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Monte Carlo Methods for Random/StochasticPDE
Jingchen Liu
Department of StatisticsColumbia University
Summer School in Monte Carlo Methods for Rare EventsBrown University, Providence RI
June 17, 2016
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Overview
I Ordinary/partial differential equations
I Imprecision and measurement error
I Stochastic and random partial differential equation
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Overview
I Ordinary/partial differential equations
I Imprecision and measurement error
I Stochastic and random partial differential equation
2 / 42
Overview
I Ordinary/partial differential equations
I Imprecision and measurement error
I Stochastic and random partial differential equation
2 / 42
Random PDE
u : D → R
I Differential equationL(u) = 0
subject to certain boundary conditions.
I Randomness L(u, ξ) = 0
I u(x , ω) : D ×Ω→ R
3 / 42
Random PDE
u : D → R
I Differential equationL(u) = 0
subject to certain boundary conditions.
I Randomness L(u, ξ) = 0
I u(x , ω) : D ×Ω→ R
3 / 42
Random PDE
u : D → R
I Differential equationL(u) = 0
subject to certain boundary conditions.
I Randomness L(u, ξ) = 0
I u(x , ω) : D ×Ω→ R
3 / 42
Interesting quantities
I The tail probabilities of functional P(Γ(u) > v)
I Some examples
Γ(u) = supx∈D
u(x), ‖u(x)‖p, supx∈D|∇u(x)|...
I Level crossing of random functions
4 / 42
Interesting quantities
I The tail probabilities of functional P(Γ(u) > v)
I Some examples
Γ(u) = supx∈D
u(x), ‖u(x)‖p, supx∈D|∇u(x)|...
I Level crossing of random functions
4 / 42
Interesting quantities
I The tail probabilities of functional P(Γ(u) > v)
I Some examples
Γ(u) = supx∈D
u(x), ‖u(x)‖p, supx∈D|∇u(x)|...
I Level crossing of random functions
4 / 42
Asymptotic regimes
I L(u, ξ(ω)) = 0⇒ uξ(x)
I Fixed noise and large deviations P(Γ(uξ) > v) as v → ∞.
I Small noise: L(u, σξ) = 0 as σ→ 0.
5 / 42
Asymptotic regimes
I L(u, ξ(ω)) = 0⇒ uξ(x)
I Fixed noise and large deviations P(Γ(uξ) > v) as v → ∞.
I Small noise: L(u, σξ) = 0 as σ→ 0.
5 / 42
Asymptotic regimes
I L(u, ξ(ω)) = 0⇒ uξ(x)
I Fixed noise and large deviations P(Γ(uξ) > v) as v → ∞.
I Small noise: L(u, σξ) = 0 as σ→ 0.
5 / 42
Korteweg-de Vries (KdV) Equation
I The wave u(x , t)
u : R× [0,T ]→ R
I Korteweg-de Vries (KdV) equation
∂tu − 6u∂xu + ∂xxxu = 0
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Korteweg-de Vries (KdV) Equation
I The wave u(x , t)
u : R× [0,T ]→ R
I Korteweg-de Vries (KdV) equation
∂tu − 6u∂xu + ∂xxxu = 0
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Wave
7 / 42
Wave
8 / 42
Wave
9 / 42
Wave
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Solitary wave
11 / 42
Solitary wave
12 / 42
Solitary wave
13 / 42
Solitary wave
u(x , t) =W(x − vt)
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Solitary wave
I ∂tu − 6u∂xu + ∂xxxu = 0
I u(x , t) =W(x − vt)
⇒ d2Wdx2
= 3W2 + vW + c0
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Solitary wave
I ∂tu − 6u∂xu + ∂xxxu = 0
I u(x , t) =W(x − vt)
⇒ d2Wdx2
= 3W2 + vW + c0
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Solitary wave
1
1The picture is published at http://kaheel7.com
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Stochastic Korteweg-de Vries (KdV) Equation
I The stochastic version
∂tu − 6u∂xu + ∂xxxu = ξ(t)
where ξ(t) is a Gaussian process.
I P(supt∈[0,T ] u(x , t) > b), for a given x
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Stochastic Korteweg-de Vries (KdV) Equation
I The stochastic version
∂tu − 6u∂xu + ∂xxxu = ξ(t)
where ξ(t) is a Gaussian process.
