A high-order numerical method for solving the 2D fourth-order reaction-diffusion equation

Haixiang Zhang; Xuehua Yang; Da Xu

doi:10.1007/s11075-018-0509-z

Numerical Algorithms

March 2019, Volume 80, Issue 3, pp 849–877 | Cite as

A high-order numerical method for solving the 2D fourth-order reaction-diffusion equation

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Authors and affiliations

Haixiang Zhang
Xuehua Yang
Da Xu

Original Paper

First Online: 26 March 2018

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Abstract

In the present work, orthogonal spline collocation (OSC) method with convergence order O(τ^3−α + h^r+ 1) is proposed for the two-dimensional (2D) fourth-order fractional reaction-diffusion equation, where τ, h, r, and α are the time-step size, space size, polynomial degree of space, and the order of the time-fractional derivative (0 < α < 1), respectively. The method is based on applying a high-order finite difference method (FDM) to approximate the time Caputo fractional derivative and employing OSC method to approximate the spatial fourth-order derivative. Using the argument developed recently by Lv and Xu (SIAM J. Sci. Comput. 38, A2699–A2724, 2016) and mathematical induction method, the optimal error estimates of proposed fully discrete OSC method are proved in detail. Then, the theoretical analysis is validated by a number of numerical experiments. To the best of our knowledge, this is the first proof on the error estimates of high-order numerical method with convergence order O(τ^3−α + h^r+ 1) for the 2D fourth-order fractional equation.

Keywords

Fourth-order fractional equation Orthogonal spline collocation Finite difference method Error estimate

Mathematics Subject Classification (2010)

65M12 65M06 65M70 35S10

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Notes

Acknowledgements

The authors thank the anonymous reviewers for their constructive comments and suggestions and Professor Graeme Fairweather for stimulating discussions and for his constant encouragement and support.

Funding information

The work is supported by the National Natural Science Foundation of China (11701168, 11601144, 11626096), Hunan Provincial Natural Science Foundation of China (2018JJ3108, 2018JJ3109, 2018JJ4062), Scientific Research Fund of Hunan Provincial Education Department (16K026,YB2016B033), China Postdoctoral Science Foundation (2016M600964), Science Challenge Project (TZ2016002).

Appendix A

Proof

Since the first step equation is easier to treat,we will only consider the case for the time stepsn ≥ 2.By using the reformulations (3.14), (3.18) and (3.19), (3.20) can be rewritten as

$$\begin{array}{@{}rcl@{}} &&(\overline{u}^{n},\chi)-\kappa\alpha_{0}\beta_{0}^{-1}({\Delta} u^{n},\chi)-\alpha_{0}\beta_{0}^{-1}({\Delta} v^{n},\chi) \\ &=&p(\overline u^{n-1},\chi) +\sum\limits_{k = 2}^{n-1}\overline{d}^{n}_{n-k}(\overline{u}^{n-k},\chi)+\overline{d}^{n}_{0}(u^{0},\chi) +\alpha_{0}\beta_{0}^{-1}(f^{n},\chi), \ \forall \chi\\ &\in& {H^{1}_{0}}({\Omega}), \ 2\leq n\leq K, \end{array} $$

(A.63)

$$\begin{array}{@{}rcl@{}} (v^{n},\psi)=-({\Delta} u^{n},\psi),\ \forall \psi\in {H^{1}_{0}}({\Omega}), \quad 2\leq n\leq K, \end{array} $$

(A.64)

where (⋅,⋅) isthe usual L²-innerproduct, ∥⋅∥ iscorresponding norm.

