Relation between Distributional and Leray-Hopf Solutions to the Navier-Stokes Equations
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1 Relation between Distributional and Leray-Hopf Solutions to the Navier-Stokes Equations Giovanni P. Galdi Department of Mechanical Engineering & Materials Science and Department of Mathematics University of Pittsburgh Prague, August
2 Preliminary Considerations Cauchy problem for the Navier-Stokes equations: } t v + v v = v p in R 3 (0, ) div v = 0 v(x, 0) = v 0 (x) x R 3.
3 Preliminary Considerations Cauchy problem for the Navier-Stokes equations: } t v + v v = v p in R 3 (0, ) div v = 0 v(x, 0) = v 0 (x) x R 3. Leray-Hopf solutions: For any v 0 L 2 σ(r 3 ) there is v L (0, T ; L 2 σ) L 2 (0, T ; H 1 ), all T > 0 lim t 0 + v(t) v 0 2 = 0 t v(x, t) v(x, τ) 2 v 0 (x) 2, R 3 0 R 3 R 3 all t [0, T ].
4 Preliminary Considerations Extra Conditions (classical)
5 Preliminary Considerations Extra Conditions (classical) v L 4 (0, T ; L 4 ) (A) Energy Equality (Prodi-Lions)
6 Preliminary Considerations Extra Conditions (classical) v L 4 (0, T ; L 4 ) (A) Energy Equality (Prodi-Lions) v L 2s s 3 (0, T ; L s ), s > 3, (B), or v C([0, T ]; L 3 ), (C) Uniqueness/Regularity (Prodi-Serrin-Ladyzhenskaya)
7 Main Results In words:
8 Main Results In words: v is any (distributional) solutions to the Navier-Stokes equations with initial data in L 2 σ (assumed in a suitable sense) ; v satisfies one of the conditions (A), (B) or (C).
9 Main Results In words: v is any (distributional) solutions to the Navier-Stokes equations with initial data in L 2 σ (assumed in a suitable sense) ; v satisfies one of the conditions (A), (B) or (C). Then v is necessarily a Leray-Hopf solution.
10 A Remark In general, classes defined by (A), (B) or (C) do not contain nor are they contained in the Leray-Hopf s.
11 A Remark In general, classes defined by (A), (B) or (C) do not contain nor are they contained in the Leray-Hopf s. In fact, if merely v {(A), (B), (C)}, both total kinetic energy and dissipation t v(x, t) 2, v(x, τ) 2, R 3 0 R 3 are infinite at all times.
12 A Remark In general, classes defined by (A), (B) or (C) do not contain nor are they contained in the Leray-Hopf s. In fact, if merely v {(A), (B), (C)}, both total kinetic energy and dissipation t v(x, t) 2, v(x, τ) 2, R 3 0 R 3 are infinite at all times. Our result implies, in particular, that this cannot happen if only the kinetic energy of the initial datum is finite.
13 Main Results: Precise Statement
14 Main Results: Precise Statement Definition. Let v 0 L 2 σ. Then v L 2 loc (R3 [0, T )) is a distributional solution (corresponding to v 0 ) if T v ( t ϕ + ϕ + v ϕ) = v 0 ϕ(0) 0 R 3 R 3 T v φ = 0, 0 R 3 for all ϕ D T := {ϕ C 0 (R 3 [0, T )) : div ϕ = 0}, and φ C 0 (R 3 (0, T )).
15 Main Results: Precise Statement Theorem 1. Let v be a distributional solution satisfying one of the following conditions: v L 4 (0, T ; L 4 ) ; v L 2s s 3 (0, T ; L s ), s > 3 ; v C([0, T ]; L 3 ). Then v is a Leray-Hopf solution.
16 Main Results: Precise Statement Theorem 2. Let v be a distributional solution satisfying one of the following conditions: v L 2s s 3 (δ, T ; L s ), s > 3 ; v C([δ, T ]; L 3 ), all small δ > 0, and also lim t 0 + R 3 (v(x, t) v 0 (x)) Φ(x)dx = 0, all Φ L 2 σ.
