I need to find the final state (i.e. the state at $t\to+\infty$) of the following ODE system: $$\begin{eqnarray*}\frac{dA}{dt}&=&-aAX\\ \frac{dB}{dt}&=&-bBX\\ \frac{dX}{dt}&=&-(aA+bB)X\end{eqnarray*}$$ where $A$, $B$, and $X$ are non-negative variables (with known initial values at $t=0$: $A(0)\ge 0$, $B(0)\ge 0$, $X(0)>0$), $a$ and $b$ are positive constants.

The ODE system appears to have no analytical solution (at least, both me and the automatic analytical ODE solvers have failed to find the solution). But is it possible to find a formula for $A(+\infty)$ and $B(+\infty)$ without having a full formula for $A(t)$ and $B(t)$?

P.S. A "physical" interpretation: the ODE describes the process of irreversible competitive binding. $A$ and $B$ are the amounts of the competing free substances, $X$ is the amount of free "receptors", $a$ and $b$ are the reaction constants ($\approx$reaction speeds). So the game is about what runs out first - the substances or the free receptors. The required $A(+\infty)$ and $B(+\infty)$ give the final ratio of bound substances.

The main pitfall here is that both situations are possible: there can be more substances than receptors ($A(0)+B(0)>X(0)$) or vice versa ($A(0)+B(0)<X(0)$).

  • $\begingroup$ Since the planes A + B - X = k are invariant, can you not reduce to the one parameter family of phase planes: $\dot{A} = -aA(A+B-k)$, $\dot{B} = -bB(a+b-k)$ and do nullcline analysis there? $\endgroup$ – Aaron Hoffman Mar 31 '16 at 21:32
  • $\begingroup$ Well, there is some kind of an "analytic" formula for the solutions, but it is implicit and I don't know whether you need it: if my sloppy calculations are not wrong, $$\int_{A(0)}^{A(t)} \frac{d\tilde{A}}{D_0 \tilde{A} - a \tilde{A}^2 - a C_0 \tilde{A}^{\frac{a+b}{a}}} = t,$$ where the constants $D_0$ and $C_0$ can be expressed in terms of $A_0, B_0, X_0$. Analogously for $B$. $\endgroup$ – Futurologist Apr 1 '16 at 0:26

EDIT: It's obvious there can be no equilibria with all $A,B,X > 0$, and $A,B,X$ are decreasing. The only possibilities are a limit where $X=0$ or a limit where $A=B=0$. As noted, $A+B-X$ is invariant, so the first occurs, with $A(\infty) + B(\infty) = A(0) + B(0) - X(0)$, if $A(0) + B(0) - X(0) \ge 0$ while the latter occurs, with $X(\infty) = -A(0) - B(0) + X(0)$, if $A(0) + B(0) - X(0) \le 0$.

For further analysis of the first case, consider $A$ and $B$ as functions of $X$. The system becomes

$$ \eqalign{\dfrac{dA}{dX} &= \dfrac{aA}{aA + bB}\cr \dfrac{dB}{dX} &= \dfrac{bB}{aA + bB}\cr}$$

This has implicit solutions $$ \eqalign{c_1 A^{b/a} &+ A - X = c_2 \cr B &= c_1 A^{b/a}\cr} $$ We can determine $c_1$ and $c_2$ from the initial conditions, and then take $X=0$ to get the limiting values.

  • $\begingroup$ That is not true, Robert: note that $\frac{dA}{dt}+\frac{dB}{dt}=\frac{dX}{dt}$; so they can be simultaneously zero only if $A(0)+B(0)=X(0)$. $\endgroup$ – Konstantin Avilov Mar 31 '16 at 21:20
  • $\begingroup$ Oops. Fixed it. $\endgroup$ – Robert Israel Mar 31 '16 at 22:07
  • $\begingroup$ Robert, could you kindly explain how you've obtained the equations for the implicit solutions from the "$\frac{d...}{dX}$" ODE system? $\endgroup$ – Konstantin Avilov Mar 31 '16 at 22:26
  • $\begingroup$ Actually it's simplest just to consider $B$ as a function of $A$. You get $\dfrac{dB}{dA} = \dfrac{bB}{aA}$, which is separable and easy to solve. $\endgroup$ – Robert Israel Mar 31 '16 at 22:30
  • $\begingroup$ Oh, thanks... It appears like I have forgotten the basic ODE tricks... My university was too long ago, and there are too many numerical ODE solvers around. =) $\endgroup$ – Konstantin Avilov Mar 31 '16 at 22:35

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