Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 1 / 33C / b / iii

Underthe chain rule gives $dot{a}_j=(a_j/2){\dot{u}}_j$. Part (ii) and $a_j>0$ therefore implyConsequentlyThe conventions $u_0=u_n=-\infty$ give $e^{u_0}=e^{u_n}=0$, exactly matching $a_0=a_n=0$.