Convergence of a system of ODE's (replicator dynamics) I have to find the points of convergence (i.e. $\lim_{t\to\infty} v^k(t)$) of the following replicator dynamics, given by a system of $4$ linear ODE's:
$$\frac{\dot{v}^k(t)}{v^k(t)}=\alpha[(Av)^k-v^TAv]\quad if\quad v^k(t)>0$$
$$\dot{v}^k(t)=0\qquad if\quad v^k(t)=0$$
given an initial $v^k(0)>0$ for $k=1,2,3,4$.
For what matters $A$ is the following matrix:
$$
A=\begin{bmatrix}
4 & 5 & 4 & 3\\
3 & 4 & 5 & 4\\
2 & 3 & 4 & 5\\
1 & 2 & 3 & 4
\end{bmatrix}
$$
I have never studied dynamical systems, and I only know how to study those of the very simple form:
$$\dot{x}=Ax$$
via the exponential matrix of $A$. So I am clueless.
Any help would be precious, also, I would be very interested in an handy reference about this kind of problems.
 A: This question can be answered using Theorem 7.2.1(b) in the chapter on replicator dynamics in the book of Hofbauer and Sigmund, Evolutionary Games and Population Dynamics.  For solutions taking values in the unit simplex in $\mathbb{R}^4$, this result says that if the solution converges, the limit must be a Nash equilibrium.  Since you know $e_1=[1,0,0,0]^T$ is the unique Nash equilibrium, it is the only possible limit.
The conclusion holds for any solution $v(t)$ in the positive orthant of $\mathbb{R}^4$, as well, for the following reasons: First,  as Futurologist showed, with $e=[1,1,1,1]^T$ one computes that $s(t)=e^Tv(t)=\sum_k v^k(t)$ satisfies
$$
\dot s(t) = (v^TAv)(1-s(t)).
$$
Since  every entry of the matrix $A$ is $1$ or larger, one has $(Av)^k>s$ for all $k$, whence $v^TAv> s^2$.  Hence $s(t)$ is strictly increasing if it is less than $1$, decreasing if greater.
If we assume $\hat v = \lim_{t\to\infty}v(t)$ exists, then from the ODEs it follows $\lim_{t\to\infty} \dot v^k(t)$ exists for each $k$, and these limits must be zero (by a mean-value-theorem argument).   For a similar reason, using the equation for $\dot s(t)$ it follows $e^T\hat v=\lim_{t\to\infty}s(t)=1$. Hence $\hat v$ is a rest point and lies in the closed unit simplex.
If $\hat v$ is not a Nash equilibrium, it follows by direct negation of the definition of such that $$
w^T A\hat v > \hat v^T A \hat v
$$
for some $w$ in the unit simplex.  The left-hand side is a weighted average of components $(A\hat v)^k$, so for some particular component we have $(A\hat v)^k > \hat v^T A \hat v$. For this component, by continuity there is a constant $\delta>0$ such that $(Av)^k-v^TAv>\delta$ for all $v$ in some neighborhood of $\hat v$.  Then the ODE's imply
$$
\frac{\dot v^k(t)}{v^k(t)} = \frac{d}{dt}\log v^k(t) \ge \alpha\delta  >0
$$
for all sufficiently large $t$.  This leads to a contradiction with the assumption that $v^k(t)\to \hat v_k \in [0,1]$.
One should note that this argument does not prove that the Nash equilibrium $e_1$ globally attracts solutions in the positive orthant. It remains possible that some kind of non-convergent behavior persists.  Complicated dynamic behavior is known to be possible in replicator dynamics in general, though perhaps it can be ruled out in this example if, say, a suitable Lyapunov function can be found.
(By the way, I was able to check that $e_1$ is indeed the only Nash equilibrium, but not very easily---if you can say how you know this, please do.)
