Adjoint operator of a partial differential operator used to do the carleman estimate

Question

I am reading Carleman estimate for parabolic equations and applications written by Masahiro Yamamoto.
Let $P$ be an operator defined by $$ Pw=\partial_tw-\Delta w+2s\nabla \varphi\cdot\nabla w+(-s\partial_t \varphi-s^2|\nabla\varphi|^2+s\Delta \varphi)w. $$ The article on page 4 says that the way is the decomposition of $P$ into the symmetric part $P_+$ and antisymmetric part $P_-$. To do this, he consider the formal ajoint operator $P^*$ to $P$: $$ (Pw,v)_{L^2(D)}=(w,P^*v)_{L^2(D)},\quad v,w\in C_0^\infty(D). $$ By integration by parts, the Green's theorem and $v,w \in C_0^\infty(D)$, he obtains $$ P^*=-\partial_t w-\Delta w -2s\nabla\varphi\cdot\nabla w-(s\Delta \varphi+s^2|\nabla\varphi|^2+s(\partial_t\varphi))w. $$ It can be seen that the term $s\Delta\varphi w$ change to $-s\Delta \varphi w$. I cannot figure out why this change happens. According to my understanding (may not be true), $$ (s\Delta \varphi w,v)=(w,s\Delta \varphi v), $$ so there must be something wrong with my thinking.

Answer 1

You are right that the part $w\mapsto s(\Delta \phi)w$ is symmetric. However, there is additionally a contribution from the part $w\mapsto 2s\nabla \phi\cdot\nabla w$: $$ \int v\nabla\phi\cdot\nabla w=-\int w\nabla\cdot(v\nabla \phi)=-\int w(\nabla v\cdot\nabla \phi+v\Delta \phi). $$ So you end up with $s(\Delta \phi)w-2s(\Delta \phi)w=-s (\Delta \phi)w$ in the adjoint.