5
$\begingroup$

Let $\phi\in C_c^\infty(\mathbb{R})$ be an even function such that $\chi_{(-1/2,1/2)}\le\phi\le \chi_{(-1,1)}$, where $\chi_{(a,b)}$ stands for the indicator function of the interval $(a,b)$. For $\lambda>0$ consider the oscillatory integral $$ I(\lambda)=\int_\mathbb{R} \phi(x)\, \exp \left(i\lambda(x+\epsilon|x|^{\sqrt{2}})\right)\, dx, $$ with some fixed (very small) positive constant $\epsilon$.

My question is: what is the asymptotic behavior of this integral as $\lambda\rightarrow \infty$? I can show, by essentially doing careful integration by parts, that the upper bound is $\lesssim \lambda^{-\sqrt{2}}$, but I wonder whether $\lambda^{-\sqrt{2}}$ is also a lower bound?

Note, that if the exponent $\sqrt{2}$ is replaced by $2k$ for some positive integer $k$, then the integral decays like $\lambda^{-M}$ for any $M>0$ due to the non-stationary phase estimate (the derivative of the function $x+\epsilon x^{2k}$ is $\gtrsim 1$).

I would appreciate any hints on how to approach this problem.

$\endgroup$
8
$\begingroup$

$\newcommand{\R}{\mathbb R}\newcommand{\de}{\delta}\newcommand{\ep}{\epsilon}\newcommand{\tI}{\tilde I}$ Take any $a\in(1,2)$ and then any nonzero $\epsilon\in(-1/a,1/a)$. Then \begin{align*} I(t)&:=\int_\mathbb{R} \phi(x)\, \exp(it(x+\epsilon|x|^a))\, dx \\ &\sim\frac{2\epsilon\,\Gamma(a+1)}{it^a}\,\sin\frac{\pi a}2 \tag{1} \end{align*} as $t\to\infty$. This asymptotics does not depend on $\phi$, as long as \begin{equation*} \phi\in C_c^\infty(\mathbb{R}) \tag{2} \end{equation*} is a (not necessarily even) function such that \begin{equation*} \chi_{(-1/2,1/2)}\le\phi\le \chi_{(-1,1)}. \tag{3} \end{equation*}

Indeed, let \begin{equation*} g(x):=x+\ep|x|^a. \end{equation*} Then $g'(x)=1+\ep ax^{[a-1]}$ for real $x\ne0$, where $x^{[c]}:=|x|^{c-1}x$. Therefore and because $|\ep|<1/a$, we see that $2\ge g'\ge1-|\ep|a>0$ on $(-1,1)$. So, there is a unique inverse $h$ of function $g$ on $(-1,1)$ such that for all $x\in(-1,1)$ and all $y\in(g(-1),g(1))$ we have \begin{equation*} y=g(x)\iff x=h(y). \tag{3.5} \end{equation*} Also, $g(-x)<0<g(x)$ for all $x\in(0,1)$.

Using now the substitution $g(x)=y$, using conditions (2)--(3) and integrating by parts, we have \begin{align*} I(t)&=\int_\R dx\,\phi(x)\, e^{itg(x)} \\ &=\int_{-1}^1 dx\,\phi(x)\, e^{itg(x)} \\ &=\int_{g(-1)}^{g(1)} dy\,h'(y)\phi(h(y))\, e^{ity} \\ &=-\frac1{it}\,(I_1+\tI_1), \tag{4} \end{align*} where \begin{align*} I_1&:=\int_{g(-1)}^{g(1)} dy\,e^{ity}h''(y)\phi(h(y)), \\ \tI_1&:=\int_{g(-1)}^{g(1)} dy\,e^{ity}h'(y)^2\phi'(h(y)). \end{align*} In view of (2)--(3), $\phi'$ is an (infinitely) smooth function supported on \begin{equation*} S:=[-1,1]\setminus(-1/2,1/2). \end{equation*} Hence, $(h')^2\,\phi'\circ h\in C^1(g(S))$. So, integrating by parts, we have \begin{equation*} \tI_1\ll\frac1t; \tag{5} \end{equation*} as usual we write $u\ll v$ to mean $|u|\le Cv$ for some real constant $C>0$. Next, in view of (3), \begin{equation*} I_1=I_2+\tI_2, \tag{6} \end{equation*} where \begin{align*} I_2&:=\int_{g(-1/2)}^{g(1/2)} dy\,e^{ity}h''(y), \\ \tI_2&:=\int_{g(S)} dy\,e^{ity}h''(y). \end{align*} Note that $h''\in C^1(g(S))$. So, integrating by parts, we have \begin{equation*} \tI_2\ll\frac1t. \tag{7} \end{equation*}

