p−1 We study a defocusing semilinear wave equation, with a power nonlinearity |u| u, deﬁned outside the unit ball of R , n ≥ 3, with Dirichlet boundary conditions. We prove that if p > n + 3 and the initial data are nonradial perturbations of large radial data, there exists a global smooth solution. The solution is unique among energy class solutions satisfying an energy inequality. The main tools used are the Penrose transform and a Strichartz estimate for the exterior linear wave equation perturbed with a large, time dependent potential. Keywords Supercritical wave equation · Exterior problem · Decay estimates · Global existence Mathematics Subject Classiﬁcation 35L05 · 58J45 · 35L20 1 Introduction Consider the Cauchy problem for the defocusing wave equation on R × R : p−1 u +|u| u = 0, u(0, x ) = u (x ), u (0, x ) = u (x ) (1.1) 0 t 1 s s−1 with Sobolev initial data (u , u ) ∈ H × H . The existence of global solutions to 0 2 this problem has been explored in considerable detail. The energy critical power for global smooth solvability is p (n) = 1+ for n ≥ 3, while p (1) = p (2) =∞. cr cr cr n−2 Global existence in the energy space is known for p ≤ p (n) [4,5,8]; regularity for cr p = p has been explicitly proved up to dimension 7 [5,7] and should hold for all cr The author is partially supported by the Project Ricerca Scientiﬁca Sapienza 2016: “Metodi di Analisi Reale e Armonica per problemi stazionari ed evolutivi”. B Piero D’Ancona dancona@mat.uniroma1.it Dipartimento di Matematica, Sapienza Università di Roma, Piazzale A. Moro 2, 00185 Rome, Italy 0123456789().: V,-vol 141 Page 2 of 29 P. D’Ancona dimensions. For supercritical nonlinearities p > p (n), very little is known, and the cr question of global existence or blow up is still open. If we restrict to spherically symmetric solutions, the difﬁculty of the problem is essentially the same. However, if a radial solution blows up, the blow up must occur at the origin. This follows at once from the Sobolev embedding for radial functions −1 |x | |u(x )| ∂u , (1.2) 2 n L (R ) which is a well known special case of the family of inequalities n − 1 1 n −σ |x | |u(x )| |D| u , + <σ < (1.3) L L |x | r p p r p r (see [3]). Here the norm L L is an L norm in the radial direction of the L norm |x | ω ω in the angular direction. We give a precise statement of this blow–up alternative result including its short proof: Proposition 1.1 Let u , u be spherically symmetric test functions, and T the supre- 0 1 mum of times τ> 0 such that a smooth solution u(t , x ) of (1.1) exists on [0,τ]× R . Assume T < ∞. Then u blows up at x = 0 as t → T , in the sense that for any R > 0, u is unbounded on the cylinder {|x | < R, t < T }. Proof By standard local existence results (e.g. [9], Chapter 5) we know that T > 0, and the solution is spherically symmetric for all t ∈[0, T ) by uniqueness of smooth solutions. Denote the energy of the solution u at time t by 1 1 2 p+1 E (t ) := ∂ u(t ) + |u(t )| dx . t ,x 2 n 2 p+1 L (R ) By conservation of energy and the radial estimate (1.2)wehave n−2 2 |x | |u(t , x )| ≤ CE (t ) = CE (0), 0 ≤ t < T . Assume by contradiction that |u(t , x )|≤ M for (t , x ) in some cylinder {|x | < R, t < T }, then by the previous bound we get 1/2 1−n/2 n |u(t , x )|≤ M + (CE (0)) R < ∞ for 0 ≤ t < T , x ∈ R and by standard arguments we can extend u to a strip [0, T ]× R for some T > T , against the assumptions. In order to obtain a global radial solution, it would be sufﬁcient to prevent blow-up at 0. This is the case if we replace the whole space with the exterior of a ball: ={x ∈ R :|x | > 1} On the supercritical defocusing NLW outside a ball Page 3 of 29 141 and consider the mixed problem on R × with Dirichlet boundary conditions p−1 u +|u| u = 0, u(0, x ) = u (x ), u (0, x ) = u (x ), u(t , ·)| = 0. 0 t 1 ∂ (1.4) In this situation the local solution can not blow up, due to the conservation of H energy and the radial estimate (1.2). Note indeed that estimate (1.2) is valid also for 1 ∞ functions in H (), since it holds on the dense subspace C ().Thisimpliesthe existence of a global radial solution for arbitrary powers p > 1. Indeed, we have the following result, where we use the notation k k + 2 n s n 1 C X = C (R ; X ), X = L (R ) or H (R ) or H (). n 1 2 Proposition 1.2 Let = R \ B(0, 1),n ≥ 2,p > 1 and let u ∈ H () ∩ H () ∩ p+1 1 L (),u ∈ H () be two radially symmetric functions. Then the mixed problem (1.4) 2 2 1 1 2 has a global solution u ∈ C L ∩ C H ∩ CH , satisfying the conservation of energy 1 1 2 p+1 E (u(t )) = E (u(0)), E (u(t )) := ∂ u(t ) + |u(t )| dx (1.5) t ,x 2 2 p+1 L () and the uniform bound ∞ + u C (u 1 , u 2 ). (1.6) 0 1 L (R ×) H L 2 2 1 1 2 If v ∈ C L ∩ C H ∩ CH is a second solution of (1.4) with the same data which is either radially symmetric in x, or locally bounded on R × , then v ≡ u. The existence of large radial solutions suggests naturally the question of stability: do nonradial perturbations of radial data give rise to global solutions? The uniform bound (1.6) is not sufﬁcient for a perturbation argument. However, we can prove an actual decay estimate, obtained by reduction to a mixed problem with moving boundary on S via the Penrose transform. Deﬁne for M > 1 the quantity C (u , u ) := u +u , N = + 1. N +1,N N ,N +1 M 0 1 0 1 0 H 0 0 (|x |≥M ) H 0 0 (|x |≥M ) (1.7) Then we have Theorem 1.3 (Decay of the radial solution) Let u be the solution constructed in Propo- sition 1.2. Assume that for some M > 1 the data satisfy p−1 n −1 p+1 (u , u ) := C (u , u )+u 2 +u 1 +x u p+1 < ∞. (1.8) 0 1 M M 0 1 0 1 0 H H L Then the following decay estimate holds n 1 1 1− − − 2 2 2 |u(t , x )|≤ C (u , u ) ·|x | t +|x | t −|x | . (1.9) 0 1 M 141 Page 4 of 29 P. D’Ancona If the initial data are smoother then regularity propagates, and in addition higher Sobolev norms remain bounded as t →∞. In order to state the regularity result, we introduce brieﬂy the nonlinear compatibility