Determination of the Optimal Integral Curves 
Author's Data: Housam Binous
Department of Chemical Engineering
National Institute of Applied Sciences and Technology
Tunis, TUNISIA
Email: binoushousam@yahoo.com

The solution to the present problem was derived by M. N. Bandyopadhyay, Control Engineering: Theory and Practice, Prentice Hall India, New Delhi, 2004, page 394.

We look for the optimal integral curve when the performance index is given by:
		
		J=(∫_0^((π/2)))(((OverDot[(Subscript[x, 1])])^2)+((OverDot[(Subscript[x, 2])])^2)+2  (Subscript[x, 1]) (Subscript[x, 2]))dt
		
The boundary conditions are:

        (Subscript[x, 1])(0)=0, (Subscript[x, 2])(0) = 
        0 ,  (Subscript[x, 1])(π/2) = -1 and (Subscript[x, 2])(π/2) = 
          1
        
We need to use the famous Euler's equation:

		δ(L/δ)x-(d/dt)(((δL)/(δ(OverDot[x]))))=0
		
	with 	L=((OverDot[(Subscript[x, 1])])^2)+((OverDot[(Subscript[x, 2])])^2)+2  (Subscript[x, 1]) (Subscript[x, 2])

L := x1'^2 + x2'^2 + 2 x1 x2

Dt[L, x1, Constants ->  {x2, x1', x2'}] /. 
      x2 ->  x2[t]

D[Dt[L,x1',Constants->{x2,x2',x1}]/.x1' -> x1'[t],{t, 1}]

Dt[L, x2, Constants ->  {x1, x1', x2'}] /. 
      x1 ->  x1[t]

D[Dt[L,x2',Constants->{x2,x1',x1}]/.x2' -> x2'[t],{t, 1}]

sol=DSolve[{2 x2[t]==2 x1''[t],2 x1[t]==2 x2''[t],x1[0] ==  0,x2[0] ==  0,x1[Pi/2] ==  -1,x2[Pi/2] ==  1},{x1[t], x2[t]},t]

Plot[Evaluate[{x1[t], x2[t]} /. sol], {t, 0, Pi/2}, 
      PlotStyle ->  {RGBColor[1, 0, 0], RGBColor[0, 0, 1]}]