research35 min studyLesson 423 in Mathematics

Dispersive decay and stationary phase

Explain decay through oscillatory cancellation, compute the free Schrödinger kernel bound, and isolate the nondegeneracy behind stationary phase.

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What you’ll understand

  • Derive the L¹-to-L∞ free Schrödinger estimate with its time exponent.
  • State the smooth compact-support and nondegenerate-phase conditions behind the leading stationary-phase formula.
  • Distinguish pointwise dispersive decay from dissipative loss of L² mass.

A wave packet can become smaller at each fixed spatial location while retaining all its total squared amplitude. Different frequencies move with different phases and cancel when recombined. Stationary phase identifies the locations where that cancellation is weakest and predicts their contribution. This mechanism differs from diffusion, which genuinely damps Fourier amplitudes. Keeping the distinction clear prevents a decreasing graph peak from being mistaken for an energy-loss theorem.

An oscillatory kernel has a useful absolute size

For the free equation i u_t plus the Laplacian of u equals zero on R to the d, Fourier inversion gives the propagator U(t)=exp(it Laplacian). For nonzero time its convolution kernel is a complex Gaussian with absolute value (4pi times |t|) to the minus d over two. For Schwartz initial data, taking absolute values inside the convolution yields an L¹-to-L∞ estimate. Density and the kernel formula extend the appropriate statement to integrable data, with the usual interpretation of the free evolution.

The estimate is singular at zero time, where the propagator is the identity and no bounded map from all L¹ data to L∞ exists. On the other hand, Plancherel gives exact unitarity on L² for every time because the Fourier multiplier has modulus one. Interpolating these two bounds gives L to the r-prime into L to the r decay with exponent d times one half minus one over r, for r between two and infinity. The domain R to the d and its spreading geometry are essential; a fixed torus has recurrent Fourier modes rather than this global decay law.

(U(t)f)(x)=1(4πit)d/2∫Rdei∣x−y∣2/(4t)f(y) dy,∥U(t)f∥∞≤(4π∣t∣)−d/2∥f∥1(U(t)f)(x)=\frac1{(4\pi it)^{d/2}}\int_{\mathbb R^d}e^{i|x-y|^2/(4t)}f(y)\,dy,\qquad \|U(t)f\|_\infty\le(4\pi|t|)^{-d/2}\|f\|_1
For t nonzero, the kernel phase has unit magnitude. The branch convention in the complex prefactor does not alter the displayed absolute-value estimate.↔ Scroll wide equations horizontally.

A nondegenerate critical point gives one quadratic contribution

Consider an oscillatory integral with a smooth real phase and a smooth compactly supported amplitude. Suppose there is exactly one critical point near the amplitude's support, it lies in the interior region under consideration, and the phase Hessian there is invertible; any other part of the support is nonstationary. For positive large time, stationary phase gives a leading term proportional to t to the minus d over two. Its coefficient uses the amplitude at the critical point, the absolute Hessian determinant, and a phase determined by the Hessian signature.

The proof locally changes coordinates to a nondegenerate quadratic phase and integrates the resulting Gaussian; outside the critical neighborhood, integration by parts gains powers of time because the phase gradient stays nonzero. These are the essential mechanisms, not a complete proof here. Degenerate Hessians, nonsmooth amplitudes, boundaries, and colliding critical points require different theorems. If the amplitude vanishes at the stationary point, the leading coefficient can vanish too, giving faster decay than the generic first term predicts.

∫eitϕ(ξ)a(ξ) dξ=(2πt)d/2eitϕ(ξ0)+iπsgn⁡(D2ϕ(ξ0))/4a(ξ0)∣det⁡D2ϕ(ξ0)∣+O(t−d/2−1)\int e^{it\phi(\xi)}a(\xi)\,d\xi=\left(\frac{2\pi}{t}\right)^{d/2}\frac{e^{it\phi(\xi_0)+i\pi\operatorname{sgn}(D^2\phi(\xi_0))/4}a(\xi_0)}{\sqrt{|\det D^2\phi(\xi_0)|}}+O(t^{-d/2-1})
The stated smooth compact-support, unique nondegenerate stationary-point assumptions and nonstationary remainder justify this standard leading expansion as t tends to positive infinity.↔ Scroll wide equations horizontally.

