Solutions of p-Laplace equation I found that for the following problem 
\begin{cases}
-\Delta_p u = 1,&x\in B_1(0)\\
u = 0,\quad &x\in\partial B_1(0)
\end{cases}
where $B_1(0)$ is the unitary ball of $\mathbb{R}^N$ and $\Delta_p u = \operatorname{div}(\|\nabla u\|^{p-2}\nabla u) $,
there exists a weak solution of the form:
$ u(x) = C_{N,p} (1-\|x\|^{\frac{p}{p-1}})$ with $C_{N,p}\in\mathbb{R}$ a constant depending just on the dimension of the ball $N$ and the value of $p>2$. 
The problem is that I cannot find the constant $C_{N,p},$ and I need it for the case $N=3$.
 A: Assume the given solution $u$ and note that it is smooth almost everywhere.
Compute the partial derivative (define $C:=C_{N,p}$)
\begin{align}
  \partial_i u
  &= -C \partial_i |x|^\frac{p}{p-1}
  \\
  &= -C \frac{p}{p-1} |x|^{\frac{p}{p-1}-1} \partial_i |x|
  \\
  &= -C \frac{p}{p-1} |x|^{\frac{p}{p-1}-1} \frac{x_i}{|x|}
  \\
  &= -C \frac{p}{p-1} |x|^{\frac{1}{p-1}-1} x_i
  \,.
\end{align}
Gather for the gradient
\begin{align}
  \nabla u
  &= -C \frac{p}{p-1} |x|^{\frac{1}{p-1}-1} x \,.
\end{align}
The norm of the gradient (assume $C>0$)
\begin{align}
  | \nabla u |
  &= C \frac{p}{p-1} |x|^\frac{1}{p-1} \,.
\end{align}
The factor needed for the $p$-Laplacian
\begin{align}
  | \nabla u |^{p-2} \nabla u
  &= \left( C \frac{p}{p-1} |x|^\frac{1}{p-1} \right)^{p-2}
  \left( -C \frac{p}{p-1} |x|^{\frac{1}{p-1}-1} x \right)
  \\
  &= - \left( C \frac{p}{p-1} \right)^{p-1} |x|^{\frac{p-2}{p-1}+\frac{1}{p-1}-1} x
  \\
  &= - \left( C \frac{p}{p-1} \right)^{p-1} x \,.
\end{align}
For the $p$-Laplacian we need to compute the divergence of the above computed factor
\begin{align}
  \Delta_p u
  &= \nabla \cdot \left( | \nabla u |^{p-2} \nabla u \right)
  \\
  &= \nabla \cdot \left( - \left( C \frac{p}{p-1} \right)^{p-1} x \right)
  \\
  &= -N \left( C \frac{p}{p-1} \right)^{p-1}
  \\
  &\overset{!}{=} -1
  \\
  \implies C &= \frac{p-1}{p} N^\frac{1}{1-p} = C_{N,p} \,.
\end{align}
Note, that there are no special cases for $p>2$.
