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Math 441 (C13) – Fall, 2008
Prof. Ward Henson
The Proof of Theorem 2.8.
The main point of Section 2.8 is to prove the existence and uniqueness theorem for first-order
differential equations, as stated in Theorem 2.8.1 on page 110. (In turn, this implies Theorem 2.4.
by a translation argument.)
The main purpose of this note is to provide the details of the proof that are not given explicitly in
the text; rather they are formulated as Problems 15–19 at the end of Section 2.8.
- Notation.
- a, b are real numbers > 0.
- R is the closed rectangle given by R = {(t, y) ∈ R
2 | |t| ≤ a and |y| ≤ b}.
- f (t, y) is a real-valued function defined on R.
- M, K are real numbers > 0 satisfying |f (t, y)| ≤ M and |
∂f ∂y (t, y)| ≤^ K^ for all (t, y)^ ∈^ R.
- Assumption. The functions f and
∂f ∂y
are continuous on R.
- Remark. Recall (from third semester calculus) that a continuous function on a closed rectangle
in R
2 is bounded. Therefore, Assumption 2 ensures that the numbers M and K exist.
The initial value problem we consider is:
(IVP) y
′ = f (t, y) y(0) = 0 on the interval [−h, h].
The result we will prove is the following:
- Theorem 2.8.1. There exists a real number h with 0 < h ≤ a such that for this value of h, the
problem (IVP) has a unique solution.
- Notation. We let h denote a real number (to be specified later) satisfying 0 < h ≤ a.
- Remark. For our proof of the uniqueness part of Theorem 2.8.1, it turns out that we can take
h to be any number satisfying 0 < h ≤ a. For our proof of the existence part, we can take h to be
any number that also satisfies h ≤ b/M. Therefore, the biggest value of h for which both parts of
our proof work is h = min(a, b/M ).
Passing from IVP to an equivalent integral equation
A key idea for our proof of Theorem 2.8.1 is to turn (IVP) into an equivalent integral equation:
- Proposition. Let y = ϕ(t) be a real valued function defined on the interval [−h, h]. Then
y = ϕ(t) is a solution to (IVP) if and only if it satisfies the following two conditions:
ϕ(t) is continuous on [−h, h]
ϕ(t) =
∫ (^) t
0
f (s, ϕ(s)) ds for all t ∈ [−h, h]
Proof. This is explained in the text in the two full paragraphs following the statement of Theorem
2.8.1 (pages 110-111).
A few basic lemmas from elementary analysis
- Lemma. If t, y 1 , y 2 ∈ R satisfy |t| ≤ a, |y 1 | ≤ b and |y 2 | ≤ b, then we have
∣ ∣f (t, y 1 )^ −^ f^ (t, y 2 )
∣ (^) ≤ K|y 1 −^ y 2 |
Proof. Let t, y 1 , y 2 be fixed as in the statement of the Lemma. By the Mean Value Theorem there
exists a real number z between y 1 and y 2 such that
f (t, y 1 ) − f (t, y 2 ) =
∂f
∂y
(t, z)(y 1 − y 2 ).
Therefore ∣ ∣f (t, y 1 )^ −^ f^ (t, y 2 )
∂f
∂y
(t, z)
∣|y 1 −^ y 2 | ≤^ K|y 1 −^ y 2 |
as claimed.
- Lemma. Let c, d be real numbers and let g 1 , g 2 be real-valued functions that are defined and
continuous on an interval containing c and d. Assume that |g 1 (s)| ≤ |g 2 (s)| holds for all s between
c and d. Then ∣ ∣ ∣
∫ (^) d
c
g 1 (s) ds
∫ (^) d
c
|g 2 (s)| ds
In particular, this holds if g 1 and g 2 are the same function.
Proof. This is a basic property of the Riemann integral; see any calculus book. We need the
absolute values outside the integral on the right side of the inequality in case d < c.
Finally, we will need the following consequences of Taylor’s Theorem for the exponential function.
- Taylor’s Theorem for e
x
. For any real numbers c > 0 and x ∈ [−c, c] and any integer n ≥ 0
∣ ∣ ∣e
x −
n ∑
j=
x
j
j!
∣ ≤^ e
c |x|
n+
(n + 1)!
An elementary argument shows that limn→∞
|x|n+ (n+1)!
= 0. Therefore Taylor’s Theorem for e
x yields
the familiar identity
e
x
∞ ∑
j=
xj
j!
for all x ∈ R. Furthermore,
∣ ∣ ∣
∑^ ∞
j=n+
x
j
j!
∑^ ∞
j=n+
|x|
j
j!
∣e
|x| −
∑^ n
j=
|x|
j
j!
∣ ≤^ e
c |x|
n+
(n + 1)!
≤ e
c c
n+
(n + 1)!
for all x ∈ [−c, c].
The proof of uniqueness
The key idea in our proof of the uniqueness part of Theorem 2.8.1 is given in the following Lemma:
- Lemma. Suppose A ≥ 0. Let U (t) be a differentiable function on the interval [0, h] such that
(a) U (0) = 0; (b) U (t) ≥ 0 for all t ∈ [0, h]; and (c) U
′ (t) ≤ A · U (t) for all t ∈ [0, h]. Then
U (t) = 0 for all t ∈ [0, h].
