Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 28, 2003, 89–97

CONTINUOUS DEPENDENCE ON OBSTACLES IN DOUBLE GLOBAL OBSTACLE PROBLEMS

Anna Olek and Kuba Szczepaniak

Technical University of Łódź, Department of Mathematics PL-90-924 Łódź, Poland; kubaszcz@ife.p.lodz.pl

Abstract. Let A: H_{0}^{1}(Ω) → H^{−1}(Ω) be Lipschitz and monotone operator; let ⟨ · , · ⟩ stand for the dual bracket between H_{0}^{1}(Ω) and H^{−1}(Ω). For given functions φ, ψ we define (under suitable assumptions) the admissible set K_{φ}^{ψ} := {v ∈ H_{0}^{1}(Ω) : φ ≤ v ≤ ψ a.e. in Ω}.

Next for sequences (φ_{n}), (ψ_{n}), converging to φ, ψ, respectively, we consider the sequence of admissible sets (K_{φ_{n}}^{ψ_{n}}) defined as K_{φ_{n}}^{ψ_{n}} := {v ∈ H_{0}^{1}(Ω) : φ_{n} ≤ v ≤ ψ_{n} a.e. in Ω}. Then for f ∈ H^{−1}(Ω) we discuss the following obstacle problems.

(P_{n}): Find u_{n} ∈ K_{φ_{n}}^{ψ_{n}} : ⟨Au_{n}, v_{n} − u_{n}⟩ ≥ ⟨f, v_{n} − u_{n}⟩ for all v_{n} ∈ K_{φ_{n}}^{ψ_{n}} and

(P): Find u ∈ K_{φ}^{ψ} : ⟨Au, v − u⟩ ≥ ⟨f, v − u⟩ for all v ∈ K_{φ}^{ψ}.

The above problems represent the so-called double global obstacle problem. See [2], [3] for existence and regularity results.

The purpose of this paper is to study convergence (in certain sense) of the solutions u_{n} of (P_{n}) providing the sequences of impediments converge to their limits. Such problems are known as “varying obstacle problems”.

We extend here the results given in [6] where the author has considered the global obstacle problem (see [2], [3] for definitions, existence and regularity results).

1. Introduction

In this paper we study continuity properties corresponding to double global obstacle problems. We consider two sequences of impediments (φ_{n}) and (ψ_{n}) converging to φ and ψ, respectively. Let u_{n} for n ∈ N denote the solution of the double global obstacle problem with the admissible set K_{φ_{n}}^{ψ_{n}} = {v ∈ H_{0}^{1}(Ω) : φ_{n} ≤ v ≤ ψ_{n} a.e. in Ω}. We analyze the convergence of the sequence (u_{n}) to the solution u of the obstacle problem with the admissible set K_{φ}^{ψ} = {v ∈ H_{0}^{1}(Ω) : φ ≤ v ≤ ψ a.e. in Ω}. Certain assumptions are imposed on the functions φ_{n} and ψ_{n}.

We present two results. In the first one we assume that the sequence (φ_{n}) converges to φ from below while (ψ_{n}) converges to ψ from above. We prove that the solutions u_{n} approach u. We also give an example illustrating that if one of the assumptions is not satisfied then the convergence of impediments does not imply convergence of the solutions to the solution of the limit problem.

2000 Mathematics Subject Classification: Primary 49J40; Secondary 35B65.

In the second one we assume that both (φ_{n}) and (ψ_{n}) converge to φ and ψ, respectively, from above. Moreover, the functions φ_{n} are convex while ψ_{n} are concave. Then the solutions u_{n} approach the solution of the limit problem.

The proofs are based on properties of H_{0}^{1}(Ω) functions, Sobolev’s embedding theorem and Minty’s lemma.

2. Notation and basic definitions

Throughout this paper we shall use the following notation and assumptions if not stated otherwise:

Ω ⊂ R^{n} bounded domain with smooth boundary ∂Ω;

H^{1}(Ω) is the Sobolev space W^{1,2}(Ω);

H_{0}^{1}(Ω) represents H^{1} closure of C_{0}^{∞}(Ω);

( · , ·) stands for the inner product in L^{2}(Ω);

Q(Ω) is the vector space of equivalent classes of quasi-continuous functions;

L_{c}^{2}(Ω) := {φ ∈ Q(Ω) : ṽ ≥ |φ| quasi-everywhere in Ω for v ∈ H_{0}^{1}(Ω), where ṽ is a quasi-continuous representative of v}.

