Stochastic Homogenization For Elliptic Nonlocal …cermics.enpc.fr/~al-hajm/ile-de-re/schwab.pdf ·...
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Stochastic Homogenization For Elliptic NonlocalEquations... And Beyond
(Ginzburg-Landau equations, Dislocations andHomogenization– Ile de Re, May 2011)
Russell Schwab
Carnegie Mellon University
26 May, 2011
The Set-UpNotation for Some Relevant Operators
• The σ/2 fractional Laplace
− (−∆)σ/2 u(x) = Cn,σ
∫Rn
(u(x + y) + u(x − y)− 2u(x)) |y |−n−σ dy
• A linear operator comparable to the fractional Laplace
L(u, x) = Cn,σ
∫Rn
(u(x + y) + u(x − y)− 2u(x)) K (x , y)dy
withλ
|y |n+σ≤ K (x , y) ≤ Λ
|y |n+σ
• A fully nonlinear operator comparable to the fractional Laplace
F (u, x) =
infα
supβ
Cn,σ
∫Rn
(u(x + y) + u(x − y)− 2u(x))Kαβ(x , y)dy.
The Set-Up
Family of Oscillatory Nonlocal Equations:
F (uε,
x
ε, ω) = 0 in D
uε = g on Rn \ D
F (u,x
ε, ω) =
infα
supβ
f αβ(
x
ε, ω) + Cn,σ
∫Rn
(u(x + y) + u(x − y)− 2u(x))Kαβ(x
ε, y , ω)dy
.(
Think of a more familiar 2nd order equation:
F (D2u,x
ε) = inf
αsupβ
f αβ(
x
ε) + aαβij (
x
ε)uxixj (x)
)
The Set-Up
Stationary Ergodic F
for all z ∈ Rn
F (u, x + z , ω) = F (u(·+ z), x , τzω)
Our F will be stationary ergodic when f αβ and Kαβ are stationary withrespect to some ergodic τz .
Translation Invariant Nonlocal Operator F
F is translation invariant if for any y ∈ Rn,
F (u, x + y) = F (u(·+ y), x).
just think of
L(u, x) = Cn,σ
∫Rn
(u(x + y) + u(x − y)− 2u(x)) K (y)dy
The Set-Up
Family of Oscillatory Nonlocal Equations:
F (uε,
x
ε) = 0 in D
uε = g on Rn \ D
Translation Invariant Limit Nonlocal Equations
F (u, x) = 0 in D
u = g on Rn \ D.
GOAL
Prove there is a unique nonlocal operator F so that uε will be very closeto u as ε→ 0. (Homogenization takes place.)
Main Theorem
Theorem (Stochastic Homogenization of Nonlocal Equations–preprint at arXiv.org)
If F is stationary ergodic and uniformly elliptic, plus technicalassumptions, then there exists a translation invariant elliptic nonlocaloperator F with the same ellipticity as F , such that for a.e. ω uε(ω)→ ulocally uniformly and u is the unique solution of
F (u, x) = 0 in D
u = g on Rn \ D.
The Set-Up– Assumptions on F
“Ellipticity”
λ
|y |n+σ≤ Kαβ(x , y) ≤ Λ
|y |n+σ
Scaling (not really necessary... just for clean presentation)
Kαβ(x , λy) = λ−n−σKαβ(x , y).
Symmetry
Kαβ(x ,−y) = Kαβ(x , y)
Recent Background– Nonlocal Elliptic Equations
Existence/Uniqueness (Barles-Chasseigne-Imbert)
Given basic assumptions on Kαβ and f αβ, there exist unique solutions tothe Dirichlet Problems F (uε, x/ε) = 0, F (u, x) = 0.
Regularity (Silvestre, Caffarelli-Silvestre)
uε are Holder continuous, depending only on λ, Λ, ‖f αβ‖∞, dimension,and g . (In particular, continuous uniformly in ε.)
Nonlocal Ellipticity– Good Definition (Caffarelli-Silvestre)
If u and v are C 1,1 at a point, x , then
M−(u − v)(x) ≤ F (u, x)− F (v , x) ≤ M+(u − v)(x).
