# asymptotic with a very degenerate stationary phase

Suppose $f(x_1,x_2)\in C^\infty_c(\mathbb R^2)$. I wonder how one may derive the asymptotic expansion of the following integral when the real paramter $\lambda\rightarrow \infty$: \begin{equation} \int_{\mathbb R^2}e^{i\lambda x_1x_2(x_1+x_2)}f(x_1,x_2) dx_1 dx_2. \end{equation} Note that the origin is the only critical point of the phase function $\phi(x_1,x_2)=x_1x_2(x_1+x_2)$, but it is a very degenerate one as the Hessian matrix $\big(\frac{\partial^2 \phi}{\partial x_i\partial x_j}\big)$ is zero at the origin. Actually, the Milnor number that describes the singularity of $\phi$ at the origin is $4$.

[EDIT: I have found an answer for this question myself. When the phase function $\phi$ is as simple as stated above, the integral's asymptotic behavior can be written down in a clean manner. The real complexity shows up when there is some perturbation in the phase, that is, when the phase is of the form $\Phi(x_1,x_2)=x_1x_2(x_1+x_2)+ax_1x_2+bx_1+cx_2$, with $a,b,c$ denoting the control parameters. In the latter case, an asymptotic $\lambda$-expansion that is uniform in $a,b,c$ is very hard to get and seems not existent in the literature yet.]

## 1 Answer

(I am not really an expert, so it should be taken with a pinch of salt.)

One may take $f=1$ and simply compute the integral using polar coordinates. (There are minor problems with convergence, but they are irrelevant.) The result is $c|\lambda|^{-2/3}$, where $c\neq 0$ is an absolute constant which I was too lazy to compute. So, in the general case the asymptotic must be $c|\lambda|^{-2/3}f(0,0)$. (To actually prove this may be a piece of work, but it certainly can be done if necessary.)

[EDIT] Now I think I can help. First, I would change coordinates to make the phase function more symmetric, $\phi(x,y)=y^3-3x^2 y$. Then, I would make use of complex variables. Let $\Sigma\subset\mathbb{C}^2$ be a surface defined by $x=(\cos t-iu\sin 2t)r,\,y=(\sin t -iu\cos 2t)r$, where $u>0$ is not very big. ($u=1/2$ will do.) Here $0\le t<2\pi$ and $r>0$.

The idea is to replace the original integral with $$\int_{\Sigma}e^{i\lambda\phi(x,y)}f(x,y)dx\wedge dy,$$ which must have the same asymptotic. This integral is convergent even if $f$ is an entire function, because on this surface $\Im \phi(x,y)>0$. The asymptotic series simply comes from the Taylor series. (To make the argument rigorous one may use a partition of unity and other standard techniques.)

• The asymptotic expansion should look like $\sum_{n=0}^\infty c_n\lambda^{-\frac{2+n}{3}}$. The precise questions are i) to prove that the expansion is of such a form and ii) to provide expressions or an algorithm for the coefficients $c_n$. The leading coefficient is $c_0=3^{-\frac{1}{6}}\Gamma(\frac{1}{3})^2f(0,0)$, which can be computed by a different method. But it is not clear how to derive other coefficients. Thanks for sharing your thought on the problem. – Chris Sep 8 '16 at 3:26
• When $u>0$, the smooth function $f$ on $\mathbb R^2$ may not be defined on the surface $\Sigma_u$ in $\mathbb C^2$. Nevertheless, one could replace $f$ by its Taylor polynomial $T_n(f)$ of degree $n$ and show that the contribution from the difference is a power of $\lambda^{-1}$ depending on $n$ in certain way. However, for $T_n(f)$, one still needs to argue that the asymptotic expansion of its integral on $\Sigma_u$ leads to the expansion of the integral on $\Sigma_0$ when $u\rightarrow 0$. Is this obvious? (As you are aware of, $Im(\phi)$ is no longer positive on $\Sigma_0$). – Chris Sep 10 '16 at 21:03
• Of course, it is not THIS straightforward. The argument may go as follows. Assume that $f=F\omega$, where $F$ is entire and $\omega$ is a "cap", i.e. it has a compact support and $\omega=1$ near the origin. Then the integral of $f$ over $\Sigma_0$ is equal to the integral of $f$ over some "intermediate" surface $\Sigma'$, which coincides with $\Sigma_0$ where $\omega\neq 0$ but coincides with $\Sigma_u$ far from the origin. On the other hand, the integral of $F$ over $\Sigma_u$ is equal to the integral of $F$ over the same $\Sigma'$, because the form is holomorphic. – Alex Gavrilov Sep 11 '16 at 6:31
• Then, the difference is an integral of $F(1-\omega)$ over $\Sigma'$, which is small because the function is smooth and zero near the origin. – Alex Gavrilov Sep 11 '16 at 6:36
• At least, this is an answer to question ii (algorithm for the coefficients). – Alex Gavrilov Sep 11 '16 at 6:43