we-solve (beta)

Econ-24 · economics · Economics

Selection and attraction of knowledge-growth traveling waves

LLM status Open unverified

Statement

GT-082 | Mean-field games | R: precise asymptotic instantiation of a published open question

Definitions and mathematical statement

Fix $\nu,\rho>0$ and a continuous increasing concave learning function $\alpha:[0,1]\to[0,\infty)$ with $\alpha(0)=0$. Let $f(t,x)$ be the probability density of log productivity $x\in\mathbb R$. In the diffusive Lucas--Moll model the value $v$, density $f$, and optimal search fraction $s_*\in[0,1]$ solve \[ \begin{split} -v_t-\nu v_{xx}+\rho v &=\max_{s\in[0,1]}\left\{(1-s)e^x+ \alpha(s)\int_x^\infty[v(t,y)-v(t,x)]f(t,y)dy\right\},\\ f_t-\nu f_{xx} &=f(t,x)\int_{-\infty}^x\alpha(s_*(t,y))f(t,y)dy -\alpha(s_*(t,x))f(t,x)\int_x^\infty f(t,y)dy, \end{split} \] Seek $C^{1,2}$ functions $v,f$ with the displayed integrals finite. Here $s_*(t,x)$ is a measurable maximizing selector, $f_0$ is a positive $C^2$ probability density with $\int e^xf_0(x)dx<\infty$, $f(0)=f_0$, $f\geq0$, and $\int f(t,x)dx=1$. The infinite-horizon boundary condition for $v$ is its value representation \[ v(t,x)=\sup_s\mathbb E_{t,x}\int_t^\infty e^{-\rho(\tau-t)}(1-s_\tau)e^{X_\tau}d\tau. \] Against fixed $f$, $X$ has diffusion $\sqrt{2\nu}\,dB$ and meeting rate $\alpha(s)$; at a meeting with $Y\sim f(\tau,\cdot)$ it becomes $\max(X,Y)$. Controls are progressively measurable search fractions, and the displayed value must be finite. This defines the backward condition instead of imposing a freely chosen $v(0)$.

A balanced-growth traveling wave has \[ f(t,x)=\varphi(x-ct),\qquad v(t,x)=e^{ct}V(x-ct),\qquad s_*(t,x)=S(x-ct),\quad \int\varphi=1. \] Its profile solves \[ \begin{split} -\nu V''+cV'+(\rho-c)V &=\max_{s\in[0,1]}\left\{(1-s)e^z+ \alpha(s)\int_z^\infty[V(y)-V(z)]\varphi(y)dy\right\},\\ -c\varphi'-\nu\varphi'' &=\varphi(z)\int_{-\infty}^z\alpha(S(y))\varphi(y)dy -\alpha(S(z))\varphi(z)\int_z^\infty\varphi(y)dy. \end{split} \] For a concrete nonempty wave regime with a sufficiently large discount, take $\alpha(s)=a\sqrt{s}$, $a>\nu$, and $\rho>a+\nu$. These imply $\rho\geq2\sqrt{\nu a}$, matching the primary paper's $\nu=\kappa^2$ convention. This is a declared subfamily, not the largest known existence regime.

Normalize profiles by $\int_{-\infty}^0\varphi=1/2$. For a solution with a unique median define $b(t)$ by $\int_{-\infty}^{b(t)}f(t,x)dx=1/2$. Characterize the initial densities $f_0$ and solution selections for which some normalized wave obeys \[ \frac{b(t)}t\longrightarrow c,\qquad \|f(t,b(t)+\cdot)-\varphi\|_{L^1(\mathbb R)}\longrightarrow0,\qquad e^{-b(t)}v(t,b(t)+\cdot)\longrightarrow V \ \text{in }C^1_{\mathrm{loc}}(\mathbb R). \] Determine how the speed and profile depend on $f_0$, including any tail assumptions. This attraction-basin formulation is editorial; arbitrary probability densities are not asserted to satisfy it.

Short English statement

Which knowledge distributions converge, after translation and value rescaling, to a balanced-growth wave, and which wave is selected?

Variants and scope boundaries

Wave existence is already known in specified regimes. Convergence in a moving quantile frame allows sublinear shifts, including possible logarithmic shifts; convergence in the rigid frame $x-ct$ is stronger. Finite-horizon terminal values and infinite-horizon value representations are different boundary conditions.

Source relationship and status

The chapter retains time-dependent convergence as open. The model, moving centering, and norms are explicit specifications; they do not certify the full parameterized claim as open in 2026.

References

[S1] Cardaliaguet and Delarue. Selected topics in mean field games, ICM2022, vol. 5; DOI: 10.4171/ICM2022/17. Locator: Section 3.3, "Traveling waves," pp. 3680--3681.

https://ems.press/content/book-chapter-files/33265

[S2] Alessio Porretta and Luca Rossi (2020). Traveling waves for a nonlocal KPP equation and mean-field game models of knowledge diffusion. arXiv:2010.10828v1.

Locators: system (2), p. 4; traveling-wave scaling (3), p. 4; Section 2, Theorem 2.2, critical-wave existence.

https://arxiv.org/pdf/2010.10828v1

Common conventions: Source mathematical conventions

Finite games. For a finite set $A$, $\Delta(A)$ is its probability simplex. Players choose independent mixed strategies in Nash equilibrium; correlation is allowed only when explicitly part of the solution concept. In a game with payoff functions $u_i$ and strategy profile $x$, an additive $\varepsilon$ Nash equilibrium satisfies \[ u_i(x)\geq u_i(x_i',x_{-i})-\varepsilon \quad\text{for every player $i$ and every admissible deviation $x_i'$}. \] For numerical approximation, payoffs are normalized as locally specified. An $\varepsilon$ payoff residual is distinct from distance at most $\varepsilon$ to the set of exact equilibria. Point convergence, convergence of empirical play, and convergence in distance to an equilibrium set are also distinct.

Computational representations. Unless stated otherwise, a complexity question takes rational data with binary encoding; polynomial time is measured in total bit length. A fixed-accuracy polynomial algorithm is not necessarily polynomial in $1/\varepsilon$, and neither is necessarily polynomial in $\log(1/\varepsilon)$. Succinct games and value, demand, or sampling oracles must declare their interfaces. Exact real solutions require an explicit representation; an unspecified list of arbitrary real numbers is not a finite computational output. A claimed algorithmic barrier is meaningful only for its specified model and reductions.

Dynamic games. Histories, information, observation, and allowable randomization are part of the model. A stationary strategy depends only on the current state; a positional strategy in a deterministic graph game chooses one action at each controlled vertex. Nash equilibrium at an initial state and equilibrium after every history are different requirements. An expectation of a pathwise $\liminf$ payoff, a $\liminf$ of expected averages, a limiting expected average, and a uniform finite-horizon equilibrium are different criteria. None is silently substituted for another.

Allocation and mechanisms. A complete allocation assigns every item exactly once unless otherwise stated. Zero-valued goods, negative values, capacity constraints, feasibility, lotteries, and copies of goods affect fairness definitions and are specified locally. Dominant-strategy incentive compatibility, Bayesian incentive compatibility, universal truthfulness, and truthfulness in expectation impose different inequalities. Whenever an objective ratio has zero denominator, its convention is stated or those instances are excluded.

Infinite-dimensional models. Expectations, integrals, stochastic equations, and optimization objectives require their local regularity, integrability, filtrations, and admissible domains. A backward--forward mean-field system is a boundary-value problem; it is not an initial-value semiflow without an additional construction. Long-time, large-population, and vanishing-noise limits require an order of limits and a convergence topology. If a minimum need not be attained, the objective is its infimum or supremum and the approximation requirement is stated separately.

Statement licence: CC-BY-4.0 · Source file

Contributions 3

Econ-24-A3 · Exploration · LLM-generated · gpt 6 astra ultra ·

GPT 6 Astra Ultra attempt on Econ-24, version 3

Knowledge-growth wave selection: finite values, automatic tail selection and sharp moment thresholds

Problem O33-23. Status: unresolved. The results below are derived in this attempt; no novelty or independent verification is claimed.

Target and scope

The target is attraction, after median centering, to a Lucas--Moll balanced-growth wave, including selection of its speed. We retain the infinite-horizon value boundary condition. Wave existence does not answer this question. The known wave theorem is Porretta--Rossi, Theorem 2.2 [PR]; the time-dependent question is discussed in [CD, Section 3.3]. We retain the necessary speed and integrability restrictions from Version 1. Version 2 adds a finite-value theorem, a uniform productivity-sensitivity bound, and a quantitative upper-tail search restriction. Version 3 proves that every admissible wave automatically selects the slow tail root, determines its exact exponential-moment cutoff and tail asymptotics, and derives a strict finite-productivity speed window. These are restrictions on existing waves and admissible solutions, not an attraction theorem.

