On sharp Strichartz inequalities in low dimensions Dirk Hundertmark - PowerPoint PPT Presentation

On sharp Strichartz inequalities in low dimensions Dirk Hundertmark University of Birmingham Gregynog May 2007 – p. 1/24

Plan The Strichartz inequality Some History Proof of sharp Strichartz The Representation Theorem Classification of maximizers: how to use rotation invariance Gregynog May 2007 – p. 2/24

The free Schrödinger evolution Gregynog May 2007 – p. 3/24

The free Schrödinger evolution on L 2 ( R d ) i∂ t u = − ∆ u With initial condition u (0 , x ) = f ( x ) ∈ L 2 ( R d ) . Gregynog May 2007 – p. 3/24

The free Schrödinger evolution on L 2 ( R d ) i∂ t u = − ∆ u With initial condition u (0 , x ) = f ( x ) ∈ L 2 ( R d ) . Solved by the unitary time evolution u ( t, x ) = ( e it ∆ f )( x ) Gregynog May 2007 – p. 3/24

The free Schrödinger evolution on L 2 ( R d ) i∂ t u = − ∆ u With initial condition u (0 , x ) = f ( x ) ∈ L 2 ( R d ) . Solved by the unitary time evolution u ( t, x ) = ( e it ∆ f )( x ) � u ( t, . ) � L 2 ( R d ) = � f � L 2 ( R d ) . Gregynog May 2007 – p. 3/24

The free Schrödinger evolution on L 2 ( R d ) i∂ t u = − ∆ u With initial condition u (0 , x ) = f ( x ) ∈ L 2 ( R d ) . Solved by the unitary time evolution u ( t, x ) = ( e it ∆ f )( x ) � u ( t, . ) � L 2 ( R d ) = � f � L 2 ( R d ) . Thus u ∈ L ∞ t L 2 as a space-time function x Gregynog May 2007 – p. 3/24

Strichartz inequality Theorem (Strichartz 1977) . For p = p ( d ) = 2 + 4 d , � � � R d dx | u ( t, x ) | p � 1 /p ≤ S d � f � L 2 � u � L p t,x = dt x R Gregynog May 2007 – p. 4/24

Strichartz inequality Theorem (Strichartz 1977) . For p = p ( d ) = 2 + 4 d , � � � R d dx | u ( t, x ) | p � 1 /p ≤ S d � f � L 2 � u � L p t,x = dt x R Note p (1) = 6 and p (2) = 4 are the only even integer exponents. Gregynog May 2007 – p. 4/24

Strichartz inequality Theorem (Strichartz 1977) . For p = p ( d ) = 2 + 4 d , � � � R d dx | u ( t, x ) | p � 1 /p ≤ S d � f � L 2 � u � L p t,x = dt x R Note p (1) = 6 and p (2) = 4 are the only even integer exponents. From a harmonic analyst point of view, Strichartz follows from Gregynog May 2007 – p. 4/24

Strichartz inequality Theorem (Strichartz 1977) . For p = p ( d ) = 2 + 4 d , � � � R d dx | u ( t, x ) | p � 1 /p ≤ S d � f � L 2 � u � L p t,x = dt x R Note p (1) = 6 and p (2) = 4 are the only even integer exponents. From a harmonic analyst point of view, Strichartz follows from � � R d e i ( xk − τt ) δ ( τ − k 2 ) ˆ u ( t, x ) = (2 π ) ( d +1) / 2 f ( k ) dkdτ, R Gregynog May 2007 – p. 4/24

Strichartz inequality Theorem (Strichartz 1977) . For p = p ( d ) = 2 + 4 d , � � � R d dx | u ( t, x ) | p � 1 /p ≤ S d � f � L 2 � u � L p t,x = dt x R Note p (1) = 6 and p (2) = 4 are the only even integer exponents. From a harmonic analyst point of view, Strichartz follows from � � R d e i ( xk − τt ) δ ( τ − k 2 ) ˆ u ( t, x ) = (2 π ) ( d +1) / 2 f ( k ) dkdτ, R the paraboloid τ = k 2 has positive Gaussian curvature in R d +1 . Now apply the Stein-Tomas theorem. Gregynog May 2007 – p. 4/24

