Two exponential-type integrators for the “good” Boussinesq equation

Abstract

We introduce two exponential-type integrators for the “good” Boussinesq equation. They are of orders one and two, respectively, and they require lower spatial regularity of the solution compared to classical exponential integrators. For the first integrator, we prove first-order convergence in $H^r$ for solutions in $H^{r+1}$ with $r>1/2$. This new integrator even converges (with lower order) in $H^r$ for solutions in $H^r$, i.e., without any additional smoothness assumptions. For the second integrator, we prove second-order convergence in $H^r$ for solutions in $H^{r+3}$ with $r>1/2$ and convergence in $L^2$ for solutions in $H^3$. Numerical results are reported to illustrate the established error estimates. The experiments clearly demonstrate that the new exponential-type integrators are favorable over classical exponential integrators for initial data with low regularity.

Appendix: Classical exponential integrators

For the purpose of a reference solution and as benchmark for comparisons, we make use of the classical exponential integrators by Gautschi and Deuflhard [10, 14, 15, 30]. In this appendix, we shortly recall these integrators and their application to the GB equation.

Applying Duhamel’s formula to (2.1), we get

$$\begin{aligned} z(t_n+s)&=\cos (s\langle \partial _x^2\rangle )z(t_n)+\frac{\sin (s\langle \partial _x^2\rangle )}{\langle \partial _x^2\rangle } z_t(t_n)\\&\quad +\frac{1}{\langle \partial _x^2\rangle }\int _0^s \sin ((s-y)\langle \partial _x^2\rangle )(z^2)_{xx}(t_n+y)\mathrm{d}y,\\ z_t(t_n+s)&=-\langle \partial _x^2\rangle \sin (s\langle \partial _x^2\rangle )z(t_n)+\cos (s\langle \partial _x^2\rangle )z_t(t_n)\\&\quad +\int _0^s \cos ((s-y)\langle \partial _x^2\rangle )(z^2)_{xx}(t_n+y)\mathrm{d}y, \end{aligned}$$

where $\langle \partial _x^2\rangle = \sqrt{\partial _x^4-\partial _x^2}$. Approximating the integral in different ways, we obtain the following three exponential integrators.

Gautschi 1 EI Setting $s=\tau$ and replacing $(z^2)_{xx}(t_n+y)$ by $(z^2)_{xx}(t_n)$ in the integrals leads to the following one-step scheme of first order (e.g., [15, Example 3.11]):

$$\begin{aligned} z^{n+1}&=\cos (\tau \langle \partial _x^2\rangle )z^n+\frac{\sin (\tau \langle \partial _x^2\rangle )}{\langle \partial _x^2\rangle } z_t^n+\frac{\tau ^2}{2}\mathrm {sinc}^2\left( \frac{\tau }{2}\langle \partial _x^2\rangle \right) ((z^n)^2)_{xx},\\ z_t^{n+1}&=-\langle \partial _x^2\rangle \sin (\tau \langle \partial _x^2\rangle )z^n+\cos (\tau \langle \partial _x^2\rangle )z_t^n+\tau \mathrm {sinc}(\tau \langle \partial _x^2\rangle )((z^n)^2)_{xx},\quad n\ge 0. \end{aligned}$$

Gautschi 2 EI Setting $s=\pm \tau$, replacing $(z^2)_{xx}(t_n\pm y)$ by $(z^2)_{xx}(t_n)$ and adding up yields the following symmetric two-step scheme of second order (e.g., [15, Example 3.11]):

$$\begin{aligned} z^{n+1}=&-z^{n-1}+2\cos (\tau \langle \partial _x^2\rangle )z^n+\tau ^2\mathrm {sinc}^2\left( \frac{\tau }{2}\langle \partial _x^2\rangle \right) ((z^n)^2)_{xx},\quad n\ge 1,\\ z^{1}=&\cos (\tau \langle \partial _x^2\rangle )z^0+\frac{\sin (\tau \langle \partial _x^2\rangle )}{\langle \partial _x^2\rangle } z_t^0+\frac{\tau ^2}{2}\mathrm {sinc}^2\left( \frac{\tau }{2}\langle \partial _x^2\rangle \right) ((z^0)^2)_{xx}. \end{aligned}$$

Deuflhard 2 EI Setting $s=\tau$ and approximating the integrals by using the trapezoidal rule, one gets the one-step scheme of second-order for $n\ge 0$ as follows (e.g., [15, Example 3.12]):

$$\begin{aligned} z^{n+1}&=\cos (\tau \langle \partial _x^2\rangle )z^n+\frac{\sin (\tau \langle \partial _x^2\rangle )}{\langle \partial _x^2\rangle } z_t^n+\frac{\tau ^2}{2}\mathrm {sinc}\left( \tau \langle \partial _x^2\rangle \right) ((z^n)^2)_{xx},\\ z_t^{n+1}&=-\langle \partial _x^2\rangle \sin (\tau \langle \partial _x^2\rangle )z^n+\cos (\tau \langle \partial _x^2\rangle )z_t^n\\&\quad +\frac{\tau }{2}\left[ \cos (\tau \langle \partial _x^2\rangle )((z^n)^2)_{xx}+ ((z^{n+1})^2)_{xx}\right] . \end{aligned}$$