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Error estimates for the third order explicit Runge-Kutta discontinuous Galerkin method for a linear hyperbolic equation in one-dimension with discontinuous initial data

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In this paper we present an error estimate for the explicit Runge-Kutta discontinuous Galerkin method to solve a linear hyperbolic equation in one dimension with discontinuous but piecewise smooth initial data. The discontinuous finite element space is made up of piecewise polynomials of arbitrary degree \(k\ge 1\), and time is advanced by the third order explicit total variation diminishing Runge-Kutta method under the standard CFL temporal-spatial condition. The \(L^2(\mathbb R \backslash \mathcal R _T)\)-norm error at the final time \(T\) is optimal in both space and time, where \(\mathcal R _T\) is the pollution region due to the initial discontinuity with the width \(\mathcal O (\sqrt{T\beta }h^{1/2}\log (1/h))\). Here \(h\) is the maximum cell length and \(\beta \) is the flowing speed. These results are independent of the time step and hold also for the semi-discrete discontinuous Galerkin method.

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Correspondence to Chi-Wang Shu.

Additional information

The research of the first author is supported by NSFC grants 10871093 and 11271187, and also supported partially by NSFC grants 10931004 and 11071116. The research of the second author is supported by NSF grant DMS-1112700 and DOE grant DE-FG02-08ER25863.

Appendix

Appendix

1.1 Proof of Lemma 3.3

The second conclusion is obviously the corollary of the first one, so we only prove the first inequality of Lemma 3.3. To that end, we just need to consider this error estimate in a given cell \(I_j\), and then summing this result over all elements.

Since the considered projection \(\mathbb W \) is linear and is exact for any piecewise polynomials \(v_h\), we have \(\mathbb W ^{\bot }(\psi v_h)=\mathbb W ^{\bot }((\psi -c_j) v_h)\), where \(c_j\) is any arbitrary constant. Starting form the general approximation property [3], or the conclusion in Lemma 3.2 (with \(\psi =1\)), we can find a bounding constant \(C>0\) independent of \(j\) and \(v_h\), such that

$$\begin{aligned}&\Vert \mathbb W ^{\bot }(\psi v_h)\Vert _{I_j}+h_j \Vert \partial _x(\mathbb W ^{\bot }(\psi v_h))\Vert _{I_j} \le Ch_j\Vert \partial _x((\psi -c_j) v_h)\Vert _{I_j} \nonumber \\&\quad \le Ch_j\Vert \partial _x\psi v_h\Vert _{I_j}+Ch_j\Vert (\psi -c_j) \partial _x v_h\Vert _{I_j}. \end{aligned}$$
(8.1)

Taking \(c_j=\psi (x_j)\) and noticing \(\gamma h^{\sigma }\ge h\ge h_j\), then we will have

$$\begin{aligned} \text {RHS of } (8.1)&\le Ch_j\Vert \partial _x\psi \Vert _{L^{\infty }(I_j)}\Vert v_h\Vert _{I_j}+ Ch_j^2\Vert \partial _x\psi \Vert _{L^{\infty }(I_j)}\Vert \partial _x v_h\Vert _{I_j} \nonumber \\&\le Ch_j\Vert \partial _x\psi \Vert _{L^{\infty }(I_j)}\Vert v_h\Vert _{I_j} \le C\gamma ^{-1}h_j^{1-\sigma }\Vert \psi \Vert _{L^{\infty }(I_j)} \Vert v_h\Vert _{I_j},\quad \qquad \end{aligned}$$
(8.2)

due to the first inverse property in Lemma 3.1 (with \(\psi =1\) there), and property (3.3b) of the weight function \(\psi (x,t)\).

Next, we multiply the quantity \(\psi ^{-1/2}(x_j)\) in the above inequality. Using property (3.3a) of the weight function gives us that

$$\begin{aligned} \Vert \mathbb W ^{\bot }(\psi v_h)\Vert _{\psi ^{-1},I_j}+h_j \Vert \partial _x(\mathbb W ^{\bot }(\psi v_h))\Vert _{I_j} \le C\gamma ^{-1} h_j^{1-\sigma }\Vert v_h\Vert _{\psi ,I_j}, \end{aligned}$$
(8.3)

the sum of which on every cells yields the first conclusion of this lemma.

