Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-FlourChapter 1: Introduction A course in 8...

Random Obstacle Problems

Lorenzo Zambotti(LPMA, Univ. Paris 6)

July 2015, Saint-Flour

Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour

Chapter 1: Introduction

A course in 8 chapters:I IntroductionI The reflecting Brownian motionI Bessel processesI The stochastic heat equationI Obstacle problemsI Integration by Parts FormulaeI The contact setI Convergence and Dirichlet Forms


Chapter 1: Introduction

A course in 8 chapters:I IntroductionI The reflecting Brownian motionI Bessel processesI The stochastic heat equationI Obstacle problemsI Integration by Parts FormulaeI The contact setI Open problems


Symmetric simple random walk

(Yi)i≥1 i.i.d. sequence with P(Yi = 1) = P(Yi = −1) = 1/2

S0 := a ∈ Z, Sn+1 = Sn + Yn+1,

(Sn)n≥0 defines a Markov chain with valued in Z.

0 2N

0 2N

a

0

Sn

a

0

Xn

1


Symmetric simple random walk with reflection

Xn+1 :=

Xn + Yn+1 if Xn > 0

1 if Xn = 0.

This defines a Markov chain with values in Z+ = 0, 1, . . .. It is easyto prove by recurrence on n that

Xn = Sn +

n∑i=1

1(Xi−1=0)(1− Yi), n ≥ 0. (1)

a

0

Xn

a

0

Zn

0 2N

0 2N

2


Absolute value of the SSRW

Let Zn := |Sn|, with S0 = a ≥ 0. Although this is not obvious at firstsight, (Zn)n≥0 is also a Markov chain:

Zn+1 =

Zn + Wn+1 if Zn > 0

1 if Zn = 0,

where Wn+1 := Yn+1(1Sn≥0 − 1Sn<0).

It is easy to see that (Wi)i≥1 is a copy of (Yi)i≥1 and then (Zn)n≥0 and(Xn)n≥0 (for the same fixed a) have the same law.

a

0

Xn

a

0

Zn

0 2N

0 2N

2


S, Z, X

0 2N

0 2N

a

0

Sn

a

0

Sn

1

a

0

Xn

a

0

Zn

0 2N

0 2N

2

a

0

Xn

a

0

Zn

0 2N

0 2N

2


Discrete interfaces

We define now a Markov chain with values in discrete paths. We fixN ∈ N and we define the state space

EN := w ∈ Z2N : w(0) = w(2N) = 0,

|w(i)− w(i− 1)| = 1, ∀ i = 1, . . . , 2N.

0 2N

0 2N

0 2N

1

Figure : A typical path in EN


The free evolution

Then we define a Markov chain with values in EN as follows: we definea map F : EN × 1, . . . , 2N − 1 → EN , F(w, j) = w ∈ EN , where

wi =

wi + 2 if i = j and wi−1 = wi+1 > wi

wi − 2 if i = j and wi−1 = wi+1 < wi

wi otherwise.

0 2N

We let (Un)n≥1 be an i.i.d. sequence of uniform random variables on1, . . . , 2N − 1; then

en+1 := F(en,Un+1), e0 ∈ EN .


The free evolution

We denote the transition matrix of (en)n≥0 by

P(x, y) = P(F(x,U1) = y).

It is easy to see that P(x, y) = P(y, x) and therefore the uniformmeasure on EN is invariant and reversible for P.

This is in fact the unique probability invariant measure of (en)n≥0 bythe following

LemmaThe Markov chain (en)n≥0 is aperiodic and irreducible.


The free evolution

Proof.Equivalence with a particle system:

EN 3 w←→ (wi − wi−1)i=1,...,2N ∈ 1,−12N ←→ •, 2N

EN ←→ CN := c ∈ •, 2N : #• = # = N.P(x, x′) > 0 iff

x←→ · · · | • | | · · · , x′ ←→ · · · | | • | · · ·

or viceversa.

Let c0 := | • | · · · | • | | · · · | | ∈ CN . From any c ∈ CN we can reachc0 with finitely many transitions having positive probability by movingall particles in c to the left starting with the leftmost one. By symmetryone can go from c0 to any c ∈ CN . Aperiodicity: P(c0, c0) > 0.


The reflected interface

Let us now add reflection to our discrete interface. We set

E+N := w ∈ EN : w(i) ≥ 0, ∀ i = 0, . . . , 2N.

0 2N

0 2N

a

0

Sn

a

0

Sn

1

Figure : A typical path in E+N

Reflection means now suppression of transitions which would let e+n

take negative values:

0 2N



The Markov evolution in E+N is defined as follows:

F+ : E+N × 1, . . . , 2N − 1 → E+

N

F+(w, j) =

F(w, j) if F(w, j) ∈ E+

Nw otherwise.

Then our E+N -valued Markov chain (e+

n )n≥0 is defined by

e+n+1 := F+(e+

n ,Un+1), e+0 ∈ E+

N .

LemmaThe Markov chain (e+

n )n≥0 is aperiodic irreducible and has a uniqueinvariant probability measure, given by the uniform probabilitymeasure on E+

N . This probability measure is furthermore reversible for(e+

n )n≥0.



Proof.We recall that the transition matrix P of (en)n≥0 is symmetric. Now wehave for x, y ∈ E+

N :

P+(x, y) = P( f+(x,U1) = y) = P(x, y) + P(x,EN \ E+N )1(y=x).

Then P+ is also symmetric and therefore the uniform measure on E+N is

invariant and reversible for P+.

E+N ←→ C+

N , set of c ∈ CN such that for all k ∈ 1, . . . , 2N the first ksites contain at least s(k) particles, where s(2i) := i, s(2i + 1) := i + 1.

Then c0 = | • | · · · | • | | · · · | | ∈ C+N and we can find a path in C+

Nfrom any c ∈ C+

N to c0 by moving particles leftwards starting with theleftmost one. Finally P+(c0, c0) > 0, so that this Markov chain isaperiodic.


An important remark

We see that the reflection for the dynamics is equivalent to aconditioning for the invariant measure.

Let |en| be the absolute value of the free interface (en)n≥0. If (en)n≥0 isstationary then the distribution of |e0| is a probability measure on E+

N

P(|e0| = w) ∝ #w′ ∈ EN : |w′| = w = 2L(w),

L(w) :=

2N∑i=1

1(wi=0), w ∈ E+N .

In other words L(w) is the number of excursions of w.

On the other hand, if (e+n )n≥0 is stationary then the law of e+

0 isuniform on E+

N , so that e+0 and |e0| have different laws.

Moreover (|en|)n≥0 is not Markovian.


Scaling limits

Let (Sn)n≥0 be the SSRW with S0 := 0 and

BNt :=

1√N

SbNtc, t ≥ 0.

By Donsker’s theorem (BNt )t≥0 =⇒ (Bt)t≥0 as N → +∞.

Under the same scaling the reflecting SSRW (Xn)n≥0 converges to thereflecting Brownian motion (ρt)t≥0 (see chapter 2).

This process is given by a stochastic differential equation

dρt = dBt + d`t, ρt ≥ 0, d`t ≥ 0,∫ ∞

0ρt d`t = 0,

I t 7→ (ρt, `t) is continuous, ρt is non-negativeI `0 = 0, `s ≤ `t for s ≤ tI supp(d`t) ⊂ t ≥ 0 : ρt = 0.

The measure d`t is the reflection term: see (1).Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour

Scaling of the free interface

Let us first consider the stationary version of (en)n≥0 and define

vN(t, x) =1√2N

eb4N2tc(b2Nxc), t ≥ 0, x ∈ [0, 1].

x0 1

1

Figure : A typical path of vN(t, ·) for any t ≥ 0 when N is large


The stochastic heat equation

As N → +∞, vN =⇒ (v(t, x), t ≥ 0, x ∈ [0, 1]), that we are going tostudy in detail in chapter 4 and is a stationary solution to a stochasticpartial differential equation (SPDE)

∂v∂t

=12∂2v∂x2 + W,

v(t, 0) = v(t, 1) = 0, t ≥ 0,

v(0, x) = v0(x), x ∈ [0, 1].

Here W is a space-time white noise, a two-parameter version of thederivative of a Brownian motion that we shall define properly later.


Scaling of the reflected interface

Let us now consider the stationary version of (e+n )n≥0 and define

uN(t, x) =1√2N

e+b4N2tc(b2Nxc), t ≥ 0, x ∈ [0, 1].

x0 1

1

Figure : A typical path of uN(t, ·) for any t ≥ 0 when N is large


A SPDE with reflection

As N → +∞, uN =⇒ (u(t, x), t ≥ 0, x ∈ [0, 1]), t, that we are going tostudy in detail from chapter 5 on and is a stationary solution to a SPDEwith reflection

∂u∂t

=12∂2u∂x2 + W + η

u(0, x) = u0(x), u(t, 0) = u(t, 1) = 0

u ≥ 0, dη ≥ 0,∫

u dη = 0.

Here (u, η) is a random pair that consists ofI a continuous non-negative functions u(t, x) ≥ 0I a Radon measure η on ]0,+∞[× ]0, 1[,

such that the support of η is contained in (t, x) : u(t, x) = 0.


The contact set

For all t ≥ 0 the typical profile of u(t, ·) is positive on ]0, 1[. Wheredoes the reflection act?

This apparent paradox is solved if we formulate the sentence moreprecisely: the correct result is that for all t ≥ 0, a.s. u(t, ·) > 0 on ]0, 1[:

∀ t > 0, P(∃ x ∈ ]0, 1[ : u(t, x) = 0) = 0.

However this does not exclude the existence, with positive probability,of exceptional times t ≥ 0 and x ∈ ]0, 1[ such that u(t, x) = 0:

P(∃ t > 0, x ∈ ]0, 1[ : u(t, x) = 0) > 0.


The contact set

The next question is: what can be said about the contact set

Z := (t, x) : t > 0, x ∈ ]0, 1[, u(t, x) = 0.After proving that with positive probability u visits 0, one can ask:I what is the typical behavior at exceptional times t ≥ 0 ?I That is, if t > 0 is such that there exists x ∈ ]0, 1[ so that

u(t, x) = 0, then how many such points x exist?

0 1

1

Figure : How many x such that u(t, x) = 0: infinitely many?


The contact set

0

1

1

Figure : How many x such that u(t, x) = 0: finitely many?

0 1

1

Figure : How many x such that u(t, x) = 0: just one? or two? or three?


Difficulties

I (u(t, x), t ≥ 0, x ∈ [0, 1]) has no standard semi-martingalestructure

I neither has (u(t, x), t ≥ 0)

I which is by the way not markovian for fixed x ∈ [0, 1]

I therefore we must study S(P)DEs without stochastic calculus

This course is about some fascinating phenomena, in search of a theory.

The dream is a stochastic calculus for infinite-dimensional processes,following the model of the classical theory (e.g. Revuz-Yor).

We shall discuss some open problems waiting for someone to solvethem.


Chapter 2: The reflecting Brownian motion

Let us meet our first obstacle problem.

Lemma (Skorokhod)Let y : R+ 7→ R be a continuous function such that y(0) ≥ 0. Thenthere exists a unique couple (z, `) of continuous functions on R+ s.t.

1. z = y + `

2. z ≥ 0

3. t 7→ `t is monotone non-decreasing, `0 = 0, and the support of theassociated measure d`t is contained in t ≥ 0 : zt = 0.

Moreover, ` are z given explicitly by

`t = sups≤t

max−ys, 0, zt = yt + `t. (2)

Remark: supp(d`) ⊂ t ≥ 0 : zt = 0 ⇐⇒∫ ∞

0zt d`t = 0.


The Skorokhod Lemma

0

z

y

`

1

Figure : An example of a triple (y, z, `) from the Skorokhod Lemma.


The Skorokhod Lemma

Proof.To prove existence, one verifies that `t := sups≤t max−ys, 0 andz := y + ` satisfy the above properties.

If (z, `) is another pair with the same properties, then z− z = `− ` is acontinuous function with bounded variation and by the chain rule

0 ≤ 12

(zt − zt)2 =

∫ t

0(zs − zs) d(`s − `s)

=

[∫ t

0zs d`s +

∫ t

0zs d`s −

∫ t

0zs d`s −

∫ t

0zs d`s

].

Since supp(d`s) ⊂ s : zs = 0 and supp(d`s) ⊂ s : zs = 0, this isequal to

−∫ t

0zs d`s −

∫ t

0zs d`s ≤ 0.

Then (zt − zt)2 = (`t − `t)

2 = 0.Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour

ODEs with reflection

Let f : R 7→ R such that

| f (x)− f (y)| ≤ L|x− y|, x, y ∈ R.

PropositionLet T ≥ 0 and w ∈ C([0,T]) with w0 ≥ 0. Then there exists a uniquecouple (ρt, `t)t∈[0,T] of continuous real functions such that

ρt = wt +

∫ t

0f (ρs) ds + `t, t ∈ [0,T]

`0 = 0,

ρt ≥ 0, d`t ≥ 0,∫ T

0ρt d`t = 0.

(3)


ODEs with reflection

Proof.Let ET := C([0,T]) with norm ‖g‖L := supt∈[0,T] e−3Lt|gt|. We defineΓ : ET 7→ ET by Γ(g) := z, where (z, `) is the solution of theSkorokhod problem of Lemma 3 associated to y defined by

yt := wt +

∫ t

0f (gs) ds, t ∈ [0,T].

