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Using these three claims, we have shown that the solution <math>\Phi(x)</math> exists. The proofs of the claims are below. |
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Using these three claims, we have shown that the solution <math>\Phi(x)</math> exists. The proofs of the claims are below. |
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Proof of Uniqueness: |
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Suppose <math>\Phi</math> and <math>\Psi</math> are both solutions. Let <math>\Chi(x) = |\Phi(x) - \Psi(x)|</math>. |
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<math>\Chi(x) = |\Phi(x) - \Psi(x)| = |\int_{x_0}^x(f(x, \Phi(x)) - f(x, \Psi(x))) dx | \leq \int_{x_0}^x k|\Phi(x) - \Psi(x)| dx</math> |
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We have that <math>\Chi \leq k \int_{x_0}^x \Chi(x) dx</math> for some constant k, which means <math>\Chi' \leq k\Chi</math>, and that <math>\Chi(x) \geq 0</math>. |
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Let <math>U(x) = e^{-kx}\int_{x_0}^x \Chi(x) dx</math>. Note that <math>U(x_0) = 0</math> as in this case we are integrating over an empty set, and that U thus defined has <math>U(x) \geq 0</math>. Then |
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<math>U'(x) = -ke^{-kx}\int_{x_0}^x\Chi(x) dx + e^{-kx} \Chi(x) = e^{-kx}(\Chi(x) - k\int_{x_0}^x\Chi(x) dx) \leq 0</math> |
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Then <math>U(x_0) = 0 \and U'(x) = 0 \implies U(x) \leq 0</math>, and <math> 0 \leq U(x) \leq 0 \implies U(x) \equiv 0 \implies \Chi(x) \equiv 0 \implies \Phi(x) \equiv \Psi(x)</math>. |
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<math>\Box</math> |
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Proof of Claim 1: |
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Proof of Claim 1: |
Disclamer: This is a student prepared note based on the lecure of Monday September 21st.
Def.
is called Lipschitz if
(a Lipschitz constant of f) such that
.
Note that any function that is Lipschitz is uniformly continuous, and that if a function f and its derivative are both continous on a compact set then f is Lipschitz.
Thm. Existence and Uniqueness Theorem for ODEs
Let
be continuous and uniformly Lipschitz relative to y. Then the equation
with
has a unique solution
where
where M is a bound of f on
.
This is proven by showing the equation
exists, given the noted assumptions.
Let
and let
.
Claim 1:
is well-defined. More precisely,
is continuous and
,
where b is as referred to above.
Claim 2: For
,
.
Claim 3: if
is a series of functions such that
, with
equal to some finite number, then
converges uniformly to some function
Using these three claims, we have shown that the solution
exists. The proofs of the claims are below.
Proof of Uniqueness:
Suppose
and
are both solutions. Let
.
We have that
for some constant k, which means
, and that
.
Let
. Note that
as in this case we are integrating over an empty set, and that U thus defined has
. Then
Then
, and
.
Proof of Claim 1:
The statement is trivially true for
. Assume the claim is true for
.
is continuous, being the integral of a continuous function.
Proof of Claim 2:
Note that the sequence
has
equal to some finite number.
Proof of Claim 3: Assigned in Homework 3, Task 1