Solving quasi-variational inequalities via their KKT conditions

Abstract

We propose to solve a general quasi-variational inequality by using its Karush–Kuhn–Tucker conditions. To this end we use a globally convergent algorithm based on a potential reduction approach. We establish global convergence results for many interesting instances of quasi-variational inequalities, vastly broadening the class of problems that can be solved with theoretical guarantees. Our numerical testings are very promising and show the practical viability of the approach.

A Appendix on monotonicity and Lipschitz properties

In this appendix we recall some well-known definitions and discuss some related results. Although the latter are also mostly well-known, in some cases we could not find in the literature the exact versions we needed. Therefore, for completeness we also report the proofs of these results.

We begin by recalling the definitions of several classes of functions.

Definition 2

Let $D \subseteq {\mathbb{R }}^n$ and $F: D \rightarrow {\mathbb{R }}^n$ be a given function. Then

• (a) $F$ is strongly monotone on $D$ with constant $\sigma$ if $\sigma >0$ and

  $$\begin{aligned} (x - y)^{\scriptscriptstyle T}\big ( F(x) - F(y) \big ) \,\ge \, \sigma \Vert x - y\Vert ^2, \quad \forall x, y \in D; \end{aligned}$$

  The largest $\sigma$ for which such a relation holds is termed the monotonicity modulus of $F$ on $D$:

  $$\begin{aligned} \sigma (D,F) \, :=\, \inf _{x\ne y, x,y \in D}\frac{(x - y)^{\scriptscriptstyle T}\big ( F(x) - F(y) \big )}{\Vert x-y\Vert ^2}. \end{aligned}$$

• (b) $F$ is co-coercive on $D$ with constant $\xi$ if $\xi >0$ and

  $$\begin{aligned} (x - y)^{\scriptscriptstyle T}\big ( F(x) - F(y) \big ) \,\ge \, \xi \Vert F(x) - F(y)\Vert ^2, \quad \forall x, y \in D; \end{aligned}$$

• (c) $F$ is Lipschitz continuous on $D$ with constant $L \ge 0$ if

  $$\begin{aligned} \Vert F(x) - F(y)\Vert \, \le \, L \Vert x - y\Vert , \quad \forall x, y \in D. \end{aligned}$$

  The smallest $L$ for which such relation holds is termed the Lipschitz modulus of $F$ on $D$:

  $$\begin{aligned} L(D,F) \, :=\, \sup _{x\ne y, x,y \in D} \frac{\Vert F(x) - F(y)\Vert }{\Vert x-y\Vert }. \end{aligned}$$

• (d) $F$ is a homeomorphism of $D$ onto $F(D)$ if $F$ is one-to-one on $D$ (that is $F(x) \ne F(y)$ whenever $x,y \in D, x \ne y$, or, in other words, $F$ has a single-valued inverse $F^{-1}$ defined on $F(D)$), and $F$ and $F^{-1}$ are continuous on $D$ and $F(D)$, respectively. $\square$

Characterizations of the Lipschitz and strong monotonicity moduli are given in the following result.

Proposition 3

Let $D \subseteq {\mathbb{R }}^n$ be an open, convex subset of ${\mathbb{R }}^n$ and let $F: D \rightarrow {\mathbb{R }}^n$ be a continuously differentiable function. Then the following statements hold:

• (a) $F$ is Lipschitz continuous on $D$ with constant $L$ if and only if $\Vert J\!F(x)\Vert \le L$ for all $x \in D$; consequently

  $$\begin{aligned} L(D,F) \, =\, \sup _{x\in D}\Vert J\!F(x)\Vert , \end{aligned}$$

  provided the $\sup$ on the right hand side is finite.

• (b) $F$ is strongly monotone on $D$ with constant $\sigma$ if and only if $h^{\scriptscriptstyle T}J\!F(x) h \ge \sigma \Vert h\Vert ^2$ for all $x \in D , \, h \in {\mathbb{R }}^n$; consequently

  $$\begin{aligned} \sigma (D,F) \, =\, \inf _{x\in D}\mu _m^s(J\!F(x)), \end{aligned}$$

  provided the $\inf$ on the right hand side is positive.

Proof

• (a) From Theorem 3.2.3 in [37], if $\Vert J\!F(x)\Vert \le L$ then $L$ is a Lipschitz constant for $F$ on $D$. Conversely, assume that

  $$\begin{aligned} \big \Vert F(x) - F(y) \big \Vert \le L \Vert x - y \Vert , \quad \forall x, y \in D \end{aligned}$$

  (35)

  holds. Applying the differential mean value theorem to each component function $F_i$ of $F$, it follows that, for any given $x, y \in D$, we can find suitable points $\xi ^{(i)} \in (x,y)$ such that

  $$\begin{aligned} F_i(x) - F_i (y) = \nabla F_i (\xi ^{(i)})^T (x-y) \quad \forall i = 1, \ldots , n. \end{aligned}$$

