18.03 Existence and Uniqueness Theorem
Jeremy Orloff
Theorem (existence and uniqueness)
∂f
Suppose f (t, y) and
(t, y) are continuous on a rectangle D as shown. Then we can
∂y
choose a smaller rectangle R (as shown) so that the IVP
dy
= f (t, y(t)), y(t0 ) = y0
dt
has a unique solution defined on [t0 − a, t0 + a] whose graph is entirely inside R.
y
D = big rectangle
y0 + b
R = middle rect.
y0
•
(t0 , y0 )
y0 − b
t
t0 − a
t0
t0 + a
The proof proceeds in a series of steps. Some of these steps are technical –I try to
give a sense of why they are true. The key steps are the definition of the contraction
map T (step (3)) and the use of T in Picard iteration (step (8)).
∂f
(1) Let M = max |f (t, y)| and L = max (t, y).
D
D
∂y
∂f
(t, c) (y2 − y1 ) (for some c
The mean value theorem ⇒ f (t, y2 ) − f (t, y1 ) =
∂y
between y1 and y2 ). ⇒ |f (t, y2 ) − f (t, y1 )| < L |y2 − y1 | (Lipshcitz condition).
(2) Choosing the rectangle R
b 1
Choose a < min( ,
). We will use this in steps (3) and (5).
M 2L
(3) The operator T
Let Y be the space of all functions y which are continuous on [t0 − a, t0 + a] and
whose graph is entirely inside R. For any y ∈ Y define
Z t
T y = z(t) = y0 +
f (s, y(s)) ds.
t0
(continued)
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18.03 Existence and Uniqueness Theorem
2
We note a number of easy facts about T .
a) T y = z(t) is well defined on [t0 −a, t0 +a]. (proof: (s, y(s)) is in R, so the integrand
f (s, y(s)) is defined and continuous.)
b) z(t) is continuous. (proof: trivial since both y and f are continuous.)
c) The graph of z(t) is entirely in R.
Z t
proof: |z(t) − y0 | = f (s, y(s)) ds ≤ M |t − t0 | ≤ M a < b. (The last inequality
t0
follows from the choice of a in step (2).)
Facts a-c show T maps the space Y into itself.
d) z 0 (t) = f (t, y(t)) (proof: fundamental theorem of calculus).
e) Definition: the function y ∈ Y is a fixed point of T means T y = y.
Claim:
y is a solution to the IVP ⇔ y is a fixed point of T .
Suppose y is a solution, i.e., y(t0 ) = y0 and y 0 (t) = f (t, y(t)). Then
Z t
T y = y0 +
f (s, y(s)) ds
t
0
Z t
= y0 +
y 0 (s) ds = y0 + y(s)|tt0 = y(t)
Proof:
t0
So y is a fixed point of T .
Z
t
Conversely, suppose y is a fixed point, then y = T y = y0 +
f (s, y(s)) ds.
t0
⇒ y(t0 ) = y0 and y 0 = f (t, y(t)). I.e. y satisfies the IVP. QED
The claim shows that proving existence and uniqueness is equivalent to proving that
T has a unique fixed point. (This is proved in (8) and (9) below.)
(4) The metric on Y
For y1 and y2 in Y define
δ(y1 , y2 ) =
max
[t0 −a,t0 +a]
|y1 (t) − y2 (t)|.
a) δ(y1 , y2 ) = 0 ⇔ y1 = y2 (proof: trivial).
b) δ satisfies the triange inequality: δ(y1 , y2 ) + δ(y2 , y3 ) ≥ δ(y1 , y3 ) (proof: not hard).
c) δ tells how to measure ’closeness’ between ’points’ of Y.
d) Y has no ’holes’.
Formally: Y is a complete metric space.
Formally: Complete means all Cauchy sequences converge.
Informally: Complete means all sequences that look like they converge really do
converge.
