Optimal Error Estimates of the Penalty Finite Element Method for the Unsteady Navier–Stokes Equations with Nonsmooth Initial Data

Abstract

In this paper, both semi-discrete and fully discrete finite element methods are analyzed for the penalized two-dimensional unsteady Navier–Stokes equations with nonsmooth initial data. First order backward Euler method is applied for the time discretization, whereas conforming finite element method is used for the spatial discretization. Optimal $L^2$ error estimates for the semi-discrete as well as the fully discrete approximations of the velocity and of the pressure are derived for realistically assumed conditions on the data. The main ingredient in the proof is the appropriate exploitation of the inverse of the penalized Stokes operator, negative norm estimates and time weighted estimates. Two numerical examples one in 2D and one in 3D are presented whose results are conforming our theoretical findings. Finally, computational experiments on benchmark problem: one on lid driven cavity problem and other on flow around a cylinder with low viscosity are discussed.

Appendix

Proof of the Lemma 5

Choose $\phi ={\textbf{u}}_\varepsilon$ in (8) and use the Cauchy–Schwarz inequality and the Poincaré inequality with Lemma 3 ($\Vert {\textbf{u}}_\varepsilon \Vert ^2 \le \frac{1}{\lambda _1}\Vert \nabla {\textbf{u}}_\varepsilon \Vert ^2\le \frac{c_0^2}{\lambda _1}\Vert A_{\varepsilon }^{\frac{1}{2}}{\textbf{u}}_\varepsilon \Vert ^2$) to find that

$$\begin{aligned} \frac{d}{dt}\Vert {\textbf{u}}_\varepsilon \Vert ^2 + \nu \Vert A_{\varepsilon }^{\frac{1}{2}}{\textbf{u}}_\varepsilon \Vert ^2 \le \frac{c_0^2}{\nu \lambda _1}\Vert \textbf{f}\Vert ^2. \end{aligned}$$

(109)

Note that the non-linear term vanishes due to (7). Now multiply by $e^{2\alpha t}$ and integrate from 0 to t to obtain

$$\begin{aligned} e^{2\alpha t}\Vert {\textbf{u}}_\varepsilon (t)\Vert ^2+(\nu -\frac{2c_0^2\alpha }{\lambda _1})\int _0^t e^{2\alpha s}\Vert A_{\varepsilon }^{\frac{1}{2}}{\textbf{u}}_\varepsilon (s)\Vert ^2ds \le \Vert \textbf{u}_{\varepsilon 0}\Vert ^2+\frac{c_0^2(e^{2\alpha t}-1)}{2\alpha \nu \lambda _1}\Vert \textbf{f}\Vert ^2_{\infty }. \end{aligned}$$

(110)

With $0< \alpha < \frac{\nu \lambda _1}{2c_0^2}$, we have $(\nu -\frac{2c_0^2\alpha }{\lambda _1})>0$. Multiply through out by $e^{-2\alpha t}$ to conclude the first proof. Now, we integrate (109) with respect to time from t to $t+T$ for any $T>0$, we have

$$\begin{aligned} \Vert {\textbf{u}}_\varepsilon (t+T)\Vert ^2+\nu \int _t^{t+T} \Vert A_{\varepsilon }^{\frac{1}{2}}{\textbf{u}}_\varepsilon (s)\Vert ^2ds \le \Vert {\textbf{u}}_\varepsilon (t)\Vert ^2 + \frac{c_0^2T}{\nu \lambda _1}\Vert \textbf{f}\Vert ^2_{\infty }. \end{aligned}$$

(111)

For the second estimate, choose $\phi =A_{\varepsilon }^{m+1}{\textbf{u}}_\varepsilon$ in (8). When $m=0$, we find that

$$\begin{aligned} \frac{1}{2}\frac{d}{dt} \Vert A_{\varepsilon }^{\frac{1}{2}}{\textbf{u}}_\varepsilon \Vert ^2 +\nu \Vert A_{\varepsilon }{\textbf{u}}_\varepsilon \Vert ^2 = (\textbf{f},A_{\varepsilon }{\textbf{u}}_\varepsilon ) -\tilde{b}({\textbf{u}}_\varepsilon ,{\textbf{u}}_\varepsilon ,A_{\varepsilon }{\textbf{u}}_\varepsilon ). \end{aligned}$$

(112)