  Setting

  $$\begin{aligned} G(\xi ) := \left( \begin{array}{ccc}&\nabla F_1 (\xi ^{(1)})^T&\\&\vdots&\\&\nabla F_n (\xi ^{(n)})^T&\end{array} \right) \in {\mathbb{R }}^{n \times n}, \end{aligned}$$

  this can be rewritten in a compact way as

  $$\begin{aligned} F(x) - F(y) = G(\xi ) (x-y). \end{aligned}$$

  (36)

  Now, let $x \in D$ be fixed, and note that

  $$\begin{aligned} G( \xi ) \rightarrow J\!F (x) \end{aligned}$$

  (37)

  for any sequence $y \rightarrow x$ in view of the continuous differentiability of $F$. We now consider a particular sequence $y = x + td$ with a fixed (but arbitrary) vector $d \in {\mathbb{R }}^n \setminus \{ 0 \}$ and a sequence $t \downarrow 0$. Then (35) and (36) together imply

  $$\begin{aligned} \big \Vert G(\xi ) t d \big \Vert = \big \Vert F(x) - F(x + td) \big \Vert \le L \Vert t d \Vert . \end{aligned}$$

  Dividing by $t$ and subsequently letting $t \downarrow 0$ (note that $\xi$ still depends on $t$), we obtain

  $$\begin{aligned} \big \Vert J\!F(x) d \big \Vert \le L \Vert d \Vert \end{aligned}$$

  in view of (37). Since $d$ was taken arbitrarily, this implies $\Vert J\!F(x) \Vert \le L$, and this inequality is true for any vector $x \in D$.

• (b) See [37, Theorem 5.4.3]. $\square$

The following result gives a relation between the Lipschitz constants etc. of a given mapping $F$ and its inverse $F^{-1}$.

Proposition 4

Let a function $F: D\rightarrow {\mathbb{R }}^n$ be given where $D$ is an open subset of ${\mathbb{R }}^n$. Assume that two positive constants $\ell$ and $L$ exist such that

$$\begin{aligned} \ell \Vert x -y\Vert \le \Vert F(x) - F(y) \Vert \le L \Vert x-y\Vert , \quad \forall x, y \in D. \end{aligned}$$

(38)

Then $F$ is a homeomorphism from $D$ to $F(D)$ (which is an open set) and

$$\begin{aligned} \frac{1}{L}\Vert a-b\Vert \le \Vert F^{-1}(a) - F^{-1}(b) \Vert \le \frac{1}{\ell } \Vert a-b\Vert , \quad \forall a, b \in F(D), \end{aligned}$$

(39)

in particular, $F$ and $F^{-1}$ are Lipschitz continuous on $D$ and $F(D)$, respectively.

Proof

The first inequality from (38) implies that $F$ is one-to-one on $D$ (therefore the inverse $F^{-1}$ exists on $F(D)$), and that, setting $a = F (x)$ and $b = F (y)$, the second inequality in (39) holds. In particular, this implies that $F^{-1}$ is Lipschitz continuous on $F(D)$, hence continuous, so that $F(D) = (F^{-1})^{-1} (D)$, being the pre-image of a continuous map of the open set $D$, is also an open set. Finally, let $a, b \in F(D)$ be arbitrarily given. Setting $x = F^{-1} (a),y = F^{-1} (b)$, we obtain from the second inequality in (38) that

$$\begin{aligned} \Vert F^{-1} (a) - F^{-1} (b) \Vert = \Vert x - y \Vert \ge \frac{1}{L} \Vert F(x) - F(y) \Vert = \frac{1}{L} \Vert a - b \Vert , \end{aligned}$$

and this completes the proof. $\square$

The next result considers a strongly monotone and Lipschitz continuous mapping and provides suitable bounds for the moduli of Lipschitz continuity and strong monotonicity of the corresponding inverse function. We stress, however, that the constant of strong monotonicity of the inverse function provided by this result is really just an estimate and typically not exact. It seems difficult to find a stronger bound in the general context discussed here. In a more specialized situation, much better results can be obtained, see Proposition 6 below.

Proposition 5

Let $D \subseteq {\mathbb{R }}^n$ be an open set and $F: D \rightarrow {\mathbb{R }}^n$ be strongly monotone with modulus $\sigma$ and Lipschitz continuous with modulus $L$ on $D$. Then $F$ is co-coercive with constant $\frac{\sigma }{L^2}$. Furthermore, it holds that the inverse $F^{-1}$ exists on $F(D)$, is Lipschitz with constant $\frac{1}{\sigma }$ and strongly monotone with constant $\frac{\sigma }{L^2}$.

Proof

We can write

$$\begin{aligned} \big ( F(x) - F(y) \big )^{\scriptscriptstyle T}(x-y) \ge \sigma \Vert x - y \Vert ^2, \quad \forall x, y \in D \end{aligned}$$

and

$$\begin{aligned} \Vert F(x) - F(y) \Vert ^2 \le L^2 \Vert x - y \Vert ^2, \quad \forall x, y \in D \end{aligned}$$

by assumption. A combination of these two inequalities yields

$$\begin{aligned} \Vert F(x) - F(y) \Vert ^2 \le L^2 \Vert x \!-\! y \Vert ^2 \le \frac{L^2}{\sigma } \big ( F(x) \!-\! F(y) \big )^{\scriptscriptstyle T}(x-y), \quad \forall x, y \in D.\qquad \end{aligned}$$

(40)

Hence $F$ is co-coercive with constant $\frac{\sigma }{L^2}$.

