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I've stumbled across an older post here trying to solve the same problem the asker of the post had. The solution that was provided stated that for a harmonic function $u$ on $\mathbb{R}^n$ we have that $\nabla u (r\cdot0) = r\nabla u (0)$ for any $r > 0$. I'm not quite sure that this is true. Here's a counter example that I found:

  • $u((x,y)) = e^x \sin(y)$ is harmonic

  • $\nabla u((x,y)) = \begin{pmatrix} e^x \sin(y) \\ e^x \cos(y) \end{pmatrix} $

  • $\nabla u((r\cdot 0,r\cdot 0)) = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \neq r\begin{pmatrix} 0 \\ 1 \end{pmatrix} = r\nabla u((0,0))$

Am I missing something or is the claim indeed wrong?

dragontrainer
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1 Answers1

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You probably got the defintion of $u_r$ wrong. $u_r(x,y)=e^{rx}\sin (ry)$ and $\nabla u_r(0,0)=(0,r)$ and not $(0,1)$

Kavi Rama Murthy
  • 311,013