# Nonconforming double set parameter finite element methods for a fourth order variational inequality with two-sided displacement obstacle

## Abstract

Two nonconforming finite elements constructed by double set parameter method are used to approximate a fourth order variational inequality with two-sided displacement obstacle. Because the exact solution does not belong to $$H^{4}_{\mathrm{loc}}(\Omega)$$ and each element space involves two sets of parameters, a series of novel approaches different from the exiting literature are developed in the procedure for presenting convergence analysis and deriving the optimal error estimates in broken energy norm.

## Introduction

Let $$\Omega\subset R^{2}$$ be a bounded convex polygon domain, $$f\in L^{2}(\Omega)$$, $$\psi_{1},\psi_{2}\in C^{2}(\Omega)\cap C(\bar{\Omega})$$, $$\psi_{1}<\psi_{2}$$ on $$\bar{\Omega}$$ and $$\psi_{1}<0<\psi_{2}$$ on Ω. Consider the following two-sided displacement obstacle problem:

$$\left \{ \textstyle\begin{array}{l} \mbox{Find } u\in K \mbox{ such that} \\ u= \arg\min_{v\in K} J(v), \end{array}\displaystyle \right .$$
(1)

where $$J(v)=\frac{1}{2}a(v,v)-(f,v)$$, $$a(w,v)=\int_{\Omega}D^{2}w:D^{2}v \,dx\,dy=\int_{\Omega}(w_{xx}v_{xx}+w_{yy}v_{yy} +2w_{xy}v_{xy} )\,dx\,dy$$, $$(f,v)=\int_{\Omega}f v \,dx\,dy$$, $$K=\{v\in H_{0}^{2}(\Omega); \psi_{1}\leq v\leq\psi_{2} \mbox{ on } \Omega\}$$.

It follows from the theory in [1, 2] that the solution of obstacle problem (1) is uniquely determined by the following fourth order variational inequality:

$$\left \{ \textstyle\begin{array}{l} \mbox{Find } u\in K \mbox{ such that} \\ a(u,v-u)\geq(f,v-u),\quad \forall v\in K. \end{array}\displaystyle \right .$$
(2)

Different from the second order displacement obstacle problems (solutions of that have the $$H^{2}$$ regularity, which allows the corresponding strong form of the variational inequality to be used in the convergence analysis of finite element methods (FEMs) ), the fourth order problem (1) has a unique solution u belonging to $$H^{3}(\Omega)\cap C^{2}(\Omega)$$ (in general $$u \notin H^{4}_{\mathrm{loc}}(\Omega)$$ even for smooth data) . This lack of the $$H^{4}$$ regularity means that the corresponding strong form of the variational inequality (2) is not available for the convergence analysis of FEMs, which leads to a cardinal difficulty in optimal order error estimates. Although FEMs for the fourth order variational inequality with one side displacement obstacle were investigated and optimal order error estimates were obtained in , the techniques, which depend greatly on the one side displacement obstacle condition, can not be applied to the two-sided case directly. The main reason is that the two-sided condition will lead to an important inequality in the error estimate invalid, which makes the convergence analysis complicated and difficult to be handled. Recently, a new unified convergence analysis for $$C^{1}$$-conforming FEs, classical nonconforming FEs, and discontinuous FEMs for problem (1) was developed in [17, 18]. In which the optimal error estimate in the energy norm with order $$O(h)$$ was established by using an auxiliary obstacle problem and an enriching operator (here and later h denotes the mesh parameter). Subsequently the idea was extended to a generalized FEM in  and a quadratic $$C^{0}$$ interior penalty method in .

It is well known that the degree of piecewise polynomials of a $$C^{1}$$-conforming element space must be very high to meet $$C^{1}$$ smoothness requirement. For example, Argyris element  with 5-degree polynomials and Bogner-Fox-Schmit element  with bicubic degree polynomials are conforming triangular and rectangular elements respectively. This phenomenon causes some computational difficulties. To reduce the order of polynomials on each element, nonconforming elements are an attractive option. Unfortunately, the degrees of freedom of classical nonconforming FEs are usually very difficult to satisfy the following two requirements simultaneously: (a) to pass through the generalized patch test or F-E-M-test (cf. [23, 24]); (b) to be simple and convenient so that the size of discrete system is small. For example, the degrees of freedom of Veubeke element  are complicated, while Zienkiewicz element [26, 27] is convergent only on three parallel direction meshes . In order to circumvent or ameliorate the above deficiency, the double set parameter method is proposed , which has two sets of parameters chosen independently. In principle, the first set is selected to meet convergence requirement, while the second set is chosen to be simple to make the total number of unknowns in the resulting discrete system as small as possible. Then the first set of parameters is discretized into linear combinations with respect to the second set (real degrees of freedom) in a certain way so as to maintain the convergence order. Up to now, several nonconforming plate elements have been successfully constructed by the double set parameter method and applied to deal with some fourth order PDEs (cf. ). However, as far as we know, there is no consideration about problem (1) with this FEM.

The main aim of this paper is to apply two nonconforming elements constructed by the double set parameter method to approximate problem (1). In this situation, enriching operators meeting the requirements in  become very difficult to be developed for each element space involves two sets of parameters. Consequently, a new approach is adopted to get the optimal error estimate, in which we skillfully use two kinds of auxiliary obstacle problems and two enriching operators $$E_{1h}$$ and $$E_{2h}$$. The first auxiliary problem proposed in  plays an important role in connecting the continuous and discrete obstacle problems. The second one introduced by ourselves can be regarded as a bridge between the discrete and the first auxiliary obstacle problems. $$E_{1h}$$ (presented by ) establishes the relationship between the triangular Morley  and Argyris element spaces, and $$E_{2h}$$ is newly constructed as a connection between the rectangular Morley  and $$Q_{4}$$ Bogner-Fox-Schmit element spaces. Finally, we derive an optimal error estimate of order $$O(h)$$ in the broken energy norm successfully by a different approach from the existing literature.

The remainder of this paper is organized as follows. In the next section, the two nonconforming finite elements constructed by the double set parameter method are described. In Section 3, we introduce two kinds of auxiliary obstacle problems and two enriching operators and obtain error estimates of order $$O(h)$$ in the energy norm.

The following standard notations for the Sobolev spaces will be used: for an integer $$m>0$$, $$H^{m}(\Omega)$$ with norm $$\|\cdot\|_{m,\Omega}$$ and semi-norm $$|\cdot|_{m,\Omega}$$, $$H^{m}(T)$$ with norm $$\|\cdot\|_{m,T}$$ and semi-norm $$|\cdot |_{m,T}$$, $$L^{2}(\Omega)$$ with norm $$\|\cdot\|_{0,\Omega}$$ and $$L^{2}(T)$$ with norm $$\|\cdot\|_{0,T}$$, respectively. Besides, let $$P_{k}(T)$$ be the space consisting of piecewise polynomials of degree k, and $$Q_{k}(T)$$ be the space of polynomials whose degrees for x, y are equal to k on element T. Throughout the paper, C denotes a positive constant independent of the mesh parameter h and may be different at each appearance.

## Two double set parameter elements

To begin with, for the sake of completeness, we introduce two nonconforming finite elements constructed by the double set parameter method in .

Assume that $$T_{h}$$ is a family of regular triangular or rectangular subdivisions of Ω .

