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) \\\

$$\begin{aligned} \\dot{x} &= x(2 - x - y) \\\\ \\dot{y} &= y(y - x) \\end{aligned}$$

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(b) \\\

$$\begin{aligned} \\dot{x} &= x - y \\\\ \\dot{y} &= x^2 - 1 \\end{aligned}$$

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(c) \\\

$$\begin{aligned} \\dot{x} &= 1 - xy \\\\ \\dot{y} &= x - y^3 \\end{aligned}$$

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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: Linearization of Autonomous Systems
*Approximating non-linear systems near equilibrium points using the Jacobian matrix.*

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

The system is given by:
\[ \begin{aligned} \dot{x} &= x(2 - x - y) \\ \dot{y} &= y(y - x) \end{aligned} \]

To find the critical (equilibrium) points, set \(\dot{x} = 0\) and \(\dot{y} = 0\):
1. \(x(2 - x - y) = 0 \implies x = 0\) or \(y = 2 - x\)
2. \(y(y - x) = 0 \implies y = 0\) or \(y = x\)

We find the intersections of these conditions:
- If \(x = 0\):
- From \(y(y - x) = 0\), we get \(y(y - 0) = 0 \implies y = 0\).
- This gives the critical point: \(P_1(0, 0)\).
- If \(y = 2 - x\):
- Substitute into \(y(y - x) = 0\):
- Case 1: \(y = 0 \implies 2 - x = 0 \implies x = 2\). This gives \(P_2(2, 0)\).
- Case 2: \(y = x \implies 2 - x = x \implies 2x = 2 \implies x = 1\), which means \(y = 1\). This gives \(P_3(1, 1)\).

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

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

Define the system functions:
\[ f(x, y) = 2x - x^2 - xy \]
\[ g(x, y) = y^2 - xy \]

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

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Step 3: Classify each critical point of system (a)

At \(P_1(0, 0)\):

\[ J(0, 0) = \begin{pmatrix} 2 & 0 \\ 0 & 0 \end{pmatrix} \]
The eigenvalues are \(\lambda_1 = 2\) and \(\lambda_2 = 0\).
Since one eigenvalue is zero, the linear system is degenerate, and linearization is inconclusive for determining stability of the non-linear system. However, looking at the linear approximation, it has a positive eigenvalue, indicating instability along the x-axis.

At \(P_2(2, 0)\):

\[ J(2, 0) = \begin{pmatrix} 2 - 4 - 0 & -2 \\ 0 & 0 - 2 \end{pmatrix} = \begin{pmatrix} -2 & -2 \\ 0 & -2 \end{pmatrix} \]
This is an upper triangular matrix, so the eigenvalues are the diagonal entries:
\[ \lambda_1 = -2, \quad \lambda_2 = -2 \]
Since both eigenvalues are real, negative, and equal, and the matrix is non-diagonalizable (it has a non-zero off-diagonal element \(-2\)), the critical point \((2, 0)\) is a **stable node** (specifically, a degenerate stable node).

At \(P_3(1, 1)\):

\[ J(1, 1) = \begin{pmatrix} 2 - 2 - 1 & -1 \\ -1 & 2 - 1 \end{pmatrix} = \begin{pmatrix} -1 & -1 \\ -1 & 1 \end{pmatrix} \]
Find the eigenvalues by solving the characteristic equation \(\det(J - \lambda I) = 0\):
\[ \det \begin{pmatrix} -1 - \lambda & -1 \\ -1 & 1 - \lambda \end{pmatrix} = 0 \]
\[ (-1 - \lambda)(1 - \lambda) - 1 = 0 \]
\[ \lambda^2 - 1 - 1 = 0 \implies \lambda^2 = 2 \implies \lambda = \pm\sqrt{2} \]
Since the eigenvalues are real and have opposite signs (\(\lambda_1 = \sqrt{2} > 0\) and \(\lambda_2 = -\sqrt{2} < 0\)), the critical point \((1, 1)\) is a **saddle point**.

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Step 4: Find the critical points of system (b)

The system is given by:
\[ \begin{aligned} \dot{x} &= x - y \\ \dot{y} &= x^2 - 1 \end{aligned} \]

Set \(\dot{x} = 0\) and \(\dot{y} = 0\):
1. \(x - y = 0 \implies y = x\)
2. \(x^2 - 1 = 0 \implies x = \pm 1\)

- If \(x = 1\), then \(y = 1\). This gives \(P_1(1, 1)\).
- If \(x = -1\), then \(y = -1\). This gives \(P_2(-1, -1)\).

