Regularity of optimal control problems with super linear growth P - - PowerPoint PPT Presentation

Regularity of optimal control problems with super linear growth

P . Cardaliaguet

• Univ. Brest

Toulon, May 18-21, 2008 Sixièmes Journées Franco-Chiliennes d’Optimisation

P . Cardaliaguet (Univ. Brest) Regularity of optimal control problems 1 / 25

Statement of the main result

Outline

1

Statement of the main result

2

Sketch of the proof of the main result

3

Sketch of proof for the reverse Hölder inequality Lemma

P . Cardaliaguet (Univ. Brest) Regularity of optimal control problems 2 / 25

Statement of the main result

Aim

We investigate the regularity of the value function u(x, t) = inf T

t

L(x(s), s, x′(s))ds + g(x(T))

• where the infimum is taken over the x(·) ∈ W 1,1([t, T], RN) such that

x(t) = x under the key assumption that L has a superlinear growth : ∃p > 1, δ ∈ (0, 1), M > 0 with δ|ξ|p − M ≤ L(x, s, ξ) ≤ 1 δ |ξ|p + M

P . Cardaliaguet (Univ. Brest) Regularity of optimal control problems 3 / 25

Statement of the main result

Aim

We investigate the regularity of the value function u(x, t) = inf T

t

L(x(s), s, x′(s))ds + g(x(T))

• where the infimum is taken over the x(·) ∈ W 1,1([t, T], RN) such that

x(t) = x under the key assumption that L has a superlinear growth : ∃p > 1, δ ∈ (0, 1), M > 0 with δ|ξ|p − M ≤ L(x, s, ξ) ≤ 1 δ |ξ|p + M

P . Cardaliaguet (Univ. Brest) Regularity of optimal control problems 3 / 25

Statement of the main result

Motivation

Stochastic homogenization of Hamilton-Jacobi equations : (HJ) uǫ

t (x, t) + H(x

ǫ , t ǫ, Duǫ(x, t)) = 0 in RN × (0, T) (Souganidis (1999), Lions-Souganidis (2005), Schwab (2008)). Recall that our value function u is solution of (HJ) − ut(x, t) + H(x, t, Du(x, t)) = 0 in RN × (0, T)

P . Cardaliaguet (Univ. Brest) Regularity of optimal control problems 4 / 25

Statement of the main result

Known results

Two types of results : Propagation of regularity : Lions (1985), Barles (1990), Rampazzo, Sartori (2000) If L and g are “sufficiently smooth", then u is Lipschitz continuous with a Lipschitz constant depending on the regularity of L and g. Weak point : not suitable for homogenization.

P . Cardaliaguet (Univ. Brest) Regularity of optimal control problems 5 / 25

Statement of the main result

Known results

Two types of results : Propagation of regularity : Lions (1985), Barles (1990), Rampazzo, Sartori (2000) If L and g are “sufficiently smooth", then u is Lipschitz continuous with a Lipschitz constant depending on the regularity of L and g. Weak point : not suitable for homogenization.

P . Cardaliaguet (Univ. Brest) Regularity of optimal control problems 5 / 25

Statement of the main result

Interior regularity/Lipschitz regularity of optimal solutions : Clarke, Vinter (1985), Ambrosio, Ascenzi, Buttazzo (1989), Dal Maso, Frankowska (2003), Quincampoix, Zlateva (2006), Frankowska, Marchini (2005), Davini (2007). Key assumption : L = L(x, x′) has a superlinear growth. Consequences :

• No Lavrentiev phenomenon.
• Lipschitz continuity of optimal solutions implies Lipschitz

continuity of the value function. Weak point : Cannot handle the time-dependant case.

P . Cardaliaguet (Univ. Brest) Regularity of optimal control problems 6 / 25

Statement of the main result

Interior regularity/Lipschitz regularity of optimal solutions : Clarke, Vinter (1985), Ambrosio, Ascenzi, Buttazzo (1989), Dal Maso, Frankowska (2003), Quincampoix, Zlateva (2006), Frankowska, Marchini (2005), Davini (2007). Key assumption : L = L(x, x′) has a superlinear growth. Consequences :

• No Lavrentiev phenomenon.
• Lipschitz continuity of optimal solutions implies Lipschitz

continuity of the value function. Weak point : Cannot handle the time-dependant case.

