Applying Moser s Iteration to the 3D Axially Symmetric Navier Stokes Equations (ASNSE)

Applying Moser s Iteration to the 3D Axially Symmetric Navier Stokes Equations (ASNSE) Advisor: Qi Zhang Department of Mathematics University of California, Riverside November 4, 2012 / Graduate Student Seminar

The Axially Symmetric Navier-Stokes Equations Introduction The ASNSE are a special case of the incompressible Navier-Stokes equations.

The Axially Symmetric Navier-Stokes Equations Introduction The ASNSE are a special case of the incompressible Navier-Stokes equations. The more general, non axis-symmetric case is given, in Cartesian coordinates, by: v (v )v p t v = 0, div v = 0 v(x, 0) L 2 (R 3 ) C (R 3 )

The Axially Symmetric Navier-Stokes Equations Introduction The ASNSE are a special case of the incompressible Navier-Stokes equations. The more general, non axis-symmetric case is given, in Cartesian coordinates, by: v (v )v p t v = 0, div v = 0 v(x, 0) L 2 (R 3 ) C (R 3 ) v = (v 1 (x, t), v 2 (x, t), v 3 (x, t)) : R 3 [0, T ] R 3 is the unknown velocity field.

The Axially Symmetric Navier-Stokes Equations Introduction The ASNSE are a special case of the incompressible Navier-Stokes equations. The more general, non axis-symmetric case is given, in Cartesian coordinates, by: v (v )v p t v = 0, div v = 0 v(x, 0) L 2 (R 3 ) C (R 3 ) v = (v 1 (x, t), v 2 (x, t), v 3 (x, t)) : R 3 [0, T ] R 3 is the unknown velocity field. p = p(x, t) is the scalar-valued pressure.

The Axially Symmetric Navier-Stokes Equations Derivation Converting to cylindrical coordinates by letting (x 1, x 2, x 3 ) = (r cos θ, r sin θ, z), we can rewrite the velocity field v as: v(x, t) = v r (r, z, t)e r + v θ (r, z, t)e θ + v z (r, z, t)e z

The Axially Symmetric Navier-Stokes Equations Derivation Converting to cylindrical coordinates by letting (x 1, x 2, x 3 ) = (r cos θ, r sin θ, z), we can rewrite the velocity field v as: v(x, t) = v r (r, z, t)e r + v θ (r, z, t)e θ + v z (r, z, t)e z where ( x1 e r = r, x ) 2 r, 0 = (cos θ, sin θ, 0) ( e θ = x 2 r, x ) 1 r, 0 = ( sin θ, cos θ, 0) e z = (0, 0, 1) are the basis vectors for R 3 in cylindrical coordinates,

The Axially Symmetric Navier-Stokes Equations Derivation Converting to cylindrical coordinates by letting (x 1, x 2, x 3 ) = (r cos θ, r sin θ, z), we can rewrite the velocity field v as: v(x, t) = v r (r, z, t)e r + v θ (r, z, t)e θ + v z (r, z, t)e z where ( x1 e r = r, x ) 2 r, 0 = (cos θ, sin θ, 0) ( e θ = x 2 r, x ) 1 r, 0 = ( sin θ, cos θ, 0) e z = (0, 0, 1) are the basis vectors for R 3 in cylindrical coordinates, and v r = v e r, v θ = v e θ, v z = v e z.

The Axially Symmetric Navier-Stokes Equations Derivation Utilizing the Chain Rule and the cylindrical gradient and Laplacian operators: = e r r + e 1 θ r θ + e z z = 2 r 2 + 1 r r + 1 2 r 2 θ 2 + 2 z 2 we can derive the ASNSE:

The Axially Symmetric Navier-Stokes Equations Derivation Utilizing the Chain Rule and the cylindrical gradient and Laplacian operators: = e r r + e 1 θ r θ + e z z = 2 r 2 + 1 r r + 1 2 r 2 θ 2 + 2 z 2 we can derive the ASNSE: ( 1 ) r ( 2 1 r 2 v r (b )v r + v 2 θ r ) v θ (b )v θ + v θv r r v z (b )v z p r v r = 0 t 1 (rv r ) + v z r r z = 0 p r v r t v r t = 0 = 0

The Axially Symmetric Navier-Stokes Equations Key Features ( 1 ) r ( 2 1 r 2 v r (b )v r + v 2 θ r ) v θ (b )v θ + v θv r r v z (b )v z p r v r = 0 t 1 (rv r ) + v z r r z = 0 p r v r t v r t = 0 = 0 The last equation is the continuity equation.

