Nonlinear Analysis 69 (2008) 1125–1137www.elsevier.com/locate/na

A power penalty method for solving a nonlinear paraboliccomplementarity problem

Song Wanga,∗, C.-S. Huangb

a School of Mathematics and Statistics, The University of Western Australia, 35 Stirling Highway, Crawley, WA 6009, Australiab Department of Applied Mathematics, National Sun Yat-sen University, Kaohsiung, Taiwan

Received 19 April 2007; accepted 5 June 2007

Abstract

In this paper we present a penalty method for solving a complementarity problem involving a second-order nonlinear parabolicdifferential operator. In this work we first rewrite the complementarity problem as a nonlinear variational inequality. Then, we definea nonlinear parabolic partial differential equation (PDE) approximating the variational inequality using a power penalty term witha penalty constant λ > 1, a power parameter k > 0 and a smoothing parameter ε. We prove that the solution to the penalizedPDE converges to that of the variational inequality in an appropriate norm at an arbitrary exponential rate of the formO([λ−k

+ ε(1 + λε1/k)]1/2). Numerical experiments, performed to verify the theoretical results, show that the computed rates ofconvergence in both λ and k are close to the theoretical ones.c© 2007 Elsevier Ltd. All rights reserved.

MSC: 65K10; 90C33; 35R45

Keywords: Nonlinear variational inequality; Nonlinear complementarity problem; Stochastic control; Free boundary; Obstacle problem; Penaltymethod

1. Introduction

Many real-world phenomena in engineering, stochastic control, physics, mechanics and economics are governedby complementarity problems involving linear and nonlinear, second-order, parabolic partial differential operators(cf., for example, [2,5,4,3,1]). These problems are of the form

∂u

∂t+ T (u(x, t)) − f (x, t) ≤ 0 (1.1)

u(x, t) − u∗(x, t) ≤ 0 (1.2)(∂u

∂t+ T (u(x, t)) − f (x, t)

)· (u(x, t) − u∗(x, t)) = 0 (1.3)

∗ Corresponding author. Tel.: +61 864883350; fax: +61 864881028.E-mail addresses: swang@maths.uwa.edu.au (S. Wang), huangcs@math.nsysu.edu.tw (C.-S. Huang).

0362-546X/$ - see front matter c© 2007 Elsevier Ltd. All rights reserved.doi:10.1016/j.na.2007.06.014

1126 S. Wang, C.-S. Huang / Nonlinear Analysis 69 (2008) 1125–1137

for (x, t) ∈ Ω × (0, T ] =: Q almost everywhere (a.e.) with the initial and boundary conditions

u(x, t) = 0, (x, t) ∈ ∂Ω × (0, T ] a.e. and u(x, 0) = u0(x), x ∈ Ω , a.e., (1.4)

where

T (u(x, t)) = −∇ · (A(x)∇u(x, t)) + G(u(x, t)), (1.5)

Ω ⊂ Rn denotes a bounded, open and connected set, n denotes a positive integer, ∂Ω denotes the boundary ofΩ , A(x) = (ai j (x)) is an n × n matrix, and G(·) : R 7→ R, f, u∗ and u0 are given functions. Because of itscomplexity and nonlinearity, analytical solutions to (1.1)–(1.4) are, in general, impossible to find except for somespecial cases, and numerical approximations to the problem are normally sought in practice. Numerical methods forlinear complementarity problem corresponding to (1.1)–(1.3) (i.e., G ≡ 0 in (1.5)) have been discussed by variousauthors. A popular approach is to reformulate the problem as a penalized differential equation using the linear penaltyterm of the form λ[uλ−u∗

]+, where λ 1 is a penalty constant, uλ denotes the solution to the penalized equation and[a]+ = maxa, 0. This method has been discussed in [2,3]. The penalty term penalizes the positive part of uλ −u∗. Ithas been proved in [2] that the solution to the penalized equation converges to the exact one in a Sobolev norm at therate ofO(λ−1/2). Power penalty methods have been used for solving conventional constrained optimization problems(cf., for example, [6,9]). In [7], the authors proposed a power penalty for a one-dimensional parabolic variationalinequality involving the Black–Scholes differential operator. The penalized equation is of the form

Luλ(x, t) + λ[uλ(x, t) − u∗(x, t)]1/k+ = 0, (1.6)

where L denotes a differential operator, λ > 1 is the penalty parameter, k > 0 is a constant and [a]+ = maxa, 0.It has been shown in [7] that the approximate solution uλ converges to the exact solution u in a proper Sobolev normat the rate of O(λ−k/2). This includes the linear penalty as the special case that k = 1. However, the penalty term in(1.6) has unbounded derivative when uλ − u∗

→ 0+, and therefore, the penalty function needs to be smoothed locallyas proposed in [7]. A power penalty method for a linear complementarity problem in Rn is also proposed in [8].

The aim of this paper is twofold. We will generalize the penalty term in (1.6) to the formλ ([uλ(x, t) − u∗

]+ + ε)1/k , where 1 ε ≥ 0 is a regularization parameter, and we extend the power penalty methodfor linear problems in one spatial dimension to the nonlinear problem (1.1)–(1.4) in multiple spatial dimensions. Therest of the paper is organized as follows.

