The pointwise error estimates of two energy-preserving fourth-order compact schemes for viscous Burgers’ equation

Abstract

A novel fourth-order three-point compact operator for the nonlinear convection term uu_x is provided in this paper. The operator makes the numerical analysis of higher-order difference schemes become possible for a wide class of nonlinear evolutionary equations under the unified framework. We take the classical viscous Burgers’ equation as an example and establish a new conservative fourth-order implicit compact difference scheme based on the method of order reduction. A detailed theoretical analysis is carried out by the discrete energy argument and mathematical induction. It is rigorously proved that the difference scheme is conservative, uniquely solvable, stable, and unconditionally convergent in discrete $L^{\infty }$-norm. The convergence order is two in time and four in space, respectively. Furthermore, we derive a three-level linearized compact difference scheme for viscous Burgers’ equation based on the proposed operator. All numerically theoretical results similar to that of the nonlinear numerical scheme are inherited completely; meanwhile, it is more time saving. Applying the compact operator to other more complex and higher-order nonlinear evolutionary equations is feasible, including Benjamin-Bona-Mahony-Burgers’ equation, Korteweg-de Vries equation, Kuramoto-Sivashinsky equation, and classification to name a few. Numerical results demonstrate that the presented schemes for Burgers’ equation can achieve second-order accuracy in time and fourth-order accuracy in space in $L^{\infty }$-norm.

Appendix. Direct proof of the convergence of the difference scheme in L 2-norm

The result and direct proof of L²-norm error estimate for the convergence of the difference scheme (3.14)–(3.18) are listed as follows.

Theorem A.1

Let {u(x,t), v(x,t)} be the solution of (3.1)–(3.4) and $\{{u_{i}^{k}},\ {v_{i}^{k}} | 0 \leqslant i \leqslant M, 0 \leqslant k \leqslant N\}$ be the solution of the difference scheme (3.14)–(3.18). Denote

$${e_{i}^{k}} = {U_{i}^{k}}-{u_{i}^{k}}, \quad {f_{i}^{k}} = {V_{i}^{k}}-{v_{i}^{k}}, \quad 0 \leqslant i \leqslant M, 0 \leqslant k \leqslant N,$$

and

$$ c_{13}= \frac12\left( \frac{2{c_{3}^{2}}}{\nu } + \nu + \frac{c_{3}}2\right), \quad c_{14} = \left( \nu + \frac1{2\nu}\right) {c_{1}^{2}}L, \quad c_{15} = \exp\left\{3c_{13}T\right\} \cdot \sqrt{\frac{c_{14}}{2c_{13}}}. $$

If $2c_{13}\tau \leqslant 1/3$, then we have:

$$ \|e^{k}\| \leqslant c_{15}(\tau^{2}+h^{4}),\quad 0 \leqslant k \leqslant N. $$

(A.1)

Proof

It is easy to know that (A.1) holds for k = 0.

Taking an inner product of (3.41) with $e^{k+\frac {1}{2}}$, we have:

$$ \begin{array}{@{}rcl@{}} && \left( \delta_{t}e^{k+\frac{1}{2}},e^{k+\frac{1}{2}}\right) + \left( \psi(U^{k+\frac{1}{2}},U^{k+\frac{1}{2}}) - \psi(u^{k+\frac{1}{2}},u^{k+\frac{1}{2}}),e^{k+\frac{1}{2}} \right) \\ &&-\frac{h^{2}}{2}\left( \psi(V^{k+\frac{1}{2}},U^{k+\frac{1}{2}})-\psi(v^{k+\frac{1}{2}},u^{k+\frac{1}{2}}),e^{k+\frac{1}{2}}\right)-\nu \left( f^{k+\frac{1}{2}},e^{k+\frac{1}{2}}\right) = \left( P^{k+\frac{1}{2}},e^{k+\frac{1}{2}}\right), \\ && \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad 0 \leqslant k \leqslant N-1. \end{array} $$

(A.2)

