Optimal control problem with cost $x(1)-\alpha\int_0^1(x(t))^2dt$ and system $x'(t)=u(t)$, $x(0)=0$ for $u$ taking values in the compact [-1,1]. Consider the optimal control problem
Minimize $$x(1)-\alpha\int_0^1(x(t))^2dt$$ with 
$$\dot{x}(t)=u(t),\hspace{0.4cm} x(0)=0.$$
The control $u$ takes values in $[-1,1]$ and $\alpha\in\mathbb R$. We want to find all locally optimal controls.
I tried using the Minimum Principle, but I was only able to prove that the control only takes values -1 and 1.
 A: Pontryagin minimum principle
The Hamiltonian formulation is
$$
I[x,λ,u]=x(1)+\int_0^1(H(x,λ,u)-λ\dot x)dt ~~\text{ where } ~~ 
H(x,λ,u)=-αx^2+λu.
$$
$\newcommand{\sign}{\operatorname{sign}}$
$\DeclareMathOperator*{\argmin}{argmin}$
The Pontryagin minimum principle then gives the equations
\begin{align}
\dot x&=\partial_λH=u,&x(0)&=0,\\
\dot λ&=-\partial_xH=2αx,&λ(1)&=1,\\[.5em]
u&=\argmin_{v\in[-1,1]}H(x,λ,v)&=\argmin_{v\in[-1,1]}(-αx^2+λu)&=-\sign(λ)
\end{align}
In consequence $\ddot λ=-2α\sign(λ)$, $\dot λ(0)=2α\dot x(0)=0$.

Solution for negative $α$
Thus for $α\le 0$ the function $λ$ is convex and growing for positive $λ(0)$, and concave and falling for negative $λ(0)$. To reach $λ(1)=1$ we need $λ\ge 0$, thus for $-α$ small enough
$$
\dot x=-1\implies x(t)=-t~~\text{ and }~~λ(t)=1+α(1-t^2).
$$
If $α<-1$ we get
\begin{cases}
 x(t)=0 , ~~ &λ(t)=0&\text{ for }t\le t_1, \text{ and }\\
 x(t)=t_1-t ,~~ &λ(t)= -α(t-t_1)^2&\text{ for }t> t_1.
\end{cases}
with $t_1=1-\sqrt{\frac1{|α|}}$.

Solution for positive $α$
For $α>0$ if $λ\ge 0$ over the full interval, then this function is concave and falling, giving again
$$
x(t)=-t,~~λ=λ(t)=1+α(1-t^2)
$$
this time without further restrictions on $α$. This is the optimum within the boundaries of the control, however other stationary points of the functional $I[x,λ,u]$ exist.