The definitions of a quasi-continuous function, properties which hold quasi-every-where and Q(Ω) can be found in [5].

It is well known (see [4]) that the sequence of solutions (u_{n}) converges strongly to u in H_{0}^{1}(Ω) provided (K_{φ_{n}}^{ψ_{n}}) is a sequence of closed, convex subsets of H_{0}^{1}(Ω) which converges to K_{φ}^{ψ} in the sense of Mosco.

The capacity problems allowed to introduce necessary and sufficient conditions for convergence in the sense of Mosco of sequences of the convex admissible sets; see [1]. Referring to the result mentioned we recall that (u_{n}) converges strongly to u in H_{0}^{1}(Ω) provided (φ_{n}) and (ψ_{n}) converge to φ, ψ in L_{c}^{2}(Ω).

Similarly as in [6] we define the class of obstacles converging from above and from below. More precisely:

Definition 1. We say that the sequence (γ_{n}) converges to γ from below (above) a.e. in Ω provided:

(i) γ_{n} ⟶_{n→∞}γ a.e. in Ω,

(ii) for all n ∈ N γ_{n} ≤ γ (γ_{n} ≥ γ) a.e. in Ω.

Definition 2. We say that the sequence (γ_{n}) converges to γ from below (above) in L^{2}(Ω) provided:

(i) γ_{n} ⟶_{n→∞}γ in L^{2}(Ω),

(ii) for all n ∈ N γ_{n} ≤ γ (γ_{n} ≥ γ) a.e. in Ω.

Throughout this paper we agree that we are given an operator A: H_{0}^{1}(Ω) → H^{−1}(Ω) which is Lipschitz and monotone i.e. there exist α, γ > 0 such that for all u, v ∈ H_{0}^{1}(Ω)

\begin{aligned}∥Au − Av∥ & ≤ γ∥u − v∥, \\ ⟨Au − Av, u − v⟩ & ≥ α∥u − v∥^{2}; \end{aligned}

f ∈ H^{−1}(Ω) and functions φ_{n}, φ, ψ_{n}, ψ ∈ L_{c}^{2}(Ω), n ∈ N.

We consider double global obstacle problems P_{n} with impediments φ_{n}, ψ_{n} given by:

P_{n}: Find u_{n} ∈ K_{φ_{n}}^{ψ_{n}} := {v ∈ H_{0}^{1}(Ω) : φ_{n} ≤ v ≤ ψ_{n} a.e. in Ω} such that

⟨Au_{n}, v_{n} − u_{n}⟩ ≥ ⟨f, v_{n} − u_{n}⟩ for all v_{n} ∈ K_{φ_{n}}^{ψ_{n}}

and the double global obstacle problem P with impediments φ, ψ defined as: P: Find u ∈ K_{φ}^{ψ} := {v ∈ H_{0}^{1}(Ω) : φ ≤ v ≤ ψ a.e. in Ω} such that

⟨Au, v − u⟩ ≥ ⟨f, v − u⟩ for all v ∈ K_{φ}^{ψ}.

3. Continuity results

Our first result consists in showing that the solution of the double global obstacle problem with impediments φ and ψ can be obtained as the limit (in the H_{0}^{1}(Ω) space) of the solutions u_{n} with impediments φ_{n}, ψ_{n} converging to φ, ψ from below and above, respectively. Observe that monotonicity of the sequences (φ_{n}) and (ψ_{n}) is not required.

The proof is based on properties of H_{0}^{1}(Ω) functions taking advantage of Sobolev’s embedding theorem and Minty’s lemma.

Theorem 1. If we assume that the sequences (φ_{n}) and (ψ_{n}) converge a.e. in Ω to φ and ψ from below and above, respectively, and φ_{n} ≤ ψ_{n} a.e. in Ω, φ_{n|∂Ω} ≤ 0, ψ_{n|∂Ω} ≥ 0 then the solutions u_{n} of the double global obstacle problem P_{n} converge strongly in H_{0}^{1}(Ω) to the solution u of the double global obstacle problem P.

Here we present an example which illustrates that convergence from above of the impediments does not necessarily imply convergence of solutions to the limit solution.

