Existence of a ground state and blow-up problem for a nonlinear Schrödinger equation with critical growth

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1 Existence of a ground state and blow-up problem for a nonlinear Schrödinger equation with critical growth Takafumi Akahori, Slim Ibrahim, Hiroaki Kikuchi and Hayato Nawa 1 Introduction In this paper, we consider the following nonlinear Schrödinger equation. i ψ t + ψ + µ ψ p 1 ψ + ψ 4 d ψ =, NLS) where ψ = ψx, t) is a complex-valued function on R d R d 3), is the Laplace operator on R d, µ > and d < p < 1 := d d ). The nonlinearity of NLS) is attractive and contains the energy-critical term ψ 4 d ψ in the scaling sense. It is known see [7, 1]) that: i) for any t R and datum ψ H 1 R d ), there exists a unique solution ψ CI max, H 1 R d )) to NLS) with ψt ) = ψ, where I max denotes the maximal interval on which ψ exists; ii) the solution ψ enjoys the following conservation laws: where ψt) L = ψ L for any t I max, 1.1) Hψt)) = Hψ ) for any t I max, 1.) Hu) := u L µ p + 1 u p+1 d L p+1 d u L. 1.3) We are interested, like in [3], in the existence of standing wave and blowup problem for NLS). Here, the standing wave means a solution to NLS) of the form ψx, t) = e i tω Qx), so that Q must satisfies the elliptic equation u + ωu µ u p 1 u u 4 d u =. 1.4) When µ =, it is well-known that the equation 1.4) has no solution for ω >. Moreover, if the subcritical perturbation is repulsive, i.e., µ <, then there is no non-trivial solution; Indeed, using the Pohozaev identity, we can verify that any solution Q H 1 R d ) 1 to 1.4) with µ < obeys that ) = ω Q L µ dp 1) 1 Q p+1, 1.5) p + 1) L p+1 which, together with ω > and p < 1, shows Q is trivial. Thus, µ > is necessary for the existence of non-trivial solution to 1.4). 1 The solution Q H 1 R d ) also belongs to L locr d ) see, e.g., Appendix B in [11]). 1

2 Our first aim is to seek a special solution to 1.4) called ground state via a certain variational problem; Q is said to be the ground state, if it is a solution to 1.4) and where S ω Q) = mins ω u): u is a solution to 1.4), 1.6) S ω u) := ω u L + Hu). 1.7) More precisely, we look for the ground state as a minimizer of the variational problem below: m ω := inf S ω u) u H 1 R d ) \, Ku) =, 1.8) where Ku) := d dλ S ωt λ u) and T λ is the L -scaling operator defined by = d λ=1 dλ HT λu) λ=1 = u 1) L µdp p + 1 u p+1 u L p+1 L, 1.9) T λ u)x) := λ d uλx) for λ >. 1.1) We can verify the following fact see [3, 8] for the proof): Proposition 1.1. Any minimizer of the variational problem for m ω becomes a ground state of 1.4). We remark that the existence of the ground state is studied by many authors see, e.g., [1,, 3]); In particular, in [1], they proved the existence of a ground state for a class of elliptic equations including 1.4) through a minimization problem different from ours. An advantage of our variational problem is that we can prove the instability of the ground state found through it. Indeed, we can prove the blowup result see Theorem 1. below). In order to find the minimizer of the variational problem for m ω, we also consider the auxiliary variational problem: m ω = inf I ω u) u H 1 R d ) \, Ku), 1.11) where I ω u) := S ω u) dp 1) Ku) 1.1) = ω u dp 1) 4 L + u 4 d )p 1) L + u L dp 1) dp 1). Note that the subcritical potential term disappears in the functional I ω. The problem 1.1) is good to treat, since the functional I ω is positive and the constraint is K which is stable under the Schwarz symmetrization. Moreover, we have the following; Proposition 1.. Assume d 3. Then, for any ω > and µ >, we have; i) m ω = m ω ii) Any minimizer of the variational problem for m ω is also a minimizer for m ω, and vice versa.

