In this paper, we consider the stochastic heat equation with the Dirichlet boundary condition, which can formally be written as (1.1) 𝜕 𝑢 𝐿 𝜕 𝑡 = 𝜕 2 𝑢 𝐿 𝜕 𝑥 2 + 𝑓 ( 𝑡 , 𝑥 , 𝑢 𝐿 ( 𝑡 , 𝑥 ) ) + 𝜎 ( 𝑡 , 𝑥 ) 𝜕 𝜇 𝜕 𝑥 , 𝑢 𝐿 ( 𝑡 , − 𝐿 ) = 𝑢 𝐿 ( 𝑡 , 𝐿 ) = 0 , 𝑢 𝐿 ( 0 , 𝑥 ) = 𝑢 0 ( 𝑥 ) , where ( 𝑡 , 𝑥 ) ∈ [ 0 , 𝑇 ] × [ − 𝐿 , 𝐿 ], and 𝜇 is a stochastic measure defined on B ( R ) (Borel 𝜎-algebra in R). For 𝜇 we assume only 𝜎-additivity in probability, assumptions for 𝑓 , 𝜎 and 𝑢 0 are given in Section 3. We study the solution to the formal equation (1.1) in the mild form (see (3.1) below).

We study the convergence as 𝐿 → ∞, and prove that 𝑢 𝐿 tends to the solution to the equation (1.2) 𝜕 𝑢 𝜕 𝑡 = 𝜕 2 𝑢 𝜕 𝑥 2 + 𝑓 ( 𝑡 , 𝑥 , 𝑢 ( 𝑡 , 𝑥 ) ) + 𝜎 ( 𝑡 , 𝑥 ) 𝜕 𝜇 𝜕 𝑥 , 𝑢 ( 0 , 𝑥 ) = 𝑢 0 ( 𝑥 ) , where ( 𝑡 , 𝑥 ) ∈ [ 0 , 𝑇 ] × R. The solution to equation (1.2) is also considered in the mild form (see (3.2) below).

As the main result of the paper, we obtain an exponential rate for the pointwise convergence 𝑢 𝐿 → 𝑢. This is a natural result that follows from the form of the Green’s functions. Zero Dirichlet conditions represent absorbing boundaries whose effect decays exponentially as the domain expands.

For an equation driven by space-time white noise, this problem was studied in [2, 3], where the Neumann and mixed boundary conditions were also considered.

The difference of the form | 𝑢 𝐿 − 𝑢 | is often called a localization error and, in different settings, has been studied in [1, 6]. The impact of localization on some computational problems in financial mathematics is considered in [8].

Theory of equations driven by stochastic measures is developed in [15]. Existence and uniqueness of mild solutions to (1.1) and (1.2), their boundedness and Hölder regularity, are proved in [18] and [17] respectively. Also, generalization of equation (1.2) is considered in [16], where the L 2 -continuity of the solution was proved. For the heat equation on a bounded domain in R 𝑑 , we can refer to [10].

Different classes of equations driven by space-time white noise are considered in detail in [4], those driven by spatially colored Gaussian noise in [5], and those driven by a Lévy process in [12].

The rest of the paper is organized as follows. Section 2 contains the basic information concerning the integrals with respect to stochastic measures. In Section 3, we formulate the problem and our main assumptions. In Section 4 we obtain an estimate for the difference of the stochastic integrals from our equations. Section 5 contains the main result of the paper.

Let L 0 = L 0 ( Ω , F , P ) be the set of all real-valued random variables defined on the complete probability space ( Ω , F , P ). Convergence in L 0 means the convergence in probability. Let X be an arbitrary set and B a 𝜎-algebra of subsets of X.

Definition 1.

A 𝜎-additive mapping 𝜇 : B → L 0 is called stochastic measure (SM).

We do not assume the moment existence or martingale properties for SM. We can say that 𝜇 is an L 0 -valued measure.

Important examples of SMs are random orthogonal measure with finite structural function (see [7, Section 5.3]), 𝛼-stable random measures defined on a 𝜎-algebra for 𝛼 ∈ ( 0 , 1 ) ∪ ( 1 , 2 ] (see [19, Chapter 3]). Many examples of the SMs on B ( R ) may be given by the Wiener-type integral (2.1) 𝜇 ( 𝐴 ) = ∫ R 1 𝐴 ( 𝑥 ) 𝑓 ( 𝑥 ) 𝑑 𝑌 𝑥 , where 𝑓 is a deterministic function, integrable on R. For instance, we can take in (2.1) any square integrable martingale 𝑌 𝑥 , or 𝑌 𝑥 = 𝑊 𝑥 𝐻 – the fractional Brownian motion with Hurst index 𝐻 > 1 / 2 (see Corollary 1.9.4 in [11]).

Other examples of SMs can be found in Section 1.2.1 [15].

For deterministic measurable functions 𝑓 : X → R, an integral of the form ∫ X 𝑓 𝑑𝜇 is defined and studied in [9, Chapter 7], [15, Chapter 1]. In particular, every bounded measurable 𝑓 is integrable with respect to any 𝜇.

Below, we will use the following statement.

Lemma 1

Let 𝜇 be an SM on ( X , B ), 𝑓 𝑘 : X → R , 𝑘 ≥ 1, be measurable functions such that 𝑓 ˆ ( 𝑥 ) = ∑ 𝑘 = 1 ∞ | 𝑓 𝑘 ( 𝑥 ) | is integrable with respect to 𝜇. Then ∑ 𝑘 = 1 ∞ ( ∫ X 𝑓 𝑘 𝑑𝜇 ) 2 < ∞ a.s.

