An elementary proof of the Voros connection formula for WKB solutions to the Airy equation with a large parameter

In [3], the Voros connection formula (cf. [12]) for the WKB solutions to Schrödinger-type ordinary differential equations is proved from the viewpoint of microlocal analysis [6]. The proof consists of two parts. In the first part, the Schrödinger-type ordinary differential equation with an analytic potential is formally transformed to the Airy equation with a large parameter near a simple turning point. The formal series appearing in the transformation can be justified by using microdifferential operators. In the second part, the Voros connection formula for the Airy equation is proved by computing the Borel transform of the WKB solutions directly in terms of the Gauss hypergeometric functions. The classical connection formulas for the hypergeometric functions yield the Voros formula for the Airy equation. Combining these two parts, we have the Voros formula for general equations. (See [7] also.)

In this article, we focus on the second part. We observe that the parameters in the hypergeometric functions used in the proof in [3, 7] are contained in the Schwarz list [11]. In other words, they are algebraic functions. Our aim is to prove the connection formula for the WKB solutions to the Airy equation by using analytic continuation of algebraic functions, not of the hypergeometric functions. This point of view gives some generalization. We find a similar structure in the Pearcey system (cf. [9]) with a large parameter. This system is a natural generalization of the Airy equation to the two-variable case. We see that the Borel transforms of the suitably normalized WKB solutions to the Pearcey system are also algebraic functions. Hence such WKB solutions are resurgent. We hope that this example provides a part of basics of the prospective theory for the exact WKB analysis of holonomic systems.

The differential equation

(1)

$$\left(-\frac{d^{2}}{dz^{2}}+z\right)w=0$$

is called the Airy equation (cf. [10]). Airy used a solution to this equation expressed by an integral, known as the “Airy function” nowadays (see § 3), effectively in his theory of the rainbow (caustics) [1]. We introduce a positive large parameter $\eta$ by setting $z=\eta^{2/3}x,\,\psi(x,\eta)=w(\eta^{2/3}x)$. Then we have a differential equation of the form

(2)

$$\left(-\frac{d^{2}}{dx^{2}}+\eta^{2}x\right)\psi=0.$$

We call this the Airy equation with the large parameter, or simply, the Airy equation. This equation plays a fundamental role in the exact WKB analysis. Let $S$ denote the logarithmic derivative of the unknown function $\psi$, namely $\displaystyle S=\frac{d}{dx}\log\psi.$ Then $S$ should satisfy the following Riccati-type equation:

(3)

$$\frac{dS}{dx}+S^{2}=\eta^{2}x.$$

This equation has a formal solution $\displaystyle S=\sum_{j=-1}^{\infty}\eta^{-j}S_{j}$ defined by the recurrence relation

(4)

$$\left\{\begin{split}\ \ S_{-1}^{2}&=x,\\ S_{j+1}&=-\frac{1}{2S_{-1}}\left(\frac{dS_{j}}{dx}+\sum_{k=0}^{j}S_{k}S_{j-k}\right)\quad(j=-1,0,1,2,3,\dots).\end{split}\right.$$

We consider the exponential of the integral of $S$:

$$\psi=\exp\left(\int Sdx\right),$$

which formally satisfies (2). We call this a WKB solution to (2). We easily see that

$$S_{-1}=x^{1/2},\ S_{0}=-\frac{1}{4x},\,\dots\,$$

and if we fix the branch of the square root, say, as $x^{1/2}>0$ for $x>0$, we have a formal solution $S^{(+)}$. Another choice of the branch also gives a formal solution $S^{(-)}$. If we set

$$S_{\rm odd}=\sum_{j=-1}^{\infty}\eta^{-2j-1}S_{2j+1}=\eta x^{1/2}-\eta^{-1}\frac{5}{32}x^{-5/2}-\eta^{-3}\frac{1105}{2048}x^{-11/2}-\cdots,$$

