Riemann integral of a random function and the parabolic equation with a general stochastic measure

0.2 MB·English

by  Vadym Radchenko

Riemann integral of a random function and the parabolic equation with a
general stochastic measure
Vadym Radchenko 1
Abstract
For stochastic parabolic equation driven by a general stochastic measure, the weak solution is
obtained. The integralof a random function in the equation is considered as a limit in probability
of Riemann integral sums. Basic properties of such integrals are studied in the paper.
1 Introduction
In this paper we consider the stochastic parabolic equation, which can formally be written as
dX(x,t) = AX(x,t)dt+f(x,t)dµ(t), X(x,0) = ξ(x), (1.1)
2
1 where (x,t) ∈ Rd×[0,T], A is a second-order strongly elliptic differential operator, and µ is a general
0
2 stochasticmeasuredefinedontheBorelσ-algebraof[0,T]. Forµweassumeσ−additivityinprobability
only, assumptions for A,f and ξ are given in Section 6. Equation (1.1) is interpreted in the weak sense
n
a (see (6.1) below). We prove existence and uniqueness of solution.
J
Weak form of (1.1) includes the integral of random function with respect to deterministic measure

9
(Jordan content). We interpret this integral as a limit in probability of Riemann integral sums. This

] definition of the integral allows to interchange the order of integration with respect to deterministic
R
and stochastic measures (Theorem 4.1), that is important for solving the equation. A large part of the
P
paper is devoted to the study of this Riemann-type stochastic integral.
.
h Parabolic stochastic partial differential equations (SPDEs) driven by the martingale measures had
t
a been introduced and discussed initially in [19]. This approach was developed in [1, 3]. Parabolic
m
SPDEs as equations in infinite dimensional space were studied in [4, 11]. In these and many other
[
papers the stochastic noise has some distributional, integrability or martingale properties. In our

1 paper, we consider very general class of possible µ on [0,T]. On the other hand, the stochastic term
v
in (1.1) is independent of u. A reason is that appropriate definition of integral of random function
2
9 with respect to µ does not exist.
7 Some motivating examples for studying SPDEs may be found in [4, Introduction], [6, section 13.2].
1
For A = ∆, equation (1.1) describes the evolution in time of the density X of some quantity such a
.
1
heat or chemical concentration in a system with random sources. In our model, the random influence
0
can be rather general.
2
1
:
v 2 Preliminaries
i
X
r Let L = L (Ω,F,P) be a set of all real-valued random variables defined on a complete probability
0 0
a
space (Ω,F,P) (equivalence classes of). Convergence in L means the convergence in probability and
0
is the convergence in the quasi-norm
kηk = inf{δ : P{|η| > δ} ≤ δ}.
Note that kη +η k ≤ kη k+kη k. The following inequality will be used in the sequel
1 2 1 2
l l
c ξ ≤ 8 max a ξ ≤ 16max ξ , |c |≤ 1, ξ ∈ L , (2.1)
k k k k k k k 0
(cid:13)X (cid:13) ak=±1(cid:13)X (cid:13) V (cid:13)X (cid:13)
(cid:13)k=1 (cid:13) (cid:13)k=1 (cid:13) (cid:13)k∈V (cid:13)
(cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13)
where the latter maximum is taken over all possible V ⊂ {1,...,l} (see [16, Theorem 3]).
Let S be an arbitrary set and B be a σ-algebra of subsets of S.
1
This research was supportedbyAlexandervon HumboldtFoundation,grant no. UKR/1074615. Theauthorwishes
to thank Prof. M. Zähle for fruitful discussions, and thehospitality of Jena Universityis gratefully acknowledged.
1
Definition 2.1. Any σ-additive mapping µ :B → L is called a stochastic measure.
0
Inotherwords, µisavectormeasurewithvaluesinL . Wedonotassumepositivityorintegrability
0
forstochasticmeasures. In[7]suchaµiscalledageneralstochasticmeasure. Inthefollowing, µalways
denotes a stochastic measure.
Examples of stochastic measures are the following. Let S = [0,T] ⊂ R , B be the σ-algebra of
+
Borel subsets of [0,T], and Y(t) be a square integrable martingale. Then µ(A) = T 1A(t)dY(t)
0
is a stochastic measure. If WH(t) is a fractional Brownian motion with Hurst indexRH > 1/2 and
f :[0,T] → R is a bounded measurable function then µ(A)= T f(t)1A(t)dWH(t) is also a stochastic
0
measure, as follows from [8, Theorem 1.1]. Some other exampRles may be found in [7, subsection 7.2].
