A Stochastic Heat Equation

A Stochastic Heat EquationRecall that F : R→ R is Lipschitz continuous ifLip(F ) := sup

−∞<x 6=y<∞

∣∣∣∣F (x)− F (y)x − y

∣∣∣∣ <∞.

Let b, σ : R→ R be nonrandom and Lipschitz continuous.We wish to solve the stochastic PDE [SPDE],u(t ,x) = u′′(t ,x) + b(u(t ,x)) + σ (u(t ,x))W (t ,x) on R+ × [0 , 1],

subject to:

I u(0) = u0 ∈ L2[0 , 1] where u0 is nonrandom;I u(t , 0) = u(t , 1) = 0 ∀t > 0.

Describes heat flow in a random environment [W ] with localinteraction/feedback [σ (u)].First, we need to give rigorous meaning to our SPDE.

July 28, 2016 1 / 17


−∞<x 6=y<∞

∣∣∣∣F (x)− F (y)x − y

∣∣∣∣ <∞.Let b, σ : R→ R be nonrandom and Lipschitz continuous.

We wish to solve the stochastic PDE [SPDE],u(t ,x) = u′′(t ,x) + b(u(t ,x)) + σ (u(t ,x))W (t ,x) on R+ × [0 , 1],

subject to:

I u(0) = u0 ∈ L2[0 , 1] where u0 is nonrandom;I u(t , 0) = u(t , 1) = 0 ∀t > 0.


July 28, 2016 1 / 17


−∞<x 6=y<∞

∣∣∣∣F (x)− F (y)x − y

∣∣∣∣ <∞.Let b, σ : R→ R be nonrandom and Lipschitz continuous.We wish to solve the stochastic PDE [SPDE],u(t ,x) = u′′(t ,x) + b(u(t ,x)) + σ (u(t ,x))W (t ,x) on R+ × [0 , 1],

subject to:

I u(0) = u0 ∈ L2[0 , 1] where u0 is nonrandom;I u(t , 0) = u(t , 1) = 0 ∀t > 0.Describes heat flow in a random environment [W ] with localinteraction/feedback [σ (u)].First, we need to give rigorous meaning to our SPDE.

July 28, 2016 1 / 17


−∞<x 6=y<∞

∣∣∣∣F (x)− F (y)x − y


subject to:I u(0) = u0 ∈ L2[0 , 1] where u0 is nonrandom;

I u(t , 0) = u(t , 1) = 0 ∀t > 0.Describes heat flow in a random environment [W ] with localinteraction/feedback [σ (u)].First, we need to give rigorous meaning to our SPDE.

July 28, 2016 1 / 17


−∞<x 6=y<∞

∣∣∣∣F (x)− F (y)x − y


subject to:I u(0) = u0 ∈ L2[0 , 1] where u0 is nonrandom;I u(t , 0) = u(t , 1) = 0 ∀t > 0.


July 28, 2016 1 / 17


−∞<x 6=y<∞

∣∣∣∣F (x)− F (y)x − y


subject to:I u(0) = u0 ∈ L2[0 , 1] where u0 is nonrandom;I u(t , 0) = u(t , 1) = 0 ∀t > 0.Describes heat flow in a random environment [W ] with localinteraction/feedback [σ (u)].

First, we need to give rigorous meaning to our SPDE.

July 28, 2016 1 / 17


−∞<x 6=y<∞

∣∣∣∣F (x)− F (y)x − y


subject to:I u(0) = u0 ∈ L2[0 , 1] where u0 is nonrandom;I u(t , 0) = u(t , 1) = 0 ∀t > 0.Describes heat flow in a random environment [W ] with localinteraction/feedback [σ (u)].First, we need to give rigorous meaning to our SPDE.

July 28, 2016 1 / 17

A Stochastic Heat Equationu = u′′ + b(u) + σ (u)W ; u(0) = u0 ∈ L2[0 , 1]; u(t , 0) = u(t , 1) = 0 ∀t > 0First pretend that W is a smooth function. Then ∃! solution ufrom general theory.

Apply Duhamel’s principle: u uniquely solves the integralequationu(t ,x) = (Ptu0)(x) + ∫[0,t]×[0,1] pt−s(x , y)b(u(s , y)) ds dy

+ ∫[0,t]×[0,1] pt−s(x , y)σ (u(s , y))W (s , y) ds dy.

Mild form of the solution.For us, the mild solution continues to make sense and is in facta weak solution.We care only about mild solutions in general.Let us study the simplest case first where b ≡ 0 and σ ≡ 1.

July 28, 2016 2 / 17

A Stochastic Heat Equationu = u′′ + b(u) + σ (u)W ; u(0) = u0 ∈ L2[0 , 1]; u(t , 0) = u(t , 1) = 0 ∀t > 0First pretend that W is a smooth function. Then ∃! solution ufrom general theory.Apply Duhamel’s principle: u uniquely solves the integralequation

u(t ,x) = (Ptu0)(x) + ∫[0,t]×[0,1] pt−s(x , y)b(u(s , y)) ds dy


Mild form of the solution.For us, the mild solution continues to make sense and is in facta weak solution.We care only about mild solutions in general.Let us study the simplest case first where b ≡ 0 and σ ≡ 1.

July 28, 2016 2 / 17




Mild form of the solution.

For us, the mild solution continues to make sense and is in facta weak solution.We care only about mild solutions in general.Let us study the simplest case first where b ≡ 0 and σ ≡ 1.

July 28, 2016 2 / 17




Mild form of the solution.For us, the mild solution continues to make sense and is in facta weak solution.

We care only about mild solutions in general.Let us study the simplest case first where b ≡ 0 and σ ≡ 1.

July 28, 2016 2 / 17




Mild form of the solution.For us, the mild solution continues to make sense and is in facta weak solution.We care only about mild solutions in general.

Let us study the simplest case first where b ≡ 0 and σ ≡ 1.