conditions, discussed in greater detail in Sect. 3 (see Deﬁnition 3.3 and also Deﬁnition 2.1). For the mixed problem on R × u = u + f (u), u(0, x ) = u , u (0, x ) = u , u(t , ·)| = 0 tt 0 t 1 ∂ we deﬁne formally the sequence of functions ψ as follows: j j −2 ψ =u(0, x )=u ,ψ =u (0, x )=u ,ψ =∂ u(0, x )=ψ + ∂ ( f (u))| , 0 0 1 t 1 j j −2 t =0 t t where in the deﬁnition of ψ we set recursively ∂ u(0, x ) = ψ (x ) for 0 ≤ k ≤ j − 2. j k Then we say that the data (u , u , f (s)) satisfy the nonlinear compatibility conditions 0 1 of order N ≥ 1if ψ ∈ H () for 0 ≤ j ≤ N . (1.10) Note that if f (s) vanishes at s = 0 of sufﬁcient order, to satisfy the nonlinear compat- ibility conditions it is sufﬁcient to assume that the initial data u , u belong to H () 0 1 for k large enough. We can now state the higher regularity result for the radial solutions. In the statement we use the norm v ∞,2;k = ∂ u 2 ∞. Y t ,x L L |α|≤k 2n Theorem 1.4 Let n ≥ 3,p > N ≥ 1,p > . Assume the radially symmetric data n−2 N +1 N p−1 u ∈ H (),u ∈ H and f (u) =|u| u satisfy the nonlinear compatibility 0 1 conditions of order N and condition (1.8). k N +1−k Then the radial solution constructed in Proposition 1.2 belongs to C H ∩ N 1 C H for all 0 ≤ k ≤ N + 1 and satisﬁes (besides (1.5)) the uniform bounds u ≤ C ((u , u ) ), ∞ 2 0 1 M L L ∇ u ≤ C (u , u ) , u , u , 1 ≤ k < p − 1. (1.11) ∞,2;k k k−1 t ,x 0 1 M 0 1 Y H H The main result of the paper is the following: Theorem 1.5 Let n ≥ 3,N ≥ n, p > n + 3, and u , u radial functions satisfying 0 1 the assumptions in Theorem 1.4. Then there exists (u , u )> 0 such that the 0 1 following holds. N +1 N p−1 Assume v ∈ H (), v ∈ H () and f (u) =|u| u satisfy the nonlinear 0 1 compatibility conditions of order N and u − v N +1 < , u − v N < 0 0 1 1 H H k N −k N −1 1 Then Problem (1.4) with data v ,v has a global solution v ∈ C H ∩ C H , 0 1 0 ≤ k ≤ N + 1. In other words, nonradial perturbations of large radial initial data in a high Sobolev norm do give rise to global smooth quasiradial solutions. Note that these solutions are unique in the class of locally bounded global solutions, but one can not in principle exclude the existence of other energy class solutions. However, one can prove a weak– strong uniqueness result, which implies in particular that the solution constructed in Theorem 1.5 is the unique energy class solution satisfying an energy inequality: On the supercritical defocusing NLW outside a ball Page 5 of 29 141 Theorem 1.6 Suppose all the assumptions in Theorem 1.5 are satisﬁed. Let I be an 1 1 2 ∞ p+1 open interval containing [0, T ] and v ∈ C (I ; H ) ∩ C (I ; L ) ∩ L (I ; L ()) a distributional solution for 0 ≤ t ≤ T to Problem (1.4) with the same initial data as v, which satisﬁes an energy inequality E (v(t )) ≤ E (v(0)) (see (1.5)). Then we have v(t ) = v(t ) for 0 ≤ t ≤ T. Remark 1.1 Similar results can be proved for other dispersive equations, notably for the nonlinear Schrödinger equation. This will be the topic of further work in preparation. The proof of Theorem 1.5 is based on perturbing the radial solution u(t , x ) with a small term w(t , x ) satisfying the equation p−1 p−1 w +|u + w| (u + w) −|u| u = 0, which can be written in the form 2 p−1 w + V (t , x )w = w · F [u,w], V (t , x ) = p|u| . To solve the last equation, we prove an energy–Strichartz estimate for the exterior problem with potential u + V (t , x )u = F , u(0, x ) = u (x ), u (0, x ) = u (x ), u(t , ·)| = 0 0 t 1 ∂ as an application of the results of [1,6,11]; note that this part of the argument holds also on arbitrary non trapping exterior domains. The Strichartz estimate for the wave equation with potential is proved by reduction to the constant coefﬁcient case, which is possible since from the previous results we know that the potential V has rapid decay in (t , x ). Finally, Theorem 1.6 is an adaptation of a weak–strong uniqueness result due to M. Struwe [12]. The plan of the paper is the following. The linear theory and the perturbed energy– Strichartz estimates are developed in Sect. 2. In Sect. 3 the global radial solution is constructed and its regularity and decay properties are studied. In the ﬁnal Sects. 4, 5 we prove the main results Theorems 1.5 and 1.6. 2 Linear theory Denote by − the Dirichlet Laplacian, that is the nonnegative selfadjoint operator 2 1 with domain H () ∩ H (), and by its nonnegative selfadjoint square root 1/2 1 = (− ) , D() = H (). 2k k We have = (− ) for integer k ≥ 0, and 2k k 2k k−1 1 D( ) = D( ) ={ f ∈ H () : f , f ,..., f ∈ H ()}, D 0 2k+1 2k+1 k 1 D( ) ={ f ∈ H () : f , f ,..., f ∈ H ()} 0 141 Page 6 of 29 P. D’Ancona so that k k j 1 D( ) ={ f ∈ H () : f ∈ H (), 0 ≤ 2 j ≤ k − 1}. We shall make repeated use of the equivalence ∇ g =g , 2 2 L () L () valid for g ∈ D() = H (). The solution of the special mixed problem u = 0, u(0, x ) = 0, u (0, x ) = u , u(t , ·)| = 0 t 1 ∂ with u ∈ H () can be represented in the form −1 S(t )u := sin(t )u (2.1) 1 1 Then the solution of the full linear mixed problem on R × u = F (t , x ) u(0, x ) = u (x ), u (0, x ) = u (x ), u(t , ·)| =0(2.2) 0 t 1 ∂ takes the form u = S(t )u + ∂ S(t )u + S(t − s)F (s)ds, (2.3) 1 t 0 where ∂ S(t )u = cos(t )u . t 0 0 We shall be concerned with several global in time estimates of S(t ) and of the solution to (2.2). We work for positive times t > 0 only, but it is clear that all results are time–reversible. Directly from the spectral theory one gets ∇ S(t )g 2 ≤g 2 , ∂ S(t )g 2 ≤g 2 , (2.4) x t L () L () L () L () −1 S(t )g 2 ≤ t g 2 , S(t )g 2 ≤ g 2 g 2n . L () L () L () L () n+2 L () (2.5) As a consequence one gets the basic energy estimate for solutions of (2.2): ∇ u ∞ 2 ∇ u 2 +u 2 +F 1 2 (2.6) t ,x x 0 1 L (0,T ;L ()) L L L (0,T ;L ()) valid for all T > 0, with a constant independent of T . Higher regularity results require compatibility conditions. Given the data (u , u , F ) 0 1 we deﬁne recursively the sequence of functions h as follows: j j −2 h = u , h = u , h = ∂ u(0, x ) = h + ∂ F (0, x ), j ≥ 2. 