I P(supt∈[0,T ] u(x , t) > b), for a given x
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Solitary wave
2
2The picture is published at http://whybecausescience.com
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Initial condition
I The equation
u(x , t = 0) = −2κ2sech2(κx)
where sech(x) = 2ex+e−x .
I If ξ(t) = 0, then
u(x , t) = −2κ2sech2κ(x − 4κ2t)
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Initial condition
I The equation
u(x , t = 0) = −2κ2sech2(κx)
where sech(x) = 2ex+e−x .
I If ξ(t) = 0, then
u(x , t) = −2κ2sech2κ(x − 4κ2t)
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Solution to the random PDE
∂tu − 6u∂xu + ∂xxxu = ξ(t)
u(x , 0) = −2κ2sech2(κx)
I The solution
u(x , t) = f (t)− 2κ2sech2
κ(x − 4κ2t) + 6κ∫ t
0f (t)dt
where f (t) =
∫ t0 ξ(s)ds
20 / 42
Hueristics
u(x , t) = f (t)− 2κ2sech2
κ(x − 4κ2t) + 6κ∫ t
0f (s)ds
I Upper bound u(x , t) ≤ f (t)
I supt u(x , t) > v ⊂ sup f (t) > v
I P(supt u(x , t) > v) ≤ P(sup f (t) > v)
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Hueristics
u(x , t) = f (t)− 2κ2sech2
κ(x − 4κ2t) + 6κ∫ t
0f (s)ds
I Upper bound u(x , t) ≤ f (t)
I supt u(x , t) > v ⊂ sup f (t) > v
I P(supt u(x , t) > v) ≤ P(sup f (t) > v)
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Hueristics
u(x , t) = f (t)− 2κ2sech2
κ(x − 4κ2t) + 6κ∫ t
0f (s)ds
I Upper bound u(x , t) ≤ f (t)
I supt u(x , t) > v ⊂ sup f (t) > v
I P(supt u(x , t) > v) ≤ P(sup f (t) > v)
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The relative difference
I Given sup f (t) > v
supt
f (t)− v = O(1/v)
andsupt
f (t)− supt
u(x , t) = O(e−εv )
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The relative difference
I The relative difference
1− e−εb ≤ P(supt u(x , t) > v)
P(supt f (t) > v)≤ 1
I Approximation
P(supt
u(x , t) > v) ∼ P(supt
f (t) > v) ∼ Cv βe−v2/2σ2
T
where σ2T = supt Var(f (t))
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The relative difference
I The relative difference
1− e−εb ≤ P(supt u(x , t) > v)
P(supt f (t) > v)≤ 1
I Approximation
P(supt
u(x , t) > v) ∼ P(supt
f (t) > v) ∼ Cv βe−v2/2σ2
T
where σ2T = supt Var(f (t))
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Simulation from the change of measure
I Simulate τ ∈ Dτ ∼ h(t)
I Simulate f (τ) according to gτ(x)
I Simulate f (t) : t 6= τ given f (τ) under P
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Simulation from the change of measure
I Simulate τ ∈ Dτ ∼ h(t)
I Simulate f (τ) according to gτ(x)
I Simulate f (t) : t 6= τ given f (τ) under P
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Simulation from the change of measure
I Simulate τ ∈ Dτ ∼ h(t)
I Simulate f (τ) according to gτ(x)
I Simulate f (t) : t 6= τ given f (τ) under P
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The change-of-measure-based analysis
I Let P be the original measure.
I The change of measure Q
dQ
dP=∫t∈D
gt(f (t))
ϕt(f (t))h(t)dt
where ϕt(x) is the marginal density of f (t), h(t) is a densityon D, and gt(x) is an alternative density.
25 / 42
The change-of-measure-based analysis
I Let P be the original measure.
I The change of measure Q
dQ
dP=∫t∈D
gt(f (t))
ϕt(f (t))h(t)dt
where ϕt(x) is the marginal density of f (t), h(t) is a densityon D, and gt(x) is an alternative density.
