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Note: The name of the title is probably not the right one, so if anyone has a better title, shoot.

Let G be a digraph with weighted edges and vertices with real values.

The values of vertices are defined recursively: $$v_{t+1} =v_t -\sum_{w\in n^+(v)}\frac{r(vw)v_t}{\sum_{w\in n^+(v)}r(vw)}+\sum_{w\in n^-(v)}\frac{r(wv)w_t}{\sum_{x\in n^+(w)}r(wx)}$$

Where $v(t)$ is the value of $v$ at time $t$, $n^+(v)$ is the set of vertices whose edges are pointing to $v$, $n^-(v)$ is the set of vertices whose edges are pointing from $v$, $r(wv)$ is the weight of the edge $wv$, and $r(wv) > 0$.

So my question is, how does one go about proving that $$\sum_{v\in V(G)}v(t) = \sum_{v\in V(G)}v(t+1).$$

I have tried proving this for a couple weeks, but honestly I don't even know where to start. Maybe I'm overlooking something trivial, or the question could be transformed into something more solvable, in any case I hope you guys can help me, thanks.

EDIT: The equation has been updated (this was the original equation before I stupidly reduced it and put it here) to account for cases mentioned in the comments and notation has been changed from $v(t)$ to $v_t$.

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    Does $w(t)$ denote the value of vertex $w$ at time $t$? – 2017-02-06
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    "The weight of the edge doesn't depends on the direction. Other than they don't necessarily need to be distinct." That's the edit. I'm new here and I didn't know you couldn't edit comments after five minutes. – 2017-02-06
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    I guess you are missing condition that $G$ is [strongly connected](http://mathworld.wolfram.com/StronglyConnectedDigraph.html). Otherwise your sum tends to zero. E. g. it is easy to see when $G$ is a digraph with 2 vertices and one arc. – 2017-02-06
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    Which sum are you talking about? – 2017-02-06

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To understand why sum remains the same the following intuition (not a formal proof!) should be useful. Let $w_t$ be the power vertex $w$ has at time $t$. What does this vertex do with its power when time goes (i. e., $t$ changes to $t + 1$)? If it has empty $N^+(w)$ then it keeps all its power and probably gets more from $N^-(w)$. If it has non-empty $N^+(w)$ then $w$ send all its power to $N^+(w)$ and each vertex $v \in N^+(w)$ gets some part of power of $w$. This part is proportional to the weight $r(wv)$ of arc $(w, v)$. In this case $w$ also may get more power from $N^-(w)$, but it will send new power later.

More formally I suggest to expand sum and group all positive summands $\frac{r(wv) w_t}{\sum_{x \in N^+(w)} r(wx)}$ by $w$. $$\sum_{v \in V(G)}v_{t + 1} = \sum_{v \in V(G)}\left(v_t - \sum_{w \in N^+(v)} \frac{r(vw) v_t}{\sum_{w \in N^+(v)} r(vw)} + \sum_{w\in N^-(v)}\frac{r(wv) w_t}{\sum_{x \in N^+(w)} r(wx)}\right)\\ = \sum_{v \in V(G)} v_t - \sum_{v \in V(G)}\sum_{w \in N^+(v)} \frac{r(vw) v_t}{\sum_{w \in N^+(v)} r(vw)} + \sum_{v \in V(G)}\sum_{w\in N^-(v)}\frac{r(wv) w_t}{\sum_{x \in N^+(w)} r(wx)}\\ = \sum_{v \in V(G)} v_t - \sum_{v \in V(G)}\sum_{w \in N^+(v)} \frac{r(vw) v_t}{\sum_{w \in N^+(v)} r(vw)} + \sum_{w \in V(G)}\sum_{v\in N^+(w)}\frac{r(wv) w_t}{\sum_{x \in N^+(w)} r(wx)}\\ = \sum_{v \in V(G)}v_t.$$

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    Yes! That's exactly my intuition too, but I just don't know how to turn this intuition into a formal proof. – 2017-02-06
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    I've edited the answer – 2017-02-06
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    How do you go from the second equation to the third, especially the third term? – 2017-02-06
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    I just change the order of summation – 2017-02-06
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    Can you somehow clarify how exactly the third term of the second line is equal to the third term of the third line? – 2017-02-06
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    For simplicity I generalize: $$\sum_{v \in V(G)}\sum_{w \in N^-(v)} f(w, v) = \sum_{(w, v) \in A(G)} f(w, v) = \sum_{w \in V(G)}\sum_{v \in N^+(w)} f(w, v).$$ – 2017-02-06
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    Again in the third line third term, you had: $$\sum_{w \in V(G)}\sum_{v\in N^+(v)}\frac{r(wv) w_t}{\sum_{x \in N^+(w)} r(wx)}$$. Did you mean: $$\sum_{w \in V(G)}\sum_{v\in N^+(w)}\frac{r(wv) w_t}{\sum_{x \in N^+(w)} r(wx)}$$. – 2017-02-06
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    Yes, surely. I've updated the answer. – 2017-02-06