to stochastic partial differential equations
In this section we will review some results on the germ Markov field (GMF) property for solutions to stochastic partial differential equations driven by a white noise which have been obtained by means of the technique of change of probability.
LetDbe a bounded domain inRk with smooth boundary, and consider a continuous stochastic process X ={Xz, z∈ D}. We will say that X is a germ Markov field (GMF) if for any ǫ > 0 and any open subset A ⊂ D, the σ-fields σ{Xz, z ∈ A} and σ{Xz, z ∈ D−Ac} are conditionally independent given theσ-field σ{Xz, z∈(∂A)ǫ}, where (∂A)ǫ denotes the ǫ-neighborhood of the boundary ofA.
We will first discuss in some detail the case of an elliptic stochastic partial differential equation with additive white noise.
(A) Elliptic stochastic partial differential equations
Let D be a bounded domain in Rk with smooth boundary, and assume k = 1,2,3. Let λk denote the Lebesgue measure on D, and set H = L2(D,B(D), λk). Consider an isonormal Gaussian processW ={W(h), h∈ H} associated with H. That is, if we set W(A) = W(1A), then W = {W(A), A∈ B(D)} is a zero-mean Gaussian process with covariance
E(W(A)W(B)) =λk(A∩B).
We want to study the equation
−∆U(x) +f(U(x)) = ˙W(x), x∈D,
U|∂D= 0. (4.30)
Let us first introduce the notion of the solution to (4.30) in the sense of distributions.
Definition 4.2.2 We will say that a continuous process U ={U(x), x∈ D} that vanishes in the boundary ofD is a solution to (4.30) if
−U,∆ϕH+f(U), ϕH=
D
ϕ(x)W(dx) for allϕ∈C∞(D) with compact support.
We will denote byG(x, y) the Green function associated with the Laplace operator ∆ with Dirichlet boundary conditions on D. That is, for any ϕ∈L2(D), the elliptic linear equation
−∆ψ(x) =ϕ(x), x∈D,
ψ|∂D= 0 (4.31)
possesses a unique solution in the Sobolev space H01(D), which can be written as
ψ(x) =
D
G(x, y)ϕ(y)dy.
We recall that H01(D) denotes the completion of C0∞(D) for the Sobolev norm ã 1,2. We will use the notation ψ = Gϕ. We recall that G is a symmetric function such that G(x,ã) is harmonic on D− {x}.
One can easily show (see [54]) thatUis a solution to the elliptic equation (4.30) if and only if it satisfies the integral equation
U(x) +
D
G(x, y)f(U(y))dy=
D
G(x, y)W(dy). (4.32) Note that the right-hand side of (4.32) is a well-defined stochastic integral because the Green function is square integrable. More precisely, we have
sup
x∈D
D
G2(x, y)dy <∞. (4.33)
In dimensionk >3 this property is no longer true, and for this reason the analysis stops at dimension three.
We will denote byU0 the solution of (4.30) forf = 0, that is, U0(x) =
D
G(x, y)W(dy). (4.34)
Using Kolmogorov’s criterion one can show (see Exercise 4.2.13) that the process {U0(x), x ∈ D} has Lipschitz paths if k = 1, H¨older continuous paths of order 1−ǫifk= 2, and H¨older continuous paths of order 38−ǫif k= 3, for anyǫ >0.
The following result was established by Buckdahn and Pardoux in [54].
Theorem 4.2.3 Let D be a bounded domain of Rk, k = 1,2,3, with a smooth boundary. Let f be a continuous and nondecreasing function. Then Eq. (4.32) possesses a unique continuous solution.
A basic ingredient in the proof of this theorem is the following inequality:
Lemma 4.2.3 There exists a constanta >0such that for anyϕ∈L2(D), Gϕ, ϕH≥aGϕ2H. (4.35) Proof: Setψ=Gϕ. Thenψsolves Eq. (4.31). Multiplying this equation byψand integrating by parts, we obtain
k i=1
∂ψ
∂xi
2
H
=ϕ, ψH.
From Poincar´e’s inequality (cf. [120, p. 157]) there exists a constanta >0 such that for anyψ∈H01(D),
k i=1
∂ψ
∂xi
2
H
≥aψ2H.
The result follows.