Theorem 8.3.1 [7] states the conditions under which the increments of a real-valued Lévy process
generate a stochastic measure.
For deterministic measurable functions g : S→ R, an integral of the form gdµ is studied in [12]
S
(see also [7, Chapter 7], [2]). The construction of this integral is standard, usRes an approximation by
simple functions and is based on results of [15, 17, 18]. In particular, every bounded measurable g is
integrable with respect to any µ. An analogue of the Lebesgue dominated convergence theorem holds
for this integral (see [7, Proposition 7.1.1] or [12, Corollary 1.2]).
For equations with stochastic measures, weak solutions of some SPDEs were obtained in [13].
Regularity properties of mild solution of the stochastic heat equation were considered in [14].
3 Riemann integral of a random function
Let B ⊂ Rd be a Jordan measurable set, and ξ : B → L be a random function. We shall say that ξ
0
has an integral on B if for any sequence of partitions
B = ∪ B , n ≥ 1, maxdiamB → 0, n → ∞, x ∈ B ,
1≤k≤kn kn kn kn kn
k
the limit in probability
p lim ξ(x )m(B )= ξ(x)dx (3.1)
n→∞ X kn kn ZB
1≤k≤kn
exists. HeremdenotestheJordan content, setsB , 1≤ k ≤ k ,areassumedtobeJordanmeasurable
kn n
and have no common interior points. By mixing of different sequences of partitions, we can prove that
the limit is independent of the choice of the sequence. For deterministic ξ, our definition is equivalent
to the definition of the standard Riemann integral in [9].
Lemma 3.1. Let ξ has an integral on B = d [a ,b ] ⊂ Rd. Then the set of values {ξ(x), x ∈ B}
k=1 k k
is bounded in probability. Q
Proof. Is analogous to the deterministic case.
For some other B ⊂ Rd, limit (3.1) can exists for unbounded ξ (for instance, in the case m(B) = 0).
We use the following
Definition 3.1. Random function ξ is called integrable on B if ξ has an integral on B and set of values
{ξ(x), x ∈ B} is bounded in probability.
Let B˜ ⊂ Rd be an unbounded set for which there exists a sequence of Jordan measurable sets B(j)
such that
B(j) ↑ B˜, ∀c> 0 ∃j : B˜ ∩{|x| ≤ c} ⊂ B(j) (3.2)
(we call B(j) the exhaustive sets). We shall say that ξ is integrable (in improper sense) on B˜, if ξ is
integrable on each B(j), and there exists the limit in probability
p lim ξ(x)dx = ξ(x)dx,
j→∞ZB(j) ZB˜
2
that is independent of choice of B(j).
All bounded subsets of Rd used in the paper are assumed to be Jordan measurable, and all un-
bounded sets are assumed to be approximable by Jordan measurable sets in the sense of (3.2). Sets in
partitions are assumed to be non-overlapping.
Obviously, if ξ has the Riemann integrable paths then ξ is integrable in our sense. Theorem 4.1
below gives other examples of integrable random functions.
Further, we establish basic properties of the integral.
Lemma 3.2. Let ξ be integrable on B. Then ξ is integrable on each A ⊂ B, and for any ε > 0 there
exists δ > 0 such that for all A ⊂ B, A= ∪ A , x ∈ A , diamA < δ, holds
1≤k≤k0 k k k k
ξ(x )m(A )− ξ(x)dx < ε.
(cid:13) X k k ZA (cid:13)
(cid:13)1≤k≤k0 (cid:13)
(cid:13) (cid:13)
Proof. Suppose the lemma were false. Then
∃ε > 0 ∀δ > 0 ∃A = ∪ A = ∪ A′, diamA , diamA′ < δ :
0 1≤k≤k0 k 1≤i≤i0 i k i
ξ(x )m(A )− ξ(x′)m(A′) ≥ ε .
k k i i 0
(cid:13) X X (cid:13)
(cid:13)1≤k≤k0 1≤i≤i0 (cid:13)
(cid:13) (cid:13)
Take an arbitrary partition
B\A = ∪ C , diamC < δ,
1≤j≤j0 j j
and add
ξ(x′′)m(C ), x′′ ∈ C ,
j j j j
X
1≤j≤j0
to each of the considered sums on A. Thus we can get two integral sums on B with arbitrary small
diameters such that the quasi-norm of their difference is greater than or equal to ε . This contradicts
0
the integrability of ξ on B.
