“…We may apply the strong Markov property of u, with respect to F , at the stopping time T n in order to see that for all integers n 0, 1] solves (1.8) starting from the random initial profile v (n) 0 (x) := v (n) (0 , x) = u(T n , x ; λ). By the very definition of the stopping time T n , and since T n is finite a.s., v (n) 0 (x) = u(T n , x ; λ) e −1 inf y∈T u(T n−1 , y ; λ) · · · e −n inf y∈T u 0 (y) =: e −n u 0 , a.s. for every x ∈ T, and with identity for some x ∈ T a.s. Because u 0 > 0 [see (1.2)], it follows from a comparison theorem [3,21,25] that v (n) (t , x) w (n) (t , x) for all t 0 and x ∈ T a.s., where w (n) solves (1.8) [for a different Brownian sheet] starting from w (n) 0 (x) := w (n) (0 , x) = e −n u 0 . In particular, for all integers n 0 and reals τ ∈ (0 , 1),…”