Urns: solving a differential system I'm working on an urn model where two urns, say A and B, are interacting in that at each step, a ball is drawn from each of them and then replaced adding a ball of the color drawn from the other urn. So if my draw vector at time $t$ is $X_t=(X_t^a,X_t^b)=(W,B)$, I'll add a black ball to urn A and a white to urn B. Now, if $Z$ represents the proportion of black balls, I end up having the following dynamics:
\begin{cases} \mathbb{E}\left[\Delta Z_a^t|\mathcal{F}_t\right]=\cfrac{1}{S_a^t+1}\left[Z_b^t-\cfrac{1}{S_a^t}\right]\\ \mathbb{E}\left[\Delta Z_b^t|\mathcal{F}_t\right] =\cfrac{1}{S_b^t+1}\left[Z_a^t-\cfrac{1}{S_b^t}\right] \end{cases}
with S being the total number of balls in the considered urn. Quite naively, I tried to solve the following differential system:
$$
\begin{cases} x'(t)=\frac{1}{t+1}\left[y(t)-\frac{1}{t}\right] \\ y'(t)=\frac{1}{t+1}\left[x(t)-\frac{1}{t}\right] \end{cases}
$$
assuming that each urn is randomly initialized with one ball. The problem is: if I solve this system I end up having functions that are not probabilities at all. I'm not proficient in stochastic calculus so I don't know whether my method is good but I need to add a constraint to my system or if I should solve SDEs rather than ODEs. Do you know how I can get around this?
EDIT: actually a mistake was hidden in the computation of the dynamics. The right system is:
$$
\begin{cases} \mathbb{E}\left[\Delta Z_a^t|\mathcal{F}_t\right]=\cfrac{1}{S_a^t+1}\left[Z_b^t-Z_a^t\right]\\ \mathbb{E}\left[\Delta Z_b^t|\mathcal{F}_t\right] =\cfrac{1}{S_b^t+1}\left[Z_a^t-Z_b^t\right] \end{cases}
$$
So the ordinary differential system associated is
$$
\begin{cases} x'(t)=\frac{1}{t+1}\left[y(t)-x(t)\right] \\ y'(t)=\frac{1}{t+1}\left[x(t)-y(t)\right] \end{cases}
$$
Which makes much more sense but seems tougher to solve.
 A: Premise
I am tackling the problem in two phases,  to use the results of the (more simple) first as a key for the second  :
 - 1st assuming that the ball extracted from one urn is placed in the other, by which the total content of each urn is constant;
 - 2nd assuming (as requested) that each ball extracted is replaced in its own urn, and a ball of same color is added to the other.
Phase 1
The extraction- replacement algorythm proposed can be seen as simulating
a discrete diffusion among the contents of the two urns.
Assume that the first urn contains $n$ white balls and a total of $N$ balls, while the second
contains $m$ white balls and a total of $M$.    
Indexing with $t$ the number of extraction, we have
$$ \bbox[lightyellow] {  
\left\{ \matrix{
  n(t + 1) = n(t) - \Delta m(t) + \Delta n(t) \hfill \cr 
  m(t + 1) = m(t) - \Delta n(t) + \Delta m(t) \hfill \cr}  \right.
 } \tag{1} $$
with $N$ and $M$ remaining constant, as well as the sum $n(t)+m(t)=n(0)+m(0)=s$, since
the number of white (and black) balls remain constant over the two urns.
Then the mechanism of extraction, swap and replace translates into
the fact that the $\Delta n$ can assume only the values $(-1,0,1)$, with probability
$$ \bbox[lightyellow] {  
\Delta n = \left\{ {\matrix{
   1 & {\left( {1 - {n \over N}} \right){m \over M} = } & {{{\left( {N - n} \right)\left( {s - n} \right)} \over {N\,M}}}  \cr 
   0 & {{n \over N}{m \over M} + \left( {1 - {n \over N}} \right)\left( {1 - {m \over M}} \right) = } 
   & {{{M\left( {N - n} \right) - N\left( {s - n} \right) + 2n\left( {s - n} \right)} \over {N\,M}}}  \cr 
   { - 1} & {\left( {1 - {m \over M}} \right){n \over N} = } & {{{n\left( {M + n - s} \right)} \over {N\,M}}}  \cr 
 } } \right.