$$S_{\rm even}=\sum_{j=0}^{\infty}\eta^{-2j}S_{2j}=-\frac{1}{4x}-\eta^{-2}\frac{15}{64}x^{-4}-\frac{1695}{1024}x^{-7}+\cdots,$$

then we have

$$S^{(\pm)}=\pm S_{\rm odd}+S_{\rm even}$$

satisfies (3) and

$$S_{\rm even}=-\frac{1}{2}\frac{d}{dx}\log S_{\rm odd}$$

holds. Hence we may take $-1/2\log S_{\rm odd}$ as a primitive of $S_{\rm even}$ and have the special WKB solutions of the form

(5)

$$\psi_{\pm}=\frac{1}{\sqrt{S_{\rm odd}}}\exp\left(\pm\int_{0}^{x}S_{\rm odd}dx\right).$$

Here the integral is defined by one half of the term-by-term contour integral of $S_{\rm odd}$ starting from $x$ on the second sheet of the Riemann surface of $\sqrt{x}$, going around the origin counterclockwise and back to the $x$ on the first sheet:

$$\int_{0}^{x}S_{\rm odd}dx=\eta\frac{2}{3}x^{3/2}+\eta^{-1}\frac{5}{48}x^{-3/2}+\eta^{-3}\frac{1105}{9216}x^{-9/2}+\cdots.$$

We observe that $S_{\rm odd}$ has the weighted homogeneity

$$S_{\rm odd}(\lambda^{2}x,\lambda^{-3}\eta)=\lambda^{-2}S_{\rm odd}(x,\eta).$$

Hence $\psi_{\pm}$ satisfies

$$\psi_{\pm}(\lambda^{2}x,\lambda^{-3}\eta)=\lambda\psi_{\pm}(x,\eta).$$

This relation and (2) imply that $\psi_{\pm}$ are formal solutions to the following system of partial differential equations in the variable $(x,\eta)$:

(6)

$$\left\{\begin{split}&\ \ \left(-\frac{\partial^{2}}{\partial x^{2}}+\eta^{2}x\right)\psi=0,\\ &\ \ \left(2x\frac{\partial}{\partial x}-3\eta\frac{\partial}{\partial\eta}-1\right)\psi=0.\end{split}\right.$$

There are two standard solutions to (1), which are called the Airy functions [10]:

(7)		$\displaystyle{\rm Ai}(z)$	$\displaystyle=\frac{1}{2\pi i}\int_{\gamma_{1}}\exp\left(-z\xi+\frac{\xi^{3}}{3}\right)d\xi,$
(8)		$\displaystyle{\rm Bi}(z)$	$\displaystyle=\frac{1}{2\pi}\int_{-\gamma_{2}+\gamma_{3}}\exp\left(-z\xi+\frac{\xi^{3}}{3}\right)d\xi.$

Here the paths $\gamma_{j}\ (j=1,2,3)$ are taken as shown in Fig. 3.1. More precisely, the arguments of three half-line asymptotes of the paths are $\pm\pi/3,\pi$.

Figure 3.1.

$\gamma_{2}$

$\gamma_{1}$

$\gamma_{3}$

We rewrite (9) by setting $t^{3}/3-xt=-y$:

(11)

$$\varphi_{j}=\int_{c_{j}}g(x,y)\exp(-y\eta)dy.$$

Here $c_{j}$ is the image of $\gamma_{j}$ by the mapping $t\mapsto y$ and

$$g(x,y)=\left.\frac{1}{x-t^{2}}\right|_{t=t(x,y)}.$$

The branch of the root $t=t(x,y)$ of the cubic equation

(12)

$$t^{3}/3-xt=-y$$

is taken suitably (see §6). We note that (11) looks like the Laplace integral defining the Borel sum of WKB solutions (see (27)) and hence $g$ is expected to have some relation with the Borel transform of WKB solutions.