July 28, 2016 2 / 17




Mild form of the solution.For us, the mild solution continues to make sense and is in facta weak solution.We care only about mild solutions in general.Let us study the simplest case first where b ≡ 0 and σ ≡ 1.July 28, 2016 2 / 17

A Stochastic Heat Equationu = u′′ + W ; u(0) ≡ 0; u(t , 0) = u(t , 1) = 0 ∀t > 0

The mild solution is, by definition,u(t ,x) = ∫[0,t]×[0,1] pt−s(x , y) dW (s , y),

provided that the Wiener integral exists; i.e.,∫ t0 ds∫ 10 dy |pt−s(x , y)|2 <∞.

Because pt−s(x , y) =∑∞n=1 en(x)en(y) exp{−n2π2(t − s)},

∫ t

0 ds∫ 1

0 dy |pt−s(x , y)|2 = ∫ t

0 ds∞∑

n=1 |en(x)|2e−2n2π2(t−s)

6 2 ∫ t

0 ds∞∑

n=1 e−2n2π2(t−s) - ∞∑n=1 n−2 <∞.

July 28, 2016 3 / 17



provided that the Wiener integral exists; i.e.,∫ t0 ds∫ 10 dy |pt−s(x , y)|2 <∞.Because pt−s(x , y) =∑∞n=1 en(x)en(y) exp{−n2π2(t − s)},

∫ t

0 ds∫ 1

0 dy |pt−s(x , y)|2 = ∫ t

0 ds∞∑

n=1 |en(x)|2e−2n2π2(t−s)

6 2 ∫ t

0 ds∞∑

n=1 e−2n2π2(t−s) - ∞∑n=1 n−2 <∞.

July 28, 2016 3 / 17



provided that the Wiener integral exists; i.e.,∫ t0 ds∫ 10 dy |pt−s(x , y)|2 <∞.Because pt−s(x , y) =∑∞n=1 en(x)en(y) exp{−n2π2(t − s)},∫ t

0 ds∫ 1

0 dy |pt−s(x , y)|2 = ∫ t

0 ds∞∑

n=1 |en(x)|2e−2n2π2(t−s)

6 2 ∫ t

0 ds∞∑

n=1 e−2n2π2(t−s) - ∞∑n=1 n−2 <∞.

July 28, 2016 3 / 17




0 ds∫ 1

0 dy |pt−s(x , y)|2 = ∫ t

0 ds∞∑

n=1 |en(x)|2e−2n2π2(t−s)

6 2 ∫ t

0 ds∞∑

n=1 e−2n2π2(t−s)

-∞∑

n=1 n−2 <∞.

July 28, 2016 3 / 17




0 ds∫ 1

0 dy |pt−s(x , y)|2 = ∫ t

0 ds∞∑

n=1 |en(x)|2e−2n2π2(t−s)

6 2 ∫ t

0 ds∞∑

n=1 e−2n2π2(t−s) - ∞∑n=1 n−2 <∞.

July 28, 2016 3 / 17

A Stochastic Heat Equationu = u′′ + W ; u(0) ≡ 0; u(t , 0) = u(t , 1) = 0 ∀t > 0Therefore the SPDE above has a mild solution u(t ,x).

That solution is the mean-zero Gaussian random fieldu(t ,x) = ∫[0,t]×[0,1] pt−s(x , y) dW (s , y).

Theorem(t ,x) 7Ï u(t ,x) is continuous a.s. [up to a modification]. In fact,u ∈ C 14−, 12− a.s.

Outline of Proof (in 2 steps).

1 One proves that for all x, y ∈ [0 , 1], s, t > 0, k > 2,E(|u(t ,x)− u(s , y)|k) 6 Ck

[|t − s|k/4 + |x − y|k/2] .

2 Appeal to a two-parameter version of the Kolmogorov continuitytheorem.

July 28, 2016 4 / 17

A Stochastic Heat Equationu = u′′ + W ; u(0) ≡ 0; u(t , 0) = u(t , 1) = 0 ∀t > 0Therefore the SPDE above has a mild solution u(t ,x).That solution is the mean-zero Gaussian random field

u(t ,x) = ∫[0,t]×[0,1] pt−s(x , y) dW (s , y).

Theorem(t ,x) 7Ï u(t ,x) is continuous a.s. [up to a modification]. In fact,u ∈ C 14−, 12− a.s.



[|t − s|k/4 + |x − y|k/2] .


July 28, 2016 4 / 17


u(t ,x) = ∫[0,t]×[0,1] pt−s(x , y) dW (s , y).Theorem(t ,x) 7Ï u(t ,x) is continuous a.s. [up to a modification]. In fact,u ∈ C 14−, 12− a.s.



[|t − s|k/4 + |x − y|k/2] .


July 28, 2016 4 / 17



Outline of Proof (in 2 steps).1 One proves that for all x, y ∈ [0 , 1], s, t > 0, k > 2,

E(|u(t ,x)− u(s , y)|k) 6 Ck

[|t − s|k/4 + |x − y|k/2] .


July 28, 2016 4 / 17



Outline of Proof (in 2 steps).1 One proves that for all x, y ∈ [0 , 1], s, t > 0, k > 2,

E(|u(t ,x)− u(s , y)|k) 6 Ck

[|t − s|k/4 + |x − y|k/2] .

2 Appeal to a two-parameter version of the Kolmogorov continuitytheorem.July 28, 2016 4 / 17



T1+2k E(|u(t , y)− u(s , y)|k)︸︷︷︸

T2.

Fact. T1 6 Ck|x − y|k/2 and T2 6 Ck|t − s|k/4.

Let us begin with the bound on T1.Idea. If Z is Gaussian with mean zero and SD σ , thenE(|Z|k) = Akσk.Therefore, it suffices to prove thatE(|u(t ,x)− u(t , y)|2) - |x − y|.

July 28, 2016 5 / 17



T1+2k E(|u(t , y)− u(s , y)|k)︸︷︷︸

T2.

Fact. T1 6 Ck|x − y|k/2 and T2 6 Ck|t − s|k/4.Let us begin with the bound on T1.