The Axially Symmetric Navier-Stokes Equations Key Features ( 1 ) r ( 2 1 r 2 v r (b )v r + v 2 θ r ) v θ (b )v θ + v θv r r v z (b )v z p r v r = 0 t 1 (rv r ) + v z r r z = 0 p r v r t v r t = 0 = 0 The last equation is the continuity equation. b(x, t) = (v r, 0, v z )

The Axially Symmetric Navier-Stokes Equations Key Features ( 1 ) r ( 2 1 r 2 v r (b )v r + v 2 θ r ) v θ (b )v θ + v θv r r v z (b )v z p r v r = 0 t 1 (rv r ) + v z r r z = 0 p r v r t v r t = 0 = 0 The last equation is the continuity equation. b(x, t) = (v r, 0, v z ) No θ derivative terms (solutions are axis-symmetric).

The Axially Symmetric Navier-Stokes Equations Key Features ( 1 ) r ( 2 1 r 2 v r (b )v r + v 2 θ r ) v θ (b )v θ + v θv r r v z (b )v z p r v r = 0 t 1 (rv r ) + v z r r z = 0 p r v r t v r t = 0 = 0 The last equation is the continuity equation. b(x, t) = (v r, 0, v z ) No θ derivative terms (solutions are axis-symmetric). No pressure term in second equation.

The Vorticity Equations Derivation We can take the cylindrical curl of the ASNSE to derive the Vorticity Equations. Let ω = curl v.

The Vorticity Equations Derivation We can take the cylindrical curl of the ASNSE to derive the Vorticity Equations. Let ω = curl v. Then ω(x, t) = ω r e r + ω θ e θ + ω z e z ω r = v θ z, ω θ = v r z v z r ω z = v θ r + v θ r

The Vorticity Equations Derivation We can take the cylindrical curl of the ASNSE to derive the Vorticity Equations. Let ω = curl v. Then ω(x, t) = ω r e r + ω θ e θ + ω z e z ω r = v θ z, ω θ = v r z v z r ω z = v θ r The corresponding vorticity equations are ( 1 ) r ( 2 1 ) r 2 ω θ (b )ω θ + 2 v θ r + v θ r v r ω r (b )ω r + ω r r + ω v r z z ω r t ω z (b )ω z + ω z v z z + ω r v θ z + ω v r θ r ω θ t v z r ω z t = 0 = 0 = 0

Preliminaries A few known results on the ASNSE Late 1960 s: Ladyzhenskaya et al showed the no-swirl case (v 0,θ = 0) has long-time existence and uniqueness of strong solutions.

Preliminaries A few known results on the ASNSE Late 1960 s: Ladyzhenskaya et al showed the no-swirl case (v 0,θ = 0) has long-time existence and uniqueness of strong solutions. 1982: Caffarelli, Kohn, Nirenberg showed blow-up could only occur on the axis of symmetry.

Preliminaries A few known results on the ASNSE Late 1960 s: Ladyzhenskaya et al showed the no-swirl case (v 0,θ = 0) has long-time existence and uniqueness of strong solutions. 1982: Caffarelli, Kohn, Nirenberg showed blow-up could only occur on the axis of symmetry. 2008: Chen, Strain, Tsai, Yau showed that if v(x, t) C then solutions are smooth. r 2 + t

Preliminaries A few known results on the ASNSE Late 1960 s: Ladyzhenskaya et al showed the no-swirl case (v 0,θ = 0) has long-time existence and uniqueness of strong solutions. 1982: Caffarelli, Kohn, Nirenberg showed blow-up could only occur on the axis of symmetry. 2008: Chen, Strain, Tsai, Yau showed that if v(x, t) C then solutions are smooth. r 2 + t 2008: Koch, Nadirashvili, Seregin, Sverák weakened the bound on v to v(x, t) C r.