P . Cardaliaguet (Univ. Brest) Regularity of optimal control problems 6 / 25

Statement of the main result

Difficulty for the time-dependent case

If L = L(x, t, ξ), one cannot expect u to be locally Lipschitz continuous because u Lipschitz continuous ⇒ optimal solutions are Lipschitz continuous. However

Proposition (Lavrentiev (1926))

There are L and initial data (x0, t0) for which optimal solutions are not Lipschitz continuous. Manià’s Example (1934) : L(x, t, ξ) = (t3 − x)2|ξ|6.

P . Cardaliaguet (Univ. Brest) Regularity of optimal control problems 7 / 25

Statement of the main result

Assumptions for the main result

We suppose that L = L(x, t, ξ) is continuous, convex w.r.t. ξ, g is bounded by some M > 0, (Superlinear growth) there are p > 1 and δ > 0 such that δ|ξ|p − M ≤ L(x, t, ξ) ≤ 1 δ |ξ|p + M ∀(x, t) ∈ RN × (0, T)

P . Cardaliaguet (Univ. Brest) Regularity of optimal control problems 8 / 25

Statement of the main result

Main result

Theorem

There are θ > p and, for any τ > 0, Kτ > 0 such that |u(x0, t0) − u(x1, t1)| ≤ Kτ

• |x0 − x1|(θ−p)/(θ−1) + |t0 − t1|(θ−p)/θ

for any x0, x1 ∈ RN, for any t0, t1 ∈ [0, T − τ], where θ = θ(M, δ, p, T) and Kτ = K(τ, M, δ, q, T).

P . Cardaliaguet (Univ. Brest) Regularity of optimal control problems 9 / 25

Sketch of the proof of the main result

Outline

1

Statement of the main result

2

Sketch of the proof of the main result

3

Sketch of proof for the reverse Hölder inequality Lemma

P . Cardaliaguet (Univ. Brest) Regularity of optimal control problems 10 / 25

Sketch of the proof of the main result

Main steps

Three main steps : Step 1 : a strange inequality satisfied by optimal solutions, Step 2 : use of the reverse Hölder inequality to get regularity of the optimal solutions, Step 3 : this regularity gives the Hölder continuity of u.

P . Cardaliaguet (Univ. Brest) Regularity of optimal control problems 11 / 25

Sketch of the proof of the main result

Step 1 : a strange inequality

Lemma

There is A = A(M, δ, T) ≥ 1 such that, for any optimal solution ¯ x starting from x0 at time t0, 1 h t0+h

t0

(α(s))pds ≤ A

• 1

h t0+h

t0

α(s)ds p ∀h ∈ [0, T − t0] where α(s) = |¯ x′(s)| + 1 Idea of proof : test the optimality of ¯ x against ˜ x(t) =

• ¯

x(t0+h)−x0 h

(t − t0) + x0 if t ∈ [t0, t0 + h] ¯ x(t)

• therwise

P . Cardaliaguet (Univ. Brest) Regularity of optimal control problems 12 / 25

Sketch of the proof of the main result

Remarks on the inequality

If α ∈ Lp(0, 1) for some p > 1, we have from the Hölder inequality :

• 1

h h |α| p ≤ 1 hp h |α|p

• h(1−1/q)p = 1

h h |α|p ∀h ∈ [0, 1] So the inequality satisfied by α(s) = |¯ x′(s)| + 1 : 1 h t0+h

t0

(α(s))pds ≤ A

• 1

h t0+h

t0

α(s)ds p ∀h ∈ [0, T − t0] is a reverse Hölder inequality.

P . Cardaliaguet (Univ. Brest) Regularity of optimal control problems 13 / 25

Sketch of the proof of the main result

Remarks on the inequality

If α ∈ Lp(0, 1) for some p > 1, we have from the Hölder inequality :

• 1

h h |α| p ≤ 1 hp h |α|p

• h(1−1/q)p = 1

h h |α|p ∀h ∈ [0, 1] So the inequality satisfied by α(s) = |¯ x′(s)| + 1 : 1 h t0+h

t0

(α(s))pds ≤ A

• 1

h t0+h

t0

α(s)ds p ∀h ∈ [0, T − t0] is a reverse Hölder inequality.