17 Main Results: Precise Statement Theorem 2. Let v be a distributional solution satisfying one of the following conditions: v L 2s s 3 (δ, T ; L s ), s > 3 ; v C([δ, T ]; L 3 ), all small δ > 0, and also lim t 0 + R 3 (v(x, t) v 0 (x)) Φ(x)dx = 0, all Φ L 2 σ. Then v is a Leray-Hopf solution.
18 Main Results: Precise Statement Theorem 3. Let v be a distributional solution satisfying v L 4 (δ, T ; L 4 ), all small δ > 0, and also lim t 0 + v(t) v 0 2 = 0.
19 Main Results: Precise Statement Theorem 3. Let v be a distributional solution satisfying v L 4 (δ, T ; L 4 ), all small δ > 0, and also lim t 0 + v(t) v 0 2 = 0. Then v is a Leray-Hopf solution.
20 Some Relevant Consequences
21 Some Relevant Consequences For example:
22 Some Relevant Consequences For example: Corollary 1. If v is a distributional solution such that v L 4 (0, T ; L 4 ), then v satisfies the energy equality, and also v C([0, T ]; L 2 ).
23 Some Relevant Consequences For example: Corollary 1. If v is a distributional solution such that v L 4 (0, T ; L 4 ), then v satisfies the energy equality, and also v C([0, T ]; L 2 ). Corollary 2. If v is a distributional solution satisfying one of the conditions v L 2s s 3 (0, T ; L s ), s > 3 ; v C([0, T ]; L 3 ), then v is the unique Leray-Hopf solution corresponding to v 0, and also v C ((0, T ] R 3 ).
24 Some Relevant Consequences Corollary 3. If v is a distributional solution such that v L 4 (δ, T ; L 4 ), all small δ > 0, and v(t) v 0 2 0, as t 0 +, then v satisfies the energy equality, and also v C([0, T ]; L 2 ).
25 Some Relevant Consequences Corollary 3. If v is a distributional solution such that v L 4 (δ, T ; L 4 ), all small δ > 0, and v(t) v 0 2 0, as t 0 +, then v satisfies the energy equality, and also v C([0, T ]; L 2 ). Corollary 4. If v is a distributional solution satisfying one of the conditions for all small δ > 0 v L 2s s 3 (δ, T ; L s ), s > 3 ; v C([0, δ, T ]; L 3 ), and v(t) v 0 weakly in L 2 σ as t 0 +, then v is a Leray-Hopf solution, so that v C ((0, T ] R 3 ).
26 Some Remarks
27 Some Remarks 1. Investigation in the class v L 4 (0, T ; L 4 ) seems to be entirely new (no previous literature);
28 Some Remarks 1. Investigation in the class v L 4 (0, T ; L 4 ) seems to be entirely new (no previous literature); 2. Investigation in either class v L 2s s 3 (0, T ; L s ), s > 3 ; v C([0, T ]; L 3 ), falls, instead, in the framework of very weak (or mild ) solutions, and has been performed extensively.
29 Some Remarks FOIAS (1961): v L 2 (0, T ; H 1 ) L r (0, T ; L s ), s>3, r > 2s s 3 ; v 0 L 2 σ.
30 Some Remarks FOIAS (1961): v L 2 (0, T ; H 1 ) L r (0, T ; L s ), s>3, r > FABES, JOHNS & RIVIERE (1972): v L 2s s 3 (0, T ; L s ), s > 3, v 0 L 2 σ L s. 2s s 3 ; v 0 L 2 σ.
31 Some Remarks FOIAS (1961): v L 2 (0, T ; H 1 ) L r (0, T ; L s ), s>3, r > FABES, JOHNS & RIVIERE (1972): v L 2s s 3 (0, T ; L s ), s > 3, v 0 L 2 σ L s. KATO (1984), GIGA (1986): 2s s 3 ; v 0 L 2 σ. v C([0, T ]; L 3 ) L 2s s 3 (0, T ; L s ), s > 3, v 0 L 2 σ L 3. ( ) lim t s 3 2s v(t) s = 0. t 0
32 Some Remarks 3. BUCKMASTER & VICOL (2017/2018), LUO (2018) show non-uniqueness in the class of distributional solutions defined by v C([0, T ]; H β ), β > 0 small, and v 0 L 2 σ. having, therefore, infinite dissipation.
