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Chapter 13: Problem 20

Most of the time, the linearization accurately predicts the behavior of a nonlinear system at an equilibrium point. There are exceptions, most notably when the the matrix \(A\) has purely imaginary eigenvalues, or when one of the eigenvalues is zero. For example, consider the system $$ \begin{aligned} &x^{\prime}=-y+x\left(x^{2}+y^{2}\right) \\ &y^{\prime}=x+y\left(x^{2}+y^{2}\right) \end{aligned} $$ Enter the system in pplane6, set the display window so that \(-1 \leq x \leq 1\) and \(-1 \leq y \leq 1\), select Arrows for the direction field, then click Proceed. In the PPLANE6 Display window, select Solutions \(\rightarrow\) Show nullclines to overlay the nullclines on the vector field and note the presence of an equilibrium point at \((0,0)\). a) Use Keyboard input to start a solution trajectory at \((0.5,0)\) and note that the origin behaves as a spiral source. If the solution takes too long to stop on its own use the Stop button. b) Select Solutions \(\rightarrow\) Find an equilibrium point and find the equilibrium point at \((0,0)\). Note that the eigenvalues of the Jacobian are purely imaginary, indicating that the linearization has a center, not a spiral source, at the origin. Display the linearization and draw some solution trajectories.

Short Answer

Expert verified
The origin is a spiral source due to higher-order terms; eigenvalues show a center.

Step by step solution

01

Analyze the System Equations

The given system of differential equations is \(x' = -y + x(x^2 + y^2)\) and \(y' = x + y(x^2 + y^2)\). These are nonlinear equations, and we initially aim to examine its behavior around potential equilibrium points.
02

Identify Equilibrium Points

To find the equilibrium points, we set \(x' = 0\) and \(y' = 0\). This gives us the equations \(-y + x(x^2 + y^2) = 0\) and \(x + y(x^2 + y^2) = 0\). Solving these equations simultaneously, we find the equilibrium point \((0, 0)\).
03

Explore the System Using PPLANE6

Load the system into PPLANE6. Set the display window with bounds \(-1 \leq x \leq 1\) and \(-1 \leq y \leq 1\). Select 'Arrows' for the direction field, then click 'Proceed'. This will allow us to visualize the direction field of the system.
04

Overlay Nullclines

In PPLANE6, go to 'Solutions' \(\rightarrow\) 'Show nullclines'. The nullclines intersect at the equilibrium point \((0, 0)\), confirming its presence visually in the vector field.
05

Investigate Solution Trajectory From Initial Point

Using 'Keyboard Input', begin a solution trajectory starting at the point \((0.5, 0)\). Observe the behavior of the trajectory as it moves away from the origin, which is hypothesized to behave as a spiral source.
06

Check for Eigenvalues of the Jacobian

Find the Jacobian matrix of the system at the equilibrium point \((0, 0)\). The Jacobian at this point would be \(\begin{pmatrix} 0 & -1 \ 1 & 0 \end{pmatrix}\). Calculate the eigenvalues of this matrix, which are \(\pm i\), indicating they are purely imaginary.
07

Understand Linearization Implications

Since the eigenvalues are purely imaginary, the linearization of the system at the equilibrium point forms a center. PPLANE6 might show a different non-linear behavior (spiral source) due to higher-order terms that the linearization does not account for.

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Key Concepts

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

Equilibrium Points
The concept of an equilibrium point in a system of differential equations is fundamental. An equilibrium point is where the system remains at rest if it is not disturbed. For a system described by \( x' = f(x, y) \) and \( y' = g(x, y) \), equilibrium points occur where both \( x' \) and \( y' \) are zero.

In the exercise's system, we set the given equations to zero: \(-y + x(x^2 + y^2) = 0\) and \(x + y(x^2 + y^2) = 0\). Solving these, we find that \((0, 0)\) is an equilibrium point.
  • This point simplifies to both \( x' = 0 \) and \( y' = 0 \), meaning the system is in balance at the origin.
  • Equilibrium points are critical as they tell us where potential sustained behavior exists in the system.
Eigenvalues
Eigenvalues are numbers that provide significant insight into the behavior of dynamical systems near equilibrium points. When analyzing a nonlinear system's stability through linear approximation, eigenvalues of the Jacobian matrix become crucial.

