Improving dispersive estimates for linear KdV

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.

• 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) Nov 13 '20 at 20:45
• @WillieWong sorry, what is the connection between the dispersive estimate and the integral estimate? Nov 13 '20 at 20:49
• $$u(t,x) = \int e^{it \xi^3} \underbrace{e^{ix\xi} \hat{u}_0(\xi)}_{a(\xi)} ~d\xi$$ and here $\Phi(\xi) = \xi^3$. Nov 13 '20 at 20:50
• Since $\hat{u}_0$ has compact support, we also have $$\|a\|_{L^1} \lesssim \|\hat{u}_0\|_{L^\infty} \lesssim \|u_0\|_{L^1}$$ Nov 13 '20 at 20:54
• @WillieWong notice the estimate in the auxiliary integral contains $a'$ rather than $a$? Nov 13 '20 at 20:56

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 $$\delta = 1/\sqrt{t}$$, which would guarantee (using that $$\eta''' = 6t > 0$$) that when you get $$\delta$$ away from $$\xi_0$$, your $$|\eta'| \geq 3 \sqrt{t}$$.