For any 𝑗 ∈ Z and all 𝑛 ≥ 0, put 𝑑 𝑘𝑛 ( 𝑗 ) = 𝑗 + 𝑘 2 − 𝑛 , 0 ≤ 𝑘 ≤ 2 𝑛 , Δ 𝑘𝑛 ( 𝑗 ) = ( 𝑑 ( 𝑘 − 1 ) 𝑛 ( 𝑗 ) , 𝑑 𝑘𝑛 ( 𝑗 ) ] , 1 ≤ 𝑘 ≤ 2 𝑛 .

The following lemma is a key tool for estimating a version of the stochastic integral. In the sequel, we assume that SM 𝜇 is defined on Borel subsets of R.

Lemma 2

Let 𝑍 be an arbitrary set, and the function 𝑞 ( 𝑧 , 𝑥 ) : 𝑍 × [ 𝑗 , 𝑗 + 1 ] → R be such that all paths 𝑞 ( 𝑧 , · ) are continuous on [ 𝑗 , 𝑗 + 1 ], A ⊂ ( 𝑗 , 𝑗 + 1 ] be a Borel set. Denote 𝑞 𝑛 ( 𝑧 , 𝑥 ) = ∑ 1 ≤ 𝑘 ≤ 2 𝑛 𝑞 ( 𝑧 , 𝑑 ( 𝑘 − 1 ) 𝑛 ( 𝑗 ) ) 1 Δ 𝑘𝑛 ( 𝑗 ) ( 𝑥 ) . Then the random function 𝜂 ( 𝑧 ) = ∫ A 𝑞 ( 𝑧 , 𝑥 ) 𝑑𝜇 ( 𝑥 ) , 𝑧 ∈ 𝑍 , has a version (2.2) 𝜂 ̃ ( 𝑧 ) = ∫ A 𝑞 0 ( 𝑧 , 𝑥 ) 𝑑𝜇 ( 𝑥 ) + ∑ 𝑛 ≥ 1 ( ∫ A 𝑞 𝑛 ( 𝑧 , 𝑥 ) 𝑑𝜇 ( 𝑥 ) − ∫ A 𝑞 𝑛 − 1 ( 𝑧 , 𝑥 ) 𝑑𝜇 ( 𝑥 ) ) such that for all 𝛽 > 0, 𝜔 ∈ Ω, 𝑧 ∈ 𝑍 (2.3) | 𝜂 ̃ ( 𝑧 ) | ≤ | 𝑞 ( 𝑧 , 𝑗 ) 𝜇 ( A ) | + { ∑ 𝑛 ≥ 1 2 𝑛𝛽 ∑ 1 ≤ 𝑘 ≤ 2 𝑛 | 𝑞 ( 𝑧 , 𝑑 𝑘𝑛 ( 𝑗 ) ) − 𝑞 ( 𝑧 , 𝑑 ( 𝑘 − 1 ) 𝑛 ( 𝑗 ) ) | 2 } 1 / 2 × { ∑ 𝑛 ≥ 1 2 − 𝑛𝛽 ∑ 1 ≤ 𝑘 ≤ 2 𝑛 | 𝜇 ( Δ 𝑘𝑛 ( 𝑗 ) ∩ A ) | 2 } 1 / 2 .

In the sequel, by 𝐶 and 𝐶 ( 𝜔 ) we denote the constants and random constants, respectively, whose exact values are not important. Notations of the kind 𝐶 𝜆 , 𝐶 𝛿 means that these constants may depend of 𝜆, 𝛿 respectively.

We consider (1.1) in the mild form (3.1) 𝑢 𝐿 ( 𝑡 , 𝑥 ) = ∫ ( − 𝐿 , 𝐿 ] 𝐺 𝐿 ( 𝑡 , 𝑥 , 𝑦 ) 𝑢 0 ( 𝑦 ) 𝑑𝑦 + ∫ 0 𝑡 ∫ ( − 𝐿 , 𝐿 ] 𝐺 𝐿 ( 𝑡 − 𝑠 , 𝑥 , 𝑦 ) 𝑓 ( 𝑠 , 𝑦 , 𝑢 𝐿 ( 𝑠 , 𝑦 ) ) 𝑑𝑦 𝑑𝑠 + ∫ ( − 𝐿 , 𝐿 ] ∫ 0 𝑡 𝐺 𝐿 ( 𝑡 − 𝑠 , 𝑥 , 𝑦 ) 𝜎 ( 𝑠 , 𝑦 ) 𝑑𝑠 𝑑𝜇 ( 𝑦 ) , where equality holds a.s. for each ( 𝑡 , 𝑥 ) ∈ [ 0 , 𝑇 ] × [ − 𝐿 , 𝐿 ]. Here 𝐺 𝐿 ( 𝑡 , 𝑥 , 𝑦 ) is the fundamental solution of the heat equation with the Dirichlet boundary condition 𝜕 𝑢 𝜕 𝑡 = 𝜕 2 𝑢 𝜕 𝑥 2 , ( 𝑡 , 𝑥 ) ∈ [ 0 , 𝑇 ] × [ − 𝐿 , 𝐿 ] , 𝑢 ( 𝑡 , − 𝐿 ) = 𝑢 ( 𝑡 , 𝐿 ) = 0 . It is known that 𝐺 𝐿 ( 𝑡 , 𝑥 , 𝑦 ) = ∑ 𝑛 = − ∞ ∞ ( 𝑝 ( 𝑡 , 𝑥 − 𝑦 + 4 𝑛𝐿 ) − 𝑝 ( 𝑡 , 𝑥 + 𝑦 + ( 4 𝑛 + 2 ) 𝐿 ) ) , where 𝑝 ( 𝑡 , 𝑥 ) = 1 4 𝜋𝑡 exp { − 𝑥 2 4 𝑡 } is a standard heat kernel (see, for example, [13, Section 8.4.2], [4, Lemma 1.4.1]).

We will refer to the following assumptions on elements of (3.1).

Assumption A1.

