# Solution of p-Laplace equation

If $$u$$ is a solution of $$\Delta_p u=0$$ weakly, then $$u^{+}$$ is also satisfies $$\Delta_p v=0$$ weakly. To prove this result, I need to prove that $$\int_{\Omega}|\nabla v|^{p-2}\nabla v.\nabla \phi\,dx=0$$ where $$v=u^{+}$$ for every $$\phi\in W_{0}^{1,p}(\Omega)$$, whereas I know that $$\int_{\Omega}|\nabla u|^{p-2}\nabla u.\nabla \phi\,dx=0\:\text{ for every }\:\phi\in W_{0}^{1,p}(\Omega).$$

• I don't believe this is true. Consider the 1D-case, where $\Delta_p u = 0$ just means that $u$ is linear. – Michał Miśkiewicz Nov 8 '18 at 14:46
• Thanks, but what about if $\Omega\subset\mathbb{R}^N$ for $N\geq 2$? – Mathlover Nov 9 '18 at 19:46
• The same. If $u$ is a counterexample in 1D, then $v(x_1,\ldots,x_n) = u(x_1)$ is a counterexample in $n$ dimensions. – Michał Miśkiewicz Nov 10 '18 at 10:11