A PDE VERSION OF THE CONSTANT VARIATION METHOD AND THE SUBCRITICALITY OF SCHRÖDINGER SYSTEMS WITH ANTISYMMETRIC POTENTIALS.

Transcription

1 A PDE VERSION OF THE CONSTANT VARIATION METHOD AND THE SUBCRITICALITY OF SCHRÖDINGER SYSTEMS WITH ANTISYMMETRIC POTENTIALS. Tristan Rivière Departement Mathematik ETH Zürich (e mail: (Homepage: riviere) Rivière-Fabes Symposium Minneapolis April 24th 2010.

2 Conservation laws for solutions to Recall from the first talk. u = Ω u For Ω L 2 (D 2, R 2 so(m)) get P W 1,2 (D 2, SO(m)) and ξ W 1,2 s.t. Observe that Perturb P : P (id + ǫ) P where Ω = P 1 ξ P + P 1 P. div(p u) = ξ P u div( ǫ P) = div((id + ǫ) ξ P) in D 2 Only Jacobians in the r. h. s.! Apply Wente and get ǫ L W 1,2 0 s.t. u = Ω u div((id + ǫ) P u) = J u where J = ǫ (id + ǫ) ξ P and hence div J = 0. Recall from the first talk. 1

3 The equation: The 1 Dimensional Analogy. u = Ω u u = Ω u. The curl in 1-D is zero: X = curl Y div X = 0 X = 0. The auxiliary equation for P : ξ := P P 1 + P Ω P 1 0 = P P 1 + P Ω P 1 In other words, P 1 satisfies (P 1 ) = Ω(P 1 ) The multiplication by P : div(p u) = ξ P u (P u ) = 0 Conclusion: the 1-D analogy is the constant variation method P u = cte u = (P 1 ) cte The 1 Dimensional Analogy. 2

4 Look now at Schrödinger Systems with Antisymmetric Potentials. v = Ω v v = Ω v 1D analogy. For m > 2, assume Ω L m 2(B m, so(n)). The system is critical under the minimal assumption v L m m 2(B m ) indeed v Lm 2 m = v L 1 = v L m m 2, loc We aim here again to exploit the antisymmetry of Ω in order to write the Schrödinger system above in conservative form and deduce the subcriticality of the system. Schrödinger Systems with Antisymmetric Potentials. 3

5 A Constant Variation Method Approach. For P W 2,m/2 (B m, SO(n)) to be fixed later one has (Pv) = P v P v 2div( P v). Hence w = Pv satisfies w = [ P P 1 + PΩP 1] w 2div( P P 1 w). Can we choose P such that P P 1 + PΩP 1 = 0? Answer : No in general! Indeed P P 1 / so(n). Instead we consider the following Auxiliary Equation for P Asym( P P 1 ) + PΩP 1 = 0 A Constant Variation Method Approach. 4

6 The resolution of the auxiliary equation. Proposition. [R. 2009] There exists δ > 0 and C > 0 such that for any Ω L m/2 (B m, so(n)) satisfying B m Ω m/2 < δ then there exists P W 2,m/2 (B m, SO(n)) solving Moreover 1 [ P P 1 P P 1 ] = PΩP 1 2 P Id W 2,m/2 C [ B m Ω m/2 For this P the equation becomes L P w = 0 where ] 2/m. L P w := w [ PP 1] 2 w + 2div( P P 1 w) The resolution of the auxiliary equation. 5

7 Construction of a Conservation Law. The idea is to look for Q : B m Gl(R n ) solving L PQ = 0 because then 0 = QL P w wl P Q = div(q w [ Q + 2 P P 1 ] w). Proposition For Ω m/2 < δ there exists a unique Q in W 2,m/2 L satisfying L P Q = 0 in Bm Q = id on B m and Q Id W 2,m/2 0 C m [ B m Ω m/2 ] 2/m. Main Ingredients of the Proof : ( P P 1 ) 2 Sym + n (R) and X R m (X t Q Q t X) 0 hence Maximum Principle implies that Q 1... Construction of a Conservation Law. 6

8 Schrödinger Systems in Conservative Form. For A := P Q one finally obtains Theorem[R. 2009] Let Ω L m/2 (B m, so(n)) then there exists A L W 2,m/2 (B m, Gl n (R)) such that A + AΩ = 0 and for any v L m/(m 2) (B m, R n ) v = Ω v div (A v A v) = 0 Recall from yesterday and from the first slide of today s talk Theorem Let Ω L 2 (B 2, R 2 so(n)) then there exists A L W 1,2 (B 2, Gl n (R)) s.t. div( A AΩ) = 0 and for u W 1,2 (B 2, R n ) one has u = Ω u div(a u + B u) = 0 where J = B := Ω A. Schrödinger Systems in Conservative Form. 7

9 Non local operators antisymmetric potentials Ω? Theorem [Da Lio, R. 2010] Let v L 2 (R, R n ) and Ω L 2 (R, so(n)) satisfying 1/4 v = Ω v Then v L loc (R). Application to the regularity of 1/2 harmonic maps. Let u Ḣ 1/2 (R, N p ), N p R n is assumed to be closed oriented and C 2, E 1/2 (u) := R 1/4 u 2 dx = Critical points to E 1/2 in Ḣ 1/2 (R, N p ) satisfy inf w Ẇ 1,2 u (R 2 +,Rn ) Non local operators antisymmetric potentials Ω? 7 R 2 + 1/2 u (ν u) = 0 in D (R) where ν C 1 (N p, Gr p (R n )) is the oriented Gauss map of N p. w 2 dx dy, Corollary [Da Lio, R. 2010] Let u Ḣ 1/2 (R, N p ) be a critical point of E 1/2. Then u is Hölder continuous.

