First-order least-squares method for the obstacle problem

Abstract

We define and analyse a least-squares finite element method for a first-order reformulation of the obstacle problem. Moreover, we derive variational inequalities that are based on similar but non-symmetric bilinear forms. A priori error estimates including the case of non-conforming convex sets are given and optimal convergence rates are shown for the lowest-order case. We provide a posteriori bounds that can be used as error indicators in an adaptive algorithm. Numerical studies are presented.

Appendices

Appendix A: Non-convexity of functional J

Recall that the functional $J(\cdot ;f,g)$ is convex if and only if

$$\begin{aligned} a_1(\varvec{u}-\varvec{v},\varvec{u}-\varvec{v}) \ge 0 \quad \text {for all } \varvec{u},\varvec{v}\in K^s := \big \{(w,{\varvec{\chi }},\nu )\in U{:}\,w\ge g, \, \nu \ge 0\big \}. \end{aligned}$$

In the following we construct a simple example that shows that the above inequality does not hold in general, thus J is not convex resp. $a(\cdot ,\cdot ) = a_1(\cdot ,\cdot )$ is not coercive.

To that end, let $u\in H_0^1(\Omega )$ denote the solution of $\Delta u = 1$ in the square domain $\Omega = (0,d)^2$. Then, $u\le 0$ in $\Omega$. Choose the obstacle as $g= u$ (or $g\le u$). Note that $\varvec{v}:= (0,0,0)\in K^s$ and that $\varvec{u}:=(u,\varvec{\sigma },\Delta u) := (u,\nabla u,\Delta u)\in K^s$. We have that

$$\begin{aligned} \Vert 1\Vert _{-1} = \sup _{0\ne v\in H_0^1(\Omega )} \frac{\langle 1,v\rangle }{\Vert \nabla v\Vert _{}} = \sup _{0\ne v\in H_0^1(\Omega )} \frac{-(\nabla u,\nabla v)}{\Vert \nabla v\Vert _{}} = \Vert \nabla u\Vert _{}. \end{aligned}$$

Therefore $\Vert 1\Vert _{-1}^2 = \Vert \nabla u\Vert _{}^2 = -\langle 1,u\rangle$. Using this we infer that

$$\begin{aligned} a_1(\varvec{u}-\varvec{v},\varvec{u}-\varvec{v}) = \Vert {{\,\mathrm{div}\,}}\varvec{\sigma }+1\Vert _{}^2 + \Vert \nabla u-\varvec{\sigma }\Vert _{}^2 + \langle 1,u\rangle = \Vert 2\Vert _{}^2 - \Vert 1\Vert _{-1}^2. \end{aligned}$$

Hence, $a_1(\varvec{u}-\varvec{v},\varvec{u}-\varvec{v}) < 0$ if and only if $\Vert 2\Vert _{} < \Vert 1\Vert _{-1}$. Clearly, $\Vert 2\Vert _{} = 2 |\Omega |^{1/2} = 2d$. We investigate the scaling of the negative order norm. Let $\widehat{\Omega }$ denote the unit square $(0,1)^2$. Then,

$$\begin{aligned} \Vert 1\Vert _{-1} = \sup _{0\ne v\in H_0^1(\Omega )} \frac{\langle 1,v\rangle }{\Vert \nabla v\Vert _{}} = |\Omega | \sup _{0\ne \widehat{v} \in H_0^1(\widehat{\Omega })} \frac{(1,\widehat{v})_{\widehat{\Omega }}}{\Vert \nabla \widehat{v}\Vert _{\widehat{\Omega }}} = |\Omega | \, \Vert 1\Vert _{-1,\widehat{\Omega }} =: |\Omega | \, C. \end{aligned}$$

Finally, $\Vert 2\Vert _{} < \Vert 1\Vert _{-1}$ if

$$\begin{aligned} 2d < C d^2 \end{aligned}$$

which holds for sufficiently large d.

Appendix B: Proof of Lemma 12

We use $T_\mathrm {C}:= T_\mathrm {C}(v)$ and $T_\mathrm {NC} := T_\mathrm {NC}(v)$. If $|T_\mathrm {NC}|=0$, then $v=0$ on T and the first inequality is trivial. Note that also $\nabla v =0$. Therefore, the second inequality is trivial as well. From now on we thus assume $|T_\mathrm {NC}|>0$. Using that $v=0$ on $T_\mathrm {C}$ and the identity

$$\begin{aligned} v(\varvec{x})-v(\varvec{y}) = (\varvec{x}-\varvec{y})\cdot \int _0^1 \nabla v (s \varvec{x}+ (1-s) \varvec{y}) \,ds \quad \text {for all }\varvec{x},\varvec{y}\in T \end{aligned}$$

