Problem 21 - Advanced Engineering Mathematics

Question

The equation $$x^{\prime}=\alpha \frac{y}{1+y} x-x=x\left(\frac{\alpha y}{1+y}-1\right)=0$$ implies that \(x=0\) or \(y=1 /(\alpha-1) .\) When \(\alpha>0, \hat{y}=1 /(\alpha-1)>0 .\) If \(x=0,\) then from the differential equation for \(y^{\prime}, y=\beta .\) On the other hand, if \(\hat{y}=1 /(\alpha-1), \hat{y} /(1+\hat{y})=1 / \alpha\) and so \(\hat{x} / \alpha-1 /(\alpha-1)+\beta=0\) It follows that $$\hat{x}=\alpha\left(\beta-\frac{1}{\alpha-1}\right)=\frac{\alpha}{\alpha-1}[(\alpha-1) \beta-1]$$ and if \(\beta(\alpha-1)>1, \hat{x}>0 .\) Therefore \((\hat{x}, \hat{y})\) is the unique critical point in the first quadrant. The Jacobian matrix is $$\mathbf{g}^{\prime}(\mathbf{X})=\left(\begin{array}{cc} \alpha \frac{y}{y+1}-1 & \frac{\alpha x}{(1+y)^{2}} \\ -\frac{y}{1+y} & \frac{-x}{(1+y)^{2}}-1 \end{array}\right)$$ and for \(\mathbf{X}=(\hat{x}, \hat{y}),\) the Jacobian can be written in the form $$\mathbf{g}^{\prime}((\hat{x}, \hat{y}))=\left(\begin{array}{cc} 0 & \frac{(\alpha-1)^{2}}{\alpha} \hat{x} \\ -\frac{1}{\alpha} & -\frac{(\alpha-1)^{2}}{\alpha^{2}}-1 \end{array}\right)$$ It follows that $$\tau=-\left[\frac{(\alpha-1)^{2}}{\alpha^{2}} \hat{x}+1\right]<0, \quad \Delta=\frac{(\alpha-1)^{2}}{\alpha^{2}} \hat{x}$$ and so \(\tau=-(\Delta+1) .\) Therefore \(\tau^{2}-4 \Delta=(\Delta+1)^{2}-4 \Delta=(\Delta-1)^{2}>0 .\) Therefore \((\hat{x}, \hat{y})\) is a stable node.

Step-by-Step Solution

Answer

\((\hat{x}, \hat{y})\) is a unique stable node in the first quadrant.

Step 1: Identify the Equilibrium Points

The differential equation given is \( x^{\prime}=x\left(\frac{\alpha y}{1+y}-1\right)=0 \). This implies that either \( x = 0 \) or \( y = \frac{1}{\alpha - 1} \) since the expression can only be zero if either of the multipliers is zero.

Step 2: Conditions for Solution Values

For the equation \( y = \frac{1}{\alpha - 1} \) to hold true, \( \alpha > 1 \) since \( \hat{y} > 0 \). If \( x = 0 \), then \( y = \beta \). On the other hand, if \( y = \frac{1}{\alpha - 1} \), a relation \( \frac{\hat{y}}{1 + \hat{y}} = \frac{1}{\alpha} \) leads to \( \hat{x} = \alpha\left(\beta - \frac{1}{\alpha - 1}\right) = \frac{\alpha}{\alpha-1}[(\alpha-1)\beta-1] \). This implies \( \hat{x} > 0 \) if \( \beta(\alpha-1) > 1 \).

Step 3: Stability Analysis Using Jacobian Matrix

The Jacobian matrix is given by \( \mathbf{g}^{\prime}(\mathbf{X})=\begin{pmatrix} \alpha \frac{y}{y+1}-1 & \frac{\alpha x}{(1+y)^{2}} \ -\frac{y}{1+y} & \frac{-x}{(1+y)^{2}}-1 \end{pmatrix} \). For \( \mathbf{X}=(\hat{x}, \hat{y}) \), it becomes \( \mathbf{g}^{\prime}((\hat{x}, \hat{y}))=\begin{pmatrix} 0 & \frac{(\alpha-1)^{2}}{\alpha} \hat{x} \ -\frac{1}{\alpha} & -\frac{(\alpha-1)^{2}}{\alpha^{2}}-1 \end{pmatrix} \).