Write $a=\max_{s\in[0,1]}\alpha(s)$ and $q(t,x)=\alpha(s_*(t,x))$. The density equation is \begin{equation}\label{wsref-0-64-65-6e-73-69-74-79} f_t=\nu f_{xx}+f(x)\int_{-\infty}^x q(y)f(y){\mathrm d} y-q(x)f(x)\int_x^\infty f(y){\mathrm d} y, \qquad 0\le q\le a. \end{equation} Time arguments inside integrals are suppressed. We assume a positive classical probability-density solution. Whenever integration by parts with an unbounded weight is used, assume the corresponding exponential moment is locally bounded in time and that the weak formulation extends to that weight by cutoff; sufficient conditions are integrability of $e^{\lambda x}(f+|f_x|)$ and vanishing weighted boundary terms. This additional moment regularity is stated, not inferred just from pointwise classical regularity.

Tail and moment equations

Proposition: Let $w(t,x)=\int_x^\infty f(t,y){\mathrm d} y$ and $A(t,x)=\int_{-\infty}^x q(t,y)f(t,y){\mathrm d} y$. If the flux at infinity vanishes, then \[ w_t=\nu w_{xx}+Aw,\qquad 0\le A\le a(1-w). \] For $\lambda>0$ with the stated moment regularity, $M_\lambda(t)=\int e^{\lambda x}f(t,x){\mathrm d} x$ satisfies \begin{align}\label{wsref-0-6d-6f-6d-65-6e-74-69-64-65-6e-74-69-74-79} M_\lambda'(t)&=\nu\lambda^2 M_\lambda(t)+I_\lambda(t),\\ I_\lambda(t)&=\int_{x<y}q(t,x)f(t,x)f(t,y)(e^{\lambda y}-e^{\lambda x}){\mathrm d} x{\mathrm d} y, \qquad 0\le I_\lambda\le aM_\lambda.\nonumber \end{align} In particular, if $b(t)$ is a median, \[ b(t)\le\frac{\log(2M_\lambda(0))}{\lambda} +\left(\nu\lambda+\frac a\lambda\right)t. \] Every existing limit $b(t)/t\to c$ therefore satisfies $c\le\inf_{\lambda\in\Lambda}(\nu\lambda+a/\lambda)$, where $\Lambda$ is any set of positive exponents for which these moment hypotheses hold.
Proof: Integrate (density) over $(x,\infty)$. In the reaction terms all ordered meetings with both states above $x$ cancel. Only meetings starting below $x$ and ending above $x$ remain, giving $Aw$. The diffusion contribution is $-\nu f_x=\nu w_{xx}$. The bound on $A$ follows from $q\le a$ and $\int_{-\infty}^x f=1-w$.

Multiplying (density) by $e^{\lambda x}$, integrating the diffusion twice by parts, and interchanging the nonnegative ordered-meeting integrals gives (momentidentity). Its upper bound follows from \[ I_\lambda\le a\int e^{\lambda y}f(t,y)\int_{-\infty}^y f(t,x){\mathrm d} x{\mathrm d} y\le aM_\lambda. \] Gronwall's inequality yields $M_\lambda(t)\le M_\lambda(0)e^{(\nu\lambda^2+a)t}$. Since at least half the mass lies above the median, $M_\lambda(t)\ge e^{\lambda b(t)}/2$. These two inequalities prove the assertion, including its infimum version. ∎

The hypotheses of the question give a first exponential moment initially. They do not supply every exponent in $\Lambda$. In particular, replacing the infimum by $2\sqrt{\nu a}$ needs the exponent $\sqrt{a/\nu}$ to be available. Even when this upper bound holds, it does not identify the selected wave, because $A$ depends on the endogenous optimal policy.

Restrictions on a possible limit wave

Theorem: Let $(\varphi,S,c)$ be a positive classical traveling density satisfying the density equation, $\int\varphi=1$, and the flux and moment conditions above. Put \[ W(z)=\int_z^\infty\varphi(y){\mathrm d} y,\quad B(z)=\int_{-\infty}^z\alpha(S(y))\varphi(y){\mathrm d} y,\quad \beta=B(+\infty). \] Then \begin{equation}\label{wsref-0-74-61-69-6c-6f-64-65} \nu W''+cW'+BW=0. \end{equation} If $\beta>0$ and the first exponential moment is finite, then $c>\nu$ and $c\ge2\sqrt{\nu\beta}$. More generally, whenever $M_\lambda(\varphi)<\infty$ with valid weighted integrations, \begin{equation}\label{wsref-0-73-70-65-65-64-6d-6f-6d-65-6e-74} c=\nu\lambda+\frac{I_\lambda(\varphi,S)}{\lambda M_\lambda(\varphi)} \le\nu\lambda+\frac a\lambda. \end{equation} If in addition $h(z)=-W'(z)/W(z)$ converges to a finite positive $\ell$ as $z\to\infty$, then \begin{equation}\label{wsref-0-64-69-73-70-65-72-73-69-6f-6e} c=\nu\ell+\frac\beta\ell. \end{equation} Finite first moment implies $\ell\ge1$; when $\ell>1$ it guarantees first-moment integrability at the right tail, whereas $\ell=1$ alone decides neither integrability nor nonintegrability.
Proof: The tail equation from Proposition Proposition with $w(t,x)=W(x-ct)$ is (tailode). A traveling density has $M_\lambda(t)=e^{\lambda ct}M_\lambda(\varphi)$, giving (speedmoment). When $\beta>0$, the set where $\alpha(S(x))\varphi(x)>0$ has positive measure. Positivity of $\varphi$ gives positive mass strictly above every such $x$, so $I_1>0$ and $c>\nu$.

Set $Y(z)=e^{cz/(2\nu)}W(z)>0$. Then \[ Y''+\left(\frac{B(z)}\nu-\frac{c^2}{4\nu^2}\right)Y=0. \] If $c^2<4\nu\beta$, the coefficient is at least some $\eta>0$ on a half-line. No positive function solves $Y''\le-\eta Y$ there: if $Y'$ becomes negative, concavity forces a zero in finite distance; if $Y'$ stays nonnegative, $Y$ stays bounded below by a positive constant and $Y''$ is uniformly negative, forcing $Y'$ to become negative. Hence $c^2\ge4\nu\beta$. The already proved $c>0$ gives the claimed lower bound.

Substitution of $h=-W'/W$ in (tailode) gives $h'=h^2-(c/\nu)h+B/\nu$. If $h\to\ell$, its right-hand side has a limit. That limit must be zero, since a derivative with a nonzero limit is incompatible with convergence of its primitive. This proves (dispersion). Finally, \[ \int_0^\infty e^z\varphi(z){\mathrm d} z=W(0)+\int_0^\infty e^zW(z){\mathrm d} z \] when finite, by integration by parts or Tonelli. If $\ell<1$, choose $\ell<r<1$; eventually $h\le r$, so $W(z)\ge C e^{-rz}$ and the last integral diverges. If $\ell>1$, choose $1<r<\ell$; eventually $h\ge r$, so $W(z)\le C e^{-rz}$ and it converges. At $\ell=1$, the tails $e^{-z}(1+z)^{-2}$ and $e^{-z}$ show the two alternatives. These last tails illustrate the integrability test; they are not asserted to be wave solutions. ∎

The published wave analysis gives stronger restrictions under its full value-profile hypotheses; Theorem Theorem is deliberately only a necessary test from the density and moment equations. It does not reprove the existence result of [PR].

Proposition (The square-root search rule): Suppose $\alpha(s)=a\sqrt{s}$ and the wave value $V$ is nondecreasing with finite meeting integral. Define \[ Q(z)=e^{-z}\int_z^\infty[V(y)-V(z)]\varphi(y){\mathrm d} y. \] Then the maximizing search fraction is uniquely \[ S(z)=\min\{1,a^2Q(z)^2/4\}. \] If $V$ is the finite infinite-horizon value and $\rho>\nu$, then $V(z)\ge e^z/(\rho-\nu)$.
Proof: Divide the objective by $e^z$ and set $r=\sqrt{s}\in[0,1]$. The maximized expression becomes $1-r^2+aQ(z)r$, a strictly concave quadratic whose maximizer is $\min\{1,aQ(z)/2\}$ because $Q\ge0$. Squaring proves the formula. Choosing zero search forever is admissible and leaves only diffusion; its expected discounted production is $e^z\int_0^\infty e^{-(\rho-\nu)t}{\mathrm d} t$. Taking the supremum proves the lower bound. ∎

Finite values and productivity sensitivity: progress beyond Version 1

Version 1 assumed that the infinite-horizon value was finite and proved only its no-search lower bound. We now verify finiteness under the catalogue's discount condition and control its dependence on the initial state. These arguments are conditional on a density flow satisfying the moment assumptions of Proposition Proposition; they do not construct a solution of the coupled density--value system.

