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The adjoint equation is $\dot{p}=-rp$, so $p(t)=p_0{e}^{-rt}$. For the interior stationary candidate from part (i), writing $s=\beta c/\alpha$ gives $p\propto c^{\alpha +\beta -1}$. Therefore it formally haswhich derives the printed formula as a stationary, generally nonoptimal candidate.

For the work-then-consume optimum for a nonconcave consumption-leisure utility, the endpoint maximization in part (i) giveswhenever consumption is chosen, and $\ell =1,c=0$ otherwise. Compare the two endpoint values: leisure and consumption are preferred whenBecause the unconstrained consumption is increasing in time, there is at most one switch: work first, then cease work and consume at an increasing rate.

Here is a global verification, not just a Pontryagin maximum principle condition. The terminal constraint is equivalent toFor any $p_0>0$, pointwise maximization of $c^{\alpha }{s}^{\beta }-p_0{e}^{-rt}(c+s-1)$ gives an upper bound on every feasible objective, after adding $p_0{x}_0$. A control attaining this pointwise maximum and using the full budget attains the bound, so it is globally optimal.

Let $a=r\alpha /(1-\alpha )$. Ifthe optimal switch time is $t_{\ast }=0$ and $C=C_0$. Otherwise choose the unique $t_{\ast }\in (0,T)$ such thatThe left side of the second equality after substitution decreases strictly from $c_{\ast }(e^{aT}-1)/a$ to zero, while the right side increases from $x_0$ to a positive number. Thus the switch exists and is unique. The control is $(\ell ,c)=(1,0)$ before $t_{\ast }$ and $(0,C{e}^{rt/(1-\alpha )})$ after it. It saturates the budget and attains the dual upper bound, completing the proof of optimality for the actual printed parameter range.