0 0 1 1 j j −2 t t (2.7) j −2 The function h is obtained by applying ∂ to the equation u = u + F. j tt t On the supercritical defocusing NLW outside a ball Page 7 of 29 141 Deﬁnition 2.1 (Linear compatibility conditions) We say that the data (u , u , F ) sat- 0 1 N +1 isfy the linear compatibility conditions of order N ≥ 1if (u , u ) ∈ H () × 0 1 N k N −k H (), F ∈ C H () for 0 ≤ k ≤ N , and h ∈ H () for 0 ≤ j ≤ N . To formulate estimates of u in a compact format we introduce a few notations. We write for short for any interval I ⊆ R and T ≥ 0 p p p p q p q q q p q q L L = L (I ; L ()), L L = L L , L L = L L . I T [0,T ] [0,∞) p q Moreover we denote the L L norm of all spacetime derivatives up to the order N by p q u p,q;N = ∂ ∂ u . (2.8) L (I ;L ()) j +|α|≤N t x When I =[0, T ] or I =[0, +∞) we write also q,r ;N q,r ;N q,r ;N q,r ;N Y = Y Y = Y . T [0,T ] [0,∞) The following result is standard, and valid for general domains with sufﬁciently −1 smooth compact boundary. We use the inequality g 2 g 2n in the for- n+2 mulation of (2.9). N +1 N Proposition 2.2 Let N ≥ 1, and assume u ∈ H (),u ∈ H () and F ∈ 0 1 k N −k C H () for 0 ≤ k ≤ N satisfy the linear compatibility conditions of order N . Then Problem (2.2) has a unique solution belonging to k N +1−k N 1 u ∈ C H ∩ C H for all 0 ≤ k ≤ N + 1. The solution satisﬁes for all T > 0 the energy estimates u ∞ u +u +F , (2.9) 2 2 2n 2n 0 1 L L L n+2 n+2 L L L ∇ u u N +1 +u N +F (2.10) ∞,2;N 1,2;N t ,x 0 1 H H Y Y T T with a constant independent of T . We next recall Strichartz estimates for the exterior problem, following [1,6,11]. These estimates are valid on the exterior of any strictly convex obstacle with smooth boundary in R , n ≥ 2. With our notations one has 1 1 1 q r S(t )g g s , s = − + (2.11) L L ˙ H 2 r q provided n ≥ 3 and 2 n−1 n−1 2(n−1) + = , 2 < q ≤∞, 2 ≤ r < . (2.12) q r 2 n−3 141 Page 8 of 29 P. D’Ancona A couple (q, r ) as in (2.12)iscalled admissible; note that the endpoint (q, r ) = 2(n−1) (2, ) is not included and it is not known if the estimate holds also in this case. n−3 One can further extend the range of indices by combining (2.11) with Sobolev embedding. We shall focus on the following special case: S(t )g q r g (2.13) L L ˙ valid for n ≥ 3 and for couples of indices of the form 2 2n q = , r = , 0 ≤ δ< 1. (2.14) δ n−2−δ Then we have: Proposition 2.3 The solution u to (2.2) saitisﬁes, for any interval I containing 0, the Sobolev–Strichartz estimate u u +u 2 +F 1 (2.15) r 0 ˙ 1 1 2 L L H L L I I provided n ≥ 3 and (q, r ) satisfy (2.14). Proof The proof is a standard application of the Christ–Kiselev Lemma to the repre- sentation (2.3) of the solution. Differentiating (2.2) with respect to t and applying the usual recursive procedure one gets, more generally, the following higher order estimates; Proposition 2.4 Assume the data (u , u , F ) of (2.2) satisfy the compatibility condi- 0 1 tions of order N ≥ 1. Then, for all (q, r ) as in (2.14) and for any interval I of length |I | 1 containing 0, the solution satisﬁes the estimates u q,r ;N u N +1 +u N +F 1,2;N . (2.16) 0 1 H H Y Y I I Proof We give a sketch of the proof. Applying ∂ to the equation, by (2.15) we get u r u 2 +u 1 +F (0, ·) 2 +F 1 2 t 0 ˙ 1 ˙ t L L H H L L I I since u (0) = u +F (0, x ) from the equation. We note that F (0, ·) 2 F 1,2;1 tt 0 provided the interval has length |I | 1, and this gives the estimate for u . In a similar way one can estimate all derivatives ∂ u. We next estimate u = u − F: tt q q q u ≤u +F . r tt r r L L L L L L I I I The u term has already been estimated. As to the second term, we note that tt F 1,2;2 F 1 F 2n L H Y ∞ I n−2 L L I On the supercritical defocusing NLW outside a ball Page 9 of 29 141 which is the endpoint δ = 0in (2.14). Moreover, F 1,2;2 F 1 +F 1 L H L H I I and by interpolation and Sobolev embedding F 1,2;2 F 2 F 2n 3/2 L H Y 2 n−3 I L L which is the other endpoint δ = 1in (2.14). This argument can be modiﬁed in the case n = 3 by using the Sobolev embedding into BM O instead of L .Again by interpolation we get F F 1,2;2 r L L for all (q, r ) as in (2.14), and we conclude u u 3 +u 2 +F 1,2;2 . r 0 1 H H L L Y Also by interpolation this covers the case N = 1of (2.16). For larger values of N one 1,2;N q,r ;m−2 proceeds in a similar way by recursion, using the embedding Y → Y just proved for all (q, r ) as in (2.14). We shall also need estimates for the exterior wave equation with a time dependent potential. We denote by u = S (t ; t )g the solution of the mixed problem with initial V 0 data at time t = t u + V (t , x )u = 0, u(t , x ) = 0, u (t , x ) = g, u(t , ·)| = 0 0 t 0 ∂ and by u = S (t ; t ) f (which is not ∂ S g) the solution of 0 t V u + V (t , x )u = 0, u(t , x ) = f , u (t , x ) = 0, u(t , ·)| = 0. 0 t 0 ∂ If V is a sufﬁciently smooth potential with good behaviour at inﬁnity, the existence and uniqueness of a solution is standard. The solution of the full problem u + V (t , x )u = F (t , x ), u(t , x ) = u , u (t , x ) = u , u(t , ·)| = 0. 0 0 t 0 1 ∂ (2.17) can be represented by Duhamel as u = S (t ; t )u + S (t ; t )u + S (t ; s)F (s)ds. (2.18) 0 0 V 0 1 V V t Note also that it is not necessary to modify the compatibility conditions, provided the potential V has a sufﬁcient regularity. Indeed the correct condition would require j −2 j j −2− j −2 h = ∂ u(0, x ) = h + ∂ V (0, x )h + ∂ F (0, x ) ∈ H j j −2 t t t =0 141 Page 10 of 29 P. D’Ancona j −2− 1 1 but the term ∂ V (0, x )h is already in H since h ∈ H , and can be omitted. 