Thus you have four real coordinates describing a phase curve: (Re[x[t]], Im[x[t]], Re[y[t]], Im[y[t]]} To plot such a curve, it seems to me that you must project it onto a lower dimension. Projecting tends to muddy the phase portrait, since the projected curves might appear to intersect, which they do not do in the actual 4D phase space.

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The constants $C_1$ and $C_2$ only determine your starting point for drawing a curve in a phase portrait. The eigenvalues $r_1$ and $r_2$ are what actually determine the major and minor axes of the ellipse. The resulting curve in phase space is the same for any pair $(C_1,C_2)$ that starts on the same ellipse. If x0= Ax and A has real eigenvalues 1 6= 2, and eigenvectors v 1, v 2, then the general solution is x(t) = C 1e 1tv 1 + C 2e 2tv 2. We can plot thephase portrait(x 2 vs. x 1) by rst drawing the \eigenvector lines". If i >0, then the solutions move away from (0;0) because lim t!1 jCe tvj= 1. If i <0, then the solutions move torward (0;0) because lim t!1

Linear Phase Portraits: Cursor Entry The phase portrait of a homogeneous linear autonomous system depends mainly upon the trace and determinant of the matrix, but there are two further degrees of freedom.

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The signs of the eigenvalues indicate the phase plane's behaviour: If the signs are opposite, the intersection of the eigenvectors is a saddle point. If the signs are both positive, the eigenvectors represent stable situations that the system diverges away from, and the intersection is an unstable node.stable. Below is the phase portrait. ˇ 2 3ˇ 2 ˇ 0 6. (15 points) For the linear system x_ = 5x+ 10y; y_ = x y nd eigenvalues, eigenvectors, and the general solution of the system. Classify the xed point and determine its stability. Sketch the phase portrait. Solution: This is linear homogeneous system and its the only FP is the origin (0;0 ...

Phase Portraits of 2-D Linear Systems with Zero Eigenvalue For each of the following systems, • Find general solutions; • skecth the phase portrait; • determine whether the equilibrium (x,y) = (0,0) is stable or unstable; • determine whether the equilibrium (x,y) = (0,0) is asymptotically stable. [1] x ′= x− 2y, y = 3x− 6y.

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eigenvalues and eigenvectors of the matrix A. Example 2. Find the eigenvalues and eigenvectors of the system in Example 1, and plot the eigenvectors on a graph. Compare with the established trajectory. The matrix []. To find the eigenvalues of the matrix we subtract λ from the diagonal entries and calculate the determinant. | These pictures are often called phase portraits. The system need not be linear. In fact, phase plane portraits are a useful tool for two-dimensional non-linear differential equations as well. In Figure Three we see four examples of phase portraits. The horizontal axis is x1 and the vertical axis is x2. Each curve represents a solution to the ...

Sliders allow manipulation of the matrix entries over. By viewing simultaneously the phase portrait and the eigenvalue plot, one can easily and directly associate phase portrait bifurcations with changes in the character of the eigenvalues.

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The phase portrait can indicate the stability of the system. The phase portrait behavior of a system of ODEs can be determined by the eigenvalues or the trace and determinant (trace = λ 1 + λ 2, determinant = λ 1 x λ 2) of the system.The phase portrait of (1) in this case is exactly the same as Figure 3, except that the direction of the arrows is reversed. Hence, the equilibrium solution x(t) = 0of (1) is an unstablenodeif both eigenvalues of A are positive. EXAMPLE: Draw the phase portrait of the linear equation x˙= Ax= " −2 −1 4 −7 # x (2) Solution: It is easily ... The phase portrait for (3) is obtained from that for (4) by applying linear transformation.>MON (see Figure 1b). A ﬁxed point of a linear system eigenvalues having different signs is called a saddle. The same term is often used to describe the corresponding phase portraits (Figure??1). Next, suppose 7 ( ) #. Then +*. As before, we ﬁrst plot ...

