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A symmetry property for polyharmonic functions vanishing on equidistant hyperplanes

Analysis of PDEs 2015-10-09 v1

Abstract

Let u(t,y)u\left( t,y\right) be a polyharmonic function of order NN defined on the strip (a,b)×Rd\left( a,b\right) \times\mathbb{R}^{d} satisfying the growth condition suptKu(t,y)o(y(1d)/2eπcy) \sup_{t\in K}\left\vert u\left( t,y\right) \right\vert \leq o\left( \left\vert y\right\vert ^{\left( 1-d\right) /2}e^{\frac{\pi}{c}\left\vert y\right\vert }\right) for y\left\vert y\right\vert \rightarrow\infty and any compact subinterval KK of (a,b)\left( a,b\right) , and suppose that u(t,y)u\left( t,y\right) vanishes on 2N12N-1 equidistant hyperplanes of the form {tj}×Rd\left\{ t_{j}\right\} \times\mathbb{R}^{d} for tj=t0+jc(a,b)t_{j}=t_{0}+jc\in\left( a,b\right) and j=(N1),...,N1.j=-\left( N-1\right) ,...,N-1. Then it is shown that u(t,y)u\left( t,y\right) is odd at t0,t_{0}, i.e. that u(t0+t,y)=u(t0t,y)u\left( t_{0}+t,y\right) =-u\left( t_{0}-t,y\right) for yRdy\in\mathbb{R}^{d}. The second main result states that uu is identically zero provided that uu satisfies the growth condition and vanishes on 2N2N equidistant hyperplanes with distance c.c.

Keywords

Cite

@article{arxiv.1510.02299,
  title  = {A symmetry property for polyharmonic functions vanishing on equidistant hyperplanes},
  author = {Ognyan Kounchev and Hermann Render},
  journal= {arXiv preprint arXiv:1510.02299},
  year   = {2015}
}
R2 v1 2026-06-22T11:15:40.809Z