Problem 22
Question
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? \(A=\left[ \begin{array}{ll}{0} & {1} \\ {1} & {0}\end{array}\right]\)
Step-by-Step Solution
Verified Answer
The equilibrium is a saddle point.
1Step 1: Identify the System of Differential Equations
We are given the system of differential equations \( \frac{d \mathbf{x}}{dt} = A \mathbf{x} \), where \( A = \begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix} \). Then we have two equations: \( \frac{dx_1}{dt} = x_2 \) and \( \frac{dx_2}{dt} = x_1 \).
2Step 2: Find the Eigenvalues of Matrix A
To determine the type of equilibrium, we first find the eigenvalues of the matrix \( A \). Solve \( \text{det}(A - \lambda I) = 0 \) where \( \lambda \) is the eigenvalue and \( I \) is the identity matrix. The determinant is calculated as \(\text{det}\begin{bmatrix}-\lambda & 1\ 1 & -\lambda\end{bmatrix} = \lambda^2 - 1\). Setting it to zero gives \( \lambda^2 = 1 \), so \( \lambda = \pm 1 \).
3Step 3: Classify the Equilibrium
The eigenvalues \( \lambda = 1 \) and \( \lambda = -1 \) are real and opposite in sign. This indicates that the equilibrium point is a saddle node. In the phase plane, trajectories will approach the equilibrium point along one eigenvector and diverge along another.
Key Concepts
Phase PlaneEigenvaluesSaddle Point
Phase Plane
Understanding the phase plane is crucial when working with systems of differential equations. It is a visual representation of possible trajectories (paths) that the system can take over time. In our system, each point in the phase plane corresponds to a set of states of the variables. For a system of two variables, like in the exercise, it's a two-dimensional plane where each axis represents one of the variables, often labeled as \( x_1 \) and \( x_2 \).
To construct the phase plane, we need to compute the trajectories of the system, which involves solving the differential equations or using numerical methods to simulate the system over time. Importantly, nullclines are also drawn in the phase plane. These are curves where the rate of change for one of the variables is zero. For each nullcline, set \( \frac{dx_1}{dt} = 0 \) or \( \frac{dx_2}{dt} = 0 \), and solve for the variables. This divides the phase plane into regions where estimations of trajectory directions can be made.
In our given exercise, the nullclines can be derived directly from the equations \( \frac{dx_1}{dt} = x_2 \) and \( \frac{dx_2}{dt} = x_1 \). The physical interpretation of the nullclines helps in predicting how the system behaves near equilibrium points.
To construct the phase plane, we need to compute the trajectories of the system, which involves solving the differential equations or using numerical methods to simulate the system over time. Importantly, nullclines are also drawn in the phase plane. These are curves where the rate of change for one of the variables is zero. For each nullcline, set \( \frac{dx_1}{dt} = 0 \) or \( \frac{dx_2}{dt} = 0 \), and solve for the variables. This divides the phase plane into regions where estimations of trajectory directions can be made.
In our given exercise, the nullclines can be derived directly from the equations \( \frac{dx_1}{dt} = x_2 \) and \( \frac{dx_2}{dt} = x_1 \). The physical interpretation of the nullclines helps in predicting how the system behaves near equilibrium points.
Eigenvalues
Eigenvalues are essential for determining the nature of equilibrium points in a system of linear differential equations. They are derived from the system's matrix, \( A \), which defines how the system behaves. In our exercise, the matrix \( A \) is \( \begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix} \).
To find the eigenvalues, solve the characteristic equation \( \text{det}(A - \lambda I) = 0 \), where \( I \) is the identity matrix and \( \lambda \) represents the eigenvalues. By plugging \( A \) into the determinant equation, we solve \( \lambda^2 - 1 = 0 \), giving us \( \lambda = \pm 1 \).
Having real and opposite eigenvalues (\( 1 \) and \( -1 \)), signifies that the system's equilibrium is a saddle point. This algebraic approach directly informs us about the phase plane's dynamics, particularly how trajectories behave around critical points.
To find the eigenvalues, solve the characteristic equation \( \text{det}(A - \lambda I) = 0 \), where \( I \) is the identity matrix and \( \lambda \) represents the eigenvalues. By plugging \( A \) into the determinant equation, we solve \( \lambda^2 - 1 = 0 \), giving us \( \lambda = \pm 1 \).
Having real and opposite eigenvalues (\( 1 \) and \( -1 \)), signifies that the system's equilibrium is a saddle point. This algebraic approach directly informs us about the phase plane's dynamics, particularly how trajectories behave around critical points.
Saddle Point
The term "saddle point" originates from the resemblance of the surface shape around the equilibrium point, similar to a horse's saddle. In the context of differential equations, a saddle point refers to a specific type of equilibrium where the trajectories in the phase plane behave in a distinct manner. When analyzing a linear system, the system's equilibrium is considered a saddle point if it possesses one stable and one unstable direction, typically when the eigenvalues of the system's matrix \( A \) are real and of opposite signs.
In our exercise, with eigenvalues \( \lambda = 1 \) and \( \lambda = -1 \), the equilibrium behaves as a saddle point. This configuration means:
In our exercise, with eigenvalues \( \lambda = 1 \) and \( \lambda = -1 \), the equilibrium behaves as a saddle point. This configuration means:
- Some trajectories will move towards the equilibrium point, only along the stable direction.
- Other trajectories will move away from the equilibrium, along the unstable direction.
- The presence of at least one trajectory that does not settle into equilibrium indicates instability around the saddle point.
Other exercises in this chapter
Problem 22
Find all equilibria and determine their local stability properties. $$x^{\prime}=x(2-x), \quad y^{\prime}=y(3-y)$$
View solution Problem 22
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}{0} & {1} \\ {-6} & {-5}
View solution Problem 23
Find all equilibria and determine their local stability properties. $$p^{\prime}=-p^{2}+q-1, \quad q^{\prime}=q(2-p-q)$$
View solution Problem 23
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} \\ {-3} & {-1
View solution