Stormer-Numerov HDG Methods for Acoustic Waves

Abstract

We introduce and analyze the first energy-conservative hybridizable discontinuous Galerkin method for the semidiscretization in space of the acoustic wave equation. We prove optimal convergence and superconvergence estimates for the semidiscrete method. We then introduce a two-step fourth-order-in-time Stormer-Numerov discretization and prove energy conservation and convergence estimates for the fully discrete method. In particular, we show that by using polynomial approximations of degree two, convergence of order four is obtained. Numerical experiments verifying that our theoretical orders of convergence are sharp are presented. We also show experiments comparing the method with dissipative methods of the same order.

An Identity Used in the Duality Argument

Lemma A.1

Suppose $\eta$ is the continuous piecewise-linear Lagrange interpolation of the values $\{\eta ^n\}_{n=0}^{N}$ and $\mu :[0,T]\rightarrow {\mathbb {R}}$ is a function. Then we have

$$\begin{aligned} \int _0^T\eta (t)\mu (t)dt&=\tfrac{k}{2}\eta (0)\mathrm I_k\mu (0)+k\sum _{n=1}^{N-1}\eta (t_n)\mathrm I_k\mu (t_n)+\tfrac{k}{2}\eta (0)\mathrm I_k\mu (0),\\ \int _0^T\dot{\eta }(t)\dot{\mu }(t)dt&=\delta _k\eta ^N\mu (T)-k\sum _{n=1}^{N-1}\mathrm D_k^2\eta ^n\mu (t_n)-\delta _k\eta ^1\mu (0). \end{aligned}$$

There result also holds for functions taking values in an inner product space, when the pointwise product is substituted by the inner product.

Proof

We have

$$\begin{aligned} \int _0^T\eta (t)\mu (t)dt&=\sum _{n=1}^{N}\Big (\eta ^n\int _{t_{n-1}}^{t_n}\tfrac{1}{k}(t-t_{n-1})\mu (t)dt+\eta ^{n-1}\int _{t_{n-1}}^{t_n}\tfrac{1}{k}(t_n-t)\mu (t)dt\Big )\\&=\eta ^N\int _{T-k}^T\tfrac{1}{k}(t-(T-k))\mu (t)dt+\sum _{n=1}^{N-1}\eta ^n\int _{t_{n-1}}^{t_n}\tfrac{1}{k}(t-t_{n-1})\mu (t)dt\\&\quad +\,\sum _{n=2}^{N}\eta ^{n-1}\int _{t_{n-1}}^{t_n}\tfrac{1}{k}(t_n-t)\mu (t)dt+\eta ^0\int _0^k\tfrac{1}{k}(k-t)\mu (t)dt. \end{aligned}$$

After using the definition of $\mathrm I_k$, we get the first identity.

To obtain the second identity, note that we have

$$\begin{aligned} \int _0^T\dot{\eta }(t)\dot{\mu }(t)dt&=\sum _{n=1}^N\int _{t_{n-1}}^{t_n}\dot{\eta }(t)\dot{\mu }(t)dt =\sum _{n=1}^N\delta _k\eta ^n(\mu (t_n)-\mu (t_{n-1})), \end{aligned}$$

and the result follows after simple rearrangements. $\square$