If someone could help me with this problem I would be greatly appreciative.
Control the system $$\dot{x}=x+u$$ From $$x(0)=0 \space to\space x(T)=2$$ Where $T\in\mathbb{R}_+$ is free
s.t. $$J=\int_{0}^{T}\frac{1}{2}u^2dt$$ is minimised.
My Approach
Let $$ \\f_0(t,x,u)=\frac{1}{2}u^2 \\f_1(t,x,u)=x+u $$
The Hamiltonian of this problem is given by: $$ \\H=-f_0+\psi \times f_1 \\=-\frac{1}{2}u^2+\psi(x+u) $$
By the PMP we wish to choose $u$ s.t. it maximises $H$, $$ \\\frac{\partial H}{\partial u}=0 \\\Rightarrow\psi=u $$
The costate equation gives us $$ \\\dot\psi= -\frac{\partial H}{\partial x} \\\Rightarrow\dot\psi=-\psi \\\Rightarrow\psi=Ae^{-t} $$
Subbing this back into the system gives $$ \\x(t)=Be^{t}-\frac{A}{2}e^{-t} \\u(t)=Ae^{-t} $$
Now along the optimal trajectory, again by the PMP, $H$ must be $0$. As this applies along any point of the trajectory, we have (after a bit of algebra) $$ \\H(t=0)=0 \\\Rightarrow A = 0 \space or \space B = 0 $$
Now this is where I am confused, if $A=0$ or $B=0$ we have
$$x(t) = -\frac{A}{2}e^{-t} \space or \space x(t) = Be^{t} $$
But given $x(0)=0$ that would imply that in both cases $x(t)=0$. Which clearly does not give the optimal solution as it will never reach $x(T)=2$.
I'm not sure if I have made some fundamental error along the way, or if the system is just not controllable, but would appreciate some guidance either way.
OK, you started well, but then something went wrong. You've found $u(t)=\psi(t)$. That's right. The equation for $\psi(t)$ is correct either. I'll just write it as $\psi(t)=\psi_0 e^{-t}$, where $\psi_0=\psi(0)$.
Then you substitute your $u(t)=\psi(t)$ into the DE for $x$, i.e., $\dot{x}=x+\psi_0e^{-t}$, and solve it to get $$x(t)=x_0e^t + \int_0^t e^{t-\tau}\psi_0 e^{-\tau}d\tau=x_0 e^t + \psi_0 \frac{e^t-e^{-t}}{2}.$$ Taking into account that $x(0)=0$ and using some hyperbolic trigonometry notation we obtain $$x(t)=\psi_0 \sinh(t).$$ It remains to sustitute the final time $T$ and solve the preceding equation to get $\psi_0=\frac{2}{\sinh(T)}$ whence you can compute the optimal control etc.
ADDED: Assume now that $T$ is free. We can compute the cost function $J=\frac{4}{e^{2T}-1}$, which attains minimum at $T\to \infty$ which is equivalent to $\psi_0=0$ and hence, $u(t)=0$. This implies that the problem does not have a solution.
It is interesting that for $T>5$ the cost $J$ becomes infinitesimally small, so one can get a practically optimal solution by setting $T$ to an arbitrary constant larger than 5. But this is a different story.