3 We state our main theorems: Theorem 1.1. Assume d 4. Then, for any ω > and µ >, there exists a minimizer of the variational problem for m ω. We also have m ω = m ω >. Theorem 1.. Assume d = 3. If µ is sufficiently small, then we have; i) the variational problem for m ω has no minimizer. ii) there is no ground state for the equation 1.4). We see from Theorem 1.1 together with Proposition 1.1 that when d 4, a ground state exists for any ω > and µ >. We give the proofs of Proposition 1. and Theorem 1.1 in Section. The proof of Theorem 1.1 is based on the idea of Brezis and Nirenberg [5]. Since we consider the variational problem different from theirs, we need a little diffrent argument in the estimate of m ω see Lemma.). We give the proof of Theorem 1. in Section 3. Next, we move to the blowup problem. Using the variational value m ω, we define the set A ω, by A ω, := u H 1 R d ) S ω u) < m ω, Ku) <. 1.13) We can see that A ω, is invariant under the flow defined by NLS) see Lemma 4.1 below). Furthermore, we have the following result: Theorem 1.3. Assume d 4, ω > and µ >. Let ψ be a solution to NLS) starting from A ω,, and let I max be the maximal interval on which ψ exists. If ψ is radially symmetric, then I max is bounded. We will prove Theorem 1.3 in Section 4. Besides the blowup result, we can also prove the scattering result; Put A ω,+ := u H 1 R d ) S ω u) < m ω, Ku) >. 1.14) Then, we have Theorem 1.4. Assume d 4, ω > and µ >. Let ψ be a solution to NLS) starting from A ω,+. Then, ψ exists globally in time and satisfies and there exists ϕ +, ϕ H 1 R d ) such that lim ψt) e i t H ϕ + = lim 1 t + ψt) A ω,+ for any t R, 1.15) t ψt) e i t H ϕ 1 =. 1.16) We will give the proof of Theorem 1.4 in a forthcoming paper. In fact, we will prove the scattering result for a more general nonlinearity. Proofs of Proposition 1. and Theorem 1.1 In this section, we prove Proposition 1. and Theorem 1.1. In the proofs, we often use the following easy facts: For any u H 1 R d ) \, there exists a unique λu) > such that > for any < λ < λu), KT λ u) = for λ = λu), < for any λ > λu). 3.1)

4 d dλ S ωt λ u) = 1 λ KT λu) for any u H 1 R d ) and λ >..) Moreover, we see from.1) and.) that for any u H 1 R d ) \, S ω T λ u) < S ω T λu) u) for any λ λu),.3) S ω T λ u) is concave for λ > λu)..4) Let us begin with the proof of Proposition 1.. Proof of Proposition 1.. We first show the claim i): m ω = m ω. Let u n be any minimizing sequence for m ω, i.e., u n is a sequence in H 1 R d ) \ such that lim I ωu n ) = m ω,.5) n Ku n ) for any n N..6) Using.1), for each n N, we can take λ n, 1] such that KT λn u n ) =. Then, we have m ω S ω T λn u n ) = I ω T λn u n ) I ω u n ) = m ω + o n 1),.7) where we have used λ n 1 to derive the second inequality. Hence, we obtain m ω m ω. On the other hand, it follows from m ω inf I ω u) = u H 1 \ K= inf S ω u) = m ω.8) u H 1 \ K= that m ω m ω. Thus, we have proved that m = m ω. Next, we shall show the claim ii). Let Q be any minimizer for m ω, i.e., Q H 1 R d ) \ with KQ) and I ω Q) = m ω. Since I ω = S ω dp 1) K see 1.1)) and m ω = m ω, it is sufficient to show that KQ) =. Suppose the contrary that KQ) <, so that there exists < λ < 1 such that KT λ Q) =. Then, we have m ω I ω T λ Q) < I ω Q) = m ω,.9) which is a contradiction. Hence, KQ) = and therefore Q is also a minimizer for m ω. Now, employing the idea of Brézis and Nirenberg [5], we prove Theorem 1.1. To this end, we introduce another variational value σ: σ := inf u L u Ḣ 1 R d ), u L = 1..1) Then, we have; Proposition.1. Assume that for ω > and µ >, we have m ω < d σ d..11) Then, there exists a minimizer of the variational problem for m ω. 4