0 0 By Duhamel we can write S (t ; s), S (t , s) as perturbations of S(t ), ∂ S(t ): V t S (t ; t ) = S(t − t ) − S(t − s)V (s, x )S (s; t )ds, (2.19) V 0 0 V 0 S (t ; t ) = ∂ S(t − t ) − S(t − s)V (s, x )S (s; t )ds. (2.20) 0 t 0 V 0 Proposition 2.5 (Perturbed energy–Strichartz estimate) Let n ≥ 3,m ≥ 1. Assume the data (u , u , F ) satisfy the compatibility conditions of order m, and that 0 1 V 1,n;m < ∞. (2.21) Then for any interval I containing t the solution of Problem (2.17) satisﬁes u ∞ 2 u 2 +u 2n +F 2n , (2.22) 0 1 L L L n+2 n+2 L L L ∇ u ∞,2;m +u q,r ;m u m+1 +u +F 1,2;m (2.23) t ,x 0 H 1 H Y Y Y I I I provided the couple (q, r ) is of the form (2.14). Proof We can assume I = I (t ) =[t , t ]; the proof for t < t is identical. Since u 0 0 solves u = F − Vu,by(2.9) we get u(t ) u +u +F + Vu ds. 2 2 2n 2n 2n 0 1 L L 1 t n+2 n+2 n+2 L L L L Noting that t t Vu 2n ds ≤ V (s) u(s) 2ds t t 0 0 n+2 and a(s) =V (s) n is integrable by (2.21), by Gronwall’s Lemma we deduce (2.22). In a similar way, by (2.10) and (2.16) we can write ∇ u ∞,2;m +u q,r ;m u m+1 +u +F 1,2;m +Vu 1,2;m . t ,x 0 1 H Y Y Y Y I (t ) I (t ) I (t ) I (t ) We have t t Vu 1,2;m b(s)u 2n ds b(s)∇ u ∞,2;m ds ∞, ;m t t Y 0 0 Y n−2 I (t ) I (s) I (s) where b(s) = ∂ ∂ V (s) j +|α|≤m x is integrable on R by (2.21). Using again Gronwall’s inequality we obtain (2.23). On the supercritical defocusing NLW outside a ball Page 11 of 29 141 3 The global radial solution This section is devoted to the proof of Proposition 1.2 and Theorems 1.3, 1.4.We begin with a few preliminary results on the mixed problem u + f (u) = 0, u(0, x ) = u , u (0, x ) = u , u(t , ·)| = 0. (3.1) 0 t 1 ∂ For data of low regularity, a solution to (3.1) is intended to be a solution of the integral equation u = S(t )u + ∂ S(t )u − S(t − s) f (u(s))ds (3.2) 1 t 0 with S(t ) as in (2.1). We will give only sketchy proofs of standard results, which are virtually identical to the corresponding ones for semilinear wave equations on R (for which we refer e.g. to Chapter 6 of [9]). Lemma 3.1 Assume f : R → R is (globally) Lipschitz with f (0) = 0. Then for any 1 2 initial data (u , u ) ∈ H × L () Problem (3.1) has a unique solution u(t , x ) in 0 1 1 1 2 CH () ∩ C L (). The solution satisﬁes the energy bound for all t > 0 u(t , ·) 1 +∂ u(t , ·) 2 ≤ C ( f , t ) u 1 +u 2 . (3.3) t 0 1 H () L () H () L () 1 1 2 Proof Apply a contraction argument in the space C ([0, T ]; H ()) ∩ C ([0, T ]; L ()) to (3.2), with T > 0 sufﬁciently small, using the energy estimates (2.4), (2.5). 1 2 The lifespan T depends only on the H × L norm of the data, thus we can iterate to a global solution. Estimate (3.3) is a byproduct of the proof. Lemma 3.2 Consider the solution u constructed in Lemma 3.1. Assume in addition 2 2 1 1 2 2 that f ∈ C ,u ∈ H () ∩ H (),u ∈ H (). Then u belongs to C L () ∩ 0 1 0 0 1 1 2 C H () ∩ CH () and solves (3.1) in both distributional and a.e. sense. If we further assume that 0 ≤ f (s)s F (s) for s ∈ R, where F (s) = f (σ )dσ , then the solution satisﬁes for all times the energy identity 1 2 1 2 E (t ) = E (0), E (t ) := |∇ u| + |u | + F (u) dx . (3.4) x t 2 2 Proof Differentiate the equation once w.r.to time and call formally v = u . Then v must solve v − v + f (u)v = 0 tt ∞ 1 where we may regard f (u) as an L coefﬁcient. As data we take v(0) = u ∈ H () and v (0) = u (0) = u(0) − f (u(0))v(0) ∈ L (). By the previous result, this t tt problem has a global solution v(t , x ), satisfying an energy estimate like u. Since u(0) + v solves the original equation (3.1), by uniqueness for (3.1) we conclude that v = u and hence u has the claimed regularity. The energy conservation follows by direct computation. Note that the assumption 0 ≤ sf (s) F (s) is sufﬁcient to prove the existence of 1 2 a global weak solution for data in H () × L (),evenif f is not Lipschitz (Segal’s 0 141 Page 12 of 29 P. D’Ancona Theorem). This can be proved like in the case of the whole space R by approximating f with a sequence of truncated Lipschitz functions and using weak compactness. The weak solution thus constructed satisﬁes then a weaker energy inequality E (t ) ≤ E (0) (proved using Fatou’s Lemma). We shall not need this variant in the sequel. For smoother data, one can prove a local existence theorem which does not require a global Lipschitz condition, similarly to the case = R . However, one must assume suitable compatibility conditions, analogous to the linear ones from Deﬁnition 2.1. Deﬁne formally a sequence of functions ψ , j ≥ 0 as follows: differentiating the equation u = u + f (u) with respect to time, set tt j j −2 ψ =u(0, x )=u ,ψ =u (0, x )=u ,ψ =∂ u(0, x )=ψ + ∂ ( f (u))| , 0 0 1 t 1 j j −2 t =0 t t where values of ∂ u(0, x ) for 0 ≤ k ≤ j − 2, required to compute ψ , are set recursively equal to ψ . For instance, ψ = u(0, x ) + f (u(0, x )) = ψ + f (ψ ), 2 0 0 ψ = ψ + f (u)u = ψ + f (ψ )ψ , 3 1 t 1 0 1 2 2 ψ = ψ + f (u)u + f (u)u = ψ + f (ψ )ψ + f (ψ )ψ 4 2 tt 2 0 0 2 t 1 and so on. Then we have: Deﬁnition 3.3 (Nonlinear compatibility conditions) We say that the data (u , u , f ) 0 1 satisfy the nonlinear compatibility conditions of order N ≥ 1if (u , u ) ∈ 0 1 N +1 N N −2 H () × H (), f ∈ C (R; R) and we have ψ ∈ H () for 0 ≤ j ≤ N . (3.5) Remark 3.1 Conditions (3.5) are implied by a number of simpler assumptions on the N −2 data. For instance, if f ∈ C and one assumes n n +N +1 +N 2 2 u ∈ H , u ∈ H (3.6) 0 1 0 0 then one checks easily that (u , u , f ) satisﬁy the nonlinear compatibility conditions 0 1 of order N . N +1 N N Lemma 3.4 (Local existence) Let N > , (u , u ) ∈ H () × H (),f ∈ C 0 1 and assume (u , u , f ) satisfy the nonlinear compatibility conditions of order N . Then 0 1 N +1 N there exists 0 < T ≤+∞, depending only on the H × H norm of (u , u ),such 0 1 k N +1−k that Problem (3.1) has a unique solution u ∈ C ([0, T ); H ()), 0 ≤ k ≤ N +1. N 1 The solution belongs to C ([0, T ); H ()). Moreover, if T is the maximal time of existence of such a solution, then either ∗ ∗ T =+∞ or u(t , ·) →∞ as t ↑ T . Proof The existence part is completely standard; it is usually proved for more general quasilinear equations, which require higher smoothness of the data; see e.g. Theo- rem 3.5 in [10], where local existence is proved for a nonlinear term of the form On the supercritical defocusing NLW outside a ball Page 13 of 29 141 α n f (t , x,∂ ∂ u) with j +|α|≤ 2, j ≤ 1 (and a regularity of order + 8 is imposed t x on the data). The proof is based on a contraction mapping argument, combined with Moser type estimates of the nonlinear term. The ﬁnal blow up alternative in the state- ment is a byproduct of the proof. 