Solutions of parametric complementarity problems monotone with respect to parameters

Abstract

In many applied problems (such as, e.g., elasto-hydrodynamic lubrication problem, some economic equilibrium problems, etc.), one of the important question is if certain complementarity problem’s solution is monotone with respect to parameters. Our paper investigates this question and provides several sufficient conditions that guarantee such a monotonicity of the solutions to linear and nonlinear complementarity problems with parameters. In the majority of cases, it is required that the principal mapping of the complementarity problem be monotone by decision variables and, vice versa, antitone with respect to parameters.

Appendix

1.1 Proof of Theorem 3

Proof

Let $u\ge v$. Consider the well-defined solutions x(u) and x(v) to problem (10) for the parameter vectors u and v, respectively. We will now show that $x_i(u)\le x_i(v)$ for every $i=1,\ldots ,n$. Split the set of indices into subsets: $I=I(u)=\{i:x_i(u)>0\}$ and $J=J(u)=\{1,\ldots ,n\} \setminus I$. Now it is to consider three different cases similar to the proof of Theorem 2.

• 1. Let i belong to $J=\{1,2,\ldots n\}\setminus I$, that is $x_i(u)= 0$. However, $x_i(v)\ge 0$ by definition, hence $x_i(u)\le x_i(v)$.

• 2. Now consider the subset $I=I(u)$. We will show that $x_i(u)\le x_i(v)$, too. For $i\in I$, the following relationships hold:

  $$\begin{aligned}&A_{II}x_I(u)+B_{II}u+\varphi _I(x(u),u)= 0; \end{aligned}$$

  (34)
  $$\begin{aligned}&A_{II}x_I(v)+B_{II}v+\varphi _I(x(v),v)\ge 0, \end{aligned}$$

  (35)

  where $A_{II},B_{II}$ are the square submatrices of the matrices A and B with the elements $a_{ij},b_{ij},i,j,\in I$, respectively; $x_I$ and $\varphi _I$ are the corresponding subvectors of the vectors x and $\varphi$, respectively. Subtract (34) from (35) and come to

  $$\begin{aligned}&A_{II}x_I(v)+B_{II}v+\varphi _I\left( x(v),v\right) -A_{II}x_I(u)-B_{II}u\\&\quad -\varphi _I\left( x(u),u\right) \ge 0; \end{aligned}$$

  hence

  $$\begin{aligned}&A_{II}\left[ x_I(v)-x_I(u)\right] +B_{II}(v-u)\nonumber \\&\quad +\left[ \varphi _I\left( x(v),v\right) -\varphi _I\left( x(u),u\right) \right] \ge 0. \end{aligned}$$

  (36)

Examine the last term of the left-hand side of inequality (36):

$$\begin{aligned}&\varphi _I\left( x(v),v\right) -\varphi _I\left( x(u),u\right) \\&\quad =\int \limits _0^1\varphi ^{\prime }_I\left( x(u)+t\left( x(v)-x(u) \right) ,u+t(v-u)\right) \\&\quad \quad \times \left[ x(v)-x(u),v-u\right] dt\\&\quad =\int \limits _0^1\varphi ^{\prime }_I\left( x(u)+t\left( x(v)-x(u) \right) ,u+t(v-u)\right) \,dt\\&\quad \quad \times \left[ x(v)-x(u),v-u\right] \\&\quad =\int \limits _0^1\varphi ^{\prime }_{I,x}\left( x(u)+t\left( x(v)-x(u)\right) , u+t(v-u)\right) \,dt\\&\quad \quad \times \left[ x(v)-x(u)\right] \\&\quad \quad +\int \limits _0^1\varphi ^{\prime }_{I,u}\left( x(u)+t\left( x(v)-x(u)\right) , u+t(v-u)\right) \,dt\cdot (v-u). \end{aligned}$$

For convenience purpose, introduce the following notation:

$$\begin{aligned} \varPhi ^{\prime }_{I,x}=\int \limits _0^1\varphi ^{\prime }_{I,x}\left( x(u)+t\left( x(v)-x(u)\right) ,u+t(v-u)\right) \,dt \end{aligned}$$

and

$$\begin{aligned} \varPhi ^{\prime }_{I,u}=\int \limits _0^1\varphi ^{\prime }_{I,u}\left( x(u)+t\left( x(v)-x(u)\right) ,u+t(v-u)\right) \,dt. \end{aligned}$$

