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arXiv:1107.0847v1 [math.AP] 5 Jul 2011 THE GLASSEY CONJECTURE WITH RADIALLY SYMMETRIC DATA KUNIO HIDANO, CHENGBO WANG, AND KAZUYOSHI YOKOYAMA Abstract. In this paper, we verify the Glassey conjecture in the radial case for all spatial dimensions. Moreover, we are able to prove the existence results with low regularity assumption on the initial data and extend the solutions to the sharp lifespan. The main idea is to exploit the trace estimates and KSS type estimates. Contents 1. Introduction 2. Sobolev type estimates 3. Space-time L2 estimates 4. Glassey conjecture when n ≥ 3 4.1. Glassey conjecture when p > pc and n ≥ 3 4.2. Glassey conjecture when p = pc and n ≥ 3 5. Glassey conjecture when p < pc and n ≥ 2 6. Glassey conjecture when n = 2, p > pc References 1 5 10 15 16 18 21 25 27 1. Introduction Let n ≥ 2, p > 1,  = ∂t2 − ∆, and a, b be constants. Consider the following nonlinear wave equations  u = a|∂t u|p + b|∇x u|p , (t, x) ∈ R × Rn (1.1) 2 1 u(0, x) = u0 (x) ∈ Hrad (Rn ), ∂t u(0, x) = u1 (x) ∈ Hrad (Rn ) . m Here Hrad stands for the space of spherically symmetric functions lying in the usual Sobolev space H m . In the 1980’s, Glassey made the conjecture that the critical exponent for the problem to admit global small solutions is 2 pc = 1 + n−1 in [4] (see also Schaeffer [16] and Rammaha [14]). The conjecture has been verified in space dimension n = 2, 3 for general data (Hidano and Tsutaya [5] and Tzvetkov Key words and phrases. Glassey conjecture, semilinear wave equations, Morawetz estimates, KSS estimates. The first author was partly supported by the Grant-in-Aid for Scientific Research (C) (No.20540165 and 23540198), Japan Society for the Promotion of Science. The second author was supported in part by NSFC 10871175 and 10911120383. 1 2 KUNIO HIDANO, CHENGBO WANG, AND KAZUYOSHI YOKOYAMA [24] independently) as well as radial data (Sideris [17] for n = 3). For higher dimension n ≥ 4, there are only negative results available (blow up with upper bound on expected sharp lifespan for p ≤ pc ) in Zhou [25]. The purpose of this paper is to verify this conjecture in the radial case for all spatial dimensions, by proving global existence for p > pc . Moreover, we are able to prove the results with low regularity assumption on the initial data and extend the solutions to the sharp lifespan (for all 1 < p < 1 + 2/(n − 2)). Before presenting our main results, let us first give a brief review of the history. The problem is scale invariant in the Sobolev space Ḣ sc with 1 n . sc = + 1 − 2 p−1 For local well-posedness of the problem, it has been intensively studied at least for p ∈ N, when the general result requires the initial data lie in H s × H s−1 for s > max(sc , (n + 5)/4) (see Ponce and Sideris [13], Tataru [23], Fang and Wang [1] and references therein). If p ≥ 3 or p = 2 with n ≥ 4, the problem is locally well-posed in H s × H s−1 for s > sc , when the initial data have radial symmetry or certain amount of angular regularity (see Fang and Wang [3] and references therein). For the long time existence of the solutions with C0∞ small data of size ǫ > 0, it is well known also for the case of p ∈ N (even for the problem of quasilinear equations). When p > pc , we have global existence. For p = pc , we have almost global existence with lifespan Tǫ which satisfies log(Tǫ ) ∼ ǫ1−p . Instead, if p < pc , we have long time existence with lifespan p−1 Tǫ ∼ ǫ− 1−(n−1)(p−1)/2 , see John and Klainerman [9], Klainerman [10], Sogge [20] and references therein. Moreover, the estimate on the lifespan Tǫ is sharp for the problem with nonlinearity |∂t u|p (see Rammaha [15] for p = 2 and n = 2, 3, Zhou [25] for p ∈ R and 1 < p ≤ pc ). There is not much work on the long time existence with low-regularity small data. 2 1 In [8], Hidano and Yokoyama proved almost global existence for small Hrad × Hrad data when p = 2 and n = 3. It was generalized to the quasilinear problem in our recent work [6]. If p ≥ 3 or p = 2 with n ≥ 4, we have global (almost global for p = 3 and n = 2) in H s with s > sc and certain angular regularity (Sterbenz [22] and Fang and Wang [3]). We will use Λi to denote the norm of the initial data, Λi := ku0 kḢ i (Rn ) + ku1 kḢ i−1 (Rn ) , i = 1, 2 . n−1 , Let ∂ = (∂√ x , ∂t ) with ∂x = (∂x1 , ∂x2 , . . . , ∂xn ), x = rω with r = |x| and ω ∈ S 2 and hri = 1 + r . Now we are ready to state our main results. The first result is the global existence theorem for p > pc and n ≥ 3. Theorem 1.1. Let n ≥ 3 and 1 + 2/(n − 1) < p < 1 + 2/(n − 2). Consider the nonlinear wave equation (1.1). For any choice of s1 , s2 such that 1/2 ≤ s1 < n/2 − 1/(p − 1) < s2 ≤ 1, there exist constants C, ǫ0 > 0, such that if 1 2 Λ1−s Λs21 + Λ1−s Λs22 ≤ ǫ0 , 1 1 GLASSEY CONJECTURE 3 then we have a unique global solution u to (1.1) satisfying 2 1 u ∈ C([0, ∞); Hrad (Rn )) ∩ C 1 ([0, ∞); Hrad (Rn )) , ′ k∂ukL∞ ([0,∞);L2 (Rn )) + kr−δ hri−1/2+δ ∂ukL2 ([0,∞)×Rn ) ≤ CΛ1 , ′ where k∂∂x ukL∞ ([0,∞);L2 (Rn )) + kr−δ hri−1/2+δ ∂∂x ukL2 ([0,∞)×Rn ) ≤ CΛ2 , 1 − (s2 − s1 )(p − 1) n − 2s2 (p − 1) , δ ′ = . 4 2 In contrast, when p = pc , we have the almost global existence. (1.2) δ= Theorem 1.2. Let n ≥ 3 and p = 1 + 2/(n − 1). Consider the nonlinear wave equation (1.1). For any choice of s such that 1/2 < s ≤ 1, there exist constants C, c, ǫ0 > 0, such that if 1/2 1/2 ǫ := Λ1 Λ2 s + Λ1−s 1 Λ2 ≤ ǫ0 , then we have a unique almost global solution u to (1.1) satisfying 2 1 u ∈ C([0, T∗ ]; Hrad (Rn )) ∩ C 1 ([0, T∗ ]; Hrad (Rn )) , k∂ukL∞([0,T∗ ];L2 (Rn )) + ǫ(p−1)/2 kr−δ hri−1/2+δ ∂ukL2 ([0,T∗ ]×Rn ) ≤ CΛ1 , k∂∂x ukL∞ ([0,T∗ ];L2 (Rn )) + ǫ(p−1)/2 kr−δ hri−1/2+δ ∂∂x ukL2 ([0,T∗ ]×Rn ) ≤ CΛ2 , where n − 2s δ= (p − 1), T∗ = exp(cǫ1−p ) . 