M−u(x) = infαβ
Lαβu(x)
and M+u(x) = sup
αβ
Lαβu(x)
.
Historical Homogenization Background
• De Giorgi, Jikov, Kozlov, Murat, Spagnolo, Tartar, Yurinskii, ...many many more
• Bensoussan-Lions-Papanicolaou 1978– Asymptotic analysis forperiodic structures
• Papanicolaou-Varadhan– Random Homogenization Linear EllipticOperators
• Lions-Papanicolaou-Varadhan (unpublished)– PeriodicHomogenization of Hamilton-Jacobi Equations
• Evans 1992– Periodic Homogenization of Fully Nonlinear PDEs
• Arisawa-Lions 1998– On Ergodic Stochastic Control
• Caffarelli-Souganidis-Wang 2005– Stochastic Homogenization OfFully Nonlinear Elliptic Equations
• (related results) Arisawa 2009– Homogenization of a Class ofIntegro-Differential Equations with Levy Operators
Motivating The “Corrector” Equation
• Expansion (formally!!!): uε(x) = u(x) + εσv(·/ε) + o(εσ)
• Scaling: L(εσv(·/ε), x) = L(v , x/ε)
• Separation of Scales: as ε→ 0, x can be considered fixed w.r.t. x/εand x/ε = y becomes a global variable
• formal generalization of Fredholm Alternative
Motivating The “Corrector” EquationJust for the fractional Laplace sufficient for the idea...
−(−∆)σ/2uε = f (x
ε)
• Plug Expansion into equation plus scaling:
−(−∆)σ/2u(x) +−(−∆)σ/2v(x
ε) = f (
x
ε)
• Global variables:
−(−∆)σ/2u(x) +−(−∆)σ/2v(y) = f (y) + F
• Need to find such a v : Fredholm says only if
F +
∫Q1
fdy − (−(−∆)σ/2u(x)) = 0
Motivating The “Corrector” EquationJust for the fractional Laplace sufficient for the idea...
−(−∆)σ/2uε = f (x
ε)
• Plug Expansion into equation plus scaling:
−(−∆)σ/2u(x) +−(−∆)σ/2v(x
ε) = f (
x
ε)
• Global variables:
−(−∆)σ/2u(x) +−(−∆)σ/2v(y) = f (y) + F
• Need to find such a v : Fredholm says only if
F +
∫Q1
fdy − (−(−∆)σ/2u(x)) = 0
Motivating The “Corrector” EquationRestate what we just did...
True Corrector... Special Case
For all u and x fixed, there exists a unique constant, F , such that thereis a bounded periodic periodic solution of
(−(−∆)σ/2u(x))− (−∆)σ/2v(y) = f (y) + F (u, x)
Formal Solution To Homogenization
−(−∆)σ/2u(x) +−(−∆)σ/2v(x
ε) = f (
x
ε)
which reads, after ε→ 0,F (u, x) = 0
Generalizing Away From Fractional Laplace
Let’s remove some unnecessary restrictions... It was all formal anyway!
• Expansion? Why one v?? Let’s try a whole sequence of v ε souε(x) = u(x) + v ε(x) + o(εσ)
• Decay! We will need a compatibility condition on the v ε... requiresupx |v ε(x)| → 0
• Fractional Laplacian? Not Special! Only used scaling (naively! notmandatory), really... upgrade to F (u, x/ε)
• Why global equation and periodic v?? How about just solve theequation in B1?
“Corrector” Equation
equation for φ + v ε
F (φ+ v ε,x
ε) =
infα
supβ
f αβ(
x
ε) +
∫Rn
(φ(x + y) + φ(x − y)− 2φ(x))Kαβ(x
ε, y)dy
+
∫Rn
(v ε(x + y) + v ε(x − y)− 2v ε(x))Kαβ(x
ε, y)dy
“frozen” operator on φ at x0
[Lαβφ(x0)](x) =
∫Rn
(φ(x0 + z) + φ(x0 − z)− 2φ(x0))Kαβ(x , z)dz
“Corrector” Equation
Analogy to 2nd order equation
aij(x
ε)(φ+ v)xixj (x) = aij(
x
ε)φxixj (x) + aij(
x
ε)vxixj (x)
and aij(xε )φxixj (x) is uniformly continuous in x .
