Fal!2001 16.3122-3 Note that we now have i(t)=Ac(t)+ Buopt(t)=Ac(t)-Buruu BuA(t with a(to)=o So combine with equation for the adjoint variable 入(t)=-Rx(t)-A入(t)=-C2RnC2x(t)-AN(t to get A B.R-1BT a(t CI R,C 入(t which of course is the hamiltonian matrix again Note that the dynamics of a(t are coup oled, but a(t)Is known initially and X(t)is known at the terminal time, since X(tf) Pie a(t This is a two point boundary value problem that is very hard to solve in general However, in this case, we can introduce a new matrix variable P(t and show that 1. A(t)=P(t)a(t 2. It is relatively easy to find P(t)Fall 2001 16.31 22—3 • Note that we now have: x˙(t) = Ax(t) + Buopt(t) = Ax(t) − BuR−1 uu Bu T λ(t) with x(t0) = x0 • So combine with equation for the adjoint variable λ˙(t) = −Rxxx(t) − AT λ(t) = −Cz TRzzCzx(t) − AT λ(t) to get: ∙ x˙(t) λ˙(t) ¸ = " A −BuR−1 uu Bu T −Cz TRzzCz −AT # ∙ x(t) λ(t) ¸ which of course is the Hamiltonian Matrix again. • Note that the dynamics of x(t) and λ(t) are coupled, but x(t) is known initially and λ(t) is known at the terminal time, since λ(tf ) = Ptfx(tf ) — This is a two point boundary value problem that is very hard to solve in general. • However, in this case, we can introduce a new matrix variable P(t) and show that: 1. λ(t) = P(t)x(t) 2. It is relatively easy to find P(t)