Lecture Notes for Math 524
Dr. Michael Y. Li
September 18, 2009
These notes are based on the lecture notes of Professor James S. Muldowney, the books of Hale,
Copple, Coddington and Levinson, and Perko. They are for the use of students in my graduate
ODE class.
Chapter 1. Existence and Uniqueness Theorems
1.1. Initial value problems
Let D be an open set in Rn+1 , and f ∈ C(D → Rn ). Consider an ODE in Rn
x0 = f (t, x).
(1)
A solution to (1) in an interval I ⊂ R is a differentiable function ϕ : I → Rn such that
ϕ0 (t) = f (t, ϕ(t)).
Example 1. The equation x0 = 1 has solutions defined in R of form
x(t) = t + c,
where c is an arbitrary constant. In general, an ODE has infinitely many solutions.
We would like to solve equation (1) subject to conditions
x(t0 ) = x0
(2)
for (t0 , x0 ) ∈ D. Equations (1) and (2) are called an Initial Value Problem, an IVP for short.
In this chapter, we investigate the following questions:
1. Existence of solutions to the IVP (1)-(2).
2. Uniqueness of solutions to the IVP (1)-(2).
3. Maximal interval of existence of a solution.
4. Continuous dependence of solutions on initial conditions and system parameters.
1
We first examine some examples.
Example 2. The IVP x0 = x, x(t0 ) = x0 has solution
x(t) = x0 e(t−t0 )
which exists for all t ∈ R and is unique.
Example 3. The ODE x0 = x2 has a solution
x(t) = −
1
,
t+c
where c can be uniquely determined by the initial condition x(t0 ) = x0 , for x0 6= 0. This solution
only exists near t0 and is unique as long as it exists. When x0 = 0 the IVP has a unique solution
x(t) = 0.
Example 4. Consider the IVP x0 = |x|1/2 , x(t0 ) = x0 . Since x0 (t) ≥ 0, x(t) is increasing.
For x(t) > 0, we have
x0
= 1, 2x1/2 = t + c,
x1/2
1/2
where c = 2x0
− t0 . A solution is given by
(
1
(t + c)2 ,
x(t) = 4 1
− 4 (t + c)2 ,
t ≥ −c,
t < −c.
This solution exists in R. When x0 = 0, we have x(t) = 0 for all t. Solutions to the IVP are not
1/2
unique over R . For instance, if x0 > 0 and c = 2x0 − t0 , you may check that the function
(
1
(t + c)2 ,
t ≥ −c,
x(t) = 4
0,
t < −c,
is another solution with the same initial condition x(t0 ) = x0 . However a solution is unique in a
nhood of (t0 , x0 ) if x0 6= 0.
1.2. An equivalent problem
The IVP (1)-(2) is equivalent to the following integral equation
Z
t
x(t) = x0 +
f (s, x(s)) ds.
t0
Namely, the IVP and the integral equation has the same set of solutions.
The following useful result is a simple version of Gronwall’s lemma.
2
(3)
Lemma 1.1 Let g(t) be a continuous real-valued function and g(t) ≥ 0 for t ∈ [t0 , b]. If, for
K > 0, C ≥ 0,
Z t
g(s) ds
g(t) ≤ C + K
t0
for t ∈ [t0 , b], then
g(t) ≤ CeK(t−t0 )
(4)
for t ∈ [t0 , b].
Proof. Let G(t) = C + K
Rt
t0
g(s)ds. Then G(t) ≥ g(t) and G(t) ≥ 0 for all t ∈ [t0 , b]. Moreover,
G0 (t) = Kg(t) ≤ KG(t).
Thus
G0 (t) − KG(t) ≤ 0
and
G0 (t)e−Kt − G(t)Ke−Kt ≤ 0,
namely
[G(t)e−Kt ]0 ≤ 0,
and thus G(t)e−Kt ≤ G(t0 )e−Kt0 = Ce−Kt0 , it follows g(t) ≤ G(t) ≤ CeK(t−t0 ) .
