why $\int_\Omega S(t) div(w) = \int_\Omega div(w)$ ? where $S(t)$ is a heat semigroup with neumann condition

Question

In this work

https://www.dropbox.com/s/sygzebrr87ma99z/cao_2015_publicado.pdf?dl=0

page 1984, The author claims that

Let $w \in (C^{\infty}_0(\Omega))^N$ then $\int_{\Omega}S(t)div(w)=\int_{\Omega}div(w)$ See,

where $div$ is the divergence operator, S(t) is a heat semigroup with Neumann condition (Analytic semigroup) and $\Omega$ is a open set of $R^N$

I understand the right side (use divergence theorem and $w=0$ in boundary) and the continuation of the proof is clear for me. I know that if div (w) = constant the statement is true. I appreciate your help.

Thanks and sorry for my English

Answer 1

The divergence term is not important here. What is important is that solutions of the heat equation with homogeneous Neumann boundary conditions preserve total heat, so that

$$\int_\Omega S(t) f \, dx = \int_\Omega f \, dx$$

for all $t$. To show this, simply note that $u(x,t)=S(t)f(x)$ solves the heat equation $u_t = \Delta u$ with $\partial u/\partial \nu=0$ on the boundary and $u(x,0) = f(x)$. Integrating by parts we have

$$\frac{d}{dt} \int_\Omega u \,dx = \int_\Omega u_t \, dx = \int_\Omega \Delta u \, dx = \int_{\partial \Omega} \frac{\partial u}{\partial \nu} \, dS = 0.$$

So

$$\int_\Omega S(t) f \, dx = \int_\Omega u(x,t) \, dx = \int_\Omega u(x,0) \, dx = \int_\Omega f(x) \, dx.$$