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The parabolic boundary values are zero. An exact solution satisfies the preceding differential inequality, so its maximum estimate with $A=0$ gives $u\le 0$. The function $v=-u$ satisfies the same equation because both the linear differential operator and the cubic term change sign. Applying the same maximum principle to $v$ gives $-u\le 0$. Hence the complete solution set isSubstitution verifies existence, and the two inequalities prove uniqueness among the assumed smooth solutions for every continuous choice of the drift coefficients.

With zero drift, choose the smooth functionIt equals $-1$ on the entire parabolic boundary, so $A=-1$. On the closed cylinder, $-1\le u\le -7/8$, whileConsequentlyNevertheless the maximum at $(0,0,1)$ isThe negative cubic term permits an interior increase while the boundary stays fixed, explaining why the sign condition in the preceding part matters.

Let $\mathcal{P}u=u_t-\Delta u+a{u}_x+b{u}_y$ on the unit disk over $0<t\le 1$. 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,At $t_0=1$ the last inequality uses the derivative from the past; at an interior time it is equality. These signs imply $\mathcal{P}u+u^3\ge M^3>0$, contradicting the given differential inequality. ThusIf $A>0$, equality at any point of $Q$ gives the same contradiction with $M=A$, so $u<A$ throughout $Q$. If $A=0$, equality is possible: $u\equiv 0$ is an example. More precisely, write the inequality as $\mathcal{P}u+cu\le 0$ with $c=u^2\ge 0$. The strong parabolic maximum principle shows that an interior zero at time $t_0>0$ forces $u\equiv 0$ 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 $t\le t_0$ and equals $-\exp (-1/(t-t_0))$ for $t>t_0$. It is smooth, nonincreasing and nonpositive, so $\mathcal{P}u+u^3\le 0$, but it becomes negative after $t_0$.