# Oscillatory integrals II: several-variables phases

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) ${\mathbb{R}}$. 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 ${\mathbb{R}^d}$ (see Exercise 18 for the sort of issues that arise in trying to deal with general open sets of ${\mathbb{R}^d}$), in this section we will study mainly objects of the form

$\displaystyle \mathcal{I}_{\psi}(\lambda) := \int_{\mathbb{R}^d} e^{i \lambda u(x)} \psi(x) dx,$

where ${\psi}$ has compact support. We have switched to ${u}$ 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 ${u}$, we say that ${x_0}$ is a critical point of ${u}$ if

$\displaystyle \nabla u(x_0) = (0,\ldots,0).$

Proposition 8 (Principle of non-stationary phase – several variables) Let ${\psi \in C^\infty_c(\mathbb{R}^d)}$ (that is, smooth and compactly supported) and let the phase ${u\in C^\infty}$ be such that ${u}$ does not have critical points in the support of ${\psi}$. Then for any ${N >0}$ we have

$\displaystyle |\mathcal{I}_\psi(\lambda)|\lesssim_{N,\psi,u} |\lambda|^{-N}.$

Proof: The proof is simply a reduction to the single variable case of Proposition 3 from part I. Indeed, by assumption there exists a ${c>0}$ such that ${|\nabla u(x)|>2c}$ for ${x}$ in the support of ${\psi}$, and therefore for any such ${x}$ there exists a small ball ${B_x}$, centered at ${x}$ and a unit vector ${\xi_x}$ such that ${\xi_x \cdot \nabla u(y) > c}$ for every ${y \in B_x}$. By compactness we can find a finite collection of such balls, ${\{B_j\}_j}$, that covers the support of ${\psi}$; with a partition of unity associated to this collection, we can therefore write ${\psi = \sum_{j} \psi_j}$ with ${\psi_j}$ supported in ${B_j}$. Since the number of balls is finite (depending only on ${\psi, u}$), it suffices to prove the claimed bound for a single ${\mathcal{I}_{\psi_j}(\lambda)}$. By a change of coordinates, we can assume ${\xi_j = (1,0, \ldots, 0)}$ and since

$\displaystyle \int_{\mathbb{R}^d}e^{i\lambda u(x)}\psi_j(x) dx = \int_{\mathbb{R}^{d-1}}\Big(\int_{\mathbb{R}} e^{i\lambda u(x_1, x_2,\ldots,x_d)} \psi_j(x_1, x_2,\ldots,x_d) dx_1\Big)dx_2 \cdots dx_d,$

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 ${B_j}$). $\Box$

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 ${u}$. 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 ${\alpha = (\alpha_1, \ldots, \alpha_d)}$ a multi-index, that is an element of ${\mathbb{N}^d}$, and by ${|\alpha|}$ the sum ${\alpha_1 + \ldots + \alpha_d}$. Then ${\partial^\alpha}$ denotes the partial derivative of order ${|\alpha|}$

$\displaystyle \partial^\alpha = \partial_{x_1}^{\alpha_1} \cdots \partial_{x_d}^{\alpha_d}.$

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 ${\psi \in C^1_c(B(0,1))}$ (that is, compactly supported in the unit ball). Assume that for every ${x}$ in the support of ${\psi}$ the phase ${u}$ satisfies

$\displaystyle |\partial^\alpha u(x)| > 1$

for a certain fixed multi-index ${\alpha}$, with ${|\alpha|\geq 1}$. Furthermore we assume ${u \in C^{|\alpha|+1}}$. Then we have

$\displaystyle |\mathcal{I}_\psi (\lambda)| \leq C_{|\alpha|,u} \big[ \|\psi\|_{L^\infty} + \|\nabla \psi\|_{L^1}\big] \cdot |\lambda|^{-1/|\alpha|},$

where the constant ${C_{|\alpha|,u}}$ depends only on ${|\alpha|}$ and on the phase ${u}$ (in particular, on the derivatives of ${u}$ up to order ${|\alpha|+1}$).

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 ${|\partial^\alpha u(x)|>\mu}$ then the inequality holds with ${|\lambda|^{-1/|\alpha|}}$ replaced by ${(\mu|\lambda|)^{-1/|\alpha|}}$.
However, there are several key differences with Corollary 6 of part I which make the above theorem weak:

1. 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;
2. as you will see from the proof, the constant also depends on the size of the support of ${\psi}$ if one removes the extra hypothesis that ${\psi \in C^1_c(B(0,1))}$, and thus the inequality does not have the nice scaling properties that it had in the single variable case;
3. the estimate is no longer tight: indeed, if we take a phase like ${u(x,y) = xy}$, so that ${|\partial_x \partial_y u| = 1}$, and an amplitude that is just a bump function supported near the origin, the theorem gives us a bound of ${O(|\lambda|^{-1/2})}$, but in reality one can prove the better estimate1 ${O(|\lambda|^{-1})}$ (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 ${k = |\alpha|}$. In order to reduce to the single variable case, we need to show that the assumption ${|\partial^\alpha u|>1}$ implies a similar lowerbound for some derivative of the form ${(\xi \cdot \nabla)^k u}$ – that is, a ${k}$-th derivative in a fixed direction, ${(\xi \cdot \nabla)f(x) = \frac{d}{dt}\big(f(x + t\xi)\big)}$, so that we can apply single variable results. Indeed, the real vector space of partial derivatives of order ${k}$ admits a basis of the above form: there exist unit vectors ${\xi^{(1)}, \ldots, \xi^{(m)}}$ (where ${m = \#\{\beta \in \mathbb{N}^d \text{ s.t. } |\beta|=k\}}$) such that any partial derivative of order ${k}$ can be expressed as a linear combination of derivatives ${(\xi^{(j)}\cdot \nabla)^k}$. Prove this in Exercise 16.
Since ${\partial^\alpha}$ can be expressed as a linear combination of ${(\xi^{(j)}\cdot \nabla)^k}$, for each ${x}$ in the support of ${\psi}$ there exists a unit vector ${\xi_x}$ such that ${|(\xi_x \cdot \nabla)^k u(x)| \gtrsim_k 1}$; moreover, since we have assumed that ${u\in C^{k+1}}$, we have that ${\|u\|_{C^{k+1}}}$ is bounded on the support of ${\psi}$, and therefore for a small ball ${B_x}$ (of radius ${\lesssim \|u\|_{C^{k+1}}^{-1}}$) centered at ${x}$ we have actually

$\displaystyle |(\xi_x \cdot \nabla)^k u(y)| \gtrsim_k 1 \qquad \text{ for all } y \in B_x$

(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 ${\psi}$ by such balls and use them to create a partition of unity that allows us to split ${\psi}$ as ${\sum_{j} \psi_j}$ with each ${\psi_j}$ supported in one of such balls. It suffices to estimate ${\mathcal{I}_{\psi_j}(\lambda)}$ separately (${\psi}$ being supported in ${B(0,1)}$, the number of such balls only depends on ${\|u\|_{C^{k+1}}}$). After a change of variables, we can assume that ${\xi = (1,0,\ldots,0)}$ for the ball under examination, and therefore we write again (with ${x' = (x_2,\ldots,x_d)}$)

$\displaystyle \int_{B(0,1)}e^{i\lambda u(x)}\psi_j(x) dx = \int_{\mathbb{R}^{d-1}\cap B(0,1)}\Big(\int e^{i\lambda u(x_1, x')} \psi_j(x_1, x') dx_1\Big)dx'.$

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

$\displaystyle \lesssim_k \Big[\|\psi\|_{L^\infty} + \int |\partial_{x_1}\psi(x_1,x')|dx_1\Big] \cdot |\lambda|^{-1/k};$

integrating in the remaining variables gives us the estimate we want. $\Box$

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

When ${|\alpha|=2}$, 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 ${x_0}$ is a critical point for phase ${u}$, we say that it is a non-degenerate critical point if

$\displaystyle \det (\mathrm{Hess}(u)(x_0)) \neq 0,$

where ${\mathrm{Hess}(u)}$ is the Hessian matrix ${\Big(\frac{\partial^2 u}{\partial x_i \partial x_j}\Big)_{ij}}$.
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 10 Assume that ${\psi \in C^{\infty}_c(\mathbb{R}^d)}$ and that the phase ${u \in C^\infty(\mathbb{R}^d)}$ has ${x_0}$ as a non-degenerate critical point. If ${\psi}$ is supported in a sufficiently small neighbourhood of ${x_0}$ (in particular, there are no other critical points of ${u}$ in it), then there exist coefficients ${a_j}$ for ${j \in \mathbb{N}}$ (each depending only on finitely many derivatives of ${u}$ and ${\psi}$ at ${x_0}$) such that

$\displaystyle \mathcal{I}_\psi(\lambda) \simeq e^{i \lambda u(x_0)} \lambda^{-d/2} \sum_{j \in \mathbb{N}} a_j \lambda^{-j}, \ \ \ \ \ (5)$

where ${\simeq}$ is in the same sense as in Theorem 7 from part I. In particular, ${|\mathcal{I}_\psi(\lambda)|\lesssim_{\phi,\psi} |\lambda|^{-d/2}}$.

When ${u(x,y) = xy}$, we see that this gives ${I_{\psi}(\lambda)= O(\lambda^{-1})}$ 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 3 The theorem above is limited to the case ${|\alpha|=2}$. Indeed, observe that in the single variable case the fact that ${\phi^{(k)}(t_0)\neq 0}$ and ${\phi(t_0)=\phi'(t_0)=\ldots=\phi^{(k-1)}(t_0)=0}$ implies that we can put the phase in the simple canonical form ${\phi(t) = t^k}$ (after a change of variable); however, in several variables there is no simple analogue of this, except in the case ${|\alpha|=2}$, in which we can put the phase ${u}$ 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 ${u}$ as well, which is sometimes a problem. It is also known that, if we consider the case ${\psi \equiv 1}$ and thus look at oscillatory integrals of the form ${\int_{\Omega} e^{i\lambda u(x)} dx}$ with ${\Omega}$ a connected (bounded) open set, the constant cannot be made independent of ${\Omega}$ (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 ${\Omega}$, and therefore we concentrate on independence on the phase ${u}$. 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.

Theorem 11 If ${u : [0,1]^2 \rightarrow \mathbb{R}}$ is such that

$\displaystyle \Big|\frac{\partial^2 u}{\partial x \partial y}(x,y)\Big|>1 \qquad \forall (x,y) \in [0,1]^2$

and furthermore

$\displaystyle \frac{\partial^3 u}{\partial^2 x \partial y}(x,y) \neq 0 \qquad \forall (x,y) \in [0,1]^2,$

we have

$\displaystyle \Big|\int_{[0,1]^2} e^{i \lambda u(x,y)}dx dy\Big|\lesssim (\log(2 + |\lambda|))^{1/2} |\lambda|^{-1/2}.$

In particular, the constant is independent of ${u}$.

Obviously, there is an analogous statement for ${\partial_x \partial_y^2 u}$ in place of ${\partial_x^2 \partial_y u}$. Notice that the assumption on this third mixed derivative is nothing more than a monotonicity assumption (which turns out to be necessary).

