12-267/Existence And Uniqueness Theorem

Disclamer: This is a student prepared note based on the lecure of Monday September 21st.

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.

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].

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].

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.

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| = |\int_{x_0}^x f(t, \Phi_{n-1}(t))dt| \leq |\int_{x_0}^x |f(t, \Phi_{n-1}(t))|dt| \leq | \int_{x_0}^x M dt | = M |x_0 - x| \leq M \delta \leq M \cdot \frac{b}{M} = b. }[/math]

[math]\displaystyle{ \Box }[/math]