In particular, z ≥ 0, d` ≥ 0, `0 = 0, and for all t ∈ [0,T]

zt := wt +

∫ t

0f (gs) ds + `t,

∫ t

0zs d`s = 0.

By the explicit representation (2) of (z, `) in terms of y we obtain that Γis a contraction in ET and it has a unique fixed point (ρt)t∈[0,T].


SDEs with reflection

CorollaryLet (Bt)t≥0 be a standard BM and x ≥ 0. Then there exists a uniquecouple (ρt, `t)t≥0 of continuous real processes such that

ρt = x + Bt +

∫ t

0f (ρs) ds + `t, t ≥ 0

`0 = 0,

ρt ≥ 0, d`t ≥ 0,∫ ∞

0ρt d`t = 0.

(4)

If f ≡ 0 we call ρ the reflecting BM.


The reflecting BM

0 t

t

1


Reflection in Rd

H. Tanaka (1979), Stochastic differential equations with reflectingboundary condition in convex regions. Hiroshima Math. J. 9, no. 1,163–177.

Let D ⊂ Rd be a convex open bounded domain. We define:I For all x ∈ ∂D a exposing hyperplane H ∈Hx(D) is a hyperplane

such that D is contained in one of the two half-spaces whoseboundary is H and H ∩ ∂D 6= ∅.

I We call n ∈ Rd an inward normal unit vector at x ∈ ∂D ifx + εn ∈ D for some ε > 0, ‖n‖ = 1 and n is perpendicular tosome H ∈Hx(D). We call Nx(D) the set of all inward unitnormal vectors at x ∈ ∂D.


Reflection in Rd

Figure : Exposing hyperplanes

(Special thanks to wikipedia)


Reflection in Rd

Theorem (Tanaka)For all w ∈ C([0, T];Rd) with w0 ∈ D there exists a unique pair (ξ, ϕ)such thatI ξ ∈ C([0,T]; D), ϕ ∈ C([0,T];Rd), ϕ(0) = 0, ξ = w + ϕ.I there exists a monotone non-decreasing ` ∈ C([0,T]) such that

ϕt =

∫ t

01(ξs∈∂D) ns d`s, t ∈ [0,T]

where ns ∈ Nξs(D) for d`-a.e. s.

Moreover the map w 7→ ξ is continuous in the sup-topology.

One can write the equation as a differential inclusion

d(ξt − wt) ∈ 1(ξt∈∂D) Nξt(D) d`t,

∫ ·01(ξs∈D) d`s = 0.


Proof of Uniqueness

Let (ξ, ϕ) and (ξ, ϕ) be two solutions. Then

d‖ξt − ξt‖2 = (ξt − ξt) · (dϕt − dϕt).

Now for d`-a.e. t

(ξt − ξt) · dϕt = 1(ξs∈∂D) (ξt − ξt) · nt d`t.

Since ξt ∈ D and nt ∈ Nξt(D), then

(ξt − ξt) · nt = −(ξt − ξt) · nt ≤ 0.

By symmetry we conclude that ‖ξt − ξt‖2 ≡ 0.


Another approach to reflection: Penalisation

Let n ≥ 1, w ∈ C([0,T]) with w0 ≥ 0 and

ρnt = wt + n

∫ t

0(ρn

s )− ds +

∫ t

0f (ρn

s ) ds, t ∈ [0,T],

where r− = (r)− := max−r, 0, r ∈ R.

Additive (deterministic!) noise and Lipschitz drift, so clearly pathwiseuniqueness and existence of solutions by a Picard iteration.

Proposition

1. if n ≤ m then ρnt ≤ ρm

t for all t ∈ [0,T].

2. ρn ↑ ρ uniformly on [0,T] as n ↑ +∞, where (ρt, `t)t≥0 is theunique solution to the equation with reflection (3). Moreover

limn↑+∞

n∫ t

0(ρn

s )− ds = `t, t ∈ [0,T].


Penalisation

0t

nt

1


Proof

We set fn : R 7→ R,

fn(x) := n x− + f (x).

Note that fn is Lipschitz-continuous on R and satisfies the one-sidedestimate:

fn(x)− fn(y) ≤ L(x− y), ∀ n ≥ 1, x > y.

For simplicity we assume that f is monotone non-increasing, so that inparticular

fn(x)− fn(y) ≤ 0, ∀ n ≥ 1, x > y.

In the lecture notes you find the argument for the general case.

We call S(n,w,T) = ρn the unique solution of

ρnt = wt +

∫ t

0fn(ρn

s ) ds, t ∈ [0,T].


Proof

Step 1 (Monotonicity). We show first thatρn = S(n,w,T) ≤ ρm = S(m,w,T) if n ≤ m. Since fn(·) ≤ fm(·)

ddt

((ρn − ρm)+

)2= 2(ρn − ρm)+ ( fn(ρn)− fm(ρm))

≤ 2(ρn − ρm)+ ( fm(ρn)− fm(ρm)) ≤ 0.

Since ρn0 = ρm

0 , we conclude that ρn ≤ ρm. Then

S(w,T) := supn

S(n,w,T) = limn→+∞

S(n,w,T)

where the supremum and the limit are taken pointwise, i.e. for allt ∈ [0,T]. It is not clear at this point whether ρ := S(w,T) is acontinuous function. This will be proved in the next steps.


Proof

Step 2 (The case of a smooth obstacle). Let w ∈ C∞([0,T]) withw0 ≥ 0. Let us fix n. Writing for simplicity x := ρn

xt = fn(xt) + wt,

ddt|xt − xt+h| = ( fn(xt)− fn(xt+h) + wt − wt+h) sign(xt − xt+h)

≤ |wt − wt+h|so that

|xt − xt+h| ≤ |x0 − xh|+∫ t

0|ws − ws+h| ds.

If we divide by h > 0 and let h ↓ 0 we obtain

|xt| ≤ |x0|+∫ t

0|ws| ds ≤ | fn(w0)|+ |w0|+

∫ t

0|ws| ds

= | f (x0)|+ |w0|+∫ t

0|ws| ds.


Proof

Thereforesupn≥1

supt∈[0,T]

|xt| ≤ C = C(T,w).

By the Arzelà-Ascoli theorem, the sequence x = ρn = S(n,w,T) inn ≥ 1 is compact in C([0,T]). Therefore ρn converges uniformly,

`nt := n

∫ t

0(ρn

s )− ds = ρnt − wt −

∫ t

0f (ρn

s ) ds −→ `t

and one shows that (ρ, `) is a solution to the SDE with reflection.


Proof

Step 3 (Continuity of the map w 7→ S(n,w,T) uniformly in n). Letnow w ∈ C([0, T]) with w0 ≥ 0. We still write x := ρn for fixed n. Set

yt := xt − wt =⇒ yt = fn(xt). (5)

Let w ∈ C([0, T]) with w0 ≥ 0, and set x = S(n,w, T), x = S(n, w, T).We denote

‖g‖∞ := sup[0,T]

|gt|, g : [0,T] 7→ R.

We set y as in (5) and y analogously after replacing (x,w) by (x, w).

Let κ := ‖w− w‖∞. Now if yt − yt − κ > 0 then

xt − xt > wt − wt + κ

≥ −‖w− w‖∞ + κ = 0.


Proof

Then since r 7→ fn(r) is monotone non-increasing,

ddt

((yt − yt − κ)+)2

= 2 (yt − yt − κ)+ ( fn(xt)− fn(xt)) ≤ 0.

Since y0 = y0, we obtain that yt − yt ≤ κ for all t ∈ [0,T], and bysymmetry

‖y− y‖∞ ≤ κ = ‖w− w‖∞.By definition y = x− w, so that

‖S(n,w,T)− S(n, w,T)‖∞ ≤ 2‖w− w‖∞.

By letting n→ +∞:

‖S(w,T)− S(w,T)‖∞ ≤ 2‖w− w‖∞.


Proof

Step 4 (Conclusion).I Let w ∈ C([0,T]), wm ∈ C∞([0,T]) such that

w0,wm0 ≥ 0, lim

m→+∞‖w− wm‖∞ = 0.

I By Step 2, S(wm,T) ∈ C([0,T]).I By Step 3 S(wm, T) converges uniformly on [0, T] to ρ = S(w, T),

so that S(w,T) ∈ C([0,T]) .I We can conclude the proof by Dini’s Theorem.


The penalised SDE

For x ∈ R and B a standard BM we set

ρnt (x) = x + Bt + n

∫ t

0(ρn

s (x))− ds +

∫ t

0f (ρn

s (x)) ds, t ≥ 0,

The infinitesimal generator of ρn is for ϕ ∈ C2c(R)

Lnϕ(x) :=12ϕ′′(x) + (nx− + f (x))ϕ′(x), x ∈ R.

Moreover ρn admits as reversible invariant measure

µn(dx) = e−n (x−)2+2F(x) dx

where F : R 7→ R is any function such that

F′(x) = f (x), x ∈ R.

Note that µn(]−∞, 0]) < +∞ for n large, but µn([0,+∞[) ≤ +∞ ingeneral.


The penalised SDE

LemmaThe measure µ is invariant and reversible for (ρt)t≥0 and

limn→+∞

∫Rϕ dµn =

∫Rϕ dµ, ∀ϕ ∈ Cc(R),

whereµ(dx) := 1(x≥0) e2F(x) dx.

Here is an important message, that we have already noticed for discreteinterfaces:

RemarkA reflection for the dynamics means a conditioning for the invariantmeasure.


Proof

Since µn is reversible for ρn, for all ϕ,ψ ∈ Cc(R)∫Rϕ(x)E(ψ(ρn

t (x)))µn(dx) =

∫Rψ(x)E(ϕ(ρn

t (x)))µn(dx).

We see that 1(x≥0) µn(dx) = 1(x≥0) µ(dx), while

µn(]−∞, 0[) =

∫]−∞,0[

e−n x2+2F(x) dx→ 0

as n→ +∞. We conclude by dominated convergence


A useful formula

LemmaLet (ρt, `t)t≥0 be the solution to the SDE with reflection (4). Then∫

R+

e2F(x) E(`t(x)) dx =t2

e2F(0), t ≥ 0. (6)

Proof. We use the notation

`nt (x) :=

∫ t

0n (ρn

s (x))− ds.

We are going to prove two formulae:

limn→+∞

∫Rµn(dx)E(`n

t (x)) =

∫R+

µ(dx)E(`t(x)), (7)

limn→+∞

∫Rµn(dx)E(`n

t (x)) =t2

e2F(0). (8)

Again, for simplicity we assume that f is monotone non-increasing. Inthe lecture notes you find the argument for the general case.


Proof

First, since µn is invariant for ρn,∫Rµn(dx)E

(∫ t

0n (ρn

s (x))− ds)

= t∫Rµn(dx) n x−

and by a simple change of variable this is equal to

t∫R

e−x2+2F(n−1/2x) x− dx→ t e2F(0)

∫R

e−x2x− dx =

t2

e2F(0) (9)

as n→ +∞. In the above limit we can use dominated convergencesince for x ≤ 0

F(x) = −∫ 0

xf (y) dy + F(0) ≤ −

∫ 0

xf (0) dy + F(0)

= −f (0) x + F(0) ≤ C(|x|+ 1)

for some C > 0. We have therefore proved (8). We must now prove (7).Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour

Proof

Since f is monotone non-increasing

`nt =

∫ t

0n (ρn

s (x))− ds = ρnt − Bt − x−

∫ t

0f (ρn

s ) ds

becomes monotone non-decreasing in n.

This result is not so trivial, since n 7→ (ρns )− is monotone

non-increasing.

Let us now use the notation

Gn(x) := E(`nt (x)) =

∫ t

0E(n (ρn

s (x))−)

ds

= E (ρnt (x))− x−

∫ t

0E ( f (ρn

s (x))) ds,

G(x) := E (`t(x)) .


Proof

By the convergence of the penalisation procedure, by monotoneconvergence and since 1(x≥0)µn(dx) = µ(dx), as n ↑ +∞∫

R+

Gn(x)µn(dx) =

∫R+

Gn(x)µ(dx) ↑∫R+

µ(dx) G(x).

For all x ≤ 0 we have ρnt (x) ≤ ρn

t (0) and ρnt (0) ≤ ρt(0). Then

Gn(x) ≤ E (ρt(0))− x−∫ t

0E ( f (ρs(0))) ds =: Ct − x,

for all x ≤ 0 and therefore as n ↑ +∞

0 ≤∫R−

Gn(x)µn(dx) =

∫R−

Gn(x) e−n x2+2F(x) dx

≤∫R−

(Ct − x) e−n x2+2C(x2+1) dx ↓ 0.

We have thus proved (7). Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour

Reflection and local time

TheoremLet (Xt)t≥0 be a real-valued continuous semimartingale with quadraticvariation (〈X〉t)t≥0. There exists a measurable process (La

t )t≥0,a∈Rsuch that a.s. for all bounded Borel ϕ : R 7→ R∫ t

0ϕ(Xs) d〈X〉s =

∫Rϕ(a) La

t da, t ≥ 0. (10)

The process (Lat )t≥0,a∈R has a modification such that a.s. (t, a) 7→ La

t iscontinuous in t and cadlag in a.

The process (Lat )t≥0 is called the local time of X at a ∈ R and (10) is

known as the occupation time formula.

For all a ∈ R the process (Lat )t≥0 is monotone non-decreasing and the

measure dLat is supported by t ≥ 0 : Xt = a.



