Penalty method with P1/P1 finite element approximation for the Stokes equations under the slip boundary condition

Abstract

We consider the P1/P1 or P1b/P1 finite element approximations to the Stokes equations in a bounded smooth domain subject to the slip boundary condition. A penalty method is applied to address the essential boundary condition $u\cdot n = g$ on $\partial \Omega$, which avoids a variational crime and simultaneously facilitates the numerical implementation. We give $O(h^{1/2} + \epsilon ^{1/2} + h/\epsilon ^{1/2})$-error estimate for velocity and pressure in the energy norm, where h and $\epsilon$ denote the discretization parameter and the penalty parameter, respectively. In the two-dimensional case, it is improved to $O(h + \epsilon ^{1/2} + h^2/\epsilon ^{1/2})$ by applying reduced-order numerical integration to the penalty term. The theoretical results are confirmed by numerical experiments.

Appendices

Appendix 1: Transformation between $\varGamma$ and $\varGamma _h$ and related estimates

Let $\Omega$ be a bounded domain in $\mathbb R^N$. Its boundary $\varGamma$ is assumed to be $C^{1,1}$-smooth, namely, there exist a system of local coordinates $\{(U_r, y_r, \varphi _r)\}_{r=1}^M$ and positive numbers $\alpha , \beta$ such that: 1) $\{U_r\}_{r=1}^M$ forms an open covering of $\varGamma$; 2) $y_r = (y_{r1}, \dots , y_{rN-1}, y_{rN}) = (y_r', y_{rN})$ is a rotated coordinate of the original one x, that is, $y_r = A_rx$ for some orthogonal transformation $A_r$; 3) $\varphi _r\in C^{1,1}(\varDelta _r)$ gives a graph representation of $\varGamma \cap U_r$, where $\varDelta _r := \{y_r' \in \mathbb R^{N-1} :\; |y_r'| < \alpha \}$, that is,

$$\begin{aligned} \varGamma \cap U_r&= \{ y_r \in \mathbb R^N \,:\; y_r' \in \varDelta _r \text { and } y_{rN} = \varphi _r(y_r') \}, \\ \Omega \cap U_r&= \{ y_r \in \mathbb R^N \,:\; y_r' \in \varDelta _r \text { and } \varphi _r(y_r') < y_{rN} < \varphi _r(y_r') + \beta \}, \\ \overline{\Omega }^c \cap U_r&= \{ y_r \in \mathbb R^N \,:\; y_r' \in \varDelta _r \text { and } \varphi _r(y_r') - \beta < y_{rN} < \varphi _r(y_r') \}. \end{aligned}$$

Since $\varGamma$ is compact, there exists $h_0>0$ such that for any $x\in \varGamma$ the open ball $B(x; h_0)$ is contained in some local coordinate neighborhood $U_r$. According to the fact that $C^{1,1}(\varDelta _r) = W^{2,\infty }(\varDelta _r)$, the derivatives of $\varphi _r$ are bounded up to second order, i.e.,

$$\begin{aligned} \Vert \varphi _r\Vert _{L^\infty (\varDelta _r)} \le C_0, \quad \Vert \nabla '\varphi _r\Vert _{L^\infty (\varDelta _r)} \le C_1, \quad \Vert \nabla '^2\varphi _r\Vert _{L^\infty (\varDelta _r)} \le C_2, \end{aligned}$$

where $C_0,\,C_1,\,C_2$ are constants independent of r, and $\nabla '$ means $\nabla _{y'}$.

The smoothness of $\varGamma$ is connected with that of the signed distance function d(x) defined by

$$\begin{aligned} d(x) = {\left\{ \begin{array}{ll} -\mathrm {dist}(x, \varGamma ) &{} \text {if }\; x\in \Omega , \\ \mathrm {dist}(x, \varGamma ) &{} \text {if }\; x\in \Omega ^c. \end{array}\right. } \end{aligned}$$

We collect several known properties on d(x) below. For the details, see e.g. [13, Section 14.6] or [8, Section 7.8]. Let $\varGamma (\delta ) := \{ x\in \mathbb R^N :\; |d(x)| < \delta \}$ be a tubular neighborhood of $\varGamma$ with width $2\delta$. Then there exists $\delta$ depending only on the curvature of $\varGamma$ such that for arbitrary $x\in \varGamma (\delta )$ the decomposition

$$\begin{aligned} x = \pi (x) + d(x)\, n(\pi (x)), \quad \pi (x)\in \varGamma , \end{aligned}$$

