# Smoothing-Strichartz estimates for the heat-Schrodinger evolution

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?

• Sure, thank you. I edited Mar 4, 2020 at 1:12

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.

• Nice. I think for the lower order term you might need some weight at the RHS Mar 4, 2020 at 8:31
• @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. Mar 4, 2020 at 14:13
• 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 Mar 4, 2020 at 14:58
• @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). Mar 4, 2020 at 16:21
• 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 $\varepsilon\to 0$. Mar 4, 2020 at 19:29