(I) Triangular element: let $$T\in T_{h}$$ be a triangle with vertices $$a_{i}(x_{i},y_{i})$$ ($$i=1,2,3$$). We denote by $$l_{i}$$, $$F_{i}$$, $$n_{i}$$, $$\lambda_{i}$$, Δ, respectively, the side opposite to $$a_{i}$$, the length of $$l_{i}$$, the unit outward normal vector on $$l_{i}$$, the area coordinates for T and the area of T. Let $$v_{i}$$, $$v_{ix}$$, $$v_{iy}$$ be the function value of v and its first derivatives at $$a_{i}$$, and $$a_{12}$$, $$a_{23}$$, $$a_{31}$$ be the midpoints of $$l_{3}$$, $$l_{1}$$, $$l_{2}$$, respectively. Furthermore, for $$i=1,2,3\ (\operatorname{mod}(3))$$, we define

\begin{aligned}& b_{i}=y_{i+1}-y_{i-1}, \qquad c_{i}=x_{i-1}-x_{i+1}, \\& r_{i}=\frac{(b_{i+1}b_{i-1}+c_{i+1}c_{i-1})}{\Delta},\qquad t_{i}= \frac{F^{2}_{i}}{\Delta}. \end{aligned}

On element T, let the shape function space be

$$P(T)=P_{2}(T)=\operatorname{Span} \{\lambda_{1}, \lambda_{2},\lambda_{3},\lambda_{1} \lambda_{2},\lambda_{2}\lambda _{3}, \lambda_{3}\lambda_{1}\},$$
(3)

and the degrees of freedom be

$$D_{1}(v)=\bigl(d_{1}(v), d_{2}(v), \ldots, d_{6}(v) \bigr)^{\prime},$$
(4)

where $$d_{i}(v)=v(a_{i})\doteq v_{i}$$, $$d_{i+3}(v)=\frac{1}{F_{i}}\int_{l_{i}}\frac{\partial v}{\partial n_{i}}\, ds$$, $$i=1, 2, 3$$.

For $$v\in P(T)$$, suppose that

$$v=\beta_{1}\lambda_{1}+\beta_{2} \lambda_{2}+\beta_{3}\lambda_{3}+\beta_{4} \lambda _{1}\lambda_{2}+\beta_{5} \lambda_{2}\lambda_{3}+\beta_{6} \lambda_{3}\lambda_{1}.$$
(5)

Substituting (5) into (4), we have $$D_{1}(v)=C_{1}b$$ with

$$C_{1}=\left ( \textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{}} 1 &0 &0 &0 & 0 & 0 \\ 0 &1 &0 &0 & 0 & 0 \\ 0 &0 &1 &0 & 0 & 0 \\ k_{1}t_{1} &k_{1}r_{3} &k_{1}r_{2} &\frac{k_{1}}{2}t_{1} &-\frac{k_{1}}{2}t_{1} & \frac {k_{1}}{2}t_{1} \\ k_{2}r_{3} &k_{2}t_{2} &k_{2}r_{1} &\frac{k_{2}}{2}t_{2} &\frac{k_{2}}{2}t_{2} & -\frac {k_{2}}{2}t_{2} \\ k_{3}r_{2} &k_{3}r_{1} &k_{3}t_{3} &-\frac{k_{3}}{2}t_{3} &\frac{k_{3}}{2}t_{3} & \frac {k_{3}}{2}t_{3} \end{array}\displaystyle \right ),$$
(6)

where $$k_{i}=-\frac{1}{2F_{i}}$$ ($$i=1,2,3$$), $$b=(\beta_{1}, \beta_{2},\ldots, \beta_{6})^{\prime}\in R^{6}$$.

It is easy to see that $$\operatorname{det} C_{1}=\frac{k_{1}k_{2}k_{3}t_{1}t_{2}t_{3}}{2}\neq0$$. Thus $$b=C_{1}^{-1}D_{1}(v)$$, i.e., there holds

$$\forall v\in P(T), \quad v=\bigl(P_{1},C_{1}^{-1}D_{1}(v) \bigr),$$
(7)

where $$P_{1}=(\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{1}\lambda_{2},\lambda_{2}\lambda _{3},\lambda_{3}\lambda_{1})^{\prime}$$.

Then the associated interpolation operator on T is defined by

$$I_{1T}: v\in H^{2}(T)\longmapsto I_{1T} v\in P(T),\quad I_{1T} v=\bigl(P_{1},C_{1}^{-1}D_{1}(v) \bigr).$$
(8)

Nodal parameters are chosen as

$$Q_{1}(v)=(v_{1},v_{1x},v_{1y},v_{2},v_{2x},v_{2y}, v_{3},v_{3x},v_{3y})^{\prime}.$$
(9)

We discrete the degrees of freedom $$D_{1}(v)$$ in terms of the nodal parameters $$Q_{1}(v)$$ as follows.

$$d_{i}(v)$$ ($$i=1,2,3$$) by the exact values of v on the three vertices of element, i.e.,

$$d_{i}(v)= v_{i},$$

$$d_{i}(v)$$ ($$i=4,5,6$$) by trapezoidal formula:

$$\left \{ \textstyle\begin{array}{l} d_{4}(v)=k_{1}({b_{1}}(v_{2x}+v_{3x})+{c_{1}}(v_{2y}+v_{3y}))+O(h^{2}|v|_{3,T}), \\ d_{5}(v)=k_{2}({b_{2}}(v_{3x}+v_{1x})+{c_{2}}(v_{3y}+v_{1y}))+O(h^{2}|v|_{3,T}), \\ d_{6}(v)=k_{3}({b_{3}}(v_{1x}+v_{2x})+{c_{3}}(v_{1y}+v_{2y}))+O(h^{2}|v|_{3,T}). \end{array}\displaystyle \right .$$

The above discretizations can be written in matrix form as

$$\forall v\in H^{3}(T), \quad D_{1}(v)=G_{1}Q_{1}(v)+E_{1}(v),$$
(10)

where $$E_{1}(v)=(0,0,0,\varepsilon(v),\varepsilon(v),\varepsilon(v))^{\prime}$$, $$\varepsilon(v)=O(h^{2}|v|_{3,T})$$, and

$$G_{1}=\left ( \textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{}} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & k_{1}b_{1} & k_{1}c_{1} & 0 & k_{1}b_{1} & k_{1}c_{1} \\ 0 & k_{2}b_{2} & k_{2}c_{2} & 0& 0 & 0 & 0 & k_{2}b_{2} & k_{2}c_{2} \\ 0 & k_{3}b_{3} & k_{3}c_{3} & 0 & k_{3}b_{3} & k_{3}c_{3} & 0 & 0 & 0 \end{array}\displaystyle \right ).$$
(11)

Note that for all $$v\in P(T)$$, $$\frac{\partial v}{\partial n}$$ is the first order polynomial on $$l_{i}$$ ($$i=1,2,3$$), and the trapezoidal rule of numerical integration is exact for linear polynomials, thus $$E_{1}(v)$$ vanishes, i.e.,

$$\forall v\in P(T),\quad D_{1}(v)=G_{1}Q_{1}(v).$$
(12)

We take the real shape function still as (5), but with

$$b=C_{1}^{-1}G_{1}Q_{1}(v),$$
(13)

then the corresponding FE space is defined by

\begin{aligned} {V}_{1h} =&\bigl\{ {v}; {v}|_{{T}}= \bigl(P_{1},C_{1}^{-1}G_{1}Q_{1}(v) \bigr), \forall T\in{{T}}_{h}, \\ &v(a)=v_{x}(a)=v_{y}(a)=0, \forall \mbox{ node } a\in \partial\Omega\bigr\} . \end{aligned}
(14)

The associated interpolation operator $$\Pi_{1h}$$ on $$V_{1h}$$ is defined by $$\Pi_{1h}|_{T}=\Pi_{1T}$$, where $$\Pi_{1T}:v\in H^{3}(T)\longmapsto\Pi_{1T} v \in P(T)$$ satisfies

$$\Pi_{1T} v=\bigl(P_{1},C_{1}^{-1}G_{1}Q_{1}(v) \bigr).$$
(15)

Obviously,

$$I_{1T} v-\Pi_{1T} v=\bigl(P_{1},C_{1}^{-1}E_{1}(v) \bigr).$$
(16)

