High Order Schemes for Hyperbolic Problems Using Globally Continuous Approximation and Avoiding Mass Matrices

Abstract

When integrating unsteady problems using globally continuous representation of the solution, as for continuous finite element methods, one faces the problem of inverting a mass matrix. In some cases, one has to recompute this mass matrix at each time steps. In some other methods that are not directly formulated by standard variational principles, it is not clear how to write an invertible mass matrix. Hence, in this paper, we show how to avoid this problem for hyperbolic systems, and we also detail the conditions under which this is possible. Analysis and simulation support our conclusions, namely that it is possible to avoid inverting mass matrices without sacrificing the accuracy of the scheme. This paper is an extension of Abgrall et al. (in: Karasözen B, Manguoglu M, Tezer-Sezgin M, Goktepe S, Ugur O (eds) Numerical mathematics and advanced applications ENUMATH 2015. Lecture notes in computational sciences and engineering, vol 112, Springer, Berlin, 2016) and Ricchiuto and Abgrall (J Comput Phys 229(16):5653–5691, 2010).

Appendices

Technical Results

Lemma A.1

Assume that K is convex and its aspect ratio is bounded by a constant C. If $v\in \mathbb {P}_r(K)$, and $v=\sum \limits _{\sigma \in K}v_\sigma \varphi _\sigma$, then

$$\begin{aligned} \sum _\sigma |v_\sigma -v_{\sigma '}|\le C_K \sum _\sigma |v(\sigma )-v({\sigma '})\big | \end{aligned}$$

where $C_K$ is the $L^\infty$ norm of the inverse of the matrix $(\varphi _\sigma (\sigma '))_{\sigma , \sigma '}$. and

$$\begin{aligned} h_K\sum _{K}|v_\sigma |\le C_K ||v||_{2,K} \end{aligned}$$

where $C_K$ only depends on K via C.

Proof

We have $v(\sigma )=\sum \limits _{\sigma '\in K}v(\sigma ')\varphi _{\sigma '}(\sigma )$, so that

$$\begin{aligned} \sum _{\sigma \in K}|v(\sigma )|\le C_1 \sum _{\sigma \in K}|v(\sigma )|^2\le C_1 ||A^{-1}||_2 \; \int _Kv^2(\mathbf {x}) d\mathbf {x}\end{aligned}$$

where A is the matrix $A=(\int _K\varphi _{\sigma '}\varphi _\sigma )_{\sigma , \sigma '\in K}$, and $C_1$ is the square root of the number of degrees of freedom in K.

By a scaling argument, $||A^{-1}||_2\le C_K h_K^{-1}$ where $C_K$ depends on the aspect ratio of K. Hence,

$$\begin{aligned} \sum _{\sigma \in K}|v_\sigma -v_{\sigma '}|\le C_K \sum _{\sigma \in K}|v(\sigma )-v(\sigma ')|\le \frac{C_K}{h_K}\int _K |v(\mathbf {x})-v(\sigma ')|d\mathbf {x}\end{aligned}$$

where $C_k$ is the $L^{\infty }$ norm of the matrix $(\varphi _{\sigma '})_{\sigma , \sigma '\in K}$. We have:

$$\begin{aligned} v(\mathbf {x})-v({\sigma '})=\int _0^1 \nabla v((1-s)\mathbf {x}+s\sigma ')\cdot (\mathbf {x}-\sigma ) ds, \end{aligned}$$

so that

$$\begin{aligned} \begin{aligned} \int _K |v(\mathbf {x})-v(\sigma ')|d\mathbf {x}&\le \int _K\int _0^1||\nabla v((1-s)\mathbf {x}+s\sigma ')|| \; ||\mathbf {x}-\sigma '|| d\mathbf {x}\\&\le h_K\; \; \int _K\bigg (\int _0^1||\nabla v((1-s)\mathbf {x}+s\sigma ')||^2 ds\bigg )^{1/2} d\mathbf {x}\\&\le h_K\bigg (\int _K\big (\int _0^1||\nabla v((1-s)\mathbf {x}+s\sigma ')||^2 ds\big ) d\mathbf {x}\bigg )^{1/2} \end{aligned} \end{aligned}$$

