Let's consider the propagator corresponding to the one-dimensional equation $$ u_t=\Lambda^\alpha u_x,\; u(x,0)=f(x) $$ where $$ \widehat{\Lambda^\alpha u}=|\xi|^\alpha\hat{u}(\xi), $$ and $-1< \alpha\leq 1$.

QUESTION 1) Do you know any good reference in the dispersive estimates for this operator? I mean inequalities of the following form $$ \|u(t)\|_{L^\infty(\mathbb{R}}\leq (1+t)^{-\theta}\|f\|_{X}, $$ for certain $\theta$ and Banach space $X$.

QUESTION 2) Same for the propagator corresponding to $$ u_t=(1-\partial_x^2)^\alpha u_x,\; u(x,0)=f(x) $$ and $-1\leq \alpha\leq 1$.

  • $\begingroup$ Yes. I already corrected. $\endgroup$
    – guacho
    Jul 2, 2017 at 6:59
  • $\begingroup$ In Q2 I wonder if you meant $|\alpha| \leq \frac12$ instead? $\alpha > \frac12$ is not a problem, but $\alpha < -\frac12$ your operator has a net negative number of derivatives on the right hand side... $\endgroup$ Jul 5, 2017 at 16:55

1 Answer 1


First note that when $\alpha = 0$ you have the linear advection equation with no decay whatsoever, in either case, as $L^\infty$ is conserved. So we focus on the situation where $\alpha \neq 0$.

Question 1

Since you have scaling homogeneity, it suffices to localize in frequency. Let $\chi_\pm$ be bump functions supported near $\xi \approx \pm 1$. Your goal is to essentially estimate $$ \int e^{i t |\xi|^\alpha \xi + i x \xi} ~\mathrm{d}\xi \approx \sum_{\lambda = -\infty}^\infty \int e^{it |\xi|^\alpha \xi + i x\xi} (\chi_+(2^\lambda \xi) + \chi_-(2^\lambda \xi)) ~\mathrm{d}\xi = \sum P^{-\lambda}_+ + P^{-\lambda}_- $$ And we have that $$ P_{\pm}^{-\lambda}(2^{\lambda(\alpha + 1)}t, 2^{\lambda} x) = \frac{1}{2^\lambda} P_{\pm}^0 $$ The Van der Corput Lemmas imply that $P^0$ has a uniform bound by $1/\sqrt{t}$ as $\alpha(\alpha+1) > 0$ if $\alpha > -1$ and $\alpha \neq 0$. So this gives the bound in terms of $X$ being a Homogeneous Besov space

$$ \| u\|_{\infty} \lesssim \frac{1}{\sqrt{t}} \|f\|_{B^{(1-\alpha)/2,1}_1} $$

Question 2

You don't have scaling homogeneity. But instead you can consider $$P_{\pm}^{\lambda} = 2^{\lambda} \int e^{i t \langle 2^{\lambda} \xi\rangle^{2\alpha} 2^{\lambda} \xi} e^{i 2^\lambda \xi x} \chi_{\pm}(\xi) ~\mathrm{d}\xi $$ for the frequency $2^{\lambda}$ piece. You see that the phase function $\eta = 2^\lambda \xi \langle 2^\lambda \xi\rangle^{2\alpha} + 2^\lambda \xi (x / t)$ is such that $$ |\eta''| \gtrsim \alpha 2^{\lambda (2\alpha -1)} |\xi|^{2\alpha - 3} \left| 1 + (1+2\alpha) 2^{2\lambda} |\xi|^2 \right|. $$ We see that when $\alpha \neq 0$ and $\alpha > -\frac12$, we have that the lower bound is by $2^{\lambda(2\alpha + 1)}$, so by Van der Corput Lemma (applied to $\tilde{t} \tilde{\eta} = t\eta$ where $\tilde{\eta} = 2^{-\lambda(2\alpha + 1)} \eta$ and $\tilde{t} = 2^{\lambda(2\alpha + 1)} t$) we get exactly the same decay estimate as in Question 1.

The regularization near $|\xi| = 0$ gives us however that, when $\alpha = -\frac12$, $|\eta''| \gtrsim 2^{-2\lambda}$. So in this case we have by Van der Corput Lemma that $$ \|u\|_{\infty} \lesssim \frac{1}{\sqrt{t}} \| f\|_{B^{2,1}_1} $$

(a loss of one derivative compared to the $\alpha > -1/2$ case)

For the small time estimate you can use $L^2$ conservation plus Sobolev to get the bound

$$ \|u\|_{\infty} \lesssim \|f\|_{W^{1,2}}$$

  • $\begingroup$ Great answer. Thank you very much. However, I'm afraid that I don't fully understand it. *I'm not really familiar to VdC Lemma. I think that you need a smooth function (and $|\xi|^\alpha$ is not) and a lower bound equal to 1 (and that is not your lower bound) *How do you get the Besov spaces? $\endgroup$
    – guacho
    Jul 6, 2017 at 14:00
  • 1
    $\begingroup$ @guacho: $|\xi|^\alpha$ is only not smooth at the origin; so after the frequency restriction you are only working with a smooth phase. (This is the same as how dispersive estimates are proven for higher dimensional wave equations.) For proving estimates, a uniform constant is "as good as 1". In the case of Question 2 the constant is not uniform in $\lambda$, so what you need to do is to absorb the constant into $t$, which gives you a frequency dependent decay constant, which is why we end up using Besov spaces. $\endgroup$ Jul 6, 2017 at 16:03
  • 1
    $\begingroup$ Similar arguments are given in my lecture notes in Chapter 3. What you want to look at is the section on "Estimates of the fundamental solution". The discussion leading up to Theorem 3.46 is similar to my answer to Question 1, the only difference being how the frequency-restricted estimates are derived (here we can use VdC). The discussion of Theorem 3.48 is similar to how I approached Question 2. $\endgroup$ Jul 6, 2017 at 16:09

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