Idea. If Z is Gaussian with mean zero and SD σ , thenE(|Z|k) = Akσk.Therefore, it suffices to prove thatE(|u(t ,x)− u(t , y)|2) - |x − y|.

July 28, 2016 5 / 17



T1+2k E(|u(t , y)− u(s , y)|k)︸︷︷︸

T2.

Fact. T1 6 Ck|x − y|k/2 and T2 6 Ck|t − s|k/4.Let us begin with the bound on T1.Idea. If Z is Gaussian with mean zero and SD σ , thenE(|Z|k) = Akσk.

Therefore, it suffices to prove thatE(|u(t ,x)− u(t , y)|2) - |x − y|.

July 28, 2016 5 / 17



T1+2k E(|u(t , y)− u(s , y)|k)︸︷︷︸

T2.

Fact. T1 6 Ck|x − y|k/2 and T2 6 Ck|t − s|k/4.Let us begin with the bound on T1.Idea. If Z is Gaussian with mean zero and SD σ , thenE(|Z|k) = Akσk.Therefore, it suffices to prove thatE(|u(t ,x)− u(t , y)|2) - |x − y|.

July 28, 2016 5 / 17

A Stochastic Heat Equationu(t ,x) = ∫[0,t]×[0,1] pt−s(x ,a) dW (s ,a)

Now,E(|u(t ,x)− u(t , y)|2) = ∫[0,t]×[0,1] |pt−s(x ,a)− pt−s(y ,a)|2 ds da

= ∫[0,t]×[0,1] |ps(x ,a)− ps(y ,a)|2 ds da.

Since ps(x ,a) =∑∞n=1 en(x)en(a)e−n2π2t ,∫ 10 |ps(x ,a)− ps(y ,a)|2 da = ∞∑

n=1 |en(x)− en(y)|2 |en(a)|2e−2n2π2s.

|en(a)| 6 √2 and |en(x)− en(y)| 6 √2 min{2 ,nπ|x − y|}.Compute.

July 28, 2016 6 / 17



= ∫[0,t]×[0,1] |ps(x ,a)− ps(y ,a)|2 ds da.


n=1 |en(x)− en(y)|2 |en(a)|2e−2n2π2s.

|en(a)| 6 √2 and |en(x)− en(y)| 6 √2 min{2 ,nπ|x − y|}.

Compute.

July 28, 2016 6 / 17



= ∫[0,t]×[0,1] |ps(x ,a)− ps(y ,a)|2 ds da.


n=1 |en(x)− en(y)|2 |en(a)|2e−2n2π2s.

|en(a)| 6 √2 and |en(x)− en(y)| 6 √2 min{2 ,nπ|x − y|}.Compute.July 28, 2016 6 / 17

A Stochastic Heat Equationu(t ,x) = ∫[0,t]×[0,1] pt−s(x ,a) dW (s ,a)WLOG 0 < s < t. We next estimate T2 = E(|u(t , y)− u(s , y)|k)by writing

T2 6 2kT2,1 + 2kT2,2,where:

I T2,1 := ∫[0,s]×[0,1] (pt−r(y ,a)− ps−r(r ,a)) dW (r ,a); andI T2,2 := ∫[s,t]×[0,1] pt−r(y ,a) dW (r ,a).By Wiener’s isometry,

E(T22,2) = ∫ t

sdr∫ 1

0 da |pt−r(y ,a)|2

= ∫ t−s

0 dr∫ 1

0 da pr(y ,a)pr(a , y) = ∫ t−s

0 p2r(y , y) dr

= ∞∑n=1

∫ t−s

0 |en(y)|2e−2n2π2r dr 6 2 ∞∑n=1

∫ t−s

0 e−2n2π2r dr.

July 28, 2016 7 / 17


T2 6 2kT2,1 + 2kT2,2,where:I T2,1 := ∫[0,s]×[0,1] (pt−r(y ,a)− ps−r(r ,a)) dW (r ,a); and

I T2,2 := ∫[s,t]×[0,1] pt−r(y ,a) dW (r ,a).By Wiener’s isometry,E(T22,2) = ∫ t

sdr∫ 1

0 da |pt−r(y ,a)|2

= ∫ t−s

0 dr∫ 1


0 p2r(y , y) dr

= ∞∑n=1

∫ t−s

0 |en(y)|2e−2n2π2r dr 6 2 ∞∑n=1

∫ t−s

0 e−2n2π2r dr.

July 28, 2016 7 / 17


T2 6 2kT2,1 + 2kT2,2,where:I T2,1 := ∫[0,s]×[0,1] (pt−r(y ,a)− ps−r(r ,a)) dW (r ,a); andI T2,2 := ∫[s,t]×[0,1] pt−r(y ,a) dW (r ,a).

By Wiener’s isometry,E(T22,2) = ∫ t

sdr∫ 1

0 da |pt−r(y ,a)|2

= ∫ t−s

0 dr∫ 1


0 p2r(y , y) dr

= ∞∑n=1

∫ t−s

0 |en(y)|2e−2n2π2r dr 6 2 ∞∑n=1

∫ t−s

0 e−2n2π2r dr.

July 28, 2016 7 / 17


T2 6 2kT2,1 + 2kT2,2,where:I T2,1 := ∫[0,s]×[0,1] (pt−r(y ,a)− ps−r(r ,a)) dW (r ,a); andI T2,2 := ∫[s,t]×[0,1] pt−r(y ,a) dW (r ,a).By Wiener’s isometry,

E(T22,2) = ∫ t

sdr∫ 1

0 da |pt−r(y ,a)|2

= ∫ t−s

0 dr∫ 1


0 p2r(y , y) dr

= ∞∑n=1

∫ t−s

0 |en(y)|2e−2n2π2r dr 6 2 ∞∑n=1

∫ t−s

0 e−2n2π2r dr.