By choosing $\chi = 2\overline {u}^{n}$,$\psi = 2\alpha _{0}\beta _{0}^{-1}{\Delta }\overline {u}^{n}$ in (A.63) and (A.64), respectively, we subtract the resulting equations to obtain

$$\begin{array}{@{}rcl@{}} &&(\overline{u}^{n},2\overline{u}^{n}) -\kappa\alpha_{0}\beta_{0}^{-1}({\Delta} u^{n},2\overline{u}^{n} ) +\alpha_{0}\beta_{0}^{-1} ({\Delta} u^{n},2{\Delta}\overline{u}^{n})\\ &=&p(\overline{u}^{n-1},2\overline{u}^{n}) +\sum\limits_{k = 2}^{n-1}\overline{d}^{n}_{n-k}(\overline{u}^{n-k},2\overline{u}^{n}) +\overline{d}^{n}_{0}(u^{0},2\overline{u}^{n}) \\ &&+\alpha_{0}\beta_{0}^{-1}(f^{n},2\overline{u}^{n}),\ \ 2\leq n\leq K. \end{array} $$

(A.65)

In virtue of the identity

$$\begin{array}{@{}rcl@{}} 2({\Delta} u^{n},{\Delta}\overline{u}^{n}) =\|{\Delta} u^{n}\|^{2}+\|{\Delta}\overline{u}^{n}\|^{2} -p^{2}\| {\Delta} u^{n-1}\|^{2}, \end{array} $$

and

$$\begin{array}{@{}rcl@{}} 2(\nabla u^{n},\nabla\overline{u}^{n} )=\|\nabla\overline{u}^{n}\|^{2}+\| \nabla u^{n}\|^{2}-p^{2}\| \nabla u^{n-1}\|^{2}. \end{array} $$

Substituting the above two equation to (A.65), using Lemma 4 and Schwartz inequality, wehave

$$\begin{array}{@{}rcl@{}} &&2\|\overline{u}^{n}\|^{2} +\kappa\alpha_{0}\beta_{0}^{-1}\|\nabla\overline{u}^{n}\|^{2} +\kappa\alpha_{0}\beta_{0}^{-1}\|\nabla u^{n}\|^{2} -p^{2}\kappa\alpha_{0}\beta_{0}^{-1}\|\nabla u^{n-1}\|^{2} \\ & & +\alpha_{0}\beta_{0}^{-1}\| {\Delta} u^{n}\|^{2} +\alpha_{0}\beta_{0}^{-1}\| {\Delta}\overline{u}^{n}\|^{2} -p^{2}\alpha_{0}\beta_{0}^{-1}\|{\Delta} u^{n-1}\|^{2} \\ &\leq& p\|\overline{u}^{n-1}\|^{2} +\sum\limits_{k = 2}^{n-1}\overline{d}^{n}_{n-k}\|\overline{u}^{n-k}\|^{2} + 2\overline{d}^{n}_{0}\|u^{0}\|^{2}+ 2\overline{d}^{n}_{0}\|(\overline{d}^{n}_{0})^{-1}\alpha_{0}\beta_{0}^{-1}f^{n}\|^{2} \\ & & +(p+\sum\limits_{k = 2}^{n-1}\overline{d}^{n}_{n-k}+\overline{d}^{n}_{0})\|\overline{u}^{n}\|^{2},\ \ 2\leq n\leq K. \end{array} $$

From (1) and (3) ofLemma 4, we attain

$$\begin{array}{@{}rcl@{}} &&\|\overline{u}^{n}\|^{2} +\kappa\alpha_{0}\beta_{0}^{-1}\|\nabla u^{n}\|^{2} +\alpha_{0}\beta_{0}^{-1}\| {\Delta} u^{n}\|^{2}\\ &\leq& p\|\overline{u}^{n-1}\|^{2} +p^{2}\kappa\alpha_{0}\beta_{0}^{-1}\|\nabla u^{n-1}\|^{2} +p^{2}\alpha_{0}\beta_{0}^{-1}\|{\Delta} u^{n-1}\|^{2} \\ & & +\sum\limits_{k = 2}^{n-1}\overline{d}^{n}_{n-k}\|\overline{u}^{n-k}\|^{2} + 2\overline{d}^{n}_{0}\|u^{0}\|^{2}+ 2\overline{d}^{n}_{0}\|(\overline{d}^{n}_{0})^{-1}\alpha_{0}\beta_{0}^{-1}f^{n}\|^{2} \\ &\leq& p\left( \|\overline{u}^{n-1}\|^{2} +\kappa\alpha_{0}\beta_{0}^{-1}\|\nabla u^{n-1}\|^{2} +\alpha_{0}\beta_{0}^{-1}\|{\Delta} u^{n-1}\|^{2}\right) \\ && +\sum\limits_{k = 2}^{n-1}\overline{d}^{n}_{n-k}\ \left( \|\overline{u}^{n-k}\|^{2} +\kappa\alpha_{0}\beta_{0}^{-1}\|\nabla u^{n-k}\|^{2} +\alpha_{0}\beta_{0}^{-1}\|{\Delta} u^{n-k}\|^{2}\right) \\ && + 2\overline{d}^{n}_{0}\|u^{0}\|^{2}+ 2\overline{d}^{n}_{0}\|(\overline{d}^{n}_{0})^{-1}\alpha_{0}\beta_{0}^{-1}f^{n}\|^{2},\ \ 2\leq n\leq K. \end{array} $$

