A two-gap contact model changes its active set with the load

The vectors $(1,1{)}^T$ and $(1,-1{)}^T$ are eigenvectors of $A$ with eigenvalues $1$ and $3$, so $A$ is symmetric positive definite. Also $A^{-1}=\frac{1}{3}\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$. For $f^{(1)}=(1,1\mathrm{/}5{)}^T$, the unconstrained solution is $g=A^{-1}{f}^{(1)}=(11\mathrm{/}15,7\mathrm{/}15{)}^T$, which has positive entries, and $Ag-f^{(1)}=0$. Thus every VI test has zero residual and the complementarity equations hold with $\rho =0$. The scaling between ${\rho }_i$ and the mortar coefficient ${\lambda }_i$ agrees with [F1].

For $f^{(2)}=(1,-1{)}^T$, take $g=(1\mathrm{/}2,0{)}^T$ and compute $Ag=(1,-1\mathrm{/}2{)}^T$, so $Ag-f^{(2)}=(0,1\mathrm{/}2{)}^T=\rho$. Both vectors are componentwise nonnegative and ${\rho }_1{g}_1=0={\rho }_2{g}_2$. For every $z\in {\mathbb{R}}_{+}^2$, $\rho \cdot (z-g)={\rho }_2{z}_2\ge 0$, so this is the VI solution with exactly the second gap active.

Let $e_1,e_2$ be the standard coordinate vectors and write $r:=Ag-f$. If $g\ge 0$, $r\ge 0$, and $r_i{g}_i=0$, then for every $z\ge 0$, $r\cdot (z-g)={\sum }_i{r}_i{z}_i-{\sum }_i{r}_i{g}_i={\sum }_i{r}_i{z}_i\ge 0$, so the complementarity equations imply the VI. Conversely, suppose $g$ solves the VI. If $g_i>0$, choose $0<t<g_i$ and test with $z=g+t{e}_i$ and $z=g-t{e}_i$; the two inequalities give $r_i\ge 0$ and $r_i\le 0$, hence $r_i=0$. If $g_i=0$, test with $z=g+t{e}_i$ for any $t>0$ to get $r_i\ge 0$. Thus $r\ge 0$ and $r_i{g}_i=0$ for both coordinates, proving equivalence. If $g,\tilde{g}$ both solve the VI for the same load, testing with each other and adding gives $-(g-\tilde{g}{)}^{T}A(g-\tilde{g})\ge 0$. The eigenvalues computed in step 1.1 are positive, so $A$ is positive definite and this forces $g=\tilde{g}$; then $\rho =Ag-f$ is also unique. Thus the displayed complementarity solutions are the unique ones for their respective loads.