Proof. From [2] (page 60) in the problem with free terminal time T we have that
H((x(T);u(T);0;(T)) = 0:
Since from Theorem 4, H((x(t);u(t);0;(t)) is a constant for 0 t T
H((x(t);u(t);0;(t)) = 0: (5.10)
5.2 Properties of optimal solutions
Proposition 11 Excluding degenerate cases, (see [5]) where the optimal time T = 0, if uis
an optimal control with corresponding trajectory (p;q)T , then at the …nal time
p(T) = q(T):
Proof. From the dynamic equation of p, it is easy to notice that the volume of primary
tumor cells is growing when p < q and is shrinking when p > q. We can therefore conclude
that growth of tumor cells stops at the time T when p(T) = q(T): Let us suppose that
we decide instead to choose our terminal time T such that p(T) < q(T): This situation
cannot possibly be optimal since prior to the terminal time p was still growing. Therefore
such condition must not be accepted. Let us now suppose that at the terminal time T,
p(T) > q(T): In this case, the tumor is shrinking but we want to reach a point of maximum
shrinkage. It’s not therefore optimal to stop when such condition occurs. In conclusion, it
is necessary to have p(T) = q(T) at the terminal time T.
Proposition 12 Extremals are normal i.e 0 = 1 and 1 and 2 cannot vanish simultane-
ously.
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