Further Exercises on the Euler-Lagrange Equation

The numerator is obvious from the expansion by the …rst column; the criticalterm of the denominator comes from the bottom row, corner and adjacent termboth of which create the appropriate term by multiplication with the top rightsubdeterminant of size 2. It now follows thatSimilarly,lim A T = 0:T !1and againB T = linear combinations of:(e( )T ; e (+)T ; 1);1( ) 2 e (+)T + :::lim B T = 0:T !1We consider a similar problem in the <strong>Exercises</strong> (11?): Exercise:For theproblem:subject toZ T0minZ Tthe trajectory is known to be of the formshow that0(x 2 + _x 2 ) dtxe 2t dt = 1 and x(0) = 1; x(T ) = 0;x(t) = Ae t + Be t + Ce 2t ;lim A T = 0:T !1Solution: The two boundary conditions and the constraint equation lead to thematrix equation:21 1 14 e T e T e 2T1 e T 1(1 e 3T 1) 3 4 e 4T )3 25 4ABC325 = 410135 :