Conservation can coexist with a falling peak

In one dimension, initial data exp(−x squared) evolve under free Schrödinger into (1+4it) to the minus one half times exp of minus x squared divided by (1+4it). Its peak magnitude is (1+16t squared) to the minus one quarter. Meanwhile its spatial width increases and its L² norm remains fixed. The lost peak is redistributed over a wider region with changing phase. This exact example separates the qualitative picture from the rough upper bound, which is not sharp at short times.

Dispersive estimates feed nonlinear analysis because they make distributed wave interactions integrable in suitable spacetime norms. A raw pointwise decay estimate can nevertheless be too singular near equal times to integrate directly in a nonlinear Duhamel formula. Strichartz estimates combine decay, unitarity, and time integration more carefully. For other dispersion relations, critical points can be degenerate and the decay exponent changes; a cubic Airy phase in one dimension has a different scaling from the quadratic Schrödinger phase. One should derive the actual phase before borrowing an exponent.

A Gaussian spreads without losing mass

For free one-dimensional Schrödinger evolution from f(x)=exp(−x²), calculate the peak magnitude and spatial-width factor at t=sqrt(3)/4. Check the squared-amplitude integral.

  1. At this time, 1+16t²=4. The peak magnitude is therefore 4^(−1/4)=1/sqrt(2), while the Gaussian width factor is sqrt(1+16t²)=2.

  2. The squared magnitude is (1+16t²)^(−1/2) times exp(−2x²/(1+16t²)). Here it becomes one half times exp(−x²/2).

  3. Integrating gives one half times sqrt(2pi)=sqrt(pi/2), the same value as the initial integral of exp(−2x²). The halved peak intensity is exactly balanced by doubled width.

The result

Peak magnitude falls to 1/sqrt(2), yet the squared L² norm stays sqrt(pi/2). Spreading and cancellation do not require dissipation.

Explore the model

Interactive lab

Interactive lab · waves

Observe cancellation before invoking a decay theorem

Pause and reset, then set both controls to phase difference 0 degrees and amplitude ratio 1. Predict the summed wave has twice either component's amplitude. For a fresh comparison, pause and reset, then set phase difference 180 degrees and ratio 1; predict cancellation to numerical roundoff. The instrument combines two fixed-frequency traveling sines. It illustrates phase cancellation, but has no spreading packet or frequency-squared Schrödinger phase, so it neither simulates the Gaussian nor proves its time-decay exponent.

Observe cancellation before invoking a decay theoremTwo waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). Current readings: Sum at x = 0: 0.866; Resultant amplitude: 1.732; Phase difference: 60°. All parameters are adjustable with the labeled controls below.−3.14−1.5701.573.14−3.4−1.701.73.4x · radiansAmplitude
SumFirst componentSecond component
Sum at x = 0
0.866
Resultant amplitude
1.732
Phase difference
60°
t = 0 s

Two waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forOscillatory cancellation is a mechanism; a dispersive decay theorem also needs the actual phase, domain, and input norm.

Check your understanding

Why can free Schrödinger peak amplitude decay while its L² norm stays constant?

Try it yourself

For the same Gaussian, solve for the positive time when its peak magnitude first equals one half. What happens to its width factor and L² norm at that time?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Raise the peak equation to the power minus four before solving for t.

Compare with the solution

The equation (1+16t²)^(−1/4)=1/2 gives 1+16t²=16, so t=sqrt(15)/4. The width factor is four, and the peak intensity is one quarter of its initial value. Their product leaves the squared L² norm unchanged at sqrt(pi/2). This is an exact identity for this Gaussian on the whole line, rather than a claim that every initial packet has the same profile or that the general dispersive upper bound is attained.

Take it with you

  • The free Schrödinger kernel gives |t|^(−d/2) L¹-to-L∞ decay away from zero time.
  • Stationary phase requires smooth amplitudes and nondegenerate critical geometry for its standard exponent.
  • Pointwise spreading can preserve L² mass and is distinct from diffusive damping.

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