Problem 19 - Advanced Engineering Mathematics

Question

Let \(\mathbf{X}=\mathbf{X}(t)\) be the solution of the plane autonomous system $$ \begin{aligned} &x^{\prime}=y \\ &y^{\prime}=-x-\left(1-x^{2}\right) y \end{aligned} $$ that satisfies \(\mathbf{X}(0)=\left(x_{0}, y_{0}\right) .\) Show that if \(x_{0}^{2}+y_{0}^{2}<1\), then \(\lim _{t \rightarrow \infty} \mathbf{X}(t)=(0,0) .\left[\right.\) Hint \(:\) Select \(r<1\) with \(x_{0}^{2}+y_{0}^{2}

Step-by-Step Solution

Answer

The solution \( \mathbf{X}(t) \) approaches \( (0,0) \) as \( t \to \infty \) if \( x_0^2 + y_0^2 < 1 \).

Step 1: Setup the given system of equations

We start with the plane autonomous system given by the differential equations: \( x' = y \) and \( y' = -x - (1-x^2)y \). Plus, we know that for initial conditions, \( \mathbf{X}(0) = (x_0, y_0) \).

Step 2: Identify the invariant region

According to the exercise, we need to identify the invariant region \( R \) defined by \( x^2 + y^2 \leq r^2 \) where \( r < 1 \) and \( x_0^2 + y_0^2 < r^2 \). This circular region constrains the motion of the system's trajectory.

Step 3: Calculate the time derivative of \( x^2 + y^2 \)

Let us compute the derivative \( \frac{d}{dt}(x^2 + y^2) = 2x x' + 2y y' \). Substitute \( x' = y \) and \( y' = -x - (1-x^2)y \) into this to get:\[ \frac{d}{dt}(x^2 + y^2) = 2x y + 2y(-x - (1-x^2)y) \].

Step 4: Simplify the derivative expression

Simplifying the expression, we have:\[ 2xy - 2xy - 2y^2 + 2x^2y^2 \], which reduces to \( -2y^2 + 2x^2y^2 \).

Step 5: Show negativity of the derivative

Notice that \( -2y^2 + 2x^2y^2 \leq 0 \) for \( x^2 \leq 1 \) since in this range \( x^2y^2 \leq y^2 \) resulting in \( -2y^2 + 2x^2y^2 \leq 0 \).

Step 6: Conclude about the invariant property

Since the derivative \( \frac{d}{dt}(x^2 + y^2) \leq 0 \), the quantity \( x^2 + y^2 \) is non-increasing. This means once \( \mathbf{X}(t) \) enters the region \( R \), it remains there as \( t \) approaches infinity.

Step 7: Apply limit conditions

With \( x^2 + y^2 \leq r^2 < 1 \), as \( t \rightarrow \infty \), the coordinates \( x(t) \) and \( y(t) \) move towards \( 0 \). The only point in \( R \) where they can possibly stabilize is \( (0,0) \).

Step 8: Final conclusion about the limit

Therefore, we determine that \( \lim_{t \to \infty} \mathbf{X}(t) = (0,0) \) for \( x_0^2 + y_0^2 < 1 \), concluding that the origin is a stable equilibrium point for \( x^2 + y^2 \leq r^2 \).

Key Concepts

Invariant Region

In the study of autonomous systems, the concept of an invariant region is crucial for understanding how solutions behave over time. An invariant region is a specified part of the phase space in which the trajectories of a dynamical system remain once they enter. In this exercise, we identified an invariant region \( R \) defined by the inequality \( x^2 + y^2 \leq r^2 \), with \( r < 1 \). The significance of this circular region is that it provides a constraint within which the system evaluates.

To determine if this circular region is indeed invariant, we differentiated \( x^2 + y^2 \) over time, using the given system of differential equations. The result showed that the derivative \( \frac{d}{dt}(x^2 + y^2) \) is non-positive within this region. This non-increasing nature means that once a solution’s initial condition \( (x_0, y_0) \) lies inside \( R \), it will stay inside as time progresses.

The invariant region is vital for limiting the system's behavior, indicating that the long-term dynamics are constrained and do not diverge beyond \( x^2 + y^2 < 1 \). It provides a foundational framework to explore further properties such as stability and equilibrium points.

Stability Analysis

Stability analysis in the context of dynamical systems is about determining how solutions behave over time, particularly when they are near an equilibrium point. In this exercise, once we have established that \( x^2 + y^2 \leq r^2 < 1 \) is an invariant region, we aimed to understand whether the solutions approach a stable state.

A key step in our analysis involved demonstrating that within the invariant region, the trajectory’s distance from the origin cannot increase. This was done by showing that the derivative \( \frac{d}{dt}(x^2 + y^2) \leq 0 \), which implies that the solutions settle into a "non-expanding" set until they reach equilibrium. Since this region is bounded and the system is autonomous, the solutions will over time either cycle around or tend toward a point.

The implication is, if \( x_0^2 + y_0^2 < 1 \), the solution remains in a domain where it systematically reduces its distance from the origin as time approaches infinity. Thus, through stability analysis, we conclude that the origin \((0,0)\) acts as a stable equilibrium where solutions are attracted over time, making it a focal point for long-term behavior.

Equilibrium Points

Equilibrium points in a dynamical system are crucial as they represent the states where the system experiences no change, meaning the derivatives of the state variables are zero. In autonomous systems like the one in this exercise, equilibrium points are determined by setting the derivatives equal to zero. Here, setting \( x' = y = 0 \) and \( y' = -x - (1-x^2)y = 0 \) identifies \((0,0)\) as the equilibrium point.

The behavior of solutions near an equilibrium point provides insight into the system's overall dynamics. Through the step-by-step solution, we found that trajectories within the invariant region gradually approached the point \((0,0)\), confirming its status as a stable equilibrium. This entails that regardless of initial conditions within the specified bounds, solutions will stabilize at the origin as time heads towards infinity.

Understanding equilibrium points helps predict the system's final steady state. In the context of this exercise, knowing that \((0,0)\) is stable and attractive further equips us to determine conditions under which the system can reliably return to or remain at rest. This forms the backbone of predictive modeling for many physical and engineering systems.