Lecture Notes in Differential Equations - Bruce E. Shapiro

225
for any φ ∈ V.
Let g, h be functions in V.
‖T g − T h‖ ∞ = sup
a≤t≤b
|T g − T h| (26.52)
∫ t
∫ t
= sup
∣ y 0 + f(x, g(x))dx − y 0 − f(x, h(x))dx
∣
a≤t≤b t 0 t 0
(26.53)
∫ t
= sup
∣ [f(x, g(x)) − f(x, h(x))] dx
∣ (26.54)
a≤t≤b t 0
Since f is continuously differentiable it is differentiable and its derivative
is continuous. Thus the derivative is bounded (otherwise it could not be
continuous on all of (a, b)). Therefore by theorem 12.3, it is Lipshitz in its
second argument. Consequently there is some K ∈ R such that
∫ t
‖T g − T h‖ ∞ ≤ L sup |g(x) − h(x)| dx (26.55)
a≤t≤b t 0
≤ K(t − t 0 ) sup |g(x)) − h(x)| (26.56)
a≤t≤b
≤ K(b − a) sup |g(x)) − h(x)| (26.57)
a≤t≤b
≤ K(b − a) ‖g − h‖ (26.58)
Since K is fixed, so long as the interval (a, b) is larger than 1/K we have
where
‖T g − T h‖ ∞ ≤ K ′ ‖g − h‖ ∞ (26.59)
K ′ = K(b − a) < 1 (26.60)
Thus T is a contraction. By the contraction mapping theorem it has a fixed
point; call this point φ. Equation 26.50 follows immediately.
This theorem means that higher order initial value problems also have
unique solutions. Why is this? Because any higher order differential equation
can be converted to a system of equations, as in the following example.
Example 26.2. Convert initial value problem
⎫
y ′′ + 4t 3 y ′ + y 3 = sin(t) ⎪⎬
y(0) = 1
⎪⎭
y ′ (0) = 1
(26.61)