Using the truncation method to prove weak solutions of Lotka-Volterra with diffusion system are nonnegative

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I'm reading Perthame's Parabolic equations in biology book and I'm confused at what he is doing in this proof.

For setup we are considering weak solution $$u\in C(\mathbb{R}^+;L^2(\mathbb{R}^d))$$ to the parabolic equation, $$\frac{\partial n}{\partial t} - \Delta n = n R(t,x),\quad \text{in }\mathbb{R}^d$$ $$n(t=0,x) = n^0(x)$$

The lemma is that given nonnegative initial data in $L^2(\mathbb{R}^d)$ and a locally bounded function $\Gamma(t)$ such that $\lvert R(t,x)\rvert \leq \Gamma(t)$ then our weak solutions are nonnegative.

The proof is trying to show that the negative part will vanish and to go about it he defines $p=-n,p_+ =\max(0,p)$ and uses a smoothing kernel $\omega_{\epsilon}$ defined as,

$$\omega_{\epsilon}(x) = \frac{1}{\epsilon^d}\omega(\frac{x}{\epsilon})$$ $$\omega \in \mathcal{D}(\mathbb{R}^d)$$ $$\omega\geq 0,\quad \int_{\mathbb{R}^d} \omega = 1$$

He also defines a truncation function $\chi_R$ as $$\chi\in \mathcal{D}(\mathbb{R}^d)\quad 0\leq \chi(\cdot)\leq 1\quad \chi_R(x)=\chi(\frac{x}{R})$$ $$\chi(x)=1\text{ for }\lvert x\rvert \leq 1\quad \chi(x)=0\text{ for }\lvert x\rvert\geq 2$$

The first step is regularizing with the smoothing kernel and getting

$$\frac{\partial}{\partial t}\omega_{\epsilon}*p-\Delta\omega_{\epsilon}*p = \omega_{\epsilon} *(p R)$$

Which is fine enough. The next step confuses me. When he uses the truncation function we get these extra terms,

$$ \frac{\partial}{\partial t}\chi_R\omega_{\epsilon}*p-\Delta[\chi_R\omega_{\epsilon}*p] = \chi_R \omega_{\epsilon} * (p R)- 2 \nabla \chi_R \nabla \omega_{\epsilon} *p - \omega_{\epsilon} * p \Delta \chi_R $$

Where did those last two terms come from? Specifically the,

$$- 2 \nabla \chi_R \nabla \omega_{\epsilon} *p - \omega_{\epsilon} * p \Delta \chi_R $$