Strauss conjecture for nontrapping obstacles

Chengbo Wang Joint work with: Hart Smith, Christopher Sogge Department of Mathematics Johns Hopkins University Baltimore, Maryland 21218 wangcbo@jhu.edu November 3, 2010

1 Problem and Background Problem History: K = History: recent works 2 our result Previous proof for n = 4 Proof for n = 3, 4 3 Minkowski estimates Obstacle case Further Problems

Problem History: K = History: recent works Problem: wave equations exterior to an obstacle K R n : compact nontrapping obstacle with smooth boundary (K {x : x 1}) (1) u(t, x) = F p (u(t, x)), (t, x) R + R n \K u(t, x) = 0, x K u t=0 = f, t u t=0 = g

Goals Problem and Background Problem History: K = History: recent works Goals: Global existence with small initial data (n = 2, 3, 4) weighted Strichartz estimates for the wave equation (n 2) generalized Strichartz estimates for the Dirichlet-wave equation (n = 2) Notation: angular mixed-norm space L q t L r x Lσ ω u σ L q t Lr x Lσ ω = R (same as L q t L r x if r = σ) ( ( ) ) r/σ q/r u(t, x ω) σ dω x n 1 d x dt 0 S n 1

Scaling analysis Problem and Background Problem History: K = History: recent works For the model equation u = u p Scaling invariance: u(t, x) u λ (t, x) = λ 2/(p 1) u(t/λ, x/λ). Homogeneous Sobolev space Invariance gives us f Ḣs = ( ) s/2 f L 2. s = s c = n 2 2 p 1.

The beginning of the story: K = Problem History: K = History: recent works John 1979: n = 3, p > p c = 1 + 2 global existence; 1 < p < p c blow up for many initial data Kato 1980: 1 < p 1 + 2 n 1 blow up for many initial data, in particular, 1 < p < for n = 1 Based on similarity of the wave equation and Schrödinger equation, Strauss 1981 made the insightful conjecture that for n 2, the critical power, p c (n), is given by the positive root of (n 1)p 2 (n + 1)p 2 = 0 In particular, p c (2) = 3 + 17, p c (3) = 1 + 2, p c (4) = 2. 2

Problem History: K = History: recent works A characterization of p c Another index of regularity (introduced in Sogge 1995 for radial existence with sharp lifespan for n = 3 and 2 p < p c ) s d = 1 2 1 p

First Wave of Study (1979-1985) Problem History: K = History: recent works Glassey 1981: n = 2, 1 p c global; Sideris 1984: blowup for all n > 3 when p p c global Glassey 81 John 79 open p = p c blow up Schaeffer 85 Schaeffer 85 open

Idea of Proof Problem and Background Problem History: K = History: recent works Exploits the positivity of the fundamental solution of the wave equation when n = 2, 3 (together with the Huygens principle of the wave equation). Idea of existence results: iteration in the space for some q(p). t n 1 2 t x q(p) u(t, x) L t L x

Problem History: K = History: recent works Second Wave of Study (1990-1997) More precise information about the solution. sharp life span for p p c global existence for high dimension

Problem History: K = History: recent works Second Wave of Study (1990-1997) More precise information about the solution. sharp life span for p p c global existence for high dimension Lifespan. If 1 p c, L ɛ (p) = exp(cɛ p(p 1) ) L ɛ (p) = An interesting observation. for p < p c s c < s d, L ɛ (p) = ɛ 1/(sc s d )

Problem History: K = History: recent works Second Wave of Study (1990-1997) Lindblad, Zhou 1990 : n 3, p p c sharp lifespan T ɛ L ɛ Zhou 1994: n = 4, p > p c global Li-Zhou 1995: n = 4, p = p c T ɛ L ɛ Lindblad-Sogge 1996: 3 n 8, T ɛ L ɛ moreover, if radial, any n, T ɛ L ɛ Georgiev-Lindblad-Sogge 1997: all dimensions n, p > p c Tataru 2001: all dimensions n, p > p c, global; lower bound T ɛ exp(cɛ (p2 1)/(3p+1) ) for p = p c

Problem History: K = History: recent works Second Wave of Study (1990-1997) Comparison of the results n = 2, 3 4 n 8 n 9 p p c 1980 s global (Z, L-S) global (G-L-S, T) p = p c sharp (Z) sharp (L-S) lower bound (T) sharp : we believe that the proved lower bound should be sharp in general. L: Lindblad, L-S: Lindblad-Sogge G-L-S: Georgiev-Lindblad-Sogge T: Tataru, Z: Zhou.