Proposition (Tanaka’s formula)For all a ∈ R and t ≥ 0

|Xt − a| = |X0 − a|+∫ t

0sgn(Xs − a) dXs + La

t

(Xt − a)+ = (X0 − a)+ +

∫ t

01(Xs>a) dXs +

12

Lat

where sgn(x) := 1(x>0) − 1(x≤0).

The solution (ρt)t≥0 to the reflected SDE (4) is a continuoussemimartingale. By Tanaka’s formula applied to (ρt)

+ = ρt

`t =12

L0t t ≥ 0, a.s.



In particular, a.s. `t is adapted to the filtration of ρ.

All this is not true for generic w in the Skorokhod Lemma. Let us givetwo instructive and simple examples for f ≡ 0:

1. if w ≡ 0, then ρ ≡ 0 and ` ≡ 0.

2. if wt = −t, then ˆt = t and ρt = 0, t ∈ [0,T].

Thenρ = ρ = 0, ` = 0, ˆt = t.

The same amount of time spent at 0 is associated with different valuesof the reflection process.

Moreover `, though always an explicit function of w by the SkorokhodLemma 3, may not be a function of ρ: indeed ρ = ρ but ` 6= ˆ.

Finally neither ρ nor ρ have local times.


The explicit law of RBM

PropositionLet (ρt)t≥0 be a reflecting Brownian motion started at x ≥ 0 and(Bt)t≥0 a standard Brownian motion in R. Then (ρt)t≥0 and(|x + Bt|)t≥0 have the same law.

Proof.By Tanaka’s formula (see Proposition 12)

|x + Bt| = x +

∫ t

0sign(x + Bs) dBs + L−x

t , t ≥ 0,

The process βt :=∫ t

0 sign(x + Bs) dBs is a standard BM by Lévy’scharacterisation theorem. By the Skorokhod lemma

L−xt = sup

s≤tmax−x− βs, 0, |x + B| = x + β + L−x,

and therefore (ρt)t≥0 = Γ(B) and (|x + Bt|)t≥0 = Γ(β).Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour

Results on the contact set

In particular t ≥ 0 : ρt = 0 and Z := t ≥ 0 : Bt = 0 have thesame law (for simplicity x = 0 here).

I a.s. Z is closed, unbounded, without isolated points and has zeroLebesgue measure.

I Let (Lt)t≥0 be the local time at 0 of B and

τt := infs > 0 : Ls > t, t ≥ 0.

Then (τt)t≥0 is a stable subordinator of index 1/2.I The set Z is a.s. the closure of the image of (τt)t≥0: a

regenerative set.I The Hausdorff dimension of Z is a.s. equal to 1/2.


Chapter 3: Bessel processes

δ-Bessel processes are solutions (ρt)t≥0 to the SDE

ρt = x +δ − 1

2

∫ t

0ρ−1

s ds + Bt, ρt ≥ 0, t ≥ 0, (11)

where (Bt)t≥0 is a BM, x ≥ 0, and δ > 1. For δ = 1 the equation is

ρt = x + `t + Bt, `0 = 0, d` ≥ 0,∫ t

0ρs d`s = 0. (12)

This is the reflecting Brownian motion. For all δ ≥ 1, the law of theδ-Bessel process is Pδx .

I Xt : C([0,+∞[) 7→ R is the canonical process, Xt(w) := wt, t ≥ 0I F 0

t := σ(Xs, s ∈ [0, t]) is the canonical filtrationI let Wx be the law of (x + Bt)t≥0.


Existence and uniqueness

Let g : ]0,+∞[ 7→ R be continuous and monotone non-increasing suchthat g is Lipschitz-continuous on [ε,+∞[ for all ε > 0.

We allow g(x) to blow up as x ↓ 0. The main example we have in mindis g(x) = Cx−α, α > 0, C > 0.

PropositionLet w ∈ C([0,T]) with w0 ≥ 0 and g : ]0,+∞[ 7→ R as above. Thenthere exists a unique couple (ρ, `) ∈ (C([0,T]))2 such that

ρ ≥ 0, g ρ ∈ L1(0,T), `0 = 0, d` ≥ 0, ρ d` = 0

ρt = wt +

∫ t

0g(ρs) ds + `t, t ∈ [0,T].


Proof

Let us start with uniqueness. If (ρ, `) and (ρ, ˆ) are two solutions, then

ρt − ρt =

∫ t

0(g(ρs)− g(ρs)) ds + `t − ˆt,

i.e. ρ− ρ is continuous with bounded variation on [0, T]. Therefore bythe chain rule

d (ρt − ρt)2 = 2 (ρt − ρt)

[(g(ρt)− g(ρt)) dt + d`t − dˆt

]≤ 0

and this proves uniqueness.


Proof

Let us show now existence. Let us set

Aε(x) := g(ε+ x+), x ∈ R,

where ε > 0 is fixed and x+ := maxx, 0. Note that Aε isLipschitz-continuous and monotone non-increasing, i.e.

Aε(x)− Aε(y) ≤ 0, ∀ ε > 0, x > y.

There is a unique continuous solution (ρε, `ε) to

ρεt = wt +

∫ t

0Aε(ρεs ) ds + `εt , t ∈ [0,T],

with the usual conditions ρε ≥ 0, d`ε ≥ 0, ρε d`ε = 0.


Proof

Now, ε 7→ Aε(x) is monotone non-increasing since g(·) is, and weexpect ε 7→ ρε to be also monotone non-increasing. Indeed, letε > ε′ > 0. Then

d((

ρεt − ρε′

t

)+)2

= 2(ρεt − ρε

′t

)+ ((Aε(ρεt )− Aε′(ρε

′t )) dt + d`εt − d`ε

′t

)≤ 2

(ρεt − ρε

′t

)+ ((Aε′(ρεt )− Aε′(ρε

′t )) dt + d`εt − d`ε

′t

)≤ 0

I the first inequality: monotonicity of ε 7→ AεI the second: monotonicity of Aε(·) and

ρεt > ρε′

t ≥ 0 =⇒ d`εt = 0 and − d`ε′

t ≤ 0.

Thereforeρε ≤ ρε′ , ∀ ε > ε′ > 0,

i.e. ε 7→ ρε is monotone non-increasing.Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour

Proof

Note now that γε := ε+ ρε satisfies

γεt = ε+ wt +

∫ t

0g(γεs ) ds + `εt , t ∈ [0,T],

with γε ≥ ε, d`ε ≥ 0, (γε − ε) d`ε = 0. In other words, γε is thesolution to a SDE with reflection at level ε, a Lipschitz non-linearityg(· ∨ ε) and driving function ε+ w; and we expect ε 7→ γε to bemonotone non-decreasing. Indeed, let ε > ε′ > 0. Then

d((

γε′

t − γεt)+)2

=

= 2(γε′

t − γεt)+ (

(g(γε′

t )− g(γεt )) dt + d`ε′

t − d`εt)≤ 0

which implies that

ε+ ρε = γε ≥ γε′ = ε′ + ρε′, ∀ ε > ε′ > 0,

i.e. ε 7→ ε+ ρε = γε is monotone non-decreasing.Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour

Proof

We obtain ε+ ρε ≥ ρε′ ≥ ρε for ε > ε′ > 0, so that (ρε) is a Cauchyfamily in the sup norm as ε ↓ 0; we denote the limit by ρ. Moreover bymonotone convergence∫ t

0g(ε+ ρεs ) ds ↑

∫ t

0g(ρs) ds, ε ↓ 0.

We also obtain that, as ε ↓ 0,

`εt = ρεt − wt −∫ t

0g(ε+ ρεs ) ds

converges to a continuous, monotone non-decreasing function `t with`0 = 0. Since ρε converges uniformly to ρ

0 =

∫ t

0ρεs d`εs →

∫ t

0ρs d`s, t ∈ [0,T].


Bessel processes

TheoremLet δ > 1 and x ≥ 0 and (Bt)t≥0 a standard BM. Then the equationdefining the δ-Bessel process enjoys pathwise uniqueness and existenceof strong solutions.

Proof. Let us set g(x) := δ−12 x−1, x > 0. By the Proposition, a.s. there

exists a unique couple (ρ, `) ∈ (C([0,T]))2 such that ρ ≥ 0,ρ−1 ∈ L1(0,T), `0 = 0, d` ≥ 0, ρ d` = 0 and

ρt = x + Bt +δ − 1

2

∫ t

0ρ−1

s ds + `t, t ∈ [0,T]. (13)

By Proposition 14 if (ρ, `) and (ρ, ˆ) with ` = ˆ≡ 0 satisfy (13), thenρ ≡ ρ, which proves pathwise uniqueness for (11).


Proof

We must now prove that the solution (ρ, `) satisfies a.s. ` ≡ 0. Let

ρεt = wt +

∫ t

0fε(ρεs ) ds + `εt ,

fε(x) :=δ − 1

2(ε+ x)−1, Fε(x) =

δ − 12

log(ε+ x).

We use (6) with e2Fε(x) = (ε+ x)δ−1 and obtain∫R+

(ε+ x)δ−1 E(`εt (x)) dx =t2εδ−1.

As ε ↓ 0 we obtain ∫R+

xδ−1 E(`t(x)) dx = 0.

Since R+ 3 x 7→ E(`t(x)) is a monotone non-increasing function, weobtain that E(`t(x)) = 0 for all x > 0 and all t ≥ 0. In order to provethat E(`t(0)) = 0, we can show that a.s. R+ 3 x 7→ `t(x) is continuous.See the lecture notes.


Squared Bessel Processes

The classical construction of Bessel processes is different: first onedefines the δ-squared Bessel process, namely the only non-negativesolution to the SDE

Zt = y + δt + 2∫ t

0

√Zs dBs, t ≥ 0, (14)

where y, δ ≥ 0. By the classical Yamada-Watanabe theorem, thisequation enjoys pathwise uniqueness, see e.g. [Revuz-Yor, TheoremIX.3.5]. One then defines the δ-Bessel ρt :=

√Zt, and shows with the

Ito formula that ρ solves (11) with x =√

y, at least for δ ≥ 2.

With equation (14) one can introduce at once the whole family ofδ-Bessel processes with δ ≥ 0, while the SDEs (11)-(12) yield thecorrect process only for δ ≥ 1: we shall discuss the case δ < 1 later.


Hitting of 0

TheoremLet T0 := inft > 0 : Xt = 0. Fix δ ≥ 1. Then we have

1. Pδx(T0 < +∞) = 1 for all x > 0 iff δ < 2.

2. Pδx(T0 < +∞) = 0 for all x > 0 iff δ ≥ 2.

Proof. By scaling and uniqueness in law, we have the followingproperty for the solution to (11): for all c > 0

the law of (cXc−2t)t≥0 under Pδx is given by Pδcx,

namely the δ-Bessel process is self-similar of index 12 . In particular,

Pδx(T0 < +∞) = Pδ1(T0 < +∞) for all x > 0.


Proof

Let δ ≥ 1 and ν := δ2 − 1 ≥ −1

2 . If we set for t ≥ 0,

At :=

∫ t

0exp(2(Bs + νs)) ds, τt := infu ≥ 0 : Au > t,

then under Pδ1 the process (Xt, t ∈ [0,T0[) has the same law as(exp(ξτt), t ∈ [0,A∞[), where we set ξt := Bt + νt. This is a particularcase of the so-called Lamperti formula.

Since Btt → 0 a.s. as t→ +∞, we obtain that a.s.

A∞ :=

∫ +∞

0exp(2(Bs + νs)) ds

is finite if ν < 0 and it is infinite if ν > 0. For the case ν = 0, see thelecture notes.


Proof

Another possible proof: via the scale function of Bessel processes

s(x) :=

−x2−δ if δ > 2

log x if δ = 2

x2−δ if δ < 2

x ≥ 0,

then under Pδx the process s(Xt∧T0)t≥0 is a local martingale; moreover,setting and Ta := inft ≥ 0 : Xt = a, we have

Pδx(Ta < Tb) =s(b)− s(x)

s(b)− s(a), 0 < a ≤ x ≤ b.

Then letting a ↓ 0 and b ↑ +∞ for x > 0, we find thatPδx(T0 < +∞) = 1 for δ < 2 and Pδx(T0 < +∞) = 0 for δ ≥ 2.


Local times at 0 for δ ∈ ]1, 2[Let (La

t )a,t≥0 be the local times a δ-Bessel process as asemi-martingale, i.e. in the sense of (10). By Tanaka’s formula (seeProposition 12)

ρt = ρ+t = ρ+

0 +

∫ t

01(ρs>0) dρs +

12L0

t = ρ0 + ρt − ρ0 +12L0

t

which implies L0t ≡ 0. However

àt := a1−δ La

t , a > 0, `0t := lim

a↓0à

t ,

yields a bi-continuous family (àt )a,t≥0 with the occupation-time

formula ∫ t

0ψ(ρs) ds =

∫R+

ψ(a) àt aδ−1 da.

I τt := infs > 0 : `0s > t is a stable subordinator of index 1− δ

2I t ≥ 0 : ρt = 0 has Hausdorff measure equal to 1− δ

2 .Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour

Diffusion local times

We have already seen the value of the scale function of Besselprocesses. Another useful concept is the speed measure

mδ(dx) := Cδ (x+)−2 δ−1δ−2 dx, δ < 2.

with Cδ > 0. We define for (Bt)t≥0 a BM with local times (Lat )t,a

At :=

∫ t

0Cδ((Bs)

+)−2 δ−1

δ−2 ds =

∫La

t mδ(da)

γt := infu ≥ 0 : Au > t.

