Linear elasticity, nearly incompressible limits and locking

Suppose $(x,\eta )$ lies in the nullspace of the augmented matrix. Then $Kx+R\eta =0$ and $R^{T}x=0$. Multiplying the first equation by $x^T$ gives $x^{T}Kx+x^{T}R\eta =0$, and the second equation makes the second term zero; hence $x^{T}Kx=0$.

For any $y\in {\mathbb{R}}^N$ and $t\in \mathbb{R}$, positive semidefiniteness gives $(x+ty{)}^{T}K(x+ty)=2t{y}^{T}Kx+t^2{y}^{T}Ky\ge 0$. If $y^{T}Kx\mathrm{\ne }0$, choosing $t$ with the opposite sign and sufficiently small magnitude makes the right side negative. Thus $y^{T}Kx=0$ for every $y$, so $Kx=0$.

Since $\ker K=\mathrm{range}R$, write $x=Ra$. Then $0=R^{T}x=R^{T}Ra$; full column rank makes $R^{T}R$ positive definite, so $a=0$ and $x=0$. The first null equation now gives $R\eta =0$, and full column rank yields $\eta =0$. The augmented square matrix is injective and therefore nonsingular by finite-dimensional rank-nullity.

Nonsingularity gives a unique solution for every right-hand side. For the right-hand side $[f;0]$, multiply $Ku+R\gamma =f$ by $R^T$. Symmetry of $K$ and $KR=0$ imply $R^{T}K=0$, so $R^{T}R\gamma =R^{T}f=0$ and positive definiteness gives $\gamma =0$. Hence $Ku=f$ and $R^{T}u=0$. Conversely, any vector satisfying those two equations with multiplier zero solves the augmented system, so its uniqueness follows from step 3.1.

Under [A1], [F1] gives $\mathrm{div}u\in L^2(D)$ and ${\int }_D\mathrm{div}u=0$. By the definition of $Q_D$ in [F2], this means $\mathrm{div}u\in Q_D$.

Multiplication by the finite scalar $-\lambda$ preserves membership in the linear subspace $Q_D$. Therefore $p=-\lambda \mathrm{div}u\in Q_D$. In the stated connected Lipschitz model domain, [F2] identifies $Q_D$ with $Q$, giving the model-space conclusion as well.

Since $\mathrm{\mid }\epsilon (v){\mathrm{\mid }}_F\le \mathrm{\mid }\mathrm{\nabla }v{\mathrm{\mid }}_F$ and $\mathrm{\mid }\mathrm{div}v\mathrm{\mid }\le \sqrt{d}\mathrm{\mid }\mathrm{\nabla }v{\mathrm{\mid }}_F$, [F4] gives $\mathrm{\mid }{\mathcal{E}}_{\lambda }(u,v)\mathrm{\mid }\le (2{\mu }_1+d\lambda )\mathrm{\parallel }u{\mathrm{\parallel }}_V\mathrm{\parallel }v{\mathrm{\parallel }}_V$. The form is real, symmetric, and bounded.

By [F3] and $\mu \ge {\mu }_0$, ${\mathcal{E}}_{\lambda }(v,v)\ge 2{\mu }_0\mathrm{\parallel }\epsilon (v){\mathrm{\parallel }}_2^2\ge {\mu }_0\mathrm{\parallel }v{\mathrm{\parallel }}_V^2$; the volumetric term is nonnegative.

By [F1]–[F2], $V$ is complete in $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_V$. Applying [F5] to the bounded functional $f$ and the form of steps 1.1–1.2 yields a unique $u\in V$.

Testing with $v=u$ gives ${\mathcal{E}}_{\lambda }(u,u)=f(u)\le \mathrm{\parallel }f{\mathrm{\parallel }}_{V^{\mathrm{\prime }}}\mathrm{\parallel }u{\mathrm{\parallel }}_V$ and, by step 1.2, ${\mu }_0\mathrm{\parallel }u{\mathrm{\parallel }}_V^2\le \mathrm{\parallel }f{\mathrm{\parallel }}_{V^{\mathrm{\prime }}}\mathrm{\parallel }u{\mathrm{\parallel }}_V$. If $u\mathrm{\ne }0$, divide by ${\mu }_0\mathrm{\parallel }u{\mathrm{\parallel }}_V$; if $u=0$, the claimed bound is immediate. Substitution into the energy identity gives the second estimate.

The form bounds in step 1.1 and the coercivity constant ${\mu }_0$ in step 1.2 satisfy the hypotheses of [F6] on every finite-dimensional conforming $V_h$. Thus its Galerkin solution is unique and the stated $M\mathrm{/}\alpha$ estimate is exactly $M=2{\mu }_1+d\lambda$, $\alpha ={\mu }_0$.

