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Chapter 9: Problem 1

(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). \(\frac{d \mathbf{x}}{d t}=\left(\begin{array}{cc}{3} & {-2} \\ {2} & {-2}\end{array}\right) \mathbf{x}\)

Short Answer

Expert verified
Based on the given system of two linear differential equations, we found the eigenvalues and eigenvectors, classified the stability of the critical point (0,0) as unstable, and sketched the trajectories in the phase plane. The eigenvalues are λ1=2 and λ2=-1, and the eigenvectors are v1= (2,1) and v2= (1,2). The critical point (0,0) is a saddle point, and the trajectories are stretched along the eigenvector corresponding to the positive eigenvalue, and compressed along the eigenvector corresponding to the negative eigenvalue.

Step by step solution

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01

Finding eigenvalues of the matrix

To find the eigenvalues, we need to find the roots of the characteristic equation given by the determinant: \(\det(A - \lambda I) = 0\), where \(A = \begin{pmatrix} 3 & -2 \\ 2 & -2 \end{pmatrix}\) and \(I\) is the identity matrix. The characteristic equation is : \[ \begin{vmatrix} 3-\lambda & -2 \\ 2 & -2-\lambda \end{vmatrix} = (3-\lambda)(-2-\lambda) - (-2)(2) = \lambda^2-\lambda-2 \] Now, we find the roots (eigenvalues) of the characteristic equation: \[ \lambda^2-\lambda-2 = 0 \]
02

Finding eigenvectors of the matrix

The eigenvalues of the characteristic equation are \(\lambda_1=2\) and \(\lambda_2=-1\). Now, we find the eigenvectors corresponding to each eigenvalue by solving the system \((A-\lambda I)v=0\). For \(\lambda_1=2\) : \[ (A - \lambda_1 I)v_1 = \begin{pmatrix} 1 & -2 \\ 2 & -4 \end{pmatrix} \begin{pmatrix} v_{11} \\ v_{12} \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \] We can see that the second row is a scalar multiple of the first row, so we only need to focus on one row: \[ v_{11} - 2 v_{12} = 0 \] We can choose \(v_{12} = 1\), which gives us \(v_{11} = 2\). So, \(v_1=\begin{pmatrix} 2 \\ 1 \end{pmatrix}\). For \(\lambda_2=-1\) : \[ (A - \lambda_2 I)v_2 = \begin{pmatrix} 4 & -2 \\ 2 & -1 \end{pmatrix} \begin{pmatrix} v_{21} \\ v_{22} \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \] Again, the second row is a scalar multiple of the first row, so we only need to focus on one row: \[ 4 v_{21} - 2 v_{22} = 0 \] We can simplify this to \(2v_{21} - v_{22} = 0\). Let \(v_{21}=1\). Then \(v_{22}=2\). So, \(v_2=\begin{pmatrix} 1 \\ 2 \end{pmatrix}\).
03

Classify the critical point and determine stability

The stability of the critical point \((0,0)\) can be determined using the eigenvalues found earlier. We have \(\lambda_1=2\) and \(\lambda_2=-1\). Since one eigenvalue is positive and the other is negative, the critical point \((0,0)\) is a saddle point and is unstable.
04

Sketching trajectories in the phase plane

To sketch the trajectories, we need to find the general solution of the system and simplify the expression. The general solution is given by: \[ \mathbf{x}(t) = c_1 e^{\lambda_1 t} \mathbf{v}_1 + c_2 e^{\lambda_2 t} \mathbf{v}_2 \] Substituting the eigenvalues and eigenvectors, we get: \[ \mathbf{x}(t) = c_1 e^{2t} \begin{pmatrix} 2 \\ 1 \end{pmatrix} + c_2 e^{-t} \begin{pmatrix} 1 \\ 2 \end{pmatrix} \] Since the system is unstable, the trajectories will move away from the origin. For the given system, the trajectories will be stretched along the eigenvector corresponding to the positive eigenvalue (\(\begin{pmatrix} 2 \\ 1 \end{pmatrix}\)), and compressed along the eigenvector corresponding to the negative eigenvalue (\(\begin{pmatrix} 1 \\ 2 \end{pmatrix}\)).

Key Concepts

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

Characteristic Equation
The characteristic equation is a crucial tool in solving systems of differential equations. It starts with finding the "eigenvalues" of a given matrix. In our scenario, the matrix is \( A = \begin{pmatrix} 3 & -2 \ 2 & -2 \end{pmatrix} \),\ which describes the system's dynamics. We use the formula \( \det(A - \lambda I) = 0 \) to derive the characteristic equation. Here, \( \lambda \) represents the eigenvalues, and \( I \) is the identity matrix. This gives us a quadratic equation:
  • \( (3-\lambda)(-2-\lambda) - (-2)(2) = \lambda^2 - \lambda - 2 \)
The roots of this polynomial equation are the sought eigenvalues. Solving \( \lambda^2 - \lambda - 2 = 0 \), you find \( \lambda_1 = 2 \) and \( \lambda_2 = -1 \). These values tell us much about the system, including its stability and phase behavior.
Stability Analysis
Understanding stability requires scrutinizing the eigenvalues obtained from the characteristic equation. Stability analysis determines how solutions to a system of differential equations behave as time passes. A critical point, such as \((0,0)\) in our problem, is classified based on its eigenvalues:
  • If all eigenvalues have negative real parts, the critical point is stable and eventually every trajectory leads to it.
  • If any eigenvalue has a positive real part, the point is unstable, meaning trajectories will drift away.
In our case, \( \lambda_1 = 2 \) is positive, and \( \lambda_2 = -1 \) is negative, indicating a saddle point. This means that the system is unstable. Trajectories will diverge away from \((0,0)\), signaling instability in the equilibrium, and thus, the system does not tend to return to this point once perturbed.
Phase Plane Trajectories
Phase plane trajectories provide a visual representation of the system's behavior over time. We derive them from solving the system’s differential equation using its general solution:
  • \( \mathbf{x}(t) = c_1 e^{2t} \begin{pmatrix} 2 \ 1 \end{pmatrix} + c_2 e^{-t} \begin{pmatrix} 1 \ 2 \end{pmatrix} \)
This expression combines the eigenvectors \( \begin{pmatrix} 2 \ 1 \end{pmatrix} \) and \( \begin{pmatrix} 1 \ 2 \end{pmatrix} \), along with the exponential growth or decay modulated by the eigenvalues \( \lambda_1 \) and \( \lambda_2 \).
The phase portrait sketches the trajectories based on these. You’ll notice that trajectories stretch along the direction of the eigenvector corresponding to \( \lambda_1 = 2 \) and compress along \( \lambda_2 = -1 \).
For the given unstable system, expect the paths to move away from the origin. The graphical representation of these trajectories showcases the directional tendencies of the system’s solutions, graphically depicting stability and instability regions.

