Optimal convergence rate for homogenization of convex Hamilton–Jacobi equations in the periodic spatial-temporal environment

Abstract

We study the optimal convergence rate for the homogenization problem of convex Hamilton–Jacobi equations when the Hamitonian is periodic with respect to spatial and time variables, and notably time-dependent. We prove a result similar to that of (Tran and Yu (2021)), which means the optimal convergence rate is also $O (ε)$ .

Keywords

Hamilton–Jacobi equations homogenization spatio-temporal periodic setting optimal convergent rate viscosity solutions

1. Introduction

1.1. Settings

For each $ε > 0$ , let $u^{ε} \in C (R^{n} \times [0, \infty))$ be an unique viscosity solution of the following Cauchy problem $\begin{matrix} (1.1) & \{\begin{array}{ll} u_{t}^{ε} + H (\frac{x}{ε}, \frac{t}{ε}, D u^{ε}) = 0 & in R^{n} \times (0, \infty), \\ u^{ε} (x, 0) = g (x) & on R^{n} . \end{array} \end{matrix}$

The Hamiltonian $H : R^{n} \times R \times R^{n} \to R$ is a given continuous function satisfying:

for each $p \in R^{n}$ , the map $(x, t) \mapsto H (x, t, p)$ is $Z^{n + 1}$ -periodic, which means $\begin{matrix} H (x, t, p) = H (x + u, t + v, p) \forall (u, v) \in Z^{n} \times Z; \end{matrix}$

there exist some positive constants $α_{0}$ , $β_{0}$ , $K_{0}$ and $m_{0} > 1$ such that $\begin{matrix} α_{0} | p |^{m_{0}} - K_{0} ⩽ H (x, t, p) ⩽ β_{0} | p |^{m_{0}} + K_{0}; \end{matrix}$

for $(x, t) \in R^{n + 1} / Z^{n + 1}$ , the map $p \mapsto H (x, t, p)$ is convex.

We assume the initial data $g \in BUC (R^{n}) \cap LIP (R^{n})$ , where $BUC (R^{n})$ and $LIP (R^{n})$ are respectively the set of bounded uniformly continuous, and Lipschitz functions, on $R^{n}$ . By those assumptions, $u^{ε}$ converges locally uniformly on $R^{n} \times [0, \infty)$ to a function u as $ε \to 0$ , and u solves the effective equation $\begin{matrix} (1.2) & \{\begin{array}{ll} u_{t} + \overline{H} (D u) = 0 & in R^{n} \times (0, \infty), \\ u (x, 0) = g (x) & on R^{n}, \end{array} \end{matrix}$ with $\overline{H}$ is defined as follows. For each $p \in R^{n}$ , there is a unique constant $\overline{H} (p)$ such that $\begin{matrix} v_{s} + H (y, s, p + D v) = \overline{H} (p) \end{matrix}$ has a continuous viscosity solution; see [5,7,9] for more details.

Condition (P2) is natural when the given Hamiltionian is time-dependent, see [5]. Also, there are some positive constants α, β, K and $m > 1$ satisfying $\begin{matrix} α | v |^{m} - K ⩽ L (x, t, v) ⩽ β | v |^{m} + K . \end{matrix}$ Here, $L (x, t, v)$ is the Legendre transform of $H (x, t, p)$ defined by $\begin{matrix} L (x, t, v) = sup_{p \in R^{n}} (p \cdot v - H (x, t, p)) \end{matrix}$ and $\frac{1}{m} + \frac{1}{m_{0}} = 1$ . The optimal control formula for solution of (1.1) is $\begin{matrix} u^{ε} (x, t) = inf_{\begin{array}{c} ε η (0) = x \\ η \in AC ([- ε^{- 1} t, 0], R^{n}) \end{array}} {g (ε η (- ε^{- 1} t)) + ε \int_{- ε^{- 1} t}^{0} L (η (s), s, \dot{η} (s) d s)} \end{matrix}$ where $AC ([- ε^{- 1} t, 0], R^{n})$ is the space of absolutely continuous curves in $R^{n}$ defined on $[- ε^{- 1} t, 0]$ .

1.2. Previous literature

We give a brief review on finding optimal convergence rate for homogenization of Hamilton–Jacobi equations.

If one consider general Hamiltonian, the best known convergence rate is $O (ε^{1 / 3})$ by Capuzzo-Dolcetta and Ishii [2]; see a detailed explanation in [9]. It only relies on classical methods, hence one needs further assumptions on Hamiltonian, such as convexity or Lipschitz property, to improve the result.

When the Hamiltonian is convex, the lower bound $u^{ε} - u ⩾ C ε$ always holds by a result of Mitake, Tran and Yu [8], and a similar upper bound is also true for $n ⩽ 2$ . Besides, if we consider multi-scaled Hamiltonian in some one-dimensional settings, Tu also showed that the convergence rate is $O (ε)$ in [11].

When $n ⩾ 3$ , by using idea from first passage percolation method, Cooperman pointed out an unconditional upper bound $| u^{ε} (x, t) - u (x, t) | ⩽ C ε log (C + t ε^{- 1})$ , see [3].

Later on, Tran and Yu established the $O (ε)$ convergence rate [10]; their proof based on a lemma on periodic metrics in Riemannian geometry by Burago [1]. This is unexpected since recent attempts at studying homogenization problems usually base on ideas from (weak) KAM theory, which originates from dynamical systems. It is likely that Burago’s result could be adapted for more complicated situations, such as time-dependent or multi-scaled convex Hamiltonians.

