Lobachevskii Journal of Mathematics

http://ljm.ksu.ru
ISSN 1818-9962

Vol. 24, 2006, 43–53 \begin{matrix}c \\ ◯ \end{matrix} Ying-Jun Jiang and Jin-Ping Zeng

Ying-Jun Jiang and Jin-Ping Zeng

L^{∞}-ERROR ESTIMATE FOR A DISCRETE TWO-SIDED OBSTACLE PROBLEM AND MULTILEVEL PROJECTIVE

ALGORITHM

(submitted by A. V. Lapin)

Abstract. We are interested in the approximation in the L^{∞}-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}(Ω) defined by

K = {v ∈ H_{0}^{1}(Ω) : Φ ≤ v ≤ Ψ in Ω},

where Ω ⊂ R^{2} is a bounded convex polygon, Φ, Ψ ∈ W^{2,s}(Ω)(s > 2) are two given functions that satisfy Φ|Ω̄ < Ψ|Ω̄ and Φ|_{∂Ω} < 0 < Ψ|_{∂Ω}. We consider the following two-sided obstacle problem: find u ∈ K such that

a(u, v − u) ≥ f(v − u), ∀v ∈ K,  (1)

China.

2000 Mathematical Subject Classification. 65N30.

Key words and phrases. Variational inequalities, L^{∞}-error estimate, Multilevel projective algorithm.

The work is supported by National Natural Science Fond 10371035 of P. R.

where a(u, v) = ∫_{Ω} ∇u · ∇vdx, f(·) = (f, ·), f ∈ L^{s}(Ω) and (·, ·) is the L^{2} inner product.

Theorem 1. Problem (1) has a unique solution u ∈ H_{0}^{1}(Ω) ∩ W^{2,s}(Ω).

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

Remark 1. u, Φ, Ψ are Hölder continuous functions by W^{2,s}(Ω) ↪ C^{0,α}(Ω) with 0 < α < 1.

A lot of results on L^{∞}-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^{∞}-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}|logh|, provided Φ, Ψ ∈ W^{2,∞}(Ω) and f ∈ L^{∞}(Ω). 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:

W_{1} = {x ∈ Ω : u(x) = Φ(x)},
W_{2} = {x ∈ Ω : u(x) = Ψ(x)}

and

W_{3} = Ω − (W_{1} ∪ W_{2}).

It’s easy to see that W_{1}, W_{2} is two disjoint closed sets and then write

r = dist(W_{1}, W_{2}) > 0.  (2)

Lemma 1. There exist three open sets W̃_{i}, i = 1, 2, 3, which have the following properties:

W_{1} ⊂ W̃_{1}, W_{2} ⊂ W̃_{2}, W̃_{3} ⊂ W_{3},
W̃_{1} ⋂ W̃_{2} = ∅

and

⋃_{i=1}^{3} W̃_{i} = Ω.

L^{∞}-ERROR ESTIMATE 45

Moreover, there is a partition of unity {θ_{i}}_{i=1}^{3} satisfying 0 ≤ θ_{i} ∈ C^{∞}(Ω̄), θ_{i} = 0 on Ω\W̃_{i} and ∑_{i=1}^{3} θ_{i} = 1.

Proof. Let Γ_{H} = {Ω′_{i}}_{i=1}^{k} be a set of coarse mesh of Ω with mesh size H < r/2 − 4δ. Here r is defined in (2), δ << r is a positive number, and Ω′_{i}, i = 1, 2, …, k, are disjoint open sets satisfy ∪_{i=1}^{k}Ω̄′_{i} = Ω̄. Then we refine Γ_{H} to get Γ_{h}, a set of fine mesh with mesh size h < δ/2. Let Ω_{i}, i = 1, 2, ⋯, k, be the enlarged subdomains of Ω′_{i} defined by

Ω_{i} = Ω′_{i} ⋃{τ ⋃((Ω̄′_{i} ⋂ τ̄)\∂Ω)|τ ∈ Γ_{h} and dist(τ, Ω′_{i}) < δ/2}.

The union of Ω_{i} cover Ω with overlap size 2δ and diam(Ω_{i}) < r/2. Following [2], let {θ̃_{i}}_{i=1}^{k} be a partition of unity satisfying θ̃_{i} ∈ C^{∞}(Ω̄), θ̃_{i} ≥ 0, θ̃_{i} = 0 in Ω\Ω_{i} and ∑_{i=1}^{k} θ̃_{i} = 1. Classify {Ω_{i}}_{i=1}^{k} into three groups: for i = 1, 2, W̃_{i} = ⋃_{Ω̄_{j} ⋂ W_{i}≠∅}{Ω_{j}}, W̃_{3} = ⋃_{Ω̄_{j} ⋂(W_{1} ⋃ W_{2})=∅}{Ω_{j}}. Let θ_{i} = ∑_{Ω_{j}⊂W̃_{i}} θ̃_{j}(i = 1, 2, 3), the lemma can be easily verified. □

Lemma 2. Let u be the solution of (1), then

−△u = f a.e. in W̃_{3},  (3)

where W̃_{3} are defined as in Lemma 1.