P . Cardaliaguet (Univ. Brest) Regularity of optimal control problems 13 / 25

Sketch of the proof of the main result

Step 2 : use of the reverse Hölder inequality

Fix A > 1 and p > 1.

Lemma (reverse Hölder inequality)

There are θ = θ(A, p) > p and C = C(A, p) > 0 such that, for any α ∈ Lp(0, 1) satisfying 1 h h |α(s)|pds ≤ A

• 1

h h |α(s)|ds p ∀h ∈ [0, 1],

• ne has

h |α(s)|ds ≤ CαLp h1−1/θ ∀h ∈ [0, 1] . Remark : This result is a weak form of Gehring’s reverse Hölder inequality.

P . Cardaliaguet (Univ. Brest) Regularity of optimal control problems 14 / 25

Sketch of the proof of the main result

Step 2 (end) : Regularity of optimal solutions

Corollary

There are θ > p and C such that, for any x0 ∈ RN and any t0 < T, if ¯ x is optimal for the initial position x0 at time t0, then t0+h

t0

|¯ x′(s)|ds ≤ C(T − t0)1/θ−1/p h1−1/θ ∀h ∈ [t0, T] Remark : this proves that optimal solutions are 1/θ−Hölder continuous.

P . Cardaliaguet (Univ. Brest) Regularity of optimal control problems 15 / 25

Sketch of the proof of the main result

Step 3 : Use of the regularity of optimal solutions

Corollary (Space regularity)

Let x0, x1 ∈ RN, t0 < T. Then u(x1, t0) − u(x0, t0) ≤ K1(T − t0)−(p−1)(θ−p)/(p(θ−1))|x1 − x0|(θ−p)/(θ−1) where K1 = K1(M, p, T, δ). Proof : Let ¯ x be optimal for x0 and consider ˜ xh(t) =

• ¯

x(t0+h)−x1 h

(t − t0) + x0 if t ∈ [t0, t0 + h] ¯ x(t)

• therwise

which is admissible for x1. Use the regularity of ¯ x and optimize w.r.t. h.

P . Cardaliaguet (Univ. Brest) Regularity of optimal control problems 16 / 25

Sketch of proof for the reverse Hölder inequality Lemma

Outline

1

Statement of the main result

2

Sketch of the proof of the main result

3

Sketch of proof for the reverse Hölder inequality Lemma

P . Cardaliaguet (Univ. Brest) Regularity of optimal control problems 17 / 25

Sketch of proof for the reverse Hölder inequality Lemma

Reverse Hölder inequality Lemma

Fix A > 1 and p > 1.

Lemma (reverse Hölder inequality)

There are θ = θ(A, p) > p and C = C(A, p) > 0 such that, for any α ∈ Lp(0, 1) satisfying 1 h h |α(s)|pds ≤ A

• 1

h h |α(s)|ds p ∀h ∈ [0, 1],

• ne has

h |α(s)|ds ≤ CαLp h1−1/θ ∀h ∈ [0, 1] .

P . Cardaliaguet (Univ. Brest) Regularity of optimal control problems 18 / 25

Sketch of proof for the reverse Hölder inequality Lemma

Gehring result

The proof of the reverse inequality Lemma could be achieved through Gehring’s result : Let p > 1 and Ω open subset of RN. Let α ∈ Lp(Ω) be such that

• Q

− |α|p ≤ A

• Q

− |α| p for any cube Q ⊂ Ω.

Lemma (Gehring (1973))

There are θ > p and C such that αLθ ≤ CαLp where θ = θ(A, p) and C = C(A, p).

