18.034, Honors Differential Equations
Prof. Jason Starr
Lecture 4: Existence & Uniqueness, Part II
Feb. 11, 2004
1. Cauchy sequences and complete metric spaces.
A metric space is complete if it “has no holes”. What does this mean? A metric space “has a
hole” if there is a sequence that “should converse”, but such that there is no limit. The
sequences which “should converse” are Cauchy sequences.
Def’n: Let (X,d) be a metric space. A sequence of elements in X, (Pi)i=0,1,…, is a Cauchy
sequences if for every ε > 0 , there is an integer M ≥ 0 such that for all pairs of integers
m, n ≥ M , the distance d(Pm , Pn ) < ε .
Thm [ ≈ Heine-Borel thm] For the Euclidean metric space (IRn, ∂
convergent if and only if it is Cauchy.
Eucl
), sequence (Pi) is
∂ Eucl ) has a name.
This property of (IRn,
Def’n: A metric space (X, ∂) is complete if for every sequence of elements in X, (Pi), the
sequence is convergent if and only if it is Cauchy.
Exercise: Prove that for any metric space, if (Pi) is convergent, then (Pi) is Cauchy.
2 Cauchy test
Let [a,b] be a bounded interval in IR , and let C([a,b], IR ) be the metric space of continuous real
valued functions on [a,b] with the metric ∂ (y 2 , y1 ) = maximum value of y 2 (t ) − y1 (t ) on [a,b].
The main theorem about this metric space is the following.
Thm [Cauchy test = Thm A.21]: Let (yi(t)) be a sequence in C([a,b], IR ). If (yi) is Cauchy, then
it converses, i.e. C([a, b], IR), ∂ is a complete metric space.
Pf: Suppose (yi) is Cauchy. Then for every t in [a,b], the sequence of real numbers (yi,(t)) i=0,1,…
, is a Cauchy sequence. Because IR is complete, (yi,(t)) converges to some real number. Call
this number y(t).
The claim is that, for every ε > 0 , there is an integer N ≥ 0 such that for all n ≥ N ,
max{(y n (t) - y(t)} < ε . To prove this, observe by the Cauchy condition that there is N ≥ 0 such
{
}
that for all m, n ≥ N , max max y m (t ) − y n (t ) <
ε
2
.
Now, for each t, because (yi,t)) converges to y(t), there exists M = M(t) ≥ 0 such that for all
m ≥ M , y m (t ) − y(t ) <
y n (t ) − y(t )
ε
. But then, for every n ≥ N (N doesn’t depend on t),
2
= (y n (t ) − y m (t )) + (y m (t ) − y(t ))
(y n (t ) − y m (t )) + (y m (t ) − y(t ))
<ε
2
(Cauchy condition)
+ε
2
(ym(t) converses to y(t))
=ε.
This proves the claim.
18.034, Honors Differential Equations
Prof. Jason Starr
Lecture 4
Page 1 of 4
If we knew that y(t) is continuous, it would follow that (yi) converses to y. So let’s prove y(t) is
continuous. Let t be any element in [a,b]. By the claim, there exists N ≥ 0 such that for all n ≥ N,
max{(y n (δ ) - y(t)} <
y n (s) −y n (t ) < ε
ε
3
. Because yn(t) is continuous, there exists z >0 such that
whenever t − s < z ,
3
Then, whenever t − s < z
y(s) − y(t ) ≤ y(s) − y n (s) + y n (s) − y n (t ) + y n (t ) − y(t )
≤ ε
(b c max{y
n
=ε
(by the triangle inequality)
3
+
}
−y < ε
)
3
ε
3
(b c y (t )cts )
n
+
ε
3
(b c max{y
n
}
−y < ε
3
)
So y(t) is continuous. Therefore y(t) is an element of C([a,b], IR ), and the sequence (yi(t))
converges to y(t).
Variation: The metric space B • (y 0 , b) is complete (defined as in lecture 3).
Proof: The only new step is to prove that the limit y(t) has graph contained in
[t 0 , t 0 + C] x [y 0 - b, y 0 + b] . But for each t, y(t) is the limit of (yi(t)). Because yi(t) is a sequence
in the closed interval [y 0 - b, y 0 + b] , its limit is also in this interval. So y 0 - b ≤ y(t) ≤ y 0 + b ,
i.e. y(t) is in B • (y 0 , b) .
(Contd. on next page… )
18.034, Honors Differential Equations
Prof. Jason Starr
Lecture 4
Page 2 of 4
3. Thm [Contraction mapping fixed point theorem, Part II]: Let (X, ∂ ) be a complete metric
space. Let ε be a number with 0 ≤ ε < 1 . Let T be an ε-contraction mapping on (X, ∂) . Then
there exists a unique fixed point of T.