The borderline case $m = 1$ corresponds to a front propagation problem, which is studied by Jing, Tran and Yu in [6]. Combining with the curve cutting lemma of Burago [1], one might also obtain the convergence rate $O (ε)$ .

1.3. Main results

Theorem 1.1.
Let $u^{ε}$ be the viscosity solution to ( 1.1 ) and u be the viscosity solution to ( 1.2 ). Then, there exists $C > 0$ depending only on H, $‖ D g ‖_{L^{\infty} (R^{n})}$ and n such that, for each $ε \in (0, 1)$ , $\begin{matrix} {‖ u^{ε} - u ‖}_{L^{\infty} (R^{n} \times (0, \infty))} ⩽ C ε . \end{matrix}$

When $n = 1$ , Proposition 4.3 of [8] confirms the optimality of our result by providing an example such that $u^{ε} ⩾ C ε$ converges to $u \equiv 0$ . Moreover, its idea could be extended to become an example in n-dimensional settings by mimicking the assumed conditions there for each coordinate of the given Hamiltonian.

The proof of Theorem 1.1 generally follows strategies of [10], but there are some new contributions. Since L depends on the time variable, one cannot apply exactly the same techniques used for the time-independent case, which requires boundedness of velocity of the minimizing curves, a key property which does not hold in general. Additionally, as one might see in the proof of Lemma 2.3, rescaling and periodic shifting processes in our setting are far more complicated. Periodic variables could only be shifted via integer vectors, hence if those previous ideas are reused, one needs to carefully take care of subcurves decomposed from the minimizers and connectors joining them.

We want to further emphasize that the condition (P1) could be relaxed, which means spatial and temporal variables could follow different periodicity. Here, we only consider the $Z^{n + 1}$ -periodic setting to reduce unnecessary complications arising in estimations of various integrals.
2. Proof of the main result

Before starting, we restate an important lemma:

Lemma 2.1 (Burago, [1]).

Let $n \in Z^{+}$ and $ξ : [0, 1] \to R^{n}$ be a continuous path. Then, there is a positive integer $k ⩽ \frac{n + 1}{2}$ and a collection of disjoint intervals ${[a_{i}, b_{i}]}_{1 ⩽ i ⩽ k}$ , which are subintervals of $[0, 1]$ , such that $\begin{matrix} \sum_{i = 1}^{k} (ξ (b_{i}) - ξ (a_{i})) = \frac{ξ (1) - ξ (0)}{2} . \end{matrix}$

Note that Burago’s lemma still holds for a path defined on $[a, b]$ by a simple rescaling; this observation will be useful later. Now for $x, y \in R^{n}$ and $t > 0$ , denote by $\begin{matrix} m (t, x, y) = inf {\int_{0}^{t} L (η (s), s, \dot{η} (s)) d s | η \in AC ([0, t], R^{n}), η (0) = x, η (t) = y} . \end{matrix}$

One might imagine $m (t, x, y)$ as the minimum cost to travel from x to y in a given time $t > 0$ . The homogenized cost is $\begin{matrix} \overline{m} (t, x, y) = lim_{k \to \infty} \frac{1}{k} m (k t, k x, k y) . \end{matrix}$

Thanks to the Hopf–Lax formula for (1.2), for any $t > 0$ , $\begin{matrix} \overline{m} (t, x, y) = t \overline{L} (\frac{y - x}{t}), \end{matrix}$ where $\overline{L}$ is the Legendre transform of the effective Hamiltonian $\overline{H}$ . For any $s > 0$ , $\begin{matrix} \overline{m} (s t, s x, s y) = s t \overline{L} (\frac{s (y - x)}{s t}) = s \overline{m} (t, x, y) . \end{matrix}$

Lemma 2.2.

With all above assumptions, the following properties hold:

m is $Z^{n}$ -periodic: for $x, y \in R^{n}$ , $w \in Z^{n}$ and $t > 0$ , $\begin{matrix} m (t, x + w, y + w) = m (t, x, y) . \end{matrix}$

For $t > 0$ and $| y | ⩽ M t$ : $\begin{matrix} m (2 t, 0, 2 y) ⩽ 2 m (t, 0, y) + C . \end{matrix}$ More generally speaking, for $p, q > 0$ , $\begin{matrix} m ((p + q) t, 0, (p + q) y) ⩽ m (p t, 0, p y) + m (q t, 0, q y) + C . \end{matrix}$

The homogenized cose $\overline{m} (t, x, y)$ exists. In particular, $\begin{matrix} \overline{m} (t, 0, y) ⩽ m (t, 0, y) + C . \end{matrix}$

Here,

C > 0

is a universal constant depending only on K, M, n, α and β.

Proof.

A minimizer $η_{1}$ of $m (t, x, y)$ induces a path $η_{2}$ from $x + w$ to $y + w$ by shifting. From the $Z^{n + 1}$ -periodicity of $L (y, t, q)$ , we deduce: $\begin{aligned} m (t, x, y) & = \int_{0}^{t} L (η_{1} (s), s, \dot{η_{1}} (s)) d s = \int_{0}^{t} L (η_{1} (s) + w, s, \dot{η_{1}} (s)) d s \\ = \int_{0}^{t} L (η_{2} (s), s, \dot{η_{2}} (s)) d s ⩾ m (t, x + w, y + w) . \end{aligned}$ The reverse inequality is obtained by shifting backward.

