 
Summary: Simplest problem of the calculus of variations with end cost
We consider a variation on the problem stated in Section 2.2. The value of the state at time T is
not prescribed, rather there is a penalty on the nal state. For notational convenience we denote
the state by q instead of x.
The problem is as follows:
Determine a function q(t), dened on the interval [0; T ], such that the integral
J(q()) = S(q(T )) +
Z T
0
F (t; q(t); _
q(t))dt (1)
is maximized (or minimized), and where q(t) in addition satises the boundary condition
q(0) = q 0 (2)
for given q 0 .
The following result is similar to Theorem 2.2.3. The dierence is that the condition x(t) = x T is
now replaced by a penalty S(x(T )).
1 Theorem
Consider the simplest problem in the calculus of variations and suppose Assumptions 2.2.1 and
2.2.2 are met. Then a necessary condition that a C 2 function q (t) maximizes (1) and satises
(2) is that q (t) is a solution of the dierential equation
