3
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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 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 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 at 7:02
  • 2
    $\begingroup$ 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 at 8:26

1 Answer 1

7
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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 at 22:14
  • 1
    $\begingroup$ It Is $2ab \leq a^2/2+2b^2$. $\endgroup$ Mar 26 at 22:19
  • $\begingroup$ Oh yes, a routine quadratic inequality. $\endgroup$ Mar 26 at 22:32
  • 2
    $\begingroup$ "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 at 3:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.