# Expansion of Green’s function and regularity of Robin’s function for
elliptic operators in divergence form

Daomin Cao and Jie Wan  Institute of Applied Mathematics, Chinese Academy of
Sciences, Beijing 100190, and University of Chinese Academy of Sciences,
Beijing 100049, P.R. China<EMAIL_ADDRESS>School of Mathematics and
Statistics, Beijing Institute of Technology, Beijing 100081, P.R. China
<EMAIL_ADDRESS>
###### Abstract.

We consider Green’s function $G_{K}$ of the elliptic operator in divergence
form $\mathcal{L}_{K}=-\text{div}(K(x)\nabla)$ on a bounded smooth domain
$\Omega\subseteq\mathbb{R}^{n}(n\geq 2)$ with zero Dirichlet boundary
condition, where $K$ is a smooth positively definite matrix-valued function on
$\Omega$. We obtain a high-order asymptotic expansion of $G_{K}(x,y)$, which
defines uniquely a regular part $H_{K}(x,y)$. Moreover, we prove that the
associated Robin’s function $R_{K}(x)=H_{K}(x,x)$ is smooth in $\Omega$,
despite the regular part $H_{K}\notin C^{1}(\Omega\times\Omega)$ in general.

Keywords: Second-order elliptic operators in divergence form; Green’s
function; Robin’s function.

##  1\. Introduction and main results

Our purpose in this article is to study the expansion of Green’s function and
regularity of Robin’s function for second-order elliptic operators in
divergence form in all dimensions. Let $\Omega\in\mathbb{R}^{n}$ be a bounded
smooth domain with $n\geq 2$. Consider the following elliptic operator:

$\mathcal{L}_{K}u=-\text{div}(K(x)\nabla u),$

where $K=(K_{i,j})_{2\times 2}$ is a positively definite matrix satisfying

  1. ($\mathcal{K}$1).

$K_{i,j}\in C^{\infty}(\overline{\Omega})$ for $1\leq i,j\leq 2;$

  2. ($\mathcal{K}$2).

There exist constants $\Lambda_{1},\Lambda_{2}>0$ such that

$\Lambda_{1}|\zeta|^{2}\leq(K(x)\zeta|\zeta)\leq\Lambda_{2}|\zeta|^{2},\ \ \ \
\forall\ x\in\Omega,\ \zeta\in\mathbb{R}^{2}.$

Let $G_{K}(x,y)$ be the Green’s function of $\mathcal{L}_{K}$ associated to
zero Dirichlet boundary condition, i.e., for any $y\in\Omega$,

$\begin{cases}\mathcal{L}_{K}G_{K}(x,y)=\delta_{y}\ \ &\text{in}\ \Omega,\\\
G_{K}(x,y)=0\ \ &\text{on}\ \partial\Omega.\end{cases}$ (1.1)

Here $\delta_{y}$ is the Dirac measure centered at $y$. Multiplying test
functions on both sides of the first equation of (1.1) and integrating on
$\Omega$, we get the equivalent characterization of Green’s function in
integral form, that is, for any $u\in C^{2}(\Omega)\cap
C^{1}_{0}(\overline{\Omega})$

$u(y)=\int_{\Omega}G_{K}(x,y)\mathcal{L}_{K}u(x)dx,\ \ \forall y\in\Omega.$
(1.2)

Here $C^{2}(\Omega)\cap C^{1}_{0}(\overline{\Omega})$ is the space of all
functions which are twice continuously differentiable and equal 0 on the
boundary $\partial\Omega$. The well-posedness of Green’s function $G_{K}$
defined by (1.1) is considered in many articles. When the dimension $n\geq 3$,
it is well known that Green’s function of $\mathcal{L}_{K}$ exists and is
unique, see [18]. Similar results is obtained in [15] when the dimension
$n=2$. More results can be found in [7, 11, 13, 14, 19, 23] for instance.

The asymptotic expansion of Green’s function of $\mathcal{L}_{K}$, meanwhile,
has also been widely concerned in recent decades. The understanding of the
expansion of Green’s function plays an essential role in many fields,
especially in the study of concentration phenomena for some fluid mechanics
models and semilinear elliptic equations, see, e.g., [1, 2, 3, 9, 12, 21, 25,
24]. For many concentration phenomena, the location of possible singularities
of solutions is always determined by the corresponding Green’s function, and
the solutions are always perturbations of Green’s function. A typical example
is the construction of concentrated vortex solutions to two dimensional, three
dimensional axisymmetric and three dimensional helical symmetric
incompressible Euler equations. For planar Euler equations, one always gets
concentrated vortex solutions by solving an elliptic problem

$-\Delta u=\lambda f(u)\ \ \text{in}\ \Omega;\ \ u=0\ \ \text{on}\
\partial\Omega,$

and proving the asymptotic behavior of solutions $u_{\lambda}$ as $\lambda$
tends to infinity, see e.g., [2, 3, 8, 20, 22]. The limiting location is
determined by the Kirchhoff-Routh function (see [17]), which is a combination
of Green’s function of $-\Delta$ and the corresponding Robin’s function. For
the construction of concentrated vortex solutions to 3D axisymmetric Euler
equations (called vortex rings) and 3D Euler equations with helical symmetry,
corresponding elliptic operators appeared in elliptic problems are
$-\frac{1}{a(x)}\text{div}(a(x)\nabla)$ and $-\text{div}(K_{H}(x)\nabla)$ for
some positive function $a$ and positively definite matrix $K_{H}$
respectively, see e.g., [4, 5, 9, 10, 12]. When constructing a family of
vortex solutions concentrating near several points, one always need to use the
asymptotic expansion of the corresponding Green’s function for these elliptic
operators. Note that the $C^{1}$ regularity of the corresponding Robin’s
function is necessary when constructing concentrated solutions by perturbation
arguments, see [3, 6, 25] for example.

The most common case is $K\equiv Id_{n\times n}$, i.e., the identity matrix of
order $n$, and then the operator $\mathcal{L}_{K}$ is the standard Laplacian.
In this case the associated Green’s function $G_{\Omega}$ of $-\Delta$ in a
domain $\Omega$ with Dirichlet condition is of the decomposition (see [13])

$G_{\Omega}(x,y)=\Phi_{0}(x-y)-H_{\Omega}(x,y),$ (1.3)

where $\Phi_{0}$ is the fundamental solution of $-\Delta$ defined by

$\Phi_{0}(x)=\begin{cases}-\frac{1}{2\pi}\log|x|,\ \ &\text{for}\ n=2;\\\
\frac{1}{n(n-2)w_{n}}|x|^{2-n},\ \ &\text{for}\ n\geq 3;\end{cases}$ (1.4)

with $w_{n}$ the Lebesgue measure of the unit ball in $\mathbb{R}^{n}$, and
$H_{\Omega}$ is the regular part of $G_{\Omega}$ satisfying for any
$y\in\Omega$

$\begin{cases}-\Delta H_{\Omega}(x,y)=0,\ \ &x\in\Omega,\\\
H_{\Omega}(x,y)=\Phi_{0}(x-y),\ \ &x\in\partial\Omega.\end{cases}$ (1.5)

Clearly $H_{\Omega}$ is a function determined by $\Omega$. Moreover,
$H_{\Omega}$ is smooth and symmetric in $\Omega\times\Omega$, i.e.,
$H_{\Omega}(x,y)=H_{\Omega}(y,x)$. Thus $R_{\Omega}(x)=H_{\Omega}(x,x)$,
called the Robin’s function, is smooth in $\Omega.$

When $K$ is a positively definite constant coefficient matrix $K_{0}$, we can
also get the expansion of associated Green’s function. Let $T_{0}$ be the
positively definite matrix satisfying

$T_{0}^{-1}T_{0}^{-t}=K_{0},$

where $T_{0}^{-t}$ is the transpose of $T_{0}^{-1}.$ Denote
$T_{0}\Omega=\\{T_{0}x\mid x\in\Omega\\}$. Using the coordinate transformation
(see Lemma A.1 in Appendix), we find that Green’s function $G_{K_{0}}$ of
$\mathcal{L}_{K_{0}}$ has the decomposition

$G_{K_{0}}(x,y)=\sqrt{\det K_{0}}^{-1}\Phi_{0}(T_{0}(x-y))-\sqrt{\det
K_{0}}^{-1}H_{T_{0}\Omega}(T_{0}x,T_{0}y)\ \ \forall x,y\in\Omega,x\neq y,$

where $H_{T_{0}\Omega}:T_{0}\Omega\times T_{0}\Omega\to\mathbb{R}$ satisfies
for any $y^{\prime}\in T_{0}\Omega$