(8.1)

is uniquely determined. Here, n is the outer unit normal field defined on $\varGamma$, which coincides with $\nabla d|_{\varGamma }$. We extend n from $\varGamma$ to $\varGamma (\delta )$ by $n(x) = \nabla d(x)$, which also agrees with $n(\pi (x))$. The fact that $\varGamma$ is $C^{1,1}$-smooth implies that $d\in C^{1,1}(\varGamma (\delta ))$, $n\in C^{0,1}(\varGamma (\delta ))$, and $\pi \in C^{0,1}(\varGamma (\delta ))$. We call $\pi : \varGamma (\delta )\rightarrow \varGamma$ the orthogonal projection onto $\varGamma$, in view of its geometrical meaning. We may assume that

$$\begin{aligned} \Vert d\Vert _{L^\infty (\varGamma (\delta ))} \le C_0, \quad \Vert \nabla d\Vert _{L^\infty (\varGamma (\delta ))} \le C_1, \quad \Vert \nabla ^2d\Vert _{L^\infty (\varGamma (\delta ))} \le C_2, \end{aligned}$$

where we re-choose the constants $C_0,\,C_1,\,C_2$ if necessary.

Now we introduce a regular family of triangulations $\{\mathcal T_h\}_{h\downarrow 0}$ of $\overline{\Omega }$ in the sense of Sect. 3.1. As before, we denote by $\mathcal S_h$ the boundary mesh inherited from $\mathcal T_h$, and we set $\overline{\Omega }_h = \cup _{T\in \mathcal T_h}T$ and $\varGamma _h = \cup _{S\in \mathcal S_h}S$. In order for $\varGamma _h$ to be compatible with the local-coordinate system $\{(U_r, y_r, \varphi _r)\}_{r=1}^M$, we assume the following:

• (1) the mesh size h is less than $\min \{h_0, 1\}$;

• (2) for every $r=1,\dots ,M$, $\varGamma _h\cap U_r$ is represented by a graph $\{(y_r', \varphi _{rh}(y_r'))\in \mathbb R^N :\; y_r'\in \varDelta _r \}$;

• (3) every vertex of $S\in \mathcal S_h$ lies on $\varGamma$.

From these we see that every $S\in \mathcal S_h$ is contained in some $U_r$ and that $\varphi _{rh}$ is a piecewise linear interpolation of $\varphi _r$. As a result of interpolation error estimates, for arbitrary r we obtain

$$\begin{aligned} \begin{aligned} \Vert \varphi _{rh}\Vert _{L^\infty (\varDelta _r)}&\le C_0, \\ \Vert \nabla '\varphi _{rh}\Vert _{L^\infty (\varDelta _r)}&\le C_1, \end{aligned} \qquad \begin{aligned} \Vert \varphi _r - \varphi _{rh}\Vert _{L^\infty (\varDelta _r)}&\le C_{0E}h^2, \\ \Vert \nabla '(\varphi _r - \varphi _{rh})\Vert _{L^\infty (\varDelta _r)}&\le C_{1E}h, \end{aligned} \end{aligned}$$

(8.2)

where we re-choose $C_0,\,C_1$ if necessary and the subscript E refers to “error”. In the following, h is made small enough to satisfy $2C_{0E}h^2< \min \{h_0, \delta \}$, which in particular ensures that $\pi$ is well-defined on $\varGamma _h$. We may assume further that $\pi (S)$ is contained in the same $U_r$ that contains S.

Based on the observations above, we see that the orthogonal projection $\pi$ maps $\varGamma _h$ into $\varGamma$. It indeed gives a homeomorphism between $\varGamma _h$ and $\varGamma$, and is element-wisely a diffeomorphism, as shown below. The representation of a function f(x) in each local coordinate $(U_r, y_r, \varphi _r)$ is defined as $\tilde{f}(y_r) := f(A_r^{-1}y_r)$. However, with some abuse of notation, we denote it simply by $f(y_r)$. Then, the local representation of the outer unit normal n associated to $\varGamma$ is given by

$$\begin{aligned} n(y_r', \varphi _r(y_r')) = \frac{1}{ K_r(y_r') } \begin{pmatrix} \nabla '\varphi _r(y_r') \\ -1 \end{pmatrix}, \qquad y_r'\in \varDelta _r, \end{aligned}$$

where $K_r(y_r') := \sqrt{1 + |\nabla '\varphi _r(y_r')|^2}$. We are ready to state the following.

Proposition 8.1

If $h>0$ is sufficiently small, then $\pi |_{\varGamma _h}$ is a homeomorphism between $\varGamma _h$ and $\varGamma$.

