Consider the heat-Schrodinger evolution $e^{(1+i)t\Delta}$, $t\geq 0$. For simplicity, suppose to work in dimension three. Due to the Strichartz estimates for the Schrodinger equation, and the smoothing effect of the heat flow, I expect that the following estimate holds true: \begin{equation}\label{tre} (1)\qquad \left\|\int_0^te^{(1+i)(t-s)\Delta}F(s)ds\right\|_{L^2((0,t),W^{1,6}(\mathbb{R}^3))}\lesssim \|F\|_{L^2((0,t),L^2(\mathbb{R}^3))} \end{equation} Nevertheless, I have not been able to prove it.

Is estimate (1) actually true? In case, is it already known in the literature?

Thank you for your suggestions.

  • $\begingroup$ Sure, thank you. I edited $\endgroup$
    – Capublanca
    Mar 4, 2020 at 1:12

1 Answer 1


I think the estimate you want does not really require Strichartz.

First, your estimate is equivalent to the following, which is a bit easier for me to think about: let $u$ be the solution to the equation $$ \partial_t u - (1 + i) \triangle u = F \tag{*}$$ with initial data $$ u(0,x) \equiv 0 $$ Then the desired estimate is $$ \| u \|_{L^2_t W^{1,6}_x} \leq \|F\|_{L^2_t L^2_x}$$

We can derive this estimate by multiplying the equation (*) by $\triangle \bar{u}$ and taking the real part, which gives $$ \partial_t u \triangle \bar{u} + \partial_t \bar{u} \triangle u - (1+i) \triangle u \triangle \bar{u} - (1-i) \triangle u \triangle\bar{u} = F \triangle \bar{u} + \bar{F} \triangle u $$ Integrate by parts in space we get $$ \partial_t \| \nabla u\|^2_{L^2_x} + 2 \|\triangle u\|_{L^2_x}^2 = - \int_{\mathbb{R}^3} F\triangle \bar{u} + \bar{F} \triangle u ~dx$$ Integrate between time 0 and time T you get (using the triviality of the data) $$ \|\nabla u(T)\|_{L^2_x}^2 + 2 \int_0^T \| \triangle u\|^2_{L^2_x} ~dt \leq \int_{[0,T]\times \mathbb{R}^3} 2 |F|\cdot |\triangle u| ~dx~dt (**)$$ Applying Young's inequality on the right, you can pull out $\epsilon \|\triangle u\|_{L^2_t L^2_x}$ which can be absorbed on the left hand side, and this reduces to

$$ \| \triangle u\|_{L^2_t L^2_x} \lesssim \| F\|_{L^2_t L^2_x} $$

Finally use elliptic estimates to bound $\|\nabla^2 u\|$ by $\|\triangle u\|$, and then Sobolev inequality in space gets you

$$ \| \nabla u\|_{L^2_t L^6_x} \lesssim \|F\|_{L^2_t L^2_x} $$

This takes care the homogeneous part of the Sobolev norm.

I don't think the inhomogeneous part holds. It doesn't have the correct scaling.

If you rescale $u(t,x) \mapsto u(\lambda^2 t, \lambda x)$ by the parabolic scaling, then $F(t,x) \mapsto \lambda^2 F(\lambda^2 t, \lambda x)$. The $L^2_t L^2_x$ norm of $F$ scales like $\lambda^{-1/2}$, but the $L^2_t L^6_x$ norm of $u$ scales like $\lambda^{-3/2}$. The term with the correct scaling should be $\|F\|_{L^1_t L^2_x}$. And the estimate

$$ \| u\|_{L^2_t L^6_x} \lesssim \|\nabla u\|_{L^2_t L^2_x} \lesssim \|F\|_{L^1_t L^2_x} $$

can be proven by an analogous manner as above but instead of multiplying by $\triangle u$, multiply by $u$.

Incidentally: both estimates also hold for the heat equation, without the Schroedinger part.

Addendum: I just noticed that your inequality is only stated for $L^2((0,T),X)$ for some Banach space $X$. Do you only care about the local estimate where your constant can depend on the length of the interval? Or do you actually want uniform estimates for all $T$? I ask because quite obviously on a fixed time interval you have that $\|F\|_{L^1_t L^2_x} \lesssim \|F\|_{L^2_t L^2_x}$, and in fact you can get the same result by noting that (**) also implies $$\|u\|_{L^\infty_t L^6_x} \lesssim \|F\|_{L^2_t L^2_x}$$ and can be easily localized to intervals.

  • $\begingroup$ Nice. I think for the lower order term you might need some weight at the RHS $\endgroup$ Mar 4, 2020 at 8:31
  • $\begingroup$ @PieroD'Ancona: I don't quite understand what you mean. Can you say a couple sentences more? What I imagine: multiplying by $u$ and IBP gives us that $$\|u(T)\|_{L^2}^2 + \| \nabla u\|_{L^2_{t,x}}^2 \leq \| F u \|_{L^1_{t,x}} $$ the right hand side bound is bounded by $\| F\|_{L^1_t L^2_x} \|u\|_{L^\infty_t L^2_x}$ which implies $$ \|u\|_{L^\infty_t L^2_x} \leq \|F\|_{L^1_t L^2_x}$$ which gives the desired result. I don't think weights are necessary. But possibly I misunderstand your comment. $\endgroup$ Mar 4, 2020 at 14:13
  • $\begingroup$ What I mean is that another way to achieve the correct scaling is, say, using a weighted norm such as $\||x|F\|_{L^2L^2}$. I think the estimate for the lower order term might be true also with this right hand side $\endgroup$ Mar 4, 2020 at 14:58
  • $\begingroup$ @PieroD'Ancona: ah, I see. For the pure Schrodinger this would be an end-point case if I am not mistaken. (It almost follows from my paper arxiv.org/abs/1701.01460v4 but it would be the endpoint case which I didn't do). $\endgroup$ Mar 4, 2020 at 16:21
  • $\begingroup$ Nice work, thank you. I was indeed interested only in the homogeneous estimates, but also the suggestion by Piero for the lower order term is very interesting. I think that, if one considers the evolution $e^{(i+\varepsilon)t\Delta}$, where $\varepsilon>0$ is a small parameter, your approach produces a divergence rate as $\varepsilon\to 0$ which can be improved by using the Strichartz estimates for the Schrodinger equation. Maybe it could be interesting (and not so trivial) to establish the optimal divergence rate as $\varepsilon\to 0$. $\endgroup$
    – Capublanca
    Mar 4, 2020 at 19:29

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