TheoremLet ρ be a δ-Bessel process, δ > 0. Then (Bγt)t≥0 and (s(ρt))t≥0 havethe same law. Moreover

`at = La2−δ

γt.

See [Rogers-Williams, 2. vol., V.47-48]Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour

Explicit computations

Let now d ∈ N and B(d)t = (B1

t , . . . ,Bdt ) a d-dimensional Brownian

motion started at 0 ∈ Rd. Then we have the following

PropositionLet a ∈ Rd. Then (|a + B(d)

t |)t≥0 has the same law as the d-Besselprocess started at x = |a|.

Proof.Since for all a ∈ Rd, P(∃t > 0 : a + B(d)

t = 0) is 1 for d = 1 and 0 ford ≥ 2, then by the Ito formula for ρt := |a + B(d)

t | for d ≥ 2

ρt = |a|+∫ t

0

d − 12

1ρs

ds +

∫ t

0

a + B(d)s

|a + B(d)s |· dB(d)

s .

The last term is a continuous martingale with quadratic variation equalto t, and by Lévy’s Theorem it is a standard Brownian motion(Bt)t≥0.


Explicit computations

PropositionLet δ ≥ 1. The transition semigroup (pδt (x, y))t≥0,x≥0,y≥0 of theδ-Bessel process is given for t > 0 and y ≥ 0 by

pδt (x, y) =1t

(yx

)νy exp

(−x2 + y2

2t

)Iν(xy

t

), x > 0,

pδt (0, y) =1

2ν tν+1Γ(ν + 1)y2ν+1 exp

(−y2

2t

),

where ν := δ2 − 1 and Iν is the modified Bessel function of index ν

Iν(z) :=∞∑

k=0

(z/2)2k+ν

k! Γ(ν + k + 1), z ≥ 0.

In particular pt(x, y) ∼ Cy2ν+1 = Cyδ−1 as y ↓ 0.


Two exercises

Let f : R 7→ R be a smooth function with bounded first derivative f ′

and consider the following SDE in R

Zt = x +

∫ t

0f (Zs) ds + Bt, t ≥ 0.

Let Qx be the law of (Zt)t≥0.

ExerciseLet F : R 7→ R such that F′ = f . With the Girsanov theorem and theIto formula show that

Qx|F 0t

= exp(

F(Xt)− F(x)− 12

∫ t

0( f ′(Xs) + f 2(Xs)) ds

)·Wx|F 0

t.

We recall that Wx is the law of (x + Bt)t≥0 and (Xt)t≥0 is the canonicalprocess.


Two exercises

We denote now by (Qx,y)y∈R a regular conditional distribution forQx[· |X1 = y], or in other words for the law of (Zt, t ∈ [0, 1]) given Z1,i.e. for all Borel set A ⊂ C([0, 1])

Qx,y(A) = Qx[A |X1 = y].

Let Wx,y be the law of the Brownian bridge from x to y over theinterval [0, 1], namely of (x(1− t) + ty + Bt − tB1)t∈[0,1].

ExerciseWe have for all a, b ∈ R

Qa,b =1

Za,bexp

(−1

2

∫ 1

0( f ′(Xs) + f 2(Xs)) ds

)·Wa,b,

where Za,b ∈ ]0,+∞[ is a normalisation constant.


Back to Bessel processes

We want to consider for c > 0 and δ = 1 + 2c > 1,

f (x) =cx1(x>0) =

δ − 12

1x1(x>0),

which is of course not smooth and therefore requires some additionalcare. Let us note that the common term in the above exercises becomes

−12

∫ 1

0( f ′(Xs) + f 2(Xs)) ds =

c− c2

2

∫ 1

0

1X2

sds.

Note that

c− c2

2= −(δ − 1)(δ − 3)

8

< 0 for δ > 3= 0 for δ = 3> 0 for δ < 3.


Absolute continuity results

We define T0 := inft > 0 : Xt = 0.PropositionLet x > 0 and δ ≥ 2. Then

Pδx |F 0t

=

(Xt∧T0

x

) δ−12

exp(−(δ − 1)(δ − 3)

8

∫ t

0

1X2

sds)·Wx|F 0

t.

Proof. If f (x) = δ−12

1x 1(x>0), then since XT0 = 0

exp(

F(Xt∧T0)− F(x)− 12

∫ t∧T0

0( f ′(Xs) + f 2(Xs)) ds

)=

=

(Xt∧T0

x

) δ−12

exp(−(δ − 1)(δ − 3)

8

∫ t

0

1X2

sds).

See the lecture notes for the full proof.


Bessel bridges

Let x, y ≥ 0. For all Borel set A ⊂ C([0, 1])

Pδx,y(A) = Pδx [A |X1 = y].

Let Wx,y the law of (x(1− t) + ty + Bt − tB1)t∈[0,1], Brownian bridgefrom x to y over the interval [0, 1].

CorollaryLet a, b > 0. Then

P3a,b =

1(T0>1)

1− exp (−2ab)·Wa,b = Wa,b( · |T0 > 1). (15)

For δ ≥ 2

Pδa,b =1

Zδa,bexp

(−(δ − 1)(δ − 3)

8

∫ 1

0

1X2

sds)· P3

a,b.


The Brownian excursion

Since a, b 7→ Pδa,b is continuous in the weak topology, we obtain

Corollary

P30,0 = lim

a↓0Wa,a( · |T0 > 1) = lim

a↓0W0,0( · | inf X ≥ −a). (16)

P30,0 is called the law of the normalised Brownian excursion.

Note that

Wa,a(T0 > 1) = W0,0(inf X ≥ −a) = 1− exp(−2a2),

W0,0(T0 > 1) = W0,0(inf X ≥ 0) = 0,

so that the above conditioning is singular.


Chapter 4: The stochastic heat equation

PropositionLetH be a separable Hilbert space. There exists a process(W(h), h ∈ H) such that h 7→ W(h) is linear, W(h) is a centered realGaussian random variable and

E(W(h) W(k)) = 〈h, k〉H, ∀ h, k ∈ H.

Proof.Let (Zi)i be a i.i.d. sequence of real standard Gaussian variables and(hi)i a complete orthonormal system inH and set

W(h) :=+∞∑i=1

〈h, hi〉H Zi, h ∈ H.

This series converges in L2(Ω).


White noise

Let now (T,B,m) be a separable measurable space, with m a σ-finitemeasure. We apply the Proposition toH := L2(T,B,m), withcanonical scalar product

〈h, k〉H :=

∫T

h(x) k(x) m(dx), h, k ∈ L2(T,B,m).

The process (W(h), h ∈ H) is called a white noise over (T,B,m).

If A ∈ B and m(A) < +∞, then 1A ∈ H and we denoteW(A) := W(1A). If A,B ∈ B with m(A) + m(B) < +∞ then

E(W(A) W(B)) = m(A ∩ B).

In particular, if m(A ∩ B) = 0, then W(A′),A′ ⊆ A andW(B′),B′ ⊆ B are independent.


White noise

0

A1

A2

A3

1

Figure : The family (W(Ai))i is Gaussian and independent since the sets Ai arepairwise disjoint. W(Ai) ∼ N (0,m(Ai)) and W(∪iAi) ∼ N (0,

∑i m(Ai)).

For all measurable set A, W(A) is the amount of noise contained in A.


Finite dimensional white noise

Let us consider first the easiest case: T = 1, . . . , d and m is thecounting measure.

In this case L2(T,B,m) = Rd and the white noise W(h) can be realisedas W(h) = 〈W, h〉Rd , where W ∼ N (0, I).


Brownian motion

Let now T = R endowed with the Borel σ-algebra and the Lebesguemeasure λ1. Then for all [a, b] and [c, d] in R

E(W([a, b]) W([c, d])) = λ1([a, b] ∩ [c, d]).

Then the process

Bt :=

W([0, t]), t ≥ 0,

W([t, 0]), t < 0.(17)

is (a modification of) a two-sided standard Brownian motion, andW(dt) is simply called white noise over R. In particular, the process(W([0, t]), t ≥ 0) is (a modification of) a standard BM.


Multi-dimensional Brownian motion

Let now T = R× 1, . . . , d endowed with the Borel σ-algebra andthe measure λ1 ⊗ m where λ1 is the Lebesgue measure and m is thecounting measure.

Then for all [a, b] and [c, d] in R and for any i, j ∈ 1, . . . , d

E(W([a, b]× i) W([c, d]× j)) = λ1([a, b] ∩ [c, d])1(i=j).

Then the process (B1t , . . . ,B

dt ), defined by

Bit :=

W([0, t]× i), t ≥ 0,

W([t, 0]× i), t < 0.

is (a modification of) a two-sided standard Brownian motion in Rd.


SDEs with additive noise

Then a standard SDE with additive noise

dXt = b(Xt) dt + dBt

can be written, if Bt = W([0, t]), as

Xt = b(Xt) + W = b(Xt) + Bt.

This is just a notation and both formulae should be interpreted as

Xt = X0 +

∫ t

0b(Xs) ds + Bt, t ≥ 0.


Brownian sheet

If T = R2+ = [0,+∞[2 endowed with the Borel σ-algebra and the

Lebesgue measure λ2, then

E(W([0, t]× [0, t′]) W([0, s]× [0, s′])) = (t ∧ s) (t′ ∧ s′)

The process (B(t, s) := W([0, t]× [0, s]), t, s ≥ 0) is called a Browniansheet and W(dt, ds) a space-time white noise. One can also use thenotations

W(dt, ds) =∂2B∂t∂s

= W(t, s).

Notice that the same construction can be done if T = Rd+: this gives a

space-time white noise with a d-dimensional space variable.


Space-time white noise and Cylindrical Brownian motion

LemmaLet (ei)i be a complete orthonormal system in L2([0,+∞[). Then

1. Let wit := W(1[0,t] ⊗ ei), t ≥ 0, i ∈ N. Then (wi)i is an iid

sequence of standard Brownian motions.

2. For all h ∈ L2([0,+∞[) and t ≥ 0

W(1[0,t] ⊗ h) =∑

i

wit 〈ei, h〉

where the equality is in L2(Ω).

A cylindrical Brownian motion in a separable Hilbert space H is

〈Wt, h〉 :=∑

i

Bit 〈ei, h〉, t ≥ 0,

where (ei)i is a complete orthonormal system in H and (Bi)i is an iidsequence of Brownian motions.


The stochastic heat equation

We want to study the stochastic PDE

∂v∂t

=12∂2v∂x2 + W,

v(t, 0) = v(t, 1) = 0, t ≥ 0,

v(0, x) = v0(x), x ∈ [0, 1]

(18)

where W(t, x) is a space-time white-noise over [0,+∞[×[0, 1].

This is the first SPDE we encounter. It is interpreted in the PDE-weaksense: for all h ∈ C2

c(0, 1) and t ≥ 0

〈vt, h〉 = 〈v0, h〉 +12

∫ t

0〈vs, h′′〉 ds +

∫ t

0

∫ 1

0h(x) W(ds, dx).


The deterministic heat equation

Let us start from the heat equation without noise:

∂v∂t

=12∂2v∂x2 ,

v(t, 0) = v(t, 1) = 0, t > 0

v(0, x) = v0(x), x ∈ [0, 1]

(19)

where v0 ∈ H := L2(0, 1). We set for all k ≥ 1:

ek(x) :=√

2 sin(kπx), x ∈ [0, 1]. (20)

Note that ekk≥1 is a complete basis of eigenvectors of the secondderivative with homogeneous Dirichlet boundary conditions:

d2

dx2 ek = −(πk)2ek, ek(0) = ek(1) = 0, k ≥ 1.


Fourier decomposition

We set

λk :=(πk)2

2, k ≥ 1,

The solution of the heat equation (19) is therefore

v(t, x) =∑k≥1

e−tλk〈ek, v0〉 ek(x), t > 0, x ∈ [0, 1].

Since |ek(x)| ≤√

2 and∑

k≥1 e−tλk km < +∞ for all m ∈ N, the aboveseries converges uniformly on [ε,+∞[×[0, 1] for all ε > 0, togetherwith all its partial derivatives in t and x. One can write more compactly,using the semigroup notation,

vt := v(t, ·) = etAv0 :=∑k≥1

e−tλk〈ek, v0〉 ek, t ≥ 0.


Fourier expansion of the SHE

Let us consider the scalar product of both terms of (18) and ek. Settingvk

t := 〈v(t, ·), ek〉 we obtain

dvkt = −(kπ)2

2vk

t dt + dBkt , vk

0 = 〈v0, ek〉

Bkt :=

∫[0,t]×[0,1]

ek(x) W(ds, dx) = W(1[0,t] ⊗ ek

).

We proved in Lemma 26 that (Bkt , t ≥ 0)k≥1 is an independent

sequence of Brownian motions. Then (vk· )k≥1 is an independent family

of O-U processes of respective parameter λk > 0, with

vkt = e−λktvk

0 +

∫ t

0e−λk(t−s) dBk

s , t ≥ 0.

An important remark is the following:∑k

1λk

=∑

k

2(πk)2 < +∞.


Fourier expansion of the SHE

LemmaFor all t ≥ 0 and v0 ∈ H = L2(0, 1) the series

Vt(v0) :=

+∞∑k=1

vkt ek = etAv0 +

+∞∑k=1

(∫ t

0e−λk(t−s) dBk

s

)ek (21)

converges in L2(Ω; H) to a well-defined r.v. with values in H.