For each $t > 0$ and $y \in R^{n}$ with $| y | ⩽ M t$ , consider $η \in AC ([0, t], R^{n})$ joining 0 and y with constant velocity. From the polynomial growth rate of $L (., q)$ , $\begin{matrix} m (t, 0, y) ⩽ \int_{0}^{t} (β \cdot \frac{| y |^{m}}{t^{m}} + K) d v ⩽ β \cdot \frac{{(M t)}^{m}}{t^{m - 1}} + K t ⩽ C t . \end{matrix}$ Similarly $- C t ⩽ m (t, 0, y)$ . Hence the theorem is true if $t ⩽ 6$ , since we will get $\begin{matrix} m (2 t, 0, 2 y) - 2 m (t, 0, y) ⩽ 2 C t + 2 C t ⩽ C . \end{matrix}$ Now only consider $t > 6$ . Let k be the greatest positive integer such that $6 k ⩽ t$ . Also, if we denote the minimizer of $m (t, 0, y)$ by η, then $\begin{aligned} \sum_{l = 0}^{k - 1} \int_{l}^{l + 6} L (η (s), s, \dot{η} (s)) d s & = \int_{0}^{t} L (η (s), s, \dot{η} (s)) d s - \int_{6 k}^{t} L (η (s), s, \dot{η} (s)) d s \\ ⩽ m (t, 0, y) - \int_{6 k}^{t} (α {| \dot{η} (s) |}^{m} - K) d s \\ ⩽ C t + (t - 6 k) K ⩽ C t + C ⩽ C t . \end{aligned}$ Then, there exists $l \in {0, 1, \dots, k - 1}$ satisfying $\begin{matrix} \int_{l}^{l + 6} L (η (s), s, \dot{η} (s)) d s ⩽ \frac{C t}{k} ⩽ C . \end{matrix}$ From our assumption on $L (., v)$ , we obtain $\begin{matrix} \int_{l}^{l + 6} (α {| \dot{η} (s) |}^{m} - K) d s ⩽ \int_{l}^{l + 6} L (η (s), s, \dot{η} (s)) d s ⩽ C, \end{matrix}$ which can be rewritten as $\begin{matrix} \int_{l}^{l + 6} {| \dot{η} (s) |}^{m} d s ⩽ C . \end{matrix}$ The idea here is straightforward: we “double” the trace of $m (t, 0, y)$ and shift it in an appropriate way such that their endpoints are as close as possible. Hence one should “move faster” on some part of the initial curve such that its velocity is bounded, to save time for connectors. See Fig. 1.

We write down this idea rigourously. Let $w \in {[0, 1]}^{n}$ such that $y - w \in Z^{n}$ . Construct a path $μ \in AC ([0, 2 t], R^{n})$ with $μ (0) = 0$ , $μ (2 t) = 2 y$ as below: $\begin{matrix} μ (s) = \{\begin{array}{ll} η (s), & 0 ⩽ s < t; \\ \frac{- s + t}{⌈ t ⌉ + 2 - t} w + y, & t ⩽ s ⩽ ⌈ t ⌉ + 2; \\ η (s - ⌈ t ⌉ - 2) + y - w, & ⌈ t ⌉ + 2 ⩽ s < ⌈ t ⌉ + l + 2; \\ η (6 (s - ⌈ t ⌉ - l - 2) + l) + y - w, & ⌈ t ⌉ + l + 2 ⩽ s < ⌈ t ⌉ + l + 3; \\ η (s - ⌈ t ⌉ + 3) + y - w, & ⌈ t ⌉ + l + 3 ⩽ s < ⌈ t ⌉ + t - 3; \\ \frac{s - ⌈ t ⌉ - t + 3}{t - ⌈ t ⌉ + 3} w + 2 y - w; & ⌈ t ⌉ + t - 3 ⩽ s ⩽ 2 t . \end{array} \end{matrix}$ For $1 ⩽ i ⩽ 6$ , denote the cost of μ on i-th time interval by $K_{i}$ . Note that $\begin{matrix} m (2 t, 0, 2 y) ⩽ \int_{0}^{2 t} L (μ (s), s, \dot{μ} (s)) d s = \sum_{i = 1}^{6} K_{i}, \end{matrix}$ so the rest is to estimate each $K_{i}$ to show that their sum is close to $2 m (t, 0, y)$ . Indeed:

Estimations for $K_{1}$ : $\begin{aligned} K_{1} & = \int_{0}^{t} L (μ (s), s, \dot{μ} (s)) d s = \int_{0}^{t} L (η_{1} (s), s, \dot{η_{1}} (s)) d s = m (t, 0, y) . \end{aligned}$

Estimations for $K_{2}$ : $\begin{aligned} K_{2} & = \int_{t}^{⌈ t ⌉ + 2} L (μ (s), s, \dot{μ} (s)) d s ⩽ \int_{t}^{⌈ t ⌉ + 2} (β {| \dot{μ} (s) |}^{m} + K) d s \\ ⩽ \int_{t}^{⌈ t ⌉ + 2} (β \cdot \frac{1}{{(⌈ t ⌉ + 2 - t)}^{m}} + K) d s \\ ⩽ \frac{β}{{(⌈ t ⌉ + 2 - t)}^{m - 1}} + K (⌈ t ⌉ + 2 - t) ⩽ \frac{β}{2^{m}} + 3 K ⩽ C . \end{aligned}$

