Stratonovich Solution for the Wave Equation

Abstract

In this article, we construct a Stratonovich solution for the stochastic wave equation in spatial dimension $d \le 2$, with time-independent noise and linear term $\sigma (u)=u$ multiplying the noise. The noise is spatially homogeneous and its spectral measure satisfies an integrability condition which is stronger than Dalang’s condition. We give a probabilistic representation for this solution, similar to the Feynman–Kac-type formula given in Dalang et al. (Trans Am Math Soc 360:4681–4703, 2008) for the solution of the stochastic wave equation with spatially homogeneous Gaussian noise, that is white in time. We also give the chaos expansion of the Stratonovich solution and we compare it with the chaos expansion of the Skorohod solution from Balan et al. (Exact asymptotics of the stochastic wave equation with time independent noise, 2020. arXiv:2007.10203).

Appendices

A Parseval-type identities

In this section, we give two Parseval-type identities which are used in this paper.

Lemma A.1

Let $d\ge 1$ be arbitrary and $\gamma :{\mathbb {R}}^d \rightarrow [0,\infty ]$ be nonnegative-definite function satisfying Assumption A. Let $\mu$ be the tempered measure on ${\mathbb {R}}^d$ such that $\gamma ={\mathcal {F}}\mu$. If $\varphi$ is a nonnegative integrable function on ${\mathbb {R}}^d$ such that $\int _{{\mathbb {R}}^{d}}|{\mathcal {F}}\varphi (\xi )|^2 \mu (\mathrm{d}\xi )<\infty$, then

$$\begin{aligned} \int _{{\mathbb {R}}^d} \varphi (x)\gamma (x)\mathrm{d}x=\int _{{\mathbb {R}}^d}{\mathcal {F}}\varphi (\xi )\mu (\mathrm{d}\xi ). \end{aligned}$$

Proof

By relation (5.37) of [11],

$$\begin{aligned} \int _{{\mathbb {R}}^d} \int _{{\mathbb {R}}^d}\gamma (x-y)\nu _1(\mathrm{d}x)\nu _2(\mathrm{d}y) =\int _{{\mathbb {R}}^d}{\mathcal {F}}\nu _1(\xi )\overline{{\mathcal {F}}\nu _2(\xi )}\mu (\mathrm{d}\xi ), \end{aligned}$$

for any finite measures $\nu _1,\nu _2$ on ${\mathbb {R}}^d$ such that $\int _{{\mathbb {R}}^d}|{\mathcal {F}}\nu _i(\xi )|^2 \mu (\mathrm{d}\xi )<\infty$ for $i=1,2$. The conclusion follows taking $\nu _1(\mathrm{d}x)=\varphi (x)\mathrm{d}x$ and $\nu _2=\delta _0$. $\square$

Lemma A.2

Let $\gamma$ and $\mu$ be as in Lemma A.1. If $\varphi$ is a nonnegative integrable function on $({\mathbb {R}}^d)^{2k}$ and $\{I_1,\ldots ,I_k\}$ is a partition of $\{1,\ldots ,2k\}$ with $I_i=\{\ell _i,m_i\}$ for $i=1,\ldots ,k$, then

$$\begin{aligned}&\int _{({\mathbb {R}}^d)^{2k}} \varphi (x_1,\ldots ,x_{2k}) \prod _{i=1}^{k}\gamma (x_{\ell _i}-x_{m_i})\mathrm{d}x_1 \ldots \mathrm{d}x_{2k}\\&\quad =\int _{({\mathbb {R}}^d)^k} {\mathcal {F}}\varphi (\xi _1,\ldots ,\xi _{2k}) \prod _{i=1}^{k}1_{\{\xi _{\ell _i}=-\xi _{m_i}=\eta _i\}}\mu (\mathrm{d}\eta _1) \ldots \mu (\mathrm{d}\eta _k), \end{aligned}$$

provided that the integral on the right-hand side is finite.

