Brief Summary on Optimal Control

Optimal Control
Prerequisites
𝒕
For 𝒙̇ (𝒕) = 𝒇(𝒕) and 𝒙(𝒕𝟏 ) = 𝒙𝟏 the unique solution is 𝒙(𝒕) = 𝒙𝟏 + ∫𝒕 𝒇(𝜽)𝒅𝜽
𝟏

point slope formula: 𝒙̇ (𝒕) = 𝒇(𝒕) and 𝒙(𝒕𝟏 ) = 𝒙𝟏 general solution is 𝑥(𝑡) = 𝑡 − 𝑡1 + 𝐾 for K a
constant
𝑑𝑥 1
separation of variables: if in the form = 𝑓(𝑥)𝑔(𝑡) goes to ∫ 𝑓(𝑥) 𝑑𝑥 = ∫ 𝑔(𝑡) 𝑑𝑡
𝑑𝑡




Standard Form
• fixed final time: 𝑡0 , 𝑡𝑓 ∈ ℝ with 𝑡𝑓 ≥ 𝑡0 , 𝑥0 ∈ ℝ𝑛 and 𝑥𝑓 ∈ ℝℓ1 +ℓ2

find p.cont 𝑢: [𝑡0 , 𝑡𝑓 ] → ℝ𝑚 and cont. & point.cont 𝑥: [𝑡0 , 𝑡𝑓 ] → ℝ𝑛 which minimize:
𝑡
𝐽(𝑢) ≔ ∫𝑡 𝑓 𝑓(𝑡, 𝑥(𝑡), 𝑢(𝑡))𝑑𝑡 + 𝜙(𝑡𝑓 , 𝑥(𝑡𝑓 )) (1)
0


subject to the constrains

𝑥̇ (𝑡) = ℎ(𝑡, 𝑥(𝑡), 𝑢(𝑡)), 𝑥(𝑡0 ) = 𝑥0
= 𝑥𝑓,𝑖 𝑖 ∈ {1, … , ℓ1 }
𝑥𝑖 (𝑡𝑓 ) {≤ 𝑥𝑓,𝑖 𝑖 ∈ {ℓ1 + 1, … , ℓ1 + ℓ2 } (2)
𝑓𝑟𝑒𝑒 𝑖 ∈ {ℓ1 + ℓ2 + 1, … , 𝑛}
𝑢(𝑡) ∈ 𝒰

• free final time: 𝑡0 ∈ ℝ with 𝑥0 ∈ ℝ𝑛 and 𝑥𝑓 ∈ ℝℓ1 +ℓ2

find 𝑡𝑓 ∈ ℝ with 𝑡𝑓 ≥ 𝑡0 , point.cont 𝑢: [𝑡0 , 𝑡𝑓 ] → ℝ𝑚 and cont. & p.cont

𝑥: [𝑡0 , 𝑡𝑓 ] → ℝ𝑛 which minimize (1) subject to (2)



How to Solve:
• CHECK MINIMISATION
• might need to define a derivative as a variable e.g. 𝒙 ≔ 𝒚̇
• n the diam of x, scalar it’ll be n=1 and vector usually n=2
• m the size of u, most often a scalar so m=1
• h from the derivative form, usually v and 𝜻, perhaps in a vector (if h=v, Filippov special
case)
• f looks at the integral, sometimes 0 (can play around with this and 𝝓)
• 𝝓 any constraints on the final time (rest means it is 0)
• t0 the time at the start (usually 0)
• tf final time, is it fixed or free?
• x0 what is x at the start?
• x(tf) what is x at the final time (if at rest, 0)