Idea of existence results Problem History: K = History: recent works Lindblad-Sogge 1996: iteration in the weighted Lebesgue space x 3 n 2p u L pq t for q(p) s.t. (same scaling as u Ḣsc ) L p x L 2n n 1 ω 1 q = 2 p 1 n 1 2 Georgive-Lindblad-Sogge 1997: iteration in the weighted Lebesgue space (1 + t 2 x 2 ) q u L p+1 R n+1 + for some q(p) 1 p(p + 1) < q(p) < n 1 n 2 p + 1..

Third Wave of Study (2008-now) Problem History: K = History: recent works Extend the classical results to the problems on manifolds 1 problems with obstacles 2 problems on asymptotically flat manifold 3 problems on Schwarzschild/Kerr background We will concentrate on the problems with obstacles. Problems on asymptotically Euclidean manifold, refer to the works of Sogge-W. arxiv:0901.0022 (radial spatial perturbation of metric: p > p c, n = 3); W.-Yu (arxiv:1009.0928): general metric p > p c, n = 3, 4. Problems on Schwarzschild background, refer to the work of Blue-Sterbenz 2006 (global for p > 3, however p c = 1 + 2)

Recent works Problem and Background Problem History: K = History: recent works Du-Zhou 2008: K =, n = 3, p = 2 T ɛ L ɛ Du-Metcalfe-Sogge-Zhou 2008: n = 4, p > p c global p = p c T ɛ A ɛ = exp(cɛ (p 1) ) Fang-W. and Hidano-Metcalfe-Smith-Sogge-Zhou 2008p: K =, 2 n 4, p > p c, global with certain angular regularity Hidano-Metcalfe-Smith-Sogge-Zhou 2008p: K =, n = 3, 4, p > p c, global (Dirichlet or Neumann BC)

Problem History: K = History: recent works Comparison of the results (with K = ) n = 2 n = 3 n = 4 n 5 p p c open HMSSZ DMSZ, HMSSZ open p = p c open Yu 09 A ɛ DMSZ A ɛ open HMSSZ: Hidano-Metcalfe-Smith-Sogge-Zhou 2008p DMSZ: Du-Metcalfe-Sogge-Zhou 2008 DZ: Du-Zhou 2008 A ɛ = exp(cɛ (p 1) ) Recall p c (4) = 2, we see that the only positive results proved in the obstacle setting are the case p 2 and n = 3, 4 with weaker lower bound of lifespan when p = p c.

Our main result Problem and Background our result Previous proof for n = 4 Proof for n = 3, 4 In the recent work with H. F. Smith and C. D. Sogge, we proved the Strauss conjecture (p > p c ) for n = 2. Theorem 1. Let n = 2, and {Z} = { 1, 2, x 1 2 x 2 1 }. Then if we fix p (p c (2), 5), there is an ε 0 > 0 so that (1) has a global solution satisfying (Z α u(t, ), t Z α u(t, )) Ḣ sc Ḣ sc 1 (R n \K), α 1, t R +, when the initial data (f, g) = (u t=0, t u t=0 ) satisfies f K = 0 and ( ) Z α f Ḣsc (R n \K) + Z α g Ḣsc 1 (R n \K) < ε, α 1 with 0 < ε < ε 0.

our result Previous proof for n = 4 Proof for n = 3, 4 Comparison of the results (with K = ) n = 2 n = 3 n = 4 n 5 p p c Smith-Sogge-W. HMSSZ DMSZ, HMSSZ open p = p c open Yu 09 A ɛ DMSZ A ɛ open HMSSZ: Hidano-Metcalfe-Smith-Sogge-Zhou 2008p DMSZ: Du-Metcalfe-Sogge-Zhou 2008 A ɛ = exp(cɛ (p 1) ) Remark. Recently, Zhou-Han 2010p proved T ɛ L ɛ even in the case K = for n 3 and 1 < p < p c.

Previous Proof for n = 4 our result Previous proof for n = 4 Proof for n = 3, 4 DMSZ 2008: use the fact that p c = 2, follow the idea for solving the quadratic equation (Keel-Smith-Sogge 2002, Metcalfe-Sogge 2006, etc) u = Q(u, u ) Using the Morawetz/KSS (Keel-Smith-Sogge) estimates x 1/2 ɛ u L 2 + (log(2 + T )) 1/2 x 1/2 u L 2 and the trace lemma C u (0) L 2 + C u L 1 t L 2 x x n 2 2 f L x L 2 ω C f Ḣ 1, n 3