Example. Let Ω = (−1; 1), A = −d^{2}/dx^{2} and f ≡ 0 in Ω. We take

φ_{n} = \{ \begin{matrix}1 & in [−\frac{1}{n + 1}; \frac{1}{n + 1}], \\ 0 & otherwise. \end{matrix}

We put

ψ_{n} = \{ \begin{matrix}\frac{1}{2} & in (−1; −\frac{2}{n + 2}] ∪ [\frac{2}{n + 2}; 1), \\ 2 & otherwise. \end{matrix}

Note that φ = 0 and ψ = \frac{1}{2} a.e. in Ω. Moreover, for all n ∈ N

(ψ_{n} ≥ φ_{n} a.e. in Ω, φ_{n|∂Ω} ≤ 0, ψ_{n|∂Ω} ≥ 0).

Next we have for all n ∈ N, (φ_{n} ≥ φ, ψ_{n} ≥ ψ).

We consider the sequence (P_{n}) of the double global obstacle problems: Find u_{n} ∈ K_{φ_{n}}^{ψ_{n}} := {v ∈ H_{0}^{1}((−1; 1)) : φ_{n} ≤ v ≤ ψ_{n} a.e. in (−1; 1)} such that:

∫_{−1}^{1} u′_{n} · (u′_{n} − v′) ≥ 0 for all v ∈ K_{φ_{n}}^{ψ_{n}}.

We know that H_{0}^{1}((−1; 1)) is embedded in C_{0}^{1,1−1/2}((−1; 1)) from the Sobolev theorem, so after an easy computation we arrive at the solutions of P_{n} given by

u_{n}(x) = \{ \begin{matrix}\frac{n + 2}{2n}x + \frac{n + 2}{2n}, & x ∈ (−1, \frac{−2}{n + 2}], \\ \frac{(n + 2)(n + 1)}{2n}x + \frac{3n + 2}{2n}, & x ∈ [\frac{−2}{n + 2}, \frac{−1}{n + 1}], \\ 1, & x ∈ [\frac{−1}{n + 1}, \frac{1}{n + 1}], \\ \frac{−(n + 2)(n + 1)}{2n}x + \frac{3n + 2}{2n}, & x ∈ [\frac{1}{n + 1}, \frac{2}{n + 2}], \\ \frac{−(n + 2)}{2n}x + \frac{n + 2}{2n}, & x ∈ [\frac{2}{n + 2}, 1). \end{matrix}

We notice that u_{n} → u^{∗} a.e. in Ω where

u^{∗} = \{ \begin{matrix}\frac{x}{2} + \frac{1}{2}, & x ∈ [−1; 0], \\ −\frac{x}{2} + \frac{1}{2}, & x ∈ [0; 1]. \end{matrix}

Now the limit obstacle problem P is:

Find u ∈ K_{φ}^{ψ} = \{v ∈ H_{0}^{1}((−1; 1)) : 0 ≤ v ≤ \frac{1}{2} a.e. in (−1; 1)\} such that ∫_{−1}^{1} u′ · (u′ − v′) ≥ 0 for all v ∈ K_{φ}^{ψ}. It is also very easy to see that u ≡ 0. Obviously (u_{n}) does not converge to u in any sense.

Now we concentrate on the case of the double obstacle problems with impediments converging from above. In order to obtain convergence of solutions we introduce some additional assumptions imposed on the functions φ_{n} and ψ_{n}. This result is obtained due to a certain approximation procedure applied to a H_{0}^{1}(Ω)-function and the theory of singular perturbation problems.

Theorem 2. Let the following be satisfied:

(1)  \begin{aligned}φ_{n}, ψ_{n}, φ, ψ & ∈ H_{0}^{1}(Ω), \\ −Δφ_{n} & ≤ 0 in H^{−1}(Ω), \\ −Δψ_{n} & ≥ 0 \end{aligned}

in H^{−1}(Ω), n ∈ N. If (φ_{n}), (ψ_{n}) approach φ, ψ, respectively, from above in L^{2}(Ω) then

u_{n} ⟶_{n→∞}u

strongly in H_{0}^{1}(Ω).

Remark. We have also managed to derive a similar result to the one given in [6] for the global double obstacle problem when the impediments φ_{n}, ψ_{n} are concave and convex, respectively.

4. Proof of Theorem 1

The existence and uniqueness of the solutions of the problems P_{n} and P follow directly from the Lions–Stampacchia theorem ([3], [5]). First we show adapting the ideas from [6] that if there exists v_{0} ∈ ⋂_{n≥1}K_{φ_{n}}^{ψ_{n}} then the following estimate holds:

∥u_{n} − v_{0}∥ ≤ \frac{1}{α}(∥f∥ + ∥Av∥).

Indeed we put v = v_{0} at every problem P_{n} to get

(2)  ⟨Au_{n}, v_{0} − u_{n}⟩ ≥ ⟨f, v_{0} − u_{n}⟩.

