Continuous solutions to a balance law L. Caravenna, OxPDE F. - - PowerPoint PPT Presentation

Introduction Sketch of the proof of the first statement

Continuous solutions to a balance law

• L. Caravenna, OxPDE

collaboration with:

• F. Bigolin, F. Serra Cassano
• 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 ut(t, x) + [u(t, x)2/2]x = g(t, x), g : R+ × R → R

• 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 ut(t, x) + f (u(t, x))x = g(t, x), g : R+ × R → R

• 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

ut(t, x) + f (u(t, x))x = g(t, x), g : R+ × R → R 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

ut(t, x) + f (u(t, x))x = g(t, x), g : R+ × R → R PDE system of ODEs If u smooth chain rule u(s, x(s)) x(t) x t

Continuous solutions to a balance law 4 / 26

Introduction Sketch of the proof of the first statement

Few recalls on the equation: characteristics

ut(t, x) + f (u(t, x))x = g(t, x), g : R+ × R → R PDE system of ODEs If u smooth chain rule d ds u(s, x(s)) = ut(s, x(s)) + ux(s, x(s)) · ˙ x(s) x(t) x t

Continuous solutions to a balance law 4 / 26

Introduction Sketch of the proof of the first statement

Few recalls on the equation: characteristics

ut(t, x) + f (u(t, x))x = g(t, x), g : R+ × R → R PDE system of ODEs If u smooth chain rule d ds u(s, x(s)) = ut(s, x(s)) + ux(s, x(s)) · ˙ x(s) If ˙ x(s) = f ′(u(s, x(s))) = ut(Ψ(s)) + f ′(u(s, x(s))) · ux(s, x(s)) = g(Ψ(s)) x(t) x t

Continuous solutions to a balance law 4 / 26

Introduction Sketch of the proof of the first statement

Few recalls on the equation: characteristics

ut(t, x) + f (u(t, x))x = g(t, x), g : R+ × R → R PDE system of ODEs If u smooth     

d ds χ(s; x0) = f ′(u(s, χ(s; x0))) d ds u(s, x(s)) = g(s, x(s))

χ(t = 0; x0) = x0 u(0, x) = u0(x) x t

Continuous solutions to a balance law 4 / 26

Introduction Sketch of the proof of the first statement

Few recalls on the equation: characteristics

ut(t, x) + f (u(t, x))x = g(t, x), g : R+ × R → R PDE system of ODEs If u smooth     

d ds χ(s; x0) = f ′(u(s, χ(s; x0))) d ds u(s, x(s)) = g(s, x(s))

χ(t = 0; x0) = x0 u(0, x) = u0(x) Straight lines when g = 0 x t

Continuous solutions to a balance law 4 / 26

Introduction Sketch of the proof of the first statement

Few recalls on the equation: characteristics

ut(t, x) + f (u(t, x))x = g(t, x), g : R+ × R → R PDE system of ODEs If u smooth z(s; x0) =

• χ(s; x0)

u(s, χ(s; x0))

• ˙

z = f ′(z))

g(z)

• z(s = 0; x) =
• x

u0(x)

• x

t

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 e.g. ut + u2 2

• x

= 0 generalized characteristics,

• def. by differential inclusions

Not the issue here

• C. M. Dafermos.

Hyperbolic Conservation Laws in Continuous Physics. Springer, 2000.

t=0 t

u(t = 0) u(t) − → Non uniqueness of distributional solutions entropy solutions Existence, stability, uniqueness clear in the scalar, 1D-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 e.g. ut + u2 2

• x

= 0 generalized characteristics,

• def. by differential inclusions

Not the issue here

• C. M. Dafermos.

Hyperbolic Conservation Laws in Continuous Physics. Springer, 2000.

t=0 t

u(t = 0) u(t) − → Non uniqueness of distributional solutions entropy solutions Existence, stability, uniqueness clear in the scalar, 1D-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 e.g. ut + u2 2

• x

= 0 generalized characteristics,

• def. by differential inclusions

Not the issue here

• C. M. Dafermos.