A use of Ladyzhenskaya’s inequality [31] ($\Vert \phi \Vert _{L^4} \le C \Vert \phi \Vert ^{\frac{1}{2}}\Vert \nabla \phi \Vert ^{\frac{1}{2}}$, and $\Vert \nabla \phi \Vert _{L^4} \le C \Vert \nabla \phi \Vert ^{\frac{1}{2}}\Vert \varDelta \phi \Vert ^{\frac{1}{2}}$) with Lemma 3, the Young’s inequalities, we bound the nonlinear term as

$$\begin{aligned} \tilde{b}({\textbf{u}}_\varepsilon ,{\textbf{u}}_\varepsilon ,A_{\varepsilon }{\textbf{u}}_\varepsilon )&\le \Vert {\textbf{u}}_\varepsilon \Vert _{L^4}\Vert \nabla {\textbf{u}}_\varepsilon \Vert _{L^4} \Vert A_{\varepsilon }{\textbf{u}}_\varepsilon \Vert \nonumber \\&\le C \Vert {\textbf{u}}_\varepsilon \Vert ^{\frac{1}{2}}\Vert A_{\varepsilon }^{\frac{1}{2}}{\textbf{u}}_\varepsilon \Vert \Vert A_{\varepsilon }{\textbf{u}}_\varepsilon \Vert ^{3/2} \nonumber \\&\le C \Vert {\textbf{u}}_\varepsilon \Vert ^2\Vert A_{\varepsilon }^{\frac{1}{2}}{\textbf{u}}_\varepsilon \Vert ^4 + \frac{\nu }{4}\Vert A_{\varepsilon }{\textbf{u}}_\varepsilon \Vert ^2. \end{aligned}$$

(113)

Substitute the above estimate in (112) to find that

$$\begin{aligned} \frac{d}{dt} (\Vert A_{\varepsilon }^{\frac{1}{2}}{\textbf{u}}_\varepsilon \Vert ^2) +\nu \Vert A_{\varepsilon }{\textbf{u}}_\varepsilon \Vert ^2 \le C \big (\Vert {\textbf{u}}_\varepsilon \Vert ^2\Vert A_{\varepsilon }^{\frac{1}{2}}{\textbf{u}}_\varepsilon \Vert ^2\big ) \Vert A_{\varepsilon }^{\frac{1}{2}}{\textbf{u}}_\varepsilon \Vert ^2 + \frac{2}{\nu }\Vert \textbf{f}\Vert ^2). \end{aligned}$$

(114)

We now apply uniform Gronwall’s Lemma (Lemma 1) in (114) and use (110) and (111) to conclude that $\Vert A_{\varepsilon }^{\frac{1}{2}}{\textbf{u}}_\varepsilon (t)\Vert ^2$ is uniformly bounded with respect to t on $[T,\infty )$. Precisely

$$\begin{aligned} \Vert A_{\varepsilon }^{\frac{1}{2}}{\textbf{u}}_\varepsilon (t)\Vert ^2 \le C, \quad \forall t\ge T. \end{aligned}$$

(115)

For $0\le t\le T$, we use the classical Gronwall’s lemma [18, 26] in (114) and obtain

$$\begin{aligned} \Vert A_{\varepsilon }^{\frac{1}{2}}{\textbf{u}}_\varepsilon (t)\Vert ^2 \le C, \quad \text {for}~~ 0\le t\le T. \end{aligned}$$

(116)

Finally, multiply (114) by $e^{2\alpha t}$ and integrate with respect to time from 0 to t and use the estimates (110), (115) and (116) to complete the second proof when $r=0$. For $r=1$, we need some intermediate estimate. First we take $\phi =e^{2\alpha t}{\textbf{u}}_{\varepsilon t}$ with $\hat{\textbf{u}}_\varepsilon =e^{\alpha t}{\textbf{u}}_\varepsilon$ in (8) to obtain

$$\begin{aligned} \frac{\nu }{2}\frac{d}{dt} \Vert A_{\varepsilon }^{\frac{1}{2}}\hat{\textbf{u}}_\varepsilon \Vert ^2+ \Vert \hat{\textbf{u}}_{\varepsilon t}\Vert ^2= \alpha \nu \Vert A_{\varepsilon h}^{\frac{1}{2}}\hat{\textbf{u}}_\varepsilon \Vert ^2 + (\hat{\textbf{f}},{\textbf{u}}_{\varepsilon t}) -e^{2\alpha t}\tilde{b}({\textbf{u}}_\varepsilon ,{\textbf{u}}_\varepsilon ,{\textbf{u}}_{\varepsilon t}). \end{aligned}$$

