Isoclines and Direction Fields; Existence and Uniqueness MATH 374 Lecture 3 Isoclines and Direction Fields; Existence and Uniqueness

1.5: The Isoclines of an Equation and Direction Fields y Here is a geometric way to solve first order ODE in the form: For each point (a,b) in the domain of f, sketch a short line segment with midpoint (a,b) and slope f(a,b). Since dy/dx is the slope of y at x, (1) says that if the graph of a solution to (1) contains the point (a,b), then the line segment through (a,b) will be tangent to the graph of the solution. The Method of Isoclines is based on this idea! y = (x); (a) = b b x a

Method of Isoclines Sketch several level curves of f(x,y), i.e. f(x,y) = c for different constants c. (Each curve is an isocline.) On the curve f(x,y) = c, put line segments with slope c. In this way we get a good idea of the structure of the direction field of (1), which is a short line segment with slope f(a,b) at each point (a,b) in the domain of f. Any solution to (1) has a graph, which at each point has a tangent line given by the direction field.

Example 1 Use the Method of Isoclines to find the direction field of dy/dx = y –x. Sketch some solutions of this differential equation. Solution: f(x,y) = c  y – x = c  y = x+c. Thus, isoclines are straight lines with slope 1 and y-intercept c. Plot isoclines for c = 2, 1, 0, and -1.

Direction Fields via Technology Software such as Mathematica or dfield can be used to draw direction fields! http://math.rice.edu/~dfield/dfpp.html

1.6: An Existence and Uniqueness Theorem Theorem 1.1 (Existence and Uniqueness): Given the first order ODE let T be the rectangular region defined by T := {(x,y)||x – x0|  a and |y-y0|  b} for a, b > 0. If f and  f/ y are continuous functions of x and y for all (x,y) in T, then there exists an interval about x0, I := {x| |x-x0|  h} for some h 2 (0,a], and a function y(x) with the properties:

1.6: An Existence and Uniqueness Theorem (continued) y = y(x) is a solution of (1) on the interval I. On interval I, y(x) satisfies the inequality |y(x) – y0|  b. At x = x0, y(x0) = y0. y(x) is unique on the interval I in the sense that it is the only function that satisfies a), b), and c).

Picture for Theorem 1.1

Example 1 In the differential equation are continuous functions for all (x,y) in the plane. Therefore the existence and uniqueness Theorem 1.1 tells us that for any point (x0, y0), there exists a unique solution to (2) on some interval about x0, passing through (x0, y0).

Example 2 Where are we sure that has a unique solution? Solution: Since are continuous away from the line y = 0, there exists a unique solution to (3) through any point (x0, y0) with y0  0.