Citation Context

...N −2 N i∂tu + ∆u + |u| u = 0 (x, t) ∈ R × R (CP) u|t=0 = u0 ∈ ˙ H1 (RN) i.e., the ˙ H 1 critical, focusing, Cauchy problem for NLS. We need two preliminary results. Lemma 2.1 (Strichartz estimate [7],=-=[14]-=-). We say that a pair of exponents (q, r) is admissible if 2 N N 2N q + r = 2 and 2 ≤ q, r ≤ ∞. Then, if 2 ≤ r ≤ N−2 (N ≥ 3) (or 2 ≤ r < ∞, N = 2 and 2 ≤ r ≤ ∞, N = 1) we have i) ii) ∣ ∫ +∞ −∞ e i(t−τ...

Citation Context

...≤ C‖v0‖L2(Rd)(1.2) where p, q are given by the scaling admissibility condition 2 p + d q = d 2 (1.3) and satisfy moreover p ≥ 2, (p, q) = (2,∞) (for the case p = 2, we refer to the paper by Keel-Tao =-=[22]-=-). It is now classical that the proof of (1.2) is a combination of functional analytic arguments and of the dispersion estimate ‖v(t)‖L∞(Rd) ≤ C |t|d/2 ‖v0‖L1(Rd),(1.4) which can be easily derived fro...

Citation Context

... , we write ( ∫ (∫ |F(x, t)| r ) q r (2.3) dx dt ||F || L q t Lr x (Rn+1 ) ≡ This norm will be shortened to L q tLr x for readability, or to Lrx,t when q = r. R We will need Strichartz-type estimates =-=[25, 15, 18]-=- involving the spaces (2.3), (2.1). We will call a pair of exponents (q, r) Schrödinger admissible for R n+1 when q, r ≥ 2, (n, q) ̸= (2, 2), and (2.4) R n 1 n n + = q 2r 4 . ) 1 q .4 J. COLLIANDER, ...

Citation Context

...d in Tataru [20] and the results follow by applying the method of the FBI transform. For the constant coefficient case we refer the reader to the survey article [6] and also to the endpoint result in =-=[10]-=-. In this first paper, in order to make the localization simpler we assume that outside a large ball the operator we consider has constant coefficients. We plan to investigate the same questions in a ...

Citation Context

... other hand, from the L 4 t L∞ x Fourier support of a(· − s)φ we have ‖a(· − s)φ‖ L 4 t L ∞ x � ‖a(· − s)φ‖X n/2−1/4,1/2+ k and the claim follows by square-summing in s. Strichartz estimate (see e.g. =-=[11]-=-) and the ∼ 2 −k/4 2 nk/2 ‖a(· − s)φ‖ L 2 t L 2 x , One can improve this bound substantially when n > 2, but we shall not need to do so here. The factor χ (4) j≥k is a gain over Bernstein’s inequality...

Citation Context

... 2 t L4 x + �u� L ∞ t L 4 x + �u� L 6 t,x + �u� L 4 t L8 x + �u� L 3 t L12 x where all spacetime norms are on I × R 4 . � �u�˙S 1 One has the following standard Strichartz estimates (see for instance =-=[16]-=-): LEMMA 2.2. Let I be a compact time interval, and let u: I × R 4 → C be a Schwartz solution to the forced Schrödinger equation M� iut + ∆u = Fm m=1 for some Schwartz functions F1, ..., FM. Then we h...

Citation Context

...nition: a pair of Lebesgue space exponents are called Schrödinger admissible for R3+1 when q, r ≥ 2, and 1 q + 3 2r = 3 4 .(1.9) Proposition 1.1 (Strichartz estimates in 3 space dimensions (See e.g. =-=[27, 28, 18, 33, 23]-=-)). Suppose that (q, r) and (q̃, r̃) are any two Schrödinger admissible pairs as in (1.9). Suppose too that φ(x, t) is a (weak) solution to the problem (i∂t +∆)φ(x, t) = F (x, t), (x, t) ∈ R 3 × [0, ...