** Method of Stationary Phase **

Proposition (non-stationary phase)Let and (and real-valued) with on , then

for every integer and .

*Proof:* Let us take the case,

In the spirit of integration by parts,

Thus, by the quotient rule,

More generally

and so

Thus

The ideas we use work for sums as well. Traditionally, dealing with sums are much harder.

Theorem (van der Corput lemma)Suppose and throughout the interval for some . If , then we additionally assume that is monotone. Then

independent of and .

Remark

- If then the phase is non-stationary; however, we cannot do better because of the end-point contributions from integration by parts (alternatively, the “amplitude function” is , which is not smooth). If , we can compute exactly and see that we cannot do better.
- To get a bound which is independent of the length of , we need to assume is monotone (or at least some additional hypothesis). Recall
We define heuristically, slow down when is positive and speed up when is negative. Doing this appropriately, we may make the right side arbitrarily large.

*Proof:* (by induction in ). For we argue as before

Thus

Now, let us do the induction step (assume the result is true for , now show for ). By hypothesis, . Let us look at . This vanishes at most at one point ; if it does not vanish on , then choose to be the end point where is least. Consider the partition of into interval and its complement, which we will write as . On , we have and is monotone. Using the inductive hypothesis,

Choosing we get , as desired.

RemarkA computation using shows that these bounds are the best that we may expect.

CorollaryLet obey the hypothesis from the van der Corput lemma and let be of bounded variation (or at least ). Considerthen

*Proof:*

Using the van der Corput lemma twice,

** An application of this lemma **

DefinitionA sequence isequi-distributediffor all continuous functions . Analogously, a curve is equi-distributed if

for all continuous functions .

Remark

- The notion is unchanged if we replace by or for arbitrary closed/open sets (Urysohn’s lemma).
- This replacement does not work for ‘s or ‘s (heuristically because they mimic the sequences too well).
- That is not uniformly continuous is implied by the fact that the curve is equi-distributed in when , because

LemmaIf is equi-distributed then it is dense.

Proof:Choose to be a bump in the open set that misses.

Theorem (Weyl’s criterion)The curve is equi-distributed in if and only if

for all .

*Proof:* The forward implication is *a fortiori* clear. For the reverse implication, trigonometric polynomials are dense.

TheoremLet be rationally independent “of constants”: that is, if with , then . Then the curve is equi-distributed in .

RemarkConversely, if where , then the curve is no equi-distributed: it lives in a co-dimension one sub-manifold.

*Proof:* By Weyl’s criterion, we need to show that for all . For fixed , we know that is a non-constant polynomial of degree , so with . Changing variables gives

where

Now, note that uniformly in . By Van der Corput’s lemma,

uniformly in . This goes to zero as .

Proposition (Multi-dimensional non-stationary phase)Fix and with on . Then

for any integer .

*Proof:* By introducing a partition of unity, we can assume that on the support of for some fixed . Now writing, with ,

Now, we apply the one dimensional result to to get with a constant that depends continuously on .

An alternative proof is to use and integrate by parts sufficiently many times. For example

Proposition (Morse Lemma)Let have a non-degenerate critical point at ; that is,Then there is a local diffeomorphism in a neighborhood of with ,

and

*Proof:* By Taylor’s formula (with integral remainder)

Note that, for sufficiently small, we can write

with a smooth, non-singular matrix with . (Eg. first make a linear change of variables to reduce to , then explicitly do the completing the square argument).

Now, set . Note that and

TheoremLet , and with a non-degenerate critical point at but otherwise, does not vanish on . Then

as .

Note that the (wait… I forgot what I wanted to say here)

RemarkOne can use a partition of unity (and spatial translations) to generalize to multiple stationary phase points.

*Proof:* By the Morse lemma, we can find some small neighborhood of where we can change variables to simplify . Now use a smooth partition of unity to write where is supported in the neighborhood of where the change of variables is defined, while vanishes in a neighborhood of . (, with . By the non-stationary phase position (and on ),

for as large as we wish. This leaves

Lets call ; it is and .

Remembering that , we see that

which has the correct asymptotics.

** Application of the theorem **

For the following discussion, we shall denote the Fourier transform (and its inverse) by

noting that .

Example (Schrödinger’s equation)Consider the long-time behavior of the solutions tofor with .

As the equation is space-translation invariant (it is also time-translation invariant, but we will not use that here), we take the spatial Fourier transform to find

with , which we take (for now) to be Schwartz . Therefore , i.e. . Taking inverse Fourier transforms, we get

and so we may apply the stationary phase theorem. Looking for stationary phase points,

Also,

So the theorem says

with the big- notation for constant and (i.e., this describes the decay if we travel along a world-line). Simplifying the above, we get

with the interpretation that and so . Note that for , the asymptotics pick up a “discrete” change

PropositionWe have, uniformly in and , the dispersive estimate,and the “Fraunhofer formula”

for every .

RemarkThe motivation for the dispersive estimate comes from the combination of our stationary phase computation and thea prioriestimate .

Proof:As a Fourier multiplier, is also just a convolution operator. To compute the convolution kernel, take a and findIn particular, the above computation gives the dispersive estimate

Furthermore, we get

Thus,

which, by the dominated convergence theorem with dominating function , goes to .

As an aside, we discuss the result in the setting of optics. If we shine a laser through a screen with an aperture, let us denote with being the horizontal and vertical axes, respectively. At the far screen we have

where is the distance from the source point to the observation point. Because and that , the dominant term is essentially the Fourier transform.

Also

cf Paraxial approximations.

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