Corollary 1.2 If g(t) ≥ 0, K > 0 satisfy
Z
t
g(t) ≤ K
g(s)ds,
t ≥ t0 .
t0
Then g(t) = 0 for t ≥ t0 .
1.3. Uniqueness under Lipschitz conditions
A function f ∈ C(D → Rn ) is said to satisfy Lipschitz condition with respect to x if there exists
(Lipschitz) constant K > 0, such that
|f (t, x1 ) − f (t, x2 )| ≤ K |x1 − x2 |
(5)
for all (t, x1 ), (t, x2 ) ∈ D. Here, | · | denote any norm in Rn .
Example 5. f (x) = sin x is Lipschitz continuous on (−∞, +∞).
Example 6. f (x) = x2 is not globally Lipschitz continuous over R. It is, however, Lipschitz
continuous over any compact subset of R. In this case, we say it is locally Lipschitz continuous
over R.
Example 7. f (x) = x12 is globally Lipschitz continuous on [α, ∞) for any α > 0, but only
locally Lipschitz continuous on (0, ∞).
Example 8. If f (x) is C 1 in an open subset U if Rn , then f is locally Lipschitz in U.
3
Theorem 1.3 (Uniqueness) Suppose
(1) f ∈ C(D → Rn ).
(2) f satisfies Lipschitz condition in D with respect to x.
Then the IVP has at most one solution in D.
More specifically, if there are two solutions ϕ1 (t) and ϕ2 (t) to the IVP, both defined for
t ∈ [a, b], such that (t, ϕi (t)) ∈ D for t ∈ [a, b] and i = 1, 2, then ϕ1 (t) = ϕ2 (t) for all t ∈ [a, b].
Proof. (Proof # 1) Suppose ϕ1 (t) and ϕ2 (t) are both solutions to the IVP. Then
Z t
f (s, ϕi (s)) ds, i = 1, 2.
ϕi (t) = x0 +
t0
and thus, if t ≥ t0 ,
Rt
|ϕ1 (t) − ϕ2 (t)| = | t0 [f (s, ϕ1 (s)) − f (s, ϕ2 (s))] ds|
Rt
≤ t0 |f (s, ϕ1 (s)) − f (s, ϕ2 (s))| ds (t ≥ t0 )
Rt
≤ t0 K |ϕ1 (s) − ϕ2 (s)| ds (Lipschitz condition)
(6)
Set l(t) = |ϕ1 (t) − ϕ2 (t)| ≥ 0. Then, if t ≥ t0 ,
Z t
l(t) ≤
Kl(s) ds,
t0
which implies l(t) = 0 for t ≥ t0 by Corollary 1.2. A similar argument can show l(t) = 0 for all
t ≤ t0 .
Proof. (Proof # 2) Set δ = maxt∈[t0 ,b] |ϕ1 (t) − ϕ2 (t)|. From (6), we have
Z t
|ϕ1 (t) − ϕ2 (t)| ≤
Kδ ds = Kδ(t − t0 ).
t0
Using (6) again, we get
Z
t
|ϕ1 (t) − ϕ2 (t)| ≤
K 2 δ(s − t0 ) ds = δK 2 (t − t0 )2 /2.
t0
Repeat the process, we get, for any integer n,
|ϕ1 (t) − ϕ2 (t)| ≤ δK n (t − t0 )n /n! → 0
as n → ∞.
This implies ϕ1 (t) = ϕ2 (t) for t ∈ [t0 , b]. (Proof for t ∈ [a, t0 ] is similar.)
Questions:
(1) Can we weaken the global Lipschitz assumption in the theorem to local Lipschitz?
(2) Can you explain the observations in Example 4 in the context of the uniqueness theorem?
Does Example 7 contradict the theorem?
4
1.4. Local existence under Lipschitz conditions: Picard’s Method of Successive
Approximation
We prove the following result.
