ǫ 2 ) | y | 2 − n ≥ v k ( y ) , . 2 Lemma 2 follows from the above.

Since v k ≤ 1, by Lemma 2, for any ǫ > 0, we have, for large k , � 2 n n − 2 ≤ ǫ. v k r 1 ≤| y |≤ δ 1 R k • Small energy implies L ∞ bound — consequence of Liouville, as showed before. Lemma 3. ∃ δ 0 > 0 and C 0 > 1 such that if 0 < u ∈ C 2 ( B 2 ), � 2 n f ( λ ( A u )) = 1 , in B 2 , n − 2 ≤ δ 0 , u B 2 then u ≤ C 0 in B 1 .

Lemma 4. ∃ C , δ 4 > 0, independent of k , such that v k ( y ) ≤ CU ( y ) , ∀ | y | ≤ δ 4 R k . Proof. ∀ 4 r 1 < r < δ 1 R k / 4, consider 1 n − 2 2 v k ( rz ) , v k ( z ) = r ˜ 4 < | z | < 4 . For large k , � � 2 n 2 n n − 2 = n − 2 ≤ ǫ := δ 0 , v k ( z ) ˜ v k ( η ) 1 r 4 < | z | < 4 4 < | η | < 4 r where δ 0 > 0 is the number in Lemma 3.

• By Lemma 3, 1 v k ( z ) ≤ C , ˜ 3 < | z | < 3 , for some universal constant C . • By local gradient estimates, 1 |∇ log ˜ v k ( z ) | ≤ C , 2 < | z | < 2 . • Thus max | z | =1 ˜ v k ( z ) ≤ min | z | =1 ˜ v k ( z ) . i.e. max | x | = r v k ( x ) ≤ C min | x | = r v k ( x ) ≤ CU ( r ) . —- used Lemma 1 for last inequality. Lemma 4 follows immediately.

Proof of Proposition 3-2. Only need to prove that there exists δ ′ and k ′ 0 such that for any k ≥ k ′ 0 , ∀ | y | ≤ δ ′ R k . v k ( y ) ≤ (1 + 2 ǫ ) U ( y ) , Suppose the contrary, passing to subsequence, ∃ | y k | = δ k R k , δ k → 0 + , but v k ( y k ) = max v k ( y ) ≥ (1 + 2 ǫ ) U ( y k ) . | y | = δ k R k Since v k → v in C 0 loc ( R n ), | y k | → ∞ . Consider rescaling of v k : | z | < δ 4 R k v k ( z ) := | y k | n − 2 v k ( | y k | z ) , ˆ | y k | → ∞ . We have | z | < δ 4 R k f k ( λ ( A ˆ v k ))( z ) := | y k | − 2 f ( λ ( A v k ))( z ) = | y k | − 2 , | y k | . Since ˆ v k ≤ C , we can apply gradient estimates to f k to obtain:

∀ 0 < α < β < ∞ , ∃ C ( α, β ) such that for large k , |∇ log ˆ v k ( z ) | ≤ C ( α, β ) , ∀ α < | z | < β. We know from Lemma 1 and the above v k ( z ) ≤ 1 + 5 ǫ | z | =1 ˆ min 4 , and v k ( z ) ≥ 1 + 3 ǫ | z | =1 ˆ max 2 . Passing to subsequence, for some 0 < v ∗ ∈ C 0 , 1 loc ( R n \ { 0 } ), loc ( R n \ { 0 } ) , ∀ 0 < α < 1 , in C 1 ,α v ∗ v k → ˆ ˆ and v ∗ satisfies in viscosity sense R n \ { 0 } . λ ( A ˆ v ∗ ) ∈ ∂ Γ ,