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

CHAPTER 1. CLASSIFYING THE PROBLEM 9
Solution. Finding a “solution” is easy enough. We can separate the variables and
integrate. It is easily verified (by direct substitution) that
(
y = 2 sin t + π )
(1.34)
2
satisfies both the differential equation and the initial condition, hence it is a solution.
It is also easily verified that y = 2 is a solution, as are functions of the form
⎧
⎪⎨ 2 sin ( t + π )
2
t < 0
y = 2 0 ≤ t ≤ φ
⎪⎩
2 sin ( t + π 2 − φ) t > φ
(1.35)
for any positive real number φ. See Figure 1.6.
Figure 1.6: There are several solutions to y ′ = √ 4 − y 2 that pass through the point
(0, 2). See Example 1.4
Since the solution is not unique, any condition that guarantees existence must be
violated. We have two such conditions: the boundedness of the partial derivative,
and the Lipshitz condition. The first implies the second, and the second implies
uniqueness. By
∂f
∂y =
−y
√
4 − y 2
(1.36)
which is unbounded at y = 2. So the first condition is violated. Of course, a violation
of the condition does not ensure non-uniqueness, all it tells us is that uniqueness
is not ensured.
What about the Lipshitz condition? Suppose that the function f(x) = √ (4−y 2 )
is Lipshitz with Lipshitz constant K > 0 on some domain D. Then for any y 1 , y 2 in
c○2007, B.E.Shapiro
Last revised: May 23, 2007
Math 582B, Spring 2007
California State University Northridge