4 For the case 1 < p < pc , we expect a long time existence of the solution. Theorem 1.3. Let n ≥ 2 and 1 < p < 1 + 2/(n − 1). Consider the nonlinear wave equation (1.1). There exist constants C, c > 0, such that we have a unique solution u to (1.1) satisfying 2 1 u ∈ C([0, T∗ ]; Hrad (Rn )) ∩ C 1 ([0, T∗ ]; Hrad (Rn )) , δ−1/2 k∂ukL∞ ([0,T∗ ];L2 (Rn )) + T∗ δ−1/2 where k∂∂x ukL∞ ([0,T∗ ];L2 (Rn )) + T∗ 1/2 kr−δ ∂ukL2 ([0,T∗ ]×Rn ) ≤ CΛ1 , kr−δ ∂∂x ukL2 ([0,T∗ ]×Rn ) ≤ CΛ2 , 1/2 2(p−1) T∗ = c(Λ1 Λ2 )− 2−(n−1)(p−1) , δ= ( (n−1)(p−1) , 2 (n−1)(p−1) , 4 1 1 < p < 1 + n−1 1 1 + n−1 ≤p<1+ 2 n−1 = pc . As can be observed from the statement, for p < pc , we do not require the smallness of the initial data, in contrast to p ≥ pc . Remark 1.1. In our Theorems, the assumptions posed on the initial data are of “multiplicative form”, which is considered as one of the main innovations in 1/2 1/2 this paper. For example, in Theorem 1.3, the quantity Λ1 Λ2 is in fact scaleinvariant, and it scales like the homogeneous Sobolev space Ḣ 3/2 . The assumptions in the other two Theorems are almost critical, which scale like Ḣ 3/2 ∩ Ḣ 3/2+ǫ for p = pc and Ḣ sc −ǫ ∩ Ḣ sc +ǫ for p > pc , with the critical scaling regularity sc = n/2 + 1 − 1/(p − 1). One of the advantages of using the “multiplicative form” is that, for p ≥ pc , even if Λ1 is not so small, we still have (almost) global solutions when Λ2 is sufficiently small. 4 KUNIO HIDANO, CHENGBO WANG, AND KAZUYOSHI YOKOYAMA Here, we would like to point out an interesting similarity between the Glassey conjecture and the Strauss conjecture. Recall that for the Strauss conjecture, where the nonlinearity is |u|p , we find similar phenomena. Besides the critical regularity sc = n/2−2/(p−1), there is one more Sobolev regularity, namely sd = 1/2−1/p (see Sogge [20] Section IV.4), as far as the radially symmetric functions are concerned. The critical exponent p = p0 for this problem to have global small solutions is given by the positive root of the equation (n − 1)p2 − (n + 1)p − 2 = 0 . There is an interesting relation between these two facts: if p > 1, we see that sc > sd if and only if p > p0 , and the sharp lifespan for 1 < p < p0 has the order ǫ1/(sc −sd ) . Interestingly enough, for the Glassey conjecture, the index 3/2 plays the same role as sd . We have sc > 3/2 if and only if p > pc for p > 1, and the sharp lifespan T∗ has also the order ǫ1/(sc −3/2) for p < pc . These observations strongly suggest that, for the equation (1.1), by adding certain amount of angular regularity if necessary, the minimal regularity for the problem to be well-posed is   3 , sc . max 2 When n = 2 and p ≥ pc = 3, it seems to us that the methods to prove the preceding theorems are not sufficient to give satisfactory results. In spite of that, we can use the generalized Strichartz estimates of Smith, Sogge and Wang [19] to prove the following global result for p > pc . Theorem 1.4. Let n = 2 and p > 3. Consider the nonlinear wave equation (1.1). There exist constants C, ǫ0 > 0, such that if 1/(p−1) ǫ := Λ1 1−1/(p−1) Λ2 ≤ ǫ0 , then we have a unique global solution u to (1.1) satisfying 2 1 2 ≤ CΛ1 , k∂∂x ukL∞ L2 ≤ CΛ2 , k∂uk p−1 ∞ ≤ Cǫ . u ∈ Ct Hrad ∩Ct1 Hrad , k∂ukL∞ L L t Lx t x t x Remark 1.2. For p = pc and n = 2, it has been proved in Fang and Wang [3] that the problem has a unique almost global solution with almost critical regularity for small data, which is not necessarily radial. A similar result for p > 3 and p ∈ N has also been obtained there. This paper is organized as follows. At the end of this section, we list our basic notation. In the next section, we give several Sobolev type estimates related with the trace estimates. In Section 3, we prove some space-time L2 estimates, which are variants of the Morawetz-KSS estimates. In Sections 4 and 5, we give the proof of the (almost) global results for n ≥ 3 (Theorems 1.1 and 1.2) and the scalesupercritical result for n ≥ 2 (Theorems 1.3), based on the results from Sections 2 and 3. In the last section, a simple proof for p > pc and n = 2 (Theorem 1.4) is provided, by using the generalized Strichartz estimates√of [19]. Notation. Let δ ∈ (0, 1/2), δ ′ < δ. We denote D = −∆ and the homogeneous Sobolev norm kukḢ s = kDs ukL2 (Rn ) . GLASSEY CONJECTURE 5 The homogeneous Sobolev space Ḣ s with s < n/2 is defined as the completion of C0∞ with respect to the semi-norm k · kḢ s . For fixed T > 0, we will use the following notation. We use k · kEi (i = 1, 2) to denote the energy norm of order i, kukE = kukE1 = k∂ukL∞([0,T ];L2 (Rn )) , kukE2 = k∂x ∂ukL∞ ([0,T ];L2 (Rn )) . We will use k · kLE to denote the local energy norm, kukLE = kukLE1 = ′ kr−δ hri−1/2+δ ∂ukL2 ([0,T ]×Rn ) ′ +kr−1−δ hri−1/2+δ ukL2 ([0,T ]×Rn )  +(log(2 + T ))−1/2 r−δ hri−1/2+δ |∂u| +   +T δ−1/2 r−δ |∂u| + |u| . r 2 n |u| r  L2 ([0,T ]×Rn ) L ([0,T ]×R ) Here, when n ≤ 2, we will assume that there are only terms about ∂u. On the basis of the space LE, we can define kukLE2 = k∂x ukLE and LE ∗ = r−δ hriδ ′ −1/2 L2t,x + (log(2 + T ))−1/2 r−δ hriδ−1/2 L2t,x + T δ−1/2 r−δ L2t,x , where h ∈ f L2t,x means that h = f g for some g ∈ L2t,x . When T = ∞, by LE norm, we mean kukLE = ′ kukLE1 = kr−δ hri−1/2+δ ∂ukL2 ([0,∞)×Rn ) ′ +kr−1−δ hri−1/2+δ ukL2 ([0,∞)×Rn )  + supT >0 (log(2 + T ))−1/2 r−δ hri−1/2+δ |∂u| + |u| r   |u| δ−1/2 −δ . |∂u| + r + supT >0 T r  L2 ([0,T ]×Rn ) L2 ([0,T ]×Rn ) 2. Sobolev type estimates In this section, we give several Sobolev type estimates related with the trace estimates. First, we state a variant of the Hardy inequality. Lemma 2.1 (Hardy’s inequality). Let n ≥ 2 and 0 ≤ s ≤ 1 (s < 1 for n = 2). Then we have (2.1) for any u ∈ H 1 . s kr−s ukL2x ≤ Ckuk1−s L2 k∂r ukL2 6 KUNIO HIDANO, CHENGBO WANG, AND KAZUYOSHI YOKOYAMA Proof. We only need to give the proof for u ∈ C0∞ . First, we prove (2.1) for s ≥ 1/2. Since s < n/2 and s ≤ 1, we have r−s u 2 L2x = Z = ≤ ≤ ≤ ≤ ∞ r−2s |u(rω)|2 rn−1 drdω 0 Z Z ∞ 1 |u(rω)|2 ∂r rn−2s drdω n − 2s Sn−1 0 Z Z ∞ 1 − ∂r (|u(rω)|2 )rn−2s drdω n − 2s Sn−1 0 Z Z ∞ 2 r1−2s |u||∂r u|rn−1 drdω n − 2s Sn−1 0 2 kr1−2s ukL2x k∂r ukL2x n − 2s 2 k(r−s |u|)(2s−1)/s |u|(1−s)/s kL2x k∂r ukL2x n − 2s 2 (2s−1)/s (1−s)/s kr−s ukL2 kukL2 k∂r ukL2x , x x n − 2s Sn−1 = Z where we have applied the Hölder inequality in the last step. This gives us the required estimates with C = (2/(n − 2s))s and s ≥ 1/2. The case s = 0 is trivial. For s ∈ (0, 1/2), we can use the result for s = 1/2 to prove the