Regolarità SBV delle soluzioni di viscosità di equazioni ...university-logo BV functions are,...

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Regolarità SBV delle soluzioni di viscositàdi equazioni di Hamilton-Jacobi in Rn.

Roger Robyr

Università di Zurigo - Istituto di Matematica.

Università di Roma "La Sapienza" – Gennaio 2011

Roger Robyr (UZH) Regolarità SBV per HJ in Rn 17 Gennaio 2011 1 / 32

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BV functions of several variables

DefinitionLet Ω ⊂ Rn be an open set. Let u ∈ L1(Ω); we say that u ∈ BV (Ω) ifthe distributional derivative of u, denoted by Du, is representable by afinite Radon measure in Ω, i.e. if∫

Ωu∂φ

∂xidx = −

∫ΩφdDiu ∀φ ∈ C∞c (Ω), i = 1, . . . ,n

for some Rn-valued measure Du = (D1u, . . . ,Dnu) in Ω. A functionu ∈ L1

loc(Ω) belongs to BVloc(Ω) if for each open set V ⊂⊂ Ω,u ∈ BV (V ).

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BV functions are, roughly speaking, functions with a bound on the totalamount of oscillations.The total amount of the oscillations can be measured by |Du|, which isthe total variation of the Radon measure Du. Then,

We can characterize the BV space as the class of u ∈ L1(Ω) functionswith finite (total) variation, |Du|(Ω) <∞.

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BV functions are, roughly speaking, functions with a bound on the totalamount of oscillations.The total amount of the oscillations can be measured by |Du|, which isthe total variation of the Radon measure Du. Then,

We can characterize the BV space as the class of u ∈ L1(Ω) functionswith finite (total) variation, |Du|(Ω) <∞.

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Functions of bounded variation in 1− d

To give an intuitive idea of BV functions let us consider theone-dimensional case.Remarks:

I If u is differentiable and u′ ∈ L1([a,b]), its total variation is

|Du|(I) =

a|u′(t)|d t . <∞

I If u(x) =

0, on ]− a,0];1, on [0,a[.

Then, |Du|(I) = 1 <∞.

I L∞(R) is not a subspace of BV (R): for instance consideru(x) = sin(1/x) on R+.

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Functions of bounded variation in 1− d

To give an intuitive idea of BV functions let us consider theone-dimensional case.Remarks:

I If u is differentiable and u′ ∈ L1([a,b]), its total variation is

|Du|(I) =

a|u′(t)|d t . <∞

I If u(x) =

0, on ]− a,0];1, on [0,a[.

Then, |Du|(I) = 1 <∞.

I L∞(R) is not a subspace of BV (R): for instance consideru(x) = sin(1/x) on R+.

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Cantor function

The Cantor function c on [0,1] is a monotone increasing continuousfunction such that

2c(x/3) = c(x) and c(x) + c(1− x) = 1.

I c ∈ BV ([0,1]), i.e. c is a function of bounded variation!I It is continuous but not absolutely continuous.I Its derivative Dc is a singular nonatomic measure, which contains

a fractal structure (the construction of c is indeed based on theclassical Cantor ternary set).

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Decomposition of the derivative

Let u ∈ BV (R).By the Radon-Nikodym Theorem we split the measure Du:

Du = Dau + Dsu = Du (Ω\S) + Du S

whereS :=

x ∈ Ω : lim

ρ↓0

|Du|(Bρ(x))

ρ=∞

I absolute continuous part Dau (with respect to the Lebesguemeasure L1)

I singular part DsuRemark: for the singular part the slope of u is equal to∞: jumps,Cantor.

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Decomposition of the singular part Dsu = Dju + Dcu

Moreover, if A denotes the set of atoms of Du, i.e. x ∈ A if and only ifDu(x) 6= 0, we have

Du = Dau + Dju + Dcu︸︷︷︸Dsu

= Du (Ω\S) + Du A + Du (S\A).

I if u ∈ BV (R) the jump set A = Ju consists of countably manypoints.