33 Some Remarks 3. BUCKMASTER & VICOL (2017/2018), LUO (2018) show non-uniqueness in the class of distributional solutions defined by v C([0, T ]; H β ), β > 0 small, and v 0 L 2 σ. having, therefore, infinite dissipation. Our results imply, in particular, that any possible non-uniqueness result in that class could only hold with β < 1/2.
34 Some Remarks 4. It is probable that the borderline assumption v L (0, T ; L 3 ) continues to ensure that the distributional solution v is in the Leray-Hopf class.
35 Some Remarks 4. It is probable that the borderline assumption v L (0, T ; L 3 ) continues to ensure that the distributional solution v is in the Leray-Hopf class. 5. Our results continue to hold in (1) bounded or (2) exterior domain (no-slip conditions), and (3) in a torus (periodic boundary conditions). The proof in cases (1), (2) becomes much more technical.
36 Basic Ideas
37 Basic Ideas Step 1. For the given distributional solution v, consider the linear Cauchy problem } t u + v u = u p in R 3 (0, T ) div u = 0 u(x, 0) = v 0 (x) x R 3.
38 Basic Ideas Step 1. For the given distributional solution v, consider the linear Cauchy problem } t u + v u = u p in R 3 (0, T ) div u = 0 u(x, 0) = v 0 (x) x R 3. By standard Galerkin method one shows the existence of a solution u L (0, T ; L 2 σ) L 2 (0, T ; H 1 ), u(t) v 0 in L 2.
39 Basic Ideas Set then for all T 0 T 0 w := v u w ( t ϕ + ϕ + v ϕ) = 0 R 3 w φ = 0, R 3 ϕ D T := {ϕ C 0 (R 3 [0, T )) : div ϕ = 0}, and φ C 0 (R 3 (0, T )).
40 Basic Ideas Step 2. Replace t ϕ + ϕ + v ϕ, ϕ D T with arbitrary f C 0 (R 3 (0, T )).
41 Basic Ideas Step 2. Replace t ϕ + ϕ + v ϕ, ϕ D T with arbitrary f C 0 (R 3 (0, T )). In which case, we get T 0 R 3 w f = 0, for all f C 0 (R 3 (0, T )), implying that v u is a Leray-Hopf solution.
42 Basic Ideas Step 2. Replace t ϕ + ϕ + v ϕ, ϕ D T with arbitrary f C 0 (R 3 (0, T )). In which case, we get T 0 R 3 w f = 0, for all f C 0 (R 3 (0, T )), implying that v u is a Leray-Hopf solution. This step can be achieved if v is in one of the classes (A), (B) or (C).
43 Basic Ideas Implementation of Step 2, under the assumption v L r (0, T ; L s ), r := 2s s 3, s > 3.
44 Basic Ideas Implementation of Step 2, under the assumption v L r (0, T ; L s ), r := 2s s 3, s > 3. Maximal regularity class (p, q (1, )): W p,q := { ψ W 1,p (0, T ; L q σ) L p (0, T ; W 2,q ) }.
45 Basic Ideas Implementation of Step 2, under the assumption v L r (0, T ; L s ), r := 2s s 3, s > 3. Maximal regularity class (p, q (1, )): W p,q := { ψ W 1,p (0, T ; L q σ) L p (0, T ; W 2,q ) }. Lemma 1. (Technical) Test space D T is dense in { ψ W p 1,q 1 W p 2,q 2 : ψ(t ) = 0 }, for any p i, q i (1, ), i = 1, 2.