Preliminaries A few known results on the ASNSE Late 1960 s: Ladyzhenskaya et al showed the no-swirl case (v 0,θ = 0) has long-time existence and uniqueness of strong solutions. 1982: Caffarelli, Kohn, Nirenberg showed blow-up could only occur on the axis of symmetry. 2008: Chen, Strain, Tsai, Yau showed that if v(x, t) C then solutions are smooth. r 2 + t 2008: Koch, Nadirashvili, Seregin, Sverák weakened the bound on v to v(x, t) C r. 2009: Burke, Zhang proved an a priori bound on ω θ and used it to derive bounds on ω r and ω z.

Preliminaries Notation and Scaling Let R > 0, S > 0, and 0 < A < B be constants.

Preliminaries Notation and Scaling Let R > 0, S > 0, and 0 < A < B be constants. Define P A,B,R = C A,B,R ( R 2, 0),

Preliminaries Notation and Scaling Let R > 0, S > 0, and 0 < A < B be constants. Define P A,B,R = C A,B,R ( R 2, 0), where C A,B,R = {(x 1, x 2, x 3 ) R 3 : AR r BR, 0 θ 2π, z BR} R 3

Preliminaries Notation and Scaling Let R > 0, S > 0, and 0 < A < B be constants. Define P A,B,R = C A,B,R ( R 2, 0), where C A,B,R = {(x 1, x 2, x 3 ) R 3 : AR r BR, 0 θ 2π, z BR} R 3 Let k > 0.

Preliminaries Notation and Scaling Let R > 0, S > 0, and 0 < A < B be constants. Define P A,B,R = C A,B,R ( R 2, 0), where C A,B,R = {(x 1, x 2, x 3 ) R 3 : AR r BR, 0 θ 2π, z BR} R 3 Let k > 0. Then (ṽ(x, t), p(x, t)) is a solution to the Navier-Stokes equations, where ṽ(x, t) = kv(kx, k 2 t) p(x, t) = k 2 p(kx, k 2 t)

Preiliminaries Notation and Scaling For a small parabolic cylinder P 1,4,k about a point (x, t) on the x 3 -axis, we make the substitution: x = x k t = t k 2 ( x, t) P 1,4,1

Preiliminaries Notation and Scaling For a small parabolic cylinder P 1,4,k about a point (x, t) on the x 3 -axis, we make the substitution: x = x k t = t k 2 ( x, t) P 1,4,1 That is, we blow up P 1,4,k to P 1,4,1 and do the analysis on scaled, hollowed-out cylinders (shells) on the scaled vorticity ω( x, t).

Preiliminaries Notation and Scaling For a small parabolic cylinder P 1,4,k about a point (x, t) on the x 3 -axis, we make the substitution: x = x k t = t k 2 ( x, t) P 1,4,1 That is, we blow up P 1,4,k to P 1,4,1 and do the analysis on scaled, hollowed-out cylinders (shells) on the scaled vorticity ω( x, t). After we are done, we shrink back to get the bound on the original vorticity ω(x, t).

Burke-Loftus, Zhang (2009) Theorem Suppose v is a smooth, axially symmetric solution of the three-dimensional Navier-Stokes equations in R 3 ( T, 0) with initial data v 0 = v(, T ) L 2 (R 3 ), and ω the vorticity. Assume further rv 0,θ L (R 3 ) and let 0 < R min{1, T 2 }. Then there exist constants, B 1 and B 2, depending only on the initial data, such that for all (x, t) P 2,3,R R 3 ( T, 0), (i) ω θ (x, t) B 1 ( x 2 1 + x 2 2 ) 5 (ii) ω r (x, t) + ω z (x, t) B 2 ( x 2 1 + x 2 2 ) 10

Proof of part (i): Step 1: Use a refined cut-off function. Let q 1 be rational and let 5 8 σ 2 < σ 1 1.