Proof. From the definition of W̃_{3} we known that Φ < u < Ψ in W̃_{3}. Let D(W̃_{3}) denote the set of infinitely differentiable functions with compact support ⊂⊂ W̃_{3}. For any ṽ ∈ D(W̃_{3}), extend it by zero to Ω\W̃_{3}, we can take positive q sufficiently small such that v = u ± qṽ ∈ K. From (1) we have

a(u, ṽ) = (f, ṽ).

What’s more,

a(u, ṽ) = ∫_{Ω} ∇u∇ṽdx = − ∫_{Ω} u△ṽdx = − ∫_{Ω} △uṽdx,

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

∫_{Ω}(−△u − f)ṽdx = ∫W̃_{3}(−△u − f)ṽdx = 0.

By the arbitrary of ṽ, we have −△u = f in W̃_{3}. □

Remark 2. Use the same technique in Lemma 2, we can also get −△u ≥ f in W̃_{1} and −△u ≤ f in W̃_{2}.

Let K_{1} = {v ∈ H_{0}^{1}(Ω) ∩ H^{2}(Ω) : v ≥ Φ}, K_{2} = {v ∈ H_{0}^{1}(Ω) ∩ H^{2}(Ω) : v ≤ Ψ}.

Remark 3. v ∈ K_{1}(or K_{2}) is Hölder continuous function by embedding theorem, therefore v is bounded.

Lemma 3. a(u, w) ≥ f(w) if w ∈ H_{0}^{1}(W̃_{1}) and u + w ∈ K_{1}, where u is the solution of (1), and w is extended by zero to Ω\W̃_{1} .

Proof. For 0 ≤ q ≤ 1, we consider the two cases: if w(x) ≥ 0, then (u+qw)(x) ≥ u(x) ≥ Φ(x); if w(x) < 0, then (u+qw)(x) ≥ (u+w)(x) ≥ Φ(x). So we have u + qw ≥ Φ for 0 ≤ q ≤ 1. Furthermore, from the definition of W̃_{1}, we know that dist(W̃_{1}, W_{2}) ≥ r/2 and it is easy to verify that u < Ψ on ¯ _{ ˜ 1} W. Since w is bounded by Remark 3, there exists a sufficiently small positive q ≤ 1 such that Ψ ≥ u + qw ∈ K. Take v = u + qw in (1), then we have a(u, w) ≥ f(w). □

We now construct two variational inequalities with one obstacle. Define

f^{(1)} = \{ \begin{matrix}−△u & x ∈ W̃_{2}, \\ f & otherwise \end{matrix}  (4)

and

f^{(2)} = \{ \begin{matrix}−△u & x ∈ W̃_{1}, \\ f & otherwise, \end{matrix}  (5)

where u ∈ W^{2,s}(Ω) is the solution of problem (1). It is obvious that f^{(1)} and f^{(2)} ∈ L^{s}(Ω). From Lemma 2 and Remark 2, we have

f^{(1)} ≤ f ≤ f^{(2)} a.e. in Ω.  (6)

Problem I: Find u_{1} ∈ K_{1} such that

a(u_{1}, v − u_{1}) ≥ f^{(1)}(v − u_{1}), ∀v ∈ K_{1}.  (7)

Problem II: Find u_{2} ∈ K_{2} such that

a(u_{2}, v − u_{2}) ≥ f^{(2)}(v − u_{2}), ∀v ∈ K_{2}.  (8)

Remark 4. Problems (7), (8) have unique solutions in W^{2,s}(Ω) (see [6]).

Lemma 4. Problems (1), (7) and (8) have the same solution.

