Question

Can be interpreted as describing the interaction of two species with population densities \(x\) and \(y .\) In each of these problems carry out the following steps. (a) Draw a direction field and describe how solutions seem to behave. (b) Find the critical points. (c) For each critical point find the corresponding linear system. Find the eigenvalues and eigenvectors of the linear system; classify each critical point as to type, and determine whether it is asymptotically stable, or unstable. (d) Sketch the trajectories in the neighborhood of each critical point. (e) Draw a phase portrait for the system. (f) Determine the limiting behavior of \(x\) and \(y\) as \(t \rightarrow \infty\) and interpret the results in terms of the populations of the two species. $$ \begin{array}{l}{d x / d t=x(1.5-0.5 y)} \\ {d y / d t=y(-0.5+x)}\end{array} $$

Short answer

Based on the analysis and solution of the given system of differential equations that represent the interaction of two species with population densities \(x\) and \(y\), we found three critical points (0, 0), (0, 3), and (1, 1). All critical points were found to be unstable, meaning the populations will not reach a stable equilibrium in the long term. Therefore, the limiting behavior of \(x\) and \(y\) as \(t \rightarrow \infty\) is dependent on the initial conditions and system dynamics, making it difficult to predict their exact behaviors. This result suggests that the populations of these two species may constantly oscillate or increase indefinitely without reaching a stable point.

Step-by-step solution

(a) Draw a direction field and describe the solution behavior.

To draw the direction field for the given system of differential equations, we can plot the slopes at various values of \(x\) and \(y\). It may be beneficial to use a computer software or online tool to visualize the direction field. The direction field will give us a visual representation of the behavior of the populations' interactions.

(b) Find the critical points.

To find the critical points, we set both \(dx/dt\) and \(dy/dt\) to zero and solve for \(x\) and \(y\). $$ \begin{cases} x(1.5-0.5y)=0 \\ y(-0.5+x)=0 \end{cases} $$ The critical points can be found by solving these equations simultaneously. We obtain the critical points (0, 0), (0, 3), and (1,1).

(c) Linearize the system and analyze the stability of critical points.

To linearize the system, we'll find the Jacobian matrix at the critical points: $$ \textbf{J}=\begin{bmatrix} \frac{\partial}{\partial x}(x(1.5-0.5 y)) & \frac{\partial}{\partial y}(x(1.5-0.5 y)) \\ \frac{\partial}{\partial x}(y(-0.5+x)) & \frac{\partial}{\partial y}(y(-0.5+x)) \end{bmatrix} = \begin{bmatrix} 1.5-0.5y & -0.5x \\ y & -0.5+x \end{bmatrix} $$ Evaluate the Jacobian at each critical point, find the eigenvalues, and classify the critical point's stability: 1. Point (0, 0): $$ \textbf{J}|_{(0,0)} = \begin{bmatrix} 1.5 & 0 \\ 0 & -0.5 \end{bmatrix} $$ Eigenvalues are the diagonals themselves, 1.5 and -0.5. Since one is positive, the point is unstable (saddle type). 2. Point (0, 3): $$ \textbf{J}|_{(0,3)} = \begin{bmatrix} -0.5 & 0 \\ 3 & 0.5 \end{bmatrix} $$ Eigenvalues are -0.75 and 0.75. Since one is positive, the point is unstable (saddle type). 3. Point (1, 1): $$ \textbf{J}|_{(1,1)} = \begin{bmatrix} 1 & -0.5 \\ 1 & 0.5 \end{bmatrix} $$ Eigenvalues are 1.25 and 0.25. Both are positive, indicating that this point is unstable (unstable node).

(d) Sketch the trajectories in the neighborhood of each critical point.

According to the calculated eigenvalues and stability types of each critical point, we can sketch the local trajectories as follows: - For (0, 0), the trajectories will move away along one direction and towards the point along the other direction (saddle type). - For (0, 3), the trajectories will move away along one direction and towards the point along the other direction (saddle type). - For (1, 1), the trajectories will move away from the point in all directions (unstable node).

(e) Draw a phase portrait for the system.

Based on the analysis and the local trajectories, we can now sketch a phase portrait for the system. The phase portrait can be drawn manually or with the help of computer software; it will show trajectories around each critical point, as well as how other trajectories behave.

(f) Determine the limiting behavior of \(x\) and \(y\) as \(t \rightarrow \infty\).

As \(t \rightarrow \infty\), the population densities \(x\) and \(y\) will approach one of the critical points. Since all critical points are unstable in this case, it is difficult to predict the exact limiting behavior. The population densities may oscillate or increase indefinitely, depending on the initial conditions and the dynamics of the system. In the context of the two species' populations, this could mean that their populations will not reach a stable equilibrium in the long term.

Key concepts

Critical Points

Critical points in a system of differential equations are particular solutions where the derivatives of all variable functions are zero. In the context of population dynamics, these points correspond to states where the population densities of the species do not change over time. To find these points in our system, we set \( \frac{dx}{dt} = 0 \) and \( \frac{dy}{dt} = 0 \) and solve for \(x\) and \(y\).

For our given system, this results in two equations: \( x(1.5 - 0.5y) = 0 \) and \( y(-0.5 + x) = 0 \). Solving these simultaneously gives us the critical points: \((0, 0)\), \((0, 3)\), and \((1, 1)\). These points represent potential equilibria in the population system, where neither species changes its population size over time.

Stability Analysis

Stability analysis helps us determine the behavior of solutions near each critical point. This analysis is done by examining the system's linearization around these points through the Jacobian matrix. The eigenvalues of this matrix reveal much about a critical point's stability.

1. For \((0, 0)\), the eigenvalues are 1.5 and -0.5. Having an eigenvalue greater than zero indicates that this point is unstable, specifically a saddle point.

2. At \((0, 3)\), the eigenvalues are -0.75 and 0.75, again indicating a saddle point due to positive and negative eigenvalues.

3. For \((1, 1)\), both eigenvalues (1.25 and 0.25) are positive, classifying this point as an unstable node. This means disturbances will cause trajectories to diverge from this point.

This analysis is crucial, as it tells us whether a system, like a population, will return to equilibrium or diverge when disturbed.

Phase Portrait

A phase portrait visually represents all possible trajectories of a system of differential equations, such as population dynamics, in the plane. It shows how solutions evolve over time and how trajectories relate to the critical points.

To create a phase portrait, we use direction fields and knowledge of the stability of critical points. In our system:

• Trajectories around \((0, 0)\) and \((0, 3)\) are influenced by their saddle point nature, moving away along one direction and approaching in another.
• At \((1, 1)\), trajectories diverge outwards uniformly due to its unstable node type.

By sketching a phase portrait, we gain insight into the interaction of the species populations over time, helping us to predict trends and behaviors.

Population Dynamics

Population dynamics refers to how densities of populations of species change over time due to interactions like competition, predation, or mutualism. In our system, the differential equations reflect these dynamics for two species.

The nature of critical points in this system suggests no stable equilibrium for the populations, as all points are unstable. As time \(t \to \infty\), populations \(x\) and \(y\) could fluctuate without reaching a steady state, heavily depending on initial conditions.

This reflects potential chaos in ecosystems with competitive or otherwise dynamic species interactions. It's critical to consider the initial state of the ecosystems and external influences, as they dramatically affect long-term population behavior.