Assume you have a simple and infinite graph. Choose $x\_{0}$ an arbitrary vertex and consider 
$$
G\_{n}:=\{x\in G:d(x\_0,x)\leq n\}
$$
with the graph metric (hop metric). Now for each pair of nodes there is a unit flow that travels through the minimum path between nodes (if there is more than one minimum path it splits equally along all such paths). The total flow in $G\_{n}$ is equal to $|G\_{n}|^2$.


Given a node $v\in G\_{n}$ we define $T\_{n}(v)$ as the total flow generated in $G\_{n}$ passing through $v$. In other words, $T\_{n}(v)$ is the sum off all the geodesic paths in $G\_{n}$ which are carrying flow and contain the node $v$. Let $M\_{n}$ be the maximum flow
$$
M\_{n}(v):=\max\_{v\in G\_{n}}{T\_{n}(v)}.
$$
For any graph $|G\_{n}|\leq M\_{n}\leq |G\_{n}|^2$. It is not very difficult to see that for $G=\mathbb{Z}^2$ 
$$
M\_{n}=O(|G\_{n}|^{3/2}).
$$



> 
> My question is: can we do better than
>  this in a planar graph? In other
>  words, does there exist a infinite
>  planar graph $G$ such that
>  $M\_{n}=o(|G\_{n}|^{3/2})$? If so, how
>  low we can go.
> 
> 
> 


Note that the same problem could be reformulated by instead of having an infinite graph $G$ having a sequence of graphs $\{G\_{n}\}$ such that $|G\_n|\to\infty$.



Asymptotic Geodesic Flow on Planar Graphs
You can't do better. Lipton and Tarjan's [planar separator theorem](http://en.wikipedia.org/wiki/Planar_separator_theorem) says that any $n$-node planar graph $G=(V,E)$ contains a set $S$ of $O(\sqrt{n})$ vertices whose removal separates the graph into components all of which have size at most $2n/3$. We can then partition $V \setminus S$ into sets $X,Y$ each containing at least $(1/3-o(1))n$ vertices; there are order $n^2$ pairs $(u,v) \in X \times Y$, and any path between such a pair $(u,v)$ contains a vertex of $S$. Since $|S|=O(\sqrt{n})$, by the pigeonhole principle it follows that some element of $S$ is in order $n^{3/2}$ paths. between $X$ and $Y$. 