In the next section, we will reformulate the complementarity problem as a variational inequality, and consider thesolvability of the resulting variational inequality. In Section 3, we will present the penalty approach. A convergenceanalysis for the penalty approach will be given in Section 4. In Section 5, we present some numerical results to confirmthe theory developed and to demonstrate the usefulness and effectiveness of the method.

Before further discussion, it is necessary to introduce some notation to be used in the rest of the paper. For an openset S ⊂ Rn and 1 ≤ p ≤ ∞, we let L p(S) = v : (

∫S |v(x)|pdx)1/p < ∞ denote the space of all p-power integrable

functions on S. The inner product on L2(S) is denoted by (·, ·)S . We use ‖ · ‖L p(S) to denote the norm on L p(S). Form = 1, 2, . . . , we let W m

p (S) denote the usual Sobolev space with the norm ‖ · ‖m,p,S . When p = 2, we simply denoteHm

2 (S) and ‖ · ‖m,2,S by Hm(S) and ‖ · ‖m,S , respectively. Let Cm(S) (respectively, Cm(S)) be the function set ofwhich a function and its derivatives of up to order k are continuous on S (respectively, S). When S = Ω , we omit thesubscript S in the above notation. We put Hm

0 (Ω) = v ∈ Hm(Ω) : v(x) = 0, x ∈ ∂Ω. For any Hilbert space H(Ω),we let L p(0, T ; H(Ω)) denote the space defined by

L p(0, T ; H(Ω)) = v(·, t) : v(·, t) ∈ H(Ω) a.e. in (0, T ); ‖v(·, t)‖H ∈ L p(0, T ),

where 1 ≤ p ≤ ∞ and ‖ · ‖H denotes the natural norm on H(Ω). The norm on this space is denoted by ‖ · ‖L p(0,T ;H),i.e.,

‖v‖L p(0,T ;H(Ω)) =

(∫ T

0‖v(·, t)‖p

H dt

)1/p

. (1.7)

Clearly, L p(0, T ; L p(Ω)) = L p(Q). Finally, we use H−1(Ω) to denote the dual space of H1(Ω) and use 〈·, ·〉 todenote the duality pair between a Hilbert space and its dual space.

S. Wang, C.-S. Huang / Nonlinear Analysis 69 (2008) 1125–1137 1127

2. The continuous problem

In this section we will formulate (1.1)–(1.4) as a variational inequality in a proper setting and discuss the propertiesof the differential operator. The latter results will ensure that the variational inequality has a unique solution in anappropriate space. We start this discussion with some assumptions on the given data of the problem.

For the given functions in (1.1)–(1.4) we assume the following assumptions hold.

A1. For i, j = 1, . . . , n, ai j ∈ L∞(Ω) ∩ H1(Ω) and there exists a positive constant a0 such that for any ξ ∈ Rn ,ξ> A(x)ξ ≥ a0|ξ |

2 a.e. in Ω .A2. G(v) ∈ L2(Ω) for every v ∈ L2(Ω), and G is monotone and Lipschitz continuous on L2(Ω), i.e., G satisfies

(G(v1) − G(v2), v1 − v2) ≥ 0 and ‖G(v1) − G(v2)‖0 ≤ α‖v1 − v2‖0 for any v1, v2 ∈ L2(Ω) and some constantα > 0.

A3. f, ∂ f∂t ∈ Lq(0, T ; Lq(Ω)), where q = k + 1 with k being the parameter appeared in the penalty term in (1.6).

A4. u∗∈ Lq(0, T ; W 2

q (Ω)), ∂u∗

∂t ∈ Lq(0, T ; Lq(Ω)), u0 ∈ W 2p(Ω) for p = 1+1/k, and u∗

≥ 0 on ∂Ω a.e. satisfyingu0(x) ≤ u∗(x, 0) a.e in Ω .

Without loss of generality, we also assume that

A5. G(0) = 0.

The case that G(0)(x, t) = G0(x, t) can be transformed into A5 by subtracting G0(x, t) from both sides of (1.1). Let

K(t) = v ∈ H10 (Ω) : v ≤ u∗(t).

It is easy to verify thatK is a convex and closed subset of H10 (Ω) for any t ∈ [0, T ]. UsingK, we define the following

problem.

Problem 2.1. Find u(t) ∈ K(t) such that, for all v ∈ K(t),(−

∂u(t)

∂t, v − u(t)

)+ B(u(t), v − u(t)) + (G(u(t)), v − u(t)) ≥ ( f (t), v − u(t)) (2.1)

a.e. in (0, T ) and u(x, 0) = u0(x) in Ω , where B(u, v) is the bilinear in u and v given by

B(u, v) = (A∇u, ∇v), u, v ∈ H10 (Ω). (2.2)

For this variational inequality problem, we have

Theorem 2.1. Problem 2.1 is the variational form corresponding to the linear complementarity problem (1.1)–(1.4).

Proof. Note that

w − u∗(t) ≤ 0 a.e. on Ω for any w ∈ K.