Using the fact that A is monotone and Lipschitz, f ∈ H^{−1}(Ω) we have

(3)  \begin{aligned}α∥u_{n} − v_{0}∥^{2} & ≤ ⟨Au_{n} − Av_{0}, u_{n} − v_{0}⟩ \\ & = ⟨Au_{n}, u_{n} − v_{0}⟩ − ⟨Av_{0}, u_{n} − v_{0}⟩ \\ & ≤ ⟨f, u_{n} − v_{0}⟩ − ⟨Av_{0}, u_{n} − v_{0}⟩ \\ & ≤ ∥f∥ ∥u_{n} − v_{0}∥ + M∥v_{0}∥ ∥u_{n} − v_{0}∥. \end{aligned}

Whence

∥u_{n} − v_{0}∥ ≤ \frac{1}{α}(∥f∥ + M∥v_{0}∥).

Since v_{0} is a priori known we get that (u_{n}) is bounded which implies that there exists a subsequence

(4)

u_{n} → u^{∗} weakly in H_{0}^{1}(Ω);
u_{n} → u^{∗} strongly in L^{2}(Ω);
u_{n} → u^{∗} a.e. in Ω

for some u^{∗} ∈ H_{0}^{1}(Ω). Since φ_{n} ≤ u_{n} ≤ ψ_{n} a.e. in Ω if we let n → ∞ we obtain that φ ≤ u^{∗} ≤ ψ a.e. in Ω and this implies that u^{∗} ∈ K_{φ}^{ψ}.

Now let us remark that for any v ∈ K_{φ}^{ψ} we get: φ_{n} ≤ φ ≤ v ≤ ψ ≤ ψ_{n} a.e. in Ω which gives that

(5)  K_{φ}^{ψ} ⊂ K_{φ_{n}}^{ψ_{n}} for all n ∈ N.

Let us note that φ^{+} − ψ^{−} ∈ K_{φ}^{ψ} (see [1], [3], [5] for definitions and details). Therefore there exists

v_{0} ∈ ⋂_{n≥1} K_{φ_{n}}^{ψ_{n}}

and

u_{n} → u^{∗} weakly in H_{0}^{1}(Ω).

From Minty’s lemma ([3]) we can identify the problem P_{n} with: u_{n} ∈ K_{φ_{n}}^{ψ_{n}} such that

⟨Av, v − u_{n}⟩ ≥ ⟨f, v − u_{n}⟩ for all v ∈ K_{φ_{n}}^{ψ_{n}}.

Having (5), we replace K_{φ_{n}}^{ψ_{n}} by K_{φ}^{ψ} and get:

⟨Av, v − u_{n}⟩ ≥ ⟨f, v − u_{n}⟩ for all v ∈ K_{φ}^{ψ}.

Now we let n → ∞ and we arrive at the following:

u^{∗} ∈ K_{φ}^{ψ} : ⟨Av, v − u^{∗}⟩ ≥ ⟨f, v − u^{∗}⟩ for all v ∈ K_{φ}^{ψ}.

Applying Minty’s lemma to the last problem we get:

u^{∗} ∈ K_{φ}^{ψ} : ⟨Au^{∗}, v − u^{∗}⟩ ≥ ⟨f, v − u^{∗}⟩ for all v ∈ K_{φ}^{ψ}.

Uniqueness of the solution of the problem P allows us to state that u^{∗} = u. So we have shown that u_{n} → u weakly in H_{0}^{1}(Ω).

In order to show strong convergence we observe that

u ∈ ⋂_{n≥1} K_{φ_{n}}^{ψ_{n}}

so we can put v_{0} = u in (3). We arrive at the following estimate:

α · ∥u_{n} − u∥^{2} ≤ ⟨f − Au, u_{n} − u⟩.

If we let n → ∞ we obtain that u_{n} → u strongly in H_{0}^{1}(Ω) which finishes the proof.

Continuous dependence on obstacles 95

5. Proof of Theorem 2 From the maximum principle we find that φ_{n} ≤ 0, ψ_{n} ≥ 0 a.e. in Ω. There-

fore 0 ∈ ⋂_{n∈N} K_{φ_{n}}^{ψ_{n}} and using the results of Theorem 1 we can estimate ∥u_{n}∥ ≤ C (where C does not depend on n).

