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Let $\mathbb{T}^3=(\mathbb{R}/\mathbb{Z})^3$ be the three-dimensional torus with sides identified. That is, I am considering the unit box $[0,1]^3$ with periodic boundary conditions.

In this case, I am (hopefully) trying to find an explicit formula for the Green function for the Laplacian $-\Delta$.

That is, what would a function $G$ on $\mathbb{T}^3$ satisfying $-\Delta G(x)=\delta^3(x)$ look like explicitly?

I am aware that on whole $\mathbb{R}^3$, $G(x)= \frac{1}{4\pi \lvert x \rvert}$.

According to standard PDE references, I think I need to find the corrector function $\phi^y(x)$ defined by \begin{equation} -\Delta_x \phi^y(x)=0 \text{ for } x \in (0,1)^3 \text{ and } \phi^y(x)=\frac{1}{4\pi \lvert x-y \rvert} \text{ for } x \in \partial \mathbb{T}^3 \end{equation} together with $y \in \mathbb{T}^3$ and $y \neq x$.

However, I have great difficulty solving the above Poisson equation, and moreover, I am not sure if the resulting $\phi^y(x)$ will even be a periodic function.

Could anyone please help me?

Edit: I forgot the fact that the periodic boundary conditions allow zero modes and the Laplacian does not have an inverse under this condition. So, I am restricting the domain of the Laplacian to $L^2$ space over $\mathbb{T}^3$ with no zero modes. Then will it make a difference?

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    $\begingroup$ With periodic boundary conditions, the Laplacian has zero modes, and therefore it doesn't have an inverse. Perhaps the application you have in mind can make use of a pseudo-inverse ... $\endgroup$
    – Michael Engelhardt
    Commented Jul 14, 2023 at 12:58
  • $\begingroup$ @MichaelEngelhardt sorry for not specifying further conditions. I am restricting the domain to the space of functions without zero mode. Will it make a difference? $\endgroup$
    – Isaac
    Commented Jul 14, 2023 at 13:11
  • $\begingroup$ What is the intention with this Green;s function? If you had a use case maybe the original problem you were going to use the Green's function for is more tractable through another means? $\endgroup$
    – Sidharth Ghoshal
    Commented Jul 14, 2023 at 13:20
  • $\begingroup$ @SidharthGhoshal Ok, I have provided more context in which this question has come out: mathoverflow.net/questions/450778/… I would deeply appreciate if you help me. $\endgroup$
    – Isaac
    Commented Jul 14, 2023 at 15:39
  • $\begingroup$ @MichaelEngelhardt Could you also help me with the above link? $\endgroup$
    – Isaac
    Commented Jul 14, 2023 at 15:40

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Well it depends on what you mean by "explicit". Let $(\varphi_k)_k \subset L^2(\mathbb{S}^1)$ be the eigenfunctions of the Laplacian on $\mathbb{S}^1$, these have an explicit form that comes by solving the relative ODE, and let $(\lambda_k)_k$ be the relative eigenvalues. Then $G_{\mathbb{S}^1}(p,q)=\sum_{k\ge 1} \frac{\varphi_k(p)\varphi_k(q)}{\lambda_k} $ and I highly doubt this has a form more explicit that this, meaning some form that does not involve an infinite sum or other similar operations. Since $\mathbb{T^3}=\mathbb{S}^1\times \mathbb{S}^1\times \mathbb{S}^1 $ you can write the Green function of $\mathbb{T}^3$ as a (triple) sum involving only the eigenfunctions $(\varphi_k)_k$ and eigenvalues $(\lambda_k)_k$, I don't think there is a form more explicit that this sum.

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    $\begingroup$ In one dimension, the sum can be carried out, yielding a quite simple form for the Green's function. In higher dimensions, it becomes more complicated. $\endgroup$
    – Michael Engelhardt
    Commented Jul 14, 2023 at 13:24
  • $\begingroup$ @MichaelEngelhardt I didn't know that you can find a closed form in one dimension, thank you for pointing out. Where can I find this explicit form? $\endgroup$
    – Michele Caselli
    Commented Jul 14, 2023 at 13:27
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    $\begingroup$ This is a very beautiful result, i'm not too familiar with green's function so forgive me if this is trivial but is it true in general that a green's function of an operator $O$ over a bounded manifold $M$ can be written as a sum over the eigenfunctions+eigenvalues of said operator over the manifold? $\endgroup$
    – Sidharth Ghoshal
    Commented Jul 14, 2023 at 13:53
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    $\begingroup$ @SidharthGhoshal sure this is always true. In the sense of distributions take the formula $u=\sum_{k\ge 1} \langle u, \varphi_k \rangle_{L^2(M)} \varphi_k$ and put $u=G(x,p)$ for fixed $p\in M$. Use the definition of eigenfunction and integrate by parts to get the Laplacian hit the Green function, you get the formula I have written for any compact manifold. $\endgroup$
    – Michele Caselli
    Commented Jul 14, 2023 at 13:58
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    $\begingroup$ @SidharthGhoshal - this is simply what is known as the spectral representation of the Green's function. $\endgroup$
    – Michael Engelhardt
    Commented Jul 14, 2023 at 14:01

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