This is the second part of a two-part post on the theory of oscillatory integrals. In the first part we studied the theory of oscillatory integrals whose phases are functions of a single variable. In this post we will instead study the case in which the phase is a function of several variables (and we integrate in all of them). Here the theory becomes weaker because these objects can indeed have a worse behaviour. We will proceed by analogy following the same footsteps as in the single-variable case.

Part I

**3. Oscillatory integrals in several variables **

In the previous section we have analysed the situation for single variable phases, that is for integrals over (intervals of) . In this section, we will instead study the higher dimensional situation, when the phase is a function of several variables. Things are unfortunately generally not as nice as in the single variable case, as you will see.

In order to avoid having to worry about connected open sets of (see Exercise 18 for the sort of issues that arise in trying to deal with general open sets of ), in this section we will study mainly objects of the form

where has compact support. We have switched to for the phase to remind the reader of the fact that it is a function of several variables now.

** 3.1. Principle of non-stationary phase – several variables **

The principle of non-stationary phase we saw in Section 2 of part I continues to hold in the several variables case.

Given a phase , we say that is a *critical point* of if

Proposition 8 (Principle of non-stationary phase – several variables)Let (that is, smooth and compactly supported) and let the phase be such that does not have critical points in the support of . Then for any we have

*Proof:* The proof is simply a reduction to the single variable case of Proposition 3 from part I. Indeed, by assumption there exists a such that for in the support of , and therefore for any such there exists a small ball , centered at and a *unit vector* such that for every . By compactness we can find a finite collection of such balls, , that covers the support of ; with a partition of unity associated to this collection, we can therefore write with supported in . Since the number of balls is finite (depending only on ), it suffices to prove the claimed bound for a single . By a change of coordinates, we can assume and since

we can conclude by applying the single-variable non-stationary phase principle (Proposition 3 from part I) to the inner integral and then integrating in the remaining variables (over the projection of ).

** 3.2. Van der Corput’s lemma in several variables **

Proceeding by analogy to what was done in Section 2 of part I, we now want to study what happens when we have a lowerbound on some (possibly mixed) derivative of the phase . We look for inequalities that share the same scaling behaviour as the single variable ones in Van der Corput’s lemma (Proposition 4 and Theorem 5 of part I).

We will denote by a multi-index, that is an element of , and by the sum . Then denotes the partial derivative of order

We have the following several-variables version of Van der Corput’s lemma (more precisely, of Corollary 6 from part I).

Theorem 9 (Van der Corput’s lemma in several variables)Let (that is, compactly supported in the unit ball). Assume that for every in the support of the phase satisfiesfor a certain fixed multi-index , with . Furthermore we assume . Then we have

where the constant depends only on and on the phase (in particular, on the derivatives of up to order ).

One can appreciate the superficial similarity between Corollary 6 from part I and Theorem 9. We reiterate that if the hypothesis on the phase becomes then the inequality holds with replaced by .

However, there are several key differences with Corollary 6 of part I which make the above theorem weak:

- unfortunately, the constant now depends on the phase too, which can sometimes be an issue when one has to deal with multiple phases at once;
- as you will see from the proof, the constant also depends on the size of the support of if one removes the extra hypothesis that , and thus the inequality does not have the nice scaling properties that it had in the single variable case;
- the estimate is no longer tight: indeed, if we take a phase like , so that , and an amplitude that is just a bump function supported near the origin, the theorem gives us a bound of , but in reality one can prove the better estimate
^{1}(see Section 3 or Exercise 19).

*Proof:* The proof will again work by reducing to the single variable case (and this is the source of the inefficiencies mentioned above).

Let . In order to reduce to the single variable case, we need to show that the assumption implies a similar lowerbound for some derivative of the form – that is, a -th derivative in a fixed direction, , so that we can apply single variable results. Indeed, the real vector space of partial derivatives of order admits a basis of the above form: there exist *unit vectors * (where ) such that any partial derivative of order can be expressed as a linear combination of derivatives . Prove this in Exercise 16.