Throughout this section assume $a>0$, as in the catalogue's concrete square-root wave regime. If $a=0$, there are no meetings: the value is exactly $e^x/(\rho-\nu)$ for $\rho>\nu$, and $\beta=0$; the strict conclusion $\beta<a$ below is not asserted in that case.

Fix a starting time $t$. Write $A_s\in[0,1]$ for a control at elapsed time $s$ to distinguish it from the constant rate bound $a$. Work on a filtered space supporting a Brownian motion $B$, a rate-$a$ Poisson process, and independent uniform marks $(U,V)$ at its points, independent of $B$. Let $F_r^{-1}$ be a measurable quantile function of the given density $f(r,\cdot)$. At a potential meeting at elapsed time $s$, accept the meeting if $U\leq\alpha(A_s)/a$ and, if accepted, replace the log state by $\max\{X_{s-},F_{t+s}^{-1}(V)\}$. Between meetings the state has diffusion $\sqrt{2\nu}\,dB_s$. Controls are predictable for this driving filtration, possibly with additional independent private randomization; in particular, they do not anticipate the current marks. This is the usual thinning construction of the controlled meeting intensity. There are finitely many potential meetings on every bounded interval, so the construction is pathwise well defined. The same construction covers progressively measurable controls after the usual predictable representative at Poisson times; it does not allow a control to inspect a meeting mark before choosing whether to search.

For this admissible class define \[ J_{t,x}(A)={\mathbb{E}}\int_0^\infty e^{-\rho s}(1-A_s)e^{X_s}\,ds, \qquad v(t,x)=\sup_A J_{t,x}(A). \] The controls are indexed by the common driving filtration, rather than restricted to a prescribed feedback function. Thus a control chosen for one starting state can be used for another starting state: any auxiliary state trajectory it uses is itself a causal function of the same driving noise. This point is needed when taking suprema in the coupling proof. The argument includes arbitrary randomized admissible deviations in this formulation.

Lemma (Productivity under bounded meeting rates): Suppose $M_1(t+s)\leq M_1(t)e^{(\nu+a)s}$ for $s\geq0$ and $M_1(t)<\infty$. For every admissible control and $s\geq0$, \[ {\mathbb{E}} e^{X_s}\leq e^{\nu s}e^x +a\int_0^s e^{\nu(s-r)}M_1(t+r)\,dr. \tag{20} \] If two processes start at $x\leq x'$ and use the same control and all the same driving noise, then $X_s\leq X'_s$ pathwise and \[ 0\leq{\mathbb{E}}(e^{X'_s}-e^{X_s})\leq e^{\nu s}(e^{x'}-e^x). \tag{21} \]
Proof: Let $Z_s=e^{X_s}$ and $\sigma=\sqrt{2\nu}$. Between meetings, $dZ_s=\nu Z_s\,ds+\sigma Z_s\,dB_s$. The continuous multiplicative factor is $L_s=e^{\sigma B_s}$, with ${\mathbb{E}} L_s=e^{\nu s}$. At a potential meeting at time $r$ the jump $\Delta Z_r$ is nonnegative and at most the mark $e^{F_{t+r}^{-1}(V)}$, even when the meeting is rejected. Variation of constants on the finitely many meeting intervals gives \[ Z_s=e^xL_s+\sum_{r\leq s}\frac{L_s}{L_r}\Delta Z_r \leq e^xL_s+ \sum_{r\leq s}\frac{L_s}{L_r}e^{F_{t+r}^{-1}(V_r)}. \] The sum is over potential meetings, so its right-hand side is independent of acceptance decisions. Tonelli's theorem and the rate-$a$ marked Poisson expectation formula apply to its nonnegative terms. The Brownian increment $B_s-B_r$ is independent of the meeting time and mark and has exponential expectation $e^{\nu(s-r)}$. Hence the expected sum is $a\int_0^s e^{\nu(s-r)}M_1(t+r)dr$, proving (20) and its finiteness. The Poisson expectation formula here can also be obtained by conditioning on the number and times of its points and integrating the independent uniform marks.

For the pair of starting states, order is preserved both by their common Brownian translation in log space and by the map $z\mapsto\max\{z,y\}$. In productivity coordinates the latter map is $z\mapsto\max\{z,e^y\}$; for $z'\geq z$ it satisfies \[ 0\leq\max\{z',e^y\}-\max\{z,e^y\}\leq z'-z. \] Acceptance is the same for the coupled processes because the control is the same. Consequently $D_s=Z'_s-Z_s$ follows the common geometric Brownian factor between meetings and has nonpositive jumps, while staying nonnegative. Iterating across the finite set of meetings yields $0\leq D_s\leq(e^{x'}-e^x)L_s$. Taking expectations proves (21). ∎

Theorem (A finite value and a uniform increment bound): Assume the moment regularity of Proposition Proposition holds on every finite time interval, $0\leq\alpha\leq a$, and $\rho>\nu+a$. Then the value defined above is finite and obeys \begin{equation}\label{wsref-0-76-61-6c-75-65-62-6f-75-6e-64-73} \frac{e^x}{\rho-\nu}\leq v(t,x) \leq\frac{e^x}{\rho-\nu} +\frac{aM_1(t)}{(\rho-\nu)(\rho-\nu-a)}. \end{equation} For every $h\geq0$, \begin{equation}\label{wsref-0-69-6e-63-72-65-6d-65-6e-74} 0\leq v(t,x+h)-v(t,x) \leq\frac{e^{x+h}-e^x}{\rho-\nu}. \end{equation} In particular $v(t,\cdot)$ is locally Lipschitz. Where it is differentiable, $0\leq v_x(t,x)\leq e^x/(\rho-\nu)$; for the classical values in the catalogue this holds at every point.
Proof: Proposition Proposition, applied from time $t$, gives the moment bound required in Lemma Lemma (Productivity under bounded meeting rates). Since $0\leq1-A_s\leq1$, (20), Tonelli's theorem, and then the moment bound give \begin{align*} J_{t,x}(A) &\leq\frac{e^x}{\rho-\nu} +a\int_0^\infty e^{-\rho s}\int_0^s e^{\nu(s-r)}M_1(t+r)\,dr\,ds\\ &=\frac{e^x}{\rho-\nu} +\frac a{\rho-\nu}\int_0^\infty e^{-\rho r}M_1(t+r)\,dr\\ &\leq\frac{e^x}{\rho-\nu} +\frac{aM_1(t)}{(\rho-\nu)(\rho-\nu-a)}. \end{align*} The bound is uniform in the control. Zero search gives the lower bound in (valuebounds). Both denominators are positive by assumption.

For a fixed control use the ordered coupling in (21). The payoff increment is nonnegative, and its expectation is at most \[ \int_0^\infty e^{-\rho s}{\mathbb{E}} D_s\,ds \leq\frac{e^{x+h}-e^x}{\rho-\nu}. \] Thus $J_{t,x}(A)\leq J_{t,x+h}(A)\leq J_{t,x}(A)+C_{x,h}$ for every control in the same admissible class. Taking suprema proves (increment); no optimizer or feedback regularity is required. Applying that bound on a compact interval proves local Lipschitz continuity, and taking difference quotients proves the derivative bound. ∎

Corollary (Search vanishes in the upper tail): Under Theorem Theorem (A finite value and a uniform increment bound), suppose $\alpha(s)=a\sqrt{s}$. For the value-based maximizing search selector define \[ Q_t(x)=e^{-x}\int_x^\infty[v(t,y)-v(t,x)]f(t,y)\,dy. \] Then \begin{align} 0\leq Q_t(x)&\leq\frac1{\rho-\nu} \int_x^\infty(e^{y-x}-1)f(t,y)\,dy,\label{wsref-0-51-74-61-69-6c}\\ s_*(t,x)&=\min\{1,a^2Q_t(x)^2/4\},\qquad \lim_{x\to\infty}s_*(t,x)=0.\nonumber \end{align} In particular every positive finite-first-moment wave in this regime satisfies $S(z)\to0$ as $z\to\infty$ and \[ \beta=\int_{\mathbb{R}} a\sqrt{S(z)}\varphi(z)\,dz<a. \] The limit is for each fixed time or wave; no uniform-in-time centered-tail estimate is asserted.
Proof: Apply (increment) to each $y\geq x$ and integrate. This proves (Qtail), including finiteness of its meeting integral. The right side is at most $e^{-x}(\rho-\nu)^{-1}\int_x^\infty e^yf(t,y)dy$, which tends to zero as $x\to\infty$ by the finite first moment. Monotonicity of the value makes $Q_t\geq0$. Maximizing in $\sqrt{s}$ as in the earlier search-rule proposition gives the stated unique selector and its limit. For a wave evaluate at time zero, where the value is $V$ and density is $\varphi$. Choose $R$ sufficiently large that $S(z)\leq1/4$ for $z\geq R$. Positivity of $\varphi$ gives $\int_R^\infty\varphi>0$, and hence \[ \beta\leq a\int_{-\infty}^R\varphi +\frac a2\int_R^\infty\varphi<a. \] ∎

