Let $\mathcal{C}^{n}$ be a Calabi model space applied to an ample line bundle $L$ over an $n-1$ dimensional compact Calabi-Yau manifold $(D,\omega_{D},\Omega_{D})$, then the Kähler form of the Calabi metric is given by

(5.8)

$$\omega_{\mathcal{C}^{n}}=\frac{n}{n+1}\sqrt{-1}\partial\bar{\partial}(-\log|\xi|^{2})^{\frac{n+1}{n}},$$

which is well-defined for $|\xi|<1$. In order to carry out separation of variables, we will study the local representation of the Laplace operator $\Delta_{\mathcal{C}^{n}}$ on $\mathcal{C}^{n}$. The authors have developed separation of variables in Section 4.1 of [HSVZ18], so we just briefly review the computations and basic estimates obtained there.

Let $\{w_{i}\}_{i=1}^{n-1}$ be some local holomorphic coordinates on $D$, and fix a local holomorphic trivialization $e_{0}$ of the line bundle $L$ with $|e_{0}|^{2}=e^{-\psi}$, where $\psi:D\to\mathbb{R}$ is a smooth function. So we get local holomorphic coordinates $(\underline{w},\zeta)\equiv(w_{1},\ldots,w_{n-1},\sigma)$ on $\mathcal{C}^{n}$ by writing a point $\xi\in\mathcal{C}^{n}$ as $\xi=\eta e_{0}(\underline{w})$, then $|\xi|^{2}=|\eta|^{2}e^{-\psi}$. We may assume $\psi(0)=1$, $d\psi(0)=0$ and $\sqrt{-1}\partial\bar{\partial}\psi=\omega_{D}$. Let $\pi:\mathcal{C}^{n}\rightarrow D$ be the obvious projection map. Denote

(5.9)

$$\rho\equiv|\xi|,$$

then we can write

(5.10)

$$w=\rho e^{\frac{\psi}{2}+\sqrt{-1}\theta},$$

where $\partial_{\theta}$ generates the natural $S^{1}$-rotation on the total space of $L$.

Now we fix some $\rho_{0}\in(0,1)$, and define $(Y^{2n-1},h_{0})$ to be the level set $\{\rho=\rho_{0}\}$ endowed with the induced Riemannian metric $h_{0}$. We denote by $\{\Lambda_{k}\}_{k=0}^{\infty}$ the spectrum of $\Delta_{h_{0}}$ with $\Lambda_{0}\equiv 0$, and let $\{\varphi_{k}\}_{k=0}^{\infty}$ be an orthonormal basis of (complex-valued) eigenfunctions which are homogeneous under the $S^{1}$ action and with

(5.11)

$$-\Delta_{h_{0}}\varphi_{k}=\Lambda_{k}\cdot\varphi_{k}.$$

From [HSVZ18], Section 4.1 we know that $\Lambda_{k}$ can be always represented as follows,

(5.12)

$$\Lambda_{k}=\frac{\lambda_{k}}{z_{0}}+\frac{nz_{0}^{n-1}\cdot j_{k}^{2}}{r_{0}^{2}}$$

such that $j_{k}\in\mathbb{N}$ and

(5.13)

$$\lambda_{k}\geq\frac{(n-1)\cdot j_{k}}{2}.$$

Notice $j_{k}$ and $\lambda_{k}$ have geometric meanings as explained in [HSVZ18], Section 4.1. Namely, $\varphi_{k}$ has weight $\pm j_{k}$ with respect to the $S^{1}$-action (notice the weight of $\bar{\varphi}_{k}$ is negative the weight of $\varphi_{k}$), and $\varphi_{k}$ corresponds to a smooth section of the induced complex line bundle over $D$, which is an eigenfunction of the $\bar{\partial}$-Hodge Laplacian with eigenvalue $\lambda_{k}$. In particular $\lambda_{0}=j_{0}=0$ and $\varphi_{0}$ is a constant. Moreover when $j_{k}=0$, $\varphi_{k}$ corresponds to an eigenfunction on $D$ and

(5.14)

$$\underline{\lambda}\equiv\inf\{\lambda_{k}>0|j_{k}=0,k\in\mathbb{Z}_{+}\}>0.$$

Now we carry out separation of variables for the Laplace equation on $\Delta_{\mathcal{C}^{n}}$. Let $u$ be a harmonic function on the model space $\mathcal{C}^{n}$, namely,

(5.15)

$$\Delta_{\mathcal{C}^{n}}u=0$$

For every fixed $z$, we can write the $L^{2}$-expansion along the fiber $Y^{2n-1}$,

(5.16)

$$u(z,\bm{y})=\sum\limits_{k=1}^{\infty}u_{k}(z)\cdot\varphi_{k}(\bm{y}).$$

The computations in [HSVZ18] tell us that for each $k\in\mathbb{N}$, $u_{k}(z)$ satisfies the differential equation

