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Econ-491-A1 · revision r1 · on Econ-491

GPT 6 Astra Ultra attempt on Econ-491

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Contents
  1. Active input constraints and an explicit infinite-time Nash trajectory certificate
  2. A constrained class satisfying the benchmark
  3. Verification of the selected constrained equilibrium
  4. Approximate saturated gains and infinite-time error
  5. Relation to the source and remaining extensions
  6. Completion Estimate
  7. References

Active input constraints and an explicit infinite-time Nash trajectory certificate

Problem ID: C73-487. Status: Solved for the precise catalogue benchmark; broader extensions remain outside scope.

A constrained class satisfying the benchmark

The catalogue asks for a stated class, a selected constrained feedback Nash equilibrium, admissible approximations, and a computable infinite-time state-error bound tending to zero. We give these objects for an arbitrary number $N\geq2$ of players sharing one scalar state. Controls are genuinely coupled through that state, and the reference equilibrium reaches the input bounds on a nonempty interval. This is a constructive resolution of the stated-class benchmark; it does not extend the source's MPC algorithm to all constrained LQ games.

Fix $a,q,r,U>0$ and set \[ \dot x=-ax+\sum_{i=1}^Nu_i,\qquad J_i=\frac12\int_{t_0}^{\infty}(qx^2+ru_i^2)\,dt, \qquad |u_i|\leq U. \] Thus $Q_i=q$, $R_{ii}=r$, and $R_{ij}=0$ for $i\ne j$, satisfying the canonical nonnegativity restrictions. Let $k$ be the unique positive solution of \begin{equation}\label{wsref-0-72-69-63} (2N-1)rk^2+2rak=q, \qquad k=\frac{\sqrt{a^2+(2N-1)q/r}-a}{2N-1}. \end{equation} Choose $M>U/k$, and put $\mathcal X=D=[-M,M]$. Define ${\operatorname{sat}}(z,U)=\max\{-U,\min\{U,z\}\}$ and $b=U/k$.

Verification of the selected constrained equilibrium

For $x\geq0$ define \[ w(x)= \begin{cases} rkx,&0\leq x\leq b,\\[2mm] \dfrac{qx^2+rU^2}{2(ax+NU)},&b\leq x\leq M, \end{cases} \qquad V(x)=\int_0^{|x|}w(s)\,ds. \] Equation (ric) gives $w(b)=rU$, so $V$ is continuously differentiable, even, and nonnegative. Its second derivative need not agree across the switching points; no $C^2$ verification assumption is imposed.
Lemma: Let $s(x)={\operatorname{sat}}(kx,U)$. At every $x\in[-M,M]$, \[ \min_{|u|\leq U}\left\{ \frac12(qx^2+ru^2)+V'(x)(-ax-(N-1)s(x)+u) \right\}=0, \] and its unique minimizer is $u=-s(x)$.
Proof: The expression is strictly convex in $u$, with unconstrained minimizer $-V'(x)/r$. For $0\leq x\leq b$, this is $-kx$ and is feasible. Substitution reduces the minimum to $x^2\{q/2+rk^2/2-rk(a+Nk)\}=0$ by (ric).

For $x\geq b$, the inequality $w(x)\geq rU$ is equivalent to \[ qx^2-2raUx-(2N-1)rU^2\geq0. \] This upward quadratic has its unique positive root at $b$, by (ric); hence the inequality holds. The constrained minimizer is therefore $-U$. Substitution gives zero by the defining formula for $w$. At zero the assertion is immediate; for negative $x$, $V'$ and $s$ are odd, and replacing $(x,u)$ by $(-x,-u)$ gives the same objective. Uniqueness follows from $r>0$. ∎

Lemma (Terminal value under every feasible deviation): If a feasible unilateral deviation has finite cost, then its state satisfies $x(t)\to0$ and $V(x(t))\to0$.
Proof: Finite cost and $q>0$ give $\int_{t_0}^{\infty}x(t)^2dt<\infty$. Feasibility and bounded inputs give $|\dot x|\leq aM+NU=:L$ almost everywhere. If $x$ did not tend to zero, some $\epsilon>0$ would occur at arbitrarily late times with $|x|\geq\epsilon$. The Lipschitz bound would then give intervals of a fixed positive length on which $|x|\geq\epsilon/2$. Select infinitely many disjoint such intervals; their squared-state integrals have an infinite sum, a contradiction. Continuity of $V$ and $V(0)=0$ prove the last assertion. ∎
Theorem: The profile $\mu_i^*(t,x)=-{\operatorname{sat}}(kx,U)$ is an admissible feedback Nash equilibrium on $D$, at every continuation time $t_0\geq0$. Its cost is $V(x_0)$ for every player. Its input constraint is active whenever $|x|>b$.
Proof: The feedbacks are globally Lipschitz and vanish at zero. Their closed loop obeys $\dot x=-ax-Ns(x)$. For $x\ne0$, $\frac{d}{dt}|x|=-a|x|-N|s(x)|\leq-a|x|$. Uniqueness prevents crossing zero, so $|x(t)|\leq|x_0|e^{-a(t-t_0)}$ and $D$ is invariant. Since $|s(x)|\leq k|x|$, all costs are finite.

