Expansion of Integrals. 4

$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\bZ}{\mathbb{Z}}$ $\newcommand{\bC}{\mathbb{C}}$ $\newcommand{\const}{\operatorname{const}}$ $\newcommand{\Hess}{\operatorname{Hess}}$ $\newcommand{\supp}{\operatorname{supp}}$ $\newcommand{\sgn}{\operatorname{sgn}}$

Oscillatory integrals. II. Multidimensional theory

  1. No stationary points
  2. Single stationary point
  3. Multiple stationary points
  4. Degenerate stationary points
  5. Complex phase

We consider \begin{equation} I(k)= \int_X e^{k \phi(x)}f(x)\,dx \label{eq-2.4.1} \end{equation} where now $X= \bR^d$ and $\phi \in C^\infty(X)$, $\phi$ is a real-valued function, $f\in C_0^\infty(X)$ which means that $f=0$ as $|x|\ge R$.

We are interested in the asymptotics of $I(k)$ as $k\to +\infty$.

No stationary points

First of all we need

Theorem 1. If $\phi$ has no stationary points on $\supp f$ then \begin{equation} I(k)= O(k^{-\infty})\qquad\text{as }\ k\to +\infty \label{eq-2.4.2} \end{equation} which means that $I(k)= O(k^{-N})$ for any $N$.

Proof. We can decompose $f=f_1+\ldots+f_d$ such that $\partial_{x_j}\phi \ne 0$ on $\supp f_j$. Then applying we can apply $1$-dimensional result.

Single stationary point

Assume now that $\phi$ has a single stationary point $\bar{x}$ on $\supp f$ and this stationary point is non-degenerate. Then $f=f_0+f_1$ where $f_0$ is supported in the small vicinity of $\bar{x}$ and $\phi$ has no stationary points on $\supp f_1$. So without any loss of the generality one can assume that $f=f_0$.

Applying Morse theory (Theorem 2.2.2) we can assume without any loss of the generality that \begin{equation} \phi (y)=\phi (\bar{x})+ \sum_{1\le j\le d} \lambda_j z_j^2 \label{eq-2.4.3} \end{equation} and applying $1$-dimensional theory (Theorem 2.3.5) we arrive to

Theorem 2. Let $\phi$ have a single stationary point $\bar{x}$ on $\supp f$ and $\phi''(\bar{x})0$ non-degenerate. Then \begin{equation} I(k) \sim e^{ik\phi (\bar{x})}\sum _{n=0}^\infty \kappa_{2n} k^{-\frac{d}{2}-n} \label{eq-2.4.4} \end{equation} in the sense that \begin{equation} |I(k)- e^{ik\phi (c)}\sum _{n=0}^{N-1} \kappa_{2n}k^{-\frac{d}{2}-n}|\le C_N k^{-N-\frac{d}{2}}. \label{eq-2.4.5} \end{equation} Here the main coefficient is \begin{equation} \kappa_0=(2\pi)^{\frac{d}{2}}|\det \phi''(\bar{x} )|^{-\frac{1}{2}} e^{i\frac{\pi}{4} \sgn \phi''(\bar{x})} f(\bar{x}) \label{eq-2.4.6} \end{equation} and $\sgn A$ is a signature of non-degenerate Hermitean matrix $A$: $\sgn A= d_+ - d_-$ where $d_\pm$ is a number of positive and negative eigenvalues of $A$ (or equivalently the dimension of positive or negative space of the quadratic form $\langle Ax,x\rangle$).

Several stationary points

Let now $\phi$ has several stationary points (or stationary points in which $\Im \phi=0$ in the framework of Theorem 3 below) on $\supp f$: $x_1,\ldots ,x_K$ each of the type considered above. Then asymptotics of $I(k)$ is given by the sum of the contributions of all these points.

