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The stochastic heat equation driven by a Gaussian noise: Markov - - PowerPoint PPT Presentation

Aug 28, 2023 480 likes •797 views

The stochastic heat equation driven by a Gaussian noise: Markov property Doyoon Kim 1 , 2 Raluca Balan 1 1 University of Ottawa 2 University of Southern California (after August 30, 2007) June 4-8, 2007 R. Balan and D. Kim (University of Ottawa)

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The stochastic heat equation driven by a Gaussian noise: Markov property

Raluca Balan1 Doyoon Kim1,2

1University of Ottawa 2University of Southern California (after August 30, 2007)

June 4-8, 2007

  • R. Balan and D. Kim (University of Ottawa)

SPDE and Markov property Large Deviations, Ann Arbor 1 / 23

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Outline

1

History of the problem

2

The Framework The noise The stochastic integral The equation and its solution

3

RKHS General characterization Bessel kernel Riesz kernel

4

Germ Markov property The necessary and sufficient condition Main result

  • R. Balan and D. Kim (University of Ottawa)

SPDE and Markov property Large Deviations, Ann Arbor 2 / 23

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History of the problem

Germ Markov property

The germ σ-field Let S be a subset in [0, T] × Rd. FS: the σ-field generated by {u(t, x) : (t, x) ∈ S} GS =

  • O open:O⊃S

FS. Definition The process {u(t, x) : (t, x) ∈ [0, T] × Rd} is germ Markov if for every precompact open set A ⊂ [0, T] × Rd, G¯

A ⊥ GAc | G∂A,

where ∂A = ¯ A ∩ Ac.

  • R. Balan and D. Kim (University of Ottawa)

SPDE and Markov property Large Deviations, Ann Arbor 3 / 23

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History of the problem

Some cases investigated

Donati-Martin and Nualart, 1994

  • −∆u + f(u) = ˙

W, x ∈ D u|∂D = 0 , where D is a bounded domain in Rd, d = 1, 2, 3. f is an affine function. Nualart and Pardoux, 1994

  • ut = uxx + f(u) + ˙

W, (t, x) ∈ [0, 1]2 u(0, x) = u0(x), 0 ≤ x ≤ 1; u(t, 0) = u(t, 1) = 0, 0 ≤ t ≤ 1. Dalang and Hou, 1997 utt = ∆u + ˙ L, where L is locally finite Lévy process.

  • R. Balan and D. Kim (University of Ottawa)

SPDE and Markov property Large Deviations, Ann Arbor 4 / 23

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History of the problem

Some cases investigated

Donati-Martin and Nualart, 1994

  • −∆u + f(u) = ˙

W, x ∈ D u|∂D = 0 , where D is a bounded domain in Rd, d = 1, 2, 3. f is an affine function. Nualart and Pardoux, 1994

  • ut = uxx + f(u) + ˙

W, (t, x) ∈ [0, 1]2 u(0, x) = u0(x), 0 ≤ x ≤ 1; u(t, 0) = u(t, 1) = 0, 0 ≤ t ≤ 1. Dalang and Hou, 1997 utt = ∆u + ˙ L, where L is locally finite Lévy process.

  • R. Balan and D. Kim (University of Ottawa)

SPDE and Markov property Large Deviations, Ann Arbor 4 / 23

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History of the problem

Some cases investigated

Donati-Martin and Nualart, 1994

  • −∆u + f(u) = ˙

W, x ∈ D u|∂D = 0 , where D is a bounded domain in Rd, d = 1, 2, 3. f is an affine function. Nualart and Pardoux, 1994

  • ut = uxx + f(u) + ˙

W, (t, x) ∈ [0, 1]2 u(0, x) = u0(x), 0 ≤ x ≤ 1; u(t, 0) = u(t, 1) = 0, 0 ≤ t ≤ 1. Dalang and Hou, 1997 utt = ∆u + ˙ L, where L is locally finite Lévy process.

  • R. Balan and D. Kim (University of Ottawa)

SPDE and Markov property Large Deviations, Ann Arbor 4 / 23

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The Framework The noise

Outline

1

History of the problem

2

The Framework The noise The stochastic integral The equation and its solution

3

RKHS General characterization Bessel kernel Riesz kernel

4

Germ Markov property The necessary and sufficient condition Main result

  • R. Balan and D. Kim (University of Ottawa)

SPDE and Markov property Large Deviations, Ann Arbor 5 / 23

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The Framework The noise

The noise

Gaussian noise with spatial correlation (Dalang, 1999) M={M(ϕ), ϕ ∈ D((0, T) × Rd)} Gaussian process with covariance E(M(ϕ)M(ψ)) = ∞