Since can be expressed as a linear combination of , for each in the support of there exists a unit vector such that ; moreover, since we have assumed that , we have that is bounded on the support of , and therefore for a small ball (of radius ) centered at we have actually

(with a worse constant than the one implicit in the previous lowerbound). As in the proof of Proposition 8, we can find a finite covering of the support of by such balls and use them to create a partition of unity that allows us to split as with each supported in one of such balls. It suffices to estimate separately ( being supported in , the number of such balls only depends on ). After a change of variables, we can assume that for the ball under examination, and therefore we write again (with )

Applying Corollary 6 from part I to the inner integral we get that it is bounded by

integrating in the remaining variables gives us the estimate we want.

** 3.3. Method of stationary phase in several variables – non-degenerate phases **

When , we can obtain better estimates than the ones given by Theorem 9, provided that we have some extra information about the phase being well-behaved. In particular, if is a critical point for phase , we say that it is a *non-degenerate* critical point if

where is the Hessian matrix .

If the phase has only non-degenerate critical points, we can assert an analogue of the single-variable stationary phase method (Theorem 7 from part I) for phases of several variables.

Theorem 10Assume that and that the phase has as a non-degenerate critical point. If is supported in a sufficiently small neighbourhood of (in particular, there are no other critical points of in it), then there exist coefficients for (each depending only on finitely many derivatives of and at ) such that

where is in the same sense as in Theorem 7 from part I. In particular, .

When , we see that this gives as claimed before.

We do not prove this theorem either (but if you have proven Theorem 7 in Exercise 13 from part I, you can prove this one as well in Exercise 20 following the same strategy with little extra effort), but rather make a remark.

Remark 3The theorem above is limited to the case . Indeed, observe that in the single variable case the fact that and implies that we can put the phase in the simple canonical form (after a change of variable); however, in several variables there is no simple analogue of this, except in the case , in which we can put the phase in a standard quadratic form – hence our limitation.

** 3.4. Estimates independent of the phase **

As observed above, one of the downsides of Van der Corput’s lemma in several variables is that the constant ends up depending on the phase as well, which is sometimes a problem. It is also known that, if we consider the case and thus look at oscillatory integrals of the form with a connected (bounded) open set, the constant cannot be made independent of (prove this in Exercise 18).

It is reasonable to look for estimates with constants that are independent of as many parameters as possible. The above observation forces us to give up independence in , and therefore we concentrate on independence on the phase . It turns out that it is possible to obtain oscillatory integral estimates with constant independent of the phase, but it seems that one has to pay the price somehow (see the log in the estimate below). Such estimates form a rich theory, but here we will only present an example to give you a taste of what kind of results and techniques are involved.

and furthermore

we have

In particular, the constant isindependent of.

Obviously, there is an analogous statement for in place of . Notice that the assumption on this third mixed derivative is nothing more than a monotonicity assumption (which turns out to be necessary).

*Proof:* Assume for convenience. By Cauchy-Schwarz and Fubini, we have

Letting we have that

which by the hypotheses on implies that is monotonic (since is single-signed) and that . By Proposition 4 from part I (the case of Van der Corput) it follows that

Thus the above is bounded by

which gives the claim.

It is not known whether the estimate holds without the logarithmic term (this is actually one of my favourite open problems).

** Exercises **

This is a (long) list of exercises meant to improve your grasp on the topic and provide you with useful ideas for the future – do the ones you like. Those that require a bit more work are marked by ‘s – they are not necessarily harder, just longer maybe. The ones unmarked are probably more important for a basic understanding though. In any case, hints are given in the next section. Since the exercises are meant to be a complement to the lectures, don’t be afraid to have a look at the hints.

Exercise 16Let denote the real vector space of partial derivatives of order , that is the space of differential operators of the form

Show that admits a basis of vectors all of the form , where the ‘s are unit vectors.

Equivalently, you can identify with the real vector space of homogeneous polynomials of degree in variables.