𝑢 0 ( 𝑦 ) = 𝑢 0 ( 𝑦 , 𝜔 ) : R × Ω → R is measurable and | 𝑢 0 ( 𝑦 ) | ≤ 𝐶 𝑢 0 ( 𝜔 ) for some random constant 𝐶 𝑢 0 ( 𝜔 ).

Assumption A2.

𝑓 ( 𝑠 , 𝑦 , 𝑣 ) : [ 0 , 𝑇 ] × R × R → R is measurable, and bounded: | 𝑓 ( 𝑠 , 𝑦 , 𝑣 ) | ≤ 𝐶 𝑓 for some constant 𝐶 𝑓 .

Assumption A3.

𝑓 ( 𝑠 , 𝑦 , 𝑣 ) is globally Lipschitz in 𝑣 , 𝑦: | 𝑓 ( 𝑠 , 𝑦 1 , 𝑣 1 ) − 𝑓 ( 𝑠 , 𝑦 2 , 𝑣 2 ) | ≤ 𝐿 𝑓 ( | 𝑣 1 − 𝑣 2 | + | 𝑦 1 − 𝑦 2 | ) , 𝑦 𝑖 , 𝑣 𝑖 ∈ R , for some constant 𝐿 𝑓 .

Assumption A4.

𝜎 ( 𝑠 , 𝑦 ) : [ 0 , 𝑇 ] × R → R is measurable, and | 𝜎 ( 𝑠 , 𝑦 ) | ≤ 𝐶 𝜎 , | 𝜎 ( 𝑠 , 𝑦 1 ) − 𝜎 ( 𝑠 , 𝑦 2 ) | ≤ 𝐿 𝜎 | 𝑦 1 − 𝑦 2 | 𝛽 ( 𝜎 ) for some 1 / 2 < 𝛽 ( 𝜎 ) < 1 and constants 𝐶 𝜎 , 𝐿 𝜎 .

Assumption A5.

| 𝑦 | 𝜏 is integrable w.r.t. 𝜇 on R for some 𝜏 > 1 / 2.

Assumptions A1–A4 and Theorem 5.1 1) [18] imply that (3.1) has a unique bounded solution.

Also, we consider (1.2) in the mild form (3.2) 𝑢 ( 𝑡 , 𝑥 ) = ∫ R 𝑝 ( 𝑡 , 𝑥 − 𝑦 ) 𝑢 0 ( 𝑦 ) 𝑑𝑦 + ∫ 0 𝑡 ∫ R 𝑝 ( 𝑡 − 𝑠 , 𝑥 − 𝑦 ) 𝑓 ( 𝑠 , 𝑦 , 𝑢 ( 𝑠 , 𝑦 ) ) 𝑑𝑦 𝑑𝑠 + ∫ R ∫ 0 𝑡 𝑝 ( 𝑡 − 𝑠 , 𝑥 − 𝑦 ) 𝜎 ( 𝑠 , 𝑦 ) 𝑑𝑠 𝑑𝜇 ( 𝑦 ) .

The existence and uniqueness of the bounded solution to (3.2) follows from Theorem 5.1 1) [17].

This paper aims to prove that 𝑢 𝐿 ( 𝑡 , 𝑥 ) → 𝑢 ( 𝑡 , 𝑥 ) , 𝐿 → ∞ , for each ( 𝑡 , 𝑥 ) ∈ [ 0 , 𝑇 ] × R. Mainly, we will consider 𝐿 = 𝑖𝛿 for some fixed 𝛿 > 0, 𝑖 ≥ 1, and obtain the estimate of | 𝑢 𝐿 ( 𝑡 , 𝑥 ) − 𝑢 ( 𝑡 , 𝑥 ) | for some versions of the solutions.

In the estimates below, we will use that (3.3) | 𝜕 𝑝 𝜕 𝑦 ( 𝑡 − 𝑠 , 𝑥 − 𝑦 ) | ≤ 𝐶 𝜆 𝑡 − 𝑠 𝑒 − 𝜆 ( 𝑥 − 𝑦 ) 2 𝑡 − 𝑠 for any 𝜆 ∈ ( 0 , 1 / 4 ) and some constant 𝐶 𝜆 .

In the sequel, we will use the notations 𝐷 𝐿 = ( − 𝐿 , 𝐿 ], 𝐷 𝐿 𝑐 = R \ ( − 𝐿 , 𝐿 ], Z ∗ = Z \ { 0 }.

Consider the stochastic terms 𝜗 𝐿 ( 𝑡 , 𝑥 ) = ∫ 𝐷 𝐿 ∫ 0 𝑡 𝐺 𝐿 ( 𝑡 − 𝑠 , 𝑥 , 𝑦 ) 𝜎 ( 𝑠 , 𝑦 ) 𝑑𝑠 𝑑𝜇 ( 𝑦 ) , 𝜗 ( 𝑡 , 𝑥 ) = ∫ R ∫ 0 𝑡 𝑝 ( 𝑡 − 𝑠 , 𝑥 − 𝑦 ) 𝜎 ( 𝑠 , 𝑦 ) 𝑑𝑠 𝑑𝜇 ( 𝑦 ) .

In the following statement, we fix any 𝛿 > 0, and consider 𝐿 = 𝑖𝛿, 𝑖 ∈ Z + .

Lemma 3.

Let Assumptions A4 and A5 hold. Then, for each 𝛿 > 0 there exist versions of 𝜗 𝐿 , 𝜗 such that for all 𝜔 ∈ Ω for each 𝑥 ∈ 𝐷 𝐿 , 𝐿 = 𝑖𝛿, 𝑖 ∈ Z + , 0 < 𝑡 ≤ 𝑇, for any 0 < 𝜆 < 1 4 holds (4.1) | 𝜗 𝐿 ( 𝑡 , 𝑥 ) − 𝜗 ( 𝑡 , 𝑥 ) | ≤ 𝐶 𝛿 ( 𝜔 ) 𝐶 𝜆 𝑒 − 𝜆 ( 𝐿 − | 𝑥 | ) 2 𝑡 .

