Proof of the existence (by a contradiction)

Transcription

1 November 6, 2013

2 ν v + (v )v + p = f in, divv = 0 in, v = h on, v velocity of the fluid, p -pressure. R n, n = 2, 3, multi-connected domain: (NS) S 2 S 1

3 Incompressibility of the fluid (divv = 0) implies the necessary compatibility condition for the solvability of problem (NS): h n ds = N j=1 Γ j h n ds = N F j = 0, j=1 n is a unit vector of the outward normal to. (F)

4 General Scheme Let A W 1,2 () be a divergence free extension diva = 0, A = h. u = v A is a weak solution of (NS) if u H() = {w W 2 1 () : divw = 0} and ν u ηdx + ((u + A) )u ηdx + = f ηdx ν A ηdx (u )A ηdx = (A )A ηdx η H(). Integral identity (4) is equivalent to an operator equation: u = Bu where B : H() H() is a compact operator.

5 In order to apply the Leray Schauder theorem we have to prove that all possible solutions of the equation u (λ) = λbu (λ) λ (0, 1] are bounded by a constant independent of λ. Equivalently, we can consider for λ [0, 1]: u (λ) ηdx+ λ ν ((u (λ) +A) )u (λ) ηdx+ λ (u (λ) )A ηdx ν = λ f ηdx λ A ηdx λν 1 (A )A ηdx η H(). Take η = u (λ) : u (λ) 2 dx + λ (u (λ) )A u (λ) dx ν = λ f u (λ) dx λ A u (λ) dx λ (A )A u (λ) dx. ν

6 Then u (λ) 2 dx C(A) + λ ν (u (λ) )u (λ) Adx. We have to prove u (λ) 2 dx C 1 (A). ( ) (i) Leray-Hopf construction is impossible. (ii) We have prove to prove (*) by getting a contradiction.

7 Main result Theorem [M. Korobkov, K. Pileckas, R. Russo, 2013] Assume that R 2 is a bounded domain with C 2 -smooth boundary. If f W 1,2 () and h W 3/2,2 ( ) satisfies condition (F), then problem (NS) admits at least one weak solution. h n ds = N j=1 Γ j h n ds = N F j = 0, j=1 (F) S 2 S 1

8 Leray s method Take the extension A W 1,2 () as a weak solution of the Stokes problem, i.e., diva = 0, A = h and ν A η dx = f η dx η H(). To prove the solvability it is sufficient to show that all possible solutions u (λ) H() of the integral identity ν u (λ) η ( dx+λ (u (λ) +A) ) u (λ) η ( dx+λ u (λ) ) A η dx ( ) = λ A η A dx η H() are uniformly bounded in H() (with respect to λ [0, 1]). Assume this is false. Then there exist sequences {λ k } [0, 1] and {u (λk) } k N = {u k } k N H() such that lim λ k = λ 0 (0, 1], k lim J k = lim u k H() =. k k

9 Leray s method The pair ( û k = 1 J k u k, p k = 1 ) p λ k Jk 2 k satisfies the following system ν k û k + ( û k ) û k + p k = f k in, divû k = 0 in, û k = h k on, where ν k = (λ k J k ) 1 ν, f k = λ kνk 2 f, h ν 2 k = λ kν k ν h. The norms û k W 1,2 () and p k W 1,q (), q [1, 2), are uniformly bounded, û k v in W 1,2 (), p k p in W 1,q (). Moreover, û k W 3,2 loc (), p k W 2,2 loc ().

10 Leray s method (i) Take in the integral identity η = J 2 k u k. Passing to the limit as k l, we get ν λ 0 = ( v ) v A dx. (C1 ) (ii) Let ϕ J0 (). Take η = J 2 k l ϕ and let k l : ( ) v v ϕ dx = 0 ϕ J 0 ().

11 Leray s method Hence, the pair ( v, p ) satisfies for almost all x the Euler equations ( v ) v + p = 0, div v = 0, and Then v = 0. p j = p(x) Γj.

12 Leray s method Multiply Euler equations by A and integrate by parts: ( ) v v A dx = p A dx = p h n ds j=1 M M = p j h n ds = p j F j Γ j If either all F j = 0 or p 1 = = p M, we get ( ) v v A dx = 0. (C2 ) This contradicts (C 1 ): ( ) ν v v A dx = > 0. (C 1 ) λ 0 j=1

13 Obtaining a contradiction For simplicity consider the case There could appear two cases (a) The maximum of Φ is attained on the boundary : max{p 0, p 1 } = ess sup Φ(x). x (b) The maximum of Φ is not attained on the boundary : max{p 0, p 1 } < ess sup Φ(x). x

14 Consider the case (a). Let p 0 < 0, p 1 = 0. We have Φ(x) 0 in. Remark. Φ k satisfies elliptic equation Φ k 1 ν k div( Φ k û k ) = ω 2 k 1 ν k f k u k, ω k = 2 û 1k 1 û 2k. If f k = 0 Φ k satisfies maximum principle.

