The Computable Differential Equation Lecture ... - Bruce E. Shapiro

CHAPTER 1. CLASSIFYING THE PROBLEM 5
is called an initial value problem. The constraint 1.18 is called an initial condition.
Figure 1.2: Illustration of a differential equation as a dynamical system. Given any
starting point an object moves as described by the differential equation. The curve
on the left shows the coordinates of an object at several time points. On the right,
the points are annotated with the direction of motion, an arrow whose direction is
specified by the components of the differential equation.
Example 1.2. Solve the initial value problem y ′ = (3 − y)/2, y(2π) = 4.
Solution. We can rearrange variables as
and integrate to obtain
Substituting the initial condition gives
which gives
2dy
3 − y
= dt (1.19)
−t = 2 + ln |y − 3| + C (1.20)
C = −2π − ln |4 − 3| = −2π (1.21)
|y − 3| = e π−t/2 (1.22)
Thus either y = 3 + e π−t/2 or y = 3 − e π−t/2 . At the initial condition, however,
y(2π) = 4, which is only obtained with the plus sign in the solution. Hence
y = 3 + e π−t/2 (1.23)
is the unique solution. The solution of the initial value problem is plotted in figure
1.3 in comparison with the one-parameter family.
We will say that an initial value problem is well posed if it meets the following
criteria:
• A solution exists;
c○2007, B.E.Shapiro
Last revised: May 23, 2007
Math 582B, Spring 2007
California State University Northridge