[Differential Equation] Existence and Uniqueness of Partial Differential Equation
Lipschiz Continuity
Definition
A function $g: \mathbb{R}^n \rightarrow \mathbb{R}$ is Lipschiz continuous if
\[\vert g(x) - g(y) \vert \leq L \vert\vert x - y \vert\vert\]for some $L > 0$ and for all $x, y \in \mathbb{D} = \mathbb{R}^n$
Example
$g(x) = x$ is Lipschiz continuous on $\mathbb{R}$ because
\[\vert g(x) - g(y) \vert = \vert x - y\vert \leq L \vert x - y \vert \leftrightarrow (L - 1) \vert x - y \vert \geq 0\]for $x, y \in \mathbb{R}$ and some $L$.
$g(x) = x^2$ is not Lipschiz continuous on $\mathbb{R}$ because there does not exist $L$ such that
\[\begin{align*} \vert g(x) - g(y) \vert = \vert x^2 - y^2 \vert = \vert x + y \vert \cdot \vert x - y \vert &\leq L \vert x - y \vert \\ \leftrightarrow \\ (L - \vert x + y \vert) \vert x - y \vert &\geq 0 \end{align*}\]for $x, y \in \mathbb{R}$.
However, $g(x) = x^2$ is Lipschiz continuous on $[0, 1]$.
Differentiability and Lipschiz Coninuity
A function $f(x)$ is said to be differentiable at a point $c \in \mathbb{R}$ if
\[\displaystyle{\lim_{x \to c}} \dfrac{f(x) - f(c)}{x - c}\]If a function $g(x)$ is differeniable, $g(x)$ is Lipschiz continuous. However although $g(x)$ is Lipschiz continuous, $g(x)$ may not be differentiable. $g(x) = \vert x \vert$ is Lipschiz continuous but is not differentiable.
\[\begin{align*} \vert g(x) - g(y) \vert = \vert \vert x \vert - \vert y \vert \vert \leq \vert x - y \vert \\ \end{align*}\]Uniqueness
If $\vert f(x, t) \vert < M$ and $\vert f(x, t) - f(y, t) < L \vert\vert x - y \vert\vert$ for all $x, y, t$, then
there exists $\epsilon > 0$ such that
\[\begin{align} \dfrac{dx}{dt} &= f(x, t) \\ x(t_0) &= x_0 \end{align}\]has a unique solution in $(t_0 - \epsilon, t_0 + \epsilon)$.
Proof
(1)의 양변을 적분하자.
\[x(T) - x(0) = \int_0^T f(x(t), t)dt\]Picard equation
\[x_1(t) = x_0 + \int_0^t f(x_0(s), s)ds \\ x_2(t) = x_0 + \int_0^t f(x_1(s), s)ds \\ \vdots \\ x_{n+1}(t) = x_0 + \int_0^t f(x_{n}(s), s)ds \\\]다음과 같이 표현할 수 있다.
\[x_n(t) = \sum_{i=1}^n \left( x_i(t) - x_{i-1}(t)\right) + x_0(t)\]다음이 수렴함을 보이자.
\[\sum_{i=1}^{\infty} \left( x_i(t) - x_{i-1}(t)\right) < \infty\] \[\begin{align*} \vert x_i(t) - x_{i-1}(t) \vert &\leq \int_0^t \vert f(x_{i-1}(s), s) - f(x_{i-2}(s), s) \vert ds \\ &\leq \int_0^t L \vert\vert x_{i-1}(s) - x_{i-2}(s) \vert\vert ds \\ &= L \int_0^t \vert\vert x_{i-1}(s) - x_{i-2}(s) \vert\vert ds \end{align*}\]