Some history Tomas 1975: Fourier restriction theorem for densities on the sphere Extended by Strichartz to some non-compact manifolds with non-vanishing Gaussian curvature. Much simplified proof by Ginibre and Velo 1985. Strichartz inequalities are at the heart of most studies of non-linear Schrödinger equations (from mid 1980 to now). Gregynog May 2007 – p. 5/24

Some questions What is � e it ∆ f � L p t,x S d = sup =? � f � L 2 f � =0 x Existence of maximizers: Are there f ∗ ∈ L 2 ( R d ) with � e it ∆ f ∗ � L p t,x S d = . � f ∗ � L 2 x What are all maximizers? Gregynog May 2007 – p. 6/24

Some questions What is � e it ∆ f � L p t,x S d = sup =? � f � L 2 f � =0 x Existence of maximizers: Are there f ∗ ∈ L 2 ( R d ) with � e it ∆ f ∗ � L p t,x S d = . � f ∗ � L 2 x What are all maximizers? all questions above turn out to be very hard to answer: Gregynog May 2007 – p. 6/24

Some questions What is � e it ∆ f � L p t,x S d = sup =? � f � L 2 f � =0 x Existence of maximizers: Are there f ∗ ∈ L 2 ( R d ) with � e it ∆ f ∗ � L p t,x S d = . � f ∗ � L 2 x What are all maximizers? all questions above turn out to be very hard to answer: The Strichartz inequality is invariant under all Galilei transformations (translations and boosts) and scaling, which is a huge non-compact group. Gregynog May 2007 – p. 6/24

Recent history Gregynog May 2007 – p. 7/24

Recent history Markus Kunze 2003: d = 1 maximizer for Strichartz inequality exist and are in L 2 ∩ L ∞ . Gregynog May 2007 – p. 7/24

Recent history Markus Kunze 2003: d = 1 maximizer for Strichartz inequality exist and are in L 2 ∩ L ∞ . Milena Stanislavova 2004: for a related functional (the dispersion management functional) all maximizers are smooth. Gregynog May 2007 – p. 7/24

Recent history Markus Kunze 2003: d = 1 maximizer for Strichartz inequality exist and are in L 2 ∩ L ∞ . Milena Stanislavova 2004: for a related functional (the dispersion management functional) all maximizers are smooth. Damiano Foschi preprint 2004: S 1 = 12 − 1 / 12 and S 2 = 2 − 1 / 2 and Gaussians are (among the) maximizers. Gregynog May 2007 – p. 7/24

The Representation Theorem Gregynog May 2007 – p. 8/24

The Representation Theorem Theorem (100DM and Vadim Zharnitsky 2006) . Let f ∈ L 2 ( R d ) . a ) If d = 1 then � � 1 | e it ∆ f ( x ) | 6 dxdt = √ 3 � f ⊗ f ⊗ f, P 1 ( f ⊗ f ⊗ f ) � L 2 ( R 3 ) 2 R R b ) If d = 2 then � � R 2 | e it ∆ f ( x ) | 4 dxdt = 1 4 � f ⊗ f, P 2 ( f ⊗ f ) � L 2 ( R 4 ) . R Gregynog May 2007 – p. 8/24

The Representation Theorem Theorem (100DM and Vadim Zharnitsky 2006) . Let f ∈ L 2 ( R d ) . a ) If d = 1 then � � 1 | e it ∆ f ( x ) | 6 dxdt = √ 3 � f ⊗ f ⊗ f, P 1 ( f ⊗ f ⊗ f ) � L 2 ( R 3 ) 2 R R b ) If d = 2 then � � R 2 | e it ∆ f ( x ) | 4 dxdt = 1 4 � f ⊗ f, P 2 ( f ⊗ f ) � L 2 ( R 4 ) . R P 1 : orthogonal projection onto functions in L 2 ( R 3 ) invariant under rotations fixing the (1 , 1 , 1) direction. Gregynog May 2007 – p. 8/24

