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.
Completion Estimate
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