1 Introduction

In this paper, we consider the ultraconvergence of FE approximation for the following elliptic problem

$$\begin{aligned} {\left\{ \begin{array}{ll} \mathcal{L}u(y)\equiv -\frac{\partial }{\partial y_{i}}\left( a_{ij}(y)\frac{\partial u}{\partial y_{j}}\right) =f(y)\quad &{}\text {in}\quad \Omega ,\\ u(y)=0\quad &{}\text {on}\quad \partial \Omega , \end{array}\right. } \end{aligned}$$
(1.1)

where \(\Omega \subset \mathfrak {R}^{n}~(n=2,3)\) is a bounded polygon \((n=2)\) or polyhedron \((n=3)\), and \(A=(a_{ij})\in (C^\infty (\Omega ))^{n^2}\) is a uniformly positive definite matrix in the sense that there exists \(\delta >0\) satisfying

$$\begin{aligned} a_{ij}\xi _{i}\xi _{j}\ge \delta \xi _{i}\xi _i, \xi \in \mathfrak {R}^n. \end{aligned}$$

Note that throughout the paper, the Einstein convention is used: the summation will be taken over all repeated indices.

The study of optimal convergence and superconverence/ultraconvergence properties has been an area of active research, see [1, 2, 49, 1315] and [1737] for an uncompleted list of references). For instances, Bank and Xu (see [5]) proved that, for linear finite element, the recovered gradient \(Q_{h}\nabla u^{h}\) converges with order \(O(h^{1+\min (1,\sigma )}|\log h|^{\frac{1}{2}})\), where \(Q_{h}\) is a global \(L^{2}\) projection, the underlying mesh \(\mathcal {T}_{h}\) is quasi-uniform and satisfies the so-called \((\alpha ,\sigma )\)- parallelogram property. Huang and Xu (see [15]) obtained that, for second-degree finite element, the recovered gradient \(Q_{h}\nabla u^{h}\) converges with order \(O(h^{2+\min (1,\sigma )}|\log h|^{\frac{1}{2}})\).

It is known that when the underlying mesh has a certain local symmetry property, the corresponding finite element solution has some natrual superconvergence properties. For instances, Schatz, Sloan and Wahlbin discovered in [22, 23, 27] that at a local symmetric vertex \(x_{0}\), when k is even, the finite element solution converges to the exact solution with order \(\mathcal{O}(h^{k+2-\varepsilon })\), where \(\varepsilon >0\) can be arbitrary small; when k is odd, the discrete gradient of the finite element solution converges to the gradient of the exact solution with order \(\mathcal{O}(h^{k+1-\varepsilon })\), where a vertex \(x_0\) is called local symmetric if there exists some radius \(d>0\) such that the underlying mesh is symmetric in the neighborhood \(B(x_0,d)=\{y:|x_0-y|\le d\}\). The natural superconvergence or even ultraconvergence of the finite element solution has been also investigated by other techniques such as the so-called weak estimate, see [6, 14, 18, 19, 36] for an uncompleted list of references.

To obtain better superconvergence/ultraconvergence result, one natural idea is to post-process the finite element solution on some local symmetric mesh, see e.g., [6, 7, 13, 1820] and references therein. Along this direction, Lin found in [19] that the gradient of a special interpolation of some odd order finite element solution on a uniform rectangular mesh converges with order \(\mathcal{O}(h^{k+2}|\ln h|\)) at an interior vertex. Zhang et al. discovered in [30, 33] that the recovered gradient (see e.g. SPR by [34, 37] and PPR by [30]) of some even order finite element solution on some uniform rectangular meshes converges with order \(\mathcal{O}(h^{k+2}|\ln h|) \) as well. M. Asadzadeh, A. Schatz, and W. Wendland showed that the discrete gradient of some extrapolated finite element solution superconverges with order \(\mathcal{O}(h^{k+1}|\ln h|)\) under suitable local symmetric mesh.

All aforementioned ultraconvergence results are valid only for elliptic equations with constant coefficients. Recall that the classical superconvergence analysis for variable-coefficient problems is done by estimating the difference between the variable coefficient bilinear form and its corresponding (piecewise) constant coefficient bilinear form. Since this difference is only a one-order-higher term, we can not obtain ultraconvergence results of the corresponding finite element solution for variable-coefficient problems. In other words, we can not use this approach to prove the ultraconvergence of the post-processed FE solution for variable-coefficient problems.

In this paper, we propose a novel local interpolation operator to post-process finite element solutions for variable-coefficient problems. Unlike the classical interpolation operator defined in [19, 20], our local operator interpolates the value of the original finite element solution at all vertices of the underlying mesh in a patch instead of interpolating all nodes in a relatively smaller-sized patch.

To prove the ultraconvergence property of our post-processed FE solution, we first investigate properties of the so-called discrete Green’s function in the whole domain \(\mathfrak {R}^n\). We found the difference between two errors, one is the FE projection error of the Green’s function with variable coefficients, another is the FE projection error of the Green’s function with constant coefficients by fixing one point value of the corresponding variable coefficients, is of order almost \(O(h^{2})\). Combining with some further nice properties for the FE projection error of the Green’s function (see Theorem 2.1 for the details), we show that for even k and translation invariant mesh, between two interior vertices \(y_{1}\) and \(y_{2}\) satisfying \(|y_{1}-y_{2}|\lesssim h\), there holds

$$\begin{aligned} |(u-u^{h})(y_1)-(u-u^{h})(y_2)|\lesssim h^{k+3}|\ln h|^{3}\Vert u\Vert _{W^{k+3,\infty }(\mathfrak {R}^{n})}, \end{aligned}$$
(1.2)

where u and \(u^h\) are the exact and finite element solutions, respectively, of the following problem

$$\begin{aligned} \mathcal{L}u(y)=f(y)\quad&\text {in}\quad \mathfrak {R}^{n},\quad u\quad \text {has a compact support}. \end{aligned}$$

The inequality (1.2) plays a critical role in the proof of our main result

$$\begin{aligned} |\nabla (u-\varPi _{2kh}^{2k}u^h)(y_0)|\lesssim h^{k+2}|\ln h|^{3}\Vert u\Vert _{W^{k+3,\infty }(\mathfrak {R}^{n})}, \end{aligned}$$
(1.3)

where k is even and \(\varPi _{2kh}^{2k}\) is our local operator which interpolates a continuous function to a polynomial of order 2k in a mesh-patch of size 2kh.

Note that the above estimate is only valid for the case that the degree k is even. To obtain the same ultra-convergence property for the case that k is odd, we need to do some special treatment. Towards this end, we first extrapolate the finite element solution to obtain

$$\begin{aligned} |(u-Pu^{h})(y_1)-(u-Pu^{h})(y_2)|\lesssim h^{k+3}|\ln h|^{3}\Vert u\Vert _{W^{k+3,\infty }(\mathfrak {R}^{n})}, \end{aligned}$$
(1.4)

where P is the extrapolation operator defined in Section 4. Subsequently, we obtain

$$\begin{aligned} |\nabla (u-\varPi _{2kh}^{2k}(P u^h))(y_0)|\lesssim h^{k+2}|\ln h|^{3}\Vert u\Vert _{W^{k+3,\infty }(\mathfrak {R}^{n})}. \end{aligned}$$
(1.5)

Based on (1.3) and (1.5) for \(\mathfrak {R}^{n}\), we establish similar estimates on the bounded region \(\Omega \) for the problem (1.1) with help of interior analysis ( see [26] et al.) and negative norm estimate (see [21]).

Other than variable coefficients, we also would like to emphasize that our results are valid for any locally symmetric mesh, particularly for simplical meshes. Comparing with the best known gradient superconvergence result for variable coefficients by Schatz-Sloan-Wahlbin [23, 26]

$$\begin{aligned} \left| \frac{\partial u(x_{0})}{\partial x_{i}}-\frac{\partial \hat{u}^{h}(x_{0},\beta )}{\partial x_{i}}\right| \lesssim h^{k+\delta ^{'}}\ln \left( \frac{1}{h}\right) ,\quad \text {if}\quad k \quad \text {is odd} \end{aligned}$$
(1.6)

and by Asadzadeh et al. [2]

$$\begin{aligned} \left| \frac{\partial u(x_{0})}{\partial x_{i}}-\frac{\partial \hat{v}^{h}(x_{0},\beta )}{\partial x_{i}}\right| \lesssim h^{k+\delta ^{'}}\ln \left( \frac{1}{h}\right) ,\quad \text {if}\quad k\ge 2 \end{aligned}$$
(1.7)

where \(0<\delta ^{'}<1\), our results raise the superocnvergence order to \(h^{k+2}|\ln h|^{3}\), an ultraconvergence result.

The rest of this paper is outlined as follows. In Sect. 2, we discuss the discrete Green’s functions in the whole domain \(\mathfrak {R}^n\) over a uniform conforming partitions. In Sects. 3 and 4, we investigate the ultraconvergence of the finite element solution for the problems in the whole domain \(\mathfrak {R}^n\) over a uniform conforming partitions, where Sect. 3 is for the even order and Sect. 4 is for the odd order finite element solution. In Sect. 5, we apply our theory to problem (1.1). The numerical experiments supporting our theory are presented in Sect. 6.

2 Discrete Green’s functions in \(\mathfrak {R}^n\)

This section is dedicated to a discussion of the finite element approximation properties of the Green’s function. For any positive definite coefficient matrix \(B=(b_{ij})_{n\times n}\), we define the associated bilinear form

$$\begin{aligned} a_B(\psi ,\phi )=\int _{\mathfrak {R}^n} b_{ij}(y)\frac{\partial \psi (y)}{\partial y_{i}} \frac{\partial \phi (y)}{\partial y_{j}}dy\quad \psi ,\phi \in H_0^{1}(\mathfrak {R}^n), \end{aligned}$$

where \(H_0^1(\mathfrak {R}^n)=\{v\in H^1(\mathfrak {R}^n)| v \text {\ has\ a compact \ support}\}\). In particular, we denote \(a(\cdot ,\cdot )=a_A(\cdot ,\cdot )\) for simplicity. Let the Green function \(G^B_{z}\) be defined by

$$\begin{aligned} a_B(G^B_{z},w)=w(z),\quad \forall w\in W^{1,q}(\mathfrak {R}^n)\cap H_{0}^{1}(\mathfrak {R}^n), \forall q>n. \end{aligned}$$
(2.1)

For a given point \(z\in \mathfrak {R}^n\), we define the shift of A by \(A_z=(a_{ij}(y+z))\) and the constant matrix \(\overline{A}_{z}=(a_{ij}(z))\). For simplicity, we denote

$$\begin{aligned} G_z=G_z^A \quad \overline{G}_z=G^{\overline{A}_z}_z. \end{aligned}$$

Let \(\mathcal T_h\) be a uniform conforming partitions of \(\mathfrak {R}^n\) and \(\mathcal N_h\) be the set of all vertices of \(\mathcal T_h\). We assume that \(\mathcal T_h\) is symmetric in the sense that each vertex \(y\in \mathcal N_h\) is a symmetric center of the mesh \(\mathcal T_h\). That is to say, for all \(z\in \mathcal N_h\), \(2z-y\in \mathcal N_h\). Let

$$\begin{aligned} S_h=\{v_h\in C(\mathfrak {R}^n): (v_h)|_e\in P_k,\forall e\in {\mathcal T}_h\} \end{aligned}$$

be the associated finite element space of degree k and let \(S_h^0=S_h\cap H_0^1(\mathfrak {R}^n)\), we introduce the finite element projector \(R_{h}^{B}:H_{0}^{1}(\mathfrak {R}^n)\rightarrow S_h^0\) for all \(\psi \in S_{h}^{0}(\mathfrak {R}^n)\) by

$$\begin{aligned} a_{B}(w-R_{h}^{B}w,\psi )=0. \end{aligned}$$
(2.2)

In particular, we denote \(R_h=R_h^{A},\quad \tilde{R}_{h}^{z}=R_{h}^{A_{z}},\quad \overline{R}_{h}^{z}=R_h^{\overline{A}_z}\).

The following approximation property

$$\begin{aligned} \Vert G^{B}_{z}-R_h^{B}G^{B}_{z}\Vert _{W^{1,\infty }(\mathfrak {R}^n\smallsetminus B(z,d))}\lesssim h^{k}d^{1-k-n}|\ln h|,\forall d>0, \end{aligned}$$
(2.3)

is shown in [24]. Usually, the estimate (2.3) can be used to derive the vertex-wise convergence or superconvergence of \(|(u-R_h u)(x_{0})|, x_0\in \mathcal N_h\), where \(u\in H_0^1(\mathfrak {R}^n)\) is the solution of

$$\begin{aligned} a(u,v)=(f,v),\forall v\in H_0^1(\mathfrak {R}^n) \end{aligned}$$
(2.4)

and \(R_hu=u^h\in S_h^0\) is the finite element solution satisfying

$$\begin{aligned} a(u^h,v_h)=(f,v_h),\forall v_h\in S_h^0. \end{aligned}$$
(2.5)

To derive the ultraconvergence of \(R_h u\), we need to discuss further approximation properties of the discrete Green’s functions \(R_hG_z\). Precisely, in this section, we shall estimate the following three quantities defined for all \(y,z\in \mathfrak {R}^n\) by

$$\begin{aligned}&\alpha (y,z)=G_{z}(y)-R_{h}G_{z}(y)-[\overline{G}_{z}(y)-\overline{R}_{h}^{z}\overline{G}_{z}(y)], \end{aligned}$$
(2.6)
$$\begin{aligned}&\alpha _{1}(y,z)=G_{0}(y)-R_{h}G_{0}(y)-[G_{z}(y+z)-(R_{h}G_{z})(y+z)], \end{aligned}$$
(2.7)
$$\begin{aligned}&\alpha _{2}(y,z)=\overline{G}_{0}(y)-\overline{R}_{h}^{0}\overline{G}_{0}(y)-[\overline{G}_{z}(y+z)-\overline{R}_{h}^{z}G_{z}(y+z)] \end{aligned}$$
(2.8)

which depend on the smoothness of variable coefficient \(A=(a_{ij})\).

We have the following Theorem.

Theorem 2.1

Let \(a_{ij}\in C^{\infty }(\mathfrak {R}^n), 1\le i,j\le n\). Then

$$\begin{aligned} \int _{\mathfrak {R}^n} |\nabla _y \alpha (y,z)| dy\lesssim h^{2}|\ln h|^3. \end{aligned}$$
(2.9)

Moreover if \(|z|\lesssim h\), then we have

$$\begin{aligned} \int _{\mathfrak {R}^n}|y||\nabla _y \alpha _{1}(y,z)|dy\lesssim h^{3}|\ln h|^3,\end{aligned}$$
(2.10)
$$\begin{aligned} \int _{\mathfrak {R}^n}|y||\nabla _y \alpha _{2}(y,z)|dy\lesssim h^{3}|\ln h|^3, \end{aligned}$$
(2.11)

and

$$\begin{aligned} \int _{\mathfrak {R}^n}|\nabla (\alpha (y,0)-\alpha (y+z,z))|dy\lesssim h^{3}|\ln h|^{3}. \end{aligned}$$
(2.12)

Before proceeding the proof of Theorem 2.1, we first introduce some lemmas.

