Minimal blowup data for potential Navier-Stokes singularities in the - PowerPoint PPT Presentation

Minimal blowup data for potential Navier-Stokes singularities in the half-space Tuan Pham Oregon State University October 29, 2018 1/18 Tuan Pham (Oregon State University) October 29, 2018 1 / 18

Cauchy problem of NSE For Ω = R 3 or R 3 + , consider  ∂ t u − ∆ u + u · ∇ u + ∇ p = f x ∈ Ω , t > 0 ,    div u = 0 x ∈ Ω , t > 0 , ( NSE ) Ω : u ( x , t ) = 0 x ∈ ∂ Ω , t > 0 ,    u ( x , 0) = u 0 x ∈ Ω . 2/18 Tuan Pham (Oregon State University) October 29, 2018 2 / 18

Cauchy problem of NSE For Ω = R 3 or R 3 + , consider  ∂ t u − ∆ u + u · ∇ u + ∇ p = f x ∈ Ω , t > 0 ,    div u = 0 x ∈ Ω , t > 0 , ( NSE ) Ω : u ( x , t ) = 0 x ∈ ∂ Ω , t > 0 ,    u ( x , 0) = u 0 x ∈ Ω . Scaling symmetry : λ u ( λ x , λ 2 t ) u ( x , t ) → λ 2 p ( λ x , λ 2 t ) p ( x , t ) → λ 3 f ( λ x , λ 2 t ) f ( x , t ) → u 0 ( x ) → λ u 0 ( λ x ) 2/18 Tuan Pham (Oregon State University) October 29, 2018 2 / 18

Cauchy problem of NSE For Ω = R 3 or R 3 + , consider  ∂ t u − ∆ u + u · ∇ u + ∇ p = f x ∈ Ω , t > 0 ,    div u = 0 x ∈ Ω , t > 0 , ( NSE ) Ω : u ( x , t ) = 0 x ∈ ∂ Ω , t > 0 ,    u ( x , 0) = u 0 x ∈ Ω . Scaling symmetry : λ u ( λ x , λ 2 t ) u ( x , t ) → λ 2 p ( λ x , λ 2 t ) p ( x , t ) → λ 3 f ( λ x , λ 2 t ) f ( x , t ) → u 0 ( x ) → λ u 0 ( λ x ) Critical spaces : u 0 ∈ L 3 , f ∈ L 5 / 3 t , x , u ∈ L 5 t , x ,. . . 2/18 Tuan Pham (Oregon State University) October 29, 2018 2 / 18

Mild Solutions • Helmholtz decomposition: g = v + ∇ φ div v = 0 , v · n | ∂ Ω = 0 3/18 Tuan Pham (Oregon State University) October 29, 2018 3 / 18

Mild Solutions • Helmholtz decomposition: g = v + ∇ φ div v = 0 , v · n | ∂ Ω = 0 ∂ t u − P ∆ u = P f − P ( u · ∇ u ) ���� A � t Mild solution: u ∈ L 5 e ( t − s ) A P ( u · ∇ u ) ds t , x , u = U + F − 0 3/18 Tuan Pham (Oregon State University) October 29, 2018 3 / 18

Mild Solutions • Helmholtz decomposition: g = v + ∇ φ div v = 0 , v · n | ∂ Ω = 0 ∂ t u − P ∆ u = P f − P ( u · ∇ u ) ���� A � t Mild solution: u ∈ L 5 e ( t − s ) A P ( u · ∇ u ) ds t , x , u = U + F − 0 Local in time, unique, regular. Characterization of finite-time blowup: T → T ∗ � u � L 5 (Ω × (0 , T )) = ∞ . lim Globally well-posed if ( u 0 , f ) is sufficiently small in critical spaces. 3/18 Tuan Pham (Oregon State University) October 29, 2018 3 / 18

Weak Solutions Suitable weak solution:  weak form,  Leray ‘34, Scheffer ‘77, local energy inequality, C–K–N ‘82, Lemari´ e-Rieusset ‘02  u ( t ) → u 0 in L 2 loc as t ↓ 0 . 4/18 Tuan Pham (Oregon State University) October 29, 2018 4 / 18