25 / 42
Choice of h and gt
I Let γ = v − 1/v
I The distribution
gt(x) = I (x > γ)ϕt(x)
P(f (t) > γ)
I The distribution
h(t) =P(f (t) > γ)∫
D P(f (t) > γ)dt
26 / 42
Choice of h and gt
I Let γ = v − 1/v
I The distribution
gt(x) = I (x > γ)ϕt(x)
P(f (t) > γ)
I The distribution
h(t) =P(f (t) > γ)∫
D P(f (t) > γ)dt
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Choice of h and gt
I Let γ = v − 1/v
I The distribution
gt(x) = I (x > γ)ϕt(x)
P(f (t) > γ)
I The distribution
h(t) =P(f (t) > γ)∫
D P(f (t) > γ)dt
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Rare-event simulation and importance sampling
I The change of measure:
dQγ
dP=
mes(Aγ)∫D P(f (t) > γ)dt
where Aγ = t : f (t) > γ.
27 / 42
Importance sampling (Li and L. 2015)
I Strongly efficient in computing the tail probabilities ofsup f (t) and sup u(x , t).
I Discretization
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Importance sampling (Li and L. 2015)
I Strongly efficient in computing the tail probabilities ofsup f (t) and sup u(x , t).
I Discretization
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Material Failure
I u(x): the shape of the material
I Ou(x): strain
I p(x): pressure
I a(x): material-specific coefficients
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Material Failure
I The partial differential equation: x ∈ D
−O · a(x)Ou(x) = p(x)
I Dirichlet boundary condition u(x) = 0 for all x ∈ ∂D.
I log a(x) = ξ(x)
I Failure probability
P(
supx∈D|∇u(x)| > b
)
30 / 42
Material Failure
I The partial differential equation: x ∈ D
−O · a(x)Ou(x) = p(x)
I Dirichlet boundary condition u(x) = 0 for all x ∈ ∂D.
I log a(x) = ξ(x)
I Failure probability
P(
supx∈D|∇u(x)| > b
)
30 / 42
Material Failure
I The partial differential equation: x ∈ D
−O · a(x)Ou(x) = p(x)
I Dirichlet boundary condition u(x) = 0 for all x ∈ ∂D.
I log a(x) = ξ(x)
I Failure probability
P(
supx∈D|∇u(x)| > b
)
30 / 42
Material Failure
I The partial differential equation: x ∈ D
−O · a(x)Ou(x) = p(x)
I Dirichlet boundary condition u(x) = 0 for all x ∈ ∂D.
I log a(x) = ξ(x)
I Failure probability
P(
supx∈D|∇u(x)| > b
)
30 / 42
Material Failure – one dimensional example
I The ordinary differential equation: x ∈ [0, 1]
(a(x)u′(x))′ = −p(x)
I Spatial variation: a(x) = eξ(x), where ξ(x) is a Gaussianprocess.
31 / 42
The failure probability
I The failure probability
P
(supx∈D|u′(x)| > b
)I The displacement u(x) depends on the process a(x).
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The failure probability
I The failure probability
P
(supx∈D|u′(x)| > b
)I The displacement u(x) depends on the process a(x).
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Material Failure – Dirichlet condition
I Dirichlet condition: u(0) = u(1) = 0
I The solution:
u(x) =∫ x0 F (y)a−1(y)dy −
∫ 10 F (y )a−1(dy )dy∫ 1
0 a−1(dy )
∫ x0 a−1(y)dy ,
where F (x) =∫ x0 p(y)dy .
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Material Failure – Dirichlet condition
I Dirichlet condition: u(0) = u(1) = 0
I The solution:
u(x) =∫ x0 F (y)a−1(y)dy −
∫ 10 F (y )a−1(dy )dy∫ 1
0 a−1(dy )
∫ x0 a−1(y)dy ,
where F (x) =∫ x0 p(y)dy .
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Material Failure – Dirichlet condition
I The strain
u′(x) = a−1(x)
(F (x)−
∫ 10 F (y)a−1(y)dy∫ 1
0 a−1(y)dy
)= a−1(x)
[F (x)− Eξ(F (Y ))
]where a−1(x) = eξ(x).
34 / 42
Theorem: approximation for continuous force (L. and Zhou2011)
I The external force p(x) is a continuously differentiablefunction.
I x∗ = arg supx p(x).
Then, we have the approximation
P( supx∈[0,1]
|u′(x)| > b)
∼ P(|u′(0)| > b) + P(|u′(1)| > b) + P( sup|x−x∗|<ε
|u′(x)| > b).