8-Let us now think about the phase portrait for this system. Since the system is 4-dimensional, the phase portrait should also be 4-dimensional. The phase portrait described by the coupled set of equations (Eqns 2) is quite complex. To get a sense of its complexity, use your data from step 7 to make a plot of ! x ú 1 vs ! x1. You will need to ...

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The eigenvalues are a pure imaginary conjugate pair. Hence, the phase portrait is that of the center. The origin is the center of closed orbits that neither spread apart nor approach each other, the essential meaning of orbital stability. Any solution starting near the origin remains near the origin. eigenvalues and eigenvectors of the matrix A. Example 2. Find the eigenvalues and eigenvectors of the system in Example 1, and plot the eigenvectors on a graph. Compare with the established trajectory. The matrix []. To find the eigenvalues of the matrix we subtract λ from the diagonal entries and calculate the determinant. |

A phase portrait is a geometric representation of the trajectories of a dynamical system in the phase plane. Each set of initial conditions is represented by a different curve, or point. Phase portraits are an invaluable tool in studying dynamical systems. They consist of a plot of typical trajectories in the state space.

The EquationTrekker package is a great package for plotting and exploring phase portraits Planar equations In first part of the course , we discussed the direction field for first order differential equations.

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negative trace such that T 2 — 4D > 0. The corresponding phase portraits are real sinks. If a l, we have a sink with repeated eigenvalues. If —1 < a < 0, we have complex eigenvalues with negative real parts. Therefore, the phase portraits are spiral sinks. If a 0, we have a degenerate case with an entire line of equilibrium points.

LINEAR PHASE PORTRAITS: MATRIX ENTRY + help The graphing window at right displays a few trajectories of the linear system x' = Ax. Below the window the name of the phase portrait is displayed. Depress the mousekey over the graphing window to display a trajectory through that point. ... Click the [eigenvalues] key to toggle display of a complex ...

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, and plot each component of the solution, as well as the phase portrait. (g)Can you design a matrix A, such that all solutions go inward to the origin [0;0]T and all the eigenvalues of A are real. (h)Now let us proceed to the case when the eigenvalues are a complex conjugate pair. Let the eigenvalues be l

stable. Below is the phase portrait. ˇ 2 3ˇ 2 ˇ 0 6. (15 points) For the linear system x_ = 5x+ 10y; y_ = x y nd eigenvalues, eigenvectors, and the general solution of the system. Classify the xed point and determine its stability. Sketch the phase portrait. Solution: This is linear homogeneous system and its the only FP is the origin (0;0 ...

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Phase Plane – A brief introduction to the phase plane and phase portraits. Real Eigenvalues – Solving systems of differential equations with real eigenvalues. Complex Eigenvalues – Solving systems of differential equations with complex eigenvalues. Repeated Eigenvalues – Solving systems of differential equations with repeated eigenvalues.

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D > 0: Real and distinct eigenvalues. D < 0: Complex conjugate pair of eigenvalues. D = 0: Double real eigenvalues. Phase portraits The solution to the system depends on the initial condition. Plot solution trajectories, starting at several initial values. Does not give a time-scale, t is not in the plots. We plot the solution trajectories, but ...

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Phase Portraits: Matrix Entry. 26.1. Phase portraits and eigenvectors. It is convenient to rep resen⎩⎪t the solutions of an autonomous system x˙ = f(x) (where x = ) by means of a phase portrait. The x, y plane is called the phase y plane (because a point in it represents the state or phase of a system). The phase portrait is a ...

