# Uniqueness of Cauchy problem

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.