Understanding the Hamiltonian function Based on this function:
$$\text{max} \int_0^2(-2tx-u^2) \, dt$$
We know that $$(1) \;-1 \leq u \leq 1, \; \; \; (2) \; \dot{x}=2u, \; \; \; (3) \; x(0)=1, \; \; \; \text{x(2) is free}$$
I can rewrite the function into a hamiltonian function:
$$H=-2tx-u^2+p2u$$
where u(t) maxizmizee H where:
\begin{equation}
u = \left\{\begin{array}{rc} 1 & p \geq 1 \\ p & -1 < p < 1 \\ -1 & p \leq -1 \end{array}\right.
\end{equation}
Now, can somebody help me understand how the last part is true, and why?
I find it hard to see the bridge between $u$ and $p$.
 A: I'd like to add some detail to the other answer posted here. 
First note that
\begin{equation}
H = -2tx - u^2 + 2pu
\end{equation}
and that we'd like to maximize $H$. There are two terms containing $u$ and hence Pontryagin's Maximum Principle tells us that at each time $t \in [0,2]$, we should choose
$u$ so that it maximizes
\begin{equation}
 -u^2 + 2pu = u(-u + 2p). 
\end{equation}
This function is concave in $u$ and we can take its derivative to find stationary points (this is point-wise in time and stationarity here is not meant in the same sense as it was used above) and concavity guarantees that any stationary point is a maximum. Doing so, we have
\begin{equation}
\frac{\partial }{\partial u} \big(-u^2 + 2pu\big) = -2u + 2p = 0.
\end{equation}
From this we see that we'd like $u = p$. The constraints on $u$ prevent us from doing this when $p \not\in [-1,1]$. In those cases, we set $u$ to be as close to $p$ as possible, which gives 
\begin{equation}
u = \left\{\begin{array}{rc} 1 & p \geq 1 \\ p & -1 < p < 1 \\ -1 & p \leq -1, \end{array}\right.,
\end{equation}
just as you have above.
A: $$
\frac{\partial H}{\partial u} = -2u + 2p \tag{1}
$$
where $u$ is the control variable and $p$ is the costate.
The optimality of $H$ requires (1)=0, where you obtain your $u_t$ expression considering its constraint.