Multiplying both sides of (1.1) by w − u∗ for an arbitrary w ∈ K and integrating the second term by parts, we obtain(−

∂u(t)

∂t, w − u∗(t)

)+ B(u(t), w − u∗(t)) + (G(u(t)), v − u∗(t)) ≥ ( f (t), w − u∗(t)), (2.3)

a.e. in (0, T ). Since K(t) is a convex subset of H10 (Ω) for any t ∈ [0, T ], we may write w as w = θv + (1 − θ)u(t),

where v ∈ K and θ ∈ [0, 1]. Therefore, (2.3) becomes(−

∂u(t)

∂t, u(t) − u∗(t)

)+

(−

∂u(t)

∂t, θ(v − u(t))

)+ B(u(t), u(t) − u∗(t))

+ (G(u(t)), v − u(t)) + B(u(t), θ(v − u(t))) + (G(u(t))), θ(v − u(t))

≥ ( f (t), u(t) − u∗(t)) + ( f (t), θ(v − u(t))). (2.4)

Integrating (1.3) and using integration by parts, we have(−

∂u(t)

∂t, u(t) − u∗(t)

)+ B(u(t), u(t) − u∗(t)) + (G(u(t)), u(t) − u∗(t)) = ( f (t), u(t) − u∗(t)).

1128 S. Wang, C.-S. Huang / Nonlinear Analysis 69 (2008) 1125–1137

Therefore, (2.4) reduces to(−

∂u(t)

∂t, θ(v − u(t))

)+ B(u(t), θ(v − u(t))) + (G(u(t)), v − u(t)) ≥ ( f (t), θ(v − u(t))).

Since θ ≥ 0, we see that the strong complementarity problem (1.1)–(1.4) can be reformulated as (2.1).

We now consider the solvability of Problem 2.1. To show this, we first prove that the operator T (·) is monotone,coercive and continuous. These are contained in the following theorem.

Lemma 2.1. For any v, w ∈ H10 (I ), there exist positive constants C and M, independent of v and w, such that

〈T (v) − T (w), v − w〉 ≥ 0, (2.5)

〈T (v), v〉 = B(v, v) + (G(v), v) ≥ C‖v‖21, (2.6)

〈T (v), w〉 = B(v, w) + (G(v), w) ≤ M‖v‖1‖w‖1. (2.7)

Proof. Let C be a generic positive constant, independent of v and w. We first note that from (2.2) and AssumptionA1 it is easy to show that, for any v ∈ H1

0 (Ω),

B(v, v) = (A∇v, ∇v) ≥ a0(∇v, ∇v) = a0‖∇v‖0 ≥ C‖v‖1. (2.8)

In the above we use the Poincare–Friedrich inequality that ‖∇v‖0 ≥ C‖v‖1 for any v ∈ H10 (Ω).

To prove (2.5), we use integration by parts, (2.8) and Assumption A2 to get

〈T (v) − T (w), v − w〉 = B(v − w, v − w) + (G(v) − G(w), v − w)

≥ C‖v − w‖1.

We thus have (2.5). When w = 0, we have T (w) = 0 because G(w) = 0 by Assumption A5. Therefore, the aboveinequality also implies (2.6).

We now prove (2.7). From Assumptions A1 and A2 we have, for any v, w ∈ H10 (Ω),

B(v, w) + (G(v), w) = (A∇v, ∇w) + (G(v), w)

≤ C(‖∇v‖0‖∇w‖0 + ‖v‖0‖w‖0)

≤ M‖v‖1‖w‖1.

This completes the proof of Lemma 2.1.

Using this lemma we have the following theorem which ensures that Problem 2.1 is uniquely solvable.

Theorem 2.2. Let Assumptions A1–A5 be fulfilled. Then there exists a unique solution to Problem 2.1.

Proof. Using Lemma 2.1 and Assumption A4 it is easy to verify that all the sufficient conditions for the uniquesolvability of Problem 2.1 listed in [11, pages. 572–575] are satisfied. Therefore, the problem has a uniquesolution.

3. The power penalty approach

Motivated by (1.6), let us consider the following nonlinear equation:

∂uλ

∂t+ T (uλ(x, t)) + λ (φ(uλ(x, t)) + ε)1/k

= f (x, t) + λε1/k, (x, t) ∈ Ω (3.1)

with the given boundary and initial conditions

uλ(x, t) = 0, (x, t) ∈ ∂Ω × (0, T ), and uλ(x, 0) = u0(x), x ∈ Ω ,

where λ > 1, ε > 0 and k ≥ 1 are parameters and

φ(v(x, t)) = [v(x, t) − u∗(x, t)]+ = maxv(x, t) − u∗(x, t), 0.

S. Wang, C.-S. Huang / Nonlinear Analysis 69 (2008) 1125–1137 1129

This is a penalized equation corresponding to (1.1)–(1.3). Note that here we assume that k ≥ 1. The case that0 < k < 1 is less interesting. This is because in this case, the convergence of uλ to u is slower than that of k ≥ 1,though the penalty term with 0 < k < 1 is smooth even when ε = 0.

In (3.1), the penalty term penalizes the positive part of uλ−u∗, as mentioned before. When λ → +∞ and ε → +0,we expect that the solution uλ to (3.1) converges to the solution u to (2.1). Clearly, the rate of convergence dependson all the parameters in the penalty term, which will be discussed in the next section. Before the discussion, we firstconsider the solvability of (3.1).

Using integration by parts, it is easy to show that the variational problem corresponding to (3.1) is

Problem 3.1. Find uλ(t) ∈ H10 (Ω) such that, for all v ∈ H1

0 (Ω),(−

∂uλ(t)

∂t, v

)+ B(uλ(t), v) + (G(uλ(t)), v) + λ((φ(uλ(t)) + ε)1/k, v) = ( f (t) + λε1/k, v) (3.2)

a.e. in (0, T ) with the initial condition uλ(x, 0) = u0(x) in Ω .