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Step 5: Linearize and classify each critical point of system (b)

Define the system functions:
\[ f(x, y) = x - y \]
\[ g(x, y) = x^2 - 1 \]

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

At \(P_1(1, 1)\):

\[ J(1, 1) = \begin{pmatrix} 1 & -1 \\ 2 & 0 \end{pmatrix} \]
The characteristic equation is:
\[ \det \begin{pmatrix} 1 - \lambda & -1 \\ 2 & -\lambda \end{pmatrix} = 0 \]
\[ -\lambda(1 - \lambda) + 2 = 0 \implies \lambda^2 - \lambda + 2 = 0 \]
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 (\(\text{Re}(\lambda) = \frac{1}{2} > 0\)), the critical point \((1, 1)\) is an **unstable spiral point**.

At \(P_2(-1, -1)\):

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

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Step 6: Find the critical points of system (c)

The system is given by:
\[ \begin{aligned} \dot{x} &= 1 - xy \\ \dot{y} &= x - y^3 \end{aligned} \]

Set \(\dot{x} = 0\) and \(\dot{y} = 0\):
1. \(1 - xy = 0 \implies xy = 1 \implies x = \frac{1}{y}\)
2. \(x - y^3 = 0 \implies x = y^3\)

Equating the two expressions for \(x\):
\[ \frac{1}{y} = y^3 \implies y^4 = 1 \]
Since we are looking for real critical points:
\[ y = \pm 1 \]
- If \(y = 1\), then \(x = 1^3 = 1\). This gives \(P_1(1, 1)\).
- If \(y = -1\), then \(x = (-1)^3 = -1\). This gives \(P_2(-1, -1)\).

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Step 7: Linearize and classify each critical point of system (c)

Define the system functions:
\[ f(x, y) = 1 - xy \]
\[ g(x, y) = x - y^3 \]

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

At \(P_1(1, 1)\):

\[ J(1, 1) = \begin{pmatrix} -1 & -1 \\ 1 & -3 \end{pmatrix} \]
The characteristic equation is:
\[ \det \begin{pmatrix} -1 - \lambda & -1 \\ 1 & -3 - \lambda \end{pmatrix} = 0 \]
\[ (-1 - \lambda)(-3 - \lambda) + 1 = 0 \]
\[ \lambda^2 + 4\lambda + 3 + 1 = 0 \implies \lambda^2 + 4\lambda + 4 = 0 \]
Factor the quadratic equation:
\[ (\lambda + 2)^2 = 0 \implies \lambda_1 = \lambda_2 = -2 \]
Since both eigenvalues are real, negative, and equal, and the matrix is non-diagonalizable, the critical point \((1, 1)\) is a **stable node** (degenerate stable node).

At \(P_2(-1, -1)\):

\[ J(-1, -1) = \begin{pmatrix} 1 & 1 \\ 1 & -3 \end{pmatrix} \]
The characteristic equation is:
\[ \det \begin{pmatrix} 1 - \lambda & 1 \\ 1 & -3 - \lambda \end{pmatrix} = 0 \]
\[ (1 - \lambda)(-3 - \lambda) - 1 = 0 \]
\[ \lambda^2 + 2\lambda - 3 - 1 = 0 \implies \lambda^2 + 2\lambda - 4 = 0 \]
Using the quadratic formula:
\[ \lambda = \frac{-2 \pm \sqrt{4 - 4(1)(-4)}}{2} = \frac{-2 \pm \sqrt{20}}{2} = -1 \pm \sqrt{5} \]
Since \(\sqrt{5} \approx 2.236\), the eigenvalues are:
\[ \lambda_1 = -1 + \sqrt{5} > 0 \]
\[ \lambda_2 = -1 - \sqrt{5} < 0 \]
Since the eigenvalues are real and have opposite signs, the critical point \((-1, -1)\) is a **saddle point**.

Answer:

System (a)

• \((0,0)\): Inconclusive via linearization (degenerate linear system with eigenvalues \(\lambda = 2, 0\)).
• \((2,0)\): Stable node
• \((1,1)\): Saddle point

System (b)

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

System (c)

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