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(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). \(\frac{d \mathbf{x}}{d t}=\left(\begin{array}{ll}{3} & {-4} \\ {1} & {-1}\end{array}\right) \mathbf{x}\)

Short Answer

Expert verified
In conclusion, we have found the eigenvalues and eigenvectors of the given matrix, classified the critical point (0,0) as an unstable node, and discussed the sketch of trajectories in the phase plane and the plot of x1 versus time t. To create accurate plots of the requested curves, use computer software with appropriate commands and functions to create the phase plane trajectories and plots of \(x_1\) versus time \(t\) for various initial conditions.

Step by step solution

01

Calculate the eigenvalues and eigenvectors

First, we need to find the eigenvalues of the given matrix: \(\left|\begin{array}{cc}{3 - \lambda} & {-4} \\ {1} & {-1 - \lambda}\end{array}\right| = 0\) Solve the characteristic equation: \((3 - \lambda)(-1 - \lambda) - (-4)(1) = \lambda^2 - 2\lambda - 3 + 4 = \lambda^2 - 2\lambda + 1 = (\lambda - 1)^2 = 0\) There is one distinct eigenvalue: \(\lambda_1 = 1\) Now let's compute the corresponding eigenvector. For \(\lambda_1\), we have: \(\left(\begin{array}{cc}{3 - 1} & {-4} \\ {1} & {-1 - 1}\end{array}\right) \left(\begin{array}{c}{v_1} \\ {v_2}\end{array}\right) = \left(\begin{array}{c}{0} \\ {0}\end{array}\right)\) Solving this linear system, we get: \(v_1 = 2v_2\) The eigenvector associated with \(\lambda_1\) can be represented as: \(\mathbf{v} = \left(\begin{array}{c}{2} \\ {1}\end{array}\right)\) So the eigenvalues and eigenvectors are: Eigenvalue: \(\lambda_1 = 1\) Eigenvector: \(\mathbf{v} = \left(\begin{array}{c}{2} \\ {1}\end{array}\right)\)
02

Classify the critical point and determine its stability

With only one eigenvalue, we can classify the critical point (0,0) as one of the following types: - Stable node: All eigenvalues have negative real parts. - Unstable node: All eigenvalues have positive real parts. - Saddle point: Eigenvalues have opposite signs. Our only eigenvalue is \(\lambda_1 = 1\), which has a positive real part. So, the critical point (0,0) is an unstable node.
03

Sketch trajectories in the phase plane and plot x1 versus time t

Since the critical point (0,0) is an unstable node, trajectories in the phase plane will move away from the origin along the direction of the eigenvector \(\mathbf{v}\). We would get several trajectories forming paths towards infinity. For graphs of \(x_1\) versus time \(t\), we can use the eigenvector and eigenvalue found earlier. In general, the solution can be written as: \(\mathbf{x}(t) = c_1 e^{\lambda_1 t} \mathbf{v}\) Where \(c_1\) is a constant determined by initial conditions. Substituting the values for \(\lambda_1\) and \(\mathbf{v}\) we get: \(\mathbf{x}(t) = c_1 e^{t} \left(\begin{array}{c}{2} \\ {1}\end{array}\right)\) To get \(x_1\) versus time \(t\), we can consider only the first component of the solution: \(x_1(t) = 2c_1 e^t\) Plotting this function for different values of \(c_1\) will give us a family of curves representing different initial conditions.
04

Use a computer to plot the curves

To plot the curves requested in part (c), use computer software such as MATLAB, Mathematica or Python (with packages like NumPy and Matplotlib). Create trajectories in the phase plane by plotting the vector field of the system and the solution given by \(\mathbf{x}(t)\). For plotting \(x_1\) versus time \(t\), use the expression \(x_1(t) = 2c_1 e^t\) for different values of \(c_1\) to obtain various curves representing different initial conditions.

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Key Concepts

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

Critical Point Classification
Understanding the behavior of dynamical systems near critical points is essential for analyzing the stability of these systems. A critical point, also known as an equilibrium point, is a point in the phase space where the system's state does not change over time. To classify these points, we look at the eigenvalues of the system's linearization around the critical point.

In the given exercise, we determined a single eigenvalue, \(\lambda_1 = 1\), which indicates the presence of a node. Since this eigenvalue is positive, it tells us that the critical point is repelling or an unstable node. It means that trajectories starting close to the point will move away from it over time. In our case, the eigenvector associated with this eigenvalue, \(\mathbf{v} = \left(\begin{array}{c}{2} \ {1}\end{array}\right)\), dictates the direction along which this repulsion takes place. This preliminary classification helps predict long-term behaviors of the system, which is crucial for both theoretical studies and practical applications.
Stability of Dynamical Systems
The stability of a dynamical system is a concept that describes how the system responds to small disturbances from its equilibrium state. Eigenvalues play a crucial role in determining the stability of the system at a critical point. For example, if all eigenvalues have negative real parts, any small disturbance will decay, and the system will return to equilibrium, indicating stability.

