Let $X=\sum_{j=1}^{n}a_{j}(x)\partial_{x_{j}}$ be a smooth vector fields defined on $\Omega\subset \mathbb{R}^n$, where $\Omega\subset \mathbb{R}^n$ is supposed to be a bounded open domain. Denote $X^{*}$ be the formal adjoint of $X$. I feel confused about the explicit formula about $X^{*}$. I get $X^{*}=-X-\text{div}X$ , but I found some papers (i.e Harnack estimates for degenerate parabolic equations modeled on the subelliptic p-Laplacian) said that $X^{*}=-X+\text{div}X$. I don't know which is correct. Here is my approach. Let $u,v\in C_{0}^{\infty}(\Omega)$, we have $$ (Xu,v)_{L^2(\Omega)}=(u,X^{*}v)_{L^2(\Omega)},$$ That is $$ \int_{\Omega}v\sum_{j=1}^{n}a_{j}(x)\partial_{x_{j}}udx=(u,X^{*}v)_{L^2(\Omega)}.$$ Observe that \begin{align} &\int_{\Omega}v\sum_{j=1}^{n}a_{j}(x)\partial_{x_{j}}udx\\ &=\sum_{j=1}^{n}\int_{\Omega}\partial_{x_{j}}(a_{j}(x)uv)dx-\sum_{j=1}^{n}\int_{\Omega}u\cdot a_{j}(x)\partial_{x_{j}}vdx-\sum_{j=1}^{n}\int_{\Omega}uv\partial_{x_{j}}a_{j}(x)dx\\ &=\sum_{j=1}^{n}\int_{\Omega}\partial_{x_{j}}(a_{j}(x)uv)dx-\int_{\Omega}u\cdot Xv-\int_{\Omega}u\cdot \text{div}X\cdot v dx\\ &=\int_{\partial\Omega} uv\cdot X\cdot \overrightarrow{\mathbf{n}}dS(x)-\int_{\Omega} u(X+\text{div}X)v dx \end{align} Here $\overrightarrow{\mathbf{n}}$ is the outward normal vector of $\partial\Omega$. Since $u,v\in C_{0}^{\infty}(\Omega)$, we have $$\int_{\Omega}v\sum_{j=1}^{n}a_{j}(x)\partial_{x_{j}}udx=-\int_{\Omega} u(X+\text{div}X)v dx$$ Which implies $$ (Xu,v)_{L^2(\Omega)}=(u,(-X-\text{div}X)v)_{L^2(\Omega)}$$.
On the other hand, I found in Wiki that https://en.wikipedia.org/wiki/Differential_operator If we denote $P=\sum_{|\alpha|\leq k}a_{\alpha}(x)D^{\alpha}$ be the $k$ order differential operator, then the adjoint of $P$ should be $$ P^{*}u=\sum_{|\alpha|\leq k}(-1)^{|\alpha|}D^{\alpha}(a_{\alpha}(x)u)$$. In particular, when we choose $P=X$, we can also get above result. Is there exist any mistake in my approach? Can someone help me with it? Thank you very much!
It's just a difference in notation: some people take $\operatorname{div}{X}$ to be $-\sum_i \partial_{x_i}a_i$ so that the Laplacian $\operatorname{div}\operatorname{grad} \phi$ is a positive operator (the "geometrical" convention).