This is a penalized variational problem corresponding to Problem 2.1. Similar to Theorem 2.2 we have the followingtheorem:

Theorem 3.1. Suppose the conditions in Theorem 2.2 are fulfilled. Then, Problem 3.1 has a unique solution for fixedε ≥ 0 and λ ≥ 1.

Proof. To prove this theorem, we consider the following equation equivalent to (3.2):(−

∂uλ(t)

∂t, v

)+ B(uλ(t), v) + (G(uλ(t)), v) + λ(Φε(uλ(t)), v) = ( f (t) + λ[ε1/k

− (φ(0) + ε)1/k], v),

where Φε(w) = (φ(w) + ε)1/k− (φ(0) + ε)1/k . Clearly,

Φε(0) = (φ(0) + ε)1/k− (φ(0) + ε)1/k

= 0.

Also, it is easy to see that Φε(w) is monotonically increasing in w because of the monotonicity of ([w − u∗]+ + ε)1/k

in w. Now, we verify the following items:(i) 〈T (v) + λΦε(v) − (T (w) + λΦε(w)), v − w〉 ≥ 0 for any v, w ∈ H1

0 (Ω).(ii) 〈T (v) + λΦε(v), v〉 ≥ C‖v‖

21 for a positive constant C , independent of v, λ, k and ε.

(iii) ‖T (v) + λΦε(v)‖H−1(Ω) ≤ C1(t) + C2‖v‖1 for a positive C1(t) ∈ L2(0, T ) and a positive constant C2,independent of v, k and ε.

The existence and uniqueness of the solution to Problem 3.1 is then guaranteed by the theoretical results in [10, pp.767–779] (also see [4, Theorem 1.37]).

Item (i) is trivial because of (2.5) and the fact that Φε(v) is monotone. To prove (ii), we use (2.6) to obtain

〈T (v) + λΦε(v), v〉 ≥ C‖v‖21 + λ(Φε(v), v) ≥ C‖v‖

21,

because (Φε(v), v) = (Φε(v) − Φε(0), v − 0) ≥ 0.We now prove (iii). Let C1 ∈ L2(0, T ) be a generic positive function of t only and C2 a generic positive constant,

independent of ε and k. From the definitions of Φε and φ(v) we have

(Φε(v), w) =

∫Ω

([v − u∗]+ + ε)1/kwdx −

∫Ω

(φ(0) + ε)1/k wdx

≤

[(∫Ω

([v − u∗]+ + ε)2/kdx

)1/2

+ C1(t)

]‖w‖0

≤ C

[(∫Ω

([v − u∗]+ + ε)2dx

)1/2k

+ C1(t)

]‖w‖0

= C(‖[v − u∗]+ + ε‖

1/k0 + C1(t))‖w‖0

≤ (C2‖v‖1/k0 + C1(t))‖w‖0

≤ (C2‖v‖0 + C1(t)) ‖w‖0,

1130 S. Wang, C.-S. Huang / Nonlinear Analysis 69 (2008) 1125–1137

since ‖v‖1/k0 ≤ max1, ‖v‖0 for k ≥ 1 and ‖u∗

‖0 ∈ L2(0, T ) by Assumption A4. Combining this with (2.7) gives

〈T (v) + λΦε(v), w〉 ≤ ‖w‖1 (C1(t) + C2‖v‖1) , v,w ∈ H10 (Ω).

Dividing both sides of this inequality by ‖w‖1 and taking the supremum with respect to w we have (iii).

Remark 3.1. We comment that Theorem 3.1 establishes the unique solvability of Problem 3.1 for any fixed ε ≥ 0.The penalty term ([v − u∗

]+ + ε)1/k is non-smooth, but has bounded upper and lower derivatives if ε > 0. However,if ε = 0, its first derivative becomes unbounded when [v − u∗

]+ → 0+.

4. Convergence of uλ

We now show that the solution to Problem 3.1 converges to that of Problem 2.1 as λ → ∞ at the rateO([λ−k

+ ε(λε1/k+ 1)]1/2

)in a proper norm. This discussion is divided into two parts. We will first establish error

bounds for [uλ − u∗]+ in proper norms, and then use these results to derive a bound for uλ − u. We begin this analysis

with the following lemma.

Lemma 4.1. Let uλ be the solution to Problem 3.1. If uλ ∈ L p(Ω), then there exists a positive constant C,independent of uλ, ε, k and λ, such that

‖[uλ − u∗]+‖L p(Q) ≤ C

(1λk + ε

), (4.1)

‖[uλ − u∗]+‖L∞(0,T ;L2(Ω)) + ‖[uλ − u∗

]+‖L2(0,T ;H10 (Ω)) ≤ C

(1λk + λε p

)1/2

, (4.2)

where k is the parameter used in (3.1) and p = 1 + 1/k.