Free and frozen variables, x and x0
Uniform continuity (Caffarelli-Silvestre)
[Lαβφ(x0)](x) is uniformly continuous in x0, independent of x and αβ
Comments on Test Functionstest functions cause a non-trivial modification for the nonlocal setting
• local operators versus nonlocal operators
“Corrector” Equation
NEW OPERATOR Fφ,x0
Fφ,x0(v ε,x
ε) = inf
αsupβ
f αβ(
x
ε) + [Lαβφ(x0)](
x
ε)
+
∫Rn
(v ε(x + y) + v ε(x − y)− 2v ε(x))Kαβ(x
ε, y)dy
New “Corrector” Equation
Fφ,x0(v ε, xε ) = F (φ, x0) in B1(x0)
v ε = 0 on Rn \ B1(x0).
“Corrector” Equation
Proposition (“Corrector” Equation)
For all smooth φ and x0 fixed, there exists a set of full measure Ωφ,x0 anda unique choice for the value of F (φ, x0) such that for ω ∈ Ωφ,x0 , thesolutions of the “corrector” equation also satisfy
limε→0
maxB1(x0)
|v ε| = 0.
(via the perturbed test function method, this proposition is equivalent tohomogenization)
Perturbed Test Function Method
Need to DetermineEffective operator
⇒All information is inoriginal operatorF (·, x/ε) = 0
Can we perturbφ to φ+ v ε
to COMPAREWITH uε???
F (φ, x0) ≥ 0 ⇒ F (φ+ v ε, x/ε) = F ≥ 0
To go BACK from comparison of φ+ v ε and uε
TO comparison of φ and u NEED
|v ε| → 0 as ε→ 0
Finding F ... Variational Problem
(Caffarelli-Sougandis-Wang... *In spirit)
Consider a generic choice of a Right Hand Side, l is fixed
Fφ,x0(v εl ,
xε ) = l in B1(x0)
v εl = 0 on Rn \ B1(x0).
How does the choice of l affect the decay of v εl ?
decay property
limε→0 maxB1(x0) |v ε| = 0 ⇐⇒ (v εl )∗ = (v εl )∗ = 0
Variational Problem
l very negative
p+(x) = (1− |x |2)2 · 1B1 is a subsolution of equation=⇒ (v εl )∗ > 0 and we missed the goal.
(Fφ,x0(v εl ,
x
ε) = l
)
Variational Problem
(Fφ,x0(v εl ,
x
ε) = l
)l very positive
p−(x) = −(|x |2 − 1)2 · 1B1 is a supersolution of equation=⇒ (v εl )∗ < 0 and we missed the goal, but in the other direction.
Can we choose an l in the middle that is “JUST RIGHT”?
Variational Problem
(Fφ,x0(v εl ,
x
ε) = l
)l very positive
p−(x) = −(|x |2 − 1)2 · 1B1 is a supersolution of equation=⇒ (v εl )∗ < 0 and we missed the goal, but in the other direction.
Can we choose an l in the middle that is “JUST RIGHT”?
Variational Problem
(Fφ,x0(v εl ,
x
ε) = l
)l very positive
p−(x) = −(|x |2 − 1)2 · 1B1 is a supersolution of equation=⇒ (v εl )∗ < 0 and we missed the goal, but in the other direction.
Can we choose an l in the middle that is “JUST RIGHT”?
Variational Problem
(l << 0
)l = ?????
(l >> 0
)
Can we choose an l in the middle that is “JUST RIGHT”?
Obstacle Problem
(Caffarelli-Sougandis-Wang) The answer is YES.
Information From Obstacle Problem
The obstacle problem gives relationship between the choice of l and thedecay of v εl .
Obstacle Problem
The Solution of The Obstacle Problem In a Set A
U lA = inf
u : Fφ,x0(u, y) ≤ l in A and u ≥ 0 in Rn
equation: U l
A is the least supersolution of Fφ,x0 = l in A
obstacle: U lA must be above the obstacle which is 0 in all of Rn
Lemma (Holder Continuity)
U lA is γ-Holder Continuous depending only on λ, Λ, ‖f αβ‖∞, φ,
dimension, and A.