Theorem 1.4 (Picard Local Existence Theorem) Suppose
(1) f ∈ C(D → Rn ).
(2) f satisfies Lipschitz condition in D with respect to x.
Then there exists α > 0 such that the IVP has a solution for t ∈ [t0 − α, t0 + α].
Proof. Let (t0 , x0 ) ∈ D. Then there exists a1 , a2 > 0 such that the closed rectangle R = {(t, x) :
|t − t0 | ≤ a1 , |x − x0 | ≤ a2 } ⊂ D ( D is open). Let
M = max{|f (t, x)|
:
(t, x) ∈ R}.
a2
We will show that α = min{a1 , M
} satisfies the conclusion in the theorem. The following method
is called Picard Method of Successive Approximation.
Consider the equivalent integral equation
Z t
x(t) = x0 +
f (s, x(s)) ds.
(7)
t0
Step 1: Set up the successive approximation
x0 (t) = x0
Rt
xn+1 (t) = x0 + t0 f (s, xn (s)) ds,
n = 0, 1, 2, ·.
(8)
Step 2: Verify (use induction) that the sequence {xn (t)} are well-defined for |t − t0 | ≤ α, and
that (t, xn (t)) ∈ R, |t − t0 | ≤ α. In fact,
Z
t
|x1 (t) − x0 | ≤
|f (s, x0 )|ds ≤ M |t − t0 | ≤ a2 ,
t0
Z
t
|xn (t) − x0 | ≤
|f (s, xn−1 (s))| ds ≤ M |t − t0 | ≤ a2 ,
n = 2, 3, · · · ,
t0
if |t − t0 | ≤ a2 /M.
Step 3: Show that {xn (t)} is a uniform Cauchy sequence for |t − t0 | ≤ α.
R
t
|xn+1 (t) − xn (t)| ≤ t0 |f (s, xn (s)) − f (s, xn−1 (s))| ds
R
t
(Lipschitz) ≤ K t0 |xn (s) − xn−1 (s)| ds, n = 1, 2, ...
and
Z t
|x1 (t) − x0 (t)| = | f (s, x0 ) ds| ≤ M |t − t0 |,
t0
5
|t − t0 | ≤ α.
Thus
|x2 (t) − x1 (t)| ≤ M K|t − t0 |2 /2!,
|t − t0 | ≤ α,
|x3 (t) − x2 (t)| ≤ M K 2 |t − t0 |3 /3!,
|t − t0 | ≤ α,
and
and, for n = 1, 2, ...
|xn+1 (t) − xn (t)| ≤
M K n |t − t0 |n+1
M (Kα)n+1
≤
.
(n + 1)!
K (n + 1)!
Thus, for 0 ≤ m ≤ n, we have
n−1
M X (Kα)k+1
|xn (t) − xm (t)| ≤
,
K
(k + 1)!
|t − t0 | ≤ α.
(9)
k=m
Step 4: The solution to the IVP is obtained as the uniform limit of {xn (t)}.
Cauchy’s criterion and (9) imply that
lim xn (t) = x(t)
t→∞
uniformly for t ∈ [t0 , t0 + α], and x(t) is continuous. In the equation (8), let t → ∞, we verify
that x(t) is a solution to the integral equation (7). Thus x(t) is differentiable and satisfies the
IVP.
Summarize the results in Sections 1.3 and 1.4, we have the following existence and uniqueness
theorem under Lipschitz conditions.
Theorem 1.5 Suppose
(1) f ∈ C(D → Rn ).
(2) f satisfies Lipschitz condition in D with respect to x.
Then there exists α > 0 such that the IVP has a unique solution for t ∈ [t0 − α, t0 + α].
1.5. The existence and uniqueness under Lipschitz conditions: a proof using fixed
point theorem
The following is an abstract version of the proof of Picard’s existence and uniqueness theorem using
a fixed point theorem.
