Posted by: Shuanglin Shao | December 6, 2009

## Kato-smoothing effect

“Kato smoothing” “is ” important in dispersive PDE.“Smoothing”, as it stands, sounds like a very good word. But why is it good? How will it be used remains as vague questions to me.

In the last week, I began to report a paper by Alazard-Burq-Zuily, “On the water waves questions with surface tension”, which contains a “local-smoothing” result: roughly speaking, solutions to 2-D water waves with surface tension in $C^0_tH^s_x$ will get $\frac 14$ regularity upgrade to $L^2_tH^{s+\frac 14}_x$, to the price of locally in space and being averaged in time. This result was first proven by Christianson-Hur-Staffilani.

A question came to me, why $1/4$? Where can I quickly see it? Soon I found out it is determined by the structure of the water-wave equation. To oversimplify the major result in the paper on the derivation of water-wave equations by means of paralinearization, it can be written as a type of dispersive equations,
$\partial_t u+iD^{3/2} u=0, \,D:=|\nabla|, \, (*)$

(I have dropped a lot of terms, for instance, only looking at a linear equation, and no flow-terms in (*); it is because I am only looking at main terms which I think reflect dispersion). (*) suggests a dispersive relation $\tau=|\xi|^{3/2}$. If one look for
$\int \int_{|x|\le 1} |D^\alpha u|^2 dxdt \lesssim \|u_0\|_{L^2}$, the maximum value of $\alpha$ turns out to be $1/4$. ($1/4$ was proved for 2D, does this suggest that it was the case in all dimensions?)

To understand this necessary condition on $\alpha$, I would like to draw an analogy with the “1/2-local smoothing” for the Schrodinger equations: for any $\epsilon >0$,

$\int\int \langle x \rangle^{-1-\epsilon} |D^{1/2} e^{it\Delta} f|^2 dx dt \le C \|f\|_2^2.$

Why $1/2$ there? and how it was proved? Then the clarifications of these matters provides a model (to me) that $1/4$ for water-wave sounds reasonable.

Let us first motivate “local-smoothing” for Schr\”odinger? It is well known that the solutions $e^{it\Delta} f$ to a free Schr\”odinger equation

$i\partial_t u+\Delta u=0, u(0,x)=f(x)$

obeys the conservation law of mass, i.e.,

$\| e^{it\Delta} f \|_{L^{\infty}_tL^2_x} =\|f\|_{L^2_x}. \,(1)$

Note that we can not add any $D^\alpha$ with $\alpha>0$ to the left hand side of (1) due to the Galilean transform (or simply by creating a bump at high frequency). One may argue that, is this no-gain-of-derivative due to we asking for too much in time by requiring a $L^\infty_t$? For instance, on a time interval, $[0,1]$, $L^\infty$-norm is stronger than any $L^q$ norm with $1\le q<\infty$? So is the following true,

$\|D^\alpha e^{it\Delta} f\|_{L^q_tL^2_x}\le C\|f\|_2,\,(2)$

for some $\alpha>0$ and $1\le q<\infty$. Unfortunately (2) does not hold for any $\alpha>0$ due to the same reason as above (one can take the same examples.).

Is there any hope that some variant of (2) holds true? By using the heuristic that solutions to a dispersive equations at high frequency travels much faster than low frequency, after waiting for a long time, only low-frequencies are left behind around spatial origin and they do not hurt positive derivatives. So if we were asking for an estimate locally in space, was it possible? The answer turns out to be Yes. We have the following estimate,

$\int \int_{|x|\le 1} |D^{1/2} e^{it\Delta}|^2 dxdt \lesssim \|f\|^2_2.\, (3)$

This is referred to as “Kato smoothing estimate" for Schr\"odinger in literature. Moreover $1/2$ is the most one can expect due to the obstruction (counterexample) we mentioned above. (3) will lead to a more general estimate, for any $\epsilon>0$

$\int \int_{\mathbb{R}^d} (1+|x|)^{-1-\epsilon} |D^{1/2} e^{it\Delta}|^2 dxdt \lesssim \|f\|^2_2.\, (4)$

The implication of (4) from (3) is easier by partitioning $\{|x|\ge 1\}$ and rescaling, and using the information that $\epsilon>0$.

We focus on proving (3). It will follow from a Plancherel argument.
Writing

$D^{1/2} e^{it\Delta} f(x)=\int_{\xi^\prime;, \tau} e^{ix^\prime\cdot \xi^\prime+it\tau}\int_{\xi_d} e^{ix_d\xi_d} \delta (\tau-|\xi|^2)|\xi|^{1/2} \widehat{f}(\xi)d\xi_d d\xi'd\tau,$

where $\delta$ is the Dirac mass, and $\xi=(\xi^\prime,\xi_d)$. We set

$F(\xi^\prime,\tau)=\int_{\xi_d} e^{ix_d\xi_d} \delta (\tau-|\xi|^2) |\xi|^{1/2}\widehat{f}(\xi)d\xi_d.$

To prove (3), it suffices to prove

$\int_{|x_d|\le 1} \int_{\mathbb{R}^d } |\widehat{F}(x^\prime,t)|^2 dx^\prime dt dx_d \le C\|f\|_2^2, \, (5)$

where $x=(x^\prime,x_d)$. Obviously by the Plancherel theorem, the left hand side of (5) is bounded by

$\int_{|x_d|\le 1} \int_{(\xi^\prime, \tau)} |F|^2 d\xi^\prime d\tau dx_d.$

Fixing $(\xi^\prime \tau)$, Cauchy-Schwarz yields

$|F|^2\le \int_{\xi_d} |\widehat{f}|^2 \delta(\tau-|\xi|^2 )d\xi_d \int_{\xi_d} |\xi|\delta(\tau-|\xi|^2) d\xi_d. \, (6)$

The second factor on the right hand side of (6) is bounded by an absolute if we restricting $\xi$ to to set $\{\xi: |\xi|\le \sqrt{d} |\xi_d|\}$; it is not hard to do so if at the beginning we aim to prove (3) under this restriction; then the general estimate follows from the triangle inequality and partition the frequency space into $laex d$ pieces.

So plugging (6) in (5) and interchanging the integration order, and using $|x_d|\le 1$, we find out the left hand side of (5) is bounded by $\|f\|_2^2$. This is exactly what we need. So we finish proving (3).

I like the previous type of argument very much; but I did not remember where it was the first place/time I saw it. So I recorded it for my own benefits.