 } \tag{2} $$
where we omitted the index $t$, and where  the probability is to be understood
as conditional, given $n$ (and $m$ for the first expression).
For $m$ we have an analoguous formula, just by exchanging the letters.
The sum of the above probabilities correctly returns $1$, and the expected value 
of $\Delta n$ results to be
$$ \bbox[lightyellow] {  
E\left( {\Delta n\;\left| n \right.} \right) = {s \over M} - \left( {{{N + M} \over {N\,M}}} \right)n
 } $$
Inserting that into the first expression in (1) and averaging over the $n$ we get
$$ \bbox[lightyellow] {  
\overline n (t + 1) = \left( {1 - {{N + M} \over {N\,M}}} \right)\overline n (t) + {s \over M}
 } $$
the solution to which is
$$ \bbox[lightyellow] {  
\overline n (t) - {{N\,s} \over {N + M}} = \left( {n(0) - {{N\,s} \over {N + M}}} \right)\left( {1 - {{N + M} \over {N\,M}}} \right)^{\,t} 
 } \tag{3} $$
which tells that the compositions of the urns converge exponentially, and with the same decay/grow ratio, to
the equilibrium value, which is as expected:
$$ \bbox[lightyellow] {  
{{\overline n (\infty )} \over N} = {{\overline m (\infty )} \over M} = {{n(0) + m(0)} \over {N + M}}
 } \tag{4} $$
That said for the average values, we shall look now to define the probability distribution for $n(t)$.
Returning to id. (2), that tells us that we can organize a Markov chain in which the 
the probabilities for $n(t)$ to attain the values ${0,1,\cdots,N}$ are collected in the vector
${\bf p}_{\,{\bf n}} (t)$ and where the $(N+1) \times (N+1)$, tri-diagonal, Transition Matrix ${\bf T^T}$
has the rows given by the probability values in (2), i.e. 
$$ \bbox[lightyellow] {  
\eqalign{
  & T^{\,T} _{\,n,\,n + \Delta n}  = p(\Delta n)\quad   \cr 
  & {\bf p}_{\,{\bf n}} (t + 1) = {\bf T}\;{\bf p}_{\,{\bf n}} (t) \cr} 
 } \tag{5} $$
(we use the column vector version of the Markov chain construction).
The matrix $\bf T$ diagonalizes, and the diagonal matrix obtained 
does not depend on $s$.
Therefore it is easy to obtain the probability distribution for
$n$ along all the steps in $t$.
We shall omit here the details of the similitude matrices, and just
refer that therefrom the distribution at the equilibrium, which is independent
from the initial conditions, is given by
$$ \bbox[lightyellow] {  
\left. {p(n)\;} \right|_{\,t = \infty }  = \left( \matrix{
  s \cr 
  n \cr}  \right)\left( \matrix{
  M + N - s \cr 
  N - n \cr}  \right)\,/\,\left( \matrix{
  M + N \cr 
  N \cr}  \right)
 } \tag{6} $$
Phase 2
With the same notation as above now we have
$$ \bbox[lightyellow] {  
\left\{ \matrix{
  N(t + 1) = N(t) + 1 = N(0) + t \hfill \cr 
  M(t + 1) = M(t) + 1 = M(0) + t \hfill \cr 
  n(t + 1) = n(t) + \Delta n(t) \hfill \cr 
  m(t + 1) = m(t) + \Delta m(t) \hfill \cr}  \right.
