Timeline for Monotonic properties of harmonic functions on graphs
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
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| Dec 13, 2010 at 0:02 | comment | added | fedja | Oh, yes, I just noticed the additional $(i,j)\in E$ requirement! Stupid me! OK, back to the drawing board then. | |
| Dec 12, 2010 at 23:59 | comment | added | fedja | Update: I ran simulations and got a clear trouble with N=9. The table of rows i+j=k from k=8 down is (after multiplying by 1000 and rounding down to an integer) (0,0,0,0,0,0,0,0,0),(19,37,53,63,63,53,37,19),(57,93,121,132,121,93,57),(114,167,200,200,167,114),(193,261,288,261,193),(299,376,376,299),(442,514,442),(652,652),(1000). Clearly, the middle of the second row is larger than the end of the third, and the function seems harmonic enough. | |
| Dec 12, 2010 at 23:33 | comment | added | Michał Oszmaniec | Thanks for your comment. When I take a limit $N\rightarrow \infty $ ( while keeping source of unit current at $(0,0)$) potential at $(0,0)$ (or resistance of the network) has a logarithmic asymptotic (when I keep $f|_{V_1^-}=0$). About level sets $x+y=const.$ - for $N\geq 4$ they do not correspond to constant potential: only for $x+y=0$, $x+y=1$ and $x+y=N-1$ you have fixed value of $f$. | |
| Dec 12, 2010 at 23:04 | comment | added | fedja | Yes, we have a singular source ($V^+$), so my intuition would be that we get essentially the renormalized function $-\log|z|$ in the quarter circle that is transferred to the triangle by the conformal map preserving the corners for large $n$. But $x+y=C$ do not correspond to $|z|=c'$ under such map. I'll try to run a few simulations myself a bit later. You may be right, but then it goes against everything I know of the Laplace equation... Note that on the boundary we have 3 adjacent nodes, not 4, and 2 of them go "away". | |
| Dec 12, 2010 at 17:32 | comment | added | Michał Oszmaniec | Indeed. It is not that clear that this will be the case in my example. You can say that I relayed on "physical intuition" when I gave this remark (conservation of the total current that flow perpendicular to $x+y=const.$ cuts) .Yet, I did try both numerics and analytical approach (I was able to get the exact expression for $f$ for this problem) - the property that interests me holds for any size of the $\Delta_N$ (in case of uniform network). As for "continuum limit" - it is singular since you put the "current source" exactly into one of the edges of a unit triangle. | |
| Dec 12, 2010 at 16:04 | comment | added | fedja | It is by no means clear to me that in your example, even if the network is uniform, you have the desired property: when you pass to the limit as $N\to\infty$, you get the usual Laplace equation inside with the Neumann boundary condition on two sides of the unit triangle $x,y>0,x+y<1$, which $1-x-y$ (the only suitable harmonic function that would depend on $x+y$ only) does not satisfy. Did you try numeric experiments with $n>10$? | |
| Dec 12, 2010 at 13:48 | comment | added | Marcin Kotowski | In general, the (unique) harmonic function $f$ is given by $f(x)=\mathbb{E}f_0(X_n)$, where $X_i$ is a random walk starting at $x$ and $X_n$ is the first vertex on the boundary hit by the walk. So if you allow general $p_{ij}$, you can have a single "super fast" path (say, with $p$ along the path equal $1-\epsilon$) to the boundary and other paths going through a region of uniform edge probabilities. If a vertex $v$ is adjacent to that "fast" path, a walk starting there may hit the boundary faster than starting from a nonadjacent vertex $w$, even if $w$ is closer to the boundary. | |
| Dec 12, 2010 at 13:30 | history | edited | Michał Oszmaniec | CC BY-SA 2.5 |
added 5 characters in body; edited body
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| Dec 12, 2010 at 13:12 | history | asked | Michał Oszmaniec | CC BY-SA 2.5 |