Lobachevskii Journal of Mathematics http://ljm.ksu.ru ISSN 1818-9962 Vol. 24, 2006, 43–53

© Ying-Jun Jiang and Jin-Ping Zeng

Abstract. We are interested in the approximation in the $L^{\infty }$-norm of variational inequalities with two-sided obstacle. We show that the order of convergence will be the same as that of variational inequalities with one obstacle. We also give multilevel projective algorithm and discuss its convergence.

1. Introduction

Let $K$ be a closed convex set in $H_0^1\left(\Omega \right)$ defined by

where $\Omega \subset R^2$ is a bounded convex polygon, $\Phi ,\Psi \in W^{2,s}\left(\Omega \right)\left(s>2\right)$ are two given functions that satisfy $\Phi {\mid }_{\bar{\Omega }}<\Psi {\mid }_{\bar{\Omega }}$ and $\Phi {\mid }_{\partial \Omega }<0<\Psi {\mid }_{\partial \Omega }$. We consider the following two-sided obstacle problem: find $u\in K$ such that

where $a\left(u,v\right)={\int }_{\Omega }\nabla u\cdot \nabla vdx$, $f\left(\cdot \right)=\left(f,\cdot \right)$, $f\in L^s\left(\Omega \right)$ and $\left(\cdot ,\cdot \right)$ is the $L^2$ inner product.

The proof is similar to variational inequalities with one obstacle(see [6]).

A lot of results on $L^{\infty }$-error estimate for the finite element approximation of obstacle problems have been obtained(see [1, 4, 5, 7, 8, 10]). In this paper, we will discuss $L^{\infty }$-error estimate for the finite element approximation of problem (1) by using the results of variational inequalities with one obstacle. Based on results in [13] we establish a rate of convergence $h^2\mid logh\mid$, provided $\Phi ,\Psi \in W^{2,\infty }\left(\Omega \right)$ and $f\in L^{\infty }\left(\Omega \right)$. This result has be established in a different way (see [4]). At last, we will present a multilevel projective algorithm (see[14]) and discuss its convergence.

2. Preliminaries

In this section, we will construct two variational inequalities with lower and upper obstacles respectively which have the same solution as problem (1). Firstly, we make some preparations.

Set three sets as follows:

and

It’s easy to see that $W_1,W_2$ is two disjoint closed sets and then write

Proof.Let ${\Gamma }_H={\left\{{\Omega }_i^{\prime }\right\}}_{i=1}^k$ be a set of coarse mesh of $\Omega$ with mesh size $H<r∕2-4\delta$. Here $r$ is defined in (2), $\delta <<r$ is a positive number, and ${\Omega }_i^{\prime },i=1,2,\dots ,k$, are disjoint open sets satisfy ${\cup }_{i=1}^k\overline{{\Omega }_i^{\prime }}=\bar{\Omega }$. Then we refine ${\Gamma }_H$ to get ${\Gamma }_h$, a set of fine mesh with mesh size $h<\delta ∕2$. Let ${\Omega }_i,i=1,2,\cdots ,k$, be the enlarged subdomains of ${\Omega }_i^{\prime }$ defined by

The union of ${\Omega }_i$ cover $\Omega$ with overlap size $2\delta$ and $diam\left({\Omega }_i\right)<r∕2$. Following [2], let ${\left\{{\tilde{\theta }}_i\right\}}_{i=1}^k$ be a partition of unity satisfying ${\tilde{\theta }}_i\in C^{\infty }\left(\bar{\Omega }\right)$, ${\tilde{\theta }}_i\ge 0$, ${\tilde{\theta }}_i=0$ in $\Omega \setminus {\Omega }_i$ and ${\sum }_{i=1}^k{\tilde{\theta }}_i=1$. Classify ${\left\{{\Omega }_i\right\}}_{i=1}^k$ into three groups: for $i=1,2$, ${\tilde{W}}_i={\bigcup }_{{\bar{\Omega }}_j\bigcap W_i\ne \varnothing }\left\{{\Omega }_j\right\},{\tilde{W}}_3={\bigcup }_{{\bar{\Omega }}_j\bigcap \left(W_1\bigcup W_2\right)=\varnothing }\left\{{\Omega }_j\right\}$. Let ${\theta }_i={\sum }_{{\Omega }_j\subset {\tilde{W}}_i}{\tilde{\theta }}_j\left(i=1,2,3\right)$, the lemma can be easily verified. $\square$

Proof.From the definition of ${\tilde{W}}_3$ we known that $\Phi <u<\Psi$ in ${\tilde{W}}_3$. Let $D\left({\tilde{W}}_3\right)$ denote the set of infinitely differentiable functions with compact support $\subset \subset {\tilde{W}}_3$. For any $\tilde{v}\in D\left({\tilde{W}}_3\right)$, extend it by zero to $\Omega \setminus {\tilde{W}}_3$, we can take positive $q$ sufficiently small such that $v=u\pm q\tilde{v}\in K$. From (1) we have

What’s more,

where the last equal follows from the definition of weak derivative. Therefore

By the arbitrary of $\tilde{v}$, we have $-△u=fin{\tilde{W}}_3$. $\square$

Let $K_1=\left\{v\in H_0^1\left(\Omega \right)\cap H^2\left(\Omega \right):v\ge \Phi \right\}$, $K_2=\left\{v\in H_0^1\left(\Omega \right)\cap H^2\left(\Omega \right):v\le \Psi \right\}$.