10 Integrability by Compensation and 2-Commutator Estimates. Theorem[Coiffman, Rochberg, Weiss ] Let m N and P a Fourier multiplier of order 0. There exists C > 0 s.t. h BMO(R m ) f L 2 (R m ) h P(f) P(h f) L 2 C h BMO f L 2 and by duality f, g L 2 (R m ) P(f) g f P g H 1 C f L 2 g L 2. where H 1 (R m ) is the real Hardy Space whose dual is BMO(R m ). Hardy estimates for Jacobians. [Coiffman, Lions, Meyer, Semmes ]. Let a W 1,2 (R m ) b W 1,2 (R m, m 2 R m ) introduce P := d d 1 P(da) = da and P( db) = 0 = da db H 1 C da L 2 db L 2. Since in 2 D. 1 (H 1 ) C 0 W 1,2 continuously we obtain Wente estimates. Integrability by Compensation and 2-Commutator Estimates. 8

11 Regularity by Compensation and 3-Commutator Estimates. Theorem.[Da Lio, R. 2009] Let m N. There exists C > 0 s.t. h BMO(R m ) and a H 1/2 (R m ) 1/4 (a 1/4 h) a 1/2 h + 1/4 a 1/4 h H 1/2 C h BMO a H 1/2 and by duality a, b H 1/2 (R m ) 1/4 ( 1/4 a b 1/4 (a b) + a 1/4 b ) H 1 C a H 1/2 b H 1/2. Moreover if one replaces BMO in the first inequality by H 1/2 the corresponding estimate holds with H 1/2 replaced by H 1. Organization of the proof of the first estimate : One uses Littlewood Paley. 1) For Low-High products one estimates each of the 3 terms separately. 2) For High-Low products a compensation arises in the grouping of the 2 first terms. The 3rd can be treated separately. 3) For products of comparable frequencies a compensation arises in the grouping of the 2 last terms. The first term can be treated separately Regularity by Compensation and 3-Commutator Estimates. 9

12 How the antisymmetry of Ω is used for non local operators? Let v L 2 (S 1, R n ) and Ω L 2 (S 1, so(n)) solving 1/4 v = Ω v We claim that v L p loc for some p > 2. Take P H1/2 (S 1, SO(n)) to be fixed later. Hence v solves 1/4 (P v) [ PΩP 1 1/4 P P 1] P v = T(P, v) where, from the 3-commutator estimate, the r.h.s. T(P, v) := 1/4 (P v) P 1/4 v + 1/4 P v H 1 plays the role of the Jacobian in the local case. Problem : A priori 1/4 P P 1 / so(n) Thus there is no hope to choose P in such a way that PΩP 1 1/4 P P 1 = 0 How the antisymmetry of Ω is used for non local operators? 10

13 Good Gauge extraction....but one can solve à la Uhlenbeck - under small Ω L 2 assumption - 1/4 P P 1 ( 1/4 P P 1 ) T = 2 PΩP 1 2 main reasons for that : A formal linearisation about id gives : P = exp(tξ) 1/4 P P 1 ( 1/4 P P 1 ) T 2t 1/4 ξ For small Ḣ 1/2 energy of P one has 1/4 P L 2 1/4 P P 1 ( 1/4 P P 1 ) T L 2 ( ) From the previous 3-commutator estimates one has for any a, b Ḣ 1/2 S(a, b) = 1/4 a b 1/4 (a b)+a 1/4 b L 2 C 1/4 a L 2 1/4 a L 2 Apply this estimate to the symmetric part of 1/4 P P 1 : 1/4 P P 1 +( 1/4 P P 1 ) T = 1/4 P P 1 1/4 (P P 1 )+P 1/4 P 1 Good Gauge extraction. 11

14 The Symmetric part of 1/4 P P 1. Take P constructed in the previous slide. We have then PΩP 1 + 1/4 P P 1 = symmetric part of 1/4 P P 1 = 1 2 ( 1/4 P P 1 + P 1/4 P 1 ) Again the 3-commutator estimates give for a, b H 1/2 (S 1 ) 1/4 S(a, b) = 1/4 ( 1/4 a b 1/4 (a b) + a 1/4 b ) H 1 (S 1 ) with estimate. Hence, since 1/4 (P P 1 ) = 0, the symmetric part of 1/4 P P 1 L 2,1 is more regular than 1/4 P P 1 L 2! The Symmetric part of 1/4 P P 1. 12

15 The Bootstrap Test. The test consists in taking Ω L 2 small and to ask if for any v L 2 1/4 v = Ω v = v = 0? For the previous choice of P H 1/2 (R, SO(n)) we then have 1/4 v = Ω v 1/4 (P v) + Sym( 1/4 P P 1 )(P v) = T(P, v) We now use the gain of regularity we obtained, T(P, V ) H 1 and 1/4 (Sym( 1/4 P P 1 )) H 1, to prove the bootstrap test. The Bootstrap Test. 13

16 Proof of the Bootstrap Test. Let φ such that g = 1/4 φ L 2,1 ( φ C 0 ) and g L 2,1 1. i) sup g L 2,1 1 S 1 φ 1/4 (P v) = sup g L 2,1 1 S 1 g P v = P v L 2, = v L 2, ii) iii) φ S 1 Sym( 1/4 P P 1 )P v φ Sym( 1/4 P P 1 ) L 2,1 Pv L 2, S 1 φ T(P, v) S = C P 2 H 1/2 v L 2, S(P, φ) v 1 1/4 S(P, φ) H 1 1/4 v BMO C P H 1/2 φ H 1/2 v L 2, i), ii) and iii) = v 2, C [ P H 1/2 + P 2 H 1/2 ] v 2, Hence small Ω L 2 = small P H 1/2 = v = 0 Proof of the Bootstrap Test. 13