Lemma 2.2

Let \(a_{ij}\in C^{\infty }(\mathfrak {R}^n), 1\le i,j\le n\) and \(E(y)=G_{z}(y)-\overline{G}_{z}(y), y\in \mathfrak {R}^n\). Then

$$\begin{aligned} \Vert E\Vert _{W^{l,\infty }(\mathfrak {R}^n\smallsetminus B(z,d))}\lesssim d^{3-n-l}|\ln d|, \quad \forall d>0. \end{aligned}$$
(2.13)

Consequently,

$$\begin{aligned} \Vert E-R_hE\Vert _{W^{1,1}(\mathfrak {R}^n)}\lesssim h^{2}|\ln h|^{2}. \end{aligned}$$
(2.14)

Proof

We first show (2.13) for the case \(l=1\). We denote \(\chi _{t}(s)=\frac{\partial E(s)}{\partial s_{t}}, t=1,\ldots ,n\). One observes that

$$\begin{aligned} \frac{\partial }{\partial s_{i}}\left( a_{ij}(s)\frac{\partial \chi _{t}(s)}{\partial s_{j}}\right) =\frac{\partial ^{2}}{\partial s_{i}\partial s_{t}}\left( a_{ij}(s)\frac{\partial E(s)}{\partial s_{j}}\right) -\frac{\partial }{\partial s_{i}}\left( \frac{\partial a_{ij}(s)}{\partial s_{t}}\frac{\partial E(s)}{\partial s_{j}}\right) . \end{aligned}$$
(2.15)

The fact that \(w(z)=a(G_z,w)=a_{\overline{A}_{z}}(\overline{G}_z,w), \forall w\in W^{1,q}(\mathfrak {R}^n)\cap H_{0}^{1}(\mathfrak {R}^n), \forall q>n\) yields that

$$\begin{aligned} \frac{\partial }{\partial s_{i}}\left( a_{ij}(s)\frac{\partial G_{z}(s)}{\partial s_{j}}\right) =\frac{\partial }{\partial s_{i}}\left( a_{ij}(z)\frac{\partial \overline{G}_z(s)}{\partial s_{j}}\right) \end{aligned}$$

and thus

$$\begin{aligned} \frac{\partial ^{2}}{\partial s_{i}\partial s_{t}}\left( a_{ij}(s) \frac{\partial E(s)}{\partial s_{j}}\right)= & {} \frac{\partial ^{2}}{\partial s_{i}\partial s_{t}}\left[ (a_{ij}(z)-a_{ij}(s))\frac{\partial \overline{G}_{z}(s)}{\partial s_{j}}\right] \nonumber \\= & {} \frac{\partial }{\partial s_{i}}\left[ (a_{ij}(z)-a_{ij}(s))\frac{\partial ^{2} \overline{G}_{z}(s)}{\partial s_{j}\partial s_{t}}\right] -\frac{\partial }{\partial s_{i}}\left[ \frac{\partial a_{ij}(s)}{\partial s_{t}}\frac{\partial \overline{G}_{z}(s)}{\partial s_{j}}\right] .\nonumber \\ \end{aligned}$$
(2.16)

Combining (2.15) and (2.16), we have

$$\begin{aligned}&\mathcal{L}\chi _{t}(s)=\frac{\partial }{\partial s_{i}}\left[ (a_{ij}(z)-a_{ij}(s))\frac{\partial ^{2} \overline{G}_{z}(s)}{\partial s_{j}\partial s_{t}}\right] \nonumber \\&-\frac{\partial }{\partial s_{i}}\left[ \frac{\partial a_{ij}(s)}{\partial s_{t}}\frac{\partial \overline{G}_{z}(s)}{\partial s_{j}}\right] -\frac{\partial }{\partial s_{i}}\left( \frac{\partial a_{ij}(s)}{\partial s_{t}}\frac{\partial E(s)}{\partial s_{j}}\right) . \end{aligned}$$
(2.17)

Let \(y\in \mathfrak {R}^{n}{\setminus } B(z,d)\). Then

$$\begin{aligned} \chi _{t}(y)&=\int _{\mathfrak {R}^n}a_{ij}(s) \frac{\partial \chi _k(s)}{\partial s_{j}}\frac{\partial G_{y}(s)}{\partial s_{i}}ds =\int _{\mathfrak {R}^n}\left[ (a_{ij}(z)-a_{ij}(s))\frac{\partial ^{2} \overline{G}_{z}(s)}{\partial s_{j}\partial s_{t}}\right] \frac{\partial G_{y}(s)}{\partial s_{i}}ds \nonumber \\&\quad - \int _{\mathfrak {R}^n}\frac{\partial a_{ij}(s)}{\partial s_{t}}\frac{\partial \overline{G}_{z}(s)}{\partial s_{j}}\frac{\partial G_{y}(s)}{\partial s_{i}}ds- \int _{\mathfrak {R}^n} \frac{\partial a_{ij}(s)}{\partial s_{t}}\frac{\partial E(s)}{\partial s_{j}}\frac{\partial G_{y}(s)}{\partial s_{i}}ds\nonumber \\&=J_{1}(y)+J_{2}(y)+J_{3}(y). \end{aligned}$$
(2.18)

We first estimate \(J_{1}(y)\). For all \(d>0\), let \(d_{0}=\max \{2d, 1\}\) and \(\Omega _{0}=B(z,d_{0})\cup B(y,d_{0})\). Then

$$\begin{aligned} J_{1}(y)&=\left( \int _{B(z,\frac{d}{2})}+\int _{B(y,\frac{d}{2})}+\int _{\Omega _{0}\smallsetminus (B(y,\frac{d}{2})\cup B(z,\frac{d}{2}))}+\int _{\mathfrak {R}^n{\setminus } \Omega _{0}}\right) \nonumber \\&\quad \times \left[ (a_{ij}(z)-a_{ij}(s))\frac{\partial ^{2} \overline{G}_{z}(s)}{\partial s_{j}\partial s_{t}}\right] \frac{\partial G_{y}(s)}{\partial s_{i}}ds \nonumber \\&=J_{1,1}(y)+J_{1,2}(y)+ J_{1,3}(y)+J_{1,4}(y). \end{aligned}$$
(2.19)

Moreover, by [11] and [12],

$$\begin{aligned} |\nabla ^{2}\overline{G}_{z}(s)|\lesssim |s-z|^{-n},\quad |\nabla G_{y}(s)|\lesssim |y-s|^{1-n} \end{aligned}$$

and the fact that \(a_{ij}\in C^\infty \) yields that

$$\begin{aligned} |a_{ij}(z)-a_{ij}(s)|\lesssim |s-z|. \end{aligned}$$

When \(y\in \mathfrak {R}^n\setminus B(z,d), s\in B(z,\frac{d}{2})\), we have \(|y-s|\ge \frac{d}{2}\) and thus

$$\begin{aligned} |J_{1,1}(y)|\lesssim \int _{B(z,d/2)} |s-z|^{1-n}|y-s|^{1-n}ds \lesssim d^{1-n}\int _{B(z,d/2)} |s-z|^{1-n} ds \lesssim d^{2-n}. \end{aligned}$$

Similarly,

$$\begin{aligned} |J_{1,2}(y)|\lesssim d^{2-n}. \end{aligned}$$

A straightforward calculation yields that

$$\begin{aligned} |J_{1,3}(y)|\lesssim d^{2-n}|\ln d|, \end{aligned}$$

and

$$\begin{aligned} |J_{1,4}|&\lesssim \int _{\mathfrak {R}^n{\setminus } \Omega _{0}} |\frac{\partial ^{2} \overline{G}_{z}(s)}{\partial s_{j}\partial s_{t}}| |\frac{\partial G_{y}(s)}{\partial s_{i}}|ds \lesssim \int _{\mathfrak {R}^n\smallsetminus \Omega _{0}}|s-z|^{-n} |y-s|^{1-n}ds\nonumber \\&\lesssim d_{0}^{1-n}\lesssim \min \{d^{2-n},1\}. \end{aligned}$$
(2.20)

Then by (2.19), we have

$$\begin{aligned} |J_{1}(y)|\lesssim d^{2-n}|\ln d|. \end{aligned}$$

Similarly,

$$\begin{aligned} |J_{2}(y)|\lesssim d^{2-n}|\ln d|,\quad \text {and}\quad |J_{3}(y)|\lesssim d^{2-n}|\ln d|. \end{aligned}$$

Then, by (2.18), we get the desired result (2.13) for the case \(l=1\). We turn now to the proof of (2.13) for \(l\ge 2\). From (2.17) it follows that, for any positive integer m,

$$\begin{aligned}&|\nabla ^{m}\mathcal{L}\chi _{t}(s)|\le \left| \nabla ^{m}\left( \frac{\partial }{\partial s_{i}}\left[ (a_{ij}(z)-a_{ij}(s))\frac{\partial ^{2} \overline{G}_{z}(s)}{\partial s_{j}\partial s_{t}}\right] \right) \right| \nonumber \\&+\left| \nabla ^{m}\left( \frac{\partial }{\partial s_{i}}\left[ \frac{\partial a_{ij}(s)}{\partial s_{t}}\frac{\partial \overline{G}_{z}(s)}{\partial s_{j}}\right] \right) \right| +\left| \nabla ^{m}\left( \frac{\partial }{\partial s_{i}}(\frac{\partial a_{ij}(s)}{\partial s_{t}}\frac{\partial E(s)}{\partial s_{j}})\right) \right| \nonumber \\&\lesssim |s-z||\nabla ^{m+3}\overline{G}_{z}(s)|+|\nabla ^{m+2}\overline{G}_{z}(s)| +|\nabla ^{m+2}G_{z}(s)| \nonumber \\&\lesssim |s-z|^{-m-n}+|s-z|^{-m-n}+|s-z|^{-m-n}\lesssim |s-z|^{-m-n}, \end{aligned}$$
(2.21)

where we have used the estimate (see [11] and [12])

$$\begin{aligned} |\nabla ^{m^{'}}\overline{G}_{z}(s)|+|\nabla ^{m^{'}}G_{z}(s)|\lesssim |s-z|^{2-m^{'}-n}, \end{aligned}$$

where \(m^{'}\) is a positive integer. Assume that the weighted Sobolev space \(\kappa _{a}^{T}(B(s,R))\) is a normed linear spaces if equipped with the norms \(\Vert \mu \Vert _{\kappa _{a}^{T}(B(s,R))}=\sum \nolimits _{|\alpha |\le T}\left( \int _{B(s,R)}\left( |D^{\alpha }\mu (x)|\rho (x,M)^{|\alpha |-a}\right) ^{2}dx\right) ^{\frac{1}{2}}\). By Theorem 6.5 (see [3]), (2.21) and (2.13) for the case \(l=1\), we obtain, for all positive integer \(m\ge 2\),

$$\begin{aligned} \Vert \chi _{t}\Vert _{\kappa _{m-2}^{m}(B(s,d/2))}&\lesssim \Vert \mathcal{L} E\Vert _{\kappa _{m-2}^{m-2}(B(s,d/2))}+ \Vert \chi _{t}\Vert _{\kappa _{-2}^{0}(B(s,d/2))}\\&\lesssim d^{m-2}d^{2-m-n+\frac{n}{2}}+d^{-2}d^{2-n+\frac{n}{2}}|\ln d| \nonumber \\ {}&\lesssim d^{-\frac{n}{2}}|\ln d|. \end{aligned}$$

This implies

$$\begin{aligned} \Vert \chi _{t}\Vert _{H^{m}(B(s,d/2))}\lesssim d^{2-m}\Vert \chi _{t}\Vert _{\kappa _{m-2}^{m}(B(s,d/2))}\lesssim d^{2-m-\frac{n}{2}}|\ln d|. \end{aligned}$$
(2.22)

By (2.22), we have

$$\begin{aligned} \Vert \chi _{t}\Vert _{W^{l-1,\infty }(B(s,d/2))}&\lesssim \Vert \chi _{t}\Vert _{W^{l+n-1,1}(B(s,d/2))}\lesssim d^{\frac{n}{2}}\Vert \chi _{t}\Vert _{H^{l+n-1}(B(s,d/2))} \nonumber \\&\lesssim d^{\frac{n}{2}}d^{3-l-n-\frac{n}{2}}|\ln d| \lesssim d^{3-l-n}|\ln d|. \end{aligned}$$
(2.23)

Consequently,

$$\begin{aligned} \Vert E\Vert _{W^{l,\infty }(\mathfrak {R}^n\smallsetminus B(z,d))}\lesssim d^{3-l-n}|\ln d|. \end{aligned}$$

This gives the desired result (2.13).

Next we show (2.14). For any \(y\in \mathfrak {R}^n\),

$$\begin{aligned} (E-R_hE)(y)= & {} a(E-R_hE, G_y) =a(E-R_hE, G_y-R_hG_y)\\= & {} a(E-I^k_hE, G_y-R_hG_y), \end{aligned}$$

where \(I_h^k\) is the standard interpolating operator from \(H^1_0(\mathfrak {R}^n)\) to \(S^0_h\). Letting

$$\begin{aligned} B_1= & {} \int _{B(z,d)}a_{ij}(s)\frac{\partial (E-I_{h}^{k}E)(s)}{\partial s_{i}}\frac{\partial (G_{y}-R_hG_{y})(s)}{\partial s_{j}}ds,\\ B_2= & {} \int _{\mathfrak {R}^n\smallsetminus B(z,d)}a_{ij}(s)\frac{\partial (E-I_{h}^{k}E)(s)}{\partial s_{i}}\frac{\partial (G_{y}-R_hG_{y})(s)}{\partial s_{j}}ds, \end{aligned}$$

we have

$$\begin{aligned} (E-R_hE)(y)=B_{1}+B_{2}. \end{aligned}$$
(2.24)

We next estimate \(B_1\) and \(B_2\) for \(y\in \mathfrak {R}^n\setminus B(z,2d)\) for some \(d\ge c_1h>h\). Since \(y\in \mathfrak {R}^n\setminus B(z,2d)\), we have \(s\in \mathfrak {R}^n\smallsetminus B(y,d)\) if \(s\in B(z,d)\). Note that

$$\begin{aligned} \mathcal{L} E=\frac{\partial }{\partial s_{i}}\left[ (a_{ij}(z)-a_{ij}(s))\frac{\partial \overline{G}_{z}(s)}{\partial s_{j}}\right] . \end{aligned}$$

Therefore, by (2.3), we have

$$\begin{aligned} |B_{1}|&\lesssim \Vert E-I_{h}^{k}E\Vert _{W^{1,1}(B(z,d))}\Vert G_{y}-R_hG_{y}\Vert _{W^{1,\infty }(\mathfrak {R}^n\smallsetminus B(y,d))} \nonumber \\&\lesssim h^{k}d^{1-k-n}|\ln h|\left[ \Vert E-I_{h}^{k}E\Vert _{W^{1,1}(B(z,h))}+h^{k}\Vert E\Vert _{W^{k+1,1}(B(z,d)\smallsetminus B(z,h))}\right] \nonumber \\&\lesssim h^{k+2}|\ln h|d^{1-k-n}, \end{aligned}$$
(2.25)

where we have used (2.3), (2.13) for the case \(l=k+1\) and the fact that \(d> h\) in the last inequality.