Lemma 3.3. Let ξ be integrable on B˜ in the improper sense, A˜ ⊂ B˜. Then ξ is integrable on A˜ (if A˜
is an unbounded set, the integral is meant in the improper sense).
Proof. Take exhaustive sets B(j) ↑ B˜, A(i) ↑ A˜. Then sets
(B(j)\A˜)∪A(i) ↑ B˜, i,j → ∞
are exhaustive too, and
ξ(x)dx = p lim ξ(x)dx+ ξ(x)dx . (3.3)
ZB˜ i,j→∞(cid:16)ZB(j)\A˜ ZA(i) (cid:17)
If plim ξ(x)dx does not exist then we can choose i, j → ∞ such that the limit in (3.3) does
i→∞ A(i)
not exist. R
Lemma 3.4. Let ξ be integrable on B. Then the set of values ξ(s)ds, A ⊂ B is bounded in
A
n o
probability. R
Proof. Suppose the lemma were false. Then
1
∃ε > 0, A ⊂ B, n ≥ 1: ξ(s)ds ≥ ε .
0 n (cid:13)n ZA (cid:13) 0
(cid:13) n (cid:13)
By Lemma 3.2, we can choose a partition B = ∪ (cid:13) B fine eno(cid:13)ugh, such that all integral sums
1≤k≤k0 k
for partitions A = ∪ (A ∩B ) will be close enough to the integrals on A . Thus, for all n,
n 1≤k≤k0 n k n
x ∈ A ∩B , we get
kn n k
1 ε
ξ(x )m(A ∩B ) ≥ 0.
(cid:13)n kn n k (cid:13) 2
(cid:13) X (cid:13)
(cid:13) 1≤k≤k0 (cid:13)
(cid:13) (cid:13)
Since the number of summands is fixed for all n, we arrive at a contradiction with boundedness of
ξ.
3
Lemma 3.5. Let ξ be integrable on B, f : B → R be a deterministic uniformly continuous on B
function. Then fξ is integrable on B.
Proof. Consider the difference of two integral sums of fξ
f(x )ξ(x )m(B )− f(x )ξ(x )m(B )
km km km in in in
(cid:13) X X (cid:13)
(cid:13)1≤k≤km 1≤i≤in (cid:13)
(cid:13) (cid:13)
= [f(x ξ(x )−f(x )ξ(x )]m(B ∩B )
km km in in km in
(cid:13) X (cid:13)
(cid:13)1≤k≤km, 1≤i≤in (cid:13)
(cid:13) (cid:13)
≤ [ξ(x )−ξ(x )]f(x )m(B ∩B )
km in in km in
(cid:13) X (cid:13)
(cid:13)1≤k≤km, 1≤i≤in (cid:13)
(cid:13) (cid:13)
+ [f(x )−f(x )]ξ(x )m(B ∩B ) = S +S .
km in km km in 1 2
(cid:13) X (cid:13)
(cid:13)1≤k≤km, 1≤i≤in (cid:13)
(cid:13) (cid:13)
From (2.1) for |f(x)| ≤ C we get
S ≤16max C [ξ(x )−ξ(x )]m(B ∩B ) , (3.4)
1 km in km in
V (cid:13) X (cid:13)
(cid:13) (k,i)∈V (cid:13)
(cid:13) (cid:13)
where the maximum is taken over all possible sets of pairs (k,i).
For example, consider
X ξ(xkm)m(Bkm∩Bin) = X ξ(xkm)h X m(Bkm∩Bin)+m(Bkm∩Bi′n)1xkm∈/(∪i:(k,i)∈VBin)i
(k,i)∈V 1≤k≤km i:(k,i)∈V
− ξ(xkm)m(Bkm∩Bi′n)1xkm∈/(∪i:(k,i)∈VBin) = I1−I2.
X
1≤k≤km
Here Bi′n is one of the sets Bin, 1 ≤ i ≤ in, that contains xkm. (If xkm lies on the border of Bi′n, we
take it only once.) I and I are integral sums and, by Lemma 3.2, they approximate the integrals
1 2
of ξ on respective sets. Therefore, for diameter small enough, I −I will be close to the integral on
1 2
∪ (B ∩B ). Similarly, ξ(x )m(B ∩B ) approximate the integral on the same set,
(k,i)∈V km in (k,i)∈V in km in
and we make the right hand sidPe of (3.4) arbitrary small by choosing the diameter.
Further, foranyα > 0,fordiametersmallenoughandB ∩B = ∅,wehave|f(x )−f(x )| < α
km in km in
in S . Inequality (2.1) implies
2
S ≤ 16max α ξ(x )m(B ∩B ) .