July 28, 2016 7 / 17



E(T22,2) = ∫ t

sdr∫ 1

0 da |pt−r(y ,a)|2= ∫ t−s

0 dr∫ 1

0 da pr(y ,a)pr(a , y)

= ∫ t−s

0 p2r(y , y) dr

= ∞∑n=1

∫ t−s

0 |en(y)|2e−2n2π2r dr 6 2 ∞∑n=1

∫ t−s

0 e−2n2π2r dr.

July 28, 2016 7 / 17



E(T22,2) = ∫ t

sdr∫ 1

0 da |pt−r(y ,a)|2= ∫ t−s

0 dr∫ 1


0 p2r(y , y) dr

= ∞∑n=1

∫ t−s

0 |en(y)|2e−2n2π2r dr 6 2 ∞∑n=1

∫ t−s

0 e−2n2π2r dr.

July 28, 2016 7 / 17



E(T22,2) = ∫ t

sdr∫ 1

0 da |pt−r(y ,a)|2= ∫ t−s

0 dr∫ 1


0 p2r(y , y) dr

= ∞∑n=1

∫ t−s

0 |en(y)|2e−2n2π2r dr

6 2 ∞∑n=1

∫ t−s

0 e−2n2π2r dr.

July 28, 2016 7 / 17



E(T22,2) = ∫ t

sdr∫ 1

0 da |pt−r(y ,a)|2= ∫ t−s

0 dr∫ 1


0 p2r(y , y) dr

= ∞∑n=1

∫ t−s

0 |en(y)|2e−2n2π2r dr 6 2 ∞∑n=1

∫ t−s

0 e−2n2π2r dr.

July 28, 2016 7 / 17

A Stochastic Heat Equationu(t ,x) = ∫[0,t]×[0,1] pt−s(x ,a) dW (s ,a)Therefore,

E(T22,2) 6 2 ∞∑n=1

∫ t−s

0 e−2n2π2r dr

= ∞∑n=1

1− e−2n2π2(t−s)n2π2

6∞∑

n=1min (1 , 2n2π2(t − s))

n2π2 · · · - (t − s)1/2.Similarly,

E(T22,1) = ∫ s

0 dr∫ 1

0 da [pt−r(y ,a)− ps−r(y ,a)]2= ∞∑

n=1∫ s

0 dr[e−n2π2(t−s+r) − e−n2π2r

]2|en(y)en(a)|2

6 2 ∞∑n=1

∫ s

0[e−n2π2(t−s+r) − e−n2π2r

]2dr.

July 28, 2016 8 / 17


E(T22,2) 6 2 ∞∑n=1

∫ t−s

0 e−2n2π2r dr = ∞∑n=1

1− e−2n2π2(t−s)n2π2

6∞∑

n=1min (1 , 2n2π2(t − s))

n2π2 · · · - (t − s)1/2.Similarly,

E(T22,1) = ∫ s

0 dr∫ 1

0 da [pt−r(y ,a)− ps−r(y ,a)]2= ∞∑

n=1∫ s


]2|en(y)en(a)|2

6 2 ∞∑n=1

∫ s


]2dr.

July 28, 2016 8 / 17


E(T22,2) 6 2 ∞∑n=1

∫ t−s

0 e−2n2π2r dr = ∞∑n=1

1− e−2n2π2(t−s)n2π2

6∞∑

n=1min (1 , 2n2π2(t − s))

n2π2

· · · - (t − s)1/2.Similarly,

E(T22,1) = ∫ s

0 dr∫ 1

0 da [pt−r(y ,a)− ps−r(y ,a)]2= ∞∑

n=1∫ s


]2|en(y)en(a)|2

6 2 ∞∑n=1

∫ s


]2dr.

July 28, 2016 8 / 17


E(T22,2) 6 2 ∞∑n=1

∫ t−s

0 e−2n2π2r dr = ∞∑n=1

1− e−2n2π2(t−s)n2π2

6∞∑

n=1min (1 , 2n2π2(t − s))

n2π2 · · · - (t − s)1/2.

Similarly,

E(T22,1) = ∫ s

0 dr∫ 1

0 da [pt−r(y ,a)− ps−r(y ,a)]2= ∞∑

n=1∫ s


]2|en(y)en(a)|2

6 2 ∞∑n=1

∫ s


]2dr.

July 28, 2016 8 / 17


E(T22,2) 6 2 ∞∑n=1

∫ t−s

0 e−2n2π2r dr = ∞∑n=1

1− e−2n2π2(t−s)n2π2

6∞∑

n=1min (1 , 2n2π2(t − s))

n2π2 · · · - (t − s)1/2.Similarly,

E(T22,1) = ∫ s

0 dr∫ 1

0 da [pt−r(y ,a)− ps−r(y ,a)]2

= ∞∑n=1

∫ s


]2|en(y)en(a)|2

6 2 ∞∑n=1

∫ s


]2dr.

July 28, 2016 8 / 17


E(T22,2) 6 2 ∞∑n=1

∫ t−s

0 e−2n2π2r dr = ∞∑n=1

1− e−2n2π2(t−s)n2π2

6∞∑

n=1min (1 , 2n2π2(t − s))

n2π2 · · · - (t − s)1/2.Similarly,

E(T22,1) = ∫ s

0 dr∫ 1

0 da [pt−r(y ,a)− ps−r(y ,a)]2= ∞∑

n=1∫ s


]2|en(y)en(a)|2

6 2 ∞∑n=1

∫ s


]2dr.

July 28, 2016 8 / 17

A Stochastic Heat Equationu(t ,x) = ∫[0,t]×[0,1] pt−s(x ,a) dW (s ,a)Thus,

E(T22,1) 6 2 ∞∑n=1

∫ s


]2dr

= 2 ∞∑n=1

∫ s

0[1− e−n2π2(t−s)]2

e−2n2π2r dr

= ∞∑n=1

[1− e−n2π2(t−s)]2 1− e−2n2π2(t−s)n2π2

6∞∑

n=1 min(1 ,n4π4(t − s)2) min (1 , 2n2π2(t − s))n2π2

· · · - (t − s)1/2.Combine T2,1+T2,2 to see thatT2 = E(|u(t , y)− u(s , y)|2) - (t − s)1/2.