estimate as follows r−s u L2x 2s =  r−1/2 |u| ≤  r−1/2 |u| = ≤ 2s 2s |u|1−2s 1/s Lx L2x |u|1−2s 2/(1−2s) Lx 1−2s r−1/2 u 2 kukL2x L  s x 2 1−s s kukL2x k∂r ukL2x . n−1 With the help of the Hardy inequality, it will be easy to prove trace estimates. Lemma 2.2 (Trace estimates). Let n ≥ 2. If 1/2 ≤ s ≤ 1 (and s < 1 for n = 2), then (2.2) 1−s s 2 ≤ Ckuk 2 k∂r uk 2 krn/2−s ukL∞ Lx L r Lω x for any u ∈ H 1 (Rn ). In particular, if s = 1/2, we have (2.3) 1/2 1/2 x x 2 ≤ Ckuk 2 k∂r uk 2 . kr(n−1)/2 ukL∞ L L r Lω GLASSEY CONJECTURE 7 Proof. We only need to give the proof for u ∈ C0∞ . The assumptions on s tell us that n − 2s > 0, 0 ≤ 2s − 1 ≤ 1 and 2s − 1 < n/2. Then by using (2.1), we see that Z Z ∞ Rn−2s ku(Rω)k2L2ω = −Rn−2s ∂r |u(rω)|2 drdω Sn−1 R Z Z ∞ rn−2s |u||∂r u|drdω ≤ 2 Sn−1 0 Z Z ∞ r1−2s |u||∂r u|rn−1 drdω = 2 Sn−1 0 1−2s ukL2x k∂r ukL2x 2−2s CkukL2 k∂r uk2s L2x , x ≤ 2kr ≤ with C independent of R > 0. This completes the proof. We will also need to use the following variant of the trace estimates for the proof of Theorem 1.3 in the case of n = 2 and 2 ≤ p < 3. Lemma 2.3. Let n ≥ 2. If s ≥ 0, then √ s−(n−1)/2 1/2 s−(n−1)/2 1/2 2 ≤ 2kr ukL2 kr ∂x ukL2 , (2.4) krs ukL∞ r Lω x x for any u such that the right hand side is finite. Proof. If u ∈ C0∞ , this inequality follows from a simple application of integration by parts and the Cauchy-Schwarz inequality, Z s 2 2s kr ukL2ω = r |u(rω)|2 dω Sn−1 Z Z ∞ 2s = −r ∂R |u(Rω)|2 dRdω Sn−1 r Z Z ∞ ≤ 2 R2s |u(Rω)||∂R u(Rω)|dRdω n−1 ZS Zr ∞ ≤ 2 R2s−(n−1) |u(Rω)||∂R u(Rω)|Rn−1 dRdω ≤ Sn−1 0 s−(n−1)/2 2kr ukL2x krs−(n−1)/2 ∂r ukL2x . Here the condition s ≥ 0 is used to control r2s by R2s . In general, if u ∈ r(n−1)/2−s L2x and ∂x u ∈ r(n−1)/2−s L2x , we only need to construct a C0∞ sequence which is convergent to u in the corresponding norm. Define ul,m (x) := ψl (x)(ρm ∗ u)(x) , R where ψl (x) = ψ(x/l), ρm (x) = mn ρ(mx), ψ, ρ ∈ C0∞ , ρ ≥ 0, Rn ρ(x)dx = 1 and ψ(x) ≡ 1 for |x| < 1. We recall the n-dimensional version of (4.2) of Lemma 4.2 in our previous paper [6], Z ρm (y) dy ≤ C|x|−α , α < n , (2.5) |x − y|α n R where the constant C is independent of m ≥ 1. We claim that there exists a function m = m(l) such that ul,m(l) → u in r(n−1)/2−s L2x as l → ∞. If it is true, then we also have (∂x u)l,m(l) → ∂x u in 8 KUNIO HIDANO, CHENGBO WANG, AND KAZUYOSHI YOKOYAMA r(n−1)/2−s L2x . Notice that ∂x ul,m = x 1 (ρm ∗ u)(x) + (∂x u)l,m (x) . (∂x ψ) l l For the first term, we see that x 1 (ρm ∗ u)(x)kL2x krs−(n−1)/2 (∂x ψ) l l Z x 1 = ρm (x − y)u(y)dy |x|s−(n−1)/2 (∂x ψ) l l Rn ≤ ≤ ≤ L2x C 1/2 k|x|s−(n−1)/2 ρ1/2 m (x − y)u(y)kL2y kρm (x − y)kL2y l C k|x|s−(n−1)/2 ρ1/2 m (x − y)u(y)kL2y L2x l C k|y|s−(n−1)/2 u(y)kL2y → 0 , l L2x as l → ∞, where we have used the inequality (2.5) with α = n − 1 − 2s and the fact that s > −1/2. This gives us the convergence of ∂x ul,m(l) to ∂x u. To complete the proof, it remains to prove the claim. Observe that ul,m − u = ψl (x) ((ρm ∗ u)(x) − u(x)) + (ψl (x) − 1)u(x) . For the second term, since rs−(n−1)/2 (ψl (x) − 1)u(x) → 0 a.e. x ∈ Rn as l → ∞, and |rs−(n−1)/2 (ψl (x) − 1)u(x)|2 ≤ C|rs−(n−1)/2 u(x)|2 ∈ L1 , we see that, by Lebesgue’s dominated convergence theorem, (ψl (x) − 1)u(x) → 0 in r(n−1)/2−s L2x as l → ∞. We only need to control the first term ψl (x) ((ρm ∗ u)(x) − u(x)). Since rs−(n−1)/2 u ∈ 2 Lx , for any ǫ > 0, there exists a continuous function g such that supp g ⊂ {x ∈ Rn : R1 ≤ |x| ≤ R2 } for some 0 < R1 < R2 < ∞, and krs−(n−1)/2 u − gkL2 ≤ ǫ . To deal with the term ψl (x) ((ρm ∗ u)(x) − u(x)), we rewrite it as follows ψl (x) ((ρm ∗ u)(x) − u(x)) = ψl (x) (ρm ∗ (u − G) + (ρm ∗ G − G) + G − u) , where G(x) := r−s+(n−1)/2 g. We easily see that krs−(n−1)/2 ψl (x)(G − u)kL2 ≤ Ckg − rs−(n−1)/2 ukL2 ≤ Cǫ . GLASSEY CONJECTURE 9 For the term involving ρm ∗ (u − G), we obtain ≤ krs−(n−1)/2 ψl (x)ρm ∗ (u − G)(x)kL2x Z C |x|s−(n−1)/2 ρm (x − y)(u − G)(y)dy Rn L2x ≤ 1/2 C k|x|s−(n−1)/2 ρ1/2 m (x − y)(u − G)(y)kL2y kρm (x − y)kL2y ≤ Ck|x|s−(n−1)/2 ρ1/2 m (x ≤ L2x − y)(u − G)(y)kL2y L2x Ck|y|s−(n−1)/2 (u − G)(y)kL2y ≤ Cǫ , where we have used the inequality (2.5) with α = n − 1 − 2s and the fact that s > −1/2. Finally, we consider the term involving (ρm ∗ G − G). Note that G is a uniformly continuous function, Z |(ρm ∗ G)(x) − G(x)| = ρm (x − y)(G(y) − G(x))dy Rn y ≤ sup |y−x|<C/m,x,y∈suppG |G(y) − G(x)| → 0 as m → ∞. Since supp ψl ⊂ {x ∈ Rn : |x| < Cl}, krs−(n−1)/2 ψl (x)((ρm ∗ G)(x) − G(x))kL2 ≤ Ckrs−(n−1)/2 ((ρm ∗ G)(x) − G(x))kL2 (|x|<Cl) ≤ Cls+1/2 sup |y−x|<C/m,x,y∈suppG |G(y) − G(x)| → 0 as m → ∞, for any fixed l. This completes the proof. As we may observe, all these estimates hold for general functions. Typically, we will apply these estimates to ∂u, which is not radial, even if u is radial. This is the main reason for us to state all the estimates above involving the L2ω norm. In this way, as we can see in the following lemma, we can easily control ∂x u and ∂r u. Lemma 2.4. Let u = u(x) be a radially symmetric function. Then −1/2 (2.6) |∂x u| = |∂r u| = An−1 k∂x ukL2ω with An−1 = |Sn−1 |. The proof is just a simple calculation. Since u is radial, we see ∂r u is radial. Further, x x ∂x u = ∂r u, |∂x u| = | ||∂r u| = |∂r u| , r r and 1/2 k∂r ukL2ω = An−1 |∂r u| . Thus, −1/2 −1/2 |∂x u| = |∂r u| = An−1 k∂r ukL2ω = An−1 k∂x ukL2ω . 10 KUNIO HIDANO, CHENGBO WANG, AND KAZUYOSHI YOKOYAMA 3. Space-time L2 estimates In this section, we prove the space-time L2 estimates, which are variants of the Morawetz-KSS estimates. Consider the wave equation  u = F, (t, x) ∈ R × Rn (3.1) u(0, x) = u0 (x), ∂t u(0, x) = u1 (x) . Lemma 3.1 (KSS type estimates). Let n ≥ 1, 0 ≤ δ < 1/2 and δ ′ < δ. For any solution u = u(t, x) to the wave equation (3.1), we have the following inequality (3.2) kukE + kukLE ≤ C(k∂x u0 kL2x + ku1 kL2x + kF kL1t L2x ) , where C is independent of T > 0 and the functions u0 ∈ H 1 , u1 ∈ L2 and F ∈ L1t L2x . This is a standard estimate now. The estimates of this type together with the application to nonlinear wave equations originate from the work of Keel, Smith and Sogge [11]. The variants with LE norm including the homogeneous weight r−δ are due to Hidano and Yokoyama [7]. Here, for completeness, we give a proof. Proof. To begin the proof, let us recall the classical local energy estimates of Smith-Sogge (Lemma 2.2 in [18]) (3.3) kβ(x)eitD f kL2 (R×Rn ) ≤ Cn,γ,β kf kḢ γ , for β ∈ C0∞ and 2γ ≤ n − 1. The inequality (3.2) follows from this inequality with γ = 0 (and γ = 1 for n ≥ 3), together with the energy estimate. First, owing to the Duhamel principle and a standard scaling argument, it is enough to prove the following six inequalities (3.4) (3.5) kr−δ eitD f kL2 ([0,1]×Rn ) ≤ Ckf kL2x , 0 ≤ δ < 1/2 , (log(2 + T ))−1/2 kr−δ hri−1/2+δ eitD f kL2 ([0,T ]×Rn ) ≤ Ckf kL2x , δ < 1/2 , ′ (3.6) kr−δ hri−1/2+δ eitD f kL2 (R×Rn ) ≤ Ckf kL2x , δ ′ < δ < 1/2 , (3.7) kr−1−δ eitD f kL2 ([0,1]×Rn ) ≤ Ckf kḢx1 , 0 ≤ δ < 1/2 , n ≥ 3 , (3.8) (log(2 + T ))−1/2 kr−1−δ hri−1/2+δ eitD f kL2 ([0,T ]×Rn ) ≤ Ckf kḢ 1 , δ < 1/2 , n ≥ 3 x (3.9) ′ kr−1−δ hri−1/2+δ eitD f kL2 (R×Rn ) ≤ Ckf kḢ 1 , δ ′ < δ < 1/2 , n ≥ 3 . x We begin by the proof of the first three inequalities for r ≤ 1 and (3.6) for r > 1. From (3.3) with γ = 0 and n ≥ 1, we see that keitD f kL2 (R×{r≤1}) ≤ Ckf kL2 . A standard scaling argument leads us to (3.10) sup 2−j/2 keitD f kL2 (R×{r≤2j }) ≤ Ckf kL2 , j∈Z GLASSEY CONJECTURE 11 and so for any δ < 1/2, kr−δ eitD f kL2 (R×{r≤1})   C 2(1/2−δ)j 2−j/2 keitD f kL2 (R×{2j−1 <r≤2j }) 2 ≤ lj:j≤0 −j/2 ≤ C sup 2 ≤ Ckf kL2 . j≤0 ke itD f kL2 (R×{2j−1 <r≤2j }) Similarly, for any δ ′ < δ < 1/2, since r ≤ hri, we obtain kr−δ hriδ ≤ ≤ ′ −1/2 itD e ′ δ −δ−1/2 itD f kL2 (R×{r≥1}) kr e f kL2 (R×{r≥1})   ′ C 2(δ −δ)j 2−j/2 keitD f kL2 (R×{2j−1 ≤r≤2j }) 2 lj:j≥1 −j/2 ≤ C sup 2 ≤ Ckf kL2 , j≥1 ke itD f kL2 (R×{2j−1 ≤r≤2j }) which is (3.6) for r > 1. It remains to prove (3.4) and (3.5) for r > 1. For (3.4), because of the assumption δ ≥ 0, we can easily get by the energy estimates kr−δ eitD f kL2 ([0,1]×{r>1}) ≤ keitD f kL2 ([0,1]×Rn ) ≤ keitD f kL∞ ([0,1];L2 (Rn )) ≤ Ckf kL2x . For (3.5) with r > 1, we consider 1 ≤ r ≤ T and r ≥ T separately. For r ≥ T , since δ − 1/2 < 0 and r ≤ hri, we obtain kr−δ hriδ−1/2 eitD f kL2 ([0,T ]×{r≥T }) ≤ kr−1/2 eitD f kL2 ([0,T ]×{r≥T }) ≤ keitD f kL∞ ([0,T ];L2 (Rn )) ≤ ≤ T −1/2 keitD f kL2 ([0,T ]×Rn ) Ckf kL2x . Now we give the estimate of (3.5) for 1 ≤ r ≤ T . By (3.10) and the elementary inequality 2[10 log(2+T )] ≥ T (where [M ] denotes the greatest integer not greater than M ), we have ≤ ≤ kr−δ hriδ−1/2 eitD f kL2 (R×{1≤r≤T }) Ckr−1/2 eitD f kL2 (R×{1≤r≤T })   C 2−j/2 keitD f kL2 (R×{2j−1 ≤r≤2j }) 2 lj:1≤j≤10 log(2+T ) 1/2 sup 2 −j/2 ≤ C(log(2 + T )) ≤ C(log(2 + T ))1/2 kf kL2x . j ke itD f kL2 (R×{2j−1 ≤r≤2j }) This completes the proof of the first three inequalities. The inequalities (3.7)-(3.9) follow from basically the same proof, by using (3.3) with γ = 1 and Hardy’s inequality. For example, to prove (3.7) for n ≥ 3, we use 12 KUNIO HIDANO, CHENGBO WANG, AND KAZUYOSHI YOKOYAMA (3.3) with γ = 1 (since 1 ≤ (n − 1)/2 for n ≥ 3), which tells us that keitD f kL2 (R×{r≤1}) ≤ Ckf kḢ 1 . A standard scaling argument leads us to (3.11) sup 2−3j/2 keitD f kL2 (R×{r≤2j }) ≤ Ckf kḢ 1 , j∈Z and so for any δ < 1/2, kr−δ−1 eitD f kL2 (R×{r≤1}) ≤   C 2(1/2−δ)j 2−3j/2 keitD f kL2 (R×{2j−1 <r≤2j }) 2 lj:j≤0 −3j/2 ≤ C sup 2 ≤ Ckf kḢ 1 . j≤0 ke itD f kL2 (R×{2j−1 <r≤2j }) For (3.7) with r > 1, since δ ≥ 0, we can easily get by the energy estimates and Hardy’s inequality (2.1) with s = 1, kr−1−δ eitD f kL2 ([0,1]×{r>1}) ≤ kr−1 eitD f kL2 ([0,1]×Rn ) ≤ kDeitD f kL∞ ([0,1];L2 (Rn )) ≤ Ckf kḢ 1 , x which completes the proof of (3.7). When n ≥ 3, we can prove the following inhomogeneous KSS type estimates with LE ∗ norm on F . Lemma 3.2 (Inhomogeneous KSS type estimates). Let n ≥ 3, 0 < δ < 1/2, and δ ′ < δ. For any solution u = u(t, x) to the wave equation (3.1), we have the following inequality (3.12) kukE + kukLE ≤ C(k∂x u0 kL2x + ku1 kL2x + kF kLE ∗ ) , where C is independent of T > 0 and the functions u0 ∈ Ḣ 1 , u1 ∈ L2 and F ∈ LE ∗ . Proof. i) Let us first consider smooth solutions. For such a case, we have the space-time L2 estimates even for certain small perturbations of the Minkowski metric (see [21], [12] and our previous work [6]). Recall that using Lemma 2.3 and (2.30) of [6], we can get Z TZ  |∂u|2 u2  (3.13) + 2+2δ dxdt T 2δ−1 2δ r 0 {x∈Rn ; 1<r<T } r Z TZ  |∂u|2 −1 u2  + log(2 + T ) + 3 dxdt r r 0 {x∈Rn ; 1<r<T } Z TZ  |∂u|2  2 u + + 3+2δ−2δ′ dxdt 1+2δ−2δ ′ r r n 0 {x∈R ; 1<r<∞} ≤ C(k∇u0 k2L2 (Rn ) + ku1 k2L2 (Rn ) )  Z TZ  |u||F | |∂u||F | + +C dxdt , hri 0 Rn GLASSEY CONJECTURE 13 for any smooth solution u to the wave equations (3.1), T > 1, δ ′ < δ and 0 < δ < 1/2. We will also need a slight variant of Lemma 2.2 of [6]. Observe that if we choose the function  k r f (r) = r+λ with k = 1 − 2δ ∈ (0, 1) and λ > 0, then the same argument as in the proof of Lemma 2.2 of [6] will tell us that λ2δ−1 (3.14) Z TZ {x∈Rn ; r<λ} 0 ≤  |∂u|2 r2δ + u2  r2+2δ dxdt C(k∇u0 k2L2 (Rn ) + ku1 k2L2 (Rn ) )  Z TZ  |u||F | |∂u||F | + 2δ +C dxdt, r (r + λ)1−2δ 0 Rn where the constant C is independent of λ > 0. We only need to check the new relations (instead of (2.15) and (2.16) there)   k(1 − k)λ2 f rk−1 f ′ ≤ − , ∆ , − f (r) ≥ (1 − k) r (λ + r)k r r3−k (λ + r)2+k and substitute these new relations to (2.12) and (2.17) there. On the basis of (3.13) and (3.14), together with the standard energy estimate ! Z Z Z T (3.15) sup t∈[0,T ] Rn |∂u|2 dx ≤ C k∂u(0)k2L2(Rn ) + 0 Rn |∂u||F |dxdt , it will be easy to prove the required estimates for the smooth solutions. Suppose T > 1 first. By applying (3.15) to the integrals over {r > T }, we see that ≤ ≤ ≤ ≤ ≤ kuk2E + kuk2LE(r>T ) Z Z TZ 2 2δ−1 sup |∂u| dx + T  |∂u|2 u2  dxdt r2δ r2+2δ Z Z  |∂u|2 −1 T u2  + log(2 + T ) + 3 dxdt r r 0 {x∈Rn ; r>T } Z TZ  |∂u|2  u2 + dxdt ′ + ′ 1+2δ−2δ 3+2δ−2δ r 0 {x∈Rn ; r>T } r Z Z TZ  u2  sup |∂u|2 + 2 dxdt |∂u|2 dx + 3T −1 r t∈[0,T ] Rn 0 {x∈Rn ; r>T } Z  2 u C sup |∂u|2 + 2 dx r n t∈[0,T ] R Z C sup |∂u|2 dx t∈[0,T ] Rn 0 {x∈Rn ; r>T } + t∈[0,T ] Rn C(k∇u0 k2L2 (Rn ) + ku1 k2L2 (Rn ) + Z 0 T Z Rn |∂u||F |dxdt), 14 KUNIO HIDANO, CHENGBO WANG, AND KAZUYOSHI