$\begin{cases}-\Delta H_{T_{0}\Omega}(x^{\prime},y^{\prime})=0,\ \
&x^{\prime}\in T_{0}\Omega,\\\
H_{T_{0}\Omega}(x^{\prime},y^{\prime})=\Phi_{0}(x^{\prime}-y^{\prime}),\ \
&x^{\prime}\in\partial T_{0}\Omega.\end{cases}$

Clearly the corresponding Robin’s function $R_{T_{0}\Omega}(x)=\sqrt{\det
K_{0}}^{-1}H_{T_{0}\Omega}(T_{0}x,T_{0}x)$ is also smooth in $\Omega.$

When $K$ is a matrix-valued function rather than a constant coefficient
matrix, the situation turns out to be less clear. If $K$ is a diagonal matrix
with the same diagonal elements, i.e., $K(\cdot)=a(\cdot)Id_{n\times n}$ for
some positive smooth function $a$, [16] constructed the expansion of Green’s
function and proved the smoothness of the corresponding Robin’s function. For
any $l\in\mathbb{N}$, it was proved in [16] that Green’s function $G_{a}$ of
$-\Delta_{a}=-\frac{1}{a(x)}\text{div}(a(x)\nabla)$ has the high-order
expansion

$G_{a}(x,y)=\Phi_{0}(x-y)+\sum_{k=1}^{n+l-2}\Phi_{k}(x-y)+H^{l}_{a}(x,y)\ \
\text{in}\ \overline{\Omega}\times\Omega,$

where $H_{a}^{l}(x,y)\in C^{l,\gamma}(\overline{\Omega}\times\Omega)$, see
also [25]. An interesting phenomena is, for any $n\geq 2$ if $\nabla a(y)\neq
0,$ then $x\mapsto H^{0}_{a}(x,y)$ does not belong to $H^{2}(\Omega)$ since
$-\Delta_{a}H^{0}_{a}(x,y)\notin L^{2}(\Omega)$. However, $H^{0}_{a}(x,x)$ is
smooth in $\Omega$, see Proposition 2.7 in [16]. For the case $K$ being a
general positively definite matrix-valued function, it follows from [5] that
Green’s function $G_{K}$ has the following structure when $n=2$ (see [5],
theorem 1.2)

$G_{K}(x,y)=\frac{\sqrt{\det K(x)}^{-1}+\sqrt{\det
K(y)}^{-1}}{2}\Phi_{0}\left(\frac{T_{x}+T_{y}}{2}(x-y)\right)+S_{K}(x,y),$
(1.6)

where $S_{K}\in C^{0,\gamma}(\Omega\times\Omega)$ for all $\gamma\in(0,1)$ and
$S_{K}(x,y)=S_{K}(y,x)$ for $x,y\in\Omega.$ See also [4]. This implies that
the regular part $S_{K}$ of $G_{K}$ is just $C^{0,\gamma}$, not smooth. Note
that (1.6) does not give us a high-order expansion of Green’s function
$G_{K}(x,y)$, nor proves whether the corresponding Robin’s function
$S_{K}(x,x)$ is smooth. For the dimension $n\geq 3$, such results are also
unknown for our knowledge. In this article, we intend to study this aspect.
For the cases that dimension $n\geq 2$, we construct a high-order asymptotic
expansion of Green’s function $G_{K}$ of (1.1), which permits us to define
uniquely a regular part $H_{K}$. Then we prove the smoothness of the
corresponding Robin’s function $R_{K}(x)=H_{K}(x,x)$. We find that $R_{K}\in
C^{\infty}(\Omega)$, although $H_{K}$ does not belong to
$C^{1}(\Omega\times\Omega)$ in general. Note that the fundamental solution of
$-\Delta$ for $n=2$ has a logarithmic term, which is different from that of
$n\geq 3$. Therefore, we get results by dealing with the odd and even cases of
$n$ separately.

Before stating our main results, let us first introduce some notations. Let
$n\geq 2$ and define $T_{x}$ a $C^{\infty}$ positively definite matrix-valued
function determined by $K$ satisfying

$T_{x}^{-1}(T_{x}^{-1})^{t}=K(x)\ \ \ \ \forall x\in\Omega.$ (1.7)

Clearly such $T_{x}$ exists and is unique. For any multi-index
$\alpha=(\alpha_{i})\in\mathbb{N}^{n}$ and $x\in\mathbb{R}^{n}$, we denote
$|\alpha|=\sum_{i=1}^{n}\alpha_{i}$ and
$x^{\alpha}=\Pi_{i=1}^{n}x_{i}^{\alpha_{i}}$. We define a linear space
$E^{n+2m}_{k+2m}$ as follows

######  Definition 1.1.

Given $k\in\mathbb{N}^{*},m\in\mathbb{N}$, denote

$E^{n+2m}_{k+2m}=\text{span}\left\\{\frac{x^{\alpha}}{r^{n+2m}}\mid|\alpha|=k+2m,\alpha\in\mathbb{N}^{n}\right\\},$

where $r=||x||$ is the classical Euclidean norm of $x$.

Our first result is on the expansion of Green’s function of (1.1) when $n$ is
odd.

######  Theorem 1.2.

Let $n\geq 3$ be odd and $\gamma\in(0,1)$ be an arbitrary constant. Then for
any $l\in\mathbb{N}$, there exists a unique $\Phi_{i}\in
E^{n+2(2i-1)}_{i+2+2(2i-1)}$ for $i=1,\cdots,n+l-2$ depending on $y\in\Omega$
and $H^{l}(x,y)=H^{l}_{y}(x)\in
C\left(\Omega,C^{l,\gamma}\left(\overline{\Omega}\right)\right)\cap
C^{l}(\Omega\times\Omega)$ such that

$G_{K}(x,y)=\sqrt{\det
K(y)}^{-1}\Phi_{0}\left(T_{y}(x-y)\right)+\sum_{i=1}^{n+l-2}\Phi_{i}\left(T_{y}(x-y)\right)+H^{l}(x,y)\
\ \text{in}\ \overline{\Omega}\times\Omega.$

######  Remark 1.3.

Note that from Theorem 1.2, each term in the expansion of Green’s function
$G_{K}$ is determined by $\Phi_{k}\left(T_{y}(x-y)\right)$, which is different
from the term $\Phi_{k}(x-y)$ in the expansion of Green’s function of
$-\Delta$ and $-\Delta_{a}=-\frac{1}{a(x)}\text{div}\left(a(x)\nabla\right)$,
see [13, 16]. However, if we choose $K$ to be $Id_{n\times n}$ and
$aId_{n\times n}$, the expansion of $G_{K}$ in Theorem 1.2 coincides with the
classical results in [13, 16].

Concerning the cases with even dimension, since the fundamental solution of
$-\Delta$ in dimension two is $-\frac{1}{2\pi}\log|x|$, we need to introduce
some notations about singular and logarithmic terms. Let $n\in
2\mathbb{N}^{*},k\in\mathbb{N}^{*},m\in\mathbb{N}$ and $E^{n+2m}_{k+2m}$ be
defined as in Definition 1.1. Let $\mathbb{R}[x]$ be the set of real
polynomials with variables $x_{i}$. When $k\geq n,$ we denote by
$E^{n+2m,s}_{k+2m}$ the singular set of $E^{n+2m}_{k+2m}$, i.e.,
$E^{n+2m,s}_{k+2m}=E^{n+2m}_{k+2m}\backslash\mathbb{R}[x]$. Denote

$L_{m}=\text{span}\\{x^{\alpha}\log
r\mid|\alpha|=m,\alpha\in\mathbb{N}^{n}\\}.$

We define $F^{n+2m}_{k+2m}$ as

$F^{n+2m}_{k+2m}=\begin{cases}E^{n+2m}_{k+2m}\ &k<n,\\\
E^{n+2m,s}_{k+2m}\oplus L_{k-n}\ &k\geq n.\end{cases}$

When the dimension $n$ is even we have the following result concerning the
expansion of Green’s function.

