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

CHAPTER 2. SUCCESSIVE APPROXIMATIONS 31
Solution.
x = x 4 + 2x 2 + x − 3
0 = x 4 + 2x 2 − 3
= (x − 1)(x + 1)(x 2 + 3)
Hence the real fixed points are x = 1 and x = −1.
A function f : R ↦→ R has a fixed point if and only if its graph intersects with the
line y = x. If there are multiple intersections, then there are multiple fixed points.
Consequently a sufficient condition is that the range of f is contained in its domain.
⇐=
Figure 2.1: A sufficient condition for a bounded continuous function to have a fixed
point is that the range be a subset of the domain. A fixed point occurs whenever
the curve of f(t) intersects the line y = t.
Theorem 2.1 (Sufficient condition for fixed point). Suppose that f(t) is a continuous
function that maps its domain into a subset of itself, i.e.,
Then f(t) has a fixed point in [a, b].
f(t) : [a, b] ↦→ S ⊂ [a, b] (2.17)
Proof. If f(a) = a or f(b) = b then there is a fixed point at either a or b. So assume
that both f(a) ≠ a and f(b) ≠ b. By assumption, f(t) : [a, b] ↦→ S ⊂ [a, b], so that
Since both f(a) ≠ a and f(b) ≠ b, this reduces to
f(a) ≥ a and f(b) ≤ b (2.18)
f(a) > a and f(b) < b (2.19)
Let g(t) = f(t) − t. Then g is continuous because f is continuous. Thus ⇐=
c○2007, B.E.Shapiro
Last revised: May 23, 2007
Math 582B, Spring 2007
California State University Northridge