Introduction Sketch of the proof of the first statement Continuous solutions to a balance law L. Caravenna, OxPDE F. Bigolin, F. Serra Cassano collaboration with: G. Alberti, S. Bianchini Hyp2012, Padova, 28th June 2012 Continuous solutions to a balance law 1 / 26
Introduction Sketch of the proof of the first statement Outline Eulerian vs Lagrangian formulation for the PDE g : R + × R → R u t ( t , x ) + [ u ( t , x ) 2 / 2] x = g ( t , x ) , 1. Introduction Characteristics for smooth solutions Continuous solutions, bounded sources Main statement 2. Sketch of the proof of the first statement Continuous solutions to a balance law 2 / 26
Introduction Sketch of the proof of the first statement Outline Eulerian vs Lagrangian formulation for the PDE g : R + × R → R u t ( t , x ) + f ( u ( t , x )) x = g ( t , x ) , 1. Introduction Characteristics for smooth solutions Continuous solutions, bounded sources Main statement 2. Sketch of the proof of the first statement Continuous solutions to a balance law 2 / 26
Introduction Sketch of the proof of the first statement Outline 1. Introduction Characteristics for smooth solutions Continuous solutions, bounded sources Main statement 2. Sketch of the proof of the first statement Continuous solutions to a balance law 3 / 26
Introduction Sketch of the proof of the first statement Few recalls on the equation: characteristics g : R + × R → R u t ( t , x ) + f ( u ( t , x )) x = g ( t , x ) , PDE � system of ODEs Continuous solutions to a balance law 4 / 26
Introduction Sketch of the proof of the first statement Few recalls on the equation: characteristics g : R + × R → R u t ( t , x ) + f ( u ( t , x )) x = g ( t , x ) , PDE � system of ODEs If u smooth chain rule t u ( s , x ( s )) x ( t ) x Continuous solutions to a balance law 4 / 26
Introduction Sketch of the proof of the first statement Few recalls on the equation: characteristics g : R + × R → R u t ( t , x ) + f ( u ( t , x )) x = g ( t , x ) , PDE � system of ODEs If u smooth chain rule d t ds u ( s , x ( s )) = u t ( s , x ( s )) + u x ( s , x ( s )) · ˙ x ( s ) x ( t ) x Continuous solutions to a balance law 4 / 26
Introduction Sketch of the proof of the first statement Few recalls on the equation: characteristics g : R + × R → R u t ( t , x ) + f ( u ( t , x )) x = g ( t , x ) , PDE � system of ODEs If u smooth chain rule d t ds u ( s , x ( s )) = u t ( s , x ( s )) + u x ( s , x ( s )) · ˙ x ( s ) x ( s ) = f ′ ( u ( s , x ( s ))) If ˙ = u t (Ψ( s )) + f ′ ( u ( s , x ( s ))) · u x ( s , x ( s )) x ( t ) x = g (Ψ( s )) Continuous solutions to a balance law 4 / 26
Introduction Sketch of the proof of the first statement Few recalls on the equation: characteristics g : R + × R → R u t ( t , x ) + f ( u ( t , x )) x = g ( t , x ) , PDE � system of ODEs If u smooth t ds χ ( s ; x 0 ) = f ′ ( u ( s , χ ( s ; x 0 ))) d d ds u ( s , x ( s )) = g ( s , x ( s )) χ ( t = 0; x 0 ) = x 0 u (0 , x ) = u 0 ( x ) x Continuous solutions to a balance law 4 / 26
Introduction Sketch of the proof of the first statement Few recalls on the equation: characteristics g : R + × R → R u t ( t , x ) + f ( u ( t , x )) x = g ( t , x ) , PDE � system of ODEs If u smooth t ds χ ( s ; x 0 ) = f ′ ( u ( s , χ ( s ; x 0 ))) d d ds u ( s , x ( s )) = g ( s , x ( s )) χ ( t = 0; x 0 ) = x 0 u (0 , x ) = u 0 ( x ) x Straight lines when g = 0 Continuous solutions to a balance law 4 / 26
Introduction Sketch of the proof of the first statement Few recalls on the equation: characteristics g : R + × R → R u t ( t , x ) + f ( u ( t , x )) x = g ( t , x ) , PDE � system of ODEs If u smooth t � χ ( s ; x 0 ) � z ( s ; x 0 ) = u ( s , χ ( s ; x 0 )) � f ′ ( z )) � � z = ˙ g ( z ) x x � � z ( s = 0; x ) = u 0 ( x ) Continuous solutions to a balance law 4 / 26
Introduction Sketch of the proof of the first statement Few recalls on the equation u may develop jumps � u 2 � e.g. u t + = 0 2 x � generalized characteristics, t def. by differential inclusions t=0 Not the issue here C. M. Dafermos. u ( t = 0) u ( t ) Hyperbolic Conservation Laws in Continuous Physics . − → Springer, 2000. Non uniqueness of distributional solutions � entropy solutions Existence, stability, uniqueness clear in the scalar, 1 D -case [Oleinik, Kruzhkov,. . . ] Continuous solutions to a balance law 5 / 26