Weak Solutions Suitable weak solution:  weak form,  Leray ‘34, Scheffer ‘77, local energy inequality, C–K–N ‘82, Lemari´ e-Rieusset ‘02  u ( t ) → u 0 in L 2 loc as t ↓ 0 . Sw-solution:  u = v + w    Seregin–Sverak v satisfies linear Stokes eq. with data ( u 0 , f ) � ∂ t w − ∆ w + ∇ π = − u · ∇ u weakly (2017)  w satisfies   energy inequality 4/18 Tuan Pham (Oregon State University) October 29, 2018 4 / 18

Weak Solutions Suitable weak solution:  weak form,  Leray ‘34, Scheffer ‘77, local energy inequality, C–K–N ‘82, Lemari´ e-Rieusset ‘02  u ( t ) → u 0 in L 2 loc as t ↓ 0 . Sw-solution:  u = v + w    Seregin–Sverak v satisfies linear Stokes eq. with data ( u 0 , f ) � ∂ t w − ∆ w + ∇ π = − u · ∇ u weakly (2017)  w satisfies   energy inequality ε -regularity criterion (C–K–N ‘82, Lin ‘98, Seregin 2002) There are two positive constants ε and C such that �� 1 � � | u ( x , t ) | ≤ C | u | 3 + | p | 3 dxdt ≤ ε = ⇒ sup r . 2 r 2 Q r / 2 ( x 0 , t 0 ) Q r ( x 0 , t 0 ) 4/18 Tuan Pham (Oregon State University) October 29, 2018 4 / 18

Minimal blowup data ρ Ω max = sup { ρ : T max ( u 0 , f ) = ∞ if � ( u 0 , f ) � X × Y < ρ } . Question If ρ Ω max is finite, does there exist a data ( u 0 , f ) ∈ X × Y with � ( u 0 , f ) � = ρ Ω max , such that the solution u of (NSE) Ω blows up in finite time ? 5/18 Tuan Pham (Oregon State University) October 29, 2018 5 / 18

Minimal blowup data ρ Ω max = sup { ρ : T max ( u 0 , f ) = ∞ if � ( u 0 , f ) � X × Y < ρ } . Question If ρ Ω max is finite, does there exist a data ( u 0 , f ) ∈ X × Y with � ( u 0 , f ) � = ρ Ω max , such that the solution u of (NSE) Ω blows up in finite time ? Affirmation for Ω = R 3 , f ≡ 0 and u 0 ∈ X X = ˙ H 1 / 2 : Rusin–Sverak 2011. X = L 3 : Jia–Sverak 2013, Gallagher–Koch–Planchon 2013. X = ˙ B − 1+3 / p (3 < p , q < ∞ ): G–K–P 2016. p , q 5/18 Tuan Pham (Oregon State University) October 29, 2018 5 / 18

Main Results Assume u 0 = 0 . � � 5 q ∗ = 3 2 − 5 t q ∗ f ∈ L q Y q = f : 2 < q < 3 , , t , x 2 q 6/18 Tuan Pham (Oregon State University) October 29, 2018 6 / 18

Main Results Assume u 0 = 0 . � � 5 q ∗ = 3 2 − 5 t q ∗ f ∈ L q Y q = f : 2 < q < 3 , , t , x 2 q Theorem 1 For Ω = R 3 and Y = Y q , minimal blowup data exists, provided that a blowup data exists. 6/18 Tuan Pham (Oregon State University) October 29, 2018 6 / 18

Main Results Assume u 0 = 0 . � � 5 q ∗ = 3 2 − 5 t q ∗ f ∈ L q Y q = f : 2 < q < 3 , , t , x 2 q Theorem 1 For Ω = R 3 and Y = Y q , minimal blowup data exists, provided that a blowup data exists. Theorem 2 (a) ρ + max ≤ ρ max , (b) If ρ + max < ρ max then there exists a minimal blowup data for Ω = R 3 + . 6/18 Tuan Pham (Oregon State University) October 29, 2018 6 / 18

Theorem 1: Sketch of proof Step 1 : Blowup happens only if there occurs a singular point. � u � Q r ( x 0 , T max ) = ∞ ∀ r > 0 . This is an application of ε -regularity criterion ! 7/18 Tuan Pham (Oregon State University) October 29, 2018 7 / 18

