| Citation: |
Yue Li, Hongjun Cao. BIFURCATION AND COMPARISON OF A DISCRETE-TIME HINDMARSH-ROSE MODEL[J]. Journal of Applied Analysis & Computation, 2023, 13(1): 34-56. doi: 10.11948/20210204
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BIFURCATION AND COMPARISON OF A DISCRETE-TIME HINDMARSH-ROSE MODEL
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Abstract
In this paper, a Hindmarsh-Rose model discretized by a nonstandard finite difference (NSFD) scheme is considered. Bifurcation behaviors are compared between the model obtained by the forward Euler scheme and the model obtained by the NSFD scheme. Through analytical and numerical comparisons, the Neimark-Sacker bifurcation of the model discretized by the NSFD method is closer to the Hopf bifurcation of the original continuous Hindmarsh-Rose model than that discretized by the forward Euler method. Moreover, due to the NSFD method's better stability and convergence, the integral step size can be chosen larger in the NSFD scheme. And much more dynamic behaviors can be obtained by using the NSFD scheme than those in the forward Euler scheme. These confirmed results can at least guarantee true available numerical results to investigate complex neuron dynamical systems.
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References
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Yue Li, Hongjun Cao. BIFURCATION AND COMPARISON OF A DISCRETE-TIME HINDMARSH-ROSE MODEL[J]. Journal of Applied Analysis & Computation, 2023, 13(1): 34-56. doi: 10.11948/20210204
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Figure 1.
(a) Bifurcation diagram of model (1.3) in
$ (a, x) $ $ b=5, c=1, d = 2, h=0.1 $ $ (0.6, 0.528) $ $ (h, x) $ $ a=1.080, b=5, c=1, d = 2 $ $ (0.6, 0.528) $ -
Figure 2.
(a) Neimark-Sacker bifurcation diagram of model (1.3) in
$ (a, x) $ $ b=1, c=2, d = 3, h=0.1 $ $ (- 0.8, -0.6) $ $ (a, x) $ $ b=1, c=2, d = 3, h=0.1 $ $ (- 0.8, -0.6) $ -
Figure 3.
(a) Neimark-Sacker bifurcation diagram of model (1.3) in
$ (h, x) $ $ a=1.2023, b=1, c=2, d = 3 $ $ (- 0.8, -0.6) $ $ (h, x) $ $ a=1.2277, b=1, c=2, d = 3 $ $ (- 0.8, -0.6) $ $ (h, a) $ $ (h, a) $ -
Figure 4.
Phase portraits for various values of
$ a $ $ a=1.207 $ $ a=1.20235 $ $ a=1.18 $ -
Figure 5.
Phase portraits for various values of
$ h $ $ h=0.855 $ $ h=1.081 $ $ h=1.299 $ $ 0.06967 $