We want to model the effect $B_H(t)$ of recent criminal activity on house $H$. First of all, let's assume that every time a house $H$ is burgled, $B_H(t)$ increases by an amount $\theta.$ This increase isn't going to last forever though: it's going to diminish over time. Write $\omega$ for the time scale over which repeat burglaries are likely to occur in a neighbourhood. Now write $\delta t$ for the length of a time step. So what we're after is a formula for $B_H(t + \delta t)$ in terms of $B_H(t)$. The formula used by Bertozzi and her colleagues is $$B_H(t + \delta t) = B_H(t)(1-\omega \delta t) + \theta E_H(t),$$ where $E_H(t)$ is the number of burglaries that happened at house $H$ at the time $t$. You can see that each of these burglaries contributes an amount of $\theta.$ Burglaries that happened before that time are accounted for by $B_H(t),$ but the amount by which they contribute is determined by the relationship of $\omega$ and the length of the time step. The larger $\omega \delta t,$ the smaller the contribution of $B_H(t)$ (we assume that $\omega \delta t$ is less than $1$). This formula only reflects how recent burglaries at house $H$ affect $B_H(t).$ But we also need to take account of the effect of recent burglaries at neighbouring houses. To do this, replace $B_H(t)$ in the equation above by $$(1-\eta)B_H(t) + \frac{\eta}{z} S_B(t),$$ where $\eta$ is a number between 0 and 1 that measures how significant the effect of crimes in adjacent houses is. The parameter $z$ is the number of houses that directly neighbour $H$ (so in a square lattice that number is four). $S_B(t)$ is the sum of the $B_J(t)$ for all the houses $J$ adjacent to $H.$ The formula now becomes $$B_H(t + \delta t) = \left( (1-\eta)B_H(t) + \frac{\eta}{z} S_B(t)\right)(1-\omega \delta t)+\theta E_H(t).$$
We want to model the effect $B_H(t)$ of recent criminal activity on house $H$. First of all, let's assume that every time a house $H$ is burgled, $B_H(t)$ increases by an amount $\theta.$ This increase isn't going to last forever though: it's going to diminish over time. Write $\omega$ for the time scale over which repeat burglaries are likely to occur in a neighbourhood. Now write $\delta t$ for the length of a time step. So what we're after is a formula for $B_H(t + \delta t)$ in terms of $B_H(t)$. The formula used by Bertozzi and her colleagues is $$B_H(t + \delta t) = B_H(t)(1-\omega \delta t) + \theta E_H(t),$$ where $E_H(t)$ is the number of burglaries that happened at house $H$ at the time $t$. You can see that each of these burglaries contributes an amount of $\theta.$ Burglaries that happened before that time are accounted for by $B_H(t),$ but the amount by which they contribute is determined by the relationship of $\omega$ and the length of the time step. The larger $\omega \delta t,$ the smaller the contribution of $B_H(t)$ (we assume that $\omega \delta t$ is less than $1$). This formula only reflects how recent burglaries at house $H$ affect $B_H(t).$ But we also need to take account of the effect of recent burglaries at neighbouring houses. To do this, replace $B_H(t)$ in the equation above by $$(1-\eta)B_H(t) + \frac{\eta}{z} S_B(t),$$ where $\eta$ is a number between 0 and 1 that measures how significant the effect of crimes in adjacent houses is. The parameter $z$ is the number of houses that directly neighbour $H$ (so in a square lattice that number is four). $S_B(t)$ is the sum of the $B_J(t)$ for all the houses $J$ adjacent to $H.$ The formula now becomes $$B_H(t + \delta t) = \left( (1-\eta)B_H(t) + \frac{\eta}{z} S_B(t)\right)(1-\omega \delta t)+\theta E_H(t).$$