Generally, optimal control problem is a functional extremum problem. Sometimes analytic solution is hard to get, so numerical methods are feasible. To be solved, it can be treated as a optimization problem and differential equations problem. Therefore, JuMP.jl and DifferentialEqutions.jl are two powerful solver.
General form of optimal control problems are:
$$min \int_{t_0}^{t_f} L[\boldsymbol{x}(t),\boldsymbol{u}(t),t]dt \\s.t. \hspace{0.4cm} \dot{\boldsymbol{x}} =
f[\boldsymbol{x}(t),\boldsymbol{u}(t),t]$$
Differential form Solution Define Hamilton function $\boldsymbol{H}$ and Lagrange multiplier vector $\boldsymbol{\lambda}$ ,
$$H[\boldsymbol{x}(t),\boldsymbol{u}(t),t] = L[\boldsymbol{x}(t),\boldsymbol{u}(t),t] + \boldsymbol{\lambda}^T(t)f[\boldsymbol{x}(t),\boldsymbol{u}(t),t]\\\
\boldsymbol{\lambda}(t) = [\lambda_1(t),\lambda_2(t),...,\lambda_n(t)]$$
Necessary conditions for the existence of functional extremum are:
$$\dot{\boldsymbol{\lambda}} =- \frac{\partial H}{\partial \boldsymbol{x}}$$
$$\frac{\partial H}{\partial \boldsymbol{u}} = 0$$
$$\dot{\boldsymbol{x}} = -\frac{\partial H}{\partial \boldsymbol{\lambda}} =f[\boldsymbol{x}(t),\boldsymbol{u}(t),t]$$
There are several different possibilities for Transversality conditions :
Condition 1: Fixed initial and final value Fixed initial and final value means:
$$\boldsymbol{x}(t_0)=\boldsymbol{x}_0,\boldsymbol{x}(t_f)=\boldsymbol{x}_{f}$$
Complete equations to be solved are:
$$\left\{\begin{matrix}
\dot{\boldsymbol{\lambda}} =- \frac{\partial H}{\partial \boldsymbol{x}} \\\
\frac{\partial H}{\partial \boldsymbol{u}} = 0\\\
\dot{\boldsymbol{x}} =f[\boldsymbol{x}(t),\boldsymbol{u}(t),t] \\\
\boldsymbol{x}(t_0)=\boldsymbol{x}_0,\boldsymbol{x}(t_f)=\boldsymbol{x}_{f}
\end{matrix}\right.$$
For example:
$$min \int_{0}^{2} u^2dt \\\
s.t. \hspace{0.4cm} \dot{\boldsymbol{x}} =
\begin{bmatrix} 0&1 \\ 0&0 \end{bmatrix}\boldsymbol{x} +\begin{bmatrix}0\\1\end{bmatrix}u \\\
\boldsymbol{x}(0)=\begin{bmatrix}1\\1\end{bmatrix},\boldsymbol{x}(2)=\begin{bmatrix}0\\0\end{bmatrix}$$
Equations to be solved are:
$$\left\{\begin{matrix}
\dot{\lambda_1} = 0\\\
\dot{\lambda_2} = -\lambda_1\\\
\dot{x_1} = x_2 \\\
\dot{x_2} = u \\\
u + \lambda_2 = 0\\\
\end{matrix}\right.$$
Boundary conditions are:
$$x_1(0)=1,x_2(0)=1,x_1(2)=0,x_2(2)=0$$
This is a typical Differential Algebraic Equations(DAE) with Boundary Value Problems(BVP) . DifferentialEqutions.jl is a powerful solver to solve this problem.
Condition 2: Fixed initial value and Free final value In this condition, consider Bolza Type :
$$min (\Phi(\boldsymbol{x}(t_f),t_f)+\int_{t_0}^{t_f} L[\boldsymbol{x}(t),\boldsymbol{u}(t),t]dt)$$
the $\Phi(\boldsymbol{x}(t_f),t_f)$ are some special requirements for end time.
