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Econ 204–Summer/Fall 2008
Diferential Equations
In this supplement, we use the methods we have developed so far to study
diFerential equations.
1
Existence and Uniqueness o± Solutions
De²nition 1
A
diferential equation
is an equation of the form
y
0
(
t
)=
F
(
y
(
t
)
,t
)whe
re
U
is an open subset of
R
n
×
R
and
F
:
U
→
R
n
.A
n
initial value problem
is a diFerential equation combined with an initial con
dition
y
(
t
0
y
0
with (
y
0
0
)
∈
U
solution
of the initial value problem is
a diFerentiable function
y
:(
a, b
)
→
R
n
such that
t
0
∈
(
a, b
),
y
(
t
0
y
0
and,
for all
t
∈
(
a, b
),
dy
dt
=
F
(
y
(
t
)
).
Theorem 2
Consider the initial value problem
y
0
(
t
F
(
y
(
t
)
)
,y
(
t
0
y
0
(1)
Let
U
be an open set in
R
n
×
R
containing
(
y
0
0
)
.
•
Suppose
F
:
U
→
R
n
is continuous. Then the initial value problem has
aso
lu
t
ion
.
•
IF, in addition,
F
is Lipschitz in
y
on U (i.e. there is a constant
K
such that For all
(
y,t
)
,
(ˆ
y, t
)
∈
U
,

F
(
)
−
F
y, t
)
≤
K

y
−
ˆ
y

), then
there is an interval
(
a, b
)
containing
t
0
such that the solution is unique
on
(
a, b
)
.
Proo±:
We will limit our proof to the case in which
F
is Lipschitz; for the
general case, see Coddington and Levinson [1].
Since
U
is open, we may
choose
r>
0 such that
R
=
{
(
):

y
−
y
0
r,

t
−
t
0
r
}⊆
U
Given the Lipschitz condition, we may assume that

F
(
)
−
F
y, t
)
K

y
−
ˆ
y

for all (
)
,
y, t
)
∈
R
.L
e
t
δ
=m
in
±
1
2
K
,
r
M
²
1
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View Full Document We claim the initial value problem has a unique solution on (
t
0
−
δ, t
0
+
δ
).
Let
C
be the space of continuous functions from [
t
0
−
δ, t
0
+
δ
]to
R
n
,
endowed with the sup norm
k
f
k
∞
=sup
{
f
(
t
)

:
t
∈
[
t
0
−
δ, t
0
+
δ
]
}
Let
S
=
{
z
∈
C
([
t
0
−
δ, t
0
+
δ
]) : (
z
(
s
)
,s
)
∈
R
for all
s
∈
[
t
0
−
δ, t
0
+
δ
]
}
S
is a closed subset of the complete metric space
C
,so
S
is a complete metric
space. Consider the function
I
:
S
→
C
deFned by
I
(
z
)(
t
)=
y
0
+
Z
t
t
0
F
(
z
(
s
)
)
ds
I
(
z
) is deFned and continuous because
F
is bounded and continuous on
R
.
Observe that if (
z
(
s
)
)
∈
R
for all
s
∈
[
t
0
−
δ, t
0
+
δ
], then

I
(
z
)(
t
)
−
y
0

=
±
±
±
±
Z
t
t
0
F
(
z
(
s
)
)
ds
±
±
±
±
≤
t
−
t
0

max
{
F
(
y,s
)

:(
)
∈
R
}
≤
δM
≤
r
so (
I
(
z
)(
t
)
,t
)
∈
R
for all
t
∈
[
t
0
−
δ, t
0
+
δ
]. Thus,
I
:
S
→
S
.
Given two functions
x, z
∈
S
and
t
∈
[
t
0
−
δ, t
0
+
δ
],

I
(
z
)(
t
)
−
I
(
x
)(
t
)

=
±
±
±
±
y
0
+
Z
t
t
0
F
(
z
(
s
)
)
ds
−
y
0
−
Z
t
t
0
F
(
x
(
s
)
)
ds
±
±
±
±
=
±
±
±
±
Z
t
t
0
(
F
(
z
(
s
)
)
−
F
(
x
(
s
)
))
ds
±
±
±
±
t
−
t
0

sup
{
F
(
z
(
s
)
)
−
F
(
x
(
s
)
)

:
s
∈
[
t
0
−
δ, t
0
+
δ
]
}
≤
δK
sup
{
z
(
s
)
−
x
(
s
)

:
s
∈
[
t
0
−
δ, t
0
+
δ
]
}
≤
k
z
−
x
k
∞
2
Therefore,
k
I
(
z
)
−
I
(
x
)
k
∞
≤
k
z
−
x
k
∞
2
I
is a contraction.
Since
S
is a
complete metric space,
I
has a unique Fxed point
y
∈
S
. Therefore, for all
t
∈
[
t
0
−
δ, t
0
+
δ
], we have
y
(
t
Z
t
t
0
F
(
y
(
s
)
)
ds
2
F
is continuous, so the Fundamental Theorem of Calculus implies that
y
0
(
t
)=
F
(
y
(
t
)
,t
)
for all
t
∈
(
t
0
−
δ, t
0
+
δ
). Since we also have
y
(
t
0
y
0
+
Z
t
0
t
0
F
(
y
(
s
)
,s
)
ds
=
y
0
y
(restricted to (
t
0
−
δ, t
0
+
δ
)) is a solution of the initial value problem.
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This note was uploaded on 08/01/2008 for the course ECON 204 taught by Professor Anderson during the Fall '08 term at University of California, Berkeley.
 Fall '08
 ANDERSON

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