For our system, we compute the Jacobian matrix at the equilibrium point \((0, 0)\). The matrix is\[\begin{pmatrix} 0 & -1 \ 1 & 0 \end{pmatrix}\]The eigenvalues of this matrix are \( \pm i \), which are purely imaginary.
  • When eigenvalues are imaginary, the system is expected to form a center near the equilibrium.
  • This indicates oscillatory behavior, but without linear growth or decay.
Linearization
Linearization is a method to approximate the behavior of a nonlinear system by a linear one around an equilibrium point. It helps in understanding the local stability and dynamic traits of the original system by simplifying the complex dynamics.

In our example, the linearization at the origin based on the system鈥檚 Jacobian \[\begin{pmatrix} 0 & -1 \ 1 & 0 \end{pmatrix}\]produces purely imaginary eigenvalues. Although this suggests the system should locally behave as a center, the true nonlinear system might exhibit more complex behaviors not captured by the simple linear model.

Therefore, while linearization is a useful first step, it does not always depict the complete picture, particularly when eigenvalues are imaginary, suggesting complex solutions.
Direction Field
A direction field, or phase portrait, is a graphical depiction of the trajectory of systems of differential equations without solving them analytically. By plotting vectors across a grid, it illustrates the flow of solutions in the system's phase space.

In our exercise, entering the system into PPLANE6 and displaying the direction field gives a visual representation of how trajectories behave around the equilibrium point.
  • The direction field provides immediate insight into the stability and behavior, such as identifying points of attraction or repulsion.
  • In this system, the direction field evidenced spiral trajectories originating from near the origin, hinting at a possible spiral source behavior.
Spiral Source
The term "spiral source" refers to a type of behavior in differential systems where trajectories spiral outwards from an equilibrium point. This occurs when the real part of an eigenvalue is positive while the imaginary part causes rotation. In our system, though linearization suggested a center, the nonlinear nature caused it to behave more like a spiral source.

During analysis and observation in PPLANE6, when starting trajectories from nearby the equilibrium point \((0.5,0)\), the system showed outward spiraling, indicating repulsion.
  • This indicates non-linear terms are significant enough to affect the trajectory path drastically and illustrate the limits of linear approximation.
  • Understanding spiral sources in systems can help in predicting pattern behaviors such as growth or oscillation in similar physical systems.

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Most popular questions from this chapter

This is another example of how a system can change as a parameter is varied. Consider the system $$ \begin{aligned} &x^{\prime}=a x+y-x\left(x^{2}+y^{2}\right) \\ &y^{\prime}=-x+a y-y\left(x^{2}+y^{2}\right) \end{aligned} $$ for \(a=0\), and \(a=\pm 0.1\). Use the display rectangle \(-1 \leq x \leq 1\), and \(-1 \leq y \leq 1\), and plot enough solutions in each case to describe behavior of the system. Describe what happens as the parameter \(a\) varies from negative values to positive values. (This is an example of a Hopf bifurcation.) Hint: Look for something completely new, something other than an equilibrium point.

In Exercises \(21-30\), perform steps i) and ii) for the linear system $$ \mathbf{x}^{\prime}=\mathbf{A x}, $$ where \(\mathbf{A}\) is the given matrix. i) Find the type of the equilibrium point at the origin. Do this without the aid of any technology or pplane6. You may, of course, check your answer using pplane. ii) Use pplane6 to plot several solution curves - enough to fully illustrate the behavior of the system. You will find it convenient to use the "linear system" choice from the Gallery menu. If the eigenvalues are real, use the "Keyboard input" option to include straight line solutions starting at \(\pm 10\) times the eigenvectors. If the equilibrium point is a saddle point, compute and plot the separatrices. $$ \left[\begin{array}{rr} -1 & 0 \\ 0 & -1 \end{array}\right] $$

In contrast to Exercises \(31-34\), consider the system $$ \begin{aligned} &x^{\prime}=y+a x^{3} \\ &y^{\prime}=-x \end{aligned} $$ for the three values 0,10 and \(-10\) of the parameter \(a\). a) Show that all three systems have the same Jacobian matrix at the origin. What type of equilibrium point at \((0,0)\) is predicted by the eigenvalues of the Jacobian? b) Use pplane6 to find evidence that will enable you to make a conjecture as to the type of the equilibrium point at \((0,0)\) in each of the three cases. c) Consider the function \(h(x, y)=x^{2}+y^{2}\). In each of the three cases, restrict \(h\) to a solution curve and differentiate the result with respect to \(t\) (Recall: \(d h / d t=(\partial h / \partial x)(d x / d t)+(\partial h / \partial y)(d y / d t))\). Can you use the result to verify the conjecture you made in part b)? Hint: Note that \(h(x, y)\) measures the distance between the origin and \((x, y)\). d) Does the Jacobian predict the behavior of the non-linear systems in this case?

