Traveling wave solutions to diffusive Holling-Tanner predator-prey models

Ching-Hui Wang; Sheng-Chen Fu

doi:10.3934/dcdsb.2021007

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2021, Volume 26, Issue 4: 2239-2255. Doi: 10.3934/dcdsb.2021007

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Traveling wave solutions to diffusive Holling-Tanner predator-prey models

Ching-Hui Wang and
Sheng-Chen Fu^,

Department of Mathematical Sciences, National Chengchi University, 64, S-2 Zhi-Nan Road, Taipei 116, Taiwan

^* Corresponding author: Sheng-Chen Fu
^* Corresponding author: Sheng-Chen Fu

Dedicated to Professor Sze-Bi Hsu

Received: September 2020

Revised: November 2020

Early access: December 2020

Published: April 2021

The second author is supported by MOST grant 109-2115-M-004-004.

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Abstract

In this paper, we first establish the existence of semi-traveling wave solutions to a diffusive generalized Holling-Tanner predator-prey model in which the functional response may depend on both the predator and prey populations. Then, by constructing the Lyapunov function, we apply the obtained result to show the existence of traveling wave solutions to the diffusive Holling-Tanner predator-prey models with various functional responses, including the Lotka-Volterra type functional response, the Holling type Ⅱ functional response and the Beddington-DeAngelis functional response.

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Mathematics Subject Classification: Primary: 35c07, 35K57; Secondary: 35K40.

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Figure 1. The solution as a function of the spatial variable x is plotted at t = 0, t = 10, t = 20 and t = 30. The initial data $ (u_0,v_0) $ is chosen so that $ u_0 = 1 $ and $ v_0 = 0.05*(1+sign(51-x))*(1+sign(x-49))/4 $. The parameter values are $ k = 1.4 $, $ b = e = 1 $, $ d = 1 $, $ r = 4 $ and $ s = 0.6 $

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Figure 2. The solution as a function of the spatial variable x is plotted at t = 0, t = 5, t = 10 and t = 20. The initial data $ (u_0,v_0) $ is chosen so that $ u_0 = 1 $ and $ v_0 = 0.05*(1+sign(51-x))*(1+sign(x-49))/4 $. The parameter values are $ k = 4 $, $ b = e = 0 $, $ d = 1 $, $ r = 2 $ and $ s = 0.5 $

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Figure 3. The solution as a function of the spatial variable x is plotted at t = 0, t = 10, t = 20 and t = 30. The initial data $ (u_0,v_0) $ is chosen so that $ u_0 = 1 $ and $ v_0 = 0.05*(1+sign(51-x))*(1+sign(x-49))/4 $. The parameter values are $ k = 10 $, $ b = 5 $, $ e = 1 $, $ d = 1 $, $ r = 4 $ and $ s = 0.6 $

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Figure 4. The solution as a function of the spatial variable x is plotted at t = 0, t = 5, t = 10 and t = 20. The initial data $ (u_0,v_0) $ is chosen so that $ u_0 = 1 $ and $ v_0 = 0.05*(1+sign(51-x))*(1+sign(x-49))/4 $. The parameter values are $ k = 10 $, $ b = e = 0 $, $ d = 1 $, $ r = 2 $ and $ s = 0.5 $

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References

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