show that the nonlinear second-order differential equation

$1 + \alpha^2 x^2)x + (\beta + \alpha^2 (x)^2)x = 0\

has a saddle point at \((0, 0)$ when $\beta

Question

show that the nonlinear second-order differential equation

$1 + \alpha^2 x^2)x + (\beta + \alpha^2 (x)^2)x = 0\

has a saddle point at \((0, 0)$ when $\beta < 0$.

the corresponding plane autonomous system is the following.
\x = y\
\y = \

the jacobian matrix is the following.

\g(x) = \begin{pmatrix} \quad & \quad \\ \quad & \quad \end{pmatrix}\

for $x = (0, 0)$, $\tau = $ and $\delta = $. since $\beta < 0$, we can conclude that $(0, 0)$ is a saddle point.

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "blank_filling_formatter", "explain_then_answer", "tutor", "mastery_adaptive_brevity" ], "needs_drawing": false, "concepts_used": [ "Phase Portrait Analysis", "Linear Systems of ODEs" ], "new_concepts": [ "Nonlinear Systems of ODEs", "Jacobian Matrix Linearization" ], "current_concepts": [ "Phase Portrait Analysis", "Linear Systems of ODEs", "Nonlinear Systems of ODEs", "Jacobian Matrix Linearization" ] } ## Convert to a plane autonomous system Using the *Linear Systems of ODEs* knowledge point $$ \begin{aligned} &x' = y \ &y' = x'' = -\frac{(\beta + \alpha^2 (x')^2)x}{1 + \alpha^2 x^2} = -\frac{(\beta + \alpha^2 y^2)x}{1 + \alpha^2 x^2} \end{aligned} $$ ## Compute the partial derivatives for the Jacobian matrix Using the *Linear Systems of ODEs* knowledge point $$ \begin{aligned} &f(x, y) = y \implies \frac{\partial f}{\partial x} = 0, \quad \frac{\partial f}{\partial y} = 1 \ &g(x, y) = -\frac{(\beta + \alpha^2 y^2)x}{1 + \alpha^2 x^2} \ &\frac{\partial g}{\partial x} = -\frac{(\beta + \alpha^2 y^2)(1 + \alpha^2 x^2) - (\beta + \alpha^2 y^2)x(2\alpha^2 x)}{(1 + \alpha^2 x^2)^2} = -\frac{(\beta + \alpha^2 y^2)(1 - \alpha^2 x^2)}{(1 + \alpha^2 x^2)^2} \ &\frac{\partial g}{\partial y} = -\frac{2\alpha^2 y x}{1 + \alpha^2 x^2} \end{aligned} $$ ## Evaluate the Jacobian matrix at the origin Using the *Phase Portrait Analysis* knowledge point $$ \begin{aligned} &\mathbf{g}'(0, 0) = \begin{pmatrix} \frac{\partial f}{\partial x}(0,0) & \frac{\partial f}{\partial y}(0,0) \ \frac{\partial g}{\partial x}(0,0) & \frac{\partial g}{\partial y}(0,0) \end{pmatrix} \ &\frac{\partial f}{\partial x}(0,0) = 0, \quad \frac{\partial f}{\partial y}(0,0) = 1 \ &\frac{\partial g}{\partial x}(0,0) = -\beta, \quad \frac{\partial g}{\partial y}(0,0) = 0 \ &\mathbf{g}'(\mathbf{X}) = \begin{pmatrix} 0 & 1 \ -\beta & 0 \end{pmatrix} \end{aligned} $$ ## Determine trace and determinant at the origin Using the *Phase Portrait Analysis* knowledge point $$ \begin{aligned} &\tau = \operatorname{tr}(\mathbf{g}'(0,0)) = 0 + 0 = 0 \ &\Delta = \det(\mathbf{g}'(0,0)) = 0(0) - 1(-\beta) = \beta \end{aligned} $$ Show that the nonlinear second-order differential equation $$(1 + \alpha^2x^2)x'' + (\beta + \alpha^2(x')^2)x = 0$$ has a saddle point at $(0, 0)$ when $\beta < 0$. The corresponding plane autonomous system is the following. $x' = y$ $y' =$ $-\frac{(\beta + \alpha^2 y^2)x}{1 + \alpha^2 x^2}$ The Jacobian matrix is the following. $$\mathbf{g}'(\mathbf{X}) = \begin{pmatrix} \text{$0$} & \text{$1$} \ \text{$-\frac{(\beta + \alpha^2 y^2)(1 - \alpha^2 x^2)}{(1 + \alpha^2 x^2)^2}$} & \text{$-\frac{2\alpha^2 x y}{1 + \alpha^2 x^2}$} \end{pmatrix}$$ For $\mathbf{X} = (0, 0)$, $\tau =$ $0$ and $\Delta =$ $\beta$. Since $\beta < 0$, we can conclude that $(0, 0)$ is a saddle point. { "subject": "Mathematics", "question_type": "Fill-in-the-blank", "knowledge_point": [ "Mathematics", "Calculus", "Nonlinear Systems of ODEs" ] }

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Question

Explanation:

Convert to a plane autonomous system

Compute the partial derivatives for the Jacobian matrix

Evaluate the Jacobian matrix at the origin

Determine trace and determinant at the origin

Answer: