Noisy differential equations with power type coefficients Samy - - PowerPoint PPT Presentation

Noisy differential equations with power type coefficients

Samy Tindel

Université de Lorraine

AMS Sectional Meeting - Las Vegas 2015 Ongoing joint work with Jorge León and David Nualart

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Outline

1

Introduction Equation under consideration Examples and heuristics

2

Results

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Outline

1

Introduction Equation under consideration Examples and heuristics

2

Results

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Outline

1

Introduction Equation under consideration Examples and heuristics

2

Results

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Typical noisy differential equation

Equation: yt = a +

d

• j=1

t

0 σj(yu) dx j u.

(1) Assumptions: x ∈ Cγ([0, τ]; Rd), with γ > 1/2 a initial data in Rm σ1, . . . , σd vector fields on Rm Classical result: If σ ∈ C

1 γ then

֒ → Existence and uniqueness result for (1)

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Equation with power type coefficient

Equation: for x ∈ Cγ([0, τ]; Rd), yt = a +

d

• j=1

t

0 σj(yu) dx j u.

(2) Further assumptions on σ: For κ ∈ (0, 1) σ(0) = 0, σ smooth away from 0 |σ(ξ2) − σ(ξ1)| |ξ2 − ξ1|κ in a neighborhood of 0. Case 1: γ + γκ > 1 ֒ → Easily handled by Young integration techniques. Case 2: γ + γκ < 1 ֒ → Case of interest for us.

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Outline

1

Introduction Equation under consideration Examples and heuristics

2

Results

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Brownian case

Feller’s diffusion: Historical process defined as Solution to dXt = βXt dt + σ√Xt ˙ Wt, where W ≡ Wiener. Obtained as limit of Galton-Watson processes.

1 2 3 4 5 2 4 6 8 12

Figure: Simulation for dXt = .02Xt dt + 2√Xt dWt on [0, 5]

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Brownian case (2)

Related question: Pathwise existence and uniqueness for dYt = |Yt|κ ˙ Wt, κ ∈ (0, 1). Yamada-Watanabe’s results: If κ ≥ 1/2, existence and pathwise uniqueness. If κ < 1/2, pathwise uniqueness fails.

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Brownian case (3)

Heuristics for critical exponent:

1

Reduction to the definition of |Yt|κ ˙ Wt as a distribution

2

˙ Wt ∈ C−1/2−ε

3

Problem to define |Yt|κ ˙ Wt: when Yt close to 0 ֒ → otherwise power function well-behaved

4

When Yt close to 0: equation becomes noiseless

(i) Morally Y ∈ C1 (ii) Morally |Y |κ ∈ Cκ instead of |Y |κ ∈ C

κ 2

(iii) Product |Yt|κ ˙ Wt well-defined if κ − 1

2 − ε > 0

Conclusion: Critical κ is 1

2

Remark: Heuristic not apparent in Yamada-Watanabe’s proof.

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Stochastic PDE case

Dawson-Watanabe process: Defined as Solution to ∂tXt(x) = 1

2∆Xt(x) +

• Xt(x) ˙

Wt(x) where ˙ W ≡ space-time white noise on R+ × R. Obtained as limit of branching Brownian particles.

Figure: A simulation by Nicolas Champagnat

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Stochastic PDE case (2)

Related question: Pathwise existence and uniqueness for ∂tYt(x) = 1 2∆Yt(x) + |Yt(x)|κ ˙ Wt, κ ∈ (0, 1). Mytnik-Mueller-Perkins’ results: If κ > 3/4, existence and pathwise uniqueness. If κ < 3/4, pathwise uniqueness fails. Criticality of 3/4 can be seen from heuristics Mytnik-Perkin’s proof is a (terrible) elaboration of heuristics.