1.2 Proof of Lemma 3.5

After an integration by parts, we have \(\mathcal H (w,\psi v)= -\sum _j (\beta \psi v, w_x)_j -\sum _j\beta \psi _{j+\frac{1}{2}}[\![w]\!]_{j+\frac{1}{2}}v^{+}_{j+\frac{1}{2}}\). Each term on the right-hand side can be easily estimated by the weighted Cauchy-Schwarz inequality and the inverse properties in Lemma 3.1, which follows

$$\begin{aligned} \sum _j\psi _{j+\frac{1}{2}} (v^{+}_{j+\frac{1}{2}})^2 \le \mu (\nu h)^{-1}\Vert v\Vert _{\psi }^2, \quad \sum _j\psi _{j+\frac{1}{2}} [\![w]\!]_{j+\frac{1}{2}}^2 \le 2\mu (\nu h)^{-1}\Vert w\Vert _{\psi }^2,\quad \qquad \end{aligned}$$
(8.4)

form the simple inequality \((a-b)^2\le 2(a^2+b^2)\), if both \(w\) and \(v\) belong to the finite element space \(V_h\). The final upper boundedness reads \(|\mathcal H (w,\psi v)|\le (1+\sqrt{2})\beta \mu (\nu h)^{-1} \Vert w\Vert _{\psi }\Vert v\Vert _{\psi }\). This completes the proof of this lemma.

1.3 Proof of Lemma 4.1

We can prove every conclusion in this lemma along the same way. First we multiply the test function \(v_h\in V_h\) on both sides of the following three equalities: the definitions (4.4a) and (4.4b), and the equality

$$\begin{aligned} u(x,t+\tau ) = \frac{1}{3}u^{(0)}(x,t)+\frac{2}{3}u^{(2)}(x,t)+\frac{2}{3}\tau u^{(2)}_{t}(x,t) +\zeta (x,t), \end{aligned}$$
(8.5)

which will be proved later. Then we transform the time derivatives into space derivatives with the application of the hyperbolic equation (4.1). Finally, fixing the time at \(t=t^n\) and integration by parts yield the conclusions of this lemma.

Now we prove Eq. (8.5) and complete the proof of this lemma. Using (4.4b) and (4.4a) successively, we will get

$$\begin{aligned}&\frac{1}{3}u^{(0)}+\frac{2}{3}u^{(2)}+\frac{2}{3}\tau u^{(2)}_{t} =\frac{5}{6}u^{(0)}+\frac{1}{6}u^{(1)}+\frac{1}{2}\tau u^{(0)}_{t}+\frac{1}{3}\tau u^{(1)}_{t} +\frac{1}{6}\tau ^2u^{(1)}_{tt}\\&\quad = u^{(0)}+\tau u^{(0)}_{t}+\frac{1}{2}\tau ^2u^{(0)}_{tt}+\frac{1}{6}\tau ^3u^{(0)}_{ttt} =u(x,t+\tau )-\zeta (x,t), \end{aligned}$$

where the last equality comes from the Taylor’s expansion in time. Here we have dropped the arguments \((x,t)\) for simplicity of notations.

1.4 Proof of Lemma 5.5

At the first step we take the test function \(v_h=\mathbb Q _h(\psi ^{n}\mathbb D _1\xi ^{n})\) in (5.1a), and obtain the explicit expression

$$\begin{aligned} \Vert \mathbb D _1\xi ^{n}\Vert _{\psi ^{n}}^2&= \tau \mathcal H (\xi ^{n},\psi ^{n}\mathbb D _1\xi ^{n})+(\mathbb D _1\xi ^{n},\mathbb Q _h^{\bot } (\psi ^{n}\mathbb D _1\xi ^{n})) \nonumber \\&\quad +(\mathbb D _1\eta ^{n},\mathbb Q _h(\psi ^{n}\mathbb D _1\xi ^{n})). \end{aligned}$$
(8.6)

This result has almost the same construction as (5.3). Each term on the right-hand side, denoted respectively by \(Y_1, Y_2\) and \(Y_3\) in order, will be estimated one by one.