Proof. For simplicity, we only prove problems (1) and (7) have the same solution. Let u be solution of (1). For any v ∈ K_{1}, v_{i} = θ_{i}(v − u) ∈ H_{0}^{1}(Ω) ⋂ H^{2}(Ω), i = 1, 2, 3, have supports ⊂ W̃_{i} respectively and the restriction of v_{i} are in H_{0}^{1}(W̃_{i}) ∩ H^{2}(W̃_{i}). Here θ_{i}, i = 1, 2, 3, are defined as in Lemma 1. Obviously, v − u = ∑_{i=1}^{3} v_{i} and

a(u, v − u) = a(u, v_{1}) + a(u, v_{2}) + a(u, v_{3}).  (9)

It’s easy verify that u + v_{1} ∈ K_{1}. So by Lemma 3 and (4) we have a(u, v_{1}) ≥ f(v_{1}) = ∫_{Ω} fv_{1}dx = ∫W̃_{1} fv_{1}dx = ∫W̃_{1} f^{(1)}v_{1}dx = ∫_{Ω} f^{(1)}v_{1}dx = f^{(1)}(v_{1}). By Lemma 2 and (4), we can directly verify that

a(u, v_{3}) = (−△u, v_{3}) = ∫_{Ω} −△uv_{3}dx = ∫W̃_{3} −△uv_{3}dx

= ∫W̃_{3} f^{(1)}v_{3}dx = ∫_{Ω} f^{(1)}v_{3}dx = f^{(1)}(v_{3}).

Similarly, a(u, v_{2}) = f^{(1)}(v_{2}). So we have

a(u, v − u) ≥ f^{(1)}(v_{1} + v_{2} + v_{3}) = f^{(1)}(v − u).

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

3. Main Results

Let a triangulation T_{h} be defined over Ω, satisfying the shape regularity and maximum angle conditions(see [1]). Let Ṽ_{h} = Ṽ_{h}(T_{h}) denote the space of continuous piecewise linear functions over T_{h}. Take V_{h} = Ṽ_{h} ∩ H_{0}^{1}(Ω). For v ∈ C^{0}(Ω̄), let π_{h}(v) ∈ Ṽ_{h} be nodal interpolation of v, that is, v = π_{h}(v) holds at each vertex. Let Φ_{h} = π_{h}(Φ), Ψ_{h} = π_{h}(Ψ). Take

K_{h} = {v ∈ V_{h} : Φ_{h} ≤ v ≤ Ψ_{h}},
K_{h}^{(1)} = {v ∈ V_{h} : v ≥ Φ_{h}}

and

K_{h}^{(2)} = {v ∈ V_{h} : v ≤ Ψ_{h}}.

The correspondent discrete problem of (1) is: find u_{h} ∈ K_{h} such that

a(u_{h}, v_{h} − u_{h}) ≥ f(v_{h} − u_{h}), ∀v_{h} ∈ K_{h}.  (10)

And the correspondent discrete problems of (7) and (8) are: find u_{h}^{(i)} ∈ K_{h}^{(i)}, i = 1, 2, such that

a(u_{h}^{(i)}, v_{h} − u_{h}^{(i)}) ≥ f^{(i)}(v_{h} − u_{h}^{(i)}), ∀v_{h} ∈ K_{h}^{(i)},  (11)

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

||u − u_{h}^{(i)}||_{∞} ≤ Ch^{μ}|logh|^{γ}, 0 ≤ μ, γ ≤ 2.  (12)

We will use the assumption (12) to estimate the error bound of ||u−u_{h}||_{∞}.

Denote {x_{i} : i = 1, 2, …, m_{h}} the interior node set and {x_{i} : i = m_{h} + 1, …, n_{h}} the boundary node set of T_{h}. Let {φ_{h}^{(i)}}_{i=1}^{n_{h}} be the nodal basis for Ṽ_{h} with φ_{h}^{(i)}(x_{j}) = δ_{ij}. P_{h} be a canonical function from Ṽ_{h} to R^{m_{h}}. Namely, for any v_{h} = ∑_{i=1}^{n_{h}} y_{i}φ_{h}^{(i)}, let y = P_{h}v_{h} with y^{T} = (y_{1}, …, y_{m_{h}}). Let A^{h} = (a_{ij})_{m_{h}×m_{h}} be the stiffness matrix given by a_{ij} = ∫_{Ω} ∇φ_{h}^{(j)}∇φ_{h}^{(i)}dx. To obtain our main results, we need the following lemmas.

Lemma 5. (see [9, 11]) If the triangulation T_{h} satisfies maximal angle condition, A^{h} is M-matrix.

Lemma 6. If the triangulation T_{h} satisfies maximal angle condition,

A^{h}e ≥ 0,

where e = (1, 1, …, 1)^{T} ∈ R^{m_{h}}.

Proof. By maximal angle condition, a(φ_{i}, φ_{j}) ≤ 0 for i ≠ j(see [3, 13]). So we have for i = 1, 2, …, m_{h},

(A^{h}e)_{i} = ∑_{j=1}^{m_{h}} a_{ij} = ∑_{j=1}^{m_{h}} a(φ_{j}, φ_{i})

≥ ∑_{j=1}^{m_{h}} a(φ_{j}, φ_{i}) + ∑_{j=m_{h}+1}^{n_{h}} a(φ_{j}, φ_{i}) = a(1, φ_{i}) = 0.

