# Weak solution of a heat equation is zero?

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 $u \equiv 0$?

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This might be more appropriate at math.stackexchange. For an answer, what about unit ball as $\Omega$ and $u(x,t)=x_1$? This solves the heat equation and averages to zero. You specified no boundary conditions. –  Joonas Ilmavirta Aug 15 at 11:37

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.