For t ∈ [0, h] we now let
U
− (t) =
0
−t
∣ϕ 1 (s)^ −^ ϕ 2 (s)
∣ (^) ds = −
−t
0
∣ϕ 1 (s)^ −^ ϕ 2 (s)
∣ (^) ds.
Note that U
− (0) = 0 and that U
− (t) ≥ 0 for all t ∈ [0, h]. Furthermore, using the Fundamental
Theorem of Calculus, the chain rule, and inequality (I) we see that
(E2)
dU
−
dt
(t) =
∣ϕ 1 (−t)^ −^ ϕ 2 (−t)
∣ (^) ≤ KU −(t)
for all t ∈ [0, h]. Again using Lemma 11 we conclude that U
− is identically 0 on [0, h]. It follows
using (E2) that ϕ 1 and ϕ 2 are identical on [−h, 0].
The proof of existence
From here to the end of this handout we prove the existence part of Theorem 2.8.1; for this argument
we fix h = min(a, b/M ).
As described in the text, we produce our solution to (IVP) as the limit of a sequence (ϕn(t) | n ≥ 0)
of functions that are defined for t ∈ [−h, h]. These functions are defined by induction on n. We
must prove that the sequence converges for all t ∈ [−h, h]; for each t we will denote this limit by
ϕ(t). Then we must prove that the limit function satisfies (IVP), which we do using Proposition
7; that is, we prove that ϕ(t) is continuous on [−h, h] and that it satisfies the integral formula in
Proposition 7. In order to do these things we need to have very good information about the rate
at which the sequence (ϕn(t) | n ≥ 0) converges to ϕ(t).
Our first step justifies the induction by which the functions ϕn(t) are defined:
- Lemma. Let χ(t) be a continuous function that maps [−h, h] into [−b, b]. Define ψ : [−h, h] → R
by
ψ(t) =
∫ (^) t
0
f (s, χ(s)) ds.
Then ψ(t) ∈ [−b, b] for all t ∈ [−h, h].
Proof. Our assumption on χ ensures that the definition of ψ makes sense, since it im-
plies that f (s, χ(s)) is defined and continuous for s ∈ [−h, h]. Using Lemma 9 we see ∣ ∣ψ(t)
∫ (^) t
0 f^ (s, χ(s))^ ds
∫ (^) t
0 M ds
∣ = M |t| ≤ M h = b
as claimed.
As in the text, we define ϕn(t) for all t ∈ [−h, h] and n ≥ 0 by induction on n, as follows:
ϕ 0 (t) = 0
ϕn+1(t) =
t
0
f (s, ϕn(s)) ds.
By induction on n we may prove, using Lemma 13, that each ϕn is defined and differentiable (hence
continuous) on [−h, h] and takes its values in [−b, b].
- Lemma. For t ∈ [−h, h] and n ≥ 0 we have
(Estimate 1)
∣ϕ n+1(t)^ −^ ϕn(t)
M
K
[
(K|t|)
n+
(n + 1)!
]
Proof. The proof is by induction on n. For the base case we have n = 0 and we see
∣ϕ 1 (t)^ −^ ϕ 0 (t)
∫ (^) t
0
f (t, 0) ds
∫ (^) t
0
M ds
∣ = M |t|.
For the induction step we assume that Estimate 1 holds for n and prove it for n + 1. This is done
as follows:
∣ ∣ϕ n+2 −^ ϕn+
∫ (^) t
0
f (s, ϕn+1(s)) − f (s, ϕn(s))
ds
∫ (^) t
0
K
∣ϕ n+1(s)^ −^ ϕn(t)
∣ (^) ds
∫ (^) t
0
K ·
M
K
[
(K|s|)
n+
(n + 1)!
]
ds
M
K
[
(K|t|)
n+
(n + 2)!
]
as claimed.
As discussed above, the series
∑∞
n=
(Kh)
n+
(n + 1)!
converges to e
Kh − 1. Using the comparison test and the absolute convergence test, this together
with Estimate 1 implies that the series
∞ ∑
n=
(ϕn+1(t) − ϕn(t))
converges for all t ∈ [−h, h]. Moreover, since ϕ 0 is identically 0, we have
N∑ − 1
n=
(ϕn+1(t) − ϕn(t)) = ϕN (t)
for all t ∈ [−h, h]. Therefore we may make the following definition of a function ϕ : [−h, h] → R.
The rest of this handout is devoted to showing that this function is a solution to (IVP).
- Definition. For t ∈ [−h, h] we define
ϕ(t) =
∑^ ∞
n=
(ϕn+1(t) − ϕn(t)) = lim n→∞
ϕn(t).
- Lemma. For t ∈ [−h, h] and n ≥ 0 we have
(Estimate 2)
∣ϕ(t) − ϕ n(t)
M e
Kh
K
[
(Kh)
n+
(n + 1)!
]
