# Space-time Poincaré inequality for solution of parabolic equation

If $u : \mathbb R^n \to \mathbb R$ is a smooth enough function then on any Euclidean $n$-ball $B_R$ of radius $R$ we have the very well-known Poincaré inequality

$$\int_{B_R} |u - \bar u|^2 \le C(R,n)\int_{B_R} |\nabla u|^2.$$

Here $\bar u = \frac{1}{|B_R|} \int_{B_R} u$ is the average of $u$.

If we suppose instead that $u : \mathbb R^{n} \times \mathbb R \to \mathbb R$ (thinking of the last coordinate as time) and attempt to get a similar estimate on the space-time cylinder $Q_R = B_R \times (-R^2, 0]$ in terms of the spatial derivatives, applying Poincaré on each time slice gives $$\int_{Q_R} |u(x,t) - \bar u(t)|^2\, dx\,dt \le C(R,n) \int_{Q_R} |\nabla u|^2$$ where $\bar u(t) = \overline{u(\cdot,t)}$ is the spatial average at time $t$ and $\nabla u$ denotes the vector of derivatives with respect to the first $n$ coordinates. In general, this $\bar u(t)$ cannot be replaced with a constant - adding large multiples of $t$ to $u$ would make the LHS blow up while leaving the RHS invariant.

Question: Does this change if we know $u$ solves a uniformly parabolic equation? That is, if $\partial_t u = \sum_{i,j=1}^n a^{ij} \partial_i \partial_j u$ with coefficients $a^{ij} \in L^\infty(\mathbb R^{n+1})$ satisfying $\lambda \delta^{ij} \le a^{ij} \le \Lambda \delta^{ij}$ ($0<\lambda\le\Lambda<\infty$), can we improve this to $$\int_{Q_R} |u(x,t) - \bar u|^2\,dx\,dt \le C(R,n,\lambda,\Lambda)\int_{Q_R} |\nabla u|^2$$ where $\bar u = \frac{1}{|Q_R|} \int_{Q_R} u$ is the space-time average?

If it helps, for the application I have in mind I also have an estimate for $\int |\nabla a^{ij}|^2\,dx\,dt$, so we can rewrite the equation in divergence form as $\partial_t u = \partial_i (a^{ij} \partial_j u) - b^j \partial_j u$ with $a^{ij}$ bounded and $b^j = \partial_i a^{ij} \in L^2(\mathbb R;L^2(\mathbb R^n))$.