5 Proof of Proposition.1. Let u n be a minimizing sequence of the variational problem for m ω. Extracting some subsequence, we may assume that I ω u n ) 1 + m ω for any n N..1) We denote the Schwarz symmetrization of u n by u n. Then, we easily see that Ku n) for any n N,.13) lim n I ωu n) = m ω,.14) sup u n H 1 < for any n N..15) n N Since u n is radially symmetric and bounded in H 1 R d ), there exists a radially symmetric function Q H 1 R d ) such that lim n u n = Q weakly in H 1 R d ),.16) lim n u n = Q strongly in L q R d ) for < q <,.17) lim n u nx) = Qx) for almost all x R d..18) This function Q is a candidate for a minimizer of the variational problem 1.8). We shall show that it is actually a minimizer. Let us begin with the non-triviality of Q. Suppose the contrary that Q is trivial. Then, it follows from.13) and.17) that u n L u L,.19) so that lim sup n Ku n) = lim sup n lim sup u n L lim inf n n u n L..) Moreover, this together with the definition of σ see.1)) gives us that so that lim sup u n L σ lim inf n n u n L σ d Hence, if Q is trivial, then we have that σ lim sup n d ) u d n L,.1) lim sup u n L..) n m ω = lim I ωu n n) lim n dp 1) 4 dp 1) u n 4 d )p 1) L + u dp 1) n L.3) d lim inf n u n L d σ d. However, this contradicts the hypothesis.11). Thus, Q is non-trivial. 5

6 We shall show that KQ) =. By the Brezis-Lieb Lemma see [4]), we have I ω u n) I ω u n Q) IQ) = o n 1)..4) Since lim n I ωu n) = m ω and I ω is positive, we see from.4) that I ω Q) m ω..5) First, we suppose that KQ) <, which together with the definition of m ω see 1.11)) and.5) implies that I ω Q) = m ω. Moreover, we can take λ, 1) such that KT λ Q) =. Then, we have m ω I ω T λ Q) < I ω Q) = m ω..6) This is a contradiction. Therefore KQ). Second, suppose the contrary that KQ) >. Then, it follows from.13) and Ku n) Ku n Q) KQ) = o n 1).7) that Ku n Q) < for any sufficiently large n N, so that there exists a unique λ n, 1) such that KT λn u n Q)) =. Then, we easily verify that m ω I ω T λn u n Q)) I ω u n Q) = I ω u n) I ω Q) + o n 1).8) = m ω I ω Q) + o n 1). Hence, we conclude I ω Q) =. However, this contradicts the fact that Q is non-trivial. Thus, KQ) =. Since Q is non-trivial and KQ) =, we have Moreover, it follows from.4) and Proposition 1. that m ω S ω Q) = I ω Q)..9) I ω Q) lim inf n I ωu n) m ω = m ω..3) Combining.9) and.3), we obtain that S ω Q) = I ω Q) = m ω. proved that Q is a minimizer for the variational problem 1.8). Thus, we have We shall give the proof of Theorem 1.1. In view of Proposition.1, it suffices to show the following lemma. Lemma.. Assume d 4. Then, for any ω > and µ >, we have m ω < d σ d..31) Proof of Lemma.. In view of Proposition 1., it suffices to show that m ω < d σ d..3) 6

7 Let W be the Talenti function W x) := dd ) 1 + x ) d Put W ε x) := ε d W ε 1 x) for ε >. It is known that..33) W W 4 d W =,.34) σ d = Wε L = W ε L for any ε >..35) We easily see that W ε belongs to L R d ) for d 5. When d = 4, W ε no longer belong to L R 4 ) and we need a cut-off function in our proof: Let b C R 4 ) be a function such that 1 if x 1, bx) =.36) if x. We define the function w ε by w ε := bwε when d = 4, W ε when d 5..37) Then, for any ε < 1, we can verify see p.35 of [13]) that w ε L = σ d + Oε d ),.38) w ε L = σ d + Oε d ),.39) w ε L = C1 ε log ε + Oε ) if d = 4, C 1 ε + Oε d ) if d 5,.4) w ε p+1 d )p+1) d = C L p+1 ε + Oε d )p+1) ),.41) where C 1 and C are positive constants independent of ε. Moreover, it follows from.1) that; For any ε, 1), there exists a unique λ ε > such that = KT λε w ε ) = λ ε w ε dp 1) 1) L µdp p + 1 λ ε w ε p+1 λ L p+1 ε w ε L..4) Combining.38).41) and.4), we obtain that λ 4 d ε dp 1) 4 d )p+1) d = 1 Cλ ε + Oε d ),.43) ε 4 where C > is some constant which is independent of ε. Since, we also find from.43) that lim λ ε = 1..44) ε d > dp 1) 7