(117)

We can estimate the nonlinear term on the right hand side of (117) similar to (113) and integrate both sides with respect to time to find that

$$\begin{aligned} \nu \Vert A_{\varepsilon }^{\frac{1}{2}}\hat{\textbf{u}}_\varepsilon \Vert ^2 + \int _0^t\Vert \hat{\textbf{u}}_{\varepsilon t}\Vert ^2 ds \le C \bigg [\int _0^t \Big (\Vert A_{\varepsilon }^{\frac{1}{2}}\hat{\textbf{u}}_\varepsilon \Vert ^2 + \Vert \hat{\textbf{f}}\Vert ^2+ \Vert A_{\varepsilon }^{\frac{1}{2}}{\textbf{u}}_\varepsilon \Vert ^2\Vert A_{\varepsilon }\hat{\textbf{u}}_\varepsilon \Vert ^2 \big ) ds\bigg ]. \end{aligned}$$

Now a use of (110) and (115) lead us to the intermediate estimate.

$$\begin{aligned} \Vert A_{\varepsilon }^{\frac{1}{2}}{\textbf{u}}_\varepsilon (t)\Vert ^2 + e^{-2\alpha t}\int _0^t e^{2\alpha s}\Vert {\textbf{u}}_{\varepsilon s}(s)\Vert ^2 ds \le C. \end{aligned}$$

(118)

We now differentiate (8) with respect to time and deduce that

$$\begin{aligned} ({\textbf{u}}_{\varepsilon tt},\phi )+\nu a_{\varepsilon }({\textbf{u}}_{\varepsilon t},\phi ) = (\textbf{f}_t,\phi ) -\tilde{b}({\textbf{u}}_{\varepsilon t},{\textbf{u}}_\varepsilon ,\phi )-\tilde{b}({\textbf{u}}_\varepsilon ,{\textbf{u}}_{\varepsilon t},\phi ), \quad \forall \phi \in \textbf{H}_0^1. \end{aligned}$$

(119)

Take $\phi =\sigma (t){\textbf{u}}_{\varepsilon tt}$ in (119) and use Lemma 6, the Cauchy–Schwarz inequality to reach at

$$\begin{aligned} \frac{d}{dt}(\sigma (t)\Vert {\textbf{u}}_{\varepsilon t}\Vert ^2) + \nu \sigma (t)\Vert A_{\varepsilon }^{\frac{1}{2}}{\textbf{u}}_{\varepsilon t}\Vert ^2 \le C e^{2\alpha t}\Vert {\textbf{u}}_{\varepsilon t}\Vert ^2 +C\sigma (t)\Big (\Vert \textbf{f}_t\Vert ^2+ \Vert {\textbf{u}}_{\varepsilon t}\Vert ^2\Vert A_{\varepsilon h}^{\frac{1}{2}}{\textbf{u}}_\varepsilon \Vert ^2\Big ). \end{aligned}$$

Integrate with respect to time and use (118), (115) and (116) to obtain

$$\begin{aligned} \tau (t)\Vert {\textbf{u}}_{\varepsilon t}(t)\Vert ^2 + \nu e^{-2\alpha t}\int _0^t \sigma (s)\Vert A_{\varepsilon }^{\frac{1}{2}}{\textbf{u}}_{\varepsilon s}(s)\Vert ^2 ds \le C. \end{aligned}$$

(120)

Now we are in position to complete the proof of the second estimate when $m=1$. For this, we set $\phi =A_{\varepsilon }{\textbf{u}}_\varepsilon$ in (8) and rewrite it and use (113) and the Cauchy–Schwarz inequality to arrive at

$$\begin{aligned} \nu \Vert A_{\varepsilon h}{\textbf{u}}_\varepsilon \Vert ^2&= (\textbf{f},A_{\varepsilon }{\textbf{u}}_\varepsilon )-({\textbf{u}}_{\varepsilon t},A_{\varepsilon }{\textbf{u}}_\varepsilon )-\tilde{b}({\textbf{u}}_\varepsilon ,{\textbf{u}}_\varepsilon ,A_{\varepsilon }{\textbf{u}}_\varepsilon )\\&\le C \big ( \Vert \textbf{f}\Vert ^2 +\Vert {\textbf{u}}_{\varepsilon t}\Vert ^2 + \Vert {\textbf{u}}_\varepsilon \Vert ^2\Vert A_{\varepsilon }^{\frac{1}{2}}{\textbf{u}}_\varepsilon \Vert ^4\big ) + \frac{\nu }{2}\Vert A_{\varepsilon }{\textbf{u}}_\varepsilon \Vert ^2. \end{aligned}$$

Multiply by $\tau (t)$ and use (110), (115), (116) and (120) to complete the second proof.

