I work on a bounded domain in $\mathbb{R}^n$. Let $u \in H^1(0,T;H^{-1})\cap L^2(0,T;H^1)$ be a solution of the heat equation:
$$\langle u', v \rangle + \int \nabla u \nabla v = 0$$
for each test function $v$. The solution has the property 
$$\int\_\Omega u(t) = 0$$
for each $t$. 


Is it possible to conclude from this information that $u \equiv 0$? 



Weak solution of a heat equation is zero?
I think the answer to your question is 'no' since the function $e^{-t}\sin (x)$ solves the heat equation on $\Omega=(-1,1)\times \mathbb R\_+$ but the integral is zero by symmetry. If you want an example where $n>1$, just add a dummy variable.


