# Estimate for the gradient $|\nabla_x u (x,t)|$ in the heat equation

heat equationpartial differential equations

Suppose $$u_0 \in C^{0}(\mathbb{R}^n) \cap L^{\infty}(\mathbb{R}^n).$$
Consider
\left\{ \begin{aligned} &u_{t} – \Delta u = 0 \ \ &&\mathbb{R}^n \times (0, +\infty),\\ & u(x,0) = u_0(x) && \mathbb{R}^n\\ \end{aligned} \right.
And let $$u$$ be the solution given in terms of the fundamental solution,
$$u(x,y) = \frac{1}{(4\pi t)^{\frac{n}{2}}}\int_{\mathbb{R}^n} e^{\frac{-|x – y|^2}{4t}}u_0(y) dy$$
Show that there is $$C > 0$$ such that
$$|\nabla_x u (x,t)| \leq \frac{C}{\sqrt t}\|u_0\|_{L^{\infty}(\mathbb{R}^n)}, \ \forall \ x \in \mathbb{R}^n, \ t > 0$$

Attempt: For the one-dimensional problem, for example,
\begin{aligned} u_x(x,t) &= \frac{1}{\sqrt 4\pi t}\int^{\infty}_{-\infty} \frac{d}{dx}e^{\frac{-|x – y|^2}{4t}}u_0(y) dy\\ & = \frac{1}{\sqrt 4\pi t}\int^{\infty}_{-\infty} -\frac{|x – y|}{2t}e^{\frac{-|x – y|^2}{4t}}u_0(y) dy\\ & \leq \frac{1}{4\sqrt \pi t }\frac{1}{\sqrt t}\|u_0\|_{L^{\infty}(\mathbb{R}^n)}\int^{\infty}_{-\infty} |x – y|e^{\frac{-|x – y|^2}{4t}} dy\\ \end{aligned}
By changing a variable, I'm finding that the integral above gives zero. Help!

You're on the right track; use translation invariance to write $$\int_{-\infty}^\infty |x-y|e^{-|x-y|^2/4t}\, dy = \int_{-\infty}^\infty |z|e^{-|z|^2/4t}\, dz.$$ Now write $$w=z/(2\sqrt{t})$$ to further write $$\int_{-\infty}^\infty |x-y|e^{-|x-y|^2/4t}\, dy = \int_{-\infty}^\infty |2\sqrt{t} w|e^{-|w|^2}\, 2\sqrt{t}dw=4t \int_{-\infty}^\infty |w|e^{-|w|^2}\, dw.$$ For higher dimensions it's the same, just remember that in that case $$dw=(2\sqrt{t})^{-n}\, dz$$.
Added: In one dimension it's easy to compute the integral: $$$$\int_{-\infty}^\infty |w|e^{-w^2}\, dw = 2\int_0^\infty we^{-w^2}\, dw = \int_0^\infty e^{-s}\, ds =1,$$$$ with $$s=w^2$$ so that $$ds=2w dw$$. In higher dimensions you can probably get something in terms of the $$\Gamma$$ function with the same change of variables.