Proof.For $0\le q\le 1$, we consider the two cases: if $w\left(x\right)\ge 0$, then $\left(u+qw\right)\left(x\right)\ge u\left(x\right)\ge \Phi \left(x\right)$; if $w\left(x\right)<0$, then $\left(u+qw\right)\left(x\right)\ge \left(u+w\right)\left(x\right)\ge \Phi \left(x\right)$. So we have $u+qw\ge \Phi$ for $0\le q\le 1$. Furthermore, from the definition of ${\tilde{W}}_1$, we know that $dist\left({\tilde{W}}_1,W_2\right)\ge r∕2$ and it is easy to verify that $u<\Psi$ on $\overline{{\tilde{W}}_1}$. Since $w$ is bounded by Remark 3, there exists a sufficiently small positive $q\le 1$ such that $\Psi \ge u+qw\in K$. Take $v=u+qw$ in (1), then we have $a\left(u,w\right)\ge f\left(w\right)$. $\square$

We now construct two variational inequalities with one obstacle. Define

and

where $u\in W^{2,s}\left(\Omega \right)$ is the solution of problem (1). It is obvious that $f^{\left(1\right)}$ and $f^{\left(2\right)}\in L^s\left(\Omega \right)$. From Lemma 2 and Remark 2, we have

Problem I: Find $u_1\in K_1$ such that

Problem II: Find $u_2\in K_2$ such that

Proof.For simplicity, we only prove problems (1) and (7) have the same solution. Let $u$ be solution of (1). For any $v\in K_1$, $v_i={\theta }_i\left(v-u\right)\in H_0^1\left(\Omega \right)\bigcap H^2\left(\Omega \right)$, $i=1,2,3$, have supports $\subset$ ${\tilde{W}}_i$ respectively and the restriction of $v_i$ are in $H_0^1\left({\tilde{W}}_i\right)\cap H^2\left({\tilde{W}}_i\right)$. Here ${\theta }_i$, $i=1,2,3$, are defined as in Lemma 1. Obviously, $v-u={\sum }_{i=1}^3{v}_i$ and

It’s easy verify that $u+v_1\in K_1$. So by Lemma 3 and (4) we have $a\left(u,v_1\right)\ge f\left(v_1\right)={\int }_{\Omega }f{v}_{1}dx={\int }_{{\tilde{W}}_1}f{v}_{1}dx={\int }_{{\tilde{W}}_1}{f}^{\left(1\right)}{v}_{1}dx={\int }_{\Omega }{f}^{\left(1\right)}{v}_{1}dx=f^{\left(1\right)}\left(v_1\right)$. By Lemma 2 and (4), we can directly verify that

Similarly, $a\left(u,v_2\right)=f^{\left(1\right)}\left(v_2\right)$. So we have

Thereby, $u$ is a solution of problem (7). By the uniqueness of the solution of problem (1) and problem (7), the lemma is proved. $\square$

3. Main Results

Let a triangulation $T_h$ be defined over $\Omega$, satisfying the shape regularity and maximum angle conditions(see [1]). Let ${\tilde{V}}_h={\tilde{V}}_h\left(T_h\right)$ denote the space of continuous piecewise linear functions over $T_h$. Take $V_h={\tilde{V}}_h\cap H_0^1\left(\Omega \right)$. For $v\in C^0\left(\bar{\Omega }\right)$, let ${\pi }_h\left(v\right)\in {\tilde{V}}_h$ be nodal interpolation of $v$, that is, $v={\pi }_h\left(v\right)$ holds at each vertex. Let ${\Phi }_h={\pi }_h\left(\Phi \right),{\Psi }_h={\pi }_h\left(\Psi \right)$. Take

and

The correspondent discrete problem of (1) is: find $u_h\in K_h$ such that

And the correspondent discrete problems of (7) and (8) are: find $u_h^{\left(i\right)}\in K_h^{\left(i\right)},i=1,2$, such that

respectively. We assume, for $i=1,2$, that

We will use the assumption (12) to estimate the error bound of $\mid \mid u-u_h\mid {\mid }_{\infty }$.

