Question

assignment 03
fixed closing date: 31 july 2026
unique number: 860536
total marks: 30

1. consider the following plane autonomous system:

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$$\begin{aligned} \\dot{x} &= y \\\\ \\dot{y} &= x - x^3 \\end{aligned}$$

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find the nature of all the singular points and sketch the phase plane diagram with the trajectories and all the isoclines.

1. classify (if possible) each critical point of the given plane autonomous systems as a stable node, an unstable node, a stable spiral point, an unstable spiral point or a saddle point.

(a) \\(\dot{x} = x(2 - x - y)\\), \\(\dot{y} = y(y - x)\\)
(b) \\(\dot{x} = x - y\\), \\(\dot{y} = x^2 - 1\\)
(c) \\(\dot{x} = 1 - xy\\), \\(\dot{y} = x - y^3\\)

1. classify (if possible) each critical point of the given second-order differential equation as a stable node, an unstable node, a stable spiral, an unstable spiral point or a saddle point.

(a) \\(\ddot{x} + \dot{x}(1 - x^2) + x - x^3 = 0\\)
(b) \\(\ddot{x} + \dot{x} + \sin(x) = 0\\)

Explanation:

🆕 New Concept Discovered: Stability of Autonomous Systems
Analyzing nonlinear systems using local linearization and Jacobian matrices.

Step 1: Find the critical points of system (a)

The system is given by:
\[ \dot{x} = x(2 - x - y) = 0 \]
\[ \dot{y} = y(y - x) = 0 \]

From the second equation, either \( y = 0 \) or \( y = x \).

- Case 1: If \( y = 0 \), substitute into the first equation:
\[ x(2 - x) = 0 \implies x = 0 \text{ or } x = 2 \]
This gives the critical points: \((0, 0)\) and \((2, 0)\).

- Case 2: If \( y = x \), substitute into the first equation:
\[ x(2 - 2x) = 0 \implies x = 0 \text{ or } x = 1 \]
This gives the critical points: \((0, 0)\) (already found) and \((1, 1)\).

The critical points are \((0, 0)\), \((2, 0)\), and \((1, 1)\).

Step 2: Linearize system (a) using the Jacobian matrix

Let \( f(x, y) = 2x - x^2 - xy \) and \( g(x, y) = y^2 - xy \).
The Jacobian matrix \( J(x, y) \) is:
\[ J(x, y) = \begin{pmatrix} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} \\ \frac{\partial g}{\partial x} & \frac{\partial g}{\partial y} \end{pmatrix} = \begin{pmatrix} 2 - 2x - y & -x \\ -y & 2y - x \end{pmatrix} \]

- **At \((0, 0)\):**
\[ J(0, 0) = \begin{pmatrix} 2 & 0 \\ 0 & 0 \end{pmatrix} \]
The eigenvalues are \(\lambda_1 = 2\) and \(\lambda_2 = 0\). Since one eigenvalue is zero, the linear test is inconclusive.

- **At \((2, 0)\):**
\[ J(2, 0) = \begin{pmatrix} -2 & -2 \\ 0 & -2 \end{pmatrix} \]
This is an upper triangular matrix, so the eigenvalues are \(\lambda_1 = -2\) and \(\lambda_2 = -2\). Since both eigenvalues are real, negative, and equal, and the matrix is non-diagonalizable (it has only one linearly independent eigenvector), this is a **stable node** (specifically, a degenerate stable node).

- **At \((1, 1)\):**
\[ J(1, 1) = \begin{pmatrix} -1 & -1 \\ -1 & 1 \end{pmatrix} \]
The characteristic equation is:
\[ \det(J - \lambda I) = (-1 - \lambda)(1 - \lambda) - 1 = \lambda^2 - 2 = 0 \implies \lambda = \pm\sqrt{2} \]
Since the eigenvalues are real and have opposite signs, this is a **saddle point**.

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Step 3: Find and classify the critical points of system (b)

The system is given by:
\[ \dot{x} = x - y = 0 \implies y = x \]
\[ \dot{y} = x^2 - 1 = 0 \implies x = \pm 1 \]

The critical points are \((1, 1)\) and \((-1, -1)\).