Proof: Assume ${\lambda>1}$ for convenience. By Cauchy-Schwarz and Fubini, we have

\displaystyle \begin{aligned} \Big|\int_{[0,1]^2} e^{i \lambda u(x,y)}dx dy\Big|^2 & \leq \int_{0}^{1} \Big|\int_{0}^{1} e^{i \lambda u(x,y)}dy\Big|^2 dx \cdot \int_{0}^{1} 1 dx \\ & = \int_{0}^{1} \int_{0}^{1} \Big(\int_{0}^{1} e^{i \lambda (u(x,y)-u(x,y'))}dx \Big) dy dy'. \end{aligned}

Letting ${\phi_{y,y'}(x):= u(x,y) - u(x,y')}$ we have that

$\displaystyle \phi_{y,y'}'(x) = \partial_x u(x,y) - \partial_x u(x,y') = \int_{y'}^{y} \partial_x \partial_y u(x,t) dt,$

which by the hypotheses on ${u}$ implies that ${\phi_{y,y'}'}$ is monotonic (since ${\phi_{y,y'}''}$ is single-signed) and that ${|\phi_{y,y'}'(x)| > |y - y'|}$. By Proposition 4 from part I (the $k=1$ case of Van der Corput) it follows that

$\displaystyle \Big|\int_{0}^{1} e^{i \lambda (u(x,y)-u(x,y'))}dx\Big| \lesssim \min\{1,(\lambda|y-y'|)^{-1}\}.$

Thus the above is bounded by

\displaystyle \begin{aligned} \int_{0}^{1} \int_{0}^{1} & \min\{1,(\lambda|y-y'|)^{-1}\} dy dy' \\ & = \lambda^{-1} \iint_{|y-y'|>\lambda^{-1}} |y-y'|^{-1} dy dy' + \iint_{|y-y'|<\lambda^{-1}} 1 dy dy' \lesssim \lambda^{-1} \log \lambda + \lambda^{-1}, \end{aligned}

which gives the claim. $\Box$

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 ${\bigstar}$‘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 16 Let ${V}$ denote the real vector space of partial derivatives of order ${k}$, that is the space of differential operators of the form

$\displaystyle \sum_{\alpha \;: |\alpha|=k} c_\alpha \partial^\alpha.$

Show that ${V}$ admits a basis of vectors all of the form ${(\xi\cdot\nabla)^k}$, where the ${\xi}$‘s are unit vectors.
Equivalently, you can identify ${V}$ with the real vector space of homogeneous polynomials of degree ${k}$ in ${d}$ variables.
[hint: Using the identification with homogeneous polynomials, it suffices to find an inner product on ${V}$ such that if a polynomial is orthogonal to all polynomials of the form ${(\xi\cdot X)^k}$, with ${X = (X_1,\ldots, X_d)}$, then it has to be the zero polynomial.]

Exercise 17 Show again the estimate

$\displaystyle |J(t)| \lesssim_d (1 + |t|)^{-(d-1)/2}$

as 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 ${J}$ in terms of an integral over the sphere, then express the half-sphere as the graph of a function.
Relatedly, recall that if ${\mu}$ is a finite measure on ${\mathbb{R}^d}$ we can define its Fourier transform ${\widehat{d\mu}}$ as

$\displaystyle \widehat{d\mu}(\xi) := \int_{\mathbb{R}^d} e^{-2\pi i x \cdot \xi} d\mu(x).$

Calculate the Fourier transform ${\widehat{d\sigma}(\xi)}$ of the spherical measure ${d\sigma}$ on the ${(d-1)}$-dimensional sphere ${\mathbb{S}^{d-1}}$ and deduce its decay properties from the above. Oscillatory integrals provide a great way to obtain such estimates.
Finally, consider the flat surface in ${\mathbb{R}^d}$ given by ${[-1,1]^{d-1} \times \{0\}}$ and let ${d\mu}$ be its surface measure. Show that there is a direction in which ${\widehat{d\mu}}$ does not decay at all. This should convince you that the source of the extra cancellation that makes ${\widehat{d\sigma}}$ decay is the curvature of the underlying surface.

Exercise 18 In this exercise you will show that in dimension greater than ${1}$ it is impossible to have estimates of the form

$\displaystyle \Big|\int_{\Omega} e^{i \lambda u(x)}dx\Big| \leq C |\lambda|^{-1/|\alpha|}$

with constant ${C}$ independent of ${\Omega}$ (here we are assuming ${|\partial^{\alpha}u|>1}$).

1. Repeat the proof you gave for i) of Exercise 10 of part I to show that the inequality above would imply

$\displaystyle |\{ x \in \Omega \text{ s.t. } |u(x)| < \lambda \}| \leq C' \lambda^{1/|\alpha|}$

with ${C'}$ independent of ${\Omega}$. Thus it will suffice to disprove the sublevel-set estimates.

2. First we show that if ${\Omega}$ is not bounded, there is no hope to have an estimate like the above. Consider ${\Omega = [-R,R]^2}$ and ${u(x,y) = \frac{1}{2}(x+y)^2}$. We have ${\partial_x \partial_y u = 1}$. Show that, however,

$\displaystyle |\{ (x,y) \in \Omega \text{ s.t. } |u(x,y)| < \lambda \}| \sim R \lambda^{1/2},$

so there is an unavoidable dependence on ${\mathrm{diam}(\Omega)}$.