July 28, 2016 9 / 17

A Stochastic Heat Equationu(t ,x) = ∫[0,t]×[0,1] pt−s(x , y) dW (s , y)

Large-time behavior of x 7Ï u(t ,x)?

Spatial covariance function: ∀t > 0 and 0 6 x, z 6 1,E [u(t ,x)u(t , z)] = ∫[0,t]×[0,1] pt−s(x , y)pt−s(z , y) ds dy

= ∫[0,t]×[0,1] ps(x , y)ps(y , z) ds dy

= ∫ t

0 p2s(x , z) ds

= ∫ t

0 ds∞∑

n=1 en(x)en(z)e−2n2π2s

t→∞−−−Ï∞∑

n=1en(x)en(z)2n2π2 .

We will simplify this sum next.

July 28, 2016 10 / 17


Large-time behavior of x 7Ï u(t ,x)?Spatial covariance function: ∀t > 0 and 0 6 x, z 6 1,E [u(t ,x)u(t , z)] = ∫[0,t]×[0,1] pt−s(x , y)pt−s(z , y) ds dy

= ∫[0,t]×[0,1] ps(x , y)ps(y , z) ds dy

= ∫ t

0 p2s(x , z) ds

= ∫ t

0 ds∞∑


t→∞−−−Ï∞∑



July 28, 2016 10 / 17



= ∫[0,t]×[0,1] ps(x , y)ps(y , z) ds dy = ∫ t

0 p2s(x , z) ds

= ∫ t

0 ds∞∑


t→∞−−−Ï∞∑



July 28, 2016 10 / 17



= ∫[0,t]×[0,1] ps(x , y)ps(y , z) ds dy = ∫ t

0 p2s(x , z) ds

= ∫ t

0 ds∞∑

n=1 en(x)en(z)e−2n2π2s t→∞−−−Ï∞∑



July 28, 2016 10 / 17



= ∫[0,t]×[0,1] ps(x , y)ps(y , z) ds dy = ∫ t

0 p2s(x , z) ds

= ∫ t

0 ds∞∑

n=1 en(x)en(z)e−2n2π2s t→∞−−−Ï∞∑


We will simplify this sum next.July 28, 2016 10 / 17

A Stochastic Heat Equation∑∞n=1 en (x)en (z)2n2π2 =?Let c0 ≡ 1 and cn(a) := √2 cos(nπa), n > 1, a ∈ [0 , 1]. Then,(

1[0,z] , cn) = √2 ∫ z

0 cos(nπa) da = √2 sin(nπz)nπ = en(z)

nπ ,

∀n > 1 and z ∈ [0 , 1].

Therefore,∞∑

n=1en(x)en(z)2n2π2 = ∞∑

n=1(1[0,x]√2 , cn

)(1[0,z]√2 , cn

)

= ∞∑n=0

(1[0,x]√2 , cn

)(1[0,z]√2 , cn

)−(1[0,x]√2 , c0

)(1[0,z]√2 , c0

)= (1[0,x]√2 ,

1[0,z]√2)− xz2

= 12 [min(x , y)− xy] .

July 28, 2016 11 / 17


1[0,z] , cn) = √2 ∫ z


nπ ,

∀n > 1 and z ∈ [0 , 1].Therefore,∞∑

n=1en(x)en(z)2n2π2 = ∞∑

n=1(1[0,x]√2 , cn

)(1[0,z]√2 , cn

)

= ∞∑n=0

(1[0,x]√2 , cn

)(1[0,z]√2 , cn

)−(1[0,x]√2 , c0

)(1[0,z]√2 , c0

)= (1[0,x]√2 ,

1[0,z]√2)− xz2

= 12 [min(x , y)− xy] .

July 28, 2016 11 / 17


1[0,z] , cn) = √2 ∫ z


nπ ,


n=1en(x)en(z)2n2π2 = ∞∑

n=1(1[0,x]√2 , cn

)(1[0,z]√2 , cn

)= ∞∑

n=0(1[0,x]√2 , cn

)(1[0,z]√2 , cn

)−(1[0,x]√2 , c0

)(1[0,z]√2 , c0

)

= (1[0,x]√2 ,1[0,z]√2

)− xz2

= 12 [min(x , y)− xy] .

July 28, 2016 11 / 17


1[0,z] , cn) = √2 ∫ z


nπ ,


n=1en(x)en(z)2n2π2 = ∞∑

n=1(1[0,x]√2 , cn

)(1[0,z]√2 , cn

)= ∞∑

n=0(1[0,x]√2 , cn

)(1[0,z]√2 , cn

)−(1[0,x]√2 , c0

)(1[0,z]√2 , c0

)= (1[0,x]√2 ,

1[0,z]√2)− xz2

= 12 [min(x , y)− xy] .

July 28, 2016 11 / 17


1[0,z] , cn) = √2 ∫ z


nπ ,


n=1en(x)en(z)2n2π2 = ∞∑

n=1(1[0,x]√2 , cn

)(1[0,z]√2 , cn

)= ∞∑

n=0(1[0,x]√2 , cn

)(1[0,z]√2 , cn

)−(1[0,x]√2 , c0

)(1[0,z]√2 , c0

)= (1[0,x]√2 ,

1[0,z]√2)− xz2 = 12 [min(x , y)− xy] .

July 28, 2016 11 / 17

A Stochastic Heat Equationu(t ,x) = ∫[0,t]×[0,1] pt−s(x , y) dW (s , y)We have shown thus far that

limt→∞

E [u(t ,x)u(t , z)] = 12 [min(x , z)− xz] .

The Brownian bridge W0(x) := [W (x)− xW (1)]/√2 is amean-zero Gaussian process withE [W0(x)W0(z)] = 12 [min(x , z)− xz] .

Therefore, it follows that u(t) t→∞===ÑW0 in the sense offinite-dimensional distributions.Tightness follows from our earlier estimateE(|u(t ,x)− u(t , y)|2) - |x − y|1/2.

