Université de Cergy-Pontoise. Insitut Universitaire de France. joint work with Frank Merle. Hatem Zaag. wave equation

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1 The blow-up rate for the critical semilinear wave equation Hatem Zaag CNRS École Normale Supérieure joint work with Frank Merle Insitut Universitaire de France Université de Cergy-Pontoise

2 utt = u + u p 1 u, u(0) = u0 et ut(0) = u1, where u(t) : x IR N u(x, t) IR, u0 H 1 loc,u (IRN ) and u1 L 2 loc,u (IRN ). v L 2 loc,u (IR N ) = sup a IR N ( x a <1 v(x) 2 dx ) 1/2. N 2 and p = pc N 1. Earlier work: 1 < p < pc (Amer. J. Math.). 1

3 Critical why? 1- When p = pc, there is a conformal invariance in the equation: if U(ξ, τ) is defined by U(ξ, τ) = ( x 2 t 2 ) N 1 2 u(x, t), ξ = x x 2 t 2, τ = t x 2 t 2, then U satisfies the same equation as u. 2- The subritical case 1 < p < pc has been solved in an earlier work (Amer. J. Math.), where major difficulties to adapt to the ciritical case appeared. The presentation is done for 1 < p pc. 2

4 CAUCHY PROBLEM IN H 1 loc,u (IRN ) L 2 loc,u (IRN ) Since p < N+2 N 2, it follows from : - the solution of the Cauchy problem in H 1 L 2 (IR N ) (Lindblad and Sogge, Shatah and Struwe) - the finite speed of propagation. 3

5 FINITE TIME BLOW-UP SOLUTIONS Existence: John, Caffarelli and Friedman, Alinhac, Kichenassamy and Litman QUESTION (was open before this work) Evaluate the norm of u, u and ut in L 2 loc,u (IRN ) near the blow-up time T. 4

6 A HINT : THE ASSOCIATED ODE utt = u p, u(t ) = + gives u(t) κ(t t) 2 p 1 where κ = κ(p) is explicitly given. 5

7 SELF-SIMILAR VARIABLES wa(y, s) = (T t) 2 p 1u(x, t), y = x a T t, s = log(t t). Equivalent problem: For all y IR N and s log T : 2 s w + p + 3 p 1 sw + 2y. sw + i,j (yiyj δi,j) 2 yiyj w + 2(p + 1) p 1 y. w = w p 1 w 2(p + 1) (p 1) 2 w, 6

8 Equivalent problem in divergence form: For all y IR N and s log T : 2 s w 1 ρ div [ρ w ρ(y. w)y] + 2(p+1) (p 1) 2 w w p 1 w = p+3 p 1 sw 2y. sw where ρ(y) = (1 y 2 ) α If p < pc N 1, then α 2 p 1 N 1 2 > 0. If p = pc, then α = 0 and ρ 1. 7

9 Th. (p pc) For any sol. u blowing up at time T, for all s log T + 1 and a IR N, wa(s) H 1 (B) + swa(s) L 2 (B) K where B = B(0, 1), K = K(N, p, u0, T ). Rk. From scaling arguments and the solution of the Cauchy Problem in H 1 L 2 (IR N ), we get for all s log T + 1, sup wa(s) H a IR N 1 (B) + swa(s) L 2 (B) ɛ0(n, p) > 0. = We are at the good scale 8

10 IN THE ORIGINAL VARIABLES Th. (p pc) For any sol. u blowing-up at time T, for any t [T (1 e 1 ), T ), u L 2 loc,u (IR N ) K(T t) 2 p 1 ut L 2 loc,u (IR N ) + u L 2 K(T t) p loc,u (IRN ) where K = K(N, p, u0, T ). 9

11 THE ARGUMENTS OF THE PROOF - Existence of a Lyapunov functional for the equation on w and energy-type estimates. - Interpolation to gain more regularity. - Gagliardo-Nirenberg type estimates. 10

12 Equivalent problem in divergence form: For all y IR N and s log T : 2 s w 1 ρ div [ρ w ρ(y. w)y] + 2(p+1) (p 1) 2 w w p 1 w = p+3 p 1 sw 2y. sw where ρ(y) = (1 y 2 ) α If p < pc N 1, then α 2 p 1 N 1 2 > 0. If p = pc, then α = 0 and ρ 1. 11