Proof

Let us construct an inverse map $\pi ^*: \varGamma \rightarrow \varGamma _h$ of $\pi |_{\varGamma _h}$ which is continuous. To this end, we fix arbitrary $x\in \varGamma$ and choose a local coordinate $(U_r, y_r, \varphi _r)$ of x such that $B(x; h_0)\subset U_r$. For simplicity, we omit the subscript r in the following. In view of the definition of $\pi (x)$ by (8.1), we introduce a segment given by

$$\begin{aligned} \begin{pmatrix} y' \\ \varphi (y') \end{pmatrix} + \frac{t}{K(y')} \begin{pmatrix} \nabla '\varphi (y') \\ -1 \end{pmatrix}, \qquad |t|\le 2C_{0E}h^2. \end{aligned}$$

To each point on this segment we associate its height H(t) with respect to the graph of $\varphi _h$, that is,

$$\begin{aligned} H(t) = \varphi (y') - \frac{t}{K(y')} - \varphi _h\left( y' + \frac{t}{K(y')} \nabla '\varphi (y') \right) . \end{aligned}$$

Then we assert that $\frac{d}{dt}H(t) < 0$ and that $H(-2C_{0E}h^2) > 0$, $H(2C_{0E}h^2) < 0$.

To prove the first assertion, letting

$$\begin{aligned} Y' := y' + \frac{t}{K(y')} \nabla '\varphi (y'), \end{aligned}$$

we have $\frac{d}{dt}H(t) = -\frac{1}{K(y')}(1 + \nabla '\varphi (y')\cdot \nabla '\varphi _h(Y'))$. One sees that

$$\begin{aligned} \nabla '\varphi _h(Y')&= \nabla '\varphi (y') + [ -\nabla '\varphi (y') + \nabla '\varphi (Y') ] + [-\nabla '\varphi (Y') + \nabla '\varphi _h(Y')] \\&=: \nabla '\varphi (y') + I_1 + I_2, \end{aligned}$$

where $I_1$ and $I_2$ satisfy

$$\begin{aligned} |I_1| \le \Vert \nabla '^2\varphi \Vert _{L^\infty (\varDelta )} |Y' - y'| \le C_2\cdot 2C_{0E}h^2, \qquad |I_2| \le C_{1E}h. \end{aligned}$$

Then it follows that

$$\begin{aligned} 1 + \nabla '\varphi (y')\cdot \nabla '\varphi _h(Y') \ge K(y')^2 - C_1(2C_2C_{0E}h^2 + C_{1E}h). \end{aligned}$$

From this we have $\frac{d}{dt}H(t) < 0$ provided $C_1(2C_2C_{0E}h^2 + C_{1E}h) < 1/2$. For the second assertion, one finds that

$$\begin{aligned} H(-2C_{0E}h^2)&= \frac{2C_{0E}h^2}{K(y')} + \varphi (y') - \varphi _h(Y') \\&= \frac{2C_{0E}h^2}{K(y')} + [ \varphi (y') - \varphi (Y') ] + [ \varphi (Y') - \varphi _h(Y') ], \end{aligned}$$

where $Y'$ is $y' - \frac{2C_{0E}h^2}{K(y')} \nabla '\varphi (y')$. By Taylor’s theorem, there exists some $\theta \in (0,1)$ such that

$$\begin{aligned} \varphi (Y') - \varphi (y')&= \nabla '\varphi (y')\cdot (Y' - y') + \frac{1}{2} (Y' - y')^T \nabla '^{2}\varphi |_{y' + \theta (Y' - y')} (Y' - y') \\&\le - \frac{2C_{0E}h^2}{K(y')} |\nabla '\varphi (y')|^2 + 2 C_2C_{0E}^2 h^4. \end{aligned}$$

By the definition of K(y) and by $|\varphi (Y') - \varphi _h(Y')| \le C_{0E}h^2$, we obtain

$$\begin{aligned} H(-2C_{0E}h^2)&\ge 2K(y')C_{0E}h^2 - 2C_2C_{0E}^2 h^4 - C_{0E} h^2 \\&\ge 2C_{0E}h^2 - 2C_2C_{0E}^2 h^4 - C_{0E} h^2 = C_{0E}h^2(1 - 2C_2C_{0E}h^2), \end{aligned}$$

which implies that $H(-2C_{0E}h^2) > 0$ provided $C_2C_{0E}h^2 \le 1/4$. In the same way, the last assertion $H(2C_{0E}h^2) < 0$ can be proved.

From these assertions we deduce that there exists a unique $t^*(x) \in [-2C_{0E}h^2, 2C_{0E}h^2]$ such that $H(t^*(x)) = 0$. Consequently, the map $\pi ^*: \varGamma \rightarrow \varGamma _h; \; x\mapsto x + t^*(x)n(x)$ is well-defined. A direct computation combined with the uniqueness of the decomposition (8.1) shows that $\pi ^*$ is the inverse of $\pi |_{\varGamma _h}$. The continuity of $\pi ^*$, especially that of $t^*$, follows from an argument similar to the proof of the implicit function theorem (see e.g. [20, Theorem 3.2.1]). $\square$

Remark 8.1

Proposition 8.1 enables us to define an exact triangulation of $\varGamma$ by $\pi (\mathcal S_h) = \{ \pi (S) \,:\, S\in \mathcal S_h \}$. In particular, we can subdivide $\varGamma$ into disjoint sets as $\varGamma = \bigcup _{S\in \mathcal S_h}\pi (S)$. Under assumption (3) this fact is obvious to see for the two-dimensional case, but not for the three-dimensional case. Furthermore, the above construction of an exact triangulation even applies to the situation where the vertices of $\varGamma _h$ do not exactly belong to $\varGamma$ but still lie closely to it so that (8.2) is fulfilled. This may be useful information and is left as a side remark.

