We will focus on $\bar{\alpha}_{1}$. The periodic version of equation (2.7) on $\mathbb{R}^{2}_{\mu_{1},\mu_{2}}\times(S^{1}\times\mathbb{R})_{\eta}$ is the measure equation

$$(\Delta_{a}\tilde{\alpha}_{1})A^{3/2}d\mu_{1}\wedge d\mu_{2}\wedge dx\wedge dy=-2\pi\sqrt{A}\int_{\mathfrak{D}_{1}}d\mu_{2}.$$

Integrating in the periodic $x$-variable from 0 to 1,

(3.7)

$$(\Delta_{a}^{\prime}\bar{\alpha}_{1})d\text{Vol}_{a}^{\prime}=-2\pi\int_{\mathfrak{D}_{1}}d\mu_{2},$$

where $\Delta_{a}^{\prime}$ is the Laplacian of the metric $g_{a}^{\prime}=a_{ij}d\mu_{i}d\mu_{j}+Ady^{2}$ on $\mathbb{R}^{2}_{\mu_{1},\mu_{2}}\times\mathbb{R}$, whose volume form is $d\text{Vol}_{a}^{\prime}=Ad\mu_{1}\wedge d\mu_{2}\wedge dy$.

Now the basic strategy is to build a function satisfying the same measure equation and then compare. For a large positive cutoff $\Lambda$, we calculate the Green representation

$$\begin{split}&-\frac{1}{4\pi}\int_{0}^{\Lambda}\frac{-2\pi}{|(\mu_{1},\mu_{2}-s,y)|_{a}^{\prime}}ds=\frac{1}{2\sqrt{a_{22}}}\sinh^{-1}\left(\frac{s}{\sqrt{\frac{A}{a_{22}^{2}}\mu_{1}^{2}+\frac{A}{a_{22}}y^{2}}}\right)|_{s=-\mu_{2}-\frac{a_{12}}{a_{22}}\mu_{1}}^{s=-\mu_{2}-\frac{a_{12}}{a_{22}}\mu_{1}+\Lambda}.\end{split}$$

If we subtract $\frac{1}{2\sqrt{a_{22}}}\log(2\Lambda)$ and take the limit $\Lambda\to\infty$, we obtain the function

$$-\frac{1}{2\sqrt{a_{22}}}\log\left(\frac{1}{\sqrt{a_{22}}}|(\mu_{1},\mu_{2},y)|_{a}^{\prime}-\mu_{2}-\frac{a_{12}}{a_{22}}\mu_{1}\right)$$

which by construction satisfies the same measure equation as (3.7).

We claim that this function differs from $\bar{\alpha}_{1}$ by a constant. By the Liouville theorem, it suffices to show that the function $\bar{\alpha}_{1}$ on $\mathbb{R}^{2}\times\mathbb{R}$ has the logarithmic growth estimate

$$\bar{\alpha}_{1}\leq CA^{-1/4}\{\log(1+A^{-3/4}\varrho)+|\log(\frac{1}{a_{22}^{-1}\mu_{1}^{2}+y^{2}})|+1\}$$

which is easy to deduce from Lemma 3.2.

Now to pin down the constant, we can evaluate $\bar{\alpha}_{1}$ for $\mu_{1}=\mu_{2}=0,y\neq 0$. Then the $\arctan$ term drops out, and

$$\begin{split}\bar{\alpha}_{1}(0,0,y)&=\frac{1}{2\sqrt{a_{22}}}\lim_{\Lambda\to+\infty}\{\int_{0}^{\Lambda}\frac{1}{\sqrt{x^{2}+y^{2}}}dx-\sum_{n=1}^{\Lambda}\frac{1}{n}\}\\ &=\frac{1}{2\sqrt{a_{22}}}\lim_{\Lambda\to\infty}\{\sinh^{-1}(\frac{\Lambda}{|y|})-\log\Lambda-\gamma_{E}\}\\ &=\frac{1}{2\sqrt{a_{22}}}(\log(\frac{2}{|y|})-\gamma_{E})\end{split}$$

Comparing the expressions give the formula for $\bar{\alpha}_{1}$. ∎

We notice in advance that $\tilde{\alpha}_{1}$ and $\bar{\alpha}_{1}$ are periodic in $\eta$, so it suffices to assume $|x|\leq 1/2$. The main idea of a variant of Cauchy’s integral test for convergence.

