Optimal control problem proving an abs min.

Question

I'm having a trouble with this optimal control problem: $$\int_0^4(\dot{x}^2+x)dt \to extr$$ $$\vert \dot{x} \vert \leqslant 1 $$ $$x(4)=0$$ I found an extreme function $\hat{x}(t)$ (Sorry for wrong notions.I don't know an english terminology for this) $$\hat{x}(t)=\begin{cases} \frac{t^2}{4}-3 , t \in[0;2] \\ t-4 , t\in[2;4] \end{cases}$$ and control $$\hat{u}(t)=\begin{cases} \frac{t}{2} , t \in[0;2] \\ 1 , t\in[2;4] \end{cases}$$ This is considered to be true (in the book it is marked as solution). But I'm stucked in proving that $\hat{x}(t)$ is abs min. What I've already done: $$B_0(\hat{x}(t)+h)-B_0(\hat{x}(t))$$ Here $h$ is perturbation. So in order to $\hat{x}(t)$ be the abs minimum this difference should be greater or equal $0$. So $$\int_0^4(\dot{h}^2+2\hat{x}'\dot{h}+h)dt$$ here the the hatch is derivative (as $\dot{x}$). $$\int_0^4hdt = h t \vert_0^4 - \int_0^4(\dot{h}t)dt$$ First term is zero ($h(4)=0$) So we have $$\int_0^4\dot{h}^2dt+\int_0^4(2\hat{x}'-t)\dot{h}dt$$ On $t\in [0;2]$ $\hat{x}'(t) = \frac{t}{2}$ So the integral over this segment is zero. Finally we have: $$\int_0^4\dot{h}^2dt+\int_2^4(2-t)\dot{h}dt$$ as $\hat{x}'(t)=1$ on $[2;4]$. And from this part I'm stucked. I don't know how to prove that the last expression is greater or equal $0$.

Answer 1

So I have found problem. $\dot{h}(t) \in [-2;0]$ Since $\vert 1+\dot{h}\vert \leqslant 1$ And the second term always greater or equal $0$. The statement about values of $\dot{h}$ always was in front of my eyes, but I thought about $h$ which is greater or equal $0$. It was horrible mistake.