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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.

However, 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.

What gives?

EDIT: 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 follow 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 $\partial_k k^2=2k$.

Cross-posted to MathOverflow

Diffusion
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  • Attaching some of your calculations would help a lot here. How did you obtain these "dispersive estimates", and what do they mean? What does $\Vert \mathrm e^{\mathrm it\Delta} u_0 \Vert_\infty$ have to do with dispersion? How did you compute the phase and group velocities? – K.defaoite May 15 '24 at 20:00
  • See my edits. They are dispersive because since the $L^\infty$ norm decays and the $L^2$ norm remains constant, the solution must spread out as time increases. – Diffusion May 16 '24 at 03:39
  • Thanks, but I still don't understand the significance of $\mathrm e^{\mathrm it~\mathscr L}$ where $\mathscr L$ is either one of the operators. – K.defaoite May 16 '24 at 15:05
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    They are solutions of $i\partial_t u = \mathscr{L}u$. – Diffusion May 16 '24 at 16:35
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    Possibly one way to think about it: The speed (e.g. group velocity) at which low frequencies travel is more relevant to dispersion here since you’re measuring dispersion against an $L^1$ norm. Near $k=0$, the group velocity for the first equation is much smaller than for the usual Schrödinger equation, which could explain the slower decay. – kieransquared May 18 '24 at 00:16

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