Proof. Assume that C is a generic positive constant, independent of uλ, ε, k and λ. Setting v = φ(uλ) in (3.2), wehave (

∂uλ

∂t, φ(uλ)

)+ B(uλ, φ(uλ)) + (G(uλ), φ(uλ)) + λ((φ + ε)1/k, φ(uλ)) = ( f + λε1/k, φ(uλ))

a.e. in (0, T ). Taking ( ∂u∗

∂t , φ(uλ)) + B(u∗, φ(uλ)) + (G(u∗), φ(uλ)) away from both sides of the above equation andusing the facts that

(uλ − u∗, φ(uλ)) = (φ(uλ), φ(uλ)) and B(uλ − u∗, φ(uλ)) = B(φ(uλ), φ(uλ)),

we have(∂φ(uλ)

∂t, φ(uλ)

)+ B(φ(uλ), φ(uλ)) + (G(uλ) − G(u∗), φ(uλ)) + λ(φ1/k(uλ), φ(uλ))

= ( f + λε1/k, φ(uλ)) −

(∂u∗

∂t, φ(uλ)

)− B(u∗, φ(uλ)) − (G(u∗), φ(uλ)). (4.3)

Note that

(G(uλ) − G(u∗), φ(uλ)) = (G(uλ) − G(u∗), φ(uλ) − φ(u∗)) ≥ 0, (4.4)

because φ(u∗) = 0 and both G and φ are monotone functions. Using integration by parts, it is easy to show that, forany v(t) satisfying v(0) = u0∫ t

0

∂φ(v(τ ))

∂tφ(v(τ))dτ =

12

[φ2(v(t)) − φ2(v(0))

]=

12φ2(v(t)). (4.5)

In the above, we used the fact that φ(v(0)) = [u0(x) − u∗(x)]+ = 0, because u0 ≤ u∗ on Ω by Assumption A4.Integrating both sides of (4.3) from 0 to t and using (2.6) in Lemma 2.1, (4.4) and (4.5), we get

S. Wang, C.-S. Huang / Nonlinear Analysis 69 (2008) 1125–1137 1131

12(φ(uλ), φ(uλ)) + C

∫ t

0‖φ(uλ)‖

21dτ + λ

∫ t

0[φ(uλ) + ε]1/kφ(uλ)dτ

≤

∫ t

0( f + λε1/k, φ(uλ))dτ −

∫ t

0

(∂u∗

∂τ, φ(uλ)

)dτ −

∫ t

0B(u∗, φ(uλ))τ +

∫ t

0(G(u∗), φ)dτ

≤ C(1 + λε1/k)

(∫ t

0‖φ(uλ)‖

pL p(Ω)

dτ

)1/p

−

∫ T

tB(u∗, φ(uλ))τ. (4.6)

In the above we used Assumptions A2–A4 and Holder’s inequality. Let us consider the last integral in (4.6). Sinceu∗(t) ∈ W 2

p(Ω) by Assumption A4, integrating by parts, we have

−

∫ t

0(A∇u∗, ∇φ(uλ))dτ =

∫ t

0(∇ · (A∇u∗), φ(uλ))dτ ≤ C

(∫ t

0‖φ(uλ)‖

pL p(Ω)

dτ

)1/p

. (4.7)

Substituting (4.7) into (4.6) gives

12(φ(uλ(t)), φ(uλ(t))) + C

∫ t

0‖φ(uλ(t))‖

21dτ + λ

∫ t

0[φ(uλ) + ε]1/kφ(uλ)dτ

≤ C(1 + λε1/k)

(∫ t

0‖φ(uλ)‖

pL p(Ω)

dτ

)1/p

.

From this bound we have

12(φ(uλ(t)), φ(uλ(t))) + C

∫ t

0‖φ(uλ(t))‖

21dτ + λ

∫ t

0φ p(uλ)dτ

≤ C(1 + λε1/k)

(∫ t

0‖φ(uλ)‖

pL p(Ω)

dτ

)1/p

, (4.8)

since 0 < φ(uλ) < φ(uλ) + ε. This implies that

λ

∫ t

0‖φ(uλ(τ ))‖L p(Ω)dτ ≤ C(1 + λε1/k)

(∫ t

0‖φ(uλ(τ ))‖

pL p(Ω)

dτ

)1/p

,

or (∫ t

0‖φ(uλ(τ ))‖

pL p(Ω)

dτ

)(p−1)/p

≤ C(λ−1+ ε1/k).

Note p − 1 = 1/k since p = 1 + 1/k. From this, it follows that(∫ t

0‖φ(uλ(τ ))‖

pL p(I )dτ

)1/p

≤ C(λ−1+ ε1/k)k

≤ C

(1λk + ε

). (4.9)

This proves (4.1).Now, from (4.8) and (4.9), we have

12(φ(uλ(t)), φ(uλ(t))) +

∫ t

0‖φ(uλ(τ ))‖2

1dτ ≤ C(1 + λε1/k)

(∫ t

0‖φ(τ)‖

pL p(I )dτ

)1/p

≤ Cλ−k(1 + λε1/k)(1 + λkε)

≤ Cλ−k(1 + λε1/k)k+1

≤ Cλ−k(1 + λk+1ε p)

≤ C

(1λk + λε p

).

From this it follows that

1132 S. Wang, C.-S. Huang / Nonlinear Analysis 69 (2008) 1125–1137

(φ(uλ(t)), φ(uλ(t)))1/2

+

(∫ t

0‖φ(uλ(τ ))‖2

1dτ

)1/2

≤ C

(1λk + λε p

)1/2

for all t ∈ (0, T ). Clearly, (4.2) follows from this inequality.

Lemma 4.1 establishes the bounds for [uλ − u∗]+ in the norms. Using these results, we are able to show that the

solution to Problem 3.1 converges to that of Problem 2.1 at a rate of order O(λ−k/2+ (λε p)1/2), as given in the next

theorem.

