Usbrth

System & Objective

I have the following open system (mass flow) with 2 compartments $A$ and $B$ and constant flow rates $a,b,alpha,beta,gamma,delta$.

I would like an expression for "the proportion of $A(t)$ at time $t$ which originated from $B(t)$", denoted $A_B(t)$.

We can assume originally no mass in $A$ came from $B$ -- i.e. $A_B(t=0) = 0$.

Approach

I assume the proportion could be expressed as a fraction,
$$ A_B(t) = fracN(t)D(t) $$
where the denominator is simply the current mass in $A$, $D(t) = A(t)$; and the numerator $N(t)$ is the absolute mass currently in $A$ which originated from $B$.

I'm struggling to derive an expression for $N(t)$.

I defined the cumulative mass which enters $A$ from $B$ as
$$ T_A_B(t) = int_0^t b B(tau) dtau $$
but we also need an expression for the cumulative mass which exits $A$ after entering from $B$, denoted $E_A_B(t)$. I would then define $N(t) = T_A_B(t) - E_A_B(t)$.

I had three approaches to defining $E_A_B(t)$... is any correct?

Cumulative Exit Attempt 1:

beginequation
beginaligned
E_A_B(t) &= int_0^t (alpha + a) T_A_B(tau) dtau \
&= int_0^t (alpha + a) left(int_0^tau b B(s) ds right) dtau
endaligned
endequation

But this double integral seems wrong and would blow up for large $t$.

Cumulative Exit Attempt 2:

beginequation
E_A_B(t) = int_0^t (alpha + a) b B(tau) dtau
endequation

But this doesn't seem right either, as the exit only depends on the current value of $B(t)$.

Cumulative Exit Attempt 3:

beginequation
E_A_B(t) = int_0^t (alpha + a) A(tau) A_B(tau) dtau
endequation

But this would result in recursive definition of $A_B(t)$, and I'm not sure how I could rearrange to isolate $A_B(t)$ from the the LHS and the RHS integral.

Overall, is this a reasonable approach? Any ideas how to derive either $E_A_B(t)$, or $A_B(t)$ by another approach altogether? I feel like I should be using Laplace.

Thanks,

I'm unsure if my answer does what you want, but the way I see it, we can write the total amount $N(t) = A_B(t) cdot A(t)$ as a differential equation:

$$
dotN(t) = b B(t) - (a + alpha) N(t)
$$
This corresponds to your case 3.

We will also need the differential equation for $A(t)$:
$$
dotA(t) = b B(t) + gamma - (a + alpha) A(t)
$$

Hence, to have $A_B(t)$ we would need to compute the quotient between these two quantities:

$$
A_B(t) = fracN(t)A(t)
$$

Note that, in order for the system to be complete, you will also need the differential equation for $B(t)$
$$
dotB(t) = a A(t) + delta - (b + beta) B(t)
$$

There are just a couple of remarks to the @Ertxiem's answer, which I believe is correct.

1. We implicitely assume that any particle that enters the compartment $B$ immediately acquires the property to be of type $B$. Otherwise, we would need to trace particles from $A$ and $B$ separately and sort out the situations when a particle $x$ leaves $A$, enters $B$ and then reenters $A$ and so on...


2. You can simplify the computations if you subtract the DE for $N$ from the DE for $A$ to get
   $$fracddt(A-N)=gamma-(a+alpha)(A-N),$$
   with the initial condition $A(0)-A_B(0)$. This DE can be solved independently of others.