I Dju = Du A is purely atomic.I Dcu = Du (S\A) is the diffusive part (i.e. without atoms)I The above decomposition is unique and the three measures

Dau,Dju,Dcu are mutually singular.We have

|Du|(Ω) = |Dau|(Ω) + |Dju|(Ω) + |Dcu|(Ω)

∫Ω\S|u ′|dt +

∑t∈A

|u(t+)− u(t−)|+ |Dcu|(Ω).

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The space SBV

DefinitionLet u ∈ BV (Ω), then u is a special function of bounded variation, andwe write u ∈ SBV (Ω), if

Dcu = 0,

i.e. if the measure Du has no Cantor part.

RemarkThe classical example of a 1-dimensional continuous function whichbelongs to BV but not to SBV is the Cantor staircase.

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The space SBV in n-dimensions

Rough intuitive picture: Lipschitz functions whose gradient ispiecewise smooth, undergoing jump discontinuities along a rectifiableset of codimension 1. We exclude that the derivative can have acomplicated fractal behaviour.

Let Ω ⊂ Rn and u ∈ BV (Ω). Then, u ∈ SBV (Ω) if Dcu = 0:

Du = Dau + Dju = ∇uLn + (u+ − u−)νuHn−1 Ju.

where (u+,u−) are the one sided limits and νu the normal to the jumpset Ju.

I 1− d : SBV (Ω) = W 1,1(Ω)⊕ u ∈ SBV : ∇u = 0, L1 a.e..I n − d : W 1,1(Ω)⊕ u ∈ SBV : ∇u = 0, Ln a.e.(SBV (Ω).

Example: u : R2\S → R, u =√ρ sin(θ/2) ∈ SBVloc(R2) where S is

the negative x-axis.

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Hamilton-Jacobi equations

Evolutionary case:

∂tu + H(Dxu) = 0, on R+ × Rn.

Stationary case:H(Dxu) = 0, x ∈ Rn.

PrincipleLooking at the classical mechanics or at optimal control problemsanother powerful principle comes into the play: the least action (orHamiltonian) principle.

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An example of stationary HJ-equation: Eikonalequation

I 1-dimensional case:|u′(x)| = 1, x ∈ (a,b) ⊂ R;u(a) = u(b) = 0, boundary data.

I n-dimensional case|∇u(x)| = 1, x ∈ Ω ⊂ Rn;u|∂Ω = 0, boundary data.

A non-smooth solution is given by the distance function:

u(x) = dist(x , ∂Ω).

The weak non-smooth solutions are not unique.

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If H ∈ C2(Rn) uniformly convex (i.e. ∃c > 0, c−1Id ≤ D2H ≤ cId) forthe Hamilton-Jacobi equation:

∂tu + H(Dxu) = 0 or H(Dxu) = 0, x ∈ Rn

we have:I Singularities appear for the gradient ∇u(x) of the solution.I Weak solutions are not unique.

But as for Conservation laws we can:I use the space BV as working space, i.e. the derivative Dxu is a

function of bounded variation.I introduce the concept of viscosity solution which single out unique

solution!

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solution!

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solution!

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solution!

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Definition (Viscosity solution)A bounded, uniformly continuous function u is called a viscositysolution of

∂tu + H(Dxu) = 0, in Ω ⊂ [0,T ]× Rn ,

provided that1. u is a viscosity subsolution: for each v ∈ C∞(Ω) such that u − v

has a maximum at (t0, x0)

vt (t0, x0) + H(Dxv(t0, x0)) ≤ 0

2. u is a viscosity supersolution: for each v ∈ C∞(Ω) such that u − vhas a minimum at (t0, x0)

vt (t0, x0) + H(Dxv(t0, x0)) ≥ 0

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Viscosity solution as vanishing viscosity limit

A viscosity solution is the uniform limit, as ε→ 0, of solutions to theparabolic problem:

uεt + H(Dxuε)− ε4uε = 0 on R+ × Rn.

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Hopf-Lax formula

The unique viscosity solution is given by:

u(x , t) := miny∈R

t · L

(x − yt

)+ u0(y)

whereL = H∗ := sup

q∈Rp · q − H(q)

is the Legendre transformation.