The main work in this proof is how to get a sharp estimate to the first term \(Y_1\). To do that, we have to abandon the conclusion in Lemma 3.5, and adopt the decomposition with respect to the generalized slope function \(\mathfrak M (\mathbb D _1\xi ^{n})\).

Note that \(\partial _x(\mathbb D _1\xi ^{n})=\mathfrak M (\mathbb D _1\xi ^{n})(x-x_j)/h_j +\partial _x(\mathfrak M (\mathbb D _1\xi ^{n}))/h_j\) in each cell \(I_j\). Expanding the spatial derivative \(\partial _x(\psi ^{n}\mathbb D _1\xi ^{n})\) in \(\mathcal H (\xi ^{n},\psi ^{n}\mathbb D _1\xi ^{n})\), and applying the weighted Cauchy-Schwarz inequality to each term there, we will obtain the inequality

$$\begin{aligned} Y_1&\le \frac{1}{2}\beta \tau \Vert \xi ^{n}\Vert _{\psi ^{n}} \Vert \partial _x(\mathfrak M (\mathbb D _1\xi ^{n}))\Vert _{\psi ^{n}}+ \nu ^{-1}\beta \tau h^{-1}\Vert \xi ^{n}\Vert _{\psi ^{n}} \Vert \mathfrak M (\mathbb D _1\xi ^{n})\Vert _{\psi ^{n}} \nonumber \\&\quad +\tau \beta \Vert \xi ^{n}\Vert _{|\partial _x\psi ^{n}|} \Vert \mathbb D _1\xi ^{n}\Vert _{|\partial _x\psi ^{n}|} +\sqrt{2}\tau \beta \Vert \!\,\xi ^{n}\,\!\Vert _{\psi ^{n},\varGamma _{\!h}} \Vert \![\![\mathbb D _1\xi ^{n}]\!]\!\Vert _{\psi ^{n},\varGamma _{\!h}}, \end{aligned}$$
(8.7)

where we have also used \(|(x-x_j)/h_j|\le 1/2\) and \(h_j^{-1}\le \nu ^{-1}h^{-1}\) in each cell. Since both \(\mathfrak M (\mathbb D _1\xi ^{n})\) and \(\xi ^{n}\) belong to \(V_h\), we use the inverse properties (3.5a) and (3.5b), in Lemma 3.1, to estimate the first and the last terms on the right-hand side of (8.7), respectively. Further, we use property (3.3b) of the weight function to cope with the third term. Together with the application of Young’s inequality, the above process yields

$$\begin{aligned} Y_1\!&\le \! C\lambda \Vert \xi ^{n}\Vert _{\psi ^{n}} (\Vert \mathfrak M (\mathbb D _1\xi ^{n})\Vert _{\psi ^{n}}\!+\!h^{\frac{1}{2}} \Vert \![\![\mathbb D _1\xi ^{n}]\!]\!\Vert _{\psi ^{n},\varGamma _{\!h}}) \!+\!C\lambda \gamma ^{-1}h^{1-\sigma }\Vert \xi ^{n}\Vert _{\psi ^{n}} \Vert \mathbb D _1\xi ^{n}\Vert _{\psi ^{n}}\\ \!&\le \! C\varepsilon \Vert \xi ^{n}\Vert _{\psi ^{n}}^2 \!+\!C\lambda ^2\varepsilon ^{-1} (\Vert \mathfrak M (\mathbb D _1\xi ^{n})\Vert _{\psi ^{n}}^2+ h\Vert \![\![\mathbb D _1\xi ^{n}]\!]\!\Vert _{\psi ^{n},\varGamma _{\!h}}^2)\\&\quad +C\lambda ^2\gamma ^{-2}h^{2-2\sigma }\Vert \xi ^{n}\Vert _{\psi ^{n}}^2 +\frac{1}{6}\Vert \mathbb D _1\xi ^{n}\Vert _{\psi ^{n}}^2, \end{aligned}$$

