Explanation Given: The plane autonomous system is x ′ = x ( 1 − x 2 − 3 y 2 ) and y ′ = y ( 3 − x 2 − 3 y 2 ) . Concept used: Stable node: The condition for a critical point to be a stable node is that τ 2 − 4 Δ > 0 , τ < 0 and Δ > 0 where, τ is the trace of the matrix and Δ is the determinant of the matrix. Saddle point: The condition for a critical point to be a saddle point is that τ 2 − 4 Δ > 0 and Δ < 0 where, τ is the trace of the matrix and Δ is the determinant of the matrix. Unstable node: The condition for a critical point to be a unstable node is that τ 2 − 4 Δ > 0 , τ > 0 and Δ > 0 where, τ is the trace of the matrix and Δ is the determinant of the matrix. Calculation: The plane autonomous system is given as follows: x ′ = x ( 1 − x 2 − 3 y 2 ) y ′ = y ( 3 − x 2 − 3 y 2 ) Consider that x ′ = P ( x , y ) and y ′ = Q ( x , y ) . The above equations are written as follows: P ( x , y ) = x ( 1 − x 2 − 3 y 2 ) Q ( x , y ) = y ( 3 − x 2 − 3 y 2 ) The critical point ( x , y ) is the point for which x ′ = 0 and y ′ = 0 . Solve the equation y ′ = 0 as follows: y ′ = Q ( x , y ) y ( 3 − x 2 − 3 y 2 ) = 0 From the above equation it is observed that either the value of y is 0 or the value of 3 − x 2 − 3 y 2 is 0 . Solve the equation x ′ = 0 as follows: x ′ = P ( x , y ) x ( 1 − x 2 − 3 y 2 ) = 0 From the above equation it is observed that either the value of x is 0 or the value of 1 − x 2 − 3 y 2 is 0 . Substitute 0 for x in 3 − x 2 − 3 y 2 = 0 . 3 − 0 − 3 y 2 = 0 3 y 2 = 3 y 2 = 1 y = ± 1 Substitute 0 for y in 1 − x 2 − 3 y 2 = 0 . 1 − x 2 − 0 = 0 1 − x 2 = 0 x 2 = 1 x = ± 1 The critical points for the given autonomous system are ( 0 , 0 ) , ( 0 , 1 ) , ( 0 , − 1 ) , ( 1 , 0 ) and ( − 1 , 0 ) . Differentiate P ( x , y ) partially with respect to x as follows: ∂ P ( x , y ) ∂ x = ∂ ( x ( 1 − x 2 − 3 y 2 ) ) ∂ x = 1 − 3 x 2 − 3 y 2 Differentiate P ( x , y ) partially with respect to y as follows: ∂ P ( x , y ) ∂ y = ∂ ( x ( 1 − x 2 − 3 y 2 ) ) ∂ y = − 6 x y Differentiate Q ( x , y ) partially with respect to x as follows: ∂ Q ( x , y ) ∂ x = ∂ ( y ( 3 − x 2 − 3 y 2 ) ) ∂ x = − 2 x y Differentiate Q ( x , y ) partially with respect to y as follows: ∂ Q ( x , y ) ∂ y = ∂ ( y ( 3 − x 2 − 3 y 2 ) ) ∂ y = 3 − x 2 − 9 y 2 The Jacobian matrix formed for the given system is as follows: g ′ ( x , y ) = ( ∂ P ∂ x ∂ P ∂ y ∂ Q ∂ x ∂ Q ∂ y ) = ( 1 − 3 x 2 − 6 x y − 2 x y 3 − x 2 − 9 y 2 ) For the critical point ( 0 , 0 ) the Jacobian matrix is as follows: g ′ ( 0 , 0 ) = ( 1 0 0 3 ) Consider the trace of the above matrix as τ . Trace of a matrix is the sum of all the elements of the main diagonal. Trace of the above matrix is calculated as follows: τ = 1 + 3 = 4 The determinant of the matrix is Jacobian matrix is calculated as follows: Δ = | 1 0 0 3 | = 3 − 0 = 3 The value of τ 2 − 4 Δ is calculated as follows: τ 2 − 4 Δ = ( 4 ) 2 − ( 4 ⋅ 3 ) = 16 − 12 = 4 The condition for a critical point to be an unstable node is that τ > 0 , τ 2 − 4 Δ > 0 and Δ > 0

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 10.3, Problem 18E

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The category of the critical point for the plane autonomous system x′=x(1−x2−3y2) and y′=y(3−x2−3y2).

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Chapter 10.3, Problem 17E

Chapter 10.3, Problem 19E

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

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The category of the critical point for the plane autonomous system x ′ = x ( 1 − x 2 − 3 y 2 ) and y ′ = y ( 3 − x 2 − 3 y 2 ) .

Solution Summary: The author explains that the critical point for the plane autonomous system is a stable node, while the saddle point is the determinant of the matrix.

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The category of the critical point for the plane autonomous system x′=x(1−x2−3y2) and y′=y(3−x2−3y2).

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