Invariant Manifolds and dispersive Hamiltonian Evolution Equations W. Schlag, http://www.math.uchicago.edu/˜schlag Boston, January 5, 2012 W. Schlag, http://www.math.uchicago.edu/˜schlag Dispersive Hamiltonian PDEs
Old-fashioned string theory How does a guitar string evolve in time? Ancient Greece: observed that musical intervals such as an octave, a fifth etc. were based on integer ratios. Post Newton: mechanistic model, use calculus and F = ma . Assume displacement u = u ( t , x ) is small. Force proportional to curvature: F = ku xx . Figure: Forces acting on pieces of string Dynamical law u tt = c 2 u xx . Write as � u = 0. This is an idealization, or model! W. Schlag, http://www.math.uchicago.edu/˜schlag Dispersive Hamiltonian PDEs
Solving for the string Cauchy problem: � u = 0 , u ( 0 ) = f , ∂ t u ( 0 ) = g d’Alembert solution: ( ∂ 2 t − c 2 ∂ 2 x ) u = ( ∂ t − c ∂ x )( ∂ t + c ∂ x ) u = 0 Reduction to first order, transport equations u t + cu x = 0 ⇔ u ( t , x ) = ϕ ( x − ct ) u t − cu x = 0 ⇔ u ( t , x ) = ψ ( x + ct ) Adjust for initial conditions, gives d’Alembert formula: � x + ct u ( t , x ) = 1 2 ( f ( x − ct ) + f ( x + ct )) + 1 g ( y ) dy 2 c x − ct If g = 0, the initial position f splits into left- and right-moving waves. W. Schlag, http://www.math.uchicago.edu/˜schlag Dispersive Hamiltonian PDEs
d’Alembert solution W. Schlag, http://www.math.uchicago.edu/˜schlag Dispersive Hamiltonian PDEs
Standing waves Clamped string: u ( t , 0 ) = u ( t , L ) = 0 for all t ≥ 0, � u = 0. Special solutions (with c = 1) with n ≥ 1 an integer � � u n ( t , x ) = sin ( π nx / L ) a n sin ( π nt / L ) + b n cos ( π nt / L ) Fourier’s claim: All solutions are superpositions of these! Ω ⊂ R d bounded domain, or compact manifold. Let − ∆ Ω ϕ n = λ 2 n ϕ n , with Dirichlet boundary condition in the former case. Then � c n , ± e ± i λ n t ϕ n ( x ) u ( t , x ) = n ≥ 0 , ± solves � u = 0 (with boundary condition). W. Schlag, http://www.math.uchicago.edu/˜schlag Dispersive Hamiltonian PDEs
Drum membranes Two-dimensional waves on a drum: u tt − ∆ u = 0 with u = 0 on the boundary. Figure: Four basic harmonics of the drum First, third pictures u ( t , r ) = cos ( t λ ) J 0 ( λ r ) , where J 0 ( λ ) = 0. Second, fourth pictures u ( t , r ) = cos ( t µ ) J m ( µ r ) cos ( m θ ) , where J m ( µ ) = 0. The J m are Bessel functions. W. Schlag, http://www.math.uchicago.edu/˜schlag Dispersive Hamiltonian PDEs
Electrification of waves Maxwell’s equations: E ( t , x ) and B ( t , x ) vector fields div E = ε − 1 0 ρ, div B = 0 curl E + ∂ t B = 0 , curl B − µ 0 ε 0 ∂ t E = µ 0 J ε 0 electric constant, µ 0 magnetic constant, ρ charge density, J current density. In vaccum ρ = 0 , J = 0. Differentiate fourth equation in time: curl B t − µ 0 ε 0 E tt = 0 ⇒ curl ( curl E ) + µ 0 ε 0 E tt = 0 ∇ ( div E ) − ∆ E + µ 0 ε 0 E tt = 0 ⇒ E tt − c 2 ∆ E = 0 Similarly B tt − c 2 ∆ B = 0. In 1861 Maxwell noted that c is the speed of light, and concluded that light should be an electromagnetic wave! Wave equation appears as a fundamental equation! Loss of Galilei invariance! W. Schlag, http://www.math.uchicago.edu/˜schlag Dispersive Hamiltonian PDEs
Visualization of EM fields Figure: E & B fields W. Schlag, http://www.math.uchicago.edu/˜schlag Dispersive Hamiltonian PDEs
Least action Principle of Least Action: Paths ( x ( t ) , ˙ x ( t )) , for t 0 ≤ t ≤ t 1 with endpoints x ( t 0 ) = x 0 , and x ( t 1 ) = x 1 fixed. The physical path determined by kinetic energy K ( x , ˙ x ) and potential energy P ( x , ˙ x ) minimizes the action: � t 1 � t 1 ( K − P )( x ( t ) , ˙ L ( x ( t ) , ˙ S := x ( t )) dt = x ( t )) dt t 0 t 0 with L the Lagrangian. In fact: equations of motion equal Euler-Lagrange equation ∂ L x + ∂ L − d ∂ x = 0 ∂ ˙ dt and the physical trajectories are the critical points of S . x 2 − U ( x ) , we obtain m ¨ For L = 1 x ( t ) = − U ′ ( x ( t )) , which is 2 m ˙ Newton’s F = ma . W. Schlag, http://www.math.uchicago.edu/˜schlag Dispersive Hamiltonian PDEs
