Variational inequalities and spatial contact discretization — Examples

The function $\psi$ is continuous and piecewise affine, hence belongs to $H^1(0,1)$; its endpoint values are $0$ and $-1\mathrm{/}2$, so [F1] applies. The displayed $u$ is continuous at $a_0$, has derivative $0$ from both sides there, and has endpoint values $u(0)=u(1)=0$. By [F2], its endpoint trace is zero and therefore $u\in H_0^1(0,1)$. On $[0,a_0]$, $u=\psi =0$; on $(a_0,1]$, $u\ge 0>\psi$. Thus $u\in K$, and equality $u=\psi$ holds exactly on $[0,a_0]$.

On $(a_0,1)$, direct differentiation gives $u^{\mathrm{\prime }\mathrm{\prime }}(x)=4-6x$; on $(0,a_0)$, $u^{\mathrm{\prime }\mathrm{\prime }}=0$. The derivative $u^{\mathrm{\prime }}$ is continuous at $a_0$, so its distributional derivative is the piecewise function just computed, with no point mass at $a_0$. For $z=v-u\in H_0^1(0,1)$, [F6] supplies a sequence $z_m\in C_c^{\mathrm{\infty }}(0,1)$ converging in $H^1$. Piecewise integration by parts for each $z_m$ has cancelling interface terms because $u^{\mathrm{\prime }}$ is continuous; passing to the limit in the $L^2$ pairings gives ${\int }_0^1{u}^{\mathrm{\prime }}{z}^{\mathrm{\prime }}=-{\int }_0^1{u}^{\mathrm{\prime }\mathrm{\prime }}z$. Since $f=-1$ on the first interval and $f=-u^{\mathrm{\prime }\mathrm{\prime }}$ on the second, this yields $a(u,v-u)-F(v-u)={\int }_0^{a_0}(v-u)dx={\int }_0^{a_0}vdx\ge 0$, where the last inequality uses $v\ge \psi =0$ almost everywhere on $(0,a_0)$. Hence $u$ solves the variational inequality. The same integration by parts against compactly supported tests gives reaction density $-u^{\mathrm{\prime }\mathrm{\prime }}-f=1$ on $(0,a_0)$ and $0$ on $(a_0,1)$; no point mass occurs at $a_0$ because $u^{\mathrm{\prime }}$ is continuous.

The function $f$ is bounded on both subintervals, hence belongs to $L^2(0,1)$. The form is symmetric and bounded since $\mathrm{\mid }a(w,z)\mathrm{\mid }\le \mathrm{\parallel }{w}^{\mathrm{\prime }}{\mathrm{\parallel }}_2\mathrm{\parallel }{z}^{\mathrm{\prime }}{\mathrm{\parallel }}_2\le \mathrm{\parallel }w{\mathrm{\parallel }}_{H^1}\mathrm{\parallel }z{\mathrm{\parallel }}_{H^1}$ by [F4]. By [F3], $a(w,w)=\mathrm{\parallel }{w}^{\mathrm{\prime }}{\mathrm{\parallel }}_2^2\ge c_P^{-2}\mathrm{\parallel }w{\mathrm{\parallel }}_{H^1}^2$ for some $c_P>0$. Thus the form is bounded, symmetric and coercive in the sense of [F5]. The load is bounded because $\mathrm{\mid }F(z)\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_2\mathrm{\parallel }z{\mathrm{\parallel }}_2\le c_P\mathrm{\parallel }f{\mathrm{\parallel }}_2\mathrm{\parallel }z{\mathrm{\parallel }}_{H^1}$ by [F3]–[F4]. If $\tilde{u}\in K$ is another solution, test its inequality with $u$ and the one for $u$ with $\tilde{u}$, then add; symmetry gives $a(u-\tilde{u},u-\tilde{u})\le 0$. Coercivity forces $u=\tilde{u}$.

The polynomial $\psi (s)=-1+4s(1-s)$ has endpoint values $-1$ and midpoint value $0$, and $\psi (s)=-(2s-1{)}^2$ shows its maximum is $0$. For $s_0,s_1\in [0,1]$ and $t\in [0,1]$, direct expansion gives $\psi ((1-t)s_0+t{s}_1)-[(1-t)\psi (s_0)+t\psi (s_1)]=4t(1-t)(s_1-s_0{)}^2\ge 0$, with strict inequality when $0<t<1$ and $s_0\mathrm{\ne }{s}_1$. Thus [F2] says $\psi$ is strictly concave, not convex, and the convex-obstacle premise is absent.

The vector field $v_h(x,y)=(-x,0)$ is affine on each triangle and is zero on ${\mathrm{\Gamma }}_D$, hence belongs to $V_h$. On the two contact vertices its normal value is $-1$, equal to the endpoint obstacle values, so both nodal inequalities hold. Along all of ${\mathrm{\Gamma }}_C$, $v_h\cdot n=-1$, while at $s=1\mathrm{/}2$ the obstacle equals $0$; continuity gives an interval around that midpoint on which $-1<\psi (s)$. Therefore the continuum contact inequality fails on a set of positive arclength and $v_h\mathrm{\notin }K$.

The affine field $u$ lies in $H^1(\mathrm{\Omega };{\mathbb{R}}^2)$ and vanishes on the left side, so $u\in V$. On the right side $u\cdot n=0=\psi$, hence $u\in K$ and has zero gap. Every affine function restricts to a continuous piecewise-affine function on the given matching mesh, so $u\in V_h$; its contact nodal inequalities hold with equality, so $u\in K_h$.

The gradient is $\mathrm{\nabla }u=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)$, so $\mathrm{div}u=0$ and $\epsilon (u)=\left(\begin{array}{cc}0& 1\mathrm{/}2\\ 1\mathrm{/}2& 0\end{array}\right)$. By [F3], $F(v)=a(u,v)$ is bounded. For every $v\in K$, $a(u,v-u)-F(v-u)=0$, so $u$ solves the continuous VI; uniqueness follows from [F4]. For every $v_h\in K_h\subseteq K$, the same identity holds, so $u$ solves the discrete VI. The inherited $H^1$ inner product makes finite-dimensional $V_h$ a Hilbert space by [F2]; the restricted form is bounded and coercive by [F3], and [F5] gives uniqueness.

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.