Theorem 4.1. Suppose that the assumptions in Lemma 4.1 are fulfilled, and let u and uλ be the solutions toProblem 2.1 and Problem 3.1, respectively. If ∂u

∂t ∈ Lk+1(Q), then there exists a constant C > 0, independent ofu, uλ and λ, such that

‖u − uλ‖L∞(0,T ;L2(Ω)) + ‖u − uλ‖L2(0,T ;H10 (Ω)) ≤ C

[1λk + ε(λε1/k

+ 1)

]1/2

, (4.10)

where λ, k and ε are the parameter used in (3.2).

Proof. We follow the notation used in the proof of Lemma 4.1. The function u − uλ can be decomposed as

u − uλ = u − u∗− (uλ − u∗)

= u − u∗+ [uλ − u∗

]− − [uλ − u∗]+

= rλ − φ, (4.11)

where

[uλ − u∗]− = − minuλ − u∗, 0 and rλ = u − u∗

+ [uλ − u∗]−. (4.12)

It is clear that([φ + ε]α − εα, [uλ − u∗

]−

)=

∫Ω

(([uλ − u∗

]+ + ε)α − εα)[uλ − u∗

]−dx ≡ 0, α ≥ 0. (4.13)

Let us consider the estimation of rλ, as the estimates for φ have been considered in Lemma 4.1. Setting v = u − rλ

in (2.1) and v = rλ in (3.2) respectively, we have(∂u

∂t, −rλ

)+ B(u, −rλ) + (G(u), −rλ)

≥ ( f, −rλ),

(∂uλ

∂t, rλ

)+ B(uλ, rλ) + (G(uλ), rλ) + λ([φ(uλ) + ε]1/k, rλ)

= ( f + λε1/k, rλ).

Adding up the above inequality and equality gives(∂(uλ − u)

∂t, rλ

)+ B(uλ − u, rλ) + (G(uλ) − G(u), rλ) + λ([φ(uλ) + ε]1/k, rλ) ≥ λε1/k(1, rλ). (4.14)

Using (4.12), we have

((φ + ε)1/k− ε1/k, rλ) = ((φ + ε)1/k

− ε1/k, u − u∗+ [uλ − u∗

]−)

= ((φ + ε)1/k− ε1/k, u − u∗) + ((φ + ε)1/k

− ε1/k, [uλ − u∗]−)︸ ︷︷ ︸

=0 by (4.13)

= ((φ + ε)1/k− ε1/k, u − u∗)

≤ 0, (4.15)

because (φ + ε)1/k− ε1/k > 0 and u − u∗

≤ 0 by (1.2). Also, from (4.11) we have

(G(uλ) − G(u), rλ) = (G(uλ) − G(u), u − uλ) + (G(uλ) − G(u), φ(uλ))

≤ (G(uλ) − G(u), φ(uλ)), (4.16)

S. Wang, C.-S. Huang / Nonlinear Analysis 69 (2008) 1125–1137 1133

because G is monotone. Therefore, multiplying (4.14) by −1 and using (4.15) and (4.16), we get(∂(u − uλ)

∂t, rλ

)+ B(u − uλ, rλ) + (G(u) − G(uλ), φ(uλ)) ≤ 0.

Using (4.11), we see that the above inequality becomes(∂rλ

∂t, rλ

)+ B(rλ, rλ) ≤

(−

∂φ

∂t, rλ

)+ B(φ, rλ) − (G(u) − G(uλ), φ(uλ)).

Integrating both sides of the above from τ = 0 to τ = t and using (4.5) and (4.11), Assumption A2 andCauchy–Schwarz inequality, we obtain

12(rλ(t), rλ(t)) +

∫ t

0B(rλ(τ ), rλ(τ ))dτ ≤

∫ t

0

(∂φ(uλ(τ ))

∂τ, rλ(τ )

)dτ +

∫ t

0B(φ(uλ(τ )), rλ(τ ))dτ

−

∫ t

0(G(u(τ )) − G(uλ(τ )), φ(uλ(τ )))dτ

≤ (φ(uλ(t)), rλ(t)) +

∫ t

0

(φ(uλ(τ )),

∂rλ(τ )

∂τ

)dτ +

∫ t

0B(φ(uλ(τ )), rλ(τ ))dτ

+ C∫ t

0‖u(τ ) − uλ(τ )‖0‖φ(uλ(τ ))‖0dτ

≤ ‖φ(uλ)‖L∞(0,T ;L2(Ω)) · ‖rλ‖L∞(0,T ;L2(Ω)) + C‖φ(uλ)‖L2(0,T ;H10 (Ω)) · ‖rλ‖L2(0,T ;H1

0 (Ω))

+ C(‖φ(uλ)‖L2(Q) + ‖rλ‖L2(Q)

)‖φ(uλ)‖L2(Q) +

∫ t

0

(φ(uλ),

∂rλ(τ )

∂τ

)dτ (4.17)

for all t ∈ (0, T ).Now, we consider the last term in (4.17). Let

Ω+= x ∈ Ω : uλ − u∗ > 0 a.e. and Ω−

= x ∈ Ω : uλ − u∗ < 0 a.e..