Automatic tail selection and the exact moment threshold

Version 2 identified the dispersion relation only under an additional assumption that the logarithmic tail slope converges. It also left the borderline first-moment case undecided by that assumption alone. The cumulative structure of $B$ in (tailode) removes both ambiguities. The next result concerns the tails of existing waves, not selection of a wave by the time-dependent initial-value and value problem.
Theorem (The slow root is forced): Suppose $W\in C^2({\mathbb{R}})$, $W>0$, $W'<0$, $W(-\infty)=1$ and $W(+\infty)=0$ satisfy (tailode). Suppose $B(z)=\int_{-\infty}^z q(y)\varphi(y)\,dy$, where $\varphi=-W'$, $0\leq q\leq a<\infty$ and $\beta=\int q\varphi>0$. Assume $c>0$ and $c^2\geq4\nu\beta$, as already proved for the admissible finite-first-moment waves in Theorem Theorem. Set \[ \ell_\pm=\frac{c\pm\sqrt{c^2-4\nu\beta}}{2\nu}. \] Then \begin{equation}\label{wsref-0-68-61-7a-61-72-64-73-65-6c-65-63-74-65-64} 0<h(z):=-\frac{W'(z)}{W(z)}\leq\ell_-,\qquad \lim_{z\to\infty}h(z)=\ell_-. \end{equation} The exact positive exponential-moment range is \begin{equation}\label{wsref-0-6d-6f-6d-65-6e-74-63-75-74-6f-66-66} \int_{\mathbb{R}} e^{\lambda z}\varphi(z)\,dz<\infty \quad\Longleftrightarrow\quad 0<\lambda<\ell_-. \end{equation} If $\ell_-<\ell_+$, there is a constant $A>0$ such that $W(z)\sim A e^{-\ell_-z}$ as $z\to\infty$. If $\ell_- =\ell_+=:\ell$, there are constants $A_1>0$ and $A_0\in{\mathbb{R}}$ such that \[ W(z)=\left(A_1z+A_0+o(1)\right)e^{-\ell z}. \]
Proof: Since $\beta>0$ and $c>0$, both roots are positive. Substituting $h=-W'/W$ in (tailode) gives \begin{equation}\label{wsref-0-72-69-63-63-61-74-69-72-6f-6f-74} h'=(h-\ell_-)(h-\ell_+)-d(z),\qquad d(z)=\frac{\beta-B(z)}\nu\geq0. \end{equation} For every fixed $R$, integration of the logarithmic derivative gives $\int_{-\infty}^R h(z)\,dz=-\log W(R)<\infty$. Consequently there are points $z_n\to-\infty$ with $h(z_n)<\ell_-$. The constant function $\ell_-$ is a supersolution of (riccatiroot): its derivative is zero whereas the right side there is $-d(z)\leq0$. Scalar comparison therefore implies $h(z)\leq\ell_-$ for $z\geq z_n$. To verify the comparison without a strict-crossing assumption, on any compact time and value interval the polynomial right side is Lipschitz in $h$; the positive part of $h-\ell_-$ satisfies the usual differential inequality $(h-\ell_-)_{+}'\leq C(h-\ell_-)_{+}$ almost everywhere until it could leave that interval. Its zero initial value and Gronwall imply it remains zero. Exhaust the finite intervals, then send $z_n\to-\infty$. This proves the global upper bound.

Fix $0<\epsilon<\ell_-$. On $0\leq h\leq\ell_--\epsilon$, \[ (h-\ell_-)(h-\ell_+) \geq\epsilon(\ell_+-\ell_-+\epsilon)>0. \] Since $d(z)\to0$, eventually $h'$ is at least half this positive constant whenever $h$ is in that interval. It must therefore enter $[\ell_--\epsilon,\ell_-]$ in finite distance and cannot exit through the lower endpoint, whose vector field points strictly upward. This holds for every $\epsilon$, proving the limit in (hazardselected).

For $\lambda>0$, Tonelli's theorem applied to $e^{\lambda z}=1+\int_0^z\lambda e^{\lambda y}dy$ for $z>0$ gives \begin{equation}\label{wsref-0-74-6f-6e-65-6c-6c-69-74-61-69-6c} \int_0^\infty e^{\lambda z}\varphi(z)\,dz =W(0)+\lambda\int_0^\infty e^{\lambda z}W(z)\,dz, \end{equation} with equality also when both sides are infinite. The negative half-line contributes at most one. If $\lambda<\ell_-$, choose $\lambda<b<\ell_-$. The established limit gives $h\geq b$ for large $z$, hence $W(z)\leq C_b e^{-bz}$; this makes (tonellitail) finite. The global upper bound on $h$ gives $W(z)\geq W(R)e^{-\ell_-(z-R)}$ for $z\geq R$. Thus (tonellitail) diverges for every $\lambda\geq\ell_-$, including equality. This proves (momentcutoff).

For the sharper asymptotics put $r=\ell_--h\geq0$ and $g=\ell_+-\ell_-$. Then \[ r'=d-gr-r^2,\qquad 0\leq d(z)=\nu^{-1}\int_z^\infty q(y)\varphi(y)dy \leq(a/\nu)W(z). \] The bound $W(z)\leq C_b e^{-bz}$ just proved for every $0<b<\ell_-$ makes $d$ integrable on a right half-line. If $g>0$, integrating $r'+gr\leq d$ from $R$ to $L$ yields \[ g\int_R^L r\leq r(R)-r(L)+\int_R^L d \leq r(R)+\int_R^\infty d. \] Hence $\int_R^\infty r<\infty$. Since $(\log(e^{\ell_-z}W(z)))'=r(z)$, the positive expression $e^{\ell_-z}W(z)$ has a finite positive limit. This proves the simple-root asymptotic.

If $g=0$, put $Y=e^{\ell z}W$. The wave equation gives $Y''=dY\geq0$, while (hazardselected) gives $Y'=rY\geq0$. Choose $0<\epsilon<\ell/2$. The bound $W(z)\leq C e^{-(\ell-\epsilon)z}$ implies \[ 0\leq Y''(z)=d(z)Y(z) \leq C' e^{-(\ell-2\epsilon)z} \] on a right half-line. Thus $Y'$ has a finite limit $A_1$, and $\int_R^\infty|Y'(z)-A_1|dz<\infty$, by integrating this exponential bound twice. It follows that $Y(z)-A_1z$ has a finite limit $A_0$. Finally $B(-\infty)=0<\beta$, so $d>0$ on some interval. There $Y''>0$, and $Y'\geq0$ everywhere; therefore $Y'$ becomes strictly positive and its nondecreasing limit $A_1$ is positive. This proves the critical asymptotic. ∎

Corollary (A strict finite-productivity speed window): Every positive wave to which Theorem Theorem applies, with $\beta>0$ and finite first exponential moment, satisfies \[ \ell_->1,\qquad \beta>\nu,\qquad 2\sqrt{\nu\beta}\leq c<\nu+\beta. \] In the square-root finite-value regime of Corollary Corollary (Search vanishes in the upper tail), these restrictions sharpen to $\nu<\beta<a$ and $c<\nu+\beta<\nu+a$.
Proof: The first assertion follows by applying (momentcutoff) at $\lambda=1$. Since $\ell_-\ell_+=\beta/\nu$ and $\ell_-\leq\ell_+$, we have $\ell_-\leq\sqrt{\beta/\nu}$, so $\beta>\nu$. The root identity gives $c=\nu\ell_-+\beta/\ell_-$. This function is strictly decreasing on $[1,\sqrt{\beta/\nu}]$ except for its zero derivative at the right endpoint. Since $\ell_->1$, its value is strictly less than $\nu+\beta$. The lower speed bound is retained from Theorem Theorem; the last assertion uses $\beta<a$ from Corollary Corollary (Search vanishes in the upper tail). ∎