since $s\mapsto \sqrt{s}$ is concave. Using Fubini, we then have

$$\begin{aligned} \int _K\big (\int _0^1||\nabla v((1{-}s)\mathbf {x}{+}s\sigma ')||^2 ds\big ) d\mathbf {x}=\int _{K\times [0,1]}||\nabla v((1-s)\mathbf {x}+s\sigma ')||^2{=}\int _K||\nabla v(\mathbf {x})||^2 \mathbf {x}\end{aligned}$$

because K is convex.

Collecting all the pieces, we get:

$$\begin{aligned} \sum _{\sigma \in K}|v_\sigma -v_{\sigma '}|\le C_K\; ||\nabla v||_{\mathcal {L}^2(K)} \end{aligned}$$

where $C_K$ only depends on the aspect ratio of K. $\square$

Non-linear RDS Scheme for Steady the Steady Problem (13)

Consider one element K. Since there is no ambiguity, the drop, for the residuals, any reference to K in the following. The total residual is defined by

$$\begin{aligned} \Phi =\int _{\partial K} \mathbf {f}(u^h)\cdot \mathbf {n}\; d\partial K, \end{aligned}$$

and we introduce the Rusanov residuals:

where $\bar{u}$ is the arithmetic average of of the $u_\sigma 's$ on K and $\alpha$ satisfies:

$$\begin{aligned} \alpha \ge \#K\; \max _{\sigma , \sigma '\in K} \bigg | \int _K \varphi _\sigma \nabla \varphi _{\sigma '}\cdot \nabla _u\mathbf {f}(u^h)\; d\mathbf {x}\bigg |. \end{aligned}$$

Here $\#K$ is the number of degrees of freedom in K. This residual can be rewritten as

$$\begin{aligned} \Phi _\sigma ^{Rus}=\sum _{\sigma '\in K} c_{\sigma \sigma '}(u_\sigma -u_{\sigma '}) \end{aligned}$$

with

$$\begin{aligned} c_{\sigma \sigma '}=\int _K \varphi _\sigma \nabla \varphi _{\sigma '}\cdot \nabla _u\mathbf {f}(u^h)\; d\mathbf {x}-\dfrac{\alpha }{\#K}. \end{aligned}$$

Under the condition above, $c_{\sigma \sigma '}\ge 0$ and hence we have a maximum principle.

The coefficients $\beta _\sigma$ introduced in the relations (22) and (23) are defined by:

$$\begin{aligned} \beta _\sigma =\dfrac{\max (0,\frac{\Phi _\sigma ^{Rus}}{\Phi })}{ \sum \limits _{\sigma '\in K} \max (0,\frac{\Phi _{\sigma '}^{Rus}}{\Phi })}. \end{aligned}$$

and can be shown to be always defined, to garanty a local maximum principle for (22) and (23), see [6].

Some Properties of Non-linear RDS Schemes

This annex is devoted to the justification of some fact stated in Sect. 4.5, namely that the $\mathcal {L}^2$ operator, i.e. for each element and each sub-time step p,

$$\begin{aligned} \int _K \psi _\sigma (V_p^{(l)}-V_0) dx+\int _K \mathcal {I}_l \Phi _\sigma ^\mathbf {x}\big (V_0), \ldots , \Phi _\sigma ^\mathbf {x}(V_{r+1}) \big ) \end{aligned}$$

can write

$$\begin{aligned} |K|\sum _{\sigma '\in K} a_{\sigma \sigma '}^K (V_{p,\sigma '}^{(l)}-V_{0,\sigma '}) +\Delta t\sum _{k=1}^{r+1} \theta _{k,r+1} \bigg \{\sum _{\sigma '\in K} c_{\sigma \sigma '}^k (V_{k,\sigma }^{(l)}-V^{(l)}_{k,\sigma '})\bigg \} \bigg ). \end{aligned}$$

with

• $a_{\sigma \sigma }^K=\gamma _\sigma ^K \frac{|K|}{\#K}$, $\gamma _\sigma ^K\in [0,1]$ and $\#K$ is the number of degrees of freedom in K

• and $c_{\sigma \sigma '}^k\ge 0$.

