Is there a cut-off function $\theta$ so that, if $\nabla \cdot \varphi = 0$, then $\nabla \cdot (\theta\varphi) = 0$

Question

Let us consider the channels $A = \mathbb R \times[-1, 1]$, $B = \mathbb R \times[-2, 2]$ and a smooth function $\boldsymbol{\varphi}: \Omega \subset \mathbb R^2 \to \mathbb R^2$, for $\Omega$ an open set containing $A$ and $B$, which has a free divergence, $\nabla \cdot \boldsymbol{\varphi} = 0$. I know that we can easily construct a $C^\infty$-function $\theta: \mathbb R^2 \to \mathbb R$ such that $$\theta(x, y) = \begin{cases} 1 & \text{if } |y| \le 1,\\ 0 & \text{if } |y| \ge 2. \end{cases}$$ Now is it possible to construct such a $\theta$ but with the additional property $\theta \boldsymbol{\varphi}$ still has a free divergence, $\nabla \cdot (\theta\boldsymbol{\varphi}) = 0~?$ I have absolutely no clue how to start, any help ?