Step 4: Calculate Trace and Determinant

The trace \( \tau \) is calculated as \( \tau=-\left[\frac{(\alpha-1)^{2}}{\alpha^{2}} \hat{x}+1\right] \), which is negative because \( \hat{x} > 0 \). The determinant \( \Delta \) is \( \Delta=\frac{(\alpha-1)^{2}}{\alpha^{2}} \hat{x} \). This provides a condition \( \tau = - (\Delta + 1) \).

Step 5: Determine Node Type

Calculate \( \tau^{2} - 4\Delta \). This results in \( (\Delta + 1)^{2} - 4\Delta = (\Delta - 1)^{2} \) which is always positive, indicating \((\hat{x}, \hat{y})\) is a stable node.

Key Concepts

Stability Analysis

Stability Analysis is an important aspect of understanding differential equations. It determines whether small perturbations in initial conditions lead to minor changes over time or cause the system to behave unpredictably.
In differential equations, stability is commonly analyzed using the eigenvalues of the Jacobian matrix at equilibrium points. If all eigenvalues have negative real parts, the system is stable at that point. This implies after a small disturbance, the system will eventually return to equilibrium.
For instance, in the given problem, the trace \( \tau \) and determinant \( \Delta \) of the Jacobian are used to assess stability. The calculations confirm that \( \tau^{2} - 4\Delta > 0 \), indicating that the equilibrium is a stable node.
Thus, analyzing stability helps predict the long-term behavior of dynamical systems, making it an invaluable tool in engineering, science, and economics.

Jacobian Matrix

The Jacobian Matrix is a mathematical tool used to study how a function changes as its input changes. In the context of systems of differential equations, it is used to examine the behavior near equilibrium points.
By taking partial derivatives of each function with respect to each variable, you construct the Jacobian matrix. This matrix plays a crucial role in linearizing the system near equilibrium points, simplifying the analysis of complex systems.
In our example, the Jacobian is calculated as \( \mathbf{g}^{\prime}((\hat{x}, \hat{y})) \). The computation involves evaluating the linearized system near critical points, facilitating an analysis of system dynamics. This matrix helps determine whether the points are attractors or repellers in the phase space.
Therefore, the Jacobian matrix is an essential part of stability analysis, allowing for the classification of equilibrium points effectively.

Equilibrium Points

Equilibrium Points in a system of differential equations are states where the system can remain indefinitely without external changes. At these points, the rate of change of the system is zero, represented by setting the derivatives to zero.
In the given system, equilibrium points are found by solving \( x^{\prime}=0 \). This implies either \( x=0 \) or \( y=\frac{1}{\alpha-1} \). These points determine the system's behavior over time and the stability conditions derived from these points dictate how perturbations will affect the system.
Equilibrium points are critical as they provide insights into potential long-term states of the system. Understanding these states helps in predicting how altering parameters like \( \alpha \) and \( \beta \) affect overall dynamics.
By identifying and analyzing equilibrium points, we can better forecast system responses to changes, providing valuable insights in various fields.

Phase Plane Analysis

Phase Plane Analysis is a graphical tool used for studying two-dimensional systems of differential equations. It simplifies the understanding of dynamic behavior by visualizing trajectories in a plane confined by two variables.
Through this method, you can visualize how the state of a system evolves over time. You often draw trajectories representing paths that solutions in the plane take given different initial conditions. These trajectories highlight stable and unstable areas, providing an intuitive feel for solving differential equations.
The intersections of trajectories are equilibrium points, and the stability of these points is crucial. Stable nodes, like the one identified at \((\hat{x}, \hat{y})\), indicate the system's state will naturally progress towards them over time.
Thus, phase plane analysis offers a compelling way to visualize and interpret the behavior of complex dynamical systems.