Suppose $u$ satisfies the problem: $$ du= \nabla u \circ \,dW(t) +|\nabla u|\,dt$$ I want to prove that $ v(x,t):= u(x+W(t),t)$ satisfies the PDE $\partial_tv=|\nabla v|$. Here '$\circ$' denotes the Stratonovich differential and $(x,t) \in \mathbb{R}^n \times (0,T)$.
Applying the chain rule in Stratonovich sense we get, $$d(u(x+W(t),t))= \nabla u (x+W(t),t)\circ dW(t) + \partial_t u(x+W(t),t)\,dt \implies d(u(x+W(t),t))-\nabla u (x-W(t),t)\circ dW(t)= \partial_t u(x+W(t),t)\,dt.$$ This gives $|\nabla v| = \partial_t v$. Are these computations correct? If we specify the initial data $u(t,0)=u_0$, can it be said that $u_0(x+W(t))$ solves $|\nabla v| = \partial_t v$?