In contrast, if the real parts of all eigenvalues are positive, as in our example with \(\lambda_1 = 1\), small disturbances grow exponentially over time, making the system unstable at the equilibrium point. The type of instability we have identified here suggests that solutions will diverge from the critical point as time progresses. This insight into the system's response forms the basis for analyzing how dynamical systems evolve over time, including in complex scenarios such as ecological systems, mechanical structures, and financial markets.
Phase Plane Trajectories
The phase plane is a graphical tool used to visualize the behavior of a two-dimensional dynamical system. Trajectories in the phase plane, also known as phase curves, represent the path that a state of the system will follow over time. These trajectories can help predict the system's behavior and identify patterns such as cycles, attractors, or chaotic movement.

By plotting several trajectories, as suggested in the exercise, we gain insight into how the system's state evolves. In systems with an unstable node, all the trajectories will diverge away from the critical point, moving outward along the direction of the eigenvectors. The visualization of trajectories is vital for analyzing non-linear systems, too, as it can reveal complex dynamics that are not immediately obvious from the system's equations.
Differential Equations
Differential equations form the mathematical backbone of dynamical systems. They describe the relationship between a function and its derivatives, reflecting how a system evolves over time based on its current state. In the context of the exercise, the linear differential equation system \(\frac{d \mathbf{x}}{d t} = \left(\begin{array}{ll}{3} & {-4} \ {1} & {-1}\end{array}\right) \mathbf{x}\) dictates how the state vector \(\mathbf{x}\) changes with time.

Solution to these equations often involves finding eigenvalues and eigenvectors, as they can greatly simplify the process. Specifically, solutions take the form \(\mathbf{x}(t) = c_1 e^{\lambda_1 t} \mathbf{v}\), which combines exponential growth or decay dictated by the eigenvalue with directionality from the eigenvector. Understanding differential equations is critical for students as they apply to a wide array of academic and professional fields, from economics to engineering.

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Most popular questions from this chapter

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 \(r_{1}+r_{2}=p\).

Each of Problems I through 6 can be interpreted as describing the interaction of two species with populations \(x\) and \(y .\) In each of these problems carry out the following steps. $$ \begin{array}{l}{\text { (a) Draw a direction field and describe how solutions seem to behave. }} \\ {\text { (b) Find the critical points. }} \\ {\text { (c) For each critical point find the corresponding linear system. Find the eigenvalues and }} \\ {\text { eigenvectors of the linear system; classify each critical point as to type, and determine }} \\ {\text { whether it is asymptotically stable, stable, or unstable. }}\end{array} $$ $$ \begin{array}{l}{\text { (d) Sketch the trajectories in the neighborhood of each critical point. }} \\ {\text { (c) Compute and plot enough trajectories of the given system to show clearly the behavior of }} \\ {\text { the solutions. }} \\ {\text { (f) Determine the limiting behavior of } x \text { and } y \text { as } t \rightarrow \infty \text { and interpret the results in terms of }} \\ {\text { the populations of the two species. }}\end{array} $$ $$ \begin{array}{l}{d x / d t=x(1.5-x-0.5 y)} \\ {d y / d t=y(2-0.5 y-1.5 x)}\end{array} $$

If \(x=r \cos \theta, y=r \sin \theta,\) show that \(y(d x / d t)-x(d y / d t)=-r^{2}(d \theta / d t)\)

(a) By solving Eq. (9) numerically show that the real part of the complex roots changes sign when \(r \cong 24.737\). (b) Show that a cubic polynomial \(x^{3}+A x^{2}+B x+C\) has one real zero and two pure imaginary zeros only if \(A B=C\). (c) By applying the result of part (b) to Eq. (9) show that the real part of the complex roots changes sign when \(r=470 / 19\).

Consider the system of equations $$ d x / d t=y-x f(x, y), \quad d y / d t=-x-y f(x, y) $$ where \(f\) is continuous and has continuous first partial derivatives. Show that if \(f(x, y)>0\) in some neighborhood of the origin, then the origin is an asymptotically stable critical point, and if \(f(x, y)<0\) in some neighborhood of the origin, then the origin is an unstable critical point. Hint: Construct a Liapunov function of the form \(c\left(x^{2}+y^{2}\right) .\)

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