Denote $\left\{x_i:i=1,2,\dots ,m_h\right\}$ the interior node set and $\left\{x_i:i=m_h+1,\dots ,n_h\right\}$ the boundary node set of $T_h$. Let ${\left\{{\varphi }_h^{\left(i\right)}\right\}}_{i=1}^{n_h}$ be the nodal basis for ${\tilde{V}}_h$ with ${\varphi }_h^{\left(i\right)}\left(x_j\right)={\delta }_{ij}$. $P_h$ be a canonical function from ${\tilde{V}}_h$ to $R^{m_h}$. Namely, for any $v_h={\sum }_{i=1}^{n_h}{y}_i{\varphi }_h^{\left(i\right)}$, let $y=P_h{v}_h$ with $y^T=\left(y_1,\dots ,y_{m_h}\right)$. Let $A^h={\left(a_{ij}\right)}_{m_h\times m_h}$ be the stiffness matrix given by $a_{ij}={\int }_{\Omega }\nabla {\varphi }_h^{\left(j\right)}\nabla {\varphi }_h^{\left(i\right)}dx$. To obtain our main results, we need the following lemmas.

Proof.By maximal angle condition, $a\left({\varphi }_i,{\varphi }_j\right)\le 0$ for $i\ne j$(see [3, 13]). So we have for $i=1,2,\dots ,m_h$,

The lemma is followed. $\square$

Take $z=P_h{u}_h$, $z^{\left(i\right)}=P_h{u}_h^{\left(i\right)}\left(i=1,2\right)$, $\phi =P_h{\Phi }_h$ and $\psi =P_h{\Psi }_h$. Discrete problems (10) and (11) are equivalent to the following three algebraic problems respectively:

where

and

By (6) we know

Proof.By the assumption (12), we have

and therefore

Define $I,J$ by

respectively. It’s easy to verify that

From (13), (14), (16) and Lemma 6, we have that

and

The lemma follows from (18), (19), (20) and Lemma 7. $\square$

The proof is similar to that of Lemma 8, we omit it here.

By Lemmas 8 and 9, we have

Then, the following theorem becomes obvious.

When $s=\infty$, [13] has obtained the estimate

Therefore by Theorem 2, we have the following theorem.

4. Multilevel Projective Algorithm

In the sequel, we assume $s=\infty$ in problem (1). We consider a sequence of regular triangulations $T_{h_k}$ of the polygonal domain $\Omega$ determined as follows. Suppose $T_{h_1}$ is given and let $T_{h_k}$, $k\ge 2$, be obtained from $T_{h_{k-1}}$ via a systematic subdivision. Edge midpoints in $T_{h_{k-1}}$ are connected by new edges to form $T_{h_k}$. Let $h_k$ be the mesh size of $T_{h_k}$ and satisfy

Let $V_{h_k}$ denote the space of continuous piecewise linear functions with respect to $T_{h_k}$ that vanish on $\partial \Omega$. Note that

Similarly, we use $m_{h_k}$ denote the number of the interior nodes of $T_{h_k}$. Next we will discuss how to solve the discrete problem (13) on the multilevel mesh.

Let $K^{\left(0\right)}$ be a vector space associated with $K_h$ defined as follows:

Let $P_{K^{\left(0\right)}}$ be the projective operator from $R^{m_h}$ into $K^{\left(0\right)}$. Therefore, for any $y\in R^{m_h}$,

where ${\left(\cdot ,\cdot \right)}_E$ is Euclidean inner product. Now we can define an operator $P{G}^h:R^{m_h}\mapsto R^{m_h}$ by

where

Here, ${\lambda }_{max}\left(A^h\right)$ and ${\lambda }_{min}\left(A^h\right)$ are the biggest eigenvalue and the smallest eigenvalue of $A^h$ respectively. Notice that ${\lambda }_{max}\left(A^h\right)=O\left(1\right)$, ${\lambda }_{min}\left(A^h\right)=O\left(h^2\right)$, and the condition number of $A^h$, $\Lambda \left(A^h\right)$, satisfies

Denote $\mid \mid \cdot \mid \mid$ the Euclidean norm of $R^{m_h}$, then we have the following theorem.

The proof of the theorem is similar to that in [14].

Now we define the intergrid transfer operators. Let $I_{h_{k-1}}^{h_k}:R^{m_{h_{k-1}}}\mapsto R^{m_{h_k}}$ defined by $I_{h_{k-1}}^{h_k}\left(y\right)=P_{h_k}\left({\sum }_{i=1}^{m_{h_{k-1}}}{y}_i{\varphi }_{h_{k-1}}^{\left(i\right)}\right)$ for $y\in R^{m_{h_{k-1}}}$. In the following, we will give the multilevel projective algorithm.

Algorithm MP:

Step 1: Solve the exact solution ${\tilde{z}}_{h_1}$ of problem (13) on the coarsest mesh;

Step 2: For $k\ge 2$, the approximate solution on the $k$th level is gotten by

where the positive integer ${\tau }_k$ is chosen such that

Here $m$ is the maximal number of triangles sharing a common vertex.

Now we give the convergence results of algorithm MP.

The proofs can be seen in [14]. We omit them here.

References

College of Mathematics and Science computing, ChangSha University of Science & Technology, Changsha, Hunan, 410076, People’s Republic of China

E-mail address: jiangyingjun@csust.edu.cn

College of Mathematics and Econometrics, Hunan University, Changsha, Hunan, 410076, People’s Republic of China

E-mail address: jpzeng@hnu.cn

Received July 14, 2006