July 28, 2016 12 / 17


limt→∞

E [u(t ,x)u(t , z)] = 12 [min(x , z)− xz] .The Brownian bridge W0(x) := [W (x)− xW (1)]/√2 is amean-zero Gaussian process with

E [W0(x)W0(z)] = 12 [min(x , z)− xz] .

Therefore, it follows that u(t) t→∞===ÑW0 in the sense offinite-dimensional distributions.Tightness follows from our earlier estimateE(|u(t ,x)− u(t , y)|2) - |x − y|1/2.

July 28, 2016 12 / 17


limt→∞


E [W0(x)W0(z)] = 12 [min(x , z)− xz] .Therefore, it follows that u(t) t→∞===ÑW0 in the sense offinite-dimensional distributions.

Tightness follows from our earlier estimateE(|u(t ,x)− u(t , y)|2) - |x − y|1/2.

July 28, 2016 12 / 17


limt→∞


E [W0(x)W0(z)] = 12 [min(x , z)− xz] .Therefore, it follows that u(t) t→∞===ÑW0 in the sense offinite-dimensional distributions.Tightness follows from our earlier estimate

E(|u(t ,x)− u(t , y)|2) - |x − y|1/2.July 28, 2016 12 / 17

A Stochastic Heat EquationTheorem (Funaki)Let u denote the solution to the SPDE

u = u′′ + W on R+ × [0 , 1],subject to:

u(0) ≡ 0;u(t , 0) = u(t , 1) = 0 ∀t > 0.

Then, u(t) C[0 ,1]===ÑW0 as t →∞.

If v solves the same SPDE subject to v(0) = f ∈ L2[0 , 1], thenv(t ,x) = (Ptf )(x) + u(t ,x).As t →∞, (Ptf )(x) =∑∞n=1(f ,en)en(x)e−n2π2t → 0 in L2[0 , 1].Therefore, v(t) L2[0 ,1]====ÑW0 as t →∞ ∀v(0) ∈ L2[0 , 1]An infinite-dimensional ergodic theorem.

July 28, 2016 13 / 17



u(0) ≡ 0;

u(t , 0) = u(t , 1) = 0 ∀t > 0.

Then, u(t) C[0 ,1]===ÑW0 as t →∞.


July 28, 2016 13 / 17



u(0) ≡ 0;u(t , 0) = u(t , 1) = 0 ∀t > 0.

Then, u(t) C[0 ,1]===ÑW0 as t →∞.


July 28, 2016 13 / 17



u(0) ≡ 0;u(t , 0) = u(t , 1) = 0 ∀t > 0.

Then, u(t) C[0 ,1]===ÑW0 as t →∞.

If v solves the same SPDE subject to v(0) = f ∈ L2[0 , 1], thenv(t ,x) = (Ptf )(x) + u(t ,x).

As t →∞, (Ptf )(x) =∑∞n=1(f ,en)en(x)e−n2π2t → 0 in L2[0 , 1].Therefore, v(t) L2[0 ,1]====ÑW0 as t →∞ ∀v(0) ∈ L2[0 , 1]An infinite-dimensional ergodic theorem.

July 28, 2016 13 / 17



u(0) ≡ 0;u(t , 0) = u(t , 1) = 0 ∀t > 0.

Then, u(t) C[0 ,1]===ÑW0 as t →∞.

If v solves the same SPDE subject to v(0) = f ∈ L2[0 , 1], thenv(t ,x) = (Ptf )(x) + u(t ,x).

As t →∞, (Ptf )(x) =∑∞n=1(f ,en)en(x)e−n2π2t → 0 in L2[0 , 1].

Therefore, v(t) L2[0 ,1]====ÑW0 as t →∞ ∀v(0) ∈ L2[0 , 1]An infinite-dimensional ergodic theorem.

July 28, 2016 13 / 17



u(0) ≡ 0;u(t , 0) = u(t , 1) = 0 ∀t > 0.

Then, u(t) C[0 ,1]===ÑW0 as t →∞.

If v solves the same SPDE subject to v(0) = f ∈ L2[0 , 1], thenv(t ,x) = (Ptf )(x) + u(t ,x).As t →∞, (Ptf )(x) =∑∞n=1(f ,en)en(x)e−n2π2t → 0 in L2[0 , 1].

Therefore, v(t) L2[0 ,1]====ÑW0 as t →∞ ∀v(0) ∈ L2[0 , 1]

An infinite-dimensional ergodic theorem.

July 28, 2016 13 / 17



u(0) ≡ 0;u(t , 0) = u(t , 1) = 0 ∀t > 0.

Then, u(t) C[0 ,1]===ÑW0 as t →∞.

If v solves the same SPDE subject to v(0) = f ∈ L2[0 , 1], thenv(t ,x) = (Ptf )(x) + u(t ,x).As t →∞, (Ptf )(x) =∑∞n=1(f ,en)en(x)e−n2π2t → 0 in L2[0 , 1].Therefore, v(t) L2[0 ,1]====ÑW0 as t →∞ ∀v(0) ∈ L2[0 , 1]

An infinite-dimensional ergodic theorem.

July 28, 2016 13 / 17



u(0) ≡ 0;u(t , 0) = u(t , 1) = 0 ∀t > 0.

Then, u(t) C[0 ,1]===ÑW0 as t →∞.


July 28, 2016 13 / 17

A Stochastic Heat Equation

July 28, 2016 14 / 17


Once again consider u = u′′ + W on (0 ,∞)× [0 , 1] withDirichlet zero boundary and initial value zero.

Since pt−s(x , y) =∑∞n=1 en(x)en(y)e−n2π2(t−s),u(t ,x) = ∞∑

n=1 en(x)∫[0,t]×[0,1] en(y)e−n2π2(t−s) dW (s , y)

= √2 ∞∑n=1 sin(nπx)∫ t

0 e−n2π2(t−s) dXn(s)︸︷︷︸=Yn(t).

dYn(t) = dXn(t)− n2π2Yn(t) dt.Yn ’s are Ornstein–Uhlenbeck processes.