Along any feasible unilateral deviation, Lemma Lemma and the chain rule for the $C^1$ function $V$ give \[ \frac12\int_{t_0}^T(qx^2+ru_i^2)dt \geq V(x_0)-V(x(T)). \] Lemma Lemma (Terminal value under every feasible deviation) implies the right side tends to $V(x_0)$. Along $\mu^*$ equality holds at each instant; its exponential decay makes the terminal value vanish, so its cost equals $V(x_0)$. This proves every required Nash inequality. All arguments are invariant under translating time, proving continuation optimality. Since $M>b$, initial states with $|x_0|>b$ yield a positive interval of saturated control by continuity. ∎

Approximate saturated gains and infinite-time error

Choose any finite nonnegative gains $\widehat k_i\geq0$, and define \[ \widehat\mu_i(t,x)=-{\operatorname{sat}}(\widehat k_i x,U),\qquad \delta=\sum_{i=1}^N|\widehat k_i-k|. \] These policies allow different switching points across players.
Theorem (Computable active-constraint certificate): Both the reference and approximate profiles keep $D$ invariant and have finite costs. Starting from the same state at time zero, \[ |x^{\widehat\mu}(t;x_0)-x^{\mu^*}(t;x_0)| \leq \delta |x_0|t e^{-at},\qquad \sup_{t\geq0}|x^{\widehat\mu}(t;x_0)-x^{\mu^*}(t;x_0)| \leq \frac{\delta |x_0|}{ae}. \] In particular, the last bound tends uniformly to zero over $x_0\in D$ as $\max_i|\widehat k_i-k|\to0$.
Proof: The approximate vector field $\widehat f(x)=-ax-\sum_i{\operatorname{sat}}(\widehat k_i x,U)$ is Lipschitz. Nonnegative gains give $d|\widehat x|/dt\leq-a|\widehat x|$ exactly as in Theorem Theorem; this proves invariance and exponential decay. Also $|\widehat\mu_i|\leq\widehat k_i|\widehat x|$, so every cost is finite.

The reference field $f(x)=-ax-N{\operatorname{sat}}(kx,U)$ satisfies \[ (x-y)(f(x)-f(y))\leq-a(x-y)^2, \qquad |\widehat f(x)-f(x)|\leq\delta|x|. \] The first inequality uses monotonicity of saturation; the second uses its one-Lipschitz property. For $z=\widehat x-x^*$, the upper right derivative of $|z|$ therefore satisfies $D^+|z|\leq-a|z|+\delta|\widehat x|$. One can justify this also at $z=0$ by using $(z^2+\epsilon^2)^{1/2}$ and taking $\epsilon\downarrow0$. Multiply the scalar comparison inequality by $e^{at}$ and integrate, using $z(0)=0$ and $|\widehat x(t)|\leq|x_0|e^{-at}$. The result is $|z(t)|\leq\delta|x_0|t e^{-at}$. Finally $\sup_{t\geq0}te^{-at}=1/(ae)$, by differentiating; $|x_0|\leq M$ gives uniform convergence. ∎

Proposition (A rational certificate from gain intervals): Suppose the game parameters, $M$, and approximate gains are rational, and a rational interval $[\ell,u]$ with $0<\ell\leq k\leq u$ is certified by evaluating the increasing quadratic in (ric) at its endpoints. Then \[ \overline\delta=\sum_i\max\{|\widehat k_i-\ell|,|\widehat k_i-u|\} \] is a rational upper bound on $\delta$. The entirely rational uniform bound $E=\overline\delta M/a$ is valid for every $x_0\in D$ and converges to zero when the interval shrinks to $k$ and the gains converge to $k$. For rational $x_0$, one may use the smaller rational bound $\overline\delta|x_0|/a$.
Proof: The quadratic is strictly increasing for $k\geq0$ and changes sign once. Interval endpoint signs therefore locate its unique positive root. Absolute distance to a point in an interval is bounded by the maximum endpoint distance. Summing and applying Theorem Theorem (Computable active-constraint certificate), with $1/e\leq1$, proves the certificate and its convergence. ∎

For a concrete rational instance, take $N=2$, $a=r=U=1$, $q=5$, and $M=2$. Then $k=1$, the reference uses saturated controls for $1<|x|\leq2$, and gains $1+h,1-h$ with $|h|\leq1$ give error at most $2|h||x_0|/e$. This example has a common controlled state and a Nash value verification, rather than merely a stable single-agent reference.

Relation to the source and remaining extensions

The source's Theorem 4 controls trajectories using gain mismatch and Hurwitz estimates. Its Section 6.3 labels the constrained example a proof of concept; Section 6.5 identifies further constrained error analysis [varga]. Those full-text locators were inspected on 9 October 2026. Our contraction proof replaces a linear matrix-exponential calculation and tolerates changing active input sets. It neither assumes nor proves that an arbitrary optimization residual bounds gain error. The supplied gain intervals are the approximation data.

Active state-boundary arcs, multidimensional coupled constraints, cross-control cost terms, and an MPC-horizon-to-gain certificate are outside this class. The state constraint here is verified invariant; it is the input constraints that actively bind. These limits do not remove the infinite-time, continuation-Nash, or active-control requirements met by the stated benchmark.

Completion Estimate

100%

This is a post hoc, heuristic estimate for the full catalogue target. The attempt supplies a coupled constrained game class with a verified continuation Nash equilibrium, active input bounds, invariant admissible approximations, and computable infinite-time trajectory certificates converging to zero. These establish every requirement of the canonical stated-class benchmark, whose scope does not require multidimensional or arbitrary MPC extensions.

References

  • [varga] B. Varga, K. Handwerker, I. Rudas and P. Galambos, Approximate Feedback Nash Equilibria in Constrained Differential Games via Model Predictive Control with Upper Bound Guarantees, arXiv:2610.04614v1 (2026), Theorem 4 and Sections 6.3, 6.5. https://arxiv.org/html/2610.04614v1.