The Representation Theorem Theorem (100DM and Vadim Zharnitsky 2006) . Let f ∈ L 2 ( R d ) . a ) If d = 1 then � � 1 | e it ∆ f ( x ) | 6 dxdt = √ 3 � f ⊗ f ⊗ f, P 1 ( f ⊗ f ⊗ f ) � L 2 ( R 3 ) 2 R R b ) If d = 2 then � � R 2 | e it ∆ f ( x ) | 4 dxdt = 1 4 � f ⊗ f, P 2 ( f ⊗ f ) � L 2 ( R 4 ) . R P 1 : orthogonal projection onto functions in L 2 ( R 3 ) invariant under rotations fixing the (1 , 1 , 1) direction. P 2 : orthogonal projection onto functions in L 2 ( R 4 ) invariant under rotations fixing the (1 , 0 , 1 , 0) and (0 , 1 , 0 , 1) directions. Gregynog May 2007 – p. 8/24

Consequences I Note that � f ⊗ f ⊗ f, f ⊗ f ⊗ f � L 2 ( R 3 ) = � f � 6 L 2 ( R ) . Gregynog May 2007 – p. 9/24

Consequences I Note that � f ⊗ f ⊗ f, f ⊗ f ⊗ f � L 2 ( R 3 ) = � f � 6 L 2 ( R ) . Since P 1 ≤ 1 L 2 ( R 3 ) , √ 3) − 1 / 6 � f � L 2 ( R ) � u � L 6 t,x ≤ (2 Gregynog May 2007 – p. 9/24

Consequences I Note that � f ⊗ f ⊗ f, f ⊗ f ⊗ f � L 2 ( R 3 ) = � f � 6 L 2 ( R ) . Since P 1 ≤ 1 L 2 ( R 3 ) , √ 3) − 1 / 6 � f � L 2 ( R ) � u � L 6 t,x ≤ (2 i.e., √ 3) − 1 / 6 = 12 − 1 / 12 . S 1 ≤ (2 and similarly, S 2 ≤ 2 − 1 / 2 . Gregynog May 2007 – p. 9/24

Consequences II Gregynog May 2007 – p. 10/24

Consequences II d = 1 : Equality in Strichartz inequality iff f ⊗ f ⊗ f ∈ Ran( P 1 ) Gregynog May 2007 – p. 10/24

Consequences II d = 1 : Equality in Strichartz inequality iff f ⊗ f ⊗ f ∈ Ran( P 1 ) i.e., iff f ⊗ f ⊗ f is invariant under rotations of R 3 which keep the (1 , 1 , 1) direction fixed. Gregynog May 2007 – p. 10/24

Consequences II d = 1 : Equality in Strichartz inequality iff f ⊗ f ⊗ f ∈ Ran( P 1 ) i.e., iff f ⊗ f ⊗ f is invariant under rotations of R 3 which keep the (1 , 1 , 1) direction fixed. If f ( x ) = e − ax 2 , then Gregynog May 2007 – p. 10/24

Consequences II d = 1 : Equality in Strichartz inequality iff f ⊗ f ⊗ f ∈ Ran( P 1 ) i.e., iff f ⊗ f ⊗ f is invariant under rotations of R 3 which keep the (1 , 1 , 1) direction fixed. If f ( x ) = e − ax 2 , then 3 ) = e − aη 2 f ⊗ f ⊗ f ( η ) = f ( η 1 ) f ( η 2 ) f ( η 3 ) = e − a ( η 2 1 + η 2 2 + η 2 is invariant under all rotations of R 3 . Gregynog May 2007 – p. 10/24

Consequences II If f ( x ) = e − ax 2 + bx , then f ⊗ f ⊗ f ( η ) = e − aη 2 + b (1 , 1 , 1) · η Gregynog May 2007 – p. 11/24

Consequences II If f ( x ) = e − ax 2 + bx , then f ⊗ f ⊗ f ( η ) = e − aη 2 + b (1 , 1 , 1) · η is invariant under rotations of R 3 keeping the (1 , 1 , 1) direction fixed. Gregynog May 2007 – p. 11/24

Consequences II If f ( x ) = e − ax 2 + bx , then f ⊗ f ⊗ f ( η ) = e − aη 2 + b (1 , 1 , 1) · η is invariant under rotations of R 3 keeping the (1 , 1 , 1) direction fixed. In particular, S 1 = 12 − 1 / 12 and analogously S 2 = 2 − 1 / 2 . Gregynog May 2007 – p. 11/24