Hyperbolic Conservation Laws in Continuous Physics. Springer, 2000.

t=0 t

u(t = 0) u(t) − → Non uniqueness of distributional solutions entropy solutions Existence, stability, uniqueness clear in the scalar, 1D-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

ut + u2 2

• x

= g, g(t, x) = 1 2 x

−∞

u2

x(y, t)dy

g may be interpreted as a control device preserving continuity

• C. M. Dafermos

Continuous solutions for balance laws Ricerche di Matematica 55 (2006), 79–91. Many authors

Continuous solutions to a balance law 6 / 26

Introduction Sketch of the proof of the first statement

An example: Hunter-Saxton equation

ut(t, x) + u2(t, x) 2

• x

= g(t, 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 Hn ≡ R2n+1 = {(x, y, z)} φ : W = {x1 = 0} ⊂ Hn → R continuous

• GraphHφ

It induces a graph quasidistance dφ 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 ◮ (n = 1) φ satisfies φy +

• φ2

2

• z = g with g continuous

Ambrosio-Serra Cassano-Vittone Franchi-Serapioni-Serra Cassano Bigolin-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 Hn ≡ R2n+1 = {(x, y, z)} φ : W = {x1 = 0} ⊂ Hn → R continuous

• GraphHφ

It induces a graph quasidistance dφ 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 φy + φ2 2

• z

= g g bounded

Citti-Manfredini-Pinamonti-Serra Cassano Franchi-Serapioni-Serra Cassano Bigolin-C-Serra Cassano

Continuous solutions to a balance law 7 / 26

Introduction Sketch of the proof of the first statement

Example

ut + u2 2

• x

=

• 1/2

x ≥ 0 −1/2 x < 0 Distributional solution: u(t, x) =

• |x|

Characteristics: ˙ x(t) = u(t, x) x t x u(t, x)

Continuous solutions to a balance law 8 / 26

Introduction Sketch of the proof of the first statement

Example

ut + u2 2

• x

=

• 1/2

x ≥ 0 −1/2 x < 0 Distributional solution: u(t, x) =

• |x|

Characteristics: ˙ x(t) = u(t, x) x t x u(t, x)

Continuous solutions to a balance law 8 / 26

Introduction Sketch of the proof of the first statement

Example

ut + u2 2

• x

=

• 1/2

x ≥ 0 −1/2 x < 0 Distributional solution: u(t, x) =

• |x|

Characteristics: ˙ x(t) = u(t, x) x t x u(t, x)

Continuous solutions to a balance law 8 / 26

Introduction Sketch of the proof of the first statement

Example

ut + u2 2

• x

=

• 1/2

x ≥ 0 −1/2 x < 0 Distributional solution: u(t, x) =

• |x|

Characteristics: ˙ x(t) = u(t, x) d dt u(t, x(t)) = g(t, x(t)) x t x u(t, x) The right representative is ˆ g(t, x) =      1/2 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: ut + f (u)x = g in D(R+ × R) Broad solution: ∃ ˆ g ∈ L∞(R+ × R) s.t. g − ˆ gL2 = 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

Introduction Sketch of the proof of the first statement

Theorem [Bigolin-C.-Serra Cassano, available on arxiv]

Let u be a continuous distributional solution of ut(t, x) + u2(t, x) 2

• x

= g(t, x), g ∈ L∞(R+ × R; R). Choosing suitably ˆ g ∈ L∞(R+ × R; R) representative of g d dt u(t, γ(t)) = ˆ g(t, γ(t)) ∀γ(t) : ˙ γ = u ◦ γ. u is moreoveover 1/2-H¨

• lder continuous

For the converse implication

Citti-Manfredini-Pinamonti-Serra Cassano

Continuous solutions to a balance law 10 / 26

Introduction Sketch of the proof of the first statement

Let f ∈ C 2(R+ × R; R)

Theorem [Alberti-Bianchini-C, forthcoming]