13 A LYAPUNOV FUNCTIONAL Antonini-Merle E(w) = B 1 2 ( sw) 2 (p + 1) + (p 1) 2w2 1 p + 1 w p+1 ρdy B ( w 2 (y. w) 2) ρdy ρ(y) = (1 y 2 ) α with α = 2 If p < pc, then α > 0. If p = pc, then α = 0 and ρ 1. If p > pc, then E is not even defined. Hence, pc is critical. p 1 N

14 Lemma 1 (Monotonicity) For all s1 and s2: (p < pc, Antonini-Merle), E(w(s2)) E(w(s1)) = 2α s2 s1 B ( sw) 2 (1 y 2 ) α 1 dyds. (p = pc: degeneracy), E(w(s2)) E(w(s1)) = s2 s1 B ( sw(σ, s)) 2 dσds. Rk. 2α(1 y 2 ) α 1 δ B as p pc. 13

15 Lemma 2 (Blow-up criterion (Antonini-Merle)) If a solution W satisfies E(W (s0)) < 0 for some s0 IR, then W blows up in finite time. BOUNDS ON E For all s log T, s2 s1 log T 0 E(w(s)) E(w( log T )) C0 where C0 = C0( u0, T ). 14

16 BOUNDS ON THE DISSIPATION OF E For all s log T, s2 s1 log T (p < pc, supported in a cylinder), s 2 s1 B ( sw) 2 (y, s)(1 y 2 ) α 1 dyds C 0 2α, (p = pc, supported in the boundary of the cylinder), s 2 s1 B ( sw(σ, s)) 2 dσds C0. This degeneracy is a major difficulty in adapting the subcritical case to the critical. 15

17 (p < pc) SUPPORT OF THE DISSIPATION IN THE (x, t) VAR. (remember w = wa): In the interior of the light cone with vertex (a, T ). t T (a,t) t_2 t_1 x a 16

18 (p = pc) SUPPORT OF THE DISSIPATION IN THE (x, t) VAR. (remember w = wa): On the EDGE of the light cone with vertex (a, T ). t T (a,t) t_2 t_1 x a 17

19 Since all is uniform in a, we move the a, and then integrate in a... t T (a,t) t_2 t_1 x a We recover an estimate in the interior of the light cone, between t1 and t2 (like the subcritical case). 18

20 More precisely (p = pc), Proposition 1 For all a IR N and s2 s1 log T, s 2 s1 B swa(y, s) 2 dyds C0. 19

21 Even better: NON CONCENTRATION OF ( sw) 2 Proposition 2 For any ball B(b, r0) B(0, 3) with r0 < 1, s 2 s1 B(b,r0) swa(y, s) 2 dyds C0r0. 20

22 With this adaptation, we concentrate (from now on) on the subcritical case. Recall Bounds on E and its dissipation: For all s log T, s2 s1 log T 0 E(w(s)) E(w( log T )) C0, s 2 s1 B ( sw) 2 (y, s)(1 y 2 ) α 1 dyds C 0 2α, where C0 = C0( u0, T ). 21

23 GOAL Prove that w, w and sw are bounded in L 2 (B(0, 1)). Since they appear in E (with a weight): E(w) = B 1 2 ( sw) 2 (p + 1) + (p 1) 2w2 1 p + 1 w p+1 ρdy B ( w 2 (y. w) 2) ρdy, it is enough to bound w p+1 ρ. Rk. We get rid of the weights through a covering argument. 22

24 Rk. Since we have an average (in time) estimate on sw, s 2 s1 B ( sw) 2 (y, s)(1 y 2 ) α 1 dyds C0, we will look for average (in time) estimates of the terms in the functional E. 23

25 CONTROL OF SPACE-TIME INTEGRALS. First, we integrate E(w) between s1 and s2 : s 2 s1 E(w(s))ds = s2 s1 B 1 2 ( sw) 2 (p + 1) + (p 1) 2w2 1 p + 1 w p+1 ρdy s 2 s1 B ( w 2 (y. w) 2) ρdy Remark that w p+1 controls all the other terms in E. 24

