= Solution
Let $m_0=\min_xu(x,0)>0$, which exists by periodicity and continuity. Fix any finite interval $[0,T]$ on which the classical solution is smooth. Choose $\epsilon>0$ with $\epsilon(1+T)<m_0/2$ and consider the strictly positive decreasing barrier $b(t)=m_0-\epsilon(1+t)$.
Initially $u>b$. If this strict inequality first failed, compactness of the spatial circle would give a first contact $(x_*,t_*)$ with $u=b>0$. At that contact, the spatial minimum of $u-b$ has $u_x=0$, $u_{xx}\geq0$, and the first-contact condition gives $(u-b)_t\leq0$. But the equation yields
$$
(u-b)_t=u^2u_{xx}+u^3+\epsilon>0,
$$
a contradiction. Therefore $u>b$ throughout the interval. Letting $\epsilon\downarrow0$ gives \b[$\boxed{u(x,t)\geq m_0>0}$] on $[0,T]$. Since $T$ was arbitrary, positivity persists for every time for which the assumed smooth solution exists. This argument is a <parabolic comparison principle> proved directly and does not assume uniform parabolicity before positivity has been established.
Back to article page