P . Cardaliaguet (Univ. Brest) Regularity of optimal control problems 19 / 25

Sketch of proof for the reverse Hölder inequality Lemma

A direct proof of the reverse inequality Lemma

Let E = {α ∈ Lp(0, 1) , α ≥ 0, α satisfies (∗) and αLp ≤ 1} where (∗) 1 h h (α(s))pds ≤ A

• 1

h h α(s)ds p ∀h ∈ [0, 1]

Claim

For any τ ∈ (0, 1] the problem ξ(τ) = max τ α(s)ds , α ∈ E

• has a unique maximum denoted ¯

ατ.

P . Cardaliaguet (Univ. Brest) Regularity of optimal control problems 20 / 25

Sketch of proof for the reverse Hölder inequality Lemma

Case of equality in (∗) : α ∈ E satisfies 1 h h (α(s))pds = A

• 1

h h α(s)ds p ∀h ∈ [0, 1] if and only if α(s) = A−1/p σ sσ−1 ∀s ∈ [0, 1] , where σ be a root of the map s → sp − A(1 − p + ps). Let γ be the smallest positive root. Then γ > 1 − 1/p.

Claim

ξ(τ) ≤ Cτ γ ∀τ ∈ [0, 1] for some C = C(p). Remark : This claim proves the Lemma with θ = 1/(1 − γ) > p.

P . Cardaliaguet (Univ. Brest) Regularity of optimal control problems 21 / 25

Sketch of proof for the reverse Hölder inequality Lemma

Case of equality in (∗) : α ∈ E satisfies 1 h h (α(s))pds = A

• 1

h h α(s)ds p ∀h ∈ [0, 1] if and only if α(s) = A−1/p σ sσ−1 ∀s ∈ [0, 1] , where σ be a root of the map s → sp − A(1 − p + ps). Let γ be the smallest positive root. Then γ > 1 − 1/p.

Claim

ξ(τ) ≤ Cτ γ ∀τ ∈ [0, 1] for some C = C(p). Remark : This claim proves the Lemma with θ = 1/(1 − γ) > p.

P . Cardaliaguet (Univ. Brest) Regularity of optimal control problems 21 / 25

Sketch of proof for the reverse Hölder inequality Lemma

Structure of the optimum

Claim

There is ¯ τ > 0 such that, ∀τ ∈ (0, ¯ τ), ¯ ατ(t) =    aτ

• n [0, τ)

bτ

• n [τ, τ1)

A−1/p γ tγ−1

• n [τ1, 1]

for some 0 < bτ ≤ aτ and τ < τ1 < 1.

P . Cardaliaguet (Univ. Brest) Regularity of optimal control problems 22 / 25

Sketch of proof for the reverse Hölder inequality Lemma

A differential equation for ξ

Claim

ξ is locally Lipschitz continous and satisfies (∗∗) (−τ)ξ′(τ) + γξ(τ) = 0 for a.e. τ ∈ (0, ¯ τ) . Proof : For λ > 0, let αλτ(s) = ¯ ατ(λs) s ≥ 0 . Then αλτ/αλτp belongs to E. So (∗ ∗ ∗) ξ τ λ

• ≥

τ/λ αλτ αλτp = λ1/p−1ξ(τ)

• 1 +

λ

1 ¯

αp

τ

1/p . with an equality for λ = 1. Deriving (∗ ∗ ∗) at λ = 1 gives (**).

P . Cardaliaguet (Univ. Brest) Regularity of optimal control problems 23 / 25

Sketch of proof for the reverse Hölder inequality Lemma

Generalizations

(work in preparation with P . Cannarsa) solutions of HJ equations (HJ) ut(x, t) + H(x, t, Du(x, t)) = 0 under a superlinear growth condition on H, but no convexity assumption, solutions of degenerate parabolic equations of the form (HJ2) ut(x, t) − Tr(A(x, t)D2u(x, t)) + H(x, t, Du(x, t)) = 0 under a superquadratic growth condition on H.

P . Cardaliaguet (Univ. Brest) Regularity of optimal control problems 24 / 25

Sketch of proof for the reverse Hölder inequality Lemma

Main open question

What happens if the growth of L is anisotropic ? δ|ξ|p − M ≤ L(x, t, ξ) ≤ 1 δ |ξ|p′ + M for some 1 < p < p′.

P . Cardaliaguet (Univ. Brest) Regularity of optimal control problems 25 / 25