Denote by T the submatrix of matrix $\varPhi ^{\prime }_{I,x}$, that consists of elements with the same subsets of indices of rows and columns: $T=\varPhi ^{\prime }_{I,x_I}$; and by H the submatrix that comprises the elements of the matrix $\varPhi ^{\prime }_{I,x}$ with the complementary subsets of indices of rows and columns, i.e., $H=\varPhi ^{\prime }_{I,x_J}$. Then

$$\begin{aligned} \varphi _I\left( x(v),v\right) -\varphi _I\left( x(u),u\right)= & {} \varPhi ^{\prime }_{I,x}\cdot \left[ x(v)-x(u)\right] + \varPhi ^{\prime }_{I,u}\cdot (v-u)\nonumber \\= & {} T\cdot \left[ x_I(v)-x_I(u)\right] +H\cdot \left[ x_J(v)-x_J(u)\right] + \varPhi ^{\prime }_{I,u}\cdot (v-u)\nonumber \\= & {} T\cdot \left[ x_I(v)-x_I(u)\right] +H\cdot x_J(v)+\varPhi ^{\prime }_{I,u}\cdot (v-u). \end{aligned}$$

(37)

By substituting (37) into (36) we get

$$\begin{aligned}&A_{II}\left[ x_I(v)-x_I(u)\right] +B_{II}(v-u)+T\cdot \left[ x_I(v)-x_I(u)\right] \\&\quad +H\cdot x_J(v)+\varPhi ^{\prime }_{I,u}\cdot (v-u)\ge 0, \end{aligned}$$

or

$$\begin{aligned}&\left( A_{II}+T\right) \cdot \left[ x_I(v)-x_I(u)\right] \ge -B_{II}(v-u)-H\cdot x_J(v)\nonumber \\&\quad - \varPhi ^{\prime }_{I,u}\cdot (v-u). \end{aligned}$$

(38)

Consider each term of the right-hand side of inequality (38) separately. If $u\ge v,$ then $-B_{II}(v-u)\ge 0$. Furthermore, as the function $\varphi$ is monotone with respect to parameters, it implies that $-\varPhi ^{\prime }_{u,I}\cdot (v-u)\ge 0$. At last, since $\varphi ^{\prime }_x$ is a positive definite M -matrix and $\varPhi ^{\prime }_{I,x_J}$ contains only non-positive non-diagonal elements, we come to $-H\cdot x_J(v)\ge 0$. Thus, we have shown the non-negativity of the right-hand side of (38); therefore

$$\begin{aligned} \left( A_{II}+T\right) \cdot \left[ x_I(v)-x_I(u)\right] \ge 0. \end{aligned}$$

(39)

First, assertion (iii) of Lemma 3 implies that the submatrix T is a positive definite M-matrix, and second, assertion (ii) of the same Lemma 3 guarantees that the sum of positive definite M-matrices $\left( A_{II}+T\right)$ is also an M-matrix. Now making use of the properties of M-matrices, left-multiply both sides of inequality (39) by the matrix $\left( A_{II}+T\right) ^{-1}$. Since all entries of the latter matrix are nonnegative, we obtain $x_I(v)-x_I(u)\ge 0$. Hence $x_I(u)\le x_I(v)$, that is, if $x_i(u)>0$ then $x_i(v)\ge x_i(u)$, which completes the proof. $\square$

1.2 Proof of Theorem 4

Proof

The proof mainly follows the steps of that of Theorem 2. In more detail, here we again consider two m-vectors of parameters u, v, $u\ge v$, and the corresponding solutions x(u) and x(v) to problem (21). We have to show that $x_i(u)\ge x_i(v)$ for each $i=1,\ldots ,n$. Similar to the proof of Theorem 2, here we partition the set of indices into the following subsets:

$$\begin{aligned} I=I(u)=\{i:x_i(u)>0\} \end{aligned}$$

and

$$\begin{aligned} J=J(u)=\{i:1,\ldots ,n\}\setminus I, \end{aligned}$$

and then consider three different cases similar to those in the proof of Theorem 2.

