1
$\begingroup$

A statement of the stationary phase method I know is the following. Suppose $\phi(x_0) = \phi'(x_0) = 0$ and $\phi''(x_0) \not = 0$. If $\psi$ is a smooth function supported in a sufficiently small neighborhood of $x_0$, then $$ \int_{\mathbb{R}} e^{i \lambda \phi(x)} \psi(x) dx \sim a_0 \lambda^{-1/2} $$ for some $a_0 \in \mathbb{C}$.

Suppose $\psi(x) = 0$ for $x> x_0 + \varepsilon$ and $x < x_0 - \varepsilon$ where $\varepsilon>0$ is sufficiently small. I was just wondering does the method still work if $\psi$ satisfies this condition but also $\psi(x) = 0$ for $x \in I$ where $I$ is an open interval $x_0 \in I$ and $I \subseteq (x_0 - \varepsilon, x_0 + \varepsilon)$? So in this case we can not say that $\psi$ is supported in a sufficiently small neighbourhood of $x_0$ as in the statement above. However, $\psi$ is $0$ outside $[x_0 - \varepsilon, x_0 + \varepsilon]$. I was wondering is this still enough to obtain the above result for some $a_0$? Any comments are appreciated. Thank you very much.

Improve this question
$\endgroup$
1
  • $\begingroup$ I think you got confused about the role of $\epsilon$ here. Its only purpose is to make sure that we have only one point of stationary phase in our interval. If you now take a $\psi$ that is zero near this only point of stationary phase, ok, then you're back in the much simpler case where you don't have any points of stationary phase, and the integral becomes small to all orders of $\lambda$ by a straightforward integration by parts. $\endgroup$
    – Christian Remling
    Commented Jan 4, 2018 at 21:00

1 Answer 1

Reset to default
3
$\begingroup$

If $\psi(x)$ vanishes at $x_0$ then the large-$\lambda$ asymptotics changes; the point of stationary phase moves away from $x_0$ and will have to be recalculated for your specific choice of $\psi$. To see what is going on, you can take $\phi(x)=(x-x_0)^2$ and $\psi(x)=(x-x_0)^{2p}$, when the large-$\lambda$ asymptotics becomes $\propto\lambda^{-p-1/2}$. If $\psi(x)$ vanishes in an interval around $x_0$, then the integral will decay exponentially rather than as a power law for large $\lambda$.

Improve this answer
$\endgroup$
7
  • $\begingroup$ I mostly agree with this answer, but as long as $\psi(x)$ vanishes to finite order at $x_0$, I would say the point of stationary phase is still $x_0$, but the exponent in the asymptotics changes in the way you describe. $\endgroup$
    – David E Speyer
    Commented Jan 4, 2018 at 14:24
  • $\begingroup$ Since $\psi$ is $0$ outside $[x_0 - \varepsilon, x_0 + \varepsilon]$, it will still have compact support, so does it matter whether $\psi (x_0)$ is $0$ or not? I don't think so. If I recall correctly, moving $\sqrt \lambda$ to the left-hand side and taking the limit $\lambda \to \infty$ gives $\psi (x_0)$ times some fixed constant depending on $\phi '' (x_0)$, so the general theory of the stationary phase method still holds. I do not understand why the OP seems to think that he is in a different setting. $\endgroup$
    – Alex M.
    Commented Jan 4, 2018 at 14:32
  • $\begingroup$ @AlexM. Because that constant would be $0$. Let's write $I(\lambda)$ for the integral. If $\psi$ vanishes to order exactly $2$, then $I(\lambda) \sim c \lambda^{-3/2}$ for some $c \neq 0$. It is true that $\lim_{\lambda \to \infty} I(\lambda)/\lambda^{-1/2}$ exists and is $0$, but it is not true that $I(\lambda) \sim 0 \lambda^{-1/2}$ because that isn't how the $\sim$ notation works. $\endgroup$
    – David E Speyer
    Commented Jan 4, 2018 at 14:39
  • $\begingroup$ The $\psi=(x-x_0)^{2p}$ example is not really relevant to the question as asked by the OP, who explicitly assumed that $\psi\equiv 0$ in a neighborhood of $x_0$ (you perhaps subconsciously wanted to make the question more interesting). The trivial answer to the question as asked is that if $\psi\equiv 0$ near $x_0$, then there are no points of stationary phase in our interval (assumed to have been chosen so small that this holds), so $I(\lambda)\lesssim \lambda^{-N}$ for all $N$. $\endgroup$
    – Christian Remling
    Commented Jan 4, 2018 at 20:56
  • $\begingroup$ By the way, exponential decay is not guaranteed. For example, if $\phi(x)=x$, so we're dealing with the Fourier transform, then exponential decay would make $\psi$ holomorphic, so can never hold for a compactly supported $\psi$. $\endgroup$
    – Christian Remling
    Commented Jan 4, 2018 at 21:05

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .