Problem 20 - Advanced Engineering Mathematics

Question

(a) \(x^{\prime}=x(-a+b y)=0\) implies that \(x=0\) or \(y=a / b .\) If \(x=0,\) then, from $$-c x y+\frac{r}{K} y(K-y)=0$$ \(y=0\) or \(K .\) Therefore (0,0) and \((0, K)\) are critical points. If \(\hat{y}=a / b,\) then $$\hat{y}\left[-c x+\frac{r}{K}(K-\hat{y})\right]=0$$ The corresponding value of \(x, x=\hat{x},\) therefore satisfies the equation \(c \hat{x}=r(K-\hat{y}) / K\). (b) The Jacobian matrix is $$\mathbf{g}^{\prime}(\mathbf{X})=\left(\begin{array}{cc} -a+b y & b x \\ -c y & -c x+\frac{r}{K}(K-2 y) \end{array}\right)$$ and so at \(\mathbf{X}_{1}=(0,0), \Delta=-a r<0 .\) For \(\mathbf{X}_{1}=(0, K), \Delta=n(K b-a)=-r b(K-a / b) .\) since we are given that \(K>a / b, \Delta<0\) in this case. Therefore (0,0) and \((0, K)\) are each saddle points. For \(\mathbf{X}_{1}=(\hat{x}, \hat{y})\) where \(\hat{y}=a / b\) and \(c \hat{x}=r(K-\hat{y}) / K,\) we can write the Jacobian matrix as $$\mathbf{g}^{\prime}((\hat{x}, \hat{y}))=\left(\begin{array}{cc} 0 & b \hat{x} \\ -c \hat{y} & -\frac{r}{K} \hat{y} \end{array}\right)$$ and so \(\tau=-r \hat{y} / K<0\) and \(\Delta=b c \hat{x} \hat{y}>0 .\) Therefore \((\hat{x}, \hat{y})\) is a stable critical point and so it is either a stable node (perhaps degenerate) or a stable spiral point. (c) Write $$\tau^{2}-4 \Delta=\frac{r^{2}}{K^{2}} \hat{y}^{2}-4 b c \hat{x} \hat{y}=\hat{y}\left[\frac{r^{2}}{K^{2}} \hat{y}-4 b c \hat{x}\right]=\hat{y}\left[\frac{r^{2}}{K^{2}} \hat{y}-4 b \frac{r}{K}(K-\hat{y})\right]$$ using $$c \hat{x}=\frac{r}{K}(K-\hat{y})=\frac{r}{K} \hat{y}\left[\left(\frac{r}{K}+4 b\right) \hat{y}-4 b K\right]$$. Therefore \(\tau^{2}-4 \Delta<0\) if and only if $$\hat{y}<\frac{4 b K}{\frac{r}{K}+4 b}=\frac{4 b K^{2}}{r+4 b K}$$ Note that $$\frac{4 b K^{2}}{r+4 b K}=\frac{4 b K}{r+4 b K} \cdot K \approx K$$ where \(K\) is large, and \(\hat{y}=a / b< K .\) Therefore \(\tau^{2}-4 \Delta< 0\) when \(K\) is large and a stable spiral point will result.

Step-by-Step Solution

Answer

The critical point \((\hat{x}, \hat{y})\) is stable, likely a stable spiral point when \(K\) is large and \(\hat{y} < \frac{4bK^2}{r + 4bK}\), confirming \(\tau^2 - 4\Delta < 0\).

Step 1: Analyzing Critical Points for x=0 or y=a/b

Given the equation \(x' = x(-a + b y) = 0\), it implies either \(x = 0\) or \(y = \frac{a}{b}\). If \(x = 0\), the equation transforms to \(-c x y + \frac{r}{K} y(K-y)=0\). Therefore, \(y = 0\) or \(y = K\), leading to the critical points \((0, 0)\) and \((0, K)\). If \(y = \frac{a}{b}\), substitute in \(\hat{y}(-c x + \frac{r}{K}(K-\hat{y})) = 0\) to find \(c\hat{x} = \frac{r}{K}(K-\hat{y})\).

Step 2: Evaluating Stability with Jacobian Matrix at (0,0) and (0,K)

The Jacobian matrix is given by \(\mathbf{g}^{\prime}(\mathbf{X})=\begin{pmatrix} -a+b y & b x \ -c y & -c x+\frac{r}{K}(K-2 y) \end{pmatrix}\). At \((0,0)\), the determinant \(\Delta = -ar < 0\) indicates a saddle point. For \((0, K)\), \(\Delta = -rb(K-\frac{a}{b})\), and since \(K > \frac{a}{b}\), \(\Delta < 0\), confirming a saddle point here as well.