Similarly,

$$\begin{aligned} |B_{2}| \lesssim h|\ln h|h^{k}\Vert E\Vert _{W^{k+1,\infty }(\mathfrak {R}^n\smallsetminus B(z,d))} \lesssim h^{k+1}|\ln h|d^{2-k-n}. \end{aligned}$$
(2.26)

Substituting (2.25) and (2.26) into (2.24) and noticing \(h<d\), we obtain

$$\begin{aligned} \Vert E-R_hE\Vert _{L^{\infty }(\mathfrak {R}^n\smallsetminus B(z,2d))}\lesssim h^{k+1}|\ln h|d^{2-k-n}. \end{aligned}$$
(2.27)

Thus

$$\begin{aligned}&\Vert E-R_hE\Vert _{W^{1,\infty }(\mathfrak {R}^n\smallsetminus B(z,2d))}\\&\quad \le \Vert E-I_{h}^{k}E\Vert _{W^{1,\infty }(\mathfrak {R}^n\smallsetminus B(z,2d))} +\Vert I_{h}^{k}E-R_hE\Vert _{W^{1,\infty }(\mathfrak {R}^n\smallsetminus B(z,2d))} \\&\quad \lesssim h^{k}\Vert E\Vert _{W^{k+1,\infty }(\mathfrak {R}^n\smallsetminus B(z,2d))}+h^{-1}\Vert E-R_hE\Vert _{L^{\infty }(\mathfrak {R}^n\smallsetminus B(z,2d))}\\&\quad \lesssim h^{k}|\ln h|d^{2-k-n}. \end{aligned}$$

This implies

$$\begin{aligned} \Vert E-R_hE\Vert _{W^{1,1}(\mathfrak {R}^n\smallsetminus B(z,2c_{1}h))}\lesssim h^{k}|\ln h|h^{2-k}|\ln h|\lesssim h^{2}|\ln h|^{2}. \end{aligned}$$
(2.28)

We turn now to the estimation of \(\Vert E-R_hE\Vert _{W^{1,1}(B(z,2c_{1}h))}\). One observes that, for all \(y\in B(z,2c_1h)\),

$$\begin{aligned} \Vert R_{h}G_{y}\Vert _{H^{1}(\mathfrak {R}^{n})}\lesssim a(R_{h}G_{y},R_{h}G_{y})^{\frac{1}{2}}\lesssim \Vert R_{h}G_{y}\Vert _{L^{\infty }(\mathfrak {R}^{n})}^{\frac{1}{2}}. \end{aligned}$$

Using the inverse estimate, we have, if \(n=2\),

$$\begin{aligned} \Vert R_{h}G_{y}\Vert _{L^{\infty }(\mathfrak {R}^{n})}&=\sup \limits _{e\in \mathcal {T}_{h}}\Vert R_{N}G_{x_{0}}\Vert _{L^{\infty }(e)} \lesssim \sup \limits _{e\in \mathcal {T}_{h}}\Vert R_{N}G_{x_{0}}\Vert _{L^{|\ln h|}(e)} |\ln h|^{\frac{-1}{2}} \\&\lesssim \Vert R_{N}G_{x_{0}}\Vert _{L^{|\ln h|}(\mathfrak {R}^{n})}|\ln h|^{\frac{-1}{2}}\lesssim |\ln h|^{\frac{-1}{2}}\Vert R_{N}G_{x_{0}}\Vert _{H^{1}(\mathfrak {R}^{n})}, \end{aligned}$$

and if \(n=3\),

$$\begin{aligned} \Vert R_{h}G_{y}\Vert _{L^{\infty }(\mathfrak {R}^{n})}&=\sup \limits _{e\in \mathcal {T}_{h}}\Vert R_{N}G_{x_{0}}\Vert _{L^{\infty }(e)} \lesssim \sup \limits _{e\in \mathcal {T}_{h}}\Vert R_{N}G_{x_{0}}\Vert _{L^{6}(e)} h^{\frac{-1}{2}} \\&\lesssim \Vert R_{N}G_{x_{0}}\Vert _{L^{6}(\mathfrak {R}^{n})}h^{\frac{-1}{2}}\lesssim h^{\frac{-1}{2}}\Vert R_{N}G_{x_{0}}\Vert _{H^{1}(\Omega )}. \end{aligned}$$

Combining the above three estimates, we have

$$\begin{aligned} \Vert R_{h}G_{y}\Vert _{L^{\infty }(\mathfrak {R}^{n})}\le h^{2-n}|\ln h|^{-1}. \end{aligned}$$
(2.29)

Furthermore, by (2.29) and the inverse estimate, we have, for \(n=2,3\),

$$\begin{aligned} \Vert R_{h}G_{y}\Vert _{W^{1,\infty }(\mathfrak {R}^{n})}\lesssim h^{-1}\Vert R_{h}G_{y}\Vert _{L^{\infty }(\mathfrak {R}^{n})}\lesssim h^{1-n}|\ln h|^{-1}. \end{aligned}$$

This implies

$$\begin{aligned} |(I_{h}^{k}E-R_hE)(y)|= & {} |a(I_{h}^{k}E-E,R_hG_{y})|\lesssim \Vert I_{h}^{k}E-E\Vert _{W^{1,1}(\mathfrak {R}^n)}\Vert R_hG_{y}\Vert _{W^{1,\infty }(\mathfrak {R}^n)} \nonumber \\&\lesssim h^{3-n}|\ln h|. \end{aligned}$$
(2.30)

Consequently,

$$\begin{aligned} \Vert E-R_hE\Vert _{W^{1,1}(B(z,2c_{1}h))}&\le \Vert E-I_{h}^{k}E\Vert _{W^{1,1}(B(z,2c_{1}h))} +\Vert I_{h}^{k}E-R_hE\Vert _{W^{1,1}(B(z,2c_{1}h))}\nonumber \\&\lesssim h^{2}|\ln h| +h^{n}h^{-1}\Vert I_{h}^{k}E-R_hE\Vert _{L^{\infty }(B(z,2c_{1}h))}\lesssim h^{2}|\ln h|. \end{aligned}$$
(2.31)

The estimate (2.14) is a direct result of (2.28) and (2.31). \(\square \)

Lemma 2.3

If all \(a_{ij}\in C^{\infty }(\mathfrak {R}^n), 1\le i,j\le n\), then

$$\begin{aligned} \Vert (R_h-\overline{R}^{z}_{h})\overline{G}_{z}\Vert _{W^{1,1}(\mathfrak {R}^n)}\lesssim & {} h^{2}|\ln h|^{3}. \end{aligned}$$
(2.32)

Proof

By the definition of \(R_h\) and \(\overline{R}^{z}_{h}\), we have that for any \(v\in S_{0}^{h}(\mathfrak {R}^n)\),

$$\begin{aligned}&\int _{\mathfrak {R}^n}a_{ij}(s)\frac{\partial (R_h\overline{G}_{z}-\overline{G}_{z})(s)}{\partial s_{i}}\frac{\partial v(s)}{\partial s_{j}}ds=0, \nonumber \\&\int _{\mathfrak {R}^n}a_{ij}(z)\frac{\partial (\overline{R}^{z}_{h}\overline{G}_{z}-\overline{G}_{z})(s)}{\partial s_{i}}\frac{\partial v(s)}{\partial s_{j}}ds =0. \end{aligned}$$

Then,

$$\begin{aligned}&\int _{\mathfrak {R}^n}a_{ij}(s)\frac{\partial ((R_h-\overline{R}^{z}_{h})\overline{G}_{z})(s)}{\partial s_{i}}\frac{\partial v(s)}{\partial s_{j}}ds \nonumber \\&=\int _{\mathfrak {R}^n}[a_{ij}(s)-a_{ij}(z)]\frac{\partial (\overline{R}^{z}_{h}\overline{G}_{z}-\overline{G}_{z})(s)}{\partial s_{i}} \frac{\partial v(s)}{\partial s_{j}}ds. \end{aligned}$$
(2.33)

Consequently,

$$\begin{aligned} \frac{\partial ((R_h-\overline{R}_{h}^{z})\overline{G}_{z})(y)}{\partial y_{l}}&=\int _{\mathfrak {R}^n}(a_{ij}(s)-a_{ij}(z)) \frac{\partial (\overline{G}_{z}-\overline{R}_{h}^{z}\overline{G}_{z})(s)}{\partial s_{i}} \frac{\partial R_h g_{l}(s)}{\partial s_{j}}ds, \end{aligned}$$
(2.34)

where for \(1\le l\le n\), \(R_{h}g_{l}\) is the discrete derivative of Green’s function defined by

$$\begin{aligned} a(R_{h}g_{l},w)=\frac{\partial w(y)}{\partial y_{l}} \quad \forall w\in H_{0}^{1}(\mathfrak {R}^n)\cap W^{2,q}(\mathfrak {R}^n),q>n. \end{aligned}$$

There exists the following estimates (see [35, 36])

$$\begin{aligned} \Vert R_hg_{l}\Vert _{W^{1,1}(\mathfrak {R}^{n})}&\lesssim |\ln h|, \Vert R_hg_{l}\Vert _{W^{1,\infty }(\mathfrak {R}^{n})}\lesssim h^{-n}|\ln h|, \Vert R_hg_{l}\Vert _{W^{1,\infty }(\mathfrak {R}^{n}\smallsetminus B(y,d))}\nonumber \\&\lesssim d^{-n}|\ln h|, \end{aligned}$$
(2.35)

where \(d>c_{0}h\). In the following, we use (2.3), (2.34), (2.35) to prove (2.32). Let \(d_{0}=c_{0}h\). Note that

$$\begin{aligned}&\int _{\mathfrak {R}^n}|(a_{ij}(s)-a_{ij}(z)) \nabla (\overline{G}_{z}-\overline{R}_{h}^{z}\overline{G}_{z})(s)|ds \nonumber \\&\quad =\int _{B(z,d_{0})}+\int _{\mathfrak {R}^n\smallsetminus B(z,d_{0})}|(a_{ij}(s)-a_{ij}(z)) \nabla (\overline{G}_{z}-\overline{R}_{h}^{z}\overline{G}_{z})(s)|ds \nonumber \\&\quad \lesssim h\Vert \overline{G}_{z}-\overline{R}_{h}^{z}\overline{G}_{z}\Vert _{W^{1,1}(\mathfrak {R}^n)}+ \int _{\mathfrak {R}^n\smallsetminus B(z,d_{0})}|s-z|\times h^{k}|s-z|^{1-k-n}|\ln h|ds\nonumber \\&\quad \lesssim h^{2}|\ln h|^{2}. \end{aligned}$$
(2.36)

Combining this estimate and (2.35), we have

$$\begin{aligned} \Vert (R_h-\overline{R}_h^{z})\overline{G}_{z}\Vert _{W^{1,\infty }(\mathfrak {R}^{n})}\lesssim h^{2}h^{-n}|\ln h|^{2}\lesssim h^{2-n}|\ln h|^{2}. \end{aligned}$$
(2.37)

We next turn to an improved estimate of \(\Vert (R_h-\overline{R}_h^{z})\overline{G}_{z}\Vert _{W^{1,\infty }(\mathfrak {R}^n\setminus B(z,d))}\) for all \(d\ge d_0=c_0h\). By (2.34), we have that for any \(y\in \mathfrak {R}^n\setminus B(z,d)\),

$$\begin{aligned}&\frac{\partial ((R_h-\overline{R}_{h}^{z})\overline{G}_{z})(y)}{\partial y_{l}} =I_{1}+I_{2}, \end{aligned}$$
(2.38)

with

$$\begin{aligned} I_1&=\int _{B(z,d/2)} (a_{ij}(s)-a_{ij}(z)) \frac{\partial (\overline{G}_{z}-\overline{R}_{h}^{z}\overline{G}_{z})(s)}{\partial s_{i}} \frac{\partial R_hg_{l}(s)}{\partial s_{j}}ds,\\ I_2&=\int _{\mathfrak {R}^n\smallsetminus B(z,d/2)}(a_{ij}(s)-a_{ij}(z)) \frac{\partial (\overline{G}_{z}-\overline{R}_{h}^{z}\overline{G}_{z})(s)}{\partial s_{i}} \frac{\partial R_hg_{l}(s)}{\partial s_{j}}ds. \end{aligned}$$

By (2.35) and (2.36), we have

$$\begin{aligned} |I_{1}|\lesssim h^{2}|\ln h|\Vert R_hg_{l}\Vert _{W^{1,\infty }(\mathfrak {R}^n\smallsetminus B(y,d/2))} \lesssim h^{2}|\ln h|^2d^{-n}. \end{aligned}$$
(2.39)

On the other hand, by the fact of \(a_{ij}\in C^\infty \) and (2.3),

$$\begin{aligned}&\left\| (a_{ij}(s)-a_{ij}(z))\frac{\partial (\overline{G}_{z}-\overline{R}_{h}^{z}\overline{G}_{z})(s)}{\partial s_{i}}\right\| _{L^{\infty }(\mathfrak {R}^n\smallsetminus B(z,d/2))}\lesssim h^{k}d^{2-k-n}|\ln h|. \end{aligned}$$

This estimate, together with (2.35), gives

$$\begin{aligned} |I_{2}|\lesssim h^{k}d^{2-k-n}|\ln h|^{2}. \end{aligned}$$
(2.40)

Inserting (2.39) and (2.40) into (2.38),

$$\begin{aligned} \Vert (R_h-\overline{R}_{h}^{z})\overline{G}_{z}\Vert _{W^{1,\infty }(\mathfrak {R}^n\smallsetminus B(z,d))}\lesssim h^{2}d^{-n}|\ln h|^{2}. \end{aligned}$$

Consequently,

$$\begin{aligned} \Vert (R_h-\overline{R}_{h}^{z})\overline{G}_{z}\Vert _{W^{1,1}(\mathfrak {R}^n\smallsetminus B(z,d_0))}\lesssim h^{2}|\ln d_0||\ln h|^{2}\lesssim h^{2}|\ln h|^{3}. \end{aligned}$$
(2.41)

As an immediate consequence of (2.37) and (2.41), we obtain

$$\begin{aligned}&\Vert (R_h-\overline{R}_{h}^{z})\overline{G}_{z}\Vert _{W^{1,1}(\mathfrak {R}^{n})}\\&\quad = \Vert (R_h-\overline{R}_{h}^{z})\overline{G}_{z}\Vert _{W^{1,1}( B(z,d_{0}))}+\Vert (R_h-\overline{R}_{h}^{z})\overline{G}_{z}\Vert _{W^{1,1}(\mathfrak {R}^n\smallsetminus B(z,d_{0}))} \nonumber \\&\quad \lesssim h^{k}h^{2-k-n}|\ln h|^{3} h^{n}+ h^{2}|\ln h|^{3}\lesssim h^{2}|\ln h|^{3} \end{aligned}$$

from which the proof is completed. \(\square \)

Based on Lemmas 2.2 and 2.3, we are now ready to prove Theorem 2.1.

Proof of Theorem 2.1

First, noticing the definition of \(\alpha \), the estimate (2.9) is a direct consequence of the Lemmas 2.3 and 2.4.