The pointwise bounds $\mathrm{\mid }\epsilon (v){\mathrm{\mid }}_F\le \mathrm{\mid }\mathrm{\nabla }v{\mathrm{\mid }}_F$ and $\mathrm{\mid }\mathrm{div}v\mathrm{\mid }\le \sqrt{d}\mathrm{\mid }\mathrm{\nabla }v{\mathrm{\mid }}_F$, together with [F9], give $\mathrm{\mid }{a}_{\mu }(u,v)\mathrm{\mid }\le 2{\mu }_1\mathrm{\parallel }u{\mathrm{\parallel }}_V\mathrm{\parallel }v{\mathrm{\parallel }}_V$ and $\mathrm{\mid }b(v,q)\mathrm{\mid }\le \sqrt{d}\mathrm{\parallel }v{\mathrm{\parallel }}_V\mathrm{\parallel }q{\mathrm{\parallel }}_Q$. By [F3], $a_{\mu }(v,v)\ge 2{\mu }_0\mathrm{\parallel }\epsilon (v){\mathrm{\parallel }}_2^2\ge {\mu }_0\mathrm{\parallel }v{\mathrm{\parallel }}_V^2$.

The space $V$ is a closed subspace of $H^1(D;{\mathbb{R}}^d)$ and its gradient norm is equivalent to the $H^1$ norm by [F8]; $Q$ is the closed kernel of the bounded integral functional on $L^2(D)$. Thus both are real Hilbert spaces under [A1]. Also $Q\mathrm{\ne }\{0\}$: two disjoint positive-measure balls inside $D$ give a nonzero difference of indicator functions with zero integral. And $V\mathrm{\ne }\{0\}$: choose a ball $B(x_0,r)$ with closure in $D$ and set $\varphi (x)=\exp (-1\mathrm{/}(r^2-\mathrm{\mid }x-x_0{\mathrm{\mid }}^2))$ inside it and zero outside. Its derivatives vanish at the sphere, so $\varphi \in C_c^{\mathrm{\infty }}(D)$ is nonzero and $(\varphi ,0,\dots ,0)\in V$.

For finite $\lambda >0$, take the primal solution from [F7] and set $p_{\lambda }=-\lambda \mathrm{div}{u}_{\lambda }$. By [F2, F10], $p_{\lambda }\in Q$. Then $a_{\mu }(u_{\lambda },v)+b(v,p_{\lambda })=a_{\mu }(u_{\lambda },v)+\lambda (\mathrm{div}{u}_{\lambda },\mathrm{div}v)=f(v)$, while $b(u_{\lambda },q)-{\lambda }^{-1}(p_{\lambda },q)=-(\mathrm{div}{u}_{\lambda },q)+(\mathrm{div}{u}_{\lambda },q)=0$. Conversely, any mixed solution has $s=\mathrm{div}u+{\lambda }^{-1}p\in Q$ by [F2]; its second equation gives $(s,q)=0$ for every $q\in Q$, hence $s=0$ by choosing $q=s$. It therefore solves the unique primal equation and has the same pressure.

At $\lambda =0$, use the primal solution and set $p_0=0$ by convention. At $\lambda =\mathrm{\infty }$, apply [F6] to $V,Q$, $a=a_{\mu }$, and $b$ with $G=0$: [F1] and step 1.1 give bounded forms, step 1.1 gives coercivity on all of $V$, [F4] gives the inf-sup bound, and step 1.2 gives real Hilbert spaces with both $V,Q$ nonzero. This yields a unique mean-zero pair $(u_{\mathrm{\infty }},p_{\mathrm{\infty }})$.

For each discrete pair, [F5] supplies a positive ${\beta }_h$ bounded below independently of $h$. For finite $\lambda >0$ or $\lambda =\mathrm{\infty }$, write $c={\lambda }^{-1}\ge 0$, with $c=0$ at infinity. A homogeneous discrete mixed solution $(w_h,r_h)$ satisfies $a_{\mu }(w_h,v_h)+b(v_h,r_h)=0$ and $b(w_h,q_h)-c(r_h,q_h)=0$. Test with $v_h=w_h$ and $q_h=r_h$ and subtract the second equality from the first to get $a_{\mu }(w_h,w_h)+c\mathrm{\parallel }{r}_h{\mathrm{\parallel }}_Q^2=0$. Step 1.1 forces $w_h=0$, and then discrete inf-sup forces $r_h=0$. The square finite-dimensional block matrix is injective, hence bijective, so the discrete system has a unique solution for every right-hand side.

