Variational inequalities and spatial contact discretization

For $j\in \{D,C\}$, restriction [F2] followed by the bounded trace map [F1] and restriction from $\mathrm{\partial }{U}_j$ to its segment ${\mathrm{\Gamma }}_j$ gives $\mathrm{\parallel }{T}_{j}v{\mathrm{\parallel }}_{L^2({\mathrm{\Gamma }}_j)}\le C_j\mathrm{\parallel }v{\mathrm{\mid }}_{U_j}{\mathrm{\parallel }}_{H^1(U_j)}\le C_j\mathrm{\parallel }v{\mathrm{\parallel }}_{H^1(\mathrm{\Omega })}$ for every $v\in H^1(\mathrm{\Omega };{\mathbb{R}}^2)$. Both maps are compositions of linear maps, so they are bounded and linear.

The space $V$ is the kernel of the bounded linear map $T_D$, hence is closed in $H^1(\mathrm{\Omega };{\mathbb{R}}^2)$; by [F3] the inherited inner product is complete on $V$, so $V$ is Hilbert.

The globally affine rigid motion $r$ is continuous on $\overline{U_D}$, so [F1] identifies $T_{D}r$ with its boundary restriction. Put $A:=a+\omega J{x}_0$ and decompose $A=A_{\parallel }J\tau +A_{\perp }$ with $A_{\perp }\cdot J\tau =0$. Since $J$ is a quarter-turn, $r(x(s))=A+\omega sJ\tau$. Integrating the two orthogonal components and completing the square in ${\int }_0^L(A_{\parallel }+\omega s{)}^{2}ds$ gives $\mathrm{\parallel }{T}_{D}r{\mathrm{\parallel }}_{L^2({\mathrm{\Gamma }}_D)}^2=L\mathrm{\mid }{A}_{\perp }{\mathrm{\mid }}^2+L(A_{\parallel }+\omega L\mathrm{/}2{)}^2+{\omega }^2{L}^3\mathrm{/}12$.

The last term of step 1.1 controls $\mathrm{\mid }\omega {\mathrm{\mid }}^2$, the first controls $\mathrm{\mid }{A}_{\perp }{\mathrm{\mid }}^2$, and the middle term together with $\mathrm{\mid }\omega \mathrm{\mid }$ controls $\mathrm{\mid }{A}_{\parallel }{\mathrm{\mid }}^2$; hence $\mathrm{\mid }A\mathrm{\mid }+\mathrm{\mid }\omega \mathrm{\mid }\le C_L\mathrm{\parallel }{T}_{D}r{\mathrm{\parallel }}_{L^2({\mathrm{\Gamma }}_D)}$. Because $a=A-\omega J{x}_0$ and $\mathrm{\Omega }$ is bounded, direct integration of $r$ and its constant gradient gives $\mathrm{\parallel }r{\mathrm{\parallel }}_{H^1(\mathrm{\Omega })}\le C_{\mathrm{\Omega }}(\mathrm{\mid }a\mathrm{\mid }+\mathrm{\mid }\omega \mathrm{\mid })\le C(\mathrm{\Omega },{\mathrm{\Gamma }}_D)\mathrm{\parallel }{T}_{D}r{\mathrm{\parallel }}_{L^2({\mathrm{\Gamma }}_D)}$. If the trace vanishes, this also forces $a=0$ and $\omega =0$.

The symmetric-part map is an orthogonal projection in the Frobenius inner product, so $\mathrm{\mid }\epsilon (w){\mathrm{\mid }}_F\le \mathrm{\mid }\mathrm{\nabla }w{\mathrm{\mid }}_F$. Also $\mathrm{\mid }\mathrm{div}w\mathrm{\mid }=\mathrm{\mid }\mathrm{tr}\mathrm{\nabla }w\mathrm{\mid }\le \sqrt{2}\mathrm{\mid }\mathrm{\nabla }w{\mathrm{\mid }}_F$ by Cauchy–Schwarz on the two diagonal entries. Thus both $\epsilon (w)$ and $\mathrm{div}w$ are in $L^2$ by [F3], so each integral defining $a(u,v)$ is finite.