Proof of part (i): Step 1: Use a refined cut-off function. Let q 1 be rational and let 5 8 σ 2 < σ 1 1. Define P(σ i ) = C(σ i ) ( σ 2 i, 0) C(σ i ) = {(r, θ, t) (5 4σ i ) < r < 4σ i, 0 θ 2π, z < 4σ i }

Proof of part (i): Step 1: Use a refined cut-off function. Let q 1 be rational and let 5 8 σ 2 < σ 1 1. Define P(σ i ) = C(σ i ) ( σ 2 i, 0) C(σ i ) = {(r, θ, t) (5 4σ i ) < r < 4σ i, 0 θ 2π, z < 4σ i } Choose ψ = φ(y)η(s) such that: supp φ C(σ i ); 0 φ 1

Proof of part (i): Step 1: Use a refined cut-off function. Let q 1 be rational and let 5 8 σ 2 < σ 1 1. Define P(σ i ) = C(σ i ) ( σ 2 i, 0) C(σ i ) = {(r, θ, t) (5 4σ i ) < r < 4σ i, 0 θ 2π, z < 4σ i } Choose ψ = φ(y)η(s) such that: and supp φ C(σ i ); 0 φ 1 supp η ( σ 2 1, 0]; 0 η 1

Proof of part (i): Step 1: Use a refined cut-off function. Let Λ = v θ L (P 1,4,1 ) rv 0,θ L (R 3 ) <

Proof of part (i): Step 1: Use a refined cut-off function. Let Λ = v θ L (P 1,4,1 ) rv 0,θ L (R 3 ) < Let Ω + = { Ω(x, t) + Λ, Ω(x, t) 0 Λ, Ω(x, t) < 0

Proof of part (i): Step 1: Use a refined cut-off function. Let Λ = v θ L (P 1,4,1 ) rv 0,θ L (R 3 ) < Let Ω + = { Ω(x, t) + Λ, Ω(x, t) 0 Λ, Ω(x, t) < 0 By assumption, Ω is smooth, therefore we can compute Ω q + (b )Ω q + + 2 r r Ω q + t Ω q + = q(q 1)Ω q 2 + Ω + 2 qωq 1 + (vθ 2) r 2 z

Proof of part (i): Step 1: Use a refined cut-off function. Let f = Ω q + and choose f ψ 2 as the test function. Then

Proof of part (i): Step 1: Use a refined cut-off function. Let f = Ω q + and choose f ψ 2 as the test function. Then f (b )f + 2 r r f t f = q(q 1)Ω q 2 + Ω + 2 qω q 1 + (vθ 2) r 2 z f

Proof of part (i): Step 1: Use a refined cut-off function. Let f = Ω q + and choose f ψ 2 as the test function. Then ( f (b )f + 2 r r f t f )f ψ 2 = (q(q 1)Ω q 2 + Ω + 2 )f ψ 2 ( qωq 1 + (vθ 2) r 2 f )f ψ2 z Multiply by f ψ 2

Proof of part (i): Step 1: Use a refined cut-off function. Let f = Ω q + and choose f ψ 2 as the test function. Then P(σ 1 ) ( f (b )f + 2 r r f t f )f ψ 2 dyds = Multiply by f ψ 2 P(σ 1 ) (q(q 1)Ω q 2 + Ω + 2 )f ψ 2 dyds P(σ 1 ) Integrate over P(σ 1 ) ( qωq 1 + (vθ 2) r 2 z f )f ψ2 dyds

Proof of part (i): Step 1: Use a refined cut-off function. Let f = Ω q + and choose f ψ 2 as the test function. Then P(σ 1 ) ( f (b )f + 2 r r f s f )f ψ 2 dyds = Multiply by f ψ 2 P(σ 1 ) (q(q 1)Ω q 2 + Ω + 2 )f ψ 2 dyds P(σ 1 ) ( qωq 1 + (vθ 2) r 2 z f )f ψ2 dyds Integrate over P(σ 1 ). Replace t by s.