Let u be a continuous distributional solution of ut(t, x) + f (u(t, x))x = g(t, x), g ∈ L∞(R+ × R). If Infl(f ) = 0 is negligible, then ∃ˆ g ∈ L∞(R+ × R) : d dt u(t, γ(t)) = ˆ g(t, γ(t)) ∀γ(t) : ˙ γ = f ′(u) ◦ γ. t → u(t, γ(t)) not generally Lipschitz without assumptions on f

Continuous solutions to a balance law 11 / 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 12 / 26

Introduction Sketch of the proof of the first statement

Some examples

Characteristics split e.g. ˙ x(t) = sgn x(t)

• |x(t)|,

ut + u2 2

• x

=

• 1/2

x ≥ 0 −1/2 x < 0 x t x u(t, x)

Continuous solutions to a balance law 13 / 26

Introduction Sketch of the proof of the first statement

Some examples

Characteristics split e.g. ˙ x(t) = sgn x(t)

• |x(t)|,

ut + u2 2

• x

=

• 1/2

x ≥ 0 −1/2 x < 0 They may also concentrate x t x u(t, x)

Continuous solutions to a balance law 13 / 26

Introduction Sketch of the proof of the first statement

Some examples

Sometimes not e.g. ˙ x(t) =

• |x(t)|,

ut + u2 2

• x

=

• 1/2

x ≥ 0 −1/2 x < 0 We are not in this case

• G. Crippa.

Lagrangian flows and the one dimensional Peano phenomenon for ODEs. JDE, to appear.

x t x u(t, x)

Continuous solutions to a balance law 13 / 26

Introduction Sketch of the proof of the first statement

Some examples

Sometimes not e.g. ˙ x(t) =

• |x(t)|,

ut + u2 2

• x

=

• 1/2

x ≥ 0 −1/2 x < 0 We are not in this case

• G. Crippa.

Lagrangian flows and the one dimensional Peano phenomenon for ODEs. JDE, to appear.

x t x u(t, x)

Continuous solutions to a balance law 13 / 26

Introduction Sketch of the proof of the first statement

Some hints of the proof

• 1. Dafermos’ computation

Given any characteristic γ, compute the balance in the region Γε = {(t, x) : t1 ≤ t ≤ t2, γ(t) ≤ x ≤ γ(t) + ε} γ(t2)+ε

γ(t2)

u(t, x)dx − γ(t1)+ε

γ(t1)

u(t, x)dx = t2

t1

γ(t)+ε

γ(t)

g(t, x)dtdx − 1 2 t2

t1

[u(t, γ(t) + ε) − u(t, γ(t))]2 dt ≥ t2

t1

γ(t)+ε

γ(t)

g(t, x)dtdx

Continuous solutions to a balance law 14 / 26

Introduction Sketch of the proof of the first statement

Some hints of the proof

• 1. Dafermos’ computation

Given any characteristic γ, compute the balance in the region Γε = {(t, x) : t1 ≤ t ≤ t2, γ(t) ≤ x ≤ γ(t) + ε} In the analogous region Γ−ε one has the opposite inequality: t2

t1

γ(t)

γ(t)−ε

g(t, x)dtdx ≤ γ(t2)+ε

γ(t2)

u(t, x)dx − γ(t1)+ε

γ(t1)

u(t, x)dx ≤ t2

t1

γ(t)+ε

γ(t)

g(t, x)dtdx

Continuous solutions to a balance law 14 / 26

Introduction Sketch of the proof of the first statement

Some hints of the proof

• 1. Dafermos’ computation

Given any characteristic γ, compute the balance in the region Γε = {(t, x) : t1 ≤ t ≤ t2, γ(t) ≤ x ≤ γ(t) + ε} When g continuous in x one can take the limit as ε ↓ 0: u(t1, γ(t2)) − u(t1, γ(t1)) = t2

t1

g(t, γ(t))dt If g only bounded |u(t1, γ(t2)) − u(t1, γ(t1))| ≤ g∞(t2 − t1) Therefore u Lipschitz along characteristics