The same stability estimate applies to any solution of $a_{\mu }(w,v)+b(v,r)=F(v)$, $b(w,q)-c(r,q)=G(q)$ when the inf-sup constant is $\beta >0$. Put $U=\mathrm{\parallel }w{\mathrm{\parallel }}_V$, $P=\mathrm{\parallel }r{\mathrm{\parallel }}_Q$, $F_0=\mathrm{\parallel }F{\mathrm{\parallel }}_{V^{\mathrm{\prime }}}$, and $G_0=\mathrm{\parallel }G{\mathrm{\parallel }}_{Q^{\mathrm{\prime }}}$. Testing with $v=w$ and $q=r$ and subtracting gives ${\mu }_0{U}^2\le F_{0}U+G_{0}P$. The first equation and inf-sup give $\beta P\le F_0+2{\mu }_{1}U$. With $S=F_0+G_0$ and $A=1+2{\mu }_1\mathrm{/}\beta$, substitution, using $F_0{G}_0\le S^2\mathrm{/}4$, yields ${\mu }_0{U}^2\le ASU+S^2\mathrm{/}(4\beta )$. Solving this quadratic gives $U\le C_{U}S$, where $C_U=(A+\sqrt{A^2+{\mu }_0\mathrm{/}\beta })\mathrm{/}(2{\mu }_0)$; then $P\le C_{P}S$ with $C_P={\beta }^{-1}(1+2{\mu }_1{C}_U)$. Thus $U+P\le C_{\ast }S$ for $C_{\ast }=C_U+C_P$, and these constants do not depend on $c$.

For $0<\lambda \le {\mu }_0$, let $P_h:Q\to Q_h$ be the $L^2$-orthogonal projection. It exists uniquely: in a basis of the finite-dimensional $Q_h$, its defining normal equations have the positive-definite Gram matrix $((q_i,q_j))$, and orthogonality gives $\mathrm{\parallel }p-P_{h}p{\mathrm{\parallel }}_Q={\inf }_{q_h\in Q_h}\mathrm{\parallel }p-q_h{\mathrm{\parallel }}_Q$. The second mixed equation gives $p_h=-\lambda P_h\mathrm{div}{u}_h$. Hence the first is the symmetric discrete equation $a_h(u_h,v_h)=f(v_h)$, where $a_h(w,v)=a_{\mu }(w,v)+\lambda (P_h\mathrm{div}w,P_h\mathrm{div}v)$. It is bounded by $M_0=2{\mu }_1+d{\mu }_0$ and coercive by ${\mu }_0$.

For any $w_h\in V_h$, subtract the equation for $a_h(w_h,\cdot )$ from the continuous primal equation. For every $v_h\in V_h$ this gives $a_h(u_h-w_h,v_h)=a_h(u-w_h,v_h)+\lambda ((I-P_h)\mathrm{div}u,(I-P_h)\mathrm{div}{v}_h)$. Since $p=-\lambda \mathrm{div}u$, $\lambda (I-P_h)\mathrm{div}u=-(I-P_h)p$; orthogonal projection gives $\mathrm{\parallel }I-P_h\mathrm{\parallel }\le 1$, while $\mathrm{\parallel }\mathrm{div}{v}_h{\mathrm{\parallel }}_2\le \sqrt{d}\mathrm{\parallel }{v}_h{\mathrm{\parallel }}_V$, so Cauchy–Schwarz bounds the remainder by $\sqrt{d}\mathrm{\parallel }p-P_{h}p{\mathrm{\parallel }}_Q\mathrm{\parallel }{v}_h{\mathrm{\parallel }}_V$. Testing with $v_h=u_h-w_h$ and using coercivity yields the displacement estimate; if $u_h=w_h$ it is immediate, and otherwise divide by ${\mu }_0\mathrm{\parallel }{u}_h-w_h{\mathrm{\parallel }}_V$ and take the infimum over $w_h$: $\mathrm{\parallel }u-u_h{\mathrm{\parallel }}_V\le (1+\frac{M_0}{{\mu }_0}){\inf }_{w_h\in V_h}\mathrm{\parallel }u-w_h{\mathrm{\parallel }}_V+\frac{\sqrt{d}}{{\mu }_0}{\inf }_{q_h\in Q_h}\mathrm{\parallel }p-q_h{\mathrm{\parallel }}_Q$. Moreover, $p-p_h=(I-P_h)p-\lambda P_h\mathrm{div}(u-u_h)$, so $\mathrm{\parallel }p-p_h{\mathrm{\parallel }}_Q\le {\inf }_{q_h\in Q_h}\mathrm{\parallel }p-q_h{\mathrm{\parallel }}_Q+{\mu }_0\sqrt{d}\mathrm{\parallel }u-u_h{\mathrm{\parallel }}_V$. Adding these bounds gives the claimed product-space best-approximation estimate with a constant depending only on $d,{\mu }_0,{\mu }_1$.

