12-267/Existence And Uniqueness Theorem
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Disclamer: This is a student prepared note based on the lecures of Friday, September 28th and Monday October 1st.
Lipschitz
Def. [math]\displaystyle{ f: \mathbb{R}_y \rightarrow \mathbb{R} }[/math] is called Lipschitz if [math]\displaystyle{ \exists \epsilon \gt 0, k \gt 0 }[/math] (a Lipschitz constant of f) such that [math]\displaystyle{ |y_1 - y_2| \lt \epsilon \implies |f(y_1) - f(y_2)| \leq k |y_1 - y_2| }[/math].
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.
Statement of Existence and Uniqueness Theorem
Thm. Existence and Uniqueness Theorem for ODEs
Let [math]\displaystyle{ f:\mathbb{R} = [x_0 - a, x_0 + a] \times [y_0 - b, y_0 + b] \rightarrow \mathbb{R} }[/math] be continuous and uniformly Lipschitz relative to y. Then the equation [math]\displaystyle{ \Phi' = f(x, \Phi) }[/math] with [math]\displaystyle{ \Phi(x_0) = y_0 }[/math] has a unique solution [math]\displaystyle{ \Phi : [x_0 - \delta, x_0 + \delta] \rightarrow \mathbb{R} }[/math] where [math]\displaystyle{ \delta = min(a, ^b/_M) }[/math] where M is a bound of f on [math]\displaystyle{ \mathbb{R} }[/math].
Proof of Existence
This is proven by showing the equation [math]\displaystyle{ \Phi(x) = y_0 + \int_{x_0}^x f(t, \Phi(t))dt }[/math] exists, given the noted assumptions.
Let [math]\displaystyle{ \Phi_0(x) = y_0 }[/math] and let [math]\displaystyle{ \Phi_n(x) = y_0 + \int_{x_0}^x f(t, \Phi_{n-1}(t))dt }[/math]. IF we can prove the following three claims, we have proven the theorem. The proofs of these claims will follow below.
Claim 1: [math]\displaystyle{ \Phi_n }[/math] is well-defined. More precisely, [math]\displaystyle{ \Phi_n }[/math] is continuous and [math]\displaystyle{ \forall x \in [x_0 - \delta, x_0 + \delta] }[/math], [math]\displaystyle{ |\Phi_n(x) - y_0| \leq b }[/math] where b is as referred to above.
Claim 2: For [math]\displaystyle{ n \geq 1 }[/math], [math]\displaystyle{ |\Phi_n(x) - \Phi_{n-1}(x)| \leq \frac{Mk^{n-1}}{n!} |x-x_0|^n }[/math].
Claim 3: if [math]\displaystyle{ \Phi_n(x) }[/math] is a series of functions such that [math]\displaystyle{ |\Phi_n(x) - \Phi_{n-1}(x)| \lt c_n }[/math], with [math]\displaystyle{ \sum_{n=1}^{\infty} c_n }[/math] equal to some finite number, then [math]\displaystyle{ \Phi_n }[/math] converges uniformly to some function [math]\displaystyle{ \Phi }[/math]
Using these three claims, we have shown that the solution [math]\displaystyle{ \Phi(x) }[/math] exists.
Proofs of Claims
Proof of Claim 1:
The statement is trivially true for [math]\displaystyle{ \Phi_0 }[/math]. Assume the claim is true for [math]\displaystyle{ \Phi_{n-1} }[/math]. [math]\displaystyle{ \Phi_n }[/math] is continuous, being the integral of a continuous function.
[math]\displaystyle{ |\Phi_n - y_0| }[/math]
[math]\displaystyle{ = |\int_{x_0}^x f(t, \Phi_{n-1}(t))dt| }[/math]
[math]\displaystyle{ \leq |\int_{x_0}^x |f(t, \Phi_{n-1}(t))|dt| }[/math]
[math]\displaystyle{ \leq | \int_{x_0}^x M dt | = M |x_0 - x| }[/math]
[math]\displaystyle{ \leq M \delta }[/math]
[math]\displaystyle{ \leq M \cdot \frac{b}{M} }[/math]
[math]\displaystyle{ = b }[/math]
[math]\displaystyle{ \Box }[/math]
Proof of Claim 2:
[math]\displaystyle{ |\Phi_n(x) - \Phi_{n-1}(x)| }[/math]
[math]\displaystyle{ = |\int_{x_0}^x f(t, \Phi_{n-1}(t))dt - \int_{x_0}^x f(t, \Phi_{n-2}(t))dt| }[/math]
[math]\displaystyle{ \leq | \int_{x_0}^x |f(t, \Phi_{n-1}(t)) - f(t, \Phi_{n-2}(t)) | dt | }[/math]
[math]\displaystyle{ \leq |\int_{x_0}^x k|\Phi_{n-1}(t) - \Phi_{n-2}(t)|dt| }[/math]
[math]\displaystyle{ \leq |\int_{x_0}^x k \frac{M k^{n-2}}{(n-1)!} |t-x_0|^{n-1}dt| }[/math]
[math]\displaystyle{ = \frac{M k^{n-1}}{(n-1)!} \int_0^{|x-x_0|} t^{n-1} dt }[/math]
[math]\displaystyle{ = \frac{M k^{n-1}}{n!} |x-x_0|^n }[/math]
[math]\displaystyle{ \Box }[/math]
Note that the sequence [math]\displaystyle{ c_n = \frac{M k^{n-1}}{n!} |x-x_0|^n }[/math] has [math]\displaystyle{ \sum_{n=1}^{\infty} c_n }[/math] equal to some finite number.
Proof of Claim 3: Assigned in Homework 3, Task 1, see page for solutions.
Proof of Uniqueness
Suppose [math]\displaystyle{ \Phi }[/math] and [math]\displaystyle{ \Psi }[/math] are both solutions. Let [math]\displaystyle{ \Chi(x) = |\Phi(x) - \Psi(x)| }[/math].
[math]\displaystyle{ \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]
We have that [math]\displaystyle{ \Chi \leq k \int_{x_0}^x \Chi(x) dx }[/math] for some constant k, which means [math]\displaystyle{ \Chi' \leq k\Chi }[/math], and that [math]\displaystyle{ \Chi(x) \geq 0 }[/math].
Let [math]\displaystyle{ U(x) = e^{-kx}\int_{x_0}^x \Chi(x) dx }[/math]. Note that [math]\displaystyle{ U(x_0) = 0 }[/math] as in this case we are integrating over an empty set, and that U thus defined has [math]\displaystyle{ U(x) \geq 0 }[/math]. Then
[math]\displaystyle{ 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]
Then [math]\displaystyle{ U(x_0) = 0 \and U'(x) = 0 \implies U(x) \leq 0 }[/math], and [math]\displaystyle{ 0 \leq U(x) \leq 0 \implies U(x) \equiv 0 \implies \Chi(x) \equiv 0 \implies \Phi(x) \equiv \Psi(x) }[/math].
[math]\displaystyle{ \Box }[/math]
