Purely quadratic autonomous first order ODE system

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I was trying to figure out the steady state (the behavior of dependent variables as t approaches infinity) of the ODE system, which describes the irreversible biological reactions. The system I am trying to solve is in 3D:

$\frac{dx_1}{dt} = -2a_1x_1^2-a_2x_1x_2-a_3x_1x_3$

$\frac{dx_2}{dt} = 2a_1x_1^2-a_2x_1x_2$

$\frac{dx_3}{dt} = a_2x_1x_2 - a_3x_1x_3$

$x_1(0) = 1, x_2(0)=x_3(0)=0$ and all $x_i \geq 0$, $a_i > 0$

It can be generalized into the form of N dimensional:

$\frac{dx_1}{dt} = -2a_1x_1^2-x_1\sum_{i=2}^N a_ix_i$

$\frac{dx_2}{dt} = x_1(2a_1x_1-a_2x_2)$

$\frac{dx_i}{dt} = x_1(a_{i-1}x_{i-1}-a_ix_i)$ for $i>2$

$x_1(0) = 1, x_i(0)$ for $i>1$ and all $x_i \geq 0$, $a_i > 0$

Basically I would like to know the value of $x_i(t \rightarrow \infty)$. Since $x_1$ is always decreasing and $x_i \geq 0$, so $x_1(t \rightarrow \infty) = 0$. I tried to set the derivatives to zero, but this led to the trivial solution $x_1(t \rightarrow \infty) = 0$, which is already known and this didn't include any information about other variables. Does anyone have any clue how to solve this?

Any kinds of help would be much appreciated! Thank you!

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A partial answer (I don't think there will be a neat analytic way of doing this).

You are right that $x_1=0$ implies all derivatives are zero, no matter the values of the other quantities. This means that the final values of $x_2, x_3$, etc are dependent on the evolution of the system from its initial condition (a different initial condition will result in different final values).

One potential avenue of attack is that if you define a new 'time-like' variable $u$ by $$ \frac{du}{dt} = x_1, \qquad u(0)=0, $$ then the system becomes linear. For your third order system: $$ \frac{d}{du} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} -2a_1 & -a_2 & -a_3 \\ 2a_1 & -a_2 & 0 \\ 0 & a_2 & -a_3 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}, \qquad \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}_{u=0} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} $$ If you could solve this system exactly, you could try to find the time $u^*$ for which $x_1=0$, then substitute that in to find $x_2(u^*)$ and $x_3(u^*)$. But unless the above matrix has some nice eigenvalues to let you do so analytically (which it doesn't as far as I can tell), this will be a numerical job.