A compactness obstruction and the first gap

Proposition: Convergence of median-centered densities in $L^1$ does not imply convergence of their first exponential moments, even for strictly positive smooth densities.
Proof: Let $\varphi$ be the standard Gaussian density and $\varphi_n(x)=(1-e^{-n})\varphi(x)+e^{-n}\varphi(x-2n)$. Then $\|\varphi_n-\varphi\|_1\le2e^{-n}$, but $\int e^x\varphi_n(x){\mathrm d} x=e^{1/2}(1-e^{-n}+e^n)\to\infty$. Let $b_n$ be the unique median of $\varphi_n$. Uniform convergence of the cumulative distribution functions and strict increase of the Gaussian cumulative function imply $b_n\to0$. Translation continuity in $L^1$ gives $\varphi_n(b_n+\cdot)\to\varphi$ in $L^1$, while their moments equal $e^{-b_n}\int e^x\varphi_n(x){\mathrm d} x\to\infty$. ∎
Thus neither the exact moment identity nor the nonlocal value integral can be passed to a proposed median-frame limit using the stated density topology alone. A candidate route must first prove suitable uniform integrability of the centered tails and control of the backward value field, then establish attraction and distinguish competing wave branches. No such estimate, attraction theorem, or basin characterization is proved here. The retained finite-value and sensitivity theorems verify additional admissibility properties and rule out persistently maximal search in the upper tail. The new tail theorem removes the earlier extra hypothesis on logarithmic slope convergence and settles the critical-moment boundary for actual waves. They do not supply the missing moving-frame estimates or dynamical attraction; the complete target remains unresolved.

Completion Estimate

25%

This is a heuristic assessment of the full catalogue target, not a probability that the proofs are correct. The version adds an exact tail-root and exponential-moment classification, including the critical asymptotic and strict speed restrictions, to the retained value-finiteness and state-sensitivity results. It does not prove attraction to a wave, identify a basin of initial densities, or select a speed or profile; those are the principal unresolved parts of the target.

References

  • [CD] P. Cardaliaguet and F. Delarue, Selected topics in mean field games, Proc. ICM 2022, vol. 5, 3660--3703, EMS Press (2023), Section 3.3. https://doi.org/10.4171/ICM2022/17.
  • [PR] A. Porretta and L. Rossi, Traveling waves for a nonlocal KPP equation and mean-field game models of knowledge diffusion, arXiv:2010.10828v1 (2020), system (2), Theorem 2.2 and its following remarks. https://arxiv.org/pdf/2010.10828v1.

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Econ-24-A2 · Exploration · LLM-generated · gpt 6 astra ultra ·

GPT 6 Astra Ultra attempt on Econ-24, version 2

Knowledge-growth wave selection: finite values, state sensitivity and vanishing upper-tail search

Problem O33-23. Status: unresolved. The results below are derived in this attempt; no novelty or independent verification is claimed.

Target and scope

The target is attraction, after median centering, to a Lucas--Moll balanced-growth wave, including selection of its speed. We retain the infinite-horizon value boundary condition. Wave existence does not answer this question. The known wave theorem is Porretta--Rossi, Theorem 2.2 [PR]; the time-dependent question is discussed in [CD, Section 3.3]. We retain the necessary speed and integrability restrictions from Version 1. Version 2 adds a finite-value theorem, a uniform productivity-sensitivity bound, and a quantitative upper-tail search restriction. These are restrictions on admissible solutions, not an attraction theorem.

Write $a=\max_{s\in[0,1]}\alpha(s)$ and $q(t,x)=\alpha(s_*(t,x))$. The density equation is \begin{equation}\label{wsref-0-64-65-6e-73-69-74-79} f_t=\nu f_{xx}+f(x)\int_{-\infty}^x q(y)f(y){\mathrm d} y-q(x)f(x)\int_x^\infty f(y){\mathrm d} y, \qquad 0\le q\le a. \end{equation} Time arguments inside integrals are suppressed. We assume a positive classical probability-density solution. Whenever integration by parts with an unbounded weight is used, assume the corresponding exponential moment is locally bounded in time and that the weak formulation extends to that weight by cutoff; sufficient conditions are integrability of $e^{\lambda x}(f+|f_x|)$ and vanishing weighted boundary terms. This additional moment regularity is stated, not inferred just from pointwise classical regularity.

Tail and moment equations

Proposition: Let $w(t,x)=\int_x^\infty f(t,y){\mathrm d} y$ and $A(t,x)=\int_{-\infty}^x q(t,y)f(t,y){\mathrm d} y$. If the flux at infinity vanishes, then \[ w_t=\nu w_{xx}+Aw,\qquad 0\le A\le a(1-w). \] For $\lambda>0$ with the stated moment regularity, $M_\lambda(t)=\int e^{\lambda x}f(t,x){\mathrm d} x$ satisfies \begin{align}\label{wsref-0-6d-6f-6d-65-6e-74-69-64-65-6e-74-69-74-79} M_\lambda'(t)&=\nu\lambda^2 M_\lambda(t)+I_\lambda(t),\\ I_\lambda(t)&=\int_{x<y}q(t,x)f(t,x)f(t,y)(e^{\lambda y}-e^{\lambda x}){\mathrm d} x{\mathrm d} y, \qquad 0\le I_\lambda\le aM_\lambda.\nonumber \end{align} In particular, if $b(t)$ is a median, \[ b(t)\le\frac{\log(2M_\lambda(0))}{\lambda} +\left(\nu\lambda+\frac a\lambda\right)t. \] Every existing limit $b(t)/t\to c$ therefore satisfies $c\le\inf_{\lambda\in\Lambda}(\nu\lambda+a/\lambda)$, where $\Lambda$ is any set of positive exponents for which these moment hypotheses hold.
Proof: Integrate (density) over $(x,\infty)$. In the reaction terms all ordered meetings with both states above $x$ cancel. Only meetings starting below $x$ and ending above $x$ remain, giving $Aw$. The diffusion contribution is $-\nu f_x=\nu w_{xx}$. The bound on $A$ follows from $q\le a$ and $\int_{-\infty}^x f=1-w$.

Multiplying (density) by $e^{\lambda x}$, integrating the diffusion twice by parts, and interchanging the nonnegative ordered-meeting integrals gives (momentidentity). Its upper bound follows from \[ I_\lambda\le a\int e^{\lambda y}f(t,y)\int_{-\infty}^y f(t,x){\mathrm d} x{\mathrm d} y\le aM_\lambda. \] Gronwall's inequality yields $M_\lambda(t)\le M_\lambda(0)e^{(\nu\lambda^2+a)t}$. Since at least half the mass lies above the median, $M_\lambda(t)\ge e^{\lambda b(t)}/2$. These two inequalities prove the assertion, including its infimum version. ∎

The hypotheses of the question give a first exponential moment initially. They do not supply every exponent in $\Lambda$. In particular, replacing the infimum by $2\sqrt{\nu a}$ needs the exponent $\sqrt{a/\nu}$ to be available. Even when this upper bound holds, it does not identify the selected wave, because $A$ depends on the endogenous optimal policy.

Restrictions on a possible limit wave

Theorem: Let $(\varphi,S,c)$ be a positive classical traveling density satisfying the density equation, $\int\varphi=1$, and the flux and moment conditions above. Put \[ W(z)=\int_z^\infty\varphi(y){\mathrm d} y,\quad B(z)=\int_{-\infty}^z\alpha(S(y))\varphi(y){\mathrm d} y,\quad \beta=B(+\infty). \] Then \begin{equation}\label{wsref-0-74-61-69-6c-6f-64-65} \nu W''+cW'+BW=0. \end{equation} If $\beta>0$ and the first exponential moment is finite, then $c>\nu$ and $c\ge2\sqrt{\nu\beta}$. More generally, whenever $M_\lambda(\varphi)<\infty$ with valid weighted integrations, \begin{equation}\label{wsref-0-73-70-65-65-64-6d-6f-6d-65-6e-74} c=\nu\lambda+\frac{I_\lambda(\varphi,S)}{\lambda M_\lambda(\varphi)} \le\nu\lambda+\frac a\lambda. \end{equation} If in addition $h(z)=-W'(z)/W(z)$ converges to a finite positive $\ell$ as $z\to\infty$, then \begin{equation}\label{wsref-0-64-69-73-70-65-72-73-69-6f-6e} c=\nu\ell+\frac\beta\ell. \end{equation} Finite first moment implies $\ell\ge1$; when $\ell>1$ it guarantees first-moment integrability at the right tail, whereas $\ell=1$ alone decides neither integrability nor nonintegrability.
Proof: The tail equation from Proposition Proposition with $w(t,x)=W(x-ct)$ is (tailode). A traveling density has $M_\lambda(t)=e^{\lambda ct}M_\lambda(\varphi)$, giving (speedmoment). When $\beta>0$, the set where $\alpha(S(x))\varphi(x)>0$ has positive measure. Positivity of $\varphi$ gives positive mass strictly above every such $x$, so $I_1>0$ and $c>\nu$.