Next we observe that S and $\pi (S)$ admit the same domain of parametrization for each $S\in \mathcal S_h$, which is important in the subsequent analysis. To describe this fact, we choose a local coordinate $(U_r, y_r, \varphi _r)$ such that $U_r \supset S\cup \pi (S)$, and introduce the projection to the base set $b_r:\mathbb R^N\rightarrow \mathbb R^{N-1}$ by $b_r(y_r) = y_r'$. The domain of parametrization is then defined to be $S' = b_r(\pi (S))$. We see that the mappings

are bijective and that $\varPhi$ is smooth on $S'$. If in addition $\varPhi _h$, especially $t^*$, is also smooth on $S'$, then $\varPhi$ and $\varPhi _h$ may be employed as smooth parametrizations for $\pi (S)$ and S respectively. The next proposition verifies that this is indeed the case.

Proposition 8.2

Under the setting above, we have

$$\begin{aligned} \Vert t^*\Vert _{L^\infty (S')} \le \tilde{C}_{0E} h^2, \qquad \Vert \nabla ' t^*\Vert _{L^\infty (S')} \le \tilde{C}_{1E} h, \end{aligned}$$

where $\tilde{C}_{0E}$ and $\tilde{C}_{1E}$ are constants depending only on N and $\varGamma$.

Proof

Since the first relation is already obtained in Proposition 8.1 with $\tilde{C}_{0E} = 2C_{0E}$, we focus on proving the second one. For notational simplicity, we omit the subscript r and also use the abbreviation $\partial _i = \frac{\partial }{\partial y_i}\,(i=1,\dots ,N)$. The fact that $t^*$ is differentiable with respect to $y'$ can be shown in a way similar to the proof of the implicit function theorem. Thereby it remains to evaluate the supremum norm of $\nabla ' t$ in $S'$, which we address in the following.

Recall that $t^*(y')$ is determined according to the equation

$$\begin{aligned} \tilde{t}(y') = \varphi (y') - \varphi _h(y' + \tilde{t}(y') \nabla '\varphi (y')), \end{aligned}$$

(8.3)

where we have set $\tilde{t}(y') := t^*(y')/K(y')$. Because $\nabla 't^* = K\nabla '\tilde{t} + \frac{\nabla '^2\varphi \nabla '\varphi }{K^2}\, t^*$, it follows that

$$\begin{aligned} \Vert \nabla 't^*\Vert _{L^\infty (S')} \le (1 + C_1) \Vert \nabla '\tilde{t}\Vert _{L^\infty (S')} + C_2C_1\tilde{C}_{0E}h^2. \end{aligned}$$

Therefore, it suffices to prove that $\Vert \nabla '\tilde{t}\Vert _{L^\infty (S')} \le Ch$; here and hereafter C denotes various constants which depends only on N and $\varGamma$.

Applying $\nabla '$ to (8.3) gives

$$\begin{aligned} \big ( 1 + \nabla '\varphi (y')\cdot \nabla '\varphi _h(Y') \big ) \nabla '\tilde{t} = \nabla '\varphi (y') - \nabla '\varphi _h(Y') - \tilde{t}(y') \nabla '^2\varphi (y') \nabla '\varphi _h(Y'), \end{aligned}$$

(8.4)

where $Y' := y' + \tilde{t}(y') \nabla '\varphi (y')$. By the same way as we estimated $I_1$ and $I_2$ in the proof of Proposition 8.1, we obtain

$$\begin{aligned} |\nabla '\varphi (y') - \nabla '\varphi _h(Y')|&\le C_2\tilde{C}_{0E}h^2 + C_{1E}h \le Ch, \\ 1 + \nabla '\varphi (y')\cdot \nabla '\varphi _h(Y')&\ge K(y')^2 - C_1(C_2\tilde{C}_{0E}h^2 + C_{1E}h) \ge \frac{1}{2}. \end{aligned}$$

Also we see that

$$\begin{aligned} |\tilde{t}(y') \nabla '^2\varphi (y') \nabla '\varphi _h(Y')| \le \tilde{C}_{0E}h^2\cdot C_2C_1 \le Ch^2. \end{aligned}$$

Combining these observations with (8.4), we deduce the desired estimate $\Vert \nabla '\tilde{t}\Vert _{L^\infty (S')} \le Ch$. $\square$

Remark 8.2

When $\varGamma \in C^{2,1}$, further differentiation of (8.4) gives us $t^*\in C^{1,1}(\pi (S))$; in fact we have $\Vert \nabla '^2 t^*\Vert _{L^\infty (S')} \le \tilde{C}_{2E}$ with some constant $\tilde{C}_{2E}$. This implies that $\pi |_{S}$ is a $C^{1,1}$-diffeomorphism between S and $\pi (S)$. However, since $\nabla '\phi _h$ is smooth only within b(S), $\pi$ is not globally a diffeomorphism.