July 28, 2016 15 / 17


Once again consider u = u′′ + W on (0 ,∞)× [0 , 1] withDirichlet zero boundary and initial value zero.Since pt−s(x , y) =∑∞n=1 en(x)en(y)e−n2π2(t−s),u(t ,x) = ∞∑

n=1 en(x) ∫[0,t]×[0,1] en(y)e−n2π2(t−s) dW (s , y)

= √2 ∞∑n=1 sin(nπx)∫ t

0 e−n2π2(t−s) dXn(s)︸︷︷︸=Yn(t).


July 28, 2016 15 / 17



n=1 en(x) ∫[0,t]×[0,1] en(y)e−n2π2(t−s) dW (s , y)= √2 ∞∑

n=1 sin(nπx)∫ t

0 e−n2π2(t−s) dXn(s)︸︷︷︸=Yn(t).


July 28, 2016 15 / 17




n=1 sin(nπx)∫ t

0 e−n2π2(t−s) dXn(s)︸︷︷︸=Yn(t).

dYn(t) = dXn(t)− n2π2Yn(t) dt.

Yn ’s are Ornstein–Uhlenbeck processes.

July 28, 2016 15 / 17




n=1 sin(nπx)∫ t

0 e−n2π2(t−s) dXn(s)︸︷︷︸=Yn(t).


July 28, 2016 15 / 17

General Remarks on SPDEsu = u′′ + WOther constant-coefficient SPDEs of interest include:

I Different boundary conditions [Neumann, mixed, . . . ].I u = Au + W , where A = generator of a nice Markov process.I u = Au + W [Stoch. Wave Eq.].I u = Au + F for a different noise model.I Etc. . . .Instead we return to the nonlinear/semilinear SPDEs of theform

u = u′′ + b(u) + σ (u)W .First-order questions:

1 Existence.2 Uniqueness.3 Local regularity.

Questions about the structure of the solution, if any ∃.

July 28, 2016 16 / 17


I Different boundary conditions [Neumann, mixed, . . . ].

I u = Au + W , where A = generator of a nice Markov process.I u = Au + W [Stoch. Wave Eq.].I u = Au + F for a different noise model.I Etc. . . .Instead we return to the nonlinear/semilinear SPDEs of theform




July 28, 2016 16 / 17


I Different boundary conditions [Neumann, mixed, . . . ].I u = Au + W , where A = generator of a nice Markov process.

I u = Au + W [Stoch. Wave Eq.].I u = Au + F for a different noise model.I Etc. . . .Instead we return to the nonlinear/semilinear SPDEs of theform




July 28, 2016 16 / 17


I Different boundary conditions [Neumann, mixed, . . . ].I u = Au + W , where A = generator of a nice Markov process.I u = Au + W [Stoch. Wave Eq.].

I u = Au + F for a different noise model.I Etc. . . .Instead we return to the nonlinear/semilinear SPDEs of theform




July 28, 2016 16 / 17


I Different boundary conditions [Neumann, mixed, . . . ].I u = Au + W , where A = generator of a nice Markov process.I u = Au + W [Stoch. Wave Eq.].I u = Au + F for a different noise model.

I Etc. . . .Instead we return to the nonlinear/semilinear SPDEs of theformu = u′′ + b(u) + σ (u)W .

First-order questions:



July 28, 2016 16 / 17


I Different boundary conditions [Neumann, mixed, . . . ].I u = Au + W , where A = generator of a nice Markov process.I u = Au + W [Stoch. Wave Eq.].I u = Au + F for a different noise model.I Etc. . . .

Instead we return to the nonlinear/semilinear SPDEs of theformu = u′′ + b(u) + σ (u)W .




July 28, 2016 16 / 17



u = u′′ + b(u) + σ (u)W .




July 28, 2016 16 / 17




1 Existence.2 Uniqueness.3 Local regularity.Questions about the structure of the solution, if any ∃.

July 28, 2016 16 / 17




1 Existence.

2 Uniqueness.3 Local regularity.Questions about the structure of the solution, if any ∃.

July 28, 2016 16 / 17




1 Existence.2 Uniqueness.

3 Local regularity.Questions about the structure of the solution, if any ∃.

July 28, 2016 16 / 17






July 28, 2016 16 / 17




1 Existence.2 Uniqueness.3 Local regularity.Questions about the structure of the solution, if any ∃.

July 28, 2016 16 / 17

General Remarks on SPDEsu = u′′ + b(u) + σ (u)W

There is also interest in these equations when b and/or σ arenon-Lipschitz continuous.

For example:

1 b(u) ≡ 0 and σ (u) =√|u| [Population genetics].2 b(u) = u − u3 and σ (u) = 1 [QFT].3 b(u) = u − u3 and σ (u) = u [Statistical mechanics].4 b(u) ≡ 0 and σ (u) = u [KPZ+random polymers].

The last Item #4 is included in the general theory to followsince b and σ are Lipschitz continuous.We will say a few things about Items #2 and #3 as well[Allen-Cahn, Φ41, Swift–Hohenberg, Ginzburg–Landau, . . . ].Item #1 is perhaps the best-studied case [super BM,Brownian density process]. Not covered here [Dawson, Dynkin,Mytnik, Perkins, Shiga, Watanabe, . . . ].

July 28, 2016 17 / 17


There is also interest in these equations when b and/or σ arenon-Lipschitz continuous.For example:

1 b(u) ≡ 0 and σ (u) =√|u| [Population genetics].2 b(u) = u − u3 and σ (u) = 1 [QFT].3 b(u) = u − u3 and σ (u) = u [Statistical mechanics].4 b(u) ≡ 0 and σ (u) = u [KPZ+random polymers].The last Item #4 is included in the general theory to followsince b and σ are Lipschitz continuous.We will say a few things about Items #2 and #3 as well[Allen-Cahn, Φ41, Swift–Hohenberg, Ginzburg–Landau, . . . ].Item #1 is perhaps the best-studied case [super BM,Brownian density process]. Not covered here [Dawson, Dynkin,Mytnik, Perkins, Shiga, Watanabe, . . . ].

