# A convergence result for a local planning problem for mean field games and rigorous proof of a Freidlin-Ventchel-type Large Deviations Principle for the 1+1 KPZ equation

Pierre-Louis Lions, Panagiotis E Souganidis

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Pierre-Louis Lions, Panagiotis E Souganidis. A convergence result for a local planning problem for mean field games and rigorous proof of a Freidlin-Ventchel-type Large Deviations Principle for the 1+1 KPZ equation. 2024. hal-05315489

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https://college-de-france.hal.science/hal-05315489v1

Preprint submitted on 15 Oct 2025

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Distributed under a Creative Commons CC0 - Public Domain Dedication 4.0 International License
A CONVERGENCE RESULT FOR A LOCAL PLANNING PROBLEM FOR MEAN FIELD GAMES AND RIGOROUS PROOF OF A FREIDLIN-VENTCHEL-TYPE LARGE DEVIATIONS PRINCIPLE FOR THE 1 + 1 KPZ EQUATION

PIERRE-LOUIS LIONS AND PANAGIOTIS E. SOUGANIDIS

Version: January 5, 2024

ABSTRACT. We prove the convergence of a viscous approximation to an one dimensional local mean field type planning problem with singular initial and terminal measures. Then we use this result to give a rigorous proof to a Freidlin-Ventchel-type Large Deviations Principle for the height of the 1 + 1 KPZ equation.

1. INTRODUCTION

Motivated by the theory of large deviations for large heights and fixed times related to the 1+1 KPZ equation, here we consider, for $\epsilon, \eta > 0$, the parabolic mean field game system

$$ egin{cases} 
\partial_t u_{\epsilon, \eta} - \epsilon \partial_x^2 u_{\epsilon, \eta} + \frac{1}{2} (\partial_x u_{\epsilon, \eta})^2 = \rho_{\epsilon, \eta} \quad \text{in } \mathbb{R} \times (0, 1), \\
\partial_t \rho_{\epsilon, \eta} + \epsilon \partial_x^2 \rho_{\epsilon, \eta} + \partial_x (\partial_x u_{\epsilon, \eta} \rho_{\epsilon, \eta}) = 0 \quad \text{in } \mathbb{R} \times (0, 1), \\
u_{\epsilon, \eta}(x, 0) = |x|^2/2\eta \quad \text{and} \quad \rho_{\epsilon, \eta}(\cdot, 1) = \delta,
\end{cases} 
\quad (1.1) $$

where $\delta$ is the Dirac mass at $x=0$, and show that, as $\epsilon, \eta \to 0$ with $\epsilon \le C\eta^\alpha$ for some $\alpha \in (0, 1)$ which can be arbitrarily small, $u_{\epsilon, \eta} \to \bar{u}$ locally uniformly in $\mathbb{R} \times (0, 1]$ and $\rho_{\epsilon, \eta} \to \bar{\rho}$ in $L^2(\mathbb{R} \times [0, 1])$ where $(\bar{u}, \bar{\rho})$ is the unique solution of the planning problem

$$ egin{cases} 
\partial_t \bar{u} + \frac{1}{2} (\partial_x \bar{u})^2 = \bar{\rho} \quad \text{in } \mathbb{R} \times (0, 1), \\
\partial_t \bar{\rho} + \partial_x (\partial_x \bar{u} \bar{\rho}) = 0 \quad \text{in } \mathbb{R} \times (0, 1), \\
\bar{\rho}(\cdot, 0) = \delta, \quad \bar{\rho}(\cdot, 1) = \delta \quad \text{and} \quad \bar{u}(0, 0) = 0,
\end{cases} 
\quad (1.2) $$

In addition, the variational problem

$$ egin{gathered} I(\epsilon, \eta) = \inf \left\{ \int_{\mathbb{R}} \rho(x, 0) \frac{x^2}{2\eta} dx + \int_0^1 \int_{\mathbb{R}} \frac{1}{2} \left[ \rho^2 + \frac{\beta^2}{\rho} \right] dx dt : \right. \\ \left. \partial_t \rho + \epsilon \partial_x^2 \rho + \partial_x \beta = 0 \in \mathbb{R} \times (0, 1) \rho(\cdot, 1) = \delta \right\}, \end{gathered} 
\quad (1.3) $$

which, as we will see below, has $\rho_{\epsilon, \eta}$ and $\beta = \sqrt{\rho_{\epsilon, \eta}} \partial_x u_{\epsilon, \eta}$ as unique minimizer, converges, as $\epsilon, \eta \to 0$ with $\epsilon \le C\eta^\alpha$ for every $\alpha > 0$, to

$$ \bar{T} = \inf \left\{ \int_0^1 \int_{\mathbb{R}} \frac{1}{2} \left[ \frac{\beta^2}{\rho} + \rho^2 \right] dx dt : \partial_t \rho + \partial_x \beta = 0 \in \mathbb{R} \times (0, 1) \rho(\cdot, 0) = \delta \quad \rho(\cdot, 1) = \delta \right\}, \quad (1.4) $$

whose unique minimizer is, at least formally, $(\bar{\rho}, \sqrt{\bar{\rho}} \partial_x \bar{u})$.

<footnote>The second author was partially supported by the National Science Foundation grants DMS-1900599 and DMS-2153822, the Office for Naval Research grant N000141712095 and the Air Force Office for Scientific Research grant FA9550-18-1-0494.</footnote>
The system (1.1) represents an one-dimensional planning mean field game problem in which the player wants to move a Dirac mass at 1 to a Dirac mass at 0. The facts that these two measures are singular and the Dirac mass at $t = 0$ is created by the “singular” limit, $\eta \to 0$, in (1.1) make the passage to the limit difficult and delicate. In addition, the convergence of the $\rho_{\varepsilon, \eta}$ requires special attention, since the estimates on $\rho_{\varepsilon, \eta}$ and $u_{\varepsilon, \eta}$ only yield convergence in $L_{\text{loc}}^2(\mathbb{R} \times (0, 1))$ which is not enough to prove the convergence of the $I(\varepsilon, \eta)$'s. The global $L^2$-convergence is established directly, using the mean field structure of (1.1) and (1.2), by estimating directly the $\|\rho_{\varepsilon, \eta} - \bar{\rho}\|_{L^2(\mathbb{R} \times (0, 1))}$, which requires a detailed understanding of the limiting behavior of some new terms. Note that we are not able to prove directly that the $I(\varepsilon, \eta)$'s converge to $\bar{I}$ despite the strong convexity of the functionals involved. Indeed, the singularities due the Dirac masses seem to prevent such a direct approach.

We remark here that (1.1) cannot be solved if we start with $u_{\varepsilon, \eta}(\cdot, 0) = \delta$, which in turn will require $\rho_{\varepsilon, \eta}(\cdot, 0) = \delta$. Indeed, having the former, will require having a Brownian bridge with a square integrable drift, which is known not to exist. Hence, it is necessary to introduce the penalization $x^2/2\eta$.

The 1 + 1 KPZ equation

$$ \partial_t h - \kappa \partial_x^2 h + l(\partial_x h)^2 = \sqrt{D} \xi \quad \text{in } \mathbb{R} \times (0, T), \qquad (1.5) $$

where $\kappa$ is the diffusivity, $\xi = \xi(x, t)$ is a Gaussian spacetime white noise with $\langle \xi(x, t) \rangle = 0$ and $\langle \xi(x_1, t_1)\xi(x_2, t_2) \rangle = \delta(x_1 - x_2)\delta(t_1 - t_2)$ and $l > 0$ and $D > 0$ are material parameters, was introduced by Kardar, Parisi, and Zhang in [22] to model the dynamics of an interface arising at the contact between a stable bulk phase with a metastable one. Assuming that the two bulk phases have no conservation laws and relax exponentially fast, it was argued in [22] that the motion of the interface is governed by (1.1). The nonlinearity arises from the asymmetry between the two phases, the second derivative reflects the interface tension, and the space-time white noise $\xi$ models the randomness in transitions from metastable to stable.

In the last twenty five years there has been major progress in the mathematical theory of the KPZ equation and the properties of its solution. Listing all papers is beyond the scope of this paper. Instead, we refer to the works by Ferrari and Spohn [15], Quastel [34], Corwin [8], Quastel and Spohn [35], Chandra and Weber [5], Corwin and Shen [12], Hairer [19] and Gubinelli, Imkeller, and Perkowski [18] and the references therein for the mathematical study of and issues related to the KPZ equation.

A topic that has been the focus of several works recently is the understanding of the one-point probability distribution $P(H, t)$ of height $H$ at a specified point in space and time. Several papers have produced exact representations of $P(H, t)$ for arbitrary times. This progress has been achieved for three classes of initial conditions (and some of their combinations and variations), namely flat interface in Calabrese and Le Doussal [3], sharp or narrow wedge in Corwin [8], Sasamoto and Spohn [36], Calabrese, Le Doussal and Rosso [4], Dotsenko [14] and Amir, Corwin and Quastel [1], and stationary interface, that is a two-sided Brownian interface pinned at a point Imamura and Samasoto [20] and Borodin, Corwin, Ferrari and Veto [7].

Large deviations of the KPZ equation have been intensively studied in the mathematics and physics communities in recent years. The results obtained so far give a fairly complete picture in the long time regime $t \to \infty$. For the narrow wedge initial data, Corwin and Ghosal [9] and Corwin and Ghosal [10] derived rigorous bounds on the one-point tail probabilities for the narrow wedge initial data and general initial data. The lower-tail rate function was derived in the physics literature (see, for example, Sasorov, Meerson and Prolhac [37], Krajenbrink, Le Doussal and Prolhac [25], Corwin, Ghosal, Krajenbrink, Le Doussal, and Tsai [11], Krajenbrink, Le Doussal and Prolhac [25] and Le Doussal [26]) and was rigorously established by Tsai [38] and Cafasso and Chayes [2]. The entire
rate function for the upper tail was derived by Le Doussal, Majumdar and Schehr [27] for narrow wedge initial data, while Das and Tsai [13] and Ghosal and Lin [17] gave rigorous proofs.

The behavior of the short, that is, fixed, time large deviations of $P(H, t)$ was derived in the physics literature by Kolokolov and Korshunov [23, 24], Meerson, Katzav and Vilenkin [31] and Kamenev, Meerson and Sasor [21]. These references used the weak-noise theory (WNT), which grew from the Martin, Siggia and Rose path integral formalism in physics [30], and the Freidlin and Wentzel large deviation theory (LDP) in mathematics [16], to study the tails of the one-point probability distribution $P(H, t)$ of height $H$. Some of the physics arguments were discussed by Lin and Tsai [28].

Here, we use the convergence of the (1.1) to (1.2) to rigorously justify in this paper the full formal physics predictions for sharp wedge initial surface.

To describe the problem, we note that, after rescaling, the (1.5) takes the form

$$ \partial_t h - \partial_x^2 h + \frac{1}{2}(\partial_x h)^2 = \sqrt{\theta} \xi \quad \text{in } \mathbb{R} \times (0,1) \quad h(x, 0) = x^2/2\eta, \qquad (1.6) $$

and the problem becomes to understand the properties of the solution as $\theta \to 0$ and $h(1, 0) \to \infty$.

Since $\xi$ is square integrable random variable, we know that, for any $\rho \in L^2(\mathbb{R} \times (0, 1))$,

$$ \mathbb{P}(\sqrt{\theta}\xi) \approx \|\rho\|_{L^2(\mathbb{R}\times(0,1))}^2 = \int_0^1 \int_{\mathbb{R}} \rho^2 dxdt. $$

In the set $\{\omega : \sqrt{\theta}\xi\}, (1.6)\)$ is approximated by

$$ \partial_t h - \partial_x^2 h + \frac{1}{2}(\partial_x h)^2 = \rho \quad \text{in } \mathbb{R} \times (0,1) \quad h(x, 0) = x^2/2\eta. \qquad (1.7) $$

The theory of large deviations suggests that the $h(0, 1) \to \infty$ and $\eta \to 0$ behavior of the solution to (1.7) is governed by the $h(0, 1) \to \infty$ and $\eta \to 0$ behavior of the functional

$$ I(\lambda, \eta) = \sup \left\{ - \int_0^1 \int_{\mathbb{R}} \rho^2 dxdt : \right. \\ \left. \partial_t h - \partial_x^2 h + \frac{1}{2}(\partial_x h)^2 = \rho \quad \text{in } \mathbb{R} \times (0,1) \quad h(x, 0) = x^2/(2\eta) \quad h(0, 1) = \lambda \right\}, \qquad (1.8) $$

with minimizers satisfying the system

$$ \begin{gathered} \partial_t u - \partial_x^2 u + \frac{1}{2}(\partial_x u)^2 = \rho \quad \text{in } \mathbb{R} \times (0,1) \quad u(x, 0) = x^2/2\eta \quad u(0, 1) = \lambda \qquad (1.9) \\ \partial_t \rho + \partial_x^2 \rho + \partial_x(\rho\partial_x u) = 0 \quad \text{in } \mathbb{R} \times (0,1) \quad \rho(\cdot, 1) = c(\lambda)\delta, \end{gathered} $$

where the constant in front of $\delta$ comes from the constraint $u(0, 1) = \lambda$.

We remark that standard variational arguments show that the problems (1.8) and (1.3) are equivalent.

Upon the rescaling (we do not change the notation to avoid introducing more unknowns) $u(x, t) = \lambda u(x/\lambda^{1/2}, t)$ and $\rho(x, t) = \lambda\rho(x/\lambda^{1/2}, t)$, (1.9) becomes

$$ \begin{gathered} \partial_t u - \frac{1}{\lambda} \partial_x^2 u + \frac{1}{2} (\partial_x u)^2 = \rho \quad \text{in } \mathbb{R} \times (0,1) \quad u(x, 0) = x^2/2\eta \quad u(0, 1) = 1 \qquad (1.10) \\ \partial_t \rho + \frac{1}{\lambda} \partial_x^2 \rho + \partial_x (\rho \partial_x u) = 0 \quad \text{in } \mathbb{R} \times (0,1) \quad \rho(\cdot, 1) = \lambda^{-3/2} c(\lambda) \delta, \end{gathered} $$

which is like (1.1) with $\varepsilon = 1/\lambda$.

We show here that the asymptotics of (1.10) follow from the $\varepsilon \to 0$ and $\eta \to 0$ behavior of the solutions to (1.1).
The paper is organized as follows. In section 2 we discuss the planning problem (1.2). In section 3 we obtain uniform in $\varepsilon$ and $\eta$ estimates for $u_{\varepsilon,\eta}$ and $\rho_{\varepsilon,\eta}$ in $\mathbb{R} \times (0,1)$. In section 4 we analyze the behavior of $u_{\varepsilon,\eta}$ and $\rho_{\varepsilon,\eta}$ near $t=0$ and $t=1$. Then in section 5 we establish the convergence of $u_{\varepsilon,\eta}$ and $\rho_{\varepsilon,\eta}$ to the unique solution of the planning problem. Finally, in section 6 we discuss the large deviations of the KPZ equation.

## 2. THE FIRST-ORDER PLANNING PROBLEM

The first-order planning problem was introduced by the first author in [29] and was analyzed for smooth data by Munoz [32] and Porreta [33]. However, there are no results available, when the initial and terminal data are singular measures. In particular, nothing was known about the system (1.2) as well as its “regularized” version

$$ egin{cases} 
\partial_t \bar{u}_\eta + \frac{1}{2}(\partial_x \bar{u}_\eta)^2 = \bar{\rho}_\eta \quad \text{in } \mathbb{R} \times (0,1), \\
\partial_t \bar{\rho}_\eta + \partial_x(\partial_x \bar{u}_\eta \bar{\rho}_\eta) = 0 \quad \text{in } \mathbb{R} \times (0,1), \\
\bar{\rho}_\eta(\cdot, 1) = \delta \quad \text{and} \quad \bar{u}_\eta(x, 0) = x^2/2\eta.
\end{cases} 
\quad (2.1) $$

These systems, that is, (1.2) and (2.1), were discussed formally in [21, 31] which put forward exact solutions and the claim (without justification) that they control the LPD behavior of the height function of the KPZ equation.

The solutions proposed in [21, 31] for (1.2) and (2.1) are respectively

$$ \bar{\rho}(x,t) = r(t) \left(1 - \left(\frac{x}{l(t)}\right)^2\right)_+ \quad \text{and} \quad \bar{\rho}_\eta(x,t) = r_\eta(t) \left(1 - \left(\frac{x}{l_\eta(t)}\right)^2\right)_+, \quad (2.2) $$

and, in the domains $\mathcal{O} = \{(x,t) : |x| \le l(t)\}$ and $\mathcal{O}_\eta = \{(x,t) : |x| \le l_\eta(t)\}$,

$$ \bar{u}(x,t) = k(t) + \frac{1}{2}a(t)x^2 \quad \text{and} \quad \bar{u}_\eta(x,t) = k_\eta(t) + \frac{1}{2}a_\eta(t)x^2. \quad (2.3) $$

For $\bar{\rho}$ and $\bar{u}$ and $\bar{\rho}_\eta$ and $\bar{u}_\eta$ to be solutions of (1.2) and (2.1) respectively, $k$, $\rho$ and $a$ and $k_\eta$, $\rho_\eta$, and $a_\eta$ must satisfy respectively the odes

$$ \dot{k} = r, \quad \dot{r} = ra \quad \text{and} \quad \dot{a} = \begin{cases} (a^2 + \frac{32}{9}r^3) & \text{if } t \in (0, t_1), \\ -(a^2 + \frac{32}{9}r^3) & \text{if } t \in (t_1, 1), \end{cases} \quad (2.4) $$

and

$$ \dot{k}_\eta = r_\eta, \quad \dot{r}_\eta = r_\eta a_\eta \quad \text{and} \quad \dot{a}_\eta = \begin{cases} (a_\eta^2 + \frac{32}{9}r_\eta^3) & \text{if } t \in (0, t_1), \\ -(a_\eta^2 + \frac{32}{9}r_\eta^3) & \text{if } t \in (t_1, 1), \end{cases} \quad (2.5) $$

where $t_1$ and $t_{\eta,1}$ are the times at which $r$ and $r_\eta$ attain their minimum values $r_1$ and $t_{\eta,1}$.