26 2nd IDENTITY We multiply the w-equation by wρ, integrate on B (s1, s2), IBP and use the definition of E to write : (p 1) 2(p + 1) s 2 s1 B w p+1 ρdy = s2 s1 E(w(s))ds + s2 s1 B ( ( sw) 2 ρ swy. wρ swwy. ρ ) dyds B w sw + p + 3 2(p 1) N w 2 ρdy s2 s1. 25

27 Proposition 3 (p pc) For all a IR N and s log T + 1, s+1 s B w p+1 ρdyds C(C0, N, p, T ). Rk. There is a weight... PROOF (p < pc) : We will control all terms on the RHS of the previous identity by C ɛ + Cɛ s2 s1 B w p+1 ρdyds and then take ɛ small. 26

28 Back to p pc Corollary 1 For all a IR N and s log T + 1, s+1 s B 1/2 ( ( swa) 2 + wa 2 + wa p+1 + wa 2) dyds C where B 1/2 B(0, 1/2), C = C(N, p, u0, T ). Rk. If p < pc, we first get estimates on B B(0, 1) with the weight ρ. If p = pc, we directly get estimates on B. 27

29 We are ready for the proof of the Thoerem that I recall here : Th. (p pc) For any sol. u blowing up at time T, for all s log T + 1 and a IR N, B ( ( swa) 2 + wa 2 + wa 2) dy K where B = B(0, 1), K = K(N, p, u0, T ). 28

30 Step 1 : (p pc) Control of B w a(y, s) 2 dy We start from s+1 s B 1/2 ( ( swa) 2 + wa 2) dyds C. We first get rid of ds, and then extend the integration in space to B. Let g(s) = wa(y, s) 2 dy 12. B We write 1/2 29

31 g L 2 (s,s+1) = ( s+1 sg 2 L 2 (s,s+1) = s+1 s s ds B w2 dyds )1 2 C B 1/2 w swdy 4 B 1/2 w 2 dy s 2 s1 ds B 1/2 ( sw) 2 dy C0. Hence, g H 1 (s, s + 1). From the Sobolev injection in one dimension, g L (s, s + 1), i.e. B 1/2 wa 2 dy C. 30

32 Now, we extend the integration to B. We take a = 0. Since b IR N, y < 1 2 w b (y, s) 2 dy C w b (y, s) = w0(y + be s, s), then (z = y + be s ) b IR N, z be s < 1 2 w0(z, s) 2 dz C + covering, this yields z <1 w 0(z, s) 2 dz C. 31

33 Step 2 : (p pc) Control of wa(s) in L r loc Proposition 4 For all s log T + 1 and a IR N, B w a(y, s) p+3 2 dy C where B = B(0, 1). Proof : Follows from w 2 and w p+1 by interpolation (H 1 L in one dimension). 32

34 Step 3 : (p pc) Control of the gradient in L 2 loc,u Lemma 3 For all s log T + 1 et a IR N, B w a p+1 C B w a p+3 γ ( 2 dy B w a 2 dy ) β, where γ(p, N) > 0 and if p < pc, β = β(p, N) [0, 1) if p = pc, β = 1. Proof : Gagliardo-Nirenberg. 33

35 Proposition 5 For all s log T + 1 and a IR N, B w a(y, s) 2 dy C. Formal proof : If all the weights were equal to 1, we would have from the functional E : B w a 2 dy C + B w a p+1 dy + Gagliardo Nirenberg B w a p+1 C B w a p+3 γ ( 2 dy B w a 2 dy ) β. If p < pc, then β < 1 and we get the conclusion. If p = pc, then β = 1. we can conclude only if B w a p+3 2 dy is small. 34

36 This is possible if we replace B by B(b, r0) for small r0. Indeed, remember the NON CONCENTRATION result: Proposition 6 For any ball B(b, r0) B(0, 3) with r0 < 1, s 2 s1 B(b,r0) swa(y, s) 2 dyds C0r0. Corollary 2 B(b,r0) w a(y, s) p+3 2 dy C0 r 0. 35

37 We then cover the unit ball B by balls of radius r0 with overlapping less than some C(N). 36