Discontinuous Galerkin Methods with Optimal $L^2$ Accuracy for One Dimensional Linear PDEs with High Order Spatial Derivatives

Abstract

In this paper, we formulate and analyze discontinuous Galerkin (DG) methods to solve several partial differential equations (PDEs) with high order spatial derivatives, including the heat equation, a third order wave equation, a fourth order equation and the linear Schrödinger equation in one dimension. Following the idea of local DG methods, we first rewrite each PDE into its first order form and then apply a general DG formulation. The numerical fluxes are introduced as linear combinations of average values of fluxes, and jumps of the solution as well as the auxiliary variables at cell interfaces. The main focus of the present work is to identify a sub-family of the numerical fluxes by choosing the coefficients in the linear combinations, so the solution and some auxiliary variables of the proposed DG methods are optimally accurate in the $L^2$ norm. In our analysis, one key component is to design some special projection operator(s), tailored for each choice of numerical fluxes in the sub-family, to eliminate those terms at cell interfaces that would otherwise contribute to the sub-optimality of the error estimates. Our theoretical findings are validated by a set of numerical examples.

Appendices

A The Proof of Theorem 4.3

In this appendix, we prove the $L^2$ stability of the DG scheme (4.6)–(4.7) with the numerical fluxes (4.5) and $\alpha =\frac{1}{2},\beta _1=\beta _2=0$, namely

$$\begin{aligned} F_p =p_h^+, \quad F_q = q_h^+, \quad F_u=u_h^-. \end{aligned}$$

(A.1)

Five energy equations will be derived first.

$\bullet$The first energy equation. With $\alpha =\frac{1}{2},\beta _1=\beta _2=0$, the first energy equation (4.9) becomes

$$\begin{aligned} 0=B(u_h,p_h,q_h;u_h,q_h,-p_h)=\frac{1}{2}\frac{d}{dt}\int _{\Omega }u_h^2dx +\frac{1}{2}\sum \limits _{j}[q_h]^2_{j-\frac{1}{2}}. \end{aligned}$$

(A.2)

$\bullet$The second energy equation. We take the test functions $v=-\frac{1}{2}(q_h)_t$, $w=\frac{1}{2}p_h$ and $z=0$ in (4.10), use the definition of $F_p$, $F_q$ and $F_u$ in (A.1), and obtain

$$\begin{aligned} 0&=B(u_h,(p_h)_t,(q_h)_t;-\frac{1}{2}(q_h)_t,\frac{1}{2}p_h,0)\nonumber \\&=-\frac{1}{2}\int _{\Omega }(u_h)_t(q_h)_t dx + \frac{1}{2}\sum \limits _{j}\left( \int _{I_j}p_h (q_h)_{tx}dx+(F_p[(q_h)_t])_{j-\frac{1}{2}}\right) \nonumber \\&\quad +\frac{1}{2}\int _{\Omega }(p_h)_t p_h dx + \frac{1}{2}\sum \limits _{j}\left( \int _{I_j}(q_h)_t(p_h)_xdx+((F_q)_t[p_h])_{j-\frac{1}{2}}\right) \nonumber \\&=\frac{1}{2}\int _{\Omega }(-(u_h)_t(q_h)_t+(p_h)_tp_h)dx+\frac{1}{2}\sum \limits _{j}\left( [p_h][(q_h)_t]\right) _{j-\frac{1}{2}}. \end{aligned}$$

(A.3)

$\bullet$The third energy equation. With $\alpha =\frac{1}{2},\beta _1=\beta _2=0$, (4.14) becomes

$$\begin{aligned} 0=&B((u_h)_t,(p_h)_t,(q_h)_t;(u_h)_t,(q_h)_t,-(p_h)_t)=\frac{1}{2}\frac{d}{dt}\int _{\Omega }(u_h)_t^2dx +\frac{1}{2}\sum \limits _{j}[(q_h)_t]^2_{j-\frac{1}{2}}. \end{aligned}$$