Then, we have φ(uλ) = 0 on Ω− and [uλ − u∗]− = 0 a.e. on Ω+. The latter also implies that ∂

∂t [uλ − u∗]− = 0 a.e.

on Ω+. From this we see that∫Ω

φ(uλ(x, t)) ·∂

∂t[uλ(x, t) − u∗(x, t)]−dx = 0

for t ∈ (0, T ) a.e. Therefore, using this equality, (4.12) and (4.1) we have∫ t

0

(φ(uλ(τ )),

∂rλ(τ )

∂τ

)dτ =

∫ t

0

(φ(uλ(τ )),

∂(u(τ ) − u∗(τ ))

∂τ

)dτ

≤ C‖φ‖L p(Ω)

(∥∥∥∥∂u

∂t

∥∥∥∥Lq (Q)

+

∥∥∥∥∂u∗

∂t

∥∥∥∥Lq (Q)

)

≤ C

(1λk + ε

),

where p = 1 + 1/k and q = k + 1 (so that 1/p + 1/q = 1). Substituting the above upper bound into (4.17) and using(2.6) and (4.2), we obtain(

‖rλ‖L∞(0,T ;L2(Ω)) + ‖rλ‖L2(0,T ;H10 (Ω))

)2≤ C

(12‖rλ‖

2L∞(0,T ;L2(Ω))

+ ‖rλ‖2L2(0,T ;H1

0 (Ω))

)≤ C

[(‖φ‖L∞(0,T ;L2(Ω)) + ‖φ‖L2(0,T ;H1

0 (Ω))

)·

(‖rλ‖L∞(0,T ;L2(Ω)) + ‖rλ‖L2(0,T ;H1

0 (Ω))

)+ ‖φ‖

2L2(Q)

+ λ−k+ ε

]≤ C

[(λ−k

+ λε p+ ε

)1/2 (‖rλ‖L∞(0,T ;L2(Ω)) + ‖rλ‖L2(0,T ;H1

0 (Ω))

)+ (λ−k

+ λε p+ ε)

]. (4.18)

1134 S. Wang, C.-S. Huang / Nonlinear Analysis 69 (2008) 1125–1137

This is of the form

y2≤ Cρ1/2 y + Cρ

that can be rewritten as(y −

12

Cρ1/2)2

≤

(C +

C2

4

)ρ.

Clearly, this implies that

y ≤ Cρ1/2.

(Note that C is a generic positive constant.) Replacing y and ρ with

‖rλ‖L∞(0,T ;L2(I )) + ‖rλ‖L2(0,T ;H10 (I )) and λ−k

+ λε p+ ε

respectively, we have from (4.18) that

‖rλ‖L∞(0,T ;L2(I )) + ‖rλ‖L2(0,T ;H10 (I )) ≤ C

[1λk + ε(λε1/k

+ 1)

]1/2

.

(Recall p = 1 + 1/k.) Finally, using (4.11), the triangle inequality, and (4.2), we have from the above inequality theestimate (4.10). This finishes the proof.

We comment that the assumption that ∂u∂t ∈ Lk+1(Ω) in Theorem 4.1 is not restrictive in practice. In fact, any

function which has a bounded partial derivative with respect to t in Ω satisfies this condition. A function v with asingular t-derivative, ∂v/∂t , is also in Lk+1(Ω) as long as

∫Ω |

∂v∂t |

k+1 < ∞.We also comment that all of the above analysis in the previous sections hold in the limiting case that ε → 0+.

Therefore, in the case we have the following corollary.

Corollary 4.1. Let the assumptions in Theorem 4.1 be fulfilled. If λε1/k≤ O(1), then, there exists a constant C > 0,

independent of λ and ε, such that

‖u − uλ‖L∞(0,T ;L2(Ω)) + ‖u − uλ‖L2(0,T ;H10 (Ω)) ≤

C

λk/2 . (4.19)

Proof. Let C be a generic positive constant, independent of λ and ε. When λε1/k≤ O(1), we have ε ≤ Cλ−k .

Estimate (4.19) then follows from this inequality and (4.10).

Remark 4.1. What Corollary 4.1 tells us is that the value of ε in (4.10) does not affect the overall convergence rateof order λ−k/2 as long as λε1/k

≤ M for a positive constant M . In the limiting case that ε → 0, the convergence rateof uλ in the norm is still of order λ−k/2. However, the derivative of the penalty function becomes unbounded when[uλ − u∗

]+ → 0+. Therefore, in computation, we need to smooth out the penalty function locally, as discussed in [7].

5. Numerical experiments

In this section we verify numerically the theoretical findings in the previous sections and demonstrate the efficiencyand usefulness of the power penalty methods by solving the following model test problems on Ω × (0, T ], whereΩ = [0, 1] × [0, 1] and T = 1. For all the tests below, we choose u(x, t) = 0 for x ∈ ∂Ω × (0, T ] and u0(x) = 0 onΩ . The right-hand side function f is set to be

f (x, y, t) = m · exp(

−0.05

(x − 1)2x2

)· exp

(−0.05

(y − 1)2 y2

)· exp

(−0.02

t

)(5.1)

for a constant m > 0. Clearly, f attains its maximum value at (0.5, 0.5) for any t ∈ [0, 1] and extends smoothly tozero on ∂Ω × (0, T ] and Ω × 0. Note that with this f , the solution u to the unconstrained PDE behaviors intuitivelyas a growing plume in t having the maximum value at the center of Ω and extending smoothly to zero on ∂Ω at anyt ∈ (0, T ].