Next for v ∈ K_{φ}^{ψ} we define its approximation v_{n} as:

(6)  v_{n} ∈ H_{0}^{1}(Ω) : −\frac{1}{n}Δv_{n} + v_{n} = (v ∨ φ_{n}) − (v ∧ ψ_{n}) + v,

where v ∨ φ_{n} = max(v, φ_{n}) and v ∧ ψ_{n} = min(v, ψ_{n}). By the theory of singular perturbation problems ([5]) (v_{n}) converges to v in H_{0}^{1}(Ω) and v_{n} → max(v, φ) − min(v, ψ) + v = v strongly in H_{0}^{1}(Ω).

Now we show that v_{n} ∈ K_{φ_{n}}^{ψ_{n}}.

First we multiply both sides of (6) by (φ_{n} − v_{n})^{+} and we integrate:

\begin{aligned}∫_{Ω}(\frac{1}{n}Δv_{n} − v_{n})(φ_{n} − v_{n})^{+} & = −\frac{1}{n} ∫_{Ω} ∇v_{n}∇(φ_{n} − v_{n})^{+} − ∫_{Ω} v_{n}(φ_{n} − v_{n}) \\ & = − ∫_{Ω}(v ∨ φ_{n} − v ∧ ψ_{n} + v)(φ_{n} − v_{n})^{+}. \end{aligned}

Next we add ∫_{Ω}(−(1/n)Δφ_{n} + φ_{n})(φ_{n} − v_{n})^{+} to both sides and obtain

− \frac{1}{n} ∫_{Ω} ∇v_{n}∇(φ_{n} − v_{n})^{+} − ∫_{Ω} v_{n}(φ_{n} − v_{n})^{+} + ∫_{Ω}(−\frac{1}{n}Δφ_{n} + φ_{n})(φ_{n} − v_{n})^{+}

= −\frac{1}{n} ∫_{Ω} ∇v_{n}∇(φ_{n} − v_{n})^{+} − \frac{1}{n} ∫_{Ω} Δφ_{n}(φ_{n} − v_{n})^{+} + ∫_{Ω}(φ_{n} − v_{n})(φ_{n} − v_{n})^{+}

= −\frac{1}{n} ∫_{Ω} ∇v_{n}∇(φ_{n} − v_{n})^{+} + \frac{1}{n} ∫_{Ω} ∇φ_{n}∇(φ_{n} − v_{n})^{+} + ∫_{Ω}(φ_{n} − v_{n})(φ_{n} − v_{n})^{+}

= ∫_{Ω} \frac{1}{n}∇(φ_{n} − v_{n})∇(φ_{n} − v_{n})^{+} + ∫_{Ω}(φ_{n} − v_{n})(φ_{n} − v_{n})^{+}

= ∫_{Ω} \frac{1}{n}|∇(φ_{n} − v_{n})^{+}|^{2} + ∫_{Ω} |(φ_{n} − v_{n})^{+}|^{2}.

On the other hand,

\begin{matrix}∫_{Ω}(−\frac{1}{n}Δφ_{n} + φ_{n} − v ∨ φ_{n} + v ∧ ψ_{n} − v)(φ_{n} − v_{n})^{+} \\ = ∫_{Ω}(−\frac{1}{n}Δφ_{n} − (v − φ_{n})^{+} − (v − ψ_{n})^{+})(φ_{n} − v_{n})^{+}. \end{matrix}

The last integral is nonpositive, since φ_{n}−(v∨φ_{n}) = −(v−φ_{n})^{+} and (v∧ψ_{n})−v = −(v − ψ_{n})^{+} . This gives us that

∫_{Ω} \frac{1}{n}|∇(φ_{n} − v_{n})^{+}|^{2} + ∫_{Ω} |(φ_{n} − v_{n})^{+}|^{2} ≤ 0

which means that φ_{n} − v_{n} ≤ 0 a.e. in Ω; so v_{n} ≥ φ_{n} a.e. in Ω.

Now we multiply both sides of (6) by (v_{n} − ψ_{n})^{+} and integrate. Thus

∫_{Ω} \frac{1}{n}∇v_{n}∇(v_{n} − ψ_{n})^{+} + ∫_{Ω} v_{n}(v_{n} − ψ_{n})^{+} = ∫_{Ω}((v ∨ φ_{n}) − (v ∧ ψ_{n}) + v)(v_{n} − ψ_{n})^{+}.