Now we give an error estimate for surface integrals on $\varGamma$ and $\varGamma _h$. Heuristically speaking, the result reads $|d\gamma - d\gamma _h| \le O(h^2)$, which may be found in the literature (see e.g. [11]). Here and hereafter, we denote the surface elements of $\varGamma$ and $\varGamma _h$ by $d\gamma$ and $d\gamma _h$, respectively.

Theorem 8.1

Let $S\in \mathcal S_h$ and f be an integrable function on S. Then we have

$$\begin{aligned} \left| \int _{\pi (S)}f\,d\gamma - \int _S f\circ \pi \, d\gamma _h \right| \le Ch^2 \int _S |f|\,d\gamma , \end{aligned}$$

where C is a constant depending only on N and $\varGamma$.

Proof

Let $(U_r, y_r, \varphi _r)$ be a local coordinate that contains $S\cup \pi (S)$. We omit the subscript r and use the abbreviation $\partial _i = \frac{\partial }{\partial y_i}\,(i=1,\dots ,N)$. We represent the surface integral using the parametrization $\varPhi$ as follows:

$$\begin{aligned} \int _{\pi (S)}f\,d\gamma = \int _{S'} f(\varPhi (y')) \sqrt{\mathrm {det}\,G}\,dy', \end{aligned}$$

where $G=(G_{ij})_{1\le i,j\le N-1}$ denotes the Riemannian metric tensor given by $G_{ij} = \partial _i\varPhi \cdot \partial _j\varPhi$ (dot means the inner product in $\mathbb R^N$). Similarly, noting that $\pi \circ \varPhi _h = \varPhi$, one obtains

$$\begin{aligned} \int _{S}f\circ \pi \,d\gamma _h = \int _{S'} f(\varPhi (y')) \sqrt{\mathrm {det}\,G_h}\,dy', \end{aligned}$$

where $G_h$ is given by $G_{h,ij} = \partial _i\varPhi _h \cdot \partial _j\varPhi _h$. Then we assert that:

$$\begin{aligned} \Vert G_h - G\Vert _{L^\infty (S')} \le Ch^2. \end{aligned}$$

To prove this, noting that $\varPhi _h = \varPhi + t^*n\circ \varPhi$, we compute each component of $G_h-G$ as follows:

$$\begin{aligned} G_{h,ij} - G_{ij}&= \partial _i\varPhi \cdot \partial _j(\varPhi _h - \varPhi ) + \partial _j\varPhi \cdot \partial _i(\varPhi _h - \varPhi ) + \partial _i(\varPhi _h - \varPhi ) \cdot \partial _j(\varPhi _h - \varPhi ) \\&= \partial _i\varPhi \cdot \partial _j(t^*n\circ \varPhi ) + \partial _j\varPhi \cdot \partial _i(t^*n\circ \varPhi ) + \partial _i(t^*n\circ \varPhi ) \cdot \partial _j(t^*n\circ \varPhi ) \\&=: I_1 + I_2 + I_3. \end{aligned}$$

For $I_1$, we notice that $\partial _i\varPhi$ is a tangent vector so that $\partial _i\varPhi \cdot n\circ \varPhi =0$. This yields

$$\begin{aligned} I_1 = \partial _i\varPhi \cdot t^*\partial _j(n\circ \varPhi ) = t^*\partial _i\varPhi \cdot (\partial _jn + \partial _j\varphi \,\partial _Nn)|_{\varPhi }, \end{aligned}$$

which is estimated by $\tilde{C}_{0E}h^2(1 + C_1)(C_2 + C_1C_2)$ thanks to Proposition 8.2. $I_2$ can be bounded in the same manner. To estimate $I_3$, we observe that

$$\begin{aligned} \partial _i(t^*n\circ \varPhi ) = (\partial _i t^*) n\circ \varPhi + t^* \big ( \partial _in + \partial _i\varphi \,\partial _Nn \big )|_{\varPhi }, \end{aligned}$$

which is bounded by $\tilde{C}_{1E}h + \tilde{C}_{0E}h^2(C_2 + C_1C_2) \le Ch$. Similarly one gets $|\partial _j(t^*n\circ \varPhi )| \le Ch$, hence it follows that $|I_3| \le Ch^2$. Therefore, $|G_{h,ij} - G_{ij}| \le Ch^2$, which proves the assertion.