[hint: Using the identification with homogeneous polynomials, it suffices to find an inner product on such that if a polynomial is orthogonal to all polynomials of the form , with , then it has to be the zero polynomial.]

Exercise 17Show again the estimateas in Exercise 7 of part I, but this time using the method of stationary phase in several variables as in Theorem 10. Use the expression for in terms of an integral over the sphere, then express the half-sphere as the graph of a function.

Relatedly, recall that if is a finite measure on we can define its Fourier transform as

Calculate the Fourier transform of the spherical measure on the -dimensional sphere and deduce its decay properties from the above. Oscillatory integrals provide a great way to obtain such estimates.

Finally, consider the flat surface in given by and let be its surface measure. Show that there is a direction in which does not decay at all. This should convince you that the source of the extra cancellation that makes decay is the curvature of the underlying surface.

Exercise 18In this exercise you will show that in dimension greater than it is impossible to have estimates of the formwith constant independent of (here we are assuming ).

- Repeat the proof you gave for i) of Exercise 10 of part I to show that the inequality above would imply
with independent of . Thus it will suffice to disprove the sublevel-set estimates.

- First we show that if is not bounded, there is no hope to have an estimate like the above. Consider and . We have . Show that, however,
so there is an unavoidable dependence on .

- Now we show that even if we take contained in a bounded set (say ), the constant will still depend on properties of . Here the trick will be to consider a set that looks a bit like a comb with teeth. In particular, define
for (this will be the -th “tooth”) and

We define the phase piecewise. Let be a smooth positive function such that on and on . Then on each tooth we prescribe to behave the same way: for , . On the rest of the comb, we prescribe: for , . Show that on and show that for sufficiently smaller than we have

so that the constant necessarily depends on .

Exercise 19() In this exercise you will prove a result that is weaker than the full asymptotics provided in Theorem 10 but which is sufficient for most applications. In particular, under the same hypotheses (in particular, the non-degeneracy of the critical point of ), you will prove the estimateThe proof is quite similar to that of Theorem 11. For starters, let satisfy the hypotheses of Theorem 10. We expand into a double integral in and change variables to by introducing . We obtain

- Verify the calculations above.
- The expression we have is of the form with an oscillatory integral. You will show that for any it is . Show that this will imply that , hence concluding the estimate we want.
- Now you will prove the claimed using essentially the same proof given for Proposition 3 in part I. The oscillatory term of is ; calculate its gradient and use this information to craft a differential operator such that .
- Your differential operator should have the form for some explicit -valued function . Show by simple Taylor expansion arguments that for any multi-index one has
- Now use integration by parts times to show that
and, using iv) above, conclude the estimate ( being the transpose of , which you will have to work out explicitely).

Exercise 20() In this exercise you will prove the method of stationary phase in several variables, that is Theorem 10, following the same strategy used in Exercise 13 of part I (which you should attempt before attempting this one). Assume .

Let , and if is a multi-index let .

- Assume the phase is purely a non-degenerate quadratic form . Take and show that (4) holds, that is
[hint: the integral factorises, then you can repeat what you did in Exercise 13.]

- Now we need to prove a result for a compactly supported . Indeed, let a smooth function compactly supported around the origin and let . You will show that, with the same purely quadratic phase as above, we have
To do so, we will need to use integration by parts at some point with respect to some , and thus it will be convenient to localise in that direction. We achieve this as follows: we roughly divide into the cones

for ; you can see that is a cone with axis the -axis. Show that

Using slightly larger cones

argue by a partition of unity that it suffices to reduce to the case where the amplitude is with a function homogeneous of degree (that is, ), smooth away from the origin, supported in and identically in a conical neighbourhood of the -axis (say, points such that ).

- This step will be the exact analogue of step ii) in Exercise 13. With phase and amplitude , split the region of integration smoothly into and the complement. For the first part use a trivial estimate, and for the second part use repeated integration by parts with respect to the differential operator
Optimizing in , conclude the estimate in ii). [hint: notice that applied to returns exactly .]