Continuous solutions to a balance law 14 / 26

Introduction Sketch of the proof of the first statement

Some hints of the proof

• 1. Dafermos’ computation

Given any characteristic γ, compute the balance in the region Γε = {(t, x) : t1 ≤ t ≤ t2, γ(t) ≤ x ≤ γ(t) + ε} When g continuous in x one can take the limit as ε ↓ 0: u(t1, γ(t2)) − u(t1, γ(t1)) = t2

t1

g(t, γ(t))dt If g only bounded |u(t1, γ(t2)) − u(t1, γ(t1))| ≤ g∞(t2 − t1) Therefore u Lipschitz along characteristics This can be done under the hypothesis on the inflection points of f Otherwise there is a counterexample

Continuous solutions to a balance law 14 / 26

Introduction Sketch of the proof of the first statement

Some hints of the proof

• 1. Dafermos’ computation

Given any characteristic γ, compute the balance in the region Γε = {(t, x) : t1 ≤ t ≤ t2, γ(t) ≤ x ≤ γ(t) + ε} When g continuous in x one can take the limit as ε ↓ 0: u(t1, γ(t2)) − u(t1, γ(t1)) = t2

t1

g(t, γ(t))dt If g only bounded |u(t1, γ(t2)) − u(t1, γ(t1))| ≤ g∞(t2 − t1) Therefore u Lipschitz along characteristics This can be done under the hypothesis on the inflection points of f Otherwise there is a counterexample

Continuous solutions to a balance law 14 / 26

Introduction Sketch of the proof of the first statement

Some hints of the proof

• 1. Dafermos’ computation

Given any characteristic γ, compute the balance in the region Γε = {(t, x) : t1 ≤ t ≤ t2, γ(t) ≤ x ≤ γ(t) + ε} When g continuous in x one can take the limit as ε ↓ 0: u(t1, γ(t2)) − u(t1, γ(t1)) = t2

t1

g(t, γ(t))dt If g only bounded |u(t1, γ(t2)) − u(t1, γ(t1))| ≤ g∞(t2 − t1) Therefore u Lipschitz along characteristics This can be done under the hypothesis on the inflection points of f Otherwise there is a counterexample

Continuous solutions to a balance law 14 / 26

Introduction Sketch of the proof of the first statement

Some hints of the proof

• 2. Monotone change of variable

Focus on R+ × R, all is local by the finite speed of propagation Define a change of variables (t, y) → (t, x) = (t, χ(t, y)) where

◮ χ is surjective ◮ for all t, y → χ(t, y) is nondecreasing ◮ for all t, d dt χ(t, χ(t, y)) = u(t, χ(t, y))

Continuous solutions to a balance law 15 / 26

Introduction Sketch of the proof of the first statement

Some hints of the proof

This change of variable is BV not Lipschitz in general

Continuous solutions to a balance law 15 / 26

Introduction Sketch of the proof of the first statement

Some hints of the proof

• 2. Monotone change of variable

Focus on R+ × R, all is local by the finite speed of propagation Define a change of variables (t, y) → (t, x) = (t, χ(t, y)) where

◮ χ is surjective ◮ for all t, y → χ(t, y) is nondecreasing ◮ for all t, d dt χ(t, χ(t, y)) = u(t, χ(t, y))

This is possible by the continuity of u, applying Peano’s theorem Let ˜ g : d dt u(t, γ(t)) = d2 dt2 χ(t, y) = ˜ g(t, χ(t, y)) in D′ Still, it is not fine.

Continuous solutions to a balance law 15 / 26

Introduction Sketch of the proof of the first statement

Some hints of the proof

• 2. Monotone change of variable

Focus on R+ × R, all is local by the finite speed of propagation Define a change of variables (t, y) → (t, x) = (t, χ(t, y)) where

◮ χ is surjective ◮ for all t, y → χ(t, y) is nondecreasing ◮ for all t, d dt χ(t, χ(t, y)) = u(t, χ(t, y))

This is possible by the continuity of u, applying Peano’s theorem Let ˜ g : d dt u(t, γ(t)) = d2 dt2 χ(t, y) = ˜ g(t, χ(t, y)) in D′ Still, it is not fine.