Applying [F2] to the two terms of [F1] and then the bounds of step 1.1 gives $\mathrm{\mid }a(u,v)\mathrm{\mid }\le 2\mu \mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_2\mathrm{\parallel }\mathrm{\nabla }v{\mathrm{\parallel }}_2+2\lambda \mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_2\mathrm{\parallel }\mathrm{\nabla }v{\mathrm{\parallel }}_2\le (2\mu +2\lambda )\mathrm{\parallel }u{\mathrm{\parallel }}_{H^1}\mathrm{\parallel }v{\mathrm{\parallel }}_{H^1}$. The integrand is symmetric in $u,v$, so $a(u,v)=a(v,u)$.

The zero field belongs to $V$, and $T_{C}0\cdot n=0\ge \psi$ almost everywhere, so $0\in K$. If $v,w\in K$ and $0\le t\le 1$, linearity of $T_C$ gives $T_C(tv+(1-t)w)\cdot n-\psi =t(T_{C}v\cdot n-\psi )+(1-t)(T_{C}w\cdot n-\psi )\ge 0$ almost everywhere, so $tv+(1-t)w\in K$.

If $g_m\in L_{+}^2({\mathrm{\Gamma }}_C)$ and $g_m\to g$ in $L^2$, then $g^{-}:=\max \{-g,0\}\le \mathrm{\mid }g-g_m\mathrm{\mid }$ almost everywhere for each $m$, hence $\mathrm{\parallel }{g}^{-}{\mathrm{\parallel }}_2\le \mathrm{\parallel }g-g_m{\mathrm{\parallel }}_2\to 0$ and $g\ge 0$ almost everywhere. Thus the cone is closed; it is convex by pointwise convexity. By [F1] and [F3], the affine map $v\mapsto T_{C}v\cdot n-\psi$ is continuous into $L^2({\mathrm{\Gamma }}_C)$, so its preimage of that closed convex cone is closed in $V$. This proves the claim.

Since $\mathrm{\Omega }$ is nonempty, open and bounded, [F0] gives $0<\mathrm{\mid }\mathrm{\Omega }\mathrm{\mid }<\mathrm{\infty }$. Hence $1\in L^2(\mathrm{\Omega })$, and [F7] with the Cauchy–Schwarz inequality makes every $L^2$ component of $v$ and its weak gradient integrable; boundedness of $\mathrm{\Omega }$ also makes $x$ and $Jx$ integrable. Thus the means in $\omega =(2\mathrm{\mid }\mathrm{\Omega }\mathrm{\mid }{)}^{-1}{\int }_{\mathrm{\Omega }}\mathrm{curl}v$ and $a=\mathrm{\mid }\mathrm{\Omega }{\mathrm{\mid }}^{-1}{\int }_{\mathrm{\Omega }}(v-\omega Jx)$ exist. Put $r(x)=a+\omega Jx$ and $z=v-r$. Then ${\int }_{\mathrm{\Omega }}z=0$, ${\int }_{\mathrm{\Omega }}\mathrm{curl}z=0$, and $\epsilon (z)=\epsilon (v)$ by [F6].

Let $q=\mathrm{curl}z\in L_0^2(\mathrm{\Omega })$ and choose $w=B_{\mathrm{\Omega }}q$ by [F1]. Define the matrix $R(w)$ rowwise by $R(w{)}_{i1}={\mathrm{\partial }}_2{w}_i$ and $R(w{)}_{i2}=-{\mathrm{\partial }}_1{w}_i$. For $w\in C_c^{\mathrm{\infty }}(\mathrm{\Omega };{\mathbb{R}}^2)$ each row of $R(w)$ has zero divergence, so distributional integration by parts gives $(\mathrm{\nabla }z,R(w){)}_{L^2}=0$. Approximate the actual $w\in H_0^1$ by compactly supported smooth fields using [F5]; since $R(w_m)\to R(w)$ in $L^2$, the same identity holds for $w$.