(A.4)

$\bullet$The fourth energy equation. Here we take the test functions $v=-p_h$, $w=(u_h)_t$ and $z=q_h$ in (4.15), use the definition of $F_p$, $F_q$ and $F_u$ in (A.1), and obtain

$$\begin{aligned} 0&=B\left( u_h,p_h,\left( q_h\right) _t;-p_h,\left( u_h\right) _t,q_h\right) \nonumber \\&=-\int _{\Omega }\left( u_h\right) _t p_h dx +\sum \limits _{j}\left( \int _{I_j}p_h\left( p_h\right) _{x}dx+\left( F_p[p_h]\right) _{j-\frac{1}{2}}\right) \nonumber \\&\quad +\int _{\Omega }p_h \left( u_h\right) _t dx + \sum \limits _{j}\left( \int _{I_j}q_h\left( u_h\right) _{tx}dx+\left( F_q[\left( u_h\right) _t]\right) _{j-\frac{1}{2}}\right) \nonumber \\&\quad + \int _{\Omega }\left( q_h\right) _tq_h dx + \sum \limits _{j}\left( \int _{I_j}\left( u_h\right) _t \left( q_h\right) _xdx+\left( \left( F_u\right) _t[q_h]\right) _{j-\frac{1}{2}}\right) \nonumber \\&=\frac{1}{2}\frac{d}{dt} \int _{\Omega }\left( q_h\right) ^2 dx+\frac{1}{2}\sum \limits _{j}[p_h]^2_{j-\frac{1}{2}}. \end{aligned}$$

(A.5)

$\bullet$The fifth energy equation. We take the time derivative of (4.15) and get

$$\begin{aligned} B((u_h)_t,(p_h)_t,(q_h)_{tt};v,w,z)=0, \quad \forall v,w,z\in {\mathcal {V}}_h^k. \end{aligned}$$

(A.6)

With the test functions in (A.6) taken as $v=-(p_h)_t$, $w=(u_h)_{tt}$ and $z=(q_h)_t$, using the definition of $F_p$, $F_q$ and $F_u$ in (A.1), we have

$$\begin{aligned} 0=B((u_h)_t,(p_h)_t,(q_h)_{tt};-(p_h)_t,(u_h)_{tt},(q_h)_t)=\int _{\Omega }(q_h)_{tt}(q_h)_t dx+\frac{1}{2}\sum \limits _{j}[(p_h)_t]^2_{j-\frac{1}{2}}. \end{aligned}$$

(A.7)

$\bullet$Proof of Theorem4.3

Proof

We sum up the energy equations (A.2)–(A.5) and (A.7), and get

$$\begin{aligned} 0&=\frac{1}{2}\frac{d}{dt}\left( \Vert u_h\Vert ^2_\Omega +\frac{1}{2}\Vert p_h\Vert ^2_\Omega +\Vert (u_h)_t\Vert ^2_\Omega +\Vert q_h\Vert ^2_\Omega +\Vert (q_h)_t\Vert ^2_\Omega \right) -\frac{1}{2}((q_h)_t,(u_h)_t)\nonumber \\&\quad +\frac{1}{2}\sum \limits _{j}\left( [q_h]^2+[(q_h)_t]^2+[(p_h)_t]^2+[p_h]^2+[p_h][(q_h)_t]\right) _{j-\frac{1}{2}}. \end{aligned}$$

(A.8)

Thus, we have

$$\begin{aligned} \frac{1}{2}\frac{d}{dt}\left( \Vert u_h\Vert ^2_\Omega +\frac{1}{2}\Vert p_h\Vert ^2_\Omega +\Vert (u_h)_t\Vert ^2_\Omega +\Vert q_h\Vert ^2_\Omega +\Vert (q_h)_t\Vert ^2_\Omega \right) \le \frac{1}{4}(\Vert (q_h)_t\Vert ^2_\Omega +\Vert (u_h)_t\Vert ^2_\Omega ). \end{aligned}$$