Proof

Without loss of generality, we will assume that $I_i=\{2i-1,2i\}$ for all $i=1,\ldots ,k$. (If not, we use the change of variables $y_{1}=x_{\ell _1},y_{2}=x_{m_1},\ldots ,y_{2k-1}=x_{\ell _k},y_{2k}=x_{m_k}$ and we let $\rho$ be the permutation of $\{1,\ldots ,2k\}$ given by $\ell _1=\rho (1),m_1=\rho (2),\ldots ,\ell _k=\rho (2k-1),m_k=\rho (2k)$. Then $y_i=x_{\rho (i)}$ for $i=1,\ldots ,2k$, and

$$\begin{aligned}&\int _{({\mathbb {R}}^d)^{2k}}\varphi (x_1,\ldots ,x_{2k}) \prod _{i=1}^{k}\gamma (x_{\ell _i}-x_{m_i}) \mathrm{d}{} \mathbf{x}\\&\quad =\int _{({\mathbb {R}}^d)^{2k}} \varphi (y_{\rho ^{-1}(1)}, \ldots y_{\rho ^{-1}(2k)}) \prod _{i=1}^{k} \gamma (y_{2i-1}-y_{2i})\mathrm{d}{} \mathbf{y}, \end{aligned}$$

where $\mathbf{x}=(x_1,\ldots ,x_{2k})$ and $\mathbf{y} =(y_1,\ldots ,y_{2k})$.) We have to prove that

$$\begin{aligned}&\int _{({\mathbb {R}}^d)^{2k}} \varphi (x_1,\ldots ,x_{2k}) \prod _{i=1}^{k}\gamma (x_{2i-1}-x_{2i})\mathrm{d}{} \mathbf{x}\\&\quad =\int _{({\mathbb {R}}^d)^k} {\mathcal {F}}\varphi (\xi _1,\ldots ,\xi _{2k}) \prod _{i=1}^{k}1_{\{\xi _{2i-1}=-\xi _{2i}=\eta _i\}} \prod _{i=1}^{k}\mu (\mathrm{d}\eta _i), \end{aligned}$$

which we will rewrite as:

$$\begin{aligned}&\int _{({\mathbb {R}}^d)^k}\varphi (x_1,x_1',\ldots ,x_k,x_k') \prod _{i=1}^{k}\gamma (x_i-x_i')\mathrm{d}{} \mathbf{x} \mathrm{d}{} \mathbf{x'}\nonumber \\&\quad =\int _{({\mathbb {R}}^d)^k} {\mathcal {F}}\varphi (\xi _1,-\xi _1,\ldots ,\xi _k,-\xi _k) \prod _{i=1}^{k}\mu (\mathrm{d}\xi _i), \end{aligned}$$

(59)

with $\mathbf{x}=(x_1,\ldots ,x_{k})$ and $\mathbf{x}' =(x_1',\ldots ,x_{k}')$.

We prove (59) by induction on $k\ge 1$. If $k=1$, then by Lemma A.1, we have:

$$\begin{aligned}&\int _{({\mathbb {R}}^d)^2}\varphi (x_1,x_1')\gamma (x_1-x_1')\mathrm{d}x_1 \mathrm{d}x_1' =\int _{{\mathbb {R}}^d} \left( \int _{{\mathbb {R}}^d} \varphi (x_1,x_1+y_1) \gamma (y_1)\mathrm{d}y_1\right) \mathrm{d}x_1\\&\quad =\int _{{\mathbb {R}}^d} \left( \int _{{\mathbb {R}}^d} {\mathcal {F}}\varphi (x_1,x_1+\cdot ) (\xi _1)\mu (\mathrm{d}\xi _1)\right) \mathrm{d}x_1\\&\quad =\int _{{\mathbb {R}}^d}\int _{{\mathbb {R}}^d}\left( \int _{{\mathbb {R}}^d} e^{-i \xi _1 \cdot (x_1'-x_1)} \varphi (x_1,x_1') \mathrm{d}x_1'\right) \mu (\mathrm{d}\xi _1)\mathrm{d}x_1\\&\quad =\int _{{\mathbb {R}}^d}\left( \int _{{\mathbb {R}}^d}\int _{{\mathbb {R}}^d} e^{-i \xi _1 \cdot (x_1-x_1')} \varphi (x_1,x_1') \mathrm{d}x_1\mathrm{d}x_1' \right) \mu (\mathrm{d}\xi _1)\\&\quad =\int _{{\mathbb {R}}^d} {\mathcal {F}}\varphi (\xi _1,-\xi _1) \mu (\mathrm{d}\xi _1), \end{aligned}$$