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SBV Regularity for scalar conservation laws.

L.AMBROSIO & C.DE LELLIS, 2004.

TheoremLet u(x) ∈ L∞(Ω) be an entropy solution of the following scalarconservation law:

Dtu(x , t) + Dx [f (u(x , t))] = 0 in an open set Ω ⊂ R2,

where f ∈ C2(R) and f ′′ > 0 (locally uniformly convex).

⇒ the entropy solutions u(x , t) ∈ SBVloc(Ω).

Moreover: there exists a set S ⊂ R at most countable such that∀τ ∈ R\S the following holds:

u(., τ) ∈ SBVloc(Ωτ ) with Ωτ := x ∈ R : (x , τ) ∈ Ω.

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SBV Regularity for scalar conservation laws.

L.AMBROSIO & C.DE LELLIS, 2004.

TheoremLet u(x) ∈ L∞(Ω) be an entropy solution of the following scalarconservation law:

Dtu(x , t) + Dx [f (u(x , t))] = 0 in an open set Ω ⊂ R2,

where f ∈ C2(R) and f ′′ > 0 (locally uniformly convex).

⇒ the entropy solutions u(x , t) ∈ SBVloc(Ω).

Moreover: there exists a set S ⊂ R at most countable such that∀τ ∈ R\S the following holds:

u(., τ) ∈ SBVloc(Ωτ ) with Ωτ := x ∈ R : (x , τ) ∈ Ω.

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Hamilton-Jacobi equations and scalar conservationlaws

If u : R+ × R→ R is a viscosity solution of a Hamilton-Jacobi equation,then ux is an entropy solution of a scalar conservation law.

In particular for these Hamilton-Jacobi equations by L.AMBROSIO &C.DE LELLIS, 2004, for viscosity solutions u ∈W 1,∞ we have that:

if H ∈ C2(R) is a uniformly convex Hamiltonian, then Dtu,Dxu ∈ SBVloc .

In particular, the "middle state" of the gradient given by the Cantorfunction will be transformed in jump.

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Hamilton-Jacobi equations in Rn

Theorem (S.Bianchini, C.De Lellis & R.Robyr)

Let u be a viscosity solution of

∂tu + H(Dxu) = 0 in Ω ⊂ [0,T ]× Rn.

coupled with a Lipschitz continuous and bounded initial datau0 : Rn → R. Moreover, assume

H ∈ C2(Rn) and c−1H Idn ≤ D2H ≤ cH Idn for some cH > 0.

and set Ωt := x ∈ Rn : (t , x) ∈ Ω. Then, the set of times

S := t : Dxu(t , .) /∈ SBVloc(Ωt )

is at most countable. In particular Dxu, ∂tu ∈ SBVloc(Ω).

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What is new?: SBV-Regularity for any dimension n ∈ N:Hamilton-Jacobi equations in Rn.

Key idea: Find a monotone geometric functional which jumps everytime that a Cantor part appears in the derivative of the solution.

Difficulty: In general the derivative Dxu(t , x) is not continuous and theprojection of the jump set (of the derivative Dxu(t , .)) on Rn × 0 hasnow a complex geometric structure.

Solution: We use the Hopf-Lax formula and we work with the regularpoints, i.e. the points for which the derivative Dxu(t , x) is continuous.

Starting Remark: Viscosity solutions are semiconcave.

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Let Ω ⊂ Rn be an open set.

DefinitionWe say that a continuous function u : Ω→ R is semiconcave if, for anyconvex K ⊂⊂ Ω, there exists CK > 0 such that

u(x + h) + u(x − h)− 2u(x) ≤ CK |h|2,

for all x ,h ∈ Rn with x , x − h, x + h ∈ K .

ExampleLet Ω = (−1,1). The function

u(x) := dist(x , ∂Ω),

is semiconcave.

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Let Ω ⊂ Rn be an open set.

DefinitionWe say that a continuous function u : Ω→ R is semiconcave if, for anyconvex K ⊂⊂ Ω, there exists CK > 0 such that

u(x + h) + u(x − h)− 2u(x) ≤ CK |h|2,

for all x ,h ∈ Rn with x , x − h, x + h ∈ K .