######  Theorem 1.4.

Let $n\in 2\mathbb{N}^{*}$ and $\gamma\in(0,1)$ be an arbitrary constant. Then
for any $l\in\mathbb{N}$, there exists a unique $\Phi_{i}\in
F^{n+2(2i-1)}_{i+2+2(2i-1)}$ for $i=1,\cdots,n+l-2$ depending on $y\in\Omega$
and $H^{l}(x,y)=H^{l}_{y}(x)\in
C\left(\Omega,C^{l,\gamma}\left(\overline{\Omega}\right)\right)\cap
C^{l}(\Omega\times\Omega)$ such that

$G_{K}(x,y)=\sqrt{\det
K(y)}^{-1}\Phi_{0}\left(T_{y}(x-y)\right)+\sum_{i=1}^{n+l-2}\Phi_{i}\left(T_{y}(x-y)\right)+H^{l}(x,y)\
\ \text{in}\ \overline{\Omega}\times\Omega.$

######  Remark 1.5.

Compare with Theorem 1.2, the only difference is $\Phi_{i}\in
F^{n+2(2i-1)}_{i+2+2(2i-1)}$, rather than $E^{n+2(2i-1)}_{i+2+2(2i-1)}$. Note
also that when taking $n=2$ and $l=0$, Theorem 1.4 yields that

$G_{K}(x,y)=\sqrt{\det K(y)}^{-1}\Phi_{0}(T_{y}(x-y))+H^{0}(x,y),$

where $H^{0}(x,y)\in C^{0,\gamma}(\overline{\Omega}\times\Omega)$. This
expansion coincides with (1.6) in [5], since in this case
$S_{K}(x,y)=H^{0}(x,y)+\sqrt{\det
K(y)}^{-1}\Phi_{0}(T_{y}(x-y))-\frac{\sqrt{\det K(x)}^{-1}+\sqrt{\det
K(y)}^{-1}}{2}\Phi_{0}\left(\frac{T_{x}+T_{y}}{2}(x-y)\right)$ belongs to
$C^{0,\gamma}(\Omega\times\Omega)$.

Our last result is on the smoothness of Robin’s function, regardless of the
parity of $n$. Let us define the regular part of $G_{K}$ and the associated
Robin’s function $R_{K}$

$H_{K}(x,y)=H^{0}(x,y)\ \ \forall x,y\in\Omega,$ $R_{K}(x)=H_{K}(x,x)\ \
\forall x\in\Omega.$

It follows from Theorem 1.2 and Theorem 1.4 that $H_{K}\in
C\left(\Omega,C^{0,\gamma}\left(\overline{\Omega}\right)\right)\cap
C(\Omega\times\Omega)$. Note that this regularity is optimal since generally
$\mathcal{L}_{K}H_{K}(\cdot,y)\notin L^{2}(\Omega)$ and thus $H_{K}(\cdot,y)$
does not belong to $H^{2}(\Omega)$ for $n\geq 2$. However, we have

######  Theorem 1.6.

Let $n\geq 2$ be an integer, $\Omega$ be a bounded smooth domain and $K$ be a
positively definite smooth matrix-valued function in $\Omega$. Then the
Robin’s function $R_{K}(\cdot)\in C^{\infty}(\Omega)$.

######  Remark 1.7.

Note that for $n=2$, it is proved in [5] that Green’s function $G_{K}$ has a
structure (1.6), where the regular part $S_{K}$ belongs to $C^{0,\gamma}$.
Since $S_{K}(x,x)=R_{K}(x)$ for any $x\in\Omega$, using Theorem 1.6, we
actually show that $S_{K}(x,x)$ is not just $C^{0,\gamma}$, but smooth indeed,
which improves the regularity results of the regular part $S_{K}(x,x)$ in [4,
5].

The paper is organized as follows. In section 2, we prove the expansion of
$G_{K}$ when $n$ is odd. By similar strategy we prove the even dimensional
case in section 3. Based on Theorem 1.2 and 1.4, we finish the proof of
Theorem 1.6 at the end of section 3.

##  2\. Odd dimensional case

Now we begin to prove the expansion of Green’s function $G_{K}$ of (1.1) when
$n\geq 3$ is odd. For any subset $A_{j}$ in a vector space $V$, we denote
$A_{0}+A_{1}+\cdots+A_{p}=\\{\sum_{j=0}^{p}v_{j}\mid v_{j}\in A_{j}\\}.$ We
write $A_{0}\oplus A_{1}\oplus\cdots\oplus A_{p}$ when the sum is direct. For
any $l\in\mathbb{N},2l\leq k+2m$, we define

$E^{n+2m}_{k+2m,l}=\text{span}\left\\{\frac{x^{\alpha}}{r^{n+2m-2l}}\mid|\alpha|=k+2m-2l,\alpha\in\mathbb{N}^{n}\right\\}.$

Note that from Definition 1.1, for any $k\in\mathbb{N}^{*}$,
$E^{n+2m}_{K+2m}\in L^{1}_{loc}(\mathbb{R}^{n})$ and for any $m\in\mathbb{N}$,
$i,j\in\mathbb{N}^{*},i\neq j$

$E^{n+2m}_{i+2m}\cap E^{n+2m}_{j+2m}=\\{0\\}.$

Moreover, for any $m,i\in\mathbb{N}^{*}$

$E^{n}_{i}\subsetneqq E^{n+2}_{i+2}\subsetneqq\cdots\cdots\subsetneqq
E^{n+2m}_{i+2m}$

and for any $k\in\mathbb{N}^{*},m,l\in\mathbb{N},2(l+1)\leq k+2m$

$E^{n+2m}_{k+2m,l+1}\subsetneqq E^{n+2m}_{k+2m,l}.$

Using algebraic structure, we first show that $E^{n+2m}_{k+2m}$ is contained
in $\Delta(E^{n+2m}_{k+2m+2})$ for any $k\in\mathbb{N}^{*},m\in\mathbb{N}$.

######  Lemma 2.1.

Let $n$ be odd. Then for any $k\in\mathbb{N}^{*},m\in\mathbb{N}$

$E^{n+2m}_{k+2m}\subseteq\Delta(E^{n+2m}_{k+2m+2}).$

That is, for any $f\in E^{n+2m}_{k+2m}$, there exists $g\in E^{n+2m}_{k+2m+2}$
such that $\Delta g=f.$

###### Proof.

Note that $E^{n+2m}_{k+2m}=\sum_{0\leq 2l\leq k+2m}E^{n+2m}_{k+2m,l}$. Given
$f(x)=\frac{x^{\alpha}}{r^{n+2m-2l}}\in E^{n+2m}_{k+2m,l}$ with
$|\alpha|=k+2m-2l\geq 0,$ let
$\bar{g}(x)=r^{2}f(x)=\frac{x^{\alpha}}{r^{n+2m-2(l+1)}}\in
E^{n+2m}_{k+2m+2,l+1}.$ Then using the fact that
$x\cdot\nabla(x^{\beta})=|\beta|x^{\beta},\forall\beta\in\mathbb{N}^{n}$, we
obtain

$\Delta\bar{g}=\Delta(\frac{x^{\alpha}}{r^{n+2m-2(l+1)}})=\frac{\Delta
x^{\alpha}}{r^{n+2m-2(l+1)}}+2(2l+2-n-2m)(|\alpha|+l-m)f.$

Since $n$ is odd, $2l+2-n-2m\neq 0.$ Since $k\geq 1,$
$|\alpha|+l-m=\frac{k+|\alpha|}{2}>0.$ Thus $2(2l+2-n-2m)(|\alpha|+l-m)\neq
0$, which implies that

$\begin{split}&\frac{1}{2(2l+2-n-2m)(|\alpha|+l-m)}\Delta\bar{g}-f\\\ &\ \
=\frac{1}{2(2l+2-n-2m)(|\alpha|+l-m)}\frac{\Delta
x^{\alpha}}{r^{n+2m-2(l+1)}}\in E^{n+2m}_{k+2m,l+1}.\end{split}$

In the case $|\alpha|\leq 1,$ we already have
$\frac{1}{2(2l+2-n-2m)(|\alpha|+l-m)}\Delta\bar{g}-f=0$, i.e.,
$\frac{1}{2(2l+2-n-2m)(|\alpha|+l-m)}\bar{g}$ is a solution. If $|\alpha|>1,$
then we conclude by decreasing induction on $|\alpha|$ since $\Delta
x^{\alpha}\in\text{span}\\{x^{\beta}\mid|\beta|=|\alpha|-2\\}.$ Therefore, we
prove the existence of $g\in E^{n+2m}_{k+2m+2}$ such that $\Delta g=f.$

∎

######  Remark 2.2.

One computes directly that $\Delta(E^{n+2m}_{k+2m+2})\subseteq
E^{n+2m+2}_{k+2m+2}.$ Thus the above lemma shows that for any
$k\in\mathbb{N}^{*},m\in\mathbb{N}$

$E^{n+2m}_{k+2m}\subseteq\Delta(E^{n+2m}_{k+2m+2})\subseteq
E^{n+2m+2}_{k+2m+2}.$

Moreover, the function $g$ constructed in Lemma 2.1 is indeed unique. Let
$g\in E^{n+2m}_{k+2m+2}$ such that $\Delta g=0.$ Since $n$ is odd, $g$ cannot
be in $E^{n+2m}_{l}\backslash\\{0\\}$ for any $l\in\mathbb{N}^{*}$. Thus the
uniqueness of $g$ holds.

