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I am stuck with problem 10 from the second chapter of Evan's pde book. It is as follows

Let $\Omega$ be the open half ball, and let $u\in C^2(\overline{\Omega})$, harmonic and 0 on the flat part of the boundary. It defines this function $$v(x)=\begin{cases}u(x)&\text{if }x_n\geq0\\ -u(x_1,\ldots,x_{n-1}, -x_n)&\text{if }x_n<0\end{cases}$$ it claims that it is harmonic. And here is where I am stuck, let $x_n<0$, $j\neq n$ and $\varphi$ the transformation that changes the sign of the last coordinate, then $$\frac{\partial v}{\partial x_j}(x)=-\frac{\partial (u\circ \varphi)}{\partial x_j}(x)=-\sum_{i=1}^n\frac{\partial u}{\partial x_i}(\varphi(x))\frac{\partial \varphi_i}{\partial x_j}=-\frac{\partial u}{x_j}(x_1, \ldots, x_{n-1}, -x_n)$$ So my question is, how can this partial derivative be continuous at $x_n=0$, given that if $x_n>0$ is $$\frac{\partial v}{\partial x_j}=\frac{\partial u}{\partial x_j}$$

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As far as I understand, your problem is that if you approach the boundary from the regions $x_n > 0$ and $x_n < 0$, $\frac{\partial v}{\partial x_j}$ takes opposite limits (if $j \neq n$).

This is true, but on the flat part of the boundary we have $u=0$ and thus $\frac{\partial u}{\partial x_j} = 0$ for $j \neq n$. Hence both limits are zero.

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  • $\begingroup$ I feel extremely dumb now. Overlooking something makes you blind I guess. $\endgroup$ – Smurf Sep 26 '16 at 10:42
  • $\begingroup$ I remember having the exact same problem in the past, so you're not the only blind one here. $\endgroup$ – Michał Miśkiewicz Oct 9 '16 at 11:13

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