Dispersive estimate for the Schrödinger equation

Remarks

For $t \ne 0$ the solution operator has an explicit kernel: $$e^{it\Delta} f(x) = (4\pi i t)^{-n/2} \int_{\R^n} e^{i \abs{x - y}^2 / (4t)} f(y) \, dy,$$ with the principal branch of the square root. The kernel has modulus $(4\pi\abs{t})^{-n/2}$ at every point, and the estimate follows at once. The formula for the kernel comes from the Fourier transform of the complex Gaussian $e^{-it\abs{\xi}^2}$, or from the heat kernel by replacing $t$ by $it$.

The solution also conserves the $L^2$ norm: $\norm{e^{it\Delta} f}_{L^2} = \norm{f}_{L^2}$, by Plancherel's theorem. Interpolation between this and the dispersive estimate with the Riesz–Thorin theorem gives, for $2 \le r \le \infty$ and $1/r + 1/r' = 1$, $$\norm{e^{it\Delta} f}_{L^r} \le (4\pi\abs{t})^{-n(1/2 - 1/r)} \norm{f}_{L^{r'}}.$$

The decay rate $\abs{t}^{-n/2}$ cannot be improved. For the Gaussian $f(x) = e^{-\abs{x}^2}$ one computes $\abs{e^{it\Delta} f(0)} = (1 + 16 t^2)^{-n/4}$, which is of exact order $\abs{t}^{-n/2}$ for large $\abs{t}$. The $L^1$ norm on the right cannot be replaced by the $L^2$ norm. For fixed $t$, the operator $e^{it\Delta}$ maps $L^2(\R^n)$ onto $L^2(\R^n)$, which contains unbounded functions, so no bound of $\norm{e^{it\Delta} f}_{L^\infty}$ by a multiple of $\norm{f}_{L^2}$ holds.

Going deeper

For a student

The dispersive estimate is the pointwise decay input for the Strichartz estimates. Combined with the $TT^*$ argument and the Hardy–Littlewood–Sobolev inequality, it gives space-time integrability of $e^{it\Delta} f$ for $L^2$ data. These bounds are the basic tool of the local and global well-posedness theory of nonlinear Schrödinger equations. The same pattern applies to other dispersive equations such as the wave and Airy equations, with different decay rates. One proves a pointwise decay bound from an oscillatory kernel and interpolates it with energy conservation.