Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 202 / 6

Fix $t>0$ and define $w(s,y)=u(t-s,|y|)$ for $0\le s\le t$. The assumed smooth extension and the Neumann boundary condition at zero make $w$ a $C^{1,2}$ function. The heat equation givesThe Itô formula therefore makes $w(s,B_s)$ a local martingale. Stop first when $|B|$ leaves a large compact interval. The exponential growth bound and the finite exponential moments of the maximum of Brownian motion on $[0,t]$ give uniform integrability, so localization and the dominated convergence theorem yieldThis is the Feynman-Kac formula for the Neumann heat problem.

Let $p_t(z)=(2\pi t{)}^{-1/2}{e}^{-z^2/(2t)}$ be the heat kernel for $u_t=\frac{1}{2}{u}_{xx}$. The symmetry of the Gaussian distribution gives the method-of-images formulaThis is the Neumann heat kernel on a half-line. Differentiation under the integral sign shows that $v\in C^{1,2}$ for $t,x>0$ and that $v_t=\frac{1}{2}{v}_{xx}$. At $x=0$, the two differentiated kernel terms cancel, so $v_x(t,0)=0$. The Gaussian approximate identity gives $v(t,x)\to f(x)$ as $t\downarrow 0$, while the dominated convergence theorem gives continuity up to $x=0$. Finally $|v(t,x)|\le \Vert f{\Vert }_{\infty }$, which is stronger than the required exponential bound. Thus $v$ satisfies every condition in the displayed boundary-value problem.

Let $\rho =t\wedge {\tau }_0\wedge {\tau }_1$ and apply the Itô formula to $u(t-s,B_s)$ for $0\le s\le \rho$. The heat equation cancels the drift, so the stopped process is a bounded martingale. The optional sampling theorem for a supermartingale gives $u(t,x)={\mathbb{E}}_x[u(t-\rho ,B_{\rho })]$. On the three mutually exclusive terminal events, the initial and Dirichlet boundary conditions identify this value asThis is the probabilistic representation of the heat equation with time-dependent Dirichlet data.