Sem#5: 8-01 (from 2020)
8-01. Problem
max∫10(2xe−t−2x˙x−˙x2)dt with
x(0)=0, x(1)=1.
The "extra" question: do sufficient conditions apply?
For each fixed t, consider 2xe−t−2xu−u2 as a function of (x,u). This has the same concavity properties as -2xu-u2 which is an indefinite quadratic form. The answer is negative.
(That does not mean that the proposed solution is not the solution - it means that our standard toolbox cannot tell. Pretty much like a non-concave function of a real variable might have a maximum too, like the function -(1-z2)2 .)
Part (a): The Euler equation:
∂F∂x=2e−t−2˙x and
∂F∂˙x=−2x−2˙x; to get the (total!) time-derivative of the latter, just add a dot to each term. The Euler equation:
2e−t−2˙x+2˙x+2¨x=0, that is,
¨x=−e−t__
Part (b): finding the solution (assuming it exists)
The Euler equation is necessary [to the precision level of this course]. Therefore, you can write as follows:
(Answer:) ¨x=−e−t ⇒ ˙x=C+e−t ⇒ x=Ct+D−e−t.
x(0)=0 yields D=1 and x(1) = 1 yields C +1 - e-1 = 1, so x(t)=e−1t+1−e−t__.