2 km km in
V (cid:13) X (cid:13)
(cid:13) (k,i)∈V (cid:13)
(cid:13) (cid:13)
Asbefore,wecanmakethesumarbitraryclosetotheintegral on∪ (B ∩B ). FromLemma3.4
(k,i)∈V km in
it follows that S → 0 as α → 0.
2
Lemma 3.6. Let ξ be integrable on B, f : B → R be a deterministic uniformly continuous on B
function, |f(x)|≤ C. Then
f(x)ξ(x)dx ≤16sup C ξ(x)dx .
(cid:13)ZB (cid:13) A⊂B(cid:13) ZA (cid:13)
(cid:13) (cid:13) (cid:13) (cid:13)
(cid:13) (cid:13) (cid:13) (cid:13)
Proof. The inequality for respective integral sums follows from (2.1). Further, we pass to the limit and
apply Lemmas 3.2 and 3.5.
Lemma 3.7. Let ξ be integrable on B, f : B → R, n ≥ 1 be a deterministic uniformly continuous
n
on B functions, sup |f (x)| → 0, n → ∞. Then
x∈B n
P
f (x)ξ(x)dx → 0, n→ ∞.
Z n
B
4
Proof. The statement follows from Lemmas 3.4 and 3.6.
Lemma 3.8. Let ξ be integrable on anunbounded set B˜ inimproper sense, f :B˜ → R be adeterministic
bounded uniformly continuous on B˜ function. Then fξ is integrable on B˜ in improper sense.
Proof. For B(j) ↑ B˜ and |f(x)| ≤ C Lemma 3.6 implies
f(x)ξ(x)dx ≤ 16 sup C ξ(x)dx . (3.5)
(cid:13)(cid:13)ZB(j)\B(i) (cid:13)(cid:13) A⊂(B(j)\B(i))(cid:13)(cid:13) ZA (cid:13)(cid:13)
(cid:13) (cid:13) (cid:13) (cid:13)
If the left hand side of (3.5) does not tend to 0 as i, j → ∞, then we can construct a sequence of
bounded sets C(j) ↑B˜ such that the sequence ξ(x)dx, j ≥ 1, is non-fundamental.
C
j
R
Lemma 3.9. Let ξ be integrable on unbounded set B˜ in improper sense, f :B˜ → R be a deterministic
n
bounded uniformly continuous on B˜ functions, sup |f (x)| = C < ∞, sup |f (x)| → 0, n →
n≥1,x∈B˜ n x∈B n
∞ for all bounded B ⊂ B˜. Then
P
f (x)ξ(x)dx → 0, n→ ∞.
Z n
B˜
Proof. Suppose the lemma is false. Applying Lemma 3.7, one can find ε > 0, subsequence f , j ≥ 1,
0 nj
and bounded disjoint sets B ⊂ (B˜∩{|x| ≥ j}) such that f (x)ξ(x)dx > ε . From Lemma 3.6it
j (cid:13) Bj nj (cid:13) 0
follows thatthere existbounded disjointsets Aj ⊂ (B˜∩{|(cid:13)(cid:13)xR|≥ j}) suchthat(cid:13)(cid:13)C A ξ(x)dx > (ε0/16).
(cid:13) j (cid:13)
This contradicts the integrability of ξ on B˜. (cid:13) R (cid:13)
(cid:13) (cid:13)
Note that the stochastic continuity of ξ does not imply the integrability.
1
Example 3.1. Consider B = [0,1], ξ (ω) = 5k1 , k ≥ 1, were P(F ) = , F are independent. Set
k Fk k k k
ξ(0) = 0, ξ(x)= ξ , 2−2k−1 ≤ x ≤ 2−2k,
k
ξ(x) = 22k+2((2−2k−1 −x)ξ +(x−2−2k−2)ξ ), 2−2k−2 ≤ x ≤ 2−2k−1.
