Houjun Liu

SU-MATH53 FEB052024

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”


Let’s consider:

\begin{equation} y’ = -y \end{equation}

and take the initial value at:

\begin{equation} y(0) = y_0 \end{equation}

we have a solution such that:

\begin{equation} y(t) = y_0e^{-t} \end{equation}

which, at \(y(10)\), we obtain:

\begin{equation} y(10) = y_0e^{-10} \end{equation}

Which brings the question: “how close should \(y_0’\) be such that \(|y’(10) - y(10)| \leq 10^{-5}\)?”

We can recast this as:

\begin{equation} |y_0’ e^{-10} - y_0 e^{-10} | < 10^{-5} \end{equation}


\begin{equation} |y_0’ - y_0| < \frac{10^{-5}}{e^{-10}} \approx \frac{1}{4} \end{equation}

If you flip it over, you will have extreme instability.


\begin{equation} \begin{cases} \dv{x}{t} = a(y-x) \\ \dv{y}{t} = (b-z)x-y \\ \dv{z}{t} = xy-cz \end{cases} \end{equation}

this seems innocuous, but no. If we set our parameters to be weirdly specific values:

\begin{equation} \begin{cases} a \approx 10 \\ b \approx 28 \\ c \approx \frac{8}{3} \end{cases} \end{equation}

These attractors spins across two separate spheres, and the number of times the system spins around a particular area is unknown. It is called…

Deterministic Chaos

Deterministic Chaos is a hard problems which there is a bounded region in which the behavior happens, but the system is bounded.

Another Example

Logistic expression:

\begin{equation} y’ = ry\qty(1-\frac{y}{k}) -h \end{equation}

You can get solutions of this form for some carrying capacity \(k\) and a constant rate of removal \(h\). You can observe that we can build a phase line of this system, and observe. This behavior is called bifurcation: when some \(h\) is high enough, our whole system dies out.

“if the finish rate is too high over other parameters, you just die out.”

You can also draw a plot, where the \(x\) axis is some parameter \(p\), and phase plot can be drawn sideways.

Cauchy Stability

Suppose \(x(t)\) satisfies:

\begin{equation} x’(t) = g(t,x(t)), x(t_0) = x_0 \end{equation}

For some interval \(t \in I\) where the IVP is satisfied; for any time interval \([t_1, t_2]\) inside \(I\) and any \(x_0’\) near to \(x_0\), the associated \(x(t_0) = x_0’\) should exist for the same interval \([t_1, t_2]\) and \(|| x’(t) - x(t) ||\) is small for \(t\).

This extends for not just initial conditions, but also parameters as well. For function parameters \(a_0\) and \(a_0’\).

Newtonian 3-body problem

\begin{equation} m_1 x_1’’ = \frac{-Gm_{1}m_2}{|x_1-x_2|^{2}}- \frac{Gm_{1}m_3}{|x_1-x_3|^{2}} \end{equation}

you will note that this expression has no close form solution, so you can’t do the Cauchy Stability thing to it.