Continuous solutions to a balance law 15 / 26

Introduction Sketch of the proof of the first statement

Some hints of the proof

• 2. Monotone change of variable

Focus on R+ × R, all is local by the finite speed of propagation Define a change of variables (t, y) → (t, x) = (t, χ(t, y)) where

◮ χ is surjective ◮ for all t, y → χ(t, y) is nondecreasing ◮ for all t, d dt χ(t, χ(t, y)) = u(t, χ(t, y))

This is possible by the continuity of u, applying Peano’s theorem Let ˜ g : d dt u(t, γ(t)) = d2 dt2 χ(t, y) = ˜ g(t, χ(t, y)) in D′ Still, it is not fine.

Continuous solutions to a balance law 15 / 26

Introduction Sketch of the proof of the first statement

Some hints of the proof

• 3. Selection

Consider the subset of {t > δ} × R × C((−δ, δ)) × [−g∞, g∞] {(t, x, γ, ξ) : γ(0) = x, ˙ γ(0) = u(t, x), ∃¨ γ(0) = ξ} One can define a Borel function ˆ g(t, x) by a selection theorem

◮ given a characteristic, it touches characteristics with different

second derivative only at most at countably many times

◮ at (t, x)-a.e. point, every characteristic through that point is

differentiable with derivative g(t, x) ˆ g provides a universal representative for the source term

Continuous solutions to a balance law 16 / 26

Introduction Sketch of the proof of the first statement

Some hints of the proof

• 3. Selection

Consider the subset of {t > δ} × R × C((−δ, δ)) × [−g∞, g∞] {(t, x, γ, ξ) : γ(0) = x, ˙ γ(0) = u(t, x), ∃¨ γ(0) = ξ} One can define a Borel function ˆ g(t, x) by a selection theorem

◮ given a characteristic, it touches characteristics with different

second derivative only at most at countably many times

◮ at (t, x)-a.e. point, every characteristic through that point is

differentiable with derivative g(t, x) ˆ g provides a universal representative for the source term

Continuous solutions to a balance law 16 / 26

Introduction Sketch of the proof of the first statement

Some hints of the proof

• 4. Approximation

Let u ∈ C(R+ × R) Let (t, χ(t, y)) be a a monotone change of variables as above: for ˜ g ∈ L∞(R+ × R) ∂tχ(t, y) = u(t, χ(t, y)), ∂2

t χ(t, y) = ˜

g(t, χ(t, y)) Then choosing a convolution kernel ρε(t, y) define χε = χ ∗ ρε smooth change of variables. Then at x = χε(t, y) define uε(t, x) = ∂tχε(t, y) gε(t, x) = ∂2

t χε(t, y)

They are smooth approximations satisfying uε

t (t, x) + [(uε(t, x))2]x

2 = gε(t, x)

Continuous solutions to a balance law 17 / 26

Introduction Sketch of the proof of the first statement

Some hints of the proof

• 5. A counterexample

Let f ∈ C 2(R) positive, strictly increasing such that N = {v : f ′(v) = f ′′(v) = 0} satisfies L1(N) > 0 Then ˜ f (v + L1(N ∩ [0, v])) = f (v) ˜ f ′(v) = f ′(f −1(˜ f (v)) The function u(t, x) = f −1(x) satisfies ut + f (u)x = 1, but on the characteristic γ(t) = ˜ f (t) d dt f −1(˜ f (t)) = L1 + f♯L1 ↾N

Continuous solutions to a balance law 18 / 26

Introduction Sketch of the proof of the first statement

Thanks for your attention!