15 Obtaining a contradiction T f - family of all connected components of level sets of f, f is continuous. Then T f is a tree. Let K T ψ with diamk > 0. Take any x K \ A and put Φ(K) = Φ(x). This definition is correct by Bernoulli s Law. Lemma 1. Let A, B T ψ, A > 0, B > 0. Consider the corresponding arc [A, B] T ψ joining A to B. Then the restriction Φ [A,B] is a continuous function. Proof. Let C i (A, B) and C i C 0 in T ψ. By construction, each C i is a connected component of the level set of ψ and the sets A, B lie in different connected components of R 2 \ C i. Therefore, diam(c i ) min(diam(a), diam(b)) > 0. By the definition of convergence in T ψ, we have sup dist(x, C 0 ) 0 as i. x C i

16 Obtaining a contradiction Let Φ(x) = c i for a.a. x C i, Φ(x) = c 0 for a.a. x C 0. Assume c i c 0. Then c i c c 0. Moreover, components C i C 0 C 0 in Hausdorf metric (Blaschke selection theorem). Let L be a line; I 0 projection of C 0 on L. Obviously, I 0 - interval. For every z I 0 denote L z the line s.t. z L z and L z L. For almost all z the function Φ Lz is absolutely continuous. Fix such z. Then C i L z for sufficiently large i. Let y i C i L z. Extract a subsequence y il y 0 C 0. Then Φ(y il ) Φ(y 0 ) = c 0. Contradiction.

17 Geometrical construction Let B 0, B 1 T ψ, B 0 Γ 0, B 1 Γ 1. Set α = min Φ(C) < 0. C [B 0,B 1 ] Let t i (0, α), t i+1 = 1 2 t i and such that Φ(C) = t i C (B 0, B 1 ) is regular

18 Geometrical construction A 0 i is an element from the set {C [B 0, B 1 ] : Φ(C) = t i } which is closest to Γ 1. A 0 i is regular cycle.

19 Geometrical construction V i is the connected component of the set \ A 0 i such that Γ 1 V i. Obviously, V i V i+1, V i = A 0 i Γ 1. Remind that t i+1 = 1 2 t i. A 0 i are regular cycles. Therefore, Φ k A 0 i Φ A 0 i = t i, and for sufficiently large k holds Φ k A 0 i < 7 8 t i, Φk A 0 i+1 > 5 8 t i, k k i.

20 Geometrical construction Take t [ 5 8 t i, 7 8 t i]. W ik (t) is the connected component of the set {x V i \ V i+1 : Φk (x) > t} such that W ik (t) A 0 i+1. Put S ik (t) = ( W ik (t)) V i \ V i+1. Then Φ k Sik (t) = t, W ik (t) = S ik (t) A 0 i+1. Since Φ k W2,loc 2 (), by the Morse-Sard theorem for almost all t [ 5 8 t i, 7 8 t i] the level set S ik (t) consists of a finite number of C 1 -cycles; moreover, Φ k is differentiable at every point x S ik (t) and Φ k (x) 0. Such t we call (k, i)-regular.

21 Geometrical construction By construction S ik (t) Φ k nds = Φ k ds < 0, S ik (t) where n outward with respect to W ik (t) normal to W ik (t).

22 Geometrical construction Γ h = {x : dist(x, Γ 1 ) = h}, h = {x : dist(x, Γ 1 ) < h}. Γ h is C 1 -smooth and H(Γ h ) C 0 h (0, δ 0 ], C 0 = 3H(Γ 1 ).

23 Obtaining a contradiction Since Φ(x) const on V i, i.e., Φ(x) 0, from the identity Φ = ω ψ it follows that V i ω 2 dx > 0. From the weak convergence ω k ω in L 2 () we get the following Lemma 2. For any i N there exists ε i > 0, δ i (0, δ 0 ) and N such that k i V i+1 \ δi ω 2 kdx > ε i k k i.

24 Obtaining a contradiction The key step is the following estimate Lemma 3. For any i N there exists k(i) N such that there holds inequality Φ k (x) ds < Ft k k(i), for almost all t [ 5 8 t i, 7 8 t i]. S ik (t) The constant F is independents of t, k and i.