To have that $\rho$ and $\rho_\eta$ are probability densities we must have, for all $t \in (0,1)$,

$$ r(t)l(t) = 3/4 \quad \text{and} \quad r_\eta(t)l_\eta(t) = 3/4. \quad (2.6) $$

Finally, to satisfy the terminal and initial conditions, that is, to have $\rho(\cdot, 1) = \rho_\eta(\cdot, 1) = \delta$, $\rho(\cdot, 0) = \delta$ and $u_\eta(x, 0) = x^2/2\eta$, we must have

$$ \lim_{t \to 0} r(t) = \lim_{t \to 1} r(t) = \lim_{t \to 1} r_\eta(t) = \infty, \quad \rho_\eta(0) = r_{\eta,0} \quad \text{and} \quad a_\eta(0) = 1/\eta. \quad (2.7) $$
The relationships among the unknown parameters $t_{\eta,1}$, $r_{\eta,1}$ and $r_{\eta,0}$ are

$$ \frac{8}{3} r_{\eta,0} \sqrt{t_{\eta} - r_{\eta,1}} = 1/\eta, \quad t_{\eta,1} + \frac{3}{16} \frac{\pi}{t_{\eta,1}^{2/3}} = 1 \text{ and } t_{\eta,1} = \frac{8}{3} \left( \frac{\sqrt{t_{\eta,0} - r\eta,1}}{r_{\eta,0}r_{\eta,1}} + \frac{\arctan \sqrt{\frac{r_{\eta,0}}{r_{\eta,1}} - 1}}{r_{\eta,1}} \right), $$

while

$$ t_1 = 1/2 \quad \text{and} \quad r_1 = (3\pi/16)^{2/3}. $$

We also remark, for future use, that, as $\eta \to 0$,

$$ \bar{\rho}_{\eta} \to \bar{\rho} \quad \text{in } L^1(\mathbb{R} \times (0,1)) \quad \text{and} \quad \bar{u}_{\eta} \to \bar{u} \quad \text{locally uniformly in } \mathbb{R} \times (0,1). \quad (2.8) $$

Tedious but straightforward calculations also yield that, as $t \to 0$,

$$ r(t) \approx \frac{1}{(4t)^{2/3}}, \quad a(t) \approx \frac{2}{3t} \quad \text{and} \quad r_{\eta} \approx \frac{1}{(4(t+\eta))^{2/3}}, \quad a_{\eta}(t) \approx \frac{2}{3(t+\eta)}, \quad (2.9) $$

and, as $t \to 1$,

$$ r(t) \approx \frac{1}{(4(1-t))^{2/3}}, \quad a(t) \approx -\frac{2}{3} \frac{1}{1-t} \quad \text{and} \quad r_{\eta} \approx \frac{1}{(4(1-t))^{2/3}}, \quad a(t)_{\eta} \approx -\frac{2}{3(1-t)}. \quad (2.10) $$

We remark, if $\rho \equiv 0$, then $a(t) = \frac{1}{2t}$. The difference in the coefficient in front of $1/t$ is a result of the presence of $\bar{\rho}$ and the singularity of $\bar{\rho}$ near $t = 0$.

It is immediate that $\bar{\rho}$ and $\bar{u}$ and $\bar{\rho}_{\eta}$ and $\bar{u}_{\eta}$ are solutions to (1.2) in $\mathcal{O}$ and (2.1) in $\mathcal{O}_{\eta}$ respectively, where the first equation is understood in the viscosity sense and the second in the sense of distributions.

To get a solution of the Hamilton-Jacobi equation in (1.2) (resp. (2.1)) in $\mathbb{R} \times (0,1)$, $\bar{u}$ (resp. $\bar{u}_{\eta}$) is extended to be a solution of

$$ \partial_t \bar{u} + \frac{1}{2}(\partial_x \bar{u})^2 = 0 \quad \text{in} \quad \mathbb{R} \times (0,1) \setminus \bar{\mathcal{O}} \quad (\text{resp.} \quad \partial_t \bar{u} + \frac{1}{2}(\partial_x \bar{u})^2 = 0 \quad \text{in} \quad \mathbb{R} \times (0,1) \setminus \bar{\mathcal{O}_{\eta}}) $$

using the method of characteristics by connecting any point $(x,t) \in \mathbb{R} \times (0,1) \setminus \bar{\mathcal{O}}$ (resp. $(x,t) \in \mathbb{R} \times (0,1) \setminus \bar{\mathcal{O}_{\eta}}$) to $\partial\mathcal{O}$ (resp. $\partial\mathcal{O}_{\eta}$) by a straight line tangent to $\partial\mathcal{O}$ (resp. $\partial\mathcal{O}_{\eta}$) at $(l(t_0), t_0)$ (resp. $(l_{\eta}(t_{\eta,0}, t_{\eta,0}))$ and then moving along $\partial\mathcal{O}$ to the origin, that is, for any $(x,t) \in \mathbb{R} \times (0,1) \setminus \bar{\mathcal{O}}$ (resp. $(x,t) \in \mathbb{R} \times (0,1) \setminus \bar{\mathcal{O}_{\eta}}$), there exists $t_0 \in (0,t)$ (resp. $t_{\eta,0} \in (0,t)$) such that $x = l(t_0) + (t - t_0)\dot{l}(t_0)$ (resp. $x = l_{\eta}(t_{\eta,0}) + (t - t_{\eta,0})\dot{l}_{\eta}(t_{\eta,0})$) and

$$ \bar{u}(x,t) = \bar{u}(l(t_0), t_0) + \frac{1}{2}\dot{l}(t_0)^2(t - t_0) = \bar{u}(l(t_0), t_0) + \frac{(x - l(t_0))^2}{2(t - t_0)}, \quad (2.11) $$

(resp.

$$ \bar{u}_{\eta}(x,t) = \bar{u}_{\eta}(l(t_{\eta,0}), t_{\eta,0}) + \frac{1}{2}\dot{l}_{\eta}(t_{\eta,0})^2(t - t_{\eta,0}) = \bar{u}_{\eta}(l_{\eta}(t_{\eta,0}), t_{\eta,0}) + \frac{(x - l_{\eta}(t_{\eta,0}))^2}{2(t - t_{\eta,0}t_0)}. \quad (2.12) $$

As we show below, the planning problem (1.2) and the system (2.1) are associated respectively with the strictly convex variational problems

$$ \bar{I} = \min \left\{ \int_{0}^{1} \int_{\mathbb{R}} \frac{1}{2} \left[ \rho^2 + \frac{\beta^2}{\rho} \right] dx dt : \rho, \beta \in \mathcal{A} \right\} \quad (2.13) $$
where

$$
\begin{equation} \tag{2.14}
\begin{aligned}
\mathcal{A} &= \left\{ \rho \in C([0, 1]; \mathcal{P}_2) \cap L^2(\mathbb{R} \times (0, 1)) : \partial_t \rho + \partial_x \beta = 0 \; \rho(\cdot, 0) = \rho(\cdot, 1) = \delta \right. \\
&\qquad \left. \beta \in L^{4/3}(\mathbb{R} \times (0, 1)) \; |\beta| \le A \rho^{1/2} \text{ for some } A \in L^2(\mathbb{R} \times (0, 1)), \frac{\beta^2}{\rho} = 0 \text{ a.s. in } \{\rho = 0\} \right\},
\end{aligned}
\end{equation}
$$

and

$$
\bar{T}_{\eta} = \min \left\{ \int_{0}^{1} \int_{\mathbb{R}} \frac{1}{2} \left[ \rho^2 + \frac{\beta^2}{\rho} \right] dxdt + \int_{\mathbb{R}} \rho(x, 0) \frac{x^2}{2\eta} dx : \rho, \beta \in \mathcal{A}_{\eta} \right\}, \quad (2.15)
$$

where

$$
\begin{equation} \tag{2.16}
\begin{aligned}
\mathcal{A}_{\eta} &= \left\{ \rho \in C([0, 1]; \mathcal{P}_2) \cap L^2(\mathbb{R} \times (0, 1)) : \partial_t \rho + \partial_x \beta = 0 \text{ in } \mathbb{R} \times (0, 1) \; \rho(\cdot, 1) = \delta \right. \\
&\qquad \left. \beta \in L^{4/3}(\mathbb{R} \times (0, 1)) \; |\beta| \le A \rho^{1/2} \text{ for some } A \in L^2(\mathbb{R} \times (0, 1)), \frac{\beta^2}{\rho} = 0 \text{ a.s. in } \{\rho = 0\} \right\},
\end{aligned}
\end{equation}
$$

which have a unique minimizers.

We remark that in both $\mathcal{A}$ and $\mathcal{A}_\eta$, $\beta \in L^{4/3}(\mathbb{R} \times (0, 1))$ since $\beta = \frac{\beta}{\sqrt{\rho}}\sqrt{\rho}$ and $\frac{\beta}{\sqrt{\rho}} \in L^2(\mathbb{R} \times (0, 1))$
and $\sqrt{\rho} \in L^4(\mathbb{R} \times (0, 1))$.

We also note that the constraints $|\beta| \le \sqrt{\rho}$ and $\frac{\beta^2}{\rho} = 0$ a.s. in $\{\rho = 0\}$ are both convex.

Indeed, if $A_1, A_2 \in L^2(\mathbb{R} \times (0, 1))$, then $\sqrt{\rho_1}A_1 + \sqrt{\rho_2}A_2 \le 2\sqrt{\rho_1 + \rho_2}\max(A_1, A_2)$, and $\max(A_1, A_2) \in L^2(\mathbb{R} \times (0, 1))$.

Also, since $\{\frac{\rho_1+\rho_2}{2} = 0\} \subset \{\rho_1 = 0\} \cup \{\rho_2 = 0\}$ and $\frac{\beta_1^2}{\rho_1+\rho_2} \le \frac{\beta_1^2}{\rho_1}$ and $\frac{\beta_2^2}{\rho_1+\rho_2} \le \frac{\beta_2^2}{\rho_2}$, it follows that
$\frac{\beta_1^2}{\rho_1+\rho_2} = 0$ in $\{\rho_1 = 0\}$ and $\frac{\beta_2^2}{\rho_1+\rho_2} = 0$ in $\{\rho_2 = 0\}$.

**Lemma 2.1.** The solutions $\bar{p}_\eta$, $\bar{u}_\eta$ and $\bar{p}$, $\bar{u}$ of (2.1) and (1.2) are respectively the unique minimizers of $\bar{T}_\eta$ and $\bar{T}$.

*Proof.* The uniqueness is an immediate consequence of the convexity of the functionals.

To show the minimizing property of $\bar{p}_\eta$, $\bar{u}_\eta$, we fix $\rho, \beta \in \mathcal{A}_\eta$ and observe that, in view of the convexity of the map $(\rho, \beta) \to \frac{1}{2}(\rho^2 + \frac{\beta^2}{\rho})$, we have, for $\bar{\beta} = \partial_x \bar{u} / \bar{p}$,

$$
\begin{align*}
& \int_0^1 \int_{\mathbb{R}} \frac{1}{2} \left[ \rho^2 + \frac{\beta^2}{\rho} \right] dxdt + \int_{\mathbb{R}} \rho(x, 0) \frac{x^2}{2\eta} \ge \int_0^1 \int_{\mathbb{R}} \frac{1}{2} \left[ \bar{\rho}^2 + \frac{\bar{\beta}^2}{\bar{\rho}} \right] dxdt + \int_{\mathbb{R}} \bar{\rho}(x, 0) \frac{x^2}{2\eta} dx \\
& \qquad + \int_0^1 \int_{\mathbb{R}} \left[ (\rho - \bar{\rho})\bar{\rho} + (\beta - \bar{\beta})\frac{\bar{\beta}}{\bar{\rho}} - \frac{1}{2}\frac{\bar{\beta}^2}{\bar{\rho}^2}(\rho - \bar{\rho}) \right] dxdt + \int_{\mathbb{R}} (\rho - \bar{\rho})(x, 0) \frac{x^2}{2\eta} dx \\
&= \int_0^1 \int_{\mathbb{R}} \left[ (\rho - \bar{\rho})\bar{\rho} + \left(\beta - \frac{\partial_x \bar{u}}{\bar{\rho}}\right)\partial_x \bar{u} - \frac{1}{2}(\rho - \bar{\rho})(\partial_x \bar{u})^2 \right] dxdt + \int_{\mathbb{R}} (\rho - \bar{\rho})(x, 0) \frac{x^2}{2\eta} dx \\
&= \int_0^1 \int_{\mathbb{R}} \left[ \partial_t \bar{u}(\rho - \bar{\rho}) + \partial_t (\rho - \bar{\rho}) \right] dxdt + \int_{\mathbb{R}} (\rho - \bar{\rho})(x, 0) \frac{x^2}{2\eta} dx \\
&= \int_0^1 \int_{\mathbb{R}} \partial_t \left[ \bar{u}(\rho - \bar{\rho}) \right] dxdt + \int_{\mathbb{R}} (\rho - \bar{\rho})(x, 0) \frac{x^2}{2\eta} dx = 0
\end{align*}
$$

The claim about the minimizing property of $\bar{p}$, $\bar{u}$ follows as the one for $\bar{p}_\eta$, $\bar{u}_\eta$ the only difference
being that now $\rho, \beta \in \mathcal{A}$ and instead of integrating in time from 0 to 1, we now integrate over $(h, 1)$
and then need to show that

$$
\lim_{h \to 0} \int_{\mathbb{R}} \bar{u}(x, r)(\rho - \bar{\rho})(x, h)dx = 0, \quad (2.17)
$$
To conclude the proof of (2.17) we first remark that, in view of the definition of $\bar{u}$, we have

$$ \lim_{h \to 0} \int_R \bar{u}(x, h) \rho(x, h) dx = \lim_{h \to 0} \int_R \rho(x, h) x^2 dx = 0, $$

the last equality being a consequence of the fact that $\rho(\cdot, 0) = \delta$.

Then, multiplying the equation satisfied by $\rho$ by $x^2$ and integrating over $\mathbb{R}$ gives

$$ \frac{d}{dt} \int_{\mathbb{R}} \rho(x, t) x^2 dx \le 2 \left( \int_{\mathbb{R}} \rho(x, t) x^2 dx \right)^{1/2} \left( \int_{\mathbb{R}} \frac{\beta^2}{\rho}(x, t) dx \right)^{1/2} $$

and, thus,

$$ \frac{d}{dt} \left( \int_{\mathbb{R}} \rho(x, t) x^2 dx \right)^{1/2} \le 2 \left( \int_{\mathbb{R}} \frac{\beta^2}{\rho}(x, t) dx \right)^{1/2}, $$

and, since $\beta/\sqrt{\rho} \in L^2(\mathbb{R} \times (0, 1))$,

$$ \int_{\mathbb{R}} \rho(x, t) x^2 dx \le Ct. $$

□

Finally, we mention that (2.13) and (3.5) can be rewritten as

$$ \bar{I} = \min \left\{ \int_{0}^{1} \int_{\mathbb{R}} \frac{1}{2} [\rho^2 + \alpha^2 \rho] dxdt : \rho, \alpha \in \mathcal{B} \right\} \quad (2.18) $$

with

$$ \mathcal{B} = \left\{ \begin{array}{l} \rho \in C([0, 1]; \mathcal{P}_2) \cap L^2(\mathbb{R} \times (0, 1)) : \partial_t \rho + \partial_x(\alpha\rho) = 0 \text{ in } \mathbb{R} \times (0, 1), \\ \rho(\cdot, 0) = \rho(\cdot, 1) = \delta, \quad \sqrt{\rho}\alpha \in L^2(\mathbb{R} \times (0, 1)) \end{array} \right\}. $$

and

$$ \bar{I} = \min \left\{ \int_{0}^{1} \int_{\mathbb{R}} \frac{1}{2} [\rho^2 + \alpha^2 \rho] dxdt + \int_{\mathbb{R}} \rho(x, 0) \frac{x^2}{2\eta} dx : \rho, \alpha \in \mathcal{B}_{\eta} \right\}, \quad (2.19) $$

with

$$ \mathcal{B}_{\eta} = \left\{ \begin{array}{l} \rho \in C([0, 1]; \mathcal{P}_2) \cap L^2(\mathbb{R} \times (0, 1)) : \\ \qquad \sqrt{\rho}\alpha \in L^2(\mathbb{R} \times (0, 1)), \partial_t \rho + \partial_x(\rho\alpha) = 0 \text{ in } \mathbb{R} \times (0, 1) \quad \rho(\cdot, 1) = \delta \end{array} \right\}. $$

In the next two propositions, we record the properties of $\bar{u}$ and $\bar{p}$ and $\bar{u}_\eta$ and $\bar{p}_\eta$ that we will use later in the paper, and we begin with the former.