In the spirit of “Appendix B” and [6], we consider the following kind of nonlinear RDS. The Galerkin residuals are defined by

$$\begin{aligned} \Phi ^{K,Gal}_\sigma =\int _{\partial K}\varphi _\sigma \bigg (\sum _{k=1}^{r+1}\theta _{k,r+1} \mathbf {f}(V_k^{(l)})\bigg )\cdot \mathbf {n}-\int _K\nabla \varphi _\sigma \bigg ((\sum _{k=1}^{r+1}\theta _{k,r+1} \mathbf {f}(V_k^{(l)})\bigg ), \end{aligned}$$

(47)

from this one writes a Rusanov residual:

$$\begin{aligned} \Phi ^{K,Rus}_\sigma =\frac{|K|}{\# K} \big ( (V_{p,\sigma }^{(l)}-V_{0,\sigma }\big )+\Delta t\Phi _\sigma ^{K, Gal}+\Delta t\alpha _K\bigg ( \big \{ \sum _{k=1}^{r+1}\theta _{k,r+1}V_k^{(l)}\big \} -\overline{V}\bigg ) \end{aligned}$$

with

$$\begin{aligned} \overline{V}=\dfrac{1}{\# K}\bigg ( \sum _{\sigma \in K} \sum _{k=1}^{r+1}\theta _{k,r+1}V_k^{(l)}\bigg ) \end{aligned}$$

and $\alpha _K$ larger than the maximum of the spectral radii of the Jacobian of the flux evaluated at the states $V_k^{(l)}$, or even larger. Then one forms

$$\begin{aligned} \Phi _\sigma ^{K,\star }=\beta _\sigma ^K\Phi _{xt}^K \end{aligned}$$

(48)

where the total residual $\Phi ^K$ is defined by

$$\begin{aligned} \Phi _{xt}^K=\int _K(V_p^{(l)}-V_0) dx+\int _{\partial K}\bigg (\sum _{k=1}^{r+1}\theta _{k,r+1} \mathbf {f}(V_k^{(l)})\bigg )\cdot \mathbf {n}\qquad \qquad \left( =\sum _{\sigma \in K}\Phi _\sigma ^{K, Gal}\right) \end{aligned}$$

and $\beta _\sigma ^K$ by

$$\begin{aligned} \beta _\sigma ^K=\dfrac{\max (0, \frac{\Phi _\sigma ^{K,Rus}}{\Phi _{xt}^K})}{\sum _{\sigma '\in K} \max (0, \frac{\Phi _\sigma ^{K,Rus}}{\Phi _{xt}^K})}. \end{aligned}$$

(49)

Let us prove now that

$$\begin{aligned} a_{\sigma \sigma }^K=\gamma _\sigma ^K \frac{|K|}{\# K} \text { with }\gamma _\sigma ^K\in [0,1]. \end{aligned}$$

(50)

and

$$\begin{aligned} c_{\sigma \sigma '}^k\ge 0. \end{aligned}$$

(51)

Proof

We note that

$$\begin{aligned} \beta _\sigma ^K\Phi _{xt}^K=\gamma _\sigma ^K \Phi _\sigma ^{K, Rus} \end{aligned}$$

with

$$\begin{aligned} \gamma _\sigma ^K=\left\{ \begin{array}{ll} 0 &{} \text {if } \max \left( 0, \frac{\Phi _\sigma ^{K,Rus}}{\Phi _{xt}^K}\right) =0\\ \dfrac{1}{\sum _{\sigma '\in K} \max \left( 0, \frac{\Phi _\sigma ^{K,Rus}}{\Phi _{xt}^K}\right) } &{}\text { else} \end{array} \right. \end{aligned}$$