we obtain that

$$\begin{aligned} \Vert v\Vert _{T}^2&= \Vert v\Vert _{T_\mathrm {NC}}^2 = \int _{T_\mathrm {NC}} v(\varvec{x})^2 \,d\varvec{x}= \frac{1}{|T_\mathrm {C}|} \int _{T_\mathrm {NC}} \int _{T_\mathrm {C}} (v(\varvec{x})-v(\varvec{y}))^2 \,d\varvec{y}\,d\varvec{x}\\&= \frac{1}{|T_\mathrm {C}|} \int _{T_\mathrm {NC}} \int _{T_\mathrm {C}} \Big ( (\varvec{x}-\varvec{y}) \cdot \int _0^1 \nabla v(s\varvec{x}+ (1-s)\varvec{y}) \,ds \Big )^2 \\&\le \frac{h_T^2}{|T_\mathrm {C}|} \int _T \int _T \int _0^1 |\nabla v(s\varvec{x}+(1-s)\varvec{y})|^2 \,ds \,d\varvec{y}\,d\varvec{x}\\&= 2 \frac{h_T^2}{|T_\mathrm {C}|} \int _T \int _{1/2}^1 \int _T |\nabla v(s\varvec{x}+(1-s)\varvec{y})|^2 \,d\varvec{x}\,ds \,d\varvec{y}, \end{aligned}$$

where in the ultimate step we have used symmetry in $\varvec{x}$ and $\varvec{y}$. With the substitution $\varvec{z}= \phi (\varvec{x}) = s\varvec{x}+ (1-s)\varvec{y}\in T$ we further get that

$$\begin{aligned} \int _T \int _{1/2}^1 \int _T |\nabla v(s\varvec{x}+(1-s)\varvec{y})|^2 \,d\varvec{x}\,ds \,d\varvec{y}&= \int _T \int _{1/2}^1 s^{-n} \int _{\phi (T)} |\nabla v(\varvec{z})|^2 \,d\varvec{z}\,ds \,d\varvec{y}\\&\le \int _T \int _{1/2}^1 s^{-n} \int _{T} |\nabla v(\varvec{z})|^2 \,d\varvec{z}\,ds \,d\varvec{y}\\ {}&= |T| \frac{1-(1/2)^{1-n}}{1-n} \Vert \nabla v\Vert _{T}^2. \end{aligned}$$

Putting altogether this proves the first inequality.

For the second inequality we use the fact that the gradient on level sets vanishes, see [13, Theorem 3.3]. This means that $\nabla v = 0$ a.e. in $T_\mathrm {C}$. Then, the same lines of proof as above (with v replaced by the components of $\nabla v$) show the second inequality, which finishes the proof.

Appendix C: Proof of Lemma 19

We consider the decompositions

$$\begin{aligned} \mathcal {J}^{k-1}&= \underbrace{\big \{j\in \mathcal {J}^{k-1}{:}\,\varvec{\lambda }_j^k > 0\big \}}_{=:\mathcal {J}_1^{k-1}} \cup \underbrace{\big \{j\in \mathcal {J}^{k-1}{:}\,\varvec{\lambda }_j^k = 0\big \}}_{=:\mathcal {J}_2^{k-1}}, \\ \mathcal {I}^{k-1}&= \underbrace{\big \{i\in \mathcal {I}^{k-1}{:}\,i\in \mathcal {N}_\omega \text { or } i\in \mathcal {N}_\gamma \text { with } \varvec{x}_i^k\ge \varvec{g}_i\big \}}_{=:\mathcal {I}_1^{k-1}} \cup \underbrace{\big \{i\in \mathcal {I}^{k-1}\cap \mathcal {N}_\gamma {:}\,\varvec{x}_i^k<\varvec{g}_i\big \}}_{=:\mathcal {I}_2^{k-1}}. \end{aligned}$$

For the decomposition of $\mathcal {J}^k$ note that from Lines 8–9 of Algorithm 1 we have that $\varvec{\lambda }_{\mathcal {N}_\gamma \cap \mathcal {I}^{k-1}} = 0$ and $\varvec{x}_{\mathcal {J}^k} = \varvec{g}_{\mathcal {J}^k}$. This yields

$$\begin{aligned} \mathcal {J}^k = \big \{j\in \mathcal {J}^{k-1}{:}\, \varvec{\lambda }_j^k>0\big \} \cup \big \{i\in \mathcal {I}^{k-1}\cap \mathcal {N}_\gamma {:}\, \varvec{x}_i^k < \varvec{g}_i\big \} = \mathcal {J}_1^{k-1} \cup \mathcal {I}_2^{k-1} \end{aligned}$$

If $\mathcal {J}^k = \mathcal {J}^{k-1}$, then, since all decompositions are disjoint,

$$\begin{aligned} \mathcal {I}_2^{k-1} = \emptyset = \mathcal {J}_2^{k-1}. \end{aligned}$$

This also means that $\widehat{\varvec{\lambda }}_j^k>0$ if and only if $\varvec{\lambda }_j^k>0$ with $\varvec{x}_j^k = \varvec{g}_j$. Thus,

$$\begin{aligned} \varvec{\lambda }_{\mathcal {N}_\gamma }^k&= \max \{ 0,\varvec{\lambda }_{\mathcal {N}_\gamma }^k-C(\varvec{x}_{\mathcal {N}_\gamma }-\varvec{g}_{\mathcal {N}_\gamma })\}, \\ \varvec{\lambda }_{\mathcal {N}_\omega }^k&= 0, \end{aligned}$$

which implies that (18) is satisfied for $\varvec{x}=\varvec{x}^k$, $\varvec{\lambda }=\varvec{\lambda }^k$ or equivalently $\varvec{x}=\varvec{x}^k$ solves (17).