Proof.Since vk

t ∼ N(

e−λktvk0,

12λk

(1− e−2λkt))

, then

E

∥∥∥∥∥m∑

k=n+1

vkt ek

∥∥∥∥∥2 =

m∑k=n+1

[e−2λkt (vk

0)2

+1

2λk(1− e−2λkt)

]→ 0

as n,m→ +∞, since λk = Ck2.


Path continuity

Until now we have considered Vt as a L2(0, 1)-valued random variable.However, if x ∈ [0, 1] is fixed, then

vn(t, x) :=

n∑k=1

vkt ek(x) =

n∑k=1

〈vt, ek〉 ek(x) ∈ R

is well defined and continuous. The deterministic part etAv0 is simple toanalyse, therefore let us suppose that v0 = 0. Then since e2

k(·) ≤ 2

E(|vn(t, x)− vm(t, x)|2

)=

m∑k=n+1

E((

vkt)2)

e2k(x)

≤ 2m∑

k=n+1

∫ t

0e−2λk(t−s) ds = 2

m∑k=n+1

1− e−2λkt

2λk→ 0

as n,m→ +∞. Therefore, there exists a well defined stochasticprocess (v(t, x), t ≥ 0, x ∈ [0, 1]), limit in L2(P) of(vn(t, x), t ≥ 0, x ∈ [0, 1]) as n→∞.


Path continuity

LemmaLet v0 = 0. For all m ∈ N there exists a constant Cm < +∞ such that

E(|v(t, x)− v(s, y)|2m

)≤ Cm

(|t − s|m2 + |x− y|m

),

for all t, s ≥ 0, x, y ∈ [0, 1].

Proof. Since (v(t, x)− v(s, y)) is a real Gaussian variable with 0 mean,in order to estimate its moments it is enough to compute the secondone, i.e. it is enough to prove that for some constant C

E(|v(t, x)− v(s, y)|2

)≤ C

(|t − s| 12 + |x− y|

).

First we have

|vn(t, x)− vn(s, y)|2 ≤ 2 |vn(t, x)− vn(s, x)|2 + 2 |vn(s, x)− vn(s, y)|2.


Proof

Now (recall that v0 = 0)

E(|vn(s, x)− vn(s, y)|2

)=

= E

∣∣∣∣∣n∑

k=1

vks (ek(x)− ek(y))

∣∣∣∣∣2

=

n∑k=1

1− e−2λks

2λk(ek(x)− ek(y))2 ≤

n∑k=1

1 ∧ (|x− y| k)2

k2

≤ 1 ∧ |x− y|+∫ ∞

1

1 ∧ (|x− y| k)2

k2 dk ≤ 3 |x− y|.


Proof

With similar computations

E(|vn(t, x)− vn(s, x)|2

)=

=

n∑k=1

e2k(x)

[(1− e−λk(t−s))2 1− e−2λks

2λk+ e−2λks 1− e−2λk(t−s)

2λk

]

≤ 2n∑

k=1

1 ∧ (|t − s| k2)

k2 ≤ 2(

1 ∧ |t − s|+∫ ∞

1

1 ∧ (|t − s| k2)

k2 dk)

≤ 6√|t − s|.

We have used the fact that

(1− e−λk(t−s))2 ≤ (1 ∧ [λk(t − s)])2 ≤ 1 ∧ [λk(t − s)].

The desired result follows if we let n→∞.


Path continuity

PropositionLet v0 = 0. There exists a continuous modification of(v(t, x), t ≥ 0, x ∈ [0, 1]), that we call v again. Moreover a.s. for allε ∈ ]0, 1[ and T < +∞

supx,y∈[0,1], t,s∈[0,T]

|v(t, x)− v(s, y)||t − s| 1−ε4 + |x− y| 1−ε2

< +∞.

This proposition is a consequence of the Kolmogorov criterion statedfor a inhomogeneous distance on Rd.


Two points of view

The solution to the stochastic heat equation can be seen:I as a continuous L2(0, 1)-valued process (Vt)t≥0, solving the SDE

dV = AV dt + dWt

where A is ∂2x with homogeneous Dirichlet boundary condition

andWt :=

∑i

Bit ei

is a cylindrical Brownian motion in L2(0, 1). This is theDa Prato-Zabczyk approach.

I as a bi-continuous real valued process (v(t, x), t ≥ 0, x ∈ [0, 1]).This is the Walsh approach.


The invariant measure

A probability measure µ on H is said to be invariant for (Vt)t≥0, see(27), if, given a r.v. Z with law µ and independent of W, the process(Vt(Z))t≥0 is stationary.

An invariant probability measure µ of (Vt)t≥0 is reversible if for allT ≥ 0, the processes (Vt(Z))t∈[0,T] and (VT−t(Z))t∈[0,T] have the samelaw.

PropositionThe law W0,0 of the brownian bridge (Bx − xB1, x ∈ [0, 1]), with B astandard BM, is the unique invariant probability measure of (Vt)t≥0.Moreover W0,0 is reversible for (Vt)t≥0.

The proof of this proposition is split into several (simple) lemmas.



First we have the Karhunen-Loève decomposition of the Brownianbridge.

LemmaLet µt,v0 the law on H = L2(0, 1) of Vt(v0), solution to (18). Thenµt,v0 =⇒ µ, law of the H-valued r.v.

β :=+∞∑k=1

1πk

Zk ek =

+∞∑k=1

1√2λk

Zk ek, (22)

where (Zk)k≥1 is an i.i.d. sequence of N (0, 1) variables and (ek)k isthe complete orthonormal system of H defined in (20).

Proof.Recall that vk

t ∼ N(

e−λktvk0,

12λk

(1− e−2λkt))

.



LemmaThe L2(0, 1)-valued r.v. β has law W0,0.

Proof.

E (βx βy) =

+∞∑k=1

1(πk)2 ek(x) ek(y) =: α(x, y), x, y ∈ [0, 1].

−∂

2α

∂x2 = δy(dx),

α(0, y) = α(1, y) = 0.

The unique solution to this equation is

α(x, y) = x ∧ y− xy x, y ∈ [0, 1]

which is the covariance function of a Brownian bridge on [0, 1].


What we have done so far

I We have seen that a reflected dynamics is associated with aconditioning of the invariant measure.

I The stochastic heat equation has the law of the Brownian bridgeas invariant measure

I The Brownian bridge conditioned to be non-negative is the3-Bessel bridge, i.e. the normalised Brownian excursion.

Therefore we expect the reflected stochastic heat equation to have thelaw of 3-Bessel bridge as invariant measure.

The next step is the construction of the solution u to this SPDE withreflection.

Then we shall see the impact that the fine properties of these Brownianprocesses have on u.


Ito formula

Since W =∑

k dBkt ek, we understand why the classical Ito calculus

runs into trouble: if ϕ ∈ C2c(R) then formally

dϕ(v(t, x)) =

(12∂2v∂x2 (t, x) dt +

∑k

dBkt ek(x)

)ϕ′(v(t, x))

+12ϕ′′(v(t, x))

∑k

〈ek, ek〉 dt

and∑

k〈ek, ek〉 = +∞.


Ito formula

In fact the two terms in red

dϕ(v(t, x)) =

(12∂2v∂x2 (t, x) dt +

∑k

dBkt ek(x)

)ϕ′(v(t, x))

+12ϕ′′(v(t, x))

∑k

〈ek, ek〉 dt

are ill-defined and compensate each other.

This is a renormalisation phenomenon.


Semilinear SPDEs

Let f : R 7→ R such that there exists a constant L > 0 with

| f (u)− f (v)| ≤ L|u− v|, ∀ u, v ∈ R.

We study the following SPDE in (t, x) ∈ R+ × [0, 1]

∂u∂t

=12∂2u∂x2 + f (u) + W

u(t, 0) = u(t, 1) = 0

u(0, x) = u0(x)

Interpreted in the PDE-weak sense

〈ut, h〉 = 〈u0, h〉 +12

∫ t

0〈us, h′′〉 ds +

∫ t

0〈 f (us), h〉 ds

+

∫ t

0

∫ 1

0h(x) W(ds, dx)

(23)


Semilinear SPDEs

We set Ut := u(t, ·), Ukt := 〈Ut, ek〉, f k

t := 〈 f (Ut), ek〉,

Bkt :=

∫[0,t]×[0,1]

ek(x) W(ds, dx) = W(1[0,t] ⊗ ek

).

Then

dUkt =

(−λk Uk

t + f kt)

dt + dBkt , Uk

0 = 〈u0, ek〉

and, by applying the Ito formula to the process t 7→ eλktUkt we obtain

Ukt = e−λktUk

0 +

∫ t

0e−λk(t−s)f k

s ds +

∫ t

0e−λk(t−s) dBk

s .

The (Bk)k are still iid BMs but the non-linearity f couples the Uk’s.


Semilinear SPDEs

If we sum again over the Fourier basis

Ut =∑

k

〈Ukt , ek〉 ek = etAu0 +

∫ t

0e(t−s)A f (Us) ds + WA(t),

where the stochastic convolution

WA(t) :=

+∞∑k=1

(∫ t

0e−λk(t−s) dBk

s

)ek

is the solution of the SHE (18) with null initial condition, see (21).

If f is Lipschitz, then we can write a fixed point argument inET := C([0,T]; H) with Γ : ET 7→ ET defined by Γ(g) := h, where

ht := etAu0 +

∫ t

0e(t−s)A f (gs) ds + WA(t), t ∈ [0,T].


Higher space dimension

One can study the stochastic PDE for t ≥ 0, x ∈ Rd

∂v∂t

=12

∆v + W,

where W(t, x) is a space-time white-noise over [0,+∞[×Rd.

The solution is well-defined as a Gaussian field, but has no continuousmodification.

In fact, its solutions live in distribution spaces: the higher thedimension, the more irregular the solution.

Non-linear SPDEs become much more difficult and one needs thetheory of Regularity Structures.

Reflection seems out of reach.


Chapter 5: Obstacle problems

Let a ≥ 0. We fix a space-time white noise W on [0,+∞[×[0, 1]. Westudy the following SPDE with reflection:

∂u∂t

=12∂2u∂x2 + f (u) + W + η

u(0, x) = u0(x), u(t, 0) = u(t, 1) = a

u ≥ 0, dη ≥ 0,∫

u dη = 0

(24)

where we assume that:

1. u0 : [0, 1] 7→ R is continuous and u0 ≥ 0.

2. f : R 7→ R is Lipschitz and bounded.


The Nualart-Pardoux equation

DefinitionA pair (u, η) is said to be a solution of equation (24) if a.s.

1. (u(t, x), t ≥ 0, x ∈ [0, 1]) is continuous

2. u ≥ 0, u(t, 0) = u(t, 1) = a for all t ≥ 0 and u(0, ·) = u0

3. η(dt, dx) is a measure on ]0,+∞[× ]0, 1[ such thatη(]0,T]× [δ, 1− δ]) < +∞ for all T, δ > 0

4. for all t ≥ 0 and h ∈ C∞c (0, 1)

〈ut, h〉 = 〈u0, h〉 +12

∫ t

0〈us, h′′〉 ds +

∫ t

0〈 f (us), h〉 ds

+

∫ t

0

∫ 1

0h(x) W(ds, dx) +

∫ t

0

∫ 1

0h(x) η(ds, dx)

(25)

5.∫

u dη = 0 or, equivalently, the support of η is contained in(t, x) : u(t, x) = 0.


Reduction to a PDE with random obstacle

Let a ≥ 0 and v be the unique solution to∂v∂t

=12∂2v∂x2 + W,

v(t, 0) = v(t, 1) = a, v(0, x) = u0(x).

(26)

Then the function z := u− v solves

∂z∂t

=12∂2z∂x2 + f (z + v) + η

z(0, x) = 0, z(t, 0) = z(t, 1) = 0

z ≥ −v, dη ≥ 0,∫

(z + v) dη = 0.

(27)

The important remark here is that equation (27) is a PDE (rather than aSPDE) with random obstacle −v.


Existence and uniqueness

TheoremLet w ∈ C([0,T]× [0, 1]) with w(0, ·) ≥ 0, w(·, 0) ≥ 0, w(·, 1) ≥ 0.Then there exists a unique pair (z, η) such thatI z ∈ C([0,T]× [0, 1]), z(0, ·) = 0, z(·, 0) = z(·, 1) = 0

I η(dt, dx) is a measure on ]0,T]× ]0, 1[ such thatη(]0,T]× [δ, 1− δ]) < +∞ for all δ > 0

I For all t ∈ [0,T] and h ∈ C∞c (0, 1)

〈zt, h〉 =12

∫ t

0〈zs, h′′〉 ds +

∫ t

0〈 f (zs + ws), h〉 ds

+

∫ t

0

∫ 1

0h(x) η(ds, dx)

(28)

I z ≥ −w,∫

(z + w) dη = 0.


Uniqueness

Let (z, η) and (z, η) two solutions. If z− z is regular enough

ddt‖z− z‖2 =

= 〈z− z, ∂xx(z− z) + 2( f (z)− f (z))〉+ 2∫ 1

0(z− z)(dη − dη)

≤ −‖∂x(z− z)‖2 + 2L‖z− z‖2 − 2∫ 1

0(z dη + z dη)

≤ 2L‖z− z‖2

where ‖ · ‖ is the norm in L2(0, 1). Since ‖z− z‖2 is zero at t = 0, weconclude that z ≡ z. This argument is not fully rigorous since z− zdoes not necessary enjoy the necessary regularity for the previouscomputations. However Nualart and Pardoux proved that this argumentcan be made rigorous.


