A new smoothing method for solving nonlinear complementarity problems

Definition 2.1

Suppose that F : ℜⁿ → ℜⁿ is locally Lipschitz function. The generalized Jacobian of F at x in the sense of Clark [42] denote by ∂F(x), then, F is said to be semismooth (or strongly semismooth) at x ∈ ℜⁿ, if F is directionally differentiable at x ∈ ℜⁿ and F(x + h) – F(x) – Vh = o(∥h∥) (or = O( ∥h∥²) ) holds for any V ∈ ∂F(x + h).

Lemma 2.1

The function Φ_τ(x) satisfies the inequality

for all x ∈ ℜⁿ and τ₁, τ₂ ≥ 0, where κ = 2n. In particular, we have

for all x ∈ ℜⁿ and τ ≥ 0.

Proof

It is obvious to hold for τ₁ = τ₂ = 0. So, we suppose at least one of the perturbation parameters is positive. By simple calculation, we can have

Obviously, for any x ∈ ℜⁿ,

where κ = 2n. □

Similar to the proof of Lemma 2.8 in [22], we have the following lemma.

Lemma 2.2

Let Φ_τ(x) be defined by (2.3). Then,

where ∂_CΦ(x) is the C-subdifferential of Φ(x), and Φτ′(x) is the Jacobian of Φ_τ(x).

Lemma 2.3

Let x ∈ ℜⁿ be arbitrary but fixed. Assume that x is not a solution of NCP. Let us define the constants

and

where β(x) := {i|x_i = F_i(x)}. Let δ > 0 be given, and define

Then

for all τ such that 0 < τ ≤ τ(x, δ).

Algorithm 3.1

Step 0 Given a starting point x⁰ ∈ ℜⁿ, choose α, σ, η, ρ, γ ∈ (0, 1), δ ∈ (0, ∞), set β₀ = ∥Φ(x₀)∥, κ = 2n,τ0=α2κβ0,μ0=∥Φτ0(x0)∥,k:=0.

1. Find a solution d^k ∈ ℜⁿ of the linear system

   Φτk′(xk)TΦτk′(xk)+μkId=−Φτk′(xk)TΦτk(xk) (3.1)

2. If the condition

   ∥Φτk(xk+dk)∥≤γ∥Φτk(xk)∥ (3.2)

   is satisfied, then set x^k+1 := x^k+d^k(we call this ‘fast step’) and go to Step 4.

3. set λ_k = ρ^m_k and x^k+1 = x^k+λ_k d^k, where m_k be the smallest nonnegative integer m such that

   Ψτk(xk+ρmdk)−Ψτk(xk)≤σρm∇Ψτk(xk)Tdk. (3.3)

4. If ∇Ψ(x^k+1) = 0, stop.

5. If

   ∥Φ(xk+1)∥≤maxηβk,α−1∥Φ(xk+1)−Φτk(xk+1)∥, (3.4)

   then set

   βk+1:=∥Φ(xk+1)∥

   and choose τ_k+1 satisfied

   0<τk+1≤minα2κβk+12,τk2,τ¯(xk+1,δβk+1), (3.5)

   where τ(⋅, ⋅) is defined in Lemma 2.4; otherwise, let β_k+1 := β_k and τ_k+1 := τ_k.

6. Set μ_k+1 = ∥Φ_τk+1(x^k+1)∥ and k := k+1. Go to Step 1.

Remark

Differently from the algorithm in [7, 15], our proposed algorithm is well-defined without the assumption that smoothing function Φ_τk(x^k) is nonsingular.

For proving that the algorithm 3.1 is well-defined and has the global convergence property, we assume that algorithm 3.1 does not terminate in a finite number of iterations.

Define the index set

It follows from Lemma 2.1 and Step 5 of the algorithm 3.1 that

which indicates ∥Φ_τk(x^k)∥ ≠ 0 for all k ∈ K. By the definition τ(⋅, ⋅) in Lemma 2.3 and the updating rule (3.5), it is not difficult to find that

holds, for any k ∈ K with k ≥ 1.