Let \( f(x, y) = x - y \) and \( g(x, y) = x^2 - 1 \).
The Jacobian matrix is:
\[ J(x, y) = \begin{pmatrix} 1 & -1 \\ 2x & 0 \end{pmatrix} \]

- **At \((1, 1)\):**
\[ J(1, 1) = \begin{pmatrix} 1 & -1 \\ 2 & 0 \end{pmatrix} \]
The characteristic equation is:
\[ \lambda^2 - \text{Tr}(J)\lambda + \det(J) = \lambda^2 - \lambda + 2 = 0 \]
Using the quadratic formula:
\[ \lambda = \frac{1 \pm \sqrt{1 - 8}}{2} = \frac{1 \pm i\sqrt{7}}{2} \]
Since the eigenvalues are complex conjugates with a positive real part (\(\text{Re}(\lambda) = \frac{1}{2} > 0\)), this is an **unstable spiral point**.

- **At \((-1, -1)\):**
\[ J(-1, -1) = \begin{pmatrix} 1 & -1 \\ -2 & 0 \end{pmatrix} \]
The characteristic equation is:
\[ \lambda^2 - \text{Tr}(J)\lambda + \det(J) = \lambda^2 - \lambda - 2 = 0 \implies (\lambda - 2)(\lambda + 1) = 0 \]
The eigenvalues are \(\lambda_1 = 2\) and \(\lambda_2 = -1\). Since the eigenvalues are real and of opposite signs, this is a **saddle point**.

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Step 4: Find and classify the critical points of system (c)

The system is given by:
\[ \dot{x} = 1 - xy = 0 \implies xy = 1 \]
\[ \dot{y} = x - y^3 = 0 \implies x = y^3 \]

Substitute \( x = y^3 \) into \( xy = 1 \):
\[ y^4 = 1 \implies y = \pm 1 \]
- If \( y = 1 \), then \( x = 1 \).
- If \( y = -1 \), then \( x = -1 \).

The critical points are \((1, 1)\) and \((-1, -1)\).

Let \( f(x, y) = 1 - xy \) and \( g(x, y) = x - y^3 \).
The Jacobian matrix is:
\[ J(x, y) = \begin{pmatrix} -y & -x \\ 1 & -3y^2 \end{pmatrix} \]

- **At \((1, 1)\):**
\[ J(1, 1) = \begin{pmatrix} -1 & -1 \\ 1 & -3 \end{pmatrix} \]
The characteristic equation is:
\[ \lambda^2 - \text{Tr}(J)\lambda + \det(J) = \lambda^2 + 4\lambda + 4 = 0 \implies (\lambda + 2)^2 = 0 \]
The eigenvalues are \(\lambda_1 = -2\) and \(\lambda_2 = -2\). Since both eigenvalues are real, negative, and equal, this is a **stable node**.

- **At \((-1, -1)\):**
\[ J(-1, -1) = \begin{pmatrix} 1 & 1 \\ 1 & -3 \end{pmatrix} \]
The characteristic equation is:
\[ \lambda^2 - \text{Tr}(J)\lambda + \det(J) = \lambda^2 + 2\lambda - 4 = 0 \]
Using the quadratic formula:
\[ \lambda = \frac{-2 \pm \sqrt{4 - 4(1)(-4)}}{2} = -1 \pm \sqrt{5} \]
Since \(\sqrt{5} > 1\), one eigenvalue is positive (\(-1 + \sqrt{5} > 0\)) and one is negative (\(-1 - \sqrt{5} < 0\)). This is a **saddle point**.

Answer:

System (a)

• \((2, 0)\): Stable node
• \((1, 1)\): Saddle point
• \((0, 0)\): Inconclusive by linearization (one zero eigenvalue)

System (b)

• \((1, 1)\): Unstable spiral point
• \((-1, -1)\): Saddle point

System (c)

• \((1, 1)\): Stable node
• \((-1, -1)\): Saddle point