YOKOYAMA where we have applied the Hardy inequality (2.1) with s = 1. For the integral over {r < T }, we use (3.13) and (3.14) with λ = 1 to get  Z TZ  |u||F | kuk2LE(r<T ) ≤ C(k∇u0 k2L2 (Rn ) + ku1 k2L2 (Rn ) ) + C |∂u||F | + dxdt. r 0 Rn Then an application of the Cauchy-Schwarz inequality yields the required estimate (3.12) for T > 1. To prove the general result for any T > 0, we only need to control the term   |u| A[u] = T δ−1/2 r−δ |∂u| + r L2 ([0,T ]×Rn ) for T ∈ (0, 1]. To control this, we only need to apply (3.14) with λ = T and (3.15) as follows:   2 |u| + kuk2E A[u]2 + kuk2E = T 2δ−1 r−δ |∂u| + r 2 n L ([0,T ]×R )  2  |u| ≤ CT 2δ−1 r−δ |∂u| + r L2 ([0,T ]×Rn :r<T ) +CT −1 |∂u| + |u| r 2 L2 ([0,T ]×Rn :r>T )   |u| ≤ CT 2δ−1 r−δ |∂u| + r +C |∂u| + ≤ |u| r + kuk2E 2 L2 ([0,T ]×Rn :r<T ) 2 2 n L∞ t ([0,T ];L (R )) + ku1 k2L2 (Rn ) )  |u||F | |∂u||F | + dxdt . r C(k∇u0 k2L2 (Rn ) Z TZ  +C 0 Rn Once again, an application of the Cauchy-Schwarz inequality gives us the required estimate (3.12) for T ≤ 1. ii) We next consider the case where u is not smooth. By Lemma 3.1, we only need to prove for the case u0 = u1 = 0. Fix T ∈ (0, ∞). Observe that for 0 < δ < 1/2, we have the Hardy inequality kr−δ ukL2 ≤ CkukḢ δ ≤ CkukH δ , which means r−δ L2x ⊂ H −δ , and so LE ∗ ⊂ L1t H −δ ([0, T ] × Rn ) if T < ∞. Thus by the standard existence and uniqueness result of the linear wave equation, we have u ∈ Ct Hx1−δ ∩ Ct1 Hx−δ ([0, T ] × Rn ). We claim that there exists a sequence of smooth functions Fk such that Fk → F in LE ∗ . If it is true, then uk are Cauchy sequence in E1 ∩ LE1 , and uk → u in Ct Hx1−δ ∩ Ct1 Hx−δ ([0, T ] × Rn ). This tells us that {uk }∞ k=1 converges to u in E1 ∩ LE1 , and so kukE1 ∩LE1 = lim kuk kE1 ∩LE1 ≤ C lim kFk kLE ∗ = CkF kLE ∗ , k→∞ which implies (3.12). k→∞ GLASSEY CONJECTURE 15 To complete the proof, it remains to prove the claim. Proof of the claim. Without loss of generality, we give the proof for F ∈ r−δ L2t,x . Let F̃ (t, x) be the zero extension of rδ F ∈ L2t,x ([0, TR] × Rn ) in R × Rn . Let φ(x) ∈ C0∞ (Rn ) be a function with the properties φ ≥ 0, Rn φ(x)dx = 1, φ = 1 near 0. We will also choose its one-dimensional counterpart ψ(t) ∈ C0∞ (R). Define φk (x) = 2kn φ(2k x) and ψk (t) = 2k ψ(2k t). Then the standard results of approximations of the identity give us F̃k = (φk (x)ψk (t)) ∗t,x F̃ → F̃ in L2t,x ([0, T ] × Rn ) , that is, Fk1 := r−δ F̃k → F in r−δ L2t,x ([0, T ] × Rn ) . Notice that Fk1 is smooth except at x = 0. It suffices to set Fk (t, x) = (1 − φ(2k x))Fk1 (t, x), which is smooth for any t, x. Indeed, by Lebesgue’s dominated convergence theorem, we see krδ (Fk − F )kL2t,x ([0,T ]×Rn ) = kF̃k (1 − φ(2k x)) − F̃ kL2t,x ([0,T ]×Rn ) ≤ k(F̃k − F̃ )(1 − φ(2k x))kL2t,x ([0,T ]×Rn ) + kφ(2k x)F̃ kL2t,x ([0,T ]×Rn ) ≤ kF̃k − F̃ kL2t,x ([0,T ]×Rn ) + kφ(2k x)F̃ kL2t,x ([0,T ]×Rn ) → 0 as k → ∞ . This completes the proof of the claim and hence that of Lemma 3.2. 4. Glassey conjecture when n ≥ 3 Now we are ready to present our proof of the Glassey conjecture for radial initial data. Let us first formulate the setup of the proof for existence and uniqueness. Define (4.1) XT : = { 1 2 u ∈ C([0, T ]; Hrad (Rn )) ∩ L∞ ([0, T ]; Hrad (Rn )) : 1 ∂t u ∈ C([0, T ]; L2rad (Rn )) ∩ L∞ ([0, T ]; Hrad (Rn )), kukE1 ∩E2 + kukLE1∩LE2 < ∞} . For R1 > 0 and R2 > 0, we next define X(R1 , R2 ; T ) := {u ∈ XT : kukEi + kukLEi ≤ Ri , i = 1, 2} . Endowed with (4.2) ρ(u, v) := ku − vkE1 + ku − vkLE1 , it is easy to check that X(R1 , R2 ; T ) is complete with the metric ρ(u, v). 2 1 For fixed (u0 , u1 ) ∈ Hrad × Hrad , we define the iteration map (4.3) Φ[u](t) := u(0) (t) + I[N [u]] , where u(0) (t) = cos(tD)u0 + D−1 sin(tD)u1 is the solution of the linear Cauchy problem, (4.4) N [u] := a|∂t u|p + b|∇x u|p , and (4.5) I[F ] := Z 0 t sin((t − s)D) F (s)ds . D 16 KUNIO HIDANO, CHENGBO WANG, AND KAZUYOSHI YOKOYAMA For the nonlinearity N [u], we have the properties |∂xα N [u]| ≤ C|∂u|p−1 |∂xα ∂u| , |α| ≤ 1, (4.6) and |N [u] − N [v]| ≤ C(|∂u|p−1 + |∂v|p−1 )|∂(u − v)| . (4.7) Notice that v = Φ[u] is defined as the solution to the following equation  v = N [u], (t, x) ∈ R × Rn (4.8) v(0, x) = u0 (x), ∂t v(0, x) = u1 (x) . We aim at showing that Φ is a contraction mapping of X(R1 , R2 ; T ), if we choose R1 , R2 and T suitably. 4.1. Glassey conjecture when p > pc and n ≥ 3. Consider the nonlinear wave equation (1.1) for p > pc and n ≥ 3. Let us begin with the estimate of the homogeneous solution, u(0) , which follows directly from the application of Lemma 3.1 to u and ∂x u with F = 0. Proposition 4.1. Let n ≥ 2. There is a positive constant C1 , independent of T > 0, such that the following estimates hold (4.9) ku(0) kE1 + ku(0) kLE1 ≤ C1 (k∂x u0 kL2 + ku1 kL2 ) , (4.10) ku(0) kE2 + ku(0) kLE2 ≤ C1 (k∂x2 u0 kL2 + k∂x u1 kL2 ) . Next, we give the estimate for the inhomogeneous part. Proposition 4.2. Let pc < p < 1 + 2/(n − 2) and n ≥ 3, u ∈ X∞ and s1 , s2 such that 1/2 ≤ s1 < n/2 − 1/(p − 1) < s2 ≤ 1. Set δ and δ ′ as in (1.2). Then there is a positive constant C2 , such that the following estimates hold s1 1−s2 s2 p−1 1 (4.11) kI[N [u]]kEi ∩LEi ≤ C2 (kuk1−s kukLEi , i = 1, 2 . E1 kukE2 + kukE1 kukE2 ) Moreover, if u, v ∈ X∞ , we have (4.12) kΦ[u] − Φ[v]kE1 ∩LE1 ≤ C3 ku − vkLE1 s1 1−s2 s2 1−s1 s1 1−s2 s2 p−1 1 ×(kuk1−s , E1 kukE2 + kukE1 kukE2 + kvkE1 kvkE2 + kvkE1 kvkE2 ) for some C3 . Proof. First, by Lemma 2.2, we have for any s ∈ [1/2, 1], 1−s s 2 ≤ Ckuk 2 k∂r uk 2 . krn/2−s ukL∞ Lx L r Lω x Fix s1 , s2 such that 1/2 ≤ s1 < n/2 − 1/(p − 1) < s2 ≤ 1. Then we have (4.13) s1 1−s2 s2 1 ku(rω)kL2ω ≤ Crs2 −n/2 hris1 −s2 (kuk1−s L2 k∂r ukL2 + kukL2 k∂r ukL2 ) . x x x By Lemma 2.4, we have for u ∈ X∞ , (4.14) s1 1−s2 s2 1 |∂u| ≤ Crs2 −n/2 hris1 −s2 (kuk1−s E1 kukE2 + kukE1 kukE2 ) . x GLASSEY CONJECTURE 17 From (4.14), (4.6) and (1.2), it is clear that, for i = 1, 2, X ′ krδ hri1/2−δ ∂xα N [u]kL2x |α|=i−1 ≤ C ′ X |α|=i−1 krδ hri1/2−δ |∂u|p−1 ∂xα ∂ukL2x s2 p−1 1 2 ≤ C(kuk1−s kuksE12 + kuk1−s E E1 kukE2 ) X1 ′ krδ+(s2 −n/2)(p−1) hri1/2−δ +(s1 −s2 )(p−1) ∂xα ∂ukL2x × |α|=i−1 s1 1−s2 s2 p−1 1 = C(kuk1−s