Houjun Liu

Non-Linear ODE

Suppose we analyze first order non-linear system:

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

We can actually turn this into an autonomous system:

\begin{equation} x_0 = t \end{equation}

\begin{equation} x_0’ = 1 \end{equation}

meaning suddenly we have an autonomous system:

\begin{equation} \begin{cases} x_0’ = 1 \\ x_1’ = F(x_0, x_1) \end{cases} \end{equation}

General strategy:

  1. Find zeros of the right side (which are the stationary solutions)
  2. Analyze near-stationary solutions through eigenvalues of the linearized Jacobian matrix: if both eigenvalues are zero
  3. Away from stationary solutions: basically guessing

Three Examples that are Hopeless to Solve

Lotha-Volterra Prey-Predictor Equation

\begin{equation} \begin{cases} x_1’ = 2x_1-x_1x_2 \\ x_2’ = x_1x_2 - 3x_2 \end{cases} \end{equation}

By default, if either \(x_1\) or \(x_2\) goes down, the system dies quickly.


\begin{equation} \begin{cases} x_1’ = r_1x_1 \qty(1- \frac{x_1 + h_{12} x_2}{k_1})\\ x_2’ = r_2x_2 \qty(1- \frac{x_2 + h_{21} x_1}{k_2}) \end{cases} \end{equation}


\begin{equation} \begin{cases} x_1’ = x_2 \\ x_2’ = -\sin x_1 - \gamma x_2 \end{cases} \end{equation}

Strategy to Analyze when its Hopeless

  1. find a stationary solutions: \(x(t) = a\): where \(x’ = F(a) = 0\) and draw them as points on the \(x_1\) and \(x_2\) plane
  2. near each equilibrium point, approximate through Linearilzation
  3. study the mesoscopic region

So, see ODE linearilzation.

Phase Portrait

Phase Portrait is a figure in the \(x_1, x_2\) plane where each solution exists as a curve on the figure.

monotone function

for linearilzation systems that are marginal (zero, negative real parts, one or more fully imaginary), we can’t use linearilzation itself to analyze the system.

Therefore, we have to use a function for which \(\dv t V(y(t)) \geq 0\) or \(\dv V(y(t)) \leq 0\) for all \(t\) called a monotone function, which could give us hints about the function’s behavior.


\begin{align} \dv t V(y(T) &= \nabla V(y(t)) \cdot y’(t) \\ &= \nabla V(y(t)) \cdot F(y(t)) \end{align}

The gradient of \(V\) is always perpendicular to the level curve of \(V\), and—when dotted with \(F\) the vector field of $y$—we obtain a value that’s either positive or negative. When positive, the angle between the vector field \(F\) and \(V\) would be less than \(\frac{\pi}{2}\), meaning the vector field point “outwards” from the level sets. Otherwise, it would be more than \(\frac{\pi}{2}\), meaning the vector field point “inwards”.

conserved function

its like a monotone function, but \(\dv{V}{t} = 0\). any solution curve would lie inside a level curve of \(V\) (parts of the level curve). Its basically the intuition of a monotone function, but the solution curves instead of pointing inwards and outwards, it just get stuck.