where \(\varepsilon \) is any small positive constant. Hence, Lemma 5.3 about the generalized slope function gives that

$$\begin{aligned} Y_1&\le C(\varepsilon +\lambda ^2\gamma ^{-2}h^{2-2\sigma })\Vert \xi ^{n}\Vert _{\psi ^{n}}^2 +C\lambda ^2\varepsilon ^{-1}h\Vert \![\![\mathbb D _1\xi ^{n}]\!]\!\Vert _{\psi ^{n},\varGamma _{\!h}}^2 +\frac{1}{6}\Vert \mathbb D _1\xi ^{n}\Vert _{\psi ^{n}}^2 \nonumber \\&\quad +C\varepsilon ^{-1}\nu ^{-2}(\Vert \mathbb D _2\xi ^{n}\Vert _{\psi ^{n}}^2 +\Vert \mathbb D _2\eta ^{n}\Vert _{\psi ^{n}}^2). \end{aligned}$$
(8.8)

By using the superconvergence results given in Lemma 3.3, the weighted Cauchy-Schwarz inequality, and Young’s inequality, we have the estimate to the remaining terms

$$\begin{aligned}&Y_2+Y_3\le C\gamma ^{-1}h^{1-\sigma }\Vert \mathbb D _1\xi ^{n}\Vert _{\psi ^{n}}^2 +C\Vert \mathbb D _1\xi ^{n}\Vert _{\psi ^{n}}\Vert \mathbb D _1\eta ^{n}\Vert _{\psi ^{n}} \le \frac{1}{3}\Vert \mathbb D _1\xi ^{n}\Vert _{\psi ^{n}}^2\nonumber \\&\quad +C\Vert \mathbb D _1\eta ^{n}\Vert _{\psi ^{n}}^2, \end{aligned}$$
(8.9)

here \(\gamma h^{\sigma -1}\) is assumed to be large enough such that \(C\gamma ^{-1}h^{1-\sigma }\le 1/6\). Finally we substitute inequalities (8.8) and (8.9) into (8.6), and solve out \(\Vert \mathbb D _1\xi ^{n}\Vert _{\psi ^{n}}^2\). This completes the proof of this lemma.

1.5 Proof of inequality (5.6)

The proof is straightforward by using an integration by parts, and is almost the same as that in [4]. A simple manipulation yields that

$$\begin{aligned} B_j^-(\mathfrak M (x))= \frac{1}{4}\psi (x_{j+\frac{1}{2}})\mathfrak M ^2(x_{j+\frac{1}{2}}) +\frac{1}{4h_j}\int \limits _{I_j}\mathfrak M ^2(x)\psi (x)\mathrm d x+\Psi , \end{aligned}$$
(8.10)

where \(\Psi =-(2h_j^2)^{-1}\int _{I_j}\mathfrak M ^2(x)\partial _x\psi (x) (x-x_{j-1/2})(x-x_j)\mathrm d x\). Property (3.3b) of the weight function implies that

$$\begin{aligned} |\Psi |\le \frac{1}{4\gamma h^{\sigma }}\int \limits _{I_j}\mathfrak M ^2(x)\psi (x)\mathrm d x\le \frac{1}{8h_j}\int \limits _{I_j}\mathfrak M ^2(x)\psi (x)\mathrm d x, \end{aligned}$$
(8.11)

if \(\gamma h^{\sigma -1}\) is assumed to be large enough, since \(|2(x-x_{j-1/2})(x-x_j)|\le h_j^2\) in each cell. Thus we have completed the proof of this lemma.

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Zhang, Q., Shu, CW. Error estimates for the third order explicit Runge-Kutta discontinuous Galerkin method for a linear hyperbolic equation in one-dimension with discontinuous initial data. Numer. Math. 126, 703–740 (2014). https://doi.org/10.1007/s00211-013-0573-1

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