Stochastic solutions for fractional wave equations

Abstract

A fractional wave equation replaces the second time derivative by a Caputo derivative of order between one and two. In this paper, we show that the fractional wave equation governs a stochastic model for wave propagation, with deterministic time replaced by the inverse of a stable subordinator whose index is one-half the order of the fractional time derivative.

Appendix

1.1 Reflection principle

The goal of this appendix is to establish the following variation of the D. André reflection principle for Brownian motion, which may also be useful in other contexts. Since this extension is not completely standard, we include its simple proof.

Theorem 6.1

(Reflection principle) Suppose that $Y_t$ is a Lévy process started at the origin, with no positive jumps, and let $S_t=\sup \{Y_u:0\le u\le t\}$. Assume that $\mathbb {P}(Y_t>0) = \mathbb {P}(Y_1 \ge 0)$, for all $t>0$. Then,

$$\begin{aligned} \mathbb {P}(S_t\ge x)&= \mathbb {P}(Y_t\ge x \,|\, Y_t\ge 0)\nonumber \\&= \frac{\mathbb {P}(Y_t\ge x)}{\mathbb {P}(Y_t\ge 0)} \end{aligned}$$

(6.1)

for all $t,x>0$.

Proof

Let $\tau _x := \inf \{u>0\,:\, Y_u > x\}$ denote the first-passage time process. Since $(Y_t)_{t\ge 0}$ has stationary independent increments, it follows that $(Y_{t+\tau _x} - Y_{\tau _x})_{t\ge 0}$ is a Lévy process, which is independent of the $\sigma$-algebra generated by $(Y_t)_{t\le \tau _x}$, and it has the same finite-dimensional distributions as $(Y_t)_{t\ge 0}$. Consequently,

$$\begin{aligned}&\mathbb {P}\left( \tau _x\le t, Y_t < Y_{\tau _x} \right) =\mathbb {P}\left( \tau _x\le t, Y_t - Y_{\tau _x} < 0\right) \nonumber \\&\quad =\mathbb {P}\left( \tau _x\le t\right) \mathbb {P}\left( Y_1<0\right) \nonumber \\&\quad =\frac{\mathbb {P}\left( Y_1<0\right) }{\mathbb {P}\left( Y_1\ge 0\right) } \mathbb {P}\left( \tau _x\le t\right) \mathbb {P}\left( Y_1\ge 0\right) \nonumber \\&\quad =\frac{\mathbb {P}\left( Y_1<0\right) }{\mathbb {P}\left( Y_1\ge 0\right) } \mathbb {P}\left( \tau _x\le t, Y_t \ge Y_{\tau _x} \right) . \end{aligned}$$

(6.2)

Observe that we have $\{\tau _x < t\} \subset \{S_t > x\} \subset \{\tau _x \le t\} \subset \{S_t\ge x\}$ for all $t$ and $x>0$. Therefore,

$$\begin{aligned}&\mathbb {P}(S_t>x) \le \mathbb {P}(\tau _x\le t) = \mathbb {P}(\tau _x\le t, Y_t\ge Y_{\tau _x}) \\&\qquad + \mathbb {P}(\tau _x\le t, Y_t < Y_{\tau _x})\\&\quad = \left( 1+ \frac{\mathbb {P}\left( Y_1<0\right) }{\mathbb {P}\left( Y_1\ge 0\right) }\right) \mathbb {P}(\tau _x\le t, Y_t\ge Y_{\tau _x})\\&\quad = \frac{1}{\mathbb {P}\left( Y_1\ge 0\right) }\,\mathbb {P}(\tau _x\le t, Y_t\ge Y_{\tau _x})\\&\quad = \frac{1}{\mathbb {P}\left( Y_t\ge 0\right) }\,\mathbb {P}(\tau _x\le t, Y_t\ge Y_{\tau _x},Y_t\ge 0)\\&\quad = \mathbb {P}(\tau _x\le t, Y_t\ge Y_{\tau _x}\,|\,Y_t\ge 0). \end{aligned}$$

On the other hand, and with similar arguments,

$$\begin{aligned}&\mathbb {P}(S_t>x) \ge \mathbb {P}(\tau _x< t) = \mathbb {P}(\tau _x < t, Y_t\ge Y_{\tau _x}) \\&\qquad + \mathbb {P}(\tau _x< t, Y_t < Y_{\tau _x})\\&\quad = \mathbb {P}(\tau _x< t, Y_t\ge Y_{\tau _x}\,|\,Y_t\ge 0)\\&\quad \ge \mathbb {P}(\tau _x< t, Y_t > Y_{\tau _x}\,|\,Y_t\ge 0). \end{aligned}$$

Therefore,

$$\begin{aligned}&\mathbb {P}(\tau _x< t, Y_t > Y_{\tau _x}\,|\,Y_t\ge 0) \le \mathbb {P}(S_t>x) \\&\quad \le \mathbb {P}(\tau _x\le t, Y_t\ge Y_{\tau _x}\,|\,Y_t\ge 0). \end{aligned}$$

Since $Y_t$ has no upward jumps, $Y_{\tau _x} = x$ and $\{\tau _x < t\}\cap \{Y_t > x\} = \{Y_t > x\}$. Therefore,

$$\begin{aligned}&\mathbb {P}(Y_t > x\,|\,Y_t\ge 0) \le \mathbb {P}(S_t>x) \\&\quad \le \mathbb {P}(Y_t\ge x\,|\,Y_t\ge 0). \end{aligned}$$