**Proposition 2.2.** (i) For any $\theta \in (0, 1)$ and $(x, t) \in \mathbb{R} \times (\theta, 1)$,

$$ \bar{u}(x, t) = \inf \left\{ \int_{t-\theta}^{t} \left[ \frac{(\dot{x}(s))^2}{2} + \bar{p}(x(s), s) \right] ds + \bar{u}(x(\theta), \theta) : x \in H^1((0, 1)) \text{ and } x(t) = x \right\}. \quad (2.20) $$

(ii) For any $(x, t) \in \mathbb{R} \times (0, 1)$,

$$ \bar{u}(x, t) = \inf \left\{ \int_{0}^{t} \left[ \frac{(\dot{x}(s))^2}{2} + \bar{p}(x(s), s) \right] ds : x \in H^1((0, 1)), x(t) = x \text{ and } x(0) = 0 \right\}. \quad (2.21) $$

(iii) For any $(x, t) \in \mathbb{R} \times (0, 1)$,

$$ \bar{u}(x, t) \ge \frac{x^2}{2t}. \quad (2.22) $$

(iv) For any $x \in \mathbb{R}$ and uniformly,

$$ \limsup_{t \to 0} \left[ \bar{u}(x, t) - \frac{x^2}{2t} \right] \le 0. \quad (2.23) $$
**Proof.** It is clear from its construction that $\bar{u}$ satisfies the first equation of (1.2) everywhere in $\mathbb{R} \times (0, 1]$. The dynamic programming principle (2.20) is then a consequence of the fact that $\bar{u}$ is a viscosity solution of the first equation of (1.2). We leave the details for this classical claim to the reader.

Note that $\partial_t \bar{u} + \frac{1}{2}(\partial_x \bar{u})^2 = \bar{\rho} \ge 0$. It follows that $\bar{u} \ge 0$ in $\mathbb{R} \times (0, 1)$ since, by construction, $\bar{u}(\cdot, t) \ge 0$ for $t \in (0, 1]$.

Next observe that, if $x \in \mathbb{R} \setminus \{0\}$ and $t$ is sufficiently small, $(x, t) \notin \bar{\mathcal{O}}$ and $\bar{u}$ is given by (2.2). Then, as $t \to 0$, it is immediate that $t_0 \to 0$ and $l(t_0) \to 0$, and, hence,

$$ \lim_{t \to 0} \bar{u}(x, t) = \infty, \qquad (2.24) $$

since

$$ \bar{u}(x, t) = \bar{u}(l(t_0), t_0) + \frac{(x - l(t_0))^2}{2(t - t_0)} \ge \frac{(x - l(t_0))^2}{2(t - t_0)} \quad \text{and} \quad \lim_{t \to 0} \frac{(x - l(t_0))^2}{2(t - t_0)} \ge \lim_{t \to 0} \frac{(x - l(t_0))^2}{2t} = \infty. $$

Using this information in (2.20), we find that, if $x_\theta$ is an approximate minimizer, then $\lim_{\theta \to 0} x_\theta(\theta) = 0$, which implies that $\bar{u}$ must actually satisfy (2.21).

The lower bound in (2.22) follows now from Jensen's inequality.

Going back to (2.21), using, for each $(x, t) \in \mathbb{R} \times (0, 1)$, the path $x(s) = \frac{x}{t}s$, denoting by $\hat{t}$ the time when the path intersects $\partial \mathcal{O}$, which may be 0, and observing that $\hat{t} \to 0$ as $t \to 0$, we find

$$ \bar{u}(x, t) \le \frac{x^2}{2t} + \int_0^{\hat{t}} \bar{\rho}(\frac{x}{t}s, s)ds, \qquad (2.25) $$

and (2.23) follows from the observation that

$$ \lim_{t \to 0} \int_0^{\hat{t}} \bar{\rho}(\frac{x}{t}s, s)ds = 0. \qquad (2.26) $$

□

The result about the properties of $\bar{u}_\eta$ and $\bar{\rho}_\eta$ is stated next. Since it is proof is almost identical with one of the previous proposition, we omit it.

**Proposition 2.3.** (i) For any $\theta > 0$ and $(x, t) \in \mathbb{R} \times (\theta, 1)$,

$$ \bar{u}_{\eta}(x, t) = \inf \left\{ \int_{t-\theta}^{t} \left[ \frac{(\dot{x}(s))^2}{2} + \bar{\rho}(x(s), s) \right] ds + \bar{u}_{\eta}(x(\theta), \theta) : x \in H^1((0, 1)), \ x(t) = x \right\}. \quad (2.27) $$

(ii) For any $(x, t) \in \mathbb{R} \times (0, 1)$,

$$ \bar{u}_{\eta}(x, t) = \inf \left\{ \int_{0}^{t} \left[ \frac{(\dot{x}(s))^2}{2} + \bar{\rho}(x(s), s) \right] ds + \frac{x(0)^2}{2\eta} : x \in H^1((0, 1)), \ x(t) = x, \ x(0) = 0 \right\}. \quad (2.28) $$

(iii) For any $(x, t) \in \mathbb{R} \times (0, 1)$,

$$ \bar{u}_{\eta}(x, t) \ge \frac{x^2}{2(t + \eta)}. \qquad (2.29) $$

(iv) Locally uniformly in $x \in \mathbb{R}$,

$$ \lim_{t \to 0} \left[ \bar{u}_{\eta}(x, t) - \frac{x^2}{2(t + \eta)} \right] = 0. \qquad (2.30) $$

The next result is the uniqueness of solutions to (1.2).

For this we need to specify the regularity of the solutions we are considering, which are consistent with the expect regularity of the limit of $u_{\varepsilon, \eta}$ and $\rho_{\varepsilon, \eta}$.
**Definition.** A pair $(u, \rho) \in C(\mathbb{R} \times (0, 1]) \times (C([0, 1]; \mathcal{P}_2) \cap L^1(\mathbb{R} \times (0, 1)))$ is a solution to (1.2), if $u$ is a viscosity solution of the Hamilton-Jacobi equation and $\rho$ is a distributional solution of the continuity equation and, in addition,

$$ (i) \quad u(x,t) \geq x^2/2t, \quad (ii) \quad \int_{\mathbb{R}} u(x,t) \bar{p}(x,t) dx \leq \int_{\mathbb{R}} \bar{p}(x,t) \frac{x^2}{2t} dx + o(1), \\ (iii) \quad u \in C^{0,\alpha}(\mathbb{R} \times (\theta, 1]) \text{ for all } \theta \in (0,1) \quad \text{and} \quad (iv) \quad \int_0^1 \int_{\mathbb{R}} \rho(\partial_x u)^2 dx dt < \infty. \tag{2.31} $$

Before we state and prove the uniqueness result for the planning problem, we state and prove the following auxiliary fact.

**Proposition 2.4.** If $\partial_t \rho + \partial_x(\rho\alpha) = 0$ in $\mathbb{R} \times (0, 1)$, $\rho(\cdot, 0) = \delta$ and $\int_0^1 \int_{\mathbb{R}} \rho\alpha^2 dx ds < \infty$, then

$$ \lim_{t \to 0} \int_{\mathbb{R}} \rho(x, t) \frac{x^2}{t} dx = 0. $$

*Proof.* Multiplying the continuity equation satisfied by $\rho$ by $x^2$ and integrating over $\mathbb{R}$ gives

$$ \begin{aligned} \frac{d}{dt} \int_{\mathbb{R}} x^2 \rho dx &= 2 \int_{\mathbb{R}} x \rho \partial_x u dx \\ &\leq 2 \left( \int_{\mathbb{R}} x^2 \rho dx \right)^{1/2} \left( \int_{\mathbb{R}} \rho (\partial_x u)^2 dx \right)^{1/2}, \end{aligned} $$

and thus

$$ \frac{d}{dt} \left( \int_{\mathbb{R}} x^2 \rho dx \right)^{1/2} \leq \left( \int_{\mathbb{R}} \rho (\partial_x u)^2 dx \right)^{1/2}. $$

It then follows, using $\int x^2 \rho(x, 0) dx = 0$, that

$$ \left( \int_{\mathbb{R}} x^2 \rho(x, t) dx \right)^{1/2} \leq t^{1/2} \left( \int_0^t \int_{\mathbb{R}} \rho(\partial_x u)^2 dx \right)^{1/2} $$

Squaring the last inequality we get

$$ \int_{\mathbb{R}} \frac{x^2}{t} \rho(x, t) dx \leq \int_{0}^{t} \int_{\mathbb{R}} \rho(\partial_x u)^2 dx $$

and, since in view of the $L^2(\mathbb{R} \times (0, 1))$- integrability of $\sqrt{\rho}\alpha$, the claim follows.

□

**Theorem 2.5.** Let $(\rho, u)$ be a solution of (1.2) satisfying (2.31). Then $\rho = \bar{p}$ and $u = \bar{u}$.

*Proof.* As it is common when studying systems with mean field games structure, the uniqueness is based on a classical computation which gives, for any $\theta \in (0, 1/2)$, the identity

$$ \begin{aligned} & \int_{\mathbb{R}} (\rho - \bar{p})(x, 1 - \theta)(u - \bar{u})(x, 1 - \theta) dx - \int_{\mathbb{R}} (\rho - \bar{p})(x, \theta)(u - \bar{u})(x, \theta) dx \\ & = \int_{\theta}^{1-\theta} \int_{\mathbb{R}} \left[ (\rho - \bar{p})^2 + \frac{1}{2}(\rho + \bar{p})(\partial_x u - \partial_x \bar{u})^2 \right] dx dt. \end{aligned} \tag{2.32} $$

Assuming we can show that

$$ \lim_{\theta \to 0} \int_{\mathbb{R}} (\rho - \bar{p})(x, 1 - \theta)(u - \bar{u})(x, 1 - \theta) dx = \lim_{\theta \to 0} \int_{\mathbb{R}} (\rho - \bar{p})(x, \theta)(u - \bar{u})(x, \theta) dx = 0, \tag{2.33} $$
it follows that

$$ \int_0^1 \int_{\mathbb{R}} [(\rho - \bar{\rho})^2 + \frac{1}{2}(\rho + \bar{\rho})(\partial_x u - \partial_x \bar{u})^2] dxdt = \lim_{\theta \to 0} \int_\theta^{1-\theta} \int_{\mathbb{R}} [(\rho - \bar{\rho})^2 + \frac{1}{2}(\rho + \bar{\rho})(\partial_x u - \partial_x \bar{u})^2] dxdt = 0, $$
which in turn implies that

$\rho = \bar{\rho} \text{ almost everywhere in } \mathbb{R} \times (0, 1) \text{ and } \partial_x u = \partial_x \bar{u} \text{ almost everywhere in } \mathcal{O}.$

It then follows from the fact that $u$ and $\bar{u}$ solve the same equation in $\mathcal{O}$ that we also have that

$\partial_t u = \partial_t \bar{u} \text{ almost everywhere in } \mathcal{O}$

and, hence, in view of the continuity of $u$ and $\bar{u}$, that, for some constant $c \in \mathbb{R}$,

$$ u - \bar{u} = c \quad \text{on } \bar{\mathcal{O}}. \tag{2.34} $$

It follows from (2.31) and Proposition 2.4 that

$$ \lim_{t \to 0} \int_{\mathbb{R}} \bar{\rho}(x, t) u(x, t) dx = \lim_{t \to 0} \int_{\mathbb{R}} \bar{\rho}(x, t) \bar{u}(x, t) dx = 0, $$

and, thus,

$$ \lim_{t \to 0} \int_{\mathbb{R}} \bar{\rho}(x, t) [u(x, t) - \bar{u}(x, t)] dx = 0. \tag{2.35} $$

Since $\bar{\rho}$ is supported in $\bar{\mathcal{O}}$ and $\int_{\mathbb{R}} \bar{\rho}(x, t) dx = 1$, it follows from (2.35) and (2.36) that $c = 0$, that is

$$ u = \bar{u} \quad \text{on } \bar{\mathcal{O}}. \tag{2.36} $$

To conclude we need to show that $u = \bar{u}$ on $\mathbb{R} \times (0, 1]$. This follows from the fact that both $u$ and $\bar{u}$ are viscosity solutions of the Hamilton-Jacobi equation of the system, and, hence, satisfy the same variational formula (see Proposition 2.2).

We know return to the $\theta \to 0$ limits in (2.33) and begin with the behavior near $t = 0$.

In view of the signs of $\int_{\mathbb{R}} \rho(x, \theta) u(x, \theta) dx$ and $\int_{\mathbb{R}} \bar{\rho}(x, \theta) \bar{u}(x, \theta) dx$, we only need to show that

$$ \lim_{\theta \to 0} \int_{\mathbb{R}} \rho(x, \theta) \bar{u}(x, \theta) dx = \lim_{\theta \to 0} \int_{\mathbb{R}} \bar{\rho}(x, \theta) u(x, \theta) dx = 0. $$

The first limit follows from (2.25), (2.26) and (2.31) (iii) while the second follows from (2.31) (ii).

We turn now to the behavior near $t = 1$, and we show that

$$ \lim_{\theta \to 0} \int_{\mathbb{R}} [(\rho - \bar{\rho})(x, 1 - \theta)\bar{u}(x, 1 - \theta) dx = 0. \tag{2.37} $$

and

$$ \lim_{\theta \to 0} \int_{\mathbb{R}} [(\rho - \bar{\rho})(x, 1 - \theta)u(x, 1 - \theta)dx = 0. \tag{2.38} $$

In view of the explicit forms of $\bar{\rho}$ and $\bar{u}$ which give immediately that

$$ \lim_{\theta \to 0} \int_{\mathbb{R}} \bar{\rho}(x, 1 - \theta) \bar{u}(x, 1 - \theta) dx = \bar{u}(0, 1), $$

to prove (2.37) we need to show that

$$ \lim_{\theta \to 0} \int_{\mathbb{R}} \rho(x, 1 - \theta) \bar{u}(x, 1 - \theta) dx = \bar{u}(0, 1). $$

The continuity of $\bar{u}$ gives that, for every $\kappa > 0$, there exists $\theta_0 > 0$ and $R_0 > 0$ such that

$$ \sup_{|x| \le R, \theta \le \theta_0} |\bar{u}(x, 1 - \theta) - \bar{u}(0, 1)| \le \kappa, $$
so to conclude we only need to show that, for every $R > 0$,

$$ \lim_{t \to 0} \int_{|x| \ge R} \rho(x, t) dx = 0. \qquad (2.39) $$

For this, after multiplying the continuity equation by $(|x| - R)_+^2$ and integrating by parts, we find

$$ \begin{aligned} \frac{d}{dt} \int_{\mathbb{R}} \rho(x,t)(|x| - R)_+^2 dx &= -2 \int_{|x| \ge R} (|x| - R)_+ \rho(x,t) \partial_x u(x,t) dx \\ &\ge -C \left( \int_R \rho(x,t)(|x| - R)_+^2 dx \right)^{1/2} \left( \int_{\mathbb{R}} \rho(x,t)(\partial_x u)^2(x,t) dx \right)^{1/2}, \end{aligned} $$

and, hence,

$$ \frac{d}{dt} \left( \int_{\mathbb{R}} \rho(x,t)(|x| - R)_+^2 dx \right)^{1/2} \ge -C \left( \int_{\mathbb{R}} \rho(x,t)(\partial_x u)^2(x,t) dx \right)^{1/2}. $$

Integrating this last inequality from $1 - \theta$ to 1 and using that, in view of the fact that $\rho(\cdot, 1) = \delta$,

$$ \int_{\mathbb{R}} \rho(x, 1)(|x| - R)_+^2 dx = 0, \text{ we find} $$

$$ \int_{\mathbb{R}} \rho(x, 1 - \theta)[(|x| - R)_+]^2 dx \le \theta \int_0^1 \int_{\mathbb{R}} \rho(x, t)(\partial_x u)^2(x, t) dx, $$

and thus, using that $\int_0^1 \int_{\mathbb{R}} \rho(x,t)(\partial_x u)^2(x,t)dx < \infty$, we get (2.39).

