# Parabolic PDE with zero Neumann condition

I'm working with the following parabolic PDE:

$$u_t = \nabla\cdot(\alpha(x)\nabla u)- \beta u\\ x\in\Omega \subset \mathbb{R}^2\\ \partial_n u = 0$$

Given the Neumann boundary condition above, if I want to integrate this PDE, will it be correct to proceed as below?

\begin{alignat}{2} \frac{\mathrm d}{\mathrm dt}\int_\Omega \! u\,\mathrm dx&=\phantom{-}\int_\Omega \! \nabla\cdot (a(x)\nabla u)\,\mathrm dx \:\:&-\int_\Omega \! \beta(x) u \,\mathrm dx \\ &=\phantom{-}\int_{\partial\Omega}\! a(x)\nabla u\cdot n \,\mathrm ds&-\int_\Omega \! \beta(x) u \, \mathrm dx \\ &=\phantom{-}\int_{\partial\Omega} a(x)\partial_n u \,\mathrm ds&-\int_\Omega \! \beta(x) u \, \mathrm dx \\ &=-\int_\Omega \! \beta(x) u \, \mathrm dx \end{alignat}

Just wanted to make sure.

• That looks right. You used Gauss's Theorem to relate the energy due to $\alpha \nabla u$ to the boundary which has zero divergence. So this tells you the effective energy is only dependent on $\beta u$ term. – gdepaul Mar 25 at 21:52