(5.17)

$$\frac{d^{2}u_{k}(z)}{dz^{2}}-(\frac{j_{k}^{2}n^{2}}{4}\cdot z^{n}+n\lambda_{k})z^{n-2}u_{k}(z)=0,\ z\geq 1.$$

We also consider the Poisson equation

(5.18)

$$\Delta_{\mathcal{C}^{n}}u=v.$$

Take the $L^{2}$-expansion of $v$ in the direction of the cross section $Y^{2n-1}$,

(5.19)

$$v(z,\bm{y})=\sum\limits_{k=1}^{\infty}\xi_{k}(z)\cdot\varphi_{k}(\bm{y}),$$

then the same procedure of separation of variables leads to an ordinary differential equation

(5.20)

$$\frac{d^{2}u_{k}(z)}{dz^{2}}-(\frac{j_{k}^{2}n^{2}}{4}\cdot z^{n}+n\lambda_{k})z^{n-2}u_{k}(z)=z^{n-1}\cdot\xi_{k}(z),\ z\geq 1.$$

Since we will study the solutions (5.16) and (5.19) in terms of the fiber-wise $L^{2}$-expansions, so there are two fundamental ingredients to analyze: First, in order to show the $L^{2}$-expansions in fact converge, we need to obtain some uniform estimates for the ODE solutions which are independent of the subscript $k\in\mathbb{N}$. The other basic aspect is to understand the asymptotics of the linearly independent solutions $\mathcal{G}_{k}(z)$ and $\mathcal{D}_{k}(z)$ as $z\to+\infty$, which in turn gives the asymptotics of the solutions (5.16) and (5.19).

Technically speaking, we will study the solutions to (5.17) and (5.20) in two different cases: $j_{k}=0$ and $j_{k}\neq 0$. The first step is to understand the solutions to homogeneous equation (5.17). Notice that, by using the change of variables $\zeta\equiv z^{n}$, (5.17) will become a homogeneous equation with linear coefficients, so that we can apply the theory of special functions to obtain some effective estimates for the solutions. Now letting

(5.21)

$\displaystyle\begin{cases}\zeta=-\log r^{2}=z^{n}\\ w_{k}(\zeta)\equiv u_{k}(z)=u_{k}(\zeta^{\frac{1}{n}}),\end{cases}$

we have

(5.22)

$$\zeta\cdot\frac{d^{2}w_{k}(\zeta)}{d\zeta^{2}}+(1-\frac{1}{n})\frac{dw_{k}(\zeta)}{d\zeta}-(\frac{j_{k}^{2}}{4}\cdot\zeta+\frac{\lambda_{k}}{n})w_{k}(\zeta)=0.$$

In the first case $j_{k}=0$, we make the transformation of the above solution $w(\zeta)$ as follows,

(5.23)

$$\begin{cases}y=2\sqrt{\frac{\lambda}{n}}\cdot\zeta^{\frac{1}{2}}\geq 0\\ w_{k}(\zeta)=\zeta^{\frac{1}{2n}}\cdot\mathcal{B}\Big(2\sqrt{\frac{\lambda}{n}}\cdot\zeta^{\frac{1}{2}}\Big),\end{cases}$$

then the function $\mathcal{B}(y)$ satisfies the modified Bessel equation,

(5.24)

$$y^{2}\cdot\frac{d^{2}\mathcal{B}(y)}{dy^{2}}+y\cdot\frac{d\mathcal{B}(y)}{dy}-(y^{2}+\frac{1}{n^{2}})\cdot\mathcal{B}(y)=0.$$

In the latter case $j_{k}\neq 0$, we make the following transformation

(5.25)

$\displaystyle\begin{cases}y=-j_{k}\cdot\zeta\leq 0\\ w_{k}(\zeta)=e^{\frac{j_{k}\cdot\zeta}{2}}\cdot\mathcal{J}(-j_{k}\cdot\zeta),\end{cases}$

then $\mathcal{J}(y)$ satisfies the confluent hypergeometric equation,

(5.26)

$$y\cdot\frac{d^{2}\mathcal{J}(y)}{dy^{2}}+(\fa-y)\cdot\frac{d\mathcal{J}(y)}{dy}-\fb\cdot\mathcal{J}(y)=0,$$

where

(5.27)

$\displaystyle\begin{cases}\fa=1-\frac{1}{n}\\ \fb=\frac{1}{2}(1-\frac{1}{n})-\frac{\lambda_{k}}{j_{k}\cdot n}.\end{cases}$

It is straightforward to see that $\alpha\in(0,1)$ and $\beta\in(-\infty,0]$.