8 Put λ ε = 1 + α ε, where lim α ε =. Then,.43) together with the expansion 1 + α ε ) q = ε qα ε + oα ε ) q > ) gives us that 1+ 4 d α ε+oα ε ) = 1 C so that 1 + dp 1) 4 α ε + oα ε ) d )p+1) d ε +Oε d ),.45) d )p+1) d α ε = C ε + Oε d ) for any sufficiently small ε >,.46) where C > is some constant which is independent of ε. When d 4, d > d d )p+1) and hence we find from.46) that d )p+1) d λ ε = 1 C ε + Oε d ) for any sufficiently small ε >..47) Using.38).41) and.47), we obtain that m ω I ω T λε w ε ) = ω w ε dp 1) 4 L + λ ε w ε 4 d )p 1) L + dp 1) dp 1) ωc 1 ε + ωc 1 ε + + dp 1) 4 dp 1) dp 1) 4 dp 1) 4 d )p 1) dp 1) λ ε w ε L λ εσ d 4 d )p 1) + λ ε σ d + Oε d ) dp 1) 1 C ε 1 C ε d d )p+1) d d )p+1) σ d σ d + oε d d )p+1) ) = d σ d 16 dd ) C ε d d )p+1) + ωc 1 ε d )p+1) d + oε ).48) Since d d )p+1) < for p > 1, we obtain m ω < d σ d. 3 Non-existence result in 3D for any sufficiently small ε >. In this section, we will give the proof of Theorem 1.. Before proving the theorem, we prepare the following lemma. Lemma 3.1. Assume d 3, ω > and µ >. Let u be a minimizer of the variational problem for m ω. Then, we have that where the implicit constant is independent of µ. u H 1 1, 3.1) 8

9 Proof of Lemma 3.1. Fix a non-trivial function χ Cc R d ) and put It is easy to see that χ L = χ L χ L ) χ 1 L χ := χ. 3.) χ L ) χ L = so that K χ) <. Hence, we obtain that χ L χ L ) χ L = χ L, 3.3) m ω I ω χ). 3.4) Note here that I ω χ) is independent of µ. Now, we take any minimizer u for m ω. Then, Proposition 1. shows that u is also a minimizer for m ω. Hence, we see from 3.4) that u H 1 I ω u) = m ω I ω χ), 3.5) where the implicit constant is independent of µ. Thus, we have completed the proof. Now, we are in a position to prove Theorem 1.. Proof of Theorem 1.. We first prove the claim i). Suppose the contrary that there exists a minimizer u of the variational problem for m ω. It follows from Proposition 1. that u is also a minimizer for m ω. Let u be the Schwarz symmetrization of u. Then, we easily see that u is a minimizer for m ω as well as u. Hence, using Proposition 1. again, we find that u is a radially symmetric solution to the elliptic equation 1.4). Moreover, Lemma 1 in [] gives us that u C R d ). Put ϕr) := ru r) r = x ). We see that ϕ satisfies the equation ϕ ϕ + µr 1 p ϕ p + r 4 ϕ 5 =, r >. 3.6) Let g W 3, [, )). Multiplying gϕ by the equation 3.6) and integrating the resulting equation, we obtain g) ϕ )) 1 g ϕ ) + 1 g ϕ p 1 + µ p + 1 r p g 1 ) p + 1 r1 p g ϕ p+1 + ) 3 r 5 g r 4 6 g ϕ 6 =. 3.7) On the other hand, multiplying the equation 3.6) by g ϕ/ and integrating the resulting equation, we obtain 1 g ϕ ) g ϕ 1 g ϕ + µ r 1 p g ϕ p r 4 ϕ 6 g =. 3.8) 9

