Decisions in Time

Let $H$ be the closed half-plane of points no farther from $E_0$ than from $P_0$ (boundary: the perpendicular bisector $x=3$). The value of the game is the distance from $T$ to $H$: for $T=(7,2)$ it is exactly $4$, attained at the foot $F=(3,2)$, which both players reach in the same time $\sqrt{13}$; for $T^{\prime }=(2,1)\in H$ it is $0$.

At every state and horizon the three values are ordered: pursuer-sees-first $\le$ simultaneous $\le$ evader-sees-first (checked exactly for all $20$ states and $T\le 10$). From the start with the pursuer at the centre and the evader on the rim: an evader that sees the pursuer's move first is never caught (value $=T$); a pursuer that sees the evader's move first captures at once (value $0$); the simultaneous value is $\frac{3}{4}$ for one period, $\frac{42}{29}$ for two, and $\approx 4.174$ for ten. From opposite rim nodes the ten-period simultaneous value is $\approx 4.996$.

Let $\Phi$ be the one-period (Shapley) operator of the simultaneous discounted game. $\Phi$ is monotone and a $\delta$-contraction, so iterating from the exact bounds ${\underline{v}}_0\equiv 0$ and ${\overline{v}}_0\equiv 1/(1-\delta )$ gives monotone rational sequences with ${\underline{v}}_n\le v\le {\overline{v}}_n$ and ${\overline{v}}_n-{\underline{v}}_n\le {\delta }^n/(1-\delta )$. At $\delta =\frac{1}{2}$, after $18$ iterations, the value from opposite rim nodes lies in $[1.672027396,1.672027826]$ and from the centre start in $[1.344055158,1.344055522]$; the width bound $2^{-17}$ is checked on the exact rationals.

Along the square law, $\alpha x^2-\beta y^2$ is constant; along the linear law, $\alpha x-\beta y$ is constant (both are polynomial identities in the vector fields, checked exactly). Consequences, at $\alpha =\beta$: an attacker of $50$ destroys defenders of $40$ and $30$ met in sequence at the price of its own annihilation — $\sqrt{50^2-40^2}=30$ survive the first battle and annihilate exactly against the second — while the combined defence of $70$ wins with $\sqrt{70^2-50^2}=\sqrt{2400}$ survivors. Under the linear law the same numbers give the defenders $20$ survivors whether split or combined. With $\alpha =4\beta$ (each unit of $x$ four times as effective), $25$ of $x$ defeat $40$ of $y$ with exactly $15$ survivors, and draw against exactly $50$: effectiveness enters the square law linearly, numbers quadratically.

In the stationary symmetric equilibrium each player quits each period with probability $q=c/(v+c)=\frac{1}{5}$, and each player's expected payoff is $0$. The standoff continues each period with probability $(1-q{)}^2=\frac{16}{25}$ and lasts $\frac{25}{9}$ periods in expectation; the prize is eventually won with probability $\frac{8}{9}$, worth $\frac{32}{9}$ in expectation, and the two players' expected fighting costs total exactly $\frac{32}{9}$: the contest dissipates the entire expected prize. With a finite horizon and a lapsing prize, backward induction gives the same $q=\frac{1}{5}$ in every period if the interior mixed equilibrium is selected at every stage; it is not the only backward-induction solution. With positive $v$ and $c$ there are exactly three stationary equilibria: the mixed one and two pure ones in which one player fights for ever and the other quits at once, with payoffs $(v,0)$. The pure equilibria need no external commitment device — they are equilibria of expectations — whereas both quitting at once (either would gain $v$ by staying) and both fighting for ever (unbounded costs against $0$ for leaving) are not equilibria. In the asymmetric variant the stationary mixed equilibrium has $q_2=c/(v_1+c)=\frac{1}{7}$, $q_1=c/(v_2+c)=\frac{1}{4}$ and still gives both players $0$: a larger prize does not raise its holder's equilibrium value, and the rival quits less often ($\frac{1}{7}$ against $\frac{1}{5}$), since its exit rate is what keeps the richer side indifferent.