Continuous solutions to a balance law 19 / 26

Introduction to intrinsic Lipschitz graphs in Hn

Outline

• 3. Introduction to intrinsic Lipschitz graphs in Hn

Carnot-Carath´ eodory spaces Graphs in the Heisenberg groups Main statement

Continuous solutions to a balance law 20 / 26

Introduction to intrinsic Lipschitz graphs in Hn

X1, . . . , Xm smooth (‘orthonormal’) horizontal vector fileds in Rn. A curve γ ∈ AC([0, 1]; RN) is horizontal if ˙ γ ∈ X1, . . . , Xm a.e. The Carnot-Carath´ eodory distance dC between P, Q ∈ RN is dC(P, B) = inf {lengthX(γ) : γ horizontal curve joining P to Q}

Continuous solutions to a balance law 21 / 26

Introduction to intrinsic Lipschitz graphs in Hn

X1, . . . , Xm smooth (‘orthonormal’) horizontal vector fileds in Rn. A curve γ ∈ AC([0, 1]; RN) is horizontal if ˙ γ ∈ X1, . . . , Xm a.e. The Carnot-Carath´ eodory distance dC between P, Q ∈ RN is dC(P, B) = inf {lengthX(γ) : γ horizontal curve joining P to Q} Sub-Riemannian Heisenberg group Hn ≡ R2n+1 = {(x, y, z)} Noncommutative Lie group: P · Q = (x, y, z) · (x′, y′, z′) is

• x + x′, y + y′, z + z′ − 1

2ℑ(x + iy, y′ − iy′)

• Horizontal vector fileds: Xj =

∂ ∂xj − 1 2yj ∂ ∂z , Yj = ∂ ∂yj + 1 2xj ∂ ∂z

Vertical vector field: Z =

∂ ∂z ≡ [Xj, Yj],

j = 1, . . . , n

Continuous solutions to a balance law 21 / 26

Introduction to intrinsic Lipschitz graphs in Hn

X1, . . . , Xm smooth (‘orthonormal’) horizontal vector fileds in Rn. A curve γ ∈ AC([0, 1]; RN) is horizontal if ˙ γ ∈ X1, . . . , Xm a.e. The Carnot-Carath´ eodory distance dC between P, Q ∈ RN is dC(P, B) = inf {lengthX(γ) : γ horizontal curve joining P to Q} Sub-Riemannian Heisenberg group Hn ≡ R2n+1 = {(x, y, z)} Noncommutative Lie group: P · Q = (x, y, z) · (x′, y′, z′) is

• x + x′, y + y′, z + z′ − 1

2ℑ(x + iy, y′ − iy′)

• Horizontal vector fileds: Xj =

∂ ∂xj − 1 2yj ∂ ∂z , Yj = ∂ ∂yj + 1 2xj ∂ ∂z

Vertical vector field: Z =

∂ ∂z ≡ [Xj, Yj],

j = 1, . . . , n dC ∼ d∞(P, Q) := max

• |x′′ + iy′′|, |z′′|1/2

, (x′′, y′′, z′′) = P−1·Q

Continuous solutions to a balance law 21 / 26

Introduction to intrinsic Lipschitz graphs in Hn

X1-graphs in Hn

Let φ : W = {x1 = 0} ⊂ Hn → R continuous n = 1 for shortness GraphHφ := {A · φ(A)e1 : A ∈ W} =

• φ(A), yA, zA − yA

2 φ(A)

• : A ∈ W
• It induces the graph quasidistance dφ on W

dφ(A, B) =

• yB − yA
• +
• zB − zA − 1

2(φ(A) + φ(B))(yB − yA)

• 1/2

[This formula for dφ is from Ambrosio-Serra Cassano-Vittone]

Continuous solutions to a balance law 22 / 26

Introduction to intrinsic Lipschitz graphs in Hn

X1-graphs in Hn

Let φ : W = {x1 = 0} ⊂ Hn → R continuous n = 1 for shortness GraphHφ := {A · φ(A)e1 : A ∈ W} =

• φ(A), yA, zA − yA

2 φ(A)

• : A ∈ W
• It induces the graph quasidistance dφ on W

dφ(A, B) =

• yB − yA
• +
• zB − zA − 1

2(φ(A) + φ(B))(yB − yA)

• 1/2

[This formula for dφ is from Ambrosio-Serra Cassano-Vittone] It induces an intrinsic differentiable structure. One has notions of intrinsic cones, intrinsic differentiability, intrinsic gradient, . . . .