Since $\sum \limits _{\sigma \in K}\Phi _\sigma ^{K,Rus}=\Phi _{xt}^K$, we have that:

$$\begin{aligned} \sum _{\sigma '\in K} \max \left( 0, \frac{\Phi _\sigma ^{K,Rus}}{\Phi _{xt}^K}\right) +\sum _{\sigma ' \in K} \max \left( 0, \frac{\Phi _\sigma ^{K,Rus}}{\Phi _{xt}^K}\right) = \sum _{\sigma '\in K} \left( \frac{\Phi _\sigma ^{K,Rus}}{\Phi _{xt}^K}\right) =1, \end{aligned}$$

so that

$$\begin{aligned} \sum _{\sigma '\in K} \max \left( 0, \frac{\Phi _\sigma ^{K,Rus}}{\Phi _{xt}^K}\right) \ge 1 \end{aligned}$$

and then $\gamma _{\sigma }^K\in [0,1].$ We get the first property (50).

The second one (51) comes from the very definition of $\beta _\sigma ^K$. $\square$

Remark C.1

In many practical applications, the residual that is considered is not (48) but

$$\begin{aligned} \Phi _\sigma ^{K,\star }=\beta _\sigma ^K\Phi _{xt}^K+\sum _{\text {edges of }K} h_K^2 \Gamma \int _e \bigg [\nabla \varphi _\sigma \bigg ]\bigg [\nabla \bigg ( \sum _{k=1}^{r+1}\theta _{k,r+1}V_k^{(l)}\bigg )\bigg ] \end{aligned}$$

(52)

with $\beta _\sigma ^K$ defined as (49). Then (50) is still true because the term

$$\begin{aligned} \int _{\partial K}\bigg (\sum _{k=1}^{r+1}\theta _{k,r+1} \mathbf {f}(V_k^{(l)})\bigg )\cdot \mathbf {n}\qquad \qquad \left( =\sum _{\sigma \in K}\Phi _\sigma ^{K, Gal}\right) \end{aligned}$$

does not contain any time increment.

In some other, we modify the definition of the Rusanov residual into

$$\begin{aligned} \Phi ^{K,Rus}_\sigma =\Phi _\sigma ^{K, Gal}+\Delta t \alpha _K\bigg ( \big \{ \sum _{k=1}^{r+1}\theta _{k,r+1}V_k^{(l)}\big \} -\overline{V}\bigg ) \end{aligned}$$

with now

$$\begin{aligned} \Phi _\sigma ^{K, Gal}= & {} \int _K(V_p^{(l)}-V_0) \varphi _\sigma dx \Delta t \int _{\partial K}\varphi _\sigma \bigg (\sum _{k=1}^{r+1}\theta _{k,r+1} \mathbf {f}(V_k^{(l)})\bigg )\cdot \mathbf {n}\nonumber \\&-\Delta t\int _K\nabla \varphi _\sigma \bigg ((\sum _{k=1}^{r+1}\theta _{k,r+1} \mathbf {f}(V_k^{(l)})\bigg ) \end{aligned}$$

and we consider

$$\begin{aligned} \Phi _\sigma ^{K,\star }=(1-\ell )\bigg \{ \Phi _\sigma ^{K, Gal}+\sum _{\text {edges of }K} h_K^2 \Gamma \int _e \bigg [\nabla \varphi _\sigma \bigg ]\bigg [\nabla \bigg ( \sum _{k=1}^{r+1}\theta _{k,r+1}V_k^{(l)}\bigg )\bigg ]\bigg \} +\ell \Phi ^{K,Rus}_\sigma \end{aligned}$$

(53)

with

$$\begin{aligned} \ell =\dfrac{|\Phi ^K|}{\sum _{\sigma \in K} |\Phi _\sigma ^{K,Rus}|}. \end{aligned}$$

(54)

Then none of the properties hold formally true but we get a maximum principle experimentally.