Using Lemma 2.1, Cauchy-Schwartz inequality, and inequality (2.4), we have:

$$ \begin{array}{@{}rcl@{}} && \left( f^{k+\frac{1}{2}},e^{k+\frac{1}{2}}\right) \\ &\leqslant & -|e^{k+\frac{1}{2}}|_{1}^{2} -\frac{h^{2}}{18}\|f^{k+\frac{1}{2}}\|^{2} + \frac{h^{2}}{12} \left( f^{k+\frac{1}{2}},Q^{k+\frac{1}{2}}\right) + \left( Q^{k+\frac{1}{2}},e^{k+\frac{1}{2}}\right) \\ &\leqslant & -\frac{h^{2}}{18}\|f^{k+\frac{1}{2}}\|^{2} + \frac{h^{2}}6\left( \frac19\|f^{k+\frac12}\|^{2} + \frac9{16}\|Q^{k+\frac{1}{2}}\|^{2}\right) + \left( \frac12\|Q^{k+\frac{1}{2}}\|^{2} + \frac12\|e^{k+\frac{1}{2}}\|^{2}\right) \\ &= & - \frac{h^{2}}{27}\|f^{k+\frac12}\|^{2} + \left( \frac{3h^{2}}{32} + \frac12\right)\|Q^{k+\frac{1}{2}}\|^{2} + \frac12\|e^{k+\frac{1}{2}}\|^{2}. \end{array} $$

(A.3)

By Lemmas 2.2, 2.3 and (3.39), computing arrives at:

$$ \begin{array}{@{}rcl@{}} && - \left( \psi(U^{k+\frac{1}{2}},U^{k+\frac{1}{2}}) - \psi(u^{k+\frac{1}{2}},u^{k+\frac{1}{2}}),e^{k+\frac{1}{2}}\right)\\ &=& -\left( \psi(e^{k+\frac{1}{2}},U^{k+\frac{1}{2}}),e^{k+\frac{1}{2}}\right)\\ &=& - \frac{h}{3} \sum\limits_{i=1}^{M-1} \left[e_{i}^{k+\frac{1}{2}}{\varDelta}_{x} U_{i}^{k+\frac{1}{2}}+ {\varDelta}_{x}(e^{k+\frac{1}{2}}U^{k+\frac{1}{2}})_{i} \right]e_{i}^{k+\frac{1}{2}}\\ &=& - \frac{h}{3} \sum\limits_{i=1}^{M-1} (e_{i}^{k+\frac{1}{2}})^{2} \cdot {\varDelta}_{x} U_{i}^{k+\frac{1}{2}}- \frac{h}{6} \sum\limits_{i=1}^{M-1} \frac{U_{i+1}^{k+\frac{1}{2}}-U_{i}^{k+\frac{1}{2}}}{h} \cdot e_{i}^{k+\frac{1}{2}}e_{i+1}^{k+\frac{1}{2}}\\ &\leqslant & \frac{c_{3}}{2}\|e^{k+\frac{1}{2}}\|^{2} \end{array} $$

and

$$ \begin{array}{@{}rcl@{}} && \frac{h^{2}}{2}\left( \psi(V^{k+\frac{1}{2}},U^{k+\frac{1}{2}})-\psi(v^{k+\frac{1}{2}},u^{k+\frac{1}{2}}),e^{k+\frac{1}{2}}\right) \\ &= & \frac{h^{2}}2\left( \psi(f^{k+\frac{1}{2}},U^{k+\frac{1}{2}}),e^{k+\frac{1}{2}}\right) \\ &= & \frac{h^{3}}6 \sum\limits_{i=1}^{M-1} \left[f_{i}^{k+\frac{1}{2}}{\varDelta}_{x}U_{i}^{k+\frac{1}{2}} +{\varDelta}_{x}(f^{k+\frac{1}{2}}U^{k+\frac{1}{2}})_{i}\right]e_{i}^{k+\frac{1}{2}} \\ &=& \frac{h^{3}}6 \sum\limits_{i=1}^{M-1} f_{i}^{k+\frac{1}{2}}e_{i}^{k+\frac{1}{2}}{\varDelta}_{x}U_{i}^{k+\frac{1}{2}} -\frac{h^{3}}6\sum\limits_{i=1}^{M-1}f_{i}^{k+\frac{1}{2}}U_{i}^{k+\frac{1}{2}}{\varDelta}_{x}e_{i}^{k+\frac{1}{2}} \\ &\leqslant & \frac{c_{3}h^{2}}{6} \|f^{k+\frac{1}{2}}\|\cdot\|e^{k+\frac{1}{2}}\| + \frac{c_{3}h^{2}}{6}\|f^{k+\frac{1}{2}}\|\cdot |e^{k+\frac{1}{2}}|_{1} \\ &\leqslant & \frac{h^{2}}6\left( \frac\nu9\|f^{k+\frac12}\|^{2} + \frac{9{c_{3}^{2}}}{4\nu}\|e^{k+\frac12}\|^{2}\right) + \frac{h^{2}}6\left( \frac\nu9\|f^{k+\frac12}\|^{2} + \frac{9{c_{3}^{2}}}{4\nu}|e^{k+\frac12}|_{1}^{2}\right) \\ &\leqslant & \frac{\nu h^{2}}{27}\|f^{k+\frac12}\|^{2} + \left( \frac{3{c_{3}^{2}}h^{2}}{8\nu} + \frac{3{c_{3}^{2}}}{2\nu }\right)\|e^{k+\frac12}\|^{2}. \end{array} $$