Next we show (2.10). Let \( E_{0}(y)=G_{z}(y+z), E_{1}(y)=G_{0}(y)-E_{0}(y), y\in \mathfrak {R}^n\). Noticing the fact that \(\tilde{R}_h^{z}E_{0}(y)=(R_{h}G_{z})(y+z)\), one observes that

$$\begin{aligned} \alpha _{1}(y,z)=(E_{1}-R_hE_{1})(y)-(R_{h}-\tilde{R}_h^{z})E_{0}(y). \end{aligned}$$
(2.42)

By the same arguments in the proof of (2.13), we have that, for all \(|z|\lesssim h\) and all \(d>0\),

$$\begin{aligned} \Vert E_{1}\Vert _{L^{\infty }(\mathfrak {R}^n\smallsetminus B(z,d))}\lesssim h d^{2-n}|\ln d|. \end{aligned}$$

Then by the same reasoning to show (2.14), we can prove

$$\begin{aligned} \int _{\mathfrak {R}^n}|\nabla (E_{1}-R_hE_{1})(y)||y|dy \lesssim h^{3}|\ln h|^{3}. \end{aligned}$$
(2.43)

Moreover, similarly to the proof of (2.32), we can show that

$$\begin{aligned} \int _{\mathfrak {R}^n}|y||\nabla ((R_{h}-\tilde{R}_h^{z})E_{0}(y))|dy\lesssim & {} h^{3}|\ln h|^{3}. \end{aligned}$$
(2.44)

Plugging the estimates (2.43) and (2.44) into the equality (2.42), we get the desired result (2.10).

By the same reasoning, we obtain the estimate (2.11) .

Next we show (2.12). Note that

$$\begin{aligned} \alpha (y,0)&=[(G_{0}-R_{h}G_{0})-(\overline{G}_{0}-\overline{R}_{h}^{0}\overline{G}_{0})](y) \\&=[(G_{0}-\overline{G}_{0})(y)-R_{h}(G_{0}-\overline{G}_{0})(y)]-(R_{h}\overline{G}_{0}-\overline{R}_{h}^{0}\overline{G}_{0})(y), \end{aligned}$$

and

$$\begin{aligned} \alpha (y+z,z)&=[(G_{z}-R_{h}G_{z })-(\overline{G}_{z}-\overline{R}_{h}^{z }\overline{G}_{z })](y+z ) \\&=[(G_{z}-\overline{G}_{z })(y+z )-R_{h}(G_{z}-\overline{G}_{z})(y+z )]-(R_{h}\overline{G}_{z }-\overline{R}_{h}^{z}\overline{G}_{z })(y+z). \end{aligned}$$

Then by letting

$$\begin{aligned} \beta _{t}(y)= & {} (G_{t}-\overline{G}_{t })(y+t),\quad \bar{\beta }_{z}(y)=\beta _{0}(y)-\beta _{z}(y),\\ \omega (y)= & {} (R_{h}-\overline{R}_{h}^{0})\overline{G}_{0}(y)-((R_{h}-\overline{R}_{h}^{z })\overline{G}_{z })(y+z ), \end{aligned}$$

and noticing the fact that \(\tilde{R}_{h}^{z}\beta _{z}(y)=R_{h}(G_{z}-\overline{G}_{z})(y+z)\), we have

$$\begin{aligned}&\displaystyle \alpha (y,0)-\alpha (y+z,z) \nonumber \\&\displaystyle =(\bar{\beta }_{z}-R_{h}\bar{\beta }_{z})(y) -(R_{h}-\tilde{R}_{h}^{z})\beta _{z}(y)- [(R_{h}\overline{G}_{0}-\overline{R}_{h}^{0}\overline{G}_{0})(y) -(R_{h}\overline{G}_{z}-\overline{R}_{h}^{z}\overline{G}_{z})(y+z)] \nonumber \\&\displaystyle =(\bar{\beta }_{z}-R_{h}\bar{\beta }_{z})(y)+\omega (y)-(R_{h}-\tilde{R}_{h}^{z})\beta _{z}(y). \end{aligned}$$
(2.45)

We next estimate the three terms of the right-hand of the above equality. First, note that

$$\begin{aligned}&\frac{\partial }{\partial y_{i}}\left( a_{ij}(y)\frac{\partial \beta _{0}(y)}{\partial y_{j}}\right) =\frac{\partial }{\partial y_{i}}\left[ (a_{ij}(0)-a_{ij}(y))\frac{\partial \overline{G}_{0}(y)}{\partial y_{j}}\right] , \\&\frac{\partial }{\partial y_{i}}\left( a_{ij}(y+z )\frac{\partial \beta _{z}(y+z )}{\partial y_{j}}\right) =\frac{\partial }{\partial y_{i}}\left[ (a_{ij}(z )-a_{ij}(y+z ))\frac{\partial \overline{G}_{z }(y+z )}{\partial y_{j}}\right] , \end{aligned}$$

we have

$$\begin{aligned} \frac{\partial }{\partial y_{i}}\left( a_{ij}(y)\frac{\partial \bar{\beta }_{z}(y)}{\partial y_{j}}\right)= & {} \frac{\partial }{\partial y_{i}}(a_{ij}(y)-a_{ij}(y+z ))\frac{\partial \beta _{z}(y+z)}{\partial y_{j}}) \nonumber \\&+ \frac{\partial }{\partial y_{i}}\left[ ((a_{ij}(0)-a_{ij}(y))-(a_{ij}(z )-a_{ij}(y+z )) \frac{\partial \overline{G}_{0}(y)}{\partial y_{j}}\right] \nonumber \\&+ \frac{\partial }{\partial y_{i}}\left[ (a_{ij}(z )-a_{ij}(y+z ))\frac{\partial (\overline{G}_{0}(y)-\overline{G}_{z }(y+z )) }{\partial y_{j}}\right] .\nonumber \\ \end{aligned}$$
(2.46)

Note that

$$\begin{aligned} |(a_{ij}(0)-a_{ij}(y))-(a_{ij}(z)-a_{ij}(y+z))|\lesssim h|y|,\quad |a_{ij}(z)-a_{ij}(y+z)|\lesssim |y|. \end{aligned}$$
(2.47)

Similarly to (2.13), it follows from (2.47) that, for all \(d>0\),

$$\begin{aligned} \Vert \beta _{z}\Vert _{W^{1,\infty }(\mathfrak {R}^n\smallsetminus B(z,d))}\lesssim d^{2-n}|\ln d|, \end{aligned}$$

and

$$\begin{aligned} \Vert \bar{\beta }_{z}\Vert _{W^{1,\infty }(\mathfrak {R}^n\smallsetminus B(z,d))}\lesssim hd^{2-n}|\ln d|. \end{aligned}$$

Furthermore, by the same arguments in the proof of (2.14), we have from the above estimate that

$$\begin{aligned} \int _{\mathfrak {R}^n}|\nabla (\bar{\beta }_{z}-R_{h}\bar{\beta }_{z})(y)|dy\lesssim h^{3}|\ln h|^{3}. \end{aligned}$$
(2.48)

We now estimate the second term of the right-hand side of (2.45). Note that, for any \(v\in S_{h}^{0}(\mathfrak {R}^n)\),

$$\begin{aligned}&\int _{\mathfrak {R}^n}a_{ij}(y)\frac{\partial (R_{h}\overline{G}_{0}(y)-\overline{R}_{h}^{0}\overline{G}_{0})(y)}{\partial y_{i}}\frac{\partial v(y)}{\partial y_{j}}dy\\&\quad =-\int _{\mathfrak {R}^n}(a_{ij}(0)-a_{ij}(y))\frac{\partial (\overline{G}_{0}-\overline{R}_{h}^{0}\overline{G}_{0})(y)}{\partial y_{i}}\frac{\partial v(y)}{\partial y_{j}}dy, \end{aligned}$$

and

$$\begin{aligned}&\int _{\mathfrak {R}^n}a_{ij}(y+z )\frac{\partial (R_{h}\overline{G}_{z}-\overline{R}_{h}^{z }\overline{G}_{z })(y+z )}{\partial y_{i}}\frac{\partial v(y)}{\partial y_{j}}dy \\&=-\int _{\mathfrak {R}^n}(a_{ij}(z )-a_{ij}(y+z))\frac{\partial (\overline{G}_{z }-\overline{R}_{h}^{z}\overline{G}_{z })(y+z )}{\partial y_{i}}\frac{\partial v(y)}{\partial y_{j}}dy. \end{aligned}$$

The above two equalities yields

$$\begin{aligned}&\int _{\mathfrak {R}^n}a_{ij}(y)\frac{\partial \omega (y) }{\partial y_{i}}\frac{\partial v(y)}{\partial y_{j}}dy\nonumber \\&\quad =-\int _{\mathfrak {R}^n}[(a_{ij}(0)-a_{ij}(y))-(a_{ij}(z)-a_{ij}(y+z))]\frac{\partial (\overline{G}_{0}-\overline{R}_{h}^{0}\overline{G}_{0})(y)}{\partial y_{i}}\frac{\partial v(y)}{\partial y_{j}}dy \nonumber \\&\quad - \int _{\mathfrak {R}^n}(a_{ij}(z )-a_{ij}(y+z))\frac{\partial \alpha _{2}(y,z)}{\partial y_{i}}\frac{\partial v(y)}{\partial y_{j}}dy. \end{aligned}$$
(2.49)

Similarly to (2.32), by (2.3), (2.10), (2.47) and (2.49), we have

$$\begin{aligned} \int _{\mathfrak {R}^n}|\nabla \omega (y)|dy\lesssim h^{3}|\ln h|^{3}. \end{aligned}$$
(2.50)

Similarly, we have

$$\begin{aligned} \int _{\mathfrak {R}^n}|\nabla (R_{h}-\tilde{R}_{h}^{z})\beta _{z}(y)|dy\lesssim h^{3}|\ln h|^{3}. \end{aligned}$$
(2.51)

Then (2.12) follows by combining (2.45), (2.48), (2.50) and (2.51). \(\square \)

3 Ultraconvergence in the case k is even

In this section, we discuss the ultraconvergence of the finite element solution of (1.1). It is known that the weak solution \(u\in H_{0}^{1}(\mathfrak {R}^n)\) of (1.1) satisfies the following variational form

$$\begin{aligned} a(u,w)=(f,w), \forall \ w\in H_{0}^{1}(\mathfrak {R}^n) \end{aligned}$$
(3.1)

and the finite element solution \(u^h\in S_h^0\) satisfies

$$\begin{aligned} a(u^h,w^h)=(f,w^h), \forall \ w^h\in S_h^0. \end{aligned}$$
(3.2)

Note that by the definition (2.2), we actually have \(u^h=R_h u\).

Let \(\tau _0\) be a parallelogram or parallelepiped constituting of the elements in the \({\mathcal {T}_h}\) such that each edge of \(\tau _0\) contains \(2k+1\) vertices of \(T_h\). We denote by \(y_0\) the center of \(\tau _0\). Note that the fact \({\mathcal {T}_h}\) yields \(y_0\in \mathcal N_h\). We introduce a \(2k-\)degree interpolation operator \(\varPi _{2kh}^{2k}\) over \(\tau _{0}\) by letting \(\varPi _{2kh}^{2k}v\in (P_{2k})^n\) satisfying

$$\begin{aligned} \varPi _{2kh}^{2k}v(y)=v(y) \quad \forall \ y\in \tau _{0}\cap \mathcal N_h. \end{aligned}$$
(3.3)

Next we present the main result of this section.

Theorem 3.1

Let \(k\ge 2\) be even. If \(a_{ij}\in C^{\infty }(\mathfrak {R}^n), 1\le i,j\le n\) and \(u\in W^{k+3,\infty }(\mathfrak {R}^n)\) with a compact support, then

$$\begin{aligned} |\nabla (u-\varPi _{2kh}^{2k}u^{h})(y_0)|\lesssim h^{k+2}|\ln h|^{3}\Vert u\Vert _{W^{k+3,\infty }(\mathfrak {R}^n)}. \end{aligned}$$
(3.4)

The rest of this section is dedicated to the proof of Theorem 3.1. Without loss of generality, we assume \(y_0=0\). Assume that \(\Delta y\in \tau _{0}\cap \mathcal N_{h}\). In the process of proving (3.4), we mainly apply the following estimate

$$\begin{aligned} |(u-R_{h}u)(0)-(u-R_{h}u)(\Delta y)|\lesssim h^{k+3}|\ln h|^{3}\Vert u\Vert _{W^{k+3,\infty }(\mathfrak {R}^n)}. \end{aligned}$$

Let \(d>0\) be a constant such that \(\mathrm{Supp} u\subset B(0,2d)\). Let \(\mu \in C^\infty _0(\mathfrak {R}^n)\) satisfy

$$\begin{aligned} \mu =1 \quad \mathrm{in}\ B(0,d). \end{aligned}$$

Moreover, let \(u^J\in \mathbb P_{k+1}\) satisfy

$$\begin{aligned} {\nabla ^{l}(u^J-u)(0)=0}, \quad 0\le |l|\le k+1 \end{aligned}$$

We split u(y) as

$$\begin{aligned} u=u_{1}+u_{2} \end{aligned}$$

with \(u_1=\mu u^J\). It is easy to check that both \(u_1, u_2\) have a compact support and \({\nabla }^{l}u_{1}(0)=0, \forall y\in B(0,d),|l|\ge k+2\) and \(u_{1}(y)=0\) and that \({\nabla }^{k+1}u_{2}(0)=0\). Next, we estimate \(|\nabla (u_1-\varPi _{2kh}^{2k}(R_hu_1))(0)|\) and \(|\nabla (u_2-\varPi _{2kh}^{2k} (R_hu_2))(0)|\) separately.

In the following, we estimate \(|(u_1-R_{h}u_1)(0)-(u_1-R_{h}u_1)(\Delta y )|\). For all \(\Delta y\in \tau _0\cap \mathcal N_h\),

$$\begin{aligned} (u_1-R_{h}u_1)(0)-(u_1-R_{h}u_1)(\Delta y)&=\left[ (u_1-\overline{R}_{h}^{0}u_1)(0)-(u_1-\overline{R}_{h}^{\Delta y}u_1)(\Delta y)\right] \nonumber \\&\quad +\left[ (\overline{R}_{h}^{0}u_1-R_{h}u_1)(0)-(\overline{R}_{h}^{\Delta y}u_1-R_{h}u_1)(\Delta y)\right] . \end{aligned}$$
(3.5)

The following Lemmas 3.2 and 3.3 estimate the two terms of the right-hand side of the above equality, respectively.

