Entropy-Stable Schemes for the Euler Equations with Far-Field and Wall Boundary Conditions

Abstract

In this paper entropy-stable numerical schemes for the Euler equations in one space dimension subject to far-field and wall boundary conditions are derived. Furthermore, a stable numerical treatment of interfaces between different grid domains is proposed. Numerical computations with second- and fourth-order accurate schemes corroborate the stability and accuracy of the proposed boundary treatment.

Appendix: Some Comments on the Proof of Theorem 2.6

Here we will further explain the balance of non-linear terms on which the proof relies. These arguments also appear in the proof of a similar theorem but for another entropy in [17].

1.1 Scalar Conservation Law

The entropy variable is $w=u$ and there is only one eigenvalue and eigenvector. Hence, the expression (2.23) becomes

$$\begin{aligned} \tilde{F}_0=F(\mathfrak{ g }) +u_{\mathfrak{ g }}((\tilde{\lambda })^--(\hat{\lambda })^+)u_{0\mathfrak{ g }} -\frac{1}{2}u_{0\mathfrak{ g }}(|\tilde{\lambda }|+|\hat{\lambda }|)u_{0\mathfrak{ g }}. \end{aligned}$$

(5.1)

Recall that $\mathfrak{ g }$ is a bounded admissible state in the sense that $f(\mathfrak{ g })$ and $\lambda (\mathfrak{ g })$ are both bounded. The numerical solution $u$ is assumed to be an admissible (but not necessarily bounded) state, i.e., $|\lambda (u)|$ can only be unbounded if $|u|$ is. In the scalar case the balance is now very simple. We split the expression into two pieces:

$$\begin{aligned} \tilde{F}_0&=F(\mathfrak{ g }) +BT_{\hat{\lambda }}+ BT_{\tilde{\lambda }}.\\ BT_{\hat{\lambda }}&= -u_{\mathfrak{ g }}(\hat{\lambda })^+u_{0\mathfrak{ g }} -\frac{1}{2}u_{0\mathfrak{ g }}|\hat{\lambda }|u_{0\mathfrak{ g }} \\ BT_{\tilde{\lambda }}&= u_{\mathfrak{ g }}(\tilde{\lambda })^-u_{0\mathfrak{ g }}-\frac{1}{2}u_{0\mathfrak{ g }}|\tilde{\lambda }|u_{0\mathfrak{ g }} \end{aligned}$$

The task is to show that both $BT_{\hat{\lambda }}$ and $BT_{\tilde{\lambda }}$ are bounded from above (but not necessarily negative). We consider $BT_{\hat{\lambda }}$. $BT_{\tilde{\lambda }}$ can be treated in the same way. We assume $\hat{\lambda }^+>0$ or else $BT_{\hat{\lambda }}$ is trivially non-positive. If $|u_{0\mathfrak{ g }}|< 2|u_{\mathfrak{ g }}|$, the right-hand side may be positive indicating a growth. However, $|u_0-u_{\mathfrak{ g }}|=|u_{0\mathfrak{ g }}|< |u_{\mathfrak{ g }}|$ implies that $u_0$ is bounded which in turn means that $\hat{\lambda }(u_0,\mathfrak{ g })$ is bounded. Hence, $BT_{\hat{\lambda }}$ is bounded although it may be positive. An upper bound is sufficient for a global bound on the entropy. The other option is that $|u_{0\mathfrak{ g }}|> 2|u_{\mathfrak{ g }}|$. Then,

$$\begin{aligned} BT_{\hat{\lambda }}=(\hat{\lambda })^+\left( -u_{\mathfrak{ g }}u_{0\mathfrak{ g }} -\frac{1}{2}u_{0\mathfrak{ g }}u_{0\mathfrak{ g }}\right) <0. \end{aligned}$$

(5.2)

Note that the size of $\hat{\lambda }$ is unimportant and it will give a bound on the entropy. We stress that the key to the proof is the observation that the eigenvalues are not independent of $u$. It is impossible that $BT_{\hat{\lambda }}>0$ and at the same time $\hat{\lambda }^+\rightarrow \infty$. This ensures an upper bound on the growth rate of the entropy and an entropy estimate is obtained.