3. Now we show that even if we take ${\Omega}$ contained in a bounded set (say ${[0,1]^2}$), the constant will still depend on properties of ${\Omega}$. Here the trick will be to consider a set ${\Omega}$ that looks a bit like a comb with ${N}$ teeth. In particular, define

$\displaystyle \Omega_j := [1/2,1] \times [j/N, (j + 1/2)/N]$

for ${j= 0, \ldots, N-1}$ (this will be the ${j}$-th “tooth”) and

$\displaystyle \Omega := [0,1/2]\times [0,1] \cup \bigcup_{j=0}^{N-1} \Omega_j.$

We define the phase ${u}$ piecewise. Let ${\varphi}$ be a smooth positive function such that ${\varphi \equiv 0}$ on ${[0,5/8]}$ and ${\varphi \equiv 1}$ on ${[3/4,1]}$. Then on each tooth we prescribe ${u}$ to behave the same way: for ${(x,y) \in \Omega_j}$, ${u(x,y) = y - j\varphi(x)/N}$. On the rest of the comb, we prescribe: for ${(x,y) \in [0,1/2]\times [0,1]}$, ${u(x,y) = y}$. Show that ${|\partial_y u| = 1}$ on ${\Omega}$ and show that for ${\lambda}$ sufficiently smaller than ${N^{-1}}$ we have

$\displaystyle |\{(x,y) \in \Omega \text{ s.t. } |u(x,y)|<\lambda\}| \gtrsim N\lambda,$

so that the constant necessarily depends on ${\Omega}$.

Exercise 19 (${\bigstar}$) 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 ${u}$), you will prove the estimate

$\displaystyle |\mathcal{I}_{\psi}(\lambda)| \lesssim_{u,\psi,d} |\lambda|^{-d/2}.$

The proof is quite similar to that of Theorem 11. For starters, let ${u,\psi}$ satisfy the hypotheses of Theorem 10. We expand ${|\mathcal{I}_\psi(\lambda)|^2}$ into a double integral in ${dx dy}$ and change variables to ${(y,z)}$ by introducing ${z = x-y}$. We obtain

$\displaystyle |\mathcal{I}_\psi(\lambda)|^2 = \int_{\mathbb{R}^d}\int_{\mathbb{R}^d} e^{i\lambda (u(y+z) - u(y))} \psi(y+z)\psi(y) dy dz.$

1. Verify the calculations above.
2. The expression we have is of the form ${\int_{\mathbb{R}^d} \mathcal{J}_\lambda(z) dz}$ with ${\mathcal{J}_\lambda}$ an oscillatory integral. You will show that for any ${N>0}$ it is ${|\mathcal{J}_\lambda(z)| \lesssim_{N} (1+\lambda|z|)^{-N}}$. Show that this will imply that ${|\mathcal{I}_\psi(\lambda)|^2\lesssim |\lambda|^{-d}}$, hence concluding the estimate we want.
3. Now you will prove the claimed ${|\mathcal{J}_\lambda(z)| \lesssim_{N} (1+\lambda|z|)^{-N}}$ using essentially the same proof given for Proposition 3 in part I. The oscillatory term of ${\mathcal{J}_\lambda (z)}$ is ${e^{i\lambda (u(y+z) - u(y))}}$; calculate its ${\nabla_y}$ gradient and use this information to craft a differential operator ${\mathsf{D}}$ such that ${\mathsf{D}(e^{i\lambda (u(y+z) - u(y))}) = e^{i\lambda (u(y+z) - u(y))}}$.
4. Your differential operator should have the form ${\mathsf{D} = (i\lambda)^{-1} \theta_z \cdot \nabla_y}$ for some explicit ${\mathbb{R}^d}$-valued function ${\theta_z(y)}$. Show by simple Taylor expansion arguments that for any multi-index ${\alpha}$ one has

$\displaystyle \|\partial^\alpha_y \theta_z (y)\| \lesssim_\alpha |z|^{-1}.$

5. Now use integration by parts ${N}$ times to show that

$\displaystyle \mathcal{J}_\lambda (z) = \int e^{i\lambda (u(y+z) - u(y))} (\mathsf{D}^{\intercal})^N (\psi(y+z)\psi(y))dy$

and, using iv) above, conclude the estimate (${\mathsf{D}^{\intercal}}$ being the transpose of ${\mathsf{D}}$, which you will have to work out explicitely).

Exercise 20 (${\bigstar \bigstar}$) 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 ${x_0 = 0}$.
Let ${x=(x_1, \ldots, x_d)}$, and if ${\alpha \in \mathbb{N}^d}$ is a multi-index let ${x^{\alpha} := x_1^{\alpha_2} \cdot \ldots \cdot x_d^{\alpha_d}}$.

1. Assume the phase is purely a non-degenerate quadratic form ${Q(x) = \sum_{j=1}^{\ell} x_j^2 - \sum_{j=\ell+1}^{d} x_j^2}$. Take ${\psi(x) = e^{-\|x\|^2} x^{\alpha}}$ and show that (4) holds, that is

$\displaystyle \int_{\mathbb{R}^d} e^{i \lambda Q(x)} e^{-\|x\|^2} x^\alpha dx \simeq \lambda^{-d/2 - |\alpha|/2} \sum_{j \in \mathbb{N}} a_j^{\ell,\alpha} \lambda^{-j}.$

[hint: the integral factorises, then you can repeat what you did in Exercise 13.]

