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Consider the initial-value problems in $d=1$ $$\begin{cases} i\partial_tu = \Delta^2 u \\ u(x,0)=u_0 \end{cases}$$ and $$\begin{cases} i\partial_t u= \Delta u \\ u(x,0)=u_0, \end{cases}$$ Solutions to these equations obey the dispersive estimates $$\|e^{-it\Delta^2}u_0\|_{L^\infty} \lesssim t^{-1/4} \|u_0\|_{L^1} \\ \|e^{-it\Delta}u_0\|_{L^\infty} \lesssim t^{-1/2} \|u_0\|_{L^1}.$$ By conservation of the $L^2$ norm, we thus expect for long times, a localized initial data will in the first case spread like $\Delta x\gtrsim t^{1/2} $ and in the second case, $\Delta x \gtrsim t$ (indeed for the latter, taking a Gaussian as our initial data, we have that $\Delta x \sim t$ for large $t$). Hence, the latter seems to be more dispersive since it can delocalize faster.

From the physical side of things, we also know that for the former, the ratio of the group velocity to the phase velocity is 2 times as large compared to the latter. So from this perspective, we would predict the former to be more dispersive insofar as we think of dispersion arising from the discrepancy between the group and the phase velocity.

What gives?

Note: The second dispersive estimate follows from computing $\mathcal{F}^{-1}(e^{-i|\xi|^2t})(x)= C t^{-1/2} e^{-i|x|^2/(4t)}$ while the first follows from this paper. The phase and group velocities for the first equation are $\omega(k)/k= k^3$ and $\partial_k k^4=4k^3$. For the second, they are $\omega(k)/k= k$ and $\partial_k k^2=2k$.

Crossposted to Math.StackExchange

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    $\begingroup$ Aren't the two equations in your title the same? $\endgroup$
    – LSpice
    Commented May 17 at 23:51
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    $\begingroup$ Right, sorry about that. It's now fixed. $\endgroup$
    – Dispersion
    Commented May 18 at 1:31
  • $\begingroup$ The short answer is that the Schroedinger equation (which is Galilean invariant) has the same dispersion at all frequencies, so that the worst case estimate given by $L^1$--$L^\infty$ is essentially homogeneous across frequencies, and translates to the spreading that you see when you look at the Gaussian initial data. But for the first equation the dispersion is not the same at all frequencies; at $k = 0$ you have a highly degenerate critical point which makes the worst case $L^1$--$L^\infty$ estimate. But localized near $k \neq 0$ the decay rate is still $t^{-1/2}$. $\endgroup$ Commented May 18 at 1:44
  • $\begingroup$ So in fact your claim that localized initial data for the first equation will "spread" like $\Delta x \approx t^{1/2}$ is I believe incorrect. $\endgroup$ Commented May 18 at 1:46
  • $\begingroup$ Also, I am not sure about your association of dispersion with the ratio between phase and group velocities. For studying wave packets and other localized initial data, what you are looking at really should be the group velocity dispersion which is the rate of change of the group velocity as you change the localized frequency, which is something like the second derivative of $\omega$ in your notation. $\endgroup$ Commented May 18 at 1:58

1 Answer 1

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Let me address this comment.

Let $\phi \in C^\infty_c(\mathbb{R})$, then the function $$ u(t,x) = \int_{\mathbb{R}} \exp(-i\xi^4 t + ix\xi) \phi(\xi) ~\mathrm{d}\xi$$ solve the first equation with initial data $u_0(x) = \int_{\mathbb{R}} \exp(ix\xi) \phi(\xi)~\mathrm{d}\xi$. Suppose now that $\phi$ is supported in the interval $[k/2,2k]$ for some $k > 0$.

Recall the standard van der Corput Lemma (versions of which can be found in most texts on oscillatory integration)

VdC Lemma Suppose $\phi, \eta$ are smooth functions. Suppose $|\eta''| \geq \lambda > 0$ on $[a,b]$, then $$ \Big| \int_a^b \exp(i\eta(\xi)) \phi(\xi) ~\mathrm{d}\xi \Big| \leq \frac{1}{\sqrt{\lambda}} \Big[ \|\phi'\|_{L^1} + 8 \|\phi\|_{L^\infty} \Big] $$

Apply this to $\eta(\xi) = -\xi^4 t + x\xi$, we see that $\eta''(\xi) = 12 t\xi^2$. On the support of $\phi$ (which is the interval $[k/2,2k]$ we have therefore $|\eta''(\xi)| > 3 |t| k^2$ and hence we see the uniform decay $$ |u(t,x)| \leq \frac{1}{\sqrt{3 |t|} k} \Big[ \|\phi'\|_{L^1} + 8 \|\phi\|_{L^\infty} \Big] $$ showing the uniform $t^{-1/2}$ decay.

The reason that for generic solutions you have slower uniform decay rate is that if the support of $\phi$ includes $0$, then $\eta''$ is no longer uniformly bounded below. The best we have is $\eta^{(4)}$ being uniformly bounded, the analogous VdC lemma reads that if $|\eta^{(4)}| \geq \lambda > 0$ on $[a,b]$, then $\Big| \int_a^b \exp(i\eta(\xi))\phi(\xi) ~\mathrm{d}\xi\Big| \leq \lambda^{-1/4} \Big[ \|\phi'\|_{L^1} + 38 \|\phi\|_{L^\infty} \Big]$.

However, as your equation is linear and you can always use frequency projectors to write $\phi = \phi_< + \phi_>$ with $\phi_<$ having frequency support within $[-1,1]$ and $\phi_>$ supported outside $[-1/2,1/2]$, this indicates that a solution to your first equation should exhibit behaviors that reflect both the $t^{-1/4}$ decay from $k \approx 0$ and $t^{-1/2}$ decay from large $k$.


Finally, in terms of the speed of delocalization: if you start with initial data that is frequency localized to $[k/2,2k]$, the above discussion indicates that for the Schrodinger equation you have a decay of order $t^{-1/2}$, while for the fourth-order equation you have a decay of order $t^{-1/2}k^{-1}$, which shows that at higher frequencies the fourth-order equation has faster decay and stronger dispersion.


For references, in no particular order:

It is particularly illustrative if you look at the Airy equation which is treated in Stein and Shakarchi; it really highlights the frequency-dependent behavior of wave packets and how the decay estimates are not naturally uniform across different frequency ranges.

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  • $\begingroup$ Thank you, this a great answer. $\endgroup$
    – Dispersion
    Commented May 27 at 17:25

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