Lemma 3.2

Under the assumptions of Theorem 3.1, we have that for all \(\Delta y\in \tau _0\cap \mathcal N_h\),

$$\begin{aligned} |(u_1-\overline{R}_{h}^{0}u_1)(0)-(u_1-\overline{R}_{h}^{\Delta y }u_1)(\Delta y )|\lesssim h^{k+3}|\ln h|^{3}\Vert u_1\Vert _{W^{k+3,\infty }(\mathfrak {R}^n)}. \end{aligned}$$
(3.6)

Proof

We first present \((u_1-\overline{R}_{h}^{0}u_1)(0)\) in its integral form. Let the linear operators \(\zeta \) and \(\zeta _{1}\) be defined for all \(v\in C^{0}(\mathfrak {R}^n)\) by

$$\begin{aligned} \zeta (v)(y)=\frac{1}{2}[v(y)+v(-y)],\quad \zeta _{1}(v)(y)=\frac{1}{2}[v(y+\Delta y )+v(\Delta y-y)], y\in \mathfrak {R}^n. \end{aligned}$$
(3.7)

Apparently,

$$\begin{aligned} \zeta (u_1)(0)=u_1(0),\quad \overline{R}_{h}^{0}\zeta (u_1)(0)=\zeta (\overline{R}_{h}^{0}u_1)(0)=\overline{R}_{h}^{0}u_1(0). \end{aligned}$$
(3.8)

Moreover, the fact \(u_1\) is a polynomial of order \(k+1\) in B(0, d) and that k is even yield

$$\begin{aligned} \nabla ^{k+1}\zeta (u_1)(0)=0, \end{aligned}$$
(3.9)

we conclude that \(\zeta (u_1)\) is a polynomial of order k in B(0, d). That is

$$\begin{aligned} \zeta (u_1)=I_{h}^{k}\zeta (u_1)\quad \mathrm{in} \quad B(0,d). \end{aligned}$$

This equality, together with (3.8), implies

$$\begin{aligned}&(u_1-\overline{R}_{h}^{0}u_1)(0)\nonumber \\&\quad =(\zeta (u_1)-\overline{R}_{h}^{0}\zeta (u_1))(0) =\int _{\mathfrak {R}^n}a_{ij}(0)\frac{\partial (\zeta (u_1)-I_{h}^{k}\zeta (u_1))(y)}{\partial y_{i}}\frac{\partial (\overline{G}_{0}-\overline{R}_{h}^{0}\overline{G}_{0})(y)}{\partial y_{j}}dy \nonumber \\&\quad =\int _{\mathfrak {R}^n\smallsetminus B(0,d/2)}a_{ij}(0)\frac{\partial (\zeta (u_1)-I_{h}^{k}\zeta (u_1))(y)}{\partial y_{i}}\frac{\partial (\overline{G}_{0}-\overline{R}_{h}^{0}\overline{G}_{0})(y)}{\partial y_{j}}dy. \end{aligned}$$
(3.10)

Next we present \((u_1-\overline{R}_{h}^{\Delta y }u_1)({\Delta y})\). We also have

$$\begin{aligned} \zeta _{1}(u_1)(0)=u_1(\Delta y ),\quad \overline{R}_{h}^{\Delta y}\zeta _{1}(u_1)(0)=\overline{R}_{h}^{\Delta y }u_1(\Delta y). \end{aligned}$$
(3.11)

Since \(|\Delta y |\lesssim h\), we have that \(y+\Delta y\in B(0,d)\) for all \(y\in B(0,d/2)\). That is, for all given \(\Delta y\in \tau _0\cap N_h\), \(\zeta _1(u)\) is also a polynomial of degree \(k+1\). On the other hand, the fact that k is even yields

$$\begin{aligned} \nabla ^{k+1}\zeta _{1}(u_1)(\Delta y)=0. \end{aligned}$$
(3.12)

Then we have

$$\begin{aligned} \zeta _{1}(u_1)-I_{h}^{k}\zeta _{1}(u_1)=0,\quad \mathrm{in}\quad B(0,d/2). \end{aligned}$$

This equality, together with (3.11), shows

$$\begin{aligned}&(u_1-\overline{R}_{h}^{\Delta y }u_1)({\Delta y} )=(\zeta _{1}(u_1)-\overline{R}_{h}^{\Delta y }\zeta _{1}(u_1))(0) \nonumber \\&\quad =\int _{\mathfrak {R}^n}a_{ij}(\Delta y)\frac{\partial (\zeta _{1}(u_1)-I_{h}^{k}\zeta _{1}(u_1))(y)}{\partial y_{i}}\frac{\partial (\overline{G}_{\Delta y }-\overline{R}_{h}^{\Delta y }\overline{G}_{\Delta y })(y)}{\partial y_{j}}dy \nonumber \\&\quad =\int _{\mathfrak {R}^n}a_{ij}(\Delta y)\frac{\partial (\zeta _{1}(u_1)-I_{h}^{k}\zeta _{1}(u_1))(y+\Delta y)}{\partial y_{i}}\frac{\partial (\overline{G}_{\Delta y}-\overline{R}_{h}^{\Delta y }\overline{G}_{\Delta y })(y+\Delta y )}{\partial y_{j}}dy \nonumber \\&\quad =\int _{\mathfrak {R}^n\smallsetminus B(0,d/2)}a_{ij}(\Delta y)\frac{\partial (\zeta _{1}(u_1)-I_{h}^{k}\zeta _{1}(u_1))(y+\Delta y)}{\partial y_{i}}\frac{\partial (\overline{G}_{\Delta y}-\overline{R}_{h}^{\Delta y }\overline{G}_{\Delta y })(y+\Delta y )}{\partial y_{j}}dy.\nonumber \\ \end{aligned}$$
(3.13)

Set \(\varsigma (y)=\frac{1}{2}(u_1(y)-u_1(y+2\Delta y ))\). One observes that

$$\begin{aligned}&\zeta (u_1)(y)-\zeta _{1}(u_1)(y+\Delta y )\nonumber \\&\quad =\frac{1}{2}(u_1(y)+u_1(-y))-\frac{1}{2}(u_1((y+\Delta y )+\Delta y )+u_1(\Delta y -(y+\Delta y ))) \nonumber \\&\quad =\frac{1}{2}(u_1(y)+u_1(-y))-\frac{1}{2}(u_1(y+2{\Delta y} )+u_1(-y))\nonumber \\&\quad =\frac{1}{2}(u_1(y)-u_1(y+2\Delta y))=\varsigma (y). \end{aligned}$$
(3.14)

Using (3.10), (3.13) and (3.14), one observes that \((u_1-\overline{R}_{h}^{0}u_1)(0)-(u_1-\overline{R}_{h}^{{\Delta y} }u_1)(\Delta y )\) can be split into

$$\begin{aligned}&(u_1-\overline{R}_{h}^{0}u_1)(0)-(u_1-\overline{R}_{h}^{\Delta y }u_1)(\Delta y ) \nonumber \\&\quad = \int _{\mathfrak {R}^n\smallsetminus B(0,d/2)}(a_{ij}(0)-a_{ij}(\Delta y ))\frac{\partial (\zeta (u_1)-I_{h}^{k}\zeta (u_1))(y)}{\partial y_{i}}\frac{\partial (\overline{G}_{0}-\overline{R}_{h}^{0}\overline{G}_{0})(y)}{\partial y_{j}}dy \nonumber \\&\qquad + \int _{\mathfrak {R}^n\smallsetminus B(0,d/2)}a_{ij}(\Delta y )\frac{\partial (\varsigma -I_{h}^{k}\varsigma )(y)}{\partial y_{i}}\frac{\partial (\overline{G}_{0}-\overline{R}_{h}^{0}\overline{G}_{0})(y)}{\partial y_{j}}dy \nonumber \\&\qquad + \int _{\mathfrak {R}^n\smallsetminus B(0,d/2)}a_{ij}(\Delta y)\frac{\partial (\zeta _{1}(u_1)-I_{h}^{k}\zeta _{1}(u_1))(y+\Delta y)}{\partial y_{i}} \frac{\partial \alpha _{2}(y,\Delta y)}{\partial y_{j}}dy \nonumber \\&\quad =W_{1}+W_{2}+W_{3}, \end{aligned}$$
(3.15)

where \(\alpha _{2}(y,z)\) is defined as (2.8). We next estimate \(W_i, i=1,2,3\) separately. To estimate \(W_{1}\), using (2.3) and the following estimate

$$\begin{aligned} |a_{ij}(\Delta y )-a_{ij}(0)|\lesssim h, \end{aligned}$$
(3.16)

we have

$$\begin{aligned} |W_{1}|\lesssim h^{2k+1}|\ln h|^{3}\Vert u_1\Vert _{W^{k+1,\infty }(\mathfrak {R}^n)}. \end{aligned}$$
(3.17)

We turn now to the estimation of \(W_{2}\). Note that

$$\begin{aligned} \Vert \nabla ^{k+1}\varsigma \Vert _{L^{\infty }(\mathfrak {R}^n)}\lesssim h\Vert u_1\Vert _{W^{k+2,\infty }(\mathfrak {R}^n)}. \end{aligned}$$
(3.18)

Inserting (2.3), (3.14) and (3.18) into (3.15), we obtain

$$\begin{aligned} |W_{2}|&= \left| \int _{\mathfrak {R}^n\smallsetminus B(0,d/2)}a_{ij}(\Delta y )\frac{\partial ( \varsigma -I_{h}^{k}\varsigma )(y)}{\partial y_{i}}\frac{\partial (\overline{G}_{0}-\overline{R}_{h}^{0}\overline{G}_{0})(y)}{\partial y_{j}}dy\right| \nonumber \\&\lesssim h^{k+1}\Vert u_1\Vert _{W^{k+2,\infty }(\mathfrak {R}^n)}\Vert \overline{G}_{0}-\overline{R}_{h}^{0}\overline{G}_{0}\Vert _{W^{1,1}(\mathfrak {R}^n\smallsetminus B(0,d/2))}\nonumber \\&\lesssim h^{2k+1}|\ln h|^{3}\Vert u_1\Vert _{W^{k+2,\infty }(\mathfrak {R}^n)}. \end{aligned}$$
(3.19)

Similarly, by (2.11), we obtain

$$\begin{aligned} |W_{3}|\lesssim h^{k+3}|\ln h|^{3}\Vert u_1\Vert _{W^{k+1,\infty }(\mathfrak {R}^n)}. \end{aligned}$$
(3.20)

Summing up (3.15), (3.17), (3.19) and (3.20), we have the desired result (3.6). \(\square \)

Lemma 3.3

Assume that \(u\in W^{k+3,\infty }(\mathfrak {R}^n)\). Then

$$\begin{aligned} |(\overline{R}_{h}^{0}u_1-R_{h}u_1)(0)-(\overline{R}_{h}^{{\Delta y} }u_1-R_{h}u_1)(\Delta y )| \lesssim h^{k+3}|\ln h|^{3}\Vert u_1\Vert _{W^{k+3,\infty }(\mathfrak {R}^n)}. \end{aligned}$$
(3.21)

Proof

Note that

$$\begin{aligned} (u_1-\overline{R}_{h}^{0})(0)=\int _{\mathfrak {R}^n}a_{ij}(0)\frac{\partial (u_1-I_{h}^{k}u_1)(y)}{\partial y_{i}}\frac{\partial (\overline{G}_{0}-\overline{R}_{h}^{0}\overline{G}_{0})(y)}{\partial y_{j}}dy, \end{aligned}$$

and

$$\begin{aligned} (u_1-R_{h}u_1)(0)=\int _{\mathfrak {R}^n}a_{ij}(y)\frac{\partial (u_1-I_{h}^{k}u_1)(y)}{\partial y_{i}}\frac{\partial (G_{0}-R_{h}G_{0})(y)}{\partial y_{j}}dy. \end{aligned}$$

We observe that \((\overline{R}_{h}^{0}u_1-R_{h}u_1)(0)\) can be split into

$$\begin{aligned} (\overline{R}_{h}^{0}u_1-R_{h}u_1)(0)= & {} (u_1-R_{h}u_1)(0)-(u_1-\overline{R}_{h}^{0}u_1)(0) \nonumber \\= & {} \int _{\mathfrak {R}^n}[a_{ij}(y)-a_{ij}(0)]\frac{\partial (u_1-I_{h}^{k}u_1)(y)}{\partial y_{i}}\frac{\partial (G_{0}-R_{h}G_{0})(y)}{\partial y_{j}}dy\nonumber \\&+ \int _{\mathfrak {R}^n}a_{ij}(0)\frac{\partial (u_1-I_{h}^{k}u_1)(y)}{\partial y_{i}}\frac{\partial \alpha (y,0)}{\partial y_{j}}dy \nonumber \\= & {} I_{1}+I_{2}, \end{aligned}$$
(3.22)

where \(\alpha (y,z)\) is defined as (2.6). Moreover, since

$$\begin{aligned}&(u_1-R_{h}u_1)(\Delta y )\\&\quad =\int _{\mathfrak {R}^n} a_{ij}(y+\Delta y )\frac{\partial (u_1-I_{h}^{k}u_1)(y+\Delta y )}{\partial y_{i}}\frac{\partial (G_{\Delta y }-R_{h}G_{\Delta y})(y+\Delta y )}{\partial y_{j}}dy, \end{aligned}$$

and

$$\begin{aligned}&(u_1-\overline{R}_{h}^{\Delta y }u_1)(\Delta y )\\&\quad =\int _{\mathfrak {R}^n}a_{ij}(\Delta y)\frac{\partial (u_1-I_{h}^{k}u_1)(y+\Delta y )}{\partial y_{i}}\frac{\partial (\overline{G}_{\Delta y }-\overline{R}_{h}^{\Delta y }\overline{G}_{\Delta y })(y+\Delta y )}{\partial y_{j}}dy. \end{aligned}$$

We obtain

$$\begin{aligned}&(\overline{R}_{h}^{\Delta y }u_1-R_{h}u_1)(\Delta y ) \nonumber \\&\quad =\int _{\mathfrak {R}^n}[a_{ij}(y+\Delta y)-a_{ij}(\Delta y )]\frac{\partial (u_1-I_{h}^{k}u_1)(y+\Delta y )}{\partial y_{i}}\frac{\partial (G_{\Delta y }-R_{h}G_{\Delta y})(y+\Delta y )}{\partial y_{j}}dy \nonumber \\&\qquad +\int _{\mathfrak {R}^n}a_{ij}(\Delta y)\frac{\partial (u_1-I_{h}^{k}u_1)(y+\Delta y )}{\partial y_{i}}\frac{\partial \alpha (y+\Delta y,\Delta y)}{\partial y_{j}}dy =J_{1}+J_{2}. \end{aligned}$$
(3.23)

Then by (3.22) and (3.23),

$$\begin{aligned} (\overline{R}_{h}^{0}u_1-R_{h}u_1)(0)-(\overline{R}_{h}^{\Delta y}u_1-R_{h}u_1)(\Delta y)=(I_{1}-J_{1})+(I_{2}-J_{2}). \end{aligned}$$
(3.24)

