A Viscosity-Independent Error Estimate of a Pressure-Stabilized Lagrange–Galerkin Scheme for the Oseen Problem

Abstract

We consider a pressure-stabilized Lagrange–Galerkin scheme for the transient Oseen problem with small viscosity. In the scheme we use the equal-order approximation of order k for both the velocity and pressure, and add a symmetric pressure stabilization term. We show an error estimate for the velocity with a constant independent of the viscosity if the exact solution is sufficiently smooth. We also show an error estimate of a discrete primitive of the pressure. Numerical examples show high accuracy of the scheme for problems with small viscosity.

A Estimates for LG Schemes

Lemma 6 is shown in [39, Lemma 5.7].

Lemma 6

Let $w^*\in W^{1,\infty }(\varOmega )^d$ and $X_1(w^*)$ be the mapping defined in (4). Under the condition $\Delta t |w^*|_{1,\infty } \le 1/4$, the estimate

$$\begin{aligned} \frac{1}{2} \le \det \biggl (\frac{\partial X_1(w^*)}{\partial x} \biggr ) \le \frac{3}{2} \end{aligned}$$

holds, where $\det (\partial X_1(w^*)/\partial x)$ is the Jacobian of $X_1(w^*)$.

Lemma 7 is shown in [35, Lemma 1].

Lemma 7

Let $w^*\in W^{1,\infty }_0(\varOmega )^d$ and $X_1(w^*)$ be the mapping defined in (4). Under the condition $\Delta t|w^*|_{1,\infty } \le 1/4$, there exists a positive constant c independent of $\Delta t$ such that for $v \in L^2(\varOmega )^d$

$$\begin{aligned} \Vert v \circ X_1(w^*) \Vert _{0}^{2} \le (1+ c |w^*|_{1,\infty } \Delta t) \Vert v\Vert _0^{2}. \end{aligned}$$

We now show an estimate for $R_1^n$ in Lemma 8, or tools for estimating $R_2^n$ and $R_3^n$ in Lemmas 9 and 10, where $R_i^n$, $i=1,2,3$, are defined in (16). Although these estimates are frequently used in the analysis of the LG method, e.g. [31, 39], we show proofs of Lemmas 8 and 10 for completeness.

Lemma 8

Suppose that $u\in Z^2$, $w \in C(W^{1,\infty }_0) \cap H^1(L^\infty )$ and $\Delta t |w|_{C(W^{1,\infty })} \le 1/4$. Then

$$\begin{aligned}&\Vert R_1^n\Vert _0 \le \sqrt{\Delta t} \left[ \sqrt{\frac{2}{3}} (\Vert w^{n-1}\Vert _{0,\infty }^2+1) \Vert u\Vert _{Z^2(t^{n-1},t^n)} + \left\| \frac{\partial w}{\partial t} \right\| _{L^2(t^{n-1}, t^n; L^\infty )} \Vert \nabla u^n\Vert _0 \right] . \end{aligned}$$

Proof

We estimate $\Vert R_1^n \Vert _0$ by dividing

$$\begin{aligned} R_{1}^n =&\left( \frac{\partial u^n}{\partial t} + (w^{n-1} \cdot \nabla )u^n -\frac{u^n-u^{n-1}\circ X_1(w^{n-1})}{\Delta t} \right) \\&+ \left( (w^{n} \cdot \nabla )u^n - (w^{n-1} \cdot \nabla )u^n \right) \\ =:&R_{11}^n + R_{12}^n. \end{aligned}$$

For $R_{11}^n$, we set

$$\begin{aligned} y(x,s):= x-sw^{n-1}(x) \Delta t, ~ t(s):=t^n-s\Delta t, \end{aligned}$$

and use Taylor’s theorem to get

$$\begin{aligned} (u^{n-1} \circ X_1(w^{n-1}))(x)&= u^n(x) - \Delta t \left( \frac{\partial u^n}{\partial t} + (w^{n-1} \cdot \nabla ) u^n \right) (x) \\&\quad + \Delta t^2 \int _0^1 (1-s) \left( \frac{\partial }{\partial t} + w^{n-1} (x) \cdot \nabla \right) ^2 u(y(x,s), t(s)) ds. \end{aligned}$$

