I have the following problem:
Let $\Omega \subset \mathbb{R}^n$ be an open and bounded subset with piecewise smooth boundary $\partial\Omega$.
$a:\Omega\to]0,\infty[$ is a smooth function.
$f:\Omega\to\mathbb{R}$ is a smooth function.
Show that the following cauchy problem has at most one classical solution.
$\cases{-\nabla\cdot(a\nabla u)=f\quad in\quad \Omega\\u=0\quad on \quad \partial\Omega}$
I think that the energy method could be useful, but i'm not sure how to use it in this case.
Follows directly from standard trick. Let $u,v$ be two classical solutions. Their difference $w=u-v$ solves the homogeneous problem $$-\nabla \cdot (a\nabla w ) = 0, \quad w = 0 \text{ on }\partial \Omega .$$
Multiply by $w$ and integrate over $\Omega$: $$ 0 = \int_\Omega -w\nabla \cdot (a\nabla w ) \overset{\tiny \substack{integration \\by\ parts} }{=} \int_\Omega \nabla w \cdot (a\nabla w) + \int_{\partial \Omega} w a\nabla w\cdot \nu \ dS \\= \int_\Omega a |\nabla w|^2 + 0 $$ so $\nabla w=0$ so $w$ is (locally) constant, and $w|_{\partial \Omega } = 0$ so $w=0$. Therefore, any two solutions coincide.