Estimations for $K_{6}$ : $\begin{aligned} K_{6} & = \int_{⌈ t ⌉ + t - 3}^{2 t} L (μ (s), s, \dot{μ} (s)) d s ⩽ \int_{⌈ t ⌉ + t - 3}^{2 t} (β {| \dot{μ} (s) |}^{m} + K) d s \\ ⩽ \int_{t}^{⌈ t ⌉ + 2} (β \cdot \frac{1}{{(t - ⌈ t ⌉ + 3)}^{m}} + K) d s = \frac{β}{{(t - ⌈ t ⌉ + 3)}^{m - 1}} + K (t - ⌈ t ⌉ + 3) \\ ⩽ \frac{β}{2^{m}} + 3 K ⩽ C . \end{aligned}$

Estimations for $K_{4}$ : $\begin{aligned} K_{4} & = \int_{⌈ t ⌉ + l + 2}^{⌈ t ⌉ + l + 3} L (μ (s), s, \dot{μ} (s)) d s \\ = \int_{⌈ t ⌉ + l + 2}^{⌈ t ⌉ + l + 3} L (η (6 (s - ⌈ t ⌉ - l - 2) + l) + y - w, s, \dot{η} (6 (s - ⌈ t ⌉ - l - 2) + l)) d s \\ = \frac{1}{6} \int_{l}^{l + 6} L (η (s), \frac{s - l}{6} + ⌈ t ⌉ + l + 2, 6 \dot{η} (s)) d s \\ ⩽ \frac{1}{6} \int_{l}^{l + 6} (β \cdot {| 6 \dot{η} (s) |}^{m} + K) d s \\ ⩽ \frac{β}{6^{m - 1}} \int_{l}^{l + 6} {| \dot{η} (s) |}^{m} d s + K ⩽ C . \end{aligned}$

Estimation for $K_{3}$ : $\begin{aligned} K_{3} & = \int_{⌈ t ⌉ + 2}^{⌈ t ⌉ + l + 2} L (μ (s), s, \dot{μ} (s)) d s \\ = \int_{⌈ t ⌉ + 2}^{⌈ t ⌉ + l + 2} L (η (s - ⌈ t ⌉ - 2) + y - w, s, \dot{η} (s - ⌈ t ⌉ - 2)) d s \\ = \int_{0}^{l} L (η (s) + y - w, s + ⌈ t ⌉ + 2, \dot{η} (s)) d s = \int_{0}^{l} L (η (s), s, \dot{η} (s)) d s . \end{aligned}$

Estimation for $K_{5}$ : $\begin{aligned} K_{5} & = \int_{⌈ t ⌉ + l + 3}^{⌈ t ⌉ + t - 3} L (μ (s), s, \dot{μ} (s)) d s \\ = \int_{⌈ t ⌉ + l + 3}^{⌈ t ⌉ + t - 3} L (η (s - ⌈ t ⌉ + 3) + y - w, s, \dot{η} (s - ⌈ t ⌉ + 3)) d s \\ = \int_{l + 6}^{t} L (η (s) + y - w, s + ⌈ t ⌉ - 3, \dot{η} (s)) d s = \int_{l + 6}^{t} L (η (s), s, \dot{η} (s)) d s . \end{aligned}$

We combine all above results to yield

\begin{aligned} m (2 t, 0, 2 y) & ⩽ \sum_{i = 1}^{6} K_{i} \\ ⩽ m (t, 0, y) + \int_{0}^{l} L (η (s), s, \dot{η} (s)) d s + \int_{l + 6}^{t} L (η (s), s, \dot{η} (s)) d s + C \\ = 2 m (t, 0, y) + C - \int_{l}^{l + 6} L (η (s), s, \dot{η} (s)) d s \\ ⩽ 2 m (t, 0, y) + C - \int_{l}^{l + 6} (α {| \dot{η} (s) |}^{m} - K) d s \\ ⩽ 2 m (t, 0, y) + C + 6 K ⩽ 2 m (t, 0, y) + C . \end{aligned}

A proof for an arbitrary choice of p, q follows exactly the same.

Fig. 1.

Joining two copies of the trace of $m (t, 0, y)$ . The blue straight lines are connectors, and the red part is the segment on which we should “move faster”.

Without loss of generality, we may assume $x = 0$ . Thanks to the previous part, $\begin{matrix} m ((p + q) t, 0, (p + q) y) ⩽ m (p t, 0, p y) + m (q t, 0, q y) + C . \end{matrix}$ Consider a function $φ : [0, \infty) \to R$ satisfying $\begin{matrix} φ (p) = m (p t, 0, p y) + C, \end{matrix}$ then φ is a subadditive function. Precisely, $\begin{matrix} φ (p + q) ⩽ φ (p) + φ (q) \end{matrix}$ for any choice of positive numbers p, q. The existence of $\overline{m} (t, x, y)$ is now affirmed thanks to Fekete’s lemma (for instance, see Lemma 11.15 at [4] for a proof): $\begin{matrix} \overline{m} (t, x, y) = lim_{k \to \infty} \frac{1}{k} m (k t, k x, k y) = inf_{k > 0} \frac{φ (k)}{k} . \end{matrix}$ From the previous part, we also deduce $\begin{matrix} m (2^{k} t, 0, 2^{k} y) ⩽ 2^{k} m (t, 0, y) + C (2^{k} - 1) \end{matrix}$ for any positive integer k, which could be rewritten as $\begin{matrix} \frac{1}{2^{k}} m (2^{k} t, 0, 2^{k} y) ⩽ m (t, 0, y) + C (1 - \frac{1}{2^{k}}) . \end{matrix}$ Now let k tend to infinity to finish the proof.