where for the second last line we used the fact that $\mu$ is symmetric. Suppose now that the statement holds for k. By applying the result for $k=1$, we obtain:

$$\begin{aligned} I&:=\int _{({\mathbb {R}}^d)^{k+1}}\varphi (x_1,x_1',\ldots ,x_{k+1},x_{k+1}') \prod _{i=1}^{k+1} \gamma (x_i-x_i')\mathrm{d}x_1 \ldots \mathrm{d}x_{k+1}\mathrm{d}x_1' \ldots \mathrm{d}x_{k+1}'\\&=\int _{({\mathbb {R}}^d)^{2k}} \left( \int _{{\mathbb {R}}^d} \psi _{\xi _{k+1}}(x_1,x_1',\ldots ,x_k,x_k') \mu (\mathrm{d}\xi _{k+1})\right) \prod _{i=1}^{k}\gamma (x_i-x_i') \mathrm{d}x_1\ldots \\&\quad \times \mathrm{d}x_{k}\mathrm{d}x_1' \ldots \mathrm{d}x_{k}', \end{aligned}$$

where $\psi _{\xi _{k+1}}(x_1,x_1',\ldots ,x_k,x_k'):={\mathcal {F}}\varphi (x_1,x_1',\ldots ,x_k,x_k',\cdot ,\cdot )(\xi _{k+1},-\xi _{k+1})$. We apply Fubini’s theorem and the induction hypothesis for the inner integral on $({\mathbb {R}}^d)^{2k}$. We obtain:

$$\begin{aligned} I=\int _{{\mathbb {R}}^d} \left( \int _{({\mathbb {R}}^d)^{k}} {\mathcal {F}}\psi _{\xi _{k+1}} (\xi _1,-\xi _1,\ldots ,\xi _k,-\xi _k) \mu (\mathrm{d}\xi _1)\ldots \mu (\mathrm{d}\xi _k)\right) \mu (\mathrm{d}\xi _{k+1}). \end{aligned}$$

The conclusion follows since ${\mathcal {F}}\psi _{\xi _{k+1}} (\xi _1,-\xi _1,\ldots ,\xi _k,-\xi _k)={\mathcal {F}}\varphi (\xi _1,-\xi _1,\ldots ,\xi _{k+1},-\xi _{k+1})$. $\square$

B Products of Wiener integrals

In this section, we develop a formula for the product of n Wiener integrals with respect to the noise W, which plays a crucial role in the present paper. This formula is a consequence of Corollary 5.1 (page 123) of [6]. We include its proof for the sake of completeness.

We recall the product formula from Malliavin calculus: for any integer $p\ge 1$, for any symmetric function $f \in {\mathcal {H}}^{\otimes p}$ and for any $g \in {\mathcal {H}}$,

$$\begin{aligned} I_p(f)I_1(g)=I_{p+1}(f \otimes g)+pI_{p-1}(f \otimes _1 g), \end{aligned}$$

(60)

where $f \otimes _{1} g$ is the first contraction of f and g, defined by:

$$\begin{aligned} (f \otimes _{1} g)(y_1,\ldots ,y_{p-1})=\langle f(\cdot ,y_1, \ldots ,y_{p-1}),g \rangle _{{\mathcal {H}}} \end{aligned}$$

(see, e.g., Theorem 2.7.10 of [12], or Proposition 1.1.3 of [13]).

We will use the fact that $I_p(f)=I_p(\widetilde{f})$ for any $f \in {\mathcal {H}}^{\otimes p}$, where $\widetilde{f}$ is the symmetrization of f. We let $f \otimes _0 g=f \otimes g$ and we denote by $f \widetilde{\otimes }_{r} g$ the symmetrization of $f \otimes _{r} g$.