Let $z\in\Omega$ be fixed and $T_{z}$ be the positively definite matrix
defined by (1.7). We define a linear space

$E^{n+2m}_{k+2m}(T_{z})=\\{f\circ T_{z}\mid f\in E^{n+2m}_{k+2m}\\}.$

Clearly, $E^{n+2m}_{i+2m}(T_{z})\cap E^{n+2m}_{j+2m}(T_{z})=\\{0\\}$ for any
$m\in\mathbb{N}$, $i,j\in\mathbb{N}^{*},i\neq j$. A direct consequence of
Lemma 2.1 is

######  Lemma 2.3.

Let $n$ be odd and $z\in\Omega$ be fixed. Then for any
$k\in\mathbb{N}^{*},m\in\mathbb{N}$

$E^{n+2m}_{k+2m}(T_{z})\subseteq-\text{div}(K(z)\nabla)(E^{n+2m}_{k+2m+2}(T_{z})).$

That is, for any $f\in E^{n+2m}_{k+2m}(T_{z})$, there exists $g\in
E^{n+2m}_{k+2m+2}(T_{z})$ such that $-\text{div}(K(z)\nabla)g=f.$

###### Proof.

For any $f\in E^{n+2m}_{k+2m}(T_{z})$, we can find $\hat{f}\in
E^{n+2m}_{k+2m}$ such that $f=\hat{f}\circ T_{z}$. Hence by Lemma 2.1, there
exists $\hat{g}\in E^{n+2m}_{k+2m+2}$ such that $-\Delta\hat{g}=\hat{f}$. Let
$g=\hat{g}\circ T_{z}$. One computes directly that $g\in
E^{n+2m}_{k+2m+2}(T_{z})$ and $-\text{div}(K(z)\nabla)g=f.$ ∎

Using Lemma 2.3, we can solve elliptic equations in divergence form with
constant coefficient matrix $K(z)$ when $f\in E^{n+2m}_{k+2m}(T_{z})$.

######  Lemma 2.4.

Let $n$ be odd and $z\in\Omega$ be fixed. Let $\Omega_{0}$ be a bounded smooth
domain in $\mathbb{R}^{n}$ containing the origin $0$ and $c$ be a smooth
function defined on $\overline{\Omega}_{0}.$ Then for any
$k\in\mathbb{N}^{*},m,l\in\mathbb{N}$, $f\in E^{n+2m}_{k+2m}(T_{z})$, there
exists

$g\in E^{n+2m}_{k+2m+2}(T_{z})\oplus
E^{n+2m}_{k+2m+3}(T_{z})\oplus\cdots\oplus E^{n+2m}_{n+2m+l}(T_{z})\oplus
C^{l,\gamma}(\overline{\Omega}_{0})$ (2.1)

such that

$-\text{div}(K(z)\nabla g)=c(x)f\ \ \text{in}\ \Omega_{0};\ \ g=0\ \
\text{on}\ \partial\Omega_{0}.$

Moreover, the expansion of $g$ is unique.

######  Remark 2.5.

If $k\geq n+l-1$, we just think of $E^{n+2m}_{k+2m+2}(T_{z})\oplus
E^{n+2m}_{k+2m+3}(T_{z})\oplus\cdots\oplus E^{n+2m}_{n+2m+l}(T_{z})$ as
$\\{0\\}.$

###### Proof.

For $x\in\mathbb{R}^{n}$, we denote the operator
$[x\cdot\nabla]=\sum_{i=1}^{n}x_{i}\partial_{i}$. It follows from the Taylor’s
formula that

$c(x)=\sum_{j=0}^{n+l-k}\frac{1}{j!}[x\cdot\nabla]^{j}c(0)+\frac{1}{(n+l-k)!}\int_{0}^{1}(1-t)^{n+l-k}[x\cdot\nabla]^{n+l-k+1}c(tx)dt.$

Thus one has

$c(x)f\in E^{n+2m}_{k+2m}(T_{z})\oplus
E^{n+2m}_{k+2m+1}(T_{z})\oplus\cdots\oplus E^{n+2m}_{n+2m+l}(T_{z})\oplus
C^{l,\gamma}(\overline{\Omega}_{0}).$

Using Lemma 2.3, we get the existence of $g_{1}\in
E^{n+2m}_{k+2m+2}(T_{z})\oplus E^{n+2m}_{k+2m+3}(T_{z})\oplus\cdots\oplus
E^{n+2m}_{n+2m+l+2}(T_{z})\oplus C^{l+2,\gamma}(\overline{\Omega}_{0})$ such
that $-\text{div}(K(z)\nabla g_{1})=f$. Since the trace of any function in
$E^{n+2m}_{j}(T_{z})$ on $\partial\Omega_{0}$ is smooth, there exists a
function $g_{2}\in C^{\infty}(\overline{\Omega}_{0})$ such that
$-\text{div}(K(z)\nabla g_{2})=0$ in $\Omega_{0}$ and $g_{2}=g_{1}$ on
$\partial\Omega_{0}$. Note that $E^{n+2m}_{n+2m+l+1}(T_{z})\oplus
E^{n+2m}_{n+2m+l+2}(T_{z})\subseteq C^{l,\gamma}(\overline{\Omega}_{0})$, we
conclude that $g=g_{1}-g_{2}$ is the desired solution. The uniqueness of the
decomposition of $g$ follows directly from the maximum principle and the fact
that the sum in (2.1) is direct.

∎

Based on Lemma 2.4, we can extend results of Lemma 2.4 to the case of
operators $-\text{div}(\tilde{K}(x)\nabla)$, when the coefficient matrix
satisfies $\tilde{K}(0)=K(z)$.

######  Proposition 2.6.

Let $n$ be odd, $z\in\Omega$, $\Omega_{0}$ and $c$ be as in Lemma 2.4. Let
$\tilde{K}\in C^{\infty}(\overline{\Omega}_{0},\mathbb{R}^{n\times n})$ be a
positively definite matrix-valued function with $\tilde{K}(0)=K(z).$ Then for
any $k\in\mathbb{N}^{*},m,l\in\mathbb{N}$, $f\in E^{n+2m}_{k+2m}(T_{z})$,
there exists

$\zeta\in E^{n+2m}_{k+2m+2}(T_{z})\oplus
E^{n+2m+4}_{k+2m+7}(T_{z})\oplus\cdots\oplus
E^{n+2m+4(n+l-k-2)}_{n+2m+l+4(n+l-k-2)}(T_{z})\oplus
C^{l,\gamma}(\overline{\Omega}_{0})$

such that

$-\text{div}(\tilde{K}(x)\nabla\zeta)=c(x)f\ \ \text{in}\ \Omega_{0};\ \
\zeta=0\ \ \text{on}\ \partial\Omega_{0}.$

Moreover, the expansion of $\zeta$ is unique.

######  Remark 2.7.

If $k\geq n+l-1$, we just think of $E^{n+2m}_{k+2m+2}(T_{z})\oplus
E^{n+2m+4}_{k+2m+7}(T_{z})\oplus\cdots\oplus
E^{n+2m+4(n+l-k-2)}_{n+2m+l+4(n+l-k-2)}(T_{z})$ as $\\{0\\}.$

###### Proof.

Consider the case $c\equiv 1$. To find a solution $\zeta$, we construct a
sequence of $\\{\zeta_{j}\\}_{j=0}^{n+l-k+1}.$ First, let $\zeta_{0}$ satisfy

$-\text{div}(\tilde{K}(0)\nabla\zeta_{0})=f\ \ \text{in}\ \Omega_{0};\ \
\zeta_{0}=0\ \ \text{on}\ \partial\Omega_{0}.$ (2.2)

It follows from Lemma 2.4 that

$\zeta_{0}\in E^{n+2m}_{k+2m+2}(T_{z})\oplus
E^{n+2m}_{k+2m+3}(T_{z})\oplus\cdots\oplus E^{n+2m}_{n+2m+l+2}(T_{z})\oplus
C^{l+2,\gamma}(\overline{\Omega}_{0}).$

Thus using Taylor’s formula, we have

$\begin{split}\text{div}((\tilde{K}(x)-\tilde{K}(0))\nabla\zeta_{0})&=\sum_{i,j=1}^{n}(\tilde{K}_{ij}(x)-\tilde{K}_{ij}(0))\partial_{ij}\zeta_{0}+\sum_{i,j=1}^{n}\partial_{i}\tilde{K}_{ij}(x)\partial_{j}\zeta_{0}\\\
&\in E^{n+2m+4}_{k+2m+5}(T_{z})\oplus
E^{n+2m+4}_{k+2m+6}(T_{z})\oplus\cdots\oplus
E^{n+2m+4}_{n+2m+l+4}(T_{z})\oplus
C^{l,\gamma}(\overline{\Omega}_{0}).\end{split}$

Let $\zeta_{1}$ satisfy

$-\text{div}(\tilde{K}(0)\nabla\zeta_{1})=\text{div}((\tilde{K}(x)-\tilde{K}(0))\nabla\zeta_{0})\
\ \text{in}\ \Omega_{0};\ \ \zeta_{1}=0\ \ \text{on}\ \partial\Omega_{0}.$