(II) Rectangular element: suppose $$T\in T_{h}$$ is a rectangular with sides parallelled to axes of coordinates. Let $$(x_{T}, y_{T})$$, $$a_{i}(x_{i},y_{i})$$ and $${l}_{i}=\overline{{a}_{i}{a}_{i+1}}$$ ($$i=1,2,3,4$$) be its center, vertices and edges, the edges length be $$2h_{x}$$ and $$2h_{y}$$, respectively. Let $$\hat{T}=[-1,1]\times[-1,1]$$ be the reference element on ξ-η plane with center point $$(0,0)$$, and its four vertices be $$\hat{a}_{1}=(-1,-1)$$, $$\hat{a}_{2}=(1,-1)$$, $$\hat{a}_{3}=(1,1)$$ and $$\hat{a}_{4}=(-1,1)$$, four edges be $$\hat{l}_{1}=\overline{\hat{a}_{1}\hat{a}_{2}}$$, $$\hat{l}_{2}=\overline{\hat{a}_{2}\hat{a}_{3}}$$, $$\hat{l}_{3}=\overline{\hat{a}_{3}\hat{a}_{4}}$$ and $$\hat{l}_{4}=\overline{\hat{a}_{4}\hat{a}_{1}}$$, $$v_{i}$$, $$v_{ix}$$, $$v_{iy}$$ be the function value of v and its first derivatives at $$a_{i}$$, respectively. Then there exists an affine mapping $$F:\hat{T}\longrightarrow T$$,

$$\left \{ \textstyle\begin{array}{l} x=x_{T}+h_{x}\xi, \\ y=y_{T}+h_{y}\eta \end{array}\displaystyle \right .$$
(17)

satisfying $$F(\hat{T})=T$$, $$v(x,y)=\hat{v}(\xi,\eta)$$.

On element T, the shape function space is taken as

$$P(T)=P_{2}(T)\cup\bigl\{ x^{3},y^{3} \bigr\} =\operatorname{Span} \{p_{1},p_{2},\ldots, p_{8}\},$$
(18)

where

\begin{aligned} \begin{aligned} &p_{1}= \frac{1}{4}(1-\xi) (1-\eta),\qquad p_{2}= \frac{1}{4}(1+\xi) (1-\eta), \\ &p_{3}= \frac {1}{4}(1+ \xi) (1+\eta),\qquad p_{4}= \frac{1}{4}(1-\xi) (1+\eta), \\ &p_{5}=1-\xi^{2}, \qquad p_{6}=1- \eta^{2},\qquad p_{7}=\xi\bigl(1-\xi^{2}\bigr), \qquad p_{8}=\eta\bigl(1-\eta^{2}\bigr). \end{aligned} \end{aligned}

The degrees of freedom are chosen as

$$D_{2}(v)=\bigl(d_{1}(v), d_{2}(v), \ldots, d_{8}(v) \bigr)^{\prime},$$
(19)

where

\begin{aligned}& d_{i}(v)=v(a_{i})\doteq v_{i}, \quad i=1,2,3,4, \\& d_{5}(v)=-\frac{h_{y}}{h_{x}}\int_{l_{1}} \frac{\partial v}{\partial n}\,ds =\int_{-1}^{1} \frac{\partial\hat{v}}{\partial\eta}(\xi,-1)\,d\xi, \\& d_{6}(v)=\frac{h_{x}}{h_{y}}\int _{l_{2}}\frac{\partial v}{\partial n}\,ds =\int_{-1}^{1} \frac{\partial\hat{v}}{\partial\xi}(1,\eta)\,d\eta, \\& d_{7}(v)=-\frac{h_{y}}{h_{x}}\int_{l_{3}} \frac{\partial v}{\partial n}\,ds =\int_{-1}^{1} \frac{\partial\hat{v}}{\partial\eta}(\xi,1)\,d\xi, \\& d_{8}(v)=\frac{h_{x}}{h_{y}}\int _{l_{4}}\frac{\partial v}{\partial n}\,ds =\int_{-1}^{1} \frac{\partial\hat{v}}{\partial\xi}(-1,\eta)\,d\eta. \end{aligned}

For $$v\in P(T)$$, it can be expressed as

$$v=\sum_{i=1}^{8} \beta_{i}p_{i}=(P_{2},b),$$
(20)

where $$b=(\beta_{1}, \beta_{2},\ldots, \beta_{8})^{\prime}\in R^{8}$$, $$P_{2}=(p_{1}, p_{2},\ldots, p_{8})^{\prime}$$.

Substituting (20) into (19), we have $$D_{2}(v)=C_{2}b$$ with

$$C_{2}=\left ( \textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{}} 1 &0 &0 &0 & 0 & 0 & 0 & 0 \\ 0 &1 &0 &0 & 0 & 0 & 0 & 0 \\ 0 &0 &1 &0 & 0 & 0 & 0 & 0 \\ 0 &0 &0 &1 & 0 & 0 & 0 & 0 \\ -\frac{1}{4} &-\frac{1}{4} &\frac{1}{4} &\frac{1}{4} & 0 & 2 & 0 & -2 \\ -\frac{1}{4} & \frac{1}{4} &\frac{1}{4} &-\frac{1}{4} & -2 & 0 & -2 &0 \\ -\frac{1}{4} &-\frac{1}{4} &\frac{1}{4} &\frac{1}{4} & 0 &-2 & 0 & -2 \\ -\frac{1}{4} & \frac{1}{4} &\frac{1}{4} &-\frac{1}{4} & 2 & 0 & -2 &0 \end{array}\displaystyle \right ).$$
(21)

It is easy to check that $$\operatorname{det} C_{2}=-64\neq0$$, thus $$b=C_{2}^{-1}D_{2}(v)$$, i.e., there holds

$$\forall v\in P(T), \quad v=\bigl(P_{2},C_{2}^{-1}D_{2}(v) \bigr).$$
(22)

The associated interpolation operator on T is defined by

$$I_{2T}: v\in H^{2}(T)\longmapsto I_{2T} v\in P(T),\quad I_{2T} v=\bigl(P_{2},C_{2}^{-1}D_{2}(v) \bigr).$$
(23)

Then we take the nodal parameters as

$$Q_{2}(v)=(v_{1},v_{1x},v_{1y},v_{2},v_{2x},v_{2y},v_{3},v_{3x},v_{3y}, v_{4},v_{4x},v_{4y})^{\prime}.$$
(24)

Discretize $$D_{2}(v)$$ into a linear combination of the nodal parameters $$Q_{2}(v)$$ as follows.

$$d_{i}(v)= v_{i}$$ ($$i=1,2,3,4$$); $$d_{i}(v)$$ ($$i=5,6,7,8$$) with the trapezoidal rule of numerical integration:

$$\left \{ \textstyle\begin{array}{l} d_{5}(v)={h_{y}}(v_{1y}+v_{2y})+O(h^{2}|v|_{3,T}), \\ d_{6}(v)={h_{x}}(v_{2x}+v_{3x})+O(h^{2}|v|_{3,T}), \\ d_{7}(v)={h_{y}}(v_{3y}+v_{4y})+O(h^{2}|v|_{3,T}), \\ d_{8}(v)={h_{x}}(v_{4x}+v_{1x})+O(h^{2}|v|_{3,T}). \end{array}\displaystyle \right .$$

Then, for $$v\in H^{3}(T)$$, the above discretizations can be written in matrix form as

$$D_{2}(v)=G_{2}Q_{2}(v)+E_{2}(v),$$
(25)

where $$E_{2}(v)=(0,0,0,0,\varepsilon(v),\varepsilon(v),\varepsilon (v),\varepsilon(v))^{\prime}$$ and

$$G_{2}=\left ( \textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{}} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & h_{y} & 0 & 0 & h_{y} & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0& h_{x} & 0 & 0 & h_{x} & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & h_{y} & 0 & 0 & h_{y} \\ 0 & h_{x} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & h_{x} & 0 \end{array}\displaystyle \right ).$$
(26)