Existence: penalisation

We introduce the following approximating problem:∂zε

∂t=

12∂2zε

∂x2 + f (zε + w) +(zε + w)−

ε

zε(0, ·) = 0, zε(t, 0) = zε(t, 1) = 0,

(29)

with ε > 0. We set

fε(r) := f (r) +(r−)

ε, r ∈ R.

We suppose as usually that f is monotone non-increasing, so that

fε(r)− fε(r′) ≤ 0, ∀ r > r′.


Monotonicity

We show that zε ≤ zε′

if ε > ε′. Since fε(·) ≤ fε′(·) and with anintegration by parts:

ddt

∥∥∥(zε − zε′)+∥∥∥2

=

= 2⟨

(zε − zε′)+,

12∂xx(zε − zε

′) + fε(zε + w)− fε′(zε

′+ w)

⟩≤ −

∥∥∥∂x(zε − zε′)+∥∥∥2

+ 2⟨

(zε − zε′)+, fε′(zε + w)− fε′(zε

′+ w)

⟩≤ 0

where ‖ · ‖ is the norm in L2(0, 1).

Since zε0 − zε′

0 = 0, we conclude that zε ≤ zε′.


The rest of the proof

Three more steps, as for one-dimensional diffusions:I The case of a smooth obstacleI Continuity of the map w 7→ zε uniformly in ε > 0I Conclusion.

The crucial point is the maximum principle of ∂2xx, i.e. the fact that

monotonicity is preserved.

In our computations it was the simple formula:

〈(zε − zε′)+, ∂xx(zε − zε

′)〉 = −

∥∥∥∂x(zε − zε′)+∥∥∥2≤ 0.


The penalised SPDE

Setting w := v and uε := zε + v, where v is defined by (26) and zε by(29), then uε solves the SPDE

∂uε

∂t=

12∂2uε

∂x2 + f (uε) +(uε)−

ε+ W

uε(0, ·) = u0, uε(t, 0) = uε(t, 1) = a.

This is a particular example of (23).

It will be useful in order to compute the invariant measure of thereflected dynamics.



We already expect the invariant measure of (ut)t≥0 to be P3a,a, law of

the 3-Bessel bridge a→ a ≥ 0 (for f = 0).

Let us consider a Brownian motion (B(d)t )t≥0 in Rd where

B(d) = (B1, . . . ,Bd) and the Bi’s are iid standard BMs.

Let V : Rd → R be a smooth function and let

dXt = ∇V(Xt) dt + dB(d)t , X0 = x ∈ Rd.

Then one can show that an invariant measure for (Xt)t≥0 is given by

exp(2V(x)) dx.

If this measure is finite on Rd, we obtain an invariant probabilitymeasure.


Gradient systems

Let us consider again a semi-linear SPDE

∂u∂t

=12∂2u∂x2 + f (u) + W.

Then W =∑

k Bk ek where (ek)k is a c.o.s. in H := L2(0, 1) such that

d2

dx2 ek = −(πk)2ek, ek(0) = ek(1) = 0, k ≥ 1,

recall (20). Question: can we find V : H 7→ R such that

du = ∇V(u) dt + dW.

Here the gradient∇ is in the Hilbert-structure of H:

〈∇V(ζ), h〉 = limε→0

1ε

(V(ζ + εh)− V(ζ)).


Gradient systems

We set A : D(A) ⊂ H = L2(0, 1)→ H

D(A) =

h ∈ H :∑k≥1

λ2k〈ek, h〉2 < +∞

,

Ah = −∑k≥1

λk 〈ek, h〉 ek, h ∈ D(A),

recall that∑

k λ−1k < +∞. If h ∈ C2

c(0, 1) then

Ah =12

d2hdx2 .

Then we set for ζ ∈ D(A)

V(ζ) :=12〈Aζ, ζ〉+ 〈F(ζ), 1〉

where F : R→ R, F′ = f and

〈F(ζ), 1〉 =

∫ 1

0F(ζx) dx.


Gradient systems

Then, if R 3 ε→ 0 and h ∈ D(A),

1ε

(〈A(ζ + εh), ζ + εh〉 − 〈Aζ, ζ〉) = 2〈Aζ, h〉+ ε〈Ah, h〉,

where we have used 〈Aζ, h〉 = 〈ζ,Ah〉, and

1ε〈F(ζ + εh)− F(ζ), 1〉 =

∫ 1

0

1ε

(F(ζx + εhx)− F(ζx)) dx

→∫ 1

0F′(ζx) hx dx = 〈 f (ζ), h〉.

Therefore〈∇V(ζ), h〉 = 〈Aζ + f (ζ), h〉,∇V(ζ) = Aζ + f (ζ).


Gradient systems: finite-dimensional approximation

Let now n ≥ 1. We note Hn the linear span of e1, . . . , en andB(n)

t :=∑n

k=1 Bkt ek. Let Πn : H 7→ Hn be the orthogonal projection

Πnh :=

n∑k=1

〈h, ek〉 ek, h ∈ H.

Then we set An : H 7→ Hn and f n : H 7→ Hn:

An := ΠnAΠn = AΠn = ΠnA, f n := Πn f Πn.

Let ζn be the solution to the SDE in Hn

ζnt = ζn

t (z) = z +

∫ t

0(Aζn

s + f n(ζns )) ds + B(n)

t , z ∈ Hn.

TheoremFix u0 ∈ H. Then a.s. (ζn

t (Πnu0))t≥0 converges to u in C([0,T]; H).


Gradient systems: finite-dimensional approximation

We define the probability measure on Hn

νn(dζn) :=1Zn

exp (2Vn(ζn)) dζn, ζn ∈ Hn,

where dζn is the n-dimensional Lebesgue measure on the Hilbert spaceHn and Zn is a normalisation constant.

Then it is well known that νn is invariant and reversible for (ζnt )t≥0,

namely if Zn has law νn and is independent of W then (ζnt (Zn))t≥0 is

stationary and for all T ≥ 0, the processes (ζnt (Zn))t∈[0,T] and

(ζnT−t(Zn))t∈[0,T] have the same law.

Can we pass to the limit as n→ +∞?


Invariant measures

Since 2V(ζ) = 〈Aζ, ζ〉+ 2〈F(ζ), 1〉

νn(dζn) :=1Zn

exp (2〈F(ζn), 1〉)N (0,Qn)(dζn)

where Qn = diag(

12λ1, . . . , 1

2λn

).

Recall now (22). Then N (0,Qn) =⇒W0,0, law of β. It is thereforesimple to show that

νn =⇒ ν :=1Z

exp (2〈F(ζ), 1〉) W0,0(dζ).

By Skorokhod’s representation theorem we can suppose that Zn

converges a.s. to Z in H.


A technical estimate

Under the usual one sided bound

f (r)− f (r′) ≤ L(r − r′), ∀ r > r′,

we have the following estimate:

‖ζnt (z)− ζn

t (z)‖ ≤ eLt‖z− z‖.

This can be proved in the usual way, denoting ζt := ζnt (z) and

ζt := ζnt (z)

ddt‖ζt − ζt‖2 = 2

⟨ζt − ζt, A

(ζt − ζt

)+ fε(ζt)− fε(ζt)

⟩≤ −

∥∥∥∂x

(ζt − ζt

)∥∥∥2+ 2

⟨ζt − ζt, f (ζt)− f (ζt)

⟩≤ 2L‖ζt − ζt‖2.



Therefore we haveI (ζn

t (Πnu0))t≥0 converges a.s. to (ut)t≥0 as n→ +∞I z 7→ ζn

t (z) is Lipschitz, uniformly in nI Zn converges a.s.I (ζn

t (Zn))t≥0 is stationary

Then it is easy to see that (ζnt (Zn))t≥0 converges a.s. to a stationary

solution of the SPDE with reflection.


The penalised invariant measure

We consider now the penalised SPDE∂uε

∂t=

12∂2uε

∂x2 + f (uε) +(uε)−

ε+ W

uε(0, ·) = u0, uε(t, 0) = uε(t, 1) = a.

The invariant measure is

νaε(dζ) :=

1Zaε

exp (2〈Fε(ζ), 1〉) Wa,a(dζ),

where Fε satisfies

Fε(0) = 0, F′ε(y) := f (y) +y−

ε= fε(y).


The penalised invariant measure

Note thatddr

(r−)2

= −2r−.

Then for a > 0

νaε(dζ) =

1Zaε

exp(

2〈F(ζ), 1〉 − 1ε〈(ζ−)2, 1〉)

Wa,a(dζ),

converges as ε ↓ 0 to

νa(dζ) :=1

Za exp (2〈F(ζ), 1〉)1K(ζ) Wa,a(dζ),

where K := u0 : [0, 1]→ R : u0 ∈ L2(0, 1), u0 ≥ 0.


a ↓ 0

We have seen in (15) that

P3a,a = Wa,a( · |T0 > 1) = Wa,a( · |K).

Thenνa(dζ) :=

1Za

exp (2〈F(ζ), 1〉) P3a,a(dζ),

and as a ↓ 0

νa(dζ) =⇒ ν0(dζ) :=1

Zaexp (2〈F(ζ), 1〉) P3

0,0(dζ).

In particular if f ≡ 0 then the invariant measure of the SPDE withreflection is simply P3

a,a.


The situation

I We have constructed the solution (u, η) to the SPDE withreflection (24)

I We have computed the (unique) probability invariant measure

νa(dζ) :=1

Zaexp (2〈F(ζ), 1〉) P3

a,a(dζ),

for a ≥ 0.

Comments:I Walsh and Da Prato-Zabczyk reconciledI In the slides we only prove results on stationary u, in the lecture

notes the general case is treatedI u is the weak limit of discrete interfaces: ask Cyril Labbé for a

confirmation.


The situation



νa(dζ) :=1

Zaexp (2〈F(ζ), 1〉) P3

a,a(dζ),

for a ≥ 0.

Comments:I Walsh and Da Prato-Zabczyk reconciled

I In the slides we only prove results on stationary u, in the lecturenotes the general case is treated

I u is the weak limit of discrete interfaces: ask Cyril Labbé for aconfirmation.


The situation



νa(dζ) :=1

Zaexp (2〈F(ζ), 1〉) P3

a,a(dζ),

for a ≥ 0.


notes the general case is treated

I u is the weak limit of discrete interfaces: ask Cyril Labbé for aconfirmation.


The situation



νa(dζ) :=1

Zaexp (2〈F(ζ), 1〉) P3

a,a(dζ),

for a ≥ 0.


notes the general case is treatedI u is the weak limit of discrete interfaces: ask Cyril Labbé for a

confirmation.


Chapter 6: Integration by parts formulae

Let (ρt, `t)t≥0 be the solution to the SDE with reflection (4)

ρt = x + Bt +

∫ t

0f (ρs) ds + `t.

Then we proved in (6) that∫R+

e2F(x) E(`t(x)) dx =t2

e2F(0), t ≥ 0.

where F′ = f .

This formula allowed us to prove that for δ > 1 the equation

ρt = x + Bt +δ − 1

2

∫ t

0ρ−1

s ds + `t

has reflection term ` ≡ 0 and the solution is a δ-Bessel process.


The Revuz measure of η

For reasons which will be clear later, it would be very useful to have asimilar formula for (u, η).

In this case, such the analogous result would be an explicit formula for∫νa(du0)E

[∫ t

0

∫ 1

0h(x)ϕ(us) η(ds, dx)

]where h : [0, 1]→ R and ϕ : H → R are bounded Borel.

The one-dimensional formula was simpler since there was no spatialvariable x, no h(x) and no ϕ, since∫ t

0ϕ(ρs) d`s = ϕ(0) `t,∫

R+

e2F(x) E[∫ t

0ϕ(ρs(x)) d`s(x)

]dx =

t2ϕ(0) e2F(0).


Integration by parts

Penalisation:

ρnt = x + Bt + n

∫ t

0(ρn

s )− ds +

∫ t

0f (ρn

s ) ds, t ≥ 0.

One can prove that

limε↓0

∫µn(dx)E

[∫ t

0ϕ(ρn

s ) n (ρns )− ds

],

=

∫µ(dx)E

[∫ t

0ϕ(ρs) d`s

].

By stationarity∫µn(dx)E

[∫ t

0ϕ(ρn

s ) n (ρns )− ds

]= t∫µn(dx)ϕ(x) n x−.

We can prove with a change of variable as in (9) that

limn→+∞

∫µn(dx)ϕ(x) n x− =

12ϕ(0) e2F(0).



The difficulty comes from

µn(dx) n x− = e−n (x−)2+2F(x) n x− dx.

However we can write

e−n (x−)2n x− =

12

ddx

e−n (x−)2.

Therefore, if ϕ ∈ C1c(R), we can integrate by parts and obtain∫

µn(dx)ϕ(x) n x− =

∫e2F(x) ϕ(x)

12

ddx

e−n (x−)2dx

= −12

∫ddx

(ϕ e2F) e−n (x−)2

dx→ −12

∫R+

ddx

(ϕ e2F) dx.

