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Linear SPDEs Driven by Stationary Random Distributions

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Abstract

Using tools from the theory of stationary random distributions developed in Itô (Mem. Coll. Sci., Univ. Kyoto, Ser. A: Math., 28:209–223, 1954) and Yaglom (Theory Probab. Appl., 2:273–320, 1957), we introduce a new class of processes which can be used as a model for the noise perturbing an SPDE. This type of noise is not necessarily Gaussian, but it includes the spatially homogeneous Gaussian noise introduced in Dalang (Electron. J. Probab. 4(6) 1999), and the fractional noise considered in Balan and Tudor (Stoch. Process. Appl., 120:2468–2494, 2010). We derive some general conditions for the existence of a random field solution of a linear SPDE with this type of noise, under some mild conditions imposed on the Green function of the differential operator which appears in this equation. This methodology is applied to the study of the heat and wave equations (possibly replacing the Laplacian by one of its fractional powers), extending in this manner the results of Balan and Tudor (Stoch. Process. Appl., 120:2468–2494, 2010) to the case H<1/2.

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Acknowledgements

The author would like to thank Davar Khoshnevisan for drawing her attention to the class of processes with stationary increments as a possible replacement for the fBm (which appears in the temporal component of the Gaussian noise of [1]), and also for pointing out a simplified proof of Lemma B.1(i).

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Correspondence to Raluca M. Balan.

Additional information

Communicated by Christian Houdré.

Research supported by a grant from the Natural Sciences and Engineering Research Council of Canada.

Appendices

Appendix A: The Stochastic Integral with Respect to \(\mathcal{M}\)

Let \(\mathcal{M}=\{\mathcal{M}(A);A \in\mathcal{R}_{d}\}\) be a complex random measure indexed by rectangles, with control measure μ, as specified by Definition 2.2.

Let \(\varphi\in L_{\mathbb{C}}^{2}(\mathbb{R}^{d},\mu)\) be a continuous function. We give below the construction of the stochastic integral of φ with respect to \(\mathcal{M}\).

Step 1. Let \(A \in\mathcal{R}_{d}\) be fixed. The stochastic integral of φ over A, with respect to \(\mathcal{M}\) is defined as the \(L_{\mathbb{C} }^{2}(\varOmega)\)-limit of \(X_{n}=\sum_{j=1}^{k_{n}}\varphi(\tau _{j,n})\mathcal{M} (A_{j,n})\), n≥1, where \(\Delta_{n}=(A_{j,n})_{1 \leq j \leq k_{n}}\) is a partition of A into sets from \(\mathcal{R}_{d}\) such that \(\| \Delta_{n}\| =\max_{j \leq k_{n}}|A_{j,n}| \to0\), and τ j,n A j,n is arbitrary. The limit exists, since the sequence (X n ) n≥1 is Cauchy: using the orthogonality and additivity of \(\mathcal{M}\), one can prove that E|X n X m |2=∫ A |φ n φ m |2 , where \(\varphi_{n}=\sum_{j=1}^{k_{n}}\varphi(\tau_{j,n})1_{A_{j,n}}\) and ∫ A |φ n |2 →∫ A |φ|2 , by the definition of the Stieltjes integral (see e.g. p. 228 of [5]). In this calculation, we used the fact that

$$E\bigl[\mathcal{M}(A)\overline{\mathcal{M}(B)}\bigr]=\mu(A \cap B), \quad\mbox {for all} \ A,B \in\mathcal{R}_d $$

which is again a consequence of the orthogonality and additivity of \(\mathcal{M}\). One can show that the limit of (X n ) n does not depend on the choice of (Δ n ) n and (τ j,n ) j,n . This limit is denoted by \(\mathcal{M}_{A}(\varphi)=\int_{A}\varphi(\tau)\mathcal{M}(d\tau)\).

This stochastic integral has the following properties:

  1. (a)

    \(\mathcal{M}_{A}(\varphi+\psi)=\mathcal{M}_{A}(\varphi )+\mathcal{M}_{A}(\psi)\) a.s.;

  2. (b)

    \(E[\mathcal{M}_{A}(\varphi)]=0\) and \(E[\mathcal {M}_{A}(\varphi)\overline {\mathcal{M}_{A}(\psi )}]=\int_{A}\varphi\overline{\psi} d\mu\).