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Most popular questions from this chapter

Consider the system $$ d x / d t=a x[1-(y / 2)], \quad d y / d t=b y[-1+(x / 3)] $$ where \(a\) and \(b\) are positive constants. Observe that this system is the same as in the example in the text if \(a=1\) and \(b=0.75 .\) Suppose the initial conditions are \(x(0)=5\) and \(y(0)=2\) (a) Let \(a=1\) and \(b=1 .\) Plot the trajectory in the phase plane and determine (or cstimate) the period of the oscillation. (b) Repeat part (a) for \(a=3\) and \(a=1 / 3,\) with \(b=1\) (c) Repeat part (a) for \(b=3\) and \(b=1 / 3,\) with \(a=1\) (d) Describe how the period and the shape of the trajectory depend on \(a\) and \(b\).

(a) Determine all critical points of the given system of equations. (b) Find the corresponding linear system near each critical point. (c) Find the eigenalues of each linear system. What conclusions can you then draw about the nonlinear system? (d) Draw a phase portrait of the nonlinear system to confirm your conclusions, or to extend them in those cases where the linear system does not provide definite information about the nonlinear system. $$ d x / d t=x-y^{2}, \quad d y / d t=y-x^{2} $$

Carry out the indicated investigations of the Lorenz equations. (a) For \(r=21\) plot \(x\) versus \(t\) for the solutions starting at the initial points \((3,8,0),\) \((5,5,5),\) and \((5,5,10) .\) Use a \(t\) interval of at least \(0 \leq t \leq 30 .\) Compare your graphs with those in Figure \(9.8 .4 .\) (b) Repeat the calculation in part (a) for \(r=22, r=23,\) and \(r=24 .\) Increase the \(t\) interval as necessary so that you can determine when each solution begins to converge to one of the critical points. Record the approximate duration of the chaotic transient in each case. Describe how this quantity depends on the value of \(r\). (c) Repeat the calculations in parts (a) and (b) for values of \(r\) slightly greater than 24 . Try to estimate the value of \(r\) for which the duration of the chaotic transient approaches infinity.

(a) Determine all critical points of the given system of equations. (b) Find the corresponding linear system near each critical point. (c) Find the eigenalues of each linear system. What conclusions can you then draw about the nonlinear system? (d) Draw a phase portrait of the nonlinear system to confirm your conclusions, or to extend them in those cases where the linear system does not provide definite information about the nonlinear system. $$ d x / d t=1-x y, \quad d y / d t=x-y^{3} $$

Consider the system (3) in Example 1 of the text. Recall that this system has an asymptotically stable critical point at (0.5,0.5) , corresponding to the stable coexistence of the two population species. Now suppose that immigration or emigration occurs at the constant rates of \(\delta a\) and \(\delta b\) for the species \(x\) and \(y,\) respectively. In this case equations ( 3 ) are replaced by $$\frac{d x}{d t}=x(1-x-y)+\delta a, \quad \frac{d y}{d t}=\frac{y}{4}(3-4 y-2 x)+\delta b$$ The question is what effect this has on the location of the stable equilibrium point. a. To find the new critical point, we must solve the equations $$\begin{aligned} x(1-x-y)+\delta a &=0 \\ \frac{y}{4}(3-4 y-2 x)+\delta b &=0 \end{aligned}$$ One way to proceed is to assume that \(x\) and \(y\) are given by power series in the parameter \(\delta ;\) thus $$x=x_{0}+x_{1} \delta+\cdots, \quad y=y_{0}+y_{1} \delta+\cdots$$ Substitute equations (44) into equations (43) and collect terms according to powers of \(\delta\). b. From the constant terms (the terms not involving \(\delta\) ), show that \(x_{0}=0.5\) and \(y_{0}=0.5,\) thus confirming that in the absence of immigration or emigration, the critical point is (0.5,0.5) . c. From the terms that are linear in \(\delta,\) show that \\[ x_{1}=4 a-4 b, \quad y_{1}=-2 a+4 b \\] d. Suppose that \(a>0\) and \(b>0\) so that immigration occurs for both species. Show that the resulting equilibrium solution may represent an increase in both populations, or an increase in one but a decrease in the other. Explain intuitively why this is a reasonable result.
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