• 1. If $S={I(u)\cap I(v)}$, then $A_{SS}x_S+\varphi _S(u)=0$, where $A_{SS}$ is a square submatrix of the matrix A with elements $a_{ij}, \ i,j\in S$. In this case, the solution of problem (21) can be represented in the explicit from: $x_S=-A^{-1}_{SS}\varphi _S(u)$, where $A^{-1}_{SS}=\left( \alpha _{ij}\right) _{i\in S,j\in S}\ge 0$.

  Rewrite $x_i(u)$ and $x_i(v)$ as follows:

  $$\begin{aligned} x_i(u)=-\sum _{j\in S}\alpha _{ij}\varphi _j(u),\ x_i(v)=-\sum _{j\in S} \alpha _{ij}\varphi _j(v). \end{aligned}$$

  Evaluate the difference:

  $$\begin{aligned} x_i(u)-x_i(v)=-\sum _{j\in S}\alpha _{ij}\left[ \varphi _j(u)-\varphi _j(v)\right] . \end{aligned}$$

  As the function $\varphi$ is antitone, it implies $\varphi _j(u)- \varphi _j(v) \le 0$, $j\in S$. Therefore, $x_i(u)-x_i(v)\ge 0$, i.e. $x_i(u)\ge x_i(v)$.

• 2. For $i\in I(u)\setminus I(v)$, one has $x_i(u)>0,\ x_i(v)=0$, which implies $x_i(u)>x_i(v)$.

• 3. Let $i\in J(u)$, that is, $x_i(u)=0$. Now we show that $x_i(v)=0$, too, i.e., $J(u)\subseteq J(v)$.

  Consider the set $L={J(u)\cap I(v)}$. Assume that $L\ne \emptyset$. The latter implies

  $$\begin{aligned}&A_{LL}x_L(v)+\varphi _L(v)=0, \end{aligned}$$

  (40)
  $$\begin{aligned}&A_{LL}x_L(u)+\varphi _L(u)\ge 0. \end{aligned}$$

  (41)

  Rewrite (40) and (41) in the following form:

  $$\begin{aligned}&\sum _{k\in L}a_{\ell k}x_k(v)+\varphi _\ell (v)=0,\ \ \ell \in L, \end{aligned}$$

  (42)
  $$\begin{aligned}&\sum _{k\in L}a_{\ell k}x_k(u)+\varphi _\ell (u)\ge 0,\ \ell \in L, \end{aligned}$$

  (43)

  respectively.

  Since $u\ge v$ and the function $\varphi$ is antitone, the inequality $\varphi _\ell (u)\le \varphi _\ell (v)$ holds. Subtract (42) from (43) to deduce

  $$\begin{aligned} \sum _{k\in L}a_{\ell k}\left[ x_k(u)-x_k(v)\right] +\left[ \varphi _\ell (u)- \varphi _\ell (v)\right] \ge 0,\ \ell \in L. \end{aligned}$$

  (44)

  By the definition of the subset L, we have $x_k(u)=0$, $\forall k\in L$. Next, as $\varphi$ is antitone, the difference $[\varphi _\ell (u)-\varphi _\ell (v)]$ is non-positive. Hence

  $$\begin{aligned} \sum _{k\in L}a_{\ell k}x_k(v)\le \left[ \varphi _\ell (u)-\varphi _\ell (v) \right] \le 0,\ \ell \in L, \end{aligned}$$

  (45)

  which implies the following inequality

  $$\begin{aligned} A_{LL}x_L(v)\le 0. \end{aligned}$$

  (46)

Since each principal submatrix $A_{LL}$ of an M-matrix A is also an M-matrix [cf., Lemma 3, assertion (iii)], we can left-multiply both sides of (46) by the nonnegative matrix $A^{-1}_{LL} \ge 0$ and obtain the following relationship: $A^{-1}_{LL}A_{LL}x_L(v)=x_L (v)\le 0$. This inequality contradicts our assumption $x_L(v)>0$. Therefore, $x_i(u)=0$ implies $x_i(v)=0$, thus $u\ge v$ implies $x(u)\ge x(v)$, which completes the proof. $\square$