Differential equations phase diagram

Question

I am trying to create a phase diagram of two differential equations. (This is an optimization problem, so I am using state(s) and control(c) variables w.r.t. time. $$\begin{align} \frac{dc}{dt}&=\frac{\delta}{n}c\\\frac{ds}{dt}&=\frac{\delta}{n}s-c,&\delta>0, n\geq2 \end{align}$$

I think I've graphed the $\frac{ds}{dt}=0$ line but I don't know how to graph the line for the first equation ($\frac{dc}{dt}=0$). I think that it should be the vertical line at $c=0$, but shouldn't there be another one? Because then how am I supposed to get 4 isosectors for the phase diagram.

This is an image with the system and what I've done so far: https://imgur.io/a/xLAGq1w $$\frac{ds}{dt}=0\text{ if }s=\frac{n}{\delta}c$$

Answer 1

First off, your system only depends on the combined value $p=\frac{\delta}{n}$.

Next, notice your control equation is about as simple as it gets: linear first order homogenous, doesn't even depend on $s$. Its solution is outright just $$c(t)=c(0)e^{pt}$$ without trying.

To get the full solution, as a system of ODEs it's also extremely simple: $$\underbrace{\begin{bmatrix}c\\s\end{bmatrix}^{'}}_{\vec{x}}=\underbrace{\begin{bmatrix}p&0\\-1&p\end{bmatrix}}_{A}\begin{bmatrix}c\\s\end{bmatrix}$$ with therefore $$\vec{x}(t)=e^{At}\vec{x}(0)$$ where $$e^{At}=\begin{bmatrix}e^{pt}&0\\-te^{pt}&e^{pt}\end{bmatrix}$$ from the matrix exponential.

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Back to the phase diagram: you were looking at $\frac{dc}{dt}=0$, which is so simple you're just not believing the solution: $$\frac{dc}{dt}=pc=0\quad\Rightarrow\quad c=0$$

Now you have your two special lines: along the line $c=0$ is vertical vectors (the horizontal component is zero), and $s=c/p$ has horizontal vectors.

You would also need to find stationary points (make both zero, that's the origin only), and any special lines that stay solutions (none, officially set $c=ms$ and solve $c'=ms'$ for some $m$).

Next, your two lines (in general, arbitrary curves) divide the space into 4 quadrants. In each quadrant (in general, arbitrary sectors of space), you get a particular combination of signs of derivatives, so $c>0,s<c/p$ means horizontally to the right and vertically downwards as an arrow.

Plotting the entire thing in Wolfram results in this graph (using $y\mapsto s,x\mapsto c,p=0.5$)