An optimal control problem with fixed final time and free final state I am attempting to solve an optimal control problem, but I am not really sure what it is asking.

Given the control system
$$\dfrac{d}{dt}\begin{bmatrix} x_1(t) \\ x_2(t) \end{bmatrix} = \begin{bmatrix} -x_1(t) \\ x_2(t) \end{bmatrix}u(t)$$
where $u(t) \in [-1,1]$ and $\mathbf{x} (t) \in \mathbb R^2$ for all $t \in [0,1]$, and $\mathbf{x}(0) =[x_1(0),x_2(0)]^T =: \mathbf{x}^0$.

*

*Prove there is an optimal controller for every $\mathbf{x}^0$ and there exists

$$\min_{u:[0,1] \to [-1,1]} \|\mathbf{x}(1)\|_2^2$$


*For every $\mathbf x^0 \in \mathbb R^2$, find an optimal controller $u:[0,1] \to [-1,1] $.


I have never taken an optimization course and only had basic education on control theory, but it all involved actual numbers and coding. The main thing I am confused on is the mathematical notation — what is the problem asking? Any advice on what the best way to approach this problem is?
 A: You do not necessarily need knowledge in optimal control. The question asks you to drive the system as close as possible to the origin in one time unit.
Now, if we look at the differential equations we see that they are decoupled and we can solve them as follows
$$\dot{x}_1=-u(t)x_1\implies x_1(t) = x_1(t=0)\exp\left[-\int_{\tau=0}^{t}u(\tau)~d\tau \right]$$
$$\dot{x}_2=u(t)x_2\implies x_2(t)=x_2(t=0)\exp\left[\int_{\tau=0}^{t}u(\tau)~d\tau \right]$$
Now, we set time to $t=1$ to obtain the position in which we terminate after one time unit.
$$x_1(1) = x_1(t=0)\exp\left[-\int_{\tau=0}^{1}u(\tau)~d\tau \right]$$
$$x_2(1)=x_2(t=0)\exp\left[\int_{\tau=0}^{1}u(\tau)~d\tau \right]$$
Instead of minimizing the distance to the origin we will minimize the square of the distance to the origin.
$$D^2(u)=x_1^2(1)+x_2^2(1)=x^2_1(0)\exp\left[-2\int_{\tau=0}^{1}u(\tau)~d\tau \right]+x^2_2(0)\exp\left[2\int_{\tau=0}^{1}u(\tau)~d\tau \right]$$
$$=\dfrac{x^2_1(0)+x^2_2(0)\left(\exp\left[2\int_{\tau=0}^{1}u(\tau)~d\tau \right]\right)^2}{\exp\left[2\int_{\tau=0}^{1}u(\tau)~d\tau \right]}$$
Now, introduce the new variable
$$\tilde{u}=\exp\left[2\int_{\tau=0}^{1}u(\tau)~d\tau \right]$$
and our goal will be to minimize
$$D^2(\tilde{u})=\dfrac{x^2_1(0)+x^2_2(0)\tilde{u}^2}{\tilde{u}}$$
by solving
$$\dfrac{d}{d\tilde{u}}D^2=0.$$
After solving this for $\tilde{u}>0$ (the solution has to be positive because the exponential function is always positive) you will have to solve
$$\tilde{u}=\exp\left[2\int_{\tau=0}^{1}u(\tau)~d\tau \right]$$
$$\implies 0.5\ln\tilde{u}=\int_{\tau=0}^{1}u(\tau)~d\tau $$
for $u(t)$ The solution of such an integral equation can be obtained by an ansatz function (e.g. constant function, linear function, ...) or by using the solution to the Fredholm integral equation. Note that a constant $u(t)=u_0$ is enough for this problem if we do not consider the input constraints.

An additional observation is that the ratio of the differential equations is given by
$$\dfrac{dx_2}{dx_1}=-\dfrac{x_2}{x_1}$$
$$\implies \dfrac{dx_2}{x_2}=-\dfrac{dx_1}{x_1}$$
$$\implies \ln x_2 - \ln x_2(0)=-\ln x_1+\ln x_1(0)$$
$$\implies x_2x_1 = x_2(0)x_1(0)$$
$$\implies x_2 = \dfrac{x_1(0)x_2(0)}{x_1}$$
These equations are hyperbolas ... What can you conclude from this? This is a constraint for the squared distance to the origin. We can reformulate the problem as a constrained optimization with Lagrange multipliers $$\text{minimize: } f(x_1,x_2)=x_1^2+x_2^2$$
$$\text{subject to: } x_1x_2 = x_1(0)x_2(0)$$
Or simply replace $x_2 = \dfrac{x_1(0)x_2(0)}{x_1}$ in $f(x_1,x_2)$ to obtain $f(x_1)$ and maximize this by standard school calculus.