Waves from a Lagrangian Let � 1 � t + |∇ u | 2 � − u 2 L ( u , ∂ t u ) := ( t , x ) dtdx (1) 2 R 1 + d t , x Substitute u = u 0 + ε v . Then � ( � u 0 )( t , x ) v ( t , x ) dtdx + O ( ε 2 ) L ( u , ∂ t u ) = L 0 + ε R 1 + d t , x where � = ∂ tt − ∆ . In other words, u 0 is a critical point of L if and only if � u 0 = 0. Significance: Underlying symmetries ⇒ invariances ⇒ Conservation laws Conservation of energy, momentum, angular momentum Lagrangian formulation has a universal character, and is flexible, versatile. W. Schlag, http://www.math.uchicago.edu/˜schlag Dispersive Hamiltonian PDEs
Wave maps Let ( M , g ) be a Riemannian manifold, and u : R 1 + d → M smooth. t , x What is a wave into M ? Lagrangian d � 1 � 2 ( −| ∂ t u | 2 | ∂ j u | 2 � L ( u , ∂ t u ) = g + dtdx g R 1 + d j = 1 t , x Critical points L ′ ( u , ∂ t u ) = 0 satisfy “manifold-valued wave equation”. M ⊂ R N imbedded, this equation is � u ⊥ T u M or � u = A ( u )( ∂ u , ∂ u ) , A being the second fundamental form. For example, M = S n − 1 , then � u = u ( −| ∂ t u | 2 + |∇ u | 2 ) Note: Nonlinear wave equation, null-form! Harmonic maps are solutions. Intrinsic formulation: D α ∂ α u = η αβ D β ∂ α u = 0, in coordinates tt + ∆ u i + Γ i jk ( u ) ∂ α u j ∂ α u k = 0 − u i η = ( − 1 , 1 , 1 , . . . , 1 ) Minkowski metric W. Schlag, http://www.math.uchicago.edu/˜schlag Dispersive Hamiltonian PDEs
Maxwell from Lagrangian To formulate electro-magnetism in a Lagrangian frame work, introduce vector potential: A = ( A 0 , A ) with B = curl A , E = ∇ A 0 − ∂ t A Define curvature tensor F αβ := ∂ α A β − ∂ β A α Maxwell’s equations: ∂ α F αβ = 0. Lagrangian: � 1 4 F αβ F αβ dtdx L = R 1 + 3 t , x Lorentz invariance: Minkowski metric [ x , y ] := η αβ x α y β = − x 0 y 0 + Σ 3 j = 1 x j y j Linear maps S : R 4 → R 4 with [ Sx , y ] = [ x , y ] for all x , y ∈ R 1 + 3 t , x are called Lorentz transforms. Note: � u = 0 ⇔ � ( u ◦ S ) = 0. ∂η α ′ ∂ξ β ∂ξ α For L : ξ = ( t , x ) �→ η = ( s , y ) , F αβ �→ ˜ F α ′ β ′ = F αβ ∂η β ′ . W. Schlag, http://www.math.uchicago.edu/˜schlag Dispersive Hamiltonian PDEs
Lorentz transformations 1 t ′ cosh α sinh α 0 0 t x ′ sinh α cosh α 0 0 x 1 1 = x ′ 0 0 1 0 x 2 2 x ′ 0 0 0 1 x 3 3 Figure: Causal structure of space-time W. Schlag, http://www.math.uchicago.edu/˜schlag Dispersive Hamiltonian PDEs
Lorentz transformations 2 Figure: Snapshots of Lorentz transforms Lorentz transforms (hyperbolic rotations) are for the d’Alembertian what Euclidean rotations are for the Laplacian. W. Schlag, http://www.math.uchicago.edu/˜schlag Dispersive Hamiltonian PDEs
Gauge invariance We obtain the same E , B fields after A �→ A + ( φ t , ∇ φ ) . B = curl ( A + ∇ φ ) = curl ( A ) E = ∇ ( A 0 + φ t ) − ∂ t ( A + ∇ φ ) = ∇ A 0 − ∂ t A Curvature F αβ invariant under such gauge transforms. Impose a gauge: ∂ α A α = 0 (Lorentz), div A = 0 (Coulomb). These pick out a unique representative in the equivalence class of vector potentials. Make Klein-Gordon equation � u − m 2 u = ∂ α ∂ α u − m 2 u = 0 gauge invariant: u �→ e i ϕ u with ϕ = ϕ ( t , x ) does not leave solutions invariant. How to modify? KG-Lagrangian is � 1 � ∂ α u ∂ α u + m 2 | u | 2 � L 0 := dtdx 2 R 1 + 3 t , x Need to replace ∂ α with D α = ∂ α − iA α . Bad choice: � 1 � D α uD α u + m 2 | u | 2 � L 1 := dtdx 2 R 1 + 3 t , x W. Schlag, http://www.math.uchicago.edu/˜schlag Dispersive Hamiltonian PDEs
Maxwell-Klein-Gordon system How is A α determined? Need to add a piece to the Lagrangian to rectify that: a “simple” and natural choice is the Maxwell Lagrangian. So obtain 2 D α uD α u + m 2 � � 1 4 F αβ F αβ + 1 2 | u | 2 � L MKG := dtdx R 1 + 3 t , x Dynamical equations, as Euler-Lagrange equation of L MKG : ∂ α F αβ = Im ( φ D β φ ) D α D α φ − m 2 φ = 0 Coupled system, Maxwell with current J β = Im ( φ D β φ ) which is determined by scalar field φ . Lorentz and U ( 1 ) gauge invariant. Maxwell-Klein-Gordon system. W. Schlag, http://www.math.uchicago.edu/˜schlag Dispersive Hamiltonian PDEs
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