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 $\partial_k k^2=2k$.
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