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.



Def. f: \mathbb{R}_y \rightarrow \mathbb{R} is called Lipschitz if \exists \epsilon > 0, k > 0 (a Lipschitz constant of f) such that |y_1 - y_2| < \epsilon \implies |f(y_1) - f(y_2)| \leq k |y_1 - y_2|.

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 f:\mathbb{R} = [x_0 - a, x_0 + a] \times [y_0 - b, y_0 + b] \rightarrow \mathbb{R} be continuous and uniformly Lipschitz relative to y. Then the equation \Phi' = f(x, \Phi) with  \Phi(x_0) = y_0 has a unique solution \Phi : [x_0 - \delta, x_0 + \delta] \rightarrow \mathbb{R} where \delta = min(a, ^b/_M) where M is a bound of f on \mathbb{R}.

Proof of Existence

This is proven by showing the equation \Phi(x) = y_0 + \int_{x_0}^x f(t, \Phi(t))dt exists, given the noted assumptions.

Let \Phi_0(x) = y_0 and let \Phi_n(x) = y_0 + \int_{x_0}^x f(t, \Phi_{n-1}(t))dt. IF we can prove the following three claims, we have proven the theorem. The proofs of these claims will follow below.

Claim 1: \Phi_n is well-defined. More precisely, \Phi_n is continuous and \forall x \in [x_0 - \delta, x_0 + \delta], |\Phi_n(x) - y_0| \leq b where b is as referred to above.

Claim 2: For n \geq 1, |\Phi_n(x) - \Phi_{n-1}(x)| \leq \frac{Mk^{n-1}}{n!} |x-x_0|^n.

Claim 3: if  \Phi_n(x) is a series of functions such that |\Phi_n(x) - \Phi_{n-1}(x)| < c_n, with \sum_{n=1}^{\infty} c_n equal to some finite number, then \Phi_n converges uniformly to some function \Phi

Using these three claims, we have shown that the solution \Phi(x) exists.

Proofs of Claims

Proof of Claim 1:

The statement is trivially true for \Phi_0. Assume the claim is true for \Phi_{n-1}. \Phi_n is continuous, being the integral of a continuous function.

|\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


Proof of Claim 2:

 |\Phi_n(x) - \Phi_{n-1}(x)|

 = |\int_{x_0}^x f(t, \Phi_{n-1}(t))dt - \int_{x_0}^x f(t, \Phi_{n-2}(t))dt|

 \leq | \int_{x_0}^x |f(t, \Phi_{n-1}(t)) - f(t, \Phi_{n-2}(t)) | dt  |

 \leq |\int_{x_0}^x k|\Phi_{n-1}(t) - \Phi_{n-2}(t)|dt|

 \leq |\int_{x_0}^x k \frac{M k^{n-2}}{(n-1)!} |t-x_0|^{n-1}dt|

 = \frac{M k^{n-1}}{(n-1)!} \int_0^{|x-x_0|} t^{n-1} dt

 = \frac{M k^{n-1}}{n!} |x-x_0|^n


Note that the sequence  c_n = \frac{M k^{n-1}}{n!} |x-x_0|^n has \sum_{n=1}^{\infty} c_n equal to some finite number.

Proof of Claim 3: Assigned in Homework 3, Task 1, see page for solutions.

Proof of Uniqueness

Suppose \Phi and \Psi are both solutions. Let \Chi(x) = |\Phi(x) - \Psi(x)|.

\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

We have that \Chi \leq k \int_{x_0}^x \Chi(x) dx for some constant k, which means \Chi' \leq k\Chi, and that \Chi(x) \geq 0.

Let U(x) = e^{-kx}\int_{x_0}^x \Chi(x) dx. Note that U(x_0) = 0 as in this case we are integrating over an empty set, and that U thus defined has U(x) \geq 0. Then

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

Then U(x_0) = 0 \and U'(x) = 0 \implies U(x) \leq 0, and  0 \leq U(x) \leq 0 \implies U(x) \equiv 0 \implies \Chi(x) \equiv 0 \implies \Phi(x) \equiv \Psi(x).