2. Now we need to prove a result for a compactly supported ${\psi}$. Indeed, let ${\eta(x)}$ a smooth function compactly supported around the origin and let ${\psi(x)=x^{\alpha} \eta(x)}$. You will show that, with the same purely quadratic phase ${Q}$ as above, we have

$\displaystyle |\mathcal{I}_\psi(\lambda)| \lesssim_{\alpha,\eta,Q} |\lambda|^{- (d + |\alpha|)/2}.$

To do so, we will need to use integration by parts at some point with respect to some ${\partial_{x_k}}$, and thus it will be convenient to localise in that direction. We achieve this as follows: we roughly divide ${\mathbb{R}^d}$ into the cones

$\displaystyle \Gamma_k := \{ x \in \mathbb{R}^d \text{ s.t. } \|x\|^2/d \leq |x_k|^2 \}$

for ${k \in \{1,\ldots,d\}}$; you can see that ${\Gamma_k}$ is a cone with axis the ${x_k}$-axis. Show that

$\displaystyle \mathbb{R}^d = \bigcup_{k=1}^{d} \Gamma_k.$

Using slightly larger cones

$\displaystyle \widetilde{\Gamma}_k := \{ x \in \mathbb{R}^d \text{ s.t. } \|x\|^2/(2d) \leq |x_k|^2 \},$

argue by a partition of unity that it suffices to reduce to the case where the amplitude is ${\psi(x)\cdot G_k(x)}$ with ${G_k}$ a function homogeneous of degree ${0}$ (that is, ${G_k(x) = G_k(x / \|x\|)}$), smooth away from the origin, supported in ${\widetilde{\Gamma}_k}$ and identically ${1}$ in a conical neighbourhood of the ${x_k}$-axis (say, points ${x}$ such that ${(9/10)\|x\|^2 \leq |x_k|^2}$).

3. This step will be the exact analogue of step ii) in Exercise 13. With phase ${Q}$ and amplitude ${x^\alpha \eta(x) G_k(x)}$, split the region of integration smoothly into ${\|x\|\lesssim \delta}$ 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

$\displaystyle D_k f(x) := \pm \frac{1}{2 i \lambda x_k} \partial_{x_k}.$

Optimizing in ${\delta}$, conclude the estimate in ii). [hint: notice that ${D_k}$ applied to ${e^{i\lambda Q}}$ returns exactly ${e^{i\lambda Q}}$.]

4. Using the same arguments as above, show that if ${g \in \mathcal{S}(\mathbb{R}^d)}$ is identically ${0}$ near the origin, then for any ${N>0}$

$\displaystyle |\mathcal{I}_g(\lambda)|\lesssim_{N,g} |\lambda|^{-N}.$

5. For a generic amplitude ${\psi}$ and phase ${Q}$ as above, write

$\displaystyle \int e^{i\lambda Q(x)} \psi(x) dx = \int e^{i\lambda Q(x)} e^{-\|x\|^2} \big(e^{\|x\|^2} \psi(x)\big) \eta(x) dx$

with ${\eta}$ compactly supported and identically ${1}$ on the support of ${\psi}$. Perform a Taylor expansion of ${e^{\|x\|^2}\psi(x)}$ to some finite degree, then use i)-iv) above to argue that (5) holds for this oscillatory integral.

6. Finally, if ${u}$ 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 ${u}$ in the form ${Q}$ above (provided ${\psi}$ 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 (${\bigstar\bigstar\bigstar}$) Oscillatory integrals are prototype solutions for many partial differential equations:

• Show that the oscillatory integral

$\displaystyle u(x,t):= \int e^{i(t|\xi|^2 + x\cdot \xi)} \psi(\xi) d\xi$

is a solution for the Schr\”{o}dinger equation ${i \partial_t u - \Delta u = 0}$.

• Show that the oscillatory integral

$\displaystyle u(x,t):= \int e^{i(t|\xi| + x\cdot \xi)} \psi(\xi) d\xi$

is a solution for the wave equation ${\partial^2_t u - \Delta u = 0}$.

• Show that the oscillatory integral

$\displaystyle u(x,t):= \int e^{i(t m \langle\xi/m\rangle + x\cdot \xi)} \psi(\xi) d\xi,$

where ${\langle \xi \rangle}$ (the so-called japanese bracket) denotes ${(1 + |\xi|^2)^{1/2}}$, is a solution for the Klein-Gordon equation ${\partial^2_t u - \Delta u + m^2 u = 0}$ (${m>0}$ is a constant – the mass).

Now consider the wave equation in ${\mathbb{R}^d}$ above. You will prove some decay estimates for a packet of waves of frequency approximately ${2^k}$.
Let ${\psi(\xi)}$ be a ${C^\infty}$ function supported in the annulus ${1<|\xi|<2}$ and radially symmetric, for simplicity. Fix an integer ${k\in \mathbb{Z}}$ and consider the solution to the wave equation

$\displaystyle \Phi_k(x,t) := \int e^{i(x\cdot \xi + t|\xi|)} \psi\Big(\frac{\xi}{2^k}\Big) d\xi,$

which is frequency localised in ${2^k < |\xi| 0}$ the above solution obeys the estimate

$\displaystyle |\Phi_k(x,t)| \lesssim_N 2^{kd} (1 + 2^k||x|-|t||)^{-N} (1 + 2^k|t|)^{-(d-1)/2}. \ \ \ \ \ (9)$

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 ${t|\xi| + x\cdot \xi}$ is stationary.