Using Lemma 2.4 again, we obtain

$\zeta_{1}\in E^{n+2m+4}_{k+2m+7}(T_{z})\oplus
E^{n+2m+4}_{k+2m+8}(T_{z})\oplus\cdots\oplus
E^{n+2m+4}_{n+2m+l+6}(T_{z})\oplus C^{l+2,\gamma}(\overline{\Omega}_{0}),$

which implies that

$\begin{split}\text{div}((\tilde{K}(x)-\tilde{K}(0))\nabla\zeta_{1})\in
E^{n+2m+8}_{k+2m+10}(T_{z})\oplus
E^{n+2m+8}_{k+2m+11}(T_{z})\oplus\cdots\oplus
E^{n+2m+8}_{n+2m+l+8}(T_{z})\oplus
C^{l,\gamma}(\overline{\Omega}_{0})\end{split}$

and so on. Thus by a recursive method we conclude that there exists
$\zeta_{j}$ for $j=1,\cdots,n+l-k+1$ with

$-\text{div}(\tilde{K}(0)\nabla\zeta_{j})=\text{div}((\tilde{K}(x)-\tilde{K}(0))\nabla\zeta_{j-1})\
\ \text{in}\ \Omega_{0};\ \ \zeta_{j}=0\ \ \text{on}\ \partial\Omega_{0}$
(2.3)

and

$\zeta_{j}\in E^{n+2m+4j}_{k+2m+5j+2}(T_{z})\oplus
E^{n+2m+4j}_{k+2m+5j+3}(T_{z})\oplus\cdots\oplus
E^{n+2m+4j}_{n+2m+4j+l+2}(T_{z})\oplus C^{l+2,\gamma}(\overline{\Omega}_{0}).$

In particular, $\zeta_{n+l-k+1}\in C^{l+2,\gamma}(\overline{\Omega}_{0})$.
Finally, we solve

$-\text{div}(\tilde{K}(x)\nabla
h)=\text{div}((\tilde{K}(x)-\tilde{K}(0))\nabla\zeta_{n+l-k+1})\ \ \text{in}\
\Omega_{0};\ \ h=0\ \ \text{on}\ \partial\Omega_{0}.$ (2.4)

By the classical theory for elliptic equations, $h\in
C^{l+2,\gamma}(\overline{\Omega}_{0}).$

Denote $\zeta=\sum_{i=0}^{n+l-k+1}\zeta_{i}+h\in
E^{n+2m}_{k+2m+2}(T_{z})\oplus E^{n+2m+4}_{k+2m+7}(T_{z})\oplus\cdots\oplus
E^{n+2m+4(n+l-k)}_{n+2m+4(n+l-k)+l+2}(T_{z})\oplus
C^{l+2,\gamma}(\overline{\Omega}_{0}).$ By (2.2), (2.3) and (2.4), one
computes directly that

$-\text{div}(\tilde{K}(x)\nabla\zeta)=f\ \ \text{in}\ \Omega_{0};\ \ \zeta=0\
\ \text{on}\ \partial\Omega_{0}.$

Note that $E^{n+2m+4(n+l-k-1)}_{n+2m+4(n+l-k-1)+l+1}(T_{z})\oplus
E^{n+2m+4(n+l-k)}_{n+2m+4(n+l-k)+l+2}(T_{z})\subseteq
C^{l,\gamma}(\overline{\Omega}_{0})$, therefore $\zeta$ is the desired
solution. For the case $c(x)$ being a smooth function, we can use the Taylor’s
expansion as that in Lemma 2.4 to get the result. The proof is thus finished.

∎

We are now able to prove the expansion of Green’s function when $n$ is odd.

Proof of Theorem 1.2: We extend $K(x):\Omega\to\mathbb{R}^{n\times n}$ to a
positively definite smooth matrix-valued function in $\mathbb{R}^{n}$. This
can be done since $\Omega$ is smooth. For any $w\in\Omega$, define
$\Omega_{w}=\\{x-w\mid x\in\Omega\\}$ and $T_{w}(\Omega)=\\{T_{w}(x)\mid
x\in\Omega\\}.$ By the positively definiteness and smoothness of $K$, we can
choose $M>0$ sufficiently large such that $T_{w}(\Omega_{w})\subseteq
B_{M}(0)$ for all $w\in\Omega.$

Now we give the expansion of $G_{K}(x,y)$. Note that $\Phi_{0}\in E^{n}_{2}.$
For fixed $y\in\Omega,$ we define $\hat{\Phi}_{0}=\sqrt{\det
K(y)}^{-1}\Phi_{0}\circ T_{y}\in E^{n}_{2}(T_{y})$. Then direct computation
yields that $-\text{div}(K(y)\nabla\hat{\Phi}_{0})=\delta_{0}$. Consider
solutions $\zeta_{y}$ of the following equations

$\begin{cases}-\text{div}(\tilde{K}_{y}(x)\nabla\zeta_{y})=\text{div}((\tilde{K}_{y}(x)-K(y))\nabla\hat{\Phi}_{0}),\
&x\in B_{M}(0),\\\ \zeta_{y}=0,\ &x\in\partial B_{M}(0),\end{cases}$ (2.5)

where the coefficient matrix $\tilde{K}_{y}(\cdot)=K(\cdot+y)$. By
assumptions, we have $\tilde{K}_{y}\in
C^{\infty}(\overline{B_{M}(0)},\mathbb{R}^{n\times n})$ and
$\tilde{K}_{y}(0)=K(y)$. Moreover, one computes directly that for any
$l\in\mathbb{N}$

$\text{div}((\tilde{K}_{y}(x)-K(y))\nabla\hat{\Phi}_{0})\in
E^{n+2}_{3}(T_{y})\oplus E^{n+2}_{4}(T_{y})\oplus\cdots\oplus
E^{n+2}_{n+2+l}(T_{y})\oplus C^{l,\gamma}(\overline{B_{M}(0)}).$

Therefore using Proposition 2.6 with $z=y$, $\Omega_{0}=B_{M}(0)$ and
$\tilde{K}=\tilde{K}_{y}$, we get the unique solution $\zeta_{y}$ of (2.5)
such that

$\begin{split}\zeta_{y}&=\sum_{i=1}^{n+l-2}\hat{\Phi}_{i}+h\\\ &\in
E^{n+2}_{5}(T_{y})\oplus E^{n+6}_{10}(T_{y})\oplus\cdots\oplus
E^{n+2+4(n+l-3)}_{n+2+4(n+l-3)+l}(T_{y})\oplus
C^{l,\gamma}(\overline{B_{M}(0)}),\end{split}$

where $\hat{\Phi}_{i}\in E^{n+2+4(i-1)}_{5i}(T_{y})$ for $i=1,\cdots,n+l-2$
and $h\in C^{l,\gamma}(\overline{B_{M}(0)})$.

Define $\eta_{y}(x)=G_{K}(x,y)-\hat{\Phi}_{0}(x-y)-\zeta_{y}(x-y)$ for any
$x\in\Omega$. It follows from the definition of $G_{K}$ that

$\mathcal{L}_{K}G_{K}(x,y)=\delta_{y},\ x\in\Omega;\ \ \ G_{K}(x,y)=0,\
x\in\partial\Omega,$

which implies that for any $x\in\Omega$

$\begin{split}\mathcal{L}_{K}\eta_{y}(x)=&\text{div}((\tilde{K}_{y}(x)-K(y))\nabla\hat{\Phi}_{0})(x-y)+\text{div}(\tilde{K}_{y}(x)\nabla\zeta_{y})(x-y)=0.\end{split}$
(2.6)

Here we have used (2.5). Note that (2.6) also holds in sense of distribution.
Thus by Weyl’s lemma, $\eta_{y}\in C^{\infty}(\overline{\Omega})$ since it is
smooth on $\partial\Omega$. To conclude, we have

$G_{K}(x,y)=\hat{\Phi}_{0}(x-y)+\sum_{i=1}^{n+l-2}\hat{\Phi}_{i}(x-y)+H^{l}(x,y),$
(2.7)

where $\hat{\Phi}_{i}\in E^{n+2+4(i-1)}_{5i}(T_{y})$ for $i=1,\cdots,n+l-2$
and $x\mapsto H^{l}(x,y)\in C^{l,\gamma}(\overline{\Omega})$. Define
$\Phi_{i}=\hat{\Phi}_{i}\circ T_{y}^{-1}\in E^{n+2+4(i-1)}_{5i}$. This
combined with (2.7) yields the expansion for $G_{K}(x,y)$

$G_{K}(x,y)=\sqrt{\det
K(y)}^{-1}\Phi_{0}(T_{y}(x-y))+\sum_{i=1}^{n+l-2}\Phi_{i}(T_{y}(x-y))+H^{l}(x,y),$

where $x\mapsto H^{l}(x,y)\in C^{l,\gamma}(\overline{\Omega})$.

