How to find $q,\beta$ such that $\nabla\cdot[\gamma\nabla u]=0\Leftrightarrow(-\Delta +q)v=0$ for some $v=\beta u $? Let $\Omega$ be an open subset of $\mathbb R^n$
. Let $\gamma\in C^1(\Omega)$ be bounded away from zero. Find $q,\beta\in C^1(\Omega)$ such that
\begin{equation*}
 \nabla\cdot[\gamma\nabla u]=0\Leftrightarrow(-\Delta +q)v=0 \text{ for some }v=\beta u 
\end{equation*}

My attempt:
$$
\begin{aligned}
0&=~ (-\Delta+q)\beta u\\
&=~-(u\Delta \beta+\beta\Delta u+2\nabla u\cdot\nabla \beta)+q\beta u\\
&=~-2\nabla u\cdot\nabla \beta+u(-\Delta \beta+q\beta)-\beta\Delta u
\end{aligned}$$
Also we have
$$
\begin{aligned}
0&=~ \nabla\cdot(\gamma\nabla u)\\
&=~\nabla u\cdot \nabla \gamma+\gamma \Delta u
\end{aligned}
$$
I do not know how should I proceed now.
 A: There's something a little bit funny about the regularity assumptions on $\beta$ and $q$ if you're working with classical solutions since you're asked to evaluate $\Delta(\beta u)$ but only assume $\beta \in C^1(\Omega)$.  I'm going to change the numerology to show the idea of the change of unknown while avoiding this issue.  Indeed, I'll assume $\gamma \in C^2(\Omega)$.
Set $\beta = \gamma^{1/2}$, which belongs to $C^2(\Omega)$ since $\gamma \in C^2(\Omega)$ and is bounded away from $0$.  Note that $\beta$ doesn't vanish, so we can define $q = \Delta \beta / \beta \in C^0(\Omega)$.
Then
$$
\nabla \beta = \frac{1}{2} \gamma^{-1/2} \nabla \gamma 
$$
and
$$
-\Delta \beta + \beta q =0.
$$
Thus if the functions $u$ and $v$ are related by $v = \beta u$, then
$$
(-\Delta + q) v = -\beta \Delta u - 2 \nabla u\cdot \nabla \beta + u(q\beta -\Delta \beta) = -\gamma^{1/2} \Delta u -\gamma^{-1/2} \nabla \gamma \cdot \nabla u \\
= -\gamma^{-1/2} \left(\gamma \Delta u + \nabla \gamma \cdot \nabla u \right)  = -\gamma^{-1/2} \nabla \cdot (\gamma \nabla u). 
$$
We immediately deduce from this that $v$ satisfies $-\Delta v + qv =0$ if and only if $u$ satisfies $\nabla\cdot(\gamma \nabla u) =0$.
