The boundary value of gradient of solution I am considering the function
\begin{equation}
\begin{cases}
-\triangle u = f(u) & \text{in }\Omega\\
u\equiv 0 & \text{on }\partial \Omega
\end{cases}
\end{equation}
where $\Omega\subset R^3$ is open bounded, smooth boundary and $f$ is smooth also. For the fact that I know the solution exist and $u\in C^2(\bar{\Omega})\cap C^\infty(\Omega)$ and $u>0$ inside $\Omega$. 
Now, my question is, can I say that $\nabla u=0$ on $\partial \Omega$? The reason I believe it is true just because $u\in C^2(\bar{\Omega})$ and hence $\nabla u$ is continuous on $\partial \Omega$ which force $\nabla u\equiv 0$ on $\partial \Omega$ b/c $\nabla u\equiv 0$ outside $\Omega$. 
In the end, can I say that $\nabla u$ is the solution of
\begin{equation}
\begin{cases}
-\triangle (\nabla u) = \nabla f(u) & \text{in }\Omega\\
\nabla u\equiv 0 & \text{on }\partial \Omega
\end{cases}
\end{equation}
?
Thank you very much for your help!
 A: I don't think it follows that $\nabla u = 0$ on $\partial \Omega.$ Consider the case when $\Omega = \{x \in \mathbb{R}^3: x_1^2 + x_2^2 + x_3^2 \leq 1\},$ a.k.a the unit ball. Let $f(u) = \lambda u,$ with $\lambda > 0.$ Then suppose we look for radially symmetric solutions to the problem 
\begin{equation}
-\Delta u = \lambda u, \mbox{ in } \Omega; u|\partial \Omega = 0,
\end{equation}
i.e., solutions of the form $u(x) = v(r),$ where $r^2 = \sum_i x_i^2,$ and $r > 0.$ 
Then it follows that $v(r)$ satisfies the ordinary differential equation 
$$
(rv(r))'' - \lambda rv(r) = 0.
$$
Thus we obtain 
$$
v(r) = \frac{c_1 e^{\lambda r} + c_2 e^{-\lambda r}}{r},
$$
for determined values of $c_1, c_2.$ Can you take it from here?
A: Short answer: No. 
Long answer:
Here's a simple counter-example. Take $f=0$ and $\Omega$ to be the unit circle. The question is now: $$\nabla^2u=0$$ on the unit circle and $u(r)=0$ for $|r|=1$. There are infinitely many solutions to this famous equation, and they are called Bessel functions of the first kind. You can see that at $r=1$ they have non-zero derivative.
The directional derivative of $u$ in the direction of the boundary clearly vanishes because $u$ is constant along $\partial \Omega$. However, in the perpendicular direction there's no reason for this to hold. 
