Explanation Given: The Lotka-Voltera model is x ′ = r 1 K 1 x ( K 1 − x − α 12 y ) and y ′ = r 2 K 2 y ( K 2 − y − α 21 x ) . Calculation: The equations of Lotka-Voltera model is given as follows: x ′ = r 1 K 1 x ( K 1 − x − α 12 y ) … ( 1 ) y ′ = r 2 K 2 y ( K 2 − y − α 21 x ) … ( 2 ) Consider P ( x , y ) = x ′ and substitute in equation ( 1 ) . P ( x , y ) = r 1 K 1 x ( K 1 − x − α 12 y ) … … ( 3 ) Consider Q ( x , y ) = y ′ and substitute in equation ( 2 ) . Q ( x , y ) = r 2 K 2 y ( K 2 − y − α 21 x ) … … ( 4 ) Differentiate equation ( 3 ) partially with respect to x . ∂ P ∂ x = r 1 K 1 ( K 1 − 2 x − α 12 y ) Differentiate equation ( 3 ) partially with respect to y . ∂ P ∂ y = − r 1 K 1 α 12 x Differentiate equation ( 4 ) partially with respect to x . ∂ Q ∂ x = − r 2 K 2 α 21 y Differentiate the both sides of the equation ( 4 ) with respect to y . ∂ Q ∂ y = r 2 K 2 ( K 2 − 2 y − α 21 x ) The Jacobian matrix is given as follows: g ′ ( x , y ) = ( ∂ P ∂ x ∂ P ∂ y ∂ Q ∂ x ∂ Q ∂ y ) Substitute the values of ∂ P ∂ x , ∂ P ∂ y , ∂ Q ∂ x and ∂ Q ∂ y in above equation. g ′ ( x , y ) = ( r 1 K 1 ( K 1 − 2 x − α 12 y ) − r 1 K 1 α 12 x − r 2 K 2 α 21 y r 2 K 2 ( K 2 − 2 y − α 21 x ) ) … … ( 5 ) At the critical point ( K 1 , 0 ) the Jacobian matrix of equation ( 5 ) is given as follows: g ′ ( K 1 , 0 ) = ( − r 1 − r 1 α 12 0 r 2 ( 1 − K 1 K 2 α 21 ) ) The determinant of the above matrix is calculated as follows: Δ = | − r 1 − r 1 α 12 0 r 2 ( 1 − K 1 K 2 α 21 ) | = − r 1 [ r 2 ( 1 − K 1 K 2 α 21 ) ] − 0 = − r 1 r 2 ( 1 − K 1 K 2 α 21 ) Consider the trace point τ and τ is equal to the sum of the diagonal elements of the above matrix. τ = − r 1 + r 2 ( 1 − K 1 K 2 α 21 ) … … ( 6 ) Assume the new coefficient as follows: c = 1 − K 1 K 2 α 21 … … ( 7 ) Substitute 1 − K 1 K 2 α 21 = c in equation ( 6 )

Differential Equations with Boundary-Value Problems (MindTap Course List)

9th Edition

ISBN: 9781305965799

Author: Dennis G. Zill

Publisher: Cengage Learning

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Chapter 10.4, Problem 13E

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To show: The point (K1,0) is a stable node when K1>K2α21 and a saddle point when K1<K2α21 in Lotka–Voltera model.

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Chapter 10.4, Problem 12E

Chapter 10.4, Problem 14E

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Differential Equations with Boundary-Value Problems (MindTap Course List)

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The point ( K 1 , 0 ) is a stable node when K 1 > K 2 α 21 and a saddle point when K 1 < K 2 α 21 in Lotka–Voltera model.

Solution Summary: The author explains that the point (K_1,0) is a stable node, and the saddle point, in Lotka–Voltera model.

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To show: The point (K1,0) is a stable node when K1>K2α21 and a saddle point when K1<K2α21 in Lotka–Voltera model.

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Chapter 10 Solutions

The point ( K 1 , 0 ) is a stable node when K 1 &gt; K 2 α 21 and a saddle point when K 1 &lt; K 2 α 21 in Lotka–Voltera model.

Solution Summary: The author explains that the point (K_1,0) is a stable node, and the saddle point, in Lotka–Voltera model.

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To show: The point (K1,0) is a stable node when K1>K2α21 and a saddle point when K1<K2α21 in Lotka–Voltera model.

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Chapter 10 Solutions

The point ( K 1 , 0 ) is a stable node when K 1 > K 2 α 21 and a saddle point when K 1 < K 2 α 21 in Lotka–Voltera model.