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Stochastic PDE case (3)

Heuristics for critical exponent:

1

Reduction to the definition of |Yt(x)|κ ˙ Wt(x) as a distribution

2

˙ W ∈ C−3/2−ε in parabolic scaling

3

Problem to define |Yt(x)|κ ˙ Wt(x): when Yt(x) close to 0 ֒ → otherwise power function well-behaved

4

When Yt(x) close to 0: equation becomes noiseless

(i) Morally Y ∈ C2 in parabolic scaling (ii) Morally |Y |κ ∈ C2κ instead of |Y |κ ∈ C

κ 2 in parabolic scaling

(iii) Product |Yt|κ ˙ Wt well-defined if 2κ − 3

2 − ε > 0

Conclusion: Critical κ is 3

4

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Stochastic PDE case (4)

Parabolic scaling: for ϕ : R × R → R set St,xϕ(s, y) = 1 δ3ϕ

s − t

δ2 , y − x δ

• Distributional exponent: F ∈ C−α in parabolic scaling if
• R×R[St,xϕ](s, y) F(s, y) dsdy ≤ cϕ δ−α

White noise irregularity: We have ˙ W ∈ C−3/2−ε since E

• ˙

W (St,xϕ)

• 2

=

• R×R
• [St,xϕ](s, y)
• 2 dsdy

= 1 δ3

• R×R |ϕ(s, y)|2 dsdy

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Back to our equation

Equation: ˙ yt = |yt|κ ˙ xt, where x ∈ Cγ Heuristics for critical exponent:

1

Reduction to the definition of |yt|κ ˙ xt as a distribution

2

˙ xt ∈ C−(1−γ)

3

Problem to define |yt|κ ˙ xt: when yt close to 0 ֒ → otherwise power function well-behaved

4

When yt close to 0: equation becomes noiseless

(i) Morally y ∈ C1 (ii) Morally |y|κ ∈ Cκ instead of |y|κ ∈ Cγ κ (iii) Product |yt|κ ˙ xt well-defined if κ − (1 − γ) > 0

Conclusion: Critical κ should be κc = 1 − γ Remark: Our story is in fact quite different!

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Outline

1

Introduction Equation under consideration Examples and heuristics

2

Results

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Setting

Equation: for x ∈ Cγ([0, τ]; Rd) with γ > 1/2, yt = a +

d

• j=1

t

0 σj(yu) dx j u.

(3) Further assumptions on σ: σ(0) = 0 |σ(ξ2) − σ(ξ1)| |ξ2 − ξ1|κ for κ ∈ (0, 1) Case of interest: γ + γκ < 1 Remark: relation γ + κ > 1 does not show up! Noisy integral: Extended Young sense ֒ → See later for details

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1-dimensional case

Additional set of assumptions: d = 1 and y is real-valued φ(ξ) =

ξ

ds σ(s) well-defined

τ

ds |φη(xs)| < ∞ for a certain η < 1 − κ

֒ → Satisfied for x ≡ fBm with H > 1/2 Under assumptions above, y = φ−1(x) solution to: yt =

t

0 σ(ys)dxs

Theorem 1. Remark: No uniqueness here (y ≡ 0 is also a solution)

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d-dimensional case: setting

Additional set of assumptions: |σj(ξ)| |ξ|κ |σj(ξ2) − σj(ξ1)| |ξ2 − ξ1|κ in a neighborhood of 0. |Dσj(ξ)| |ξ|−(1−κ). A stopping time: Set t∗ = inf {t; yt = 0} ∧ τ

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d-dimensional case: result

Under assumptions of previous slide ֒ → There exists a solution y to equation (3) in following sense: (i) For t ∈ [0, t∗] we have yt = a +

d

• j=1

t

0 σj(yu) dx j u

(ii) On [t∗, τ], we have y ≡ 0 (iii) Integral understood in extended Young sense Theorem 2.

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Extended Young integral

Consider: x and y in Cγ with γ > 1/2 κ such that γ + γκ < 1 σ of the form σ(ξ) ≍ |ξ|κ Assume y satisfies: |y|−1 ∈ Lβ([0, τ]), with β ≥ 1 − (γ + γκ) γ2 Then following integral well-defined as limit of Riemann sums:

t

0 σ(ys) dxs

Proposition 3.

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Sketch of the proof

Strategy for existence result for (3): Proceed by regularization σ → σn and y → y n y n ∈ Cγ Main step: prove uniform bounds on

t∗

n

0 |y n s |−βds

To prove these bounds: use regularity increase as |ys| close to 0 ֒ → Back to Mytnik-Perkins’ heuristics

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