Continuous solutions to a balance law 22 / 26

Introduction to intrinsic Lipschitz graphs in Hn

Intrinsic Regular X1-graphs in Hn

Let φ : ω ⊂ W = {x1 = 0} ⊂ Hn → R continuous GraphHφ := {A · φ(A)e1 : A ∈ ω} A W-linear functional L : W → R is a group homeomorphism s.t. L(rx2, . . . , ryn, r2z) = rL(x2, . . . , yn, z) for r > 0 φ is ∇φ-differentiable at A0 ∈ W if ∃ a W-linear L : W → R s.t. lim

A→A0

φ(A) − φ(A0) − L(A−1 · A) dφ(A0, A) = 0. (1) Roughly, L represented as scalar product with horizontal variables, the intrinsic gradient ∇φφ given by this representation

Continuous solutions to a balance law 23 / 26

Introduction to intrinsic Lipschitz graphs in Hn

Intrinsic Regular X1-graphs in Hn

Let φ : ω ⊂ W = {x1 = 0} ⊂ Hn → R continuous GraphHφ := {A · φ(A)e1 : A ∈ ω} The graph is intrinsic regular if

◮ φ is uniformly intrinsic differentiable ◮ it is the level set of Φ : H → R which is C 1 along all Xi, Yi

and s.t. X1Φ = 0 (Intrinsic Implicit Function Theorem)

◮ (n = 1) φ satisfies φy +

• φ2

2

• z = g with g : ω → R continuous

In this case, φ is uniformly intrinsic differentiable and ∇φφ = g

Ambrosio-Serra Cassano-Vittone Franchi-Serapioni-Serra Cassano Bigolin-Serra Cassano

Continuous solutions to a balance law 23 / 26

Introduction to intrinsic Lipschitz graphs in Hn

Intrinsic Regular X1-graphs in Hn

Let φ : ω ⊂ W = {x1 = 0} ⊂ Hn → R continuous GraphHφ := {A · φ(A)e1 : A ∈ ω} The graph is intrinsic regular if

◮ φ is uniformly intrinsic differentiable ◮ it is the level set of Φ : H → R which is C 1 along all Xi, Yi

and s.t. X1Φ = 0 (Intrinsic Implicit Function Theorem)

◮ (n = 1) φ satisfies φy +

• φ2

2

• z = g with g : ω → R continuous

In this case, φ is uniformly intrinsic differentiable and ∇φφ = g

Ambrosio-Serra Cassano-Vittone Franchi-Serapioni-Serra Cassano Bigolin-Serra Cassano

Continuous solutions to a balance law 23 / 26

Introduction to intrinsic Lipschitz graphs in Hn

Intrinsic Regular X1-graphs in Hn

Let φ : ω ⊂ W = {x1 = 0} ⊂ Hn → R continuous GraphHφ := {A · φ(A)e1 : A ∈ ω} The graph is intrinsic regular if

◮ φ is uniformly intrinsic differentiable ◮ it is the level set of Φ : H → R which is C 1 along all Xi, Yi

and s.t. X1Φ = 0 (Intrinsic Implicit Function Theorem)

◮ (n = 1) φ satisfies φy +

• φ2

2

• z = g with g : ω → R continuous

In this case, φ is uniformly intrinsic differentiable and ∇φφ = g

Ambrosio-Serra Cassano-Vittone Franchi-Serapioni-Serra Cassano Bigolin-Serra Cassano

They can be very irregular from the Euclidean point of view

Kirchheim-Serra Cassano

Continuous solutions to a balance law 23 / 26

Introduction to intrinsic Lipschitz graphs in Hn

Intrinsic Regular X1-graphs in Hn

Let φ : ω ⊂ W = {x1 = 0} ⊂ Hn → R continuous GraphHφ := {A · φ(A)e1 : A ∈ ω} The graph is intrinsic regular if

◮ φ is uniformly intrinsic differentiable ◮ it is the level set of Φ : H → R which is C 1 along all Xi, Yi

and s.t. X1Φ = 0 (Intrinsic Implicit Function Theorem)