Proof of the well-posedness of the discrete solution of problem (66)

We can rewritten (66) as

$$\begin{aligned} (\textbf{U}_{\varepsilon }^n,\phi _h)+\nu k a_{\varepsilon }(\textbf{U}_{\varepsilon }^n,\phi _h)+k\tilde{b}(\textbf{U}_{\varepsilon }^n,\textbf{U}_{\varepsilon }^n,\phi _h)= k(\textbf{f}^n,\phi _h) +(\textbf{U}_{\varepsilon }^{n-1},\phi _h),~~~\forall \phi _h\in \textbf{H}_h. \end{aligned}$$

Consider a function $F:\textbf{H}_h\rightarrow \textbf{H}_h$ such that

$$\begin{aligned} (F(\textbf{v}),\phi _h) = (\textbf{v},\phi _h)+\nu k a_{\varepsilon }(\textbf{v},\phi _h)+k\tilde{b}(\textbf{v},\textbf{v},\phi _h)- k(\textbf{f}^n,\phi _h) -(\textbf{U}_{\varepsilon }^{n-1},\phi _h),~~~\forall \phi _h\in \textbf{H}_h. \end{aligned}$$

Clearly, F is continuous. Then, a use of pointcaré inequality and inverse hypothesis yields

$$\begin{aligned} (F(\textbf{U}_{\varepsilon }^n),\textbf{U}_{\varepsilon }^n)&= (\textbf{U}_{\varepsilon }^n,\textbf{U}_{\varepsilon }^n)+\nu k a_{\varepsilon }(\textbf{U}_{\varepsilon }^n,\textbf{U}_{\varepsilon }^n)+k\tilde{b}(\textbf{U}_{\varepsilon }^n,\textbf{U}_{\varepsilon }^n,\textbf{U}_{\varepsilon }^n)- k(\textbf{f}^n,\textbf{U}_{\varepsilon }^n) -(\textbf{U}_{\varepsilon }^{n-1},\textbf{U}_{\varepsilon }^n)\\&\ge \Vert \textbf{U}_{\varepsilon }^n\Vert ^2+\nu k \Vert A_{\varepsilon h}^{\frac{1}{2}} \textbf{U}_{\varepsilon }^n\Vert ^2 - k\Vert \textbf{f}^n\Vert \Vert \textbf{U}_{\varepsilon }^n\Vert -\Vert \textbf{U}_{\varepsilon }^{n-1}\Vert \Vert \textbf{U}_{\varepsilon }^n\Vert \\&\ge \Big (\big (1+\frac{\nu k\lambda _1}{c^2}\big ) \Vert \textbf{U}_{\varepsilon }^n\Vert - \big (k\Vert \textbf{f}^n\Vert +\Vert \textbf{U}_{\varepsilon }^{n-1}\Vert \big )\Big )\Vert \textbf{U}_{\varepsilon }^n\Vert . \end{aligned}$$

Now, choose $\textbf{U}_{\varepsilon }^n\in \textbf{H}_h$ such that

$$\begin{aligned} \textbf{U}_{\varepsilon }^n = \frac{2\big (k\Vert \textbf{f}^n\Vert +\Vert \textbf{U}_{\varepsilon }^{n-1}\Vert \big )}{(\big (1+\frac{\nu k\lambda _1}{c^2}\big )}= \alpha _1. \end{aligned}$$

If either $\Vert \textbf{f}^n\Vert \ne 0$ or $\Vert \textbf{U}_{\varepsilon }^{n-1}\Vert \ne 0$, then $\alpha _1>0$, which implies that there exists $\textbf{U}_{\varepsilon }^*\in \textbf{H}_h$ such that $\Vert \textbf{U}_{\varepsilon }^*\Vert \le \alpha _1$ and $F(\textbf{U}_{\varepsilon }^*)=0$.