Set $G=\mathrm{\nabla }z-R(w)$. In two dimensions $J:G=G_{21}-G_{12}=\mathrm{curl}z-\mathrm{div}w=q-q=0$, so the off-diagonal entries of $G$ agree and $G$ is symmetric. Write $S=(\mathrm{\nabla }z-(\mathrm{\nabla }z{)}^T)\mathrm{/}2$, which is skew-symmetric. Symmetric and skew-symmetric matrices are orthogonal in the Frobenius inner product, so $(S,G{)}_{L^2}=0$ and $(\mathrm{\nabla }z,G)=(\epsilon (z),G)$. By step 2.1, $(\mathrm{\nabla }z,G)=\mathrm{\parallel }\mathrm{\nabla }z{\mathrm{\parallel }}_2^2-(\mathrm{\nabla }z,R(w))=\mathrm{\parallel }\mathrm{\nabla }z{\mathrm{\parallel }}_2^2$. The entries of $R(w)$ are the entries of $\mathrm{\nabla }w$ permuted and sign-changed, so $\mathrm{\parallel }R(w){\mathrm{\parallel }}_2=\mathrm{\parallel }\mathrm{\nabla }w{\mathrm{\parallel }}_2$. Also $\mathrm{\mid }q{\mathrm{\mid }}^2=\mathrm{\mid }{\mathrm{\partial }}_1{z}_2-{\mathrm{\partial }}_2{z}_1{\mathrm{\mid }}^2\le 2(\mathrm{\mid }{\mathrm{\partial }}_1{z}_2{\mathrm{\mid }}^2+\mathrm{\mid }{\mathrm{\partial }}_2{z}_1{\mathrm{\mid }}^2)\le 2\mathrm{\mid }\mathrm{\nabla }z{\mathrm{\mid }}_F^2$ almost everywhere; integration gives $\mathrm{\parallel }q{\mathrm{\parallel }}_2\le \sqrt{2}\mathrm{\parallel }\mathrm{\nabla }z{\mathrm{\parallel }}_2$. Thus [F1] and the triangle inequality imply $\mathrm{\parallel }G{\mathrm{\parallel }}_2\le \mathrm{\parallel }\mathrm{\nabla }z{\mathrm{\parallel }}_2+\mathrm{\parallel }\mathrm{\nabla }w{\mathrm{\parallel }}_2\le (1+\sqrt{2}{C}_{\mathrm{d}\mathrm{i}\mathrm{v}})\mathrm{\parallel }\mathrm{\nabla }z{\mathrm{\parallel }}_2$. Therefore $\mathrm{\parallel }\mathrm{\nabla }z{\mathrm{\parallel }}_2^2\le \mathrm{\parallel }\epsilon (z){\mathrm{\parallel }}_2\mathrm{\parallel }G{\mathrm{\parallel }}_2\le (1+\sqrt{2}{C}_{\mathrm{d}\mathrm{i}\mathrm{v}})\mathrm{\parallel }\epsilon (z){\mathrm{\parallel }}_2\mathrm{\parallel }\mathrm{\nabla }z{\mathrm{\parallel }}_2$. If $\mathrm{\nabla }z=0$ the gradient estimate is immediate; otherwise divide by $\mathrm{\parallel }\mathrm{\nabla }z{\mathrm{\parallel }}_2$. Then [F2] and ${\int }_{\mathrm{\Omega }}z=0$ give $\mathrm{\parallel }z{\mathrm{\parallel }}_2\le C_P\mathrm{\parallel }\mathrm{\nabla }z{\mathrm{\parallel }}_2$, hence $\mathrm{\parallel }z{\mathrm{\parallel }}_{H^1}\le \sqrt{1+C_P^2}\mathrm{\parallel }\mathrm{\nabla }z{\mathrm{\parallel }}_2\le C_0\mathrm{\parallel }\epsilon (v){\mathrm{\parallel }}_2$.