If in addition \(\mathcal{M}\) is symmetric, then the following property also holds:

  1. (c)

    \(\mathcal{M}_{A}^{-}(\varphi)=\overline{\mathcal{M}}_{A}(\varphi )\), where \(\mathcal{M}_{A}^{-}(B)=\mathcal{M}_{A}(-B)\) and \(\overline{\mathcal{M}}_{A}(B)=\overline{\mathcal{M}_{A}(B)}\).

Step 2. The stochastic integral of φ with respect to \(\mathcal{M} \) is defined as the \(L_{\mathbb{C}}^{2}(\varOmega)\)-limit of \(Y_{n}=\mathcal{M} _{A_{n}}(\varphi )\), n≥1, where the sequence \((A_{n})_{n} \subset\mathcal{R}_{d}\) is chosen such that A n A n+1 for all n, and ⋃ n A n =ℝd. The sequence (Y n ) n is Cauchy, since \(E|Y_{n}-Y_{m}|^{2}=\int_{A_{m} \backslash A_{n}}|\varphi|^{2} d\mu\) for any m>n, and \(\int _{A_{n}}|\varphi|^{2} d\mu\to\int_{\mathbb{R}^{d}}|\varphi|^{2} d\mu\) by the monotone convergence theorem. Since the limit of (Y n ) n does not depend on the choice of (A n ) n , we denote it by \(\mathcal{M}(\varphi)=\int_{\mathbb {R}^{d}}\varphi(\tau )\mathcal{M} (d\tau)\). This stochastic integral enjoys properties similar to (a)–(c) above. In particular,

$$ E\bigl[\mathcal{M}(\varphi)\overline{\mathcal {M}(\psi)}\bigr ]=\int _{\mathbb{R}^{d}}\varphi\overline{\psi}d\mu. $$
(55)

Moreover, if \(\mathcal{M}\) is symmetric, then \(\mathcal{M}(\varphi)\) is a real-valued random variable for any function φ such that \(\overline {\varphi (\tau)}=\varphi(-\tau)\) for all τ∈ℝd. This follows since

$$ \overline{\mathcal{M}(\varphi)}=\int_{\mathbb{R}^d} \overline{\varphi(\tau)} \overline{\mathcal{M}}(d\tau)=\int _{\mathbb{R}^d} \varphi(-\tau)\mathcal{M}^{-}(d\tau)=\mathcal{M}(\varphi), $$
(56)

where we used from property (c) for the second equality above.

Appendix B: Some Elementary Estimates

Lemma B.1

Let γ∈(−1,1) be arbitrary. (i) For any a>0, we have:

$$\int_{\mathbb{R}}\frac{1}{\tau^2+a^2}|\tau|^{-\gamma}d \tau=C_{\gamma} a^{-\gamma-1}, $$

where C γ =∫(s 2+1)−1 s γ ds. (ii) For any a>1, we have:

$$C_{\gamma}^{(1)} a^{-\gamma-1} \leq\int_{\mathbb{R}} \frac{1}{\tau^2+a^2}\bigl(\tau^2+1\bigr)^{-\gamma/2}d\tau\leq C_{\gamma}^{(2)} a^{-\gamma-1}, $$

where \(C_{\gamma}^{(1)}\) and \(C_{\gamma}^{(2)}\) are some positive constants depending on γ.

Proof

(i) The conclusion follows by the change of variable τ/a=τ′.

(ii) We denote by I 1 and I 2 the integrals over the regions |τ|≤a, respectively |τ|≥a. We use the notation f(τ)∼g(τ) if c 1 g(τ)≤f(τ)≤c 2 g(τ) for any τ∈ℝ, for some constants c 1,c 2>0. When |τ|≤a,

$$\bigl(\tau^2+a^2\bigr)^{-1} \bigl( \tau^2+1\bigr)^{-\gamma/2} \sim a^{-2} \bigl( \tau^2+1\bigr)^{-\gamma /2} \sim a^{-2} ( \tau+1)^{-\gamma}, $$

and hence I 1 is bounded below and above by some constants multiplied by:

$$a^{-2}\int_{0}^{a}( \tau+1)^{-\gamma}d\tau=a^{-2} \frac{1}{1-\gamma } \bigl[(a+1)^{1-\gamma}-1\bigr] \sim a^{-2}(a+1)^{1-\gamma} \sim a^{-\gamma-1}. $$