We first estimate \(I_{1}-J_{1}\). Let \(\alpha _{1}(y,z)\) be defined by (2.7). We have that

$$\begin{aligned}&I_{1}-J_{1}\nonumber \\&\quad =\int _{\mathfrak {R}^n}[(a_{ij}(y)-a_{ij}(0))-(a_{ij}(y+\Delta y )-a_{ij}(\Delta y ))]\frac{\partial (u_1-I_{h}^{k}u_1)(y)}{\partial y_{i}}\frac{\partial (G_{0}-R_{h}G_{0})(y)}{\partial y_{j}}dy \nonumber \\&\qquad +\int _{\mathfrak {R}^n}(a_{ij}(y+\Delta y )-a_{ij}(\Delta y))\frac{\partial [(u_1-I_{h}^{k}u_1)(y)-(u_1-I_{h}^{k}u_1)(y+\Delta y)]}{\partial y_{i}}\frac{\partial (G_{0}-R_{h}G_{0})(y)}{\partial y_{j}}dy\nonumber \\&\qquad +\int _{\mathfrak {R}^n}(a_{ij}(y+\Delta y)-a_{ij}(\Delta y ))\frac{\partial (u_1-I_{h}^{k}u_1)(y+\Delta y )}{\partial y_{i}}\frac{\partial \alpha _{1}(y,\Delta y)}{\partial y_{j}}dy \nonumber \\&\quad =Z_{1}^{1}+Z_{1}^{2}+Z_{1}^{3}. \end{aligned}$$
(3.25)

We first estimate \(Z_{1}^{1}\). Note that

$$\begin{aligned} |(a_{ij}(y)-a_{ij}(0))-(a_{ij}(y+\Delta y )-a_{ij}(\Delta y ))|&\lesssim h|y|,\quad \nonumber \\ |a_{ij}(y+\Delta y )-a_{ij}(\Delta y )|&\lesssim |y|. \end{aligned}$$
(3.26)

Combining (2.3) and (3.26) gives

$$\begin{aligned} |Z_{1}^{1}|&\lesssim h^{k+1}\Vert u_1\Vert _{W^{k+1,\infty }(\mathfrak {R}^n)} \int _{\mathfrak {R}^n}|y||\nabla (G_{0}-R_{h}G_{0})(y)|dy\nonumber \\&\lesssim h^{k+3}|\ln h|^{3}\Vert u_1\Vert _{W^{k+1,\infty }(\mathfrak {R}^n)}. \end{aligned}$$
(3.27)

We turn now to the estimation of \(Z_{1}^{2}\). Set \(\varsigma _{1}(y)=u_1(y)-u_1(y+\Delta y )\). One observes that

$$\begin{aligned} \Vert \varsigma _{1}-I_{h}^{k}\varsigma _{1}\Vert _{W^{1,\infty }(\mathfrak {R}^n)} \lesssim h^{k+1}\Vert u_1\Vert _{W^{k+2,\infty }(\mathfrak {R}^n)}. \end{aligned}$$
(3.28)

By (2.3), (3.26) and (3.28), we arrive at

$$\begin{aligned} |Z_{1}^{2}|\lesssim h^{k+3}|\ln h|^{3}\Vert u_1\Vert _{W^{k+3,\infty }(\mathfrak {R}^n)}. \end{aligned}$$

Note that the combination of (2.10) and (3.26) implies

$$\begin{aligned} |Z_{1}^{3}|\lesssim h^{k+3}|\ln h|^{3}\Vert u_1\Vert _{W^{k+3,\infty }(\mathfrak {R}^n)}. \end{aligned}$$

Inserting the above two estimates and (3.27) into (3.25) yields

$$\begin{aligned} |I_{1}-J_{1}|\lesssim h^{k+3}|\ln h|^{3}\Vert u_1\Vert _{W^{k+3,\infty }(\mathfrak {R}^n)}. \end{aligned}$$
(3.29)

Next we estimate \(I_{2}-J_{2}\). One observes that \(I_{2}-J_{2}\) can be decomposed into

$$\begin{aligned} I_{2}-J_{2}= & {} \int _{\mathfrak {R}^n}(a_{ij}(0)-a_{ij}(\Delta y))\frac{\partial (u_1-I_{h}^{k}u_1)(y)}{\partial y_{i}}\frac{\partial \alpha (y,0)}{\partial y_{j}}dy \nonumber \\&+\int _{\mathfrak {R}^n}a_{ij}(\Delta y ) \frac{\partial [(u_1-I_{h}^{k}u_1)(y)-(u_1-I_{h}^{k}u_1)(y+\Delta y)]}{\partial y_{i}}\frac{\partial \alpha (y,0)}{\partial y_{j}}dy \nonumber \\&+\int _{\mathfrak {R}^n}a_{ij}(\Delta y)\frac{\partial (u_1-I_{h}^{k}u_1)(y+\Delta y )}{\partial y_{i}}\frac{\partial ( \alpha (y,0)- \alpha (y+\Delta y,\Delta y))}{\partial y_{j}}dy \nonumber \\= & {} Z_{2}^{1}+Z_{2}^{2}+Z_{2}^{3}. \end{aligned}$$
(3.30)

We need estimate the three items of the right-hand side. Summing up (2.9), (2.12), (3.26) and (3.28), we also get, for \(l=1,2,3\),

$$\begin{aligned} |Z_{2}^{l}|\lesssim h^{k+3}|\ln h|^{3}\Vert u_1\Vert _{W^{k+3,\infty }(\mathfrak {R}^n)}. \end{aligned}$$

Substituting the above estimate into (3.30), we obtain

$$\begin{aligned} |I_{2}-J_{2}|\lesssim h^{k+3}|\ln h|^{3}\Vert u_1\Vert _{W^{k+3,\infty }(\mathfrak {R}^n)}. \end{aligned}$$
(3.31)

The desired result (3.21) follows from (3.24), (3.29) and (3.31). \(\square \)

We are now in a perfect position to give an estimate for \(\nabla (u_1-\varPi _{2kh}^{2k}(R_{h}u_1))(0)\).

Theorem 3.4

Under the assumptions of Theorem 3.1, we have

$$\begin{aligned} |\nabla (u_1-\varPi _{2kh}^{2k}(R_{h}u_1))(0)|\lesssim h^{k+2}|\ln h|^{3}\Vert u_1\Vert _{W^{k+3,\infty }(\mathfrak {R}^n)}. \end{aligned}$$
(3.32)

Proof

The estimate of \(|\nabla (u_1-\varPi _{2kh}^{2k}(R_{h}u_1))(0)|\) can be reduced to the boundedness of \(|(u_1-R_{h}u_1)(0)-(u_1-R_{h}u_1)(\Delta y )|\) for all \(\Delta y\in \tau _0\cap \mathcal N_h\). In fact,

$$\begin{aligned} |\nabla (u_1-\varPi _{2kh}^{2k}(R_{h}u_1))(0)|&\le |\nabla (u_1-\varPi _{2kh}^{2k}(I_{h}u_1))(0)|\\&\quad +|\nabla \varPi _{2kh}^{2k}(R_{h}u_1-(I_{h}u_1))(0)|, \end{aligned}$$

where \(I_hu_1\in S_h\) is the interpolation of \(u_1\). We first estimate \(\nabla (u_1-\varPi _{2kh}^{2k}(I_{h}u_1))(0)\). One has

$$\begin{aligned} |\nabla (u_1-\varPi _{2kh}^{2k}(I_{h}u_1))(0)|\lesssim h^{2k}\Vert u_1\Vert _{W^{2k+1,\infty }(\mathfrak {R}^{n})}\lesssim h^{2k}\Vert u_1\Vert _{W^{2k+1,\infty }(\mathfrak {R}^{n})}. \end{aligned}$$
(3.33)

Next we estimate \(\nabla \varPi _{2kh}^{2k}(R_{h}u_1-I_{h}u_1)(0)\). Combining (3.6) and (3.21) gives

$$\begin{aligned} |(u_1-R_{h}u_1)(0)-(u_1-R_{h}u_1)(\Delta y )|\lesssim h^{k+3}|\ln h|^{3}\Vert u_1\Vert _{W^{k+3,\infty }(\mathfrak {R}^n)}. \end{aligned}$$

Set \(\psi (y)=(R_{h}u_1-I_{h}u_1)(y)-(R_{h}u_1-I_{h}u_1)(0)\) for all \(y\in \tau _{0}\). Then, by the inverse estimate,

$$\begin{aligned} |\nabla \varPi _{2kh}^{2k}(R_{h}u_1-I_{h}u_1)(0)|= & {} |\nabla \varPi _{2kh}^{2k}\psi (0)| \lesssim h^{-1}|\psi \Vert _{L^{\infty }(\tau _{0})}\nonumber \\\lesssim & {} h^{-1}\max _{\Delta y\in \tau _{0}} |\psi (\Delta y)| \nonumber \\\lesssim & {} h^{-1}\max _{\Delta y\in \tau _{0}}|(u_1-R_{h}u_1)(0)-(u_1-R_{h}u_1)(\Delta y )|\nonumber \\\lesssim & {} h^{k+2}|\ln h|^{3}\Vert u_1\Vert _{W^{k+3,\infty }(\mathfrak {R}^n)}. \end{aligned}$$
(3.34)

Then (3.32) can be obtained by combining the estimates (3.33) and (3.34). \(\square \)

Next, we turn to the estimation of \(|\nabla (u_2-\varPi _{2kh}^{2k}(R_{h}u_2))(0)|\).

Theorem 3.5

Under the same assumptions of Theorem 3.1,

$$\begin{aligned} |\nabla (u_2-\varPi _{2kh}^{2k}(R_{h}u_2))(0)|\lesssim h^{k+2}|\ln h|^{3}\Vert u_2\Vert _{W^{k+3,\infty }(\mathfrak {R}^n)}. \end{aligned}$$
(3.35)

Proof

By the same arguments in the proof of Theorem 3.4, the estimate of \(|\nabla (u_2-\varPi _{2kh}^{2k}(R_{h}u_2))(0)|\) can be reduced to the boundedness of \(|(u_2-R_{h}u_2)(0)-(u_2-R_{h}u_2)(\Delta y )|\) for all \(\Delta y\in \tau _0\cap \mathcal N_h\). We decompose

$$\begin{aligned}&(u_2-R_{h}u_2)(0)-(u_2-R_{h}u_2)(\Delta y) \nonumber \\&=[(u_2-\overline{R}_{h}^{0}u_2)(0)-(u_2-\overline{R}_{h}^{\Delta y}u_2)(\Delta y)]\nonumber \\&\qquad +[(\overline{R}_{h}^{0}u_2-R_{h}u_2)(0)-(\overline{R}_{h}^{\Delta y}u_2-R_{h}u_2)(\Delta y)], \end{aligned}$$
(3.36)

and we will estimate the above two terms separately. We first present \((u_2-\overline{R}_{h}^{0}u_2)(0)-(u_2-\overline{R}_{h}^{\Delta y}u_2)(\Delta y)\) in its integral form. Similarly to Theorem 3.1, we have

$$\begin{aligned} ( u_2-\overline{R}_{h}^{0}u_2)(0) =\int _{\mathfrak {R}^n}a_{ij}(0)\frac{\partial ( u_2-I_{h}^{k} u_2)(y)}{\partial y_{i}}\frac{\partial (\overline{G}_{0}-\overline{R}_{h}^{0}G_{0})(y)}{\partial y_{j}}dy, \end{aligned}$$

and

$$\begin{aligned}&( u_2- \overline{R}_{h}^{\Delta y}u_2)(\Delta y )\\&\quad =\int _{\mathfrak {R}^n}a_{ij}(\Delta y)\frac{\partial ( u_2-I_{h}^{k} u_2)(y+\Delta y)}{\partial y_{i}}\frac{\partial (\overline{G}_{\Delta y}-\overline{R}_{h}^{\Delta y}\overline{G}_{\Delta y})(y+\Delta y )}{\partial y_{j}}dy. \end{aligned}$$

Then

$$\begin{aligned}&( u_2-\overline{R}_{h}^{0}u_2)(0)-( u_2-\overline{R}_{h}^{\Delta y}u_2)(\Delta y ) \nonumber \\&\quad =\int _{\mathfrak {R}^n}[a_{ij}(0)-a_{ij}(\Delta y)]\frac{\partial ( u_2-I_{h}^{k}u_2)(y)}{\partial y_{i}}\frac{\partial (\overline{G}_{0}-\overline{R}_{h}^{0}\overline{G}_{0})(y)}{\partial y_{j}}dy \nonumber \\&\qquad + \int _{\mathfrak {R}^n}a_{ij}(\Delta y)\frac{\partial [( u_2-I_{h}^{k} u_2)(y)-(u_2-I_{h}^{k} u_2)(y+\Delta y )]}{\partial y_{i}}\frac{\partial (\overline{G}_{0}-\overline{R}_{h}^{0}\overline{G}_{0})(y)}{\partial y_{j}}dy \nonumber \\&\qquad +\int _{\mathfrak {R}^n}a_{ij}(\Delta y) \frac{\partial ( u_2-I_{h}^{k} u_2)(y+\Delta y )}{\partial y_{i}}\frac{\partial \alpha _{2}(y,\Delta y)}{\partial y_{j}}dy \nonumber \\&\quad =K_{1}+K_{2}+K_{3}. \end{aligned}$$
(3.37)

Noticing the facts that \(\nabla ^{k+1}u_2(0)=0, \mathrm{Supp} u_2\subset B(0,2d)\), (2.3) and (3.26), we have

$$\begin{aligned} |K_{1}|&\lesssim h^{k+1}\Vert u_2\Vert _{W^{k+2,\infty }(\mathfrak {R}^n)}\int _{B(0,2d)}|y||\nabla (\overline{G}_{0}-\overline{R}_{h}^{0}\overline{G}_{0}) (y)|dy\nonumber \\&\lesssim h^{k+3}|\ln h|\Vert u_2\Vert _{W^{k+2,\infty }(\mathfrak {R}^n)}. \end{aligned}$$
(3.38)

By \(\nabla ^{k+1}u_2(0)=0\) and (2.10), we have

$$\begin{aligned} |K_{3}|&\lesssim h^{k}\Vert u_2\Vert _{W^{k+2,\infty }(\mathfrak {R}^n)}\int _{B(0,2d)}|y| |\nabla _{y}\alpha _{2}(y,\Delta y)|dy \nonumber \\&\lesssim h^{k+3}|\ln h|^{3}\Vert u_2\Vert _{W^{k+2,\infty }(\mathfrak {R}^n)}. \end{aligned}$$
(3.39)

We next estimate \(K_{2}\). For any \(y\in \mathfrak {R}^{n}\),

$$\begin{aligned} \overline{G}_{0}(y)=\overline{G}_{0}(-y),\qquad \overline{R}_{h}^{0}\overline{G}_{0}(y)=\overline{R}_{h}^{0}\overline{G}_{0}(-y). \end{aligned}$$

This equality implies

$$\begin{aligned} \frac{\partial (\overline{G}_{0}-\overline{R}_{h}^{0}\overline{G}_{0})(-y)}{\partial y_{j}}=-\frac{\partial (\overline{G}_{0}-\overline{R}_{h}^{0}\overline{G}_{0})(y)}{\partial y_{j}}. \end{aligned}$$

Then, we have

$$\begin{aligned} K_{2}&=\int _{\mathfrak {R}^n}a_{ij}(\Delta y)\frac{\partial [( u_2-I_{h}^{k} u_2)(-y)-(u_2-I_{h}^{k} u_2)(-y+\Delta y )]}{\partial y_{i}}\frac{\partial (\overline{G}_{0}-\overline{R}_{h}^{0}\overline{G}_{0})(-y)}{\partial y_{j}}dy \nonumber \\&=-\int _{\mathfrak {R}^n}a_{ij}(\Delta y)\frac{\partial [( u_2-I_{h}^{k} u_2)(-y)-(u_2-I_{h}^{k} u_2)(-y+\Delta y )]}{\partial y_{i}}\frac{\partial (\overline{G}_{0}-\overline{R}_{h}^{0}\overline{G}_{0})(y)}{\partial y_{j}}dy. \end{aligned}$$
(3.40)