We are going to reformulate the above existence and uniqueness theorem in an alternative way. Consider the Banach space
B={ω∈C(D), ω|∂D= 0},
equipped with the supremum norm, and the transformation T : B → B given by
T(ω)(x) =ω(x) +
D
G(x, y)f(ω(y))dy. (4.36) Note that{U(x), x∈D} is a solution to (4.32) if and only if
T(U(x)) =U0(x).
Then Theorem 4.2.3 is a consequence of the following result.
Lemma 4.2.4 Let f be a continuous and nondecreasing function. Then the transformation T given by (4.36) is bijective.
Proof: Let us first show that T is one to one. Let u, v ∈ B such that T(u) =T(v). Then
u−v+G[f(u)−f(v)] = 0. (4.37) Multiplying this equation byf(u)−f(v), we obtain
u−v, f(u)−f(v)H+G[f(u)−f(v)], f(u)−f(v)H= 0.
Using the fact thatf is nondecreasing, and Lemma 4.2.3, it follows that aG[f(u)−f(v)]2H ≤0.
By (4.37) this is equivalent toau−v2H≤0, sou=vandT is one to one.
In order to show thatT is onto, we will assume thatf is bounded. The general case would be obtained by a truncation argument. Let v ∈ B, and let {vn, n ∈ N} be a sequence of functions in C2(D), with compact support inD, such thatv−vn∞ tends to zero asntends to infinity. Set hn = −∆vn. It follows from Lions ([199], Theorem 2.1, p. 171) that the elliptic partial differential equation
−∆un+f(un) =hn
un|∂D= 0 admits a unique solutionun ∈H01(D). Then,
un+G[f(un)] =Ghn=vn, (4.38) that is,T(un) =vn. We now prove thatun is a Cauchy sequence inL2(D).
Multiplying the equation
un−um+G[f(un)−f(um)] =vn−vm
byf(un)−f(um), and using Lemma 4.2.3 and the monotonicity property off, we get
aG[f(un)−f(um)]2H ≤ vn−vm, f(un)−f(um)H, which implies, using the above equation,
aun−umH ≤ vn−vm, f(un)−f(um) + 2a(un−um)H. Since {vn} is a Cauchy sequence in L2(D) and f is bounded, {un} is a Cauchy sequence in L2(D). Define u = limun. Then f(un) converges to f(u) in L2(D). Taking the limit in (4.38), we obtain
u+G[f(u)] =v.
Thusu∈B (f is bounded) andT(u) =v.
Let us now discuss the germ Markov field property of the processU(x).
First we will show that the Gaussian processU0(x) verifies the germ Markov field property. To do this we shall use a criterion expressed in terms of the reproducing kernel Hilbert space (RKHS) H associated to U0. Let H1 ⊂ L2(Ω) be the Gaussian space (i.e., the first chaos) generated by W. An element v ∈B belongs to the RKHS Hiff there exists a random variable X ∈ H1such that
v(x) =E[XU0(x)],
for all x∈ D, i.e., iff there exists φ∈ L2(D) such that v =Gφ. In other words,H={v∈B : ∆v∈L2(D)}, andv1, v2H=∆v1,∆v2H.
We now have the following result (see Pitt [285] and K¨unsch [177]).
Proposition 4.2.1 A continuous Gaussian fieldU ={U(x), x∈D} pos- sesses the germ Markov field property iff its RKHS H ⊂B is local in the sense that it satisfies the following two properties:
(i) Whenever u,v in Hhave disjoint supports,u, vH= 0.
(ii) If v ∈ H is of the form v = v1+v2 with v1, v2 ∈ B with disjoint supports, thenv1, v2∈ H.
The RKHS associated to the process U0 verifies conditions (i) and (ii), and this implies the germ Markov field property of U0. Concerning the processU, one can prove the following result.
Theorem 4.2.4 Assume that f is a C2 function such thatf′ >0 andf′ has linear growth. Then the solution {U(x), x∈D} of the elliptic equation (4.30) has the germ Markov property if and only iff′′= 0.
This theorem has been proved by Donati-Martin and Nualart in [82]. In dimension one, Eq. (4.30) is a second-order stochastic differential equation studied by Nualart and Pardoux in [251]. In that case the germ σ-field corresponding to the boundary points {s, t} is generated by the variables {Xs,X˙s, Xt,X˙t}, and the theorem holds even if the function f depends on Xt and ˙Xt (assuming in that case more regularity on f). The main difference between one and several parameters is that in dimension one, one can explicitly compute the Carleman-Fredholm determinant of Du.
