Sensitivity to Initial Conditions + Parameters.

ODE Existence and Uniqueness

We can recast all high order systems into a first-order vector-valued system. So, for any system:

\begin{equation} x’ = g(t,x, a) \end{equation}

if \(g\) is differentiable across \(t,x\) and \(a\), the IVP given by \(x’ = g(t,x,a)\) and \(x(0) = x_0\), has the property has that:

1. the ODE has a solution \(x(t_0) = x_0\) for any \(t_0\), and any two solutions on the interval coincide as the same solution
2. The only way for a solution to fail to extend temporally is due to the bounds’ \(||x(t)||\) becomes unbounded as \(t\) approaches the endpoints
3. On any interval \(t_0 \leq t \leq T\) the solution \(y_{a,y_0}\) depends continuously on \(a, y_0\), “if I look at my solution sometime later, it would be a non-discontinuous change on the choice of initial condition”

Example

Let’s consider:

Non-Linear ODE

Still Non-Linear ODE

How would we solve equations like:

\begin{equation} \begin{cases} y’’ - 2xy’ + 2\lambda y = 0 \\ y’’ - xy = 0 \end{cases} \end{equation}

Taylor Series

Its time to have a blast from the past! Taylor Series time.

\begin{equation} p_{n}(x) = \sum_{i=0}^{n} \frac{f^{(n)}(0) x^{n}}{n!} \end{equation}

Taylor’s Theorem with Remainder gives us that, at some \(n\), \(|f(x) - p_{n}(x)|\) is bounded.

\begin{equation} |x(t+h) - (x(t) + h x’(t))| \leq Ch \end{equation}

Insight: if your derivatives are bounded, then at high values of \(j\) we have \(\frac{f^{(j)}\qty(0)}{n!}\) tends eventually towards zero as \(n\) increases.