I'm interested in the limits $$\lim_{t\rightarrow\infty}A\left(t\right),\lim_{t\rightarrow\infty}B\left(t\right),$$ of the system $$\begin{cases} \frac{d}{dt}A\left(t\right)=p_{A}\left(1-A-B\right)\\ \frac{d}{dt}B\left(t\right)=\left(p_{B}+q_{B}\right)\left(1-A-B\right)-q_{B}\left(1-B\right)e^{-\left(p_{A}+p_{B}\right)t} \end{cases}$$ with initial conditions $A(0)=B(0)=0$, where $p_{A},p_{B},q_{B}>0$ are constants.
Since most of what I know about critical points of dynamical systems concerns autonomous ones, I thought about investigating the following autonomous system: $$\begin{cases} \frac{d}{dt}A\left(t\right)=p_{A}\left(1-A-B\right)\\ \frac{d}{dt}B\left(t\right)=\left(p_{B}+q_{B}\right)\left(1-A-B\right)-q_{B}\left(1-B\right)C\left(t\right)\\ \frac{d}{dt}C\left(t\right)=-\left(p_{A}+p_{B}\right)C\left(t\right) \end{cases}$$ where $C(0)=1$. Setting the left hand sides to zero results in the conditions $$A+B =1,C =0,$$ which means there's an infinite number of critical points, which is the set $$\left\{ \left(a,1-a,0\right)|a\in\mathbb{R}\right\}.$$
My question is, is there some way to find the critical point towards the solution converges to?
My attempts at solving said system have failed (other than when $q_{B}=0$, which is a special but simpler case), and I'm hoping there's some other method of finding the limits, that don't require completely solving the system.
Any thoughts would be appreciated on the matter,
Thank you in advance!
You need to ascertain, first, whether the system converges to an equilibrium for the regime of parameters and initial condition proposed. Assuming that it converges, it might be hard to obtain a closed form expression for this limit as a function of the vector of parameters and initial condition.
Below, I prove that it attains equilibrium (Claim 3).
Remark. I will write $x(t)$ instead of $A(t)$ and $y(t)$ instead of $B(t)$.
Claim 1.[Invariant set] $\mathcal{I}=\left\{(x,y)\in\mathbb{R}^2\,:\,(x\geq 0)\,\wedge\,(x+y\leq 1)\right\}$ is invariant w.r.t. the underlying dynamics.
This is quite straightforward to check.
Claim 1 is relevant to proving Claim 2.
Claim 2. [Attractor] Any orbit with initial condition in $\mathcal{I}$ is attracted to the line $\mathcal{A}=\left\{(x,y)\in\mathbb{R}^2\,:\,x+y=1\right\}$. That is, $d(z(t),\mathcal{A})\overset{t\rightarrow +\infty}\longrightarrow 0$ whenever $z(0)=(x(0),y(0))\in \mathcal{I}$ and $z(t)=(x(t),y(t))$ is solution to the ODE.
Proof. Indeed, define $w(t)\overset{\Delta}=(1-x(t)-y(t))$, and you have
$$\dot{w}(t)=-\widetilde{p} w(t)-q_B(1-y(t))e^{-(p_A+p_B)t}\leq -(\widetilde{p}+q_Be^{-(p_A+p_B)t})w(t)\leq -\widetilde{p}w(t),$$
where $\widetilde{p}=p_A+p_B+q_B$; the first inequality above holds since $x(t)\geq 0$ for all $t$ and the second inequality holds since $w(t)\geq 0$ for all $t$. These are the case as we are choosing $(x(0),y(0))\in \mathcal{I}$ and $\mathcal{I}$ is invariant (here is where the invariance is important).
Now, you get that $w(t)\leq e^{-\widetilde{p}t}$, in view of Gronwall's inequality, i.e., $w(t)\overset{t\rightarrow +\infty}\longrightarrow 0$ exponentially fast, which implies that $\mathcal{A}$ is an attractor with Basin of attraction containing $\mathcal{I}$.
Claims 1 and 2 yield the following.
Claim 3. [System attains an equilibrium] Let $z(0)\in \mathcal{I}$. Then, $z(t)\overset{t\rightarrow +\infty}\longrightarrow v$ for some $v\in \mathcal{A}$.
Proof. Indeed, remark that $\dot{x}(t)=p_A w(t)\geq 0$ since $\mathcal{I}$ is invariant (Claim 1). This implies that $x(t)$ is monotonous increasing for solutions with initial condition at $\mathcal{I}$. Claim 2 implies $\dot{x}(t)\longrightarrow 0$. To grant $x(t)\longrightarrow a$ for some $a$, we just need to guarantee that $\dot{x}(t)$ is integrable, i.e., $\int_{0}^{\infty} \dot{x}(s)ds<\infty$. But, this holds since $w(t)$ is bounded by an exponential as shown in Claim 2 (hence $\dot{x}(t)$ is bounded by an exponential, as well). Now, since $d(z(t),\mathcal{A})\longrightarrow 0$ and $x(t)\longrightarrow a$ for some $a$, we have that $y(t)\longrightarrow b$ for some $b$ with $(a,b)\in \mathcal{A}$.