ExampleLet Ω = (−1,1). The function

u(x) := dist(x , ∂Ω),

is semiconcave.

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DefinitionLet u : Ω→ R be a measurable function. The set ∂u(x), called thesuperdifferential of u at point x ∈ Ω, is defined as

∂u(x) :=

p ∈ Rn : lim supy→x

u(y)− u(x)− p · (y − x)

|y − x |≤ 0

ExampleLet Ω = (−1,1) and consider the semiconcave functionu(x) := dist(x , ∂Ω), thus

I ∂u(x) = 1 for x ∈ (−1,0) and ∂u(x) = −1 for x ∈ (0,1).I ∂u(x) = [−1,1] for x = 0.

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DefinitionLet u : Ω→ R be a measurable function. The set ∂u(x), called thesuperdifferential of u at point x ∈ Ω, is defined as

∂u(x) :=

p ∈ Rn : lim supy→x

u(y)− u(x)− p · (y − x)

|y − x |≤ 0

ExampleLet Ω = (−1,1) and consider the semiconcave functionu(x) := dist(x , ∂Ω), thus

I ∂u(x) = 1 for x ∈ (−1,0) and ∂u(x) = −1 for x ∈ (0,1).I ∂u(x) = [−1,1] for x = 0.

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By this Proposition:I To every semiconcave function we can associate a concave

function!I Concave functions have monotonicity properties.

Proposition

Let Ω ⊂ Rn be open and K ⊂ Ω a compact convex set. Let u : Ω→ Rbe a semiconcave function with semiconcavity constant CK ≥ 0. Then,the function

u : x 7→ u(x)− CK

2|x |2 is concave in K .

In particular, for any given x , y ∈ K , p ∈ ∂u(x) and q ∈ ∂u(y) we havethat

〈q − p, y − x〉 ≤ 0.

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Following the work of G.ALBERTI & L.AMBROSIO, 1999

Definition

Let B : Rn → Rn be a multifunction, then1. B is a monotone function if

〈y1 − y2, x1 − x2〉 ≤ 0 ∀xi ∈ Rn, yi ∈ B(xi), i = 1,2.

2. A monotone function B is called maximal when it is maximal withrespect to the inclusion in the class of monotone functions, i.e. ifthe following implication holds:

A(x) ⊃ B(x) for all x ,A monotone ⇒ A = B.

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Helpful remarks:

Let u be the unique viscosity solution of the Hamilton-Jacobi equation(with H ∈ C2(Rn) uniformly convex).

I u is semiconcave and ∂u(x) = Du(x) for a.e. x , and at anypoint where ∂u is single-valued, Du is continuous.

I To every semiconcave function we can associate a concavefunction!

I The supergradient ∂u of a concave function is a maximalmonotone function.

I Hille-Yosida Approximation: for every ε and for every maximalmonotone function B on Rn there exists a 1/ε-Lipschitz maximalmonotone function Bε := (εId − B−1)−1.

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Helpful remarks:

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Helpful remarks:

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Helpful remarks:

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We recall the Hopf-Lax formula:

u(x , t) := miny∈R

t · L

(x − yt

)+ u0(y)

which is the unique (semiconcave) viscosity solution.

We have:I ∂u = Du(x) is single-valued, if x is a continuous point for the

derivative: then we have a unique backward characteristicI ∂u is not single-valued, if x is a discontinuous point for the

derivative: several backward characteristic with bounded speed.

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Definition of the geometric functional

We consider the projection of the points for which the derivativeDxu(t , x) is single valued (or continuous):

for such a regular point (t , x) ∈ t × Rn the minimum in the Hopf–Laxformula is obtained for a unique y ∈ 0 × Rn and

χt ,0(x) := x − t DH(Dxu(t , x)) = y .

The function χt ,0 is defined a.e. on Ωt := x ∈ Rn : (t , x) ∈ Ω and wedefine the nonincreasing functional:

F (t) := |χt ,0(Ωt )|.

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χt ,0(x) := x − t DH(Dxu(t , x)) = y .

