I'm trying to find asymptotics for an oscillatory integral on $\mathbb{S}^{n-1}$, for which my advisor said I should use stationary phase arguments. The particular, he claims that:

If $\lambda\gg 1$, then $$I(\lambda,x) = \int_{\mathbb{S}^{n-1}}(x\cdot y)e^{i\lambda(x\cdot y)}\,d\sigma(y),$$ is $\mathcal{O}(|x|(\lambda |x|)^{-\frac{n-1}{2}})$ when $|x|\geq \lambda^{-1}$.

When I hear stationary phase, I think of working with operators of the form $$ L = \frac{\nabla_y(x\cdot y)}{i\lambda |\nabla_y (x\cdot y)|^2} \cdot \nabla_y, $$ since then $L^N[e^{i\lambda (x\cdot y)}] = e^{i\lambda (x\cdot y)}$, for any $N \geq 1$, and I can use integration by parts to move these operators over to the $(x\cdot y)$ term. However, wouldn't that require that the integral be defined over an $n$-dimensional region, rather than an $(n-1)$-dimensional surface?

I wouldn't be struggling so much if the gradient $\nabla_y$ could be taken in Cartesian coordinates. But we have a surface integral in $d\sigma(y)$, meaning that we would need to parametrize our surface with $n-1$ parameters, say $\omega = (\omega_1, \ldots, \omega_{n-1}) \in \Omega \subset \mathbb{R}^{n-1}$, with any derivates now being taken in these new variables. Specifically $\nabla_y$ would generate an $n$-vector, while $\nabla_{\omega}$ would generate an $n-1$-vector. So how should I go about applying the same kind of stationary phase arguments to this new integral? $$ I(\lambda, x) = \int_{\Omega} (x \cdot y(\omega)) e^{i\lambda (x\cdot y(\omega))} \,dV(\omega) $$ Now that $y$ depends on $\omega$, the gradients $\nabla_\omega (x\cdot y(\omega))$ become much trickier to get a grasp on. I'm particularly struggling trying to argue how we can find regions where $|\nabla_\omega (x\cdot y(\omega))| > 0$, so that our $L$ type operators are properly defined.

Am I going about this all wrong? If I can get the big-oh asymptotics I mentioned above in some other way, it doesn't really matter. I just need to prove these results as a lemma to something bigger. Any help is much appreciated!


You have $ I(\lambda, x)=x\cdot\int_{\mathbb S^{n-1}} ye^{i \lambda x\cdot y} d\sigma(y)=x\cdot J(x,\lambda) $ and you claim that for $\vert x\vert \lambda \ge 1$, you have $$ J(x,\lambda)=O((\vert x\vert \lambda)^{-\frac{n-1}{2}}). $$ Indeed, using coordinate charts and a finite partition of unity, you are reduced to the case where $$ J(x,\lambda)=\int_{\mathbb R^{n-1}} a(z) e^{i\lambda (x'\cdot z+ x_n\sqrt{1- \vert z\vert^2})}dz, \quad\text{$a\in C^\infty_0(\mathbb R^{n-1})$ supported near $0$, $x=(x', x_n)\in \mathbb R^{n-1}\times \mathbb R$}. $$ Let us set $\phi(x,z)=x'\cdot z+ x_n\sqrt{1- \vert z\vert^2}$. We have $$ \frac{\partial \phi}{\partial z}= x'-(1- \vert z\vert^2)^{-1/2} z x_n, $$ which vanishes at $z=0$ when $x'=0$. Then you calculate the Hessian of $\phi$ at $z=0$ and get $$ \phi''_{zz}=-x_n. $$ The stationary phase in $n-1$ dimensions gives the sought estimate with $O((\vert x_n\vert \lambda)^{-\frac{n-1}{2}})$ (note that you know that $\vert x_n\vert \lambda \ge 1$, since you are near $x'=0$).

| cite | improve this answer | |
  • $\begingroup$ I'm still reading through this carefully, but thank you! I have a couple small clarification questions, though, so if you don't mind I'll just enumerate them here in a series of separate comments: $\endgroup$ – Patch Mar 25 at 6:20
  • $\begingroup$ 1) By bringing the integral inside of the dot product, you are effectively making $\mathbf{J}(x,\lambda)$ a vector-valued function, correct? But isn't your cutoff function, $a(z)$ a scalar-valued function? Should I just be interpreting $a(z)$ as one scalar component of $\mathbf{J}(x,\lambda)$? If not, then wouldn't this seem to contradict the definition for $\mathbf{J}(x,\lambda)$ you had earlier? $\endgroup$ – Patch Mar 25 at 6:21
  • $\begingroup$ 2) Assuming we have clarified the issue with $\mathbf{J}(x,\lambda)$ above, lets call $J_k(x, \lambda)$ the $k$-th component of the vector. Then I can see how stationary phase around $(x,z) = (0,0)$ gives us the $(|x_n| \lambda)^{(n-1)/2}$ behavior, but this would only be asymptotic behavior in each coordinate. Moreover, every $J_k$ would satisfy the same big-oh bounds involving the absolute value $|x_n|$ only; so how do I put these all together to get a uniform bound involving the norm, $|x|$? $\endgroup$ – Patch Mar 25 at 6:21
  • 1
    $\begingroup$ @Patch (1) $J$ is indeed vector-valued, but you can handle each coordinate separately. (2) You start with $x$ such that $\vert x\vert \lambda\ge 1$; then you look at the stationary phase method with an integral near a given point $y_0$ of the sphere. You check if you have a stationary point: if not you get a better estimate with $O((\vert x\vert \lambda)^{-N})$ for all $N$. If you land on a stationary point, you check the Hessian. Note that the choice of $y_0$ in the answer above is $e_n$. $\endgroup$ – Bazin Mar 25 at 9:53
  • $\begingroup$ Thanks, again!! $\endgroup$ – Patch Mar 25 at 16:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.