Let \(\mu (y)=(u_{2}(y)-u_{2}(y+\Delta y))-(u_{2}(-y)-u_{2}(-y+\Delta y))\). Combining (3.37) and (3.40) gives

$$\begin{aligned} 2K_{2}=\int _{\mathfrak {R}^n}a_{ij}(\Delta y)\frac{\partial (\mu -I^k_h\mu )(y)}{\partial y_{i}}\frac{\partial (\overline{G}_{0}-\overline{R}_{h}^{0}\overline{G}_{0})(y)}{\partial y_{j}}dy. \end{aligned}$$
(3.41)

Again by \(\nabla ^{k+1}u_{2}(0)=0\), we have

$$\begin{aligned} | \nabla ^{k+1}\mu (y)|\lesssim h|y|\Vert u_2\Vert _{W^{k+3,\infty }(\mathfrak {R}^n)}. \end{aligned}$$
(3.42)

Plugging (2.3) and (3.42) into (3.41) and noticing that \(\mathrm{Supp} u_2\subset B(0,2d)\), we have

$$\begin{aligned} |K_{2}|\lesssim h^{k+3}|\ln h|^{3}\Vert u_2\Vert _{W^{k+2,\infty }(\mathfrak {R}^n)}. \end{aligned}$$
(3.43)

Combining (3.38), (3.39) and (3.43), we have that

$$\begin{aligned} |(u_2-R_{h}u_2)(\Delta y )-(u_2- R_{h}u_2)(0)|\lesssim h^{k+3}|\ln h|^{3}\Vert u_2\Vert _{W^{k+2,\infty }(\mathfrak {R}^n)}. \end{aligned}$$

Similarly to (3.21), we have

$$\begin{aligned} |(\overline{R}_{h}^{0}u_2-R_{h}u_2)(0)-(\overline{R}_{h}^{{\Delta y} }u_2-R_{h}u_2)(\Delta y )| \lesssim h^{k+3}|\ln h|^{3}\Vert u_2\Vert _{W^{k+3,\infty }(\mathfrak {R}^n)}. \end{aligned}$$

Therefore,

$$\begin{aligned} |(u_2-R_{h}u_2)(0)-(u_2-R_{h}u_2)(\Delta y)|\lesssim h^{k+3}|\ln h|^{3}\Vert u_2\Vert _{W^{k+2,\infty }(\mathfrak {R}^n)}. \end{aligned}$$
(3.44)

By the same arguments in the proof of (3.32), using (3.44), we get the desired result (3.35).

Finally, we are ready to prove Theorem 3.1.

Proof of Theorem 3.1

By the definition of \(u_1\) and \(u_2\), we have that

$$\begin{aligned} \Vert u_1\Vert _{W^{k+3,\infty }(\mathfrak {R}^n)}, \Vert u_2\Vert _{W^{k+3,\infty }(\mathfrak {R}^n)}\lesssim \Vert u\Vert _{W^{k+3,\infty }(\mathfrak {R}^n)}. \end{aligned}$$

Then (3.4) is an immediate consequence of Theorems 3.4 and 3.5.

4 Ultraconvergence in the case k is odd

When k is odd, the equality (3.9) is not necessary valid. Therefore, the reasoning in the previous section can not be generalized to an arbitrary integer k. In other words, the inequality (3.4) is not necessary valid. To obtain the similar result for the case k is odd, we first need to extrapolate the finite element solution. Precisely, let \(\mathcal T_{h/2}\) be obtained by decomposing each element of \(\mathcal T_h\) into \(2^{n}\) equal-sized elements. We assume that \(\mathcal {T}_{h/2}\) is symmetric in the sense that each vertex \(y\in \mathcal N_{h/2}\) is a symmetric center of the mesh \(\mathcal T_{h/2}\). We denote by \(u_{h/2}\) the finite element solution corresponding to the mesh \(\mathcal T_{h/2}\). We define the Richardson extrapolating function \(Pu^h\in S_{h}\) by letting

$$\begin{aligned} Pu^{h}(y)=\frac{2^{k+1}u^{h/2}(y)-u^{h}(y)}{2^{k+1}-1}, y\in \mathcal N_h. \end{aligned}$$
(4.1)

Theorem 4.1

Assume that \(k\ge 3\) is odd. Let \(a_{ij}\in C^{\infty }(\mathfrak {R}^n), 1\le i,j\le n\) and \(u\in W^{k+3,\infty }(\mathfrak {R}^{n})\). Then

$$\begin{aligned} |\nabla (u-\varPi _{2kh}^{2k}Pu^{h})(y_0)|\lesssim h^{k+2}|\ln h|^{3}\Vert u\Vert _{W^{k+3,\infty }(\mathfrak {R}^{n})}. \end{aligned}$$
(4.2)

As in the previous section, we also decompose \(u=u_1+u_2\) and estimate the errors for \(u_1\) and \(u_2\) separately.

Lemma 4.2

Under the assumptions of Theorem 4.1,

$$\begin{aligned} |(u_{1}-P(\overline{R}_{h}^{0}u_{1}))(0)-(u_{1}-P(\overline{R}_{h}^{\Delta y }u_{1}))({\Delta y} )|\lesssim h^{k+3}|\ln h|^{3}\Vert u\Vert _{W^{k+3,\infty }(\mathfrak {R}^n)}. \end{aligned}$$
(4.3)

Proof

We first present \((u_{1}-P(\overline{R}_{h}^{0}u_{1}))(0)\) in its integral form. One observes that

$$\begin{aligned} (u_{1}-\overline{R}_{h}^{0}u_{1})(0)=\int _{\mathfrak {R}^n}a_{ij}(0)\frac{\partial (u_{1}-I_{h}^{k}u_{1})(y)}{\partial y_{i}} \frac{\partial (\overline{G}_{0}-\overline{R}_{h}^0\overline{G}_{0})(y)}{\partial y_{j}}dy. \end{aligned}$$
(4.4)

Set

$$\begin{aligned} \chi (y)=2^{k+1}u_{1}(\frac{y}{2}),\quad \chi _{1}(y)=u_{1}(y)-\chi (y). \end{aligned}$$
(4.5)

Similarly to (4.4), by (4.5), we have

$$\begin{aligned} 2^{k+1}(u_{1}-\overline{R}_{h/2}^{0}u_{1})(0)&=(\chi -\overline{R}_{h}^{0}\chi )(0)\nonumber \\&= \int _{\mathfrak {R}^n}a_{ij}(0)\frac{\partial (\chi -I_{h}^{k}\chi )(y)}{\partial y_{i}} \frac{\partial (\overline{G}_{0}-\overline{R}_{h}^0\overline{G}_{0})(y)}{\partial y_{j}}dy. \end{aligned}$$
(4.6)

Since \(u_1\) is a polynomial of order \(k+1\) in B(0, d). One observes that \(\chi _{1}\) is a polynomial of order k in B(0, d). By (4.4) and (4.6), we have

$$\begin{aligned}&(u_{1}-P(\overline{R}_{h}^{0}u_{1}))(0)\\&\quad =\frac{1}{2^{k+1}-1}[2^{k+1}(u_{1}-\overline{R}_{h/2}^{0}u_{1})(0)-(u_{1}-\overline{R}_{h}^{0}u_{1})(0)] \\&\quad =\frac{-1}{2^{k+1}-1}\int _{\mathfrak {R}^n}a_{ij}(0)\frac{\partial (\chi _{1}-I_{h}^{k}\chi _{1})(y)}{\partial y_{i}} \frac{\partial (\overline{G}_{0}-\overline{R}_{h}^0\overline{G}_{0})(y)}{\partial y_{j}}dy \\&\quad =\frac{-1}{2^{k+1}-1}\int _{\mathfrak {R}^n\smallsetminus B(0,d/2)}a_{ij}(0)\frac{\partial (\chi _{1}-I_{h}^{k}\chi _{1})(y)}{\partial y_{i}} \frac{\partial (\overline{G}_{0}-\overline{R}_{h}^0\overline{G}_{0})(y)}{\partial y_{j}}dy. \end{aligned}$$

Similarly, we have

$$\begin{aligned}&(u_{1}-P\overline{R}_{h}^{\Delta y }u_{1})(\Delta y ) \\&\quad = \frac{-1}{2^{k+1}-1}\int _{\mathfrak {R}^n\smallsetminus B(0,d/2)} a_{ij}({\Delta y} )\frac{\partial (\chi _{3}- I_{h}^{k}\chi _{3}) (y+\Delta y)}{\partial y_{i}}\frac{\partial (\overline{G}_{\Delta y}-\overline{R}_{h}^{\Delta y }\overline{G}_{\Delta y })(y+\Delta y )}{\partial y_{j}}dy, \end{aligned}$$

where \(\chi _{2}(y)=2^{k+1} u_{1}(\frac{y+\Delta y }{2}), \chi _{3}(y)=u_{1}(y)-\chi _{2}(y)\). Combining the above two estimates, we obtain

$$\begin{aligned}&(1-2^{k+1})[(u_{1}-P(\overline{R}_{h}^{0}u_{1}))(0)-(u_{1}-P(\overline{R}_{h}^{\Delta y}u_{1}))(\Delta y)] \nonumber \\&\quad =\int _{\mathfrak {R}^n\smallsetminus B(0,d/2)}(a_{ij}(0)-a_{ij}(\Delta y))\frac{\partial (\chi _{1}- I_{h}^{k}\chi _{1})(y)}{\partial y_{i}}\frac{\partial (\overline{G}_{0}-\overline{R}_{h}^0\overline{G}_{0})(y)}{\partial y_{j}}dy \nonumber \\&\qquad +\int _{\mathfrak {R}^n\smallsetminus B(0,d/2)}a_{ij}(\Delta y )\frac{\partial (\chi _{4}-I_{h}^{k}\chi _{4})(y)}{\partial y_{i}}\frac{\partial (\overline{G}_{0}-\overline{R}_{h}^0\overline{G}_{0})(y)}{\partial y_{j}}dy \nonumber \\&\qquad +\int _{\mathfrak {R}^n\smallsetminus B(0,d/2)}a_{ij}(\Delta y )\frac{\partial (\chi _{3}- I_{h}^{k}\chi _{3}) (y+\Delta y )}{\partial y_{i}} \frac{\partial \alpha _{2}(y,\Delta y)}{\partial y_{j}}dy \nonumber \\&\quad =S_{1}+S_{2}+S_{3}, \end{aligned}$$
(4.7)

where \(\chi _{4}(y)=\chi _{1}(y)-\chi _{3}(y+\Delta y)\) and \(\alpha _{2}(y,z)\) is defined as (2.8).

Next we estimate \(S_i, i=1,2,3.\) First, noticing (2.3) and (3.16), we obtain

$$\begin{aligned} |S_{1}|\lesssim h^{2k+1}|\ln h|\Vert u\Vert _{W^{k+1,\infty }(\mathfrak {R}^n)}. \end{aligned}$$
(4.8)

Secondly, the estimate (2.3) and the fact that

$$\begin{aligned} | \nabla ^{k+1}\chi _{4}(y)| \lesssim h\Vert u\Vert _{W^{k+2,\infty }(\mathfrak {R}^n)} \end{aligned}$$

yield

$$\begin{aligned} |S_{2}|\lesssim h^{2k+1}\Vert u\Vert _{W^{k+2,\infty }(\mathfrak {R}^n)}. \end{aligned}$$
(4.9)

Finally, by (2.11),

$$\begin{aligned}&|S_{3}|\lesssim h^{k}\Vert u\Vert _{W^{k+1,\infty }(\mathfrak {R}^n)}\int _{\mathfrak {R}^n\smallsetminus B(0,d/2)}|\nabla _y\alpha _{2}(y,\Delta y)|dy \lesssim h^{k+3}|\ln h|^{3}\Vert u\Vert _{W^{k+1,\infty }(\mathfrak {R}^n)}. \end{aligned}$$
(4.10)

Then (4.3) follows by substituting (4.7), (4.8), (4.9) into (4.10). \(\square \)

Based on Lemmas 3.3, 4.2 and Theorem 3.4, we are now in a position to show Theorem 4.1.

Proof of Theorem 4.1

A straightforward calculation yields that

$$\begin{aligned}&(u-P(R_{h}u))(0)-(u-P(R_{h}u))(\Delta y )=Y_{1}+Y_{2}+Y_{3}, \end{aligned}$$
(4.11)

with

$$\begin{aligned} Y_1= & {} [(u_{1}-P(\overline{R}_{h}^{0}u_{1}))(0)-(u_{1}-P(\overline{R}_{h}^{\Delta y }u_{1}))(\Delta y )], \\ Y_2= & {} [(P(\overline{R}_{h}^{0}u_{1})-P(R_{h}u_{1}))(0)-(P(\overline{R}_{h}^{\Delta y}u_{1})-P(R_{h}u_{1}))(\Delta y)], \\ Y_3= & {} [(u_{2}-P(R_{h}u_{2}))(0)-(u_{2}-P(R_{h}u_{2}))(\Delta y)]. \end{aligned}$$

First by (4.3),

$$\begin{aligned} |Y_{1}|\lesssim h^{k+3}|\ln h|^{3}\Vert u\Vert _{W^{k+3,\infty }(\mathfrak {R}^n)}. \end{aligned}$$

Secondly, by Lemma 3.3,

$$\begin{aligned} |Y_{2}|&\le \left| (\overline{R}_{h}^{0}u_{1}-R_{h}u_{1})(0)-(\overline{R}_{h}^{\Delta y}u_{1}-R_{h}u_{1})(\Delta y )\right| \\&\ \ \ + \left| (\overline{R}_{h/2}^{0}u_{1}-R_{h/2}u_{1})(0)-(\overline{R}_{h/2}^{\Delta y}u_{1}-R_{h/2}u_{1})(\Delta y )\right| \\&\lesssim h^{k+3}|\ln h|^{3}\Vert u\Vert _{W^{k+3,\infty }(\mathfrak {R}^n)}. \end{aligned}$$

Finally, it follows from (3.35) that

$$\begin{aligned} |Y_{3}|&\lesssim |(u_{2}-R_{h}u_{2})(\Delta y )-(u_{2}-R_{h}u_{2})(0 )|\\&\ \ \ +|(u_{2}-R_{h/2}u_{2})(\Delta y)-(u_{2}-R_{h/2}u_{2})(0)| \\&\lesssim h^{k+3}|\ln h|^{3}\Vert u\Vert _{W^{k+3,\infty }(\mathfrak {R}^n)}. \end{aligned}$$

Inserting the above three estimates into the equality (4.11), we have

$$\begin{aligned} |(u-P(R_{h}u)(\Delta y)-(u-P(R_{h}u))(0)| \lesssim h^{k+3}|\ln h|^{3}\Vert u\Vert _{W^{k+3,\infty }(\mathfrak {R}^n)}. \end{aligned}$$

