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doi: 10.3934/dcdss.2020259
Stability and bifurcation analysis in a delay-induced predator-prey model with Michaelis-Menten type predator harvesting
Ming Liu , Dongpo Hu and Fanwei Meng ,
School of Mathematical Sciences, Qufu Normal University, Qufu 273165, Shandong, China
*Corresponding author: Fanwei Meng
Received May 2019 Revised September 2019 Published February 2020
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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.
Keywords: Predator-prey model, Michaelis-Menten type harvesting, time delays, stability, Hopf bifurcation.
Mathematics Subject Classification: Primary: 92D25; Secondary: 34K18, 34K20, 34K60.
Citation: Ming Liu, Dongpo Hu, Fanwei Meng. Stability and bifurcation analysis in a delay-induced predator-prey model with Michaelis-Menten type predator harvesting. Discrete & Continuous Dynamical Systems - S, doi: 10.3934/dcdss.2020259
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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)$
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) $
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)$
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)$
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)$
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)$
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)$
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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