By Proposition 4 we know that $F$ is a homeomorphism from $D$ to $F(D)$ and $F^{-1}$ is Lipschitz continuous with constant $\frac{1}{\sigma }$. Finally writing $a=F(x), b=F(y)$ in (40) gives

$$\begin{aligned} \big ( F^{-1}(a) - F^{-1}(b) \big )^{\scriptscriptstyle T}(a-b) \ge \frac{\sigma }{L^2} \Vert a - b \Vert ^2, \quad \forall a, b \in F(D). \end{aligned}$$

This completes the proof. $\square$

The following result gives an exact estimate of the Lipschitz and strong monotonicity moduli of the inverse of a function under the assumption that the mapping $F$ itself is a gradient mapping, i.e. that $F = \nabla f$ for a differentiable real-valued function $f$.

Proposition 6

Let $D \subseteq {\mathbb{R }}^n$ be an open convex set and $F: D\rightarrow {\mathbb{R }}^n$ be a gradient mapping. Assume that $F$ is strongly monotone with modulus $\sigma$ and Lipschitz continuous with modulus $L$ on $D$. Then the inverse function $F^{-1}$ exists on $F(D)$, is Lipschitz with modulus $\frac{1}{\sigma }$ and strongly monotone with modulus $\frac{1}{L}$.

Proof

The result can easily be derived from the Baillon–Haddad Theorem, see [1], when $D={\mathbb{R }}^n$. We give here a direct proof which is valid also when $D \ne {\mathbb{R }}^n$. In view of Proposition 5, we only have to verify the statement that $F^{-1}$ is strongly monotone with constant $\frac{1}{L}$.

To this end, first consider a symmetric positive definite matrix $A \in {\mathbb{R }}^{n \times n}$, let $A^{1/2}$ be the corresponding (unique) symmetric positive definite square root of $A$ so that $A^{1/2} A^{1/2} = A$, and let $A^{-1/2}$ be the inverse of $A^{1/2}$. Then the symmetry of $A^{1/2}$ together with the Cauchy–Schwarz inequality implies

$$\begin{aligned} \Vert d \Vert ^2 = d^{\scriptscriptstyle T}d = d^{\scriptscriptstyle T}A^{1/2} A^{-1/2} d \le \Vert A^{1/2} d \Vert \cdot \Vert A^{-1/2} d \Vert , \quad \forall d \in {\mathbb{R }}^n. \end{aligned}$$

Squaring both sides shows that

$$\begin{aligned} \Vert d \Vert ^4 \le \big ( d^{\scriptscriptstyle T}A d \big ) \big ( d^{\scriptscriptstyle T}A^{-1} d \big ), \quad \forall d \in {\mathbb{R }}^n \end{aligned}$$

(41)

holds. Since $F$ is strongly monotone, the Jacobian $J\!F(x)$ is positive definite for all $x \in D$; furthermore, since $F$ is a gradient mapping, this Jacobian is also symmetric. Hence we can apply inequality (41) to the matrix $A := J\!F(x)$ and obtain

$$\begin{aligned} \Vert d \Vert ^4&\le \big ( d^{\scriptscriptstyle T}J\!F(x) d \big ) \big ( d^{\scriptscriptstyle T}J\!F(x)^{-1} d \big ) \\&\le \big ( d^{\scriptscriptstyle T}J\!F(x)^{-1} d \big ) \Vert d \Vert ^2 \Vert J\!F(x) \Vert \\&\le \big ( d^{\scriptscriptstyle T}J\!F(x)^{-1} d \big ) L \Vert d \Vert ^2, \qquad \forall d \in {\mathbb{R }}^n, \end{aligned}$$

where the second inequality uses the Cauchy–Schwarz inequality once again, and the third inequality takes into account Proposition 3. This implies

$$\begin{aligned} \frac{1}{L} d^{\scriptscriptstyle T}d = \frac{1}{L} \Vert d \Vert ^2 \le \big ( d^{\scriptscriptstyle T}J\!F(x)^{-1} d \big ), \quad \forall d \in {\mathbb{R }}^n \forall x \in D. \end{aligned}$$

Since $J\!F(x)^{-1} = J\!F^{-1}(y)$ for $y = F(x)$ by the Inverse Function Theorem, this gives

$$\begin{aligned} \frac{1}{L} d^{\scriptscriptstyle T}d \le d^{\scriptscriptstyle T}J\!F^{-1}(y) d, \quad \forall d \in {\mathbb{R }}^n \forall y \in F(D). \end{aligned}$$

By a well-known result, see [37, Theorem 5.4.3] this is equivalent to saying that $F^{-1}$ is strongly monotone on $F(D)$ with constant $1/L$. $\square$

The sharper result from Proposition 6 regarding the modulus of strong monotonicity does, in general, not hold for non-gradient mappings, see the corresponding discussion and (counter-) example at the end of Sect. 3.2.