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 ∈ 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 ∈ 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 |y| 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

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