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One-Dimensional Stationary Mean-Field Games with Local Coupling

Abstract

A standard assumption in mean-field game (MFG) theory is that the coupling between the Hamilton–Jacobi equation and the transport equation is monotonically non-decreasing in the density of the population. In many cases, this assumption implies the existence and uniqueness of solutions. Here, we drop that assumption and construct explicit solutions for one-dimensional MFGs. These solutions exhibit phenomena not present in monotonically increasing MFGs: low-regularity, non-uniqueness, and the formation of regions with no agents.

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Corresponding author

Correspondence to Diogo A. Gomes.

Additional information

D. Gomes, L. Nurbekyan and M. Prazeres were partially supported by KAUST baseline and start-up funds.

Appendix

Appendix

Proof of Proposition 6.1

We have that \(u_x+p=\frac{j}{m}\) a.e.. Thus, the second equation in (1.1) holds in the sense of distributions. Next, we observe that u is differentiable for all \(x\ne x_i\) and that the first equation in (1.1) is satisfied in the classical sense at those points. Thus, we just need to check the viscosity condition at \(x=x_i\).

There are two possible cases:

  1. 1.

    \(m(x_i^-)>m(x_i^+)\). In this case, \(m^*(x_i)=m(x_i^-). \) Moreover,

    $$\begin{aligned} u_x(x_i^-)=j/m(x_i^-)-p<j/m(x_i^+)-p=u_x(x_i^+). \end{aligned}$$

    Hence, there is no smooth function touching u from above; u can only be touched from below. Therefore, we need to check that, for any \(\phi \) touching u from below at \(x_i\), we have

    $$\begin{aligned} \frac{(\phi _x(x_i)+p)^2}{2}+V(x_i)+m(x_i^-)-{\overline{H}}\geqslant 0. \end{aligned}$$

    Because (1.1) is satisfied at \(x\ne x_i\) in the classical sense, we have that

    $$\begin{aligned} \frac{(u_x(x_i^\pm )+p)^2}{2}+V(x_i)+m(x_i^\pm )-{\overline{H}}=0. \end{aligned}$$

    Because \(\phi \) touches u from below and \(j>0\), we have

    $$\begin{aligned} 0<u_x(x_i^-)+p\leqslant \phi _x(x_i)+p\leqslant u_x(x_i^+)+p. \end{aligned}$$

    Hence,

    $$\begin{aligned}&\frac{(\phi _x(x_i)+p)^2}{2}+V(x_i)+m(x_i^-)-{\overline{H}}\\&\geqslant \frac{(u_x(x_i^-)+p)^2}{2}+V(x_i)+m(x_i^-)-{\overline{H}}= 0. \end{aligned}$$
  2. 2.

    \(m(x_i-)<m(x_i+)\). In this case, \(m_*(x_i)=m(x_i^-)\) and

    $$\begin{aligned} u_x(x_i^-)=j/m(x_i^-)-p>j/m(x_i^+)-p=u_x(x_i^+). \end{aligned}$$

    Hence, there is no smooth function touching u from below – only from above. Therefore, for any \(\phi \) touching u from above at \(x_i\), we have

    $$\begin{aligned} \frac{(\phi _x(x_i)+p)^2}{2}+V(x_i)+m(x_i^-)-{\overline{H}}\leqslant 0. \end{aligned}$$
    (11.4)

    Because (1.1) holds in the classical sense for \(x\ne x_i,\) we have that

    $$\begin{aligned} \frac{(u_x(x_i^\pm )+p)^2}{2}+V(x_i)+m(x_i^\pm )-{\overline{H}}=0. \end{aligned}$$

    Because \(\phi \) touches u from above, we have \(0<u_x(x_i^+)+p\leqslant \phi _x(x_i)+p\leqslant u_x(x_i^-)+p. \) Hence, (11.4) holds.

\(\square \)

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Gomes, D.A., Nurbekyan, L. & Prazeres, M. One-Dimensional Stationary Mean-Field Games with Local Coupling. Dyn Games Appl 8, 315–351 (2018). https://doi.org/10.1007/s13235-017-0223-9

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Keywords

  • Mean-field games
  • Stationary problems
  • Dynamic games

Mathematics Subject Classification

  • 91A13
  • 91A25