Let $\Omega \subset \mathbf{R^2}$ be open and $a,b,c \in \mathbf{C(\Omega)}$ with $c<0$ on $\Omega$.Show that if $v\in \mathbf{C(\Omega)}$, and $\Delta u + a(x,y)u_x + b(x,y)u_y+c(x,y)=0$ then $u$ does not has a local maximun in $\Omega$.
If $u$ has a local maximun in $\Omega$, then $u$ is constant, $u_x=u_y=0$ and $\Delta u =-c(x,y) > 0 \dots$
How can I continue?
I'm assuming our OP Carlos Andres Henao Acevedo intended $u \in \mathbf C^2(\Omega)$, the space of twice continuously differentiable functions on $\Omega$, since otherwise $\Delta u = \nabla^2 u = \nabla \cdot \nabla u$ does not make much sense. These things being said:
If $u$ has a local maximum $p \in \Omega$, then
$u_x(p) = u_y(p) = 0; \tag 1$
thus
$\nabla^2 u(p) + c(p) = 0, \tag 2$
whence
$\dfrac{\partial^2 u(p)}{\partial x^2} + \dfrac{\partial^2 u(p)}{\partial y^2} = \nabla^2 u(p) = -c(p) > 0; \tag 3$
this implies the trace of the matrix
$H_u(p) = \begin{bmatrix} \dfrac{\partial^2 u(p)}{\partial x^2} & \dfrac{\partial^2 u(p)}{\partial x \partial y} \\ \dfrac {\partial^2 u(p)}{\partial y \partial x} & \dfrac{\partial^2 u(p)}{\partial y^2} \end{bmatrix}, \tag 4$
which is the sum of its eigenvalues, is positive; thus $H_u(p)$ has a positive eigenvalue $\mu$ at $p$, which in turn implies that $p$ cannot be a local maximum for $u$, since $u(x, y)$ will increase in the direction of $\vec v$, the eigenvector corresponding to $\mu$:
$H_u(p) \vec v = \mu \vec v. \tag 5$
Nota Bene: I am assuming without proof my assertions of the last sentence or so, that $u(x, y)$ increases in the $\vec v$ direction; this may be proved by changing coordinates so that the tangent line to one of them is parallel to $\vec v$; but this is a lot of work and I don't have time to type it up at the moment. End of Note.