Proof.

For 𝑥 ∈ 𝐷 𝐿 , we will estimate 𝜗 𝐿 ( 𝑡 , 𝑥 ) − 𝜗 ( 𝑡 , 𝑥 ) = ∫ 𝐷 𝐿 ∫ 0 𝑡 𝐺 𝐿 ( 𝑡 − 𝑠 , 𝑥 , 𝑦 ) 𝜎 ( 𝑠 , 𝑦 ) 𝑑𝑠 𝑑𝜇 ( 𝑦 ) − ∫ R ∫ 0 𝑡 𝑝 ( 𝑡 − 𝑠 , 𝑥 − 𝑦 ) 𝜎 ( 𝑠 , 𝑦 ) 𝑑𝑠 𝑑𝜇 ( 𝑦 ) = ∫ 𝐷 𝐿 ∫ 0 𝑡 ( 𝐺 𝐿 ( 𝑡 − 𝑠 , 𝑥 , 𝑦 ) − 𝑝 ( 𝑡 − 𝑠 , 𝑥 − 𝑦 ) ) 𝜎 ( 𝑠 , 𝑦 ) 𝑑𝑠 𝑑𝜇 ( 𝑦 ) − ∫ 𝐷 𝐿 𝑐 ∫ 0 𝑡 𝑝 ( 𝑡 − 𝑠 , 𝑥 − 𝑦 ) 𝜎 ( 𝑠 , 𝑦 ) 𝑑𝑠 𝑑𝜇 ( 𝑦 ) : = 𝐼 1 − 𝐼 2 . First, consider 𝐼 1 . For 𝑥 , 𝑦 ∈ 𝐷 𝐿 , (4.2) 𝐺 𝐿 ( 𝑡 − 𝑠 , 𝑥 , 𝑦 ) − 𝑝 ( 𝑡 − 𝑠 , 𝑥 − 𝑦 ) = ∑ 𝑛 ∈ Z ∗ 𝑝 ( 𝑡 − 𝑠 , 𝑥 − 𝑦 + 4 𝑛𝐿 ) − ∑ 𝑛 ∈ Z ∗ 𝑝 ( 𝑡 − 𝑠 , 𝑥 + 𝑦 + ( 4 𝑛 + 2 ) 𝐿 ) − 𝑝 ( 𝑡 − 𝑠 , 𝑥 + 𝑦 + 2 𝐿 ) .

For 𝑥 , 𝑦 ∈ 𝐷 𝐿 we have that | 𝑥 + 𝑦 + 2 𝐿 | ≥ 𝐿 − | 𝑥 |.

We will take 𝑖 large enough such that | 𝑥 − 𝑦 + 4 𝑛𝐿 | , | 𝑥 + 𝑦 + ( 4 𝑛 + 2 ) 𝐿 | > 𝑇 ≥ 𝑡 − 𝑠 , 𝑛 ∈ Z ∗ here. (By Lemma 4.2 [17], 𝜗 is bounded. By (4.9) of [18], each 𝜗 𝑖𝛿 is bounded. Thus, excluding the finite number of 𝑖 does not influence (4.1).)

For fixed 𝑡 , 𝑥 set (4.3) 𝑞 ( 𝑧 , 𝑦 ) = ∫ 0 𝑡 𝑝 ( 𝑡 − 𝑠 , 𝑥 − 𝑦 ) 𝜎 ( 𝑠 , 𝑦 ) 𝑑𝑠 , 𝑧 = ( 𝑡 , 𝑥 ) , 𝑦 ∈ 𝐷 𝐿 . Our considerations will imply that this function is continuous in 𝑦; thus, the conditions of Lemma 2 hold (for this purpose, we can use estimate (4.6) below).

Consider 𝑞 ( 𝑧 , 𝑦 + ℎ ) − 𝑞 ( 𝑧 , 𝑦 ) = ∫ 0 𝑡 𝑝 ( 𝑡 − 𝑠 , 𝑥 − 𝑦 ) ( 𝜎 ( 𝑠 , 𝑦 + ℎ ) − 𝜎 ( 𝑠 , 𝑦 ) ) 𝑑𝑠 + ∫ 0 𝑡 ( 𝑝 ( 𝑡 − 𝑠 , 𝑥 − 𝑦 − ℎ ) − 𝑝 ( 𝑡 − 𝑠 , 𝑥 − 𝑦 ) ) 𝜎 ( 𝑠 , 𝑦 + ℎ ) 𝑑𝑠 : = 𝐼 11 + 𝐼 12 .