Let (B, | · |) be a Banach space. Let
V = {ϕ ∈ B : |ϕ| ≤ b}
be a ball in B with radius b.
Lemma 1.6 Suppose that T : V → B is a transformation that satisfies:
(a) |T (ϕ)| ≤ b for ϕ ∈ V, i.e., T (V ) ⊂ V.
6
(b) ∃ 0 < K1 < 1, such that |T (ϕ) − T (ψ)| ≤ K1 |ϕ − ψ|, for ϕ, ψ ∈ V.
Then, there exists a unique ϕ ∈ V such that T (ϕ) = ϕ, namely, T has a unique fixed point in
V.
Proof. Exercise. (Hint: use Picard successive approximation.)
A mapping that satisfies assumption (b) is called a contraction, and the fixed point theorem
in the lemma is also called the contraction mapping theorem. In the following, we prove Picard’s
theorem using this fixed point result.
Let B = C[t0 − a1 , t0 + a1 ] and V = {x ∈ B : |x − x0 | ≤ a2 }. Define T using the integral
equation (3) as follows:
Z t
f (s, ψ(s)) ds.
T (ψ)(t) = x0 +
t0
Choose a1 small enough so that a1 K < 1, where K is the Lipschitz constant of f in D, then
we can show that T satisfies
(a) ϕ ∈ B implies T (ϕ) ∈ B.
(b) |T (ϕ) − T (ψ)| ≤ a1 K |ϕ − ψ|, for |ϕ|, |ψ| ∈ B.
Then by the previous lemma, T has a unique fixed point ϕ ∈ B and x = ϕ(t) is a solution to the
integral equation (3), and hence a solution to the IVP.
1.6. Peano existence theorem
In Theorem 1.5, if the assumption (2) on the Lipschitz condition is removed, the conclusion still
holds, except for the uniqueness.
Theorem 1.7 Suppose that f ∈ C(D → Rn ). Then the IVP has a solution which exists on a
maximal interval (ω− , ω+ ).
Note: ω− and ω+ depend on t0 , x0 in general.
To establish this result, we need some preliminary propositions.
Let F be a family of real-valued (or Rn valued) functions on a closed interval I ⊂ R.
(a) F is uniformly bounded if there exists M > 0 such that
|f (t)| ≤ M,
∀f ∈ F,
∀ t ∈ I.
(b) F is equi-continuous if for any > 0, there exists δ > 0 such that t1 , t2 ∈ I, |t1 − t2 | <
δ, f ∈ F imply
|f (t1 ) − f (t2 )| < .
7
Proposition 1.8 (Arzelà-Ascoli Theorem) Suppose that the family F is uniformly bounded
and equi-continuous. Then for any sequence {fk } ⊂ F, there exists a subsequence that is uniformly
convergent on I.
A more abstract way to state the Arzelà-Ascoli Theorem is: a subset F of the Banach space
C[a, b] is precompact iff F satisfies properties (a) and (b).
Proof. Let {r1 , r2 , · · · } = I ∩ Q (rationals in I ). Consider the sequence in R :
{fk (r1 )} = {f1 (r1 ), f2 (r1 ), f3 (r1 ), · · · }.
(10)
It is bounded and therefore has a convergent subsequence:
{f11 (r1 ), f12 (r1 ), f13 (r1 ), · · · }.
The sequence
{f11 (r2 ), f12 (r2 ), f13 (r2 ), · · · }.
is bounded and thus has a convergent subsequence
{f21 (r2 ), f22 (r2 , f23 (r2 ), · · · }.
The subsequence {f2k (r)} is now convergent for r = r1 , r2 . Proceeding by induction, we construct
{fmk }, m = 1, 2, 3, · · · where
• {fmk } is a subsequence of {f(m−1)k }, m = 2, 3, 4, · · · , and
• {fmk (r)} = {fm1 (r), fm2 (r), fm2 (r), · · · } is convergent (as k → ∞ ) for r = r1 , r2 , · · · , rm .