(continued)
18.03 Existence and Uniqueness Theorem
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1
1
1
On the real line R the sequence 1 + , 1 + , 1 + , . . . converges to 1.
2
3
4
However, in R − {1} the sequence doesn’t converge. This is because R − {1} has a
’hole’ at 1.
X
For us it is enough to know that if the sequence y0 , y1 , . . . satisfies
δ(yn+1 , yn ) < ∞
then the sequence converges, i.e., lim yn = y exists.
Analogy:
n→∞
(Picture: each step from y0 to y1 to y2 etc gets smaller in such a way that the total
distance is finite.)
Completeness is not hard to show. It does require a careful ’ − δ’ proof.
1
δ(T y1 , T y2 ) ≤ δ(y1 , y2 ).
2
Proof:
Z t
|T y1 (t) − T y2 (t)| = f (s, y1 (s)) − f (s, y2 (s)) ds
Z tt0
≤
|f (s, y1 (s)) − f (s, y2 (s))| ds
t0Z
t
≤ L
|y1 (s) − y2 (s)| ds
(Lipschitz condition)
t0
Z t
≤ Lδ(y1 , y2 )
ds
(pull out max(y1 (s) − y2 (s)))
(5) Claim:
t0
1
= Lδ(y1 , y2 )(t − t0 ) ≤ δ(y1 , y2 ) L · a < δ(y1 , y2 )
2
The last inequality uses the choice of a in step (2).
QED
Note: since T shrinks distances it is called a contraction mapping.
By way of analogy let A : [0, 1] → [0, 1] by the formula A(x) = 21 (1 − x). It’s easy to
see |A(x1 ) − A(x2 )| = 12 |x1 − x2 |, so A is a contraction mapping. If you start with
the whole interval [0, 1] and repeatedly apply A you get a sequence of intervals [0, 1],
[0, 1/2], [1/4, 1/2], [1/4, 3/8]. These intervals get smaller and smaller, shrinking down
to the fixed point x = 1/3.
(6) Claim: T has at most one fixed point.
Proof:
Suppose there were two different fixed points y1 and y2 . Then since
T yj = yj we get δ(T y1 , T y2 ) = δ(y1 , y2 ). But, this contradicts (5) where we saw
1
δ(T y1 , T y2 ) ≤ δ(y1 , y2 ).
2
(7) If the sequence y0 , y1 , y2 , . . . converges to y then T y0 , T y1 , T y2 , . . . converges to
T y.
Formally: T is a continuous map of Y to itself.
(continued)
18.03 Existence and Uniqueness Theorem
(8) Picard itereation
Start with y0 (t) = y0 . Let y1 = T y0 , y2 = T y1 , . . . , yn+1 = T yn = T n y0 .
Claim:
the sequence y0 , y1 , . . . converges.
1
Proof: δ(y2 , y1 ) = δ(T y1 , T y0 ) ≤ δ(y1 , y0 )
2
1
1
Likewise, δ(y3 , y2 ) = δ(T y2 , T y1 ) ≤ δ(y2 , y1 ) ≤ δ(y1 , y0 )
2
4
n
1
Generally, δ(yn+1 , yn ) ≤
δ(y1 , y0 )
2
∞
∞ n
X
X
1
.
⇒
δ(yn+1 , yn ) ≤ δ(y1 , y0 )
2
n=0
0
Since this last sum converges the completeness of Y proves the claim.
(9) Take the sequence from (8) and let lim yn = y.
n→∞
Claim: y is a fixed point of T .
Proof: Since y = lim T n y0 we have T y = lim T n+1 y0 = y. QED
Example: (Picard iteration) Consider the IVP y 0 = y, y(0) = 1.
Picard iteration gives y0 (t) = 1.
Z t
1 ds = 1 + t.
y1 (t) = y0 +
0
Z t
t2
1 + t ds = 1 + t + .
y2 (t) = y0 +
2
0
⇒ In this case Picard iteration leads to the power series for et .
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