Using the fact that $|\frac{\partial^{2}\alpha_{1}}{\partial x^{2}}|\leq C\frac{a_{22}}{(\mu_{1}^{2}+a_{22}|\eta|^{2})^{3/2}}$, and the mean value type inequality

$$f(0)-\int_{-1/2}^{1/2}f(s)ds\leq C\sup_{|s|\leq 1/2}|f^{\prime\prime}(s)|,$$

we deduce that for $|\eta|^{2}+\frac{\mu_{1}^{2}}{a_{22}}\gtrsim 1$,

$$|\alpha_{1}(\mu_{1},\mu_{2},\eta)-\int_{x-1/2}^{x+1/2}\alpha_{1}(\mu_{1},\mu_{2},s+\sqrt{-1}y)ds|\leq CA^{-1/4}(|\eta|^{2}+\frac{\mu_{1}^{2}}{a_{22}})^{-3/2}$$

Thus for $|x|\leq 1/2$,

$$\begin{split}&|\sum_{n\in\mathbb{Z}\setminus\{0\}}\{\alpha_{1}(\mu_{1},\mu_{2},\eta+n)-\int_{x-1/2}^{x+1/2}\alpha_{1}(\mu_{1},\mu_{2},s+n+\sqrt{-1}y)ds\}|\\ &\leq CA^{-1/4}\sum_{|n|\neq 0}(|\eta+n|^{2}+\frac{\mu_{1}^{2}}{a_{22}})^{-3/2}\leq CA^{-1/4},\end{split}$$

and the sum converges to zero as $\mu_{1}^{2}+|\eta|^{2}\to\infty$.

In particular if $y^{2}+a_{22}^{-1}\mu_{1}^{2}\gtrsim 1$, then adding the above two inequalities already implies the bound

$$\begin{split}|\tilde{\alpha}_{1}-\bar{\alpha}_{1}|&=|\sum_{n\in\mathbb{Z}}\{\alpha_{1}(\mu_{1},\mu_{2},\eta+n)-\int_{x-1/2}^{x+1/2}\alpha_{1}(\mu_{1},\mu_{2},s+n+\sqrt{-1}y)ds\}|\leq CA^{-1/4},\end{split}$$

and that $|\tilde{\alpha}_{1}-\bar{\alpha}_{1}|$ converges to zero as $\mu_{1}^{2}+|\eta|^{2}\to\infty$.

If however $y^{2}+a_{22}^{-1}\mu_{1}^{2}\ll 1$ but $\mu_{2}\lesssim-A^{1/4}$, then we can make

$$\frac{a_{22}\mu_{2}+a_{12}\mu_{1}}{A^{1/2}\sqrt{\mu_{1}^{2}+a_{22}|\eta|^{2}}}\lesssim-1,$$

and the Taylor expansion of $\arctan$ will ensure $|\alpha_{1}(\mu_{1},\mu_{2},\eta)|\leq\frac{C}{-\mu_{2}}$, so

$$|\alpha_{1}(\mu_{1},\mu_{2},\eta)-\int_{x-1/2}^{x+1/2}\alpha_{1}(\mu_{1},\mu_{2},s+\sqrt{-1}y)ds|\leq C\mu_{2}^{-1}\leq CA^{-1/4},$$

from which we again deduce $|\tilde{\alpha}_{1}-\bar{\alpha}_{1}|\leq CA^{-1/4}$. ∎

We focus on the region $\{\text{dist}_{g_{a}^{\prime}}(\cdot,\mathfrak{D}_{1})\gtrsim A^{1/2}\}$. The key idea is that $\tilde{\alpha}_{1}-\bar{\alpha}_{1}$ is $\Delta_{a}$-harmonic , bounded and has no zero Fourier mode in the $S^{1}$ direction defined by the $x$-variable, so the exponential decay follows from Fourier analysis. We remark that similar ideas have appeared in the recent paper [13].