We now prove the regularity of $H^{l}$, i.e., $H^{l}(x,y)=H^{l}_{y}(x)\in
C\left(\Omega,C^{l,\gamma}\left(\overline{\Omega}\right)\right)\cap
C^{l}(\Omega\times\Omega)$. Indeed, from the above construction of $G_{K}$, we
know that $H^{l}(x,y)$ satisfies for all $y\in\Omega$

$\begin{split}\mathcal{L}_{K}H^{l}(x,y)=&\mathcal{L}_{K}G_{K}(x,y)-\mathcal{L}_{K}\hat{\Phi}_{0}(x-y)-\mathcal{L}_{K}\left(\sum_{i=1}^{n+l-2}\hat{\Phi}_{i}(x-y)\right)\\\
=&\text{div}\left(\left(K(x)-K(y)\right)\nabla\hat{\Phi}_{0}(x-y)\right)+\text{div}\left(K(x)\nabla\left(\sum_{i=1}^{n+l-2}\hat{\Phi}_{i}(x-y)\right)\right)\\\
\in&E^{n+2+4(n+l-3)}_{n+1+4(n+l-3)+l}(T_{y})(x-y)\oplus
E^{n+2+4(n+l-3)}_{n+2+4(n+l-3)+l}(T_{y})(x-y)\oplus
C^{l,\gamma}(\overline{B_{M}(0)})(x-y).\end{split}$ (2.8)

Here we denote $E^{n+2+4(n+l-3)}_{n+1+4(n+l-3)+l}(T_{y})(x-y)=\\{f(x-y)\mid
f\in E^{n+2+4(n+l-3)}_{n+1+4(n+l-3)+l}(T_{y})\\}$ and
$C^{l,\gamma}(\overline{B_{M}(0)})(x-y)=\\{f(x-y)\mid f\in
C^{l,\gamma}(\overline{B_{M}(0)})\\}$. Moreover, it follows from
$G_{K}(\cdot,y)=0$ on $\partial\Omega$ that

$H^{l}(x,y)=-\hat{\Phi}_{0}(x-y)-\sum_{i=1}^{n+l-2}\hat{\Phi}_{i}(x-y)\ \ \ \
x\in\partial\Omega.$

By the continuity of the right-hand side of (2.8) and the boundary condition
with respect to $y$ in $C^{l-2,\gamma}(\Omega)$ and
$C^{l,\gamma}(\partial\Omega)$, respectively, we can get $H^{l}(x,y)\in
C(\Omega,C^{l,\gamma}(\overline{\Omega}))$ and thus $\frac{\partial
H^{l}(x,y)}{\partial x},\frac{\partial^{2}H^{l}(x,y)}{\partial
x^{2}},\cdots,\frac{\partial^{l}H^{l}(x,y)}{\partial x^{l}}\in
C(\Omega\times\Omega)$.

Similarly, taking $\nabla_{y}$ to both sides of (2.8), we can check that
$\nabla_{y}H^{l}(x,y)\in C(\Omega,C^{l-1,\gamma}(\overline{\Omega}))$, which
implies that $\frac{\partial H^{l}(x,y)}{\partial
y},\frac{\partial^{2}H^{l}(x,y)}{\partial y\partial
x},\cdots,\frac{\partial^{l}H^{l}(x,y)}{\partial y\partial x^{l-1}}\in
C(\Omega\times\Omega)$. Moreover, we can get that for any integer $k,m$ with
$0\leq k\leq m\leq l$,

$\frac{\partial^{m}H^{l}(x,y)}{\partial y^{k}\partial x^{m-k}}\in
C(\Omega\times\Omega).$

Thus we conclude that $H^{l}(x,y)$ is a $C^{l}$ function over
$\Omega\times\Omega$.

∎

Note that the regular part of $G_{K}$ and the associated Robin’s function
$R_{K}$ are defined by $H_{K}(x,y)=H^{0}(x,y)$ and $R_{K}(x)=H_{K}(x,x)$ for
any $x,y\in\Omega$. Therefore by Theorem 1.2, for any $l\in\mathbb{N}^{*}$

$H_{K}(x,y)=H^{0}(x,y)=\sum_{i=n-1}^{n+l-2}\Phi_{i}(T_{y}(x-y))+H^{l}(x,y),\ \
\forall x,y\in\Omega.$ (2.9)

Now we can prove Theorem 1.6 when $n$ is odd.

######  Proposition 2.8.

Let $n$ be odd, $\Omega$ and $K$ be as in Theorem 1.6. Then the Robin’s
function $R_{K}(\cdot)\in C^{\infty}(\Omega)$.

###### Proof.

For any $l\in\mathbb{N}^{*}$, it follows from (2.9) that $R_{K}(x)=H^{l}(x,x)$
since $\Phi_{i}(0)=0$ for any $i\geq n-1$. So $R_{K}\in C^{l}(\Omega)$. By the
arbitrariness of $l$, we get the result. ∎

##  3\. Even dimensional case: proof of Theorems 1.4 and 1.6

Now we prove the even dimensional case. Let $n\in 2\mathbb{N}^{*}$. Clearly
from the definition of $E^{n+2m,s}_{k+2m}$, we have
$E^{n+2m}_{k+2m}=E^{n+2m,s}_{k+2m}\oplus\mathbb{R}[x]$ for any $k\geq
n,m\in\mathbb{N}.$ For $L_{m}$, we first have

######  Lemma 3.1.

For any $m\in\mathbb{N},$ $L_{m}\subseteq\Delta(L_{m+2}\oplus\mathbb{R}[x])$.

###### Proof.

We prove it by a direct algebraic construction. Let $l\in\mathbb{N},$ $2l\leq
m,$ and $L_{m,l}=\text{span}\\{x^{\alpha}r^{2l}\log r\mid|\alpha|=m-2l\\}$.
Then $L_{m}=\sum_{0\leq 2l\leq m}L_{m,l}.$ For $f=x^{\alpha}r^{2l}\log r\in
L_{m,l}$, let $g_{1}=r^{2}f\in L_{m+2,l+1}$, then one computes directly that

$\Delta g_{1}=r^{2l+2}\log
r\Delta(x^{\alpha})+(2|\alpha|+n+4l+2)x^{\alpha}r^{2l}+(2l+2)(2l+|\alpha|+n)f.$

Let $g_{2}$ be a polynomial with order $m+2$ satisfying $\Delta
g_{2}=(2|\alpha|+n+4l+2)x^{\alpha}r^{2l}.$ Since $(2l+2)(2l+|\alpha|+n)>0$, we
find that there exists $g_{3}=\frac{1}{(2l+2)(2l+|\alpha|+n)}(g_{1}-g_{2})\in
L_{m+2}\oplus\mathbb{R}[x]$ such that

$\Delta g_{3}-f=\frac{1}{(2l+2)(2l+|\alpha|+n)}r^{2l+2}\log
r\Delta(x^{\alpha})\in L_{m,l+1}.$

Thus we can get the result by decreasing induction on $|\alpha|.$ ∎

Let us define

$\tilde{E}^{n+2m}_{k+2m}=\begin{cases}F^{n+2m}_{k+2m}\ &k<n,\\\
F^{n+2m}_{k+2m}\oplus\mathbb{R}[x]\ &k\geq n.\end{cases}$

Similar to Lemma 2.1, using the definition of $F^{n+2m}_{k+2m}$ and
$\tilde{E}^{n+2m}_{k+2m}$, we get

######  Lemma 3.2.

There holds

$\tilde{E}^{n+2m}_{k+2m}\subseteq\Delta(\tilde{E}^{n+2m}_{k+2m+2})\ \ \forall
k\in\mathbb{N}^{*},m\in\mathbb{N}.$

###### Proof.

If $k<n-2$, then $\tilde{E}^{n+2m}_{k+2m}=E^{n+2m}_{k+2m}$,
$\tilde{E}^{n+2m}_{k+2m+2}=E^{n+2m}_{k+2m+2}$. The results follows from Lemma
2.1.

When $k=n-2$ or $n-1$, let $f\in E^{n+2m}_{k+2m,l}$ with $0\leq
l\leq\frac{n-2+2m}{2}$. If $l<\frac{n-2+2m}{2}$, we can repeat the
construction in Lemma 2.1 with decreasing induction on $|\alpha|$ to get the
results. The only one case we need to pay attention to is $2l=n-2+2m$, i.e.,
$f=\frac{x^{\alpha}}{r^{2}}$ with $|\alpha|=k-n+2\leq 1.$ Let
$g=x^{\alpha}\log r\in L_{k-n+2}\subseteq\tilde{E}^{n+2m}_{k+2m+2}$. One
computes directly that $\Delta g=(2|\alpha|+n-2)f.$ Note that
$2|\alpha|+n-2>0$ for $n\geq 4.$ For $n=2,$ only the case $|\alpha|=1$ occurs
since $k\geq 1$. Thus $2|\alpha|+n-2\neq 0$ and we get the results.