The lemma is followed. □

Lemma 7. [12] Let sets I ⊂ {1, 2, …, m_{h}}, J = {1, 2, …, m_{h}}\I. For any y, ȳ ∈ R^{m_{h}}, if y_{I} ≤ ȳ_{I} and (A^{h}y)_{J} ≤ (A^{h}ȳ)_{J}, we have that

y ≤ ȳ,

where A^{h} is the stiffness matrix given before.

Take z = P_{h}u_{h}, z^{(i)} = P_{h}u_{h}^{(i)}(i = 1, 2), ϕ = P_{h}Φ_{h} and ψ = P_{h}Ψ_{h}. Discrete problems (10) and (11) are equivalent to the following three algebraic problems respectively:

\{ \begin{matrix}(A^{h}z)_{i} = g_{i} & if ϕ_{i} < z_{i} < ψ_{i}, \\ (A^{h}z)_{i} ≥ g_{i} & if z_{i} = ϕ_{i}, \\ (A^{h}z)_{i} ≤ g_{i} & if z_{i} = ψ_{i}, \\ ϕ ≤ z ≤ ψ; \end{matrix}  (13)

\{ \begin{matrix}(A^{h}z^{(1)})_{i} = g_{i}^{(1)} & if z_{i}^{(1)} > ϕ_{i}, \\ (A^{h}z^{(1)})_{i} ≥ g_{i}^{(1)} & if z_{i}^{(1)} = ϕ_{i}, \\ z^{(1)} ≥ ϕ; \end{matrix}  (14)

\{ \begin{matrix}(A^{h}z^{(2)})_{i} = g_{i}^{(2)} & if z_{i}^{(2)} < ψ_{i}, \\ (A^{h}z^{(2)})_{i} ≤ g_{i}^{(2)} & if z_{i}^{(2)} = ψ_{i}, \\ z^{(2)} ≤ ψ, \end{matrix}  (15)

L^{∞}-ERROR ESTIMATE 49

where

g = ((f, φ_{h}^{(1)}), …, (f, φ_{h}^{(m_{h})}))^{T}

and

g^{(i)} = ((f^{(i)}, φ_{h}^{(1)}), …, (f^{(i)}, φ_{h}^{(m_{h})}))^{T} , i = 1, 2.

By (6) we know

g^{(1)} ≤ g ≤ g^{(2)}.  (16)

Lemma 8. If the assumption (12) holds, we have

z^{(1)} − Ch^{μ}|logh|^{γ}e ≤ z,  (17)

where e = (1, 1, …, 1)^{T} ∈ R^{m_{h}}. Proof. By the assumption (12), we have

u_{h}^{(1)} − Ch^{μ}|logh|^{γ} ≤ u ≤ Ψ,

and therefore

z^{(1)} − Ch^{μ}|logh|^{γ}e ≤ ψ.

Define I, J by

I = {i | z_{i}^{(1)} = ϕ_{i}} ∪ {i | z_{i} = ψ_{i}},
J = {1, 2, …, m_{h}}\I,

respectively. It’s easy to verify that

z_{j}^{(1)} − Ch^{μ}|logh|^{γ} ≤ z_{j}, j ∈ I.  (18)

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

(A^{h}z)_{J} ≥ g_{J}  (19)

and

A(z^{(1)} − Ch^{μ}|logh|^{γ}e)_{J} ≤ (Az^{(1)})_{J} = g_{J}^{(1)} ≤ g_{J}.  (20)

The lemma follows from (18), (19), (20) and Lemma 7. □ Lemma 9. If the assumption (12) holds, then

z^{(2)} + Ch^{μ}|logh|^{γ}e ≥ z.

The proof is similar to that of Lemma 8, we omit it here. By Lemmas 8 and 9, we have

u_{h}^{(1)} − Ch^{μ}|logh|^{γ} ≤ u_{h} ≤ u_{h}^{(2)} + Ch^{μ}|logh|^{γ}.

Then, the following theorem becomes obvious.

Theorem 2. If the assumption (12) holds, we have

||u − u_{h}||_{∞} ≤ Ch^{μ}|logh|^{γ}.

When s = ∞, [13] has obtained the estimate

||u − u_{h}^{(i)}||_{∞} ≤ Ch^{2}|logh|, for i = 1, 2.

Therefore by Theorem 2, we have the following theorem.

