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Consider the equation

$$\partial_t u(t,x) = -\partial_{xxx} u(t,x)$$ for $x \in \mathbb R.$

It is well-known that in general we have

$$\Vert u(t) \Vert_{L^{\infty}} \le C t^{-1/3} \Vert u_0 \Vert_{L^1}.$$

I now read in some lecture notes, without proper argument, that for $u_0$ such that $\operatorname{supp} (\hat u_0) \subset [\tfrac{1}{2},2] \cup [-2,-\tfrac{1}{2}]$ we have the improved estimate

$$\Vert u(t) \Vert_{L^{\infty}} \le C t^{-1/2} \Vert u_0 \Vert_{L^1}.$$

My question is: Why is that true?

Remark: Before that statement there is the remark that the integral operator

$$I(t)=\int_{\mathbb R} e^{it \Phi(x)} a(x) \ dx$$ with $\Phi'' \ge c >0$

satisfies the estimate

$$I(t)=\int_{\mathbb R} e^{it \Phi(x)} a(x) \ dx \le Ct^{-1/2} \Vert a' \Vert_{L^1}.$$

However, I am not sure if one is supposed to follow from the other.

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  • $\begingroup$ Yes, it is supposed to follow. This is basically the higher derivative Van Der Corput Lemma. en.wikipedia.org/wiki/Van_der_Corput_lemma_(harmonic_analysis) $\endgroup$ Nov 13 '20 at 20:45
  • $\begingroup$ @WillieWong sorry, what is the connection between the dispersive estimate and the integral estimate? $\endgroup$
    – Martinique
    Nov 13 '20 at 20:49
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    $\begingroup$ $$ u(t,x) = \int e^{it \xi^3} \underbrace{e^{ix\xi} \hat{u}_0(\xi)}_{a(\xi)} ~d\xi $$ and here $\Phi(\xi) = \xi^3$. $\endgroup$ Nov 13 '20 at 20:50
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    $\begingroup$ Since $\hat{u}_0$ has compact support, we also have $$ \|a\|_{L^1} \lesssim \|\hat{u}_0\|_{L^\infty} \lesssim \|u_0\|_{L^1}$$ $\endgroup$ Nov 13 '20 at 20:54
  • $\begingroup$ @WillieWong notice the estimate in the auxiliary integral contains $a'$ rather than $a$? $\endgroup$
    – Martinique
    Nov 13 '20 at 20:56
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Let $\chi$ be a smooth function with support on $[-3,-1/3] \cup [1/3,3]$ and equals 1 on $[-2,-1/2] \cup [1/2,2]$.

You can write $$ u(t,x) = \int e^{it \xi^3}e^{ix\xi} \chi(\xi) \hat{u}_0(\xi) ~d\xi $$ if $u_0$ has Fourier support as you demanded. If you write $$ V(t,x) = \int e^{it\xi^3}e^{ix\xi} \chi(\xi) ~d\xi$$ this shows $u(t,x) = \int V(t,x-y) u_0(y)~dy$ and so it suffices to show that $ \|V(t,x)|_{L^\infty_x} \lesssim t^{-1/2}$.

For convenience write $\eta(\xi) = t\xi^3 + x \xi$. We have that $\eta'(\xi_0) = 0 \iff \xi_0 = \pm\sqrt{ -x/t}$ (provided $t$ and $x$ are such that the quantity under the square root is positive).

Suppose for convenience $t > 0$ (the $t < 0$ case follows the same way). And similarly we will only look at the integral over $[1/3,3]$, the other is similar.

On this domain, we have $\eta'' = 6 t \xi > 2t$ and so $\eta'$ is monotonic.

Let $\delta = \frac{1}{\sqrt{t}}$. Split the integral for $V$ into three pieces:

$$ \int_{1/3}^{\xi_0 - \delta} + \int_{\xi_0-\delta}^{\xi_0+\delta} + \int_{\xi_0+\delta}^3 e^{i\eta}\chi ~d\xi $$

On the first and third piece we have that $|\eta'| > 2 \sqrt{t}$. So the standard Van der Corput argument gives that the integral is bounded by

$$ \frac{1}{2\sqrt{t}} \left( \| \chi'\|_{L^1} + 3\|\chi\|_{L^\infty} \right) $$

In the middle term we are integrating over an interval of width $2\delta$ so we have the integral is bounded by

$$ \frac{2}{\sqrt{t}} \|\chi\|_{L^\infty} $$

Since $\chi$ is a fixed function, we can just absorb them into constants. And have that $|V(t,x)| \lesssim t^{-1/2}$ as claimed for $t >0$.


The key step is the lower bound on $\eta''$, which is available since we restricted away from $\xi = 0$. This ensures that if you go a distance $\delta$ away from the critical point you pick up a lower bound on $|\eta'|$ so the usual stationary phase argument gives you a bound by $\frac{1}{\inf |\eta'|}$.

Without this restriction $\eta'$ may stay small for a wider range of $\xi$. This would be the case when $x = 0$ (which in our case is not a problem because the corresponding $\xi_0$ is outside of the domain of integration). In this case instead of $\delta = 1/\sqrt{t}$ you need to choose $\delta = 1/\sqrt[3]{t}$, which would guarantee (using that $\eta''' = 6t > 0$) that when you get $\delta$ away from $\xi_0$, your $|\eta'| \geq 3 \sqrt[3]{t}$.

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