Note that for all $$v\in P(T)$$, $$\frac{\partial v}{\partial n}$$ is the first order polynomial of x on $$l_{1}$$ and $$l_{3}$$, and of y on $$l_{2}$$ and $$l_{4}$$, thus

$$D_{2}(v)=G_{2}Q_{2}(v).$$
(27)

We take the real shape function still as (20), but with

$$b=C_{2}^{-1}G_{2}Q_{2}(v),$$
(28)

then the corresponding FE space is defined by

\begin{aligned} {V}_{2h} =&\bigl\{ {v}; {v}|_{{T}}= \bigl(P_{2},C_{2}^{-1}G_{2}Q_{2}(v) \bigr), \forall T\in{{T}}_{h}, \\ & v(a)=v_{x}(a)=v_{y}(a)=0, \forall \mbox{ node } a\in \partial\Omega\bigr\} . \end{aligned}
(29)

The associated interpolation operator $$\Pi_{2h}$$ on $$V_{2h}$$ is defined by $$\Pi_{2h}|_{T}=\Pi_{2T}$$, where $$\Pi_{2T}:v\in H^{3}(T)\longmapsto\Pi_{2T} v \in P(T)$$ satisfies

$$\Pi_{2T} v=\bigl(P_{2},C_{2}^{-1}G_{2}Q_{2}(v) \bigr).$$
(30)

Obviously,

$$I_{2T} v-\Pi_{2T} v=\bigl(P_{2},C_{2}^{-1}E_{2}(v) \bigr).$$
(31)

In addition, define $$\|\cdot \|_{h}= (\sum_{T\in {{T}}_{h}}{|\cdot|^{2}_{2,T}})^{\frac{1}{2}}$$, it can be checked that $$\|\cdot \|_{h}$$ is a norm over $${V}_{kh}$$.

## Error estimates

Consider the double set parameter FE approximation of (1):

$$\left \{ \textstyle\begin{array}{l} \mbox{Find } u_{ kh}\in K_{kh} \mbox{ such that} \\ u_{ kh}= \arg\min_{v\in K_{kh}} J_{ h}(v), \end{array}\displaystyle \right .$$
(32)

where $$K_{kh}=\{v\in V_{kh}; \psi_{1}(a)\leq v(a)\leq\psi_{2}(a), \forall a \in D_{h} \}$$, $$k=1,2$$, $$D_{h}$$ denotes the set of the vertices of $$T\in T_{h}$$, for any $$v,w\in K_{h}$$, $$J_{ h}(v)=\frac{1}{2}a_{ h}(v,v)-(f,v)$$, $$a_{h}(w,v)=\sum_{T\in T_{h}}\int_{T} D^{2}w:D^{2}v \,dx\,dy$$.

Since $$a_{h}(\cdot,\cdot)$$ is symmetric, bounded and coercive on $$V_{kh}$$, and $$K_{kh}\subset V_{kh}$$, the obstacle problem (32) has a unique solution $$u_{k h}$$ determined by the following discrete variational inequality:

$$\left \{ \textstyle\begin{array}{l} \mbox{Find } u_{k h}\in K_{kh} \mbox{ such that} \\ a_{h}(u_{k h},v-u_{k h})\geq(f,v-u_{k h}),\quad \forall v\in K_{kh}. \end{array}\displaystyle \right .$$
(33)

In order to obtain optimal error estimates, we firstly introduce two kinds of auxiliary obstacle problems as follows.

(i) An auxiliary obstacle problem which is built as a bridge between the continuous problem (1) and the discrete obstacle problem (32):

$$\left \{ \textstyle\begin{array}{l} \mbox{Find } \tilde{u}_{h}\in\tilde{K}_{h} \mbox{ such that} \\ \tilde{u}_{h}= \arg\min_{v\in\tilde{K}_{h}} J(v), \end{array}\displaystyle \right .$$
(34)

where $$\tilde{K}_{h}=\{v\in H_{0}^{2}(\Omega); \psi_{1}(a)\leq v(a)\leq \psi_{2}(a), \forall a \in D_{h} \}$$.

Equation (34) has a unique solution $$\tilde{u}_{h}$$ determined by

$$\left \{ \textstyle\begin{array}{l} \mbox{Find } \tilde{u}_{h}\in\tilde{K}_{h} \mbox{ such that} \\ a(\tilde{u}_{h},v-\tilde{u}_{h})\geq(f,v-\tilde{u}_{h}),\quad \forall v\in \tilde{K}_{h}. \end{array}\displaystyle \right .$$
(35)

It was shown in  that the solution u of (1) and $$\tilde{u}_{h}$$ has the following estimate:

$$|u-\tilde{u}_{h}|_{2,\Omega}\leq Ch,$$
(36)

and there exists $$h_{0}>0$$ such that for $$h \leq h_{0}$$,

$$\hat{u}_{h}=\tilde{u}_{h}+ \delta_{h,1}\phi_{1}-\delta_{h,2}\phi_{2}\in K,$$
(37)

where $$\phi_{i}\in C_{0}^{\infty}(\bar{\Omega})$$ and $$\delta_{h,i}$$ satisfy

$$\delta_{h,i}\leq Ch^{2},\quad i=1,2.$$

(ii) Let $${\bar{V}}_{kh}$$ ($$k=1,2$$) be the triangular and rectangular Morley element spaces respectively, i.e., $${\bar{V}}_{kh}=\{{v},{v}|_{{T}}=(P_{k},C_{k}^{-1}D_{k}(v)),\int_{l}[\frac {\partial v}{\partial n}]\,ds=0, l\subset\partial T, \forall T\in{{T}}_{h}, v(a)=0, \forall \mbox{ node } a\in \partial\Omega\}$$, where $$[v]$$ is the jump of v across the edge l, and $$[v]=v$$ if $$l\subset\partial\Omega$$. Let $$I_{kh}$$ be $$I_{kh}|_{T}=I_{kT}$$. Then the second kind of auxiliary obstacle problem is introduced to establish a relationship between the discrete obstacle problem (32) and the first auxiliary obstacle problem (34):

$$\left \{ \textstyle\begin{array}{l} \mbox{Find } \bar{{u}}_{kh}\in\bar{K}_{kh} \mbox{ such that} \\ \bar{u}_{kh}= \arg\min_{v\in\bar{K}_{kh}} J_{h}(v), \end{array}\displaystyle \right .$$
(38)

where $$\bar{K}_{kh}=\{v\in\bar{V}_{kh}; \psi_{1}(a)\leq v(a)\leq \psi_{2}(a), \forall a \in D_{h} \}$$.

Equation (38) has a unique solution $$\bar{u}_{kh}$$ determined by the following discrete variational inequality:

$$\left \{ \textstyle\begin{array}{l} \mbox{Find } \bar{u}_{kh}\in\bar{K}_{kh} \mbox{ such that} \\ a_{h}(\bar{u}_{kh},v-\bar{u}_{kh})\geq(f,v-\bar{u}_{kh}),\quad \forall v\in \bar{K}_{kh}. \end{array}\displaystyle \right .$$
(39)