Therefore, the sequence {fkk (r)} is convergent for every rational r ∈ I. (Is the convergence
uniform for r ∈ I ?) We will show that {fkk (t)} is convergent for every t ∈ I. (How do we get
uniform convergence?)
Let > 0. Choose δ > 0 such that t1 , t2 ∈ I, |t1 − t2 | < δ, f ∈ F imply
|f (t1 ) − f (t2 )| < /3.
(11)
Choose r1 , r2 , · · · , rN such that {(ri − δ, ri + δ) : i = 1, · · · , N } covers I ( I compact). There
exists H such that for m, n ≥ H,
|fmm (ri ) − fnn (ri )| ≤ /3,
i = 1, 2, · · · , N.
(12)
If t ∈ I, choose ri such that |ri − t| < δ. Then m, n ≥ H implies
|fmm (t) − fnn (t)|
≤ |fmm (t) − fmm (ri )| + |fmm (ri ) − fnn (ri )| + |fnn (ri ) − fnn (t)|
< /3 + /3 + /3 = by (11), (12).
Therefore, {fkk } is a subsequence of {fk } and is uniformly convergent on I (Cauchy criterion).
8
Proposition 1.9 (Peano Local Existence Theorem) Suppose that f ∈ C(D → Rn ). Then
the IVP has a solution which exists in a neighborhood of t0 .
Proof. (Proof # 1, Tonelli). We show the existence of a solution on a right half-neighborhood of
t0 . (left half-neighborhood is similar)
For > 0, let
x (t) = x0 , t0 − ≤ t ≤ t0 ,
Z t−
(13)
x (t) = x0 +
f (s, x (s)) ds, t0 < t
t0 −
Let a1 , a2 , M, α be as in the Picard Local Existence Theorem (Theorem1.4). Then x (t) is well
defined by (13) for t0 ≤ t ≤ t0 + α, since
Z t−
|x (t) − x0 | = |
f (s, x (s)) ds| ≤ M |t − t0 | ≤ M a1 ≤ a2 .
t0 −
This also shows that {x : 0 < } is uniformly bounded on [t0 , t0 + α]. The set of functions is
also equi-continuous, since
Z t1 −
f (s, x (s)) ds| ≤ M |t1 − t2 | ≤ η, if |t1 − t2 | < η/M.
|x (t1 ) − x (t2 )| = |
t2 −
By Arzelà-Ascoli Theorem, there exists sequence k → 0+ , such that {xk } is uniformly convergent
on [t0 , t0 + α]. Therefore (13) implies that the limit function x(t) satisfies
Z t
x(t) = x0 +
f (s, x(s)) ds
t0
and x(t) is a solution of the IVP on [t0 , t0 + α].
Proof. (Proof #2, Peano). Construct an -approximate solution (Euler polygon) as follows:
Let ti = t0 + i, i = 0, 1, 2, · · · , n. ( n ≤ α/. )
If ti ≤ t ≤ ti+1 , let
x (t) = xi + (t − ti )f (ti , xi ),
where xi = x (ti ), i = 1, 2, · · · , n.
Thus x (t) satisfies an integral equations
Z t
x (t) = x0 +
F (s) ds,
t0
where F (s) = f (ti , xi ), ti ≤ s ≤ ti+1 , i = 1, 2, · · · , n, is a step function.
Note
|x1 − x0 | = |f (t0 , x0 )|(t1 − t0 ) ≤ M (t1 − t0 ),
|xi+1 − x0 | = |x (ti+1 ) − x0 | ≤ |x (ti+1 ) − x (ti )| + · · · + |x (t1 ) − x0 |
≤ M (ti+1 − t0 ) ≤ a2 ,
(since t0 ≤ ti ≤ t0 + α)
Thus (use induction) (ti , xi ) ∈ R for i = 1, 2, · · · , n.