July 28, 2016 17 / 17


There is also interest in these equations when b and/or σ arenon-Lipschitz continuous.For example:1 b(u) ≡ 0 and σ (u) =√|u| [Population genetics].

2 b(u) = u − u3 and σ (u) = 1 [QFT].3 b(u) = u − u3 and σ (u) = u [Statistical mechanics].4 b(u) ≡ 0 and σ (u) = u [KPZ+random polymers].The last Item #4 is included in the general theory to followsince b and σ are Lipschitz continuous.We will say a few things about Items #2 and #3 as well[Allen-Cahn, Φ41, Swift–Hohenberg, Ginzburg–Landau, . . . ].Item #1 is perhaps the best-studied case [super BM,Brownian density process]. Not covered here [Dawson, Dynkin,Mytnik, Perkins, Shiga, Watanabe, . . . ].

July 28, 2016 17 / 17


There is also interest in these equations when b and/or σ arenon-Lipschitz continuous.For example:1 b(u) ≡ 0 and σ (u) =√|u| [Population genetics].2 b(u) = u − u3 and σ (u) = 1 [QFT].

3 b(u) = u − u3 and σ (u) = u [Statistical mechanics].4 b(u) ≡ 0 and σ (u) = u [KPZ+random polymers].The last Item #4 is included in the general theory to followsince b and σ are Lipschitz continuous.We will say a few things about Items #2 and #3 as well[Allen-Cahn, Φ41, Swift–Hohenberg, Ginzburg–Landau, . . . ].Item #1 is perhaps the best-studied case [super BM,Brownian density process]. Not covered here [Dawson, Dynkin,Mytnik, Perkins, Shiga, Watanabe, . . . ].

July 28, 2016 17 / 17


There is also interest in these equations when b and/or σ arenon-Lipschitz continuous.For example:1 b(u) ≡ 0 and σ (u) =√|u| [Population genetics].2 b(u) = u − u3 and σ (u) = 1 [QFT].3 b(u) = u − u3 and σ (u) = u [Statistical mechanics].

4 b(u) ≡ 0 and σ (u) = u [KPZ+random polymers].The last Item #4 is included in the general theory to followsince b and σ are Lipschitz continuous.We will say a few things about Items #2 and #3 as well[Allen-Cahn, Φ41, Swift–Hohenberg, Ginzburg–Landau, . . . ].Item #1 is perhaps the best-studied case [super BM,Brownian density process]. Not covered here [Dawson, Dynkin,Mytnik, Perkins, Shiga, Watanabe, . . . ].

July 28, 2016 17 / 17


There is also interest in these equations when b and/or σ arenon-Lipschitz continuous.For example:1 b(u) ≡ 0 and σ (u) =√|u| [Population genetics].2 b(u) = u − u3 and σ (u) = 1 [QFT].3 b(u) = u − u3 and σ (u) = u [Statistical mechanics].4 b(u) ≡ 0 and σ (u) = u [KPZ+random polymers].

The last Item #4 is included in the general theory to followsince b and σ are Lipschitz continuous.We will say a few things about Items #2 and #3 as well[Allen-Cahn, Φ41, Swift–Hohenberg, Ginzburg–Landau, . . . ].Item #1 is perhaps the best-studied case [super BM,Brownian density process]. Not covered here [Dawson, Dynkin,Mytnik, Perkins, Shiga, Watanabe, . . . ].

July 28, 2016 17 / 17


There is also interest in these equations when b and/or σ arenon-Lipschitz continuous.For example:1 b(u) ≡ 0 and σ (u) =√|u| [Population genetics].2 b(u) = u − u3 and σ (u) = 1 [QFT].3 b(u) = u − u3 and σ (u) = u [Statistical mechanics].4 b(u) ≡ 0 and σ (u) = u [KPZ+random polymers].The last Item #4 is included in the general theory to followsince b and σ are Lipschitz continuous.

We will say a few things about Items #2 and #3 as well[Allen-Cahn, Φ41, Swift–Hohenberg, Ginzburg–Landau, . . . ].Item #1 is perhaps the best-studied case [super BM,Brownian density process]. Not covered here [Dawson, Dynkin,Mytnik, Perkins, Shiga, Watanabe, . . . ].

July 28, 2016 17 / 17


There is also interest in these equations when b and/or σ arenon-Lipschitz continuous.For example:1 b(u) ≡ 0 and σ (u) =√|u| [Population genetics].2 b(u) = u − u3 and σ (u) = 1 [QFT].3 b(u) = u − u3 and σ (u) = u [Statistical mechanics].4 b(u) ≡ 0 and σ (u) = u [KPZ+random polymers].The last Item #4 is included in the general theory to followsince b and σ are Lipschitz continuous.We will say a few things about Items #2 and #3 as well[Allen-Cahn, Φ41, Swift–Hohenberg, Ginzburg–Landau, . . . ].

Item #1 is perhaps the best-studied case [super BM,Brownian density process]. Not covered here [Dawson, Dynkin,Mytnik, Perkins, Shiga, Watanabe, . . . ].

July 28, 2016 17 / 17


There is also interest in these equations when b and/or σ arenon-Lipschitz continuous.For example:1 b(u) ≡ 0 and σ (u) =√|u| [Population genetics].2 b(u) = u − u3 and σ (u) = 1 [QFT].3 b(u) = u − u3 and σ (u) = u [Statistical mechanics].4 b(u) ≡ 0 and σ (u) = u [KPZ+random polymers].The last Item #4 is included in the general theory to followsince b and σ are Lipschitz continuous.We will say a few things about Items #2 and #3 as well[Allen-Cahn, Φ41, Swift–Hohenberg, Ginzburg–Landau, . . . ].Item #1 is perhaps the best-studied case [super BM,Brownian density process]. Not covered here [Dawson, Dynkin,Mytnik, Perkins, Shiga, Watanabe, . . . ].

July 28, 2016 17 / 17

A Stochastic Heat Equation

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