Problem 32 - Advanced Engineering Mathematics

Question

Use the phase-plane method to show that the solution to the nonlinear second- order differential equation \(x^{\prime \prime}+2 x-x^{2}=0\) that satisfies \(x(0)=1\) and \(x^{\prime}(0)=0\) is periodic.

Step-by-Step Solution

Answer

The solution is periodic as the initial conditions place the trajectory on a closed orbit around a center in the phase plane.

Step 1: Convert to First-Order System

To analyze the equation using the phase-plane method, convert the second-order differential equation to a first-order system. Set \( y = x' \). Then we have the system: \[ x' = y \]\[ y' = -2x + x^2 \]

Step 2: Analyze the Phase Plane

The system is now defined by the equations \(x' = y\) and \(y' = -2x + x^2\). The critical points occur where \(x' = 0\) and \(y' = 0\). Solve \(y = 0\) and \(-2x + x^2 = 0\) to find critical points: \[ x(x-2) = 0 \] Thus, \(x = 0\) or \(x = 2\). This gives critical points at \((0, 0)\) and \((2, 0)\).

Step 3: Determine Nature of Critical Points

To ascertain the nature of the critical points, consider the linear stability of the system around the critical points using the Jacobian matrix: \[ J = \begin{bmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \end{bmatrix} = \begin{bmatrix} 0 & 1 \ -2 + 2x & 0 \end{bmatrix} \]Evaluate the Jacobian at each critical point:- At \((0,0)\): \[ J = \begin{bmatrix} 0 & 1 \ -2 & 0 \end{bmatrix} \]The eigenvalues are \(\pm i\sqrt{2}\), indicating a center (suggesting periodic solutions).- At \((2,0)\):\[ J = \begin{bmatrix} 0 & 1 \ 2 & 0 \end{bmatrix} \]The eigenvalues are \(\pm i\sqrt{2}\), also indicating a center.

Step 4: Analyze Initial Conditions and Periodicity

Given the initial conditions \(x(0) = 1\) and \(x'(0) = 0\), plot this point on the phase plane. This point falls on a trajectory around one of the centers found previously. Centers imply closed, periodic orbits in the phase plane for linearized systems, hence the solution is periodic.

Key Concepts

Nonlinear Differential Equations

Nonlinear differential equations involve relationships where the dependent variable and its derivatives appear with powers greater than one or in functions such as trigonometric or exponential functions. These equations are often complex and may not have straightforward solutions like linear differential equations. In the context of our exercise, the nonlinear second-order differential equation is given by:

• \( x'' + 2x - x^2 = 0 \)

Unlike linear equations, which have solutions that are straightforward combinations of functions, nonlinear equations can present challenging dynamics including chaos, bifurcations, or multiple periodic solutions.
To solve such equations, especially when looking for behavior like periodicity, we often use qualitative methods like phase-plane analysis, which provides insights into the system's behavior over time without finding explicit solutions.

First-Order System

A first-order system in differential equations refers to equations that involve first derivatives only. In our problem, we convert the second-order differential equation into a first-order system by introducing a new variable. This conversion process is essential because it simplifies the problem and makes it more tractable using phase-plane methods.
To do this, we set:

• \( y = x' \)

This creates the system:

• \( x' = y \)
• \( y' = -2x + x^2 \)

Analyzing this system helps us understand the trajectory of the solution in a two-dimensional plane defined by \(x\) and \(y\). It allows us to visualize how solutions evolve over time, which is especially useful for identifying periodic behaviors.

Critical Points

Critical points in the phase-plane method occur where both first-order derivatives, \(x'\) and \(y'\), simultaneously equal zero. These points are important because they describe the equilibrium states or steady conditions of the system, where the system isn't changing in time.For the system:

• \( x' = y \)
• \( y' = -2x + x^2 \)

Solving \(y = 0\) and \(-2x + x^2 = 0\), we find the critical points:

• \( x(x-2) = 0 \)
• Resulting in the points: \( (0, 0) \) and \( (2, 0) \)

These points are analyzed to determine their nature, whether they represent stable nodes, spirals, centers or other types of behavior in the phase plane. Understanding critical points provides insight into the potential long-term behavior of the system's trajectories.

Jacobian Matrix

The Jacobian matrix is a crucial tool in analyzing the stability of critical points within the system. It contains partial derivatives that help us understand how the system dynamics change near these points, which sheds light on their stability and nature.
For our first-order system:

• \( x' = y \)
• \( y' = -2x + x^2 \)

The Jacobian matrix is defined as:\[ J = \begin{bmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \end{bmatrix} = \begin{bmatrix} 0 & 1 \ -2 + 2x & 0 \end{bmatrix}\]Evaluating J at each critical point gives us information about the nature of the equilibrium:

• At \((0,0)\): Eigenvalues \(\pm i\sqrt{2}\) suggest a center, indicating possible periodic solutions.
• At \((2,0)\): Again, eigenvalues \(\pm i\sqrt{2}\) confirm the observation of a center.

These analyses are crucial for visualizing the qualitative behavior of the system and predicting the possible periodic or oscillatory solutions.