Writing, for brevity, $x$ for $h(y)$ (cf. (3.5)), we have \begin{equation*} h'(y)=\frac1{1+\ep ah(y)^{[a-1]}}=\frac1{1+\ep ax^{[a-1]}} \end{equation*} and hence \begin{align*} h''(y)&=-\frac{\ep a(a-1)|x|^{a-2}}{(1+\ep ax^{[a-1]})^2},\\ h'''(y)&\ll|x|^{a-3}+|x|^{2a-4}\ll|x|^{a-3}\ll|y|^{a-3} \tag{8} \end{align*} for $|x|\le1/2$, that is, for $y\in[g(-1/2),g(1/2)]$. Next, for $y=g(x)\in[g(-1/2),g(1/2)]$ we have $y=x(1+\ep x^{[a-1]})$ and hence \begin{align*} x=h(y)&=\frac y{1+\ep x^{[a-1]}} \\ &=\frac y{1+\ep y^{[a-1]}(1+O(|y|^{a-1}))} \\ &=\frac y{1+\ep y^{[a-1]}} \, (1+O(|y|^{2a-2}) \\ &=y\, (1+O(|y|^{a-1})) \end{align*} and \begin{align*} h''(y)&=-\frac{\ep a(a-1)|h(y)|^{a-2}}{(1+\ep ah(y)^{[a-1]})^2}\\ &=-\ep a(a-1)|y|^{a-2}[1+O(|y|^{a-1})] \\ &=-\ep a(a-1)|y|^{a-2}+O(|y|^{2a-3}). \tag{9} \end{align*} Further,
\begin{equation*} I_2=I_3+\tI_3, \tag{10} \end{equation*} where \begin{align*} I_3&:=\int_{|y|\le\de} dy\,e^{ity}h''(y), \\ \tI_3&:=\int_{[g(-1/2),g(1/2)]\setminus[-\de,\de]} dy\,e^{ity}h''(y), \\ \de&:=t^{-3/4}. \end{align*} Integrating by parts and using (8) and (9), we have \begin{align*} \tI_3&\ll\frac{|h''(\de)|+|h''(-\de)|+O(1)}t +\frac1t\, \int_{|y|>\de} dy\,|y|^{a-3}, \\ &\ll \frac{\de^{a-2}}t=t^{1/2-3a/4}=o(t^{1-a}). \end{align*} Using (9) again, we have \begin{equation*} I_3=-\ep a(a-1)I_4+O(\tI_4), \tag{11} \end{equation*} where \begin{align*} I_4&:=\int_{|y|\le\de} dy\,e^{ity}|y|^{a-2}, \\ \tI_4&:=\int_{|y|\le\de} dy\,|y|^{2a-3}. \end{align*} Next, \begin{equation*} \tI_4\ll\de^{2a-2}=t^{-(2a-2)3/4}=o(t^{1-a}), \tag{12} \end{equation*} \begin{align*} I_4&=t^{1-a}\int_{|z|\le t\de} dz\,e^{iz}|z|^{a-2} \\ &=t^{1-a}\int_{|z|\le t^{1/4}} dz\,e^{iz}|z|^{a-2} \\ &\sim t^{1-a}\int_\R dz\,e^{iz}|z|^{a-2} =2 t^{1-a}\Gamma(a-1)\sin\frac{\pi a}2. \tag{13} \end{align*}

Collecting the pieces (4)--(7) and (10)--(13), we get the result.

$\endgroup$
9
  • $\begingroup$ That's a wonderful answer! Thanks so much, Professor Pinelis! $\endgroup$
    – Tony419
    Jan 3 at 17:14
  • $\begingroup$ I do not see the $i$ at the denominator in (1). $I(t)$ is real if $\phi$ is even and this is in any case true in the asymptotics. Am I wrong? $\endgroup$ Jan 3 at 22:18
  • $\begingroup$ @GiorgioMetafune : Thank you for this sanity check. However, I do not see an argument for $I(t)$ to be real even when $\phi$ is even (pun not intended). That would be the case if $g$ were odd. But $g$ is odd only if $\epsilon=0$, in which case the expression after $\sim$ in (1) is $0$ and hence real, even though (1) then loses meaning. Also, I do not see a mistake (do you?) in this proof, except that I now see that I forgot to require $\epsilon$ to be nonzero -- going to fix this now. $\endgroup$ Jan 3 at 22:44
  • $\begingroup$ Yes, you are right, no need to be real. But then I do not see why it is purely imaginary! I get almost the same but with $\exp{i a \pi/2}$ instead of $i^{-1}\sin{\pi a/2}$. I did the computations on $(0, \infty)$ instead of the whole line, but this should only mean a factor 2. $\endgroup$ Jan 3 at 23:21
  • $\begingroup$ @GiorgioMetafune : (i) The complete explanation of why it is purely imaginary is the entire proof, which is pretty long. A short explanation may be this: If $\epsilon=0$, then $g$ is odd and hence $I(t)$ is real, and then $g$ is also smooth and hence $I(t)\to0$ faster than any (say negative) power of $t$. Now, $\epsilon$ is the measure of both the non-oddness and non-smoothness of $g$. Therefore and because $\epsilon$ comes with the factor $i$, it seems natural to expect $I(t)$ to be asymptotically purely imaginary. $\endgroup$ Jan 4 at 0:35

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.