The following proposition shows the well-definiteness of the proposed algorithm.

Proposition 3.1

Suppose that x_k is the sequence generated by algorithm 3.1. Then Algorithm 3.1 is well-defined.

Proof

1. If d^k = 0. Obviously, there exists a finite nonnegative m_k such that (3.3) always holds.

2. If d^k ≠ 0. From the above analysis, we know that μ_k = ∥Φ_τk(x^k)∥ is positive, then Φτk′(xk)TΦτk′(xk)+μkI is always positive definite. Since, ∇Ψτk(xk)=Φτk′(xk)TΦτk(xk). By the construction of Algorithm 3.1, we have

   ∇Ψτk(xk)Tdk=−(dk)T(Φτk′(xk)TΦτk′(xk)+μkI)(dk)<0, (3.9)

   moreover

   Ψτk(xk+λdk)−Ψτk(xk)=λ∇Ψτk(xk)Tdk+o(λ)=σλ∇Ψτk(xk)Tdk+(1−σ)λ∇Ψτk(xk)Tdk+o(λ). (3.10)

   It follows from (3.9) and σ ∈ (0, 1) that

   (1−σ)λ∇Ψτk(xk)Tdk<0. (3.11)

   By (3.10) and (3.11), we can obtain the following inequality

   Ψτk(xk+λdk)−Ψτk(xk)≤σλ∇Ψτk(xk)Tdk.

   So, there exists a finite nonnegative integer m_k such that

   Ψτk(xk+ρmkdk)−Ψτk(xk)≤σρmk∇Ψτk(xk)Tdk.

   Therefore, Algorithm 3.1 is well-defined. □

Theorem 4.1

Suppose that {x^k} is a sequence generated by algorithm 3.1. If there exists at least an accumulation point in the sequence {x^k}, then the index set K defined by (3.6) is infinite, and

Proof

Assume that x^* is an arbitrary accumulation point of the sequence {x^k}, and {x^k}_{k ∈ K} be a subsequence converging to x^*. We first show that the index set K is infinite. By contradiction, we assume that the set K is finite. Let k̂ be the largest number in K, then for any k ≥ k̂, wehave τ_k = τ_k̂ and β_k = β_k̂.

Denote

then for all k ≥ k̂, we have

and

In the following paragraphs, we will show that the following equation holds.

Now, we consider two cases for the sequence {μ_k}:

case 1: If μ_k → 0(k ∈ K), then, we can obtain ∥Φ_τ̂(x^k)∥ → 0 (k ∈ K). It follows from the boundness of {∥Φτ^′(xk)∥} that

which implies ∇Ψ_τ̂(x^*) = 0.

case 2: In this case, there exists a constant ξ such that μ_k ≥ ξ > 0 for all k ∈ K. Next, we show that ∇Ψ_τ̂(x^*) = 0. Suppose to the contrary that ∇Ψ_τ̂(x^*) ≠ 0.

It follows from (3.1), we have

and

From the continuity of Φ_τ̂(x^k) and the upper semicontinuity of the generalized Jacobian, we deduce that for a constant δ > 0,

We now note that

In fact, if {∥d^k∥}_K → 0, by (4.5) and (4.7), we can deduce {∥Φτ^′(xk)TΦτ^(xk)∥}K→0, thus contradicting the assumption ∇Ψ_τ̂(x^*) ≠ 0. On the other hand, we denote the bound of {∥Φτ^′(xk)TΦτ^(xk)∥} as M, it follows from (4.6) and (4.8) that

Since μ_k = ∥Φ_τ̂(x^k)∥ > 0, for all k ∈ K, Φτ^′ (x^k)^TΦ_τ̂(x^k)+μ_kI is positive definite, this together with (3.1) and (4.9), implies that

