Consider the heat flow $e^{t\Delta}$, $t>0$, on an Euclidean domain (say $\mathbb{R}^3$). I expect, in analogy with the Strichartz estimates for the Schrodinger equation, that the following estimates hold: $$\Vert e^{t\Delta}f\Vert_{L^s(\mathbb{R}^+,L^p(\mathbb{R}^3))}\lesssim \Vert f\Vert_{L^2(\mathbb{R}^3)},$$ where $(s,p)$ is a Strichartz pair. Is this fact actually true? In such a case, there exists a reference where it is stated explicitly?
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
4
Sure. Actually you have in addition the maximum principle (i.e. an $L^\infty$--$L^\infty$ estimate) which combined with the others gives a full triangle of indices for the Strichartz type estimates in your question. Even more, the strong decay of the symbol $e^{-t|\xi|^2}$ can be exploited for some gains in regularity. A short account is in the book by Wang, Huo, Hao, Guo "Harmonic analysis method for nonlinear evolution equations" (around p.35) link
answered May 22, 2017 at 10:52
-
$\begingroup$ Thank you very much Piero! I have another doubt: is it also true that $e^{t\Delta}$ maps continuosly $L^{\infty}((0,T),L^2(\mathbb{R}^3))$ into $L^{s}((0,T),L^p(\mathbb{R}^3))$? For the Schrodinger flow it should not be true, but maybe it works in the heat case. $\endgroup$ Commented May 22, 2017 at 13:18
-
$\begingroup$ You mean the non homogeneous equation $u_t-\Delta u=F$? you certainly can apply non homogeneous Strichartz estimates, so $u$ in $L^pL^q$ is estimated by $F$ in $L^1L^2$. For bounded intervals this is controlled by $L^\infty L^2$. $\endgroup$ Commented May 22, 2017 at 20:00
-
$\begingroup$ I understand the non homogeneous case, thank you. However i'm wonder if also the bound $\Vert e^{t\Delta}F(t,\cdot)\Vert_{L^sL^p}\lesssim \Vert F \Vert_{L^{\infty}L^2}$ is true, at least for finite time intervals and for $F\in\mathcal{C}(\mathbb{R}^+,L^2)$. $\endgroup$ Commented May 23, 2017 at 3:43
-
$\begingroup$ That's what I said. Inhomogeneous Strichartz takes you to $L^1L^2$ which is estimated by $L^\infty L^2$. But of course the constant in this inequality is unbounded as $T\to \infty$ (it is exactly $T$) $\endgroup$ Commented May 23, 2017 at 20:18