A note on a priori $\mathbf {L^p}$-error estimates for the obstacle problem

Abstract

This paper is concerned with a priori error estimates for the piecewise linear finite element approximation of the classical obstacle problem. We demonstrate by means of two one-dimensional counterexamples that the $L^2$-error between the exact solution u and the finite element approximation $u_h$ is typically not of order two even if the exact solution is in $H^2(\varOmega )$ and an estimate of the form $\Vert u - u_h\Vert _{H^1} \le {Ch}$ holds true. This shows that the classical Aubin–Nitsche trick which yields a doubling of the order of convergence when passing over from the $H^1$- to the $L^2$-norm cannot be generalized to the obstacle problem.

Appendix: Unilateral FE-approximation in one dimension

In this section, we construct the unilateral finite element approximations that are needed in the proof of Proposition 8. The underlying analysis essentially goes back to Mosco and Strang, cf. [8, 14]. For the convenience of the reader, we shortly recall the arguments in the following.

Theorem A1

Let $\varOmega$ be an open bounded interval and assume that $\{\mathcal {T}_h\}_{0< h < h_0}$ is a family of partitions of $\varOmega$ such that $ch< \mathrm {diam}\,T < \min (Ch, \mathrm {diam}\,\varOmega )$ holds for all $T \in \mathcal {T}_h$ and all $0< h < h_0$ with constants $c, C > 0$ independent of h. Let

$$\begin{aligned} V_h^0 := H_0^1(\varOmega ) \cap \{ v \in C(\overline{\varOmega }) : v |_{{T}} \text { is affine for all cells } T \in \mathcal {T}_{h}\} \end{aligned}$$

and suppose that $z \in H_0^1(\varOmega ) \cap W^{2,q}(\varOmega )$, $1< q < \infty$, is a given function. Then, there exist constants $C_1, C_2$ and $C_3$ independent of h and functions $\{z_h\}_{0< h < h_0}$ satisfying $z \le z_h \in V_h^0$ for all $0< h < h_0$ such that

$$\begin{aligned} \Vert z - z_h\Vert _{L^q} \le C_1 h^2 |z|_{W^{2,q}},\quad \Vert z - z_h\Vert _{W^{1,q}} \le C_2 h |z|_{W^{2,q}} \end{aligned}$$

(20)

and

$$\begin{aligned} \Vert z - z_h\Vert _{L^\infty } \le C_3 h^{2 - 1/q}|z|_{W^{2,q}} \end{aligned}$$

(21)

holds for all $0< h < h_0$.

Proof

In what follows, we will always identify z with its $C^1(\overline{\varOmega })$-representative. To prove Theorem A1, we consider for an arbitrary but fixed $0< h < h_0$ the optimization problem

$$\begin{aligned} \text {min} \sum _{x \in \mathcal {C}_h} v_h(x)\quad \text {s.t. } z \le v_h \text { in } \varOmega \text{ and } v_h \in V_h^0 , \end{aligned}$$

(22)

where $\mathcal {C}_h$ denotes the set of all vertices of the partition $\mathcal {T}_h$. Using standard techniques from finite-dimensional optimization, it is easy to see that (22) admits at least one global minimum $z_h$. From the definition of (22), it follows that the function values $z_h(x)$, $x \in \mathcal {C}_h$, of such a minimum cannot be decreased without violating the constraint $z \le z_h$. This implies that for every node $x \in \mathcal {C}_h {\setminus } \partial \varOmega$ with adjacent mesh cells $T_l = [x_l, x]$ and $T_r = [x, x_r]$ one of the following has to be true:

• (a) It holds $z_h(x) = z(x)$.

• (b) There exists an $a \in [x_l, x_r] {\setminus } \{x\}$ such that $z_h(a) = z(a)$ and $z_h'(a) = z'(a)$. If $a \in \{x_l, x_r\}$, we mean the left (resp., right) limit of the derivative here.

If (b) is the case, then the fundamental theorem of calculus yields

$$\begin{aligned} z_h(x) - z(x)&= \left| \int _a^x z''(t)(x-t) \mathrm {d}t \right| \\&\le \left( \frac{q-1}{2q-1} \right) ^{\frac{q-1}{q}} (Ch)^{2 - \frac{1}{q}} \left( \max _{T \in \mathcal {T}_h : x \in T} |z|_{W^{2,q}(T)} \right) \end{aligned}$$

and if (a) is true (or $x \in \partial \varOmega$), it trivially holds $z_h(x) - z(x) = 0$. Thus, we obtain that $z_h$ satisfies

$$\begin{aligned} 0&\le z_h(x) - I_h z(x)\nonumber \\&\le \left( \frac{q-1}{2q-1} \right) ^{\frac{q-1}{q}} (Ch)^{2 - \frac{1}{q}} \left( \max _{T \in \mathcal {T}_h : x \in T} |z|_{W^{2,q}(T)} \right) \quad \forall \, x \in \mathcal {C}_h\text {.} \end{aligned}$$

(23)

Here, $I_h: H_0^1(\varOmega ) \rightarrow V_h^0$ again denotes the Lagrange interpolation operator. From (23) and the piecewise linearity of the functions in $V_h^0$, it readily follows

$$\begin{aligned} \Vert z_h - I_h z\Vert _{L^\infty (\varOmega )} \le \left( \frac{q-1}{2q-1} \right) ^{\frac{q-1}{q}} (Ch)^{2 - \frac{1}{q}} |z|_{W^{2,q}(\varOmega )}. \end{aligned}$$

(24)

Combining (24) with the triangle inequality and standard error estimates for the Lagrange interpolant yields (21). Further, we obtain from (23) that

$$\begin{aligned} \Vert z_h - I_h z\Vert _{L^q(T)} \le \left( \frac{q-1}{2q-1} \right) ^{\frac{q-1}{q}} (Ch)^{2} \left( \max _{\tilde{T} \in \mathcal {T}_h : T \cap \tilde{T} \ne \emptyset } |z|_{W^{2,q}(\tilde{T})}\right) \quad \forall \, T \in \mathcal {T}_h\text {.} \end{aligned}$$

Summation over all $T \in \mathcal {T}_h$ now yields

$$\begin{aligned} \Vert z_h - I_h z\Vert _{L^q(\varOmega )} \le 3^{\frac{1}{q}}\left( \frac{q-1}{2q-1} \right) ^{\frac{q-1}{q}} (Ch)^{2} |z|_{W^{2,q}(\varOmega )}\text {.} \end{aligned}$$

Using again the triangle inequality and standard interpolation error estimates, we obtain the first estimate in (20). To prove the $W^{1,q}$-estimate, note that

$$\begin{aligned} \Vert z_h' - I_h' z\Vert _{L^\infty (T)} \le \frac{2}{c\,h} \Vert z_h - I_h z\Vert _{L^\infty (T)}\quad \forall \,T \in \mathcal {T}_h. \end{aligned}$$

Proceeding as in case of the $L^q$-error now gives the second estimate in (20). $\square$

It should be noted that in higher dimensions, it is still possible to prove $L^\infty$-error estimates for unilateral approximations provided the function z under consideration possesses enough regularity. We refer to [2] for details.