(A.9)

We integrate (A.9) with respect to time over [0, T], and obtain

$$\begin{aligned}&\Vert u_h(T)\Vert ^2_\Omega +\frac{1}{2}\Vert p_h(T)\Vert ^2_\Omega +\Vert (u_h)_t(T)\Vert ^2_\Omega +\Vert q_h(T)\Vert ^2_\Omega +\Vert (q_h)_t(T)\Vert ^2_\Omega \nonumber \\&\quad \le \frac{1}{2}\int _0^T(\Vert (q_h)_t(t) \Vert ^2_\Omega +\Vert (u_h)_t(t) \Vert ^2_\Omega )dt\nonumber \\&\qquad +\left( \Vert u_h(0)\Vert ^2_\Omega +\frac{1}{2}\Vert p_h(0)\Vert ^2_\Omega +\Vert (u_h)_t(0)\Vert ^2_\Omega +\Vert q_h(0)\Vert ^2_\Omega +\Vert (q_h)_t(0)\Vert ^2_\Omega \right) . \end{aligned}$$

(A.10)

From the third energy equation (A.4) and the fifth energy equation (A.7), we have

$$\begin{aligned} \Vert (u_h)_t(t)\Vert ^2_\Omega \le \Vert (u_h)_t(0)\Vert ^2_\Omega , \quad \Vert (q_h)_t(t)\Vert ^2_\Omega \le \Vert (q_h)_t(0)\Vert ^2_\Omega , \quad \forall t\ge 0. \end{aligned}$$

(A.11)

Therefore, we can obtain the $L^2$ stability in Theorem 4.3. $\square$

B The Derivation of the Conditions (4.18)

In this appendix, we will give the derivation of the conditions (4.18) that are to ensure the matrices $S_1$ and $S_2$ in (4.22) to be positive semi-definite. For this, we use the following sufficient and necessary condition for an $n\times n$ matrix to be positive semi-definite: all the principal minors$D_k$are nonnegative, $k=1, \ldots n$. Here, $D_k$ is formed by deleting any$n-k$ rows and the corresponding columns. Additionally, we require the relation $\alpha ^2+\beta _1\beta _2=\frac{1}{4}$, which helps with simplifying the conditions and is also needed for optimal accuracy.

For the matrix $S_1$, the first principal minors are

$$\begin{aligned} D_{1,1}=-\beta _1,\;D_{1,2}=-\beta _2,\;D_{1,3}=\frac{1}{2}; \end{aligned}$$

(B.1)

the second principal minors are

$$\begin{aligned} D_{2,1}=\begin{vmatrix} -\beta _2&\frac{\alpha +0.5}{2} \\ \frac{\alpha +0.5}{2}&\frac{1}{2} \end{vmatrix} ,\; D_{2,2}=\begin{vmatrix} -\beta _1&\frac{\beta _1}{2} \\ \frac{\beta _1}{2}&\frac{1}{2} \end{vmatrix} ,\; D_{2,3}=\begin{vmatrix} -\beta _1&0 \\ 0&-\beta _2 \end{vmatrix} ; \end{aligned}$$

(B.2)

and the third principal minor is

$$\begin{aligned} D_3=|S_1|=\frac{1}{2}\beta _1\beta _2+\frac{1}{4}\beta _1^2\beta _2+\frac{1}{4}\beta _1{\left( \alpha +\frac{1}{2}\right) ^2}. \end{aligned}$$

(B.3)

Let the first principal minors be nonnegative, we have $\beta _1\le 0$ and $\beta _2\le 0$. From the second principal minors being nonnegative, we obtain

$$\begin{aligned} 2\beta _2+\left( \alpha +\frac{1}{2}\right) ^2\le 0,\;\;\beta _1(2+\beta _1)\le 0,\; \;\beta _1\beta _2\ge 0. \end{aligned}$$

