Suggest edit — Non-Linear ODE

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title: "Non-Linear ODE"
source: https://www.jemoka.com/posts/kbhnon_linear_ode/
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Suppose we analyze first order non-linear system:
\begin{equation} x&rsquo; = 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&rsquo; = 1 \end{equation}
meaning suddenly we have an autonomous system:
\begin{equation} \begin{cases} x_0&rsquo; = 1 \\ x_1&rsquo; = F(x_0, x_1) \end{cases} \end{equation}
General strategy:
Find zeros of the right side (which are the stationary solutions) Analyze near-stationary solutions through eigenvalues of the linearized Jacobian matrix: if both eigenvalues are zero Away from stationary solutions: basically guessing Three Examples that are Hopeless to Solve Lotha-Volterra Prey-Predictor Equation \begin{equation} \begin{cases} x_1&rsquo; = 2x_1-x_1x_2 \\ x_2&rsquo; = x_1x_2 - 3x_2 \end{cases} \end{equation}
By default, if either $x_1$ or $x_2$ goes down, the system dies quickly.
Example \begin{equation} \begin{cases} x_1&rsquo; = r_1x_1 \qty(1- \frac{x_1 + h_{12} x_2}{k_1})\\ x_2&rsquo; = r_2x_2 \qty(1- \frac{x_2 + h_{21} x_1}{k_2}) \end{cases} \end{equation}
Example \begin{equation} \begin{cases} x_1&rsquo; = x_2 \\ x_2&rsquo; = -\sin x_1 - \gamma x_2 \end{cases} \end{equation}
Strategy to Analyze when its Hopeless find a stationary solutions: $x(t) = a$: where $x&rsquo; = F(a) = 0$ and draw them as points on the $x_1$ and $x_2$ plane near each equilibrium point, approximate through Linearilzation 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&rsquo;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&rsquo;s behavior.
Meaning:
\begin{align} \dv t V(y(T) &amp;= \nabla V(y(t)) \cdot y&rsquo;(t) \\ &amp;= \nabla V(y(t)) \cdot F(y(t)) \end{align}
The gradient of $V$ is always perpendicular to the level curve of $V$, and&mdash;when dotted with $F$ the vector field of $y$&mdash;we obtain a value that&rsquo;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 &ldquo;outwards&rdquo; from the level sets. Otherwise, it would be more than $\frac{\pi}{2}$, meaning the vector field point &ldquo;inwards&rdquo;.
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.