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
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 Open problems
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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 .
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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 .
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
The reflected interface
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
The reflected interface
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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?
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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?
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
The Skorokhod Lemma
0
z
y
`
1
Figure : An example of a triple (y, z, `) from the Skorokhod Lemma.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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)
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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].
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
The reflecting BM
0 t
t
1
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
Reflection in Rd
Figure : Exposing hyperplanes
(Special thanks to wikipedia)
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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].
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
Penalisation
0t
nt
1
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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].
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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‖∞.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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)) .
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
Reflection and local time
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
Reflection and local time
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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].
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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].
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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).
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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
`at := a1−δ La
t , a > 0, `0t := lim
a↓0`a
t ,
yields a bi-continuous family (`at )a,t≥0 with the occupation-time
formula ∫ t
0ψ(ρs) ds =
∫R+
ψ(a) `at 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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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).
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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).
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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).
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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 < +∞.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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→∞.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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|.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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→∞.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
The invariant measure
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))
.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
The invariant measure
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].
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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〉 = +∞.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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)
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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].
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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′.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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′.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
The invariant measure
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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)).
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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).
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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).
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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→ +∞?
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
The invariant measure
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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).
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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 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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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 treated
I u is the weak limit of discrete interfaces: ask Cyril Labbé for aconfirmation.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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).
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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).
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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).
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
Integration by parts
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).
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
Integration by parts
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).
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
Integration by parts
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
Integration by parts
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?
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
Integration by parts
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?
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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)).
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
The boundary measure
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).
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
The boundary measure
0 1 0 1
0 r 1
1
Figure : A typical path under the boundary measure and the action of Tr.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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).
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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).
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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).
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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
F(α⊕ β) P+i−1(dα) P+
N−i(dβ).
This concludes the proof.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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+
F(α⊕ β) P+i−1(dα) P+
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)] .
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
The boundary measure
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
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
The Revuz measure of η
TheoremFor all bounded Borel ϕ : H 7→ R and h ∈ Cc(0, 1)∫
νa(du0)E[∫ t
0
∫ 1
0h(x)ϕ(us) η(ds, dx)
]=
=t
2Za
∫ 1
0dr h(r) γ(r, a)
∫ϕ(ζ) e2F(ζ) Σa(r, dζ).
where Σa(r, ·) := P3a,a[ · |Xr = 0].
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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).
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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〉
))
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
The contact set
(S)
••
•• ••
•
•
••
••
••
••
•
•
••
••
•••
•
••
(rt, t)
0 1
t
1
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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
0h(x)ϕ(us) η(ds, dx)
]=
=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[.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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).
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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].
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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).
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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→∞.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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).
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
Integration by parts and Bessel processes
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
Integration by parts and Bessel processes
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(`a
t − `0t ) aδ−2 da
where the family of local times `a is defined by the occupation timeformula ∫ t
0φ(ρs) ds =
∫R+
φ(a) `at aδ−1 da.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
Integration by parts and Bessel processes
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(`a
t − `0t ) aδ−2 da
where the family of local times `a is defined by the occupation timeformula ∫ t
0φ(ρs) ds =
∫R+
φ(a) `at aδ−1 da.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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).
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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).
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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).
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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).
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
Integration by parts and SPDEs
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
Integration by parts and SPDEs
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.
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
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
Lorenzo Zambotti (LPMA, Univ. Paris 6) July 2015, Saint-Flour
Integration by parts and SPDEs
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-Flour
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