10 Subtracting 3.8) from 3.7), we have the identity g) ϕ )) = µ g 4g )ϕ p 1 p + 1 g p + 3 p + 1) rg r p ϕ p g rg r 5 ϕ ) We take g such that g r) 4g r) =, r >, It is easy to verify that g) =. 3.1) gr) := 1 e r Besides, we see from an elementary calculus that gr) rg r) = 1 e r is a solution to 3.1). 3.11) re r for any r >, 3.1) gr) rg r)) = re r > for any r >. 3.13) Taking this function, we see from the formula 3.9) that p 1 = µ p + 1 g p + 3 p + 1) rg r p ϕ p We shall show that lim inf µ g rg r 5 ϕ ) p 1 p + 1 g p + 3 p + 1) rg r p ϕ p+1 1, 3.15) g rg r 5 ϕ 6 1, 3.16) where the implicit constants are independent of µ. Once we obtain these estimates, taking µ in 3.9), we derive a contradiction. Thus, the claim i) holds. We first prove 3.15). It is easy to verify that the right-hand side of 3.15) is bounded by 1 e r )r u r) p+1 dr + e r r u r) p+1 dr r u r) p+1 dr u p+1 L p+1 R 3 ) u p+1 H 1, 3.17) where the implicit constants are independent of µ. Combining 3.17) with Lemma 3.1, we obtain the estimate 3.15). Next, we prove 3.16). It follows from 3.1), 3.13) and the mean value theorem that g rg r 5 ϕ 6 = g rg r u 6 r u 6 u 6 L 6, 3.18) 1

11 where the implicit constants are independent of µ. Here, using the Sobolev embedding, Ku) =, the Hölder inequality and Lemma 3.1, we obtain that Hence, we find that u L 6 3p 1) u L u + u 6 L 6 L6 for all < µ ) 1 u 6 L6 for all < µ 1. 3.) The estimate 3.18) together with 3.) gives 3.16). We shall prove the claim ii). Let Q be a ground state of the equation 1.4). We see from 1.6) and Proposition 1.1 that Q is also a minimizer for m ω. However, this contradicts the claim i). Thus, ii) follows. 4 Blowup result In this section, we prove Theorem 1.3. The key lemma is the following. Lemma 4.1. Assume d 4, ω > and µ >. Let ψ be a solution to NLS) starting from A ω,, and let I max be the maximal interval where ψ exists. Then, we have ψt) A ω, for any t I max, 4.1) sup Kψt)) <. 4.) t I max Proof of Lemma 4.1. Since the action S ω ψt)) is conserved with respect to t, we have S ω ψt)) < m ω for any t I max. Thus, it suffices for 4.1) to show that Kψt)) < for any t I max. 4.3) Suppose the contrary that Kψt)) > for some t I max. Then, it follows from the continuity of ψt) in H 1 R d ) that there exists t I max such that Kψt )) =. Then, we see from the definition of m ω that S ω ψt )) m ω, which is a contradiction. Thus, 4.3) holds. Next, we shall prove 4.). Since Kψt)) < for any t I max, we see from.1); For any t I max, there exists < λt) < 1 such that KT λt) ψt)) =. 4.4) This together with the definition of m ω shows that S ω T λt) ψt)) m ω. 4.5) Moreover, it follows from the concavity.4) together with.), 4.3) and 4.5) that S ω ψt)) > S ω T λt) ψt)) + 1 λt)) d dλ S ωt λ ψt)) = S ω T λt) ψt)) + 1 λt))kψt)) > m ω + Kψt)). λ=1 Fix t I max. Then, we have from 4.6) and the conservation law of S ω that 4.6) Kψt)) < m ω + S ω ψt)) = m ω + S ω ψt )). 4.7) Hence, 4.) follows from the hypothesis S ω ψt )) < m ω. 11

12 Let us move to the proof of Theorem 1.3. Proof of Theorem 1.3. The proof is based on the idea of [9, 1]. We introduce the generalized version of virial identity; Let ρ be a smooth function on R such that ρx) = ρ4 x) for all x R, 4.8) ρx) for all x R, 4.9) ρx) dx = 1, 4.1) R supp ρ 1, 3), 4.11) ρ x) for all x <. 4.1) Put wr) := r r ) x W R x) := R w r s)ρs) ds for r, 4.13) R for x R d and R >. 4.14) Then, for any solution ψ CI max, H 1 R d )) to NLS) and t I max, putting ψ := ψt ), we have the identity W R ψt) R d t t = W R ψ + t t )I W R ψ ψ + Kψt )) dt dt R d R d t t t t x ) R ρr) dr ψt ) + 4 x x R ρ x R x ψt ) dt dt + t t t t t t R d R d d x R ρr) dr + x R ) x ρ R µ p 1 ψt ) p+1 + ψt ) dt dt p + 1 d 1 4 t t t t R d W R ψt ) dt dt for any t I max and R >. 4.15) We easily verify that W R x) 4W R x) for any R > and x R d 4.16) W R L R, 4.17) W R L R, 4.18) W R L 1 R ρ W, R). 4.19) 1