  • Rd
  • Rd ϕ(t, x)f(x − y)ψ(t, y) dx dy dt

= T

  • Rd Fϕ(t, ξ)Fψ(t, ξ) µ(dξ)dt := ϕ, ψ0

Here f = Fµ, where µ is a tempered measure on Rd Riesz kernel f(x) = cα,d|x|−α Bessel kernel f(x) = cα ∞

0 s(α−d)/2−1e−s−|x|2/(4s)ds

Heat kernel f(x) = cα,de−|x|2/(4α) Poisson kernel f(x) = cα,d(|x|2 + α2)−(d+1)/2

  • R. Balan and D. Kim (University of Ottawa)

SPDE and Markov property Large Deviations, Ann Arbor 6 / 23

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The Framework The noise

The noise

Gaussian noise with spatial correlation (Dalang, 1999) M={M(ϕ), ϕ ∈ D((0, T) × Rd)} Gaussian process with covariance E(M(ϕ)M(ψ)) = ∞

  • Rd
  • Rd ϕ(t, x)f(x − y)ψ(t, y) dx dy dt

= T

  • Rd Fϕ(t, ξ)Fψ(t, ξ) µ(dξ)dt := ϕ, ψ0

Here f = Fµ, where µ is a tempered measure on Rd Riesz kernel f(x) = cα,d|x|−α Bessel kernel f(x) = cα ∞

0 s(α−d)/2−1e−s−|x|2/(4s)ds

Heat kernel f(x) = cα,de−|x|2/(4α) Poisson kernel f(x) = cα,d(|x|2 + α2)−(d+1)/2

  • R. Balan and D. Kim (University of Ottawa)

SPDE and Markov property Large Deviations, Ann Arbor 6 / 23

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The Framework The stochastic integral

Outline

1

History of the problem

2

The Framework The noise The stochastic integral The equation and its solution

3

RKHS General characterization Bessel kernel Riesz kernel

4

Germ Markov property The necessary and sufficient condition Main result

  • R. Balan and D. Kim (University of Ottawa)

SPDE and Markov property Large Deviations, Ann Arbor 7 / 23

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The Framework The stochastic integral

Stochastic integral with respect to M

Space of deterministic integrands P(d) is the completion of D((0, T) × Rd) with respect to ·, ·0 (This is a space of distributions in x !) Stochastic integral M(ϕ) = T

  • Rd ϕ(s, x)M(ds, dx)

is defined an an isometry ϕ → M(ϕ) between P(d) and the Gaussian space HM: EM(ϕ)M(ψ) = ϕ, ψ0, ∀ϕ, ψ ∈ P(d)

  • R. Balan and D. Kim (University of Ottawa)

SPDE and Markov property Large Deviations, Ann Arbor 8 / 23

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The Framework The equation and its solution

Outline

1

History of the problem

2

The Framework The noise The stochastic integral The equation and its solution

3

RKHS General characterization Bessel kernel Riesz kernel

4

Germ Markov property The necessary and sufficient condition Main result

  • R. Balan and D. Kim (University of Ottawa)

SPDE and Markov property Large Deviations, Ann Arbor 9 / 23

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The Framework The equation and its solution

Stochastic heat equation

Stochastic heat equation driven by ˙ M

  • ut = ∆u + ˙

M in [0, T] × Rd u(0, x) = 0 . Mild solution u(t, x) = t

  • Rd G(t − s, x − y) M(ds, dy),

where G(t, x) = (4πt)−d/2e−|x|2/(4t), t > 0, x ∈ Rd Remark: G(t − ·, x − ·) ∈ P(d) if and only if

  • Rd(1 + |ξ|2)−1 µ(dξ) < ∞.
  • R. Balan and D. Kim (University of Ottawa)

SPDE and Markov property Large Deviations, Ann Arbor 10 / 23

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The Framework The equation and its solution

Stochastic heat equation

Stochastic heat equation driven by ˙ M

  • ut = ∆u + ˙

M in [0, T] × Rd u(0, x) = 0 . Mild solution u(t, x) = t

  • Rd G(t − s, x − y) M(ds, dy),

where G(t, x) = (4πt)−d/2e−|x|2/(4t), t > 0, x ∈ Rd Remark: G(t − ·, x − ·) ∈ P(d) if and only if

  • Rd(1 + |ξ|2)−1 µ(dξ) < ∞.
  • R. Balan and D. Kim (University of Ottawa)

SPDE and Markov property Large Deviations, Ann Arbor 10 / 23

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The Framework The equation and its solution

Stochastic heat equation

Stochastic heat equation driven by ˙ M

  • ut = ∆u + ˙

M in [0, T] × Rd u(0, x) = 0 . Mild solution u(t, x) = t

  • Rd G(t − s, x − y) M(ds, dy),

where G(t, x) = (4πt)−d/2e−|x|2/(4t), t > 0, x ∈ Rd Remark: G(t − ·, x − ·) ∈ P(d) if and only if