We will use the following simple estimate for 𝑏 / 𝑡 ≥ 1. (4.4) ∫ 0 𝑡 1 𝑟 𝑒 − 𝑏 𝑟 𝑑𝑟 = 𝑏 / 𝑟 = 𝑧 ∫ 𝑏 / 𝑡 ∞ 1 𝑧 𝑒 − 𝑧 𝑑𝑧 ≤ ∫ 𝑏 / 𝑡 ∞ 𝑒 − 𝑧 𝑑𝑧 = 𝑒 − 𝑏 / 𝑡 . We get for | 𝑥 − 𝑦 | , | 𝑥 − 𝑦 − ℎ | ≥ 𝑘𝐿 − | 𝑥 | > 𝑇, ℎ < 𝐿 | 𝐼 12 | ≤ A4 𝐶 𝜎 ∫ 0 𝑡 | 𝑝 ( 𝑡 − 𝑠 , 𝑥 − ℎ − 𝑦 ) − 𝑝 ( 𝑡 − 𝑠 , 𝑥 − 𝑦 ) | 𝑑𝑠 ≤ 𝐶 𝜎 ∫ 0 𝑡 | ∫ 𝑥 − ℎ 𝑥 𝜕 𝑝 ( 𝑡 − 𝑠 , 𝑣 − 𝑦 ) 𝜕 𝑣 𝑑𝑣 | 𝑑𝑠 ≤ (3.3) 𝐶 𝜎 ∫ 0 𝑡 ( 𝐶 𝜆 𝑡 − 𝑠 ∫ 𝑥 − ℎ 𝑥 𝑒 − 𝜆 | 𝑣 − 𝑦 | 2 𝑡 − 𝑠 𝑑𝑣 ) 𝑑𝑠 = 𝑡 − 𝑠 = 𝑟 𝐶 𝜎 𝐶 𝜆 ∫ 𝑥 − ℎ 𝑥 𝑑𝑣 ∫ 0 𝑡 1 𝑟 𝑒 − 𝜆 | 𝑣 − 𝑦 | 2 𝑟 𝑑𝑟 ≤ (4.4) 𝐶 𝜎 𝐶 𝜆 ∫ 𝑥 − ℎ 𝑥 𝑒 − 𝜆 | 𝑣 − 𝑦 | 2 𝑡 𝑑𝑣 ≤ 𝐶 𝜎 𝐶 𝜆 | ℎ | 𝑒 − 𝜆 ( 𝑘𝐿 − | 𝑥 | ) 2 𝑡 .

Also, (4.5) | 𝐼 11 | ≤ ∫ 0 𝑡 𝑝 ( 𝑡 − 𝑠 , 𝑥 − 𝑦 ) | 𝜎 ( 𝑠 , 𝑦 + ℎ ) − 𝜎 ( 𝑠 , 𝑦 ) | 𝑑𝑠 ≤ A4 𝐶 𝐿 𝜎 ℎ 𝛽 ( 𝜎 ) ∫ 0 𝑡 1 𝑡 − 𝑠 𝑒 − | 𝑥 − 𝑦 | 2 4 ( 𝑡 − 𝑠 ) 𝑑𝑠 ≤ 𝐶 ℎ 𝛽 ( 𝜎 ) ∫ 0 𝑡 1 𝑡 − 𝑠 𝑒 − ( 𝑘𝐿 − | 𝑥 | ) 2 4 ( 𝑡 − 𝑠 ) 𝑑𝑠 ≤ 𝐶 ℎ 𝛽 ( 𝜎 ) 𝑒 − ( 𝑘𝐿 − | 𝑥 | ) 2 4 𝑡 ∫ 0 𝑡 1 𝑡 − 𝑠 𝑑𝑠 ≤ 𝐶 ℎ 𝛽 ( 𝜎 ) 𝑒 − ( 𝑘𝐿 − | 𝑥 | ) 2 4 𝑡 . (The constants 𝐶 𝜆 and 𝐶 may also depend on 𝜎, though this plays no role in our considerations.) Thus, for any 0 < 𝜆 < 1 4 (4.6) | 𝑞 ( 𝑧 , 𝑦 + ℎ ) − 𝑞 ( 𝑧 , 𝑦 ) | ≤ 𝐶 𝜆 ℎ 𝛽 ( 𝜎 ) 𝑒 − 𝜆 ( 𝑘𝐿 − | 𝑥 | ) 2 𝑡 . Similarly, consider 𝑝 ( 𝑡 − 𝑠 , 𝑥 + 𝑦 + ( 4 𝑛 + 2 ) 𝐿 ), 𝑛 ∈ Z ∗ , 𝑝 ( 𝑡 − 𝑠 , 𝑥 + 𝑦 + 2 𝐿 ), and for 𝑞 ̃ ( 𝑧 , 𝑦 ) = ∫ 0 𝑡 ( ∑ 𝑛 ∈ Z ∗ 𝑝 ( 𝑡 − 𝑠 , 𝑥 − 𝑦 + 4 𝑛𝐿 ) − ∑ 𝑛 ∈ Z ∗ 𝑝 ( 𝑡 − 𝑠 , 𝑥 + 𝑦 + ( 4 𝑛 + 2 ) 𝐿 ) − 𝑝 ( 𝑡 − 𝑠 , 𝑥 + 𝑦 + 2 𝐿 ) ) 𝜎 ( 𝑠 , 𝑦 ) 𝑑𝑠 we get (4.7) | 𝑞 ̃ ( 𝑧 , 𝑦 + ℎ ) − 𝑞 ̃ ( 𝑧 , 𝑦 ) | ≤ 𝐶 𝐶 𝜆 ℎ 𝛽 ( 𝜎 ) 𝑒 − 𝜆 ( 𝐿 − | 𝑥 | ) 2 𝑡 , 𝑧 = ( 𝑡 , 𝑥 ) , 𝑦 , 𝑦 + ℎ ∈ 𝐷 𝐿 . As in (4.5), we obtain | 𝑞 ( 𝑧 , 𝑦 ) | ≤ 𝐶 𝐶 𝜆 𝑒 − 𝜆 ( 𝑘𝐿 − | 𝑥 | ) 2 𝑡 . Therefore, (4.8) | 𝑞 ̃ ( 𝑧 , 𝑦 ) | ≤ 𝐶 𝐶 𝜆 𝑒 − 𝜆 ( 𝐿 − | 𝑥 | ) 2 𝑡 , 𝑧 = ( 𝑡 , 𝑥 ) , 𝑦 ∈ 𝐷 𝐿 .