Now we use the following crude estimate for perturbation of determinants (cf. [17, equation (3.13)]): if A and B are $N\times N$ matrices such that $|A_{ij}| \le a$ and $|B_{ij}|\le b$ for all i, j, then

$$\begin{aligned} |\mathrm {det}\,(A+B) - \mathrm {det}\,A| \le N!N(a+b)^{N-1}b. \end{aligned}$$

Combining this with the assertion above and also with $\sqrt{\alpha }- \sqrt{\beta }= (\alpha - \beta )/(\sqrt{\alpha }+ \sqrt{\beta })$, we obtain

$$\begin{aligned} \Vert \sqrt{\mathrm {det}\,G} - \sqrt{\mathrm {det}\,G_h}\Vert _{L^\infty (S')} \le Ch^2. \end{aligned}$$

In addition, note that $\sqrt{\mathrm {det}\,G} = \sqrt{1 + |\nabla '\varphi |^2} \ge 1$. Consequently,

$$\begin{aligned} \left| \int _{\pi (S)}f\,d\gamma - \int _S f\circ \pi \, d\gamma _h \right| \le Ch^2\,\int _{S'} |f(\varPhi (y'))| \sqrt{\mathrm {det}\,G} \,dy' = Ch^2 \int _{\pi (S)}|f|\,d\gamma , \end{aligned}$$

which proves the theorem. $\square$

Remark 8.3

Adding up the results of the theorem for all $S\in \mathcal S_h$ yields

$$\begin{aligned} \left| \int _\varGamma f\,d\gamma - \int _{\varGamma _h} f\circ \pi \, d\gamma _h \right| \le Ch^2 \int _\varGamma |f|\,d\gamma . \end{aligned}$$

It also follows that $|\int _{\varGamma _h} f\circ \pi \, d\gamma _h| \le C\int _\varGamma |f|\,d\gamma$. Choosing in particular $|f|^p$ as the integrand gives $\Vert f\circ \pi \Vert _{L^p(\varGamma _h)} \le C^{1/p}\Vert f\Vert _{L^p(\varGamma )}$ for $p\in [1,\infty ]$.

Let f be a smooth function given on $\varGamma$. Then its transformation to $\varGamma _h$ is defined by $f\circ \pi$. However, if f is extended to a neighborhood of $\varGamma$, e.g. to $\varGamma (\delta )$, then we may also consider f’s natural trace on $\varGamma _h$. The next theorem provides error estimation of these two quantities.

Theorem 8.2

Let $f\in W^{1,p}(\varGamma (\delta _1))$, where $p\in [1,\infty ]$ and $\delta _1 \in [\tilde{C}_{0E}h^2, 2\tilde{C}_{0E}h^2]$. Then,

$$\begin{aligned} \Vert f - f\circ \pi \Vert _{L^p(\varGamma _h)} \le C\delta _1^{1-1/p}\Vert f\Vert _{W^{1,p}(\varGamma (\delta _1))}, \end{aligned}$$

where C is a constant depending only on $p,\,N,\,\Omega$.

Proof

Since $\varGamma (\delta _1) = \bigcup _{S\in \mathcal S_h} \pi (S,\delta _1)$, where $\pi (S, \delta _1) = \{ x\in \varGamma (\delta _1) : \pi (x)\in \pi (S) \}$ denotes a tubular neighborhood of S, it suffices to prove that

$$\begin{aligned} \int _S |f - f\circ \pi |^p\,d\gamma \le C\delta _1^{p - 1} \int _{\pi (S, \delta _1)} |\nabla f|^p\,dx \qquad \forall S\in \mathcal S_h. \end{aligned}$$

(8.5)

To this end, using the notation in Theorem 8.1, we estimate the left-hand side of (8.5) by

$$\begin{aligned} \int _S |f - f\circ \pi |^p\,d\gamma&= \int _{S'} |f\circ \varPhi _h - f\circ \varPhi |^p \sqrt{\mathrm {det}\,G_h}\,dy' \\&\le C \int _{S'} |f\circ \varPhi _h - f\circ \varPhi |^p\,dy'. \end{aligned}$$

Here, for fixed $y'\in S'$ we have

$$\begin{aligned} f(\varPhi _h(y')) - f(\varPhi (y'))&= \int _0^1 \frac{d}{ds} f\big ( \varPhi (y') + s(\varPhi _h(y') - \varPhi (y')) \big )\,ds \\&= \int _0^1 \frac{d}{ds} f\big ( \varPhi (y') + s\, t^*(y') n(\varPhi (y')) \big )\,ds \\&= \int _0^1 t^*(y') n(\varPhi (y')) \cdot \nabla f\big ( \varPhi (y') + s\, t^*(y') n(\varPhi (y')) \big )\,ds \\&= \int _0^{t^*(y')} n(\varPhi (y')) \cdot \nabla f\big ( \varPhi (y') + t n(\varPhi (y')) \big )\,dt. \end{aligned}$$