- Using the same arguments as above, show that if is identically near the origin, then for any
- For a generic amplitude and phase as above, write
with compactly supported and identically on the support of . Perform a Taylor expansion of to some finite degree, then use i)-iv) above to argue that (5) holds for this oscillatory integral.

- Finally, if is a generic phase satisfying the hypotheses of Theorem 10, argue by Morse’s Lemma that there is a smooth change of variables that lets us put in the form above (provided has small enough support). Therefore, conclude by v). [If you do not know about Morse’s lemma, it says more or less precisely what we claimed; look it up, it is a very convenient tool.]

Exercise 21() Oscillatory integrals are prototype solutions for many partial differential equations:

- Show that the oscillatory integral
is a solution for the

Schr\”{o}dinger equation.- Show that the oscillatory integral
is a solution for the

wave equation.- Show that the oscillatory integral
where (the so-called

japanese bracket) denotes , is a solution for theKlein-Gordon equation( is a constant – the mass).Now consider the wave equation in above. You will prove some decay estimates for a packet of waves of frequency approximately .

Let be a function supported in the annulus and radially symmetric, for simplicity. Fix an integer and consider the solution to the wave equationwhich is frequency localised in the above solution obeys the estimate

The proof will boil down to a number of applications of the non-stationary phase principle, which will reduce matters to estimating the contribution of the critical region of integration where the phase is stationary.

- As a warm-up, give a physical interpretation of (9) when is large. For the sake of the interpretation, replace the fast-decaying term by the characteristic function and draw its support in a space-time diagram. Show that the estimate is compatible with the energy conservation law for waves.
- Show by rescaling that it suffices to prove (9) for the case .
- Show that the symmetries of the problem allow one to assume that and with .
- Now you have an oscillatory integral with phase . Show by using the non-stationary phase principle in that when one has for any . This proves (9) in the small time regime. Notice that the phase is not of the form and therefore one cannot use the non-stationary phase principle (Proposition 3 from part I) out of the box; however, the proof adapts easily.
- Assume then that . We concentrate on proving the localization in . Show, again by non-stationary phase principle in , that if or then one has for any . Show that this proves (9) in this regime.
- Assume therefore that and that . Show that the phase has critical points only if and the set of such critical points is contained in . We expect that the main contribution to the integral defining comes from these critical points; therefore, smoothly excise a thin cone with axis , say , from the support of and show by non-stationary phase principle in that the remaining integral (outside the cone) is dominated by (which is acceptable since it is dominated by (9) in this regime).

- Show by non-stationary phase principle in that one can do away with the portion of the cone where . Indeed, in that region ; show that this contribution is bounded by , which is again acceptable in this regime.
- Finally, we come to the critical region (the portion of the cone where and ). To compensate for this, observe that there is oscillation in the directions as well, and thus you should apply a non-stationary phase argument both in and . Show therefore that this contribution is bounded exactly by (9). [If it helps, consider first the totally critical case . ]
- For the remaining sub-regions, show that for each the quantity is positive and has size (recall ). Use once again the non-stationary phase principle in to estimate the contribution of each region. Finally, sum up in and show that the result is controlled by (9) again.

** Hints **

** Hints for Exercise 16:** Try with inner product , where if then . What does it mean that ?

** Hints for Exercise 17:** Decompose the sphere smoothly in a bounded number of pieces (see Exercise 20 part ii) for an idea how). Do a change of variables: a point is given by where and . Performing the change of variables, one has that is a sum of integrals of the form

where is a smooth bump function supported around the origin and gives the surface measure of in the -coordinates. Justify all of the above, check the Hessian of and apply Theorem 10 directly.

** Hints for Exercise 18:**

- just solve for .
- Show that for close to the function is constant in and periodic in with period . Thus concentrate on a single tooth and show that for these the set of in this tooth such that has measure bounded from below by .

