Uniqueness and Existance — Jemoka Knowledge Base

Questions of Uniqueness and Existance are important elements in Differential Equations. Here’s a very general form of a differential equations. First, here’s the: function behavior tests continuity Weakest statement. A function is continuous if and only if:

\begin{equation} \lim_{x \to y} f(x) =f(y) \end{equation}

Lipschitz Condition Stronger statement. The Lipschitz Condition is a stronger test of Continuity such that:

\begin{equation} || F(t,x)-F(t,y)|| \leq L|| x- y|| \end{equation}

for all t \in I, x,y \in \omega, with L \in (0,\infty), named “Lipschitz Constant”, in the dependent variable x. Reshaping this into linear one-dimensional function, we have that:

\begin{equation} \left | \frac{F(t,x)-F(t,y)}{x-y} \right | \leq L < \infty \end{equation}

The important thing here is that its the same L of convergence \forall t. However, L may not be stable—in can oscillate Differentiable We finally have the strongest statement.

\begin{equation} \lim_{x \to y} \frac{f(x)-f(y)}{x-y} = C \end{equation}

To make something Differentiable, it has to not only converge but converge to a constant C. Existence and Uniqueness Check for differential equation Assume some F:I \times \omega \to \mathbb{R}^{n} (a function F whose domain is in some space I \times \omega) is bounded and continuous and satisfies the Lipschitz Condition, and let x_{0} \in \omega, then, there exists T_{0} > 0 and a unique solution for x(t) that touches x_{0} to the standard First-Order Differential Equation \dv{x}{t} = F(t,x), x(t_{0}) = t_{0} for some |t-t_{0}| < T_{0}. To actually check that F satisfies Lipschitz Condition, we pretty much usually just go and take the partial derivative w.r.t. x (dependent variable, yes its x) of F on x, which—if exists on some bound—satisfies the Lipschitz condition on that bound. Proof So we started at:

\begin{equation} \dv{x}{t} = F(t,x), x(t_{0}) = x_{0} \end{equation}

We can separate this expression and integrate:

\begin{align} & \dv{x}{t} = F(t,x) \\ \Rightarrow\ & \dd{x} = F(t,x)\dd{t} \\ \Rightarrow\ & \int_{x_{0)}}^{x(t)} \dd{x} = \int_{t_{0}}^{t} F(s,x(s)) \dd{s} \\ \Rightarrow\ & x(t)-x_{0}= \int_{t_{0}}^{t} F(s,x(s)) \dd{s} \end{align}

At this point, if F is seperable, we can then seperate it out by \dd{t} and taking the right integral. However, we are only interested in existance and uniquness, so we will do something named… Picard Integration Picard Integration is a inductive iteration scheme which leverages the Lipschitz Condition to show that a function integral converges. Begin with the result that all First-Order Differential Equations have shape (after forcibly separating):

\begin{equation} x(t)-x_{0}= \int_{t_{0}}^{t} F(s,x(s)) \dd{s} \end{equation}

We hope that the inductive sequence:

\begin{equation} x_{n+1}(t) = x_{0} + \int_{t_{0}}^{t} F(s,x_{n}(s)) \dd{s} \end{equation}

converges to the same result above (that is, the functions x_{n}(s) stop varying and therefore we converge to a solution x(s) to show existance. This is hard! Here’s a digression/example: if we fix a time t=10: we hope to say that:

\begin{equation} \lim_{n \to \infty } G_{n}(10) = G(10) \end{equation}

\forall \epsilon > 0, \exists M < \infty, \forall n>M,

\begin{equation} |G_{n}(10)-G(10)| < \epsilon \end{equation}

Now, the thing is, for the integral above to converge uniformly, we hope that M stays fixed \forall t (that all of the domain converges at once after the same under of the iterations. Taking the original expression, and applying the following page of algebra to it: Finally, we then apply the Lipschitz Condition because our setup is that F satisfies the Lipschitz Condition, we have that:

\begin{equation} ||x_{n+1}(t)-x_{n}(t)|| \leq L\int_{x_{0}}^{t} ||x_{n}(s)-x_{n-1}(s)||ds \end{equation}