We work with a Calabi model space $\mathcal{C}$ with a smooth divisor $D$ defined as in Section 3. The Calabi metric is given by

(4.3)

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

which is well-defined for $|\xi|_{h}<1$. In order to carry out separation of variables, we will study the local representation of the Laplace operator $\Delta_{\mathcal{C}}$ on $\mathcal{C}$.

We choose local holomorphic coordinates $\underline{z}=\{z_{i}\}_{i=1}^{n-1}$ on the smooth divisor $D$, 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 $(\underline{z},w)\equiv(z_{1},\ldots,z_{n-1},w)$ on $\mathcal{C}$ by writing a point $\xi\in\mathcal{C}$ as $\xi=we_{0}(\underline{z})$. Then $|\xi|_{h}^{2}=|w|^{2}e^{-\psi}$. We may assume $\psi(0)=1$, $d\psi(0)=0$ and $\sqrt{-1}\partial\bar{\partial}\psi=\omega_{0}$. Let $\pi:\mathcal{C}\rightarrow D$ be the obvious projection map. Then we obtain

(4.4)

$$\omega_{\mathcal{C}}=(-\log|\xi|^{2}_{h})^{\frac{1}{n}}\omega_{D}+\frac{1}{n}(-\log|\xi|_{h}^{2})^{\frac{1}{n}-1}\sqrt{-1}(\frac{dw}{w}-\partial\psi)\wedge(\frac{d\bar{w}}{\bar{w}}-\bar{\partial}\psi).$$

Let $u$ be a $C^{2}$-function in the Calabi space $\mathcal{C}$, the Laplacian at points in the fiber $\pi^{-1}(0)$ is given by

(4.5)

$$\Delta_{\mathcal{C}}u=(-\log|\xi|^{2}_{h})^{-\frac{1}{n}}\sum_{i=1}^{n-1}\frac{\partial^{2}u}{\partial z_{i}\partial\bar{z}_{i}}+n(-\log|\xi|_{h}^{2})^{-\frac{1}{n}+1}|w|^{2}\frac{\partial^{2}u}{\partial w\partial\bar{w}}.$$

Now denote $\varrho\equiv|\xi|_{h}$, then we can write

(4.6)

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

where $\partial_{\theta}$ generates the natural $S^{1}$-rotation on the total space of $L$. Then it is straightforward to check that

(4.7)

$$\frac{\partial\varrho}{\partial w}=\frac{\varrho}{2w},\ \frac{\partial\theta}{\partial w}=\frac{1}{2\sqrt{-1}w},\frac{\partial\varrho}{\partial z_{i}}=-\frac{1}{2}\varrho\partial_{z_{i}}\psi,\frac{\partial\theta}{\partial z_{i}}=0$$

and

(4.8)

$$|w|^{2}\frac{\partial^{2}}{\partial w\partial\bar{w}}u=\frac{1}{4}(\varrho^{2}u_{\varrho\varrho}+\varrho u_{\varrho}+u_{\theta\theta}).$$

For fixed $r_{0}\in(0,1)$, the level set $Y^{2n-1}\equiv\{\varrho=r_{0}\}$ is equipped with the induced Riemannian metric given by

(4.9)

$$h_{0}=(-\log r_{0}^{2})^{\frac{1}{n}}g_{D}+\frac{1}{n}(-\log r_{0}^{2})^{\frac{1}{n}-1}r_{0}^{2}(d\theta-\frac{1}{2}d^{c}\psi)\otimes(d\theta-\frac{1}{2}d^{c}\psi).$$

Now we consider a smooth function $\phi\in C^{\infty}(Y^{2n-1})$ with

(4.10)

$$\mathcal{L}_{\partial_{\theta}}\phi=\sqrt{-1}j\phi$$

for some integer $j$. Replacing $\phi$ by $\bar{\phi}$ if necessary we may assume $j\geq 0$. Then $\phi$ is induced by a smooth section $\hat{\phi}$ of $(L^{*})^{\otimes j}$. Precisely, if we locally write $\hat{\phi}(\underline{z})=\Phi(\underline{z})(e_{0}(\underline{z})^{*})^{\otimes j}$, then

(4.11)

$$\phi(\underline{z},w)=w^{j}\Phi(\underline{z})|_{\varrho=r_{0}}=r_{0}^{j}e^{j(\frac{\psi}{2}+\sqrt{-1}\theta)}\Phi(\underline{z}).$$

Now let $\hat{\phi}$ be a non-zero eigen-section of the $\bar{\partial}$-Laplace operator, i.e.