13 Now, let ψ be a solution to NLS) starting from A ω,. We see from Lemma 4.1 that ε := sup t I max Kψt)) >. 4.) Our aim is to show that the maximal existence interval I max is bounded. We divide the proof into three steps. Step 1. We claim that; There exists m > and R > such that for any R R, m < inf vx) v Hrad 1 dx Rd ), K R v) ϵ 4,, x R v L ψ L where H 1 rad Rd ) is the set of radially symmetric functions in H 1 R d ), and K R v) := R d R d d x R x R ρr) dr + 4 x R ρr) dr + x R ) x ρ v R ) x ρ R µ p 1 p + 1 v p+1 + d v. 4.1) 4.) Let R > be a sufficiently large constant to be chosen later, and let v be a function such that v H 1 rad Rd ), 4.3) K R v) ϵ 4, 4.4) v L ψ L. 4.5) Then, we see from 4.4) that ϵ x 4 + R d R ρr) dr + 4 x x R ρ R K R v) + R d = R d d x R x R ρr) dr + x R ) v ρr) dr + 4 x x R ρ R ) x ρ R ) v µ p 1 p + 1 v p+1 + d v. 4.6) To estimate the right-hand side of 4.6), we employ the following inequality see Lemma in [6]); Assume that d 1. Let κ be a non-negative and radially symmetric function 13

14 in C 1 R d ) with x d 1) max x κ x, L R d ). Then, we have κ 1 u L u 1 L x d 1) κ x u x 1 L + x d 1) max x κ, x u 1 L L 1 4.7) for any u H 1 rad Rd ), where the implicit constant depends only on d. Now, put so that Then, we can verify that κ 1 x) := κ x) := d x R x R K R v) = κ 1 v R d ρr) dr + 4 x R ρr) dr + x R R d κ x ρ R x ρ R ), 4.8) ), 4.9) µ p 1 p + 1 v p+1 + ) d v. 4.3) supp κ j x R for j = 1,, 4.31) κ j L 1 for j = 1,, 4.3) sup max x κ x), 1 x R d x R, 4.33) κ 1 x) 1 inf x R κ x). 4.34) The inequality 4.7) together with 4.3), 4.33) and 4.5) yields that κ R d d v 1 4 d 3 κ v d L κ R d v d 1 d v d κ L x R) v + 1 d v v L L L R d 1) d ψ L R d 1) d ) v L x R) d κ v L where the implicit constant depends only on d and ρ. inequality ab ar r + br R d + ψ L R, 4.35) Moreover, applying the Young r with 1 r + 1 r = 1) to the right-hand side of 4.35), we obtain that R d κ d v v L x R) κ v L + ψ L R d 1) d ) d d 3 + ψ L R. 4.36) 14

15 Similarly, we have R d κ d v p+1 v L x R) κ v L + If R is so large that ψ L R d 1)p 1) R d ) 4 5 p + ψ L R dp 1) + ψ L R d 1) d ψ L R d 1)p 1) then 4.6) together with 4.36) and 4.38) shows that ) κ 1 C v L x R) κ v ε 8 ) 4 5 p + ψ L R dp 1). 4.37) ) d d 3 + ψ L R ε, 4.38) 4.39) for some constant C > depending only on d, µ, p and ρ the notation A ε means that the quantity A is much smaller than ε ). Hence, we conclude that which together with 4.34) gives us the desired result. κ 1 x) inf x R κ x) < C v L x R), 4.4) Step. Let m and R be constants found in 4.1), and let T max := sup I max. Then, our next claim is that sup ψx, t) dx m 4.41) t [t,t max ) x R for any R > R satisfying the following properties: W R L ψ L ε, 4.4) 1 R x R ψ x) dx < m, 4.43) 1 + 1ε ψ L ) R d W R ψ < m. 4.44) Here, we remark that it is possible to take R satisfying 4.44) see [9]). In order to prove 4.41), we introduce T R := sup T > t sup ψx, t) dx m for R >, 4.45) t t<t x R and prove that T R = T max for any R > satisfying 4.4) 4.44). It follows from 4.43) together with the continuity of ψt) in L R d ) that T R > t. We suppose the contrary that T R < T max. Then, we have ψx, T R ) dx = m. 4.46) x R Hence, we see from the definition of m see 4.1)) together with the mass conservation law 1.1) that ε 4 KR ψt R )). 4.47) 15