We observe that the characteristic variety of the system (14) is the conormal bundle of the curve $4x^{3}-9y^{2}=0$ and hence the system is holonomic. The second equation of (14) implies that the unknown function can be written in the form

$$g(x,y)=\frac{1}{x}h\left(\frac{y}{x^{3/2}}\right)$$

by using a function $h$ of one variable. It is easy to find a second-order ordinary differential equation for $h$ by using the first equation. This implies the rank of the holonomic system equals 2. Hence we have

Let $\psi_{\pm,B}$ denote the Borel transform of $\psi_{\pm}$. Explicitly, $\psi_{\pm}$ have the forms

	$\displaystyle\psi_{\pm}$	$\displaystyle=\eta^{-\frac{1}{2}}x^{-\frac{1}{4}}\left(1-\frac{5}{32}\eta^{-2}x^{-3}-\frac{1105}{2048}\eta^{-4}x^{-6}-\cdots\right)^{-\frac{1}{2}}$
		$\displaystyle\hskip 28.45274pt\times\exp\left(\pm\left(\eta\frac{2}{3}x^{\frac{3}{2}}+\eta^{-1}\frac{5}{48}x^{-\frac{3}{2}}+\eta^{-3}\frac{1105}{9216}x^{-\frac{9}{2}}+\cdots\right)\right).$

Setting

$$A=-\frac{5}{32}x^{-3}-\frac{1105}{2048}\eta^{-2}x^{-6}-\cdots$$

and

$$B=\frac{5}{48}x^{-\frac{3}{2}}+\eta^{-2}\frac{1105}{9216}x^{-\frac{9}{2}}+\cdots,$$

we may rewrite them as

	$\displaystyle\eta^{-\frac{1}{2}}x^{-\frac{1}{4}}(1+\eta^{-2}A(x,\eta))^{-\frac{1}{2}}\exp\left(\pm\eta\frac{2}{3}x^{\frac{3}{2}}\right)\exp(\pm\eta^{-1}B(x,\eta))$
	$\displaystyle=\eta^{-\frac{1}{2}}x^{-\frac{1}{4}}\exp\left(\pm\eta\frac{2}{3}x^{\frac{3}{2}}\right)\left(1-\frac{1}{2}\eta^{-2}A-\frac{3}{8}\eta^{-4}A^{2}+\cdots\right)$
	$\displaystyle\hskip 142.26378pt\times\left(1\pm\eta^{-1}B+\eta^{-2}\frac{1}{2}B^{2}\pm\cdots\right)$
	$\displaystyle=\eta^{-\frac{1}{2}}x^{-\frac{1}{4}}\exp\left(\pm\eta\frac{2}{3}x^{\frac{3}{2}}\right)(1+b_{1}^{\pm}\eta^{-1}x^{-\frac{3}{2}}+b_{2}^{\pm}(\eta^{-1}x^{-\frac{3}{2}})^{2}+\cdots)$

with some constants $b_{j}^{\pm}$. Then $\psi_{\pm,B}$ can be written in the form

(16)		$\displaystyle\psi_{\pm,B}(x,y)$	$\displaystyle=\frac{1}{x\sqrt{\pi}}\left(\frac{y}{x^{\frac{3}{2}}}\pm\frac{2}{3}\right)^{-\frac{1}{2}}\left\{1+\frac{\sqrt{\pi}{b}_{1}^{\pm}}{\Gamma(\frac{3}{2})}\left(\frac{y}{x^{\frac{3}{2}}}\pm\frac{2}{3}\right)+\cdots\right.$
		$\displaystyle\hskip 99.58464pt\cdots+\left.\frac{\sqrt{\pi}{b}_{j}^{\pm}}{\Gamma(j+\frac{1}{2})}\left(\frac{y}{x^{\frac{3}{2}}}\pm\frac{2}{3}\right)^{j}+\cdots\right\}.$

We introduce a new variable $\displaystyle s=\frac{3y}{4x^{\frac{3}{2}}}+\frac{1}{2}$. Then we may rewrite $\psi_{\pm,B}$ as follows:

