If $u$ is a solution of $F(x) \cdot \nabla u = h(u) $ and $F(x) \cdot v(x)>0 \forall x \in \partial U$ then $u\equiv 0$ 
Let $U\subseteq \mathbb{R}^n$ open and bounded; $0 \in U $ and $\partial U$ is smooth. Let $u \in C^1(\bar{U})$ be a solution of the equation
$$F(x) \cdot \nabla u = h(u),$$
with $F:\bar{U} \to \mathbb{R}^n$ continuous satisfying $F(x)\cdot v(x)>0, \forall x \in \partial U$, where $v(x)$ is the normal unit vector exterior to $\partial U$. If $h$ is a decreasing function and $h(0) = 0$ then $u \equiv 0$.

I have to solve this (yes, it is my homework, I signed up to some classes and I don't have the prerequisites and so I'm lost in almost everything I have to do. Please help me learn. I might need to know something very basic but I'm missing that piece of information).
Now, given thet $\partial U$ is smooth I can use the divergence theorem on this. So far I have this:
$$\int_U \nabla\cdot F(x) dx = \int_{\partial U} F(x) \cdot v(x) dx >0,$$
As $h$ is decreasing and $h(0)=0$ I know $h(x)\leq 0 \forall x \in \mathbb{R}$ and so
$$0\geq\int_U h(u(x))dx = \int_U F(x)\cdot \nabla u(x) dx = \int_U \nabla\cdot [F(x)u(x)]dx - \int_U [\nabla\cdot F(x)]u(x)dx$$
So $$\int_U \nabla \cdot [F(x)u(x)]dx \leq \int_U[\nabla\cdot F(x)]u(x)dx$$
How do I carry on? I get that I should find something on the lines of: "on one hand, on $\partial U$ we have that $F(x)\cdot v(x)$ is always positive, on the other $h(u(x))$ is never positive, so $u$ must be $0$ everyhere".
But I don't know how to do so. Thanks in advance.

Tips provided by the professor today (haven't tried it yet, but I'll post it here already in case someone wants to give it a try as well).
This is just my recollection of what he said in class, it is in no way formally written:
The function u is $C^1(\bar{U})$ and therefore has a maximum (and minimum) value somewhere in $\bar{U}$. Now we have two cases, either the maximum (and minimum) is inside $U$ or is in the boundary $\partial U$.
We'll talk about the maximum but the result follows similarly for the minimum.
If u has it's maximum in $x \in U$ then we have that, at the maximum the gradient of $u$ is $0$. and so
$$h(u(x)) = F(x)\cdot \nabla u(x)  = F(x) \cdot 0 =0$$
as the function $h$ is decreasing, and $h(0) = 0$, we have that $u(x)$ must be $0$, as it is the only place where $h$ assumes the value $0$.
and so the maximum of the function $u(x)=0$. Similarly we find that it's minimum is also $0$, so $u \equiv 0$.
If on the other hand the maximum(and minimum) is on the boundary, then we have a problem. The professor sugested using "Lagrange multiplier" to solve this part.
I have no clue what a lagrange multiplier is, but I'll read and after I learn it I believe I'll be able to solve the rest.
 A: I think I have a solution. Fix me if I am wrong.
There are two different options:

*

*$u$ has it's maximum in $x_{\max} \in U$ we obtain that $u(x_{\max}) = 0$


*$u$ has it's maximum in $x_{\max} \in \partial U$.
Consider the second case:
As far as $\partial U$ is smooth we can choose such function $g(x) \in C^1(\bar{U})$ so that $\partial U = \{x \, |\, g(x) = 0\}$ and the normal vector $\nabla g(x)$ to $\partial U$ will be external hence $\nabla g(x) = \alpha v(x)$ where $\alpha > 0$. So now we can apply Lagrange necessary conditions for a constrained extrema:
$$\exists \lambda: \,  \nabla u(x_{\max}) = \lambda \nabla g(x_{\max})$$
By multiplying left and right part of the last equation by $F(x_{\max})$ we will get:
$$ F(x_{\max}) \nabla u(x_{\max}) = \lambda F(x_{\max}) \nabla g(x_{\max})$$
$$ h(u(x_{\max})) = \lambda \alpha F(x_{\max}) v(x_{\max})$$
By the condition of the problem $F(x_{\max}) v(x_{\max}) > 0$
Let $\lambda \geq 0$ then: $h(u(x_{\max})) > 0 $ which is impossible because $h$ is decreasing and $h(0) = 0$ hence $h(x) < 0$.
So $\lambda < 0$. Which actually mean that $\nabla u(x_{\max})$ is interior normal of the $\partial U$.
Hence if $h > 0$ we obtain that vector ($x_{\max} + h \nabla u(x_{\max})) \in U$. Since $x_{\max}$ is maximum of $u$:
$$ \frac{u(x_{\max} + h \nabla u(x_{\max})) - u(x_{\max})}{h} \leq 0$$
Going to the limit we will get the directional derivative of $u$ at $x_{\max}$ along the vector $\nabla u(x_{\max})$
$$ 0 \leq \nabla u(x_{\max}) \cdot \nabla u(x_{\max}) = \lim_{x \to 0+} \frac{u(x_{\max} + h \nabla u(x_{\max})) - u(x_{\max})}{h} \leq 0$$
Which means that $\nabla u(x_{\max}) = 0$ which is impossible. Thus we got that $x_{\max} \in U \implies u(x_{\max}) = 0$.
Doing similarity for $x_{\min}$ we conclude that:

*

*if $x_{\min} \in U$ we obtain that $u(x_{\min}) = 0$

*if $x_{\min} \in \partial U$ we obtain that $h(u(x_{\min})) = \lambda \alpha F(x_{\min}) v(x_{\min}) < 0$. But since $h$ is decreasing we get $u(x_{\min}) \geq 0 = u(x_{\max})$.

It both cases we get $u(x_{\min}) = u(x_{\max}) = 0$ which implies $u \equiv 0$.