Previous Proof for n = 4 our result Previous proof for n = 4 Proof for n = 3, 4 By duality and Duhamel s principle, we can essentially obtain for u(0) = u t (0) = 0 x 1/2 ɛ u L 2 + (log(2 + T )) 1/2 x 1/2 u L 2 C x n 2 2 u L 1 t L 1 x L2 ω enough to show T ɛ exp(c/ɛ) for p = 2 ( n 2 2 = 2 1 2 ). In fact, if (f, g) of size ɛ, first iteration: M = (log(2 + T )) 1/2 x 1/2 u L 2 Cɛ However, the norm of the forcing term ( u = u 2 ) x n 2 2 u L 1 t L 1 = (log(2 + T x L2 ))M2 ω (log(2 + T ))M 2 M T ɛ exp(c/ɛ)

Previous Proof for n = 4 our result Previous proof for n = 4 Proof for n = 3, 4 Instead, if we consider p > p c = 2, weighted Sobolev inequality h L (R 1 x R+1) R n 1 2 Z α h L 2 (R 2 x R+2) α n+2 2 Comes from local Sobolev (in polar coordinates) f L ([R 1,R+1] S n 1 ) C α n+2 2 Z α f L 2 ([R 2,R+2] S n 1 ) and the volume element dx = r n 1 drdω.

our result Previous proof for n = 4 Proof for n = 3, 4 Idea of Proof for n = 2, 3, 4 with K = The method of DMSZ can also be applied to n 5 and p = 2 (regular F p (u)), n = 3 and p > 5/2. (Far from p c except for n = 4) Need more general method for p > p c Fang-W. and HMSSZ 2008p: weighted Strichartz estimates, for 2 q and u = 0 x n 2 n+1 q γ u L q t Lq x L2 C u (0) Ḣγ 1, ω By duality and Duhamel s principle, 1 2 1 q < γ < n 2 1 q x n 2 n+1 p γ u L p t Lp x L2 C u (0) Ḣγ 1 + C x n 2 +1 γ u ω L 1 t L 1 x L2 ω for 2 p, 1 2 1 p < γ < n 2 1 p, 1 2 < 1 γ < n 2

our result Previous proof for n = 4 Proof for n = 3, 4 Idea of Proof for n = 2, 3, 4 with K = If u = u p, we can iterate in x n 2 n+1 p γ u L p t Lp x L2 ω with γ = n 2 2 p 1 (from scaling). By weighted Strichartz estimates, Condition for iteration: 2 p, 1 2 1 p < γ < 1 2

our result Previous proof for n = 4 Proof for n = 3, 4 Idea of Proof for n = 3, 4 with K = If K =, need to prove similar weighted Strichartz estimates of type x n 2 n+1 q γ u L q t Lq x L2 C u (0) Ḣγ 1 θ

our result Previous proof for n = 4 Proof for n = 3, 4 Idea of Proof for n = 3, 4 with K = If K =, need to prove similar weighted Strichartz estimates of type x n 2 n+1 q γ u L q t Lq x L2 C u (0) Ḣγ 1 θ Theorem 2[HMSSZ, abstract Strichartz estimates] Minkowski L q t X estimates u L q t X C u (0) Ḣγ 1 + local (in space) energy decay u L 2 C u (0) t,x: x 2 L 2 +local (in time) L q t X estimates L q t X estimates, if q > 2 and γ [ 3 n 2, n 1 2 ].

Radial Strichartz Problem and Background Minkowski estimates Obstacle case Further Problems To remedy the difficulty, let us give the motivation of our result for n = 2.

Radial Strichartz Problem and Background Minkowski estimates Obstacle case Further Problems To remedy the difficulty, let us give the motivation of our result for n = 2. Fang-W. 2006: If n = 2, we have radial Strichartz estimates u L q t Lr x C u (0) Ḣγ 1 1 q < 1 2 1 r or(q, r) = (, 2), (q, r) (, ), γ = 1 1 q 2 r.

Minkowski estimates Obstacle case Further Problems 1/q 1/2 1/q=1/2-2/r, γ = 1/2 1/q = 1 γ 2/r, 0 < γ < 1/2 γ = 0 (energy) 1/4 1/2 1/r Figure: Minkowski radial estimates

Proof of radial Strauss conjecture Minkowski estimates Obstacle case Further Problems Key observation: Let q = p 1 2, by radial Strichartz estimates, we have u L qp t L p C u x (0) Ḣγ 1 + u L q t L1 x when γ = 1 2 p 1 and p c p c Iteration in L qp L p will give us the proof in the radial case. Recall the positivity of the fundamental solution when n = 2, we can also remove radial assumption (iterate in L qp t L p x L θ ), by adding certain angular regularity on the initial data.