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The general case is very similar to this example. Indeed, assume that a system has 0 and as eigenvalues. Hence if is an eigenvector associated to 0 and an eigenvector associated to , then the general solution is We have two cases, whether or . If , then is an equilibrium point. The left plot is a temporal representation of the system's development, with time \(t\) being represented on the horizontal axis. The right plot is a phase plane (or phase space or state space) portrait of the system. Method 1: Calculate by hands with phase plane analysis. First, find the eigenvalues of the characteristic equation: $$ \begin{aligned} &\lambda^{2}+1=0\\ &s_{1,2}=\pm i \end{aligned} $$ And we know that with such pole distribution, the phase portrait should look like: phase portrait w.r.t pole distribution

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b. Linearize the system and compute the eigenvalues about all the equilibrium points. c. Classify the types of the equilibrium points on a phase plane and plot the phase portraits of the nonlinear system. Solution: a. Let x 1 (t) = (t) and x 2 (t) = _(t), then the state-space representation is x_ 1 x_ 2 = x 2 cx 2 2sinx 1 + 1 The equilibrium ...

This is because there are additional symmetries (generated by an antiholomorphic involution of the complex phase space) that are not preserved by the holonomy. However, for the elliptic singularities (with the pair of nonreal eigenvalues) the germ of a self-map, called the Poincare monodromy, is well defined.

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40.3 Phase Portraits for Imprecisely Known Systems Note that the basic nature of a phase portrait for a 2×2 constant matrix system x′ = Ax depends strongly on the whether the real and imaginary parts of the eigenvalues of A are positive, negative or zero. This can be a signiﬁcant issue when the matrix A is only approximately known, such as

Sliders allow manipulation of the matrix entries over. By viewing simultaneously the phase portrait and the eigenvalue plot, one can easily and directly associate phase portrait bifurcations with changes in the character of the eigenvalues.

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(b) When a 6= ±1, then D < 0, so the phase portrait is a saddle. If a = −1, then T = −1 and D = 0, so the eigenvalues are 0 and −1; the phase portrait has a line of equilibria and all other solutions are straight lines converging to the line of equilibria. If a = 1, then T = 1 and D = 0, MATH 2700 Syllabus . Text: Blanchard, Devaney, Hall, Differential Equations, third edition . SECTION . NUMBER of 50 minute periods

phase_plot. With transfer function in s-domain. % Phase portrait plot for SECOND and THIRD order ODE % sys is the system transfer function (in s-domain) % % intial_values is ithe initial states of th system (vector of nx1) % where n is the order of the system % % range is the minimum and the maximum boundary for the states % e.g: 3rd order system, with states: x1, x2, and x3 % [x1_min x1_max ...

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The three compartment model for lead in the body considered in Lead is simplified to give a two-compartment (and thus 2x2) system with "nice" eigenvalues and eigenvectors. The solution of the system with initial conditions is graphed and, if desired, shown as a trajectory in the phase plane. Lead2: 3.4 Feb 02, 2005 · It then allows you to find their equilibrium points and plot trajectories as well as a number of other fun things. For example, you can Jacobian linearize a system around and equilibrium point and it will give you the linear phase portrait as well as the eigenvalues of the linearlized system and a set of normalized eigenvectors. terms of the form eλjt where fλjg is the set of eigenvalues of the Jacobian. The eigenvalues of the Jacobian are, in general, complex numbers. Let λj = µj +iνj, where µj and νj are, respectively, the real and imaginary parts of the eigenvalue. Each of the exponential terms in the expansion can therefore be writ-ten eλjt =eµjteiνjt:

(d)Use numerical simulation to plot the phase portrait in the original (x 1;x 2) coordinates and su-perimpose the lines y = 0 and z = 0 on the same plot. Discuss whether the phase portrait is consistent with the properties of the center manifold discussed in class. 2.Strogatz, Problem 3.7.3 (attached). 3.Strogatz, Problem 3.7.4 (attached).

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Eigenvalue leaves through +1 Vertical Bifurcation of Equilibria 0<λ2<λ1<1 0 <λ2 <1=λ1 0 <λ2 <1<λ1 Example = 1 1/2 a 0 A Axum: As a is tuned from 0.9 to 1.1, an eigenvalue leaves the unit circle through +1. λ=1/2, a Phase Portrait: The zero equilibrium changes from a stable node to a saddle. At a = 1, there exists a line of neutrally stable equilibria.