Theorem B.1

For any $n\ge 2$ and for any functions $f_1,\ldots ,f_n\in {\mathcal {H}}$,

$$\begin{aligned} \prod _{j=1}^{n}I_1(f_j)=\sum _{k=0}^{\lfloor n/2 \rfloor } \sum _{\begin{array}{c} J \subset [n]\\ |J|=n-2k \end{array}} \sum _{\begin{array}{c} \{I_1,\ldots ,I_k\} \ \mathrm{partition} \ \mathrm{of} \ J^c \\ I_i=\{\ell _i,m_i\} \forall i=1,\ldots ,k \end{array}}I_{n-2k}\left( \bigotimes _{j\in J}f_j\right) \prod _{i=1}^{k}\langle f_{\ell _i},f_{m_i}\rangle _{{\mathcal {H}}}, \end{aligned}$$

(61)

with the convention that $\otimes _{j \in \emptyset } f_j=1$ and $I_0(1)=1$.

Proof

We use induction on n. The case $n=2$ is clear.

Suppose that the result is true for n. Assume that n is even, $n=2K$ for $K\in {\mathbb {Z}}_{+}$. The case when n is odd is similar and will be omitted.

We multiply relation (61) by $I_{n+1}(f_{n+1})$. To evaluate the product $I_{n-2k}(\bigotimes _{j\in J}f_j)I_1(f_{n+1})$, we distinguish 2 cases:

• (a) if $k=K$, then $J=\emptyset$ and $I_{n-2k}(\bigotimes _{j\in J}f_j)I_1(f_{n+1})=I_1(f_{n+1})$;

• (b) if $k \le K-1$, then by (60),

  $$\begin{aligned} I_{n-2k}\left( \bigotimes _{j\in J}f_j\right) I_{1}(f_{n+1})= & {} I_{n+1-2k} \Big (\big (\widetilde{\bigotimes }_{j\in J} f_j \big ) \otimes f_{n+1} \Big )\\&+(n-2k)I_{n-2k-1} \Big (\big (\widetilde{\bigotimes }_{j\in J} f_j \big ) \otimes _1 f_{n+1} \Big ). \end{aligned}$$

  Note that $I_{n+1-2k}\Big (\big (\widetilde{\bigotimes }_{j\in J} f_j \big )\otimes f_{n+1} \Big )=I_{n+1-2k}\Big (\bigotimes _{j\in J} f_j \otimes f_{n+1} \Big )$ since $\big (\widetilde{\bigotimes }_{j\in J} f_j \big )\otimes f_{n+1}$ and $\bigotimes _{j\in J} f_j \otimes f_{n+1}$ have the same symmetrizations. Moreover, $\big (\widetilde{\bigotimes }_{j \in J} f_j \big ) \otimes _1 f_{n+1}=\frac{1}{n-2k} \sum _{j\in J} \langle f_{j},f_{n+1} \rangle _{{\mathcal {H}}} \widetilde{\bigotimes }_{\ell \in J \backslash \{j\}} f_{\ell }$ and hence,

  $$\begin{aligned} I_{n-2k-1}\Big ( \big (\widetilde{\bigotimes }_{j\in J}f_j \big ) \otimes _1 f_{n+1}\Big )=\frac{1}{n-2k}\sum _{j \in J} \langle f_{j},f_{n+1} \rangle _{{\mathcal {H}}} I_{n-2k-1} \Big ( \bigotimes _{\ell \in J - \{j\}} f_{\ell } \Big ). \end{aligned}$$

