# Unique solutions to the heat equation on $\mathbb{R}^3$

Pierre-Gilles Lemarie-Rieusset, The Navier-Stokes Problem in the 21st Century treats the heat equation on $$\mathbb{R}^3$$ for time $$t\geq 0$$, and proves uniqueness of suitably smooth solutions by a kind of energy argument. For a solution $$u(t, x)$$ with $$u(0,x)=0$$ he looks at the integral over all $$x\in \mathbb{R}^3$$
$$\begin{equation}\int |u(t,x)|^2 e^{-|x|}\,dx.\end{equation}$$ Using the equation $$u_t=\Delta u$$, calculations very like other energy proofs show $$\begin{equation} \frac{d}{dt}\int |u(t,x)|^2 e^{-|x|}\,dx = -2\int |\vec{\nabla} u|^2 e^{-|x|}\,dx\ + 2\int u e^{-|x|} \sum_{j=1}^{3}\frac{x_j}{|x|}\partial_j u\,dx. \end{equation}$$ Those calculations begin with integration by parts and proceed by pretty straightforward calculus. Lemarie-Rieusset immediately concludes $$\begin{equation} \frac{d}{dt}\int |u(t,x)|^2 e^{-|x|}\,dx \leq \frac{1}{2}\int |u(t,x)|^2 e^{-|x|}\,dx. \end{equation}$$ But I do not see how.

• There must be some information missing. This can't be true in general simply because of the units: By rescaling $t$, one can always make it false. Mar 26 at 6:03
• @MichaelEngelhardt I believe the required information is just that $u$ solves the heat equation, which is in the question. You cannot rescale $t$, preserving that equation, without rescaling $x$ also. Mar 26 at 6:21
• Ah yes, and it's that $e^{-|x|}$ factor that prevents one from simply undoing the $x$-scaling in the integral. Mar 26 at 7:02
• In the last integral where $u$ and $\nabla u$ appear, use Cauchy-Schwartz with respect to the weight $e^{-|x|}$ and $2ab \leq a^2/2+2b^2$. Mar 26 at 8:26

Sorry, maybe my previous comment was not clear enough. You have $$\frac{d}{dt} \int |u|^2 e^{-|x|}\, dx=-2\int |\nabla u|^2 e^{-|x|}\, dx +2 \int u \nabla u \cdot \frac{x}{|x|}e^{-|x|}\, dx.$$ Now use $$2|u \nabla u| \leq \frac 12 |u|^2+2|\nabla u|^2$$ to estimate the last integral.
• Now I see how to get it from $2|u \nabla u| \leq \frac 12 |u|^2+2|\nabla u|^2$ (using Cauchy-Schwarz as you said). But how do you get that inequality? Mar 26 at 22:14
• It Is $2ab \leq a^2/2+2b^2$. Mar 26 at 22:19
• "routine quadratic inequality" A.K.A. $\varepsilon$-Cauchy-Schwarz: In analysis (of PDEs?) we use all the time that $|ab|\leq \varepsilon |a|^2+\frac{1}{\varepsilon}|b|^2$ for any $\varepsilon$ of one's liking or choosing. This is typically used to "reabsorb" the $|a|^2$ in the left-hand side of some inequality, just as in this thread. Mar 27 at 3:10