Show
9
(i) |x (t) − x0 | ≤ M (t − t0 ),
(ii) |x (t) − x (s)| ≤ M |t − s|,
t0 ≤ t ≤ t0 + α;
t0 ≤ t, s ≤ t0 + α;
(iii) There is a sequence k → 0+ such that {xk (t)} is uniformly convergent on [t0 , t0 + α] to a
function x(t);
(iv) Fk (t) converges uniformly on [t0 , t0 + α] to f (t, x(t));
(v) x0 (t) = f (t, x(t)),
x(t0 ) = x0 .
The rest of the section concerns the maximal interval of existence.
Proposition 1.10 Suppose
(i) f ∈ C(D → Rn );
(ii) K ⊂ D compact;
(iii) (t, x(t)) ∈ K, t0 ≤ t < ω < ∞ for some solution x(t) of the IVP.
Then x1 = limt→ω− x(t) exists and (ω, x1 ) ∈ K. Furthermore, [t0 , ω) is not a right-maximal
interval of existence for x(t).
Proof. To show x1 = limt→ω− x(t) exists, we show that x(t) is uniformly continuous on [t0 , ω).
From the integral equation we have
Z t2
|x(t1 ) − x(t2 )| = |
f (t, x(t))dt| ≤ M |t1 − t2 |
t1
for all t1 , t2 ∈ [t0 , ω), where M = max {|f (t, x)| : (t, x) ∈ K}. The Peano Existence Theorem
(Theorem1.9) implies that x(t) can be continued to the right of ω.
Exercise. Suppose that x1 (t) is a solution of the IVP
x0 (t) = f (t, x),
x(t0 ) = x0 ,
in [t0 , a) and that limt→a x1 (t) = x1 exists and that (a, x1 ) ∈ D. Let x2 (t) be a solution to the
IVP
x0 = f (t, x), x(a) = x1
in [a, b). Define x(t) = x1 (t) for t ∈ [t0 , a) and x(t) = x2 (t) for t ∈ [a, b) Show that x(t) is a
solution of the IVP
x0 = f (t, x), x(t0 ) = x0
in [t0 , b).
Proposition 1.11 Suppose
(i) f ∈ C(D → Rn );
10
(ii) x(t) is a solution of the IVP defined for [t0 , ω) ;
(iii) ∃ t0 < tk < ω, limk→∞ tk = ω and limk→∞ x(tk ) = x1 ;
(iv) (ω, x1 ) ∈ D.
Then x1 = limt→ω− x(t) and [t0 , ω) is not a right-maximal interval of existence for x(t).
Proof. Let > 0 be sufficiently small that J ⊂ D, where J = {(t, x) : |t−ω| ≤ , |x−x1 | ≤ }.
Let M = max{|f (t, x)| : (t, x) ∈ J }. Then ∃ N 3 0 < ω − tN < /2M and |x(tN ) − x1 | < /2.
We claim that
|x(t) − x(tN )| < M (ω − tN ) < /2, tN ≤ t < ω.
(14)
Then, tN ≤ t < ω implies
|x(t) − x1 | ≤ |x(t) − x(tN )| + |x(tN ) − x1 | < /2 + /2 = .
and thus limt→ω− x(t) = x1 . Moreover, (ω, x1 ) ∈ D implies that [t0 , ω) is not the right-maximal
interval of existence.
To show claim (14): suppose it is false. Then there exists a least T > tN such that
|x(T ) − x(tN )| = M (ω − tN ) < /2.
(15)
Therefore,
|x(t) − x1 | ≤ |x(t) − x(tN )| + |x(tN ) − x1 | < /2 + /2 = ,
for tN ≤ t < T. Thus (t, x(t)) ∈ J , tN ≤ t < T. Therefore
Z
T
|x(T ) − x(tN )| = |
Z
T
f (s, x(s))ds| ≤
tN
|f (s, x(s))|ds
(16)
tN
≤ M (T − tN ) < M (ω − tN )
(17)
which contradicts (15), proving the claim. (Question: why can’t we use (refBBB) to prove (14)
directly?)