Next, in order to establish a connection between the auxiliary obstacle problems (34) and (38), we present two enriching operators $$E_{kh}: v\in\bar{V}_{kh}\longmapsto E_{kh} v\in {\tilde{V}}_{kh}$$ ($$k=1,2$$) as follows:

$$\left \{ \textstyle\begin{array}{l} E_{1h} v(a)=v(a), \\ \frac{\partial(E_{1h} v)}{\partial n}(m)=\frac{\partial v}{\partial n}(m), \\ \partial^{\alpha}(E_{1h}v)(a)=\frac{1}{|\Upsilon_{a}|}\sum_{T\in\Upsilon_{a}}\partial^{\alpha}v|_{T}(a), \quad |\alpha|=1, \\ \partial^{\alpha}(E_{1h}v)(a)=0, \quad |\alpha|=2 \end{array}\displaystyle \right .$$
(40)

and

$$\left \{ \textstyle\begin{array}{l} E_{2h} v(a)=v(a), \\ \frac{\partial(E_{2h} v)}{\partial n}(m)=\frac{\partial v}{\partial n}(m), \\ E_{2h} v(c)=v(c), \\ E_{2h} v(m)=\frac{1}{|\Upsilon_{m}|}\sum_{T\in\Upsilon _{m}}v|_{T}(m), \\ \partial^{\alpha}(E_{2h}v)(a)=\frac{1}{|\Upsilon_{a}|}\sum_{T\in\Upsilon_{a}}\partial^{\alpha}v|_{T}(a),\quad |\alpha|=1, \\ \partial^{\alpha}(E_{2h}v)(a)=0, \quad \alpha=(1,1), \end{array}\displaystyle \right .$$
(41)

where $${\tilde{V}}_{1h}$$ and $${\tilde{V}}_{2h}$$ are the Argyris and $$Q_{4}$$ Bogner-Fox-Schmit element spaces associated with $$T_{h}$$ , respectively, $$m\in M_{h}$$, $$c\in C_{h}$$, $$a \in D_{h}$$, $$M_{h}$$ and $$C_{h}$$ are the sets of midpoints of edges and centers of elements in $$T_{h}$$, $$\Upsilon_{m}$$ and $$\Upsilon_{a}$$ are the sets of elements in $$T_{h}$$ sharing m as a common midpoint of edge and a as a common vertex, respectively, $$|\Upsilon_{m}|$$ and $$|\Upsilon_{a}|$$ are the numbers of elements in $$\Upsilon_{m}$$ and $$\Upsilon_{a}$$.

Because $${\tilde{V}}_{1h}$$ and $${\tilde{V}}_{2h}$$ are $$C^{1}$$-conforming spaces, $${\tilde{V}}_{kh} \subset H_{0}^{2}(\Omega)$$, thus for all $$v\in\bar{K}_{kh}$$, there holds $$E_{kh} v\in \tilde{K}_{h}$$. Furthermore, we have the following.

### Lemma 1

For all $$u\in H^{3}(\Omega)$$, $$v\in \bar{V}_{kh}$$ ($$k=1,2$$), the following estimates hold:

\begin{aligned}& \| v-E_{kh} v\|_{0,\Omega}\leq Ch^{2}\|v \|_{h}, \end{aligned}
(42)
\begin{aligned}& \biggl(\sum_{T\in T_{h}}| v-E_{kh} v|^{2}_{1,T}\biggr)^{\frac{1}{2}}\leq Ch\|v\|_{h}, \end{aligned}
(43)
\begin{aligned}& | E_{kh} v|_{2,\Omega}\leq C\|v\|_{h}, \end{aligned}
(44)
\begin{aligned}& \| u-E_{kh} I_{kh}u\|_{0,\Omega}+h| u-E_{kh} I_{kh}u|_{1,\Omega}+h^{2}| u-E_{kh} I_{kh}u|_{2,\Omega}\leq Ch^{3}|u|_{3,\Omega}. \end{aligned}
(45)

### Proof

Because the above properties of operator $$E_{1h}$$ were proven in , we only need to present the proof for $$E_{2h}$$.

In fact, from (41), we have

\begin{aligned} v|_{T}-(E_{2h}v)|_{T} =& \sum _{i=1}^{4}\bigl(v|_{T}-(E_{2h}v)|_{T} \bigr) (m_{i})q_{i} \\ &{} +\sum_{i=1}^{4}\sum _{|\alpha|=1}\partial^{\alpha}\bigl(v|_{T}-(E_{2h}v)|_{T} \bigr) (a_{i})q_{\alpha,i} \\ &{} +\sum_{i=1}^{4}\frac{\partial ^{2}(v|_{T}-(E_{2h}v)|_{T})}{\partial x\, \partial y}(a_{i})q_{2,i}, \end{aligned}
(46)

where $$q_{i}$$, $$q_{\alpha,i}$$ and $$q_{2,i}$$ are the nodal basis functions corresponding to the nodal parameters $$v(m_{i})$$, $$\partial^{\alpha}v(a_{i})$$ and $$\frac{\partial^{2} v}{\partial x\, \partial y}(a_{i})$$ of the $$Q_{4}$$ Bogner-Fox-Schmit element respectively.

Then there holds

\begin{aligned} \bigl\Vert v-(E_{2h}v)\bigr\Vert _{0,T} \leq& \sum_{i=1}^{4}\bigl\vert \bigl(v|_{T}-(E_{2h}v)|_{T}\bigr) (m_{i})\bigr\vert \|q_{i}\|_{0,T} \\ &{}+ \sum_{i=1}^{4}\sum _{|\alpha|=1}\bigl\vert \partial^{\alpha}\bigl(v|_{T}-(E_{2h}v)|_{T}\bigr) (a_{i})\bigr\vert \|q_{\alpha,i}\|_{0,T} \\ &{}+ \sum_{i=1}^{4}\biggl\vert \frac{\partial ^{2}(v|_{T}-(E_{2h}v)|_{T})}{\partial x\, \partial y}(a_{i})\biggr\vert \|q_{2,i} \|_{0,T} \\ \doteq& I_{1}+I_{2}+I_{3}. \end{aligned}
(47)

Now we start to estimate $$I_{j}$$ one by one for $$j=1,2,3$$.

For $$I_{1}$$, let $$l_{i}\subset\Omega$$, it follows from (41) that

$$\bigl\vert \bigl(v|_{T}-(E_{2h}v)|_{T}\bigr) (m_{i})\bigr\vert =\biggl\vert \frac{1}{2}\sum _{T'\in\Upsilon _{m_{i}}}(v|_{T}-v|_{T'}) (m_{i})\biggr\vert =\frac{1}{2}\bigl\vert (v|_{T}-v|_{T'}) (m_{i})\bigr\vert .$$

Denote $$(v|_{T}-v|_{T'})=\{v\}$$, note that $$\{v\}$$ is the third order polynomial of x on $$l_{1}$$ and $$l_{3}$$, and of y on $$l_{2}$$ and $$l_{4}$$. Taking $$l_{1}$$ for example and using Taylor expansion, we have

\begin{aligned} \{v\}(x) =&\{v\}\bigl(x^{\ast}\bigr)+\bigl(x-x^{\ast}\bigr)\{v\}_{x}\bigl(x^{\ast}\bigr) \\ &{}+\frac{1}{2}\bigl(x-x^{\ast}\bigr)^{2}\{v \}_{xx}\bigl(x^{\ast}\bigr)+\frac{1}{6} \bigl(x-x^{\ast}\bigr)^{3}\{v\} _{xxx} \bigl(x^{\ast}\bigr), \end{aligned}
(48)

where $$x^{\ast}=\frac{x_{1}+x_{2}}{2}$$, $$x_{1}$$ and $$x_{2}$$ are x-axis coordinate components of the midpoints and two endpoints of $$l_{1}$$, respectively.