Because $|t^*(y')| \le \tilde{C}_{0E}h^2 \le \delta _1$, it follows that

$$\begin{aligned} |f(\varPhi _h(y')) - f(\varPhi (y'))|&\le \int _{-\delta _1}^{\delta _1} \big | \nabla f\big ( \varPhi (y') + tn(\varPhi (y')) \big ) \big |\,dt \\&\le (2\delta _1)^{1-1/p} \left( \int _{-\delta _1}^{\delta _1} \big | \nabla f\big ( \varPhi (y') + tn(\varPhi (y')) \big ) \big |^p\,dt \right) ^{1/p}, \end{aligned}$$

where we have used Hölder’s inequality. Consequently,

$$\begin{aligned} \int _S |f - f\circ \pi |^p\,d\gamma \le C\delta _1^{p-1} \int _{S'\times [-\delta _1,\delta _1]} \big | \nabla f\big ( \varPhi (y') + tn(\varPhi (y')) \big ) \big |^p\,dy'dt. \end{aligned}$$

(8.6)

On the other hand, we observe that the N-dimensional transformation

$$\begin{aligned} \Psi : S'\times [-\delta _1, \delta _1] \rightarrow \pi (S, \delta _1); \quad (y', t) \mapsto \varPhi (y') + tn(\varPhi (y')) \end{aligned}$$

(8.7)

is bijective and smooth. Application of this transformation to the right-hand side of (8.5) leads to

$$\begin{aligned} \int _{\pi (S, \delta _1)} |\nabla f|^p\,dx = \int _{S'\times [-\delta _1, \delta _1]} \big | \nabla f\big ( \varPhi (y') + tn(\varPhi (y')) \big ) \big |^p\, |\mathrm {det}\, J| \,dy'dt, \end{aligned}$$

where $J = (\partial _1\varPhi + t\partial _1(n\circ \varPhi ), \cdots , \partial _{N-1}\varPhi + t\partial _{N-1}(n\circ \varPhi ), n\circ \varPhi )$ denotes the Jacobi matrix of $\Psi$. Letting $\tilde{J} := (\partial _1\varPhi , \cdots , \partial _{N-1}\varPhi , n\circ \varPhi )$, we find that

$$\begin{aligned} \Vert J - \tilde{J}\Vert _{L^\infty (S'\times [-\delta _1, \delta _1])} \le C\delta _1, \end{aligned}$$

because $|t\partial _i(n\circ \varPhi )| = |t (\partial _in + \partial _i\varphi \partial _Nn)_{|\varPhi }| \le \delta _1(C_2 + C_1C_2)$. This implies

$$\begin{aligned} \Vert \mathrm {det}\, J - \mathrm {det}\,\tilde{J}\Vert _{L^\infty (S'\times [-\delta _1, \delta _1])} \le C\delta _1, \end{aligned}$$

which combined with $\mathrm {det}\,\tilde{J} = K(y') \ge 1$ yields $\mathrm {det}\,J \ge 1/2$ if h is sufficiently small. Therefore,

$$\begin{aligned} \int _{\pi (S, \delta _1)} |\nabla f|^p\,dx \ge \frac{1}{2} \int _{S'\times [-\delta _1, \delta _1]} \big | \nabla f\big ( \varPhi (y') + tn(\varPhi (y')) \big ) \big |^p \,dy'dt. \end{aligned}$$

(8.8)

The desired estimate (8.5) is now a consequence of (8.6) and (8.8). This completes the proof. $\square$

Finally, we show that the $L^p$-norm in a tubular neighborhood can be bounded in terms of its width. Such estimate is stated e.g. in [35, Lemma 2.1] or in [28, equation (3.6)]. However, since we could not find a full proof of this fact (especially for $N=3$) in the literature, we present it here.

Theorem 8.3

Under the same assumptions as in Theorem 8.2, we have

$$\begin{aligned} \Vert f\Vert _{L^p(\varGamma (\delta _1))} \le C (\delta _1 \Vert \nabla f\Vert _{L^p(\varGamma (\delta _1))} + \delta _1^{1/p} \Vert f\Vert _{L^p(\varGamma )}), \end{aligned}$$

where C is a constant depending only on $p,\,N,\,\Omega$.

Proof

We adopt the same notation as in the proofs of Theorems 8.1 and 8.2. Then it suffices to prove that

$$\begin{aligned} \int _{\pi (S,\delta _1)} |f|^p\,dy \le C \left( \delta _1^p \int _{\pi (S,\delta _1)} |\nabla f|^p\,dy + \delta _1 \int _{\pi (S)} |f|^p\,d\gamma \right) \qquad \forall S\in \mathcal S_h. \end{aligned}$$

