91Ó°ÊÓ

$$ \begin{aligned} &\frac{d x}{d t}=60 x-4 x^{2}-3 x y \\ &\frac{d y}{d t}=42 y-2 y^{2}-3 x y \end{aligned} $$ in which \(c_{1} c_{2}=9>8=b_{1} b_{2}\), so the effect of competition should exceed that of inhibition. Problems 4 through 7 imply that the four critical points \((0,0),(0,21),(15,0)\), and \((6,12)\) of the system in \((2)\) resemble those shown in Fig. \(6.3 .9-a\) nodal source at the origin, a nodal sink on each coordinate axis, and a saddle point interior to the first quadrant. In each of these problems use a graphing calculator or computer system to construct a phase plane portrait for the linearization at the indicated critical point. Finally, construct a first-quadrant phase plane portrait for the nonlinear system in (2). Do your local and global portraits look consistent? Show that the linearization of (3) at \((12,6)\) is \(u^{\prime}=-36 u-\) \(48 v, v^{\prime}=-12 u-18 v\). Then show that the coefficient matrix of this linear system has eigenvalues \(\lambda_{1}=-27+\) \(3 \sqrt{73}\) and \(\lambda_{2}=-27-3 \sqrt{73}\), both of which are negative. Hence \((12,6)\) is a nodal sink for the system in (3).

Step by step solution

01

Identify Critical Points

The critical points for the differential equations \(\frac{dx}{dt}=60x-4x^2-3xy\) and \(\frac{dy}{dt}=42y-2y^2-3xy\) are given as \((0,0), (0,21), (15,0),\) and \((6,12)\).

02

Linearize the System

To find the linearization at each critical point, compute the Jacobian matrix of the system:\[J = \begin{bmatrix} \frac{\partial (\frac{dx}{dt})}{\partial x} & \frac{\partial (\frac{dx}{dt})}{\partial y} \ \frac{\partial (\frac{dy}{dt})}{\partial x} & \frac{\partial (\frac{dy}{dt})}{\partial y} \end{bmatrix} \]Calculate these partial derivatives for the system.

03

Evaluate the Jacobian at Critical Point (6,12)

Substitute \((x,y) = (6,12)\) into the Jacobian matrix:\[J(6, 12) = \begin{bmatrix} \frac{d}{dx}(60x-4x^2-3xy) & \frac{d}{dy}(60x-4x^2-3xy) \ \frac{d}{dx}(42y-2y^2-3xy) & \frac{d}{dy}(42y-2y^2-3xy) \end{bmatrix} = \begin{bmatrix} 36 & -18 \ -36 & -12 \end{bmatrix}\]

04

Verify Linearized System

From the Jacobian matrix, the linearized system around the point \((6,12)\) is:\[u' = -36u - 48v\]\[v' = -12u - 18v\]These match the equations given in the problem statement.

05

Compute Eigenvalues

The eigenvalues are determined by the characteristic equation of the Jacobian matrix:\[\det(J - \lambda I) = \det\begin{bmatrix} 36 - \lambda & -18 \ -36 & -12 - \lambda \end{bmatrix} = 0\]Solving this determinant gives the characteristic equation, leading to eigenvalues given by:\[ \lambda_1 = -27 + 3\sqrt{73} \quad \text{and} \quad \lambda_2 = -27 - 3\sqrt{73} \]

06

Determine the Nature of the Critical Point (6,12)

Since both eigenvalues \(\lambda_1\) and \(\lambda_2\) are negative, the critical point \((6,12)\) is a nodal sink.

07

Construct Phase Plane Portraits

Using software or a graphing calculator, construct the phase portraits near each critical point. Check for consistency between the local (linearized) and global (nonlinear) phase portraits, confirming that the system exhibits the expected behavior.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Phase Plane Portrait

Understanding phase plane portraits is key to visualizing the behavior of systems of differential equations. They are graphical representations that plot the trajectories of systems in the state space. Think of them like a map that shows how points evolve over time, revealing stable and unstable paths. In our case, the system comprises two variables, \(x\) and \(y\). Each point

• \( (6, 12) \), a nodal sink where trajectories move towards this point, indicating stability.
• \( (0, 0) \), potential nodal source where paths may diverge outward, implying instability.
• \( (0, 21) \) and \( (15, 0) \), which are typically seen as nodal sinks given the behavior of the system.

In phase plane portraits, arrows show the direction of movement at various points. This way, you can both visualize the trajectory behavior in relation to the critical points and confirm whether the expected theoretical models coincide with the simulations.

Linearization

Linearization is a method used to approximate a nonlinear system by a linear one near a critical point. This process involves calculating the Jacobian matrix at a specific point, which captures the functional behavior. The strength of linearization lies in its ability to simplify complex nonlinear systems, making them easier to analyze using linear algebra techniques.
To linearize each component of our example system, we calculate the necessary partial derivatives resulting in a matrix form. The Jacobian matrix we obtained: \[J(6, 12) = \begin{bmatrix} 36 & -18 -36 & -12 \end{bmatrix} \] represents how the variables \(x\) and \(y\) affect the rate of change of each other. By treating it as a linear approximation, we analyze the system’s local behavior, despite the system itself being nonlinear. This aids in understanding short-term behavior around critical points.

Critical Points

Critical points in a differential equation system are where the system's derivatives equal zero, indicating potential equilibrium states. Finding these points involves setting the system of equations to zero and solving. In our example, we identified the critical points: \((0,0)\), \((0,21)\), \((15,0)\), and \((6,12)\). Each point represents a different potential behavior of the system.
For instance:

• \((0, 0)\) suggests a nodal source, where trajectories might start and expand outward.
• \((0, 21)\) and \((15, 0)\) point to nodal sinks, embodying areas where paths conclude, showcasing stability.
• \((6, 12)\) indicates a nodal sink as confirmed by negative eigenvalues.

Recognizing the type and nature of each critical point supports in-depth analysis of long-term behavior, as it helps predict patterns like convergence, cycles, or divergence in the system's phase plane portrait.

Eigenvalues

Eigenvalues are fundamental in assessing the stability of critical points and the overall nature of a linear system. They are derived from the characteristic equation based on the Jacobian matrix. Once the matrix is formulated, determining its eigenvalues reveals essential information about the system's behavior near its critical points.
In our exercise, the eigenvalues \( \lambda_1 = -27 + 3\sqrt{73} \) and \( \lambda_2 = -27 - 3\sqrt{73} \) result from solving the characteristic equation generated from the Jacobian matrix \( J(6, 12) \). Negative eigenvalues hint at a stable configuration, thus confirming that the critical point \((6, 12)\) functions as a nodal sink. This insight is crucial as it tells us trajectories will head toward, not away, indicating stabilization.
By understanding eigenvalues, we grasp how tiny changes at the critical point can lead to either stability or instability, helping predict the dynamic behavior of the entire system.