If we integrate by parts again, we find now

limn→+∞

∫µn(dx)ϕ(x) n x− =

12ϕ(0) e2F(0), ∀ϕ ∈ C1

c(R).



Can we repeat the same trick for (u, η)?

This means integrating by parts w.r.t. to

P3a,a = Wa,a( · |T0 > 1) = Wa,a( · |K)

where K = u0 : [0, 1]→ R : u0 ∈ L2(0, 1), u0 ≥ 0.



Consider a regular bounded open set O ⊂ Rd. Then the classicalGauss-Green formula states that for all h ∈ Rd∫

O(∂hϕ) ρ dx = −

∫Oϕ∂hρ

ρρ dx −

∫∂Oϕ 〈n, h〉 ρ dσ

I ϕ, ρ ∈ C1b(O) with λ ≤ ρ ≤ λ−1, λ ∈ ]0, 1] is a constant,

I n is the inward-pointing normal vector to the boundary ∂OI σ is the surface measure on ∂OI ∂hϕ is the directional derivative of ϕ along hI ∂h log ρ = (∂hρ)/ρ.

For us, Wa,a = ρ dx, K = O.

What is the analog of ρ dσ? and of n? and of ∂O?


An example

Let µ = N (0, 12λ). Then∫

ϕ′ dµ =

∫2λ xϕ(x)µ(dx)

∫R+

ϕ′ dµ =

∫R+

λ xϕ(x)µ(dx)−√λ

πϕ(0).

Here µ(dx) = ρ(x) dx =√

λπ exp(−λ x2) dx,

∂ρ

ρ(x) = −2λ x.


Integration by parts on the Gaussian measure

∫H∂hϕ dW0,0 = −

∫H〈ζ, h′′〉ϕ(ζ) W0,0(dζ)

I 〈 · , · 〉 is the scalar product in H := L2(0, 1)

I ϕ : H 7→ R is bounded and smoothI h ∈ C2

c(0, 1), h′′ is the second derivative of hI ∂hϕ is the directional derivative of ϕ along h:

∂hϕ(ζ) = limε→0

1ε

(ϕ(ζ + εh)− ϕ(ζ)).


The boundary measure

P3a,a [∂hϕ] =−P3

a,a[ϕ(X) 〈X, h′′〉

]−∫ 1

0dr h(r) γ(r, a) P3

a,a [ϕ(X) |Xr = 0] .

where γ(r, a) ≥ 0 is an explicit function of r ∈ ]0, 1[, a ≥ 0.

∫O

(∂hϕ) ρ dx = −∫

Oϕ∂hρ

ρρ dx −

∫∂Oϕ 〈n, h〉 ρ dσ



By the Markov property and the Brownian scaling of P3a,a

Pδa,b [ϕ(X) |Xr = c] =

∫ϕ(Tr(k1, k2)) Pδa,c(dk1) Pδc,b(dk2)

where for all r ∈ ]0, 1[ we have the scaling-plus-concatenation mapTr : L2(0, 1)× L2(0, 1) 7→ L2(0, 1),

[Tr(k1, k2)](τ) :=

:=√

r k1

(τr

)1(τ≤r) +

√1− r k2

(τ − r1− r

)1(r<τ≤1).



0 1 0 1

0 r 1

1

Figure : A typical path under the boundary measure and the action of Tr.


A random walk model

Let (Yi)i∈N be i.i.d. with Yi ∼ N (0, 1). For n ∈ N and y ∈ R we set

Sn := Y1 + · · ·+ Yn, gt(y) =1√2πt

exp(−y2

2t

).

PN is the law of (S1, . . . , SN) under the conditioning SN+1 = 0. Forall Borel set A ⊆ RN with φ0 := φN+1 := 0

PN(A) = limε↓0

E[1A(S1, . . . , SN)1[0,ε](SN+1)

]P(SN+1 ∈ [0, ε])

=1

CN

∫RN1A(φ) exp −HN(φ) dφ

where

HN(φ) :=12

N+1∑i=1

(φi − φi−1)2.


A conditioned random walk

We set for N ≥ 1, Ω+N := RN

+ = [0,∞[N and

dP+N =

1ZN

exp −HN(φ)1(φ∈Ω+N ) dφ

where ZN is a normalisation constant. In particular

P+N = PN( · |Ω+

N ).

We define the Brownian scaling

RN 3 φ 7→ ΛN(φ)(x) :=1√NφdNxe, x ∈ [0, 1],

PN := Λ∗N(PN), P+N := Λ∗N(P+

N ).

PropositionAs N → +∞, PN converges weakly to W0,0 and P+

N to P30,0.


Integration by parts in finite dimension

LemmaFor all F ∈ C1

b(RN) and c ∈ RN with c0 := cN+1 := 0∫RN

N∑i=1

ci∂F∂φi

dPN = −∫RN

N∑i=1

ci (φi+1 + φi−1 − 2φi) F(φ) PN(dφ)

and passing to the limit N → +∞ under Brownian scaling∫H∂hϕ dW0,0 = −

∫H〈ζ, h′′〉ϕ(ζ) W0,0(dζ)

for all h ∈ C2c(0, 1) and ϕ ∈ C1

b(H).


Integration by parts in finite dimension

LemmaFor all F ∈ C1

b(RN) and c ∈ RN with c0 := cN+1 := 0∫RN+

N∑i=1

ci∂F∂φi

dP+N =

= −∫RN+

N∑i=1

ci (φi+1 + φi−1 − 2φi) F(φ) P+N (dφ)

−N∑

i=1

ciZi−1 ZN−i

ZN

∫Ri−1+ ×R

N−i+

F(α⊕ β) P+i−1(dα) P+

N−i(dβ)

where α⊕ β := (α1, . . . , αi−1, 0, β1, . . . , βN−i).


Proof

The point is that the boundary of RN+ is

∂(RN

+

)=

N⋃i=1

Ri−1+ × 0 × RN−i

+ .

For all i ∈ 1, . . . ,N we can apply a one-dimensional IbPF fixing φj

for j 6= i∫R+

∂F∂φi

dφi = −∫R+

(φi+1 + φi−1 − 2φi) F(φ) dφi−

− 1ZN

F(ψi) exp−HN(ψi)

where ψi = (φ1, . . . , φi−1, 0, φi+1, . . . , φN).


Proof

Now

HN(φ1, . . . , φi−1, 0, φi+1, . . . , φN) =

= Hi−1(φ1, . . . , φi−1) + HN−i(φi+1, . . . , φN).

and therefore

1ZN

∫R1,...,N\i+

F(ψi) exp−HN(ψi)

dψ =

=1

ZN

∫Ri+×R

N−i+

F(α⊕ β) exp −HN(α⊕ β) dα dβ

=Zi−1 ZN−i

ZN

∫Ω+

i−1×Ω+N−i


N−i(dβ).

This concludes the proof.


N → +∞

∫RN+

N∑i=1

ci∂F∂φi

dP+N =

= −∫RN+

N∑i=1

ci (φi+1 + φi−1 − 2φi) F(φ) P+N (dφ)

−N∑

i=1

ciZi−1 ZN−i

ZN

∫Ri−1+ ×R

N−i+


N−i(dβ)

Under Brownian rescaling as N → +∞:

P30,0 [∂hϕ] =−P3

0,0[ϕ(X) 〈X, h′′〉

]−∫ 1

0dr h(r) γ(r, 0) P3

0,0 ⊗ P30,0 [ϕ(Tr)] .



where for all r ∈ ]0, 1[, Tr : L2(0, 1)× L2(0, 1) 7→ L2(0, 1),

[Tr(k1, k2)](τ) :=

:=√

r k1

(τr

)1(τ≤r) +

√1− r k2

(τ − r1− r

)1(r<τ≤1).

0 1 0 1

0 r 1

1


The Revuz measure of η

TheoremFor all bounded Borel ϕ : H 7→ R and h ∈ Cc(0, 1)∫

νa(du0)E[∫ t

0

∫ 1


]=

=t

2Za

∫ 1

0dr h(r) γ(r, a)

∫ϕ(ζ) e2F(ζ) Σa(r, dζ).

where Σa(r, ·) := P3a,a[ · |Xr = 0].


Proof.

We set

η(ds, h) :=

∫ 1

0h(x) η(ds, dx),

ηε(ds, h) :=

∫ 1

0h(x)

1ε

(uε(s, x))− dx ds =1ε〈h, (uεs )−〉 ds,

We need to prove two formulae:

limε↓0

∫νaε(du0)E

[∫ t

0ϕ(uεs ) ηε(ds, h)

]=

=

∫νa(du0)E

[∫ t

0ϕ(us) η(ds, h)

],

limε↓0

∫νaε(du0)E

[∫ t

0ϕ(uεs ) ηε(ds, h)

]=

=1

2Za

∫ 1

0dr h(r) γ(r, a)

∫ϕ(ζ) e2F(ζ) Σa(r, dζ).


Proof.

Let ϕ ∈ C1b(H). By an IbPF∫

νaε(dζ)ϕ(ζ) 〈h, 1

εζ−〉 =

=1

Zaε

∫Wa,a(dζ)ϕ(ζ) 〈h, 1

εζ−〉 exp

(〈2F(ζ)− 1

ε(ζ−)2, 1〉

)=

1Zaε

∫Wa,a(dζ)ϕ(ζ) exp (2〈F(ζ), 1〉)

⟨h,

12∇ exp

(−1ε〈(ζ−)2, 1〉

)⟩= − 1

Zaε

∫Wa,a(dζ)

(〈∇ϕ(ζ), h〉+ ϕ(ζ)

(12〈h′′, ζ〉+ 〈 f (ζ), h〉

))·

· exp(〈2F(ζ)− 1

ε(ζ−)2, 1〉

)= −

∫νaε(dζ)

(〈∇ϕ(ζ), h〉+ ϕ(ζ)

(12〈h′′, ζ〉+ 〈 f (ζ), h〉

))


Proof.

Then, letting ε ↓ 0∫νaε(dζ)ϕ(ζ) 〈h, 1

εζ−〉 →

→ −∫νa(dζ)

(〈∇ϕ(ζ), h〉+ ϕ(ζ)

(12〈h′′, ζ〉+ 〈 f (ζ), h〉

))=

12Za

∫ 1

0dr h(r) γ(r, a)

∫ϕ(ζ) e2F(ζ) Σa(r, dζ)

where we have integrated by parts again in the last equality.


The contact set

We denote by π : [0,+∞[×[0, 1] 7→ [0,+∞[ the projection (t, x) 7→ t,and for a set S ⊂ [0,+∞[×[0, 1] we write

St := x ∈ [0, 1] : (t, x) ∈ S, t ≥ 0.

TheoremLet (u, η) be the stationary solution to equation (24). Let us denote by

C := (t, x) : u(t, x) = 0, t > 0, x ∈ ]0, 1[

the contact set and let us recall that the support of η is contained in C .Then a.s. the set π(C ) has zero Lebesgue measure and there exists ameasurable set S ⊂ C such that

1. η(C \ S) = 0

2. for all t > 0, either St = ∅ or St = rt, with rt ∈ ]0, 1[.

3. if St = rt, then u(t, x) > 0 for all x ∈ ]0, 1[ \rt andu(t, rt) = 0.


The contact set

(S)

••

•• ••

•

•

••

••

••

••

•

•

••

••

•••

•

••

(rt, t)

0 1

t

1


Proof

Let I1, I2 ⊂ ]0, 1[ be closed intervals with I1 ∩ I2 = ∅. Letϕ : C([0, 1])→ R be defined by

ϕ(ζ) := 1(infI1 ζ=infI2 ζ=0), ζ ∈ C([0, 1]).

Then for all h ≥ 0∫νa(du0)E

[∫ t

0

∫ 1


]=

=t

2Za

∫ 1

0dr h(r) γ(r, a)

∫ϕ(ζ) e2F(ζ) Σa(r, dζ)

= 0.

If we consider I1 = [q1, q2], I2 = [q3, q4] with qi ∈ Q then a.s. forη(ds× [0, 1])-a.s. s, u(s, ·) has only one zero over ]0, 1[.


SPDEs with repulsion from 0

We study now the SPDE

∂u∂t

=12∂2u∂x2 +

cu3 + W

u(t, 0) = u(t, 1) = a, t ≥ 0

u(0, x) = u0(x), x ∈ [0, 1]

where a ≥ 0 and c > 0 are fixed, u0 ∈ C([0, 1]) ∩ K and we search forsolutions u ≥ 0.


SPDEs with repulsion

TheoremLet a ≥ 0, c > 0 and u0 ∈ C([0, 1]) ∩ K. Then there exists a uniquecontinuous u : [0,+∞[×[0, 1] 7→ [0,+∞[ such that

1. u−3 ∈ L1loc([0,+∞[× ]0, 1[)

2. A.s. for all t ≥ 0 and h ∈ C∞c (0, 1)

〈ut, h〉 = 〈u0, h〉+12

∫ t

0〈us, h′′〉 ds +

∫ 1

0h(x) W(ds, dx)

+ c∫ t

0

∫ 1

0h(x) u−3(s, x) ds dx.

(30)

If δ > 3 is such that c = (δ−3)(δ−1)8 , then the only invariant probability

measure of (30) is Pδa,a.


Proof

The proof is based on the ideas we used in order to construct Besselprocesses from SDEs with reflection. Let

∂uε

∂t=

12∂2uε

∂x2 +c

(ε+ uε)3 + W + ηε

uε(t, 0) = uε(t, 1) = a, uε(0, x) = u0(x)

uε ≥ 0, dηε ≥ 0,∫

uε dηε = 0

where ε > 0.

