By Milota J.

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**Additional resources for Invitation to mathematical control theory**

**Example text**

If xε is a solution to (2,3,1) corresponding to uε then xε = x(t), t ∈ [t0 , τ − ε) t = x(τ − ε) + f (s, xε (s), v)ds, t ∈ [τ − ε, τ ] τ −ε t = xε (τ ) + f (s, xε (s), u(s))ds, t ∈ (τ, t1 ]. τ With help of derivation with respect to initial conditions we get that z(t) := ∂xε |ε=0 ∂ε satisfies z(t) = 0, 0≤t<τ (2,3,2) t = f (τ, x(τ ), v) − f (τ, x(τ ), u(τ )) + τ f2 (s, x(s), u(s))z(s)ds, t ∈ (τ, t1 ). Assume that a cost functional φ has continuous partial derivatives and let w be a solution of the adjoint equation w(t) ˙ = − (f2 )∗ (t, x(t), u(t))w(t) w(t1 ) = − φ (x(t1 )).

Let Q ∈ Lsa (X), R ∈ Lsa (U ) and let Q be non-negative and R be positive-definite. Assume that (A, B, Q) is open-loop stable and J(x0 , u) is given by (1,7,1). Then for any x0 ∈ X there is uˆ ∈ L2 (R+ , U ) such that J(x0 , uˆ) = inf u∈L2 (R2 ,U ) J(x0 , U ). Moreover, this uˆ is given by uˆ(t) = −R−1 B ∗ P xˆ(t), (1,7,3) x˙ =[A − BR−1 B ∗ P ]x x(0) =x0 (1,7,4) where xˆ is a solution to Here P is as in Propositon 1,56. Proof. The operator P solves the equation (1,6,6) on any interval [0, T ] with the initial condition P (0) = P.

An easy computation shows that 1 α(t, x, λ) = − R−1 B ∗ (t)λ 2 (2,3,14) is a unique minimum in (2,3,13). e. ∇x V (t, x) = 2P (t)x. Assuming (2,3,15) we get a feedback in the form F (t, x) = −R−1 B ∗ (t)P (t)x. If P is continuous on [t0 , t1 ] then this feedback is admissible, since ϕ is a solution to y˙ =[A(t) − B(t)B ∗ P (t)]y y(τ ) =x. If we substitute ϕ for x and F for u in (2,3,8) we find that P (t) := P (t1 − t) satisfies the Riccati differential equation (1,6,5) and the value function V is continuously differentiable.

### Invitation to mathematical control theory by Milota J.

by George

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