Monotonicity and Stationarity of Obstacle Problem
If A ⊂ B, then U lA ≤ U l
B . For z ∈ Rn, U lA+z(x , ω) = U l
A(x − z , τzω)
Obstacle Problem
The Solution of The Obstacle Problem In a Set A
U lA = inf
u : Fφ,x0(u, y) ≤ l in A and u ≥ 0 in Rn
equation: U l
A is the least supersolution of Fφ,x0 = l in A
obstacle: U lA must be above the obstacle which is 0 in all of Rn
Lemma (Holder Continuity)
U lA is γ-Holder Continuous depending only on λ, Λ, ‖f αβ‖∞, φ,
dimension, and A.
Monotonicity and Stationarity of Obstacle Problem
If A ⊂ B, then U lA ≤ U l
B . For z ∈ Rn, U lA+z(x , ω) = U l
A(x − z , τzω)
Obstacle Problem
NOTATION
Rescaled Solution
uε,l = inf
u : Fφ,x0(u,y
ε) ≤ l in Q1 and u ≥ 0 in Rn
.
Solution in Q1 and Solution in Q1/ε
uε,l(x) = εσU lQ1/ε
(x
ε)
Contact Set
Kε = U lQ1/ε
= 0 and kε = uε,l = 0
Obstacle Problem- Relationship to v εlEquation for uε,l
Fφ,x0(uε,l ,x
ε) ≤ C (F )1kε + l and uε,l = 0 in Rn \ Q1
Equation for uε,l − v εl
M−(uε,l − v εl ) ≤ Fφ,x0(uε,l ,x
ε)− Fφ,x0(v εl ,
x
ε) ≤ C (F )1kε
Obstacle Problem- Contact Limits
Subadditive Quantity
Q 7→ |K (Q, ω)|
is a subadditive and stationary (with respect to τ) mapping on therectangles, Q, contained in Rn
(Dal Maso - Modica relationship...)
Subadditive Ergodic Theorem
there exists a constant, K , and a set of full measure, Ωφ,x0 , such that
1
|tQ||K (tQ, ω)| → K as t →∞
(Obstacle function scaling!!!!!!!)
Technical Requirement To Conclude
Comparison With Measurable Ingredients
Suppose that 0 ≤ f ε ≤ 1, |f ε > 0| → 0 as ε→ 0, andM−(w ε) ≤ f ε in Q1
w ε = 0 in Rn \ Q1.
Then |w ε| → 0 as ε→ 0.
Conclusion of “Corrector” Equation
K characterizes limits of v εl
if K > 0, then (v εl )∗ ≤ 0 and if K = 0 then (v ε)∗ ≥ 0
The Good Choice of F
F (φ, x0) = sup
l : K = 0
A Very Basic QuestionWhen does comparison with measurable ingredients hold? Even forLINEAR EQUATIONS!
Equations with measurable ingredients (supersolutions)
Lw ε ≤ f ε in Q1
w ε = 0 in Rn \ Q1.
Lw(x) =
∫(w(x + y) + w(x − y)− 2w(x))K (x , y)dy
(K bounded measurable, comparable to |y |−n−σ)
limit as |f ε > 0| → 0????? (simplest case 0 ≤ f ε ≤ 1)
‖w ε‖∞ → 0??????
Recent Results... Aleksandrov-Bakelman-Pucci type results
ABP type estimates– Joint with N. Guillen, preprint at arXiv.org
M−w ≤ f in D
w ≥ 0 in Rn \ D,
then
− infD
w ≤ C (n, σ)
λ(diam(D))(‖f ‖L∞(K(w)))
(2−σ)/2(‖f ‖Ln(K(w)))2/σ.
Only A Restricted M−!!!!
M−(w , x) = inftrace(A)≥λ
∫(w(x + y) + w(x − y)− 2w(x))
yTAy
|y |n+σ+2dy
Recent Results... Aleksandrov-Bakelman-Pucci type results
why the special form
K (x , y) =yTA(x)y
|y |n+σ+2
???????
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