1. Quizz 8
Let C[0, 1] be the space of real valued continuous functions on [0, 1]. Since [0, 1] is closed and bounded in R, it is sequentially compact in R. We have shown in class that C[0, 1] is a closed subspace of B[0, 1], the space of bounded real valued functions on [0, 1]. (In class, we prove a more general case.) Hence C[0, 1] with the supnorm k · k∞ is a real Banach space.
In this exercise, let us try to solve the the initial value problem dx
dt = f (t, x), x(0) = x0
via Picard iteration method/Banach contraction mapping principle.
(1) Solve for the initial value problem
x0(t) = tx(t), x(0) = 1
using following steps. In this case, f (t, x) = tx for t ∈ [0, 1] and x ∈ R (a) For any h ∈ C[0, 1], define
T (h)(t) = 1 + Z t
0
sh(s)ds, t ∈ [0, 1].
Prove that T (h) ∈ C[0, 1].
Proof. Let g : [0, 1] → R be the function defined by g(t) = th(t) for 0 ≤ t ≤ 1.
Then g is continuous. Let G : [0, 1] → R be the function defined by G(t) =
Z t 0
g(s)ds, t ∈ [0, 1].
Since g ∈ C[0, 1], G is continuously differentiable on [0, 1] by the Fundamental Theorem of calculus. Since G is continuous on [0, 1], T (h)(t) = 1 + G(t) is also
continuous on [0, 1].
(b) Define a map T : C[0, 1] → C[0, 1] by h 7→ T (h) where T (h) is defined above.
Prove that T is a contraction mapping.
Proof. Let h1, h2∈ C[0, 1]. For any t ∈ [0, 1], T (h1)(t) = 1 +
Z t
0
sh1(s)ds, T (h2)(t) = 1 +
Z t 0
sh2(s)ds.
For any t ∈ [0, 1]
T (h1)(t) − T (h2)(t) = Z t
0
s(h1(s) − h2(s))ds.
Hence for any t ∈ [0, 1],
|T (h1)(t) − T (h2)(t)| ≤ Z t
0
s|h1(s) − h2(s)|ds ≤ kh1− h2k∞ Z t
0
sds
= kh1− h2k∞t2 2 ≤ 1
2kh1− h2k∞. Therefore, we see that
kT (h1) − T (h2)k∞≤ 1
2kh1− h2k∞.
1
2
This proves that T is a contraction mapping.
(c) Define x0(t) = 1 for all t ∈ [0, 1] and
xn(t) = T (xn−1)(t), t ∈ [0, 1].
Find xn for all n ≥ 1.
Proof. We see that x1(t) = 1 +
Z t 0
sds = 1 + t2
2, t ∈ [0, 1].
Inductively, one can see that xn(t) =
n
X
k=0
1 k!
t2 2
k
, t ∈ [0, 1].
(d) Prove that (xn) forms a Cauchy sequence in (C[0, 1], k · k∞).
Proof. By the property of T, for any n ≥ 1, kxn+1− xnk∞= kT (xn) − T (xn−1)k∞≤ 1
2kxn− xn−1k∞. By induction, for any n ≥ 1,
kxn+1− xnk∞≤ 1
2nkx1− x0k∞.
From here, we leave it to the reader to verify that (xn) is a Cauchy sequence in
(C[0, 1], k · k∞).
(e) Solve for x via (xn).
Proof. By definition, x = limn→∞xn. Hence for any t ∈ [0, 1], x(t) =
∞
X
n=0
1 n!
t2 2
n
= et22 .
(2) Solve for the initial value problem:
x0(t) = 2t(1 + x(t)), x(0) = 0.
In this case, f (t, x) = 2t(1 + x).
(a) Find b > 0 such that the map
T : C[0, b] → C[0, b], h 7→ T (h) is a contraction mapping, where
T (h)(t) = Z t
0
2s(1 + h(s))ds (b) Define x0(t) = 0 and xn(t) by
xn= T (xn−1).
Find (xn).
(c) Solve for the initial value problem by (xn).
3
(3) Let f : R2 → R be a continuous function. Suppose g, h : [a, b] → R are continuous.
Define k : [a, b] → R by
k(x) = f (g(x), h(x)), x ∈ [a, b].
Prove that k is continuous on [a, b].