35 / 42
Exact asymptotic approximation for continuous body force
I Let p(x∗)r−1er−12 = b. Then,
P( sup|x−x∗|<ε
|u′(x)| > 0) ∼ κ∗ × r−1/2 exp−r2/2.
I Let H0r0−1/2er0 = b. Then,
P(|u′(0)| > b) = κ0 × r0−1e−r0
2/2
I Let H1r1−1er1 = b. Then,
P(|u′(1)| > b) = κ1 × r1−1e−r1
2/2
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Exact asymptotic approximation for continuous body force
I Let p(x∗)r−1er−12 = b. Then,
P( sup|x−x∗|<ε
|u′(x)| > 0) ∼ κ∗ × r−1/2 exp−r2/2.
I Let H0r0−1/2er0 = b. Then,
P(|u′(0)| > b) = κ0 × r0−1e−r0
2/2
I Let H1r1−1er1 = b. Then,
P(|u′(1)| > b) = κ1 × r1−1e−r1
2/2
36 / 42
Exact asymptotic approximation for continuous body force
I Let p(x∗)r−1er−12 = b. Then,
P( sup|x−x∗|<ε
|u′(x)| > 0) ∼ κ∗ × r−1/2 exp−r2/2.
I Let H0r0−1/2er0 = b. Then,
P(|u′(0)| > b) = κ0 × r0−1e−r0
2/2
I Let H1r1−1er1 = b. Then,
P(|u′(1)| > b) = κ1 × r1−1e−r1
2/2
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Key components of the conditional distribution
I Where does the break occur or arg sup u′(x) =?I Where does ξ(x) attain it maximum?
I At what level?
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Choice of h and gt
I The distribution h(t): mixture of three components
I The distribution gt(x).
38 / 42
Choice of h and gt
I The distribution h(t): mixture of three components
I The distribution gt(x).
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High-dimensional PDE (L. et al. 2014)
−O · a(x)Ou(x) = p(x)
I Random index h(τ) ∼ Uniform(D)
I The distributiongτ(x) ∼ N(lτ, 1)
39 / 42
High-dimensional PDE (L. et al. 2014)
−O · a(x)Ou(x) = p(x)
I Random index h(τ) ∼ Uniform(D)
I The distributiongτ(x) ∼ N(lτ, 1)
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The choice of lτ
I Given that ξ(τ) = lτ, the conditional field is
ξ(x) = lτ × C (x − τ) + r(x − τ)
where E [r(x)] = 0.
I Set r(x) ≡ 0 for the asymptotic analysis
I Choose ξ(x) = lτC (x − τ) such that sup |∇u(x)| = v .
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The choice of lτ
I Given that ξ(τ) = lτ, the conditional field is
ξ(x) = lτ × C (x − τ) + r(x − τ)
where E [r(x)] = 0.
I Set r(x) ≡ 0 for the asymptotic analysis
I Choose ξ(x) = lτC (x − τ) such that sup |∇u(x)| = v .
40 / 42
The choice of lτ
I Given that ξ(τ) = lτ, the conditional field is
ξ(x) = lτ × C (x − τ) + r(x − τ)
where E [r(x)] = 0.
I Set r(x) ≡ 0 for the asymptotic analysis
I Choose ξ(x) = lτC (x − τ) such that sup |∇u(x)| = v .
40 / 42
The computation of lt
1. Set the initial guess l(0)τ = log v .
2. For l(n)τ , set
ξ(t) = l(n)τ C (t − τ)
and compute u(n)
l(n+1)τ = l
(n)τ − log sup |∇u(n)|+ log v .
3. Repeat step 2 until convergence.
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The computation of lt
1. Set the initial guess l(0)τ = log v .
2. For l(n)τ , set
ξ(t) = l(n)τ C (t − τ)
and compute u(n)
l(n+1)τ = l
(n)τ − log sup |∇u(n)|+ log v .
3. Repeat step 2 until convergence.
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The computation of lt
1. Set the initial guess l(0)τ = log v .
2. For l(n)τ , set
ξ(t) = l(n)τ C (t − τ)
and compute u(n)
l(n+1)τ = l
(n)τ − log sup |∇u(n)|+ log v .
3. Repeat step 2 until convergence.
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Visualizing lt
Figure: Contour plot of the excursion level function lτ with homogeneousexternal force p ≡ 1 (left panel) and the discontinuous external forcep(x) = sign(x < 0.5) (right panel) for b = 4.
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