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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.

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  • $\begingroup$ 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. $\endgroup$ Mar 26, 2022 at 6:03
  • $\begingroup$ @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. $\endgroup$ Mar 26, 2022 at 6:21
  • $\begingroup$ Ah yes, and it's that $e^{-|x|}$ factor that prevents one from simply undoing the $x$-scaling in the integral. $\endgroup$ Mar 26, 2022 at 7:02
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    $\begingroup$ In the last integral where $u$ and $\nabla u$ appear, use Cauchy-Schwartz with respect to the weight $e^{-|x|}$ and $2ab \leq a^2/2+2b^2$. $\endgroup$ Mar 26, 2022 at 8:26

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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.

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  • $\begingroup$ 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? $\endgroup$ Mar 26, 2022 at 22:14
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    $\begingroup$ It Is $2ab \leq a^2/2+2b^2$. $\endgroup$ Mar 26, 2022 at 22:19
  • $\begingroup$ Oh yes, a routine quadratic inequality. $\endgroup$ Mar 26, 2022 at 22:32
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    $\begingroup$ "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. $\endgroup$ Mar 27, 2022 at 3:10

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