16 Combining the generalized virial identity 4.15) with 4.47) and 4.4), we obtain W R ψt R ) < W R ψ + T R I W R ψ ψ T R R d R d R d ε + T R 8 ε + T R 8 ε = W R ψ + 1 R d ε W R ψ ψ R d ε T R I W R ψ ψ 4 ε R d ) 1 + 1ε W R L W R ψ. R d ) 4.48) Hence, we see from 4.44) that R d W R ψt R ) < R m. 4.49) Since W R x) R for x > R, 4.49) gives us that ψx, T R ) dx = 1 R R ψx, T R ) dx 1 R R W R ψt r ) < m, 4.5) d x R x R which contradicts 4.46). Thus, we have proved T R = T max and therefore 4.41) holds. Similarly, putting T min := inf I max, we have sup ψx, t) dx m 4.51) t T min,t ] x R for any R > R satisfying 4.4) 4.44). Step 3 We complete the proof of Theorem 1.3. Take R > R satisfying 4.4) 4.44). Then, it follows from the definition of m together with 4.41) and 4.51) that ϵ 4 KR ψt)) for any t I max. 4.5) Hence, we see from the generalized virial identity 4.15) together with 4.5) and 4.4) that Rd W R ψt) W R ψ + ti W R ψ ψ ε t for any t I max. 4.53) R d R d Supposing sup I max = + or inf I max =, we have from 4.53) that which is a contradiction. Thus, I max is bounded. R d W R ψt) < for some t I max, 4.54) 16

17 References [1] Alves, C. O., Souto, M.A.S., and Montenegro, M., Existence of a ground state solution for a nonlinear scalar field equation with critical growth. Calc. Var. [] Berestycki, H. and Lions, P.-L. Nonlinear scalar field equations. I. Existence of a ground state. Arch. Rational Mech. Anal ), [3] Berestycki, H. and Cazenave, T., Instabilité des états stationnaires dans les équations de Schrödinger et de Klein-Gordon non linéaires. C. R. Acad. Sci. Paris Sér. I Math ), [4] Brézis, H. and Lieb, E.H., A relation between pointwise convergence of functions and convergence of functionals. Proc. Amer. Math. Soc ) [5] Brezis, H. and Nirenberg, L., Positive solutions of nonlinear elliptic equations involving critical Sobolev exponents, Comm. Pure Appl. Math ), [6] Cazenave, T., Semilinear Schrödinger equations. Courant Lecture Notes in Mathematics, 1. New York university, Courant Institute of Mathematical Sciences, New York 3). [7] Cazenave, T. and Weissler, F.B., The Cauchy problem for the critical nonlinear Schrödinger equation in H s, Nonlinear Anal ), [8] Le Coz, S., A note on Berestycki-Cazenave s classical instability result for nonlinear Schrödinger equations. Adv. Nonlinear Stud. 8 8), [9] Nawa, H., Asymptotic and Limiting Profiles of Blowup Solutions of the Nonlinear Schrödinger Equations with Critical Power, Comm. Pure Appl. Math ), [1] Ogawa, T. and Tsutsumi, Y., Blow-up of H 1 -solution for the nonlinear Schrödinger equation. J. Differential Equations ) [11] Struwe, M. Variational methods. Applications to nonlinear partial differential equations and Hamiltonian systems. Fourth edition. Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Springer-Verlag, Berlin. [1] Tao, T. and Visan, M., Stability of energy-critical nonlinear Schrödinger equations in high dimensions, Electron. J. Differential Equations 118 5), 1 8. [13] Willem, M., Minimax theorems, Birkhäuser, Takafumi Akahori, Faculty of Engineering Shizuoka University, Jyohoku 3-5-1, Hamamatsu, , Japan ttakaho@ipc.shizuoka.ac.jp Slim Ibrahim, Department of Mathematics and Statistics University of Victoria 17

18 Victoria, British Columbia Hiroaki Kikuchi, School of Information Environment Tokyo Denki University, Inzai, Chiba 7-138, Japan Hayato Nawa, Division of Mathematical Science, Department of System Innovation Graduate School of Engineering Science Osaka University, Toyonaka , Japan 18

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