Theorem 3. If s = ∞ in problem (1), we have

||u − u_{h}||_{∞} ≤ Ch^{2}|logh|.

4. Multilevel Projective Algorithm

In the sequel, we assume s = ∞ in problem (1). We consider a sequence of regular triangulations T_{h_{k}} of the polygonal domain Ω determined as follows. Suppose T_{h_{1}} is given and let T_{h_{k}}, k ≥ 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

h_{k} = 2^{−(k−1)}h_{1}, k = 1, 2, ….

Let V_{h_{k}} denote the space of continuous piecewise linear functions with respect to T_{h_{k}} that vanish on ∂Ω. Note that

T_{h_{k−1}} ⊂ T_{h_{k}} ⇒ V_{h_{k−1}} ⊂ V_{h_{k}}.

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^{(0)} be a vector space associated with K_{h} defined as follows:

K^{(0)} = {y ∈ R^{m_{h}}|ϕ ≤ y ≤ ψ}.

Let P_{K^{(0)}} be the projective operator from R^{m_{h}} into K^{(0)}. Therefore, for any y ∈ R^{m_{h}},

(y − P_{K^{(0)}}y, ȳ − P_{K^{(0)}}y)_{E} ≤ 0, for all ȳ ∈ K^{(0)},

where (·, ·)_{E} is Euclidean inner product. Now we can define an operator P G^{h} : R^{m_{h}} ↦ R^{m_{h}} by

P G^{h}(y) = P_{K^{(0)}}(y − ρ_{h}(A^{h}y − g)),

where

ρ_{h} = \frac{1}{λ_{max}(A^{h})}.

Here, λ_{max}(A^{h}) and λ_{min}(A^{h}) are the biggest eigenvalue and the smallest eigenvalue of A^{h} respectively. Notice that λ_{max}(A^{h}) = O(1), λ_{min}(A^{h}) = O(h^{2}), and the condition number of A^{h}, Λ(A^{h}), satisfies

Λ(A^{h}) = \frac{λ_{max}}{λ_{min}} = O(\frac{1}{h^{2}} ) = O(m_{h}).

Denote || · || the Euclidean norm of R^{m_{h}}, then we have the following theorem.

Theorem 4. z ∈ K^{(0)} solves the discrete problem (13), then for any y ∈ R^{m_{h}},

||P G^{h}(y) − z|| ≤ (1 − \frac{1}{Λ(A^{h})})||y − z||,

||(P G^{h})^{l}(y) − z|| ≤ (1 − \frac{1}{Λ(A^{h})})^{l}||y − z||,

here l is a positive integer.

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}}} ↦ R^{m_{h_{k}}} defined by I_{h_{k−1}}^{h_{k}}(y) = P_{h_{k}}(∑_{i=1}^{m_{h_{k−1}}} y_{i}φ_{h_{k−1}}^{(i)}) for y ∈ R^{m_{h_{k−1}}}. In the following, we will give the multilevel projective algorithm. Algorithm MP:

Step 1: Solve the exact solution z̃_{h_{1}} of problem (13) on the coarsest mesh;

Step 2: For k ≥ 2, the approximate solution on the kth level is gotten by

z̃_{h_{k}} = (P G^{h_{k}})^{τ_{k}}I_{h_{k−1}}^{h_{k}}(z̃_{h_{k−1}}),

where the positive integer τ_{k} is chosen such that

(1 − \frac{1}{Λ(A^{h_{k}})})^{τ_{k}} ≤ \frac{1}{5 + 2\sqrt{1 + m/2}}.

Here m is the maximal number of triangles sharing a common vertex. Now we give the convergence results of algorithm MP.

Theorem 5. If z̃_{h_{k}}, (k = 1, …), is generated by algorithm MP, we have

||z_{h_{k}} − z̃_{h_{k}}|| ≤ Ch|logh|,

where z_{h_{k}} ∈ K^{h_{k}} is the exact solution of problem (13) on k-th level.

Theorem 6. If z̃_{h_{k}}, (k = 1, …), is generated by algorithm MP, we have

||u_{h_{k}} − Q_{h_{k}}(z̃_{h_{k}})||_{L^{2}(Ω)} ≤ Ch^{2}|logh|,

52 YING-JUN JIANG AND JIN-PING ZENG

where u_{h_{k}} ∈ V_{h_{k}} is the exact solution of problem (10) on k-th level, Q_{h_{k}} : R^{m_{h_{k}}} ↦ V_{h_{k}}, is defined by

Q_{h_{k}}(y) = ∑_{i=1}^{m_{h_{k}}} y_{i}φ_{h_{k}}^{(i)}, for any y ∈ R^{m_{h_{k}}} .

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

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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