We turn now to the proof of (2.38) and show that

$$ \lim_{\eta \to 0} \int_{\mathbb{R}} \rho(x, 1 - \eta) u(x, 1 - \eta) dx = u(0, 1), \qquad (2.40) $$

and

$$ \lim_{\eta \to 0} \int_{\mathbb{R}} \bar{\rho}(x, 1 - \eta) u(x, 1 - \eta) dx = u(0, 1). \qquad (2.41) $$

For (2.40), we use again the identity

$$ \frac{d}{dt} \int_{\mathbb{R}} \rho(x,t) u(x,t) dx = \int_{\mathbb{R}} \left[ \rho^2(x,t) + \frac{1}{2} \rho(x,t) (\partial_x u)^2(x,t) \right] dx, $$

and the continuity of $u$ near $t = 1$ to get

$$ \begin{aligned} u(0, 1) - \int_{\mathbb{R}} \rho(x, 1 - \theta) u(x, 1 - \theta) dx &= \int_{\mathbb{R}} \rho(x, 1) u(x, 1) dx - \int_{\mathbb{R}} \rho(x, 1 - \theta) u(x, 1 - \theta) dx \\ &= \int_{1-\theta}^{1} \int_{\mathbb{R}} \left[ \rho^2(x,t) + \frac{1}{2} \rho(x,t) (\partial_x u)^2(x,t) \right] dx dt, \end{aligned} $$

and we conclude from the fact that

$$ \lim_{\theta \to 0} \int_{1-\theta}^{1} \int_{\mathbb{R}} \left[ \rho^2(x,t) + \frac{1}{2} \rho(x,t) (\partial_x u)^2(x,t) \right] dx dt = 0. $$

For the proof of (2.41), using the form of $\bar{\rho}$, we find

$$ \begin{aligned} \int_R \bar{\rho}(x,t) u(x,t) dx - u(0,1) &= \left[ \int_R \bar{\rho}(x,t) [u(x,t) - u(0,1)] dx \right. \\ &= \left. \int_R (1 - |x|^2)_+ [u(l(t)x,t) - u(0,1)] dx \right]. \end{aligned} $$
It then follows from the continuity of $u$ that

$$ \lim_{t \to 1} \int_{\mathbb{R}} (1 - |x|^2)_+ (u(l(t)x, t) - u(0, 1)) dx = 0. \quad \square $$

### 3. UNIFORM ESTIMATES FOR THE SOLUTION TO THE APPROXIMATE PLANNING PROBLEM

We work with solutions to (1.1) and prove here a number of estimates for the solution $(u_{\varepsilon,\eta}, \rho_{\varepsilon,\eta})$ of (1.1), which are independent of $\eta$ and $\varepsilon$. These bounds yield the necessary compactness to pass in the $\varepsilon, \eta \to 0$ limit to show convergence to solutions of (1.2) in $\mathbb{R} \times (0, 1)$.

For definiteness, we recall the basic existence and uniqueness result of solutions of the system (1.1).

**Proposition 3.1.** For each $\varepsilon$ and $\eta$, (1.1) has a unique solution $(u_{\varepsilon,\eta}, \rho_{\varepsilon,\eta}) \in C(\mathbb{R} \times [0, 1]) \times C([0, 1]; \mathcal{P})$ which is classical in $\mathbb{R} \times (0, 1)$.

We turn now to the variational characterization of the solution as the minimizer of the functional

$$ I_{\varepsilon,\varepsilon}(\rho, \beta) = \int_{\mathbb{R}} \rho(x, 0) \frac{x^2}{2\eta} dx + \int_{0}^{1} \int_{\mathbb{R}} \left[ \rho^2(x, t) + \frac{1}{2} \frac{\beta^2(x, t)}{\rho(x, t)} \right] dxdt, \quad (3.1) $$

where $\rho$ and $\beta$ are such that

$$ \partial_t \rho + \partial_x \beta + \varepsilon \partial_x^2 \rho = 0 \quad \text{in} \quad \mathbb{R} \times (0, 1), \quad \rho(\cdot, 1) = \delta. \quad (3.2) $$

Since both $I_{\varepsilon,\varepsilon}$ and the constraint are strictly convex, there exists a unique minimizer $(\rho_{\varepsilon,\eta}, \beta_{\varepsilon,\eta})$ of

$$ I_{\varepsilon,\eta} = \inf \{ I_{\varepsilon,\eta}(\rho, \beta) : \rho, \beta \text{ satisfying (3.2)} \}. \quad (3.3) $$

It also well known (see [29]) that $(u_{\varepsilon,\eta}, \rho_{\varepsilon,\eta})$ with $u_{\varepsilon,\eta} = \beta_{\varepsilon,\eta}/\rho_{\varepsilon,\eta}$ is the solution of (1.1).

Finally, we remark that the minimization problem can also be written as

$$ I_{\varepsilon,\eta} = \inf \left\{ \int_{\mathbb{R}} \rho(x, 0) \frac{x^2}{2\eta} dx + \int_{0}^{1} \int_{\mathbb{R}} \left[ \rho^2(x, t) + \frac{1}{2} \rho \alpha^2 \right] dxdt : \right. \\ \left. \partial_t \rho + \partial_x(\rho\alpha) + \varepsilon \partial_x^2 \rho = 0 \quad \text{in} \quad \mathbb{R} \times (0, 1), \quad \rho(\cdot, 1) = \delta \right\}. \quad (3.4) $$

It is clear that, if we could prove that, as $(\varepsilon, \eta) \to 0$, $I_{\varepsilon,\eta} \to \bar{I}$, where

$$ \bar{I} = \min \left\{ \int_{0}^{1} \int_{\mathbb{R}} \left[ \rho^2 + \frac{1}{2} \alpha^2 \rho \right] dxdt : \partial_t \rho + \partial_x(\alpha\rho) = 0 \quad \rho(\cdot, 0) = \rho(\cdot, 1) = \delta \right\}, \quad (3.5) $$

we would have that the minimizer $(u_{\varepsilon,\eta}, \rho_{\varepsilon,\eta})$ converges to the unique minimizer of $\bar{I}$.

Here, we are not able to provide a direct proof of the convergence of the functional, which, of course, follows after we have established the convergence of the solutions.

We begin with a result about a uniform upper bound on $I_{\varepsilon,\eta}$ as well as the easy part of what would have been a convergence result.

**Proposition 3.2.** For every $\alpha \in (0, 1)$ there exists $C > 0$, which is independent of $\varepsilon$ and $\eta$, such that, if $\varepsilon \le C\eta^\alpha$, then

$$ (i) \sup \{ I_{\varepsilon,\eta} : \varepsilon \le \eta^\alpha \} \le C \quad \text{and} \quad (ii) \liminf_{(\varepsilon,\eta) \to (0,0)} \varepsilon \le C\eta^\alpha \quad I_{\varepsilon,\eta} \ge \bar{I}. \quad (3.6) $$

*Proof.* To prove (3.6)(i), we construct an admissible pair $(\rho_{\varepsilon,\eta}, \beta_{\varepsilon,\eta})$ such that, for some uniform $C > 0$,

$$ I_{\varepsilon,\eta}(\rho_{\varepsilon,\eta}, \beta_{\varepsilon,\eta}) \le C. $$

Let $p$ the fundamental solution of the heat equation in $\mathbb{R} \times (0, 1)$, that is, $p(x, t) = \exp(-x^2/4t)/\sqrt{2t}$.
In $\mathbb{R} \times (1/2, 1)$, we take $\rho(x, t) = p(x, b(t))$ with $b$ a smooth decreasing function and $b(1) = 0$. It is immediate that

$$0 = \partial_t \rho - b' \partial_x^2 \rho = \partial_t \rho + \epsilon \partial_x^2 \rho - (b' + \epsilon) \partial_x (\rho \partial_x (\log \rho)) = \partial_t \rho + \epsilon \partial_x^2 \rho + \partial_x (\rho a) \text{ with } a(x, t) = \frac{b'(t) + \epsilon}{2b(t)} x.$$

We also have

$$\int_{1/2}^{1} \int_{\mathbb{R}} \rho^2 dx dt \approx \int_{1/2}^{1} \frac{1}{\sqrt{b}} dt \quad \text{and} \quad \int_{1/2}^{1} \int_{\mathbb{R}} \rho a^2 dx dt \approx \int_{1/2}^{1} \frac{(b' + \epsilon)^2}{b} dt.$$

We choose $b(t) = \epsilon(1-t) + (1-t)^\theta$ for $\theta \in (1, 2)$. Then, for some $C > 0$,

$$\int_{1/2}^{1} \frac{1}{\sqrt{b}} dt \le \int_{0}^{1/2} t^{-\theta/2} \le C \quad \text{and} \quad \int_{1/2}^{1} \frac{(b' + \epsilon)^2}{b} dt \le \int_{0}^{1/2} t^{\theta-2} dt \le C.$$

Finally,

$$\rho(x, 1/2) = \frac{1}{2b_{\epsilon}} \exp[-\frac{x^2}{4b_{\epsilon}}] \text{ with } b_{\epsilon} = \epsilon/2 + 1/2^{\theta}.$$

In $(0, 1/2)$, we take $\rho(x, t) = p(x, \gamma + c(t))$ with $\gamma > 0$ and $c$ an increasing smooth function with $c(0) = 0$. It follows as above that

$$\partial_t \rho + \epsilon \partial_x^2 \rho + \partial_x (\rho a) = 0 \text{ in } \mathbb{R} \times (0, 1) \text{ with } a(x, t) = \frac{c'(t) + \epsilon}{\gamma + c(t)} x.$$

Moreover,

$$\frac{1}{\eta} \int_{R} x^2 \rho(x, 0) dx = \frac{1}{\eta} \int_{R} x^2 \frac{1}{\sqrt{\gamma}} \exp(-x^2/\gamma) dx \approx \frac{\gamma}{\eta} = 1 \text{ if } \gamma = \eta.$$

and

$$\int_{0}^{1/2} \int_{\mathbb{R}} \rho^2 dx dt \approx \int_{0}^{1/2} \frac{1}{\sqrt{\gamma + c(t)}} dt \quad \text{and} \quad \int_{0}^{1/2} \int_{\mathbb{R}} \rho a^2 dx dt \approx \int_{0}^{1/2} \frac{(c'(t) + \epsilon)^2}{\gamma + c(t)} dt.$$

We choose $c(t) = c_0 t^\theta$ for the same $\theta \in (1, 2)$ as in the construction of $\rho$ in $(1/2, 1)$.

For $\rho$ to be continuous at $t = 1/2$, we choose $c_0$ so that $\eta + c_0/2^\theta = \epsilon/2 + 1/2^\theta$, and note that $c_0 \to 1$ in the limit $\eta, \epsilon \to 0$.

Moreover, for some $C > 0$,

$$\int_{0}^{1/2} \frac{1}{\sqrt{\eta + c(t)}} dt = \int_{0}^{1/2} \frac{1}{\sqrt{\eta + c_0 t^\theta}} dt \le \int_{0}^{1/2} \frac{1}{t^{\theta/2}} dt \le C$$

and

$$\int_{0}^{1/2} \frac{(c'(t) + \epsilon)^2}{\gamma + c(t)} dt = \int_{0}^{1/2} \frac{(c_0 \theta t^{\theta-1} + \epsilon)^2}{\gamma + c_0 t^{\theta}} dt \le 2 \int_{0}^{1/2} \frac{(c_0 \theta t^{\theta-1})^2}{\gamma + c_0 t^{\theta}} dt + 2 \int_{0}^{1/2} \frac{\epsilon^2}{\gamma + c_0 t^{\theta}} dt.$$

It is clear that

$$\int_{0}^{1/2} \frac{(c_0 \theta t^{\theta-1})^2}{\eta + c_0 t^{\theta}} dt \le C,$$

while

$$\int_{0}^{1/2} \frac{\epsilon^2}{\eta + c_0 t^{\theta}} dt \le \frac{\epsilon^2}{\eta^{1-1/\theta}} \int_{0}^{\infty} \frac{1}{1 + c_0 s^{\theta}} ds \le C \frac{\epsilon^2}{\eta^{1-1/\theta}}.$$

For this last integral to be finite we need to choose $\epsilon^2 \le \eta^\alpha$ with $\alpha = \frac{\theta-1}{\theta} \in (0, 1)$.
To prove (3.6)(ii), let $(u_{\varepsilon,\eta}, \rho_{\varepsilon,\eta})$ be the minimizer of the strictly convex functional $I_{\varepsilon,\eta}$, that is, for $\beta_{\varepsilon,\eta} = \rho_{\varepsilon,\eta} \partial_x u_{\varepsilon,\eta}$,

$$I_{\varepsilon,\eta} = I_{\varepsilon,\eta}(\rho_{\varepsilon,\eta}, \beta_{\varepsilon,\eta}) = \int_{\mathbb{R}} \rho_{\varepsilon,\eta}(x,0) \frac{x^2}{2\eta} dxdt + \int_{0}^{1} \int_{\mathbb{R}} \left[ \rho_{\varepsilon,\eta}^2 + \frac{1}{2} \frac{\beta_{\varepsilon,\eta}^2}{\rho_{\varepsilon,\eta}} \right] dxdt \quad (3.7)$$

and

$$\partial_t \rho_{\varepsilon,\eta} + \partial_x \beta_{\varepsilon,\eta} + \varepsilon \partial_x^2 \rho_{\varepsilon,\eta} = 0 \quad \text{in } \mathbb{R} \times (0,1) \quad \rho_{\varepsilon,\eta}(\cdot, 1) = \delta. \quad (3.8)$$

It follows from (3.6)(i) that, for some $C \in (0, \infty)$,

$$\int_{0}^{1} \int_{\mathbb{R}} \rho_{\varepsilon, \eta}^{2} dxdt \leq C \quad \text{and} \quad \int_{0}^{1} \int_{\mathbb{R}} \frac{1}{2} \frac{\beta_{\varepsilon, \eta}^{2}}{\rho_{\varepsilon, \eta}} dxdt \leq C. \quad (3.9)$$

We also know that, for any measurable subset $A$ of $\mathbb{R} \times (0,1)$,

$$\iint_{A} |\beta_{\varepsilon, \eta}| dxdt \leq 2C,$$

which is an immediate consequence of (3.20) and the elementary inequality

$$|\beta| = \frac{|\beta|\rho}{\rho} \leq \frac{\beta^2}{2\rho} + \frac{\rho}{2}.$$

It follows that the $\rho_{\varepsilon,\eta}$'s converge weakly in $L^2(\mathbb{R} \times (0,1))$ to some $\rho$, and the $\beta_{\varepsilon,\eta}$'s and $\beta_{\varepsilon,\eta}^2/\rho_{\varepsilon,\eta}$'s weakly in $L^1(\mathbb{R} \times (0,1))$ to some $\beta$ and $\beta^2/\rho$.

It is now clear that

$$\liminf_{\varepsilon \to 0, \eta \to 0, \varepsilon \leq \eta^\alpha} I(\rho_{\varepsilon, \eta}, \beta_{\varepsilon, \eta}) \geq I(\rho, \beta)$$

and

$$\partial_t \rho + \partial_x \beta = 0 \quad \text{in } \mathbb{R} \times (0,1) \quad \text{and} \quad \rho(\cdot, 1) = \delta,$$

and, hence, the claim.

□

The first basic estimates for $u_{\varepsilon,\eta}$ and $\rho_{\varepsilon,\eta}$ are stated and proved in the next proposition.

**Proposition 3.3.** There exists $C > 0$ such that, for $\varepsilon \leq C\eta^\alpha$,

$$\int_{0}^{1} \int_{\mathbb{R}} [\rho_{\varepsilon, \eta}^{2}(x,t) + \frac{1}{2}\rho_{\varepsilon, \eta}(x,t)u_{\varepsilon, \eta}^{2}(x,t)]dxdt + \int_{\mathbb{R}} \rho_{\varepsilon, \eta}(x,0)\frac{x^2}{2\eta}dx \\ + \sup_{t \in [0,1]} \left[ \int_{\mathbb{R}} \rho_{\varepsilon, \eta}(x,t)u_{\varepsilon, \eta}(x,t)dx + \int_{\mathbb{R}} \rho_{\varepsilon, \eta}(x,t)\frac{x^2}{2(\eta+t)}dx \right] \leq C. \quad (3.10)$$

*Proof.* Throughout the proof we omit the explicit dependence of $u_{\varepsilon,\eta}$ and $\rho_{\varepsilon,\eta}$ on $\varepsilon, \eta$ and simply write $u$ and $\rho$.

The first three bounds are immediate from Proposition 3.2.

The uniform in $t$, $\varepsilon$ and $\eta$ bound on $\int_{\mathbb{R}} \rho(x,t)u(x,t)dx$ follows from the first two bounds and the identity

$$\frac{d}{dt} \int_{\mathbb{R}} \rho(x,t)u(x,t)dx = \int_{\mathbb{R}} \left[ \frac{1}{2} \rho(x,t) (\partial_x u)^2(x,t) dx + \rho^2(x,t) dx \right]. \quad (3.11)$$

Finally, for the bound on $\int_{\mathbb{R}} \rho(x,t)x^2dx$ we observe that, since $\rho \geq 0$, $u$ is a supersolution of

$$w_t - \varepsilon \partial_x^2 w + \frac{1}{2} (\partial_x w)^2 = 0 \quad \text{and} \quad w(x,0) = x^2/2\eta,$$
whose solution is

$$w(x,t) = \frac{x^2}{2(\eta + t)} + \varepsilon(\ln(\eta + t) - \ln \eta). \qquad (3.12)$$

It follows that

$$u \ge w \text{ in } \mathbb{R} \times [0, 1]. \qquad (3.13)$$

The claim follows from the bound on $\sup_{t \in (0,1)} \int_{\mathbb{R}} \rho(x,t)u(x,t)dx$ and the assumption that $\varepsilon \le C\eta^\alpha$. $\square$

To show the convergence of $\rho_{\varepsilon,\eta}$ and $u_{\varepsilon,\eta}$, we need several estimates involving derivatives, which we formulate and prove in the propositions below.