By monotonicity arguments, we see that ε 7→ uε is monotonenon-increasing, while ε 7→ ε+ uε is monotone non-decreasing.


Proof

Then, as ε ↓ 0, uε converges uniformly to u and for all non-negativeh ∈ C([0, 1]) and t ≥ 0∫ t

0

∫ 1

0

c(ε+ uε(s, x))3 h(x) dx ds ↑

∫ t

0

∫ 1

0

c(u(s, x))3 h(x) dx ds

as ε ↓ 0. Moreover ηε ↓ η and (u, η) solves

∂u∂t

=12∂2u∂x2 +

cu3 + W + η

u(t, 0) = u(t, 1) = a, u(0, x) = u0(x)

u ≥ 0, dη ≥ 0,∫

u dη = 0, u−3 ∈ L1loc

We need to prove now that η = 0.


Proof

Note first that the invariant measure of uε is

νε,a := C(ε, a) exp(−(δ − 1)(δ − 3)

8

∫ 1

0

1(ε+ Xτ )2 dτ

)P3

a,a.

which converges to Pδa,a as ε ↓ 0. By the IbPF for ϕ ≡ 1∫νε,a(du0)E

[∫ t

0

∫ 1

0h(x) ηε(ds, dx)

]=

= t C(ε, a)

∫ 1

0dr h(r) γ(r, a)

∫e2Fε(ζ) Σa(r, dζ).


Proof

Now for r ∈ ]0, 1[ and c = (δ−1)(δ−3)8 > 0∫

e2Fε(ζ) Σa(r, dζ) =

= E3a,a

[exp

(−c∫ 1

0

1(ε+ Xτ )2 dτ

) ∣∣∣∣ Xr = 0]

↓ E3a,a

[exp

(−c∫ 1

0

1X2τ

dτ) ∣∣∣∣ Xr = 0

]= 0

by the law of the iterated logarithm or an explicit computation.


Hitting of zero

We have now functions u = uδ for δ ≥ 3, stationary solutions toequations with reflection (δ = 3) or repulsion from 0 (δ > 3).

One of the main results of this course is the following

Theorem (Dalang, Mueller, Z.)Let δ ≥ 3. If k ∈ N satisfies

k >4

δ − 2,

the probability that there exist t > 0 and x1, . . . , xk ∈ [0, 1] such that0 < x1 < · · · < xk < 1 and u(t, xi) = 0 for all i = 1, . . . , k, is zero.


Hitting of zero

In particular, setting for δ ≥ 3

ζ(δ) := supk : ∃ (t, x1, . . . , xk) ∈ ]0, 1]× ]0, 1[, u(t, xi) = 0

thenI for δ = 3, a.s. ζ(δ) ≤ 4I for δ ∈ ]3, 3 + 1/3], a.s. ζ(δ) ≤ 3I for δ ∈ ]3 + 1/3, 4], a.s. ζ(δ) ≤ 2I for δ ∈ ]4, 6], a.s. ζ(δ) ≤ 1I for δ > 6, a.s. ζ(δ) = 0.

In any case ζ(δ) ≤ 4 a.s. for all δ ≥ 3. The behavior at the transitionpoints δ ∈ 3, 3 + 1/3, 4, 6 might be non-optimal. Indeed, weconjecture that a.s.

ζ(3) ≤ 3, ζ(3 + 1/3) ≤ 2, ζ(4) ≤ 1, ζ(6) = 0.


Hölder properties

LemmaLet δ ≥ 3. For all β ∈ (0, 1/2) and T > 0, there exists a finite randomvariable γ ≥ 0 such that

|u(t, x)− u(t, y)| ≤ γ |x− y|β, x, y ∈ [0, 1], T ≥ t ≥ 0,

and

u(t, x)− u(s, x) ≥ − γ (t − s)β/2, T ≥ t ≥ s ≥ 0, x ∈ [0, 1].


Proof of the Theorem

For all qi : i = 1, . . . , 2k ⊂ Q with 0 < q1 < · · · < q2k < 1, wedefine

Q := [0, 1]×k∏

i=1

[q2i−1, q2i]

and the random set

A := (t, x1, . . . , xk) ∈ Q : u(t, xi) = 0, i = 1, . . . , k.

Then the claim follows if P(A 6= ∅) = 0 for all such (qi)i.


Proof

Since k > 4δ−2 , we can fix α ∈ ]0, 1[ such that

4 + 2k − αδk < 0.

For such α, we define the random set

An := (t, x1, . . . , xk) ∈ Q : u(t, xi) ≤ 2−αn, i = 1, . . . , k.

For all n ∈ N, let

Gn := (j 2−4n, i1 2−2n, . . . , ik 2−2n) : j, i1, . . . , ik ∈ Z,

and consider the events

Kn := An ∩ Gn 6= ∅, Ln := A 6= ∅, An ∩ Gn = ∅.

Since A ⊂ An a.s.,A 6= ∅ ⊆ Kn ∪ Ln.


0 1

1

•••

x1 x2 x3

t

22n

24n

1

Figure : An element (t, x1, x2, x3) of A with k = 3, namely u(t, xi) = 0,i = 1, 2, 3. Either u ≤ 2−αn at k points (s, y1), . . . , (s, yk) ∈ Gn (event Kn) oru has a large oscillation in one of the three rectangles containing the (t, xi)’s(event Ln).


Proof

In order to prove that P(A 6= ∅) = 0, we will show that theprobabilities of Kn and Ln tend to 0 as n→∞.

Step 1. By definition, on the event Ln, there exists(t, x) ∈ [0, 1]× ]0, 1[ such that u(t, x) = 0 but An ∩ Gn = ∅. Inparticular, on Ln, there exists (s, y) ∈ Gn such that

u(s, y) > 2−αn, 0 < t − s ≤ 2−4n, |x− y| ≤ 2−2n.

Let β ∈ ]α/2, 1/2[. Then on Ln

2−αn < u(s, y) = [u(s, y)− u(t, y)] + [u(t, y)− u(t, x)]

≤ γ(

(t − s)β2 + |y− x|β

)≤ γ 2−2βn+1.

Therefore,

P(Ln) ≤ P(2−αn < γ 2−2βn+1) = P(γ > 2(2β−α)n−1) → 0

as n→∞, since 2β > α and γ is a.s. finite.Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour

Proof

Step 2. We set In := Gn ∩ Q. Then, by definition,

P(Kn) = P(∃(t, x1, . . . , xk) ∈ In : u(t, xi) ≤ 2−αn, i = 1, . . . , k

).

Let Jn := (x1, . . . , xk) : (0, x1, . . . , xk) ∈ In. Then

P(Kn) ≤24n∑j=1

∑(x1,...,xk)∈Jn

P(u(j 2−4n, xi) ≤ 2−αn, i = 1, . . . , k

)= 24n

∑(x1,...,xk)∈Jn

Pδa,a(Xxi ≤ 2−αn, i = 1, . . . , k

),

(31)

since we have chosen u to be stationary and therefore, for any t ≥ 0, ut

has distribution Pδa,a.


Proof

For ε > 0 and x0 := 0, y0 := a

Pδa,a (Xxi ≤ ε, i = 1, . . . , k)

=

∫[0,ε)k

[k∏

i=1

pxi−xi−1(yi−1, yi)

]p1−xk(yk, a)

p1(a, a)dy1 · · · dyk,

Since pδt (x, y) ≤ K yδ−1, for all (xi)i=1,...,k ∈ Jn

Pδa,a (Xxi ≤ ε, i = 1, . . . , k) ≤ C[∫ ε

0yδ−1 dy

]k

≤ C εδk, ε > 0.

Therefore, by (31), since Jn has at most 22kn elements

P(Kn) ≤ C 24n 22kn (2−αn)δk = C 2(4+2k−αδk)n −→ 0

as n→∞.


Hitting of zero

Theorem (Dalang, Mueller, Z.)

(a) For all δ ∈ [3, 5], with positive probability, there exist t > 0 andx ∈ (0, 1) such that ut(x) = 0.

(b) For δ = 3, with positive probability there exist t > 0 andx1, x2, x3 ⊂ (0, 1), x1 < x2 < x3, such that ut(xi) = 0,i = 1, 2, 3.

We conjecture that

1. ζ(δ) <4

δ − 2a.s. for all δ ≥ 3.

2. P(ζ(δ) = 1) > 0 for δ ∈ [4, 6), P(ζ(δ) = 2) > 0 forδ ∈ [10/3, 4), and P(ζ(δ) = 3) > 0 for δ ∈ [3, 10/3).


Integration by parts and Bessel processes

For δ = 1 we have for all ϕ ∈ C1c(R)∫

R+

ϕ′(x) dx = −ϕ(0).

The associated dynamics isρt = x + Bt + `t, ρ0 = x, `0 = 0,

ρt ≥ 0, d`t ≥ 0,∫ ∞

0ρt d`t = 0.



For δ > 1 we have for all ϕ ∈ C1c(R)∫

R+

ϕ′(x) xδ−1 dx = −∫Rϕ(x)

δ − 1x

xδ−1 dx.

The associated dynamics is

ρt = x +δ − 1

2

∫ t

0

1ρs

ds + Bt, ρt ≥ 0

limit as ε ↓ 0 of

ρεt = x +δ − 1

2

∫ t

0

1ε+ ρεs

ds + Bt + `εt .

There is a family of local times `a defined by the occupation timeformula ∫ t

0φ(ρs) ds =

∫R+

φ(a) `at aδ−1 da.



For δ < 1 we have for all ϕ ∈ C1c(R)∫

R+

ϕ′(x) xδ−1 dx = −∫R

(ϕ(x)− ϕ(0))δ − 1

xxδ−1 dx.

The associated dynamics is the limit as ε ↓ 0 of

ρεt = x +δ − 1

2

∫ t

0

1ε+ ρεs

ds + Bt + `εt .

The limit is equal to

ρt = ρ0 + Bt +δ − 1

2

∫ +∞

0(à

t − `0t ) aδ−2 da

where the family of local times à is defined by the occupation timeformula ∫ t

0φ(ρs) ds =

∫R+

φ(a) àt aδ−1 da.


Integration by parts and SPDEs

For δ = 3 we have

P3a,a [∂hϕ] =−P3

a,a[ϕ(X) 〈X, h′′〉

]−∫ 1

0dr h(r) γ(r, a) P3

a,a [ϕ(X) |Xr = 0] .

and the dynamics is

∂u∂t

=12∂2u∂x2 + W + η

u(0, x) = u0(x), u(t, 0) = u(t, 1) = a

u ≥ 0, dη ≥ 0,∫

u dη = 0


Local times

Theorem

1. For all t ≥ 0: η([0, t], dx) = η([0, t], x) dx.

2. For all t ≥ 0:

η([0, t], x) =34

limε↓0

1ε3

∫ t

01[0,ε](u(s, x)) ds.

3. There exists a family of local times (`a(·, x))a∈[0,∞), x∈(0,1), of usuch that for all bounded Borel ϕ : R+ → R:∫ t

0ϕ(u(s, x)) ds =

∫ ∞0

ϕ(a) `a(t, x) a2 da, t ≥ 0.

4. For all t ≥ 0: η([0, t], x) =14`0(t, x).



For δ > 3 we have

Eδa,a [∂hϕ(X)] = − Eδa,a[ϕ(X) 〈X, h′′〉

]− (δ − 1)(δ − 3)

4Eδa,a

[ϕ(X) 〈X−3, h〉

].

and the dynamics is∂u∂t

=12∂2u∂x2 +

(δ − 1)(δ − 3)

8u−3 + W

u(0, x) = u0(x), u(t, 0) = u(t, 1) = a

limit of

∂uε

∂t=

12∂2uε

∂x2 +(δ − 1)(δ − 3)

8(uε + ε)−3 + W + ηε.



For δ ∈ [2, 3[ we set

∂uε

∂t=

12∂2uε

∂x2 +(δ − 1)(δ − 3)

8(uε + ε)−3 + W + ηε.

and as ε ↓ 0 we have again a renormalisation effect.

Open problems:I IbPF ?I dynamics ?

The non-linearity has the wrong sign.


δ ∈ [2, 3[

Eδa,a [∂hϕ(X)] = −Eδa,a[ϕ(X) 〈X, h′′〉

]− κ(δ)

∫ 1

0dr h(r)

∫ +∞

0

pδr (a, c)pδ1−r(c, a)

pδ1(a, a)·

· 1c3

[Eδa,a(ϕ(X) |Xr = c)− Eδa,a(ϕ(X) |Xr = 0)

]cδ−1 dc

and the SPDE

∂u∂t

=12∂2u∂x2 +

∂

∂tκ(δ)

2

∫ +∞

0

`c(t, x)− `0(t, x)

c3 cδ−1 dc + W

u(t, 0) = u(t, 1) = a, u(0, x) = u0(x)∫ t0 ψ(u(s, x)) ds =

∫ +∞0 ψ(c) `c(t, x) cδ−1 dc



For δ < 2 the situation is even more complicated since 0 is hit by thestationary profile.

The case δ = 1 is the most intriguing since it is the limit ofhomogeneous pinning models.

There is an IbPF for δ = 1 but the form of the dynamics is hard even toconjecture.


Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-FlourChapter 1: Introduction A course in 8...

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Transcript of Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-FlourChapter 1: Introduction A course in 8...