(4.12)

$$\Delta_{\bar{\partial}}\hat{\phi}=\hat{\lambda}\hat{\phi}.$$

By Kodaira-Nakano formula $\Delta_{\bar{\partial}}=\Delta_{\partial}+j(n-1)$, so we have $\hat{\lambda}\geq j(n-1)$. By a direct calculation, we get that on $\pi^{-1}(0)$,

(4.13)

$$\sum_{i=1}^{n-1}\frac{\partial^{2}\phi}{\partial z_{i}\partial\bar{z}_{i}}=\Big(-\hat{\lambda}+\frac{j(n-1)}{2}\Big)\phi.$$

Moreover, by the local expression of $h_{0}$ as in (4.9), one can directly check that on $\pi^{-1}(0)\cap Y^{2n-1}$,

(4.14)

$\displaystyle\begin{split}\Delta_{h_{0}}\phi&=(-\log r_{0}^{2})^{-\frac{1}{n}}\sum_{i}\frac{\partial^{2}\phi}{\partial z_{i}\partial\bar{z}_{i}}+n(-\log r_{0}^{2})^{-\frac{1}{n}+1}r_{0}^{-2}\phi_{\theta\theta}\\ &=\Big((-\log r_{0}^{2})^{-\frac{1}{n}}(-\hat{\lambda}+\frac{j(n-1)}{2})-j^{2}n(-\log r_{0}^{2})^{-\frac{1}{n}+1}r_{0}^{-2}\Big)\phi.\end{split}$

Now suppose a smooth function $u(\varrho,z)$ on the Calabi space $\mathcal{C}$ is of the form $u\equiv f(\varrho)\phi(y)$, where $\phi$ is a function on $Y^{2n-1}$ satisfying (4.10) and (4.13). In polar coordinates, we obtain

	$\displaystyle\Delta_{\mathcal{C}}u$	$\displaystyle=\phi(y)\cdot\Big((-\log\varrho^{2})^{-\frac{1}{n}}((-\hat{\lambda}+\frac{j(n-1)}{2})f-\frac{n-1}{2}{\varrho}f_{\varrho})$
		$\displaystyle\ \ \ \ +\frac{n}{4}(-\log{\varrho}^{2})^{1-\frac{1}{n}}({\varrho}^{2}f_{{\varrho}{\varrho}}+{\varrho}f_{\varrho}-j^{2}f)\Big)$
		$\displaystyle=\phi(y)\cdot(-\log\varrho^{2})^{-\frac{1}{n}}\Big(\frac{n}{4}(-\log{\varrho}^{2})({\varrho}^{2}f_{{\varrho}{\varrho}}+{\varrho}f_{\varrho}-j^{2}f)$
(4.15)			$\displaystyle\ \ \ \ -\frac{n-1}{2}{\varrho}f_{\varrho}-(\hat{\lambda}-\frac{j(n-1)}{2})f\Big).$

Notice this formula is now independent of the choice of local holomorphic coordinates. So $u$ is harmonic if and only if

(4.16)

$$\frac{n}{4}(-\log{\varrho}^{2})(\varrho^{2}f_{{\varrho}{\varrho}}+\varrho f_{\varrho}-j^{2}f)-\frac{n-1}{2}{\varrho}f_{\varrho}-(\hat{\lambda}-\frac{j(n-1)}{2})f=0.$$

