Some standard and nonstandard finite difference schemes for a reaction–diffusion–chemotaxis model

Abstract

Abstract Two standard and two nonstandard finite difference schemes are constructed to solve a basic reaction–diffusion–chemotaxis model, for which no exact solution is known. The continuous model involves a system of nonlinear coupled partial differential equations subject to some specified initial and boundary conditions. It is not possible to obtain theoretically the stability region of the two standard finite difference schemes. Through running some numerical experiments, we deduce heuristically that these classical methods give reasonable solutions when the temporal step size k k is chosen such that k ≤ 0.25 k\le 0.25 with the spatial step size h h fixed at h = 1.0 h=1.0 (first novelty of this work). We observe that the standard finite difference schemes are not always positivity preserving, and this is why we consider nonstandard finite difference schemes. Two nonstandard methods abbreviated as NSFD1 and NSFD2 from Chapwanya et al. are considered. NSFD1 was not used by Chapwanya et al. to generate results for the basic reaction–diffusion–chemotaxis model. We find that NSFD1 preserves positivity of the continuous model if some criteria are satisfied, namely, ϕ ( k ) [ ψ ( h ) ] 2 = 1 2 γ ≤ 1 2 σ + β \frac{\phi \left(k)}{{\left[\psi \left(h)]}^{2}}=\frac{1}{2\gamma }\le \frac{1}{2\sigma +\beta } and β ≤ σ \beta \le \sigma , and this is the second novelty of this work. Chapwanya et al. modified NSFD1 to obtain NSFD2, which is positivity preserving if R = ϕ ( k ) [ ψ ( h ) ] 2 = 1 2 γ R=\frac{\phi \left(k)}{{\left[\psi \left(h)]}^{2}}=\frac{1}{2\gamma } and 2 σ R ≤ 1 2\sigma R\le 1 , that is σ ≤ γ \sigma \le \gamma , and they presented some results. For the third highlight of this work, we show that NSFD2 is not always consistent and prove that consistency can be achieved if β → 0 \beta \to 0 and k h 2 → 0 \frac{k}{{h}^{2}}\to 0 . Fourthly, we show numerically that the rate of convergence in time of the four methods for case 2 is approximately one.