Proof of part (i): Step 1: Use a refined cut-off function. Next, integrate the first term and the time derivative term by parts, utilize the product rule, and use the cut-off function to arrive at:

Proof of part (i): Step 1: Use a refined cut-off function. Next, integrate the first term and the time derivative term by parts, utilize the product rule, and use the cut-off function to arrive at: (f ψ) 2 dyds + 1 f 2 (y, 0)φ 2 (y) dy 2 P(σ 1 ) C(σ 1 )

Proof of part (i): Step 1: Use a refined cut-off function. Next, integrate the first term and the time derivative term by parts, utilize the product rule, and use the cut-off function to arrive at: (f ψ) 2 dyds + 1 f 2 (y, 0)φ 2 (y) dy P(σ 1 ) 2 C(σ 1 ) b f (f ψ 2 ) dyds P(σ 1 )

Proof of part (i): Step 1: Use a refined cut-off function. Next, integrate the first term and the time derivative term by parts, utilize the product rule, and use the cut-off function to arrive at: (f ψ) 2 dyds + 1 f 2 (y, 0)φ 2 (y) dy P(σ 1 ) 2 C(σ 1 ) b f (f ψ 2 ) dyds + (η s η + ψ 2 )f 2 dyds P(σ 1 ) P(σ 1 )

Proof of part (i): Step 1: Use a refined cut-off function. Next, integrate the first term and the time derivative term by parts, utilize the product rule, and use the cut-off function to arrive at: (f ψ) 2 dyds + 1 f 2 (y, 0)φ 2 (y) dy P(σ 1 ) 2 C(σ 1 ) b f (f ψ 2 ) dyds + (η s η + ψ 2 )f 2 dyds P(σ 1 ) + P(σ 1 ) 2 r r f (f ψ 2 ) dyds P(σ 1 )

Proof of part (i): Step 1: Use a refined cut-off function. Next, integrate the first term and the time derivative term by parts, utilize the product rule, and use the cut-off function to arrive at: (f ψ) 2 dyds + 1 f 2 (y, 0)φ 2 (y) dy P(σ 1 ) 2 C(σ 1 ) b f (f ψ 2 ) dyds + (η s η + ψ 2 )f 2 dyds P(σ 1 ) + P(σ 1 ) 2 r r f (f ψ 2 ) dyds + P(σ 1 ) P(σ 1 ) qω 2q 1 + (vθ 2) r 2 z ψ2 dyds

Proof of part (i): Step 1: Use a refined cut-off function. Next, integrate the first term and the time derivative term by parts, utilize the product rule, and use the cut-off function to arrive at: (f ψ) 2 dyds + 1 f 2 (y, 0)φ 2 (y) dy P(σ 1 ) 2 C(σ 1 ) b f (f ψ 2 ) dyds + (η s η + ψ 2 )f 2 dyds P(σ 1 ) + P(σ 1 ) 2 r r f (f ψ 2 ) dyds + P(σ 1 ) P(σ 1 ) qω 2q 1 + (vθ 2) r 2 z ψ2 dyds := T 1 + T 2 + T 3 + T 4 = (Step 2) + (Step 3) + (Step 4) + (Step 5) of the proof.

Proof of part (i) After the Estimates are done. From these steps, and after converting f back to Ω q + we get the "reversed Hölder s Inequality": [ Ω 2qγ Cq 2 ] γ + dyds (σ 1 σ 2 ) 4 Ω 2q + dyds P(σ 2 ) P(σ 1 )

Proof of part (i) After the Estimates are done. From these steps, and after converting f back to Ω q + we get the "reversed Hölder s Inequality": [ Ω 2qγ Cq 2 ] γ + dyds (σ 1 σ 2 ) 4 Ω 2q + dyds P(σ 2 ) which generalizes to P(σ i+1 ) Ω 2γi+1 + dyds [c i+2 γ 2i P(σ 1 ) Ω 2γi + dyds P(σ i ) ] γ