Denote $z=(-\log\varrho^{2})^{\frac{1}{n}}$, then we get

(4.17)

$$f_{zz}-(n(\hat{\lambda}-\frac{j(n-1)}{2})+\frac{j^{2}n^{2}}{4}z^{n})z^{n-2}f=0.$$

In this section, we will also analyze the Poisson equation

(4.18)

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

Suppose now $v\equiv\zeta(\varrho)\cdot\phi(y)$, then the same separation of variables gives the following ODE

(4.19)

$$f_{zz}-(n(\hat{\lambda}-\frac{j(n-1)}{2})+\frac{j^{2}n^{2}}{4}z^{n})z^{n-2}f=z^{n-1}\cdot\zeta.$$

We remark that a similar separation of variables was carried out in [KK10], but we will need stronger estimates on solutions in order to prove Theorem 4.3.

For our application we focus on the case $n=2$. So the corresponding ODEs become

(4.20)

$$f_{zz}-(\lambda+j^{2}z^{2})f=0$$

and

(4.21)

$$f_{zz}-(\lambda+j^{2}z^{2})f=z\cdot\zeta,$$

where

(4.22)

$$\lambda\equiv 2\hat{\lambda}-j\geq j.$$

We have assumed $j\geq 0$ in the above discussion, but notice that the Laplace operator is a real operator, so the ODEs we get for $j$ and $-j$ are the same. Denote $z_{0}\equiv(-\log r_{0}^{2})^{\frac{1}{2}}$, then we notice that each eigenvalue of $\Delta_{h_{0}}$ can be represented by

(4.23)

$$\Lambda=\frac{\lambda}{2z_{0}}+\frac{2z_{0}\cdot j^{2}}{r_{0}^{2}}.$$

With the above computations, we are ready to set up the ODE system. Now we fix some $r_{0}\in(0,1)$, and define $(Y^{3},h_{0})$ to be the level set $\{r=r_{0}\}$ endowed with the induced Riemannian metric $h_{0}$. The above computations tell us that the eigenvalues of $Y^{3}$ is given by linear combinations of $\hat{\lambda}$ and $j$. Below we will parametrize our summation in terms of eigenvalues of $(Y^{3},h_{0})$ (counted with multiplicity), but we shall keep in mind that we have further split the eigenspaces of $\Delta_{h_{0}}$ according to the $S^{1}$ action hence an eigenvalue is naturally written in terms of a linear combination of $\hat{\lambda}$ and $j$.

We denote by $\{\Lambda_{k}\}_{k=1}^{\infty}$ the spectrum of $\Delta_{h_{0}}$ and let $\{\varphi_{k}\}_{k=1}^{\infty}$ be the eigenfunctions which are homogeneous under the $S^{1}$ action and with

(4.24)

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

In the above notations, one can compute that in the case $n=2$,

(4.25)

$$\Lambda_{k}=\frac{\lambda_{k}}{2z_{0}}+\frac{2z_{0}\cdot j_{k}^{2}}{r_{0}^{2}}.$$

First, we carry out separation of variables for harmonic functions on $\Delta_{\mathcal{C}}$. Let $u$ be a harmonic function on the model space $\mathcal{C}$, namely,

(4.26)

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

For every fixed $z$, we can write the $L^{2}$-expansion along the fiber $Y^{3}$,

(4.27)

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

The above computations tell us that for each $k\in\mathbb{Z}_{+}$, there are numbers $j_{k}\in\mathbb{N}$ and $\lambda_{k}\geq nj_{k}$ such that the function $u_{k}(z)$ satisfies the differential equation

(4.28)

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

We also consider the Poisson equation

(4.29)

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

Take the $L^{2}$-expansion of $v$ in the direction of the cross section $Y^{3}$,

(4.30)

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

then the same procedure of separation of variables leads to a differential equation

(4.31)

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

We end this subsection by giving a model example of the fiber $Y^{3}$.

In the following subsections, we will analyze the convergence and regularity issues of the formal solutions (4.28) and (4.31).