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Chapter 9: Problem 2

(a) Find the eigenvalues and eigenvectors. (b) Classify the critical point \((0,0)\) as to type and determine whether it is stable, asymptotically stable, or unstable. (c) Sketch several trajectories in the phase plane and also sketch some typical graphs of \(x_{1}\) versus \(t .\) (d) Use a computer to plot accurately the curves requested in part (c). \(\frac{d \mathbf{x}}{d t}=\left(\begin{array}{rr}{5} & {-1} \\ {3} & {1}\end{array}\right) \mathbf{x}\)

Short Answer

Expert verified
The eigenvalues of the given matrix are λ₁ = 2 and λ₂ = 4. The origin (0,0) represents an unstable improper node.

Step by step solution

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01

Finding eigenvalues and eigenvectors

To find the eigenvalues and eigenvectors, we need to solve the following equation: \(|A - \lambda I|=0\), where \(I\) is the identity matrix, in this case: \(\left|\begin{array}{cc}{5-\lambda} & {-1} \\ {3}& {1-\lambda}\end{array}\right| = 0\) \((5-\lambda)(1-\lambda) - (- 1)(3) = 0\) \(\lambda^{2} -6\lambda + 8 = (\lambda-2)(\lambda-4)=0\) The eigenvalues are \(\lambda_1=2\) and \(\lambda_2=4\). Now, we need to find the eigenvectors for each eigenvalue: For \(\lambda_1 = 2\): \((A - 2I)v_1 = 0\) \(\left(\begin{array}{cc}{3} & {-1} \\ {3}& {-1}\end{array}\right) \begin{pmatrix}v_1^1\\v_1^2\end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix}\) Solving for the eigenvector \(v_1\): \(v_1^1 = v_1^2\) Let \(v_1^1 = 1\), then \(v_1=\begin{pmatrix}1\\1\end{pmatrix}\). For \(\lambda_2 = 4\): \((A - 4I)v_2 = 0\) \(\left(\begin{array}{cc}{1} & {-1} \\ {3}& {-3}\end{array}\right) \begin{pmatrix}v_2^1\\v_2^2\end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix}\) Solving for the eigenvector \(v_2\): \(v_2^1 = v_2^2\) Let \(v_2^1 = 1\), then \(v_2=\begin{pmatrix}1\\1\end{pmatrix}\).
02

Classifying the critical point and determining stability

Let's analyze the eigenvalues to classify the critical point and determine its stability: - Both eigenvalues are positive and real. - Since \(\lambda_2>\lambda_1>0\), the critical point (0, 0) is an unstable improper node.
03

Sketching trajectories and x₁ vs t graphs

Using the eigenvalues and eigenvectors, we can write the solution to the system as: \(\mathbf{x}(t) = c_1e^{\lambda_1 t}\mathbf{v}_1 + c_2e^{\lambda_2 t}\mathbf{v}_2 = c_1e^{2t}\begin{pmatrix}1\\1\end{pmatrix} + c_2e^{4t}\begin{pmatrix}1\\1\end{pmatrix}\) Now let's discuss the qualitative behavior of the trajectories and sketch them in the phase plane: 1. The eigenvectors indicate that the system moves away from the origin in the direction of the eigenvectors. 2. Since both eigenvalues are positive, the trajectories will move away from the origin, confirming the instability of the improper node. To sketch the x₁ vs t graphs, we can consider the x₁ component of the solution: \(x_1(t) = c_1e^{2t} + c_2e^{4t}\) The qualitative behavior of the \(x_1\) component depends on the coefficients \(c_1\) and \(c_2\).
04

Using a computer for accurate plots

To create accurate plots of the trajectories and the x₁ vs t graphs, you can use software such as MATLAB, Mathematica, or Python's libraries like NumPy and Matplotlib. 1. Choose initial conditions for the trajectories. This will help determine \(c_1\) and \(c_2\) for the general solution. 2. Solve the system of ODEs with the chosen initial conditions and plot the trajectories in the phase plane. 3. Observe the behavior of x₁ as a function of time for the chosen trajectories and plot the x₁ vs t graphs.