Since $$v|_{T}$$ and $$v|_{T'}$$ agree at the two endpoints of $$l_{1}$$, i.e., $$\{v\}(x_{1})=\{v\}(x_{2})=0$$, there holds

$$\{v\}\bigl(x^{\ast}\bigr)=-\frac{1}{8}(x_{1}-x_{2})^{2} \{v\}_{xx}\bigl(x^{\ast}\bigr),$$

which together with the standard inverse estimate yields

$$\bigl\vert \{v\}(m_{1})\bigr\vert \leq \frac{h_{x}^{2}}{2}\bigl(\vert v\vert _{2,\infty,T}+|v|_{2,\infty,T'} \bigr)\leq C{h}|v|_{2,T}+Ch|v|_{2,T'}.$$
(49)

On the other hand, if $$l_{i}\subset\partial\Omega$$, (41) implies

$$\bigl(v|_{T}-(E_{2h}v)|_{T}\bigr) (m_{i})=0,$$
(50)

then in view of (49) and (50), employing the fact $$\|q_{i}\|_{0,T}\leq Ch$$, we have

$$I_{1}\leq Ch^{2}\bigl(\vert v\vert _{2,T}+|v|_{2,T'} \bigr).$$

As to $$I_{2}$$, it follows from (41) and a standard inverse estimate that

\begin{aligned}& \sum_{|\alpha|=1}\bigl\vert \partial^{\alpha}\bigl(v|_{T}-(E_{2h}v)|_{T} \bigr) (a_{i})\bigr\vert \\& \quad =\frac{1}{|\Upsilon_{a_{i}}|}\sum_{T'\in \Upsilon_{a_{i}}}\sum _{|\alpha|=1}\bigl\vert \partial^{\alpha}(v|_{T}-v|_{T'}) (a_{i})\bigr\vert \\& \quad \leq C\biggl(\sum_{T'\in \Upsilon_{a_{i}}}\sum _{|\alpha|=1}\bigl\vert \partial^{\alpha}(v|_{T}-v|_{T'}) (a_{i})\bigr\vert ^{2}\biggr)^{\frac{1}{2}} \\& \quad \leq C\biggl(\sum_{l\in \Phi_{a_{i}}}\frac{1}{|l|}\sum _{|\alpha|=1}\bigl\Vert \partial^{\alpha}(v|_{T'}-v|_{T''})\bigr\Vert ^{2}_{0,l} \biggr)^{\frac{1}{2}} \\& \quad \leq C\biggl(\sum_{l\in \Phi_{a_{i}}}\frac{1}{|l|} \biggl(\biggl\Vert \frac{\partial(v|_{T'}-v|_{T''})}{\partial s}\biggr\Vert ^{2}_{0,l}+ \biggl\Vert \frac{\partial(v|_{T'}-v|_{T''})}{\partial n}\biggr\Vert ^{2}_{0,l} \biggr)\biggr)^{\frac{1}{2}}, \end{aligned}
(51)

where $$\Phi_{a_{i}}$$ is the set of edges in $$T_{h}$$ sharing $$a_{i}$$ as a common vertex, $${T'}, {T''}\in\Upsilon_{a_{i}}$$ satisfy $${T'}\cap{T''}=l$$, $$\frac{\partial }{\partial n}$$ and $$\frac{\partial }{\partial s}$$ denotes the outward normal derivative and tangential derivative along l, respectively. Here and later, $$|l|$$ denotes the length of l.

Since $$v|_{T'}$$ and $$v|_{T''}$$ agree at the two endpoints of l, we get $$\int_{l}\frac{\partial(v|_{T'}-v|_{T''})}{\partial s}\,ds=0$$, thus

\begin{aligned}& \biggl\Vert \frac{\partial(v|_{T'}-v|_{T''})}{\partial s}\biggr\Vert _{0,l} \\& \quad =\biggl\Vert \frac{\partial(v|_{T'}-v|_{T''})}{\partial s}-\frac{1}{|l|}\biggl(\int _{l}\frac{\partial(v|_{T'}-v|_{T''})}{\partial s}\,ds\biggr)\biggr\Vert _{0,l} \\& \quad \leq Ch^{\frac{1}{2}}\biggl\vert \frac{\partial(v|_{T'}-v|_{T''})}{\partial s}\biggr\vert _{1,{T'}\cup{T''}} \leq Ch^{\frac{1}{2}}\bigl(\vert v\vert _{2,{T'}}+|v|_{2,{T''}}\bigr). \end{aligned}
(52)

Similarly, $$\frac{\partial(v|_{T'}-v|_{T''})}{\partial n}(m)=0$$ leads to $$\int_{l}\frac{\partial(v|_{T'}-v|_{T''})}{\partial n}\,ds=0$$ and

$$\biggl\Vert \frac{\partial(v|_{T'}-v|_{T''})}{\partial n}\biggr\Vert _{0,l}\leq Ch^{\frac{1}{2}}\bigl(\vert v\vert _{2,T'}+|v|_{2,T''} \bigr).$$
(53)

Then substituting (52) and (53) into (51) yields

$$\sum_{|\alpha|=1}\bigl\vert \partial^{\alpha}\bigl(v|_{T}-(E_{2h}v)|_{T} \bigr) (a_{i})\bigr\vert \leq C\biggl(\sum _{{T'}\in\Upsilon_{a_{i}}}|v|^{2}_{2,T'}\biggr)^{\frac{1}{2}},$$
(54)

which together with $$\|q_{\alpha,i}\|_{0,T}\leq Ch^{2}$$ ($$|\alpha|=1$$) gives

$$I_{2}\leq Ch^{2}\biggl(\sum_{{T'}\in \Upsilon_{a_{i}}}|v|^{2}_{2,T'} \biggr)^{\frac{1}{2}}.$$

In order to estimate $$I_{3}$$, employing (41) and the standard inverse estimate, we have

$$\biggl\vert \frac{\partial^{2}(v|_{T}-(E_{2h}v)|_{T})}{\partial x\, \partial y}(a_{i})\biggr\vert \leq\biggl\vert \frac{\partial^{2} (v|_{T})}{\partial x\, \partial y}(a_{i})\biggr\vert \leq|v|_{2,\infty,T} \leq Ch^{-1}|v|_{2,T},$$
(55)

which in conjunction with $$\|q_{2,i}\|_{0,T}\leq Ch^{3}$$ implies

$$I_{3}\leq Ch^{2}|v|_{2,T}.$$

Therefore, (42) follows by (47) and estimates of $$I_{1}$$, $$I_{2}$$ and $$I_{3}$$ directly.

Furthermore, applying (42), the standard inverse estimate and triangle inequality, we obtain the desired results (43) and (44) immediately.

At last, it follows from (23) and (41) that

$$(E_{2h} I_{2h}u) (a_{i})=(I_{2h}u) (a_{i})=u(a_{i}),\qquad \frac{\partial(E_{2h} I_{2h}u)}{\partial n}(m_{i})= \frac{\partial(I_{2h}u)}{\partial n}(m_{i})=\frac{\partial u}{\partial n}(m_{i}),$$

i.e., for all $$u|_{T}\in P(T)$$, there holds $$u|_{T}=E_{2h} I_{2h} u|_{T}$$, then (45) can be obtained by the interpolation theorem. The proof is completed. □

Now, we are ready to present the convergence analysis of double set parameter nonconforming FEM for (1).

### Theorem 1

Assume that u and $$u_{kh}$$ are the solutions of (1) and (32), respectively, $$u\in H^{3}(\Omega)\cap H_{0}^{2}(\Omega)$$, then we have

$$\|u-u_{kh}\|_{h}\leq Ch, \quad k=1,2.$$
(56)

### Proof

Using the discrete variational inequality (33), we get

\begin{aligned} \|\Pi_{kh}u-u_{kh}\|_{h}^{2} \leq& a_{h}(\Pi_{kh}u-u_{kh},\Pi_{kh}u-u_{kh}) \\ =&a_{h}(\Pi_{kh}u-u,\Pi_{kh}u-u_{kh}) \\ &{} +a_{h}(u,\Pi_{kh}u-u_{kh})-a_{h}(u_{kh}, \Pi_{kh}u-u_{kh}) \\ \leq& C\|\Pi_{kh}u-u\|_{h}\|\Pi_{kh}u-u_{kh} \|_{h} \\ &{} +a_{h}(u,\Pi_{kh}u-u_{kh})-(f, \Pi_{kh}u-u_{kh}) \\ \leq& C\|\Pi_{kh}u-u\|^{2}_{h}+\frac{1}{2} \|\Pi_{kh}u-u_{kh}\|^{2}_{h} \\ &{} +\bigl[a_{h}(u,\Pi_{kh}u-u_{kh})-(f, \Pi_{kh}u-u_{kh})\bigr], \end{aligned}

which implies

$$\|\Pi_{kh}u-u_{kh}\|_{h}^{2} \leq C\|\Pi_{kh}u-u\|_{h}^{2} +\bigl[a_{h}(u,\Pi_{kh}u-u_{kh})-(f, \Pi_{kh}u-u_{kh})\bigr].$$
(57)

