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Randomly, I came across an interesting note on Stochastic Solutions of the Wave Equation. It states the following:

Let $D\subset\mathbb{R}^d$ be bounded and open, $f:\mathbb{R}\times \partial D\rightarrow \mathbb{R}$ be measurable and bounded, and $B_t$ be a $d$-dimensional Brownian motion with $B_0=x\in D$. Let $X\sim \text{Cauchy}(0,1)$ and $Z\sim \mathcal{N}(0,1)$, where $B_t$, $Z$, and $X$ are independent. Let $\tau = \text{argmin}_t \{ B_t\in\partial D \}$ and $t\in\mathbb{R}_+$. Then: $$ u(t,x) = \mathbb{E}_x\left[ f(tX + \sqrt{\tau}Z, B_\tau) \right] $$ satisfies $\partial^2_t u = \Delta u $.

My main motivation was being curious about stochastic solutions to the wave equation, since the heat/diffusion equation and Laplace equation have well-known probabilistic versions of their solutions.

For this particular formulation, I have two questions:

  1. What if $D$ has no boundary? Is there any way to modify this formulation to work?

  2. The note shows that this formulation cannot generate all solutions, including if one wishes to specify initial conditions for $u$ and $u_t$. But is there not a way to choose $f$ in such a way to satisfy some fixed $u(0,x)$ and $u_t(0,x)$? I'm not sure how to go about that; any simple demonstration of this would be appreciated.

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  • $\begingroup$ what is $\partial D$ ? $\endgroup$ – Arnaud Mégret Mar 14 '17 at 15:33
  • $\begingroup$ @ArnaudMégret Usually $\partial$ means boundary (if it doesn't mean partial derivative of course). $\endgroup$ – Ian Mar 14 '17 at 15:53
  • $\begingroup$ @ArnaudMégret As Ian said, it means the boundary of the set $D$ in this case. $\endgroup$ – user3658307 Mar 14 '17 at 19:16

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