1. As a warm-up, give a physical interpretation of (9) when ${t}$ is large. For the sake of the interpretation, replace the fast-decaying term ${(1 + 2^k||x|-|t||)^{-N}}$ by the characteristic function ${\mathbf{1}_{[-1,1]}(2^k||x|-|t||)}$ and draw its support in a space-time diagram. Show that the estimate is compatible with the energy conservation law for waves.
2. Show by rescaling that it suffices to prove (9) for the case ${k=0}$.
3. Show that the symmetries of the problem allow one to assume that ${t>0}$ and ${x = (x_1, 0, \ldots, 0)}$ with ${x_1 \geq 0}$.
4. Now you have an oscillatory integral with phase ${\phi(\xi) = t |\xi| + x_1 \xi_1}$. Show by using the non-stationary phase principle in ${\xi_1}$ that when ${t\lesssim 1}$ one has ${|\Phi_0(x_1,t)| \lesssim_N (1 + x_1)^{-N}}$ for any ${N>0}$. This proves (9) in the small time regime. Notice that the phase is not of the form ${\lambda \phi}$ and therefore one cannot use the non-stationary phase principle (Proposition 3 from part I) out of the box; however, the proof adapts easily.
5. Assume then that ${t \gtrsim 1}$. We concentrate on proving the localization in ${|x|-|t|}$. Show, again by non-stationary phase principle in ${\xi_1}$, that if ${t > 4x_1}$ or ${\frac{1}{4} x_1 > t}$ then one has ${|\Phi_0(x_1,t)| \lesssim_N (1 + |x_1 - t|)^{-N}}$ for any ${N>0}$. Show that this proves (9) in this regime.
6. Assume therefore that ${t \gtrsim 1}$ and that ${4x_1> t > \frac{1}{4} x_1}$. Show that the phase ${\phi(\xi) = t |\xi| + x_1 \xi_1}$ has critical points only if ${t=x_1}$ and the set of such critical points is contained in ${(-\infty,0] \mathbf{e}_1}$. We expect that the main contribution to the integral defining ${\Phi_0}$ comes from these critical points; therefore, smoothly excise a thin cone with axis ${\mathbb{R} \mathbf{e}_1}$, say ${\Gamma := \{\xi = (\xi_1,\xi') \in \mathbb{R} \times \mathbb{R}^{d-1} \text{ s.t. } |\xi_1| > (99/100) |\xi| \}}$, from the support of ${\psi}$ and show by non-stationary phase principle in ${\xi'}$ that the remaining integral (outside the cone) is dominated by ${O_N((1 + t)^{-N})}$ (which is acceptable since it is dominated by (9) in this regime).

7. Show by non-stationary phase principle in ${\xi_1}$ that one can do away with the portion of the cone ${\Gamma}$ where ${\xi_1 > 0}$. Indeed, in that region ${|\nabla \phi| \gtrsim x_1}$; show that this contribution is bounded by ${O_N((1 + x_1)^{-N})}$, which is again acceptable in this regime.
8. Finally, we come to the critical region (the portion of the cone where ${|\xi|\sim 1}$ and ${\xi_1 2}$). To compensate for this, observe that there is oscillation in the ${\xi'}$ directions as well, and thus you should apply a non-stationary phase argument both in ${\xi_1}$ and ${\xi'}$. Show therefore that this contribution is bounded exactly by (9). [If it helps, consider first the totally critical case ${x_1 = t}$. ]
9. For the remaining sub-regions, show that for each ${j}$ the quantity ${\partial_{\xi_1} \phi(\xi) - (x_1 - t)}$ is positive and has size ${\sim 2^{2j}}$ (recall ${2^{2j} \lesssim t}$). Use once again the non-stationary phase principle in ${\xi_1, \xi'}$ to estimate the contribution of each region. Finally, sum up in ${j}$ and show that the result is controlled by (9) again.

Hints

Hints for Exercise 16: Try with inner product ${\langle P, Q\rangle := [P(\nabla)]Q}$, where if ${P(X) = \sum_{\alpha : |\alpha|=k} a_\alpha X^\alpha}$ then ${P(\nabla) = \sum_{\alpha : |\alpha|=k} a_\alpha \partial^\alpha}$. What does it mean that ${\langle (\xi\cdot X)^k, Q\rangle = 0}$?

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 ${\omega \in \mathbb{S}^{d-1}}$ is given by ${\omega=(\underline{x},\phi(\underline{x}))}$ where ${\underline{x} \in \mathbb{R}^{d-1}}$ and ${\phi(\underline{x}) = \big(1 - x_1^2 - \ldots - x_{d-1}^2)^{1/2}}$. Performing the change of variables, one has that ${\int_{\mathbb{S}^{d-1}} \exp(-2\pi i t \omega_1) d\sigma(\omega)}$ is a sum of integrals of the form

$\displaystyle \int_{\mathbb{R}^{d-1}} e^{-2\pi i t \phi(\underline{x})} S(\underline{x})\psi(\underline{x}) d\underline{x},$

where ${\psi}$ is a smooth bump function supported around the origin and ${S(\underline{x}) d\underline{x}}$ gives the surface measure of ${\mathbb{S}^{d-1}}$ in the ${\underline{x}}$-coordinates. Justify all of the above, check the Hessian of ${\phi}$ and apply Theorem 10 directly.

Hints for Exercise 18:

1. just solve ${|u(x,y)| < \lambda}$ for ${x,y}$.
2. Show that for ${x}$ close to ${1}$ the function ${u(x,y)}$ is constant in ${x}$ and periodic in ${y}$ with period ${1/N}$. Thus concentrate on a single tooth and show that for these ${x}$ the set of ${y}$ in this tooth such that ${|u(x,y)|<\lambda}$ has measure bounded from below by ${\gtrsim \lambda}$.

