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Vector Fields and Dynamical Systems Jim Lambers MAT 605 Fall Semester 2015-16 Lecture 3 Notes These notes correspond to Section 1.2 in the text. Vector Fields and Dynamical Systems Since a dynamical system prescribes the tangent vector of the solution, it is natural to identify a dynamical system with a vector field, that expresses the tangent vector as a function of the independent and dependent variables of the system. This facilitates the study of the system and its solutions from a geometric perspective. n n Definition 1 (Vector Field) A vector field is a function X : D ⊆ R ! R . For each x 2 D, we write X(x) = X1(x);X2(x);:::;Xn(x) : The functions Xi, i = 1; 2; : : : ; n, are called the component functions of X. 3 4 Example 1 The vector field X(x1; x2) = (x1x2; x1 + x1x2) corresponds to the dynamical system 0 x1 = x1x2 0 3 4 x2 = x1 + x1x2: 1 2 3 4 The component functions of X are X (x1; x2) = x1x2 and X (x1; x2) = x1 + x1x2. 2 Now, we can state a more precise definition of a solution of a dynamical system. n Definition 2 (Solutions of Autonomous Systems) A curve in R is a function n r : I ! R , where I ⊆ R is an interval. If r is differentiable, we say that r is a differentiable curve. n n Let X : D ⊆ R ! R be a vector field. A solution of the autonomous dynami- cal system x0 = X(x); also known as a solution curve, integral curve (of X) or streamline, is a differen- tiable curve r : I ! D such that r0(t) = X(r(t)); t 2 I: Vector fields and their associated integral curves have physical interpretations in various appli- cations. When a dynamical system x0 = X(x) describes the flow of a fluid, the vector field X is called the velocity field of the fluid, and the integral curves are more often referred to as streamlines or flow lines. If X is a force field, such as for electrostatic force or gravitational force, then the integral curves represent force field lines. 1 Example 2 Consider the dynamical system 1 x0 = (x + x ) (1) 1 2 1 2 1 x0 = − x : (2) 2 2 2 This system corresponds to the vector field 1 1 X(x ; x ) = (x + x ); − x : 1 2 2 1 2 2 2 This system can be solved analytically. From the second equation, we obtain −t=2 x2(t) = b2e ; where b2 is an arbitrary constant. We then solve the inhomogeneous equation 1 1 x0 − x = b e−t=2: 1 2 1 2 2 Using the integrating factor µ(t) = e−t=2, we obtain 1 x (t) = b et=2 − b e−t=2; 1 1 2 2 where b1 is also an arbitrary constant. In some cases, the solution curves can be easier to understand if they can be expressed in non-parametric form. To eliminate the parameter, we solve for t in terms of x2 to obtain t = −2 ln(x2=b2). Substituting this into x1 yields 1 x = b b =x − x ; 1 1 2 2 2 2 which can be rearranged to obtain 2 2x1x2 + x2 = 2b1b2; which can be rewritten as 2 2 (x1 + x2) − x1 = 2b1b2: It can be seen from this equation that the solution curves are hyperbolas, which can be seen in a plot of X, which is shown in Figure 1. It can also be seen from the plot of X that there are straight-line solutions. First, if b2 = 0, then x2 = 0, which corresponds to the solution curve on the x-axis. On the other hand, if b1 = 0; b2 6= 0, then we have 2x1 + x2 = 0, which corresponds to the straight-line solution y = −2x. These straight-line solutions are highlighted in Figure 1. 2 Note that we visualize a two-dimensional vector field X by using x as the initial point of the vector X(x), rather than the origin as is usually done when plotting vectors. That is, we treat x as a position vector, and X(x) as a velocity vector. In plots of vector fields, X(x) is normalized and scaled for visual clarity. Example 3 We revisit the dynamical system x0 = sinh y y0 = − sin x that has fixed points at (kπ; 0) for each integer k. A plot of the associated vector field X(x; y) = (sinh y; − sin x) is shown in Figure 2. We see that near the fixed points for even k, the solution curves are orbital, whereas for jyj sufficiently large, they approach horizontal lines. 2 2 Figure 1: Velocity field plot for the system (1), (2), with straight-line solutions shown in red Figure 2: Plot of vector field X(x; y) = (sinh y; − sin x) Exercises Section 1.2: Exercises 1b, 3 3.
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