Control $\dot x = x+u, u=u(t)$, from $x(0)=3$ to $x(t_1)=2$, s.t $J = \int_0^{t_1} x^2+ux+u^2 \: dt$ is minimized Control $\dot x = x+u, u=u(t)$ unconstrained, from $x(0)=3$ to $x(t_1)=2$, s.t $J = \int_0^{t_1} x^2+ux+u^2 \: dt$ is minimized. Find, the optimal control $u^*$, optimal path $x^*$  and $t_1$
Attempt
Using PMP, $H = -(x^2+ux+u^2) + \psi (x+u)$
Costate eqns:
$\displaystyle \dot \psi = -\frac{\partial H}{\partial x}= 2x + u -\psi$
Maximize $H$ as a fn of $u$:
\begin{align}
    0 &= \frac{\partial H}{\partial u} = -x -2u + \psi
    \Longrightarrow u = \frac{1}{2}(\psi - x)
\end{align}
which is a max since $\frac{\partial^2 H}{\partial u^2} =-2$
$\Longrightarrow \dot \psi = 2x + u -\psi = 2x + \frac{1}{2}(\psi - x) - \psi = \frac{3}{2}x -\frac{1}{2}\psi$ 
and $\dot x = x+u = {x+\psi \over 2}$
Solving the ode system gives $x = Ae^t+Be^{-t}$ and $\psi =Ae^t-3Be^{-t}$
$H(t_1)=0 \Longrightarrow x = 2\pm \sqrt{-12 + 4\psi +\psi^2}$
If $x = 2 - \sqrt{-12+4\psi(0)+\psi(0)^2}$
For $t=0$
\begin{align*}
x(0) &= 3 = 2 - \sqrt{-12+4\psi(0)+\psi(0)^2} = A+B \\
    \Longrightarrow - 1 &= \sqrt{-12+4(A-3(3-A))+(A-3(3-A))^2} \quad \text{is impossible}
\end{align*}
so no solution for $x=2 - \sqrt{-12+4\psi(0)+\psi(0)^2}$
If $x= 2 +\sqrt{-12+4\psi(0)+\psi(0)^2}$
For $t=0 \Longrightarrow A= \frac{1}{4}(7\pm\sqrt{17})$
If $A= \frac{1}{4}(7+\sqrt{17}), B = 3-A$, at $x(t_1)=2=Ae^{t_1}+Be^{-t_1} \Longrightarrow t_1 <0$
If $A= \frac{1}{4}(7-\sqrt{17}), B = 3-A$ there is no solution for $t_1$

The system can't be controlled from $x(0)=3$ to $x(t_1)=2$ for $t_1>0$

 A: You have to find a stationary point of 
$$
L(x,ψ,u,T) = \int_0^T[H(x,ψ,u)-ψ\dot x]dt
$$
For the variation of this functional with two final times $T$ and $T+δT>T$ use some smooth continuation of $u$. Then $x(T)=(x+δx)(T+δT)$. The variation can now be computed, ignoring terms quadratic or higher in the increments, as \begin{align}
\delta L&=L(x+δx,ψ+δψ,u+δu,T+δT)-L(x,ψ,u,T)\\
&=\int_0^{T+δT}[H_u(x,ψ,u)δu]dt+ \int_0^{T+δT}[H_ψ(x,ψ,u)-\dot x]δψ\,dt
\\&~ ~ ~ ~+\int_0^{T+δT}[H_x(x,ψ,u)+\dot ψ]δx\,dt-[δx(T+δT)ψ(T+δT)-δx(0)ψ(0)]
\\&~ ~ ~ ~+[H(x(T),ψ(T),u(T))-ψ(T)\dot x(T)]δT
\end{align}
The last term compensates for the additional integral over $[T,T+δT]$ that is not present in $L(x,ψ,u,T)$ but used in all the other linearizations.
Now with the fixed boundary conditions for $x$ you get $δx(0)=0$. 
On the other end the situation is a little more involved. Taylor-expanding at $T$ gives $$δx(T+δT)=x_f-x(T+δT)=-\dot x(T)δT+O(δT^2).$$
The other terms but the first give the usual dynamic of Pontryagin's maximum principle, the time variation of the end point gives the additional condition
$$
0=H(x(T),ψ(T),u(T))=−(x(T)^2+u(T)x(T)+u(T)^2)+ψ(T)(x(T)+u(T))
$$
From the local optimality condition for the control $u$ it follows $ψ=x+2u$, so that this last condition is equal
$$
0=−(x(T)^2+u(T)x(T)+u(T)^2)+(x(T)+2u(T))(x(T)+u(T))\\=2u(T)x(T)+u(T)^2=u(T)(2x(T)+u(T))
$$
Following your solution, this means $AB=0$, so that either $x(T)=Ae^t$ or $x(T)=Be^{-t}$. As the boundary values are falling, $A=0$, $B=3$, $e^T=\frac32$.