Corollary 1.12 Assume that D = (a, b) × Rn , t0 ∈ (a, b), and (ω− , ω+ ) is a maximal interval
of existence for a solution of x0 = f (t, x). Then
ω− = a
or
ω+ = b
or
lim |x(t)| = ∞,
+
t→ω−
lim |x(t)| = ∞.
−
t→ω+
Proof. Suppose that ω+ < b. Then the uniform continuity of x(t) in [t0 , ω+ ), now a finite interval,
implies that limt→ω− |x(t)| exists (though may not be finite). Suppose that limt→ω− |x(t)| <
+
+
∞. Then the uniform continuity of x(t) again implies that there exists x1 ∈ Rn such that
limt→ω− x(t) = x1 , and (ω+ , x1 ) ∈ D = (a, b) × Rn . Therefore x(t) can be extended to the
+
11
closed interval [t0 , ω+ ], contradicting the assumption that [t0 , ω+ ) is the right maximal interval
of existence. The proof for the case of ω− can be proved similarly.
In Summary: If f ∈ C(D → Rn ), then the IVP has a solution and every solution x(t) of the
IVP exists on a maximal interval (ω− , ω+ ) which depends on the solution. Moreover,
lim d((t, x(t)), ∂D) = 0,
t→ω±
(18)
where d(x, A) = inf a∈A |x − a|.
To see the relation (18), suppose that limt→ω+ d((t, x(t)), ∂D) 6= 0. Then there exist δ0 > 0 and
tn → ω+ such that limn→∞ d((t, x(tn )), ∂D) ≥ δ0 . First of all, this relation implies that x(tn ) is
bounded, since otherwise, D must be unbounded and {∞} ⊂ ∂D, and there exists a subsequence
tk such that limk→∞ |x(tk )| = ∞. This implies that limk→∞ d((t, x(tk )), ∂D) = 0 < δ0 . Now
|x(tn )|, being bounded, must have a convergent subsequence, x(tm ) such that tm → ω+ and
x(tm ) → x1 , and thus d((ω+ , x1 ), ∂D) ≥ δ0 . This implies that (ω+ , x1 ) ∈ D, and the solution
can then be further continued. This contradiction shows that (18) must hold on a maximal interval.
However, for a solution x(t) on its maximal interval of existence (ω− , ω+ ), the limit limt→ω± x(t)
may not exist, as the following example shows.
Example. Consider
x0 (t) =
1
1
cos .
2
t
t
The function f (t, x) = t12 cos 1t is continuous in D = (0, ∞) × R. A solution x(t) = c − sin 1t has
a left maximal interval (0, t0 ). We see that limt→0+ d((t, x(t)), ∂D) → 0, but limt→0+ x(t) does
not exist.
1.7. Continuous dependence on initial conditions and parameters
Consider the IVP
x0 = f (t, x, µ) x(t0 ) = x0 ,
(19)
where t ∈ R, x ∈ Rn , µ ∈ Rm . We make the following assumptions:
(i) f ∈ C(G → Rn ), G open in Rn+m+1 .
(ii) Solutions to the IVP is unique for each (t0 , x0 , µ) ∈ G.
Denote the solution to the IVP as x = x(t; t0 , x0 , µ).
We will show that x(t; t0 , x0 , µ) is continuous in its domain.
Theorem 1.13 Assume that x(t; t0 , x0 , µ0 ) exists for t ∈ [a, b]. Then, ∀ > 0, ∃ δ > 0 such
that
|t∗0 − t0 | < δ, |x∗0 − x0 | < δ, |µ∗0 − µ0 | < δ
imply
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(1) x(t; t∗0 , x∗0 , µ∗0 ) exists for t ∈ [a, b].
