Math 2.0

Calculus of variations and optimal control; optimization Definition

Hamilton–Jacobi–Isaacs equation

Builds on Differential gameHamilton–Jacobi equationHamilton–Jacobi–Bellman equationViscosity solution

Definition

Let A⊆ℝkA \subseteq \R^{k} and B⊆ℝlB \subseteq \R^{l} be nonempty compact sets and let T>0T > 0. Let f ⁣:ℝn×A×B→ℝnf \colon \R^n \times A \times B \to \R^n, h ⁣:ℝn×A×B→ℝh \colon \R^n \times A \times B \to \R and g ⁣:ℝn→ℝg \colon \R^n \to \R be continuous. Consider the two-player zero-sum differential game with dynamics x′=f(x,a,b)x' = f(x, a, b) and payoff ∫tTh(x(s),a(s),b(s)) ds+g(x(T))\int_t^T h(x(s), a(s), b(s)) \, ds + g(x(T)). The player with control a(s)∈Aa(s) \in A minimises the payoff, and the player with control b(s)∈Bb(s) \in B maximises it. The upper and lower Hamiltonians are, for x,p∈ℝnx, p \in \R^n, H+(x,p)=min⁡a∈Amax⁡b∈B{f(x,a,b)⋅p+h(x,a,b)},H−(x,p)=max⁡b∈Bmin⁡a∈A{f(x,a,b)⋅p+h(x,a,b)}.H^+(x, p) = \min_{a \in A} \max_{b \in B} \left\{ f(x, a, b) \cdot p + h(x, a, b) \right\}, \qquad H^-(x, p) = \max_{b \in B} \min_{a \in A} \left\{ f(x, a, b) \cdot p + h(x, a, b) \right\}. The Hamilton–Jacobi–Isaacs equations are the terminal value problems Vt+H±(x,∇xV)=0in (0,T)×ℝn,V(T,x)=g(x),V_t + H^\pm(x, \nabla_x V) = 0 \quad \text{in } (0, T) \times \R^n, \qquad V(T, x) = g(x), for an unknown function V ⁣:[0,T]×ℝn→ℝV \colon [0, T] \times \R^n \to \R, one with H+H^+ and one with H−H^-. The game satisfies Isaacs' condition if H+=H−H^+ = H^-, and then the two equations coincide.

Remarks

Always H−≤H+H^- \le H^+. If BB has one point, both equal min⁡a{f⋅p+h}\min_{a} \{ f \cdot p + h \} and the equation is the Hamilton–Jacobi–Bellman equation, a special Hamilton–Jacobi equation. Assume also that ff, hh and gg are bounded and Lipschitz in xx. Then the upper and lower values of the game, defined with nonanticipating strategies, are the viscosity solutions of these equations. Under Isaacs' condition the two values are equal, so the game has a value.

Example: if f(x,a,b)=f1(x,a)+f2(x,b)f(x, a, b) = f_1(x, a) + f_2(x, b) and h(x,a,b)=h1(x,a)+h2(x,b)h(x, a, b) = h_1(x, a) + h_2(x, b), then the expression splits into a part that depends only on aa and a part that depends only on bb. So min⁡amax⁡b\min_a \max_b and max⁡bmin⁡a\max_b \min_a are both the sum of min⁡a\min_a of the first part and max⁡b\max_b of the second, and Isaacs' condition holds.

Non-example: n=1n = 1, A=B=[−1,1]A = B = [-1, 1], f(x,a,b)=(a−b)2f(x, a, b) = (a - b)^2 and h=0h = 0. For p=1p = 1, max⁡b(a−b)2=(1+∣a∣)2\max_b (a - b)^2 = (1 + \abs{a})^2, so H+(x,1)=1H^+(x, 1) = 1, while min⁡a(a−b)2=0\min_a (a - b)^2 = 0, so H−(x,1)=0H^-(x, 1) = 0. Isaacs' condition fails, and the two equations are different.

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  1. Differential game concept
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  5. Hamilton–Jacobi–Isaacs equation definition

Further reading

  1. Web https://en.wikipedia.org/wiki/Differential_game
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