Theoren34.Letx∈Rn,u∈R,g:Rn×Rh×R→Rn,f:Rn×RkxR→R.For problem J(ao, at, to)=max/f([z(t), u(t),t]dt st.i(t)=gz(t), u(t), t r(to)=r0,x()≥0 define the hamiltonian as H=f(x,,t)+A(t)·9(x,,t) Under certain differentiability conditions, if u* is a solution, then there exists A to, TRn such that u is a solution of Ha=0, 入=-Hx with transversality conditions imAx=0,A(T)≥0.■ heorem35.Letx∈Rn,u∈R,g:Rn×R×R→Rn,f:Rn×Rk×R→R.Por problem J(o,IT, to)=max/f[a(t), u(t)]e-o(t-to) dt st.i(t)=gz(t), u(t), t define the hamiltonian as H=f(, a)+A(t).g(, u, t) Under certain differentiability conditions. if u* is a solution then there exists a to,T]-Rn such that u is a solution of 入=6入-H with transversality condition lim入re 0,A()≥0Theorem 3.4. Let x ∈ Rn, u ∈ Rk, g : Rn × Rk × R → Rn, f : Rn × Rk × R → R. For problem J(x0, xT , t0) ≡ maxu ] T t0 f[x(t), u(t), t] dt s.t. x˙(t) = g[x(t), u(t), t] x(t0) = x0, x(T) ≥ 0 define the Hamiltonian as H = f(x, u, t) + λ(t) · g(x, u, t). Under certain differentiability conditions, if u∗ is a solution, then there exists λ : [t0, T] → Rn such that u∗ is a solution of Hu = 0, (3.14) λ˙ = −Hx, (3.15) with transversality conditions: lim t→T λx = 0, λ(T) ≥ 0.  (3.16) Theorem 3.5. Let x ∈ Rn, u ∈ Rk, g : Rn × Rk × R → Rn, f : Rn × Rk × R → R. For problem J(x0, xT , t0) ≡ maxu ] T t0 f[x(t), u(t)]e−θ(t−t0) dt s.t. x˙(t) = g[x(t), u(t), t] x(t0) = x0, x(T) ≥ 0 define the Hamiltonian as H = f(x, u) + λ(t) · g(x, u, t). Under certain differentiability conditions, if u∗ is a solution, then there exists λ : [t0, T] → Rn such that u∗ is a solution of Hu = 0, λ˙ = θλ − Hx, with transversality condition lim t→T λxe−θt = 0, λ(T) ≥ 0.  3—6