Proof of part (i) After the Estimates are done. From these steps, and after converting f back to Ω q + we get the "reversed Hölder s Inequality": [ Ω 2qγ Cq 2 ] γ + dyds (σ 1 σ 2 ) 4 Ω 2q + dyds P(σ 2 ) which generalizes to P(σ i+1 ) γ > 1, σ i σ i 1 = τ i Ω 2γi+1 + dyds [c i+2 γ 2i P(σ 1 ) Ω 2γi + dyds P(σ i ) ] γ

Proof of part (i) After the Estimates are done. From these steps, and after converting f back to Ω q + we get the "reversed Hölder s Inequality": [ Ω 2qγ Cq 2 ] γ + dyds (σ 1 σ 2 ) 4 Ω 2q + dyds P(σ 2 ) which generalizes to P(σ i+1 ) γ > 1, σ i σ i 1 = τ i Ω 2γi+1 + dyds Let τ i = 2 i 2, q = γ i. [c i+2 γ 2i P(σ 1 ) Ω 2γi + dyds P(σ i ) ] γ

Proof of part (i) Estimate on Solutions via Moser s Iteration After i iterations of taking the 1 γ -th power and using the previous inequality, this is the result: ( Ω 2γi+1 + dyds P(σ i+1 ) ) 1 γ i+1 c γ j γ (j+1)γ j+1 P(σ 0 ) Ω 2 +dyds

Proof of part (i) Estimate on Solutions via Moser s Iteration After i iterations of taking the 1 γ -th power and using the previous inequality, this is the result: ( Ω 2γi+1 + dyds P(σ i+1 ) ) 1 γ i+1 c γ j γ (j+1)γ j+1 P(σ 0 ) Ω 2 +dyds SEND i, and all the series converge, and we have at last (noting that this argument also holds for Ω and Ω = Ω + Ω ) sup Ω 2 C Ω 2 dyds P 2,3,1 P 1,4,1

Proof of part (i) Estimate on ω θ via rescaling. The previous estimate is really a statement about the scaled vorticity Ω( x, t) = ω θ( x, t) r.

Proof of part (i) Estimate on ω θ via rescaling. The previous estimate is really a statement about the scaled vorticity Ω( x, t) = ω θ( x, t) r. Back-substitute ω θ, x = x k, and t = t, then rescale (shrink k 2 down) to get the result: sup k 4 ωθ 2 (x, t) B 1 (x,t) P 2,3,k k 6

Proof of part (i) Estimate on ω θ via rescaling. The previous estimate is really a statement about the scaled vorticity Ω( x, t) = ω θ( x, t) r. Back-substitute ω θ, x = x k, and t = t, then rescale (shrink k 2 down) to get the result: sup k 4 ωθ 2 (x, t) B 1 (x,t) P 2,3,k k 6 Therefore ω θ (x, t) L (P 2,3,k ) B 1 k 10

Thank you!

(32) References 1. O.A. Ladyzhenskaya, Unique global solvability of the three-dimensional Cuachy problem for the Navier-Stokes equations in the presence of axial symmetry (1968) 2. D. Chae and J. Lee, On the regularity of the axisymmetric solutions of the Navier-Stokes equations (2002) 3. L. Caffarelli, R. Kohn, and L.Nirenberg, Partial regularity of suitable weak solutions of the Navier-Stokes equations (1982) 4. Chiun-Chuan Chen, Robert M. Strain, Tai-Peng Tsai, and Horng-Tzer Yau, Lower bound on the blow-up rate of the axisymmetric Navier-Stokes equations (2008) 5. G. Koch, N. Nadirashvili, G. Seregin, and V. Sverak, Liouville thereoms for the Navier-Stokes equations and applications (2008)

(32) References 6. G. Seregin and V. Sverak, On Type I singularities of the local axi-symmetric solutions of the Navier-Stokes equations (2008) 7. Jennifer Burke and Qi S. Zhang, A priori bounds for the vorticity of axis symmetric solutions to the Navier-Stokes equations(2010) 8. Qi S. Zhang and Zheng Lei, Improved a priori bounds for the axially symmetric solutions to the Navier-Stokes equations (Preliminary)(2011)