For the case $k\geq n,$ if $f\in L_{k-n}\oplus\mathbb{R}[x]$, the result is
shown by Lemma 3.1. Let $f\in E^{n+2m}_{k+2m,l}$ for some $0\leq 2l\leq k+2m$.
If $2l<n-2+2m$, we can get the result by decreasing induction on $|\alpha|$.
If $2l\geq n+2m$, then $f$ is a polynomial with order $k-n$. So there exists a
polynomial $g\in\mathbb{R}[x]\subseteq\tilde{E}^{n+2m}_{k+2m+2}$ such that
$\Delta g=f.$ It remains to solve the case $2l=n-2+2m$, i.e.,
$f=\frac{x^{\alpha}}{r^{2}}$ with $|\alpha|=k-n+2.$ Let $g_{0}=x^{\alpha}\log
r\in L_{k+2-n}.$ Then

$\Delta g_{0}=(2|\alpha|+n-2)\frac{x^{\alpha}}{r^{2}}+\log
r\times\Delta(x^{\alpha}),$

which implies that $\Delta(\frac{1}{2|\alpha|+n-2}g_{0})-f\in L_{k-n}$. Since
$L_{k-n}\subseteq\Delta(L_{k-n+2}\oplus\mathbb{R}[x])$, we get the existence
of $g\in\mathbb{R}[x]\subseteq\tilde{E}^{n+2m}_{k+2m+2}$ such that $\Delta
g=f.$

∎

Let $z\in\Omega$ be fixed and $T_{z}$ be the positively definite matrix
defined by (1.7). We define

$F^{n+2m}_{k+2m}(T_{z})=\\{f\circ T_{z}\mid f\in F^{n+2m}_{k+2m}\\};\ \
\tilde{E}^{n+2m}_{k+2m}(T_{z})=\\{f\circ T_{z}\mid
f\in\tilde{E}^{n+2m}_{k+2m}\\}.$

A direct consequence of Lemma 3.2 is

######  Lemma 3.3.

Let $n$ be even and $z\in\Omega$ be fixed. Then for any
$k\in\mathbb{N}^{*},m\in\mathbb{N}$

$\tilde{E}^{n+2m}_{k+2m}(T_{z})\subseteq-\text{div}(K(z)\nabla)(\tilde{E}^{n+2m}_{k+2m+2}(T_{z})).$

Based on Lemma 3.3, we get the existence of solutions to elliptic equations in
divergence form with constant coefficient matrix $K(z)$ when $f\in
F^{n+2m}_{k+2m}(T_{z})$.

######  Lemma 3.4.

Let $n$ be even and $z\in\Omega$ be fixed. Let $\Omega_{0}$ be a bounded
smooth domain in $\mathbb{R}^{n}$ containing the origin $0$ and $c$ be a
smooth function defined on $\overline{\Omega}_{0}.$ Then for any
$k\in\mathbb{N}^{*},m,l\in\mathbb{N}$, $f\in F^{n+2m}_{k+2m}(T_{z})$, there
exists

$g\in F^{n+2m}_{k+2m+2}(T_{z})\oplus
F^{n+2m}_{k+2m+3}(T_{z})\oplus\cdots\oplus F^{n+2m}_{n+2m+l}(T_{z})\oplus
C^{l,\gamma}(\overline{\Omega}_{0})$

such that

$-\text{div}(K(z)\nabla g)=c(x)f\ \ \text{in}\ \Omega_{0};\ \ g=0\ \
\text{on}\ \partial\Omega_{0}.$

Moreover, the expansion of $g$ is unique.

###### Proof.

Using Taylor’s formula, we have

$c(x)f\in F^{n+2m}_{k+2m}(T_{z})\oplus
F^{n+2m}_{k+2m+1}(T_{z})\oplus\cdots\oplus F^{n+2m}_{n+2m+l}(T_{z})\oplus
C^{l,\gamma}(\overline{\Omega}_{0}).$

By Lemma 3.3, we get the existence of
$\bar{g}_{1}\in\tilde{E}^{n+2m}_{k+2m+2}(T_{z})\oplus\tilde{E}^{n+2m}_{k+2m+3}(T_{z})\oplus\cdots\oplus\tilde{E}^{n+2m}_{n+2m+l+2}(T_{z})\oplus
C^{l+2,\gamma}(\overline{\Omega}_{0})$ such that
$-\text{div}(K(z)\nabla\bar{g}_{1})=f$. So $\bar{g}_{1}\in
F^{n+2m}_{k+2m+2}(T_{z})\oplus F^{n+2m}_{k+2m+3}(T_{z})\oplus\cdots\oplus
F^{n+2m}_{n+2m+l+2}(T_{z})\oplus C^{l+2,\gamma}(\overline{\Omega}_{0})$. Let
$\bar{g}_{2}\in C^{\infty}(\overline{\Omega}_{0})$ be a function satisfying
$-\text{div}(K(z)\nabla\bar{g}_{2})=0$ in $\Omega_{0}$ and
$\bar{g}_{2}=\bar{g}_{1}$ on $\partial\Omega_{0}$. Since
$F^{n+2m}_{n+2m+l+1}(T_{z})\oplus F^{n+2m}_{n+2m+l+2}(T_{z})\subseteq
C^{l,\gamma}(\overline{\Omega}_{0})$, we conclude that
$g=\bar{g}_{1}-\bar{g}_{2}$ is the desired solution.

∎

Then, we further extend results of Lemma 3.4 to elliptic operators in
divergence form, which is close to $-\text{div}(K(z)\nabla)$ near the origin
$0$ when $n$ is even.

######  Proposition 3.5.

Let $n$ be even, $z\in\Omega$, $\Omega_{0}$ and $c$ be as in Lemma 3.4. Let
$\tilde{K}\in C^{\infty}(\overline{\Omega}_{0},\mathbb{R}^{n\times n})$ be a
positively definite matrix-valued function with $\tilde{K}(0)=K(z).$ Then for
any $k\in\mathbb{N}^{*},m,l\in\mathbb{N}$, $f\in F^{n+2m}_{k+2m}(T_{z})$,
there exists

$\xi\in F^{n+2m}_{k+2m+2}(T_{z})\oplus
F^{n+2m+4}_{k+2m+7}(T_{z})\oplus\cdots\oplus
F^{n+2m+4(n+l-k-2)}_{n+2m+l+4(n+l-k-2)}(T_{z})\oplus
C^{l,\gamma}(\overline{\Omega}_{0})$

such that

$-\text{div}(\tilde{K}(x)\nabla\xi)=c(x)f\ \ \text{in}\ \Omega_{0};\ \ \xi=0\
\ \text{on}\ \partial\Omega_{0}.$

Moreover, the expansion of $\zeta$ is unique.

###### Proof.

The idea is similar to the proof of Proposition 2.6 and we prove it here just
for completeness. Consider the case $c\equiv 1$. To find a solution $\xi$, we
construct a sequence of $\\{\xi_{j}\\}_{j=0}^{n+l-k+1}.$ First, let $\xi_{0}$
satisfy

$-\text{div}(\tilde{K}(0)\nabla\xi_{0})=f\ \ \text{in}\ \Omega_{0};\ \
\xi_{0}=0\ \ \text{on}\ \partial\Omega_{0}.$

It follows from Lemma 3.4 that

$\xi_{0}\in F^{n+2m}_{k+2m+2}(T_{z})\oplus
F^{n+2m}_{k+2m+3}(T_{z})\oplus\cdots\oplus F^{n+2m}_{n+2m+l+2}\oplus
C^{l+2,\gamma}(\overline{\Omega}_{0}),$

from which we deduce,

$\begin{split}\text{div}((\tilde{K}(x)-\tilde{K}(0))\nabla\xi_{0})&=\sum_{i,j=1}^{n}(\tilde{K}_{ij}(x)-\tilde{K}_{ij}(0))\partial_{ij}\xi_{0}+\sum_{i,j=1}^{n}\partial_{i}\tilde{K}_{ij}(x)\partial_{j}\xi_{0}\\\
&\in F^{n+2m+4}_{k+2m+5}(T_{z})\oplus
F^{n+2m+4}_{k+2m+6}(T_{z})\oplus\cdots\oplus
F^{n+2m+4}_{n+2m+l+4}(T_{z})\oplus
C^{l,\gamma}(\overline{\Omega}_{0}).\end{split}$

Let $\xi_{1}$ satisfy

$-\text{div}(\tilde{K}(0)\nabla\xi_{1})=\text{div}((\tilde{K}(x)-\tilde{K}(0))\nabla\xi_{0})\
\ \text{in}\ \Omega_{0};\ \ \xi_{1}=0\ \ \text{on}\ \partial\Omega_{0}.$