We perform Fourier decomposition in the $S^{1}$ direction

$$\tilde{\alpha}_{1}-\bar{\alpha}_{1}=\sum_{n\neq 0}h_{n}(\mu_{1},\mu_{2},y)e^{2\pi inx},$$

Parseval identity combined with Lemma 3.4 shows

$$\sum_{n}|h_{n}|^{2}=\int_{0}^{1}|\tilde{\alpha}_{1}-\bar{\alpha}_{1}|^{2}dx\leq CA^{-1/2}.$$

Now $\Delta_{a}$-harmonicity translates into the 3-dimensional Helmholtz equations:

$$\Delta_{a}^{\prime}h_{n}-4\pi^{2}n^{2}A^{-1}h_{n}=0.$$

The remaining task is conceptually speaking to estimate the Dirichlet Green’s function for the Helmholtz equation on the noncompact 3-dimensional domain $\{\text{dist}_{g_{a}^{\prime}}(\cdot,\mathfrak{D}_{1})\gtrsim A^{1/2}\}$. In practice, building an upper barrier for the Green’s function suffices for our purpose.

Recall $\varrho=|(\mu_{1},\mu_{2},y)|_{a}^{\prime}$ is the distance function for the Euclidean metric $g_{a}^{\prime}$ on $\mathbb{R}^{2}_{\mu_{1},\mu_{2}}\times\mathbb{R}_{y}$. By simple direct computation, for any $\kappa>0$,

$$(\Delta_{a}^{\prime}-4\pi^{2}n^{2}A^{-1})e^{-\kappa\varrho}\leq(\kappa^{2}-4\pi^{2}n^{2}A^{-1})e^{-\kappa\varrho},$$

so for $k_{n}=2\pi|n|A^{-1/2}$, the function $e^{-k_{n}\varrho}$ is a supersolution of the Helmholtz equation. Now we build a barrier function

$$h_{n}^{\prime}(\mu_{1},\mu_{2},y)=A^{-1/2}\int_{0}^{\infty}e^{-k_{n}|(\mu_{1},\mu_{2}-s,y)|_{a}^{\prime}}ds,$$

whose singularity lies on $\mathfrak{D}_{1}$. Since $h_{n}^{\prime}$ is a positive superposition of supersolutions, it must be itself a supersolution. Other basic properties are:

• •

  On $\{\text{dist}_{g_{a}^{\prime}}(\cdot,\mathfrak{D}_{1})\geq A^{1/2}\}$, using the saddle point method for Laplace type integrals

  $$0\leq h_{n}^{\prime}\leq CA^{-3/4}\text{dist}_{g_{a}^{\prime}}(\cdot,\mathfrak{D}_{1})\exp(-k_{n}\text{dist}_{g_{a}^{\prime}}(\cdot,\mathfrak{D}_{1})).$$

• •

  On the boundary of $\{\text{dist}_{g_{a}^{\prime}}(\cdot,\mathfrak{D}_{1})\geq A^{1/2}\}$, we have $h_{n}^{\prime}\geq\frac{A^{-1/4}}{C|n|}$.

Since $|h_{n}|\leq CA^{-1/4}$ by the Parseval identity, the comparison principle implies

$$\begin{split}h_{n}\leq C|n|h_{n}^{\prime}\leq CA^{-3/4}|n|\text{dist}_{g_{a}^{\prime}}(\cdot,\mathfrak{D}_{1})\exp(-k_{n}\text{dist}_{g_{a}^{\prime}}(\cdot,\mathfrak{D}_{1})).\end{split}$$

Thus on $\{\text{dist}_{g_{a}^{\prime}}(\cdot,\mathfrak{D}_{1})\geq A^{1/2}\}$, the desired bound on $|\tilde{\alpha}_{1}-\bar{\alpha}_{1}|$ follows by summing over these estimates over $n$. It is worth commenting that we expect the exponential decay rate to be sharp. ∎

Clear from $\alpha_{1}(\mu_{1},\mu_{2},\eta)+{\alpha}_{1}(-\mu_{1},-\mu_{2},-\eta)=\frac{1}{2\sqrt{\mu_{1}^{2}+a_{22}|\eta|^{2}}}.$ ∎