the following afirmation is true ?
Consider $\Omega $ a bounded and smooth domain . Let $u \in W^{1,p} ( \Omega)$ ( p>1). Supose that $u \geq 0$. let $\alpha >1$ . Then $\nabla u ^{\alpha} = \alpha u^{\alpha -1} \nabla u$ ( in the weak sense ). If this fact is true, its gonna help me a lot ..
Thank you (my english is terrible ,sorry)
On
Let $\Omega=B_1(0)\subset\mathbb{R}^4$ and define $u(x)=\frac{1}{|x|}$ in $\Omega$.This implies that $|\nabla u(x)|=\frac{1}{|x|^2}$. Let $p>1$ and let's calculate $\displaystyle\int_\Omega\frac{1}{|x|^{2p}}dx$. Write $x=r\omega$, where $r\in (0,1)$ and $|\omega|=1$. Note that \begin{eqnarray} \int_\Omega \frac{1}{|x|^{2p}}dx &=& \int_0^1\int_{S_1(0)}\frac{r^3}{r^{2p}}d\omega dr = C \, \int_0^1 r^{3-2\,p} \, d r\nonumber \\ \end{eqnarray}
where $S_1(0)$ is the boundary of $B_1(0)$ and $C>0$. From the last equality we have that for $p\in (1,2)$, $u\in W^{1,p}(\Omega)$. On the other hand, $u^q$ does not belong to $L_{loc}^1(\Omega)$ for $q \ge 4$, so it is nonsense to talk about it's weak derivative.
If $u \in C^\infty(\Omega)$, your assertion holds.
Assert that $W^{1,p}(\Omega) \hookrightarrow L^{p'/(\alpha-1)}(\Omega)$ for $1/p + 1/p' = 1$ (this is anyhow needed in order that $u^{\alpha-1} \, \nabla u \in L^1(\Omega)$). Now, let $u \in W^{1,p}(\Omega)$ be given and $u_n \to u$ in $W^{1,p}(\Omega)$ for some $u_n \in C^\infty$. For any test function $v \in C_0^\infty(\Omega)$ you have $$ \int_\Omega u^\alpha \, \nabla v \, \mathrm{d} x \leftarrow \int_\Omega u_n^\alpha \, \nabla v \, \mathrm{d} x = -\int_\Omega \alpha u_n^{\alpha-1} \, \nabla u_n \, v \, \mathrm{d} x \to -\int_\Omega \alpha u^{\alpha-1} \, \nabla u \, v \, \mathrm{d} x, $$ since $u_n \to u$ in $L^{p'/(\alpha-1)}(\Omega)$, hence $u_n^{\alpha-1} \to u^{\alpha-1}$ in $L^{p'}(\Omega)$ and $u_n^\alpha \to u^\alpha$ in $L^1(\Omega)$ and $u_n^{\alpha-1} \, \nabla u_n \to u^{\alpha-1} \, \nabla u$ in $L^1(\Omega)$.
This shows that your assertion holds for all $u \in W^{1,p}(\Omega)$.