Next, we consider two cases for the sequence {inf_kλ_k}:

1. If inf_{k ∈ K}λ_k = λ^* > 0 for all k ≥ k̂. By (3.3) and (4.10), we can obtain that

   Ψτ^(xk+1)−Ψτ^(xk)≤σλk∇Ψτ^(xk)Tdk≤σλ∗∇Ψτ^(xk)Tdk<0. (4.11)

   Using Ψ_τ̂(x^k+1) − Ψ_τ̂(x^k) → 0 yields that

   {∇Ψτ^(xk)Tdk}K→0. (4.12)

   By (3.1), we have

   ∇Ψτ^(xk)Tdk=−∇Ψτ^(xk)T(Φτ^′(xk)TΦτ^′(xk)+μkI)−1∇Ψτ^(xk)  ∀k. (4.13)

   Since ∂Φ_τ̂(x) is a nonempty compact set for any x ∈ ℜⁿ, {Φτ^′}K is bounded. Without loss of generality, assume that {Φτ^′}K →Φ_*. Considering that the set-valued mapping x → ∂Φ_τ̂(x) is closed and {x^k}_K → x^*, we have Φ_* ∈ ∂Φ_τ̂(x^*). In addition, since Φ_τ̂(x^*) ≠ 0, we have μ_k → μ_* with μ_* = ∥Φ_τ̂(x^*)∥ > 0. Thus, {Φτ^′(xk)TΦτ^′(xk)+μkI}K→Φ∗TΦ∗+μ∗I≻0. This together with (4.13) and the continuity of ∇Ψ_τ̂, implies that ∇Ψ_τ̂(x^k)^T d^k has a nonzero limit as k → + ∞, which contradicts (4.12). Hence ∇Ψ_τ̂(x^*) = 0.

2. If inf_{k ∈ K}λ_k = 0. In this case, without loss of generality, we assume that {λ_k}_{k ∈ K} → 0. By (4.9), there exists a subsequence {d^k}_{k ∈ K} → d ≠ 0.

   From the line search rule (3.3), it is clear that for all k ≥ k̂,

   Ψτ^(xk+ρmk−1dk)−Ψτ^(xk)ρmk−1>σ∇Ψτ^(xk)Tdk. (4.14)

   On K, by taking the limit k → ∞, we obtain from (4.14),

   ∇Ψτ^(x∗)Td¯≥σ∇Ψτ^(x∗)Td¯. (4.15)

   Since, σ ∈ (0, 1), then, by (4.15) we have ∇Ψ_τ̂(x^*)^T d ≥ 0. On the other hand, d is the solution of (Φ∗TΦ∗+μ∗I)d¯=−∇Ψτ^(x∗), which implies ∇Ψ_τ̂(x^*)^Td < 0, (since (Φ∗TΦ∗+μ∗I)≻0.) Thus, we get a contradiction. Then, ∇Ψ_τ̂(x^*) = 0.

   To sum up by case (1) and case (2), we have ∇Ψ_τ̂(x^*) = 0.

   On the other hand, in view of Lemma 2.1 and (4.3), we have Ψ_τ̂(x^k) → Ψ_τ̂(x^*) = 0, then, there exists k͠ ≥ k̂ such that for all k ∈ L with k ≥ k͠,

   ∥Φτ^(xk)∥≤(1−α)ηβ^.

   In view of (4.1) and (4.2), we deduce that for all k ∈ K with k ≥ k͠,

   ∥Φτ^(xk)∥≤(1−α)∥Φ(xk)∥≤(1−α)(∥Φτ^(xk)∥+∥q(xk)∥),

   i.e.,

   ∥Φτ^(xk)∥≤(1−1α)∥q(xk)∥.

   Therefore,

   ∥Φ(xk)∥≤∥Φτ^(xk)∥+∥q(xk)∥<∥q(xk)∥α

   which in contradiction to (4.1). Hence the set K is infinite.

   By the updating rule of τ_k, we have {τ_k} → 0. Similar to the proof of Proposition 4.2 in [8], we deduce that

   ∥Φ(xk)∥≤rj(1+α)∥Φ(x0)∥, as kj≤k<kj+1. (4.16)

   where r=max{12,η}.

   Since the set K is infinite, it follows from (3.8) and (4.16) that

   limk→∞Φτk(xk)=0, limk→∞Φ(xk)=0.

   □

   As a consequence of Theorem 4.1, we can obtain the following global convergence result.

Theorem 4.2

Let {x^k} be a sequence generated by Algorithm 3.1, then every accumulation point of the sequence {x^k} is a solution of the Problem(1.1).

In order to obtain the local convergent result, we introduce the following lemma.

Lemma 4.1

Let {x^k}_{k ∈ K} converge to x^*. If V ∈ ∂_CΦ(x^*) is nonsigular, then there exist two constants M₁ > 0, M₂ > 0 and k̂ such that for all k ∈ K with k ≥ k̂,

Proof

By the condition, we know that the matrix V is nonsingular. Notice that ∂_CΦ(x) is compact for all x ∈ ℜⁿ. Therefore, there exists V_k ∈ ∂_CΦ(x^k) such that

Combining with (3.8), we can deduce that for all k ∈ K,

By theorem 4.1, we have {∥Φ(x^k)∥} → 0. This together with the compactness and the upper semicontinuity of ∂_CΦ(x^*), we obtain from the above inequality that the matrices Φτk′(xk) and Φτk′(xk)TΦτk′(xk) are nonsingular and there exist M₁ > 0, M₂ > 0 such that (4.17) holds. □

The following lemma can be seen in Ref. [44].

Lemma 4.2

Assume that A, B ∈ ℜ^{n × n} and A is nonsingular. If ∥A⁻¹B∥ < 1, then A+B is nonsingular and satisfies

From Lemmas 4.1 and 4.2, we can prove that the following lemmas hold.

Lemma 4.3

Assume that the conditions of Lemma 4.1 hold. Then, for μk≤12M2,

Theorem 4.3

Assume that {x^k} is a sequence generated by Algorithm 3.1. If x^* is an accumulation point of the sequence {x^k}, and for all V ∈ ∂_CΦ(x^*) are nonsinglar. Then the sequence {x^k} converges to x^* quadratically.

Proof

It follows from Theorem 4.1 that x^* is a solution of Φ(x) = 0. Notice that ∂_BΦ(x^*) ⊆ ∂_CΦ(x^*). From Proposition 2.5 in [45], it follows that there exists a neighbourhood of x^* such that x^* is the unique solution.

Since the sequence {x^k} has an accumulation point x^*, there exists a subsequence {x^k}_{k ∈ K} that converges to x^*. By Lemma 4.1, there exist M₁ > 0, M₂ > 0 and k ∈ K,

Further, from Lemma 2.1, (3.5) and the Lipschitz continuity of Φ(x), we deduce that

Then, it follows from (4.21) and Lemmas 4.2, 4.3 that

Therefore, for all k > k̂ and k ∈ K, we have

By Proposition 2.3 in [16], we known that Φ(x) is strongly semismooth, combining with Theorem 3.2 in [41], we have

where Vki denotes the i-th row of V_k. Hence, From (4.19)-(4.23), we can deduce

The above equation together with (4.21) indicate that

It follows from triangular inequality and (3.8) that

The above equations (4.25) and (4.26) imply that

which means that there exists k ≥ k̂ such that k ∈ K and for any k ≥ k,

That is, x^k+1 = x^k+d^k for all k ≥ k and k ∈ K. This together with (4.24) implies that {x^k} converges to x^* quadratically. □

Example 5.1

Mathiesen Problem. This test problem was used by Jiang and Qi [46] with four variables, which was also tested by Pang and Gabriel [47]. Let