Difficulty for obstacle case Minkowski estimates Obstacle case Further Problems We can also have L qp L p estimates in the general case. (Sterbenz 2005 (n 4), Fang-W. 2008p (n 2)) for some b > 0. u L qp t L p x C (1 2 θ )b u (0) Ḣγ 1 1 We do have L qp t L p x Minkowski estimates; 2 We can not have L qp t L p x estimates in the obstacle case, due to the example in HMSSZ.

L q t L r x L2 θ estimates Problem and Background Minkowski estimates Obstacle case Further Problems Theorem 3. If n = 2, we have generalized Strichartz estimates u L q t Lr x L2 θ C u (0) Ḣγ 1 1 q < 1 2 1 r or(q, r) = (, 2), (q, r) (, ), γ = 1 1 q 2 r. In contrast to the L q t L r x estimates, No angular regularity is required for the initial data. The proof is basically elementary, it is based on Littlewood-Paley decomposition, Fourier transform, Fourier series (in the angle of the Fourier variable), Bessel functions and change of variables. The same set of estimates are also valid for higher dimension (n 2, including local-in-time estimates), see Fang-W., Jiang-W.-Yu.

Minkowski estimates Obstacle case Further Problems 1/q 1/2 1/q=1/2-2/r, γ = 1/2 1/q = 1 γ 2/r, 0 < γ < 1/2 γ = 0 (energy) 1/4 1/2 1/r Figure: Minkowski L q L r x L2 θ

Adaption in the obstacle case Minkowski estimates Obstacle case Further Problems As in HMSSZ, due to technical difficulties in using the rotational vector fields near K, we shall modify the Lebesgue spaces near K slightly. Specifically, assuming K {x : x 1}, given 0 γ < 1, we define h Xr,γ = h L sγ ( x <2) + h L r x L 2 θ ( x >2), with γ = 2(1/2 1/s γ). Recall (Smith-Sogge 2000) for φ C 0 and γ (n 2)/2 and the energy estimate φe it f L 2 t H γ C f Ḣ γ φe it f L t Ḣ C f γ Ḣγ we have for q 2 and γ [0, 1/2], e it f L q t Lsγ ( x <2) φeit f L q t Ḣγ C f Ḣγ

Adaption in the obstacle case Minkowski estimates Obstacle case Further Problems Combine with the L q t L r x L2 θ Minkowski estimates estimates, we get the global (in time) u L q t Xr,γ C u (0) Ḣγ 1 γ = 1 1 q 2 r [0, 1/2], q > 2. Vainberg 1975 (also Burq 2000): if K = is nontrapping, we have local (in space) energy estimates for Dirichlet BCs, but not for Neumann BCs. finite speed of propagation+sobolev+local energy Local (in time) L q t X estimates

What s next? Problem and Background Minkowski estimates Obstacle case Further Problems As in the radial K = case, need L q t X r,γ for 0 γ < 1. 1/q 1/2 1/q=1/2-2/r, γ = 1/2 γ = 0 (energy) 1/4 1/2 1/r

What s next? Problem and Background Minkowski estimates Obstacle case Further Problems As in the radial K = case, need L q t X r,γ for 0 γ < 1. Recall, however, we have energy estimates and so by Sobolev u L t L sγ x (R + K c ) C f Ḣ γ (K c ) + C g Ḣ γ 1 (K c ). 1/q 1/2 1/q=1/2-2/r, γ = 1/2 γ = 0 (energy) 1/4 1/2 1/r

L q t X r,γ for 0 γ < 1 Problem and Background Minkowski estimates Obstacle case Further Problems Interpolation L q t X r,γ for 0 γ < 1 except r =. 1/q 1/2 1/q=1/2-2/r, γ = 1/2 1/q = 1 γ 2/r, 0 < γ < 1/2 γ = 0 (energy) 1/4 1/2 1/r

Strauss conjecture for n = 2 Minkowski estimates Obstacle case Further Problems Modification of the radial proof Proof of Strauss conjecture for n = 2

Further Problems Problem and Background Minkowski estimates Obstacle case Further Problems The problem for 1 < p p c. K = and 1 < p p c, sharp lifespan obtained for n 3 or p < p c, and expected sharp lower bound for n 4 and p = p c (Lindblad, Zhou, Lindblad-Sogge, Zhou-Han) K =, sharp lifespan obtained for n = 3 and 2 p < p c (Du-Zhou, Yu), certain non-sharp lower bound for p = p c and n = 3, 4 (DMSZ, Yu) p = p c? 1 < p < p c for n = 2? 1 < p < 2 for n = 3, 4? The corresponding problem for n 5 (extremely difficult, p c < 2).

Minkowski estimates Obstacle case Further Problems Thanks!