 } $$
which, for the average values, gives
$$ \bbox[lightyellow] {  
\left\{ \matrix{
  \overline n (t + 1) = \overline n (t) + {{\overline m (t)} \over {M(0) + t}} \hfill \cr 
  \overline m (t + 1) = \overline m (t) + {{\overline n (t)} \over {N(0) + t}} \hfill \cr}  \right.\quad  \Rightarrow \quad \left( {\matrix{
   {\overline n (t + 1)}  \cr 
   {\overline m (t + 1)}  \cr 
 } } \right) = \left( {\matrix{
   1 & {1/\left( {M(0) + t} \right)}  \cr 
   {1/\left( {N(0) + t} \right)} & 1  \cr 
 } } \right)\left( {\matrix{
   {\overline n (t)}  \cr 
   {\overline m (t)}  \cr 
 } } \right)
 } \tag{7} $$
Let's change the variables, taking the sum and the difference of the ratios
$$ \bbox[lightyellow] {  
\left( {\matrix{
   {x(t)}  \cr 
   {y(t)}  \cr 
 } } \right) = \left( {\matrix{
   {1/\left( {N(0) + t} \right)} & {1/\left( {M(0) + t} \right)}  \cr 
   {1/\left( {N(0) + t} \right)} & { - \,1/\left( {M(0) + t} \right)}  \cr 
 } } \right)\left( {\matrix{
   {\overline n (t)}  \cr 
   {\overline m (t)}  \cr 
 } } \right)
 } \tag{8} $$
we reach to
$$ \bbox[lightyellow] {  
\eqalign{
  & \left( {\matrix{
   {x(t + 1)}  \cr 
   {y(t + 1)}  \cr 
 } } \right) = \left( {\matrix{
   1 & {{{N(0) - M(0)} \over {\left( {N(0) + t + 1} \right)\left( {M(0) + t + 1} \right)}}}  \cr 
   0 & {1 - {1 \over {\left( {N(0) + t + 1} \right)}} - {1 \over {\left( {M(0) + t + 1} \right)}}}  \cr 
 } } \right)\left( {\matrix{
   {x(t)}  \cr 
   {y(t)}  \cr 
 } } \right) =   \cr 
  &  = \left( {\matrix{
   1 & {{{2Q} \over {\left( {P + Q + t + 1} \right)\left( {P - Q + t + 1} \right)}}}  \cr 
   0 & {{{\left( {P + t + \sqrt {Q^{\,2}  + 1} } \right)\left( {P + t - \sqrt {Q^{\,2}  + 1} } \right)} \over {\left( {P + Q + t + 1} \right)\left( {P - Q + t + 1} \right)}}}  \cr 
 } } \right)\left( {\matrix{
   {x(t)}  \cr 
   {y(t)}  \cr 
 } } \right) \cr} 
 } \tag{9} $$
where $P=(N(0)+M(0))/2$ and $Q=(N(0)-M(0))/2$.
This matrix is Upper Triangular and the recurrence can be solved in hypergeometric terms, giving
$$ \bbox[lightyellow] {  
\eqalign{
  & y(t) = C\;{{\Gamma \left( {P + \sqrt {Q^{\,2}  + 1}  + t} \right)\;\Gamma \left( {P - \sqrt {Q^{\,2}  + 1}  + t} \right)} \over 
  {\Gamma \left( {P + \,Q + t + 1} \right)\;\Gamma \left( {P - \,Q + t + 1} \right)}} =   \cr 
  &  = C\;\left( {P + \,Q + t + 1} \right)^{\,\overline {\, - \;\left( {1 - \left( {\sqrt {Q^{\,2}  + 1}  - Q} \right)} \right)\,} } 
  \;\left( {P - \,Q + t + 1} \right)^{\,\overline {\, - \;\left( {1 + \left( {\sqrt {Q^{\,2}  + 1}  - Q} \right)} \right)\,} }   \cr 
  & C = {{\Gamma \left( {P + \,Q + 1} \right)\;\Gamma \left( {P - \,Q + 1} \right)} \over {\Gamma \left( {P + \sqrt {Q^{\,2}  + 1} } \right)\;
  \Gamma \left( {P - \sqrt {Q^{\,2}  + 1} } \right)}}y(0)  \cr 
  & x(t) = \left\{ {\matrix{
   {x(0)} & {Q = 0}  \cr 
   {x(0) - {{P^{\,2}  - P - Q^{\,2} } \over Q}y(0) + {{\left( {P + t} \right)^{\,2}  - \left( {P + t} \right) - Q^{\,2} } \over Q}y(t)} & {Q \ne 0}  \cr 
 } } \right. \cr} 
 } \tag{10.a} $$
From the expression for $y(t)$ in terms of Rising Factorials,
applying Stirling approximation, we obtain
$$ \bbox[lightyellow] {  
\left\{ \matrix{
  y(t) \propto {C \over {\left( {P + t + 1} \right)^{\,2}  - \,Q^{\,2} }} \propto 0\quad \left| {\,t \to \infty } \right. \hfill \cr 
  x(t) \propto x(0) + \left[ {0 \ne Q} \right]\left( { - {{P^{\,2}  - P - Q^{\,2} } \over Q}y(0) + {C \over Q}} \right)\quad \left| {\,t \to \infty } 
  \right. \hfill \cr}  \right.