By Lemma 3.4, $\xi_{1}\in F^{n+2m+4}_{k+2m+7}(T_{z})\oplus
F^{n+2m+4}_{k+2m+8}(T_{z})\oplus\cdots\oplus
F^{n+2m+4}_{n+2m+l+6}(T_{z})\oplus C^{l+2,\gamma}(\overline{\Omega}_{0}).$
This yields that $\text{div}((\tilde{K}(x)-\tilde{K}(0))\nabla\xi_{1})\in
F^{n+2m+8}_{k+2m+10}(T_{z})\oplus
F^{n+2m+8}_{k+2m+11}(T_{z})\oplus\cdots\oplus
F^{n+2m+8}_{n+2m+l+8}(T_{z})\oplus C^{l,\gamma}(\overline{\Omega}_{0})$ and so
on. Hence we can find $\xi_{j}$ for $j=1,\cdots,n+l-k+1$ satisfying

$-\text{div}(\tilde{K}(0)\nabla\xi_{j})=\text{div}((\tilde{K}(x)-\tilde{K}(0))\nabla\xi_{j-1})\
\ \text{in}\ \Omega_{0};\ \ \xi_{j}=0\ \ \text{on}\ \partial\Omega_{0}$

and

$\xi_{j}\in F^{n+2m+4j}_{k+2m+5j+2}(T_{z})\oplus
F^{n+2m+4j}_{k+2m+5j+3}(T_{z})\oplus\cdots\oplus
F^{n+2m+4j}_{n+2m+4j+l+2}(T_{z})\oplus C^{l+2,\gamma}(\overline{\Omega}_{0}).$

In particular, $\xi_{n+l-k+1}\in C^{l+2,\gamma}(\overline{\Omega}_{0})$.
Finally, we solve $h\in C^{l+2,\gamma}(\overline{\Omega}_{0})$ satisfying

$-\text{div}(\tilde{K}(x)\nabla
h)=\text{div}((\tilde{K}(x)-\tilde{K}(0))\nabla\xi_{n+l-k+1})\ \ \text{in}\
\Omega_{0};\ \ h=0\ \ \text{on}\ \partial\Omega_{0}.$

Denote $\xi=\sum_{i=0}^{n+l-k+1}\xi_{i}+h\in F^{n+2m}_{k+2m+2}(T_{z})\oplus
F^{n+2m+4}_{k+2m+7}(T_{z})\oplus\cdots\oplus
F^{n+2m+4(n+l-k)}_{n+2m+4(n+l-k)+l+2}(T_{z})\oplus
C^{l+2,\gamma}(\overline{\Omega}_{0}).$ One computes directly that

$-\text{div}(\tilde{K}(x)\nabla\xi)=f\ \ \text{in}\ \Omega_{0};\ \ \zeta=0\ \
\text{on}\ \partial\Omega_{0}.$

Thus $\xi$ is the desired solution. For the case $c$ being a smooth function,
we can use the same Taylor’s expansion as in Lemma 3.4 to get the result.

∎

Proof of Theorem 1.4: Based on Proposition 3.5, we can use similar method as
that in Theorem 1.2 to get the expansion of Green’s function $G_{K}$ of the
operator $\mathcal{L}_{K}$ in case that $n$ is even, just by replacing
$E^{n+2m}_{k+2m}$ with $F^{n+2m}_{k+2m}$. We omit the proof here. ∎

Similar to the odd dimensional case, we define the regular part of $G_{K}$ by
$H_{K}(x,y)=H^{0}(x,y)$ for any $x,y\in\Omega$ and Robin’s function
$R_{K}(x)=H_{K}(x,x)$ for any $x\in\Omega.$ By Theorem 1.4, for any
$l\in\mathbb{N}^{*}$

$H_{K}(x,y)=\sum_{i=n-1}^{n+l-2}\Phi_{i}(T_{y}(x-y))+H^{l}(x,y),\ \ \forall
x,y\in\Omega.$ (3.1)

Since $\Phi_{i}(0)=0$ for any $i\geq n-1$, we can get that $R_{K}$ is smooth
when $n$ is even, which is summarized as follows.

######  Proposition 3.6.

Let $n$ be even, $\Omega$ and $K$ be as in Theorem 1.6. Then Robin’s function
$R_{K}(\cdot)\in C^{\infty}(\Omega)$.

Proof of Theorem 1.6: Using Propositions 2.8 and 3.6, we finish the proof of
Theorem 1.6. ∎

######  Remark 3.7.

We give a final remark that the smoothness of $K$ and $\Omega$ is indeed used
to obtain the smoothness of Robin’s function $R_{K}$. If we just need to get
$R_{K}\in C^{l}$ for some $l$, the assumptions on $K$ and $\Omega$ can be
weakened.

## Appendix

In this Appendix, we give proof of the expansion of Green’s function of
$\mathcal{L}_{K}$ when the coefficient matrix $K$ is a constant coefficient
matrix $K_{0}$.

######  Lemma A.1.

Let $T_{0}$ be the positively definite matrix satisfying

$T_{0}^{-1}T_{0}^{-t}=K_{0}.$

Denote $T_{0}\Omega=\\{T_{0}x\mid x\in\Omega\\}$. Let $G_{K_{0}}$ be the
Green’s function of $\mathcal{L}_{K_{0}}$ in $\Omega$ with zero Dirichlet
boundary condition. Then

$G_{K_{0}}(x,y)=\sqrt{\det K_{0}}^{-1}\Phi_{0}(T_{0}(x-y))-\sqrt{\det
K_{0}}^{-1}H_{T_{0}\Omega}(T_{0}x,T_{0}y)\ \ \forall x,y\in\Omega,x\neq y,$
(3.2)

where $H_{T_{0}\Omega}:T_{0}\Omega\times T_{0}\Omega\to\mathbb{R}$ is the
regular part satisfying for any $y^{\prime}\in T_{0}\Omega$

$\begin{cases}-\Delta H_{T_{0}\Omega}(x^{\prime},y^{\prime})=0,\ \
&x^{\prime}\in T_{0}\Omega,\\\
H_{T_{0}\Omega}(x^{\prime},y^{\prime})=\Phi_{0}(x^{\prime}-y^{\prime}),\ \
&x^{\prime}\in\partial T_{0}\Omega.\end{cases}$ (3.3)

###### Proof.

For any $u\in C^{2}(\Omega)\cap C^{1}_{0}(\overline{\Omega})$ with
$f=\mathcal{L}_{K_{0}}u$, let
$\bar{u}(x^{\prime})=u(T_{0}^{-1}x^{\prime})=u(x)$ and
$\bar{f}(x^{\prime})=f(T_{0}^{-1}x^{\prime})=f(x)$ for any $x^{\prime}\in
T_{0}\Omega$. Here we set $x^{\prime}=T_{0}x$. Then one computes directly that
$-\Delta\bar{u}=\bar{f}$ in $T_{0}\Omega$ and $\bar{u}\in
C^{2}(T_{0}\Omega)\cap C^{1}_{0}(\overline{T_{0}\Omega})$. From (1.3), we have

$\bar{u}(y^{\prime})=\int_{T_{0}\Omega}\left(\Phi_{0}(x^{\prime}-y^{\prime})-H_{T_{0}\Omega}(x^{\prime},y^{\prime})\right)\bar{f}(x^{\prime})dx^{\prime},\
\ \forall y^{\prime}\in T_{0}\Omega,$ (3.4)

where $H_{T_{0}\Omega}:T_{0}\Omega\times T_{0}\Omega\to\mathbb{R}$ satisfies
(3.3). Using coordinate transformation, (3.4) yields

$\begin{split}u(y)=&\int_{\Omega}\left(\Phi_{0}(T_{0}x-T_{0}y)-H_{T_{0}\Omega}(T_{0}x,T_{0}y)\right)f(x)\cdot\det
T_{0}dx\\\ =&\sqrt{\det
K_{0}}^{-1}\int_{\Omega}\left(\Phi_{0}(T_{0}x-T_{0}y)-H_{T_{0}\Omega}(T_{0}x,T_{0}y)\right)f(x)dx,\
\ \forall y\in\Omega.\end{split}$

This implies that $G_{K_{0}}$ has decomposition (3.2). The proof is thus
complete. ∎

### Acknowledgments:

D. Cao was supported by National Key R&D Program (2022YFA1005602) and NNSF of
China(grant No.12371212). J. Wan was supported by NNSF of China (grant No.
12101045).

### Conflict of interest statement

On behalf of all authors, the corresponding author states that there is no
conflict of interest.

### Data availability statement

All data generated or analysed during this study are included in this
published article and its supplementary information files.

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