(17)		$\displaystyle\psi_{+,B}$	$\displaystyle=\frac{\sqrt{3}}{2\sqrt{\pi}x}s^{-\frac{1}{2}}\left(1+\tilde{b}_{1}^{+}\frac{4}{3}s+\cdots\right),$
(18)		$\displaystyle\psi_{-,B}$	$\displaystyle=\frac{\sqrt{3}}{2\sqrt{\pi}x}(s-1)^{-\frac{1}{2}}\left(1+\tilde{b}_{1}^{-}\frac{4}{3}(s-1)+\cdots\right).$

Here we take the branches as $s^{1/2}>0$ if $s>0$ and $(s-1)^{1/2}>0$ if $s>1$.

On the other hand, it follows from the definition of the Borel transform that $\psi_{\pm,B}$ satisfies the formal Borel transform of (6):

(19)

$$\left\{\begin{split}\ &\left(-\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}x\right)\psi_{\pm,B}=0\\ &\left(2x\frac{\partial}{\partial x}-3\frac{\partial}{\partial y}(-y)-1\right)\psi_{\pm,B}=0,\end{split}\right.$$

or, equivalently,

(20)

$$\left\{\ \begin{split}\ \ &\left(-\frac{\partial^{2}}{\partial x^{2}}+x\frac{\partial^{2}}{\partial y^{2}}\right)\psi_{\pm,B}=0,\\ \ \ &\left(2x\frac{\partial}{\partial x}+3y\frac{\partial}{\partial y}+2\right)\psi_{\pm,B}=0.\end{split}\right.$$

This system coincides with (15). Hence Theorem 3.3 yields

We set

$$g=\frac{X}{x\,s^{1/2}(1-s)^{1/2}}$$

in (13). Then we have the following cubic equation for $X$:

(22)

$$16X^{3}-3X-s^{1/2}(1-s)^{1/2}=0.$$

If $s=0$, we have $X=0,\,\pm\!\sqrt{3}/4$ and we find three roots $X_{1},X_{2},X_{3}$ of (22) near $s=0$ with expansions

(23)

$$\left\{\begin{split}\ X_{1}&=\frac{\sqrt{3}}{4}+\frac{1}{6}s^{1/2}-\frac{1}{6\sqrt{3}}s+\cdots,\\ X_{2}&=-\frac{\sqrt{3}}{4}+\frac{1}{6}s^{1/2}+\frac{1}{6\sqrt{3}}s+\cdots,\\ X_{3}&=-\frac{1}{3}s^{1/2}-\frac{5}{162}s^{3/2}-\cdots.\end{split}\right.$$

We take the analytic continuation of $X_{j}$ to $s=1$. If $s=1$, we also have $X=0,\,\pm\!\sqrt{3}/4$. For $s=1/2$, we have a double root $X=-1/4$ and a simple root $X=1/2$. The expansion of the branches of $X$ which merge at $s=1/2$ are

(24)

$$X=-\frac{1}{4}\pm\frac{1}{2\sqrt{3}}\left(s-\frac{1}{2}\right)+\frac{1}{18}\left(s-\frac{1}{2}\right)^{2}\pm\cdots.$$

The graphs of $Y=16X^{3}-3X-s^{1/2}(1-s)^{1/2}$ for $s=0,\,1/2,\,1$ are shown in Figures 4.1–4.3. Since the coefficients of $(s-1/2)$ of (24) do not vanish, two roots $X_{2}$ and $X_{3}$ pass each other at $s=1/2$ and the coefficients of the leading terms of the expansions of $(X_{1},X_{2},X_{3})$ at $s=1$ are $(\sqrt{3}/4,0,-\sqrt{3}/4)$.