3.1 Proof of Proposition 1.2 p−1 Fix M > 0 and deﬁne f (s) = min{|s|, M } s. Then f is Lipschitz and the M M problem u + f (u) = 0, u(0, x ) = u , u (0, x ) = u , u(t , ·)| = 0 (3.7) M 0 t 1 ∂ 2 2 1 1 2 has a global, unique, radially symmetric solution u ∈ C L ()∩C H ()∩CH () satisfying the bound (3.4) with F = F = f . Combining (3.4) with (1.2) we get M M p+1 −1 2 |x | |u(t , x )|≤ C K , K := u +u 2 +u (3.8) 0 0 ˙ 1 1 0 L p+1 for some universal constant C . Since |x|≥ 1on , this gives |u(t , x )|≤ C K and if we choose M = C K + 1 we see that f (u(t , x )) = f (u(t , x )), i.e., u(t , x ) is 0 M a global solution of the untruncated problem p−1 u +|u| u = 0, u(0, x ) = u , u (0, x ) = u , u(t , ·)| = 0. (3.9) 0 t 1 ∂ The same argument guarantees also uniqueness of radially symmetric solutions. More 2 ∞ generally, two solutions in H ∩L with the same data must coincide, as an elementary argument based on classical energy estimates shows. Since energy estimates are valid also on domains of dependence, by ﬁnite speed of propagation, the same arguent 2 ∞ shows uniqueness of solutions in H ∩ L . In particular, a locally bounded solution loc with radial data must coincide with the radial solution. The proof of Proposition 1.2 is concluded. 3.2 Proof of Theorem 1.3 We now prove the decay estimate (1.9), using the Penrose transform. We recall its deﬁnition. Describe the sphere S using coordinates (α, θ ) with α ∈ (0,π) and θ ∈ n−1 n−1 n−1 n 2 S ,as S = (0,π) × S . Denoting with dθ the metric of S , the metric α n−1 α,θ θ θ on S can then be written as 2 2 2 dα + (sin α) dθ . n−1 n−1 n + Similarly, on R = R × S , use polar coordinates (r,θ) with r ∈ (0, +∞) and x r θ n−1 2 2 2 θ ∈ S , so that the euclidean metric can be written dr + r dθ . Then we can n−1 S 141 Page 14 of 29 P. D’Ancona deﬁne the Penrose map : R × R → R × S as t x T : (t , r,θ) → (T ,α,θ) where T = arctan(t + r ) + arctan(t − r ), α = arctan(t + r ) − arctan(t − r ). The map (t , r ) → (T ,α) takes the quadrant {(t , r ) : t ≥ 0, r ≥ 0} to the triangle {(T ,α) : T ≥ 0, 0 ≤ α< π − T } + n so that maps R × R to the positive half of the Einstein diamond + n−1 E ={(T ,α,θ) : T ≥ 0, 0 ≤ α< π − T,θ ∈ S } The boundary |x|= 1 i.e. r = 1 is mapped by to a curve described parametrically by (T ,α) = (arctan(t + 1) + arctan(t − 1), arctan(t + 1) − arctan(t − 1)) (3.10) for t ≥ 0. We denote this curve by α = (T ), T ∈[0,π) or T = γ(α), α ∈[0,π) −1 (i.e., γ = ). One has the explicit (but not particularly useful) formulas π − (T ) = + arcsin(2 cos T ), γ (α) = arccos(sin α − cos α). It is easy to check that γ (α) < −1. (3.11) We denote by ω the conformal factor 2 1/2 ω = cos T + cos α = , s= (1 + s ) . t + r t − r + + Note that ω> 0on E .The inverseof (deﬁned on E ) can be written −1 −1 −1 (t , r,θ) = (T ,α,θ) = (ω sin T ,ω sin α, θ ). We deﬁne a new function U (T ,α,θ) via n−1 u(t , rθ) = ω U ◦ (t , r,θ). (3.12) On the supercritical defocusing NLW outside a ball Page 15 of 29 141 Since u, U are independent of θ, we shall write simply U (T ,α). Commuting with gives n−1 n+3 (n−1) 2 2 n n u = (ω U ◦ ) = ω ( U + U ) ◦ R ×R t R ×S T 4 α,θ so that U is a solution of the equation (n−1) ν p−1 n−1 n+3 U + U + ω |U | U = 0,ν = p − (3.13) R×S 4 2 2 on the subset E of R × S given by the conditions E := {(T , a) : 0 ≤ T <π, (T ) ≤ α< π − T } which is the image of R × via . We introduce also the notation D ={(α, θ ) ∈ S : (T ) ≤ α< π − T }, 0 ≤ T <π for the slice of (R × ) at time T . Note that in coordinates, Eq. (3.13) reads (n−1) 2 2 cos α ν p−1 ∂ U − ∂ U − (n − 1) ∂ U + U + ω |U | U = 0. T sin α 4 We plan to extend the solution beyond the line T + α = π, i.e., in the region where ω< 0. Thus we consider the following extended equation on (T ,α) ∈[0,π ] : (n−1) 2 2 cos α ν p−1 ∂ U − ∂ U − (n − 1) ∂ U + U + ω |U | U = 0 (3.14) T α sin α 4 where we have replaced ω by ω if T + α ≤ π, ω := 0if T + α> π. The solution U satisﬁes the identity 2 2 2 |∂ U | +|∂ U | ω (n−1) n−1 T α p+1 2 ∂ (sin α) + |U | + |U | = 2 p+1 8 ν−1 n−1 ν ω n−1 p+1 = ∂ (sin α) ∂ U ∂ U − (sin α) (sin T )|U | . α α T p+1 (3.15) We can now extend U to a larger domain in the cylinder R × S . Recall that the data u , u satisfy C (u , u )< ∞ with C (u , u ) as in (1.7). Thus if we ﬁx a 0 1 M 0 1 M 0 1 smooth cutoff function χ(x ) equal to 0 for |x|≤ M +1 and equal to 1 for |x|≥ M +2, we have χu +χu < ∞. N +1,N N ,N +1 0 0 0 1 0 0 H H 141 Page 16 of 29 P. D’Ancona Denote by U , U the functions obtained by applying the transformation (3.12)to 0 1 χu ,χu respectively (with t = 0). By the ﬁrst Lemma in Section 4 of [2]wehave 0 1 then U N +1 +U N < ∞. n n 0 0 1 0 H (S ) H (S ) In order to solve (3.13) locally via the energy method we require that the coefﬁcient ν ν N ω be sufﬁciently smooth i.e. ω ∈ C . This is true as soon as n−1 n+3 11 ν = p − > N which is implied by p > . 