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Chapter 10: Problem 16

Sketch several solution curves in the phase plane of the system of differential equations \(d \mathbf{x} / d t=A \mathbf{x}\) using the given eigenvalues and eigenvectors of \(A .\) \(\lambda_{1}=1, \quad \lambda_{2}=-1 ; \quad \mathbf{v}_{1}=\left[ \begin{array}{l}{3} \\ {2}\end{array}\right] \quad \mathbf{v}_{2}=\left[ \begin{array}{r}{-4} \\ {1}\end{array}\right]\)

Short Answer

Expert verified
Plot eigenvectors; curves depart from \( \mathbf{v}_2 \) and follow \( \mathbf{v}_1 \). Origin is a saddle point.

Step by step solution

01

Identify the Type of Critical Point

The system is given by \( \frac{d \mathbf{x}}{d t} = A \mathbf{x} \) with eigenvalues \( \lambda_1 = 1 \) and \( \lambda_2 = -1 \). Here, since \( \lambda_1 > 0 \) and \( \lambda_2 < 0 \), the origin acts as a saddle point in the phase plane. This indicates trajectories will display unstable behavior along the direction corresponding to \( \lambda_1 \) and stable behavior along the direction corresponding to \( \lambda_2 \).
02

Sketch Eigenvectors in the Phase Plane

Plot the eigenvector \( \mathbf{v}_1 = \begin{bmatrix} 3 \ 2 \end{bmatrix} \) corresponding to \( \lambda_1 = 1 \) and the eigenvector \( \mathbf{v}_2 = \begin{bmatrix} -4 \ 1 \end{bmatrix} \) corresponding to \( \lambda_2 = -1 \) in the phase plane. \( \mathbf{v}_1 \) indicates the direction of unstable behavior, while \( \mathbf{v}_2 \) indicates the direction of stable behavior.
03

Sketch Solution Curves

The solution curves on the phase plane follow the direction of the eigenvectors. Solutions exponentially move away from the origin along the \( \mathbf{v}_1 \) direction (unstable manifold) and decay towards the origin along the \( \mathbf{v}_2 \) direction (stable manifold). Sketch curves starting from various initial conditions, approaching the stable line \( \mathbf{v}_2 \) and emanating along \( \mathbf{v}_1 \) as they move away from the origin.
04

Analyze Intersections and Curve Trajectories

Since the eigenvectors are not orthogonal, note that the trajectories will not just switch directly between \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \). Instead, they will initially accelerate away from the origin along the positive \( \lambda_1 \) direction and decay along the negative \( \lambda_2 \) direction, creating a hyperbolic pattern in the phase plane.

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Key Concepts

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

Phase Plane
The phase plane is a powerful visual tool in the study of differential equations, especially when dealing with a system in two dimensions. Imagine a coordinate plane where each point represents a state of the system, marked by values of two variables. In our scenario with a system matrix \( A \), we explore how solutions evolve over time by plotting them on this plane. This results in a graphical depiction of trajectories, illustrating how each state transitions into the next.
  • Each trajectory symbolizes a solution to the differential equations.
  • The origin often represents an equilibrium point, where the system doesn't change over time if started there.
  • The behavior of trajectories is governed by the nature of the eigenvalues and eigenvectors.
By examining the phase plane, we visualize stable and unstable behavior, which is particularly useful for understanding long-term dynamics and stability of the system.
Eigenvalues
Eigenvalues play a pivotal role in determining the system's behavior portrayed in the phase plane. For any square matrix \( A \), eigenvalues analyze how transformations 'stretch' different vectors in different ways. In the context of our differential system, they decide the dynamics at the critical point, often the origin.
  • Positive eigenvalues, like \( \lambda_1 = 1 \), indicate growth or an unstable direction.
  • Negative eigenvalues, such as \( \lambda_2 = -1 \), suggest decay or stability.
  • The combination of these values can reveal the presence of saddle points, centers, spirals, or nodes.
In our exercise, the contrasting signs of \( \lambda_1 \) and \( \lambda_2 \) disclose that our system's origin is a saddle point, setting the stage for the trajectories' behaviors near this point.
Saddle Point
The saddle point classification arises from a critical nature of eigenvalues, where one is positive and the other negative, as observed in our solution. A saddle point is characterized by its stable and unstable directions in the phase plane. This mixed behavior makes the system unpredictable when slightly perturbed, as a small deviation can lead to dramatically different outcomes.
  • Unstable directions are linked with positive eigenvalues, where solutions diverge away from the point.
  • Stable directions correspond to negative eigenvalues, drawing solutions towards the point.
  • The phase plane shows these as manifolds intersecting at the saddle, creating what looks like a saddle shape.
Thus, the origin in our system displays saddle-like behavior, indicative of the trajectories diverging along the direction of \( \mathbf{v}_1 \) and converging along \( \mathbf{v}_2 \).
Eigenvectors
Eigenvectors are indispensable companions to eigenvalues, providing the directions along which these eigenvalue-induced transformations occur. For our differential equation system, they illustrate the dynamics in the phase plane. When you sketch an eigenvector, you're essentially drawing a roadmap for the solution trajectories.
  • Eigenvectors corresponding to positive eigenvalues show the direction of instability, in our case, \( \mathbf{v}_1 = \begin{bmatrix} 3 \ 2 \end{bmatrix} \).
  • Those associated with negative eigenvalues indicate the stable path, here represented by \( \mathbf{v}_2 = \begin{bmatrix} -4 \ 1 \end{bmatrix} \).
  • The trajectories most strongly aligned with these directions dominate the behavior near the origin.
Visualizing these vectors in the phase plane helps us predict how the solutions to the differential equations evolve over time, both towards and away from stability.

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Most popular questions from this chapter

Given the system of differential equations \(d \mathbf{x} / d t=A \mathbf{x}\) , construct the phase plane, including the nullclines. Does the equilibrium look like a saddle, a node, or a spiral? \(A=\left[ \begin{array}{ll}{1} & {1} \\ {0} & {1}\end{array}\right]\)

$$\begin{array}{l}{\text { A model for self-reproducing resources with limited }} \\ {\text { growth is obtained by choosing } f(R)=r R(1-R / K)} \\ {g(R, C)=b R C, \text { and } h(C)=\mu C . \text { Suppose } r=2, K=5} \\ {b=1, \varepsilon=1, \text { and } \mu=1}\end{array}$$

Write each system of linear differential equations in matrix notation. \(d x / d t=5 y, \quad d y / d t=2 x-y\)

Find all equilibria. Then find the linearization near each equilibrium. $$\begin{array}{l}{\frac{d x_{1}}{d t}=e^{-x_{1}}\left(x_{1}-x_{2}\right)} \\\ {\frac{d x_{2}}{d t}=x_{1}-x_{2}^{2}+2 x_{1} x_{2}}\end{array}$$

Solve the initial value problem \(d \mathbf{x} / d t=A \mathbf{x}\) with \(\mathbf{x}(0)=\mathbf{x}_{0} .\) \(A=\left[ \begin{array}{rr}{-1} & {-2} \\ {2} & {-2}\end{array}\right] \quad \mathbf{x}_{0}=\left[ \begin{array}{l}{1} \\ {5}\end{array}\right]\)
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