(A.4)

Substituting (A.3)–(A.4) into (A.2) and combining with (3.8) and (3.9), we can obtain:

$$ \begin{array}{@{}rcl@{}} & & \frac{1}{2\tau}(\|e^{k+1}\|^{2}-\|e^{k}\|^{2})+ \frac{\nu h^{2}}{27}\|f^{k+\frac{1}{2}}\|^{2} \\ &\leqslant & \left( \frac{3\nu h^{2}}{32} + \frac\nu2\right)\|Q^{k+\frac{1}{2}}\|^{2} + \frac\nu2\|e^{k+\frac{1}{2}}\|^{2} + \frac{c_{3}}2\|e^{k+\frac12}\|^{2} \\ & &+ \frac{\nu h^{2}}{27}\|f^{k+\frac12}\|^{2} + \left( \frac{3{c_{3}^{2}}h^{2}}{8\nu} + \frac{3{c_{3}^{2}}}{2\nu }\right)\|e^{k+\frac12}\|^{2}+(P^{k+\frac{1}{2}},e^{k+\frac{1}{2}}), \quad 0 \leqslant k \leqslant N-1 \end{array} $$

or

$$ \begin{array}{@{}rcl@{}} && \frac{1}{2\tau}\left( \|e^{k+1}\|^{2}-\|e^{k}\|^{2}\right) \\ &\leqslant & \nu\|Q^{k+\frac{1}{2}}\|^{2} + \left( \frac{2{c_{3}^{2}}}{\nu } + \nu + \frac{c_{3}}2\right)\|e^{k+\frac12}\|^{2}+ \frac1{2\nu}\|P^{k+\frac{1}{2}}\|^{2} \\ &\leqslant & c_{13}\left( \|e^{k+1}\|^{2} + \|e^{k}\|^{2}\right) + c_{14}(\tau^{2}+h^{4})^{2}, \quad 0 \leqslant k \leqslant N-1. \end{array} $$

Then we have

$$(1-2c_{13}\tau)\|e^{k+1}\|^{2} \leqslant (1+2c_{13}\tau) \|e^{k}\|^{2} + 2 c_{14}\tau(\tau^{2}+h^{4})^{2}, \quad 0 \leqslant k \leqslant N-1.$$

When $2c_{13}\tau \leqslant \frac {1}{3}$, we have

$$\|e^{k+1}\|^{2} \leqslant (1+6c_{13}\tau) \|e^{k}\|^{2} + 3c_{14}\tau(\tau^{2}+h^{4})^{2}, \quad 0 \leqslant k \leqslant N-1.$$

Using Gronwall inequality, equations (3.43) and (3.44), we can obtain

$$\|e^{k+1}\|^{2} \leqslant \exp\left\{6c_{13}k\tau\right\} \cdot \frac{c_{14}}{2c_{13}} \cdot (\tau^{2}+h^{4})^{2} \leqslant c_{15}^{2}(\tau^{2}+h^{4})^{2}, \quad 0 \leqslant k \leqslant N-1.$$

This completes the proof. □