We take 𝜏 > 1 / 2 from A5 and some 𝛽 > 0 such that 𝛽 + 1 − 2 𝛽 ( 𝜎 ) < 0. From (2.3), for the version (2.2) of the stochastic integral (taking intervals [ 𝑗𝛿 , ( 𝑗 + 1 ) 𝛿 ] instead of [ 𝑗 , 𝑗 + 1 ] and 𝑑 𝑘𝑛 ( 𝑗 ) = ( 𝑗 + 𝑘 2 − 𝑛 ) 𝛿), we get (4.9) | 𝐼 1 | = | ∫ 𝐷 𝐿 ∫ 0 𝑡 ( 𝐺 𝐿 ( 𝑡 − 𝑠 , 𝑥 , 𝑦 ) − 𝑝 ( 𝑡 − 𝑠 , 𝑥 − 𝑦 ) ) 𝜎 ( 𝑠 , 𝑦 ) 𝑑𝑠 𝑑𝜇 ( 𝑦 ) | = | ∫ 𝐷 𝐿 𝑞 ̃ ( 𝑧 , 𝑦 ) 𝑑𝜂 ( 𝑦 ) | ≤ ∑ 𝑗 ∈ Z | ∫ ( 𝑗𝛿 , ( 𝑗 + 1 ) 𝛿 ] ∩ 𝐷 𝐿 𝑞 ̃ ( 𝑧 , 𝑦 ) 𝑑𝜇 ( 𝑦 ) | ≤ (2.3) , 𝐿 = 𝑖𝛿 ∑ 𝑗 : ( 𝑗𝛿 , ( 𝑗 + 1 ) 𝛿 ] ⊂ 𝐷 𝐿 | 𝑞 ̃ ( 𝑧 , 𝑗𝛿 ) 𝜇 ( ( 𝑗𝛿 , ( 𝑗 + 1 ) 𝛿 ] ) | + { ∑ 𝑛 ≥ 1 2 𝑛𝛽 ∑ 1 ≤ 𝑘 ≤ 2 𝑛 | 𝑞 ̃ ( 𝑧 , 𝑑 𝑘𝑛 ( 𝑗 ) ) − 𝑞 ̃ ( 𝑧 , 𝑑 ( 𝑘 − 1 ) 𝑛 ( 𝑗 ) ) | 2 } 1 / 2 × { ∑ 𝑛 ≥ 1 2 − 𝑛𝛽 ∑ 1 ≤ 𝑘 ≤ 2 𝑛 | 𝜇 ( Δ 𝑘𝑛 ( 𝑗 ) ) | 2 } 1 / 2 ≤ (4.8) , (4.7) 𝐶 𝐶 𝜆 𝑒 − 𝜆 ( 𝐿 − | 𝑥 | ) 2 𝑡 [ ∑ 𝑗 : ( 𝑗𝛿 , ( 𝑗 + 1 ) 𝛿 ] ⊂ 𝐷 𝐿 | 𝜇 ( 𝑗𝛿 , ( 𝑗 + 1 ) 𝛿 ] ) | + ∑ 𝑗 : ( 𝑗𝛿 , ( 𝑗 + 1 ) 𝛿 ] ⊂ 𝐷 𝐿 { ∑ 𝑛 ≥ 1 2 𝑛 ( 𝛽 + 1 − 2 𝛽 ( 𝜎 ) ) ∑ 1 ≤ 𝑘 ≤ 2 𝑛 | 𝜇 ( Δ 𝑘𝑛 ( 𝑗 ) ) | 2 } 1 / 2 ] ≤ 𝐶 𝐶 𝜆 𝑒 − 𝜆 ( 𝐿 − | 𝑥 | ) 2 𝑡 [ ( ∑ 𝑗 ∈ Z ( | 𝑗 | + 1 ) 2 𝜏 ( 𝜇 ( ( 𝑗𝛿 , ( 𝑗 + 1 ) 𝛿 ] ) ) 2 ) 1 / 2 ( ∑ 𝑗 ∈ Z ( | 𝑗 | + 1 ) − 2 𝜏 ) 1 / 2 + ( ∑ 𝑗 ∈ Z ( | 𝑗 | + 1 ) 2 𝜏 ∑ 𝑛 ≥ 1 2 𝑛 ( 𝛽 + 1 − 2 𝛽 ( 𝜎 ) ) ∑ 1 ≤ 𝑘 ≤ 2 𝑛 | 𝜇 ( Δ 𝑘𝑛 ( 𝑗 ) ) | 2 ) 1 / 2 ( ∑ 𝑗 ∈ Z ( | 𝑗 | + 1 ) − 2 𝜏 ) 1 / 2 ] .

Here, the sums with 𝜇 have a form ∑ 𝑙 = 1 ∞ ( ∫ R 𝑓 𝑙 𝑑𝜇 ) 2 , where { 𝑓 𝑙 ( 𝑦 ) , 𝑙 ≥ 1 } = { ( | 𝑗 | + 1 ) 𝜏 1 ( 𝑗𝛿 , ( 𝑗 + 1 ) 𝛿 ] ( 𝑦 ) , 𝑗 ∈ Z } , { 𝑓 𝑙 ( 𝑦 ) , 𝑙 ≥ 1 } = { ( | 𝑗 | + 1 ) 𝜏 2 𝑛 ( 𝛽 + 1 − 2 𝛽 ( 𝜎 ) ) / 2 1 Δ 𝑘𝑛 ( 𝑗 ) ( 𝑦 ) , 𝑗 ∈ Z , 𝑛 ≥ 1 , 1 ≤ 𝑘 ≤ 2 𝑛 } . We have ∑ 𝑙 = 1 ∞ | 𝑓 𝑙 ( 𝑦 ) | ≤ 𝐶 ( | 𝑦 | + 1 ) 𝜏 in both cases. From Lemma 1 it follows that ∑ 𝑙 = 1 ∞ ( ∫ X 𝑓 𝑙 𝑑𝜇 ) 2 < + ∞ a.s. We take a set Ω 1 ⊂ Ω such that for 𝜔 ∈ Ω 1 , this series is finite for both sequences { 𝑓 𝑙 }. Thus P ( Ω 1 ) = 1 and (4.10) | 𝐼 1 | ≤ 𝐶 𝛿 ( 𝜔 ) 𝐶 𝜆 𝑒 − 𝜆 ( 𝐿 − | 𝑥 | ) 2 𝑡 holds for all 𝐿 = 𝑖𝛿, 𝑖 ∈ Z + , and 𝜔 ∈ Ω 1 . The set Ω 1 depends only on 𝜇, 𝛿 and 𝛽.