(2) supa≤t≤b |x(t; t∗0 , x∗0 , µ∗0 ) − x(t; t0 , x0 , µ0 )| < .
W.o.l.g., we may drop the dependence on µ. The IVP
x0
= f (t, x, µ)
x = x(t; t0 , x0 , µ)
x(t0 ) = x0 ,
is equivalent to the IVP




x0
y0
x(t0 )



y(t0 )
= f (t, x, y)
=
0
=
x0
=
µ,
x = x(t; t0 , x0 , µ)
y =
y0 (= µ).
Proof of Theorem 1.13. Equivalently, we prove the following. Let
(tk , xk ) → (t0 , x0 ),
as k → ∞.
Then for k > K,
(1) x(t; tk , xk ) exists for t ∈ [a, b].
(2) limk→∞ supa≤t≤b |x(t; tk , xk ) − x(t; t0 , x0 )| = 0, or equivalently, x(t; , tk , xk ) → x(t, t0 , x0 ) as
k → ∞ uniformly for t ∈ [a, b].
Step 1. Choose D0 ⊂ D0 ⊂ D1 ⊂ D1 ⊂ D such that (t, x(t)) ∈ D0 for t ∈ [a, b]. Set
M>
sup
|f (t, x)|.
(t,x)∈D1
Choose α such that
Rα (t0 , x0 ) = {(t, x) : |t − t0 | ≤ α, |x − x0 | ≤ M α} ⊂ D1 .
Then x(t; t0 , x0 ) is defined for [t0 − α, t0 + α], and ∃K > 0, such that k > K implies (tk , xk ) ∈
Rα (t0 , x0 ) and
Rα (tk , xk ) ⊂ D1 ,
and thus xk (t) = x(t; tk , xk ) is defined for t ∈ [t0 − α0 , t0 + α0 ] for some 0 < α0 ≤ α. From the
integral equation
Z t
xk (t) = xk +
f (s, xk (s)) ds
tk
we conclude that {xk (t)} is uniformly bounded and equi-continuous in [t0 − α0 , t0 + α0 ]. ArzelàAscoli Theorem implies that any subsequence of {xk (t)} has a subsequence uniformly convergent
to the same limit x(t, t0 , x0 ) (uniqueness of solutions). This implies that
x(t; tk , xk ) → x(t; t0 , x0 )
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as k → ∞
uniformly in [t0 − α0 , t0 + α0 ].
Step 2. Suppose [a, b] 6⊂ [t0 − α0 , t0 + α0 ]. Let t10 = t0 + α0 , x10 = x(t10 ). Choose α1 such that
Rα1 (t10 , x10 ) ⊂ D1 .
Repeating the preceding argument, we conclude: ∃ K1 > 0 such that k > K1 implies x(t; tk , xk )
is defined for t ∈ [t0 , t0 + α0 + α1 ], and
x(t; tk , xk ) → x(t; t0 , x0 )
as k → ∞
uniformly in [t0 , t0 + α0 + α1 ]. Proceed like this, after finite number of steps, we obtain [a1 , b1 ] ⊃
[a, b] and K > 0 such that k > K implies
(1) x(t; tk , xk ) exists for t ∈ [a1 , b1 ].
(2) limk→∞ supa1 ≤t≤b1 |x(t; tk , xk ) − x(t; t0 , x0 )| = 0.
Remark. The result also includes the continuous dependence on the vector field f (t, x). Suppose
xk (t) solves the IVP
x0 = fk (t, x), x(tk ) = xk ,
and, as k → ∞, (tk , xk ) → (t0 , x0 ), fk (t, x) → f (t, x) on any compact subset of D ⊂ Rn+1 .
Then, for k > K > 0, x(t; tk , xk , fk ) exists in [a, b] and
lim sup |x(t; tk , xk , fk ) − x(t; t0 , x0 , f )| = 0.
k→∞ a≤t≤b
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