I am stack at solving the following non linear system
$$\dot{x}= x(1-x-y-z)-ax$$ $$\dot{y}= y(1-x-y-z)-ay$$ $$\dot{z}= z(1-x-y-z)-az$$
where $a<1$ And the empty space 1 − x − y − z.
And I can only find its trivial solution $(0,0,0)$
How can I find its non trivial steady state solution?
The definition of steady state is the point $(x^*,y^*,z^*)$ such that $\dot{x}=0$ , $\dot{y}=0$ , $\dot{z}=0$
That is I need to find the solution of the following system
$$0= x(1-x-y-z)-ax$$ $$0= y(1-x-y-z)-ay$$ $$0= z(1-x-y-z)-az$$
Thanks a lot
Let $w=x+y+z$. Then we have that
$$\dot{w} = \dot{x} + \dot{y} + \dot{z} = w(1-w)-aw$$
$$\implies \int dt = \int\frac{dw}{w(1-a-w)} = \frac{1}{1-a}\int \frac{1}{w}+\frac{1}{1-a-w}dw$$
by separation of variables. Solving, we get
$$(1-a)(t+C) = \log\left(\frac{w}{1-a-w}\right) \implies w(t) = \frac{1-a}{1+Ce^{(a-1)t}}$$
We can plug this back in to get the following system of equations:
$$\begin{cases} \dot{x} = (1-a)\left(1-\frac{1}{1+Ce^{(a-1)t}}\right)x \\ \dot{y} = (1-a)\left(1-\frac{1}{1+Ce^{(a-1)t}}\right)y \\ \dot{z} = (1-a)\left(1-\frac{1}{1+Ce^{(a-1)t}}\right)z \\ \end{cases}$$
which is now fully decoupled. We can solve them by using the fact that
$$\dot{x}(t) = f(t)x(t) \implies x(t) = C\exp\left(\int f(t) dt\right)$$
so we get
$$x(t) = C_x \exp\left((1-a)\int 1 - \frac{e^{(1-a)t}}{e^{(1-a)t}+C}dt \right) = C_x \frac{e^{(1-a)t}}{e^{(1-a)t}+C} = \frac{C_x}{1-a}w(t)$$
with equivalent answers for $y$ and $z$.
To find nontrivial steady state solutions, notice that the $w$ equation was in the form
$$\dot{w} = (1-a-w)w$$
so either $w=0$ or $1-a-w=0$. The steady state solutions are all constants such that
$$w(t) = 1-a$$
or in other words, $x(t)+y(t)+z(t) = 1-a$