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


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.

  • $\begingroup$ 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 Sep 2 '17 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 Dachiraju Sep 2 '17 at 9:04
  • 1
    $\begingroup$ 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 Sep 2 '17 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 Dachiraju Sep 2 '17 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 Sep 2 '17 at 12:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.