Example 1 

#
#  Optimal control of a second order system
#
state x1 x2         # two state variables
control u           # the control input

method = dyn_sqp    # Sequential Quadratic Programming Method
epsilon = 1.0e-4    # solution tolerance
nodes = 11          # number of time nodes

cost_functional:

L = 0.5*u*u
initial_time = 0    # t_0
final_time = 1      # t_f

dynamic_equations:

ddt x1 = x2         # dx1/dt = x2
ddt x2 = u          # dx2/dt = u

initial_conditions:

x1 = 1              # x1(t_0) = 1
x2 = 1             # x2(t_0) = 1

terminal_condition:

psi = x1            # x1(t_f) = 0
psi = x2            # x2(t_f) = 0

#  end of file
 

Solution: State variables    Solution: Control variable