= Solution
With zero drift, choose the <smooth function>
$$
\boxed{u(x,y,t)=-1+\frac{t}{8}(1-x^2-y^2).}
$$
It equals $-1$ on the entire <parabolic boundary>, so $A=-1$. On the closed cylinder, $-1\leq u\leq-7/8$, while
$$
u_t-\Delta u=\frac{1-x^2-y^2}{8}+\frac t2\leq\frac58.
$$
Consequently
$$
u_t-\Delta u+u^3\leq\frac58-\left(\frac78\right)^3
=-\frac{23}{512}<0.
$$
Nevertheless the maximum at $(0,0,1)$ is
$$
\boxed{\max_{\overline Q}u=-\frac78>-1=A.}
$$
The negative cubic term permits an interior increase while the boundary stays fixed, explaining why the sign condition in the preceding part matters.
Back to article page
