Let $\Omega\subset\mathbb{R}^{N}$ be smooth and bounded. Let $u\in H^{1}_{0}(\Omega)$ be a weak solution of, \begin{align} Lu+\mu u:=-\text{div}(A\nabla u+\textbf{a}u)+(\textbf{b}\cdot\nabla u+mu)+\mu u&=f\in L^{2}(\Omega)\\ u&=0\text{ on }\partial\Omega, \end{align} where $m\in L^{\infty}(\Omega)$, $m\geq 0$ and $m(x)>0$ for some $x\in\Omega$. Also $A\in L^{\infty}(\Omega)^{N\times N}$ and is positive semi-definite, $\textbf{a},\textbf{b},\in L^{\infty}(\Omega)^{N}$ and $\mu>0$.
I want to show that $u$ is monotonic as a function of $m$. That is, if $m_{1}<m_{2}$, then $u_{1}>u_{2}$, where $u_{1}$ and $u_{2}$ are the respective solutions of $Lu+\mu u=f$ with $m:=m_{1}$ and $m:=m_{2}$.
Solution
I have been able to show that, \begin{align} a(u,u)+\mu\|u\|_{2}^{2}\geq p\|u\|_{H^{1}}^{2}, \end{align} where $p>0$. This was done by choosing $\mu$ sufficiently large so that the coefficient of the $\|u\|_{2}^{2}$ component of the $\|u\|_{H^{1}}^{2}$ on the RHS is positive. So it is clear to me that there exists $\mu>0$ sufficiently large so that the above inequality holds.
To show that $u$ is monotonic as a function of $m$ I look at the variational form. Let $u_{d}=u_{1}-u_{2}$ and choose $u_{d}^{-}=u_{1}-u_{2}^{-}:=\max(-(u_{1}-u_{2}),0)$ to be my test function, giving, \begin{align} \int_{\Omega_{-}}-A|\nabla u_{d}|^{2}-(\textbf{a}+\textbf{b})u_{d}\cdot\nabla u_{d}-\mu u_{d}^{2}-m_{2} u_{d}^{2}dx>0, \end{align} where $\Omega_{-}:=\{x\in\Omega\,|\,u_{d}<0\}$. Now assume $\mu>0$ is sufficiently large such that, \begin{align} \textbf{a}u\cdot\nabla v+\textbf{b}v\cdot\nabla u+\mu uv\geq 0\quad\forall v\in H^{1}(\Omega), \end{align} then we have, \begin{align} 0&\leq\int_{\Omega_{-}}\alpha|\nabla u_{d}|^{2}dx\leq \int_{\Omega_{-}}A|\nabla u_{d}|^{2}dx\quad\text{positive semi-definiteness of }A,\\ 0&\leq\int_{\Omega_{-}}(\textbf{a}+\textbf{b})u_{d}\cdot\nabla u_{d}+\mu u_{d}^{2}dx,\\ 0&\leq\int_{\Omega_{-}}m_{2}u_{d}^{2}dx. \end{align} So, \begin{align} \int_{\Omega_{-}}A|\nabla u_{d}|^{2}+(\textbf{a}+\textbf{b})u_{d}\cdot\nabla u_{d}+\mu u_{d}^{2}+m_{2}u_{d}^{2}dx<0, \end{align} however this implies that the sum of non-negative terms is negative, which is not possible. Therefore $u_{d}=u_{1}-u_{2}\geq 0$. From the variational form though, \begin{align} \int_{\Omega}(A\nabla u_{d}+\textbf{a}u_{d})\cdot\nabla v)dx+\int_{\Omega}(\textbf{b}v\cdot\nabla u_{d}+m_{2}u_{d}v)dx+\int_{\Omega}\mu u_{d}vdx=0\not>0, \end{align} if $u_{d}=0$. Therefore $u_{1}-u_{2}>0$. Hence if $m_{1}<m_{2}$ then $u_{1}>u_{2}$.
Concerns
My only concern is the assumption that there exists $\mu>0$ sufficiently large such that, \begin{align} \textbf{a}u\cdot\nabla v+\textbf{b}v\cdot\nabla u+\mu uv\geq 0\quad\forall v\in H^{1}(\Omega). \end{align} I clearly need it to complete the question, but I do not feel confident that this is a safe assumption. In the sense that I don't see why I can make such a claim. In the proof of the coercivity of the bilinear form it was clear, as I was picking $\mu>0$ large enough so that it is bigger than another constant, where as here I need $\mu>0$ large enough so that it dominates other terms dependent on $u,v,\textbf{a}$ and $\textbf{b}$.
Does anyone have any help for me?