46 Basic Ideas Recall that (r := 2s/(s 3)) w := v u L r (0, T ; L s ) + L (0, T ; L 2 σ).
47 Basic Ideas Recall that (r := 2s/(s 3)) w := v u L r (0, T ; L s ) + L (0, T ; L 2 σ). Lemma 2. (Embeddings + Hölder) The linear form T ψ W 1,2 W r,s v ψ w R 0 R 3 is continuous (r := r/(r 1), s := s/(s 1).)
48 Basic Ideas As a consequence, for all implies T 0 T 0 w ( t ϕ + ϕ + v ϕ) = 0 R 3 ϕ D T, R 3 w ( t ψ + ψ + v ψ) = 0 for all ψ W 1,2 W r,s, with ψ(t ) = 0.
49 Basic Ideas As a consequence, for all implies for all T 0 T 0 R 3 w ( t ϕ + ϕ + v ϕ) = 0 R 3 w ϕ D T, =f C 0 (R 3 (0, T )) {}}{ ( t ψ + ψ + v ψ) = 0 ψ W 1,2 W r,s, with ψ(t ) = 0.
50 Basic Ideas Lemma 3. Let α L r (0, T ; L s σ) L r,s, r := 2s s 3. For any F C0 (R 3 (0, T )), the problem } t Ψ α Ψ = Ψ p + F in R 3 (0, T ), div Ψ = 0 Ψ(x, 0) = 0 x R 3, has at least one solution Ψ W 1,2 W r,s,.
51 Basic Ideas Sketch of the Proof. It is based on two a priori estimates, we can formally derive.
52 Basic Ideas Sketch of the Proof. It is based on two a priori estimates, we can formally derive. Estimate in W r,s. For λ > 0, set ζ = e λt Ψ, ρ = e λt p, and G = e λt F + α ζ.
53 Basic Ideas Sketch of the Proof. It is based on two a priori estimates, we can formally derive. Estimate in W r,s. For λ > 0, set ζ = e λt Ψ, ρ = e λt p, and G = e λt F + α ζ. We get the Stokes problem } t ζ + λ ζ = ζ ρ + G in R 3 (0, T ), div ζ = 0 ζ(x, 0) = 0 x R 3,
54 Basic Ideas Maximal regularity (λ large ): ζ W r,s + λ ζ L r,s c G L r,s c ( α ζ L r,s + F L r,s ).
55 Basic Ideas Maximal regularity (λ large ): ζ W r,s + λ ζ L r,s c G L r,s c ( α ζ L r,s + F L r,s ). Split α = α 1 + α 2 with α 1 L r,s < ε, α 2 L,. We get (Hölder + embeddings + Ehrling) α ζ L r,s ε ζ L r r 2, s s 2 + α 2 L, ζ L r,s c ε ζ W r,s + c ε α 2 L, ζ L r,s.
56 Basic Ideas Maximal regularity (λ large ): ζ W r,s + λ ζ L r,s c G L r,s c ( α ζ L r,s + F L r,s ). Split α = α 1 + α 2 with α 1 L r,s < ε, α 2 L,. We get (Hölder + embeddings + Ehrling) α ζ L r,s ε ζ L r r 2, s s 2 + α 2 L, ζ L r,s c ε ζ W r,s + c ε α 2 L, ζ L r,s. Therefore: ζ W r,s c F L r,s.
57 Basic Ideas Estimate in W 1,2. It is entirely classical.
58 Basic Ideas Estimate in W 1,2. It is entirely classical. Test the original equation with Ψ, Ψ, and t Ψ: 1 d 2 dt Ψ Ψ 2 2 = F Ψ R 3 1 d 2 dt Ψ Ψ 2 2 = (α Ψ F ) Ψ R 3 1 d 2 dt Ψ t Ψ 2 2 = (α Ψ F ) t Ψ R 3
59 Basic Ideas Estimate in W 1,2. It is entirely classical. Test the original equation with Ψ, Ψ, and t Ψ: 1 d 2 dt Ψ Ψ 2 2 = F Ψ R 3 1 d 2 dt Ψ Ψ 2 2 = (α Ψ F ) Ψ R 3 1 d 2 dt Ψ t Ψ 2 2 = (α Ψ F ) t Ψ R 3 Assumption on α + Sobolev lead to Ψ W 1,2 c F L 2,2. QED
