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).

A crucial step in applying the method of separation of variables is to prove the $C^{0}$-regularity of a formal solution obtained from the above separation of variables. Specifically, in our context, to prove such a $C^{0}$-regularity result, first we need to obtain some effective estimates for the fundamental solutions to the linear differential equation (see (4.28))

(4.35)

$$\frac{d^{2}u(z)}{dz^{2}}-(j^{2}z^{2}+\lambda)u(z)=0,$$

which arises from the harmonic functions on the Calabi manifold $(\mathcal{C},g_{\mathcal{C}})$. In our context, we always require

(4.36)

$$j\geq 0,\ h\geq 0,\ z>1.$$

There are two different cases to analyze.

The first case is much simpler, i.e. $j=0$ and the ODE becomes

(4.37)

$$\frac{d^{2}u(z)}{dz^{2}}=\lambda\cdot u(z).$$

Further, if $\lambda=0$, the solutions to (4.37) are linear. If $\lambda>0$, the above equation has two linearly independent solutions $e^{\sqrt{\lambda}z}$ and $e^{-\sqrt{\lambda}z}$. All the required estimates in this case are standard and straightforward. Geometrically, the ODE analysis for (4.37) arises naturally from the flat cylindrical geometry and corresponding gluing constructions.

So in our case, we only focus on the case $j\in\mathbb{Z}_{+}$ which is substantially much more technically involved. In the case $j\in\mathbb{Z}_{+}$, we have already shown in Section 4.1 that $\lambda$ and $j$ satisfy the relation

(4.38)

$$\lambda\geq j\geq 1.$$

Hence for each pair of $\lambda$ and $j$ satisfying the above, we choose $h\geq 0$ such that

(4.39)

$$\lambda=(2h+1)j.$$

From now on, we focus on the differential equation for every $j\in\mathbb{Z}_{+}$ and $h\geq 0$,

(4.40)

$$\frac{d^{2}u(z)}{dz^{2}}=j(jz^{2}+2h+1)u(z).$$

We will simplify the above equation by the following transformations. Let

(4.41)

$$y=\sqrt{j}z,\ V(y)=u\Big(\frac{y}{\sqrt{j}}\Big),$$

then $V(y)$ satisfies

(4.42)

$$\frac{d^{2}V(y)}{dy^{2}}=(y^{2}+(2h+1))V(y).$$

Further, we make the transformation

(4.43)

$$V(y)=e^{-\frac{y^{2}}{2}}Q(y),$$

then $Q$ sovles the differential equation

(4.44)

$$\frac{d^{2}Q(y)}{dy^{2}}-2y\frac{dQ(y)}{dy}-2(h+1)Q(y)=0.$$

Notice that equation (4.44) is invariant under the change of variables $y\mapsto-y$. Given $y>1$ and $h\geq 0$, we define the following exponential integral

(4.45)

$$H_{-h-1}(y)\equiv\int_{0}^{\infty}e^{-t^{2}-2ty}t^{h}dt.$$

Straightforward calculations show that for each given $h\geq 0$, the functions $H_{-h-1}(y)$ and $H_{-h-1}(-y)$ are linearly independent solutions to (4.44). In fact, the above solutions coincide with the usual Hermite functions up to a constant (see [Leb72] for more details). Eventually, we obtain two solutions to (4.40),

(4.46)

$$\mathcal{F}(z)=e^{-\frac{jz^{2}}{2}}H_{-h-1}(-\sqrt{j}z)=e^{-\frac{y^{2}}{2}}\int_{0}^{\infty}e^{-t^{2}+2ty+h\log t}dt$$

and

(4.47)

$$\mathcal{U}(z)=e^{-\frac{jz^{2}}{2}}H_{-h-1}(\sqrt{j}z)=e^{-\frac{y^{2}}{2}}\int_{0}^{\infty}e^{-t^{2}-2ty+h\log t}dt.$$

The lemma below shows that $\mathcal{F}$ and $\mathcal{U}$ are two linearly independent solutions.

The regularity of the formal solutions obtained from the above separation of variables requires very precise uniform estimates for the fundamental solutions $\mathcal{F}$ and $\mathcal{U}$. We will use the Laplace Method, which is inspired by [SS16] in a different context. Again we denote $y=\sqrt{j}z$ and define

(4.51)

$\displaystyle\begin{split}F(t)&\equiv-t^{2}+2ty+h\log t\\ U(t)&\equiv-t^{2}-2ty+h\log t.\end{split}$

Straightforward computations tell us that both $F$ and $U$ are strictly concave when $h\geq 0$. For fixed $y$, let $t_{0}$ and $s_{0}$ be the unique (positive) critical points of $F$ and $U$ respectively. It is straightforward that

(4.52)

$\displaystyle\begin{split}t_{0}&=\frac{y}{2}+\sqrt{\frac{h^{2}}{2}+\frac{y^{2}}{4}}\\ s_{0}&=-\frac{y}{2}+\sqrt{\frac{h^{2}}{2}+\frac{y^{2}}{4}}.\end{split}$

Apply the same method as in Lemma 4.6, we have the following asymptotic property of $\mathcal{F}$ and $\mathcal{U}$.