E1 kukE2 + kukE1 kukE2 ) X |α|=i−1 ′ kr−δ hri−1/2+δ ∂xα ∂ukL2x . It is easy to check that δ and δ ′ satisfy 0 < δ < 1/2 and δ ′ < δ. Now applying Lemma 3.2 to ∂xα u with |α| ≤ 1 and u0 = u1 = 0, we have for i = 1, 2, kI[N [u]]kEi ∩LEi X ′ krδ hri1/2−δ ∂xα N [u]kL2 ([0,∞);L2x ) ≤ C |α|=i−1 1−s2 s2 p−1 s1 1 ≤ C(kuk1−s E1 kukE2 + kukE1 kukE2 ) ≤ s1 1 C(kuk1−s E1 kukE2 + X |α|=i−1 s2 p−1 2 kuk1−s kukLEi E1 kukE2 ) ′ kr−δ hri−1/2+δ ∂xα ∂ukL2 ([0,∞);L2x ) . This proves (4.11). A similar argument with (4.7) instead of (4.6) will yield (4.12). With these two Propositions 4.1 and 4.2 in hand, it will be easy to show Theorem 1.1. Setting Λi := ku0 kḢ i (Rn ) + ku1 kḢ i−1 (Rn ) , i = 1, 2 , we find by Propositions 4.1 and 4.2 that the mapping Φ, defined by (4.3), is a contraction mapping from X(2C1 Λ1 , 2C1 Λ2 ; T ) into itself, for any T > 0 provided that (4.15) and (4.16) 1 2 C2 (2C1 )p−1 (Λ1−s Λs21 + Λ1−s Λs22 )p−1 ≤ 1/2 , 1 1 1 2 C3 (4C1 )p−1 (Λ1−s Λs21 + Λ1−s Λs22 )p−1 ≤ 1/2 . 1 1 Define a positive constant C0 by −(p−1) C0 = max(2C3 (4C1 )p−1 , 2C2 (2C1 )p−1 ) . Then we see that when (4.17) 1 2 Λ1−s Λs21 + Λ1−s Λs22 ≤ C0 , 1 1 the map Φ is a contraction mapping of X(2C1 Λ1 , 2C1 Λ2 ; T ) for any T > 0, the global in time unique fixed point u ∈ X(2C1 Λ1 , 2C1 Λ2 ; ∞) is the solution which we seek. To complete the proof of Theroem 1.1, we also need to establish the regularity of u, i.e., (4.18) ∂ti u ∈ C([0, ∞); H 2−i (Rn )), i = 0, 1, 18 KUNIO HIDANO, CHENGBO WANG, AND KAZUYOSHI YOKOYAMA and the uniqueness of the solution u. First, for the problem of regularity, it suffices to show ∂ti u ∈ C([0, ∞); Ḣ 2−i (Rn )), i = 0, 1 . In fact, using the inequalities (4.11), (4.15) and the fact that u ∈ LE2 , we have k∂∂x (u(T ) − u(0))kL2x = k∂∂x (Φ[u](T ) − u(0))kL2x ≤ k∂∂x I[N [u]](T )kL2x + k∂∂x (u(0) (T ) − u(0))kL2x s1 1−s2 s2 p−1 1 ≤ C2 (kuk1−s E1 kukE2 + kukE1 kukE2 ) ′ ×kr−δ hri−1/2+δ ∂∂x ukL2 ([0,T ];L2x ) + o(1) ′ ≤ kr−δ hri−1/2+δ ∂∂x ukL2 ([0,T ];L2x ) + o(1) = o(1) as T → 0+. This proves the continuity at t = 0. Recall that our solution satisfies u = Φ[u], which tells us that we can also view u as the solution to the linear wave equation v = N [u](t0 + t) with initial data (u(t0 ), ∂t u(t0 )) at any other time t0 ∈ (0, ∞). Then a similar argument will give us the continuity at any t ∈ [0, ∞). Now, we turn to the proof of uniqueness. Assume there exists another solution v ∈ X∞ ∩ Ct H 2 ∩ Ct1 H 1 , with the same initial data. Recall that u, v ∈ Ct H 2 ∩ Ct1 H 1 . If we restrict these solutions to small enough time interval [0, T ], owing to ∂∂xα (u − v)(0) = 0, we have X X (k∂∂xα (u − v)kC([0,T ];L2x) + k∂∂xα ukC([0,T ];L2x) ) k∂∂xα vkC([0,T ];L2x) ≤ |α|=i−1 |α|=i−1 ≤ o(1) + 2C1 Λi , as T → 0+. Using the inequality (4.12), we see that ′ kr−δ hri−1/2+δ ∂(u − v)kL2 ([0,T ];L2x ) = ≤ ≤ ′ kr−δ hri−1/2+δ ∂(Φ[u] − Φ[v])kL2 ([0,T ];L2x ) ′ C3 kr−δ hri−1/2+δ ∂(u − v)kL2 ([0,T ];L2x ) s1 1−s2 s2 1−s1 s1 1−s2 s2 p−1 1 ×(kuk1−s E1 kukE2 + kukE1 kukE2 + kvkE1 kvkE2 + kvkE1 kvkE2 ) 3 −δ −1/2+δ′ kr hri ∂(u − v)kL2 ([0,T ];L2x ) , 4 provided T > 0 is small enough, where we have used (4.17) and (4.16). By this we arrive at the conclusion that u = v for t ∈ [0, T ], which shows the uniqueness. This completes the proof of Theorem 1.1. 4.2. Glassey conjecture when p = pc and n ≥ 3. Consider (1.1) for p = pc and n ≥ 3. The estimate of the homogeneous solution, u(0) , is given by Proposition 4.1. We only need to give the estimate for the inhomogeneous part. Proposition 4.3. Let p = pc and n ≥ 3, u ∈ XT and s ∈ (1/2, 1]. Define (4.19) δ= n − 2s (p − 1) . 4 GLASSEY CONJECTURE 19 Then there is a positive constant C4 , independent of T > 0, such that the following estimates hold (4.20) 1/2 1/2 s p−1 kukLEi , i = 1, 2 . kI[N [u]]kEi∩LEi ≤ C4 log(2 + T )(kukE1 kukE2 + kuk1−s E1 kukE2 ) Moreover, we have (4.21) kΦ[u] − Φ[v]kE1 ∩LE1 ≤ C5 log(2 + T )ku − vkLE1 1/2 1/2 1/2 1/2 1−s s p−1 s , ×(kukE1 kukE2 + kuk1−s E1 kukE2 + kvkE1 kvkE2 + kvkE1 kvkE2 ) for some C5 . Proof. First, by (4.13) with s1 = 1/2 and s2 = s, we have (4.22) 1/2 1/2 x x s ku(rω)kL2ω ≤ Crs−n/2 hri1/2−s (kukL2 k∂r ukL2 + kuk1−s L2 k∂r ukL2x ) . x By Lemma 2.4, we have for u ∈ XT , 1/2 1/2 s |∂u| ≤ Crs−n/2 hri1/2−s (kukE1 kukE2 + kuk1−s E1 kukE2 ) . (4.23) From (4.23) and (4.6), it is clear that, for i = 1, 2, X krδ hri1/2−δ ∂xα N [u]kL2x |α|=i−1 ≤ C X |α|=i−1 1/2 krδ hri1/2−δ |∂u|p−1 ∂xα ∂ukL2x 1/2 s p−1 ≤ C(kukE1 kukE2 + kuk1−s E1 kukE2 ) X krδ+(s−n/2)(p−1) hri1/2−δ+(1/2−s)(p−1) ∂xα ∂ukL2x × |α|=i−1 1/2 1/2 s p−1 = C(kukE1 kukE2 + kuk1−s E1 kukE2 ) X |α|=i−1 kr−δ hri−1/2+δ ∂xα ∂ukL2x . Now applying Lemma 3.2 to ∂xα u with |α| ≤ 1 and u0 = u1 = 0, we have for i = 1, 2, kI[N [u]]kEi ∩LEi ≤ C(log(2 + T ))1/2 X |α|=i−1 krδ hri1/2−δ ∂xα N [u]kL2([0,T ];L2x ) 1/2 1/2 s p−1 ≤ C(log(2 + T )) (kukE1 kukE2 + kuk1−s E1 kukE2 ) X kr−δ hri−1/2+δ ∂xα ∂ukL2 ([0,T ];L2x ) × 1/2 |α|=i−1 1/2 1/2 s p−1 ≤ C log(2 + T )(kukE1 kukE2 + kuk1−s kukLEi . E1 kukE2 ) This proves (4.20). A similar argument with (4.7) instead of (4.6) will yield (4.21). Here, for later use, we record the following inequality which is a direct consequence 20 KUNIO HIDANO, CHENGBO WANG, AND KAZUYOSHI YOKOYAMA of the last one, sup k∂∂x I[N [u]](t)kL2x (Rn ) (4.24) t∈[0,T ] ≤ C4 (log(2 + T ))1/2 kr−δ hri−1/2+δ ∂x ∂ukL2 ([0,T ];L2x ) 1/2 1/2 s p−1 . ×(kukE1 kukE2 + kuk1−s E1 kukE2 ) With these two Propositions 4.1 and 4.3 in hand, the proof of Theorem 1.2 proceeds similarly to that of Theorem 1.1. With Λi := ku0 kḢ i (Rn ) + ku1 kḢ i−1 (Rn ) , i = 1, 2 , we find by Propositions 4.1 and 4.3 that the mapping Φ, defined by (4.3), is a contraction mapping from X(2C1 Λ1 , 2C1 Λ2 ; T ) into itself, for any T > 0 provided that (4.25) 1/2 1/2 s p−1 + Λ1−s ≤ 1/2 , 1 Λ2 ) 1/2 1/2 s p−1 + Λ1−s ≤ 1/2 . 