Then employing (57) and a triangle inequality yields

\begin{aligned} \|u-u_{kh}\|_{h}^{2} \leq& C\|u- \Pi_{kh}u\|_{h}^{2}+\bigl[a_{h}(u, \Pi_{kh}u-u_{kh})-(f,\Pi_{kh}u-u_{kh}) \bigr] \\ \doteq&\triangle_{1}+\triangle_{2}. \end{aligned}
(58)

Notice that

$$\triangle_{1}=\|u-\Pi_{kh}u\|_{h}^{2} \leq \|u-I_{kh}u\|_{h}^{2}+\|I_{kh}u- \Pi_{kh}u\|_{h}^{2}.$$

Then it follows from the interpolation theorem that

$$\|u-I_{kh}u\|_{h}^{2}\leq Ch^{2}|u|_{3,\Omega}^{2}.$$

On the other hand, applying (16) and (31) yields

$$|I_{kh}u-\Pi_{kh}u|_{2,T}=\bigl\vert \bigl(P_{k},C_{k}^{-1}E_{k}(u)\bigr) \bigr\vert _{2,T}\leq C|P_{k}|_{2,T}\bigl\vert \varepsilon(u)\bigr\vert ,$$

which together with $$|P_{k}|_{2,T}\leq Ch^{-1}$$ and $$|\varepsilon(u)|\leq Ch^{2}|u|_{3,T}$$ implies

$$\|I_{kh}u-\Pi_{kh}u\|_{h}^{2} \leq Ch^{2}|u|_{3,\Omega}^{2}.$$
(59)

Immediately we get

$$\triangle_{1}\leq Ch^{2}|u|_{3,\Omega}^{2}.$$

Next we focus on the estimate of $$\triangle_{2}$$, which is the key difficulty in convergence analysis.

Note that

\begin{aligned} \triangle_{2} =&a_{h}(u, \Pi_{kh}u-u_{kh})-(f,\Pi_{kh}u-u_{kh}) \\ =&\bigl[a_{h}(u,\Pi_{kh}u-I_{kh}u)-(f, \Pi_{kh}u-I_{kh}u)\bigr] \\ &{} +\bigl[a_{h}(u,I_{kh}u-\bar{u}_{kh})-(f,I_{kh}u- \bar{u}_{kh})\bigr] \\ &{} +\bigl[a_{h}(u,\bar{u}_{kh}-u_{kh})-(f, \bar{u}_{kh}-u_{kh})\bigr] \\ \doteq& N_{1}+N_{2}+N_{3}. \end{aligned}
(60)

Now we start to estimate $$N_{j}$$ one by one for $$j=1,2,3$$.

For $$N_{1}$$, applying Green’s formula gives

\begin{aligned} N_{1} =&\bigl[a_{h}(u,\Pi_{kh}u-I_{kh}u)-(f, \Pi_{kh}u-I_{kh}u)\bigr] \\ =& -\sum_{T\in T_{h}}\int_{T} \nabla \triangle u\nabla(\Pi _{kh}u-I_{kh}u)\,dx\,dy \\ &{} -\sum_{T\in T_{h}}\int_{T} f ( \Pi_{kh}u-I_{kh}u)\,dx\,dy \\ &{} +\sum_{T\in T_{h}}\int_{\partial T} \biggl(\triangle u-\frac{\partial^{2}u}{\partial s^{2}}\biggr)\frac{\partial(\Pi_{kh}u-I_{kh}u)}{\partial n}\,ds \\ &{} +\sum_{T\in T_{h}}\int_{\partial T} \frac{\partial ^{2}u}{\partial s\, \partial n }\frac{\partial(\Pi_{kh}u-I_{kh}u)}{\partial s}\,ds \\ \doteq& M_{1}+M_{2}+M_{3}+M_{4}. \end{aligned}
(61)

Then it follows from $$|P_{k}|_{1,T}\leq C$$, $$\|P_{k}\|_{0,T}\leq Ch$$ and the Schwarz inequality that

\begin{aligned} M_{1} \leq&\biggl\vert -\sum _{T\in T_{h}}\int_{T}\nabla\triangle u\nabla( \Pi_{kh}u-I_{kh}u)\,dx\,dy\biggr\vert \\ \leq& \sum_{T\in T_{h}} |u|_{3,T}| \Pi_{kh}u-I_{kh}u|_{1,T}\leq \sum _{T\in T_{h}} C|u|_{3,T}|P_{k}|_{1,T} \bigl\vert \varepsilon(u)\bigr\vert \\ \leq& \sum_{T\in T_{h}} Ch^{2}_{T}|u|^{2}_{3,T} \leq Ch^{2}|u|^{2}_{3,\Omega} \end{aligned}
(62)

and

\begin{aligned} M_{2} \leq&\biggl\vert -\sum _{T\in T_{h}}\int_{T} f (\Pi_{kh}u-I_{kh}u) \,dx\,dy\biggr\vert \\ \leq& \sum_{T\in T_{h}} \|f\|_{0,T}\| \Pi_{kh}u-I_{kh}u\|_{0,T}\leq \sum _{T\in T_{h}}C\|f\|_{0,T}\|P_{k}\|_{0,T} \bigl\vert \varepsilon(u)\bigr\vert \\ \leq& \sum_{T\in T_{h}}Ch^{3}_{T}|u|_{3,T} \|f\|_{0,T} \leq Ch^{3} |u|_{3,\Omega}\|f\|_{0,\Omega}. \end{aligned}
(63)

At the same time, for all $$l\subset \partial T\cap T'$$, $$T,T'\in T_{h}$$, employing the definitions of $$\Pi _{kh}$$ and $$I_{kh}$$ yields

$$\int_{l} \biggl[\frac{\partial(\Pi_{kh}u-I_{kh}u)}{\partial n}\biggr]\,ds =\int_{l} \biggl[\frac{\partial\Pi_{kh}u}{\partial n}\biggr]\,ds-\int _{l} \biggl[\frac{\partial I_{kh}u}{\partial n}\biggr]\,ds=0$$
(64)

and

$$\int_{l} \biggl[\frac{\partial(\Pi_{kh}u-I_{kh}u)}{\partial s}\biggr]\,ds =\int_{l} \biggl[\frac{\partial\Pi_{kh}u}{\partial s}\biggr]\,ds-\int _{l} \biggl[\frac{\partial I_{kh}u}{\partial s}\biggr]\,ds=0.$$
(65)

Then let $$P_{0}v= \frac{1}{|l|} \int_{l}v\, ds$$, applying to (64) and (59), we have

\begin{aligned} M_{3} =& \sum_{T\in T_{h}}\sum _{l\in\partial T}\int_{l} \biggl(\triangle u-\frac{\partial^{2}u}{\partial s^{2}}\biggr) \frac{\partial(\Pi_{kh}u-I_{kh}u)}{\partial n}\,ds \\ =& \sum_{T\in T_{h}}\sum_{l\in\partial T} \int_{l}\biggl(\biggl(\triangle u-\frac{\partial^{2}u}{\partial s^{2}} \biggr)-P_{0}\biggl(\triangle u-\frac{\partial^{2}u}{\partial s^{2}}\biggr)\biggr) \\ &{}\times \biggl(\frac{\partial(\Pi_{kh}u-I_{kh}u)}{\partial n}-P_{0} \frac{\partial(\Pi_{kh}u-I_{kh}u)}{\partial n}\biggr) \,ds \\ \leq& Ch|u|_{3,\Omega}\|\Pi_{kh}u-I_{kh}u \|_{h}\leq Ch^{2}|u|^{2}_{3,\Omega}. \end{aligned}
(66)

