Problem 16
Question
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 \(A .\) \(\lambda_{1}=1, \quad \lambda_{2}=-1 ; \quad \mathbf{v}_{1}=\left[ \begin{array}{l}{3} \\ {2}\end{array}\right] \quad \mathbf{v}_{2}=\left[ \begin{array}{r}{-4} \\ {1}\end{array}\right]\)
Step-by-Step Solution
Verified Answer
Plot eigenvectors; curves depart from \( \mathbf{v}_2 \) and follow \( \mathbf{v}_1 \). Origin is a saddle point.
1Step 1: 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 \( \lambda_2 \).
2Step 2: 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.
3Step 3: 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.
4Step 4: 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.
Key Concepts
Phase PlaneEigenvaluesSaddle PointEigenvectors
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.
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.
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.
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.
Other exercises in this chapter
Problem 15
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
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A Jacobian matrix and two equlibria are given. Determine if each is locally stable, unstable, or if the analysis is inconclusive. $$\begin{array}{l}{J=\left[ \b
View solution Problem 16
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
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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}{-\frac{3}{2}} & {\frac{
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