Set $Y(z)=e^{cz/(2\nu)}W(z)>0$. Then \[ Y''+\left(\frac{B(z)}\nu-\frac{c^2}{4\nu^2}\right)Y=0. \] If $c^2<4\nu\beta$, the coefficient is at least some $\eta>0$ on a half-line. No positive function solves $Y''\le-\eta Y$ there: if $Y'$ becomes negative, concavity forces a zero in finite distance; if $Y'$ stays nonnegative, $Y$ stays bounded below by a positive constant and $Y''$ is uniformly negative, forcing $Y'$ to become negative. Hence $c^2\ge4\nu\beta$. The already proved $c>0$ gives the claimed lower bound.

Substitution of $h=-W'/W$ in (tailode) gives $h'=h^2-(c/\nu)h+B/\nu$. If $h\to\ell$, its right-hand side has a limit. That limit must be zero, since a derivative with a nonzero limit is incompatible with convergence of its primitive. This proves (dispersion). Finally, \[ \int_0^\infty e^z\varphi(z){\mathrm d} z=W(0)+\int_0^\infty e^zW(z){\mathrm d} z \] when finite, by integration by parts or Tonelli. If $\ell<1$, choose $\ell<r<1$; eventually $h\le r$, so $W(z)\ge C e^{-rz}$ and the last integral diverges. If $\ell>1$, choose $1<r<\ell$; eventually $h\ge r$, so $W(z)\le C e^{-rz}$ and it converges. At $\ell=1$, the tails $e^{-z}(1+z)^{-2}$ and $e^{-z}$ show the two alternatives. These last tails illustrate the integrability test; they are not asserted to be wave solutions. ∎

The published wave analysis gives stronger restrictions under its full value-profile hypotheses; Theorem Theorem is deliberately only a necessary test from the density and moment equations. It does not reprove the existence result of [PR].

Proposition (The square-root search rule): Suppose $\alpha(s)=a\sqrt{s}$ and the wave value $V$ is nondecreasing with finite meeting integral. Define \[ Q(z)=e^{-z}\int_z^\infty[V(y)-V(z)]\varphi(y){\mathrm d} y. \] Then the maximizing search fraction is uniquely \[ S(z)=\min\{1,a^2Q(z)^2/4\}. \] If $V$ is the finite infinite-horizon value and $\rho>\nu$, then $V(z)\ge e^z/(\rho-\nu)$.
Proof: Divide the objective by $e^z$ and set $r=\sqrt{s}\in[0,1]$. The maximized expression becomes $1-r^2+aQ(z)r$, a strictly concave quadratic whose maximizer is $\min\{1,aQ(z)/2\}$ because $Q\ge0$. Squaring proves the formula. Choosing zero search forever is admissible and leaves only diffusion; its expected discounted production is $e^z\int_0^\infty e^{-(\rho-\nu)t}{\mathrm d} t$. Taking the supremum proves the lower bound. ∎

Finite values and productivity sensitivity: progress beyond Version 1

Version 1 assumed that the infinite-horizon value was finite and proved only its no-search lower bound. We now verify finiteness under the catalogue's discount condition and control its dependence on the initial state. These arguments are conditional on a density flow satisfying the moment assumptions of Proposition Proposition; they do not construct a solution of the coupled density--value system.

Throughout this section assume $a>0$, as in the catalogue's concrete square-root wave regime. If $a=0$, there are no meetings: the value is exactly $e^x/(\rho-\nu)$ for $\rho>\nu$, and $\beta=0$; the strict conclusion $\beta<a$ below is not asserted in that case.

Fix a starting time $t$. Write $A_s\in[0,1]$ for a control at elapsed time $s$ to distinguish it from the constant rate bound $a$. Work on a filtered space supporting a Brownian motion $B$, a rate-$a$ Poisson process, and independent uniform marks $(U,V)$ at its points, independent of $B$. Let $F_r^{-1}$ be a measurable quantile function of the given density $f(r,\cdot)$. At a potential meeting at elapsed time $s$, accept the meeting if $U\leq\alpha(A_s)/a$ and, if accepted, replace the log state by $\max\{X_{s-},F_{t+s}^{-1}(V)\}$. Between meetings the state has diffusion $\sqrt{2\nu}\,dB_s$. Controls are predictable for this driving filtration, possibly with additional independent private randomization; in particular, they do not anticipate the current marks. This is the usual thinning construction of the controlled meeting intensity. There are finitely many potential meetings on every bounded interval, so the construction is pathwise well defined. The same construction covers progressively measurable controls after the usual predictable representative at Poisson times; it does not allow a control to inspect a meeting mark before choosing whether to search.

For this admissible class define \[ J_{t,x}(A)={\mathbb{E}}\int_0^\infty e^{-\rho s}(1-A_s)e^{X_s}\,ds, \qquad v(t,x)=\sup_A J_{t,x}(A). \] The controls are indexed by the common driving filtration, rather than restricted to a prescribed feedback function. Thus a control chosen for one starting state can be used for another starting state: any auxiliary state trajectory it uses is itself a causal function of the same driving noise. This point is needed when taking suprema in the coupling proof. The argument includes arbitrary randomized admissible deviations in this formulation.

Lemma (Productivity under bounded meeting rates): Suppose $M_1(t+s)\leq M_1(t)e^{(\nu+a)s}$ for $s\geq0$ and $M_1(t)<\infty$. For every admissible control and $s\geq0$, \[ {\mathbb{E}} e^{X_s}\leq e^{\nu s}e^x +a\int_0^s e^{\nu(s-r)}M_1(t+r)\,dr. \tag{20} \] If two processes start at $x\leq x'$ and use the same control and all the same driving noise, then $X_s\leq X'_s$ pathwise and \[ 0\leq{\mathbb{E}}(e^{X'_s}-e^{X_s})\leq e^{\nu s}(e^{x'}-e^x). \tag{21} \]
Proof: Let $Z_s=e^{X_s}$ and $\sigma=\sqrt{2\nu}$. Between meetings, $dZ_s=\nu Z_s\,ds+\sigma Z_s\,dB_s$. The continuous multiplicative factor is $L_s=e^{\sigma B_s}$, with ${\mathbb{E}} L_s=e^{\nu s}$. At a potential meeting at time $r$ the jump $\Delta Z_r$ is nonnegative and at most the mark $e^{F_{t+r}^{-1}(V)}$, even when the meeting is rejected. Variation of constants on the finitely many meeting intervals gives \[ Z_s=e^xL_s+\sum_{r\leq s}\frac{L_s}{L_r}\Delta Z_r \leq e^xL_s+ \sum_{r\leq s}\frac{L_s}{L_r}e^{F_{t+r}^{-1}(V_r)}. \] The sum is over potential meetings, so its right-hand side is independent of acceptance decisions. Tonelli's theorem and the rate-$a$ marked Poisson expectation formula apply to its nonnegative terms. The Brownian increment $B_s-B_r$ is independent of the meeting time and mark and has exponential expectation $e^{\nu(s-r)}$. Hence the expected sum is $a\int_0^s e^{\nu(s-r)}M_1(t+r)dr$, proving (20) and its finiteness. The Poisson expectation formula here can also be obtained by conditioning on the number and times of its points and integrating the independent uniform marks.