1.2 System of Conservation Laws

Next, we turn to a system such as the Euler equations. Again, we make the assumption that $\mathfrak{ g }$ is an admissible state. (Specifically, for the Euler equations $\mathfrak{ g }$ must be a bounded state and bounded away from $\rho =0$.) Furthermore, we also assume that the states $u$ generated by the scheme are admissible in the sense that $\rho >0$. Both these assumptions are standard. Hence, $|\lambda ^i(\mathfrak{ g })|$ are all bounded. Furthermore, if $|w_{0\mathfrak{ g }}|\le 2|w_{\mathfrak{ g }}|$, then $w_0$ is a bounded admissible state and $\hat{\lambda }^i(w_0,\mathfrak{ g })$ and $\tilde{\lambda }^i(w_0,\mathfrak{ g })$ are bounded for all $i$. So, the first part of the scalar argument applies to this situation as well. Namely, for states close to $w_{\mathfrak{ g }}$ the entropy might grow but only at a bounded rate since the eigenvalues remain bounded.

Similarly, if $|w_{0\mathfrak{ g }}^i|>2|w_{\mathfrak{ g }}^i|$ for all $i$ then the boundary terms will be negative capping the growth just like in the scalar case.

However, for a system we must also consider when $w_{0\mathfrak{ g }}^i$ remains small for one or more $i$ but grows in other directions. For the sake of the argument, assume that $|w_{0\mathfrak{ g }}^1|<2|w_{\mathfrak{ g }}^1|$ while all the other directions satisfy $|w_{0\mathfrak{ g }}^i|>2|w_{\mathfrak{ g }}^i|, i\ne 1$. This would render $BT_1$ positive which may cause a growth of the entropy. ($BT_{2N}<0$ under these assumptions and one might expect those terms to dominate $BT_1$. Such an argument would require some assumptions relating the different eigenvalues. We use another argument.) Here,

$$\begin{aligned} BT_{1}&= -w_{\mathfrak{ g }}^1\hat{\lambda }^1 {}^+w_{0\mathfrak{ g }}^1 -\frac{1}{2}w_{0\mathfrak{ g }}^1|\hat{\lambda }^1|w_{0\mathfrak{ g }}^1>0 \end{aligned}$$

(5.3)

If $\hat{\lambda }^1$ remains bounded the growth is bounded and stability obtained. ($BT_1>0$ but since $|w_{0\mathfrak{ g }}^1|,|w_{\mathfrak{ g }}^1|, |\hat{\lambda }^1|$ are all bounded we have $BT_1<constant$ which is sufficient for stability.) The only remaining question is if $w_{0\mathfrak{ g }}^1$ can be small while $\hat{\lambda }^1$ grows indefinitely. This would require that $\hat{\lambda }^1$ does not affect $w^1_{0\mathfrak{ g }}$ or else it would cause $w^1_{0\mathfrak{ g }}$ to grow which would make $BT_1$ negative and the growth would be capped.

The final part of the proof demonstrates that this can not happen for a conservation law. If it could happen, it would certainly also happen locally. Hence, we can freeze the coefficients and study the local growth pattern.

By freezing the coefficients the dependency between the solution and the eigenvalues can be determined. Equation (2.26) is of the form

$$\begin{aligned} Av_t + \Lambda v_x=0. \end{aligned}$$

Since $A=A^T>0$ it is inevitable that $\lambda ^i$ will affect $v_i$. We will spell out the argument for a $2\times 2$ system. Say that $v_2$ is small but $\lambda _2$ is growing. The growth of $\lambda _2$ will cause $v_1$ to grow but not $v_2$, for if it did we would obtain a bound. Consequently, $[A]_{22}=0$. Hence, the assumption that $A>0$, i.e. $[A]_{22}>0$, is contradicted. (We remark that $v_1$ must not affect $v_2$ either, which results in a decoupling of the equations.)

This argument has the general meaning that the coupling is strong in a conservation law and in particular, the growth of a particular eigenvalue affects the corresponding component. Hence, the scalar argument for boundedness translates to a system and $\tilde{F}_0\le Constant$. Therefore, $U(t)\le Constant$.