Theorem 1: Sketch of proof Step 1 : Blowup happens only if there occurs a singular point. � u � Q r ( x 0 , T max ) = ∞ ∀ r > 0 . This is an application of ε -regularity criterion ! Step 2 : Set up minimizing sequence. � f k � ↓ ρ max ( u k , p k ) is mild solution with data f k , singular at ( x k , t k ). 7/18 Tuan Pham (Oregon State University) October 29, 2018 7 / 18

Theorem 1: Sketch of proof Step 3 : Normalize ( x k , t k ) to (0,1) by translation/scaling symmetry. � x − x k � √ , t u k ( x , t ) λ k u k t k → , λ k = λ 2 λ k k . . . 8/18 Tuan Pham (Oregon State University) October 29, 2018 8 / 18

Theorem 1: Sketch of proof Step 3 : Normalize ( x k , t k ) to (0,1) by translation/scaling symmetry. � x − x k � √ , t u k ( x , t ) λ k u k t k → , λ k = λ 2 λ k k . . . Step 4 : Compactness Theorem (Seregin–Sverak ‘17, Lin ‘98). u k → u in L 3 loc p k ⇀ p in L 3 / 2 loc f k ⇀ f in Y q ( u , p ) is sw-solution with data f . 8/18 Tuan Pham (Oregon State University) October 29, 2018 8 / 18

Theorem 1: Sketch of proof Step 5 : z 0 = (0 , 1) is a singularity of u ? �� � � 1 | u k | 3 + | p k | 3 dxdt > ε ∀ r > 0 , k = 1 , 2 , . . . 2 r 2 Q r ( z 0 ) 9/18 Tuan Pham (Oregon State University) October 29, 2018 9 / 18

Theorem 1: Sketch of proof Step 5 : z 0 = (0 , 1) is a singularity of u ? �� � � 1 | u k | 3 + | p k | 3 dxdt > ε ∀ r > 0 , k = 1 , 2 , . . . 2 r 2 Q r ( z 0 ) Want: � 1 | p k − p | 3 2 dxdt → 0 as k → ∞ , r → 0 r 2 Q r ( z 0 ) 9/18 Tuan Pham (Oregon State University) October 29, 2018 9 / 18

Theorem 1: Sketch of proof Step 5 : z 0 = (0 , 1) is a singularity of u ? �� � � 1 | u k | 3 + | p k | 3 dxdt > ε ∀ r > 0 , k = 1 , 2 , . . . 2 r 2 Q r ( z 0 ) Want: � 1 | p k − p | 3 2 dxdt → 0 as k → ∞ , r → 0 r 2 Q r ( z 0 ) Interior pressure decomposition: � � � � � � p k = R i R j u k i u k u k i u k u k i u k = R i R j + R i R j j (1 − χ ) j χ j � �� � � �� � p k p k 1 → p 1 2 harmonic in B 1 9/18 Tuan Pham (Oregon State University) October 29, 2018 9 / 18

Theorem 1: Sketch of proof Step 5 : z 0 = (0 , 1) is a singularity of u ? �� � � 1 | u k | 3 + | p k | 3 dxdt > ε ∀ r > 0 , k = 1 , 2 , . . . 2 r 2 Q r ( z 0 ) Want: � 1 | p k − p | 3 2 dxdt → 0 as k → ∞ , r → 0 r 2 Q r ( z 0 ) Interior pressure decomposition: � � � � � � p k = R i R j u k i u k u k i u k u k i u k = R i R j + R i R j j (1 − χ ) j χ j � �� � � �� � p k p k 1 → p 1 2 harmonic in B 1 Step 6 : u must blow up ! k →∞ � f k � = ρ max ρ max ≤ � f � ≤ lim inf 9/18 Tuan Pham (Oregon State University) October 29, 2018 9 / 18

Theorem 2 (b): Sketch of proof Normalize ( x k , t k ) to ((0 , 0 , d k ) , 1) by translation and scaling. � x − x k � √ , t u k ( x , t ) λ k u k → λ k = t k , λ 2 λ k k . . . 10/18 Tuan Pham (Oregon State University) October 29, 2018 10 / 18