To this end, using the transformation $\Psi$ given in (8.7) we express the left-hand side as

$$\begin{aligned} \int _{\pi (S,\delta _1)} |f|^p\,dy&= \int _{S'\times [-\delta _1,\delta _1]} |f(\Psi (y', t))|^p|\mathrm {det}\,J(y',t)|\,dy'dt \\&\le C \int _{S'\times [-\delta _1,\delta _1]} \big ( |f(\Psi (y', t)) - f(\varPhi (y'))|^p + |f(\varPhi (y'))|^p \big ) \,dy'dt \\&=: I_1 + I_2. \end{aligned}$$

For $I_1$, we see from the same argument as before that

$$\begin{aligned} |f(\Psi (y', t)) - f(\varPhi (y'))|^p \le (2\delta _1)^{p-1} \int _{-\delta _1}^{\delta _1} \big | \nabla f\big ( \varPhi (y') + sn(\varPhi (y')) \big ) \big |^p\,ds, \end{aligned}$$

which yields

$$\begin{aligned} |I_1| \le C\delta _1^p \int _{S'\times [-\delta _1,\delta _1]} \big | \nabla f(\Psi (y',s)) \big |^p |\mathrm {det}\,J(y',s)|\,dy'ds = C\delta _1^p \int _{\pi (S,\delta _1)} |\nabla f|^p\,dy. \end{aligned}$$

For $I_2$, it follows that

$$\begin{aligned} |I_2|= & {} 2C\delta _1 \int _{S'} |f(\varPhi (y'))|^p\,dy' \le 2C\delta _1 \int _{S'} |f(\varPhi (y'))|^p \sqrt{\mathrm {det}\,G}\,dy'\\= & {} 2C\delta _1 \int _{\pi (S)} |f|^p\,d\gamma . \end{aligned}$$

We have thus obtained the desired estimate, which completes the proof. $\square$

Appendix 2: Error of n and $n_h$

Let us prove that $|n\circ \pi - n_h| \le O(h)$ on $\varGamma _h$ and also that, when $N=2$, it is improved to $O(h^2)$ if the consideration is restricted to the midpoint of edges.

Lemma 9.1

Let n and $n_h$ be the outer unit normals to $\varGamma$ and $\varGamma _h$ respectively. Then there holds

$$\begin{aligned} \Vert n\circ \pi - n_h\Vert _{L^\infty (\varGamma _h)} \le Ch. \end{aligned}$$

(9.1)

If in addition $N=2$, $\varGamma \in C^{2,1}$, and $m_S$ denotes the midpoint of $S\in \mathcal S_h$, then

$$\begin{aligned} \sup _{S\in \mathcal S_h}|n\circ \pi (m_S) - n_h(m_S)| \le Ch^2. \end{aligned}$$

(9.2)

Here, C is a constant depending only on N and $\varGamma$.

Proof

Let $S\in \mathcal S_h$ be arbitrary and let $(U_r, y_r, \phi _r)$ be a local coordinate that contains $S\cup \pi (S)$. We omit the subscript r in the following. One sees that n and $n_h$ are represented as

$$\begin{aligned} n(y', \varphi (y'))= & {} \frac{1}{\sqrt{1 + |\nabla '\varphi |^2}} \begin{pmatrix}\nabla '\varphi \\ -1\end{pmatrix}, \\ n_h(y', \varphi _h(y'))= & {} \frac{1}{\sqrt{1 + |\nabla '\varphi _h|^2}} \begin{pmatrix}\nabla '\varphi _h \\ -1\end{pmatrix}, \quad y'\in \varDelta . \end{aligned}$$

A direct computation gives

$$\begin{aligned} |n(y', \varphi (y')) - n_h(y', \varphi _h(y'))| \le 2|\nabla '(\varphi (y') - \varphi _h(y'))| \le 2C_{1E}h. \end{aligned}$$

(9.3)

This combined with the observation that

$$\begin{aligned} |n\circ \pi (y', \varphi _h(y')) - n(y', \varphi (y'))|&= |n\circ \pi (y', \varphi _h(y')) - n\circ \pi (y', \varphi (y'))| \\&\le C_1|\pi (y', \varphi _h(y')) - \pi (y', \varphi (y'))| \\&\le C_1 \Vert \nabla \pi \Vert _{L^\infty (\varGamma (\delta ))} |\varphi _h(y') - \varphi (y')| \\&\le C_1 \Vert \nabla \pi \Vert _{L^\infty (\varGamma (\delta ))} C_{0E}h^2 \end{aligned}$$

proves (9.1). When $N=2$ and $\varGamma \in C^{2,1}$, by using Taylor expansion, we find that $m_S$ is a point of super-convergence such that $|\frac{d\varphi _h}{dy_1} (b(m_S)) - \frac{d\varphi }{dy_1} (b(m_S))| \le Ch^2$, which improves (9.3) to $O(h^2)$. Thus (9.2) is proved. $\square$

Remark 9.1

In the case $N=3$, it is known that the barycenter of a triangle is not a point of super-convergence for the derivative of linear interpolations; see [34, p. 1930]. For this reason, (9.2) holds only for $N=2$.