Sem#6: that text problem

Problem: $\displaystyle \max_{u(t)\in [0,3]}\int_{2020}^{2021}e^{-\delta t}u(t)\: dt$ where $\dot x=-u,\quad x(2020)=5,\quad x(2021)\geq 4.$

• State the conditions from the maximum principle.
• Show that if these conditions hold, then $x^*(2021)=4$.
• Was the optimal control determined by your usual first-order condition?

Let $H(t,x,u,p)=e^{-\delta t}u-pu$.

First question: the conditions.

The conditions are as follows:

1.   u* maximizes (e ^-δt-p) u over u in [0,3]
2. $\ \dot p=0$ with p(2021) ≥ 0, and p(2021)=0 if x*(2021)>4. 
3. $\ \dot x^*=-u^*$ with x*(2020)=5 and x*(2021) ≥4

[as mentioned, #3 is kinda voluntary to include this as per the question formulation, but it has to hold and you need it].

Second question: we end at 4. 

So above we have the conditions. What can we extract from them? 

1. We know that $u^*=\begin{cases}3\quad\text{if }e^{-\delta t}>p\\0\quad\text{if }e^{-\delta t}<p\\\text{((?) not yet any info if }e^{-\delta t}=p\text{)}\end{cases}$
2. So p(t) = a constant P, nonnegative (b/c the transversality condition) and zero if x*(2021)>4.
   From this it follows that e ^-δt-p = e ^-δt- [constant], is a strictly decreasing function. As such, it crosses zero at most once. There the "(?)" issue above is irrelevant: it occurs at at most one point, and that will not affect any integral. 

We are now asked to prove that x*(2021) = 4. What the problem aims at, is to get a contradiction from the contrary assumption: Assume for contradiction that x*(2021) > 4; then p(2021) = 0, hence the constant P is = 0, and thus e ^-δt-p > 0 always - but that leads to u* = 3 constantly, and that leads to x*(2021)=5-3=2, a contradiction. 

If you spotted that, you are done with this question.

What if you didn't see that argument?

If not, here is how to nest the loose threads out: 
By now we know that there are three possible ways u* can develop, depending on the decreasing function e ^-δt - P:

• If that function is always positive - or even, if it hits zero at the righthand endpoint - then u* is =3 on the entire interval [2020, 2021).
  Then x*(t) = x*(0) - 3(t-2020). But that is not admissible: it gives x*(2021) = 2 < 4.  
  Therefore, e ^-δt - P must somewhere be negative.
  • That leads to P>0, hence p(2021)>0, and that implies  x*(2021) = 4. 
    If you didn't spot that, go on! Argue as follows: 
    If on the other extreme e ^-δt - P is negative for all t>2020, then x* is constant and thus = 5.  But then P=0 and thus the assumed-negative e ^-δt - P is > 0. Contradiction! 
    So there must be a crossing. Then P must be positive (being equal to exp(something)), and again ... 
    But even if you don't spot it by then, you can actually compute everything: At the crossing, u* changes from 3 to 0, and afterwards, x is constant. If that constant is >4, we get p(2021)=0.  

Did you use the usual or any FOC to get u*?

It is a vague question. If you differentiate wrt u and put =0, you get e ^-δt - p, right? And p is not the function e ^-δt. That points to a "no", don't you agree?  

But: if your idea of first-order condition for max of a single variable on a closed interval [a,b] is $h'(u)\begin{cases}\leq 0 & \text{if}\quad u=a\\\geq 0 & \text{if}\quad u=b\\ = 0 & \text{otherwise}\end{cases}$ then the answer is "yes".