Complete equations to be solved are:
$$\left\{\begin{matrix}
\dot{\boldsymbol{\lambda}} =- \frac{\partial H}{\partial \boldsymbol{x}} \\\
\frac{\partial H}{\partial \boldsymbol{u}} = 0\\\
\dot{\boldsymbol{x}} =f[\boldsymbol{x}(t),\boldsymbol{u}(t),t] \\\
\boldsymbol{x}(t_0)=\boldsymbol{x}_0\\\boldsymbol{\lambda}(t_f)=\frac{\partial \boldsymbol{\Phi}}{\partial \boldsymbol{x}}|_{t=t_f}
\end{matrix}\right.$$
For example:
$$min \int_{0}^{2} u^2dt \\\
s.t. \hspace{0.4cm} \dot{\boldsymbol{x}} =
\begin{bmatrix} 0&1 \\ 0&0 \end{bmatrix}\boldsymbol{x} +\begin{bmatrix}0\\1\end{bmatrix}u \\\
\boldsymbol{x}(0)=\begin{bmatrix}1\\1\end{bmatrix},\boldsymbol{x}(2)=\begin{bmatrix}0\\free\end{bmatrix}$$
Equations to be solved are:
$$\left\{\begin{matrix}
\dot{\lambda_1} = 0\\\
\dot{\lambda_2} = -\lambda_1\\\
\dot{x_1} = x_2 \\\
\dot{x_2} = u \\\
u + \lambda_2 = 0\\\
\end{matrix}\right.$$
Boundary conditions are:
$$x_1(0)=1,x_2(0)=1,x_1(2)=0,\lambda_2(2)=\frac{\partial \Phi}{\partial x_2(1)}=0$$
This is also a DAE with BVP.
There still some other conditions which DifferentialEquations can't apply to, but JuMP can solve them with a Optimization form.
Essentially, Optimal Control Problem is a optimization problem.Optimization methods can solve it by discretizing the continuous function.
From
$$min \int_{t_0}^{t_f} L[\boldsymbol{x}(t),\boldsymbol{u}(t),t]dt \\s.t. \hspace{0.4cm} \dot{\boldsymbol{x}} =
f[\boldsymbol{x}(t),\boldsymbol{u}(t),t]$$
to
$$min \sum_{i=1}^{n} L(\boldsymbol{x}_i,\boldsymbol{u}_i,t_i) \\s.t. \hspace{0.4cm} \boldsymbol{x}_{i+1} =\boldsymbol{x}_{i}+f(\boldsymbol{x}_i,\boldsymbol{u}_i,t_i)*dt$$
Actually, discretization method above is Euler method or we can use Backward Euler method :
$$min \sum_{i=1}^{n} L(\boldsymbol{x}_i,\boldsymbol{u}_i,t_i) \\s.t. \hspace{0.4cm} \boldsymbol{x}_{i+1} =\boldsymbol{x}_{i}+f(\boldsymbol{x}_{i+1},\boldsymbol{u}_{i+1},t_{i+1})*dt$$
There are still lots of discretization methods we can choose like Trapezoidal Method , Simpson Method , Adams method and so on.
For example in Condition 1 , we can get:
$$min \sum_{i=1}^{100} u_i^2dt \\\
s.t. \hspace{0.4cm} \boldsymbol{x}_{i+1} = \boldsymbol{x}_i +
(\begin{bmatrix} 0&1 \\ 0&0 \end{bmatrix}\boldsymbol{x}_i +\begin{bmatrix}0\\1\end{bmatrix}u_i) * 0.02 \\\
\boldsymbol{x}_1=\begin{bmatrix}1\\1\end{bmatrix},\boldsymbol{x}_{100}=\begin{bmatrix}0\\0\end{bmatrix}$$
With the Optimization form, Condition 3 can be solved.
Condition 3: Fixed initial value and final value with constrain $$min \int_{0}^{2} u^2dt \\\
s.t. \hspace{0.4cm} \dot{\boldsymbol{x}} =
\begin{bmatrix} 0&1 \\ 0&0 \end{bmatrix}\boldsymbol{x} +\begin{bmatrix}0\\1\end{bmatrix}u \\\
\boldsymbol{x}(0)=\begin{bmatrix}1\\1\end{bmatrix}\\\
x_1(2)+x_2(2) =0$$
With the Optimization form, we just add a different constraint:
$$min \sum_{i=1}^{100} u_i^2dt \\\
s.t. \hspace{0.4cm} \boldsymbol{x}_{i+1} = \boldsymbol{x}_i +
(\begin{bmatrix} 0&1 \\ 0&0 \end{bmatrix}\boldsymbol{x}_i +\begin{bmatrix}0\\1\end{bmatrix}u_i) * 0.02 \\\
\boldsymbol{x}_1=\begin{bmatrix}1\\1\end{bmatrix}\\\
x_{1,100}+x_{2,100}=0$$
Awesome!