= Solution
Let $\mathcal Pu=u_t-\Delta u+a u_x+b u_y$ on the <unit disk> over $0<t\leq1$. Smoothness is understood to include continuity on the closed cylinder. If $u$ exceeded its nonnegative <parabolic boundary> maximum $A$, it would attain a positive global maximum $M>A$ at some spatial interior point at time $t_0>0$. There,
$$
\nabla u=0,\qquad \Delta u\leq0,\qquad u_t\geq0.
$$
At $t_0=1$ the last inequality uses the derivative from the past; at an interior time it is equality. These signs imply $\mathcal Pu+u^3\geq M^3>0$, contradicting the given <differential inequality>. Thus
$$
\boxed{u\leq A\quad\text{throughout }Q.}
$$
If $A>0$, equality at any point of $Q$ gives the same contradiction with $M=A$, so $u<A$ throughout $Q$. If $A=0$, equality is possible: $u\equiv0$ is an example. More precisely, write the inequality as $\mathcal Pu+c u\leq0$ with $c=u^2\geq0$. The <strong parabolic maximum principle> shows that an interior zero at time $t_0>0$ forces $u\equiv0$ on $B\times[0,t_0]$.
That conclusion need not extend into the future for a subsolution. For $t_0<1$, take a spatially constant function which is zero for $t\leq t_0$ and equals $-\exp(-1/(t-t_0))$ for $t>t_0$. It is smooth, nonincreasing and nonpositive, so $\mathcal Pu+u^3\leq0$, but it becomes negative after $t_0$.
Back to article page