For $v\in V$, [F3] gives $T_{D}r=-T_{D}z$. Apply [F4] and the trace bound [F3] to obtain $\mathrm{\parallel }r{\mathrm{\parallel }}_{H^1}\le C_D\mathrm{\parallel }{T}_{D}z{\mathrm{\parallel }}_{L^2({\mathrm{\Gamma }}_D)}\le C_D{C}_T\mathrm{\parallel }z{\mathrm{\parallel }}_{H^1}$. Since $v=z+r$, step 3.1 yields $\mathrm{\parallel }v{\mathrm{\parallel }}_{H^1}\le \mathrm{\parallel }z{\mathrm{\parallel }}_{H^1}+\mathrm{\parallel }r{\mathrm{\parallel }}_{H^1}\le (1+C_D{C}_T)C_0\mathrm{\parallel }\epsilon (v){\mathrm{\parallel }}_2$, proving the assertion.

The set $V_h$ is finite dimensional. By [F3], $K_h$ is the intersection of finitely many closed affine halfspaces in $V_h$, so it is closed and convex. The zero field is in $V_h$, and $0\ge I_h\psi (a)=\psi (a)$ at every contact vertex since $\psi \le 0$; hence $0\in K_h$ and $K_h$ is nonempty.

Fix a contact edge $e=[a,b]$. For $v_h\in K_h$, the function $q_h:=v_h\cdot n-I_h\psi$ is affine on $e$ and is nonnegative at $a,b$ by [F3]. Thus $q_h((1-t)a+tb)=(1-t)q_h(a)+t{q}_h(b)\ge 0$ throughout $e$. Convexity [F5] gives $I_h\psi ((1-t)a+tb)\ge \psi ((1-t)a+tb)$, so $v_h\cdot n\ge \psi$ on every contact edge. Since $v_h\in C(\bar{\mathrm{\Omega }})$, [F4] identifies its contact trace with this restriction. Therefore $v_h\in K$ and $K_h\subseteq K$.

Let $v\in K\cap H^2\cap C(\bar{\mathrm{\Omega }})$. Its continuous gap $T_{C}v\cdot n-\psi$ is nonnegative almost everywhere by [F1], hence everywhere: if it were negative at one point, continuity would make it negative on a subsegment of positive length. Define $I_{h}v$ cell by cell as the affine polynomial matching $v$ at the three vertices. On a shared edge the two affine traces agree at both endpoints, so they agree along the whole edge. For every test function compactly supported in $\mathrm{\Omega }$, integration by parts on each triangle cancels the internal-edge boundary terms because these traces agree; the remaining piecewise constant gradients are in $L^2(\mathrm{\Omega })$. Hence $I_{h}v\in H^1(\mathrm{\Omega };{\mathbb{R}}^2)$ and is globally continuous. Its boundary values on ${\mathrm{\Gamma }}_D$ are zero: $v\in V$ means $T_{D}v=0$ almost everywhere, and [F4] identifies this trace with the continuous boundary restriction, which therefore vanishes everywhere on the segment. Since the mesh resolves ${\mathrm{\Gamma }}_D$, the affine boundary traces of $I_{h}v$ vanish there, so $I_{h}v\in V_h$. At every contact node, $I_{h}v(a)=v(a)$ and [F4] identifies the contact trace with the continuous restriction; the continuous gap is nonnegative, so the nodal constraints hold and $I_{h}v\in K_h$.

