Existence/Uniqueness
Recall the theorem:
Theorem
Given the IVP
y
0
=
f
(
x, y
)
,
y
(
x
0
) =
y
0
,
suppose
f
and
∂f/∂y
are continuous in the rectangle
R
given by
x
0
≤
x
≤
x
0
+
a

y

y
0
 ≤
b.
Compute the values
M
=
max
(
x,y
)
in
R

f
(
x, y
)

,
α
= min
a,
b
M
.
Then the Picard Iterates converge to a unique solution on the interval
[
x
0
, x
0
+
α
]
.
The next example shows how that the interval predicted by the theorem will be smaller than the
actual interval of existence of the solution:
I
Example
What is the largest interval of existence predicted by the above theorem for the solution to the IVP
y
0
= 1 +
y
2
,
y
(0) = 0?
Solution
Consider the rectangle
R
:
0
≤
x
≤
a

y
 ≤
b.
We compute
M
=
max
(
x,y
)
in
R

f
(
x, y
)

=
max
(
x,y
)
in
R
1 +
y
2
= 1 +
b
2
and then
α
= min
a,
b
1 +
b
2
.
Since
1
/
(1+
b
2
)
has a maximum value of
1
/
2
,
α
can be
at most
1
/
2
(and it attains this value,
for example pick
a
= 1
and
b
= 1
).
Then the largest interval of existence guaranteed by
the theorem is
[0
,
1
/
2]
, when in fact the solution
y
(
x
) = tan(
x
)
exists on the larger interval
(

π/
2
, π/
2)
.
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 '08
 staff
 Existence, Philosophy of life, Rectangle, unique solution

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