Linear elasticity, nearly incompressible limits and locking — Examples

On each nondegenerate simplex, an affine map is uniquely determined by its values at the $d+1$ affinely independent vertices, so its nodal P1 interpolant equals it. On each shared face the affine traces are determined by the same face-vertex values, so they agree and define a global continuous member of $S_h$.

Since $\mathrm{\Omega }$ is skew-symmetric, $\epsilon (r)=(\mathrm{\Omega }+{\mathrm{\Omega }}^T)\mathrm{/}2=0$ and $\mathrm{div}r=\mathrm{tr}\mathrm{\Omega }=0$. Thus the stiffness bilinear form pairs $r$ to zero with every test basis function, which is exactly $Kr=0$ in coefficient coordinates. By [F1], every complete gauge basis must span these null vectors, although other null modes are not excluded here.

The gradient of $u=(0,x)$ is $\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)$, so $\epsilon (u)=\left(\begin{array}{cc}0& 1\mathrm{/}2\\ 1\mathrm{/}2& 0\end{array}\right)$ and $\mathrm{div}u=0$. Hence [F1] gives ${\sigma }_{\lambda }(u)=\left(\begin{array}{cc}0& \mu \\ \mu & 0\end{array}\right)$, independent of $\lambda$.

The stress is constant, so its row-wise divergence is zero. The displacement vanishes at $x=0$. Multiplication by the outward unit normals $(1,0)$, $(0,1)$, and $(0,-1)$ on the right, top, and bottom edges gives the three displayed tractions.

Set $\eta (t)=\exp (-1\mathrm{/}((t-1\mathrm{/}4)(3\mathrm{/}4-t)))$ for $1\mathrm{/}4<t<3\mathrm{/}4$ and $\eta (t)=0$ otherwise. At each endpoint every derivative from the interior is a finite sum of reciprocal-distance powers multiplied by $\exp (-c\mathrm{/}\text{distance})$ for some $c>0$, hence tends to zero; so $\eta \in C_c^{\mathrm{\infty }}(0,1)$. Put $\psi (x,y)=\eta (x)\eta (y)$ and $u=({\mathrm{\partial }}_y\psi ,-{\mathrm{\partial }}_x\psi )$. Equality of mixed derivatives gives $\mathrm{div}u=0$. Since $\eta (1\mathrm{/}2)>0$ and $\eta (1\mathrm{/}4)=0$, the mean value theorem gives $y\in (1\mathrm{/}4,1\mathrm{/}2)$ with ${\eta }^{\mathrm{\prime }}(y)>0$; hence $u_1(1\mathrm{/}2,y)=\eta (1\mathrm{/}2){\eta }^{\mathrm{\prime }}(y)>0$. Thus $u\mathrm{\ne }0$ and is compactly supported in $D$.

Let $f=-\mathrm{div}(2\epsilon (u))\in C_c^{\mathrm{\infty }}(D;{\mathbb{R}}^2)$. For $v\in C_c^{\mathrm{\infty }}(D;{\mathbb{R}}^2)$, integration by parts and symmetry of $2\epsilon (u)$ give $2(\epsilon (u),\epsilon (v))=(f,v)$. By [A1, F5], choose such tests converging to any given $v\in V$ in $H^1$; [F4] makes both pairings continuous under this convergence, so the identity holds for every $v\in V$.

A P1 vector field in this mesh vanishes at every boundary vertex and is determined by its center value $z=(a,b)$. On the bottom triangle the edge vectors from the center to its corners are $(-1\mathrm{/}2,-1\mathrm{/}2)$ and $(1\mathrm{/}2,-1\mathrm{/}2)$; its affine gradient $G$ maps both to $-z$. Subtracting gives the first column of $G$ zero, and adding gives the second column $(2a,2b)$, so divergence is $2b$. On the right triangle the edge vectors are $(1\mathrm{/}2,-1\mathrm{/}2)$ and $(1\mathrm{/}2,1\mathrm{/}2)$; subtraction gives the second column zero and addition the first column $(-2a,-2b)$, hence divergence is $-2a$. On the top triangle the edge vectors are $(1\mathrm{/}2,1\mathrm{/}2)$ and $(-1\mathrm{/}2,1\mathrm{/}2)$; subtraction gives the first column zero and addition the second column $(-2a,-2b)$, hence divergence is $-2b$. On the left triangle the edge vectors are $(-1\mathrm{/}2,1\mathrm{/}2)$ and $(-1\mathrm{/}2,-1\mathrm{/}2)$; subtraction gives the second column zero and addition the first column $(2a,2b)$, hence divergence is $2a$. If all four divergences vanish, then $a=b=0$, so the elementwise divergence-free subspace is exactly $\{0\}$.

Since $\mathrm{div}u=0$, the volumetric term vanishes and step 1.2 gives ${\mathcal{E}}_{\lambda }(u,v)=f(v)$ for every $v\in V$ and every finite $\lambda \ge 0$. By [F1], this identifies the same nonzero $u$ as the unique continuous solution for each such $\lambda$.

Let $u_{h,\lambda }$ be the P1 Galerkin solution, which exists uniquely by [F1]. Testing with $u_{h,\lambda }$, using [F6] with $\mu =1$, and applying [F4] gives $\mathrm{\parallel }{u}_{h,\lambda }{\mathrm{\parallel }}_V\le \mathrm{\parallel }f{\mathrm{\parallel }}_{V^{\mathrm{\prime }}}$ and $\lambda \mathrm{\parallel }\mathrm{div}{u}_{h,\lambda }{\mathrm{\parallel }}_2^2\le \mathrm{\parallel }f{\mathrm{\parallel }}_{V^{\mathrm{\prime }}}^2$. Thus along every sequence ${\lambda }_n\to \mathrm{\infty }$ the discrete solutions are bounded in the finite-dimensional P1 space and have a convergent subsequence; the divergence of its limit is zero because the divergence norms tend to zero and convergence in $H^1$ implies convergence of divergence in $L^2$. The limit is zero by step 1.3. If the full family did not converge to zero, a sequence bounded away from zero would have such a zero-convergent subsequence, a contradiction. Hence $u_{h,\lambda }\to 0$ and the reverse triangle inequality gives $\mathrm{\parallel }u-u_{h,\lambda }{\mathrm{\parallel }}_V\to \mathrm{\parallel }u{\mathrm{\parallel }}_V>0$.

The mini mixed comparison is exactly the scoped result [F2] under the spaces [F3]; its estimate keeps the best-approximation term and its constant is independent of $\lambda$ on this mesh family. No conclusion about a P1 rate as $h\to 0$ is used.