**Proposition 3.4.** For every $\theta \in (0, 1/2)$ and uniformly for $\varepsilon$ and $\eta$ such that $\varepsilon \le C\eta^\alpha$, the $u_{\varepsilon,\eta}$'s are bounded in $L^\infty((\theta, 1); L_{\text{loc}}^1(\mathbb{R}))$, the $\partial_x u_{\varepsilon,\eta}$'s are bounded in $L^2((\eta, 1); L_{\text{loc}}^2(\mathbb{R}))$, and, hence, the $u_{\varepsilon,\eta}$'s are bounded in $L^2((\eta, 1); H_{\text{loc}}^1(\mathbb{R}))$.

*Proof.* To simplify the writing, in what follows we drop, when possible, the dependence on $\varepsilon$ and $\eta$ and we write $u$ and $\rho$. Moreover, we denote by $C$ constants may change from line to line, are uniform in $\varepsilon$ and $\eta$ and may depend on $\theta$.

Note that the reason we obtain bounds for $t \in (\theta, 1)$ instead of $(0, 1)$ is in order to have estimates that do not depend on $\eta$.

To set things up, we first observe that, since $t \in (\theta, 1)$, it follows from Proposition 3.3 that

$$\sup_{t \ge \theta} \int_{\mathbb{R}} \rho(x,t) |x|^2 dx \le C,$$

and, hence, there exists some $R_0 > 0$ such that, for all $R \ge R_0$ and $t \in (\theta, 1)$,

$$\int_{|x| \ge R} \rho(x,t) |x|^2 dx dt \le \frac{1}{2} \quad \text{and} \quad \int_{|x| \le R} \rho(x,t) |x|^2 dx dt \ge \frac{1}{2}. \qquad (3.14)$$

The proof consists of two parts. In the first, we provide, for $R \ge R_0 \ge 1$, a bound on $\int_{-R}^R u(x,t)dx$ by a quantity that depends on $\int_h^1 \int_{-R}^R u_x^2(x,t)dxdt$, which we bound in the second step.

For every $R \ge R_0$, let

$$\langle u \rangle(t) = \frac{1}{2R} \int_{-R}^{R} u(x,t)dx,$$

and observe that, in view of the lower bound on $u$ in (3.13),

$$\langle u \rangle(t) \ge R^2/6(\eta + t).$$
Using (3.14), Poincare's inequality and the bound on $\int_0^1 \int_{\mathbb{R}} \rho(x,t)u(x,t)dxdt$, we find that, for all $R \ge R_0$,

$$ 
\begin{aligned}
\frac{1}{2}\langle u \rangle(t) &\le \int_{|x|\le R} \rho(x,t)\langle u \rangle(t)dx \\
&\le \int_{|x|\le R} \rho(x,t)|u - \langle u \rangle(t)|dx + \int_0^1 \int_{|x|\le R} \rho(x,t)u(x,t)dxdt \\
&\le C + \left( \int_{|x|\le R} \rho^2(x,t)dx \right)^{1/2} \left( \int_{|x|\le R} |u(x,t) - \langle u \rangle(t)|^2dx \right)^{1/2} \\
&\le C \left( 1 + \left( \int_{|x|\le R} (\partial_x u)^2(x,t)dx \right)^{1/2} \left( \int_{\mathbb{R}} \rho^2(x,t)dx \right)^{1/2} \right) \\
&= C \left( 1 + \phi^{1/2}(t) \int_{|x|\le R} (\partial_x u)^2(x,t)dxdt \right)^{1/2},
\end{aligned}
$$

where $\phi(t) = \int_{\mathbb{R}} \rho^2(x,t)dx$ and $\phi \in L^1((0,1))$.

Since $\rho$ and $u$ are even in $x$, throughout the argument below all integrals are taken over $[0, R]$ instead of $[-R, R]$.

Hence,

$$ \langle u \rangle(t) = \frac{1}{R} \int_0^R u(x,t)dx, \quad \int_{|x|\le R} (\partial_x u)^2(x,t)dxdt = 2 \int_0^R (\partial_x u)^2(x,t)dxdt, $$

and

$$ \langle u \rangle(t) \le C \left(1 + \phi(t)^{1/2} \left(\int_0^R (\partial_x u)^2(x,t)dx\right)^{1/2}\right). \quad (3.15) $$

Let

$$ F(R, \theta) = \int_{\theta}^{1} \int_{0}^{R} (\partial_x u)^2(x, t) dx dt $$

Integrating the $u$-equation in (1.1) over $[0, R] \times [h, 1]$, using that $R_0 \ge 1$, $u \ge 0$, $\int_{\mathbb{R}} \rho(x,t)dx = 1$ and (3.15) give, after some elementary rearrangement of terms,

$$ 
\begin{aligned}
F(R, \theta) &\le 2 \left( 1 + \int_0^R u(x, \theta) dx + \varepsilon \int_\theta^1 \partial_x u(R, s) ds \right) \\
&\le 2R \left( 1 + \langle u \rangle(\theta) + \varepsilon \int_\theta^1 \partial_x u(R, s) ds \right) \\
&\le C R \left( 1 + \left( \int_\theta^1 (\partial_x u)^2(R, t) dt \right)^{1/2} + \phi(\theta)^{1/2} \left( \int_0^R (\partial_x u)^2(x, \theta) dx \right)^{1/2} \right) \\
&\le C R \left( 1 + \left( \frac{\partial}{\partial R} F(R, \theta) \right)^{1/2} + \phi(\theta)^{1/2} \left( - \frac{\partial}{\partial \theta} F(R, \theta) \right)^{1/2} \right),
\end{aligned}
$$

and, finally,

$$ F(R, h)^2 \le C R^2 \left( 1 + \frac{\partial}{\partial R} F(R, h) - \phi(\theta) \frac{\partial}{\partial h} F(R, h) \right). \quad (3.16) $$

We now fix $\bar{R} \ge R_0$ and $\theta_0$, write $F_0 = F(\bar{R}, \theta_0)$, and, for $s > 0$, we seek $t_0 > 0$ such that the odes

$$ \dot{R} = 1 \quad \text{and} \quad \dot{\theta} = -\phi(\theta) \quad \text{in} \quad [0, t_0] \quad \text{and} \quad R(t_0) = \bar{R} \quad \text{and} \quad \theta(t_0) = \theta_0, \quad (3.17) $$

have a solution. We do this at the end of the ongoing proof.
Assuming for the moment that (3.17) has solution, we observe that (3.16) yields that, for some $C > 0$, either

$$ \text{either } F \le \bar{C} \text{ or } \frac{\partial}{\partial s} F(R(s), \theta(s)) \ge \frac{1}{C} F^2. $$

The second alternative implies

$$ \frac{1}{F(\bar{R}, \theta_0)} - \frac{1}{F(R(s), \theta(s))} \ge \frac{s}{C}, $$

and, finally,

$$ F(\bar{R}, \theta_0) \le \frac{C}{\theta_0}. \qquad (3.18) $$

We now return to solving (3.17), the main issue being the existence of an interval of existence so that $\theta(t_0) = \theta_0$.

Let $a = \phi \in L^1(0, 1)$, assume without loss of generality that $\int_0^1 a(z)dz \ge 1$ and rewrite the $\theta$ equation as $\frac{d\theta}{a} = -ds$, which leads to

$$ \int_{\theta}^{\theta_0} \frac{1}{a(z)} dz = s. \qquad (3.19) $$

Since, in view of Jensen's inequality,

$$ \frac{1}{\theta_0 - \theta} \int_{\theta}^{\theta_0} \frac{1}{a(z)} dz \ge \left( \frac{1}{\theta_0 - \theta} \int_{\theta}^{\theta_0} a(z) dz \right)^{-1}, $$

it follows from (3.19) that

$$ s \ge (\theta_0 - \theta)^2 \left( \int_{\theta}^{\theta_0} a(z) dz \right)^{-1}. $$

Letting $\theta \to 0$, we see that, to have $\theta(t_0) = \theta_0$, it suffices to choose $t_0 \ge \theta_0^2 / \int_{\theta}^{\theta_0} a(z)dz$, which is possible in view of the assumption that $\int_0^1 a(z)dz \ge 1$.

Finally, we choose $R(0) = \bar{R} - t_0$.

Having established that the $\partial_x u_{\varepsilon, \eta}$'s are bounded in $L^2((\theta, 1); L_{\text{loc}}^2(\mathbb{R}))$, we now turn to the bound on $u_{\varepsilon, \eta}$ in $L^\infty((\eta, 1); L_{\text{loc}}^1(\mathbb{R}))$, which will follow from (3.15).

Fix $R' > R \ge R_0$. Since $\int_{\theta/2}^1 \int_0^{R'} (\partial_x u_x)^2(x, s) dx ds$ is bounded independently of $\varepsilon$ and $\eta$, there exists $R'' \in (R, R')$ such that

$$ \int_{\theta/2}^{1} (\partial_x u)^2(R'', s) ds \quad \text{is bounded independently of } \varepsilon \text{ and } \eta. \qquad (3.20) $$

We also know from the first estimate in this proof and the fact that $\int_0^1 \int_{\mathbb{R}} \rho^2 dx ds \le C$ that

$$ \int_{\theta/2}^{\theta} \int_{0}^{R''} [(\partial_x u)^2(x, s) + \rho^2(x, s)] dx ds \quad \text{is bounded uniformly in } \varepsilon \text{ and } \eta. $$
It follows that there exists some $h \in [\theta/2, \theta]$ such that

$$ \int_0^{R''} \rho^2(x, h)dx \le C \quad \text{and} \quad \int_0^{R''} (\partial_x u)^2(x, h)dx \text{ are bounded uniformly in } \varepsilon \text{ and } \eta \quad (3.21) $$

which together with (3.15) imply that

$$ \int_0^{R''} u(x, h)dx \text{ is bounded independently of } \varepsilon \text{ and } \eta. \quad (3.22) $$

Integrating the $u$ equation over $[0, R''] \times [h, t]$ we find

$$ \int_0^{R''} u(x, t)dx \le 1 + \int_0^{R''} u(x, h)dx + \varepsilon \left( \int_h^1 (\partial_x u)^2(R'', s)ds \right)^{1/2}. \quad (3.23) $$

In view of the above observation and (3.20) we know that the right hand side of (3.23) is bounded independently of $\varepsilon$ and $\eta$, so the same is true for $\int_0^{R''} u(x, t)dx$, and, hence, $\int_0^R u(x, t)dx$.

We prove next a new uniform estimate on $\rho_{\varepsilon,\eta}$, which, given that their limit will be Dirac masses at $t=0$ and $t=1$, can be only local in time.

**Proposition 3.5.** For $\varepsilon$ and $\eta$ such that $\varepsilon \le C\eta^\alpha$ and any $\theta \in (0, 1/2)$, the $\partial_x\rho_{\varepsilon,\eta}$'s and $\sqrt{\rho_{\varepsilon,\eta}}\partial_x^2 u_{\varepsilon,\eta}$'s are bounded uniformly in $\varepsilon$ and $\eta$ in $L^2(\theta, 1-\theta; L^2(\mathbb{R}))$, and the $\rho_{\varepsilon,\eta}$'s are bounded uniformly in $\varepsilon$ and $\eta$ in $L^2(\theta, 1-\theta; H^1(\mathbb{R}))$.

*Proof.* As in the previous proof, we omit the dependence of $u_{\varepsilon,\eta}$ and $\rho_{\varepsilon,\eta}$ on $\varepsilon$ and $\eta$.

A simple computation, which can be justified by considering approximate problems with regularized $\rho(\cdot, 1)$ and then passing to the limit, leads, for each $t \in (0, 1)$, to the identity

$$ \frac{d}{dt} \int_{\mathbb{R}} \partial_x \rho(x, t) \partial_x u(x, t) dx dt = \int_{\mathbb{R}} \left[ (\partial_x \rho)^2(x, t) + \rho(x, t) (\partial_x^2 u)^2(x, t) \right] dx dt. \quad (3.24) $$

Next we localize the estimate in time staying away from 0 and 1. For this, we consider a smooth cut-off function $\chi$ supported in $(\theta, 1-\theta)$ and multiply (3.24) by $\chi^2$.

Integrating in time we get

$$ \int_0^1 \chi^2(t) \frac{d}{dt} \int_{\mathbb{R}} \partial_x \rho(x, t) \partial_x u(x, t) dx dt = \int_0^1 \int_{\mathbb{R}} \chi^2(t) \left[ (\partial_x \rho)^2(x, t) + \rho(x, t) (\partial_x^2 u)^2(x, t) \right] dx dt, $$

and, hence,

$$ \begin{align*} \int_0^1 \int_{\mathbb{R}} \chi^2(t) \left[ (\partial_x \rho)^2(x, t) + \rho(x, t) (\partial_x^2 u)^2(x, t) \right] dx dt &= 2 \int_0^1 \int_{\mathbb{R}} \chi(t) \chi'(t) \partial_x \rho(x, t) \partial_x u(x, t) dx dt \\ &= -2 \int_0^1 \int_{\mathbb{R}} \chi(t) \chi'(t) \rho(x, t) \partial_x^2 u(x, t) dx dt \\ &\le \frac{1}{2} \int_0^1 \int_{\mathbb{R}} \chi(t) \chi'(t) \rho(x, t) (\partial_x^2 u)^2(x, t) dx dt + 2 \int_0^1 \int_{\mathbb{R}} (\chi')^2(t) \rho^2(x, t) dx dt, \end{align*} $$

which leads to the claim in view of the fact that

$$ \int_0^1 \int_{\mathbb{R}} (\chi')^2(t) \rho^2(x, t) dx dt \le C \int_0^1 \int_{\mathbb{R}} \rho^2(x, t) dx dt. $$

□
Next we obtain uniform in $\varepsilon$ and $\eta$ estimates on $\rho_{\varepsilon,\eta}$ and $u_{\varepsilon,\eta}$ describing their behavior $t = 1$. These estimates as well as the ones about the behavior near $t = 0$ are needed in order to obtain the convergence of the full family of solutions of (1.1).

The first result is about a surprising estimate in view of the fact that $\rho_{\varepsilon,\eta}(\cdot, 1) = \delta$.

**Proposition 3.6.** For each $\theta \in (0, 1)$, there exists an, independent of $\varepsilon$ and $\eta$, $C > 0$ such that, for all $\alpha \in (0, 1/2)$,

$$ \int_{\theta}^{1} [\rho_{\varepsilon,\eta}(\cdot, t)]_{C^{0,\alpha}} dt \leq C. \qquad (3.25) $$

*Proof.* As before, when there is no confusion, we drop the dependence on $\varepsilon$ and $\eta$.

In the proof we work with $\theta = 1/2$ and leave it up to the reader to adjust the argument for $\theta \in (0, 1)$.

The starting point is the identity (3.24) which can be rewritten as

$$ \frac{d}{dt} \int_{\mathbb{R}} -\rho(x,t) \partial_x^2 u(x,t) dx = \int_{\mathbb{R}} [(\partial_x \rho)^2(x,t) + \rho(x,t)(\partial_x^2 u)^2(x,t)] dx. \qquad (3.26) $$

It follows that

$$ \frac{d}{dt} \int_{\mathbb{R}} -\rho(x,t) \partial_x^2 u(x,t) dx \geq \frac{1}{2} \left( - \int_{\mathbb{R}} \rho(x,t) \partial_x^2 u(x,t) dx \right)^2, $$

and, hence, there exists $C > 0$ such that, for all $t \in (1/2, 1)$,

$$ \int_{\mathbb{R}} -\rho(x,t) \partial_x^2 u(x,t) dx \leq \frac{C}{1-t}. \qquad (3.27) $$

It follows from Proposition 3.3 and Proposition 3.5 that there exists $C > 0$ such that, for each $\varepsilon$ and $\eta$, there exists $t_{\varepsilon,\eta} \in (1/4, 1/2)$ such that

$$ \int_{\mathbb{R}} [(\partial_x \rho_{\varepsilon,\eta})^2(x, t_{\varepsilon,\eta}) + (\partial_x u_{\varepsilon,\eta})^2(x, t_{\varepsilon,\eta})] dx \leq C. $$

and, hence,

$$ \int_{\mathbb{R}} \partial_x \rho_{\varepsilon,\eta}(x, t_{\varepsilon,\eta}) \partial_x u_{\varepsilon,\eta}(x, t) dx \leq C. \qquad (3.28) $$

Integrating (3.26) over $(t_{\varepsilon,\eta}, t)$ we find

$$ \int_{1/2}^{t} \int_{\mathbb{R}} (\partial_x \rho)^2(x,t) dx \leq \int_{t_{\varepsilon,\eta}}^{t} \int_{\mathbb{R}} (\partial_x \rho)^2(x,t) dx \leq \int_{\mathbb{R}} \partial_x \rho(x, t_{\varepsilon,\eta}) \partial_x u(x, t_{\varepsilon,\eta}) dx - \int_{\mathbb{R}} \partial_x \rho(x, t) \partial_x u(x, t) dx, $$

and, in view of (3.27) and (3.28), there exists some other $C > 0$,

$$ \int_{1/2}^{t} \int_{\mathbb{R}} (\partial_x \rho)^2(x,t) dx \leq \frac{C}{1-t}. \qquad (3.29) $$

Next, we recall that, for every $\alpha \in (0, 1/2)$, there exists $\beta \in (0, 1)$ such that

$$ [\rho(\cdot, t)]_{C^{0,\alpha}}^{2+\beta} \leq C \int_{\mathbb{R}} (\partial_x \rho)^2(x,t) dx, $$

which together with (3.29) imply that

$$ \int_{1/2}^{t} [\rho(\cdot, s)]_{C^{0,\alpha}}^{2+\beta} ds \leq \frac{C}{1-t}. $$
Then, an elementary analysis lemma which we state and prove after the end of the ongoing proof yields (3.25). □

We state and prove next the auxiliary fact we used at the end of the proof above. For simplicity we change $t$ to $1-t$.