S. Wang, C.-S. Huang / Nonlinear Analysis 69 (2008) 1125–1137 1135

Fig. 5.1. Computed solution for Example 1.

Table 5.1Errors for k = 2

λ Error Ratio

5 1.80e−310 9.50e−3 1.8920 4.87e−3 1.95

Example 1. In this example, we choose a11 = a22 = 1, a12 = a21 = 0 and G(u) = u2. The parameter ε is chosen tobe ε = 10−8. Also, we choose u∗

= −0.05(x − 1) + 0.05 and m = 1.5 in (5.1).

To solve this problem numerically, we divide Ω into a uniform mesh with mesh size h in both the x- and they-directions. The time interval is also divided into a uniform mesh with the mesh size ∆t . Some standard finitedifference schemes are used for the discretization of (3.1). More specifically, the central difference discretization isapplied to the operator T in (1.5) and Crank–Nicolson time-stepping scheme is used for the time derivative term in(1.5). Note that the resulting discretized equation is a system of nonlinear algebraic equations due to the nonlinearityof the G term in (1.5) and λ[φ(uλ(x, t)) + ε]1/k term in (3.1). We apply the classical Newton method with dampingto this nonlinear system. The damping parameter and stopping criterion for the Newton’s method are chosen to be 0.1and 10−10 respectively.

To visualize the solution to the problem, we plot, in Fig. 5.1, the numerical solution uλ with λ = 100 and k = 4on the mesh with h = 1/20 = ∆t at t = 1, with and without the constraint u∗. It is easy to see that the solution uλ isconstrained by u∗.

Let us now examine the convergence rate (4.19) of the proposed power penalty scheme. A major difficulty inaccurately estimating the numerical rates of convergence in either λ or k is that the exact solutions u and uλ usedin (4.19) are both unknown for Example 1. Therefore, numerical solutions from the above discretization need to beused as the reference (or ‘exact’) solutions which involve discretization errors. These discretization errors sometimesdominate the errors from the penalty approach, as can be seen later in this section. In what follows, we shall use, fora fixed mesh, the computed solution with λ = 100 and k = 8 as the approximated exact solution ue,h,∆t .

Table 5.1 is a list of the computed errors ‖ue,h,∆t −uλ,k,h,∆t‖L2(0,T ;H10 (Ω)) for various values of λ and for k = 2 on

the uniform mesh with h = 1/100 = ∆t . From the table we see that the ratios of two consecutive errors are close to2 which is the theoretical convergence rate in (4.19) when k = 2 and the value of λ is progressively doubled. (Recallthat the discretization error needs to be factored in.)

To further test the convergence in λ of (4.19), we also solve the problem on the same mesh for k = 3 and thecomputed errors are listed in Table 5.2. The theoretical rate is 23/2

∼ 2.82 when the value of λ is progressively

1136 S. Wang, C.-S. Huang / Nonlinear Analysis 69 (2008) 1125–1137

Table 5.2Errors for k = 3

λ Error Ratio

2 1.41e−34 6.14e−4 2.308 3.48e−4 1.76

Table 5.3Errors for λ = 10

k Error Ratio

2 9.50e−43 3.06e−4 3.104 9.95e−4 3.07

Fig. 5.2. Computed solution for Example 2.

doubled. However, the computed rates are only 2.30 and 1.76. This is mainly because the discretization errordominates the total one when k is large.

In order to study the k-convergence, we keep λ = 10 fixed but calculate the ratios of the errors in the numericalsolutions for a sequence of k. These computed errors and their ratios are listed in Table 5.3. From (4.19) we seethat when λ = 10 if the difference between two consecutive values of k is 1, the theoretical ratio of the errors is√

10 ∼ 3.16. Thus, our computed ratios are close to this theoretical value.

Example 2. The functions and coefficients in (3.1) are chosen to be a11 = 3 + sin(2πy), a22 = 3 + sin(2πx),a12 = a21 = 0, G(u) = u2, u∗

= 0.2(x + 2 sin(2πx)) + 0.3, and f is given in (5.1) with m = 500. The parametersare ε = 10−8, λ = 3000, and k = 8.

This problem is solved using the same discretization scheme as the one for Example 1 on a uniform partition withh = 1/20 = ∆t . In Fig. 5.2, we plot the computed uλ at t = 1 with and without u∗. It is easy to see that the functionuλ is constrained by u∗.

6. Conclusion

In this paper we proposed and analyzed a nonlinear penalty method for the solution of the nonlinear, time-dependentcomplementarity problem. The method is to formulate the corresponding variational inequality as a nonlinear PDE.

S. Wang, C.-S. Huang / Nonlinear Analysis 69 (2008) 1125–1137 1137

We have shown that the solution to the PDE converges to that of the nonlinear variational inequality at an arbitraryrate depending on the choice of parameters in the penalty term. Numerical experiments were performed to confirmthe theoretical results, and the numerical results show that the computed convergence rates in both λ and k are inreasonably agreement with the theoretical ones.

Acknowledgments

We would like to thank one of the anonymous referees for several suggestions for the improvement of this paper.This work is partially supported by the Discovery Grant No. DP0344330 from the Australian Research Council. Partof the work was developed when SW was visiting the National Sun Yat-sen University, and the support from the NSFof Taiwan is gratefully acknowledged.

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