Unformatted text preview: essive
approximations to the solution of [1] by repeatedly applying the operator . Start with an
initial guess
to generate the sequence
,
Example.
Let , where . Apply to and from [3] to get .
Apply to to get ).
This procedure generates successive approximations to the solution 10
Chapter 3
.■
Our text presents a constructive proof of the Existence and Uniqueness theorem by analyzing
the convergence of Picard iterates. The proof is longer than that using the Contraction
Mapping theorem, but it has the advantage of dropping the restriction
. If is not Lipschitz, a solution to [1] may exist but fail to be unique. Example (The Leaky Bucket) (Strogatz, Nonlinear Dynamics and Chaos)
Sketch partially full bucket with water running out of hole
Symbols:
height of water remaining at time .
crosssection of bucket.
area of hole
velocity of water passing through hole
density of water
Suppose that in a short time interval the height of the water in the bucket decreases b y
The mass of the water that escaped is
. Assume that the potential energy of
entirely converted into kinetic energy of escaping water. .
is Solve for
,
where the minus sign reflects the fact that the escaping water is falling. By conservation of
volume
.
Combine the last two equations to obtain
where the constant . 11
Chapter 3
Iw ub u h ‘b ckw d ’v the right hand side is not Lipschitz at b for we obtain . Notice that . Now suppose we are taking a walk and that at
we notice an empty bucket with a hole in
hb
d
w
ud
h. L ’ u
u
d gk
d
wh
the water ran out by integrating backwards in time! Consider the initial value problem
, . [4] Our initial condition corresponds to the fact that we found the bucket empty . We notice
immediately that one solution is
,
that is, there never was any water in the bucket. Undaunted, we try separating variables
,
and integrate, applying the initial condition, to find
.
This means it is possible that the water has just finished running out of the bucket. Finally we
notice that we can combine these two solutions to obtain for any
:
.
The initial value problem [4] admits an infinite family of solutions! sketch solutions for
and
. ■ 3.4 Dependence on Initial Conditions and Parameters
Here we modify section 3.3 eq. [1] by considering
,
for some close to [1]
. Write the solution 12
Chapter 3
,
showing the dependence on the initial condition. Operator has initial condition : .
Lemma 3.12 (Neighborhood Existence). Suppose that for a given
satisfies a Lipschitz condition with constant , and
Then the family of solutions of [1] exists for each there is a such that
. and is unique for providing
Proof. Very similar to that of theorem 3.10. Steps (A) and (C) are essentially identical. Step
(B), which shows
is modified slightly. We now have
.
Use the triangle inequality and the maximum value of :
.
T...
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 Fall '14
 MarcEvans
 Metric space, Lipschitz

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