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Math

Posted about 1 month ago

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`Determine the stability of the critical point $(0,0)$ for the system of differential equations $dot{x} = x^2 - y$, $dot{y} = y^2 - x$.`

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Answer from Sia

Posted about 1 month ago

Solution by Steps

step 1

We start with the system of differential equations: $dot{x} = x^2 - y$, $dot{y} = y^2 - x$. We need to find the critical points by setting $dot{x} = 0$ and $dot{y} = 0$. This gives us the equations: $x^2 - y = 0$ and $y^2 - x = 0$

step 2

Solving these equations simultaneously at the critical point $(0,0)$, we substitute $x = 0$ and $y = 0$: $0^2 - 0 = 0$ and $0^2 - 0 = 0$. Thus, $(0,0)$ is indeed a critical point

step 3

To determine the stability of the critical point, we analyze the Jacobian matrix of the system. The Jacobian matrix $J$ is given by: $J = \begin{pmatrix} \frac{\partial dot{x}}{\partial x} & \frac{\partial dot{x}}{\partial y} \\ \frac{\partial dot{y}}{\partial x} & \frac{\partial dot{y}}{\partial y} \end{pmatrix}$. Calculating the partial derivatives at $(0,0)$, we find: $J(0,0) = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}$

step 4

The eigenvalues of the Jacobian matrix are found by solving the characteristic equation: $det(J - \lambda I) = 0$. This leads to the equation: $\lambda^2 + 1 = 0$, giving us eigenvalues $\lambda = i$ and $\lambda = -i$. Since the eigenvalues are purely imaginary, the critical point $(0,0)$ is a center and is stable

Answer

The critical point $(0,0)$ is stable and behaves like a center.

Key Concept

Stability of critical points in dynamical systems is determined by the eigenvalues of the Jacobian matrix.

Explanation

The eigenvalues being purely imaginary indicates that trajectories near the critical point will oscillate around it, confirming its stability.

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