k+1 k
TakingallpossiblefiniteunionsA = ∪ [2−2k−1,2−2k],weseethatthevalues ξ(x)dxarenotbounded
k A
in probability. By Lemma 3.4, ξ is not integrable on [0,1]. R
4 Interchange of the order of integration
Theorem 4.1. Let µ be a stochastic measure on (S,B), B ⊂ Rd be a bounded set. Assume that
h(x,s) : B×S → R is a measurable deterministic function which is Riemann integrable on B for each
fixed s, and |h(x,s)| ≤ g(s), where g :S → R is integrable on S with respect to dµ(s). Then the random
function ξ(x) = h(x,s)dµ(s) is integrable on B, and
S
R
dx h(x,s)dµ(s) = dµ(s) h(x,s)dx. (4.1)
Z Z Z Z
B S S B
Proof. From the inequality |h(x,s)| ≤ g(s) and (2.1) it follows that values of ξ are bounded in proba-
bility (see Lemma 1.1 and Theorem 1.3 [12]). Integral sums of ξ(x)dx have the form
B
R
m(B ) h(x ,s)dµ(s) = g (s)dµ(s),
kn Z kn Z n
X S S
1≤k≤kn
g (s)= h(x ,s)m(B ) → h(x,s)dx.
n kn kn Z
X B
1≤k≤kn
Boundedness condition of h and the analogue of the Lebesgue theorem [7, Proposition 7.1.1] for the
integral with respect to dµ(s) imply the statement.
5
Corollary 4.1. Let µ be a stochastic measure on (S,B), B˜ ⊂ Rd be an unbounded set. Assume that
h(x,s) : B˜ × S → R is a measurable deterministic function which is Riemann integrable on B˜ in
improper sense for each fixed s, and |h(x,s)| ≤ g(s), |h(x,s)|dx = g (s), where g, g : S → R are
B˜ 1 1
integrable on S with respect to dµ(s). Then the randoRm function ξ(x) = h(x,s)dµ(s) is integrable
S
on B˜ in improper sense, and R
dx h(x,s)dµ(s) = dµ(s) h(x,s)dx. (4.2)
Z Z Z Z
B˜ S S B˜
Proof. For bounded sets B(j) ↑ B˜, Theorem 4.1 implies
dx h(x,s)dµ(s) = dµ(s) h(x,s)dx.
Z Z Z Z
B(j) S S B(j)
Further, we use the analogue of the Lebesgue theorem and integrability of g .
1
Theorem 4.2. Let B ⊂ Rd, S ⊂ Rm be a bounded sets, random function ξ(x,s) : B × S → L be
0
integrable on B×S with respect to dx×ds and be integrable on S with respect to ds for each fixed x.
Then
ξ(x,s)dx×ds = dx ξ(x,s)ds. (4.3)
Z Z Z
B×S B S
Proof. Integral sums of integral with respect to dx in (4.3) has the form
m(B ) ξ(x ,s)ds. (4.4)
k Z k
X S
1≤k≤k0
Each integral in (4.4) may be approximated by sums of the form m(S )ξ(x ,s ). Thus, the
1≤i≤i0 i k i
sums P
m(B )m(S )ξ(x ,s ).
k i k i
X X
1≤k≤k01≤i≤i0
will approximate the right hand side of (4.4). But they are the integral sums for the integral with
respecttodx×ds in(4.3),andwillbeclosetothelefthandsideof (4.3)forsufficiently smalldiameters
of B ×S .
k i
Corollary 4.2. Let S ⊂ Rm be a bounded set, B˜ ⊂ Rd be an unbounded set. Assume that the random
function ξ(x,s) : B˜ × S → L is integrable on B˜ × S with respect to dx × ds in improper sense, is
0
integrable on B˜ with respect to dx in improper sense for each fixed s, and is integrable on S with respect
to ds for each fixed x. Then
ξ(x,s)dx×ds = ds ξ(x,s)dx = dx ξ(x,s)ds. (4.5)
Z Z Z Z Z
B˜×S S B˜ B˜ S
Proof. Consider exhaustive sets B(j) ↑B˜. For the first of the repeated integrals (4.5), the integral sums
has the form
m(S ) ξ(x,s )dx (4.6)
i Z i
X B˜
1≤i≤i0
The integrals in (4.6) can be approximated by ξ(x,s )dx, and the lastintegral is the limit of sums
B(j) i
R
m(B(j))ξ(x(j),s ).
k k i
X
1≤k≤k0
If integral sums (4.6) does not converge, then we can construct a non-convergent sequence of sums
m(S )m(B(j))ξ(x(j),s ),
i k k i
X X
1≤i≤i01≤k≤k0
6
and this contradicts the integrability of ξ on S×B˜.
Further, by Theorem 4.2, for each j we have
ξ(x,s)dx×ds = dx ξ(x,s)ds.
Z Z Z
B(j)×S B(j) S
The left hand side has the limit in probability as j → ∞. Hence, the right hand side has the limit,
and the second equality of (4.5) holds.
5 Integration by parts
To solve the parabolic stochastic equation, we need the following two lemmas.