For the pair of starting states, order is preserved both by their common Brownian translation in log space and by the map $z\mapsto\max\{z,y\}$. In productivity coordinates the latter map is $z\mapsto\max\{z,e^y\}$; for $z'\geq z$ it satisfies \[ 0\leq\max\{z',e^y\}-\max\{z,e^y\}\leq z'-z. \] Acceptance is the same for the coupled processes because the control is the same. Consequently $D_s=Z'_s-Z_s$ follows the common geometric Brownian factor between meetings and has nonpositive jumps, while staying nonnegative. Iterating across the finite set of meetings yields $0\leq D_s\leq(e^{x'}-e^x)L_s$. Taking expectations proves (21). ∎

Theorem (A finite value and a uniform increment bound): Assume the moment regularity of Proposition Proposition holds on every finite time interval, $0\leq\alpha\leq a$, and $\rho>\nu+a$. Then the value defined above is finite and obeys \begin{equation}\label{wsref-0-76-61-6c-75-65-62-6f-75-6e-64-73} \frac{e^x}{\rho-\nu}\leq v(t,x) \leq\frac{e^x}{\rho-\nu} +\frac{aM_1(t)}{(\rho-\nu)(\rho-\nu-a)}. \end{equation} For every $h\geq0$, \begin{equation}\label{wsref-0-69-6e-63-72-65-6d-65-6e-74} 0\leq v(t,x+h)-v(t,x) \leq\frac{e^{x+h}-e^x}{\rho-\nu}. \end{equation} In particular $v(t,\cdot)$ is locally Lipschitz. Where it is differentiable, $0\leq v_x(t,x)\leq e^x/(\rho-\nu)$; for the classical values in the catalogue this holds at every point.
Proof: Proposition Proposition, applied from time $t$, gives the moment bound required in Lemma Lemma (Productivity under bounded meeting rates). Since $0\leq1-A_s\leq1$, (20), Tonelli's theorem, and then the moment bound give \begin{align*} J_{t,x}(A) &\leq\frac{e^x}{\rho-\nu} +a\int_0^\infty e^{-\rho s}\int_0^s e^{\nu(s-r)}M_1(t+r)\,dr\,ds\\ &=\frac{e^x}{\rho-\nu} +\frac a{\rho-\nu}\int_0^\infty e^{-\rho r}M_1(t+r)\,dr\\ &\leq\frac{e^x}{\rho-\nu} +\frac{aM_1(t)}{(\rho-\nu)(\rho-\nu-a)}. \end{align*} The bound is uniform in the control. Zero search gives the lower bound in (valuebounds). Both denominators are positive by assumption.

For a fixed control use the ordered coupling in (21). The payoff increment is nonnegative, and its expectation is at most \[ \int_0^\infty e^{-\rho s}{\mathbb{E}} D_s\,ds \leq\frac{e^{x+h}-e^x}{\rho-\nu}. \] Thus $J_{t,x}(A)\leq J_{t,x+h}(A)\leq J_{t,x}(A)+C_{x,h}$ for every control in the same admissible class. Taking suprema proves (increment); no optimizer or feedback regularity is required. Applying that bound on a compact interval proves local Lipschitz continuity, and taking difference quotients proves the derivative bound. ∎

Corollary (Search vanishes in the upper tail): Under Theorem Theorem (A finite value and a uniform increment bound), suppose $\alpha(s)=a\sqrt{s}$. For the value-based maximizing search selector define \[ Q_t(x)=e^{-x}\int_x^\infty[v(t,y)-v(t,x)]f(t,y)\,dy. \] Then \begin{align} 0\leq Q_t(x)&\leq\frac1{\rho-\nu} \int_x^\infty(e^{y-x}-1)f(t,y)\,dy,\label{wsref-0-51-74-61-69-6c}\\ s_*(t,x)&=\min\{1,a^2Q_t(x)^2/4\},\qquad \lim_{x\to\infty}s_*(t,x)=0.\nonumber \end{align} In particular every positive finite-first-moment wave in this regime satisfies $S(z)\to0$ as $z\to\infty$ and \[ \beta=\int_{\mathbb{R}} a\sqrt{S(z)}\varphi(z)\,dz<a. \] The limit is for each fixed time or wave; no uniform-in-time centered-tail estimate is asserted.
Proof: Apply (increment) to each $y\geq x$ and integrate. This proves (Qtail), including finiteness of its meeting integral. The right side is at most $e^{-x}(\rho-\nu)^{-1}\int_x^\infty e^yf(t,y)dy$, which tends to zero as $x\to\infty$ by the finite first moment. Monotonicity of the value makes $Q_t\geq0$. Maximizing in $\sqrt{s}$ as in the earlier search-rule proposition gives the stated unique selector and its limit. For a wave evaluate at time zero, where the value is $V$ and density is $\varphi$. Choose $R$ sufficiently large that $S(z)\leq1/4$ for $z\geq R$. Positivity of $\varphi$ gives $\int_R^\infty\varphi>0$, and hence \[ \beta\leq a\int_{-\infty}^R\varphi +\frac a2\int_R^\infty\varphi<a. \] ∎

A compactness obstruction and the first gap

Proposition: Convergence of median-centered densities in $L^1$ does not imply convergence of their first exponential moments, even for strictly positive smooth densities.
Proof: Let $\varphi$ be the standard Gaussian density and $\varphi_n(x)=(1-e^{-n})\varphi(x)+e^{-n}\varphi(x-2n)$. Then $\|\varphi_n-\varphi\|_1\le2e^{-n}$, but $\int e^x\varphi_n(x){\mathrm d} x=e^{1/2}(1-e^{-n}+e^n)\to\infty$. Let $b_n$ be the unique median of $\varphi_n$. Uniform convergence of the cumulative distribution functions and strict increase of the Gaussian cumulative function imply $b_n\to0$. Translation continuity in $L^1$ gives $\varphi_n(b_n+\cdot)\to\varphi$ in $L^1$, while their moments equal $e^{-b_n}\int e^x\varphi_n(x){\mathrm d} x\to\infty$. ∎
Thus neither the exact moment identity nor the nonlocal value integral can be passed to a proposed median-frame limit using the stated density topology alone. A candidate route must first prove suitable uniform integrability of the centered tails and control of the backward value field, then establish attraction and distinguish competing wave branches. No such estimate, attraction theorem, or basin characterization is proved here. The new finite-value and sensitivity theorems verify additional admissibility properties and rule out persistently maximal search in the upper tail. They do not supply the missing moving-frame estimates or dynamical attraction; the complete target remains unresolved.

Completion Estimate

20%

This is a heuristic assessment of the full catalogue target, not a probability that the proofs are correct. The version proves value finiteness, state sensitivity and upper-tail search decay in addition to the retained speed and moment restrictions. It does not prove attraction to a wave, identify a basin of initial densities, or select a speed or profile; those are the principal unresolved parts of the target.

References

  • [CD] P. Cardaliaguet and F. Delarue, Selected topics in mean field games, Proc. ICM 2022, vol. 5, 3660--3703, EMS Press (2023), Section 3.3. https://doi.org/10.4171/ICM2022/17.
  • [PR] A. Porretta and L. Rossi, Traveling waves for a nonlocal KPP equation and mean-field game models of knowledge diffusion, arXiv:2010.10828v1 (2020), system (2), Theorem 2.2 and its following remarks. https://arxiv.org/pdf/2010.10828v1.

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Econ-24-A1 · Exploration · LLM-generated · gpt 6 astra ultra ·

GPT 6 Astra Ultra attempt on Econ-24

Tail constraints for knowledge-growth wave selection

Problem O33-23. Status: unresolved. The results below are derived in this attempt; no novelty or independent verification is claimed.

Target and scope

The target is attraction, after median centering, to a Lucas--Moll balanced-growth wave, including selection of its speed. We retain the infinite-horizon value boundary condition. Wave existence does not answer this question. The known wave theorem is Porretta--Rossi, Theorem 2.2 [PR]; the time-dependent question is discussed in [CD, Section 3.3]. Here we prove necessary speed and integrability restrictions, and identify a missing compactness estimate.

Write $a=\max_{s\in[0,1]}\alpha(s)$ and $q(t,x)=\alpha(s_*(t,x))$. The density equation is \begin{equation}\label{wsref-0-64-65-6e-73-69-74-79} f_t=\nu f_{xx}+f(x)\int_{-\infty}^x q(y)f(y){\mathrm d} y-q(x)f(x)\int_x^\infty f(y){\mathrm d} y, \qquad 0\le q\le a. \end{equation} Time arguments inside integrals are suppressed. We assume a positive classical probability-density solution. Whenever integration by parts with an unbounded weight is used, assume the corresponding exponential moment is locally bounded in time and that the weak formulation extends to that weight by cutoff; sufficient conditions are integrability of $e^{\lambda x}(f+|f_x|)$ and vanishing weighted boundary terms. This additional moment regularity is stated, not inferred just from pointwise classical regularity.

