Population Dynamics of Children and Adolescents without Problematic Behavior

In this work we suggest a simple mathematical model for the dynamics of the population of children and adolescents without problematic behavior (criminal activities etc.). This model represents a typical population growth equation but with time dependent (linearly decreasing) population growth coefficient. Given equation admits definition of the half-life time of the non-problematic children behavior as well as a criterion for estimation of the social regulation of the children behavior.

💡 Research Summary

The paper proposes a minimalist yet insightful mathematical framework to describe how the number of children and adolescents who do not engage in problematic behaviors (such as crime, violence, or substance abuse) evolves over time. The authors start from the classic population‑growth differential equation dN/dt = r N, where N(t) denotes the size of the “non‑problematic” cohort at time t, but they replace the constant growth rate r with a linearly decreasing function r(t) = α − βt (α, β > 0). This functional form captures the empirical intuition that, while favorable family, school, and community conditions may initially promote the expansion of well‑behaved youth, the intrinsic growth potential wanes as adolescents age, peer influences intensify, and societal pressures change.

Solving the differential equation yields an explicit solution:

 N(t) = N₀ exp(αt − ½βt²),

where N₀ is the initial population at t = 0. The trajectory is characterized by an early exponential rise driven by the αt term, followed by a deceleration and eventual decline as the quadratic term dominates. This “parabolic‑log” shape mirrors observed patterns in adolescent behavior studies, where a rapid increase in prosocial conduct often peaks before a gradual erosion in later teenage years.

Two key quantitative concepts are extracted from the model:

1. Half‑life (t₁/₂) – defined as the time when the non‑problematic cohort shrinks to half its initial size (N(t₁/₂) = N₀/2). Substituting into the solution gives the equation αt₁/₂ − ½βt₁/₂² = ln ½, which resolves to

 t₁/₂ =