**Lemma 3.7.** Let $a: (0,1) \to [0, \infty)$ and $m > 2$ be such that, for $h, h' \in (0, 1/2)$ with $h < h'$, $\int_h^{h'} a^m(s)ds \le C/(h' - h)$. Then $\int_0^{1/2} a(s)ds \le C$.

*Proof.* Observe that

$$ \int_0^{1/2} a(s)ds = \sum_{n \ge 0} \int_{1/2^{n+1}}^{1/2^n} a(s)ds \le \sum_{n \ge 0} \left[ \left( \int_{1/2^{n+1}}^{1/2^n} a^m(s)ds \right)^{1/m} \left( 2^{-(n+1)/m} \right)^{m-1/m} \right] \\ \le C \sum_{n \ge 0} 2^{-(n+1)/m} 2^{-(n+1)((m-1)/m)} = C \sum_{n \ge 0} 2^{-n((m-1)/m)}, $$

and $\sum_{n \ge 0} 2^{-n((m-1)/m)} < \infty$ if $\frac{m-1}{m} > 1/m$, that is, $m > 2$.

□

The next proposition provides $L^\infty$- bounds for $u_{\varepsilon, \eta}$ and improves their available regularity up to and including $t = 1$.

**Proposition 3.8.** As $\varepsilon, \eta \to 0$ while $\varepsilon \le C\eta^\alpha$ and every $\theta \in (0, 1)$, the $u_{\varepsilon, \eta}$'s are bounded in $L^\infty((\theta, 1]; L_{\text{loc}}^\infty(\mathbb{R}))$. Moreover, for some uniform $C > 0$,

$$ \sup_{t \in (\theta, 1]} \|u_{\varepsilon, \eta}(\cdot, t)\|_{C^{0, \alpha}_{\text{loc}}} \le \|u_{\varepsilon, \eta}(\cdot, \theta)\|_{C^{0, \alpha}_{\text{loc}}} + C. \quad (3.30) $$

Moreover, they converge along subsequences and locally uniformly to a viscosity solution of the Hamilton-Jacobi equation $\partial_t u + \frac{1}{2}(\partial_x u)^2 = \rho$ in $\mathbb{R} \times (0, 1)$ satisfying $(2.31)$.

*Proof.* In view of Proposition 3.6 and the comment at the beginning of its proof, (3.30) follows from the classical parabolic theory once we establish an $L_{\text{loc}}^\infty(\mathbb{R})$ bound on $u_{\varepsilon, \eta}(\cdot, \theta)\|.$

Proposition 3.5 yields that, for every $\theta \in (0, 1)$, the $u_{\varepsilon, \eta}$'s are uniformly bounded in $L^2((\theta, 1); H_{\text{loc}}^1(\mathbb{R})) \cap L^\infty((\theta, 1); L_{\text{loc}}^1(\mathbb{R}))$.

Thus, for all $\theta \in (0, 1/2)$, there exists $C_\theta > 0$ such that

$$ \sup_{t \in (\theta, 1-\theta)} \int_{-R}^{R} u(x,t)dx + \int_{\theta}^{1-\theta} \int_{\mathbb{R}} (\partial_x u)^2(x,t)dxdt \le C_{\theta}R, \quad (3.31) $$

which implies that, for some $t_{\varepsilon, \eta}^{\theta, 0} \in (\theta/2, \theta)$ and $x_{\varepsilon, \eta}^{\theta, 0} \in \mathbb{R}$,

$$ \int_{-R}^{R} u(x, t_{\varepsilon, \eta}^{\theta, 0}) dx + \int_{-R}^{R} (\partial_x u)^2(x, t_{\varepsilon, \eta}^{\theta, 0}) \le C_{\theta}(R+1), \quad (3.32) $$

and

$$ u(x_{\varepsilon, \eta}^{\theta, 0}, t_{\varepsilon, \eta}^{\theta, 0}) \le C_{\theta}. \quad (3.33) $$

It also follows from (3.31), that, for $m > 1$,

$$ \int_{\theta}^{1-\theta} \int_{\mathbb{R}} \frac{(\partial_x u)^2(x,t)}{(1+x^2)^{m/2}} dxdt \le C_{\theta}, $$

which also yields some $t_{\varepsilon, \eta}^{\theta, 1} \in (\theta/2, \theta)$ such that

$$ \int_{\mathbb{R}} \frac{u(x, t_{\varepsilon, \eta}^{\theta, 1}) + (\partial_x u)^2(x, t_{\varepsilon, \eta}^{\theta, 1})}{(1+x^2)^{m/2}} dx \le C_{\theta}. \quad (3.34) $$
Finally, since the $\rho_{\varepsilon,\eta}$'s are bounded in $L^2((t_{\varepsilon,\eta}^{\theta,0}, 1-\theta); L^\infty(\mathbb{R}))$, we observe that

$$u(x,t) \leq v(x,t) + \int_{t_0}^{t} \|\rho(\cdot, s)\|_{\infty} ds,$$

where $v$ solves

$$\partial_t v - \varepsilon \partial_x^2 v + \frac{1}{2} (\partial_x v)^2 = 0 \quad \text{in } \mathbb{R} \times (t_{\varepsilon,\eta}^{\theta,0}, 1-\theta) \quad v(\cdot, t_{\varepsilon,\eta}^{\theta,0}) = u(\cdot, t_{\varepsilon,\eta}^{\theta,0}).$$

Below, to keep the writing simpler we write $u_0$ for $u(\cdot, t_{\varepsilon,\eta}^{\theta,0})$.

It follows that, for each progressively measurable $\alpha : (t_{\varepsilon,\eta}^{\theta,0}, 1-\theta) \to \mathbb{R}$,

$$u(x,t) \leq \mathbb{E} \left[ \int_{t_{\lambda,0}}^{t} \frac{1}{2} |\alpha_s|^2 + u_0(X_t) \right], \quad (3.35)$$

where $dX_s = \alpha_s ds + \sqrt{2\varepsilon}dW_s$ and $X_{t_{\varepsilon,\eta}^{\theta,0}} = x$.

We choose next the control $\alpha = \frac{x_0-x}{t-t_{\varepsilon,\eta}^{\theta,0}}$. Then $dX_s = x_0 + \sqrt{2\varepsilon}dW_s$ and (3.35) yields, for $\tau = t-t_{\varepsilon,\eta}^{\theta,0}$,

$$u(x,t) \leq \frac{1}{2} \frac{(x-x_0)^2}{t-t_{\varepsilon,\eta}^{\theta,0}} + \mathbb{E}[u_0(x_0 + \sqrt{2\varepsilon}W_{\tau})]. \quad (3.36)$$

To conclude we estimate the last term of the right hand side of the inequality above. It follows from (3.36) that

$$\begin{aligned} \mathbb{E}[u_0(x_0 + \sqrt{2\varepsilon}(W_t - W_{t_{\varepsilon,\eta}^{\theta,0}}))] &= \frac{1}{\sqrt{2\varepsilon\tau}} \int_{\mathbb{R}} u_0(x_{\varepsilon,\eta}^{\theta,0} + y) e^{-\frac{y^2}{2\varepsilon\tau}} dy \\ &\leq u_0(x_{\varepsilon,\eta}^{\theta,0}) + \frac{1}{\sqrt{2\varepsilon\tau}} \int_{|y| \leq 1} |u_0(x_{\varepsilon,\eta}^{\theta,0} + y) - u_0(x_{\varepsilon,\eta}^{\theta,0})| e^{-\frac{y^2}{2\varepsilon\tau}} dy. \end{aligned} \quad (3.37)$$

Since $\int_{\theta}^{1} (\partial_x u)^2(x, t_{\varepsilon,\eta}^{\theta,0}) dx$ is bounded, we have, for some uniform $C_{\theta} > 0$,

$$\sup_{|y| \leq 1} |u_0(x_{\varepsilon,\eta}^{\theta,0} + y) - u_0(x_{\varepsilon,\eta}^{\theta,0})| \leq C_{\theta}.$$

Moreover, as it was shown above, for some $m > 1$, $\int_{\mathbb{R}} \frac{u_0(z)}{(1+|z|^2)^{m/2}} dz \leq C_{\theta}$.

It then follows that

$$\frac{1}{\sqrt{2\varepsilon\tau}} \int_{|y| \leq 1} |u_0(x_{\varepsilon,\eta}^{\theta,0} + y) - u_0(x_{\varepsilon,\eta}^{\theta,0})| e^{-\frac{y^2}{2\varepsilon\tau}} dy \leq C$$

and

$$\frac{1}{\sqrt{2\nu\tau}} \int_{|y| \geq 1} u_0(x_{\varepsilon,\eta}^{\theta,0} + y) e^{-\frac{y^2}{2\nu\tau}} dy \leq \int_{|y| \geq 1} \frac{u_0(x_{\varepsilon,\eta}^{\theta,0} + y)}{(1+|y|^2)^{m/2}} \sup_{|z| \geq 1} [|z|^m e^{\frac{-|z|^2}{\sqrt{2\nu\tau}}} \frac{e^{-\frac{y^2}{2\nu\tau}}}{\sqrt{2\nu\tau}} dy \leq C_{\theta},$$

and, hence, for some uniform $C > 0$,

$$u(x,t) \leq \frac{(x - x_{\varepsilon,\eta}^{\theta,0})^2}{2(t - t_{\varepsilon,\eta}^{\theta,0})} + C_{\theta}. \quad (3.38)$$

This upper bound and the strong convergence of the $\rho_{\varepsilon,\eta}$' in $L^2((0,T); C(\mathbb{R}))$ imply that the $u_{\varepsilon,\eta}^{\theta,0}$'s converge locally uniformly to a viscosity solution of

$$\partial_t u + \frac{1}{2} (\partial_x u)^2 = \rho. \quad (3.39)$$
The convergence is established in compact subsets of $(0, 1)$. To extend it on compact subsets of $\mathbb{R} \times (0, 1]$, It remains to show that (3.38) holds on $\mathbb{R} \times (0, 1]$, which is possible in view of Proposition 3.6. $\square$

<figure><img src="image_2.png" /></figure>

The proof above extends to

$$u(x, t) \leq \frac{(x - x_{\varepsilon, \eta}^{\theta, 0})^2}{2(t - t_{\varepsilon, \eta}^{\theta, 0})} + C + F(x, t),$$

where, as $t \to 1$, $F(\cdot, t) \to 0$ in $L^2$ which is not enough to conclude.

Next we investigate the behavior of $u_{\varepsilon, \eta}$ and $\rho_{\varepsilon, \eta}$ near $t = 0$.

**Proposition 3.9.** There exists $C > 0$ such that, for all $\varepsilon, \eta \to 0$ with $\varepsilon \log \eta \to 0$ and for any nonnegative compactly supported $\chi : \mathbb{R} \to [0, \infty)$, which is smooth in its support $[-a, a]$, there exists, an independent of $\varepsilon$ and $\eta$, $e : \mathbb{R} \to \mathbb{R}$ such that $\lim_{t \to 0} e(t) = 0$ and

$$\limsup_{\varepsilon \to 0, \eta \to 0, \varepsilon \log \eta \to 0} \left[ \int_R u_{\varepsilon, \eta}(x, t) \frac{1}{t^{2/3}} \chi\left(\frac{x}{t^{2/3}}\right) dx \right] \leq \int_R \frac{x^2}{2t} \frac{1}{t^{2/3}} \chi\left(\frac{x}{t^{2/3}}\right) dx + e(t). \quad (3.40)$$

*Proof.* Like in previous proofs we write $u$ and $\rho$ in place of $u_{\varepsilon, \eta}$ and $\rho_{\varepsilon, \eta}$.

Recall (3.12) and observe that $\lim_{t \to 0} \varepsilon(\log(t + \eta) - \log \eta) = 0$ uniformly on $\varepsilon$ and $\eta$ in view of the assumption on $\varepsilon$ and $\eta$.

Set

$$u(x, t) = \frac{x^2}{2(t + \eta)} + \varepsilon(\log(t + \eta) - \log \eta) + v(x, t).$$

It is immediate that

$$\partial_t v - \varepsilon \partial_x^2 v + \frac{1}{2} (\partial_x v)^2 + \frac{x}{t+\eta} \partial_x v = 0 \text{ and } v \ge 0 \text{ in } \mathbb{R} \times (0,1) \text{ and } v(\cdot, 0) = 0. \quad (3.41)$$

Given $\chi$ as in the statement, set $\chi_{\eta}(x, t) = \frac{1}{(t + \eta)^{2/3}} \chi(\frac{x}{(t + \eta)^{2/3}})$.

Then

$$\begin{aligned} & \frac{d}{dt} \int_{\mathbb{R}} v \chi_{\eta} dx - \int_{\mathbb{R}} v \partial_t \chi_{\eta} dx + \frac{1}{2} \int_{\mathbb{R}} (\partial_x v)^2 \chi_{\eta} dx + \int_{\mathbb{R}} \frac{x}{t+\eta} \partial_x v \chi_{\eta} dx \\ & - \varepsilon \int_{\mathbb{R}} \partial_x^2 v \chi_{\eta} dx = \int_{\mathbb{R}} \rho \chi_{\eta} \le \frac{C}{(t+\eta)^{2/3}}, \end{aligned} \quad (3.42)$$

the last inequality following from the facts that $\chi$ is bounded and $\int_R \rho(x, t) dx = 1$.

Next we analyze several of the terms in the left hand side of (3.42) beginning with the last one.

Integrating by parts and using that $|\chi'| \le C$ and $|\chi''| \le C$ in $[-a, a]$, we find

$$\begin{aligned} & \left| \int_{\mathbb{R}} \partial_x^2 v \chi_{\eta} dx \right| \le C \frac{1}{(t+\eta)^{4/3}} \|v(\cdot, x)\|_{\infty} + \left| \int_{\mathbb{R}} v \partial_x^2 \chi_{\eta} dx \right| \\ & \le C \left[ \frac{1}{(t+\eta)^{4/3}} \|v(\cdot, t)\|_{\infty} + \left| \int_{-a(t+\eta)^{4/3}}^{a(t+\eta)^{4/3}} v \partial_x^2 \chi_{\eta} dx \right| \right] \\ & \le C \frac{1}{(t+\eta)^{4/3}} \|v(\cdot, t)\|_{\infty}. \end{aligned}$$
It also follows from (3.41) that

$$ \frac{d}{dt} \int_R vdx - \frac{1}{t+\eta} \int_R vdx + \int_R (\partial_x v)^2 = 1, $$

and, thus, since $\int_{\mathbb{R}} \rho dx = 1$,

$$ \frac{d}{dt} \left( \frac{1}{t+\eta} \int_R vdx + \frac{1}{t+\eta} \int_R (\partial_x v)^2 dx \right) = \frac{1}{t+\eta}, $$

which then yields

$$ \int_{\mathbb{R}} v(x,t)dx \le (t+\eta)\log\frac{t+\eta}{\eta}, \quad (3.43) $$

and

$$ \int_0^t \frac{1}{s+\eta} \int_{\mathbb{R}} (\partial_x v)^2(x,s) dx ds \le \log \frac{t+\eta}{\eta}. \quad (3.44) $$

Using next the imbedding $\|w\| \le c\|w\|^{1/3}\|\partial_x w\|^{2/3}$, we find from (3.43) that, for each $t \in (0, 1)$,

$$ \|v(\cdot, t)\|_{\infty} \le c \left( (t+\eta)\log\frac{t+\eta}{\eta} \right)^{1/3} \|\partial_x v(\cdot, t)\|^{2/3}, $$

and, therefore, in view of (3.44),

$$ \int_0^t \frac{1}{(s+\eta)^2} \frac{1}{\log \frac{s+\eta}{\eta}} \|v(\cdot, s)\|^3 ds \le c \log \frac{t+\eta}{\eta}. $$

Then,

$$ \begin{aligned} \varepsilon \int_0^t \frac{1}{(s+\eta)^{4/3}} \|v(\cdot, s)\|_{\infty} ds &= \varepsilon \int_0^t \left[ \frac{1}{(s+\eta)^{2/3}} \frac{1}{(\log \frac{s+\eta}{\eta})^{1/3}} \|v(\cdot, s)\| \right] \left[ \frac{1}{(s+\eta)^{2/3}} (\log \frac{s+\eta}{\eta})^{1/3} \right] ds \\ &\le c\varepsilon \left(\log \frac{t+\eta}{\eta}\right)^{1/3} \left(\int_0^t \frac{1}{s+\eta} \left(\log \frac{s+\eta}{\eta}\right)^{1/2} ds\right)^{2/3} \\ &\le c\varepsilon \left(\log \frac{t+\eta}{\eta}\right)^{2/3} \le c\varepsilon \log \frac{t+\eta}{\eta}, \end{aligned} $$

and, thus,

$$ \limsup_{\varepsilon, \eta, \varepsilon \log \eta \to 0} \left[ \varepsilon \int_0^t \int_{\mathbb{R}} \partial_x^2 v \chi_\eta dx ds \right] \le 0. \quad (3.45) $$