Similar to the work done in [81] and [82] we will give a proof for the case k= 2 ork= 3.
Proof of Theorem 4.2.4: The proof follows the same lines as the proof of Theorem 2.1. We will indicate the main steps of the argument.
Step 1: We will work on the abstract Wiener space (B, H, à), whereàis the law ofU0, and the continuous injectioni:H→Bis defined as follows:
i(h)(x) =
D
G(x, y)h(y)dy.
From Lemma 4.2.3 we deduce that i is one to one, and from Eq. (4.33) we see that iis continuous. The image i(H) is densely included in B. We identifyH andH∗, and in this wayB∗ can be viewed as a dense subset of H, the inclusion map being given by
α→
D
G(y,ã)α(dy) =G∗α.
Finally, for anyα∈B∗ we have
B
eiα,ωà(dω) = E
&
exp(i
D
U0(x)α(dx)) '
= E
&
exp(i
D
D
G(x, y)dWyα(dx)) '
= exp
−1 2
D
D
G(x, y)α(dx) 2
dy
= exp
−1
2G∗α2H
,
which implies that (B, H, à) is an abstract Wiener space. Note thati(H) coincides with the RKHS Hintroduced before, and that U0(x) =ω(x) is now the canonical process in the space (B,B(B), à).
We are interested in the germ Markov field property of the process U(x) =T−1(U0)(x). Letνbe the probability onB defined byà=ν◦T−1. That is,ν is the law ofU.
Step 2: Let us show that the transformationT verifies the hypotheses of Theorem 4.1.2. We already know from Lemma 4.2.4 thatT is bijective.
Notice that we can write
T(ω) =ω+i(f(ω)), so we have to show that:
(i) the mapping ω → i(f(ω)) from B to H isH-continuously differen- tiable;
(ii) the mappingIH+Du(ω) :H →H is invertible for allω∈B, where Du(ω) is the Hilbert-Schmidt operator given by the kernel
Du(ω)(x, y) =f′(ω(x))G(x, y).
Property (i) is obvious and to prove (ii), from the Fredholm alternative, it suffices to check that−1 is not an eigenvalue ofDu(ω). Leth∈H be an element such that
h(x) +f′(ω(x))
D
G(x, y)h(y)dy= 0.
Multiplying this equality by f′h(x)(ω(x)) and integrating overD, we obtain
D
h2(x)
f′(ω(x))dx+h, GhH= 0.
From Lemma 4.2.3, h, GhH≥aGh2H, thusGhH= 0 and h= 0.
Therefore, by Theorem 4.1.2 we obtain dν
dà =|det2(I+Du)|exp(−δ(u)−12u2H). (4.39) SetL= exp(−δ(u)−12u2).
Step 3: For a fixed domain A with smooth boundary Γ and such that A⊂D, we denote
Fi=σ{U0(x), x∈A}, Fe=σ{U0(x), x∈D−A}, and
F0=∩ǫ>0σ{U0(x), x∈(∂A)ǫ}. Consider the factorizationL=LiLe, where
Li= exp
−δ(u1A)−1
2u1A2H
and
Le= exp
−δ(u1D−A)−1
2u1D−A2H
.
We claim that Ji is Fi-measurable andJe isFe-measurable. This follows from the fact that the Skorohod integrals
δ(u1A) =
A
f(U0(x))W(dx) and
δ(u1D−A) =
D−A
f(U0(x))W(dx) areFi-measurable and Fe-measurable, respectively (see [81]).
Step 4: From Step 3 it follows that iff′′= 0, the Radon-Nikodym density given by (4.39) can be expressed as the product of two factors, one being Fi-measurable, and the second one beingFe-measurable. This factorization implies the germ Markov field property ofX underà.
Step 5: Suppose conversely thatU possesses the germ Markov property under à. By the same arguments as in the proof of Theorem 2.1 we can
show that for any nonnegative random variable ξ that is Fi-measurable, the quotient
Gξ = E[ξΦ| Fe] E[Φ| Fe]
is F0-measurable, where Φ = det2(I+f′(U0(x))G(x, y)). Observe that Φ≥0 because the eigenvalues of the kernelf′(U0(x))G(x, y)) are positive.
