Show that the interior of a certain triangle is a positive invariant set of a differential system Consider the following autonomous system $x'(t) = f(x(t)),$ written explicitly:
$$\left\{
\begin{array}{ll} 
x_1'(t) = a(1-b) - c \cdot x_1(t) \cdot x_2(t) - a \cdot x_1(t) \\
x_2'(t) = c \cdot x_1(t) \cdot x_2(t) - (d+a) \cdot x_2(t),
\end{array} 
\right.$$ where $a,b,c,d \in \mathbb{R}$ are some constants with:
$$a, c, d > 0 \text{ and } b \in [0,1]. $$
How would we show that the interior of the triangle with vertices in $(0,0), (0,1)$ and $(1,0)$ is a positive invariant set for our system?
More specifically, the set we consider is
$$S = \{(x,y) \in \mathbb{R}^2 \ \mid \ x, y \in (0,1) \text{ and } x+y < 1\}, $$
and we want to show that it is a positive invariant set, i.e. that the following implication is true:
$$(x_1(0), x_2(0)) \in S \implies (x_1(t), x_2(t)) \in S, \forall t > 0. $$
I don't really know how to approach this problem. I thought about considering the behaviour of the right hand side (i.e. of $f$) along the boundary of the triangle. However, I don't really know what to conclude from there, as we are not considering the entire closed triangle (we consider only its interior). Also, I cannot consider the boundary of the triangle as a differentiable curve, as it is only a continuous curve (or a union of three smooth curves).
 A: You should indeed determine the behavior of $f$ along the boundary of the triangle, but in a different manner. You should compute the the inner product of $f$ with the normal vector to the boundary of the triangle (i.e. normal on each side, because it's not continuous).
A more detailed explanation can be found here: Showing that a nonlinear system is positively invariant on a subset of $\mathbb{R}^2$
A: Using a public domain software here is how one can visualize with arrows and integral curves the behavior of the system (case $a=4, b=\tfrac34, c=d=1$):

Let us show how the method referenced to by @C_M works.
This method, based on the dot product with the normal vectors to the sides of the triangle, deserves to be  explained in one of the cases.
Let us detail the less evident one: the case of the dot product of $\binom{x'_1}{x'_2}$ with the normal to the hypotenuse directed towards the inside of the triangle, i.e.,  $\binom{-1}{-1}$ :
Let us consider a point $(x_1(t),x_2(t))$ on this hypotenuse i.e.  verifying
$$x_1(t)+x_2(t)=1 \ \text{with} \ 0<x_1(t)<1, \ \ 0<x_2(t)<1.$$
We have to show that for such a point, the dot product of the two vectors defined above is always positive or in an equivalent way that, for any $t$:
$$x'_1(t)+x'_2(t)<0\tag{1}$$
Adding together the two equations of the initial differential system, we get:
$$x'_1(t)+x'_2(t)=a(1-b-\underbrace{(x_1(t)+x_2(t))}_{= 1, \  \text{due to (1)}})-dx_2=-ab-dx_2\tag{2}$$
a negative quantity, which establishes (1).
The two other cases with normal vectors $\binom{1}{0}$ and $\binom{0}{1}$ can be treated in a similar way.
Remark 1: The point of convergence of the system is $(1-b,0)$.
Remark 2: This differential system looks like a Lotka-Volterra system (classical preys-predators interaction), but it is in fact different. One can be convinced of this fact by making the change of functions:
$$x_1=X_1+(1-b), \ \text{while keeping} \ x_2=X_2$$
leading to the elimination of the constant term $a(1-b)$ (this amounts to  bringing back the fixed point to $0$). But the resulting system would have the form :
$$\begin{cases}X'_1=AX_1X_2+BX_1+CX_2\\X'_2=CX_1X_2+DX_2\\\end{cases}$$
which isn't the form of a Lotka-Volterra system (unless $C:=-c(1-b)=0$ which isn't possible).