□

Next, we prove the superadditive property.

Lemma 2.3.

Given any constant $M > 0$ , if $| y | ⩽ M t$ then $\begin{matrix} 2 m (t, 0, y) ⩽ m (2 t, 0, 2 y) + C . \end{matrix}$ In particular, $\begin{matrix} m (t, 0, y) ⩽ \overline{m} (t, 0, y) + C . \end{matrix}$ Here, $C > 0$ is a universal constant depending only on K, M, n, α and β.

Proof.

By similar arguments in the proof of Lemma 2.2, we get $\begin{matrix} - C t ⩽ m (t, 0, y), m (2 t, 0, 2 y) ⩽ C t . \end{matrix}$ Hence if $t < 4 {(n + 4)}^{2}$ , the superadditive property is obvious since $\begin{matrix} 2 m (t, 0, y) - m (2 t, 0, 2 y) ⩽ 2 C t + C t ⩽ C . \end{matrix}$ Now only consider $t > 4 {(n + 4)}^{2}$ . Let η be a minimizer for $m (2 t, 0, 2 y)$ . Note that $η \in AC ([0, 2 t], R^{n})$ , also $η (0) = 0$ and $η (2 t) = 2 y$ . By using Burago’s lemma for continuous curve $ζ : [0, 2 t] \to R^{n + 1}$ satisfying $ζ (s) = (η (s), s)$ , there exist a positive integer $k ⩽ \frac{n + 2}{2}$ and a collection of disjoint subintervals ${[a_{i}, b_{i}]}_{1 ⩽ i ⩽ k}$ of $[0, 2 t]$ such that $\begin{matrix} \sum_{i = 1}^{k} (η (b_{i}) - η (a_{i})) = \frac{η (2 y) - η (0)}{2} = \frac{2 y}{2} = y, and \sum_{i = 1}^{k} (b_{i} - a_{i}) = \frac{2 t - 0}{2} = t . \end{matrix}$

For convenience, let $d_{0} = 0$ . Inductively define ${c_{i}}$ and ${d_{i}}$ for $1 ⩽ i ⩽ k$ as follow:

for each $1 ⩽ i ⩽ k$ , let $c_{i}$ be the smallest number such that $c_{i} - a_{i} \in Z$ and $c_{i} - d_{i - 1} ⩾ 1$ ; note that from this definition, $1 ⩽ c_{i} - d_{i - 1} < 2$ ;

after defining $c_{i}$ , choose $d_{i}$ satisfying $d_{i} - c_{i} = b_{i} - a_{i}$ .

Fig. 2.

Time-scaling process with integer vectors.

A visualization for that scaling process is demonstrated in Fig. 2. Those difference will be time-difference for connectors between corresponding curve segments. Moreover, their time endpoints should be far enough so that the velocity of connectors would not blow up. Now on each interval $[c_{i}, d_{i}]$ , we proceed periodic shifting by defining paths $μ_{i} \in AC ([c_{i}, d_{i}], R^{n})$ satisfying:

$μ_{1} (c_{i}) \in {[0, 1]}^{n}$ ;

there exists $w_{i} \in Z^{n}$ such that $\begin{matrix} μ_{i} (s + c_{i} - a_{i}) - η (s) = w_{i} for each s \in [a_{i}, b_{i}], \end{matrix}$ which leads to $\begin{matrix} \sum_{i = 1}^{k} (μ_{i} (d_{i}) - μ_{i} (c_{i})) = \sum_{i = 1}^{k} (η (b_{i}) - η (a_{i})) = y; \end{matrix}$

for $1 ⩽ i ⩽ k - 1$ , $μ_{i} (d_{i})$ and $μ_{i + 1} (c_{i + 1})$ land in the same unit cube, so $\begin{matrix} | μ_{i} (d_{i}) - μ_{i + 1} (c_{i + 1}) | ⩽ \sqrt{n} . \end{matrix}$ In other words, the spatial endpoints are as close as possible.

From the above definition and the periodicity of L, we have $\begin{aligned} \int_{c_{i}}^{d_{i}} L (μ_{i} (s), s, \dot{μ_{i}} (s)) d s & = \int_{c_{i}}^{d_{i}} L (μ_{i} (s), s + a_{i} - c_{i}, \dot{μ_{i}} (s)) d s \\ = \int_{c_{i}}^{d_{i}} L (η (s + a_{i} - c_{i}), s + a_{i} - c_{i}, \dot{η} (s + a_{i} - c_{i})) d s \\ = \int_{a_{i}}^{b_{i}} L (η (s), s, \dot{η} (s)) d s . \end{aligned}$

