PierreGilles LemarieRieusset, The NavierStokes 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. LemarieRieusset 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. $\endgroup$– Michael EngelhardtMar 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. $\endgroup$– Colin McLartyMar 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. $\endgroup$– Michael EngelhardtMar 26, 2022 at 7:02

2$\begingroup$ In the last integral where $u$ and $\nabla u$ appear, use CauchySchwartz with respect to the weight $e^{x}$ and $2ab \leq a^2/2+2b^2$. $\endgroup$– Giorgio MetafuneMar 26, 2022 at 8:26
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1 Answer
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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 $2u \nabla u \leq \frac 12 u^2+2\nabla u^2$ to estimate the last integral.

$\begingroup$ Now I see how to get it from $2u \nabla u \leq \frac 12 u^2+2\nabla u^2$ (using CauchySchwarz as you said). But how do you get that inequality? $\endgroup$ Mar 26, 2022 at 22:14

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$\begingroup$ Oh yes, a routine quadratic inequality. $\endgroup$ Mar 26, 2022 at 22:32

2$\begingroup$ "routine quadratic inequality" A.K.A. $\varepsilon$CauchySchwarz: 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 lefthand side of some inequality, just as in this thread. $\endgroup$ Mar 27, 2022 at 3:10