Tail and moment equations

Proposition: Let $w(t,x)=\int_x^\infty f(t,y){\mathrm d} y$ and $A(t,x)=\int_{-\infty}^x q(t,y)f(t,y){\mathrm d} y$. If the flux at infinity vanishes, then \[ w_t=\nu w_{xx}+Aw,\qquad 0\le A\le a(1-w). \] For $\lambda>0$ with the stated moment regularity, $M_\lambda(t)=\int e^{\lambda x}f(t,x){\mathrm d} x$ satisfies \begin{align}\label{wsref-0-6d-6f-6d-65-6e-74-69-64-65-6e-74-69-74-79} M_\lambda'(t)&=\nu\lambda^2 M_\lambda(t)+I_\lambda(t),\\ I_\lambda(t)&=\int_{x<y}q(t,x)f(t,x)f(t,y)(e^{\lambda y}-e^{\lambda x}){\mathrm d} x{\mathrm d} y, \qquad 0\le I_\lambda\le aM_\lambda.\nonumber \end{align} In particular, if $b(t)$ is a median, \[ b(t)\le\frac{\log(2M_\lambda(0))}{\lambda} +\left(\nu\lambda+\frac a\lambda\right)t. \] Every existing limit $b(t)/t\to c$ therefore satisfies $c\le\inf_{\lambda\in\Lambda}(\nu\lambda+a/\lambda)$, where $\Lambda$ is any set of positive exponents for which these moment hypotheses hold.
Proof: Integrate (density) over $(x,\infty)$. In the reaction terms all ordered meetings with both states above $x$ cancel. Only meetings starting below $x$ and ending above $x$ remain, giving $Aw$. The diffusion contribution is $-\nu f_x=\nu w_{xx}$. The bound on $A$ follows from $q\le a$ and $\int_{-\infty}^x f=1-w$.

Multiplying (density) by $e^{\lambda x}$, integrating the diffusion twice by parts, and interchanging the nonnegative ordered-meeting integrals gives (momentidentity). Its upper bound follows from \[ I_\lambda\le a\int e^{\lambda y}f(t,y)\int_{-\infty}^y f(t,x){\mathrm d} x{\mathrm d} y\le aM_\lambda. \] Gronwall's inequality yields $M_\lambda(t)\le M_\lambda(0)e^{(\nu\lambda^2+a)t}$. Since at least half the mass lies above the median, $M_\lambda(t)\ge e^{\lambda b(t)}/2$. These two inequalities prove the assertion, including its infimum version. ∎

The hypotheses of the question give a first exponential moment initially. They do not supply every exponent in $\Lambda$. In particular, replacing the infimum by $2\sqrt{\nu a}$ needs the exponent $\sqrt{a/\nu}$ to be available. Even when this upper bound holds, it does not identify the selected wave, because $A$ depends on the endogenous optimal policy.

Restrictions on a possible limit wave

Theorem: Let $(\varphi,S,c)$ be a positive classical traveling density satisfying the density equation, $\int\varphi=1$, and the flux and moment conditions above. Put \[ W(z)=\int_z^\infty\varphi(y){\mathrm d} y,\quad B(z)=\int_{-\infty}^z\alpha(S(y))\varphi(y){\mathrm d} y,\quad \beta=B(+\infty). \] Then \begin{equation}\label{wsref-0-74-61-69-6c-6f-64-65} \nu W''+cW'+BW=0. \end{equation} If $\beta>0$ and the first exponential moment is finite, then $c>\nu$ and $c\ge2\sqrt{\nu\beta}$. More generally, whenever $M_\lambda(\varphi)<\infty$ with valid weighted integrations, \begin{equation}\label{wsref-0-73-70-65-65-64-6d-6f-6d-65-6e-74} c=\nu\lambda+\frac{I_\lambda(\varphi,S)}{\lambda M_\lambda(\varphi)} \le\nu\lambda+\frac a\lambda. \end{equation} If in addition $h(z)=-W'(z)/W(z)$ converges to a finite positive $\ell$ as $z\to\infty$, then \begin{equation}\label{wsref-0-64-69-73-70-65-72-73-69-6f-6e} c=\nu\ell+\frac\beta\ell. \end{equation} Finite first moment implies $\ell\ge1$; when $\ell>1$ it guarantees first-moment integrability at the right tail, whereas $\ell=1$ alone decides neither integrability nor nonintegrability.
Proof: The tail equation from Proposition Proposition with $w(t,x)=W(x-ct)$ is (tailode). A traveling density has $M_\lambda(t)=e^{\lambda ct}M_\lambda(\varphi)$, giving (speedmoment). When $\beta>0$, the set where $\alpha(S(x))\varphi(x)>0$ has positive measure. Positivity of $\varphi$ gives positive mass strictly above every such $x$, so $I_1>0$ and $c>\nu$.

Set $Y(z)=e^{cz/(2\nu)}W(z)>0$. Then \[ Y''+\left(\frac{B(z)}\nu-\frac{c^2}{4\nu^2}\right)Y=0. \] If $c^2<4\nu\beta$, the coefficient is at least some $\eta>0$ on a half-line. No positive function solves $Y''\le-\eta Y$ there: if $Y'$ becomes negative, concavity forces a zero in finite distance; if $Y'$ stays nonnegative, $Y$ stays bounded below by a positive constant and $Y''$ is uniformly negative, forcing $Y'$ to become negative. Hence $c^2\ge4\nu\beta$. The already proved $c>0$ gives the claimed lower bound.

Substitution of $h=-W'/W$ in (tailode) gives $h'=h^2-(c/\nu)h+B/\nu$. If $h\to\ell$, its right-hand side has a limit. That limit must be zero, since a derivative with a nonzero limit is incompatible with convergence of its primitive. This proves (dispersion). Finally, \[ \int_0^\infty e^z\varphi(z){\mathrm d} z=W(0)+\int_0^\infty e^zW(z){\mathrm d} z \] when finite, by integration by parts or Tonelli. If $\ell<1$, choose $\ell<r<1$; eventually $h\le r$, so $W(z)\ge C e^{-rz}$ and the last integral diverges. If $\ell>1$, choose $1<r<\ell$; eventually $h\ge r$, so $W(z)\le C e^{-rz}$ and it converges. At $\ell=1$, the tails $e^{-z}(1+z)^{-2}$ and $e^{-z}$ show the two alternatives. These last tails illustrate the integrability test; they are not asserted to be wave solutions. ∎

The published wave analysis gives stronger restrictions under its full value-profile hypotheses; Theorem Theorem is deliberately only a necessary test from the density and moment equations. It does not reprove the existence result of [PR].

Proposition (The square-root search rule): Suppose $\alpha(s)=a\sqrt{s}$ and the wave value $V$ is nondecreasing with finite meeting integral. Define \[ Q(z)=e^{-z}\int_z^\infty[V(y)-V(z)]\varphi(y){\mathrm d} y. \] Then the maximizing search fraction is uniquely \[ S(z)=\min\{1,a^2Q(z)^2/4\}. \] If $V$ is the finite infinite-horizon value and $\rho>\nu$, then $V(z)\ge e^z/(\rho-\nu)$.
Proof: Divide the objective by $e^z$ and set $r=\sqrt{s}\in[0,1]$. The maximized expression becomes $1-r^2+aQ(z)r$, a strictly concave quadratic whose maximizer is $\min\{1,aQ(z)/2\}$ because $Q\ge0$. Squaring proves the formula. Choosing zero search forever is admissible and leaves only diffusion; its expected discounted production is $e^z\int_0^\infty e^{-(\rho-\nu)t}{\mathrm d} t$. Taking the supremum proves the lower bound. ∎

A compactness obstruction and the first gap

Proposition: Convergence of median-centered densities in $L^1$ does not imply convergence of their first exponential moments, even for strictly positive smooth densities.
Proof: Let $\varphi$ be the standard Gaussian density and $\varphi_n(x)=(1-e^{-n})\varphi(x)+e^{-n}\varphi(x-2n)$. Then $\|\varphi_n-\varphi\|_1\le2e^{-n}$, but $\int e^x\varphi_n(x){\mathrm d} x=e^{1/2}(1-e^{-n}+e^n)\to\infty$. Let $b_n$ be the unique median of $\varphi_n$. Uniform convergence of the cumulative distribution functions and strict increase of the Gaussian cumulative function imply $b_n\to0$. Translation continuity in $L^1$ gives $\varphi_n(b_n+\cdot)\to\varphi$ in $L^1$, while their moments equal $e^{-b_n}\int e^x\varphi_n(x){\mathrm d} x\to\infty$. ∎
Thus neither the exact moment identity nor the nonlocal value integral can be passed to a proposed median-frame limit using the stated density topology alone. A candidate route must first prove suitable uniform integrability of the centered tails and control of the backward value field, then establish attraction and distinguish competing wave branches. No such estimate, attraction theorem, or basin characterization is proved here. All conclusions are necessary restrictions or conditional identities; the complete target remains unresolved.

References

  • [CD] P. Cardaliaguet and F. Delarue, Selected topics in mean field games, Proc. ICM 2022, vol. 5, 3660--3703, EMS Press (2023), Section 3.3. https://doi.org/10.4171/ICM2022/17.
  • [PR] A. Porretta and L. Rossi, Traveling waves for a nonlocal KPP equation and mean-field game models of knowledge diffusion, arXiv:2010.10828v1 (2020), system (2), Theorem 2.2 and its following remarks. https://arxiv.org/pdf/2010.10828v1.

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