Step 3: Checking Stability for (\hat{x}, \hat{y}) with y=a/b

With \(\hat{y} = \frac{a}{b}\) and \(c\hat{x} = \frac{r}{K}(K-\hat{y})\), the Jacobian at \((\hat{x}, \hat{y})\) simplifies to \(\mathbf{g}^{\prime}((\hat{x}, \hat{y}))=\begin{pmatrix} 0 & b\hat{x} \ -c\hat{y} & -\frac{r}{K}\hat{y} \end{pmatrix}\). Here, \(\tau = -\frac{r}{K}\hat{y} < 0\) and \(\Delta = b c \hat{x} \hat{y} > 0\), confirming it's a stable critical point, either a stable node or a stable spiral point.

Step 4: Analyzing Condition for Stable Spiral Point

For \(\tau^2 - 4\Delta\), substitute to get \(\tau^2 - 4\Delta = \hat{y}\left[\frac{r^2}{K^2}\hat{y} - 4b\frac{r}{K}(K - \hat{y})\right]\). For \(\tau^2 - 4\Delta < 0\), we require \(\hat{y} < \frac{4bK^2}{r + 4bK}\). Since \(K\) is large and \(\hat{y} = \frac{a}{b}\), \(\tau^2 - 4\Delta < 0\) suggests a stable spiral point will occur.

Key Concepts

Critical Points

In the study of dynamical systems, identifying critical points is a crucial step. Critical points are where the system doesn't change—in other words, they are points where the derivatives are zero. For the given system, the critical points are where \\[ x' = x(-a + by) = 0 \] \leading to either \(x = 0\) or \(y = \frac{a}{b}\). \

• When \(x = 0\), another equation determines efforts in finding \(y = 0\) or \(y = K\), resulting in critical points, like \((0, 0)\) and \((0, K)\).
• With \(y = \frac{a}{b}\), you solve for a corresponding \(x\), giving another type of critical point \((\hat{x}, \hat{y})\).

\These points become the focus for further analysis to understand system behavior at those locations.

Jacobian Matrix

The Jacobian matrix is a mathematical tool that helps analyze the stability of critical points in a system of differential equations. It consists of first-order derivatives of a vector function, representing how a change in the system's state variables affects its dynamics. \For a system described by functions \( f(x, y) \) and \( g(x, y) \), the Jacobian matrix is: \\[\mathbf{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}\] \Applying this to our example yields & \[\mathbf{g}^{\prime}(\mathbf{X})=\begin{pmatrix} \-a+by & b x \ -cy & -c x+\frac{r}{K}(K-2y) \end{pmatrix}\] \Evaluating this at each critical point gives us insights into their nature and stability.

Stability Analysis

Stability analysis of a dynamical system involves determining whether the critical points are stable or not. It tells us if small perturbations in the system will die out or grow over time. \ ### Determinants and Traces In mathematical terms, this analysis often involves calculating the determinant \( \Delta \) and the trace \( \tau \) of the Jacobian matrix at each critical point.

• If \( \Delta < 0 \), the critical point is a saddle point, indicating instability.
• For \( \Delta > 0 \) and \( \tau < 0 \), the point is stable—it could be either a stable node or a stable spiral.

\For example, at \((0, 0)\) and \((0, K)\), both points are classified as saddle points due to their \( \Delta < 0 \). Meanwhile, the point \((\hat{x}, \hat{y})\) has a positive \( \Delta \) and a negative \( \tau \), indicating stability.

Saddle Point

A saddle point in a dynamical system occurs at a critical point where the stability is such that the system will be stable along one dimension and unstable along another. ### Visualizing a Saddle PointImagine a horse saddle: it curves upwards in one direction and downwards in another, which is akin to how trajectories behave near a saddle point in a system.

• These are typically characterized by \( \Delta < 0 \), meaning the determinant of the Jacobian evaluated at the point is negative.
• It's crucial to identify these as they often indicate points of transition or bifurcation within a system.

\In our example, the critical points \((0, 0)\) and \((0, K)\) were found to be saddle points due to the negative determinant at each point.

Stable Spiral Point

A stable spiral point is a type of stable critical point where trajectories spiral towards the point as time progresses. \### Understanding Stable Spiral DynamicsThis behavior is identified by:

• \( \Delta > 0 \)
• \( \tau < 0 \)
• \( \tau^2 - 4\Delta < 0 \) indicates the complex eigenvalues, characteristic of spiral dynamics.

\In practical settings, a stable spiral point means the system will return to equilibrium not in a straightforward manner, but with oscillations that gradually reduce in magnitude. \The point \((\hat{x}, \hat{y})\) in the exercise holds this property under certain conditions, implying it acts like a damping oscillator that ultimately stabilizes at the equilibrium.