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Consider $u$ solving the following heat equation \begin{cases} \partial_t u = \Delta u + f & \text{in } \mathbb R^2 \times \mathbb R^+, \\ u(\cdot, 0) = 0 & \text{in } \mathbb R^2, \end{cases} with $f \in L^\infty(\mathbb R^2 \times \mathbb R^+).$ I was wondering if there was a way to prove that for any $R > 0$, we could show the following estimate $$\|\nabla u\|_{L^\infty(Q_{R/2}(t_0, x_0))} \le \frac{1}{R}\left(\|f\|_{L^\infty(Q_{R}(t_0, x_0))} + \|u\|_{L^\infty(Q_{R}(t_0, x_0))}\right),$$ where the cylinder $Q_R$ is defined by $$Q_R(t_0, x_0) = \{(t,x) ~|~ |x - x_0| \le R, ~ t_0 - R^2 \le t \le t_0\}.$$ This type of interior estimate for the derivative is well known for the Poisson problem, i.e. $\Delta u = f$, but does it still hold when we consider the heat equation ? I saw in Evan's book that we had such inequality when $f = 0$, but the proof strongly uses the mean value formula which is, as far as I know, valid only when $f = 0$. Is there any way to solve this question for the heat equation ?

Falcon
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  • $f$ being merely $L^{\infty}$ makes a direct application of the Green function formula difficult: $u(x,t) = \int_{\mathbb{R}^2} G(x-y, t) f(y) dy$ where $G(\cdot, t)$ is the Gaussian kernel at $t$. Since you only care about a local (in time) and interior estimate, I suggest mollify the solution to be more tamable and start working from there. – Shuhao Cao Sep 09 '24 at 20:01
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    @Falcon Could you please provide a ref. for the Poisson equation? The estimate seems not to be invariant under dilation $x\to a x$. – Andrew Jan 31 '25 at 20:26

1 Answers1

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I think that your claim is false. You can check this by sending $R\to\infty$. In this limit, you recover the domain $Q_\infty = (0,t_0)\times \mathbb R^2$. You can compute a weak solution $u\in L^\infty(Q_\infty)$ by convolution with the Green's function: $$ u = G*f \\ G = \frac{\exp(-x^2/4t)}{4\pi t} $$ Since $G\in L^1(Q_\infty)$, with $f\in L^\infty(Q_\infty)$, the convolution is well defined and $u\in L^\infty(Q_\infty)$ by Young's inequality.

Your result would therefore conclude that: $$ \|\nabla u\|_\infty=0 $$ since all the norms have finite limits.

Alternatively, you can see the issue simply by dimensional analysis. $f$ and $u$ differ by units of time, so it does not make sense to add their $L^\infty$ norm.

Here is a correct estimate for $Q_\infty$. You can compute: $$ \nabla u = \int \nabla G(x-\xi,t-\tau)f(\xi,\tau)d^2\xi d\tau $$ since $\nabla G(x-\xi,t-\tau)\in L^1$ with: $$ \|\nabla G(x-\xi,t-\tau)\|_1 = \frac{\sqrt{\pi t_0}}4 $$ you can deduce: $$ \|\nabla u\|_\infty\leq \frac{\sqrt{\pi t_0}}4\|f\|_\infty $$ Notice that everything is sound dimensionally. Perhaps you can generalize this approach with finite $R$. I think that in front of $\|f\|_\infty$, you are missing a factor proportional to $R\sqrt{t_0}$ to match this result and solve the units mismatch.

LPZ
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