  • Rd(1 + |ξ|2)−1 µ(dξ) < ∞.
  • R. Balan and D. Kim (University of Ottawa)

SPDE and Markov property Large Deviations, Ann Arbor 10 / 23

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RKHS General characterization

Outline

1

History of the problem

2

The Framework The noise The stochastic integral The equation and its solution

3

RKHS General characterization Bessel kernel Riesz kernel

4

Germ Markov property The necessary and sufficient condition Main result

  • R. Balan and D. Kim (University of Ottawa)

SPDE and Markov property Large Deviations, Ann Arbor 11 / 23

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RKHS General characterization

The Reproducing kernel Hilbert space Hu

Row of isometries P(d) → HM = Hu → Hu ϕ → M(ϕ) = Y → hY(t, x) = E(Yu(t, x)) Hu = span of {u(t, x) : (t, x) ∈ [0, T] × Rd} in L2(Ω) HM = {M(ϕ); ϕ ∈ P(d) } Definition of Hu: Hu = {h(t, x) = E(M(ϕ)u(t, x)) : ϕ ∈ P(d) } and h, gHu = E(M(ϕ)M(ψ)) = ϕ, ψ0,

  • R. Balan and D. Kim (University of Ottawa)

SPDE and Markov property Large Deviations, Ann Arbor 12 / 23

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RKHS General characterization

The Reproducing kernel Hilbert space Hu

Row of isometries P(d) → HM = Hu → Hu ϕ → M(ϕ) = Y → hY(t, x) = E(Yu(t, x)) Hu = span of {u(t, x) : (t, x) ∈ [0, T] × Rd} in L2(Ω) HM = {M(ϕ); ϕ ∈ P(d) } Definition of Hu: Hu = {h(t, x) = E(M(ϕ)u(t, x)) : ϕ ∈ P(d) } and h, gHu = E(M(ϕ)M(ψ)) = ϕ, ψ0,

  • R. Balan and D. Kim (University of Ottawa)

SPDE and Markov property Large Deviations, Ann Arbor 12 / 23

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RKHS General characterization

Attempt to characterize the elements of Hu

Formal calculation h(t, x) = EM(ϕ)u(t, x) = EM(ϕ)M(G(t − ·, x − ·)) = ϕ, G(t − ·, x − ·)0 = t

  • Rd
  • Rd G(t − s, x − y)f(y − z)ϕ(s, z) dy dz ds

= t

  • Rd G(t − s, x − y)ϕ1(s, y) dy,

where ϕ1(s, y) =

  • Rd ϕ(s, z)f(y − z) dz.

Intuitively, h should be a solution of:

  • ht = ∆h + ϕ1

in (0, T) × Rd h(0, x) = 0 .

  • R. Balan and D. Kim (University of Ottawa)

SPDE and Markov property Large Deviations, Ann Arbor 13 / 23

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RKHS General characterization

Attempt to characterize the elements of Hu

Formal calculation h(t, x) = EM(ϕ)u(t, x) = EM(ϕ)M(G(t − ·, x − ·)) = ϕ, G(t − ·, x − ·)0 = t

  • Rd
  • Rd G(t − s, x − y)f(y − z)ϕ(s, z) dy dz ds

= t

  • Rd G(t − s, x − y)ϕ1(s, y) dy,

where ϕ1(s, y) =

  • Rd ϕ(s, z)f(y − z) dz.

Intuitively, h should be a solution of:

  • ht = ∆h + ϕ1

in (0, T) × Rd h(0, x) = 0 .

  • R. Balan and D. Kim (University of Ottawa)

SPDE and Markov property Large Deviations, Ann Arbor 13 / 23

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RKHS Bessel kernel

Outline

1

History of the problem

2

The Framework The noise The stochastic integral The equation and its solution

3

RKHS General characterization Bessel kernel Riesz kernel

4

Germ Markov property The necessary and sufficient condition Main result

  • R. Balan and D. Kim (University of Ottawa)

SPDE and Markov property Large Deviations, Ann Arbor 14 / 23

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RKHS Bessel kernel

Characterization of Hu (Bessel kernel of order α)

Spaces of Bessel potentials Hγ

2 (Rd) = {(1 − ∆)−γ/2g; g ∈ L2(Rd)},

γ ∈ R Isometry P(d) ⊂ L2((0, T), H−α/2

2

(Rd)) → L2((0, T), Hα/2

2

(Rd)) ϕ → ϕ1 = (1 − ∆)−α/2ϕ Theorem A Let h(t, x) = EM(ϕ)u(t, x), ϕ ∈ P(d) . Then h is the unique solution in L2((0, T), Hα/2+2

2

(Rd)) of ht = ∆h + ϕ1, h(0, x) = 0.