Combining this estimate and the inverse inequality, we get the desired result (4.2) by the same arguments in the proof of (3.32). \(\square \)

5 Ultraconvergence in a bounded domain

In this section, we apply the previous theory to the problem (1.1). Let \(\mathcal T_{h}^{\Omega }\) be a quasi-uniform conforming partition of \(\Omega \) satisfying the following property: there exists a parallelogram or parallelepipe \( \tau \subset \Omega \) such that : 1) \(\tau \) is the union of some elements in \(\mathcal T_h^\Omega ,\) 2) \(\tau \) has a size \(h_\tau \simeq 1,\) 3) \(\mathcal T_{h}^\Omega \cap \tau =\mathcal T_{h}\cap \tau \). Here \(\mathcal T_{h}\) is a mesh in the whole domain \(\mathfrak {R}^n\) defined in Sect. 2, \(\mathcal T_h^\Omega \cap \tau =\{\tau '\in \mathcal T_h^\Omega :\tau '\subset \tau \}\) and \(\mathcal T_h\cap \tau =\{\tau '\in \mathcal T_h:\tau '\subset \tau \}\). Let

$$\begin{aligned} S_h(\Omega )=\{v_h\in C(\Omega ): v_h|_{\tau '}\in P_k,\forall \tau '\in {\mathcal T}_h^{\Omega }\} \end{aligned}$$

be the associated standard finite element space of degree k and let \(S_h^0(\Omega )=S_h(\Omega )\cap H_0^1(\Omega )\), we introduce the finite element projector \(R_{h}^{\Omega }:H_{0}^{1}(\Omega )\rightarrow S_h^0(\Omega )\) for all \(\psi \in S_{h}^{0}(\Omega )\) by

$$\begin{aligned} a^{\Omega }(w-R_{h}^{\Omega }w,\psi )=0, \end{aligned}$$
(5.1)

where the associated bilinear form is defined by

$$\begin{aligned} a^{\Omega }(\phi ,\psi )=\int _{\Omega } a_{ij}(y)\frac{\partial \phi (y)}{\partial y_{i}} \frac{\partial \psi (y)}{\partial y_{j}}dy\quad \phi ,\psi \in H^{1}(\Omega ). \end{aligned}$$

Based on Theorem 3.1, we have the following result.

Theorem 5.1

Let \(a_{ij}\in C^{\infty }(\Omega ), 1\le i,j\le n\) and \(\varPi _{2kh}^{2k}\) be defined as in Section 3, and \(\mathcal N_h^{\tau }\) be the set of all vertices of \(\overline{\mathcal T}_{h}^\Omega \cap \tau \). Let \(k\ge 2\) be an even. Assume that \(y_{0}\in \mathcal N_{h}^{\Omega }\) is away from the boundary of \(\tau \) with a fixed distance and \(u\in W^{k+3,\infty }(\tau )\), then

$$\begin{aligned} |\nabla (u-\varPi _{2kh}^{2k}(R_{h}^{\Omega }u))(y_{0})|\lesssim h^{k+2}|\ln h|^{3}\Vert u\Vert _{W^{k+3,\infty }(\tau )}+\Vert u-R_{h}^{\Omega }u\Vert _{W^{1-k,2}(\Omega )}. \end{aligned}$$
(5.2)

Proof

Assume that \(d\simeq 1\) satisfies \(B(y_{0},2d)\subset \tau \). Let \(\phi \in C^{\infty }(\mathfrak {R}^{n})\) satisfy \(0\le \phi \le 1\), and \(\phi =1\) in B\((y_0,d)\), \(\phi =0\) in \( \mathfrak {R}^{n}\smallsetminus B(y_0,2d)\), and \(\Vert \phi \Vert _{W^{k+3,\infty }(\mathfrak {R}^{n})}\lesssim 1\). Let \(\upsilon (y)=\phi (y)u(y)\). One observes that \(\nabla (u-\varPi _{2kh}^{2k}(R_{h}^{\Omega }u))(y_0)\) can be decomposed into

$$\begin{aligned}&\nabla (u-\varPi _{2kh}^{2k}(R_{h}^{\Omega }u))(y_0)\nonumber \\&\quad =\nabla (\upsilon -\varPi _{2kh}^{2k}(R_{h}\upsilon ))(y_0) +\nabla (\varPi _{2kh}^{2k}(R_{h}\upsilon )-\varPi _{2kh}^{2k}(R_{h}^{\Omega }u))(y_0). \end{aligned}$$
(5.3)

Using (3.4), we have

$$\begin{aligned} |\nabla (\upsilon -\varPi _{2kh}^{2k}(R_{h}\upsilon ))(y_0)| \lesssim h^{k+2}|\ln h|^{3}\Vert \upsilon \Vert _{W^{k+3,\infty }(\mathfrak {R}^{n})} \lesssim h^{k+2}|\ln h|^{3}\Vert u\Vert _{W^{k+3,\infty }(\tau )}. \end{aligned}$$
(5.4)

We turn now to the estimation of \(\nabla \varPi ^{2k}_{2kh}(R_{h}\upsilon -R_{h}^{\Omega }u)(y_0)\). We denote that \(S_{h}^{0}(B(y_{0},d))=\{v^h\in C(\Omega ):v^h|_e\in P_k,\forall e\in \overline{\mathcal T}_{h}^\Omega \} \cap H_{0}^{1}(B(y_{0},d))\). Note that \(\overline{\mathcal T}_{h}^\Omega \cap \tau =\overline{\mathcal T}_{h}\cap \tau \), we have, for any \(w\in S_{h}^{0}(B(y_{0},d))\),

$$\begin{aligned} a(R_{h} \upsilon -R_{h}^{\Omega }u,w)&=a(R_{h} \upsilon - \upsilon ,w)+a( \upsilon -u,w)+a(u-R_{h}^{\Omega }u,w)\nonumber \\&=0+0+0=0. \end{aligned}$$
(5.5)

Then by the arguments of Schatz et al. in [22, 23], we have

$$\begin{aligned} \Vert R_{h}\upsilon -R_{h}^{\Omega }u\Vert _{W^{1,\infty }(B(y_{0},\frac{d}{2}))}\lesssim \Vert u-R_{h}^{\Omega }u\Vert _{W^{1-k,2}(B(y_{0},d))}\lesssim \Vert u-R_{h}^{\Omega }u\Vert _{W^{1-k,2}(\Omega )}. \end{aligned}$$

Furthermore, we get

$$\begin{aligned} |\nabla (\varPi _{2kh}^{2k}(R_{h}\upsilon )-\varPi _{2kh}^{2k}(R_{h}^{\Omega }u))(y_0)|\lesssim \Vert u-R_{h}^{\Omega }u\Vert _{W^{1-k,2}(\Omega )}. \end{aligned}$$
(5.6)

Inserting the estimates (5.4) and (5.6) into the equality (5.3), we get the desired result (5.2). \(\square \)

Next we consider the case that k is odd. Let \(\overline{\mathcal T}_{h/2}^{\Omega }\) be a quasi-uniform partition obtained by decomposing each element of \(\overline{\mathcal T}_h^{\Omega }\) into \(2^{n}\) elements. Furthermore, we assume that \(\overline{\mathcal T}_{h/2}^{\Omega }\cap \tau =\overline{\mathcal T}_{h/2}\cap \tau \) where \(\overline{\mathcal T}_{h/2}\) is defined as in Section 4. We denote by \(R_{h/2}^{\Omega }u\) the finite element solution corresponding to the mesh \(\overline{\mathcal T}_{h/2}^{\Omega }\). We define the Richardson extrapolating function \(P(R_{h/2}^{\Omega }u)\in S_{h}(\Omega )\) by letting

$$\begin{aligned} P(R_{h}^{\Omega }u)(y)=\frac{2^{k+1}R_{h/2}^{\Omega }u(y)-R_{h}^{\Omega }u(y)}{2^{k+1}-1}, y\in \mathcal N_h^{\tau }. \end{aligned}$$
(5.7)

Similarly to Theorem 5.1, we have the following result.

Theorem 5.2

Under the assumptions of Theorem 5.1, if \(k\ge 3\) is odd, then

$$\begin{aligned} |\nabla (u-\varPi _{2kh}^{2k}(P(R_{h}^{\Omega }u)))(y_{0})|\lesssim h^{k+2}|\ln h|^{3}\Vert u\Vert _{W^{k+3,\infty }(\tau )}+\Vert u-R_{h}^{\Omega }u\Vert _{W^{1-k,2}(\Omega )}. \end{aligned}$$
(5.8)

Based on Theorems 5.1 and 5.2, we have the following corollary.

Corollary 5.3

Under the assumptions of Theorem 5.1, if k is even, then

$$\begin{aligned} |\nabla (u-\varPi _{2kh}^{2k}(R_{h}^{\Omega }u))(y_{0})|\lesssim h^{k+2}|\ln h|^{3}\Vert u\Vert _{W^{k+3,\infty }(\Omega )}, \end{aligned}$$
(5.9)

and, if \(k\ge 3\) is odd, then

$$\begin{aligned} |\nabla (u-\varPi _{2kh}^{2k}(P(R_{h}^{\Omega }u)))(y_0)|\lesssim h^{k+2}|\ln h|^{3}\Vert u\Vert _{W^{k+3,\infty }(\Omega )}. \end{aligned}$$
(5.10)

Proof

Recall a classical result [21]

$$\begin{aligned} \Vert u-R_{h}^{\Omega }u\Vert _{W^{1-k,2}(\Omega )}\lesssim h^{k+l}\Vert u\Vert _{W^{k+l,2}(\Omega )}, \end{aligned}$$

when \(u\in W^{k+l,2}(\Omega )\) for any \(2\le l\le k\). Then (5.9) and (5.10) follows from (5.2) and (5.8), respectively.

6 Numerical examples

We consider (1.1) with \(\Omega =[0,1]^2\) and the coefficients

$$\begin{aligned}&a_{11}(x_{1},x_{2})=1+x_{1},\quad a_{22}(x_{1},x_{2})=1+x_{2},\\&a_{12}(x_{1},x_{2})=a_{21}(x_{1},x_{2})=0.25(x_{1}+x_{2}). \end{aligned}$$

The problem admits the exact solution

$$\begin{aligned} u(x)=16x_{1}(1-x_{1})x_{2}(1-x_{2})e^{x_{1}+x_{2}}. \end{aligned}$$

We will validate (5.9) and (5.10) with numerical experiments. For simplicity, the underlying mesh is chosen as a uniform one which consists of equal-sized isosceles right-angled triangles.

Since the estimates (5.9) and (5.10) are only valid for an interior vertex \(y_0\), without loss of generality, we test our results in the following vertices set

$$\begin{aligned} \mathcal N^0_h=\mathcal N_h^{\Omega }\cap [\frac{1}{4},\frac{3}{4}]\times [\frac{1}{4},\frac{3}{4}]. \end{aligned}$$

Correspondingly, we define the discrete norm \(\quad \Vert v\Vert _{\infty ,h}=\max _{x_{j}\in \mathcal N_h^0}|v(x_{j})|\).

We will test our estimates for different orders \(k=2,3,4\). Note that once k and the mesh size h are given, the corresponding finite element solution \(R_h^\Omega u\) can be computed with the standard finite element method. Let \(y_0\in \mathcal N_h^0, \Delta y_{1}=(h,0)\) and \(\Delta y_{2}=(0,h)\). One observes that \(\varPi _{2kh}^{2k}R_{h}^{\Omega }u(y)\) is a polynomial of degree 2k along each direction \(y_{i},i=1,2\).

In the following, we explain how to compute \(\varPi _{2kh}^{2k}R_{h}^{\Omega }u(y_0)\) and \(\varPi _{2kh}^{2k}(PR_{h}^{\Omega })u(y_0)\).

When \(k=2\), we have

$$\begin{aligned}&\frac{\partial \varPi _{2kh}^{2k}R_{h}^{\Omega }u(y_{0})}{\partial y_{i}}\nonumber \\&\quad = \frac{8[R_{h}^{\Omega }u(y_{0}+\Delta y_{i})-R_{h}^{\Omega }u(y_{0}-\Delta y_{i})]-[R_{h}^{\Omega }u(y_{0}+2\Delta y_{i})-R_{h}^{\Omega }u(y_{0}-2\Delta y_{i})]}{12 h}. \end{aligned}$$
(6.1)
Table 1 \(k=2\)
Table 2 \(k=4\)
Table 3 \(k=3\)

When \(k=4\), we use

$$\begin{aligned}&\frac{\partial \varPi _{2kh}^{2k}R_{h}^{\Omega }u(y_{0})}{\partial y_{i}}\nonumber \\&\quad \simeq \frac{0.8[R_{h}^{\Omega }u(y_{0}+\Delta y_{i})-R_{h}^{\Omega }u(y_{0}-\Delta y_{i})]-0.2[R_{h}^{\Omega }u(y_{0}+2\Delta y_{i})-R_{h}^{\Omega }u(y_{0}-2\Delta y_{i})]}{ h} \nonumber \\&\qquad +\frac{0.038095238095[R_{h}^{\Omega }u(y_{0}+3\Delta y_{i})-R_{h}^{\Omega }u(y_{0}-3\Delta y_{i})]}{ h} \nonumber \\&\qquad -\frac{0.003571428571[R_{h}^{\Omega }u(y_{0}+4\Delta y_{i})-R_{h}^{\Omega }u(y_{0}-4\Delta y_{i})]}{ h}. \end{aligned}$$
(6.2)

to compute \(\frac{\partial \varPi _{2kh}^{2k}R_{h}^{\Omega }u(y_{0})}{\partial y_{i}}\).

When \(k=3\), we first use (5.7) to compute \(P(R_h^\Omega u)(y_0)\) and we obtain

$$\begin{aligned}&\frac{\partial \varPi _{2kh}^{2k}PR_{h}^{\Omega }u(y_{0})}{\partial y_{i}}\nonumber \\&\quad =\frac{45[R_{h}^{\Omega }u(y_{0}+\Delta y_{i})-R_{h}^{\Omega }u(y_{0}-\Delta y_{i})]-9[R_{h}^{\Omega }u(y_{0}+2\Delta y_{i})-R_{h}^{\Omega }u(y_{0}-2\Delta y_{i})]}{60 h} \nonumber \\&\qquad +\frac{R_{h}^{\Omega }u(y_{0}+3\Delta y_{i})-R_{h}^{\Omega }u(y_{0}-3\Delta y_{i})}{60 h}. \end{aligned}$$
(6.3)

Depicted in Tables 1, 2, 3 are our numerical ultraconvergence results corresponding to the finite element degree \(k=2,4,3\) respectively.

From Tables 1, 2, 3, we observe that the gradients of the post-processed FE solutions approximate the gradient of u with orders \(\mathcal{O}(h^4)\), \(\mathcal{O}(h^6)\) and \(\mathcal{O}(h^5)\), respectively, which validate the estimates (5.9) and (5.10). Moreover, it is interesting to find that the hidden constant is independent of the mesh size h which indicates that maybe the ‘lnh’ appeared in the right-hand side of the estimates (5.9) and (5.10) can be removed.