Lemma 5.1. Let a random function ξ(u) : [0,s] → L be integrable on [0,s]. Then η(u) = uξ(v)dv
0 0
is integrable on [0,s], and R
s u s
du ξ(v)dv = (s−v)ξ(v)dv.
Z Z Z
0 0 0
Proof. By Lemma 3.5, the function (s−v)ξ(v) is integrable, by Lemma 3.2 η(u) is well defined. The
s
integral sum of η(u)du has the form
0
R
uk
m(B ) ξ(v)dv, u ∈ B . (5.1)
k Z k k
X 0
1≤k≤k0
We can take a new partition [0,s] = ∪ C such that each integral uk ξ(v)dv be close enough to
1≤i≤i0 i 0
integral sum with this partition (Lemma 3.2). Thus we can approximateR(5.1) arbitrary closely by the
sum
m(B ) m(C ∩[0,u ])ξ(v ), v ∈ C . (5.2)
k i k i i i
X X
1≤k≤k0 1≤i≤i0
s
For (s−v)ξ(v)dv, take the integral sum
0
R
m(C )(s−v )ξ(v ). (5.3)
i i i
X
1≤i≤i0
The difference of (5.3) and (5.2) is equal to
ξ(v )[m(C )(s−v )−m(C ) m(B )− m(B )m(C ∩[0,u ])]. (5.4)
i i i i k k i k
1≤Xi≤i0 k: XCi<Bk k: CXi∩Bk6=∅
Notation C < B means that v < u for all v ∈C ,u∈ B . We have
i k i k
0 ≤ (s−v )− m(B ) ≤ maxdiamC +maxdiamB .
i k i k
k: XCi<Bk i k
The last sum of (5.4) is not greater than
m(C ) m(B ) ≤ m(C )(maxdiamC +2maxdiamB ).
i k i i k
k: CXi∩Bk6=∅ i k
Therefore, value (5.4) may be written in the form ξ(v )m(C )α , where α → 0 as
1≤i≤i0 i i i i
diamC , diamB → 0. From (2.1) we obtain P
i k
ξ(v )m(C )α ≤ 16max max|α | ξ(v )m(C ) . (5.5)
i i i i i i
(cid:13) X (cid:13) V (cid:13) i X (cid:13)
(cid:13)1≤i≤i0 (cid:13) (cid:13) i∈V (cid:13)
(cid:13) (cid:13) (cid:13) (cid:13)
The sums ξ(v )m(C ) are close to respective integrals for diamC small enough (Lemma 3.2) and
i∈V i i i
values of inPtegrals are bounded inprobability (Lemma 3.4). Therefore, the lefthand side of (5.5)tends
to zero as max |α | → 0.
i i
7
Lemma 5.2. Let a random function ξ(u) : [0,s] → L be integrable on [0,s], f ∈ C(1)([0,s]) be a
0
deterministic function. Then
s s s u
f(s) ξ(u)du = f(u)ξ(u)du+ f′(u)du ξ(v)dv. (5.6)
Z Z Z Z
0 0 0 0
Proof. From Lemmas 3.5 and 5.1 it follows that the random functions ζ (u) = f(u)ξ(u), ζ (u) =
1 2
f′(u) uξ(v)dv are integrable on [0,s]. First, let us show that for 0 = u < u < ··· < u = s, α =
0 0 1 k0
max R|u −u |, we have
k k k−1
uk P s u
(f(u )−f(u )) ξ(v)dv → f′(u)du ξ(v)dv, α → 0. (5.7)
k k−1 Z Z Z
X 0 0 0
1≤k≤k0
Applying the Lagrange formula and integrability of ζ , for some u˜ ∈ (u ,u ) we obtain
2 k k−1 k
uk uk
(f(u )−f(u )) ξ(v)dv = f′(u˜ )(u −u ) ξ(v)dv
k k−1 Z k k k−1 Z
X 0 X 0
1≤k≤k0 1≤k≤k0
u˜k uk
= f′(u˜ )(u −u ) ξ(v)dv+ f′(u˜ )(u −u ) ξ(v)dv,
k k k−1 Z k k k−1 Z
1≤Xk≤k0 0 1≤Xk≤k0 u˜k
u˜k P s u
f′(u˜ )(u −u ) ξ(v)dv → f′(u)du ξ(v)dv, α → 0.
k k k−1 Z Z Z
X 0 0 0
1≤k≤k0
For C = max |f′(u)|, from (2.1) we have
1 u
uk uk
f′(u˜ )(u −u ) ξ(v)dv ≤ 16max C α ξ(v)dv ≤ 16sup C α ξ(v)dv .