If we define M as a least integer such that $\begin{matrix} M > \sum_{i = 0}^{k - 1} (c_{i + 1} - d_{i}), \end{matrix}$ then $\begin{matrix} M ⩽ \sum_{i = 0}^{k - 1} (c_{i + 1} - d_{i}) + 1 ⩽ 2 k + 1 ⩽ n + 3 ⩽ 4 n^{2} < t . \end{matrix}$ Now we point out an index $1 ⩽ j ⩽ k$ and a subinterval $[l, l + 3 M]$ of $[c_{j}, d_{j}]$ satisfying $\begin{matrix} \int_{l}^{l + 3 M} L (μ_{j} (s), s, \dot{μ_{j}} (s)) d s ⩽ C \end{matrix}$ to reuse our time-saving idea. First, there exists j such that $d_{j} - c_{j} > 3 M$ and $\begin{matrix} \int_{c_{j}}^{d_{j}} L (μ_{j} (s), s, \dot{μ_{j}} (s)) d s ⩽ C t . \end{matrix}$ Indeed, the existence of j follows from $\begin{aligned} max_{1 ⩽ i ⩽ k} (d_{i} - c_{i}) & ⩾ \frac{1}{k} \sum_{i = 1}^{k} (d_{i} - c_{i}) = \frac{1}{k} \sum_{i = 1}^{k} (b_{i} - a_{i}) = \frac{t}{k} \\ ⩾ \frac{2 t}{n + 2} ⩾ \frac{8 {(n + 4)}^{2}}{n + 2} > 3 (n + 3) ⩾ 3 M . \end{aligned}$ And by defining $b_{0} = 0$ and $a_{k + 1} = 2 t$ , we obtain $\begin{aligned} \int_{c_{j}}^{d_{j}} L (μ_{i} (s), s, \dot{μ_{i}} (s)) d s \\ = \int_{a_{j}}^{b_{j}} L (η (s), s, \dot{η} (s)) d s \\ = \int_{0}^{2 t} L (η (s), s, \dot{η} (s)) d s - \int_{0}^{a_{j}} L (η (s), s, \dot{η} (s)) d s - \int_{b_{j}}^{2 t} L (η (s), s, \dot{η} (s)) d s \\ ⩽ m (2 t, 0, 2 y) - \int_{0}^{a_{j}} (α {| \dot{η} (s) |}^{m} - K) d s - \int_{b_{j}}^{2 t} (α {| \dot{η} (s) |}^{m} - K) d s \\ ⩽ m (2 t, 0, 2 y) + K (a_{j} + 2 t - b_{j}) ⩽ C t + 2 K t ⩽ C t . \end{aligned}$ Again by subdividing $[c_{i}, d_{i}]$ into consecutive segments with length $3 M$ and possibly a redundant part shorter than $3 M$ , then using arguments in the proof of Lemma 2.2, there exists a subinterval $[l, l + 3 M]$ of $[c_{j}, d_{j}]$ satisfying $\begin{matrix} \int_{l}^{l + 3 M} L (μ_{i} (s), s, \dot{μ_{i}} (s)) d s ⩽ C t \cdot \frac{3 M}{t} ⩽ C . \end{matrix}$ From this we easily deduce $\begin{matrix} - C ⩽ \int_{l}^{l + 3 M} {| \dot{η} (s) |}^{m} d s ⩽ C . \end{matrix}$

That boundedness saves a time interval length $2 M$ for straight connectors. Now the essential step is to construct a path $ζ \in AC ([0, t], R^{n})$ such that $\begin{matrix} \int_{0}^{t} L (ζ (s), s, \dot{ζ} (s)) d s ⩽ \sum_{i = 1}^{k} \int_{a_{i}}^{b_{i}} L (η (s), s, \dot{η} (s)) d s + C . \end{matrix}$ We define ζ as follows:

for $0 ⩽ s < c_{1}$ , $\begin{matrix} ζ (s) = \frac{μ_{1} (c_{1}) s}{c_{1}}; \end{matrix}$

Fig. 3.

Gluing ${μ_{i}}$ to obtain a path from 0 to y. The red and blue segments are straight connectors, which use redundant time obtained by rescaling the green segment.

if $1 ⩽ i < j$ , then $\begin{matrix} ζ (s) = \{\begin{array}{ll} μ_{i} (s), & c_{i} ⩽ s < d_{i}, \\ \frac{(μ_{i + 1} (c_{i + 1}) - μ_{i} (d_{i})) (s - d_{i})}{c_{i + 1} - d_{i}} + μ_{i} (d_{i}), & d_{i} ⩽ s < c_{i + 1}; \end{array} \end{matrix}$

if $i = j$ , then $\begin{matrix} ζ (s) = \{\begin{array}{ll} μ_{j} (s), & c_{i} ⩽ s < l, \\ μ_{j} (3 s - 2 l), & l ⩽ s < l + M, \\ μ_{j} (s + 2 M), & l + M ⩽ s < d_{j} - 2 M, \\ \frac{(μ_{j + 1} (c_{j + 1}) - μ_{j} (d_{j})) (s + 2 M - d_{j})}{c_{j + 1} - d_{j}} + μ_{j} (d_{j}), & d_{j} - 2 M ⩽ s < c_{j + 1} - 2 M; \end{array} \end{matrix}$

if $j < i < k$ , then $\begin{matrix} ζ (s) = \{\begin{array}{ll} μ_{i} (s + 2 M), & c_{i} - 2 M ⩽ s < d_{i} - 2 M, \\ \frac{(μ_{i + 1} (c_{i + 1}) - μ_{i} (d_{i})) (s + 2 M - d_{i})}{c_{i + 1} - d_{i}} + μ_{i} (d_{i}), & d_{i} - 2 M ⩽ s < c_{i + 1} - 2 M; \end{array} \end{matrix}$

if $i = k$ , then $\begin{matrix} ζ (s) = \{\begin{array}{ll} μ_{i} (s + 2 M), & c_{k} - 2 M ⩽ s < d_{k} - 2 M, \\ \frac{(y - μ_{k} (d_{k})) (s - d_{k} + 2 M)}{t - d_{k} + 2 M} + μ_{k} (d_{k}), & d_{k} - 2 M ⩽ s ⩽ t . \end{array} \end{matrix}$

