Stability and bifurcation analysis in a delay-induced predator-prey model with Michaelis-Menten type predator harvesting

Ming Liu; Dongpo Hu; Fanwei Meng

doi:10.3934/dcdss.2020259

Article Contents

2021, Volume 14, Issue 9: 3197-3222. Doi: 10.3934/dcdss.2020259

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Stability and bifurcation analysis in a delay-induced predator-prey model with Michaelis-Menten type predator harvesting

School of Mathematical Sciences, Qufu Normal University, Qufu 273165, Shandong, China

^*Corresponding author: Fanwei Meng
^*Corresponding author: Fanwei Meng

Received: April 30, 2019

Revised: August 31, 2019

Early access: February 2020

Published: September 2021

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Abstract

The present paper considers a delay-induced predator-prey model with Michaelis-Menten type predator harvesting. The existence of the nontrivial positive equilibria is discussed, and some sufficient conditions for locally asymptotically stability of one of the positive equilibria are developed. Meanwhile, the existence of Hopf bifurcation is discussed by choosing time delays as the bifurcation parameters. Furthermore, the direction of Hopf bifurcation and the stability of the bifurcated periodic solutions are determined by the normal form theory and the center manifold theorem for functional differential equations. Finally, some numerical simulations are carried out to support the analytical results.

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Mathematics Subject Classification: Primary: 92D25; Secondary: 34K18, 34K20, 34K60.

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Figure 1. The diagram (a) shows the time series of $ x(t) $, $ y(t) $ and the diagram (b) shows the phase portrait of model (3) with $ \tau_1 = \tau_2 = 0 $. The positive equilibrium point $ E_2(0.68, 0.32) $ is locally asymptotically stable. Here the initial value is $ (0.8, 0.6)$

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Figure 2. The diagram (a) shows the time series of $ x(t) $, $ y(t) $ and the diagram (b) shows the phase portrait of model (3) with $ \tau_1 = 0 $, $ \tau_2 = 2.8 < \tau_{20} = 2.91 $. The positive equilibrium point $ E_2(0.48, 0.52) $ is locally asymptotically stable. Here the initial value is $ (0.5,0.5)$

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Figure 3. The diagram (a) shows the time series of $ x(t) $, $ y(t) $ and the diagram (b) shows the phase portrait of model (3) with $ \tau_1 = 0 $, $ \tau_2 = 2.92 > \tau_{20} = 2.91 $. The positive equilibrium point $ E_2(0.48, 0.52) $ is unstable. Here the initial value is $ (0.5,0.5) $

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Figure 4. The diagram (a) shows the time series of $ x(t) $, $ y(t) $ and the diagram (b) shows the phase portrait of model (3) with $ \tau_1 = 2.0 < \tau_{30} = 2.11 $, $ \tau_2 = 0 $. The positive equilibrium point $ E_2(0.53, 0.47) $ is locally asymptotically stable. Here the initial value is $ (0.5,0.5)$

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Figure 5. The diagram (a) shows the time series of $ x(t) $, $ y(t) $ and the diagram (b) shows the phase portrait of model (3) with $ \tau_1 = 2.1185 > \tau_{30} = 2.11 $, $ \tau_2 = 0 $. The positive equilibrium point $ E_2(0.53, 0.47) $ is unstable. Here the initial value is $ (0.5,0.5)$

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Figure 6. The diagram (a) shows the time series of $ x(t) $, $ y(t) $ and the diagram (b) shows the phase portrait of model (3) with $ \tau_1 = \tau_2 = 1.85 < \tau_{40} = 1.92 $. The positive equilibrium point $ E_2(0.53, 0.47) $ is locally asymptotically stable. Here the initial value is $ (0.55,0.6)$

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Figure 7. The diagram (a) shows the time series of $ x(t) $, $ y(t) $ and the diagram (b) shows the phase portrait of model (3) with $ \tau_1 = \tau_2 = 1.926 > \tau_{40} = 1.92 $. The positive equilibrium point $ E_2(0.53, 0.47) $ is unstable. Here the initial value is $ (0.55,0.6)$

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Figure 8. The diagram (a) shows the time series of $ x(t) $, $ y(t) $ and the diagram (b) shows the phase portrait of model (3) with $ \tau_1 = 3 < \tau_{50} = 4.90 $, $ \tau_2 = 1.8 $. The positive equilibrium point $ E_2(0.48, 0.52) $ is locally asymptotically stable. Here the initial value is $ (0.55,0.6)$

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Figure 9. The diagram (a) shows the time series of $ x(t) $, $ y(t) $ and the diagram (b) shows the phase portrait of model (3) with $ \tau_1 = 6 > \tau_{50} = 4.90 $, $ \tau_2 = 1.8 $. The positive equilibrium point $ E_2(0.48, 0.52) $ is unstable. Here the initial value is $ (0.55,0.6)$

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