We can now use standard approximation techniques:

$$\begin{aligned} \{X \ge x\}&= \bigcap _n\big \{X> x-1/n\big \} \\&= \bigcap _n\big \{X\ge x-1/n\big \} \end{aligned}$$

and

$$\begin{aligned} \{X > x\} = \bigcup _n\{X > x+1/n\} \end{aligned}$$

to get

$$\begin{aligned} \mathbb {P}(Y_t\ge x \,|\,Y_t\ge 0)&= \lim _{n\rightarrow \infty } \mathbb {P}(Y_t > x - 1/n \,|\,Y_t\ge 0)\\&\le \lim _{n\rightarrow \infty } \mathbb {P}(S_t > x - 1/n)\\&= \mathbb {P}(S_t\ge x) \end{aligned}$$

and

$$\begin{aligned} \mathbb {P}(S_t\ge x)&= \lim _{n\rightarrow \infty }\mathbb {P}(S_t > x-1/n)\\&\le \lim _{n\rightarrow \infty }\mathbb {P}(Y_t \ge x-1/n\,|\, Y_t\ge 0)\\&= \mathbb {P}(Y_t \ge x \,|\, Y_t\ge 0) \end{aligned}$$

which proves $\mathbb {P}(S_t\ge x) = \mathbb {P}(Y_t\ge x\,|\, Y_t\ge 0)$.$\square$

Remark 6.2

If $(Y_t)_{t\ge 0}$ is a Brownian motion, then (6.1) becomes the classical reflection principle: $\mathbb {P}(Y_t\ge 0)=1/2$, so that (6.1) is equivalent to $\mathbb {P}(S_t\ge x) = 2\mathbb {P}(Y_t\ge ~x)$.

The proof of Theorem 6.1 relies essentially on local symmetry and the strong Markov property. Let $Y_t$ be a strong Markov process with càdlàg paths and transition function $p_t(z,\mathrm{d}y)=\mathbb {P}^z(Y_t\in \mathrm{d}y)$. Write $\tau _x^z = \inf \{u>0\,:\, Y_u-z > x\}$ for the first passage time above the level $x+z$ for the process $Y_t$ started at $z$; observe that, in general, $Y_{\tau _x^z} \ge x+z$. We can use the strong Markov property in (6.2) to get for any starting point $z$

$$\begin{aligned}&\mathbb {P}^z\left( \tau _x^z\le t, Y_t < Y_{\tau _x^z} \right) =\mathbb {P}^z\left( \tau _x^z\le t, Y_t - Y_{\tau _x^z} < 0\right) \\&\quad = \int \limits _{\{ \tau _x^z\le t\}} \mathbb {P}^{Y_{\tau _x^z}(\omega )}\big (Y_{t- \tau _x^z(\omega )} - Y_0 < 0\big )\, \mathbb {P}^z(\mathrm{d}\omega ). \end{aligned}$$

If we assume, in addition, some local “symmetry,” i.e., that for some constant $c\in (0,\infty )$ we have

$$\begin{aligned} \frac{\mathbb {P}^z(Y_t - z < 0)}{\mathbb {P}^z(Y_t - z \ge 0)} = c \quad \text {for all } t>0,\;z\in \mathbb {R}, \end{aligned}$$

(6.3)

then we get $\mathbb {P}^{Y_{\tau _x^z}(\omega )}\big (Y_{t-\tau _x^z(\omega )} - Y_0 < 0\big ) = c \,\mathbb {P}^{Y_{\tau _x^z}(\omega )}\big (Y_{t-\tau _x^z(\omega )} - Y_0 \ge 0\big )$ and, with a similar argument,

$$\begin{aligned} \mathbb {P}^z\left( \tau _x^z\le t, Y_t < Y_{\tau _x^z} \right) = c\,\mathbb {P}^z\left( \tau _x^z\le t, Y_t\ge Y_{\tau _x^z}\right) . \end{aligned}$$

This means that we can follow the lines of the proof of Theorem 6.1 to derive the following general result.

Theorem 6.3

(Markov reflection principle) Suppose $(Y_t,\mathbb {P}^z)$ is a strong Markov process satisfying the local symmetry condition (6.3). Set $S_t = \sup \{Y_u-Y_0 \,:\, 0\le u\le t\}$. Then, we have for all $t,x>0$ and $z\in \mathbb {R}$

$$\begin{aligned} \mathbb {P}^z(S_t>x)&\le \mathbb {P}^z (S_t\ge x, Y_t\ge Y_{ \tau _x^z}\,|\,Y_t\ge z)\\&\le \mathbb {P}^z (Y_t-z\ge x\,|\,Y_t\ge z). \end{aligned}$$

(6.4)

If $Y_t$ has only non-positive jumps, then $Y_{ \tau _x^z}=x+z$ a.s., and we get for all $t,x>0$ and $z\in \mathbb {R}$

$$\begin{aligned} \mathbb {P}^z(S_t\ge x) = \mathbb {P}^z(Y_t-z\ge x\,|\,Y_t\ge z). \end{aligned}$$

(6.5)

Remark 6.4

It is also possible to prove (6.1) using relation (3.6) in Alili and Chaumont [1].