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 $\mathcal{U}$.

A key technical point of this section is to construct a well-behaved solution of the Poisson equation

(4.88)

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

by applying separation of variables and the uniform estimate on the ODE solutions. For this purpose, we need the following monotonicity.