(cid:13)(cid:13)1≤Xk≤k0 k k k−1 Zu˜k (cid:13)(cid:13) V (cid:13)(cid:13) 1 kX∈V Zu˜k (cid:13)(cid:13) A (cid:13)(cid:13) 1 ZA (cid:13)(cid:13)
(cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13)
From Lemma 3.4 it follows that the last value tends to 0 as α→ 0. Therefore, (5.7) is proved.
Integrability of ζ implies
1
uk
f(u ) ξ(v)dv = f(u )ξ(u )(u −u )
k−1 Z k−1 k−1 k k−1
1≤Xk≤k0 uk−1 1≤Xk≤k0
uk
+ f(u ) (ξ(v)−ξ(u ))dv,
k−1 Z k−1
1≤Xk≤k0 uk−1
s
P
f(u )ξ(u )(u −u ) → f(u)ξ(u)du, α → 0.
k−1 k−1 k k−1 Z
X 0
1≤k≤k0
For C = max |f(u)|, from (2.1) we get
0 u
uk uk
f(u ) (ξ(v)−ξ(u ))dv ≤ 16max C (ξ(v)−ξ(u ))dv
(cid:13)(cid:13)1≤Xk≤k0 k−1 Zuk−1 k−1 (cid:13)(cid:13) V (cid:13)(cid:13) 0kX∈V Zuk−1 k−1 (cid:13)(cid:13)
(cid:13) (cid:13) (cid:13) (cid:13)
= 16max C ( ξ(v)dv− ξ(u )(u −u )) .
V (cid:13)(cid:13) 0 Z∪k∈V[uk−1,uk] kX∈V k−1 k k−1 (cid:13)(cid:13)
(cid:13) (cid:13)
By Lemma 3.2, the last value tends to 0 as α → 0.
Further, we take the obvious equality
s uk uk
f(s) ξ(v)dv = (f(u )−f(u )) ξ(v)dv+ f(u ) ξ(v)dv
Z k k−1 Z k−1 Z
0 1≤Xk≤k0 0 1≤Xk≤k0 uk−1
and pass to the limit as α → 0.
8
6 Parabolic equation with a general stochastic measure
Consider the differential operator
∂2g(x) ∂g(x)
Ag(x) = a (x) + b (x) +c(x)g(x),
ij i
∂x ∂x ∂x
X i j X i
1≤i,j≤d 1≤i≤d
where g : Rd → R and a = a . Suppose that A is strongly elliptic in Rd (see (4.5) [5]).
ij ji
∂a ∂2a ∂b
Assumption 1. Allfunctionsa , b , c, ij, ij , i areboundedandHöldercontinuous in Rd.
ij i
∂x ∂x ∂x ∂x
i i j i
From now on let µ be a stochastic measure on Borel subsets of [0,T].
We will study the equation
dX(x,t) = AX(x,t)dt+f(x,t)dµ(t), X(x,0) = ξ(x), (1.1)
where X :Rd×[0,T] → L is an unknown random function.
0
We consider (1.1) in the weak sense, i.e.
X(x,t)ϕ(x)dx = ξ(x)ϕ(x)dx
Z Z
Rd Rd
t
+ A∗ϕ(x)dx X(x,s)ds+ dµ(s) f(x,s)ϕ(x)dx (6.1)
Z Z Z Z
Rd 0 [0,t] Rd
for all test functions ϕ ∈ S(Rd) (rapidly decreasing Schwartz functions from C∞(Rd)). For each
fixed t ∈ [0,T] equality (6.1) holds a.s. Integrals of random functions with respect to dx and ds are
considered in Riemann sense (see section 3), and A∗ denotes the adjoint operator of A.
Assumption 2. ξ :Rd → L issuchthatξ(·,ω) iscontinuous andbounded inRd foreachfixedω ∈ Ω.
0
Assumption 3. f : Rd ×[0,T] → R is Borel measurable, sup |x|−k|f(x,t)| → 0, |x| → ∞, for some
t
k > 0, f(x,·) is continuous and bounded in Rd for each fixed t ∈ [0,T].
By Theorem 1 §4 [5], under Assumption 1, the equation ∂g/∂t = Ag has a fundamental solu-
tion p(x,y,t − s) (recall that coefficients of A do not depend on t). The following estimate is well
known:
|p(x,y,t)| ≤C t−d/2exp{−C |x−y|2/t}
1 2
(see, for example, (4.16) [5]). Consider the semigroup
S(t)g(x) = p(x,y,t)g(y)dy, t > 0, S(0)g(x) = g(x).