Similarly, (65) and (59) imply

$$M_{4}\leq Ch^{2}|u|^{2}_{3,\Omega}.$$
(67)

Therefore, substituting (62)-(63) and (66)-(67) into (61) yields

$$N_{1}\leq Ch^{2}|u|_{3,\Omega}\bigl( \vert u\vert _{3,\Omega}+\|f\|_{0,\Omega}\bigr).$$
(68)

As to $$N_{2}$$, it follows from (40) and (41) that

$$\int_{l} \biggl[\frac{\partial(v-E_{kh}v)}{\partial n}\biggr]\,ds= \int _{l} \biggl[\frac{\partial(v-E_{kh}v)}{\partial s}\biggr]\,ds=0, \quad \forall v \in\bar{V}_{kh},$$

which together with (43) and (44) in Lemma 1 yields

\begin{aligned} a_{h}(u,v-E_{kh}v) =& -\sum_{T\in T_{h}}\int_{T} \nabla\triangle u\nabla(v-E_{kh}v)\,dx\,dy \\ & {}+\sum_{T\in T_{h}}\int_{\partial T} \biggl(\triangle u-\frac{\partial^{2}u}{\partial s^{2}}\biggr)\frac{\partial(v-E_{kh}v)}{\partial n}\,ds \\ &{}+\sum_{T\in T_{h}}\int_{\partial T} \frac{\partial ^{2}u}{\partial s \, \partial n }\frac{\partial(v-E_{kh}v)}{\partial s}\,ds \\ =& -\sum_{T\in T_{h}}\int_{T} \nabla\triangle u\nabla (v-E_{kh}v)\,dx\,dy \\ & {}+\sum_{T\in T_{h}}\sum _{l\in\partial T}\int_{l} \biggl(\biggl(\triangle u- \frac{\partial^{2}u}{\partial s^{2}}\biggr)-P_{0}\biggl(\triangle u-\frac{\partial^{2}u}{\partial s^{2}} \biggr)\biggr) \\ & {}\times\biggl(\frac{\partial(v-E_{kh}v)}{\partial n}-P_{0} \frac{\partial(v-E_{kh}v)}{\partial n} \biggr)\,ds \\ & {}+\sum_{T\in T_{h}}\sum _{l\in\partial T}\int_{l}\biggl( \frac{\partial^{2}u}{\partial s\, \partial n }-P_{0}\frac{\partial^{2}u}{\partial s\, \partial n }\biggr) \\ & {}\times\biggl(\frac{\partial(v-E_{kh}v)}{\partial s}-P_{0}\frac{\partial(v-E_{kh}v)}{\partial s} \biggr)\,ds \\ \leq&|u|_{3,\Omega}\biggl(\sum_{T\in T_{h}}|v-E_{kh}v|^{2}_{1,T} \biggr)^{\frac{1}{2}}+Ch|u|_{3,\Omega}\|v-E_{kh}v\|_{h} \\ \leq& Ch|u|_{3,\Omega}\|v\|_{h}. \end{aligned}
(69)

Then using inequalities (2), (35) and (39), the definition of $${E}_{kh}$$, Lemma 1, (36), (69), the fact that $$\hat{u}_{h}\in K$$, $$\delta_{h,i}\leq Ch^{2}$$, and a similar argument as the one in , we can obtain

$$\bigl[a_{h}(u,I_{kh}u-\bar{u}_{kh})-(f,I_{kh}u- \bar{u}_{kh})\bigr]\leq Ch\|I_{kh}u-\bar{u}_{kh} \|_{h}+Ch^{2}$$
(70)

and

$$\|u-\bar{u}_{kh}\|_{h}\leq Ch.$$
(71)

Hence employing the interpolation theorem and a triangle inequality implies

\begin{aligned} \begin{aligned}[b] N_{2}&=\bigl[a_{h}(u,I_{kh}u- \bar{u}_{kh})-(f,I_{kh}u-\bar{u}_{kh})\bigr] \\ & \leq Ch\bigl(\Vert I_{kh}u-u\Vert _{h}+\|u- \bar{u}_{kh}\|_{h}+h\bigr) \leq Ch^{2}. \end{aligned} \end{aligned}
(72)

For $$N_{3}$$, from the construction of $$V_{kh}$$, (12) and (27), we know that

$${u_{kh}}|_{{T}}=\bigl(P_{k},C_{k}^{-1}G_{k}Q_{k}(u_{kh}) \bigr)=\bigl(P_{k},C_{k}^{-1}D_{k}(u_{kh}) \bigr),$$
(73)

which along with (7) and (22) implies $$u_{kh}\in \bar{V}_{kh}$$.

Then employing (39) and (71) yields

\begin{aligned} N_{3} =&a_{h}(u,\bar{u}_{kh}-{u}_{kh})-(f, \bar{u}_{kh}-{u}_{kh}) \\ \leq& a_{h}(u,\bar{u}_{kh}-{u}_{kh})-a_{h}( \bar{u}_{kh},\bar {u}_{kh}-{u}_{kh}) \\ =&a_{h}(u-\bar{u}_{kh},\bar{u}_{kh}-u)+a_{h}(u- \bar{u}_{kh},u-{u}_{kh}) \\ \leq& -\|u-\bar{u}_{kh}\|_{h}^{2}+\|u- \bar{u}_{kh}\|_{h}\|u-{u}_{kh}\|_{h} \\ \leq&\|u-\bar{u}_{kh}\|_{h}\|u-{u}_{kh} \|_{h} \leq \frac{1}{2} \|u-\bar{u}_{kh} \|_{h}^{2}+\frac{1}{2}\|u-{u}_{kh} \|_{h}^{2} \\ \leq& Ch^{2}+\frac{1}{2}\|u-{u}_{kh} \|_{h}^{2}. \end{aligned}
(74)

At last, substituting (68), (72) and (74) into (60) yields

$$\triangle_{2}\leq Ch^{2}+\frac{1}{2} \|u-{u}_{kh}\|_{h}^{2}.$$

Therefore the desired result (56) follows from the estimates of $$\triangle_{1}$$ and $$\triangle_{2}$$ immediately. The proof is completed. □

### Remark 1

With the help of two kinds of auxiliary obstacle problems (34) and (38), and two enriching operators $$E_{1h}$$ and $$E_{2h}$$, we successfully deduce the optimal error estimates in broken energy norm of the two double set parameter nonconforming element approximations to problem (1). From the proofs of Theorem 1, one can check that the analysis approaches of this paper are indeed very different from [17, 18], and the results presented herein are also valid to the one-obstacle problem discussed in .

### Remark 2

It should be pointed out that (12), (27), (16) and (31) play an important role in the convergence analysis. Unfortunately, not all nonconforming elements constructed by the double set parameter method satisfy these properties [29, 3336]. This means that it is not an easy thing to derive the optimal error estimates of double set parameter nonconforming element approximation to problem (1).

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3. Brezzi, F, Hager, WW, Raviart, PA: Error estimates for the finite element solution of variational inequalities. Numer. Math. 28(4), 431-443 (1977)

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## Acknowledgements

The authors are supported by the National Natural Science Foundation of China (Grant Nos. 10971203; 11271340), the Foundation of He’nan Educational Committee (Nos. 13B110144; 13A110005), Research Fund for the Doctoral Program of Higher Education of China (No. 20094101110006).

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Correspondence to Lifang Pei.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

PL carried out the theoretical studies and drafted the manuscript. SD participated in the design of the study and improved the final version. All authors read and approved the final draft.

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