Problem 1
In Exercises \(1-4\), without the aid of technology, using only your algebra skills, sketch the nullclines and find the equilibrium point(s) of the assigned system. Indicate the flow of the the vector field along each nullcline, similar to that shown in Figure 13.1. Check your result with pplane6. If the Symbolic Toolbox is available, use the solve command to find the equilibrium point(s). $$ \begin{aligned} &x^{\prime}=x+2 y \\ &y^{\prime}=2 x-y \end{aligned} $$
Problem 2
In Exercises \(1-4\), without the aid of technology, using only your algebra skills, sketch the nullclines and find the equilibrium point(s) of the assigned system. Indicate the flow of the the vector field along each nullcline, similar to that shown in Figure 13.1. Check your result with pplane6. If the Symbolic Toolbox is available, use the solve command to find the equilibrium point(s). $$ \begin{aligned} &x^{\prime}=x+2 y-4 \\ &y^{\prime}=2 x-y-2 \end{aligned} $$
Problem 3
In Exercises \(1-4\), without the aid of technology, using only your algebra skills, sketch the nullclines and find the equilibrium point(s) of the assigned system. Indicate the flow of the the vector field along each nullcline, similar to that shown in Figure 13.1. Check your result with pplane6. If the Symbolic Toolbox is available, use the solve command to find the equilibrium point(s). $$ \begin{aligned} &x^{\prime}=2 x-y+3 \\ &y^{\prime}=y-x^{2} \end{aligned} $$
Problem 4
In Exercises \(1-4\), without the aid of technology, using only your algebra skills, sketch the nullclines and find the equilibrium point(s) of the assigned system. Indicate the flow of the the vector field along each nullcline, similar to that shown in Figure 13.1. Check your result with pplane6. If the Symbolic Toolbox is available, use the solve command to find the equilibrium point(s). $$ \begin{aligned} &x^{\prime}=2 x+y \\ &y^{\prime}=4 x+2 y \end{aligned} $$
Problem 6
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. $$ \begin{aligned} &x^{\prime}=6 x-y \\ &y^{\prime}=-3 y \end{aligned} $$
Problem 8
What condition on \(\lambda\) will ensure that the straight line solution \(\mathbf{x}(t)=e^{\lambda t} \mathbf{v}\) moves toward the equilibrium point at the origin as time increases? What condition ensures that the straight line solution will move away from the equilibrium point as time increases?
Problem 9
If the eigenvalues and eigenvectors of a planar, autonomous, linear system are complex, then there are no straight line solutions. Use \([\mathrm{v}, \mathrm{e}]=\mathrm{eig}(\mathrm{A})\) to demonstrate that the eigenvalues and eigenvectors of the system \(x^{\prime}=2.9 x+2.0 y, y^{\prime}=-5.0 x-3.1 y\) are complex, then use pplane6 to show that solution trajectories spiral in the phase plane.
Problem 11
If a system has two distinct negative eigenvalues, then both straight line solutions will decay to the origin with the passage of time. Consequently, all solutions will decay to the origin. Enter the system, \(x^{\prime}=-4 x+y, y^{\prime}=-2 x-y\), in pplane6 and plot the straight line solutions. Plot several more solutions and note that they also decay to the origin. Select Solutions \(\rightarrow\) Find an equilibrium point, find the equilibrium point at the origin, then read its classification from the PPLANE6 Equilibrium point data window.
Problem 13
If a system has one negative and one positive eigenvalue, then one straight line solution moves toward the origin and the other moves away. Consequently, general solutions (being linear combinations of straight line solutions) must do the same thing. Enter the system \(x^{\prime}=9 x-14 y, y^{\prime}=7 x-12 y\), in pplane6 and plot the straight line solutions. Plot several more solutions and note that they move toward the origin only to move away at the last moment. Select Solutions \(\rightarrow\) Find an equilibrium point, find the equilibrium point at the origin, then read its classification from the PPLANE6 Equilibrium point data window.
Problem 17
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?