1 Λ2 ) C4 log(2 + T )(2C1 )p−1 (Λ1 Λ2 and (4.26) C5 log(2 + T )(4C1 )p−1 (Λ1 Λ2 Define a positive constant C6 by C6−1 = max(2C5 (4C1 )p−1 , 2C4 (2C1 )p−1 ) , and set T∗ according to 1/2 1/2 log(2 + T∗ )(Λ1 Λ2 s p−1 + Λ1−s = C6 , 1 Λ2 ) which is possible in general only if 1/2 1/2 ǫ = Λ1 Λ2 s + Λ1−s 1 Λ2 ≪ 1 . That is T∗ = exp(C6 ǫ1−p ) − 2, ǫ ≪ 1 . (4.27) Since Φ is a contraction mapping in X(2C1 Λ1 , 2C1 Λ2 ; T∗ ), the unique fixed point u ∈ X(2C1 Λ1 , 2C1 Λ2 ; T∗ ) is the solution which we seek. To complete the proof of Theroem 1.2, we need also to establish the regularity of u, i.e., (4.28) ∂ti u ∈ C([0, T∗ ]; H 2−i (Rn )), i = 0, 1, and the uniqueness of the solution u. First, for the problem of regularity, it suffices to show ∂ti u ∈ C([0, T∗ ]; Ḣ 2−i (Rn )), i = 0, 1 . Indeed, since u ∈ LE2 (T∗ ), we know that kr−δ hri−1/2+δ ∂x ∂ukL2([0,T∗ ];L2x ) < ∞ , and so lim kr−δ hri−1/2+δ ∂x ∂ukL2 ([0,T ];L2x ) = 0 . T →0+ GLASSEY CONJECTURE 21 Using the inequality (4.24) and (4.25), we have = ≤ ≤ ≤ k∂∂x (u(T ) − u(0))kL2x k∂∂x (Φ[u](T ) − u(0))kL2x k∂∂x I[N [u]](T )kL2x + k∂∂x (u(0) (T ) − u(0))kL2x C4 (log(2 + T ))1/2 kr−δ hri−1/2+δ ∂x ∂ukL2 ([0,T ];L2x ) 1/2 1/2 s p−1 + o(1) ×(kukE1 kukE2 + kuk1−s E1 kukE2 ) kr−δ hri−1/2+δ ∂x ∂ukL2 ([0,T ];L2x ) + o(1) = o(1) as T → 0+. This proves the continuity at t = 0. A similar argument will give us the continuity at any t ∈ [0, T∗ ]. Now, we turn to the proof of uniqueness. Assume that there exists another solution v ∈ XT∗ ∩ Ct H 2 ∩ Ct1 H 1 , with the same initial data. Recall that u, v ∈ Ct H 2 ∩ Ct1 H 1 . If we restrict these solutions to small enough time interval [0, T ], owing to ∂∂xα (u − v)(0) = 0, we have X X (k∂∂xα (u − v)kC([0,T ];L2x) + k∂∂xα ukC([0,T ];L2x) ) k∂∂xα vkC([0,T ];L2x) ≤ |α|=i−1 |α|=i−1 ≤ o(1) + 2C1 Λi , as T → 0+. Using the inequality (4.12), we see that kr−δ hri−1/2+δ ∂(u − v)kL2 ([0,T ];L2x ) = ≤ kr−δ hri−1/2+δ ∂(Φ[u] − Φ[v])kL2 ([0,T ];L2x ) C5 log(2 + T )kr−δ hri−1/2+δ ∂(u − v)kL2 ([0,T ];L2x ) 1/2 1/2 1/2 1/2 1−s s p−1 s ×(kukE1 kukE2 + kuk1−s E1 kukE2 + kvkE1 kvkE2 + kvkE1 kvkE2 ) 3 −δ −1/2+δ ≤ kr hri ∂(u − v)kL2 ([0,T ];L2x) , 4 provided T > 0 is small enough, where we have used (4.26). By this, we conclude that u = v for t ∈ [0, T ], which shows the uniqueness. This completes the proof of Theorem 1.2. 5. Glassey conjecture when p < pc and n ≥ 2 In this section, we aim at giving the proof of Theorem 1.3 for n ≥ 2. As we will see, the argument in the previous section can be adapted to the scale-supercritical case p < pc , for n ≥ 3. The argument in the previous section does not apply when n = 2, owing to the fact that current techniques do not yield the inhomogeneous KSS type estimates (3.12) for n = 2. Alternatively, applying the homogeneous estimates in Lemma 3.1 gives us the proof. In this section, by δ in LE norm, we mean ( (n−1)(p−1) 1 , 1 < p < 1 + n−1 2 (5.1) δ= (n−1)(p−1) 1 2 , 1 + n−1 ≤ p < 1 + n−1 = pc . 4 Note that 0 < δ < 1/2. We aim at showing that Φ is a contraction mapping of X(R1 , R2 ; T ), if we choose R1 , R2 and T suitably. As before, the estimate of the homogeneous solution, u(0) , 22 KUNIO HIDANO, CHENGBO WANG, AND KAZUYOSHI YOKOYAMA is given by Proposition 4.1. We only need to obtain a similar estimate for the inhomogeneous part. Proposition 5.1. Let 1 < p < pc , u ∈ XT and δ as in (5.1). Then there is a positive constant C7 , independent of T > 0, such that the following estimates hold (5.2) kI[N [u]]kEi + kI[N [u]]kLEi ≤ C7 T 1−(n−1)(p−1)/2 (kukE1 ∩LE1 kukE2∩LE2 )(p−1)/2 kukLEi , i = 1, 2 . Moreover, we have (5.3) ≤ kΦ[u] − Φ[v]kE1 ∩LE1 C8 T 1−(n−1)(p−1)/2 (kukE1 ∩LE1 kukE2 ∩LE2 + kvkE1 ∩LE1 kvkE2 ∩LE2 )(p−1)/2 ×ku − vkLE1 for some C8 , independent of T > 0. Remark 5.1. From the proof of (5.3), we can extract the following estimates. If 1 < p < min(pc , 2), then kr−δ ∂(Φ[u] − Φ[v])kL2 ([0,T ];L2x ) (5.4) ≤ C8 T 1−(n−1)(p−1)/2 (kukE1 kukE2 + kvkE1 kvkE2 )(p−1)/2 ×kr−δ ∂(u − v)kL2 ([0,T ];L2x ) . If 2 ≤ p < 3 and n = 2, kr−δ (Φ[u] − Φ[v])kL2 ([0,T ];L2x ) (5.5) ≤ C8 T (3−p)/4 (kukE1 kukE2 + kvkE1 kvkE2 )(p−2)/2 × kr−δ ∂ukL2 ([0,T ];L2x ) kr−δ ∂∂x ukL2 ([0,T ];L2x ) 1/2 + kr−δ ∂vkL2 ([0,T ];L2x ) kr−δ ∂∂x vkL2 ([0,T ];L2x ) ×kr−δ ∂(u − v)kL2 ([0,T ];L2x ) . Remark 5.2. From the proof of (5.2) with i = 2, we can extract the following estimates. If 1 < p < min(pc , 2), then (5.6) k∂∂x (Φ[u](T ) − u(0))kL2x ≤ k∂∂x I[N [u]](T )kL2x + k∂∂x (u(0) (T ) − u(0))kL2x ≤ C7 T 1−(n−1)(p−1)/2 (kukE1 kukE2 )(p−1)/2 T δ−1/2 kr−δ ∂∂x ukL2 ([0,T ];L2x ) + o(1) as T → 0+. If 2 ≤ p < 3 and n = 2, (5.7) k∂∂x (Φ[u](T ) − u(0))kL2x ≤ k∂∂x I[N [u]](T )kL2x + k∂∂x (u(0) (T ) − u(0))kL2x 1/2 ≤ C7 (kukE1 kukE2 )(p−2)/2 kr−(p−1)/4 ∂ukL2 ([0,T ];L2 ) x 3/2 ×kr−(p−1)/4 ∂∂x ukL2 ([0,T ];L2 ) + o(1) x as T → 0+. Proof. We will deal with three different cases: 1 < p < 1 + 1/(n − 1) when n ≥ 2; 2 ≤ p < 3 when n = 2; and 1 + 1/(n − 1) ≤ p < pc when n ≥ 3. GLASSEY CONJECTURE 23 Case i) 1 < p < 1 + 1/(n − 1) with δ = (n − 1)(p − 1)/2. First, by (2.3) and Lemma 2.4, we have for u ∈ XT , |∂u| ≤ Cr−(n−1)/2 (kukE1 kukE2 )1/2 . (5.8) By (5.8) and (4.6), it is clear that, for i = 1, 2, X X k|∂u|p−1 ∂xα ∂ukL2x k∂xα N [u]kL2x ≤ C |α|=i−1 |α|=i−1 ≤ C(kukE1 kukE2 )(p−1)/2 = (p−1)/2 C(kukE1 kukE2 ) X |α|=i−1 X |α|=i−1 kr−(n−1)(p−1)/2 ∂xα ∂ukL2x kr−δ ∂xα ∂ukL2x . Now applying Lemma 3.1 to ∂xα u with |α| ≤ 1 and u0 = u1 = 0, we have for i = 1, 2, ≤ kI[N [u]]kEi∩LEi X k∂xα N [u]kL1 ([0,T ];L2x ) C ≤ CT 1/2 ≤ CT 1/2 ≤ CT 1−δ (kukE1 kukE2 )(p−1)/2 T δ−1/2 ≤ |α|=i−1 X |α|=i−1 CT k∂xα N [u]kL2 ([0,T ];L2x) (kukE1 kukE2 )(p−1)/2 1−δ (p−1)/2 (kukE1 kukE2 ) X |α|=i−1 kr−δ ∂xα ∂ukL2 ([0,T ];L2x ) X |α|=i−1 kr−δ ∂xα ∂ukL2 ([0,T ];L2x ) kukLEi . This proves (5.2). A similar argument with (4.7) instead of (4.6) will yield (5.3). Case ii) 1 + 1/(n − 1) ≤ p < pc , n ≥ 3 and δ = (n − 1)(p − 1)/4. In this case, we may use Lemma 3.2 instead. Applying it to ∂xα u with |α| ≤ 1 and u0 = u1 = 0, we have for i = 1, 2, ≤ ≤ ≤ ≤ kI[N [u]]kEi + kI[N [u]]kLEi X krδ ∂xα N [u]kL2 ([0,T ];L2x ) CT 1/2−δ |α|=i−1 CT 1/2−δ CT 1−2δ CT 1−2δ (kukE1 kukE2 )(p−1)/2 (p−1)/2 (kukE1 kukE2 ) (p−1)/2 (kukE1 kukE2 ) X |α|=i−1 kr−(n−1)(p−1)/2 rδ ∂xα ∂ukL2 ([0,T ];L2x ) T δ−1/2 X |α|=i−1 kr−δ ∂xα ∂ukL2 ([0,T ];L2x) kukLEi , where we have used (5.8) and (4.6). Using (4.7) instead of (4.6), (5.3) follows similarly. Case iii) 2 ≤ p < 3, n = 2 and δ = (p − 1)/4. Notice that Lemma 2.3 with s = (3 − p)/4 > 0 gives us √ −(p−1)/4 