Hints for Exercise 19:

1. Just perform the integration ${\int_{\mathbb{R}^d} (1 + \lambda |z|)^{-N}dz}$ with ${N}$ large. If you prefer, you can split ${\mathbb{R}^d}$ into dyadic annuli ${\{z \in \mathbb{R}^d \; : \; 2^{j}<|z|\leq 2^{j+1}\}}$, estimate the integral over each annulus and sum in ${j \in \mathbb{Z}}$.
2. ${\nabla_y e^{i\lambda(u(y+z) - u(y))} = i \lambda e^{i\lambda(u(y+z) - u(y))} \nabla_y(u(y+z) - u(y))}$, so take the inner product of this vector by vector ${(i\lambda)^{-1}\frac{\nabla_y(u(y+z) - u(y))}{|\nabla_y(u(y+z) - u(y))|^2}}$ to get the pure exponential factor back.
3. From the previous point, ${\theta_z(y)}$ is the vector ${\frac{\nabla_y(u(y+z) - u(y))}{|\nabla_y(u(y+z) - u(y))|^2}}$. For the numerator, observe that by hypothesis we can control ${|\partial_y^{\beta}u|\lesssim_{\beta,u,\psi} 1}$ for any multi-index ${\beta}$; show by the Fundamental Theorem of Calculus that ${\partial_y^\alpha (u(y+z) - u(y)) = \int_{0}^{1} \nabla_y(\partial_y^\alpha u(y+tz))\cdot z dt}$ and deduce that ${|\partial_y^\alpha (u(y+z) - u(y))|\lesssim_{\alpha,u,\psi} |z|}$. For the denominator, use the hypothesis that ${\psi}$ has small support to argue that by Taylor expansion ${\nabla_y(u(y+z) - u(y))}$ is approximately ${[\mathrm{Hess}(u)](y) z}$, and use the hypothesis on ${\mathrm{Hess}(u)}$ to argue that this is ${\gtrsim_{u,\psi} |z|}$ uniformly in ${y}$. Finally, combine the above information and the generalized Leibniz’s rule to estimate ${\partial_y^{\alpha} \theta_z}$.
4. Show that ${\mathsf{D}^{\intercal}f = (i\lambda)^{-1} \nabla \cdot (f \theta_z)}$ (that is, the divergence of the ${\mathbb{R}^d}$-valued function ${f\theta_z}$). Let ${\Psi(y,z) := \psi(y+z)\psi(y)}$ for simplicity. Show by induction that ${(\mathsf{D}^{\intercal})^N \Psi}$ contains only terms of the form ${\partial_{j_0}^{\alpha_0}\Psi \cdot \partial_{j_1}^{\alpha_1}\theta_z \cdot \ldots \cdot \partial_{j_N}^{\alpha_N}\theta_z}$ with ${\alpha_0 + \alpha_1 + \ldots + \alpha_N = N}$, and deduce by iv) that ${|(\mathsf{D}^{\intercal})^N \Psi|}$ is thus controlled by a sum of terms of the form ${|z|^{-N} |\partial_{j_0}^{\alpha_0}\Psi|}$.

Hints for Exercise 20:

1. The integral factorises into a product of integrals that are exactly like the ones in i) of Exercise 13 of part I.
2. The decomposition of ${\mathbb{R}^d}$ into cones follows from a trivial pigeonholing argument. Building the partition of unity ${1 = \sum_{k=1}^{d} G_k(x)}$ is also a standard exercise: simply take non-negative bump functions ${\rho_k}$ on the sphere ${\mathbb{S}^{d-1}}$ with support contained in ${\tilde{\Gamma}_k \cap \mathbb{S}^{d-1}}$ and identically ${1}$ on ${\Gamma_k}$, then define ${G_k (x) := \rho_k(x/\|x\|) \big(\sum_{j=1}^{d} \rho_j(x/\|x\|)\big)^{-1}}$ and show this is ${C^\infty}$ and satisfies the other conditions.
3. 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 ${\int |x^{\alpha}| \phi(x/\delta) dx \lesssim \delta^{d + |\alpha|}}$. Then you will end up having to estimate the integral of ${(D_k^{\intercal})^N(x^\alpha \omega_\delta (x) G_k(x))}$, where ${\omega_\delta(x)= \eta(x) (1 - \varphi(x/\delta))}$. You can again appeal to induction to prove that the terms of this expression are of the form ${x_k^{-(2N - j)}\partial_k^j(x^\alpha \omega_\delta (x) G_k(x))}$, then apply Leibniz’s rule to expand the ${\partial_k^j}$ derivative and conclude that since all the terms contain factors of ${\partial_k^\ell G_k}$ they are supported in ${\tilde{\Gamma}_k}$, where crucially it holds ${|x_k| \sim \|x\|}$. Performing the integration will give a total contribution of ${O(\lambda^{-N} \delta^{-(2N - |\alpha|)+d} )}$, which combined with the trivial estimate and optimised in ${\delta}$ will give you the desired estimate ${|\mathcal{I}_{x^\alpha \eta}(\lambda)| \lesssim |\lambda|^{-(d + |\alpha|)/2}}$.
4. Again, repeat the second part of the proof you gave for step iii) above, but with ${\delta \sim 1}$.
5. See the hints for part iv) of Exercise 13 of part I and follow them closely.
6. Morse’s lemma says precisely what we want: if ${f \in C^\infty(\mathbb{R}^d)}$ is such that ${f(0) = 0}$, ${\nabla f (0)= 0}$ but ${\det (\mathrm{Hess}(f)) \neq 0}$ (where ${\mathrm{Hess}(f)}$ denotes the Hessian of ${f}$), then there exists a neighbourhood ${U}$ of ${0}$ and a ${C^\infty}$-diffeomorphism ${\Phi : \mathbb{R}^d \rightarrow \mathbb{R}^d}$ such that ${\Phi(0) = 0}$ and ${f \circ \Phi^{-1} (y) = x_1^2 + \ldots + x_k^2 - x_{k+1}^2 - \ldots - x_d^2}$ for all ${y \in \Phi(U)}$. Moreover, ${(k,d-k)}$ is the signature of ${\mathrm{Hess}(f)}$.
Use the change of variable provided by the diffeomorphism ${\Phi}$ to turn the phase ${u(x)}$ into a quadratic phase and conclude by appealing to the previous steps.