Now we add

∫_{Ω}(\frac{1}{n}Δψ_{n} − ψ_{n})(v_{n} − ψ_{n})^{+}

to both sides of the above equality and transform, using the Stampacchia result, in order to get

∫_{Ω}(\frac{1}{n}∇(v_{n} − ψ_{n})∇(v_{n} − ψ_{n})^{+} + (v_{n} − ψ_{n})(v_{n} − ψ_{n})^{+})

= ∫_{Ω} \frac{1}{n}|∇(v_{n} − ψ_{n})^{+}|^{2} + ∫_{Ω} |(v_{n} − ψ_{n})^{+}|^{2}

= ∫_{Ω}(\frac{1}{n}Δψ_{n} − ψ_{n} + (v ∨ φ_{n}) − (v ∧ ψ_{n}) + v)(v_{n} − ψ_{n})^{+}.

Having (1) we deduce that φ_{n} ≤ ψ_{n} a.e. in Ω. Moreover, since ψ_{n} ≥ ψ for all n ∈ N we note that the first factor in the last integral is nonpositive.

Therefore

∫_{Ω}(\frac{1}{n}|∇(v_{n} − ψ_{n})^{+}|^{2} + |(v_{n} − ψ_{n})^{+}|^{2}) ≤ 0

which implies that v_{n} ≤ ψ_{n} a.e. in Ω. This finishes the argumentation demanded for showing that v_{n} ∈ K_{φ_{n}}^{ψ_{n}}.

Applying Minty’s lemma to P_{n} we get

(7)  ⟨Aw, w − u_{n}⟩ ≥ ⟨f, w − u_{n}⟩ for all w ∈ K_{φ_{n}}^{ψ_{n}}.

Taking into account the above results we deal with the approximation (v_{n}) of v ∈ K_{φ}^{ψ}. Therefore we can replace w = v_{n} in (7) and we have:

⟨Av_{n}, v_{n} − u_{n}⟩ ≥ ⟨f, v_{n} − u_{n}⟩.

If we let n → ∞ we arrive at the following:

⟨Av, v − u^{∗}⟩ ≥ ⟨f, v − u^{∗}⟩ for all v ∈ K_{φ}^{ψ},

where u^{∗} was introduced during the proof of Theorem 1.

Since u_{n} ∈ K_{φ_{n}}^{ψ_{n}} we have φ_{n} ≤ u_{n} ≤ ψ_{n} a.e. in Ω. Letting n → ∞, we obtain, using (4), that φ ≤ u^{∗} ≤ ψ a.e. in Ω, which indicates that u^{∗} ∈ K_{φ}^{ψ}. Using Minty’s lemma again we transform the last variational inequality into ⟨Au^{∗}, v − u^{∗}⟩ ≥ ⟨f, v − u^{∗}⟩ for all v ∈ K_{φ}^{ψ}.

Using the uniqueness of the solution of P we have u = u^{∗}. This implies that u_{n} → u weakly in H_{0}^{1}(Ω).

To get the strong convergence we can take an approximation (ū_{n}) of u. By coerciveness of A we evaluate

α∥u_{n} − ū_{n}∥^{2} ≤ ⟨Au_{n} − Aū_{n}, u_{n} − ū_{n}⟩.

Next, since we have ⟨Au_{n}, ū_{n} − u_{n}⟩ ≥ ⟨f, ū_{n} − u_{n}⟩, we estimate

α∥u_{n} − ū_{n}∥^{2} ≤ ⟨Au_{n} − Aū_{n}, u_{n} − ū_{n}⟩
= ⟨Au_{n}, u_{n} − ū_{n}⟩ + ⟨−Aū_{n}, u_{n} − ū_{n}⟩
≤ ⟨f, u_{n} − ū_{n}⟩ + ⟨−Aū_{n}, u_{n} − ū_{n}⟩
= ⟨f − Aū_{n}, u_{n} − ū_{n}⟩.

So if we let n → ∞ we have that u_{n} → u strongly in H_{0}^{1}(Ω), which completes the proof.

References

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[2] Chipot, M.: Variational Inequalities and Flow Through Porous Media. - Springer-Verlag, 1984.

[3] Kinderlehrer, D., and G. Stampacchia: An Introductions to Variational Inequalities and their Applications. - Academic Press, 1980.

[4] Mosco, U.: Convergence of convex sets and solutions of variational inequalities. - Adv. Math. 3, 1969, 510–585.

[5] Rodriguez, J.F.: Obstacle Problems in Mathematical Physics. - North Holland, 1987.

[6] Toyoizumi, H.: Continuous dependence on obstacles in variational inequalities. - Funkcial. Ekvac. 34, 1991, 103–115.

Received 18 December 2001