Problem 6 - Advanced Engineering Mathematics

Question

Show that the given plane autonomous system (or second-order differential equation) has no periodic solutions. $$ \begin{aligned} &x^{r}=2 x+y^{2} \\ &y^{r}=x y-y \end{aligned} $$

Step-by-Step Solution

Answer

The system has no periodic solutions as the potential function \(V(x, y)\) does not remain constant along trajectories.

Step 1: Identify the system of equations

The given set of equations is a system of first-order autonomous differential equations:\[\begin{aligned} x^{r} &= 2 x + y^{2} \y^{r} &= x y - y\end{aligned}\]The goal is to determine if this system has periodic solutions, i.e., if there exist functions \(x(t)\) and \(y(t)\) that repeat values over constant intervals.

Step 2: Analyze potential energy function

In some cases, constructing a potential energy-like function can help identify periodic behavior. Consider the potential candidate \( V(x, y) = x^2 + y^2 \). Evaluating the change in \(V\) over time by computing its derivative:\[\frac{dV}{dt} = \frac{d}{dt}(x^2 + y^2) = 2x\cdot x^r + 2y\cdot y^r\]Inject the derivatives into this expression:

Step 3: Substitute derivatives from the system

Substitute the expressions for \(x^r\) and \(y^r\) into the derivative of the potential function:\[\frac{dV}{dt} = 2x(2x + y^2) + 2y(xy - y)\]Carry out the algebraic simplification:\[= 4x^2 + 2xy^2 + 2xy^2 - 2y^2 = 4x^2 + 4xy^2 - 2y^2\]

Step 4: Evaluate the derivative of V

Since \(\frac{dV}{dt} = 4x^2 + 4xy^2 - 2y^2\), analyze the terms. Note that this expression is not identically zero. Thus, \(V(x,y)\) is not constant along solutions. It increases or decreases depending on the values of \(x\) and \(y\), suggesting no closed orbits (periodic solutions) exist.

Step 5: Conclude the absence of periodic solutions

Since \(\frac{dV}{dt} \) is not equal to zero for all \(x\) and \(y\), \(V(x, y)\) does not remain constant, which indicates that the system lacks closed orbits in its phase plane. Therefore, the system has no periodic solutions.

Key Concepts

First-Order Differential Equations

First-order differential equations form the foundation for understanding changes in various systems. They involve functions that depend on a single variable, typically time. In this context, the differential equation dictates how the system evolves over time. Autonomous systems are a special type where the equation does not explicitly depend on the independent variable, making them simpler to analyze. For example, in our given exercise, the first-order equations are:

• \( x^{r} = 2x + y^2 \)
• \( y^{r} = xy - y \)

These equations describe how the variables \( x \) and \( y \) change over time. The system is autonomous because the time variable does not appear explicitly in these equations. This simplifies the analysis, as the long-term behavior of the system only depends on the current state, not on when it is evaluated.

Periodic Solutions

Periodic solutions are solutions to differential equations that repeat themselves over time. In simple terms, they lead to cycles or orbits where the system returns to the same state after some interval. Imagine the sine wave: it goes up and down in a regular, predictable cycle. For a system to have a periodic solution, it should consistently revisit the same values repeatedly.In the given exercise, the goal was to determine whether the system of equations had periodic solutions. By evaluating a potential energy-like function, \( V(x, y) = x^2 + y^2 \), and its derivative \( \frac{dV}{dt} \), we concluded there are no periodic solutions. Since \( \frac{dV}{dt} \) is not equal to zero always, it suggests that the energy is not constant, and hence the system does not have paths or cycles that repeat over time. Therefore, closed orbits do not exist, which rules out periodic solutions for the system.

Phase Plane Analysis

Phase plane analysis is a technique used to visualize the behavior of dynamical systems. It turns the equations into trajectories in a plane defined by the variables (in this case, \( x \) and \( y \)). Each point in this plane corresponds to a possible state of the system. As the system evolves over time, it traces a path through the plane, called a trajectory.In analyzing our system, plotting the trajectories in the phase plane allows us to see whether paths close onto themselves, indicating periodic behavior. However, the analysis showed that \( \frac{dV}{dt} = 4x^2 + 4xy^2 - 2y^2 \) is generally non-zero. This suggests that trajectories diverge or converge but do not form closed loops. Therefore, no periodic orbits exist in the phase plane, supporting the conclusion that the system does not have periodic solutions.