See Fig. 3 for a visualization. The rest is just calculation. Indeed,

Keep in mind that $d_{0} = 0$ , then for $0 ⩽ i < k$ : $\begin{aligned} \int_{d_{i}}^{c_{i + 1}} L (ζ (s), s, \dot{ζ} (s)) d s & ⩽ \int_{d_{i}}^{c_{i + 1}} (β {| \dot{ζ} (s) |}^{m} + K) d s \\ ⩽ β \int_{d_{i}}^{c_{i + 1}} {| \dot{ζ} (s) |}^{m} d s + K (c_{i + 1} - d_{i}) \\ ⩽ β \int_{d_{i}}^{c_{i + 1}} max_{1 ⩽ i < k} {{| \frac{μ_{1} (c_{1})}{c_{1}} |}^{m}, {| \frac{μ_{i + 1} (c_{i + 1}) - μ_{i} (d_{i})}{c_{i + 1} - d_{i}} |}^{m}} d s + 2 K \\ ⩽ β (c_{i + 1} - d_{i}) {(\frac{\sqrt{n}}{2})}^{m} + C ⩽ 2 β {(\frac{\sqrt{n}}{2})}^{m} + C ⩽ C . \end{aligned}$

The cost of ζ on $[d_{k} - 2 M, t]$ is bounded by a constant. Indeed: $\begin{aligned} \int_{d_{k} - 2 M}^{t} L (ζ (s), s, \dot{ζ} (s)) d s & ⩽ \int_{d_{k} - 2 M}^{t} (β {| \dot{ζ} (s) |}^{m} + K) d s \\ ⩽ β \int_{d_{k} - 2 M}^{t} \frac{| y - μ_{k} (d_{k}) |^{m}}{{(t - d_{k} + 2 M)}^{m}} d s + K M \\ ⩽ \frac{β}{{(t - d_{k} + 2 M)}^{m}} \int_{t - M}^{t} {| \sum_{i = 1}^{k} (μ_{i} (d_{i}) - μ_{i} (c_{i})) - μ_{k} (d_{k}) |}^{m} d s + C \\ ⩽ C \int_{t - M}^{t} {(\sum_{i = 0}^{k - 1} | μ_{i} (d_{i}) - μ_{i + 1} (c_{i + 1}) |)}^{m} d s + C \\ ⩽ C \int_{t - M}^{t} {(k \sqrt{n})}^{m} d s + C = C k^{m} {(\sqrt{n})}^{m} + C ⩽ C . \end{aligned}$

When $1 ⩽ i < j$ and s lies outside $[c_{i}, d_{i}]$ , we have $\begin{matrix} \int_{c_{i}}^{d_{i}} L (ζ (s), s, \dot{ζ} (s)) d s = \int_{c_{i}}^{d_{i}} L (μ_{i} (s), s, \dot{μ_{i}} (s)) d s = \int_{a_{i}}^{b_{i}} L (η (s), s, \dot{η} (s)) d s . \end{matrix}$

When $j < i ⩽ k$ and s lies in $[c_{i} - 2 M, d_{i} - 2 M]$ , we have $\begin{aligned} \int_{c_{i} - 2 M}^{d_{i} - 2 M} L (ζ (s), s, \dot{ζ} (s)) d s & = \int_{c_{i} - 2 M}^{d_{i} - 2 M} L (μ_{i} (s + 2 M), s, \dot{μ_{i}} (s + 2 M)) d s \\ = \int_{c_{i}}^{d_{i}} L (μ_{i} (s), s - 2 M, \dot{μ_{i}} (s)) d s \\ = \int_{c_{i}}^{d_{i}} L (μ_{i} (s), s, \dot{μ_{i}} (s)) d s = \int_{a_{i}}^{b_{i}} L (η (s), s, \dot{η} (s)) d s . \end{aligned}$

When $i = j$ , we obtain, $\begin{matrix} \int_{c_{i}}^{l} L (ζ (s), s, \dot{ζ} (s)) d s = \int_{c_{i}}^{l} L (μ_{i} (s), s, \dot{μ_{i}} (s)) d s \end{matrix}$ and also $\begin{aligned} \int_{l + M}^{d_{j} - 2 M} L (ζ (s), s, \dot{ζ} (s)) d s & = \int_{l + M}^{d_{j} - 2 M} L (μ_{i} (s + 2 M), s, \dot{μ_{i}} (s + 2 M)) d s \\ = \int_{l + 3 M}^{d_{j}} L (μ_{i} (s), s - 2 M, \dot{μ_{i}} (s)) d s \\ = \int_{l + 3 M}^{d_{j}} L (μ_{i} (s), s, \dot{μ_{i}} (s)) d s . \end{aligned}$