25 Obtaining a contradiction We receive the required contradiction using the Coarea formula. For i and k k(i) put E i = S ik (t). t [ 5 8 t i, 7 8 t i] By the Coarea formula for any integrable function g : E i R holds the equality E i g Φ k dx = 7 8 t i 5 8 t i S ik (t) g(x) dh 1 (x) dt. In particular, taking g = Φ k and using Lemma 3, we obtain E i Φ k 2 dx = 7 8 t i 5 8 t i S ik (t) Φ k dh 1 (x)dt 7 8 t i 5 8 t i Ft dt = F t 2 i.

26 Obtaining a contradiction Now, taking g = 1 in the Coarea formula and using the Hölder inequality we get 7 8 t i H 1( S ik (t) ) dt = Φ k dx 5 8 t i E i ( ) 1 Φ k 2 2 ( dx meas(ei ) ) 1 2 ( F t i meas(ei ) ) 1 2. E i By construction, for almost all t [ 5 8 t i, 7 8 t i] the set S ik (t) is a smooth cycle and S ik (t) separates A 0 i from A 0 i+1. Thus, each set S ik (t) separates Γ 0 from Γ 1. In particular, H 1 (S ik (t)) H 1 (Γ 1 ). Hence, So, it holds 7 8 t i 5 8 t i H 1( S ik (t) ) dt 1 4 H1 (Γ 1 )t i. 1 4 H1 (Γ 1 )t i F t i ( meas(ei ) ) 1 2,

27 Obtaining a contradiction or 1 4 H1 (Γ 1 ) F ( meas(e i ) ) 1 2. Since meas(e i ) meas ( V i \ V i+1 ) 0 as i, we obtain a contradiction!!!.

28 Proof of Lemma 3 Lemma 3. For any i N there exists k(i) N such that there holds inequality Φ k (x) ds < Ft k k(i) for almost all t [ 5 8 t i, 7 8 t i]. S ik (t) The constant F is independents of t, k and i. Proof. Fix i N and assume k k i : Φ k A 0 i < 7 8 t i, Φk A 0 i+1 > 5 8 t i, k k i. Take a sufficiently small σ > 0 and choose the parameter δ σ (0, δ i ] small enough to satisfy the following conditions: δσ A 0 i = δσ A 0 i+1 =, Γ h Φ 2 ds < 1 3 σ2 h (0, δ σ ], (1)

29 Proof of Lemma σ2 < Γ h Φ 2 k ds Φ 2 k ds < 1 Γ h 3 σ2 (2) h, h (0, δ σ ] k N. This estimate follows from the fact that for any q (1, 2) the norms Φ k W 1,q () are uniformly bounded. Hence, the norms Φ k Φ k L 6 5 () are uniformly bounded and Γ h Φ 2 k ds Γ h Φ 2 k ds 2 h \ h Φ k Φ k dx ( ) 5 2 Φ k Φ k 6/5 dx 6 meas(h \ h ) as h, h 0. h \ h

30 Proof of Lemma 3 From the weak convergence Φ k Φ in the space W 1,q (), q (1, 2), it follows that Φ k Γh Φ Γh as k for almost all h (0, δ σ ). Therefore, from (1) (2) we see that there exists k N such that Γ h Φ2 k ds < σ 2 h (0, δ σ ] k k. (3) For a function g W 2,2 () and for an arbitrary C 1 -cycle S we have by the Stokes theorem g n ds = g l ds = 0, S where l is the tangent vector to S. Since we have Φ k = ν k ω k + ω k û k + f k, S S Φ k n ds = S ω k û k n ds.

31 Proof of Lemma 3 Fix a sufficiently small ε > 0. For a given sufficiently large k k we make a special procedure to find a number h k (0, δ σ ) such that Φ k n ds = ω k û Γ hk Γ hk k n ds < ε, (4) û k 2 ds < C(ε)νk 2, (5) Γ hk where the constant C(ε) is independent of k and σ. Define a sequence of numbers 0 = h 0 < h 1 < h 2 <... by the recurrent formulas û k û k dx = νk 2, (6) U j where U j = {x : dist(x, Γ 1 ) (h j 1, h j )}.

32 Proof of Lemma 3 Since û k 2 ds = (λ kν k ) 2 ν 2 h 2 L 2 ( ), where λ k (0, 1], from (6) we get by induction that Γ h û k 2 ds Cjν 2 k h (h j 1, h j ), (7) where C is independent of k, j, σ. Therefore, Then νk 2 = û k û k dx U j Thus, we have U j û k 2 dx (h j h j 1 )Cjν 2 k. (8) ( ) 1 (h j h j 1 )Cjνk 2 û k 2 2 dx. U j νk 2 C j û k 2 dx. (9) h j h j 1 U j