Proof: We already proved there is at most one point. The real content is that there exists a fixed
point.
Let p0 be any point of (X, ∂ ). Define p1 = T(p 0 ) , define p 2 = T(p1 ) , etc. In other words, define
a sequence (pi ) i = 0,1,…. of elements in X inductively by T(pi ) = p i + 1 .
Denote by D the distance D = ∂(p1 , p 0 ) . The claim is that for all i=0,1,…, ∂ (pi + 1 , pi ) ≤ ε i • D .
This is proved by induction on i, the base case being done. If ∂(pi + 1 , pi ) ≤ ε i • D , then
∂ (pi + 2 , pi + 1 ) = ∂ (T(pi+1),T(pi)) (T(pi + 1 ), T(pi )) ≤ ε • ∂ (pi + 1 , pi )
i
≤ ε •ε •D
=ε
i + 21
(b/c T is ε-contracting)
(by hypothesis)
•D
This proves the induction step, thus ∂ (pi + 1 , pi ) ≤ ε i • D for all i.
The claim is that (pi) is a Cauchy sequence. Let ε ' > 0 be any number. Let N be an integer such
that
ε N .D
< ε ' , i.e.
1− ε
(1 − ε )ε '
N > log
D
log(ε )
.
Then for all m ≥ n ≥ N ,
∂ (p m , pn )
≤ ∂ (pn+1 , pn ) + ∂(pn+2 , pn-1 ) + … .. + ∂(pm , p m-1 )
≤ ε n • D + ε n + 1 + 1 • D + ε n + 2 + 2 • D + ….. + ε m-1 • D
≤ ε n • D + ε n + 1 + 1 • D + ε n + 2 + 2 • D + ….. = ε n • D •
1
.
1−ε
ε N .D
< ε ' for n ≥ N . ∂ (p m , pn ) < ε ' .
1−ε
So (pi) is a Cauchy sequence.
Since
Because (X, ∂ ) is a complete metric space, the X Cauchy sequence (pi) converses to an element
p of X. Let ε ' > 0 be a number. There exists N such that for all m ≥ n ≥ N , ∂ (p m , pn ) < ε
Also, there is n ≥ N such that ∂ (p, pn ) <
ε
3
3
.
.
Thus, ∂ (p, T(p)) ≤ ∂ (p, pn ) + ∂ (pn , pn + 1 ) + ∂ (pn + 1 , T(p)) .
But ∂ (pn + 1 , T(p)) = ∂ (T(pn ), T(p)) < ∂ (pn , p) .
So ∂ (pn + 1 , T(p)) ≤ 2 • ∂ (p, pn ) + ∂ (pn + pn + 1 ) < 2 • ε
3
Since ∂ (p, T(p)) (p,T(p)) is less than ε for all ε > 0 ,
a fixed point of T.
+ε
3
=ε.
∂ (p, T(p)) = 0 . Therefore
p = T(p) , i.e. p is
4. Existence of a solution to the IVP. Let R, D, f, M, L, a, b and c be as in Lecture 3.
Thm: There exists a differentiable function y(t) defined on [t0,t0+c] such that
(1) y(t 0 ) = y 0
18.034, Honors Differential Equations
Prof. Jason Starr
Lecture 4
Page 3 of 4
(2) y' (t) = f(t, y(t))
(3) y(t ) − y 0 ≤ b
Proof: As proved in Lecture 3, on the metric space B • (y 0 , b) of continuous functions y(t) on
[t0,t0+c] such that y(t ) − y 0 ≤ b , the mapping T(y) = z ,
t
z(t) := y 0 +
∫ f (s, y(s))ds .
t0
is a 1/2 – contraction mapping. By the Cauchy test, B • (y 0 , b) is a complete metric space. By
the contraction mapping fixed point theorem, part II, there exists a continuous function y(t) in
B • (y 0 , b) such that y(t) = T(y(t)) , i.e.
t
y(t) = y 0 +
∫ f (s, y(s))ds .
t0
By the fundamental theorem of calculus, y(t) is differentiable and y' (t) = f(s, y(s)) . Moreover,
t0
y(t 0 ) = y 0 +
∫ f (s, y(s))ds = y
0
+ 0 = y0 .
t0
Finally since y(t) is in B • (y 0 , b) , y (t ) − y 0 ≤ b for all t in [t0,t0+c].
18.034, Honors Differential Equations
Prof. Jason Starr
Lecture 4
Page 4 of 4