In 𝐼 2 we have | 𝑥 − 𝑦 | ≥ 𝐿 − | 𝑥 |. Again we consider 𝑞 ( 𝑧 , 𝑦 ) from (4.3), now with | 𝑦 | ≥ 𝐿. Repeating the consideration from (4.3) to (4.6), we get here | 𝑞 ( 𝑧 , 𝑦 + ℎ ) − 𝑞 ( 𝑧 , 𝑦 ) | ≤ 𝐶 𝜆 ℎ 𝛽 ( 𝜎 ) 𝑒 − 𝜆 ( 𝐿 − | 𝑥 | ) 2 𝑡 , | 𝑞 ( 𝑧 , 𝑦 ) | ≤ 𝐶 𝐶 𝜆 𝑒 − 𝜆 ( 𝑘𝐿 − | 𝑥 | ) 2 𝑡 , | 𝑦 | ≥ 𝐿 . Similarly to (4.9), we get (4.11) | 𝐼 2 | = | ∫ 𝐷 𝐿 𝑐 ∫ 0 𝑡 𝑝 ( 𝑡 − 𝑠 , 𝑥 − 𝑦 ) 𝜎 ( 𝑠 , 𝑦 ) 𝑑𝑠 𝑑𝜇 ( 𝑦 ) | = | ∫ 𝐷 𝐿 𝑐 𝑞 ( 𝑧 , 𝑦 ) 𝑑𝜂 ( 𝑦 ) | ≤ ∑ 𝑗 ∈ Z | ∫ ( 𝑗𝛿 , ( 𝑗 + 1 ) 𝛿 ] ∩ 𝐷 𝐿 𝑐 𝑞 ( 𝑧 , 𝑦 ) 𝑑𝜇 ( 𝑦 ) | ≤ (2.3) , 𝐿 = 𝑖𝛿 ∑ 𝑗 : ( 𝑗𝛿 , ( 𝑗 + 1 ) 𝛿 ] ⊂ 𝐷 𝐿 𝑐 | 𝑞 ( 𝑧 , 𝑗𝛿 ) 𝜇 ( ( 𝑗𝛿 , ( 𝑗 + 1 ) 𝛿 ] ) | + { ∑ 𝑛 ≥ 1 2 𝑛𝛽 ∑ 1 ≤ 𝑘 ≤ 2 𝑛 | 𝑞 ( 𝑧 , 𝑑 𝑘𝑛 ( 𝑗 ) ) − 𝑞 ( 𝑧 , 𝑑 ( 𝑘 − 1 ) 𝑛 ( 𝑗 ) ) | 2 } 1 / 2 × { ∑ 𝑛 ≥ 1 2 − 𝑛𝛽 ∑ 1 ≤ 𝑘 ≤ 2 𝑛 | 𝜇 ( Δ 𝑘𝑛 ( 𝑗 ) ) | 2 } 1 / 2 ≤ (4.8) , (4.7) , 𝐿 = 𝑖𝛿 𝐶 𝐶 𝜆 𝑒 − 𝜆 ( 𝐿 − | 𝑥 | ) 2 𝑡 [ ∑ 𝑗 : ( 𝑗𝛿 , ( 𝑗 + 1 ) 𝛿 ] ⊂ 𝐷 𝐿 𝑐 | 𝜇 ( 𝑗𝛿 , ( 𝑗 + 1 ) 𝛿 ] ) | + ∑ 𝑗 : ( 𝑗𝛿 , ( 𝑗 + 1 ) 𝛿 ] ⊂ 𝐷 𝐿 𝑐 { ∑ 𝑛 ≥ 1 2 𝑛 ( 𝛽 + 1 − 2 𝛽 ( 𝜎 ) ) ∑ 1 ≤ 𝑘 ≤ 2 𝑛 | 𝜇 ( Δ 𝑘𝑛 ( 𝑗 ) ) | 2 } 1 / 2 ] ≤ 𝐶 𝐶 𝜆 𝑒 − 𝜆 ( 𝐿 − | 𝑥 | ) 2 𝑡 [ ( ∑ 𝑗 ∈ Z ( | 𝑗 | + 1 ) 2 𝜏 ( 𝜇 ( ( 𝑗𝛿 , ( 𝑗 + 1 ) 𝛿 ] ) ) 2 ) 1 / 2 ( ∑ 𝑗 ∈ Z ( | 𝑗 | + 1 ) − 2 𝜏 ) 1 / 2 + ( ∑ 𝑗 ∈ Z ( | 𝑗 | + 1 ) 2 𝜏 ∑ 𝑛 ≥ 1 2 𝑛 ( 𝛽 + 1 − 2 𝛽 ( 𝜎 ) ) ∑ 1 ≤ 𝑘 ≤ 2 𝑛 | 𝜇 ( Δ 𝑘𝑛 ( 𝑗 ) ) | 2 ) 1 / 2 ( ∑ 𝑗 ∈ Z ( | 𝑗 | + 1 ) − 2 𝜏 ) 1 / 2 ] .

As in (4.10), we get that (4.12) | 𝐼 2 | ≤ 𝐶 𝛿 ( 𝜔 ) 𝐶 𝜆 𝑒 − 𝜆 ( 𝐿 − | 𝑥 | ) 2 𝑡 , (4.10) and (4.12) imply that (4.1) holds. □

Remark 1.