33 Proof of Lemma 3 Define h j for j = 1,..., j max, where j max is the first index which satisfies one of the following two conditions: STOP CASE 1. h jmax 1 < δ σ, h jmax δ σ. STOP CASE 2. Cj max U jmax û k 2 dx < ε. By construction, U j û k 2 dx 1 Cj ε for every j < j max. Hence, 2 û k 2 ε ( 1 1 ) dx > C ε ln(j max 1). C 2 j max 1 U 1 U jmax 1 So, for both cases we have the following uniform estimate with C independent of k and σ. j max 1 + exp( 1 C ε ) (10)

34 Proof of Lemma 3 Assume that Stop case 1 take place. Then δσ U 1 U jmax and by construction (see (6): U j û k û k dx = νk 2, and (7): Γ h û k 2 ds Cjνk 2 h (h j 1, h j ) ) we have δσ û k û k dx j max ν 2 k, (11) Γ h û k 2 ds Cj max ν 2 k h (0, δ σ ]. (12) From (11) it follows that there exists h k (0, δ σ ) such that û k û k ds 1 j max νk 2. δ Γ hk σ

35 Proof of Lemma 3 Then, taking into account that j max does not depend on σ and k (see (10) j max 1 + exp( 1 C ε ) ), and that ν k 0 as k, we obtain the required estimates (4) (5) for sufficiently large k: Φ k n ds = ω k û Γ hk Γ hk k n ds < ε, (4) Γ hk û k 2 ds < C(ε)ν 2 k, (5)

36 Now, let Stop case 2 arises. By definition of this case and by (9) ( ν 2 k h j h j 1 1 h jmax h jmax 1 C j U j û k 2 dx ), we obtain U jmax û k û k dx = Cj max U jmax û k 2 dx < ε. ν 2 k h jmax h jmax 1 Therefore, there exists h k (h jmax 1, h jmax ) such that (4) holds. Estimate (5) follows again from (7) and the fact that j max depends on ε only. So, for any sufficiently large k we have proved the existence of h k (0, δ σ ) such that (4) (5) hold.

37 Proof of Lemma 3 Now, for (k, i)-regular value t [ 5 8 t i, 7 8 t i] consider the domain i hk (t) = W ik (t) V i+1 \ hk. By construction, i hk (t) = Γ hk S ik (t).

38 Proof of Lemma 3 Integrating the equation Φ k = ω 2 k + 1 ν k div( Φ k û k ) 1 ν k f k u k over the domain i hk (t), we have + 1 ν k = S ik (t) i hk (t) S ik (t) where F = 1 ν F 0. Φ k n ds + Φ k n ds = ω k 2 dx Γ hk i hk (t) Φ k û k n ds + 1 ν k Φk û k n ds 1 ν Γ hk k ω k 2 1 dx tλ k F+ Φk û k n ds 1 ν k ν Γ hk k i hk (t) i hk (t) f k u k dx f k u k dx, (13)

39 Proof of Lemma 3 Since and (see (4)), we get S ik (t) S ik (t) Φ k nds = Φ k ds S ik (t) Γ hk Φ k nds ε Φ k ds tf + ε ω k 2 dx i hk (t) + 1 ) 1 2 Φ2 ν k ds k (Γ hk ( ) 1 û k ds f k u k dx ν Γ hk k i hk (t) with F = F.

40 Proof of Lemma 3 By definition 1 ν k f k L 2 () = λ kν k f ν 2 L 2 () 0 as k. Therefore, 1 f k u k dx ε for sufficiently large k. ν k i hk (t) Using inequalities and Γ h Φ k 2 ds σ 2, h (0, δ σ ], k k, (see (3), (5)), we obtain Γ hk û k 2 ds C(ε)ν 2 k,

41 Proof of Lemma 3 S ik (t) Φ k ds tf σ C(ε) i hk (t) tf + 2ε + σ C(ε) ω k 2 dx, V i+1 \ δi where C(ε) is independent of k and σ. By Lemma 2, V i+1 \ δi ω 2 kdx > ε i k k i, δ i (0, δ 0 ). ω 2 k dx Choosing ε = 1 6 ε 1 i, σ = ε i, and a sufficiently large k, we 3 C(ε) obtain 2ε + σ C(ε) V i+1 \ δi ωk 2 dx 0. Therefore, Φ k ds tf. S ik (t)

42 The case (b). (b) The maximum of Φ is not attained on the boundary : max{p 0, p 1 } < ess sup Φ(x). x We do not exclude the case ess sup Φ(x) = +. We can assume that x max{p 0, p 1 } < 0 < ess sup Φ(x). x Denote σ = max{p 0, p 1 } < 0. Lemma 4. There exists F T ψ such that diamf > 0, F = and Φ(F) > σ.

43 Fix F and consider the behavior of Φ on the arcs [B 0, F] and [B 1, F]. All other considerations are similar to above. The role of B 1 is played now by F.

44 Axially symmetric 3D - case