When |τ|≥a,

$$\bigl(\tau^2+a^2\bigr)^{-1} \bigl( \tau^2+1\bigr)^{-\gamma/2} \sim\tau^{-2} \bigl( \tau^2+1\bigr)^{-\gamma/2} \sim\tau^{-2} ( \tau+1)^{-\gamma} \sim\tau^{-2-\gamma}, $$

and hence I 2 is bounded below and above by some constants multiplied by:

$$\int_{a}^{\infty}\tau^{-2-\gamma}d\tau= \frac{1}{\gamma+1}a^{-\gamma-1}. $$

 □

Lemma B.2

Suppose that η satisfies either (C1) or (C2). Then for any K>0 there exists a constant M K >0 such that

$$\frac{1}{a} \int_{0}^{\infty} \frac{1}{\tau^2+a^2}\eta(\tau)d\tau\leq M_K \frac{\eta(a)}{a^2}, \quad\mbox{\textit{for any}} \ a \geq K. $$

Proof

If (C1) holds, then

If (C2) holds, then

 □

Appendix C: Verification of Conditions (Hyperbolic Case)

Lemma C.1

(i) Let η(τ)=|τ|γ with γ∈(−1,1). Then η satisfies (C). If γ∈(0,1), then η satisfies (C1). If γ∈(−1,0), then η satisfies (C2).

(ii) Let η(τ)=(1+τ 2)γ/2 with γ>−1. Then η satisfies (C). If γ∈(0,1), then η satisfies (C1). If γ∈(−1,0), then η satisfies (C2).

Proof

(i)(C) is clearly satisfied since η(λτ)/η(τ)=λ γ=:C λ for any τ>0.

If γ∈(0,1), (C1) holds since for any a>0,

$$\int_0^a \eta(\tau)d\tau=\int _0^a \tau^{-\gamma}d\tau= \frac{1}{1-\gamma }a^{1-\gamma}=\frac{1}{1-\gamma}a \eta(a). $$

If γ∈(−1,0), (C2) holds since for any a>0,

$$\int_{a}^{\infty} \tau^{-2} \eta(\tau)d \tau=\int_{a}^{\infty}\tau^{-2-\gamma}d\tau= \frac{1}{\gamma +1}a^{-1-\gamma}= \frac{1}{\gamma+1} a^{-1}\eta(a). $$

(ii) We first check (C). If γ>0, then the inequality

$$ \frac{\eta(\lambda \tau)}{\eta(\tau)}= \biggl(\frac{1+\tau^2}{1+\lambda^2 \tau^2} \biggr)^{\gamma/2} \leq C_{\lambda} $$
(57)

is equivalent to \(1+\lambda^{2} \tau^{2} \geq C_{\lambda}^{-2/\gamma}(1+\tau^{2})\). We choose C λ ={min(1,λ 2)}γ/2.

If γ<0, then (57) is equivalent to \(1+\lambda^{2} \tau^{2} \leq C_{\lambda}^{-2/\gamma}(1+\tau^{2})\). We choose C λ ={max(1,λ 2)}γ/2. If γ∈(0,1) then (C1) holds since for any aK,

$$\int_0^a \bigl(1+\tau^2 \bigr)^{-\gamma/2}d\tau\leq\int_0^a \tau^{-\gamma}d\tau=\frac{a}{1-\gamma} \biggl(\frac{1}{a^2} \biggr)^{\gamma/2} \leq\frac {a}{1-\gamma} \biggl(\frac{C_K}{1+a^2} \biggr)^{\gamma/2}, $$

where C K =1+K −2. If γ∈(−1,0) then (C2) holds since for any aK,

 □

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Balan, R.M. Linear SPDEs Driven by Stationary Random Distributions. J Fourier Anal Appl 18, 1113–1145 (2012). https://doi.org/10.1007/s00041-012-9240-7

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