Let SM 𝜇 be such that 𝜇 ( ( 𝑎 , 𝑏 ] ) ≤ 𝐶 ( 𝜔 ), 𝑎 , 𝑏 ∈ R. Then we can obtain (4.1) for arbitrary 𝐿 ∈ R + , and get | 𝜗 𝐿 ( 𝑡 , 𝑥 ) − 𝜗 ( 𝑡 , 𝑥 ) | ≤ 𝐶 ( 𝜔 ) 𝐶 𝜆 𝑒 − 𝜆 ( 𝐿 − | 𝑥 | ) 2 𝑡 , where 𝐶 ( 𝜔 ) is independent of 𝐿. To prove this, we can fix 𝛿 = 1, and note that in this case in (4.9) we get additional summands | 𝜇 ( ( 𝑗 0 , 𝐿 ] ∩ 𝐷 𝐿 ) | + | 𝜇 ( ( − 𝐿 , − 𝑗 0 ] ) | + ∑ 𝑛 ≥ 1 2 − 𝑛𝛽 | 𝜇 ( ( 𝑑 𝑘 𝑛 𝑛 ( 𝑗 0 ) , 𝐿 ] ) | + ∑ 𝑛 ≥ 1 2 − 𝑛𝛽 | 𝜇 ( ( − 𝐿 , 𝑑 𝑘 𝑛 𝑛 ( − 𝑗 0 ) ] ) | for some 𝑗 0 , 𝑘 𝑛 that depend on 𝐿 (and similar summands in (4.11)). If 𝜇 ( ( 𝑎 , 𝑏 ] ) ≤ 𝐶 ( 𝜔 ), then these sums are uniformly bounded.

Theorem 1.

Let Assumptions A1–A5 hold, 𝑢 𝐿 is the solution to (3.1), 𝑢 is the solution to (3.2). Then, for each 𝛿 > 0 there exist versions of 𝑢 𝐿 , 𝑢 such that for all 𝜔 ∈ Ω for each 𝑥 ∈ 𝐷 𝐿 , 𝐿 = 𝑖𝛿, 𝑖 ∈ Z + , 0 < 𝑡 ≤ 𝑇, for any 0 < 𝜆 < 1 4 holds (5.1) | 𝑢 𝐿 ( 𝑡 , 𝑥 ) − 𝑢 ( 𝑡 , 𝑥 ) | ≤ 𝐶 𝛿 ( 𝜔 ) 𝐶 𝜆 𝑒 − 𝜆 ( 𝐿 − | 𝑥 | ) 2 𝑡 .

Proof.

Consider 𝑢 𝐿 ( 𝑡 , 𝑥 ) − 𝑢 ( 𝑡 , 𝑥 ) = ( ∫ 𝐷 𝐿 𝐺 𝐿 ( 𝑡 , 𝑥 , 𝑦 ) 𝑢 0 ( 𝑦 ) 𝑑𝑦 − ∫ R 𝑝 ( 𝑡 , 𝑥 − 𝑦 ) 𝑢 0 ( 𝑦 ) 𝑑𝑦 ) + ( ∫ 0 𝑡 ∫ 𝐷 𝐿 𝐺 𝐿 ( 𝑡 − 𝑠 , 𝑥 , 𝑦 ) 𝑓 ( 𝑠 , 𝑦 , 𝑢 𝐿 ( 𝑠 , 𝑦 ) ) 𝑑𝑦 𝑑𝑠 − ∫ 0 𝑡 ∫ R 𝑝 ( 𝑡 − 𝑠 , 𝑥 − 𝑦 ) 𝑓 ( 𝑠 , 𝑦 , 𝑢 ( 𝑠 , 𝑦 ) ) 𝑑𝑦 𝑑𝑠 ) + ( ∫ 𝐷 𝐿 ∫ 0 𝑡 𝐺 𝐿 ( 𝑡 − 𝑠 , 𝑥 , 𝑦 ) 𝜎 ( 𝑠 , 𝑦 ) 𝑑𝑠 𝑑𝜇 ( 𝑦 ) − ∫ R ∫ 0 𝑡 𝑝 ( 𝑡 − 𝑠 , 𝑥 − 𝑦 ) 𝜎 ( 𝑠 , 𝑦 ) 𝑑𝑠 𝑑𝜇 ( 𝑦 ) ) : = 𝐽 1 + 𝐽 2 + 𝐽 3 , by Lemma 3, for some versions of the integrals, holds | 𝐽 3 | ≤ 𝐶 𝛿 ( 𝜔 ) 𝐶 𝜆 𝑒 − 𝜆 ( 𝐿 − | 𝑥 | ) 2 𝑡 .

For 𝐽 1 , we have | 𝐽 1 | ≤ ∫ 𝐷 𝐿 | 𝐺 𝐿 ( 𝑡 , 𝑥 , 𝑦 ) − 𝑝 ( 𝑡 , 𝑥 − 𝑦 ) | | 𝑢 0 ( 𝑦 ) | 𝑑𝑦 + ∫ 𝐷 𝐿 𝑐 𝑝 ( 𝑡 , 𝑥 − 𝑦 ) | 𝑢 0 ( 𝑦 ) | 𝑑𝑦 ≤ A1 𝐶 𝑢 0 ( 𝜔 ) ∫ 𝐷 𝐿 | 𝐺 𝐿 ( 𝑡 , 𝑥 , 𝑦 ) − 𝑝 ( 𝑡 , 𝑥 − 𝑦 ) | 𝑑𝑦 + 𝐶 𝑢 0 ( 𝜔 ) ∫ 𝐷 𝐿 𝑐 𝑝 ( 𝑡 , 𝑥 − 𝑦 ) 𝑑𝑦 : = 𝐽 11 + 𝐽 12 .