Introduction Sketch of the proof of the first statement Few recalls on the equation u may develop jumps � u 2 � e.g. u t + = 0 2 x � generalized characteristics, t def. by differential inclusions t=0 Not the issue here C. M. Dafermos. u ( t = 0) u ( t ) Hyperbolic Conservation Laws in Continuous Physics . − → Springer, 2000. Non uniqueness of distributional solutions � entropy solutions Existence, stability, uniqueness clear in the scalar, 1 D -case [Oleinik, Kruzhkov,. . . ] Continuous solutions to a balance law 5 / 26
Introduction Sketch of the proof of the first statement Few recalls on the equation u may develop jumps � u 2 � e.g. u t + = 0 2 x � generalized characteristics, t def. by differential inclusions t=0 Not the issue here C. M. Dafermos. u ( t = 0) u ( t ) Hyperbolic Conservation Laws in Continuous Physics . − → Springer, 2000. Non uniqueness of distributional solutions � entropy solutions Existence, stability, uniqueness clear in the scalar, 1 D -case [Oleinik, Kruzhkov,. . . ] Continuous solutions to a balance law 5 / 26
Introduction Sketch of the proof of the first statement An example: Hunter-Saxton equation � x � u 2 � g ( t , x ) = 1 u 2 u t + = g , x ( y , t ) dy 2 2 −∞ x g may be interpreted as a control device preserving continuity C. M. Dafermos Many authors Continuous solutions for balance laws Ricerche di Matematica 55 (2006), 79–91. Continuous solutions to a balance law 6 / 26
Introduction Sketch of the proof of the first statement An example: Hunter-Saxton equation � u 2 ( t , x ) � u t ( t , x ) + = g ( t , x ) 2 x We restrict from now on to the particular case u continuous, g bounded u continuous is a low regularity for this equation Not enough for the theory on ODEs with rough coefficients Continuous solutions to a balance law 6 / 26
Introduction Sketch of the proof of the first statement Motivation Sub-Riemannian Heisenberg group H n ≡ R 2 n +1 = { ( x , y , z ) } φ : W = { x 1 = 0 } ⊂ H n → R continuous Graph H φ � It induces a graph quasidistance d φ and an intrinsic differentiable structure on W . One has notions of intrinsic cones, intrinsic differentiability, intrinsic gradient, . . . . The graph is intrinsic regular if ◮ φ is uniformly intrinsic differentiable � φ 2 � ◮ ( n = 1) φ satisfies φ y + z = g with g continuous 2 Ambrosio-Serra Cassano-Vittone Bigolin-Serra Cassano Franchi-Serapioni-Serra Cassano Very irregular from the Euclidean point of view Kirchheim-Serra Cassano Continuous solutions to a balance law 7 / 26
Introduction Sketch of the proof of the first statement Motivation Sub-Riemannian Heisenberg group H n ≡ R 2 n +1 = { ( x , y , z ) } φ : W = { x 1 = 0 } ⊂ H n → R continuous � Graph H φ It induces a graph quasidistance d φ and an intrinsic differentiable structure on W . One has notions of intrinsic cones, intrinsic differentiability, intrinsic gradient, . . . . n = 1 for simplicity of presentation The graph is intrinsic Lipschitz if φ satisfies � φ 2 � φ y + = g g bounded 2 z Citti-Manfredini-Pinamonti-Serra Cassano Bigolin-C-Serra Cassano Franchi-Serapioni-Serra Cassano Continuous solutions to a balance law 7 / 26
Introduction Sketch of the proof of the first statement Example t � � u 2 � 1 / 2 x ≥ 0 u t + = 2 − 1 / 2 x < 0 x Distributional solution: x � u ( t , x ) = | x | Characteristics: u ( t , x ) x ( t ) = u ( t , x ) ˙ x Continuous solutions to a balance law 8 / 26
Introduction Sketch of the proof of the first statement Example t � � u 2 � 1 / 2 x ≥ 0 u t + = 2 − 1 / 2 x < 0 x Distributional solution: x � u ( t , x ) = | x | Characteristics: u ( t , x ) x ( t ) = u ( t , x ) ˙ x Continuous solutions to a balance law 8 / 26
Introduction Sketch of the proof of the first statement Example t � � u 2 � 1 / 2 x ≥ 0 u t + = 2 − 1 / 2 x < 0 x Distributional solution: x � u ( t , x ) = | x | Characteristics: u ( t , x ) x ( t ) = u ( t , x ) ˙ x Continuous solutions to a balance law 8 / 26
Introduction Sketch of the proof of the first statement Example t � � u 2 � 1 / 2 x ≥ 0 u t + = 2 − 1 / 2 x < 0 x Distributional solution: x � u ( t , x ) = | x | Characteristics: u ( t , x ) x ( t ) = u ( t , x ) ˙ x d dt u ( t , x ( t )) = g ( t , x ( t )) 1 / 2 x > 0 The right representative is ˆ g ( t , x ) = 0 x = 0 − 1 / 2 x < 0 Continuous solutions to a balance law 8 / 26
Introduction Sketch of the proof of the first statement Two notions of continuous solutions Let f ∈ C 2 ( R + × R ) and g ∈ L ∞ ( R + × R ). Let u ∈ C ( R + × R ) Distributional solution: in D ( R + × R ) u t + f ( u ) x = g g ∈ L ∞ ( R + × R ) s.t. � g − ˆ Broad solution: ∃ ˆ g � L 2 = 0 and d dt u ( t , γ ( t )) = ˆ g ( t , γ ( t )) ∀ γ ( t ) : ˙ γ = u ◦ γ. Continuous solutions to a balance law 9 / 26
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