2 2 2 Then a standard local existence result guarantees the existence of a local solution U to Eq. (3.13) with data U , U on some strip [0,δ) × S . The lifespan δ, which can be 0 1 assumed 1, depends only on δ = δ(C (u , u ), n, p) (3.16) M 0 1 where C (u , u ) was deﬁned in (1.7). Comparing U with the solution U constructed M 0 1 above, and noting that Eq. (3.13) has ﬁnite speed of propagation equal to 1, by local uniqueness we see that U , U must coincide on the forward dependence domain ema- nating from the set T = 0, 2arctan(M + 2)<α < π. Thus we can glue the two solutions, at least for 0 < T <δ/2, and we obtain an extended solution of (3.13), which we denote again U (T ,α), deﬁned on the larger domain + δ n E ∪ ([0, ) × S ). We next prove that the energy of U , deﬁned as 2 2 ν 2 π |∂ U | +|∂ U | n−1 ω p+1 (n−1) 2 T α E (T ) = (sin α) + |U | + |U | dα (T ) 2 p+1 8 remains bounded for 0 ≤ T <δ/2. To this end we integrate identity (3.15)onaslice T < T < T ,(T)<α < π, 1 2 for arbitrary times 0 ≤ T ≤ T <δ/2. Dropping negative terms at the RHS, we are 1 2 left with the inequality n−1 1 2 2 E (T ) − E (T ) ≤ (sin α) ν ∂ U ∂ U − ν |∂ U | +|∂ U | (3.17) 2 1 α α T T T α T =γ(α) where ν ,ν are the components of the exterior normal to the curve T = γ(α), i.e., α T 1 γ (α) ν =− ,ν = . T α 2 2 1 + γ (α) 1 + γ (α) On the supercritical defocusing NLW outside a ball Page 17 of 29 141 The Dirichlet condition U (γ (α), α) = 0 along the curve implies (∂ U + γ (α)∂ U )| = 0. α T T =γ(α) Thus the RHS of (3.17) is equal to 1 − γ (α) n−1 2 = (sin α) |∂ U | . 2 1 + γ (α) Recalling (3.11), we see that the RHS of (3.17) is negative, and we conclude that the energy is nonincreasing as claimed: E (T ) ≤ E (0) for all 0 ≤ T <δ/2. (3.18) Now, consider the set, which is a dependence domain for (3.13) (keep in mind that the speed of propagation for (3.13) is exactly 1): δ δ D ={(T ,α) : ≤ T <π, (T)<α < π − T + }. 3 4 We have already extended U to the part of D in the time strip δ/3 < T <δ/2, and our next goal is to prove that U can be extended to a bounded solution of (3.13)on the whole set D. Clearly, it is sufﬁcient to prove an a priori L bound of the solution on this domain in order to achieve the result via a continuation argument. To this end we prove an energy estimate similar to the previous one, but now we integrate identity (3.15) over the slice T < T < T ,(T)<α < π − T + δ/4, 1 2 where δ/3 ≤ T < T <π are ﬁxed, and we denote by F (T ) the energy 1 2 2 2 2 π −T +δ/4 |∂ U | +|∂ U | ω (n−1) n−1 T α p+1 2 F (T ) := (sin α) + |U | + |U | dα. (T ) 2 p+1 8 After integration of (3.15), the terms on the side α = π − T + δ/4giveanegative contribution at the RHS which can be dropped, as in the standard energy estimate, since the speed of propagation is 1. Proceeding as before, we are left with the inequality n−1 2 2 F (T ) − F (T ) ≤ (sin α) ν ∂ U ∂ U − ν |∂ U | +|∂ U | 2 1 α α T T T α T =γ(α) and again the RHS here is negative thanks to (3.11). We conclude that the energy F (T ) is nonincreasing: F (T ) ≤ F (δ/3) for all δ/3 ≤ T <π. Since F (δ/3) ≤ E (δ/3),by(3.18) we conclude F (T ) ≤ E (0) for all δ/3 ≤ T <π. (3.19) 141 Page 18 of 29 P. D’Ancona To proceed, we need a Lemma: Lemma 3.5 Let I ⊆ (0, +∞) be a bounded interval and n ≥ 3. Then for any V ∈ H (I ) we have 1 1 −1 n−1 2 −1 n−1 2 2 2 2 sup s |V (s)| ( s |V (s)| ds) +|I | ( s |V (s)| ds) (3.20) I I s∈I with a constant independent of I and V . Proof of the Lemma Pick any points α, β ∈ I with β ≥ α/3. We ﬁrst prove that n n −1 −1 n−1 2 2 2 2 α |V (α)| s |V | + β |V (β)|. (3.21) We have two cases: either α ≤ β or α ≥ β ≥ α/3. If α ≤ β, we write β β β n−1 2 1−n |V (α)|≤ |V |+|V (β)|≤ s |V | ( s ) +|V (β)|. α α α 1−n 2−n Using the inequality s ds ≤ α and recalling that α ≤ β we obtain (3.21). If α ≥ β ≥ α/3, we have in a similar way α α α n−1 2 1−n |V (α)|≤ |V |+|V (β)|≤ s |V | ( s ) +|V (β)|. β β β 1−n 2−n Now s ds ≤ β and we get n n −1 n−1 2 2 −1 2 2 β |V (α)|≤ s |V | + β |V (β)| and recalling that β ≥ α/3 we obtain again (3.21). Next, split I in thirds I = I ∪ I ∪ I (with I at the left and I at the right). If 1 2 3 1 3 α ∈ I ∪ I ,pick β ∈ I arbitrary and apply (3.21) to get 1 2 3 n n −1 n−1 2 −1 2 2 α |V (α)| s |V | + inf β |V (β)|. β∈I I 3 Since 1 n 1 n n−1 2 −1 −1 2 2 2 2 ( s |V | ) ≥ inf β |V (β)|· ( sds) |I | inf β |V (β)| I I 3 3 I I 3 3 2 2 (b−a) b −a (indeed sds = ≥ ) our claim (3.20) is proved for the points in I ∪ I . 1 2 a 2 2 On the other hand, if α ∈ I ,wepick β ∈ I arbitrary, we apply (3.21), and we get 3 2 n n −1 n−1 2 2 −1 2 2 α |V (α)| s |V | + inf β |V (β)| β∈I and the same argument gives again (3.20). On the supercritical defocusing NLW outside a ball Page 19 of 29 141 We apply (3.20)to U (T ,α) at a ﬁxed T ∈ (δ/3,π) on the interval I =[(T ), π − T + δ/4]⊆ (0, K]= (0,π − ]; note that |I|≥ δ/4. We get π −T +δ/4 −1 −1 n−1 2 2 sup α |U (T ,α)| δ α (|∂ U | +|U | )dα . (T ) For α ∈[0,π − ] we have sin(π − δ/12) −1 sin α ≥ · α ⇒ α δ sin α, π − δ/12 thus substituting in the previous inequality we get, for all α ∈[(T ), π − T + δ/4], n n+1 π −T +δ/4 −1 − n−1 2 2 2 2 (sin α) |U (T ,α)| δ (sin α) (|∂ U | +|U | )dα . (T ) Recalling the deﬁnition of F (T ) and (3.19) we conclude n n+1 −1 − 1/2 2 2 (sin α) |U (T ,α)| δ E (0) . (3.22) We now convert (3.22) into an estimate for u(t , x ). Since sin α = ωr, we obtain n n 1−n n 1 1 −1 −1 −1 2 2 2 2 2 2 (sin α) |U (T ,α)| (ωr ) ω |u(t , x )|= r t + r t − r |u(t , x )|. On the other hand, a change of variable shows that −1 U (0, ·) = 2r u 2 2 n L (S ) L () where S denotes the image of {t = 0}× via the Penrose transform, p−1 √ p−1 1−n n −1 p+1 p+1 U (0, ·) = 2 r u , p+1 n 0 L () L (S ) −1 ∂ U (0, ·) 2 n = 2 r u 2 T 1 L () L (S ) and −1 ∂ U (0, ·) r u 2 +r ∇ u 2 . α 2 n 0 x 0 L () L () L (S ) This gives the estimate p+1 p−1 n −1 1/2 −1 2 p+1 E (0) x u 2 +x u +x (|∇u |+|u |) 2 (3.23) 0 0 0 1 L p+1 L and recalling (3.22) (and the dependence of δ in (3.16)), we conclude the proof of Theorem 1.3. 