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We consider a discretization of the Laplace operator on $\mathbb Z^2$, https://en.wikipedia.org/wiki/Discrete_Laplace_operator

Then, it is natural to consider its fundamental solution $u$, i.e. $|u(x)|\leq C \ln|x|,\Delta u = 1$ at $(0,0)$ and $\Delta u=0$ elsewhere. I am sure that somewhere it is proven that the discrete derivatives of $u$ behave exactly as we expect them : $|u'(x)|\leq \frac{1}{|x|}, |u''(x)|\leq c\frac{1}{|x|^2}$. But I can not find anything like that in the Internet. Could you provide me with a reference, please? Any information about $c$?

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    $\begingroup$ May be here:proxy.math.rsu.ru/cgi-bin/library/… $\endgroup$
    – Sergei
    Oct 11, 2015 at 9:19
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    $\begingroup$ Try the discrete dbar and its adjoint: both should have homogeneity $-1$ and their convolution should provide the fundamental solution of the Laplace operator. $\endgroup$
    – Bazin
    Oct 14, 2015 at 20:34
  • $\begingroup$ Try "The Laplacian and ¯∂ operators on critical planar graphs" -Kenyon $\endgroup$
    – AHusain
    Jan 14, 2016 at 23:24

3 Answers 3

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For nearest neighbor Laplacian, you can find a short self-contained proof of the formula $$u(x)=\frac1{2\pi}\log |x|+c+O\left(\frac1{|x|^2}\right)$$ at these lecture notes, pages 6-7. It is a simple exercise in Laplace's method to work out further coefficients of the asymptotic expansion, if necessary. The proof there is not the shortest possible; the easiest way would be to compare the double integral formula defining $G_0$ on page 6 with its counterpart for continuous Laplacian. But then you do not get the beautiful formula (4.4) along the way.

"Random Walk: a modern introduction" by Lawler and Limic does the same in much greater generality (not necessarily nearest neighbour), using local CLT with error bounds.

Historically, the first sources usually cited for the above formula are

Stohr, A., [50]: "Uber einige lineare partielle Differenzengleichungen mit konstanten Koefizienten," I, II, and Ill, Mathematische Nachrichten 3 (1950),

or even

W. H. McCrea and F. J. W. Whipple, Random paths in two and three dimensions, Proc. Roy. Soc. Edinburgh 60 (1940),

although the latter apparently proves a slightly weaker result (I have no access to it at the moment to check).

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  • $\begingroup$ Hi, the link to the lecture notes is no longer valid. Could you share these lecture notes? $\endgroup$
    – Leo
    Mar 1 at 22:18
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Formula (71) in Basic Properties Of Discrete Analytic Functions, R. J. Duffin, Duke Math. J. Volume 23, Number 2 (1956), 335-363 (warning: it is not the fist paper where it appears) gives asymptotics $$ 2\pi u(x,y)=\log 4r+\gamma+\cos(4\varphi)/6r^2+O(r^{-4}), $$ where $(x,y)=(r\cos \varphi,r\sin \varphi)$. Remainder term remains the same for discrete derivatives (they are the same as discrete differences like $u(x+1,y)-u(x,y)$, yes?), which in turn are genuine derivatives in intermediate points. Taking derivative of these functions allows to get much more than you ask about.

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Isn't it $$ T(x,y)=\frac{1}{4π}\sum_{(k,l)\in \mathbb Z^2}\ln\bigl((x-k)^2+(y-l)^2\bigr)? $$ Reading the objections below, I note that $T$ is indeed well defined as a distribution on $\mathbb R^2$. Take $\phi\in C_c^0(\mathbb R^2)$, we may consider the bracket of duality $$ \langle T,\phi\rangle=\frac{1}{4π}\sum_{(k,l)\in \mathbb Z^2}\iint \phi(x,y)\ln\bigl((x-k)^2+(y-l)^2\bigr) dx dy, $$ since the sum is finite (thanks to the compact support of $\phi$) and $\mathbb R^2\ni (x,y)\mapsto\ln\bigl((x-k)^2+(y-l)^2\bigr)$ is locally integrable.

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  • $\begingroup$ No, it isn't. This is undefined for various reasons ($(k,l)=(x,y)$, plus the sum is badly divergent), and even if it weren't, then it would be constant. $\endgroup$ Jan 16, 2016 at 19:19
  • $\begingroup$ @Christian Remling I did edit my answer. $\endgroup$
    – Bazin
    Jan 17, 2016 at 19:41
  • $\begingroup$ I still don't think this works. For starters, the fundamental solution is a function of $(x,y)\in\mathbb Z^2$, not $\in\mathbb R^2$. (By the way, I didn't downvote.) $\endgroup$ Jan 17, 2016 at 20:46

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