(A.66)

Next we will prove the following estimates (A.67) by the mathematical induction.

$$\begin{array}{@{}rcl@{}} &&\|\overline{u}^{n}\|^{2} +\kappa\alpha_{0}\beta_{0}^{-1}\|\nabla u^{n}\|^{2} +\alpha_{0}\beta_{0}^{-1}\| {\Delta} u^{n}\|^{2}\\ &\leq& 2(\|u^{0}\|^{2}+ \|(\overline{d}^{n}_{0})^{-1}\alpha_{0}\beta_{0}^{-1}f^{n}\|^{2}),\ \ 2\leq n\leq K. \end{array} $$

(A.67)

It is easy to check that (A.67) hold for the casen = 2. Assuming the estimate(A.67) is true for n = 3, 4,⋯ , i − 1, we want to prove that it also true for n = i,and deduce from (A.66)

$$\begin{array}{@{}rcl@{}} &&\|\overline{u}^{i}\|^{2} +\kappa\alpha_{0}\beta_{0}^{-1}\|\nabla u^{i}\|^{2} +\alpha_{0}\beta_{0}^{-1}\| {\Delta} u^{i}\|^{2} \\ &\leq& 2(p+\sum\limits_{k = 2}^{i-1}\overline{d}^{i}_{i-k}+\overline{d}^{i}_{0}) (\|u^{0}\|^{2}+ \|(\overline{d}^{i}_{0})^{-1}\alpha_{0}\beta_{0}^{-1}f^{i}\|^{2}) \\ &\leq& 2(\|u^{0}\|^{2}+ \|(\overline{d}^{i}_{0})^{-1}\alpha_{0}\beta_{0}^{-1}f^{i}\|^{2}). \end{array} $$

Thus the estimate (A.67) is proven.

From Lemma 4(1) and (4), we have

$$\begin{array}{@{}rcl@{}} \|(\overline{d}^{i}_{0})^{-1}\alpha_{0}\beta_{0}^{-1}f^{i}\| \leq \frac{i^{\alpha}\tau^{\alpha}{\Gamma}(2-\alpha)}{(2-\alpha)(1-\alpha)}\| f^{i}\|. \end{array} $$

(A.68)

Through a recombination of the terms in (A.67) and (A.68), we obtain

$$\begin{array}{@{}rcl@{}} &&\|\overline{u}^{n}\|^{2} +\kappa\alpha_{0}\beta_{0}^{-1}\|\nabla u^{n}\|^{2} +\alpha_{0}\beta_{0}^{-1}\| {\Delta} u^{n}\|^{2}\\ &\leq& 2(\|u^{0}\|^{2}+\frac{T^{2\alpha}({\Gamma}(2-\alpha))^{2}}{(2-\alpha)^{2}(1-\alpha)^{2}} \|f^{n}\|^{2}),\ \ 2\leq n\leq K. \end{array} $$

(A.69)

Finally, since $\overline {u}^{n}=u^{n}-pu^{n-1}$, wenow turn to estimate ∥uⁿ∥². Applying the triangle inequality and (A.69) yields

$$\begin{array}{@{}rcl@{}} \|u^{n}\|=\|\overline{u}^{n}+pu^{n-1}\| &\leq &\|\overline{u}^{n}\|+p\|u^{n-1}\|\\ &\leq& \sqrt{2} \left( \|u^{0}\|+\frac{T^{\alpha}{\Gamma}(2-\alpha)}{(2-\alpha)(1-\alpha)}\| f^{n}\|\right)+p\|u^{n-1}\|, \end{array} $$