  It follows that

  $$\begin{aligned} A&:=\prod _{i=1}^{n+1}I_1(f_i) =\sum _{k=0}^{K-1} \sum _{\begin{array}{c} J \subset [n]\\ |J|=n-2k \end{array}} \sum _{\begin{array}{c} \{I_1,\ldots ,I_k\} \ \mathrm{partition} \ \mathrm{of} \ [n]-J\\ I_i=\{\ell _i,m_i\} \forall i=1,\ldots ,k \end{array}}I_{n+1-2k}\Big (\bigotimes _{j\in J} f_j \otimes f_{n+1} \Big ) \\&\quad \prod _{i=1}^{k} \langle f_{\ell _i},f_{m_i}\rangle _{{\mathcal {H}}}\\&\quad +\sum _{k=0}^{K-1} \sum _{\begin{array}{c} J \subset [n]\\ |J|=n-2k \end{array}} \sum _{j \in J} \sum _{\begin{array}{c} \{I_1,\ldots ,I_k\} \ \mathrm{partition} \ \mathrm{of} \ [n]-J\\ I_i=\{\ell _i,m_i\} \forall i=1,\ldots ,k \end{array}} I_{n-2k-1}\Big ( \bigotimes _{\ell \in J - \{j\}} f_{j} \Big )\\&\quad \prod _{i=1}^{k}\langle f_{\ell _i},f_{m_i}\rangle _{{\mathcal {H}}} \langle f_{j},f_{n+1} \rangle _{{\mathcal {H}}} \\&\quad +\sum _{\begin{array}{c} (\{I_1,\ldots ,I_k\} \ \mathrm{partition} \ \mathrm{of} \ [n]-J\\ I_i=\{\ell _i,m_i\} \forall i=1,\ldots ,k \end{array}} I_1(f_{n+1}) \prod _{i=1}^{K}\langle f_{\ell _i},f_{m_i}\rangle _{{\mathcal {H}}} =:A_1+A_2+A_3. \end{aligned}$$

We have to prove that $A=B$, where

$$\begin{aligned} B=\sum _{k=0}^{K} \sum _{\begin{array}{c} J' \subset [n+1]\\ |J'|=n+1-2k \end{array}} \sum _{\begin{array}{c} \{I_1,\ldots ,I_k\} \ \mathrm{partition} \ \mathrm{of} \ [n+1]-J' \\ I_i=\{\ell _i,m_i\} \forall i=1,\ldots ,k \end{array}}I_{n+1-2k} \left( \bigotimes _{j\in J'}f_j\right) \prod _{i=1}^{k} \langle f_{\ell _i},f_{m_i}\rangle _{{\mathcal {H}}}. \end{aligned}$$

We split B using $k \le K-1$ and $k=K$: (if $k=K$ then $|J'|=n+1-2K=1$)

$$\begin{aligned} B&=\sum _{k=0}^{K-1} \sum _{\begin{array}{c} J' \subset [n+1]\\ |J'|=n+1-2k \end{array}} \sum _{\begin{array}{c} \{I_1,\ldots ,I_k\} \ \mathrm{partition} \ \mathrm{of} \ [n+1]-J' \\ I_i=\{\ell _i,m_i\} \forall i=1,\ldots ,k \end{array}}I_{n+1-2k} \left( \bigotimes _{j\in J'}f_j\right) \prod _{i=1}^{k} \langle f_{\ell _i},f_{m_i}\rangle _{{\mathcal {H}}}\\&\quad +\sum _{j=1}^{n} \sum _{\begin{array}{c} \{I_1,\ldots ,I_K\} \ \mathrm{partition} \ \mathrm{of} \ [n+1]-J' \\ I_i=\{\ell _i,m_i\} \forall i=1,\ldots ,K \end{array}}I_1(f_j) \prod _{i=1}^{K}\langle f_{\ell _i}, f_{m_i}\rangle _{{\mathcal {H}}}=:B_1+B_2. \end{aligned}$$

We split $A_2$ using $k\le K-2$ and $k=K-1$: (if $k=K-1$ then $n-2k-1=1$)