141 Page 20 of 29 P. D’Ancona 3.3 Proof of Theorem 1.4 Let u be the solution given by Proposition 1.2 and Theorem 1.3, and v be the local solution given by Lemma 3.4, maximally extended to a lifespan [0, T ). Since the data are radial, v is also radial, and by the uniqueness part of Proposition 1.2 we see ∗ ∗ that u ≡ v. In particular, v is bounded as t ↑ T , hence T =+∞. This proves the regularity of the radial solution. It remains to prove the uniform bounds (1.11). For the L norm we have −1 p u(t , ·) 2 ≤u 2 + u 2 + |u| 2n ds. 0 1 L L L n+2 Then we can write, if n ≥ 3, n+2 n+2 n+2 n+2 p− p− p n−2 n−2 n−2 n−2 |u| 2n ≤u u u ∇ u ∞ ∞ x 2n 2 L L L () n+2 L n−2 by Hölder and Sobolev embedding, and we note the consequence of (1.9) (also valid only if n ≥ 3) −1 u C ((u , u ) )t (3.24) L 0 1 M and the conservation of energy (1.5). Summing up, we obtain n+2 n+2 p− n+2 n−2 −p+ 2(n−2) n−2 u(t , ·) 2 u 2 +u 2n +(u , u ) E (0) s ds 0 1 0 1 L L M 0 n+2 n+2 2n where the integral converges since p > 1 + = . Noting that n−2 n−2 E (0) ≤ C ((u , u ) ), u x u (u , u ) , 2n 2 0 1 M 1 1 0 1 M n+2 we get the uniform bound u(t , ·) 2 ≤ C ((u , u ) ). (3.25) 0 1 M Estimate (1.11)for k = 1 is a consequence of the energy conservation E (t ) = E (0) and of (3.25). For k > 1 we proceed by induction on k. By the energy estimate (2.10) we have ∇ u u +u +|u| . ∞,2;k k+1 k 1,2;k t ,x 0 1 H H Y Y T T We apply Gagliardo–Nirenberg estimates to estimate the last term. Handling separately p−1 j the highest order terms ∼u ∂ u k− j , we get t H L j ≤k T j α p |u| = ∂ ∂ |u| dt 1,2;k 2 t x L Y 0 j +|α|≤k T p−1 p−k (u +u ) 1 + ∂ u dt ·u ∞,2;k ∞ ∞ L 0 L L ≤k−1 On the supercritical defocusing NLW outside a ball Page 21 of 29 141 provided p > k + 1. Since u is radial in x we have ∂ u u ∞,2;k ≤k−1 p−k which is bounded by the induction hypothesis; on the other hand u is integrable since p > k + 1by (3.24). Thus the right hand side is ﬁnite and this proves (1.11). 4 Global quasiradial solutions This section is devoted to the proof of Theorem 1.5. Let u be the global radial solution with initial data (u , u ) given by Proposition 1.2 0 1 and Theorem 1.4, and let v be the local solution with data (v ,v ) given by Lemma 0 1 3.4. Denote by w = v − u the difference of the two solutions, which satisﬁes the equation p−1 p−1 w +|u + w| (u + w) −|u| u = 0. (4.1) We can write p−1 p−1 p−1 2 |u + w| (u + w) −|u| u = p|u| w − w · F [u,w] where p−3 F [u,w]=−p(p − 1) |u + σw| (u + σw)(1 − σ)dσ (4.2) so that w satisﬁes 2 p−1 w + V (t , x )w = w · F [u,w], V (t , x ) = p|u| . For j ≤ p − 3 we can write j j p−3−μ j μ 2 1 n,∞ n,∞ ∂ V n,1 u u ∂ u ... ∂ u (4.3) ∞ W W t W 2 t t μ=0 L where j + ··· + j = j. By Sobolev embedding and energy estimates (1.11)wehave 1 μ n,∞ ∂ u u ∞,2;N ≤ C (u , u ) , u N , u N −1 W 0 1 M 0 1 t Y H H provided N > n + + j, and this implies, recalling (3.24), j +3−p ∂ V (t ) n,1 t (4.4) t W provided (u , u ) +u +u < ∞, N > n + j . (4.5) N N −1 0 1 M 0 1 H H 2 141 Page 22 of 29 P. D’Ancona Now let m ≥ 1 to be chosen and assume u , u satisfy (4.5) with 0 1 N > m + n while w ,w satisfy the compatibility conditions of order m. We see that if p > m + 4 0 1 the assumptions of Proposition 2.5 are satisﬁed and the estimates (2.22), (2.23)are valid. Consider now the problem w+V (t , x )w = w ·F [u,w],w(0, x ) = w ,w (0, x ) = w ,w(t , ·)| = 0 0 t 1 ∂ (4.6) p−1 where w = v − u , j = 0, 1, and V (t , x ) = p|u| . Deﬁne the spacetime norm j j j M (T ) =w ∞,2;m+1 +w q,r ;m Y Y where the couple (q, r ), satisfying (2.14), will be made precise below. Our goal is to prove the a priori estimate 2 p M (T ) + M (T ) + M (T ) (4.7) where is a suitable norm of the data (w ,w ). 0 1 Applying (2.22), (2.23)to(4.6) we get M (T ) w +w +w F [u,w] (4.8) m+1 2n 2n 0 1 1,2;m 1 n+2 n+2 H ∩L Y ∩L L T I We must estimate the last term. We begin by noticing that p−2 p−2 2 2 w F [u,w] w u +w 2n ∞ ∞ 4n L L n+2 L n+2 and if we assume m + 1 > so that w w m+1, we get n+2 n−2 2 n n 2 2 w(t ) ≤w w w ≤ M (T ) 4n ∞ m+1 L H n+2 for 0 ≤ t ≤ T . If we assume in addition m > and use the embedding w m,r w , we can write p−2 q p−2−q q p−q−2 w ≤w w ≤w M (T ) ∞ m,r m,r m+1 L W W −1 provided p > q + 2. Since |u(t , ·)| t (recall (3.24)), we obtain 2 2 2−p p−q w F [u,w] M (T ) t + M (T ) w 2n m,r n+2 L On the supercritical defocusing NLW outside a ball Page 23 of 29 141 and integrating in time we conclude 2 2 p w F [u,w] 2n ds M (T ) + M (T ) . (4.9) n+2 L L We next estimate 2 α 2 w F [u,w] = ∂ ∂ (w F [u,w]) . 1,2;m 1 2 j +|α|≤m t x L L Recalling the expression of F in (4.2), we may expand the derivative as a ﬁnite sum j 1 α 2 ∂ ∂ (w F [u,w]) = G · W W U · ··· · U dσ 1 2 1 ν t x where 0 ≤ ν ≤ m and, assuming p ≥ m + 2, h α h α 1 1 2 2 • W = ∂ ∂ w and W = ∂ ∂ w 1 x 2 x t t j β k k • U = ∂ ∂ (u + σw) k x • h + h + j + ··· + j = j, α + α + β + ··· + β = α, and j +|α|≤ m 1 2 1 ν 1 2 1 ν • it is not restrictive to assume that j +|β |≥ j +|β | for all k and that h +|α |≥ ν ν k k 2 2 h +|α | 1 1 p−ν−2 p−ν−2 ∞ ∞ ∞ • G satisﬁes |G| |u + σw| so that G (u +w ) . L L L 2 2 We take the L norm in x of each product, and we estimate it with the L norm of the derivative of highest order, times the L norm of the remaining factors; it may happen that the highest order derivative falls on w or on u + σw. Thus we need to distinguish two cases: (1) The highest order derivative is on u + σw, that is to say j +|β |≥ h +|α |. ν ν 2 2 Then we estimate ∞ ∞ ∞ ∞ ∞ GW W U ... U 2 ≤G W W U ... U U 2 . 1 2 1 ν L 1 L 2 L 1 L ν−1 L ν L L We have for 0 ≤ t ≤ T U 2 M (T ). Moreover for i <ν we have j +|α |≤ , hence if we assume m > n − 2wehave i i by Sobolev embedding U ∞ u +w ≤ C + M (T ), ∞,2;m+1 ∞,2;m+1 i L m+1 Y Y T T where in the last step we used the energy estimates (1.11)for u, with C = C ((u , u ) , u , u m ). m+1 m+1 0 1 M 0 1 H In a similar way W M (T ), i = 1, 2 i L Finally, by (3.24) and Sobolev embedding p−ν−2 −1 q p−ν−q−2 G ∞ (u ∞ +w ∞ ) (t +w m,r ) (1 + M (T )) L L L W 141 Page 24 of 29 P. D’Ancona provided m > and p ≥ m + q + 2. Summing up we have, for m > max{n − 2, n/r }, p ≥ m + q + 2, 0 ≤ t ≤ T GW W U ... U 1 2 1 ν −1 q p−ν−q−2 2 ν (t +w m,r ) (1 + M (T )) M (T ) (C + M (T )) . W m+1 We now integrate in t on [0, T ] to get 2 p−q−2 −q GW W U ... U M (T ) (1 + M (T )) (t +w )dt 1 2 m,r 1 2 1 ν L L 2 p−2 M (T ) (1 + M (T )) (4.10) with an implicit constant depending on (u , u ) , u , u m . m+1 0 1 M 0 1 H (2) The highest order derivative is on w, that is to say j +|β |≤ h +|α |.Inthis ν ν 2 2 case we estimate GW W U ... U ≤G ∞ W ∞ W U ∞ ... U ∞ U ∞ . 