60 Basic Ideas Proof of Leray-Hopf property.
61 Basic Ideas Proof of Leray-Hopf property. For a given f C 0 (R 3 (0, T )) we use the previous lemma with F (t) := f(t t), α(t) := v(t t)
62 Basic Ideas Proof of Leray-Hopf property. For a given f C 0 (R 3 (0, T )) we use the previous lemma with F (t) := f(t t), α(t) := v(t t) Thus, Ψ(T t) ψ(t) W 1,2 W r,s, and solves the final-value problem } t ψ + v ψ + ψ = p + f in R 3 (0, T ), div ψ = 0 ψ(x, T ) = 0 x R 3,
63 Basic Ideas Recalling that w := v u satisfies T 0 R 3 w ( t ψ + ψ + v ψ) = 0 for all ψ W 1,2 W r,s, with ψ(t ) = 0, we get T 0 R 3 (v u) f = 0, for all f C 0 (R 3 (0, T )) that is v u. QED
64 Basic Ideas Sketch of the proof under the alternate assumption v L 2s s 3 (δ, T ; L s ), s > 3 ; v v 0 weakly in L 2.
65 Basic Ideas Sketch of the proof under the alternate assumption v L 2s s 3 (δ, T ; L s ), s > 3 ; v v 0 weakly in L 2. Let { 1 if t 2δ χ δ (t) = 0 if t δ, χ δ c δ.
66 Basic Ideas Sketch of the proof under the alternate assumption v L 2s s 3 (δ, T ; L s ), s > 3 ; v v 0 weakly in L 2. Let χ δ (t) = { 1 if t 2δ 0 if t δ, χ δ c δ. One can show (w := v u) T δ χ δ R 3 w ( t ψ + ψ + v ψ) = for all ψ W 1,2 W r,s, ψ(t ) = 0. 2δ χ δ δ R 3 w ψ,
67 Basic Ideas Reasoning as in the case δ = 0, T 2δ χ δ w f = χ δ R δ w ψ, f C0 (R 3 (0, T )). 3 δ R 3
68 Basic Ideas Reasoning as in the case δ = 0, T 2δ χ δ w f = χ δ R δ w ψ, f C0 (R 3 (0, T )). 3 δ R 3 Letting δ 0, T 0 R 3 w f = lim δ 0 2δ χ δ δ R 3 w ψ
69 Basic Ideas Reasoning as in the case δ = 0, T 2δ χ δ w f = χ δ R δ w ψ, f C0 (R 3 (0, T )). 3 δ R 3 Letting δ 0, because T 0 w f = lim R 3 δ 0 = 0 2δ χ δ δ R 3 w ψ χ δ δ 1, w 0 weakly in L 2, ψ C([0, T ]; L 2 ). QED
70 Entirely similar arguments can be employed to draw the same conclusions if either or v L 4 (0, T ; L 4 ) [resp. v L 4 (δ, T ; L 4 )], v C([0, T ]; L 3 ) [resp. v C([δ, T ]; L 3 )]. with the mentioned conditions on the attainability of the initial datum v 0.
71 Entirely similar arguments can be employed to draw the same conclusions if either or v L 4 (0, T ; L 4 ) [resp. v L 4 (δ, T ; L 4 )], v C([0, T ]; L 3 ) [resp. v C([δ, T ]; L 3 )]. with the mentioned conditions on the attainability of the initial datum v 0. The case remains open. v L (0, T ; L 3 )
72 Entirely similar arguments can be employed to draw the same conclusions if either or v L 4 (0, T ; L 4 ) [resp. v L 4 (δ, T ; L 4 )], v C([0, T ]; L 3 ) [resp. v C([δ, T ]; L 3 )]. with the mentioned conditions on the attainability of the initial datum v 0. The case v L (0, T ; L 3 ) remains open. Thank you
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