  • R. Balan and D. Kim (University of Ottawa)

SPDE and Markov property Large Deviations, Ann Arbor 15 / 23

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RKHS Riesz kernel

Outline

1

History of the problem

2

The Framework The noise The stochastic integral The equation and its solution

3

RKHS General characterization Bessel kernel Riesz kernel

4

Germ Markov property The necessary and sufficient condition Main result

  • R. Balan and D. Kim (University of Ottawa)

SPDE and Markov property Large Deviations, Ann Arbor 16 / 23

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RKHS Riesz kernel

Characterization of Hu (Riesz kernel of order α = 4k)

Spaces of Riesz potentials K α/2(Rd) = {(−∆)−α/4g; g ∈ L2(Rd)} ⊂ Lq(Rd) where 1/q = 1/2 − α/(2d). K −α/2

2

(Rd) := {ϕ ∈ S′(Rd); Fϕ is a function,

  • Rd |Fϕ(ξ)|2|ξ|−αdξ < ∞}

= completion of S(Rd) w.r.t. · K −α/2

2

(Rd)

Maps: P(d) → L2((0, T) × Rd) → L2((0, T), K α/2

2

(Rd)) ϕ → ϕ0 = (−∆)−α/4ϕ → ϕ1 = (−∆)−α/4ϕ0

  • R. Balan and D. Kim (University of Ottawa)

SPDE and Markov property Large Deviations, Ann Arbor 17 / 23

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RKHS Riesz kernel

Characterization of Hu (Riesz kernel of order α)

Theorem B Let h(t, x) = EM(ϕ)u(t, x), ϕ ∈ P(d) . Then h is the unique solution in W 1,2

2,q ((0, T) × Rd) of

ht = ∆h + ϕ1, h(0, x) = 0, where W 1,2

2,q ((0, T) × Rd) is the space of functions u such that

u, ut, uxi, uxixj are in L2((0, T), Lq(Rd)).

  • R. Balan and D. Kim (University of Ottawa)

SPDE and Markov property Large Deviations, Ann Arbor 18 / 23

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Germ Markov property The necessary and sufficient condition

Outline

1

History of the problem

2

The Framework The noise The stochastic integral The equation and its solution

3

RKHS General characterization Bessel kernel Riesz kernel

4

Germ Markov property The necessary and sufficient condition Main result

  • R. Balan and D. Kim (University of Ottawa)

SPDE and Markov property Large Deviations, Ann Arbor 19 / 23

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Germ Markov property The necessary and sufficient condition

A fundamental result

  • H. Künsch, 1979

The Gaussian process u is germ Markov if and only if the following two conditions are satisfied: If h, g ∈ Hu are such that (supp h) ∩ (supp g) = ∅ and supp h is compact, then h, gHu = 0. If ζ = h + g ∈ Hu, where h and g are such that (supp h) ∩ (supp g) = ∅ and supp h is compact, then h, g ∈ Hu.

  • R. Balan and D. Kim (University of Ottawa)

SPDE and Markov property Large Deviations, Ann Arbor 20 / 23

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Germ Markov property Main result

Outline

1

History of the problem

2

The Framework The noise The stochastic integral The equation and its solution

3

RKHS General characterization Bessel kernel Riesz kernel

4

Germ Markov property The necessary and sufficient condition Main result

  • R. Balan and D. Kim (University of Ottawa)

SPDE and Markov property Large Deviations, Ann Arbor 21 / 23

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Germ Markov property Main result

Theorem

  • B. and Kim, 2006

The process solution u is germ Markov if: (i) f is the Bessel kernel of order α = 2k, k ∈ Z+; or (ii) f is the Riesz kernel of order α = 4k, k ∈ Z+ Idea of the Proof: h(t, x) = E(M(ϕ)u(t, x)), g(t, x) = E(M(ψ)u(t, x)), and (supp h) ∩ (supp g) = ∅.

  • ht = ∆h + ϕ1

h(0, x) = 0 ,

  • gt = ∆g + ψ1

g(0, x) = 0 . We need to prove that h, gHu = 0.

  • R. Balan and D. Kim (University of Ottawa)

SPDE and Markov property Large Deviations, Ann Arbor 22 / 23

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Germ Markov property Main result

Idea of the Proof: (cont’d)

h, gHu = ϕ, ψ0 = T

  • Rd
  • Rd ψ(s, x)f(x − y)ϕ(s, y) dx dy ds

= T

  • Rd ψ(s, x)ϕ1(s, x) dx ds.

Note that supp ϕ1 ⊂ supp h, supp ψ ⊂ supp ψ1 ⊂ supp g. since: (i) if f is the Bessel kernel of order α = 2k ψ(t, ·) = (1 − ∆)kψ1(t, ·) (ii) if f is the Riesz kernel of order α = 4k ψ(t, ·) = (−∆)2kψ1(t, ·)

  • R. Balan and D. Kim (University of Ottawa)

SPDE and Markov property Large Deviations, Ann Arbor 23 / 23