◮ (n = 1) φ satisfies φy +

• φ2

2

• z = g with g : ω → R continuous

In this case, φ is uniformly intrinsic differentiable and ∇φφ = g

Ambrosio-Serra Cassano-Vittone Franchi-Serapioni-Serra Cassano Bigolin-Serra Cassano

They can be very irregular from the Euclidean point of view

Kirchheim-Serra Cassano

Continuous solutions to a balance law 23 / 26

Introduction to intrinsic Lipschitz graphs in Hn

Intrinsic Regular X1-graphs in Hn

Let φ : ω ⊂ W = {x1 = 0} ⊂ Hn → R continuous GraphHφ := {A · φ(A)e1 : A ∈ ω} The graph is intrinsic regular if

◮ φ is uniformly intrinsic differentiable ◮ it is the level set of Φ : H → R which is C 1 along all Xi, Yi

and s.t. X1Φ = 0 (Intrinsic Implicit Function Theorem)

◮ (n = 1) φ satisfies φy +

• φ2

2

• z = g with g : ω → R continuous

In this case, φ is uniformly intrinsic differentiable and ∇φφ = g

Ambrosio-Serra Cassano-Vittone Franchi-Serapioni-Serra Cassano Bigolin-Serra Cassano

They can be very irregular from the Euclidean point of view

Kirchheim-Serra Cassano

Continuous solutions to a balance law 23 / 26

Introduction to intrinsic Lipschitz graphs in Hn

Intrinsic Lipschitz X1-graphs in Hn

Let φ : ω ⊂ W = {x1 = 0} ⊂ Hn → R continuous GraphHφ := {A · φ(A)e1 : A ∈ ω} C(0, α) := {||(0, x2, . . . , yn, z)||∞ ≤ α ||(x1, 0, 0)||∞}(x,y,z)∈Hn The graph is intrinsic Lipschitz if

◮ ‘translating’ an intrinsic cone C(0, α) at any point of the

graph, it intersects the graph only in the vertex

◮ ∃L > 0 tale che |φ(A) − φ(B)| ≤ Ldφ(A, B)

∀A, B ∈ ω

◮ ‘by approximation’ with intrinsic regular graphs

In this case, φ is almost everywhere intrinsically differentiable

Franchi-Serapioni-Serra Cassano

Continuous solutions to a balance law 24 / 26

Introduction to intrinsic Lipschitz graphs in Hn

Intrinsic Lipschitz X1-graphs in Hn

Let φ : ω ⊂ W = {x1 = 0} ⊂ Hn → R continuous GraphHφ := {A · φ(A)e1 : A ∈ ω} C(0, α) := {||(0, x2, . . . , yn, z)||∞ ≤ α ||(x1, 0, 0)||∞}(x,y,z)∈Hn The graph is intrinsic Lipschitz if

◮ ‘translating’ an intrinsic cone C(0, α) at any point of the

graph, it intersects the graph only in the vertex

◮ ∃L > 0 tale che |φ(A) − φ(B)| ≤ Ldφ(A, B)

∀A, B ∈ ω

◮ ‘by approximation’ with intrinsic regular graphs

In this case, φ is almost everywhere intrinsically differentiable

Franchi-Serapioni-Serra Cassano

Local concepts

Continuous solutions to a balance law 24 / 26

Introduction to intrinsic Lipschitz graphs in Hn

Intrinsic Lipschitz X1-graphs in Hn

Let φ : ω ⊂ W = {x1 = 0} ⊂ Hn → R continuous GraphHφ := {A · φ(A)e1 : A ∈ ω}

Theorem (for shortness written here for n = 1)

The graph is intrinsic Lipschitz if and only if φ satisfies φy + φ2 2

• z

= g with g : ω → R bounded. g = ∇φφ, the intrinsic gradient of φ.

Citti-Manfredini-Pinamonti-Serra Cassano Bigolin-C.-Serra Cassano

We also have the analogous statement for n > 1

Continuous solutions to a balance law 25 / 26