2 2 1 2 1 ν L 1 L 2 1 L ν−1 L ν L L L Proceeding in a similar way, we obtain again (4.10). Summing up, recalling (4.8), we have proved 2 p M (T ) w +w + M (T ) + M (T ) . (4.11) m+1 2n 0 1 n+2 H ∩L The implicit constant depends on (u , u ) +u +u m . The conditions m+1 0 1 M 0 1 H on the parameters are p ≥ m + q + 2, m > , m > n − 2 and (q, r ) satisfy (2.14), that is to say 2 n δ δ ∈ (0, 1), p ≥ m + + 2, m > n − 2, m > − 1 − . δ 2 2 All the constraints are satisﬁed if m = n − 1, p > n + 3 (4.12) with δ< 1 chosen accordingly. Recall also that in order to apply Proposition 2.5 we N N −1 assumed p > m + 4 = n + 3 and that u , u are in H × H with compatibility 0 1 3 5 conditions of order N , where N > m + n = n − 1. 2 2 To conclude the proof it is then sufﬁcient to apply a standard continuation argument; if the norm w +w m+1 2n 0 1 n+2 H ∩L On the supercritical defocusing NLW outside a ball Page 25 of 29 141 of the initial data is sufﬁciently small with respect to the hidden constant (u , u ) +u m+1 +u 0 1 M 0 1 H then the quantity M (T ) must remain ﬁnite as T →+∞ and global existence is proved. 5 Weak–strong uniqueness Following [12], we prove a more general stability result which implies Theorem 1.6 as a special case. We use the notation 2 2 p+1 |∂ u| +|∇ u| |u| t x E (u) = E (u(t )) = ( + )dx, ={|x | > 1} 2 p+1 for the energy of a solution u(t , x ) at time t of the Cauchy problem p−1 u +|u| u = 0, u(t , ·)| = 0. (5.1) Theorem 5.1 Let I be an open interval containing [0, T ],T > 0. Let u,v be two distributional solutions to (5.1) on I × such that 2 1 1 2 2 ∞ u ∈ C (I ; H ()) ∩ C (I ; H ()) ∩ C (I ; L ()), u, u ∈ L (I × ), 1 1 2 ∞ p+1 v ∈ C (I ; H ()) ∩ C (I ; L ()) ∩ L (I ; L ()). Assume in addition that v satisﬁes an energy inequality E (v(t )) ≤ E (v(0)). Then the difference w = v − u satisﬁes the energy estimate Ct 2 E (w(t )) ≤ Ce (E (w(0)) +w(0) ), t ∈[0, T ] L () where C is a constant depending on C = C (p, T , |u|+|u | ). (5.2) t L ([0,T ]×) Proof We shall need the following easy estimate of the L norm of w = v − u: t t 2 2 2 2 w(t ) ≤ 2 w ds + 2w(0) ≤ 2T E (w(s))ds + 2w(0) . (5.3) 2 2 2 2 0 0 L L L L The difference w = v − u satisﬁes in the sense of distributions p−1 p−1 w +|u + v| (u + w) −|u| u = 0. We expand E (v) = E (u) + A(t ) + B(t ) 141 Page 26 of 29 P. D’Ancona where 2 2 p+1 p+1 |∂ w| +|∇ w| |u + w| −|u| t x p−1 A(t ) = dx + ( −|u| uw)dx 2 p + 1 p−1 B(t ) = (u w +∇u ·∇w +|u| uw)dx . t t From the energy inequality for v, and the conservation of energy for the smooth solution u,wehave 0 ≤ E (v(0)) − E (v(t )) = A(0) − A(t ) + B(0) − B(t ). (5.4) We get easily 1 σ p+1 p+1 |u+w| −|u| p−1 p−1 2 −|u| uw = p |u + τw| dτ dσ |w| p+1 0 0 2−p 2 p+1 p−1 2 ≥ |w| − |u| |w| p+1 2 which implies p p−1 −p 2 A(t ) ≥ 2 E (w(t )) − C w , C = u . 2 L ([0,T ]×) Recalling (5.3), this gives −p 2 A(t ) ≥ 2 E (w(t )) − C E (w(s))ds − C w(0) for some C = C (T , u ). On the other hand L ([0,T ]×) p+1 p+1 |u+w| −|u| p−1 p+1 2 −|u| uw ≤ C (p, u )(|w| +|w| ) p+1 which implies A(0) ≤ CE (w(0)) + C w(0) and in conclusion −p 2 A(0) − A(t ) ≤−2 E (w(t )) + C E (w(s))ds + C w(0) , (5.5) with C = C (p, T , u ∞ ). In order to estimate B(t ), we need the following remark. Assume the function 1 1 2 W (t , x ) ∈ C ([0, T ]; H ()) ∩ C ([0, T ]; L ()) (5.6) satisﬁes in D ((0, T ) × ),for some q ∈ (1, ∞), W = F (t , x ), F ∈ L ((0, T ) × ). (5.7) On the supercritical defocusing NLW outside a ball Page 27 of 29 141 ∞ ∞ Then for all χ(t ) ∈ C ((0, T )) and U (t , x ) ∈ C (R × ) we have the identity c c χ (t )(U W +∇U ·∇W )dxdt =− χ(t )(W U + U F )dxdt . (5.8) t t t t Now assume 2 1 1 2 2 ∞ U ∈ C (I ; H ()) ∩ C (I ; H ()) ∩ C (I ; L ()), U , U ∈ L (I × ) (5.9) 2 q on an open interval I ⊃[0, T ], so that U ∈ C (I ; L ()) and U ∈ L (I × ).We can approximate U with a sequence U ∈ C (I × ) in such a way that ∂ U → U , ∇U →∇U , U → U in L ((0, T ) × ) and ∂ U → U in L ((0, T ) × ) as → 0 (e.g., extend U as 0 on R \ , apply a radial change of variables u(t , x ) = −1 u(t,(1− )x ), truncate with a smooth cutoff ψ(|x|− ) where ψ = 1for |x|≤ 1 and ψ = 0for |x|≥ 2, and ﬁnally convolve with a delta sequence of the form ρ (x )σ (t )). Applying (5.8)to U and letting → 0 we obtain that (5.8) holds for any functions U , W satisfying (5.6), (5.7), (5.9) and any χ ∈ C ((0, T )). Now, consider a sequence of test functions χ (t ) ∈ C ((0, T )), non negative, such that χ ↑ 1 pointwise, t ∈ (0, T ]. We can write k [0,t ] p−1 B(t ) = B(0) + lim χ (t )(u w +∇u ·∇w +|u| uw)dxdt . t t p−1 p−1 Applying formula (5.8) with U = u, W = w, F =|u| u −|u + v| (u + w) (with q = (p + 1)/p) and using the equations for u,w we obtain the identity p−1 p−1 B(t ) − B(0) = lim χ (s)(|u| uw − Fu ) − lim χ (s)∂ (|u| uw) k t t k t p−1 p−1 p−1 = lim χ (s)(|u + v| (u + w) −|u| u − p|u| w)u dxds. k t We have p−1 p−1 p−1 |u + v| (u + w) −|u| u − p|u| w 1 σ p−2 2 = p(p − 1) |u + τw| (u + τw)dτ dσ |w| 0 0 which implies p−1 p−1 p−1 p+1 2 |u + v| (u + w) −|u| u − p|u| w ≤ C (p, u ∞ )(|w| +|w| ). Thus we get B(0)−B(t ) ≤ C [E (w(s))+w(s) ]ds, C = C (p, |u|+|u | ). t L ([0,T ]×) L () 141 Page 28 of 29 P. D’Ancona Using (5.3) we conclude B(0) − B(t ) ≤ C E (w(s))ds + C w(0) for some C as in (5.2). Plugging (5.5), (5.2)in (5.4) we arrive at E (w(t )) ≤ C E (w(s))ds + CE (w(0)) + C w(0) 0 L with C as in (5.2), and by Gronwall’s Lemma we conclude the proof. Conﬂict of interest The author declare that they have no conﬂict of interest. Funding Open access funding provided by Universitá degli Studi di Roma La Sapienza within the CRUI- CARE Agreement. Data availibility Data sharing not applicable to this article as no datasets were generated or analysed during the current study. Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. 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Analysis and Mathematical Physics – Springer Journals
Published: Jul 21, 2021
Keywords: Supercritical wave equation; Exterior problem; Decay estimates; Global existence; 35L05; 58J45; 35L20
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