In Exercises \(5-7\), find the eigenvalues and eigenvectors with the eig and null commands, as demonstrated in Example 4 of Chapter 12. You may find format rat helpful. Then enter the system into pplane6, and draw the straight line solutions. For example, if one eigenvector happens to be \(\mathbf{v}=[1,-2]^{T}\), use the Keyboard input window to start straight line solutions at \((1,-2)\) and \((-1,2)\). Perform a similar task for the other eigenvector. Finally, the straight line solutions in these exercises divide the phase plane into four regions. Use your mouse to start several solution trajectories in each region. $$ \begin{aligned} &x^{\prime}=6 x-y \\ &y^{\prime}=-3 y \end{aligned} $$

It is a nice exercise to classify linear systems based on their position in the trace-determinant plane. Consider the matrix $$ A=\left[\begin{array}{ll} a & b \\ c & d \end{array}\right] $$ a) Show that the characteristic polynomial of the matrix \(A\) is \(p(\lambda)=\lambda^{2}-T \lambda+D\), where \(T=a+d\) is the trace of \(A\) and \(D=\operatorname{det}(A)=a d-b c\) is the determinant of \(A\). b) We know that the characteristic polynomial factors as \(p(\lambda)=\left(\lambda-\lambda_{1}\right)\left(\lambda-\lambda_{2}\right)\), where \(\lambda_{1}\) and \(\lambda_{2}\) are the eigenvalues. Use this and the result of part (a) to show that the product of the eigenvalues is equal to the determinant of matrix A. Note: This is a useful fact. For example, if the determinant is negative, then you must have one positive and one negative eigenvalue, indicating a saddle equilibrium point. Also, show that the sum of the eigenvalues equals the trace of matrix \(A\). c) Show that the eigenvalues of matrix \(A\) are given by the formula $$ \lambda=\frac{T \pm \sqrt{T^{2}-4 D}}{2} $$ Note that there are three possible scenarios. If \(T^{2}-4 D<0\), then there are two complex eigenvalues. If \(T^{2}-4 D>0\), there are two real eigenvalues. Finally, if \(T^{2}-4 D=0\), then there is one repeated eigenvalue of algebraic multiplicity two. d) Draw a pair of axes on a piece of poster board. Label the vertical axis \(D\) and the horizontal axis \(T\). Sketch the graph of \(T^{2}-4 D=0\) on your poster board. The axes and the parabola defined by \(T^{2}-4 D=0\) divide the trace- determinant plane into six distinct regions, as shown in Figure 13.17. e) You can classify any matrix \(A\) by its location in the trace-determinant plane. For example, if $$ A=\left[\begin{array}{rr} 1 & 2 \\ -3 & 2 \end{array}\right] $$ then \(T=3\) and \(D=8\), so the point \((T, D)\) is located in the first quadrant. Furthermore, \((3)^{2}-4(8)<\) 0 , placing the point \((3,8)\) above the parabola \(T^{2}-4 D=0\). Finally, if you substitute \(T=3\) and \(D=8\) into the formula \(\lambda=\left(T \pm \sqrt{T^{2}-4 D}\right) / 2\), then you get eigenvalues that are complex with a positive real part, making the equilibrium point of the system \(\mathbf{x}^{\prime}=A \mathbf{x}\) a spiral source. Use pplane6 to generate a phase portrait of this particular system and attach the printout to the poster board at the point \((3,8)\). f) Linear systems possess a small number of distinctive phase portraits. Each of these is graphically different from the others, but each corresponds to the pair of eigenvalues and their multiplicities. For each case, use pplane 6 to construct a phase portrait, and attach a printout at its appropriate point \((T, D)\) in your poster board trace-determinant plane. Hint: There are degenerate cases on the axes and the parabola. For example, you can find degenerate cases on the parabola in the first quadrant that separate nodal sources from spiral sources. There are also a number of interesting degenerate cases at the origin of the trace-determinant plane. One final note: We have intentionally used the words "small number of distinctive cases"' so as to spur argument amongst our readers when working on this activity. What do you think is the correct number?
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