$\displaystyle-\frac{\sqrt{3}}{4}$                     $\displaystyle\frac{\sqrt{3}}{4}$             $X_{2}=X_{3}$               $\displaystyle\frac{1}{2}$        $\displaystyle-\frac{\sqrt{3}}{4}$                       $\displaystyle\frac{\sqrt{3}}{4}$

$X_{2}$       $X_{3}$              $X_{1}$              $\displaystyle-\frac{1}{4}$                     $X_{1}$         $X_{3}$       $X_{2}$             $X_{1}$

$O$                                       $O$                                       $O$

Fig. 4.1. $s=0$                   Fig. 4.2. $s=1/2$                    Fig. 4.3. $s=1$

Note that if $x\neq 0$, then $\pm 2/3x^{3/2}$ are singularities of square root-type of $\psi_{+,B}$ and of $\psi_{-,B}$ in $y$-variable.

The Stokes curve (cf. [7]) of (2) is defined by

$${\rm Im}\int_{0}^{x}\sqrt{x}\,dx=0.$$

It consists of three half-lines

$$\arg x=0,\pm\frac{2}{3}\pi.$$

Let ${\mathcal{R}}_{\rm I}$ and ${\mathcal{R}}_{\rm II}$ denote the sectors (two of the Stokes regions)

$$\{\,x\in{\mathbb{C}}\,|\,-2/3\pi<\arg x<0\,\}\quad\mbox{and}\quad\{\,x\in{\mathbb{C}}\,|\,0<\arg x<2/3\pi\,\},$$

respectively. Let $\ell_{\pm}(x)$ be half-lines

$$\left\{\left.\,\mp\frac{2}{3}x^{3/2}+t\,\right|\,t\geq 0\,\right\}$$

with the positive orientation in the $y$-plane and we set

(27)

$$\Psi_{\pm}^{J}=\int_{\ell_{\pm}(x)}\psi_{\pm,B}(x,y)\exp(-y\eta)dy\quad\mbox{for}\ \ x\in{\mathcal{R}}_{J}\quad(J={\rm I,II}).$$

Figures 5.1 and 5.2 show ${\mathcal{R}}_{J}$ and $\ell_{\pm}(x)$ for $x\in{\mathcal{R}}_{J}$ ($J={\rm I,II}$), respectively. The wavy lines designate the branch cuts for $\psi_{+,B}$ in $y$-plane. Theorem 4.2 implies that these functions are well-defined and each $\Psi_{\pm}^{J}$ gives the Borel sum of $\psi_{\pm}$ in ${\mathcal{R}}_{J}$ ($J={\rm I,II}$), respectively.

$\frac{2}{3}x^{3/2}$                                 $\frac{2}{3}x^{3/2}$

$\ell_{-}(x)$

$-\frac{2}{3}x^{3/2}$                                $-\frac{2}{3}x^{3/2}$

${\mathcal{R}}_{\rm II}$                                        $\ell_{+}(x)$

Fig. 5.2.1. $x\in{\mathcal{R}}_{\rm II}$                  Fig. 5.2.2. $x\in{\mathcal{R}}_{\rm II}$

$O$

${\mathcal{R}}_{\rm I}$                                         $\ell_{+}(x)$

$-\frac{2}{3}x^{3/2}$                                $-\frac{2}{3}x^{3/2}$

$\frac{2}{3}x^{3/2}$                                 $\frac{2}{3}x^{3/2}$

$\ell_{-}(x)$

Fig. 5.1. Stokes regions           Fig. 5.2.3. $x\in{\mathcal{R}}_{\rm I}$                   Fig. 5.2.4. $x\in{\mathcal{R}}_{\rm I}$

$\tilde{\ell}_{+}(x)$            $\frac{2}{3}x^{3/2}$                                                 $\Gamma$

​​​$-\frac{2}{3}x^{3/2}$                                                                                       $\ell_{+}(x)$

Fig. 5.3.                                          Fig. 5.4.

In this section, we relate the Airy functions and the WKB solutions. We employ the same notation as in the previous sections.