Past exam of the differential equations and applications course of the University of Oxford / 2011 / B / 5 / a / Solution

Assume a classical solution continuous up to the relevant boundary. At an interior positive maximum, $u_t=0$ and $u_{xx}\le 0$, hence $Lu=u_t-u_{xx}+u>0$, a contradiction. At a positive maximum on the top edge $t=\tau$, away from the spatial endpoints, the left time derivative satisfies $u_t\ge 0$, while again $u_{xx}\le 0$; the same contradiction applies. Thus a positive maximum can occur only on the initial or lateral parabolic boundary.

For two solutions with the same data, their difference $w$ satisfies $Lw=0$ and vanishes on that parabolic boundary. The weak parabolic maximum principle prohibits a positive maximum of $w$ and also of $-w$. Therefore $w=0$, proving at most one solution. The positive reaction term is what gives the immediate strict contradiction at a positive maximum.