Using the property of the Bochner integral, we then have

$$\begin{aligned} \Vert R_{11}^n\Vert _0&\le \Delta t \int _0^1 \biggl \Vert (1-s) \left( \frac{\partial }{\partial t} + w^{n-1} (\cdot ) \cdot \nabla \right) ^2 u ( y(\cdot , s), t(s)) \biggr \Vert _0 ds\\&\le \Delta t \biggl ( \int _0^1 (1-s)^2 dx \biggr )^{1/2} \biggl ( \int _0^1 \biggl \Vert \left( \frac{\partial }{\partial t} + w^{n-1} (\cdot ) \cdot \nabla \right) ^2 u ( y(\cdot , s), t(s)) \biggr \Vert _0^2 ds \biggr )^{1/2} \\&\le \sqrt{2/3} \sqrt{\Delta t} (\Vert w^{n-1}\Vert _{0,\infty }^2+1) \Vert u\Vert _{Z^2(t^{n-1},t^n)}. \end{aligned}$$

where we have used the transformation of independent variables from x to y and s to t, and the estimate $|\det (\partial x/\partial y)| \le 2$ by virtue of Lemma 6. It is easy to show

$$\begin{aligned} \Vert R_{12}^n\Vert _0 \le \sqrt{\Delta t} \left\| \frac{\partial w}{\partial t} \right\| _{L^2(t^{n-1}, t^n; L^\infty )} \Vert \nabla u^n\Vert _0. \end{aligned}$$

Combining the two estimate, we have the conclusion. $\square$

Lemma 9 is a direct consequence of [1, Lemma 4.5] and Lemma 6.

Lemma 9

Let $1\le q <\infty$, $1\le p \le \infty$, $1/p+1/p'=1$ and $w_i\in W_0^{1,\infty }(\varOmega )^d$, $i=1,2$. Under the condition $\Delta t |w_i|_{1,\infty }\le 1/4$, it holds that, for $v \in W^{1,qp'}(\varOmega )^d$,

$$\begin{aligned} \Vert v \circ X_1(w_1) - v \circ X_1(w_2)\Vert _{0,q} \le 2^{1/(qp')} \Delta t \Vert w_1-w_2\Vert _{0,pq}\Vert \nabla v \Vert _{0,qp'}, \end{aligned}$$

where $X_1(\cdot )$ is defined in (4).

Lemma 10

Suppose that $v \in H^1(H^1)$, $w^*\in W^{1,\infty }_0(\varOmega )^d$, and $\Delta t |w^*|_{1,\infty } \le 1/4$. Then

$$\begin{aligned}&\left\| v^n - v^{n-1} \circ X_1(w^*) \right\| _0 \le \sqrt{2 \Delta t }\biggl ( \left\| \frac{\partial v}{\partial t} \right\| _{L^2(t^{n-1}, t^n ;L^2)} + \Vert w^*\Vert _{0,\infty } \Vert \nabla v\Vert _{L^2(t^{n-1}, t^n; L^2)} \biggr ), \end{aligned}$$

where $X_1(\cdot )$ is defined in (4).

Proof

Similar to the proof of Lemma 8, by defining

$$\begin{aligned} y(x,s):= x-sw^{*}(x) \Delta t, ~ t(s):=t^n-s\Delta t, \end{aligned}$$

and by using the property of the Bochner integral, we have the estimate

$$\begin{aligned} \left\| v^n - v^{n-1} \circ X_1(w^*) \right\| _0 \le \Delta t \int _0^1 \left\| \left( \frac{\partial }{\partial t} + (w^{*}(\cdot ) \cdot \nabla ) \right) v (y(\cdot ,s), t(s)) \right\| _0 ds. \end{aligned}$$

The conclusion follows from the transformation of the independent variables from x to y and s to t, and the estimate $|\det (\partial x/\partial y)| \le 2$ by virtue of Lemma 6. $\square$

Lemma 11 is an extension of [17, Lemma 1] obtained in the 1D case.

Lemma 11

Let $w^*\in W_0^{1,\infty }(\varOmega )^d$. Under the condition $\Delta t |w^*|_{1,\infty }\le 1/4$, there exists a positive constant c independent of $\Delta t$ such that, for $v \in L^2(\varOmega )^d$,

$$\begin{aligned} \Vert v-v\circ X_1(w)\Vert _{-1} \le c \Delta t \Vert w\Vert _{1,\infty } \Vert v\Vert _{0}, \end{aligned}$$

where $X_1(\cdot )$ is defined in (4) and $\Vert \cdot \Vert _{-1}$ is the norm in $H^{-1}$.