Question

classify (if possible) each critical point of the given plane autonomous system as a stable node, a stable spiral point, an unstable spiral point, an unstable node, or a saddle point. (order your answers from smallest to largest \\(x\\), then from smallest to largest \\(y\\).)

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\

$$\begin{aligned} x &= 2x - y^2 \\\\ y &= -y + xy \\end{aligned}$$

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Explanation:

Find the critical points

Using the Nonlinear Systems of ODEs knowledge point

$$
LATEXBLOCK0
$$

If \(y = 0\):

$$ 2x - 0^2 = 0 \implies x = 0 \implies (0, 0) $$

If \(x = 1\):

$$ 2(1) - y^2 = 0 \implies y^2 = 2 \implies y = \pm\sqrt{2} \implies (1, -\sqrt{2}),\ (1, \sqrt{2}) $$

Ordered from smallest \(x\), then smallest \(y\):

$$ (0, 0),\ (1, -\sqrt{2}),\ (1, \sqrt{2}) $$

Find the Jacobian matrix

Using the Jacobian Matrix Linearization knowledge point

$$ J(x, y) =
LATEXBLOCK1
=
LATEXBLOCK2
$$

Classify the critical point (0, 0)

Using the Phase Portrait Analysis knowledge point

$$ J(0, 0) =
LATEXBLOCK3
$$

$$ \lambda_1 = 2 > 0,\quad \lambda_2 = -1 < 0 \implies \text{saddle point} $$

Classify the critical point (1, -sqrt(2))

Using the Phase Portrait Analysis knowledge point

$$ J(1, -\sqrt{2}) =
LATEXBLOCK4
$$

$$ \text{tr}(J) = 2,\quad \det(J) = 0 - (2\sqrt{2})(-\sqrt{2}) = 4 $$

$$ \lambda^2 - 2\lambda + 4 = 0 \implies \lambda = 1 \pm i\sqrt{3} $$

$$ \text{Re}(\lambda) = 1 > 0 \implies \text{unstable spiral point} $$

Classify the critical point (1, sqrt(2))

Using the Phase Portrait Analysis knowledge point

$$ J(1, \sqrt{2}) =
LATEXBLOCK5
$$

$$ \text{tr}(J) = 2,\quad \det(J) = 0 - (-2\sqrt{2})(\sqrt{2}) = 4 $$

$$ \lambda^2 - 2\lambda + 4 = 0 \implies \lambda = 1 \pm i\sqrt{3} $$

$$ \text{Re}(\lambda) = 1 > 0 \implies \text{unstable spiral point} $$

Answer:

\((x, y)\)	Conclusion
\((1, -\sqrt{2})\)	unstable spiral point
\((1, \sqrt{2})\)	unstable spiral point