We continue with the second term in (3.42) and obtain the following sequence of equalities and inequalities.
$$
\begin{align*}
\int_{\mathbb{R}} v \partial_t \chi_{\eta} &= -\frac{2}{3} \frac{1}{(t+\eta)} \int_{\mathbb{R}} v \chi_{\eta} dx - \frac{2}{3} \frac{1}{(t+\eta)^{2/3}} \int_{\mathbb{R}} v \chi' \left( \frac{x}{(t+\eta)^{2/3}} \right) \frac{x}{(t+\eta)^{5/3}} dx \\
&= -\frac{2}{3} \frac{1}{(t+\eta)} \left[ \int_{\mathbb{R}} v \chi_{\eta} dx + \int_{\mathbb{R}} v \partial_x \chi_{\eta} x dx \right] \\
&= -\frac{2}{3} \frac{1}{(t+\eta)} \left[ \int_{\mathbb{R}} v \chi_{\eta} dx - \int_{\mathbb{R}} v \chi_{\eta} dx - \int_{\mathbb{R}} \partial_x v \chi_{\eta} x dx \right] \\
&= \frac{2}{3} \frac{1}{(t+\eta)} \int_{\mathbb{R}} \partial_x v \chi_{\eta} x dx \\
&\le \frac{2}{3} \left[ \frac{1}{8} \int_{\mathbb{R}} (\partial_x v)^2 \chi_{\eta} dx + 2 \int_{\mathbb{R}} \chi_{\eta} \left( \frac{x}{(t+\eta)^{2/3}} \right) \left( \frac{x^2}{(t+\eta)^2} \right) dx \right] \\
&\le \frac{2}{3} \left[ \frac{1}{8} \int_{\mathbb{R}} (\partial_x v)^2 \chi_{\eta} dx + 2 \frac{1}{(t+\eta)^{2/3}} \int_{\mathbb{R}} \chi_{\eta} \left( \frac{x}{(t+\eta)^{2/3}} \right) \left( \frac{x}{(t+\eta)^{2/3}} \right)^2 dx \right] \\
&\le \frac{2}{3} \left[ \frac{1}{8} \int_{\mathbb{R}} (\partial_x v)^2 \chi_{\eta} dx + 2C \frac{1}{(t+\eta)^{2/3}} \right].
\end{align*}
$$

Finally, since

$$
\left| \int_{\mathbb{R}} \frac{x}{t+\eta} \partial_x v \chi_{\eta} dx \right| \le \frac{1}{8} \int_{\mathbb{R}} (\partial_x v)^2 \chi_{\eta} dx + 2 \int_{\mathbb{R}} \frac{x^2}{(t+\eta)^2} \chi_{\eta} dx \le \frac{1}{8} \int_{\mathbb{R}} (\partial_x v)^2 \chi_{\eta} dx + C \frac{1}{(t+\eta)^{2/3}},
$$

we find

$$
\int_{\mathbb{R}} (v\chi_{\eta})(x,t)dx \le C \left[ \int_{0}^{t} \frac{1}{(s+\eta)^{2/3}}ds + \varepsilon \left| \int_{0}^{t} \int_{\mathbb{R}} \partial_{x}^{2}v\chi_{\eta}dxd\right| \right],
$$

and, hence,

$$
\limsup_{\epsilon, \eta, \epsilon \log \eta \to 0} \int_{\mathbb{R}} v \chi_{\eta}(x,t) dx \le C t^{1/3}, \quad (3.46)
$$

and the claim follows.

□

4. THE CONVERGENCE

We are now ready to prove the main result about the convergence of $(u_{\varepsilon,\eta}, \rho_{\varepsilon,\eta})$ to $(\bar{u}, \rho)$.

Although it is possible to show, based on the interior estimates, the local uniform convergence in $\mathbb{R} \times (0, 1]$ of the $u_{\varepsilon,\eta}$'s to $\bar{u}$, the arguments so far do not allow to prove the convergence of the $\rho_{\varepsilon,\eta}$'s to $\bar{p}$ in $L^2(\mathbb{R} \times (0, 1))$ instead of only $L^2_{\text{loc}}(0, 1); L^2(\mathbb{R})$.

We develop here a different argument comparing directly $(u_{\varepsilon,\eta}, \rho_{\varepsilon,\eta})$ and $(\bar{u}, \rho)$ using the classical mean field games identity. We need, however, to deal with the fact that $\bar{p}$ is only Lipschitz continuous and not twice differentiable.

We also mention that we can complete the proof assuming that $\epsilon \log \eta \to 0$. However, since we need to have $\epsilon \le C\eta^\alpha$ for some $\alpha \in (0,1)$ to obtain the a priori bounds on $\int_0^1 \int \rho_{\epsilon,\eta}^2 dxdt$ and $\int_0^1 \int \rho_{\epsilon,\eta} (\partial_x u_{\epsilon,\eta})^2 dxdt$, we state next the main result under this assumption.
**Theorem 4.1.** As $\varepsilon \to 0$ and $\eta \to 0$, while $\varepsilon \le A\eta^\alpha$ for some $A > 0$,

$$u_{\varepsilon, \eta} \to \bar{u} \text{ locally uniformly in } \mathbb{R} \times (0, 1], \quad \rho_{\varepsilon, \eta} \to \bar{\rho} \text{ in } L^2(\mathbb{R} \times (0, 1)) \text{ and } I_{\varepsilon, \eta} \to \bar{I}.$$

*Proof.* The proof of the convergence is based on comparing $(u_{\varepsilon, \eta}, \rho_{\varepsilon, \eta})$ and the solution $(\bar{u}_{\eta}, \bar{\rho}_{\eta})$ of the approximate first-order (2.1), for which know that, as $\eta \to 0$, $\bar{u}_{\eta} \to \bar{u}$ locally uniformly in $\mathbb{R} \times (0, 1]$ and $\bar{\rho}_{\eta} \to \bar{\rho}$ in $L^2(\mathbb{R} \times (0, 1))$.

The standard computation for mean field type systems gives

$$\begin{aligned} & \frac{d}{dt} \int_{\mathbb{R}} (u_{\varepsilon, \eta} - \bar{u}_{\eta}) (\rho_{\varepsilon, \eta} - \bar{\rho}_{\eta}) dx \\ & = \int_{\mathbb{R}} (|\rho_{\varepsilon, \eta} - \bar{\rho}_{\eta}|^2 + \frac{\rho_{\varepsilon, \eta} + \bar{\rho}_{\eta}}{2} |\partial_x u_{\varepsilon, \eta} - \partial_x \bar{u}_{\eta}|^2) dx dt + \varepsilon \int_{\mathbb{R}} (\partial_x^2 \rho (u_{\varepsilon, \eta} - \bar{u}_{\eta}) - \partial_x^2 u (\rho_{\varepsilon, \eta} - \bar{\rho}_{\eta})) dx, \end{aligned}$$

and, since $u_{\varepsilon, \eta}$ is smooth, after integrating by parts

$$\begin{aligned} & \frac{d}{dt} \int_{\mathbb{R}} (u_{\varepsilon, \eta} - \bar{u}_{\eta}) (\rho_{\varepsilon, \eta} - \bar{\rho}_{\eta}) (x, t) dx \\ & = \int_{\mathbb{R}} (|\rho_{\varepsilon, \eta} - \bar{\rho}_{\eta}|^2 + \frac{\rho_{\varepsilon, \eta} + \bar{\rho}_{\eta}}{2} |\partial_x u_{\varepsilon, \eta} - \partial_x \bar{u}_{\eta}|^2) dx dt + \varepsilon \int_{\mathbb{R}} (\partial_x^2 u_{\varepsilon, \eta} \bar{\rho}_{\eta} - \rho_{\varepsilon, \eta} \partial_x^2 \bar{u}_{\eta}) dx. \end{aligned} \quad (4.1)$$

Integrating (4.1) over $t \in (0, 1)$ – recall that both $u_{\varepsilon, \eta}$ and $\bar{u}_{\eta}$ are continuous in $\mathbb{R} \times [0, 1]$ so that integration at $t = 0$ does not create any problem – we get

$$\begin{aligned} & \int_{\mathbb{R}} (u_{\varepsilon, \eta} - \bar{u}_{\eta}) (\rho_{\varepsilon, \eta} - \bar{\rho}_{\eta}) (x, 1) dx - \int_{\mathbb{R}} (u_{\varepsilon, \eta} - \bar{u}_{\eta}) (\rho_{\varepsilon, \eta} - \bar{\rho}_{\eta}) (x, 0) dx \\ & = \int_{0}^{1} \int_{\mathbb{R}} (|\rho_{\varepsilon, \eta} - \bar{\rho}_{\eta}|^2 + \frac{\rho_{\varepsilon, \eta} + \bar{\rho}_{\eta}}{2} |\partial_x u_{\varepsilon, \eta} - \partial_x \bar{u}_{\eta}|^2) dx dt + \varepsilon \int_{0}^{1} \int_{\mathbb{R}} (\partial_x^2 u_{\varepsilon, \eta} \bar{\rho}_{\eta} - \rho_{\varepsilon, \eta} \partial_x^2 \bar{u}_{\eta}) dx, \end{aligned} \quad (4.2)$$

and, thus,

$$\begin{aligned} & \int_{0}^{1} \int_{\mathbb{R}} (|\rho_{\varepsilon, \eta} - \bar{\rho}_{\eta}|^2 + \frac{\rho_{\varepsilon, \eta} + \bar{\rho}_{\eta}}{2} |\partial_x u_{\varepsilon, \eta} - \partial_x \bar{u}_{\eta}|^2) dx dt \\ & = \int_{\mathbb{R}} (u_{\varepsilon, \eta} - \bar{u}_{\eta}) (\rho_{\varepsilon, \eta} - \bar{\rho}_{\eta}) (x, 1) dx - \int_{\mathbb{R}} (u_{\varepsilon, \eta} - \bar{u}_{\eta}) (\rho - \bar{\rho}_{\eta}) (x, 0) dx \quad (4.3) \\ & \quad + \varepsilon \int_{0}^{1} \int_{\mathbb{R}} (\rho_{\varepsilon, \eta} \partial_x^2 \bar{u}_{\eta} - \partial_x^2 u_{\varepsilon, \eta} \bar{\rho}_{\eta}) dx. \end{aligned}$$

The claim about the convergence of the $u_{\varepsilon, \eta}$'s and $\rho_{\varepsilon, \eta}$'s will follow, if we show that the right hand side of (4.3) tends to 0 as $\varepsilon, \eta \to 0$ while $\varepsilon \le A\eta^\alpha$, and we proceed to prove this.

First we observe that, since $\rho_{\varepsilon, \eta}(\cdot, 1) = \bar{\rho}_{\eta}(\cdot, 1) = \delta$ and $u_{\varepsilon, \eta}(x, 0) = \bar{u}_{\eta}(x, 0) = x^2/(2\eta)$,

$$\int_{\mathbb{R}} (u_{\varepsilon, \eta} - \bar{u}_{\eta}) (\rho_{\varepsilon, \eta} - \bar{\rho}_{\eta}) (x, 1) dx - \int_{\mathbb{R}} (u_{\varepsilon, \eta} - \bar{u}_{\eta}) (\rho_{\varepsilon, \eta} - \bar{\rho}_{\eta}) (x, 0) dx = 0. \quad (4.4)$$

Thus, to conclude we must show that

$$\lim_{\varepsilon \to 0, \eta \to 0, \varepsilon \le A\eta^\alpha} \varepsilon \int_0^1 \int_{\mathbb{R}} (\rho_{\varepsilon, \eta} \partial_x^2 \bar{u}_{\eta} - \partial_x^2 u_{\varepsilon, \eta} \bar{\rho}_{\eta}) dx = 0. \quad (4.5)$$

In view of the construction of $\bar{u}_{\eta}$, we know that $\partial_x^2 \bar{u}_{\eta}(x, t) \approx 1/t+\eta$, and, hence, since $\int_{\mathbb{R}} \rho_{\varepsilon, \eta}(x, t) dx = 1$,

$$\varepsilon \int_{0}^{1} \int_{\mathbb{R}} \rho_{\varepsilon, \eta}(x, t) \partial_{x}^{2} \bar{u}_{\eta}(x, t) dx dt \approx \varepsilon \int_{0}^{1} \frac{1}{t+\eta} dt = \varepsilon [\log(1+\eta) - \log\eta], \quad (4.6)$$
and

$$ \lim_{\epsilon \to 0, \eta \to 0, \epsilon \le A\eta^\alpha} \epsilon[\log(1+\eta) - \log\eta] = 0. \quad (4.7) $$

We turn now to the term $\epsilon \int_0^1 \int_{\mathbb{R}} \partial_x^2 u_{\epsilon, \eta}(x, t) \bar{p}_\eta(x, t) dx dt$. Its analysis is more complicated consisting on splitting the time integration on integrals over $(0, h)$, $(h, 1-h)$ and $(1-h, 1)$ and using different arguments for each one.

We begin with the integral over $(1-h, 1)$ and recall that $\bar{p}_\eta$'s are supported in $[-r_{\eta,0}, r_{\eta,0}] \times [0, 1]$, where $r_{\eta,0}$ is the minimum of $r_\eta$. Moreover, in view of Proposition 3.4, the $u_{\epsilon, \eta}$'s are uniformly bounded in $L^2((\theta, 1); H^1(J_\eta))$ with $J_\eta = [-r_{\eta,0}, r_{\eta,0}]$ and $\theta \in (0, 1)$.

In view of the uniform bound of the $u$'s in $L^2((1/4, 1/2); H^1(J_\eta))$, there exists some $t_{\epsilon, \eta} \in (1/4, 1/2)$ such that, for some $C > 0$,

$$ \|\partial_x u_{\epsilon, \eta}(\cdot, t_{\epsilon, \eta})\|^2 \le C. \quad (4.8) $$

Next we multiply the $u_{\epsilon, \eta}$ equation by $\epsilon \partial_x^2 u_{\epsilon, \eta}$ and integrate over $\mathbb{R} \times (t_{\epsilon, \eta}, 1)$ to get

$$ \begin{aligned} & \frac{\epsilon}{2} \int_{\mathbb{R}} (\partial_x u_{\epsilon, \eta})^2(x, 1) dx + \|\epsilon \partial_x^2 u_{\epsilon, \eta}\|_{L^2(J_\eta \times (t_{\epsilon, \eta}, 1))}^2 \\ & \le \|\rho_{\epsilon, \eta}\|_{L^2(\mathbb{R} \times (1/2, 1))} \|\epsilon \partial_x^2 u_{\epsilon, \eta}\|_{L(J_\eta \times (t_{\epsilon, \eta}, 1))} + \frac{\epsilon}{2} \int_{J_\eta} (\partial_x u_{\epsilon, \eta})^2(x, t_{\epsilon, \eta}) dx. \end{aligned} $$

It then follows from (4.8) that, for some $C > 0$,

$$ \|\epsilon \partial_x^2 u_{\epsilon, \eta}\|_{L^2(J_\eta \times (1/2, 1))}^2 \le \|\epsilon \partial_x^2 u_{\epsilon, \eta}\|_{L(J_\eta \times (t_{\epsilon, \eta}, 1))} \le C. $$

Then

$$ \begin{aligned} \epsilon \left| \int_{1-h}^{1} \int_{\mathbb{R}} \bar{p}_{\eta} \partial_x^2 u_{\epsilon, \eta} dx dt \right| & \le \left( \int_{1-h}^{1} \int_{\mathbb{R}} \bar{p}_{\eta}^2 dx dt \right)^{1/2} dt \right)^{1/2} \left( \int_{1-h}^{1} \int_{\mathbb{R}} \epsilon^2 (\partial_x^2 u_{\epsilon, \eta})^2 dx dt \right)^{1/2} \\ & \le C \left( \int_{1-h}^{1} \int_{\mathbb{R}} \bar{p}_{\epsilon, \eta}^2 dx dt \right)^{1/2} = o(1), \end{aligned} $$

the last claim following from the fact that the $\bar{p}$'s are uniformly bounded in $L^2(\mathbb{R} \times (0, 1))$.

For the integral over $(h, 1-h)$ we observe that, since $u_{\epsilon, \eta}$ is smooth and $\bar{p}_\eta$ is Lipschitz and has compact support,

$$ \begin{aligned} \epsilon \left| \int_{1-h}^{1} \int_{\mathbb{R}} \partial_x^2 u_{\epsilon, \eta}(x, t) \bar{p}_\eta(x, t) dx dt \right| &= \epsilon \left| \int_{1-h}^{1} \int_{\mathbb{R}} \partial_x u_{\epsilon, \eta}(x, t) \partial_x \bar{p}_\eta(x, t) dx dt \right| \\ &\le \epsilon C \left( \int_{h}^{1-h} \|u_{\epsilon, \eta}(\cdot, t)\|_{H^1(J_\eta)}^2 dt + \int_{h}^{1-h} \int_{\mathbb{R}} (\partial_x \bar{p}_\eta)^2 dx dt \right), \end{aligned} $$

and, hence,

$$ \lim_{\epsilon \to 0} \epsilon \int_{1-h}^{1} \int_{\mathbb{R}} \partial_x^2 u_{\epsilon, \eta}(x, t) \bar{p}_\eta(x, t) dx dt = 0 \quad \text{uniformly in } h \in (0, 1/2), \quad (4.9) $$

since, in view of Proposition 3.4 and Proposition 3.5, for each $\theta \in (0, 1/2)$, $u_{\epsilon, \eta} \in L^2((\theta, 1); H_{\text{loc}}^1)$ and $\bar{p}_\eta \in L^2((\theta, 1-\theta); H^1)$.