Degenerate stationary points

The degenerate stationary points of the function of several variables can be of the very different types. What is more: these functions usually depend on extra parameters $\phi=\phi(x; y)$ where we integrate with respect to $x$ only and want remainder estimate uniform with respect to $y$. In this case we using the above approach eliminate integration with respect to "some of $x$".

Complex phase

Theorem 3. Assume now that $\phi$ is a complex-valued function, $\Im\phi \ge 0$ and there exists a single point $\bar{x}$ such that $\Im \phi (\bar{x})=0$, $\nabla \phi (\bar{x})=0$ and $\phi''(\bar{x})$ is non-degenerate.

Then decomposition (\ref{eq-2.4.4}) holds.

Proof. Observe, that $\Re i\phi = -\Im \phi \le 0$. Also observe that if $\Im \phi>0$ on $\supp f$ then $I(k) =O(e^{-\epsilon k})$. Further, if $\nabla \phi \ne 0$ on $\supp f$ then $I(k) =O(k^{-\infty})$.

Thus we need to consider a small vicinity of $\bar{x}$. Without any loss of the generality one can assume that $\bar{x}=0$ and $\phi(0)=0$.

Let $Q(x)=\frac{1}{2}\langle \phi''(0)x,x\rangle$ be a quadratic part of $\phi(x)$, and $S(x)=\phi(x)-Q(x)$. Then \begin{equation*} e^{ikS (x)}\sim \sum_{n\ge 0} \frac{1}{n!} (iS(x))^n \sim \sum _{m\ge 0} k^m \sum_{\alpha:|\alpha|\ge 3m} c_{m,\alpha} x^\alpha \end{equation*} in the sense that the remainder is $O(|x|^{3N}k^N)$ and since $f(x)$ could be decomposed into asymptotic Taylor series similarly \begin{equation*} e^{ikS (x)}f(x)\sim \sum _{m\ge 0} k^m \sum_{\alpha:|\alpha|\ge 3m} c'_{m,\alpha} x^\alpha \end{equation*} and then \begin{multline} I(k)\sim \sum _{m\ge 0} k^m \sum_{\alpha:|\alpha|\ge 3m} c'_{m,\alpha} \int e^{ikQ(x)}x^\alpha\,dx \sim\\ \sum _{m\ge 0} \sum_{\alpha:|\alpha|\ge 3m} c'_{m,\alpha} k^{m- \frac{d}{2}-\frac{|\alpha|}{2}} \int e^{iQ(x)}x^\alpha\,dx \label{eq-2.4.7} \end{multline} and terms with odd $|\alpha|$ vanish. One can prove (\ref{eq-2.4.7}) by integration by parts using that $|\nabla \phi|\asymp |x|$ since $\phi''$ is non-degenerate.

This proves decomposition (\ref{eq-2.4.4}).

Remark 1. One can prove that \begin{equation} \kappa_0 = (2\pi)^{\frac{d}{2}} \bigl(\det (-i\phi ''(\bar{x})^{\frac{1}{2}}\bigr)^{-1}= \pm (2\pi)^{\frac{d}{2}} \bigl(\det (-i\phi ''(\bar{x})\bigr)^{-\frac{1}{2}}. \label{eq-2.4.8} \end{equation} Since $\Im \phi''(\bar{x})$ is non-negative definite matrix, $-i\phi''(\bar{x})$ is a sectorial matrix in the sense that its spectrum belongs to ${z\in \bC:\, \Re z\ge 0, \, z\ne 0}$ and since it cannot belong $(-\infty,0]$ the square root of it $(-i\phi ''(\bar{x})^{\frac{1}{2}}$ is properly defined. However $(\det (-i\phi ''(\bar{x})\bigr)^{\frac{1}{2}}$ is defined up to a factor $\pm 1$.

Remark 2. If $f= O(|x-\bar{x}|^l)$ then $\kappa_n=0$ as $n=\lceil l/2\rceil$.

$\Leftarrow$  $\Uparrow$  $\Rightarrow$