Example of Suitable Runge–Kutta Method

Consider the problem

$$\begin{aligned} \dfrac{dy}{dt}=f(y) \end{aligned}$$

wit $y_0=y(0)$. The Runge–Kutta method of [26] is:

$$\begin{aligned} \begin{aligned} y^{(0)}&=u^n\\ y^{(1)}&=y^{(0)}+\Delta t f(y^{(0)})\\ y^{(2)}&=y^{(1)}+\frac{\Delta t}{2}\bigg ( f(y^{(0)})+ f(y^{(1)})\bigg ) \end{aligned} \end{aligned}$$

(55)

We first show that this is indeed a DeC method.

Defining

$$\begin{aligned} \mathcal {L}^2(Y)=Y-Y_0+\frac{\Delta t}{2}\big (f(Y_0)+f(Y))\big ) \end{aligned}$$

and

$$\begin{aligned} \mathcal {L}^{(1)}(Y)=Y-Y_0+\Delta t f(Y_0), \end{aligned}$$

The first iteration leads to

$$\begin{aligned} \begin{aligned} y^{(1)}-y^{(0)}+\Delta t f(y^{(0)})=&\bigg (y^{(0)}-y^{(0)}+\Delta t f(y^{(0)})\bigg )\\&-\bigg ( y^{(0)} -y^{(0)}+\frac{\Delta t}{2}\bigg ( f(y^{(0)})+ f(y^{(0)})\bigg ) \end{aligned} \end{aligned}$$

i.e.

$$\begin{aligned} y^{(1)}=y^{(0)}+\Delta t f(y^{(0)}). \end{aligned}$$

The second iteration gives:

$$\begin{aligned} \begin{aligned} y^{(2)}-y^{(0)}+\Delta t f(y^{(0)})&=\bigg (y^{(1)}-y^{(0)}+\Delta t f(y^{(0)})\bigg )\\&-\bigg ( y^{(1)}-y^{(0)}+\frac{\Delta t}{2}\bigg ( f(y^{(0)})+ f(y^{(1)})\bigg ) \end{aligned} \end{aligned}$$

i.e.

$$\begin{aligned} y^{(2)}=y^{(1)}+ \frac{\Delta t}{2}\bigg ( f(y^{(0)})+ f(y^{(1)})\bigg ). \end{aligned}$$

From this, let us construct an example of RK scheme which is third order, using this method as a building block. We know the exact solution satisfies:

$$\begin{aligned} y(t_{n+1})=y(t_n))+\int _{t_n}^{t_{n+1}} f(y(s)) ds. \end{aligned}$$

Using Simpson formula, an approximation is:

$$\begin{aligned} y_{n+1}=y_n+\dfrac{\Delta }{6}\bigg ( f(y_n)+4 f(y_{n+1/2})+f(y_{n+1})\bigg ). \end{aligned}$$

To get a third odred approximation of this third order approximation of the exact solution, we need second order approximations of $y_{n+1/2}\approx y(t_n+\frac{\Delta t}{2})$ and $y_{n+1}\approx y(t_{n+1})$.

To do so, we can use the method (55) to extimate $y_{n+1}$. And using the intermediate steps at $t_n+\frac{\Delta t}{4}$ and $t_n+\frac{\Delta t}{2}=t_{n+1/2}$, with (55) we get a second order approximation of $y_{n+1/2}$.

Note we can save one iteration (or one step) by first evaluating $y_{n+1/2}$ with (55) and using the second step of (55) to get a second order approximation of $y_{n+1}$.

We have used this method for convextion problems, and we get similar results as with the method presented here, which is more systematic.

This can be extended as follows: one first start by a quadrature formula of order k of $\int _{t_n}^{t_{n+1}}f(y(s))$ and we iteratively get approximations of order $k-1$ of the quadrature points by induction.