F (t) := |χt ,0(Ωt )|.

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χt ,0(x) := x − t DH(Dxu(t , x)) = y .

F (t) := |χt ,0(Ωt )|.

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3 steps:

I Step 1: By the dynamic programming principle we prove that thefunctional is nonincreasing.

I Step 2a: For every t ∈ R+, there exists Borel set E ⊂ Ωt with:

|E | = 0, and |D2c ut |(Ωt \ E) = 0.

I Step 2b: For every (t , x) ∈ E the function χt ,0(x) is single valued(since Dxu continuous on E).

I Step 2c: χt ,0(E) ∩ χt+δ,0(Ωt+δ) = ∅.I Step 3: When |D2

c ut | 6= 0, the functional jumps

F (t) > F (t + δ).

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3 steps:

|E | = 0, and |D2c ut |(Ωt \ E) = 0.

F (t) > F (t + δ).

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3 steps:

|E | = 0, and |D2c ut |(Ωt \ E) = 0.

F (t) > F (t + δ).

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Remarks:I the function χt ,0 is injective, i.e. for x 6= y ⇒ χt ,0(x) ∩ χt ,0(y) = ∅.I we approximate the maximal monotone functions ∂u = ∂u − CK x

with Lipschitz continuous function to prove some estimates. TheLipschitz regularity is important to use the area formula.

I Key estimate: For t ∈]0,T ], any δ ∈ [0, t ] and any Borel setE ⊂ Ωt := (s, x) ∈ Ω : s = t we have:

|Xt ,δ(E)| ≥ (t − δ)n

tn |Xt ,0(E)|;

where for t < s we define Xt ,s(x) := x − (t − s)DH(∂ut (x)) andut : Rn → Rn with ut (x) := u(t , x).

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|Xt ,δ(E)| ≥ (t − δ)n

tn |Xt ,0(E)|;

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|Xt ,δ(E)| ≥ (t − δ)n

tn |Xt ,0(E)|;

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On the functionals:

We observe that in the one-dimensional case, the functional defined inthe proof of Ambrosio & De Lellis for the scalar conservation laws andfor the Hamilton-Jacobi equations are similar:

F HJ(.) is exactly the "complementary" functional of F CL(.).

I F HJ(.) measures the projection of regular points.I F CL(.) measures the projection of singular points.

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Overview: results on the SBV regularity forHamilton-Jacobi equations:

I 1997 - P.CANNARSA, A. MENNUCCI & C.SINESTRARI.Under strong regularity assumption: for a fixed integer R letu0 ∈W 1,∞(Rn) ∩ CR+1(Rn) and consider:

Dtu(x , t) + H(Dxu(x , t), x , t) = 0, t ∈ R+, x ∈ Rn;u(x ,0) = u0(x), x ∈ Rn.

where H(., x , t) is strictly convex. Then, Du belongs to the classSBVloc , i.e D2u is a measure with no Cantor part.

I 2004 - L.AMBROSIO & C.DE LELLIS.2-dimensional case: Let H ∈ C2(R2) be locally uniformly convexand let u ∈W 1,∞(Ω) be a viscosity solution of H(Du) = 0. ThenDu ∈ SBVloc(Ω).

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I 2007 - R.ROBYR.Let H(p, x , t) ∈ C2(R× R× R+) be locally uniformly convex in p,i.e. DppH > 0. If u ∈W 1,∞(Ω) is a viscosity solution of

Dtu(x , t) + H(Dxu(x , t), x , t) = 0, (1)

then Du ∈ SBVloc(Ω).I 2010 - S.BIANCHINI, C.DE LELLIS & R.ROBYR

n-dimensional case: Under the convexity assumption that

H ∈ C2(Rn) and c−1H Idn ≤ D2H ≤ cH Idn for some cH > 0,

the gradient of any viscosity solution u of

H(Dxu) = 0 in Ω ⊂ Rn,

belongs to SBVloc(Ω).

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Thank you!!!!!

Regolarità SBV delle soluzioni di viscosità di equazioni ...university-logo BV functions are,...

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