1/2 1/2 2 ≤ kr−(p−3)/4 ukL∞ 2kr ukL2 (R2 ) kr−(p−1)/4 ∂x ukL2 (R2 ) . L r ω x x 24 KUNIO HIDANO, CHENGBO WANG, AND KAZUYOSHI YOKOYAMA Then for i = 1, 2, we obtain by using Lemma 3.1 ≤ kI[N [u]]kEi + kI[N [u]]kLEi X k∂xα N [u]kL1 ([0,T ];L2x (R2 )) C ≤ C(kukE1 kukE2 )(p−2)/2 ≤ (p−2)/2 |α|=i−1 X |α|=i−1 kr−(p−2)/2 ∂u∂xα ∂ukL1 ([0,T ];L2x (R2 )) kr−(p−3)/4 ∂ukL2 ([0,T ];L∞ C(kukE1 kukE2 ) x ) X −(p−1)/4 α kr ∂x ∂ukL2 ([0,T ];L2x ) × |α|=i−1 ≤ 2 C(kukE1 kukE2 )(p−2)/2 kr−(p−3)/4 ∂ukL2 ([0,T ];L∞ r Lω ) X kr−(p−1)/4 ∂xα ∂ukL2 ([0,T ];L2x ) × |α|=i−1 ≤ CT 1−2δ (kukE1 kukE2 )(p−2)/2 (kukLE1 kukLE2 )1/2 kukLEi , where we have used (4.6) and Lemma 2.4. The estimate (5.3) follows from the similar arguments by using (4.7) instead of (4.6). This completes the proof. With these two Propositions 4.1 and 5.1 in hand, it will be easy to show Theorem 1.3. Setting Λi := ku0 kḢ i (Rn ) + ku1 kḢ i−1 (Rn ) , i = 1, 2 , we find by Propositions 4.1 and 5.1 that the mapping Φ, defined by (4.3), is a contraction mapping from X(2C1 Λ1 , 2C1 Λ2 ; T ) into itself, provided that C7 T 1−(n−1)(p−1)/2 (2C1 )p−1 (Λ1 Λ2 )(p−1)/2 ≤ 1/2 and C8 T 1−(n−1)(p−1)/2 (4C1 )p−1 (Λ1 Λ2 )(p−1)/2 ≤ 1/2 . Define a positive constant C9 by −(1−(n−1)(p−1)/2) C9 = max(2C8 (4C1 )p−1 , 2C7 (2C1 )p−1 ) , and set T∗ according to −(1−(n−1)(p−1)/2) C9 1−(n−1)(p−1)/2 T∗ (Λ1 Λ2 )(p−1)/2 = 1 , that is (5.9) p−1 T∗ = C9 (Λ1 Λ2 )− 2−(n−1)(p−1) . Since Φ is a contraction mapping of X(2C1 Λ1 , 2C1 Λ2 ; T∗ ), the unique fixed point u ∈ X(2C1 Λ1 , 2C1 Λ2 ; T∗ ) is the solution which we seek. To complete the proof of Theroem 1.3, we also need to establish the uniqueness of u in XT∗ , and the regularity of u, i.e., (5.10) ∂ti u ∈ C([0, T∗ ]; H 2−i (Rn )), i = 0, 1. First, for the proof of uniqueness, assume that there exists another solution v ∈ XT∗ , with the same initial data. Recall the estimates (5.4) and (5.5). If we restrict these GLASSEY CONJECTURE 25 solutions to small enough 0 < T < T∗ , we have kr−δ ∂(u − v)kL2 ([0,T ];L2x ) = kr−δ ∂(Φ[u] − Φ[v])kL2 ([0,T ];L2x ) 1 −δ kr ∂(u − v)kL2 ([0,T ];L2x ) . ≤ 2 Combining this with the fact that u and v share the same initial data, we conclude that u = v for t ∈ [0, T ], which shows the uniqueness. By Remark 5.2, 1 − (n − 1)(p − 1)/2 + δ − 1/2 > 0, and the fact that r−δ ∂∂x u ∈ 2 L ([0, T∗ ]; L2x ) (and so kr−δ ∂∂x ukL2 ([0,T ];L2x ) = o(1) as T → 0+), we see that ∂∂x u(T ) converges to ∂∂x u(0) in L2x . This tells us that the continuity at t = 0. A similar argument will give us the continuity at any t ∈ [0, T∗ ]. This completes the proof of Theorem 1.3. 6. Glassey conjecture when n = 2, p > pc For this case, it seems not enough for us to prove global results by applying the KSS type estimates, mainly because we do not have the favorable inhomogeneous estimates as (3.12) in Lemma 3.2. Instead, we want to present a proof based on the recent generalized Strichartz estimates of Smith, Sogge and Wang [19] (with the previous radial estimates in Fang and Wang [2]). Lemma 6.1 (Generalized Strichartz estimates). Let n = 2 and q ∈ (2, ∞). For any solution u = u(t, x) to the wave equation (3.1), we have the following inequality with s = 1 − 1/q, (6.1) 2 2 k∂ukLq ([0,∞);L∞ ≤ Cq (k∂x u0 kḢ s + ku1 kḢ s + kF kL1 Ḣ s ) , r Lω (R )) x x t x where C is independent of the functions u0 , u1 and F . With these estimates, we are able to present a simple proof of Theorem 1.4. Let Λi := ku0 kḢ i (Rn ) + ku1 kḢ i−1 (Rn ) , i = 1, 2 . By using (4.6) and the energy estimates, we have (6.2) 2 k∂Φ[u]kL∞ t Lx ≤ CΛ1 + CkN [u]kL1t L2x ≤ 2 , k∂ukL∞ CΛ1 + Ck∂ukp−1 t Lx Lp−1 L∞ ≤ CΛ2 + Ck∂x N [u]kL1t L2x t x and (6.3) 2 k∂∂x Φ[u]kL∞ t Lx ≤ 2 . CΛ2 + Ck∂ukp−1 k∂∂x ukL∞ t Lx Lp−1 L∞ t x Recall the convex inequality kf kḢ 1−θ ≤ kf kθL2 kf k1−θ , θ ∈ [0, 1], Ḣ 1 26 KUNIO HIDANO, CHENGBO WANG, AND KAZUYOSHI YOKOYAMA together with (4.6), Lemma 6.1 and Lemma 2.4. We see that for p > 3, (6.4) k∂Φ[u]kLp−1L∞ t ≤ ≤ ≤ x k∂Φ[u]kLp−1L∞ L2 t r ω Ck∂u(0)kḢ 1−1/(p−1) + CkN [u]kL1 Ḣ 1−1/(p−1) t 1/(p−1) 1−1/(p−1) CΛ1 Λ2 + 1−1/(p−1) 1/(p−1) Ck∂ukp−1 k∂ukL∞L2 k∂∂x ukL∞ L2 Lp−1 L∞ x x t t t x . Moreover, we have (6.5) ≤ ≤ 2 k∂(Φ[u] − Φ[v])kL∞ t Lx CkN [u] − N [v]kL1t L2x 2 . C(k∂ukLp−1L∞ + k∂vkLp−1 L∞ )p−1 k∂(u − v)kL∞ t Lx t t x x Let ǫ0 > 0 be the number such that C(4Cǫ0 )p−1 = 1/2 . If 1/(p−1) ǫ = Λ1 1−1/(p−1) Λ2 ≤ ǫ0 , then we see that Φ is a contraction mapping in Y (2CΛ1 , 2CΛ2 , 2Cǫ). Here the complete space Y (R1 , R2 , R3 ) is defined as Y (R1 , R2 , R3 ) = 1 2 ≤ R1 , {u ∈ Ct Hrad ∩ Ct1 L2rad ; k∂ukL∞ t Lx 2 ≤ R2 , k∂uk p−1 ∞ ≤ R3 } k∂∂x ukL∞ L L t Lx t x 2. with the metric ρ(u, v) = k∂(u − v)kL∞ t Lx To prove the regularity, we only need to show the continuity at t = 0. For that, since ∂u ∈ Lp−1 ([0, ∞); L∞ x ), we have k∂∂x (u(t) − u(0))kL2x ≤ k∂∂x I[N [u]](t)kL2x + k∂∂x (u(0) (t) − u(0))kL2x ≤ ∞ 2 Ck∂ukp−1 Lp−1 ([0,t];L∞ ) k∂∂x ukLt Lx + o(1) = o(1) ≤ k∂x N [u]kL1 ([0,t];L2x ) + o(1) x as t → 0+. This tells us that u ∈ Ct Ḣ 2 ∩ Ct1 Ḣ 1 . For uniqueness, suppose that there exists another solution v ∈ Y ∩Ct H 2 ∩Ct1 H 1 , with the same initial data. Using the inequality (6.5), we see that = ≤ ≤ k∂(u − v)kCt ([0,T ];L2x) k∂(Φ[u] − Φ[v])kCt ([0,T ];L2x ) C(k∂ukLp−1([0,T ];L∞ ) + k∂vkLp−1 ([0,T ];L∞ ) )p−1 k∂(u − v)kCt ([0,T ];L2x ) t x t x o(1)k∂(u − v)kCt ([0,T ];L2x ) as T → 0+. 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Tzvetkov, Existence of global solutions to nonlinear massless Dirac system and wave equation with small data, Tsukuba J. Math. 22 (1998), no. 1, 193–211. [25] Y. Zhou, Blow up of solutions to the Cauchy problem for nonlinear wave equations, Chinese Ann. Math. Ser. B, 22 (2001), no. 3, 275–280. 28 KUNIO HIDANO, CHENGBO WANG, AND KAZUYOSHI YOKOYAMA Department of Mathematics, Faculty of Education, Mie University, 1577 Kurimamachiya-cho, Tsu, Mie 514-8507, Japan E-mail address: hidano@edu.mie-u.ac.jp Department of Mathematics, Zhejiang University, Hangzhou 310027, China E-mail address: wangcbo@gmail.com Hokkaido Institute of Technology, 7-15-4-1 Maeda, Teine-ku, Sapporo, Hokkaido 006-8585, Japan E-mail address: yokoyama@hit.ac.jp