Testing the continuous primal equation with $u$ and the equivalent discrete $a_h$ equation with $u_h$, coercivity gives $\mathrm{\parallel }u{\mathrm{\parallel }}_V,\mathrm{\parallel }{u}_h{\mathrm{\parallel }}_V\le {\mu }_0^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}_{V^{\mathrm{\prime }}}$. The pressure identities and $\lambda \le {\mu }_0$ then give $\mathrm{\parallel }p{\mathrm{\parallel }}_Q\le \lambda \sqrt{d}\mathrm{\parallel }u{\mathrm{\parallel }}_V\le \sqrt{d}\mathrm{\parallel }f{\mathrm{\parallel }}_{V^{\mathrm{\prime }}}$ and $\mathrm{\parallel }{p}_h{\mathrm{\parallel }}_Q\le \lambda \sqrt{d}\mathrm{\parallel }{u}_h{\mathrm{\parallel }}_V\le \sqrt{d}\mathrm{\parallel }f{\mathrm{\parallel }}_{V^{\mathrm{\prime }}}$.

For $\lambda =0$, [F7] gives the solution bounds and the displacement estimate with constant $2{\mu }_1\mathrm{/}{\mu }_0$; $p_0=p_{h,0}=0$. For $0<\lambda \le {\mu }_0$, step 4.2 gives uniform solution bounds and step 4.1 the uniform best-approximation estimate. For $\lambda \ge {\mu }_0$ or $\lambda =\mathrm{\infty }$, step 2.3 with the continuous inf-sup bound [F4] and discrete bound [F5] gives uniform solution bounds. For the same upper range, $c={\lambda }^{-1}\le {\mu }_0^{-1}$ and $\mathrm{\mid }{\mathcal{A}}_{\lambda }((u,p),(v,q))\mathrm{\mid }\le 2{\mu }_1\mathrm{\parallel }u{\mathrm{\parallel }}_V\mathrm{\parallel }v{\mathrm{\parallel }}_V+\sqrt{d}\mathrm{\parallel }v{\mathrm{\parallel }}_V\mathrm{\parallel }p{\mathrm{\parallel }}_Q+\sqrt{d}\mathrm{\parallel }u{\mathrm{\parallel }}_V\mathrm{\parallel }q{\mathrm{\parallel }}_Q+{\mu }_0^{-1}\mathrm{\parallel }p{\mathrm{\parallel }}_Q\mathrm{\parallel }q{\mathrm{\parallel }}_Q$, so its product-space operator norm is at most $M=2{\mu }_1+2\sqrt{d}+{\mu }_0^{-1}$, independently of $\lambda$.

Write $\mathrm{\parallel }(v,q){\mathrm{\parallel }}_{V\times Q}=\mathrm{\parallel }v{\mathrm{\parallel }}_V+\mathrm{\parallel }q{\mathrm{\parallel }}_Q$. Let $x=(u,p)$ be the continuous solution, $x_h=(u_h,p_h)$ its Galerkin solution, and fix $y_h=(v_h,q_h)\in V_h\times Q_h$. Galerkin orthogonality gives ${\mathcal{A}}_{\lambda }(x_h-y_h,z_h)={\mathcal{A}}_{\lambda }(x-y_h,z_h)$ for every $z_h=(w_h,r_h)$. Thus $x_h-y_h$ solves the discrete mixed system with data $F_h(w_h)={\mathcal{A}}_{\lambda }(x-y_h,(w_h,0))$ and $G_h(r_h)={\mathcal{A}}_{\lambda }(x-y_h,(0,r_h))$. The bound on ${\mathcal{A}}_{\lambda }$ gives $\mathrm{\parallel }{F}_h\mathrm{\parallel }+\mathrm{\parallel }{G}_h\mathrm{\parallel }\le 2M\mathrm{\parallel }x-y_h{\mathrm{\parallel }}_{V\times Q}$. Applying the stability bound of step 2.3 with $\beta \ge \min (C_{\mathrm{d}\mathrm{i}\mathrm{v}}(D{)}^{-1},{\beta }_{\ast })>0$, then the triangle inequality and the infimum over $y_h$, gives the required estimate with a constant at most $1+2M{C}_{\ast }$. Combining this with steps 1.3–2.1, 2.2, 4.1–4.2, and 5.1 covers every $\lambda \in [0,\mathrm{\infty }]$.