Hints for Exercise 21:

1. At time ${t}$ the wave solution is concentrated in a shell of width ${\sim 1}$ and radius ${\sim t}$, and it has amplitude ${\sim t^{-(d-1)/2}}$. Multiply the square of the amplitude by the volume occupied by the wave to get an estimate for the energy.
2. The gradient of the phase is ${\nabla \phi(\xi) = t\frac{\xi}{|\xi|} + x_1 \mathbf{e}_1}$, so if ${t \lesssim 1}$ and ${x_1 \gtrsim 1}$ one has ${\partial_{\xi_1} \phi}$ is bounded from below by ${\gtrsim x_1}$. Show that for higher derivatives one has ${|\partial_{\xi_1}^k \phi| \lesssim 1}$. Apply the integration by parts argument of Proposition 3 from part I (the non-stationary phase principle) to the integral in ${\xi_1}$ with respect to the differential operator ${D_1 = (i \partial_{\xi_1}\phi(\xi))^{-1} \partial_{\xi_1}}$. Show by induction in ${N}$ that ${|(D_1^{\intercal})^N \psi| \lesssim_{N,\psi} x_1^{-N}}$ (by now you should know how to proceed: show that ${(D_1^{\intercal})^N \psi}$ consists of terms of the form ${(\partial_{\xi_1} \phi)^{-\ell}\partial_{\xi_1}^{\beta_1} \psi \prod_{k=2}^{m}(\partial_{\xi_1}^{k}\phi)^{\beta_k}}$ where ${\ell\geq N}$…). Finish by integrating in the remaining variables ${\xi_2, \ldots, \xi_d}$.
3. Show that one always has ${|\partial_{\xi_1} \phi(\xi)|\gtrsim |x_1 - t|\gtrsim t}$ in this regime. Then repeat the analysis done in iv), but be more careful this time: now you can only say (show this) that ${|\partial_{\xi_1}^k \phi| \lesssim t}$. In order not to lose the decay, you will have to show by induction that the terms of the form ${(\partial_{\xi_1} \phi)^{-\ell}\partial_{\xi_1}^{\beta_1} \psi \prod_{k=2}^{m}(\partial_{\xi_1}^{k}\phi)^{\beta_k}}$ appearing in the expansion of ${(D_1^{\intercal})^N \psi}$ actually satisfy ${\ell - \sum_{k=1}^{m} k \beta_k = 0}$ and ${\beta_1 + \sum_{k=2}^{m} (k-1) \beta_k = N}$, thus yielding the correct decay of ${|x_1 - t|^{-N}}$.
4. 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 ${\xi_1}$ component one has ${|\nabla \phi(\xi)| \gtrsim t |\xi'|/|\xi|}$ and outside the cone one has ${|\xi'| \gtrsim |\xi|}$. Upon a further smooth partition of unity into cones with axes the coordinate directions ${\xi_2, \ldots, \xi_d}$, one can reduce to the case where ${|\xi_j| \gtrsim |\xi|}$ and thus ${|\partial_{\xi_j} \phi(\xi)| \gtrsim t}$. Then apply the same argument as in v) above with differential operator ${D_j = (i \partial_{\xi_j}\phi(\xi))^{-1} \partial_{\xi_j}}$ to conclude.
5. Same reasoning as in vi).
6. You are allowed to use Taylor expansion of ${|\xi|= \xi_1 ( 1 + (|\xi'|^2 / \xi_1^2))^{1/2}}$ in terms of ${|\xi'|^2 / \xi_1^2}$, both for estimating the difference in the phases and for estimating ${\partial_{\xi_1}\phi}$. The non-stationary phase principle in both ${\xi_1, \xi_j}$ (after once again using a smooth partition of unity to reduce to the case where ${|\xi_j| \gtrsim |\xi|}$) will result in having to estimate (for example) ${\int |(D_j^{\intercal} D_1^{\intercal})^N \psi(\xi)| d\xi}$. This is quite excruciating, but you can again use an inductive argument to show that the integrand consists entirely of terms of the form ${ (\partial_j \phi)^{-\ell} (\partial_1 \phi)^{-\ell'} \cdot \partial_j^{\alpha}\partial_1^{\beta}\psi \cdot \prod_{k+k'\geq 2} (\partial_j^{k}\partial_1^{k'}\phi)^{\gamma(k,k')}}$ where the various parameters involved satisfy certain favourable contraints. Then one can see that ${|\partial_1 \phi| \gtrsim |x_1 - t|}$, ${|\partial_1^2 \phi|\lesssim 1}$ (this because of the restriction to ${|\xi'|\lesssim t^{-1/2}}$), ${|\partial_j \phi| \sim t}$ and in general ${|\partial_j^{k}\partial_1^{k'}\phi|\lesssim t}$ for all ${k + k'\geq 2}$. Using these facts and the induction one can verify that ${|(D_j^{\intercal} D_1^{\intercal})^N \psi(\xi)|\lesssim |x_1 - t|^{-N}}$. The factor of ${t^{-(d-1)/2}}$ will come from the fact that the region of integration is approximately a cylinder of length ${1}$ and radius ${t^{-1/2}}$.
7. This part is a repetition of part viii) above essentially, except for the fact that now we can assume ${\xi_j \sim 2^{j} t^{-1/2}}$ 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 ${\int_{\Omega} \exp(i\lambda xy) dx dy}$ where ${\Omega = \{(x,y) \; : \; |x|+|y|\leq 1\}}$ (hint: rotate coordinates and use single variable Van der Corput’s lemma). [go back]