Fix a contact vertex $a_i$. On a nondegenerate triangle $K$ with vertices $b_0,b_1,b_2$, the matrix with columns $b_1-b_0$ and $b_2-b_0$ is invertible by [F1]. For $x\in K$, write $(s,t)$ for its image under the inverse affine coordinate map $x=b_0+s(b_1-b_0)+t(b_2-b_0)$, and define ${\lambda }_0=1-s-t$, ${\lambda }_1=s$, ${\lambda }_2=t$. These affine functions take the Kronecker values at the three vertices. Define ${\varphi }_i{\mathrm{\mid }}_K={\lambda }_j$ if $a_i=b_j$, and ${\varphi }_i{\mathrm{\mid }}_K=0$ if $a_i$ is not a vertex of $K$. This finite cellwise construction gives a piecewise-affine scalar function with value $1$ at $a_i$ and $0$ at every other mesh vertex.

On any common facet, the two affine restrictions of ${\varphi }_i$ agree at every facet vertex: the value is $1$ exactly at $a_i$ if that vertex lies on the facet, and is $0$ at the other vertices. Affine functions on an edge are determined by their two endpoint values, so the restrictions agree on every common edge; at a shared vertex they both have the same prescribed nodal value. Thus ${\varphi }_i$ is globally continuous and belongs to the conforming piecewise-affine space [F4]. By [F5], ${\varphi }_i\in H^1(\mathrm{\Omega })$.

No contact vertex lies on ${\mathrm{\Gamma }}_D$ by [F6]. Since ${\mathrm{\Gamma }}_D$ is a union of boundary edges, ${\varphi }_i$ vanishes at both endpoints of each such edge and, being affine on the edge, vanishes on the entire segment. The vector field $w_i:={\varphi }_{i}n$ is continuous on $\bar{\mathrm{\Omega }}$ and belongs to $H^1(\mathrm{\Omega };{\mathbb{R}}^2)$ because ${\varphi }_i\in H^1(\mathrm{\Omega })$ by step 2.1. Its ordinary boundary restriction vanishes on ${\mathrm{\Gamma }}_D$, so the vector-valued trace agreement in [F3] gives $T_D{w}_i=0$; hence $w_i\in V_h$ by [F2]. At each contact vertex, $B_j(w_i)={\varphi }_i(a_j)(n\cdot n)={\delta }_{ij}$ by [F6]. For any $y=(y_0,\dots ,y_N)\in {\mathbb{R}}^{N+1}$, the finite sum $w={\sum }_i{y}_i{w}_i$ lies in $V_h$ and satisfies $B_j(w)=y_j$ for every $j$. Hence $B$ is onto.

By [F1], $V$ is a real Hilbert space and $K$ is nonempty, closed and convex. The elasticity form is bounded by [F2] and coercive by [F3], with positive constant $\alpha =2\mu \mathrm{/}{c}_K^2$. Thus every hypothesis of [F4] holds, and [F4] gives exactly one $u\in K$ satisfying $a(u,v-u)\ge F(v-u)$ for all $v\in K$.

Apply [F5] to the same $V$, $K$ and $a$, with the two bounded loads $F_1,F_2$. Its hypotheses are the Hilbert, closed-convex, boundedness, and coercivity properties verified in step 1.1; hence $\alpha \mathrm{\parallel }{u}_1-u_2{\mathrm{\parallel }}_V\le \mathrm{\parallel }{F}_1-F_2{\mathrm{\parallel }}_{V^{\ast }}$.

On one edge, direct integration of [F1]–[F2] gives ${\int }_e{\chi }_L{\varphi }_L=h_k$, ${\int }_e{\chi }_L{\varphi }_R=0$, ${\int }_e{\chi }_R{\varphi }_L=0$, and ${\int }_e{\chi }_R{\varphi }_R=h_k$. Summing over edges, only the edges incident to node $a_i$ contribute to $\int {\chi }_i{\varphi }_i$, and no edge contributes to $\int {\chi }_i{\varphi }_j$ for $i\mathrm{\ne }j$; hence the global pairing is $d_i{\delta }_{ij}$. This also proves linear independence of the ${\chi }_i$.