(A.70)

using arguments similar to (A.70), we have

$$\begin{array}{@{}rcl@{}} \|u^{n}\| &\leq& \sqrt{2} \left( \|u^{0}\|+\frac{T^{\alpha}{\Gamma}(2-\alpha)}{(2-\alpha)(1-\alpha)}\| f^{n}\|\right) \\ && +p\left( \sqrt{2} (\|u^{0}\|+\frac{T^{\alpha}{\Gamma}(2-\alpha)}{(2-\alpha)(1-\alpha)}\|f^{n-1}\|)+p\|u^{n-2}\|\right) \\ &\leq& \sqrt{2}\left( 1+p+p^{2}+\cdots+p^{n}\right) \left( \|u^{0}\|+ \frac{T^{\alpha}{\Gamma}(2-\alpha)}{(2-\alpha)(1-\alpha)}\max\limits_{0\leq j\leq n}\|f^{j}\|\right) \\ &\leq& 3\sqrt{2} \left( \|u^{0}\|+ \frac{T^{\alpha}{\Gamma}(2-\alpha)}{(2-\alpha)(1-\alpha)}\max\limits_{0\leq j\leq n}\|f^{j}\|\right). \end{array} $$

Combining the above estimate with (A.69) gives (3.21). The proof is completed.□

Appendix B

Proof

Firstly, by using the definition ${\Delta }\overline {\xi }^{n}={\Delta }\xi ^{n}-p{\Delta }\xi ^{n-1}$, we have

$$\begin{array}{@{}rcl@{}} \langle {\Delta}\xi^{n},2{\Delta}\overline{\xi}^{n}\rangle &=&\langle {\Delta}\xi^{n},2{\Delta}\xi^{n}-2p{\Delta}\xi^{n-1}\rangle \\ &=&2\langle {\Delta}\xi^{n},{\Delta}\xi^{n}\rangle -2p\langle {\Delta}\xi^{n},{\Delta}\xi^{n-1}\rangle, \end{array} $$

then,

$$\begin{array}{@{}rcl@{}} 2p\langle {\Delta}\xi^{n},{\Delta}\xi^{n-1}\rangle = 2\langle {\Delta}\xi^{n},{\Delta}\xi^{n}\rangle-\langle {\Delta}\xi^{n},2{\Delta}\overline{\xi}^{n}\rangle, \end{array} $$

(B.71)

Also,

$$\begin{array}{@{}rcl@{}} \langle {\Delta}\xi^{n},2{\Delta}\overline{\xi}^{n}\rangle &=&\langle {\Delta}\overline{\xi}^{n}+p{\Delta}\xi^{n-1},2{\Delta}\overline{\xi}^{n}\rangle\\ &=&2\langle {\Delta}\overline{\xi}^{n},{\Delta}\overline{\xi}^{n}\rangle + 2p\langle {\Delta}\xi^{n-1},{\Delta}\xi^{n}-p{\Delta}\xi^{n-1}\rangle \\ &=&2\langle {\Delta}\overline{\xi}^{n},{\Delta}\overline{\xi}^{n}\rangle + 2p\langle {\Delta}\xi^{n-1},{\Delta}\xi^{n}\rangle-2p^{2}\langle {\Delta}\xi^{n-1},{\Delta}\xi^{n-1}\rangle. \end{array} $$

(B.72)

Substituting (B.71) into (B.72) to get (4.48), the proof is completed. □

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Authors and Affiliations

Haixiang Zhang
- 1
- 2
Xuehua Yang
- 1
- 2
Email author
Da Xu
- 3
Email author

1.School of ScienceHunan University of TechnologyZhuzhouChina
2.National Key Laboratory of Science and Technology on Computational PhysicsInstitute of Applied Physics and Computational MathematicsBeijingChina
3.Department of MathematicsHunan Normal UniversityChangshaChina

A high-order numerical method for solving the 2D fourth-order reaction-diffusion equation

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