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In the last few decades, lots of work on first eigenfunction of $p$-Laplace with Dirichlet and other boundary conditions. But I couldn't find much on periodic boundary conditions. I have computed the first eigenfunction for $p=3$, square domain in $\Bbb R^2$. I have shown it as a grey scale image below. One period I tiled in the plane ($4\times 4$ tiles) to illustrate its meeting periodic boundary conditions.

enter image description here

I wonder such a simple structure not have a closed form expression. Can we guess anything on its closed form expression? Any reference to work in this direction (especially in dimension greater than $1$.)

Color picture : enter image description here

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Let $g_p(x)$ be the first eigenfunction of $p$-Laplacian for the domain $(0,1)$ under periodic boundry conitions. Then, for all $p\ge(d+1)$ it can be shown that $$u(\vec{x}) = g_p( \vec{x}\cdot\vec{z}), z \in \mathbb{Z}^d,$$ is an eigenfunction for the $p$-Laplacian for the domain $(0,1)^d$ under periodic boundary conditions.

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  • $\begingroup$ It is mentioned in loc.cit. that $\sin_3$ is the eigenfunction of the 3-Laplacian, so maybe the non-trivial part of the question is to show that this is the eigenfunction with the smallest eigenvalue. $\endgroup$
    – Fan Zheng
    Commented Sep 2, 2017 at 8:34
  • $\begingroup$ @FanZheng : That was for the case 1 dimension. For 2 dimensions or greater, there hasnt been an analytic form discovered for any non trivial eigenfunctions till date, let alone first. $\endgroup$
    – Rajesh D
    Commented Sep 2, 2017 at 9:04
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    $\begingroup$ which brings me to this question: does it work if I just lift the 1-dimensional eigenfunction to the torus, say $\sin_3(x)$ regarded as a function of two variables? $\endgroup$
    – Fan Zheng
    Commented Sep 2, 2017 at 9:15
  • $\begingroup$ @FanZheng : I am at a loss. I really don't know what you are saying, and also I don't know differential geometry. So I cannot get any interpretation of what you are saying and its intent. If its simple enough, you may explain. From my point of view, I am interested only square domains in the Euclidean space $\mathbb{R}^d$. So probably it may help me if you can explain a bit. $\endgroup$
    – Rajesh D
    Commented Sep 2, 2017 at 12:02
  • $\begingroup$ OK. Maybe I have used a little fancy terminology, but I'm really trying to say: Way don't you try $u(x,y)=\sin_3x$ (with appropriate scaling)? BTW, it seems that you missed a factor of $\sqrt2$ or something in the above computation. If that is the case, then $\sin_3x$ would have a smaller eigenvalue. $\endgroup$
    – Fan Zheng
    Commented Sep 2, 2017 at 12:07

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