Lecture Notes in Differential Equations - Bruce E. Shapiro

224 LESSON 26. GENERAL EXISTENCE THEORY*
Hence
‖T u − u‖ ≤ ‖T u − v n+1 ‖ + ‖v n+1 − u‖ (26.42)
= ‖T u − T v n ‖ + ‖u − v n+1 ‖ (26.43)
≤ K‖u − v n ‖ + ‖u − v n+1 ‖
(because T is a contraction)
(26.44)
≤ ‖u − v n ‖ + ‖u − v n+1 ‖ (because K < 1)
(26.45)
= 2‖u − v n ‖ (26.46)
< δ (26.47)
This is a contradiction. Hence u must be a fixed point of T .
To prove uniqueness, suppose that there is another fixed point w ≠ u.
Then ‖w − u‖ > 0 (otherwise they are equal). But
‖u − w‖ = ‖T u − T w‖ ≤ K‖u − w‖ < ‖u − w‖ (26.48)
which is impossible and hence and contradiction.
Thus u is the unique fixed point of T .
Theorem 26.11. Fundamental Existence Theorem. Let D ∈ R 2 be
convex and suppose that f is continuously differentiable on D. Then the
initial value problem
y ′ = f(t, y), y(t 0 ) = y 0 (26.49)
has a unique solution φ(t) in the sense that φ ′ (t) = f(t, φ(y)), φ(t 0 ) = y 0 .
Proof. We begin by observing that φ is a solution of equation 26.49 if and
only if it is a solution of
φ(t) = y 0 +
Our goal will be to prove 26.50.
∫ t
t 0
f(x, φ(x))dx (26.50)
Let V be the set of all continuous integrable functions on an interval (a, b)
that contains t 0 . Then V is a complete normed vector space with the supnorm
as norm, as we have already seen. mn
Define the linear operator T on V by
T (φ) = y 0 +
∫ t
t 0
f(s, φ(s))ds (26.51)