Hence, $\begin{aligned} \int_{c_{i}}^{l} L (ζ (s), s, \dot{ζ} (s)) d s + \int_{l + M}^{d_{j} - 2 M} L (ζ (s), s, \dot{ζ} (s)) d s \\ = \int_{c_{i}}^{d_{j}} L (μ_{i} (s), s, \dot{μ_{i}} (s)) d s - \int_{l}^{l + 3 M} L (μ_{i} (s), s, \dot{μ_{i}} (s)) d s \\ ⩽ \int_{c_{i}}^{d_{j}} L (μ_{i} (s), s, \dot{μ_{i}} (s)) d s + C . \end{aligned}$ Moreover, since velocity of ζ is bounded on $[l, l + 3 M]$ : $\begin{aligned} \int_{l}^{l + M} L (ζ (s), s, \dot{ζ} (s)) d s & = \int_{l}^{l + M} L (μ_{j} (3 s - 2 l), s, 3 \dot{μ_{j}} (3 s - 2 l)) d s \\ = \frac{1}{3} \int_{l}^{l + 3 M} L (μ_{j} (s), \frac{s + 2 l}{3}, 3 \dot{μ_{j}} (s)) d s \\ ⩽ \frac{1}{3} \int_{l}^{l + 3 M} (β {| \dot{ζ} (s) |}^{m} + K) d s \\ = \frac{β}{3} \int_{l}^{l + 3 M} {| \dot{ζ} (s) |}^{m} d s + K M ⩽ C . \end{aligned}$

Therefore by decomposing ζ and summing by parts, $\begin{matrix} \int_{0}^{t} L (ζ (s), s, \dot{ζ} (s)) d s ⩽ \sum_{i = 1}^{k} \int_{a_{i}}^{b_{i}} L (η (s), s, \dot{η} (s)) d s + C \end{matrix}$ as desired. From here, we conclude $\begin{matrix} (2.1) & m (t, 0, y) ⩽ \sum_{i = 1}^{k} \int_{a_{i}}^{b_{i}} L (η (s), s, \dot{η} (s)) d s + C . \end{matrix}$

Replicating all previous arguments for a collection of $k + 1$ disjoint subintervals ${[u_{i}, v_{i}]}_{1 ⩽ i ⩽ k + 1}$ of $[0, 2 t]$ , which is defined by $\begin{matrix} [u_{i}, v_{i}] = \{\begin{array}{ll} [0, a_{i}], & i = 1, \\ [b_{i - 1}, a_{i}], & 2 ⩽ i ⩽ k, \\ [b_{k}, 2 t], & i = k + 1, \end{array} \end{matrix}$ and also satisfies $\begin{matrix} \sum_{i = 1}^{k} (η (v_{i}) - η (u_{i})) = \frac{η (2 y) - η (0)}{2} = \frac{2 y}{2} = y, and \sum_{i = 1}^{k} (v_{i} - u_{i}) = \frac{2 t - 0}{2} = t, \end{matrix}$ the below estimation also holds: $\begin{matrix} (2.2) & m (t, 0, y) ⩽ \sum_{i = 1}^{k + 1} \int_{u_{i}}^{v_{i}} L (η (s), s, \dot{η} (s)) d s + C . \end{matrix}$

Adding (2.1) and (2.2) to finish the proof. □

Now we finalize the proof of Theorem 1.1.

Proof of Theorem 1.1.

Thanks to Lemma 2.2.ii) and Lemma 2.3, for $t > 0$ and $y \in R^{n}$ such that $| y | ⩽ C t$ , $\begin{matrix} | m (t, 0, y) - \overline{m} (t, 0, y) | ⩽ C t . \end{matrix}$

By scaling and translation, it suffices to prove the result for $(x, t) = (0, 1)$ . Since $g \in LIP (R^{n})$ , $\begin{matrix} | g (x) | ⩽ | g (x) - g (0) | + | g (0) | ⩽ C (| x | + 1) \end{matrix}$ for some constant C. Recall the optimal control formula: $\begin{matrix} u^{ε} (0, 1) = inf_{\begin{array}{c} η (0) = 0 \\ η \in AC ([- ε^{- 1}, 0], R^{n}) \end{array}} {g (ε η (- ε^{- 1})) + ε \int_{- ε^{- 1}}^{0} L (η (s), s, \dot{η} (s)) d s} . \end{matrix}$ From the boundedness of L and Jensen’s inequality, $\begin{aligned} ε \int_{- ε^{- 1}}^{0} L (η (s), s, \dot{η} (s) d s) & ⩾ ε \int_{- ε^{- 1}}^{0} (α {| \dot{μ} (s) |}^{m} - K) d s \\ ⩾ α {| ε \int_{- ε^{- 1}}^{0} \dot{η} (s) d s |}^{m} - K \\ = α ε^{m} {| η (- ε^{- 1}) |}^{m} - K . \end{aligned}$

Thus, for some constant $C > 0$ depending only on L, $‖ D g ‖_{L^{\infty} (R^{n})}$ and n, $\begin{aligned} u^{ε} (0, 1) & = inf_{\begin{array}{c} η (0) = 0 \\ ε | η (- ε^{- 1}) | ⩽ C \end{array}} {g (ε η (- ε^{- 1})) + ε \int_{- ε^{- 1}}^{0} L (η (s), s, \dot{η} (s)) d s} \\ = inf_{| y | ⩽ C} (g (y) + ε m (ε^{- 1}, - ε^{- 1} y, 0)) \\ = inf_{| y | ⩽ C} (g (y) + \overline{m} (1, 0, - y)) + O (ε) \\ = u (0, 1) + O (ε) . \end{aligned}$ Hence the proof is completed. □

Footnotes

Acknowledgements

This work is partly supported by grant U2022-01 at the University of Science, Vietnam National University Ho Chi Minh City.

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