(B.4)

Let the third principal minor $D_3$ be nonnegative, with $\beta _1\le 0$ and the assumption $\alpha ^2+\beta _1\beta _2=\frac{1}{4}$, we get

$$\begin{aligned} 2\beta _2+\beta _1\beta _2+(\alpha +\frac{1}{2})^2 =2\beta _2+\alpha +\frac{1}{2}\le 0. \end{aligned}$$

(B.5)

We observe that, with $\beta _1\beta _2\ge 0$, the inequality (B.5) will automatically ensure the first inequality in (B.4). Combining all we have so far, the following conditions are derived to ensure $S_1$ be positive semi-definite

$$\begin{aligned} -2\le \beta _1\le 0,\;\;\beta _2\le 0,\;\;2\beta _2+\alpha \le -\frac{1}{2},\;\;\alpha ^2+\beta _1\beta _2=\frac{1}{4}. \end{aligned}$$

(B.6)

Fig. 1

The region of $\beta _1,\beta _2$ in the condition (4.18) for the stability. Left: $\alpha =-\sqrt{\frac{1}{4}-\beta _1\beta _2}$; Right: $\alpha =-\frac{1}{4}$

For the matrix $S_2$, we follow the similar analysis as for $S_1$. By requiring all the first and the second principal minors being nonnegative, we get

$$\begin{aligned}&\beta _1+\alpha \le 0, \quad \beta _2\le 0, \end{aligned}$$

(B.7)

$$\begin{aligned}&8(\beta _1+\alpha )+\left( \frac{1}{2}-\alpha \right) ^2\le 0,\; \; \beta _2(8+\beta _2)\le 0,\; \;(\alpha +\beta _1)\beta _2-\frac{1}{4}\beta _2^2\ge 0, \end{aligned}$$

(B.8)

Let the third order principal minor of $S_2$ be nonnegative, also with $\beta _2\ge 0$, we obtain

$$\begin{aligned}&8(\beta _1+\alpha )+\left( \frac{1}{2}-\alpha \right) ^2+\beta _2\left( \frac{1}{2}-\alpha \right) +\beta _2(\beta _1+\alpha )-2\beta _2\nonumber \\&\quad =8(\beta _1+\alpha )+\left( \frac{1}{2}-\alpha \right) ^2+\beta _2(\beta _1-\frac{3}{2}) \le 0. \end{aligned}$$

(B.9)

Using $\beta _2\le 0$ and assuming $\beta _1\le 0$, one can see that (B.9) implies the first inequality in (B.8), which on the other hand ensures the first inequality in (B.7). Combining (B.7)–(B.9) with $\alpha ^2+\beta _1\beta _2=\frac{1}{4}$, we have the conditions for $S_2$ as

$$\begin{aligned} \alpha ^2+\beta _1\beta _2=\frac{1}{4},\;\beta _1\le 0,\;-8\le \beta _2\le 0,\;4(\beta _1+\alpha )\le \beta _2,\;8\beta _1+7\alpha -\frac{3}{2}\beta _2\le -\frac{1}{2}. \end{aligned}$$

(B.10)

Finally, we reach the conditions in (4.18) by putting (B.6) and (B.10) together. To get some idea about these conditions in (4.18), we present two plots in Fig. 1. In the left figure, we plot those pairs $(\beta _1,\beta _2)$ such that with the respective $\alpha =-\sqrt{\frac{1}{4}-\beta _1\beta _2}$, the conditions in (4.18) are all satisfied. In the right figure, we fix $\alpha =-\frac{1}{4}$, and plot $(\beta _1,\beta _2)$ satisfying the conditions in (4.18). Note that such $(\beta _1,\beta _2)$ form part of the parabola: $\beta _1\beta _2=\frac{1}{4}-\alpha ^2=\frac{3}{16}$, see the solid line in red.