To conclude, we need to investigate the behavior of $\epsilon \int_0^h \int_{\mathbb{R}} \partial_x^2 u_{\epsilon, \eta} \bar{p}_\eta dx dt$, and for this we recall that, if

$$ u_{\epsilon, \eta}(x, t) = v_{\epsilon, \eta}(x, t) + \frac{x^2}{2(t+\eta)} + \epsilon \log\left(\frac{t+\eta}{\eta}\right), $$
then

$$ \partial_t v_{\varepsilon, \eta} + \frac{1}{2} (\partial_x v_{\varepsilon, \eta})^2 - \varepsilon \partial_x^2 v_{\varepsilon, \eta} + \frac{x}{t+\eta} \partial_x v_{\varepsilon, \eta} = \rho_{\varepsilon, \eta} \quad \text{and} \quad v_{\varepsilon, \eta}(\cdot, 0) = 0 \quad \text{and} \quad v_{\varepsilon, \eta} \ge 0. $$

Then,

$$ \begin{aligned} \varepsilon \int_0^h \int_{\mathbb{R}} \partial_x^2 u_{\varepsilon, \eta} \bar{p}_\eta dxdt &= \varepsilon \int_0^h \int_{\mathbb{R}} \partial_x^2 v_{\varepsilon, \eta} \bar{p}_\eta dxdt + \varepsilon \int_0^h \int_{\mathbb{R}} \frac{1}{t+\eta} \bar{p}_\eta dxdt \\ &= \varepsilon \int_0^h \int_{\mathbb{R}} \partial_x^2 v_{\varepsilon, \eta} \bar{p}_\eta dxdt + \log\left(\frac{t+\eta}{\eta}\right). \end{aligned} $$

To finish, we need to estimate $\varepsilon \int_0^h \int_{\mathbb{R}} \partial_x^2 v_{\varepsilon, \eta} \bar{p}_\eta dxdt$.

Recalling that $\bar{p}_\eta = r_\eta(t)(1 - \frac{x^2}{l_\eta^2(t)})_+$, we find

$$ \begin{aligned} \varepsilon \int_0^h \int_{\mathbb{R}} \partial_x^2 v_{\varepsilon, \eta} \bar{p}_\eta dxdt &= \varepsilon \int_0^h \int_{\mathbb{R}} \partial_x v_{\varepsilon, \eta} \partial_x \bar{p}_\eta dxdt \\ &= \varepsilon \int_0^h \left[ -2v_{\varepsilon, \eta}(l_\eta(t), t) \frac{r_\eta(t)}{l_\eta(t)} + 2v_{\varepsilon, \eta}(-l_\eta(t), t) \frac{r_\eta(t)}{l_\eta(t)} \right] dt + 2\varepsilon \int_0^h \frac{r_\eta(t)}{l_\eta^2(t)} \int_{\{|x| \le l_\eta(t)\}} v_{\varepsilon, \eta}(x,t) dxdt, \end{aligned} $$

and, hence, in view of the asymptotic behavior of the $r_\eta$'s and $l_\eta$'s near $t = 0$,

$$ |\varepsilon \int_0^h \int_{\mathbb{R}} \partial_x^2 v_{\varepsilon, \eta} \bar{p}_\eta dxdt| \le C \int_0^h \frac{1}{(t+\eta)^{4/3}} \|v_{\varepsilon, \eta}(\cdot, t)\|_{\infty} dt. $$

Finally, we need to estimate the right hand side of this last inequality, which have already done in the proof of Proposition 3.9.

It follows from the equation satisfied by $v_{\varepsilon, \eta}$ that

$$ \frac{d}{dt} \left( \frac{1}{t+\eta} \int_{\mathbb{R}} v_{\varepsilon, \eta}(x,t) dx \right) + \frac{1}{2(t+\eta)} \int_{\mathbb{R}} (\partial_x v_{\varepsilon, \eta})^2(x,t) dx = \frac{1}{t+\eta}, $$

and, hence,

$$ \int_{\mathbb{R}} v_{\varepsilon, \eta}(x,t) dx \le (t+\eta) \log\left(\frac{t+\eta}{\eta}\right), \quad (4.10) $$

and

$$ \int_0^t \frac{1}{s+\eta} \int_{\mathbb{R}} (\partial_x v_{\varepsilon, \eta})^2(x,s) dxds \le \log\left(\frac{t+\eta}{\eta}\right). \quad (4.11) $$

Using next the imbedding $\|w\| \le c\|w\|^{1/3}\|\partial_x w\|^{2/3}$, we find from (4.10) that, for each $t \in (0, 1)$,

$$ \|v_{\varepsilon, \eta}(\cdot, t)\|_{\infty} \le c \left( (t+\eta) \log \frac{t+\eta}{\eta} \right)^{1/3} \|\partial_x v_{\varepsilon, \eta}(\cdot, t)\|^{2/3}, $$

and, therefore, in view of (4.11),

$$ \int_0^t \frac{1}{(s+\eta)^2} \frac{1}{\log \frac{s+\eta}{\eta}} \|v_{\varepsilon, \eta}(\cdot, s)\|^3 ds \le c \log \frac{t+\eta}{\eta}. $$
Then, for some $c > 0$,

$$
\begin{align*}
& \varepsilon \int_0^t \frac{1}{(s+\eta)^{4/3}} \|v_{\varepsilon,\eta}(\cdot, s)\|_{\infty} ds \\
&= \varepsilon \int_0^t \left[ \frac{1}{(s+\eta)^{2/3}} \frac{1}{(\log \frac{s+\eta}{\eta})^{1/3}} \|v_{\varepsilon,\eta}(\cdot, s)\| \right] \left[ \frac{1}{(s+\eta)^{2/3}} (\log \frac{s+\eta}{\eta})^{1/3} \right] ds \\
&\le c\varepsilon \left(\log \frac{t+\eta}{\eta}\right)^{1/3} \left(\int_0^t \frac{1}{s+\eta} \left(\log \frac{s+\eta}{\eta}\right)^{1/2} ds\right)^{2/3} \\
&\le c\varepsilon \left(\log \frac{t+\eta}{\eta}\right)^{2/3} \le c\varepsilon \log \frac{t+\eta}{\eta}.
\end{align*}
$$

Summarizing the above arguments we find that

$$
|\varepsilon \int_0^h \int_{\mathbb{R}} \partial_x^2 u_{\varepsilon, \eta} \bar{p}_{\eta} dxdt| \leq C \varepsilon \log\left(\frac{h+\eta}{\eta}\right),
$$

and, since $\varepsilon \le A\eta^\alpha$,

$$
\lim_{\varepsilon \to 0, \eta \to 0, \eta \le A\eta^\alpha} |\varepsilon \int_0^h \int_{\mathbb{R}} \partial_x^2 u_{\varepsilon, \eta} \bar{p}_\eta dxdt| \le \lim_{\varepsilon \to 0, \eta \to 0, \eta \le A\eta^\alpha} C\varepsilon \log\left(\frac{h+\eta}{\eta}\right) = 0.
$$

The final item to show is the convergence of the $I_{\varepsilon, \eta}$'s. This follows from the facts that,

$$
\lim_{\varepsilon \to 0, \eta \to 0, \eta \le A\eta^\alpha} \int_0^1 \int_{\mathbb{R}} (\rho_{\varepsilon, \eta} - \bar{p}_\eta)^2 dxdt = 0 \quad \text{and} \quad \lim_{\eta \to 0} \int_0^1 \int_{\mathbb{R}} (\bar{p}_\eta - \bar{p})^2 dxdt = 0,
$$

and the observation that

$$
\lim_{\varepsilon \to 0, \eta \to 0, \eta \le A\eta^\alpha} \int_0^1 \int_{\mathbb{R}} (\rho_{\varepsilon, \eta} + \bar{p}_\eta) \left( \partial_x u_{\varepsilon, \eta} - \partial_x \bar{u}_\eta \right)^2 dxdt = 0
$$

and

$$
\lim_{\eta \to 0} \int_0^1 \int_{\mathbb{R}} \left( \bar{p}_{\eta} \left( \partial_x \bar{u}_{\eta} \right)^2 - \bar{p} \left( \partial_x \bar{u} \right)^2 \right) dxdt = 0
$$

yield that

$$
\lim_{\varepsilon \to 0, \eta \to 0, \eta \le A\eta^\alpha} \int_0^1 \int_{\mathbb{R}} \left( \rho_{\varepsilon, \eta} \left( \partial_x u_{\varepsilon, \eta} \right)^2 - \bar{p} \left( \partial_x \bar{u} \right)^2 \right) dxdt = 0.
$$

□

5. THE KPZ ASYMPTOTICS

As it was discussed in the introduction the original motivation of this work was to provide a complete and rigorous proof to the large deviations-type (LDP) approximation for the KPZ equation for large heights and sharp-wedge initial condition, a topic that has attracted considerable attention in the math physics community.

We begin with a short and informal introduction to the problem.

We consider the following KPZ problem

$$
\partial_t h - \partial_x^2 h + \frac{1}{2} (\partial_x h)^2 = \sqrt{\theta} \xi \quad \text{in } \mathbb{R} \times (0, 1) \quad h(x, 0) = x^2/2L. \qquad (5.1)
$$

For any $\rho \in L^2(\mathbb{R} \times (0, T))$, it is known that

$$
\mathbb{P}(\sqrt{\theta}\xi \approx \rho) \approx \|\rho\|_2^2 = \int_0^1 \int_{\mathbb{R}} \rho^2(x,t) \, dxdt.
$$
Then, in $\{\sqrt{\theta}\xi \approx \rho\}$, the height $h$ is approximated by

$$ \partial_t h - \partial_x^2 h + \frac{1}{2}(\partial_x h)^2 = \rho \quad \text{in } \mathbb{R} \times (0, 1] \quad h(x, 0) = |x|^2/L. \qquad (5.2) $$

The LDP theory says that, for large heights and sharp wedge initial condition, $h_\theta$ satisfies a large deviation principle if, as $\lambda \to \infty$ and $L \to 0$,

$$ h \approx \exp[-I(\lambda, L)] = \exp[-I + o(1)], $$

where

$$ \begin{aligned} I(\lambda, L) &= \sup \left\{ -\frac{1}{2} \|\rho\|_2^2 \right. \\ &= -\frac{1}{2} \int_0^1 \int_{\mathbb{R}} (\partial_t h - \frac{1}{2} \partial_{xx} h + \frac{1}{2} (\partial_x h)^2)^2 dx dt : h \text{ solves (5.2) and } h(0, 1) = \lambda \right\}. \end{aligned} \qquad (5.3) $$

A variational argument shows that a minimizer in (5.3) solves the system

$$ \left\{ \begin{array}{l} \partial_t u^{\lambda,L} - \partial_x^2 u^{\lambda,L} + \frac{1}{2} (\partial_x u^{\lambda,L})^2 = \rho^{\lambda,L} \quad \text{in } \mathbb{R} \times (0, 1], \\ \partial_t \rho^{\lambda,L} + \partial_x^2 \rho^{\lambda,L} + \partial_x (\partial_x u^{\lambda,L} \rho^{\lambda,L}) = 0 \quad \text{in } \mathbb{R} \times [0, 1), \\ \rho^{\lambda,L}(\cdot, 1) = c(\lambda)\delta \quad u^{\lambda,L}(0, 1) = \lambda \quad u^{\lambda,L}(x, 0) = |x|^2/L. \end{array} \right. \qquad (5.4) $$

The terminal value of $\rho^{\lambda,L}$ is due to the constraint that $u^{\lambda,L}(0, 1) = \lambda$ and $c(\lambda)$ is an appropriate constant.

The limiting behavior of $I(\lambda, L)$ as $\lambda \to \infty$ and $L \to 0$ is related to the behavior of the solution of the KPZ equation for $\lambda \to \infty$ and $L \to 0$.

To understand this behavior it is convenient to introduce the rescaling

$$ u_{\lambda,L}(x,t) = \frac{1}{\lambda} u^{\lambda,L}(\lambda^{1/2} x, t) \quad \text{and} \quad \rho_{\lambda,L}(x,t) = \frac{1}{\lambda} \rho^{\lambda,L}(\lambda^{1/2} x, t), $$

which leads to the forward-backwards system

$$ \begin{aligned} \partial_t u_{\lambda,L} - \partial_x^2 u_{\lambda,L} + \frac{1}{2} (\partial_x u_{\lambda,L})^2 &= \rho \lambda, L \quad \text{in } \mathbb{R} \times (0, 1), \\ \partial_t \rho_{\lambda,L} + \partial_x^2 \rho_{\lambda,L} + \partial_x (u_{\lambda,L} \rho_{\lambda,L}) &= 0 \quad \text{in } \mathbb{R} \times (0, 1), \end{aligned} \qquad (5.5) $$

$$ u_{\lambda,L}(x,0) = |x|^2/L, \quad \rho_{\lambda,L}(\cdot, 1) = \lambda^{-2/3} c(\lambda) \delta \quad \text{and} \quad u_{\lambda,L}(0,1) = 1. $$

We show next how to obtain the. $\lambda \to \infty$, $L \to 0$ behavior of $u_{\lambda,L}$ and $\rho_{\lambda,L}$ using what was already established in the paper.

We choose $a > 0$ such that the “physics solution” $\bar{u}_a$, which corresponds to (1.2) with $\bar{p}(\cdot, 1) = a\delta$, has the property that $\bar{u}_a(0, 1) = 1$. This can be accomplished by the appropriate choices of $r, k$ and $a$. To simplify the notation, in what follows we assume that $a = 1$.

We consider now, for $\varepsilon, \eta > 0$ such that $\varepsilon \le A\eta^\alpha$ the solution $(u_{\varepsilon,\eta}, \rho_{\varepsilon,\eta})$ of (1.1), and recall that we have shown that $(u_{\varepsilon,\eta}, \rho_{\varepsilon,\eta})$ converges in the appropriate sense to the solution $(\bar{u}, \bar{p})$ of (1.2).

Let $A(\varepsilon, \eta) = u_{\varepsilon,\eta}(0, 1)$. It follows from the convergence result that

$$ \lim_{\varepsilon \to 0, \eta \to 0, \varepsilon \le A\eta^\alpha} A(\varepsilon, \eta) = 1. \qquad (5.6) $$

Scaling (1.1) by $\mu > 0$ changes $(u_{\varepsilon,\eta}, \rho_{\varepsilon,\eta})$ to a new solution $(u_{\varepsilon',\eta}, \rho_{\varepsilon',\eta})$ for $\varepsilon' = \varepsilon\mu^\alpha$ with $\rho_{\varepsilon,\eta}(\cdot, 1) = \mu^\beta \delta$ and $u_{\varepsilon',\eta}(0, 1) = A(\varepsilon, \eta)\mu^\gamma$.
We choose $\mu = A(\varepsilon, \eta)^{-1/\gamma} \to 1$ as $\varepsilon \to 0$, $\eta \to 0$, $\varepsilon \le A\eta^\alpha$. This gives a solution to the KPZ system for $\eta > 0$, $\varepsilon' = \varepsilon\mu^\alpha = A(\varepsilon, \eta)^{-\alpha/\gamma}$; note that $\lim_{\varepsilon \to 0, \eta \to 0, \varepsilon \le A\eta^\alpha} \varepsilon'/\varepsilon = 1$.

Finally, since obviously $A(\varepsilon, \eta)$ is continuous in $\varepsilon$ for each $\eta$, to obtain a solution of (1.1), we choose $\varepsilon = \varepsilon'r$ and want to solve for a solution such that $A(\varepsilon'r, \eta)^{-\alpha/\gamma} = 1$.

Observe next that, for any $\theta, \theta' \in (0, 1)$,

$$ \lim_{\varepsilon' \to 0, \eta \to 0, \varepsilon' \le A\eta^\alpha} \theta A(\varepsilon'\theta, \eta) = \theta < 1 \quad \text{and} \quad \lim_{\varepsilon' \to 0, \eta \to 0, \varepsilon' \le A\eta^\alpha} \frac{1}{\theta'} A(\varepsilon'\frac{1}{\theta'}, \eta) = \frac{1}{\theta'} > 1. $$

Then the claimed existence follows from a standard continuity argument.

Note that

$$ \rho(\cdot, 1) = \mu^\beta \delta = A(\varepsilon, \eta)^{-\beta/\gamma} \delta \to c_0 \delta $$

which implies that

$$ \lim_{\lambda \to \infty} \lambda^{-2/3} c(\lambda) \to c_0. $$

**Statements and Declarations.** The authors state that there are no conflicts of interest. Data sharing is not applicable to this article as no datasets were generated or analyzed during the current study.

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COLLEGE DE FRANCE, 11 PL. MARCELIN BERTHELOT, 75231 PARIS, FRANCE

DEPARTMENT OF MATHEMATICS, UNIVERSITY OF CHICAGO, CHICAGO, ILLINOIS 60637, USA