Key Concepts

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

Critical Point Classification
Understanding the nature of a critical point, which is a point where the system of differential equations does not change, is essential in the study of dynamical systems. Classification depends on the eigenvalues of the matrix associated with the system. In the given exercise, since both eigenvalues are positive, the critical point at the origin (0,0) is classified as an unstable improper node. This means trajectories move away from this point, and any small disturbance can cause the long-term behavior to deviate significantly from the equilibrium.
Stability Analysis
Stability analysis in the context of differential equations involves determining whether solutions will remain close to an equilibrium point over time. An equilibrium point can be stable, asymptotically stable, or unstable. The eigenvalues obtained from the exercise, are ideal for such analysis. Since both are positive, the point (0,0) is classified as unstable. There is no tendency for the system to return to equilibrium if perturbed, indicating that solutions are sensitive to initial conditions.
Phase Plane Analysis
Phase plane analysis provides a visual representation of trajectories of a system of differential equations in two-dimensional space. The exercise asks for sketches that represent these trajectories. The characteristics and direction of these trajectories are determined by the eigenvalues and eigenvectors. With both eigenvalues being positive, the sketches will show trajectories moving away from the critical point in the direction of the eigenvectors, illustrating the system's instability as time progresses.
Differential Equations
Differential equations describe relationships involving rates of change. The given exercise demonstrates a system of linear ordinary differential equations (ODEs), represented in matrix form. The solution to this system informs how variables change over time, which is fundamental to predicting system behavior. By understanding the solution to these differential equations, one can determine the position and velocity of a system at any given time, a crucial concept in fields ranging from physics to economics.
System of ODEs
A system of ODEs consists of multiple differential equations involving several dependent variables and their derivatives with respect to one independent variable, often time. In the exercise, the system is represented in matrix form, which simplifies solving and analyzing the system. The eigenvalues and eigenvectors found are part of the solution, which can be expressed as a sum of exponentials scaled by the initial conditions. This form helps to describe exponential growth or decay in multi-dimensional systems, a principle that can be observed in natural phenomena and financial growth models alike.

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Most popular questions from this chapter

Prove that if a trajectory starts at a noncritical point of the system $$ d x / d t=F(x, y), \quad d y / d t=G(x, y) $$ then it cannot reach a critical point \(\left(x_{0}, y_{0}\right)\) in a finite length of time. Hint: Assume the contrary, that is, assume that the solution \(x=\phi(t), y=\psi(t)\) satisfies \(\phi(a)=x_{0}, \psi(a)=y_{0}\). Then use the fact that \(x=x_{0}, y=y_{0}\) is a solution of the given system satisfying the initial condition \(x=x_{0}, y=y_{0}\) at \(t=a\).

Consider the linear system $$ d x / d t=a_{11} x+a_{12} y, \quad d y / d t=a_{21} x+a_{22} y $$ where \(a_{11}, \ldots, a_{22}\) are real constants. Let \(p=a_{11}+a_{22}, q=a_{11} a_{22}-a_{12} a_{21},\) and \(\Delta=\) \(p^{2}-4 q\). Show that the critical point \((0,0)\) is a (a) Node if \(q>0\) and \(\Delta \geq 0\) (b) Saddle point if \(q<0\); (c) Spiral point if \(p \neq 0\) and \(\Delta<0\); (d) Center if \(p=0\) and \(q>0\). Hint: These conclusions can be obtained by studying the eigenvalues \(r_{1}\) and \(r_{2}\). It may also be helpful to establish, and then to use, the relations \(r_{1} r_{2}=q\) and \(r_{1}+r_{2}=p\).

Assuming that the trajectory corresponding to a solution \(x=\phi(t), y=\psi(t),-\infty0\) such that \(\phi\left(t_{0}+T\right)=x_{0}, \psi\left(t_{0}+T\right)=y_{0} .\) Show that \(x=\Phi(t)=\phi(t+T)\) and \(y=\Psi(t)=\psi(t+T)\) is a solution and then use the existence and uniqueness theorem to show that \(\Phi(t)=\phi(t)\) and \(\Psi(t)=\psi(t)\) for all \(t .\)

an autonomous system is expressed in polar coordinates. Determine all periodic solutions, all limit cycles, and determine their stability characteristics. $$ d r / d t=r(1-r)(r-2), \quad d \theta / d t=-1 $$

The motion of a certain undamped pendulum is described by the equations $$ d x / d t=y, \quad d y / d t=-4 \sin x $$ If the pendulum is set in motion with an angular displacement \(A\) and no initial velocity, then the initial conditions are \(x(0)=A, y(0)=0\) (a) Let \(A=0.25\) and plot \(x\) versus \(t\). From the graph estimate the amplitude \(R\) and period \(T\) of the resulting motion of the pendulum. (b) Repeat part (a) for \(A=0.5,1.0,1.5,\) and \(2.0 .\) (c) How do the amplitude and period of the pendulum's motion depend on the initial position \(A^{7}\) Draw a graph to show each of these relationships. Can you say anything about the limiting value of the period as \(A \rightarrow 0 ?\) (d) Let \(A=4\) and plot \(x\) versus \(t\) Explain why this graph differs from those in parts (a) and (b). For what value of \(A\) does the transition take place?
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