I have $n$ vectors $v_1(t), \dots, v_n(t)$. Time is divided into discrete rounds. Initially, all vectors have length $\leq 1$. The vectors at the next time step $t+1$ can be calculated as follows:
\begin{align*} v_1(t+1) &= \frac{v_1(t)}{2\|v_1(t)\|} + \frac{1}{2}v_2(t)\\ v_i(t+1) &= \frac{1}{2}v_{i-1}(t) + \frac{1}{2}v_{i+1}(t) \\ v_n(t+1) &= \frac{1}{2}v_{n-1}(t) + \frac{v_n(t)}{2\|v_n(t)\|}\\ \end{align*} , where $||v_i(t)||$ denotes the Euclidean norm of $v_i(t)$. It turns out that this procedure converges for $t \rightarrow \infty $ and in the end all vectors are the same. However, I cannot predict the limit vector. Does anyone have an idea?
Here's my idea...
It reminds me a discrete version of the heat equation for a vector valued function on the interval $(1,2,...,n)$, with Von Neumann conditions imposing the norm of the function at the boundary to be $1$.
I think this since the discrete version of the laplacian operator at one point of a graph-like domain is just the mean taken on all the adiacent vertexes.
That said we can understand that the limit will have norm equal to $1$, but the exact value seems harder to determine. We may also note that all vectors will have the same limit value if the norm we use in the recursion is not the $\infty$-norm, otherwhise they may take different values.