$$\begin{aligned} A_2&=\sum _{k=0}^{K-2} \sum _{\begin{array}{c} J \subset [n]\\ |J|=n-2k \end{array}} \sum _{j \in J} \sum _{\begin{array}{c} \{I_1,\ldots ,I_k\} \ \mathrm{partition} \ \mathrm{of} \ [n]-J\\ I_i=\{\ell _i,m_i\} \forall i=1,\ldots ,k \end{array}} I_{n-2k-1}\Big ( \bigotimes _{\ell \in J - \{j\}} f_{j} \Big )\\&\quad \prod _{i=1}^{k}\langle f_{\ell _i},f_{m_i}\rangle _{{\mathcal {H}}} \langle f_{j},f_{n+1} \rangle _{{\mathcal {H}}}\\&\quad +\sum _{\begin{array}{c} J \subset [n] \\ |J|=2 \end{array}} \sum _{\begin{array}{c} \{I_1,\ldots ,I_{K-1}\} \ \mathrm{partition} \ \mathrm{of} \ [n]-J\\ I_i=\{\ell _i,m_i\} \forall i=1,\ldots ,K-1 \end{array}} \sum _{j\in J} I_1\big (\bigotimes _{\ell \in J-\{j\}}f_{\ell }\big ) \langle f_j,f_{n+1} \rangle _{{\mathcal {H}}} \\&\quad \prod _{i=1}^{K-1} \langle f_{\ell _i}, f_{m_1}\rangle _{{\mathcal {H}}} :=A_2'+A_2''. \end{aligned}$$

We notice that:

$$\begin{aligned} (a) \ A_1+A_2'=B_1 \quad \text{ and } \quad (b) \ A_2''+A_3=B_2. \end{aligned}$$

• (a) We use the decomposition $B_1=B_1'+B_1''$, where $B_1'$ and $B_1''$ correspond to the cases $n+1 \in J'$, respectively $n+1 \not \in J'$. We observe that $A_1=B_1''$ and $A_2'=B_1'$.

• (b) We use the decomposition $B_2=B_2'+B_2''$, where $B_2'$ and $B_2''$ correspond to the cases $j=n+1$, respectively $j\le n$. We observe that $A_3=B_2'$ and $A_2''=B_2''$.

$\square$

C Semigroup-type property of G

In this section, we prove a property of G which was used in Step 3.(a) of the proof of Theorem 3.2. We denote $G_t(x)=G(t,x)$.

Lemma C.1

(Lemma 4.3 of [3]) Suppose that $d=2$. Let $q \in (\frac{1}{2},1)$ be arbitrary. For any $0<r<t$ and $x,z \in {\mathbb {R}}^d$

$$\begin{aligned} \int _r^t \big (G_{t-s}^{2q}*G_{s-r}^{2q}\big )^{1/q}(x-z) \mathrm{d}s \le A_q (t-r)^{\frac{1}{q}-1}G_{t-r}^{2-\frac{1}{q}}(x-z), \end{aligned}$$

where $A_q>0$ is a constant depending on q.

As an immediate consequence of the previous lemma, we obtain the following semigroup-type property of G.

Lemma C.2

For any $0<r<t$ and $x,z\in {\mathbb {R}}^d$,

$$\begin{aligned} \int _r^t \int _{{\mathbb {R}}^d} G_{t-s}(x-y)G_{s-r}(y-z)\mathrm{d}y\mathrm{d}s \le C_t^{(1)} G_{t-r}(x-z), \end{aligned}$$

where $C_t^{(1)} =Ct^2$ and C is a positive constant.

Proof

If $d=1$, this follows by relation (2.6) of [14].

Suppose that $d=2$. Let $q \in (\frac{1}{2},1)$ be arbitrary. Note that $G_t(x) \le C_{t,q} G_t^{2q}(x)$, where $C_{t,q}=(2\pi t)^{2q-1}$. By Hölder’s inequality and Lemma C.1,

$$\begin{aligned}&\int _r^t\int _{{\mathbb {R}}^d}G_{t-s}(x-y)G_{s-r}(y-z)\mathrm{d}y\mathrm{d}r \le C_{t,q}^2 \int _r^t \big ( G_t^{2q}*G_{s}^{2q}\big ) (x-z)\mathrm{d}r \\&\quad \le C_{t,q}^2 (t-r)^{1-q} \left[ \int _r^t \big ( G_t^{2q}*G_{s}^{2q}\big )^{1/q}(x-z) \mathrm{d}s\right] ^{q} \le C_{t,q}^2 (t-r)^{2(1-q)} \\&\quad \quad \times A_q^q G_{t-r}^{2q-1}(x-z). \end{aligned}$$

The conclusion follows using the fact that $G_{t-r}^{2q-1}(x-z) \le (2\pi t)^{2-2q}G_{t-r}(x-z)$. $\square$