= Solution
Use the minimization convention for <Pontryagin maximum principle>: $H=c(x,t)+\lambda^Ta(x,u)$, 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 $\lambda=\lambda_2$. Then
$$
H=z+u(3-\lambda z),\qquad
\dot z=-uz,\qquad\dot\lambda=-1+u\lambda.
$$
The 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 $0<t_*<T$.
Before it, $z=z(0)e^{-t}$; after it, $\lambda=T-t$ and $z=z(0)e^{-t_*}$. Setting $s(t_*)=0$ gives
$$
\boxed{u(t)=
\begin{cases}1,&t<t_*,\\0,&t>t_*,\end{cases}
\qquad3e^{t_*}=z(0)(T-t_*)}.
$$
The 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 $a=\int_0^Tu\,dt$, the cumulative effort by time $t$ is at most $\min(t,a)$. Since $z(t)=z(0)e^{-\int_0^tu}$, the best arrangement for that fixed effort is to advertise immediately at full strength for time $a$. Its cost is
$$
J(a)=z(0)(1-e^{-a})+z(0)(T-a)e^{-a}+3a.
$$
Its <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 $t_*$. This verifies a global optimum, not only a necessary extremal.
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