 } \tag{10.b} $$
The formulas above check with computations, as per the attached example.

It remains now to tackle with the distribution ...
A: Wow, thanks for such a comprehensive answer! I'm still trying to figure out how I can relate this type of result to my work. I guess a major difference is that you have the constraint $n(t)+m(t)=n(0)+m(0)=s, \forall t$ and $N$ and $M$ fixed. Actually, not having this is my major issue, as when solving my differential system, I can't find a closed-form solution unless I manage to have a time-independent transition matrix when diagonalizing. This is the case if and only if each urn is initialized with the same number of balls, which is a huge loss of generality as the initial number of balls characterizes the speed of convergence. I guess this is an inherent limitation to the model I use, which may not be the most suitable. I did not mention it but my goal is to extend this model to general network structures. For the moment I can't, apart from the case of regular graphs. Maybe I should look for smoother forms of updating, e.g. randomly picking a color from my neighbours draws, so that at least the $\Delta S_i^t$ is constant over $i$ and $t$. Anyway, it forced me to have a deep look at SDE and stochastic approximation theory and the answers helped a lot :)
A: Wow, I must say I'm impressed by your results! I managed to find quite similar formulas solving the mean-flow ODE system, which was requiring this assumption that, in your notation, $M(0)=N(0)$ which is not satisfactory, though it yields something very similar in flavour. From the dynamics we easily see that we will have fast-pace convergence between the two urns (rate determined by the relative initial total number of balls, 1 in my case times $\frac{s^2}{(s+t)^2}$) and limit average proportion of the average of initial proportions. That is:
$$
\begin{cases} x(t)=\cfrac{a^2\left[Z_a^0-Z_b^0\right]}{2(t+a)^2}+\cfrac{1}{2}\left[Z_b^0+Z_a^0\right] \\  \\y(t)=\cfrac{a^2\left[Z_b^0-Z_a^0\right]}{2(t+a)^2}+\cfrac{1}{2}\left[Z_b^0+Z_a^0\right] \end{cases}
$$
where $x$ and $y$ are the mean-flow proportions and $a$ equals to your $M(0)=N(0)$.
Though I did not manage to bound the fluctuations better than triviality.
From recent papers in this field, I saw that the set of limit point for our system should be contained in the set of limit points from the ODE system, granted it's gradient-like in the closed simplex. So I guess computing a Lyapunov function may help a lot finding limit points in a more general setting (which seems to be how they usually sove this kind of animal in probability theory) while being an elegant way to solve this all. But I'm mostly unexperienced in that domain so it's taking me a lot of time. What I have looks like:
$$
 Z_a^{t+1}-Z_a^t=\cfrac{1}{S_a^0+t+1}\left[\left(Z_b^t-Z_a^t\right)+\left(X_b^{t+1}-\mathbb{E}\left[X_b^{t+1}|\mathcal{F}_n\right]\right)\right]
$$
$$
\Leftrightarrow Z_a^{t+1}-Z_a^t=\gamma_t\left[F(Z^t)+u_t\right]
$$
Where $F$ is a smooth enough drift-mapping and $u$ a martingale difference. Now i need to find a Strict Lyapunov function to identify the equilibria and check for stability.
I'm making some progress though, I hope I'll have nice results to share soon! Thanks a lot again for your help! I did not know the method you use to solve the recurrence, I'm going to check how it works in details.