** Hints for Exercise 19:**

- Just perform the integration with large. If you prefer, you can split into dyadic annuli , estimate the integral over each annulus and sum in .
- , so take the inner product of this vector by vector to get the pure exponential factor back.
- From the previous point, is the vector . For the numerator, observe that by hypothesis we can control for any multi-index ; show by the Fundamental Theorem of Calculus that and deduce that . For the denominator, use the hypothesis that has small support to argue that by Taylor expansion is approximately , and use the hypothesis on to argue that this is uniformly in . Finally, combine the above information and the generalized Leibniz’s rule to estimate .
- Show that (that is, the divergence of the -valued function ). Let for simplicity. Show by induction that contains only terms of the form with , and deduce by iv) that is thus controlled by a sum of terms of the form .

** Hints for Exercise 20:**

- The integral factorises into a product of integrals that are exactly like the ones in i) of Exercise 13 of part I.
- The decomposition of into cones follows from a trivial pigeonholing argument. Building the partition of unity is also a standard exercise: simply take non-negative bump functions on the sphere with support contained in and identically on , then define and show this is and satisfies the other conditions.
- See the hints for ii) of Exercise 13 of part I, for this is extremely similar. The trivial estimate near the origin will give you a contribution . Then you will end up having to estimate the integral of , where . You can again appeal to induction to prove that the terms of this expression are of the form , then apply Leibniz’s rule to expand the derivative and conclude that since all the terms contain factors of they are supported in , where crucially it holds . Performing the integration will give a total contribution of , which combined with the trivial estimate and optimised in will give you the desired estimate .
- Again, repeat the second part of the proof you gave for step iii) above, but with .
- See the hints for part iv) of Exercise 13 of part I and follow them closely.
- Morse’s lemma says precisely what we want: if is such that , but (where denotes the Hessian of ), then there exists a neighbourhood of and a -diffeomorphism such that and for all . Moreover, is the signature of .

Use the change of variable provided by the diffeomorphism to turn the phase into a quadratic phase and conclude by appealing to the previous steps.

** Hints for Exercise 21:**

- At time the wave solution is concentrated in a shell of width and radius , and it has amplitude . Multiply the square of the amplitude by the volume occupied by the wave to get an estimate for the energy.
- The gradient of the phase is , so if and one has is bounded from below by . Show that for higher derivatives one has . Apply the integration by parts argument of Proposition 3 from part I (the non-stationary phase principle) to the integral in with respect to the differential operator . Show by induction in that (by now you should know how to proceed: show that consists of terms of the form where …). Finish by integrating in the remaining variables .
- Show that one always has in this regime. Then repeat the analysis done in iv), but be more careful this time: now you can only say (show this) that . In order not to lose the decay, you will have to show by induction that the terms of the form appearing in the expansion of actually satisfy and , thus yielding the correct decay of .
- For the smooth excision of a cone, refer to the hints for parts ii)-iii) of Exercise 20. For the estimate, observe that ignoring the component one has and outside the cone one has . Upon a further smooth partition of unity into cones with axes the coordinate directions , one can reduce to the case where and thus . Then apply the same argument as in v) above with differential operator to conclude.
- Same reasoning as in vi).
- You are allowed to use Taylor expansion of in terms of , both for estimating the difference in the phases and for estimating . The non-stationary phase principle in both (after once again using a smooth partition of unity to reduce to the case where ) will result in having to estimate (for example) . This is quite
**excruciating**, but you can again use an inductive argument to show that the integrand consists entirely of terms of the form where the various parameters involved satisfy certain favourable contraints. Then one can see that , (this because of the restriction to ), and in general for all . Using these facts and the induction one can verify that . The factor of will come from the fact that the region of integration is approximately a cylinder of length and radius . - This part is a repetition of part viii) above essentially, except for the fact that now we can assume instead and consequently one has to update the upper- and lower-bounds on the derivatives of the phase accordingly.

** Footnotes: **

^{1}: For a suggestive calculation: estimate where (hint: rotate coordinates and use single variable Van der Corput’s lemma). [go back]