Z
Rd
Theorem 2 §4 [5] implies that for any continuous bounded g
t
S(t)g(x) = g(x)+A [S(s)g(x)]ds. (6.2)
Z
0
Theorem 6.1. Suppose Assumptions 1–3 hold. Then the random function
X(x,t) = S(t)ξ(x)+ [S(t−s)f(x,s)]dµ(s) (6.3)
Z
[0,t]
is the solution of (6.1).
In addition, suppose the operator A is self-adjoint, X(x,t) satisfies (6.1), is integrable on Rd×[0,T]
with respect to dx×dt, is integrable on Rd with respect to dx for each fixed t, and is integrable on [0,T]
with respect to dt for each fixed x. Then X(x,t) is given by (6.3).
9
Proof. From (6.2)itfollows that forX (x,t) = S(t)ξ(x) andf = 0equality (6.1)holds. For X (x,t) =
1 2
[S(t−s)f(x,s)]dµ(s) we have
[0,t]
R
t
A∗ϕ(x)dx X (s)ds+ dµ(s) f(x,s)ϕ(x)dx
Z Z 2 Z Z
Rd 0 [0,t] Rd
t
= A∗ϕ(x)dx ds [S(s−u)f(x,u)]dµ(u)+ dµ(s) f(x,s)ϕ(x)dx
Z Z Z Z Z
Rd 0 [0,s] [0,t] Rd
t
(4=.1) A∗ϕ(x)dx dµ(u) [S(s−u)f(x,u)]ds+ dµ(s) f(x,s)ϕ(x)dx
Z Z Z Z Z
Rd [0,t] u [0,t] Rd
t
(4=.2) dµ(u) A∗ϕ(x)dx [S(s−u)f(x,u)]ds+ dµ(s) f(x,s)ϕ(x)dx
Z Z Z Z Z
[0,t] Rd u [0,t] Rd
t
= dµ(u) ϕ(x)dxA [S(s−u)f(x,u)]ds+ dµ(s) f(x,s)ϕ(x)dx
Z Z Z Z Z
[0,t] Rd u [0,t] Rd
(6.2)
= dµ(u) ϕ(x)dx([S(t−u)f(x,u)]−f(x,u))+ dµ(s) f(x,s)ϕ(x)dx
Z Z Z Z
[0,t] Rd [0,t] Rd
(4.2)
= dµ(u) ϕ(x)[S(t−u)f(x,u)]dx = ϕ(x)dx [S(t−s)f(x,s)]dµ(s)
Z Z Z Z
[0,t] Rd Rd [0,t]
= X (x,t)ϕ(x)dx.
Z 2
Rd
Therefore (6.1) holds for X = X +X .
1 2
Finally, we will prove the uniqueness of the solution. Section 4 implies that random function X
given by (6.3) is integrable. Thus, it is enough to prove that the equation
t
X(x,t)ϕ(x)dx = A∗ϕ(x)dx X(x,s)ds (6.4)
Z Z Z
Rd Rd 0
has only the zero solution provided that A= A∗.
For ϕ ∈ S(Rd) and 0 < s < t set ψ (x) = S(t−s)ϕ(x). Then
t,s
∂
ψ ∈ S(Rd), Aψ + ψ = 0, ψ → ϕ
t,s t,s t,s t,s
∂s
uniformly on any bounded subset of Rd as t ↓s (see (4.13) [5]), and we get
s
(6.4)
X(x,t)ψ (x)dx = Aψ (x)dx X(x,u)du
Z t,s Z t,s Z
Rd Rd 0
(5.6) s s ∂ u
= dx Aψ (x)X(x,u)du+ dx A ψ (x)du X(x,v)dv
ZRd Z0 t,u ZRd Z0 ∂u t,u Z0
(4.5) s s ∂ u
= dx Aψ (x)X(x,u)du+ du A ψ (x)dx X(x,v)dv
ZRd Z0 t,u Z0 ZRd ∂u t,u Z0
(6.4) s s ∂
= dx Aψ (x)X(x,u)du+ du ψ (x)X(x,u)dx
ZRd Z0 t,u Z0 ZRd ∂u t,u
(4.5) s ∂
= dx Aψ (x)+ ψ (x) X(x,u)du = 0.
ZRd Z0 (cid:16) t,u ∂u t,u (cid:17)
Passing to the limit as t ↓s and applying Lemma 3.9, we arrive at
X(x,s)ϕ(x)dx = 0.
Z
Rd
10