Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 4 / 29I (Optimization and Control)

Use the minimization convention for Pontryagin maximum principle: $H=c(x,t)+{\lambda }^{T}a(x,u)$, with $\dot{x}=H_{\lambda }$, $\dot{\lambda }=-H_x$, and $u$ minimizing $H$ over its allowed set. The initial state is fixed; since terminal time is fixed and terminal state is free, the transversality condition is $\lambda (T)=\nabla K(x(T))$. No free-terminal-time condition $H(T)=0$ is imposed.

For the advertising problem, ${\lambda }_1(T)={\lambda }_2(T)=0$ and ${\dot{\lambda }}_1=0$, so ${\lambda }_1=0$. Write $z=x_2>0$ and $\lambda ={\lambda }_2$. ThenThe switching function $s=3-\lambda z$ satisfies $\dot{s}=z>0$, while $s(T)=3>0$. Thus there is at most one switch, from $u=1$ to $u=0$, and no singular interval. If $u$ were always zero, then $\lambda (t)=T-t$ and $s(0)=3-z(0)T<0$, contradicting Hamiltonian minimization. Hence a switch occurs at some $0<t_{\ast }<T$.

Before it, $z=z(0)e^{-t}$; after it, $\lambda =T-t$ and $z=z(0)e^{-t_{\ast }}$. Setting $s(t_{\ast })=0$ givesThe left side increases and the right side decreases, so the switch is unique.

Sufficiency can also be proved directly. For any control with total effort $a={\int }_0^{T}udt$, the cumulative effort by time $t$ is at most $\min (t,a)$. Since $z(t)=z(0)e^{-{\int }_0^{t}u}$, the best arrangement for that fixed effort is to advertise immediately at full strength for time $a$. Its cost isIts derivative is $3-z(0)(T-a)e^{-a}$ and its second derivative is $z(0)(T-a+1)e^{-a}>0$ on $[0,T]$. The unique minimizer is exactly the preceding $t_{\ast }$. This verifies a global optimum, not only a necessary extremal.