Sem#5: 8-01 (from 2020)

 

8-01. Problem  LaTeX: \displaystyle \max\int_0^1\Big(2xe^{-t}-2x\dot x-\dot x^2\Big)dtmax10(2xet2x˙x˙x2)dt with LaTeX: x(0)=0,\ x(1)=1x(0)=0, x(1)=1.

The "extra" question: do sufficient conditions apply? 

For each fixed t, consider LaTeX: 2xe^{-t}-2xu-u^22xet2xuu2 as a function of (x,u). This has the same concavity properties as -2xu-uwhich 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: 

LaTeX: \frac{\partial F}{\partial x}=2e^{-t}-2\dot xFx=2et2˙x and LaTeX: \frac{\partial F}{\partial \dot x}=-2x-2\dot xF˙x=2x2˙x; to get the (total!) time-derivative of the latter, just add a dot to each term. The Euler equation: 
LaTeX: 2e^{-t}-2\dot x+2\dot x+2\ddot x=02et2˙x+2˙x+2¨x=0, that is, LaTeX: \underline{\underline{\ddot x=-e^{-t}}}¨x=et__

 

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:) LaTeX: \ddot x=-e^{-t}\ \Rightarrow\ \dot x=C+e^{-t}\ \Rightarrow\ x=Ct+D-e^{-t}¨x=et  ˙x=C+et  x=Ct+Det.
x(0)=0 yields D=1 and x(1) = 1 yields C +1 - e-1 = 1, so
LaTeX: \underline{\underline{x(t)=e^{-1}t+1-e^{-t}}}x(t)=e1t+1et__.