Step 6: The next step will be to translate the above measurability prop- erty into an analytical condition. Fix ǫ >0 such that A−ǫ =A−Γǫ and A+ǫ = (D−A)−Γǫare nonempty sets. We have thatGξisσ{U0(x), x∈Γǫ}- measurable. If we assume that ξis a smooth random variable, then Gξ is in D1,2loc, and by Lemma 1.3.3 we obtain that
DGξ ∈ G(x,ã), x∈ΓǫH.
This implies that for any functionφ∈C0∞(D−Γǫ) we haveφ,∆DGξH = 0 a.s. Suppose that φ ∈ C0∞(A+ǫ). In that case we have in addition that φ,∆DξH= 0, becauseξisFi-measurable. Consequently, for such a func- tionφwe get
E[ξφ,∆DΦH | Fe]E[Φ| Fe] =E[ξΦ| Fe]E[φ,∆DφH| Fe].
The above equality holds true for any bounded and nonnegative random variableξ, therefore, we obtain that
1
Φφ,∆DΦH (4.40)
isFe-measurable.
Step 7: The derivative of the random variable Φ can be computed using the expression for the derivative of the Carleman-Fredholm determinant.
We have
DzΦ = ΦT
((I+f′(U0(x))G(x, y))−1−I)Dz[f′(U0(x))G(x, y)]
. So, from (4.40) we get that
T
((I+f′(U0(x))G(x, y))−1−I)φ,∆Dã[f′(U0(x))G(x, y)]H isFe-measurable. Note that
Dz[f′(U0(x))G(x, y)] =f′′(U0(x))G(x, z)G(x, y) and
∆φ, G(x,ã)H =φ.
Thus, we have that T
((I+f′(U0(x))G(x, y))−1−I)(φ(x)f′′(U0(x))G(x, y))
isFe-measurable, and we conclude that for anyx∈A+ǫ, f′′(U0(x))
D
K(y, x)G(x, y)dy
is Fe-measurable, where K(x, y) = ((I+Du)−1−I)(x, y). Suppose now that there exists a pointb∈Rsuch thatf′′(b)= 0. Thenf′′will be nonzero in some intervalJ. Set
A={ω∈B:f′′(U0(x))∈J}. The setAhas nonzero probability, it belongs toFe, and
1A
D
K(y, x)G(x, y)dy
is Fe-measurable. Applying again Lemma 1.3.3 and using the same argu- ments as above, we obtain that on the setA
G(I+Du)−1[f′′(U0(x1))ψ(x1)G(x1, x2)](I+Du)−1
(x, x) = 0 for any functionψ∈C0∞(A−ǫ). So we get
1{f′′(U0(x))∈J}1{f′′(U0(x1))∈J}G(I+Du)−1(x, x1)G(I+Du)−1(x1, x) = 0 for all x, x1 such that x ∈ A+ǫ and x1 ∈ A−ǫ. Notice that the operator G(I+Du)−1 has a singularity in the diagonal of the same type asG. So from the above equality we get
1{f′′(U0(x))∈J}= 0,
which is not possible.
(B) Parabolic stochastic partial differential equations Consider the following equation studied in Section 2.4.2:
2 ∂u
∂t −∂∂x2u2 =f(u(t, x)) +∂∂t∂x2W, (t, x)∈[0, T]×[0,1], u(t,0) =u(t,1) = 0, 0≤t≤T.
We will impose two different types of boundary conditions:
(B.1) u(0, x) =u0(x),
(B.2) u(0, x) =u(1, x), 0≤x≤1.
In case (B.1) we are given a initial condition u0 ∈ C([0,1]) such that u0(0) = u0(1) = 0, and in case (B.2) we impose a periodic boundary condition in time. Under some hypotheses on the functionf there exists a unique continuous solution of the corresponding integral equation:
For (B.1) a sufficient condition is thatf is Lipschitz (see Theorem 2.4.3).
For (B.2) (see [253]) we require that there exists a constant 0 < c <2 such that
(z−y)(f(z)−f(y))≤c(z−y)2 for ally, z∈R.
From the point of view of the Markov property of the solution, the be- havior of these equations is completely different. In case (B.1) the GMF property always holds. On the other hand, assuming that f(z) is of class Cb2 and that the boundary condition (B.2) holds, then the solution uhas the GMF property if and only iff′′ = 0. These results have been proved under more general assumptions on the functionf in [253].