# Report on the 12th Annual USA Junior Mathematical Olympiad

Béla Bajnok and Evan Chen

(September 10, 2021)

The USA Junior Mathematical Olympiad (USAJMO) is the final round in the
American Mathematics Competitions series for high school students in grades 10
or below, organized each year by the Mathematical Association of America. The
competition follows the style of the International Mathematics Olympiad: it
consists of three problems each on two consecutive days, with an allowed time
of four and a half hours both days.

The 12th annual USAJMO was given on Tuesday, April 13, 2021 and Wednesday,
April 14, 2021. This year, 259 students were invited to take the USAJMO and,
as in 2020, the competition was administered online. The names of winners and
honorable mentions, as well as more information on the American Mathematics
Competitions program, can be found on the site https://www.maa.org/math-
competitions. Below we present the problems and solutions of the competition;
a similar article for the USA Mathematical Olympiad (USAMO) can be found in
Mathematics Magazine.

The problems of the USAJMO are chosen – from a large collection of proposals
submitted for this purpose – by the USAMO/USAJMO Editorial Board, whose co-
editors-in-chief this year were Evan Chen and Jennifer Iglesias, with
associate editors Ankan Bhattacharya, John Berman, Zuming Feng, Sherry Gong,
Alison Miller, Maria Monks Gillespie, and Alex Zhai. This year’s problems were
created by Ankan Bhattacharya, Vincent Huang, Mohsen Jamaali, Carl
Schildkraut, Brandon Wang, and Alex Zhai.

The solutions presented here are those of the present authors, relying in part
on the submissions of the problem authors and members of the editorial board.
Each problem was worth 7 points; the nine-tuple
$(n;a_{7},a_{6},a_{5},a_{4},a_{3},a_{2},a_{1},a_{0})$ states the number of
students who submitted a paper for the relevant problem, followed by the
numbers who scored $7,6,\dots,0$ points, respectively.

Problem 1 $(236;83,23,4,0,5,9,26,86)$; proposed by Vincent Huang. Let
$\mathbb{N}$ denote the set of positive integers. Find all functions
$f\colon\mathbb{N}\to\mathbb{N}$ such that for all positive integers $a$ and
$b$,

$f(a^{2}+b^{2})=f(a)f(b)\quad\text{and}\quad f(a^{2})=f(a)^{2}.$

Solution. It is easy to see that the constant function $f\equiv 1$ satisfies
the two given properties; we will use induction to prove that there are no
others. Since, according to the second condition, $f(1^{2})=f(1)^{2}$, we see
that $f(1)=1$; and now from the first condition we get
$f(2)=f(1^{2}+1^{2})=f(1)^{2}=1$ as well. For the inductive step, we let
$n\geq 3$ be any integer, and suppose that for every positive integer $m<n$ we
have $f(m)=1$; we need to prove that $f(n)=1$ holds also.

We start with the identity (familiar from Pythagorean triples) that for all
positive integers $u$ and $v$ with $u>v$ we have

$(u^{2}-v^{2})^{2}+(2uv)^{2}=(u^{2}+v^{2})^{2},$

and therefore, using the two given conditions, we find that

$f(u^{2}-v^{2})f(2uv)=f\left((u^{2}+v^{2})^{2}\right)=\left(f(u^{2}+v^{2})\right)^{2}=\left(f(u)f(v)\right)^{2}.$

In particular, if $u>v$ are positive integers for which $f(u)=f(v)=1$, then
$f(u^{2}-v^{2})=1$ and $f(2uv)=1$.

We consider two cases: when $n$ is odd and when $n$ is even. If $n=2k+1$ for
some positive integer $k$, then, since $f(k+1)=f(k)=1$ by our inductive
assumption, we get $f((k+1)^{2}-k^{2})=f(n)=1$.

Similarly, when $n=2k$, we have $f(k)=f(1)=1$, and thus $f(2k)=f(n)=1$. Our
proof is now complete.

Problem 2 $(176;66,1,1,0,0,0,8,100)$; proposed by Ankan Bhattacharya.
Rectangles $BCC_{1}B_{2}$, $CAA_{1}C_{2}$, and $ABB_{1}A_{2}$ are erected
outside an acute triangle $ABC$. Suppose that

$\angle BC_{1}C+\angle CA_{1}A+\angle AB_{1}B=180^{\circ}.$

Prove that lines $B_{1}C_{2}$, $C_{1}A_{2}$, and $A_{1}B_{2}$ are concurrent.

Solution. Let $\omega_{A}$, $\omega_{B}$, and $\omega_{C}$ be the
circumcircles of rectangles $BCC_{1}B_{2}$, $CAA_{1}C_{2}$, and
$ABB_{1}A_{2}$, respectively.

Figure 1: The main circles used in Problem 2

Define $P$ to be the foot of the altitude from $A$ to $B_{1}C_{2}$.

Observe that, since $\angle ACC_{2}$ and $\angle APC_{2}$ are both
$90^{\circ}$, $P$ lies on $\omega_{B}$ by the inscribed angle theorem and,
therefore, $\angle APC$ and $\angle CA_{1}A$ are supplementary angles. A
similar argument shows that $P$ lies on $\omega_{C}$ and thus $\angle APB$ and
$\angle AB_{1}B$ are supplementary angles as well.

But then

$\angle BPC=360^{\circ}-(\angle APC+\angle APB)=\angle CA_{1}A+\angle
AB_{1}B,$

which, by the given equation, yields

$\angle BPC=180^{\circ}-\angle BC_{1}C,$

and thus $P$ lies on $\omega_{A}$ as well. Therefore, $P$ is the (unique)
common point of all three circles.

Similar arguments would prove that the feet of the altitudes from $B$ and $C$
to $C_{1}A_{2}$ and $A_{1}B_{2}$, respectively, are on each of the three
circles, and thus must coincide with $P$. But then lines $B_{1}C_{2}$,
$C_{1}A_{2}$, and $A_{1}B_{2}$ are concurrent, as claimed.

Problem 3 $(159;13,11,2,0,1,6,26,100)$; proposed by Alex Zhai. An equilateral
triangle $\Delta$ of side length $L>0$ is given. Suppose that $n$ equilateral
triangles with side length $1$ and with non-overlapping interiors are drawn
inside $\Delta$, such that each unit equilateral triangle has sides parallel
to $\Delta$, but with opposite orientation. (An example with $n=2$ is drawn in
Figure 2.)

Figure 2: An example for Problem 3 with two triangles

Prove that $n\leq\frac{2}{3}L^{2}.$

Solution. For each triangle, we draw a green regular hexagon of side length
$1/2$ as shown in Figure 3. Because the edges of $\Delta$ slope upwards at the
same angle as the edges in the upper half of the green hexagons, it follows
that the green hexagons lie entirely inside $\Delta$.

Figure 3: The key hexagon in the solution of Problem 3

Our main claim is that the green hexagons are pairwise disjoint. To see this,
anchor each triangle by the midpoint of the top side. By fixing one anchor and
moving another anchor as close as possible so that the triangles do not
overlap, we find that the boundary of the allowable region for the second
anchor is a hexagon of side length $1$ centered at the first anchor. This is
equivalent to the hexagons not overlapping.

Now since each hexagon has area $\frac{3\sqrt{3}}{8}$ and lies inside
$\Delta$, and the total area of $\Delta$ is $\frac{\sqrt{3}}{4}L^{2}$, we
conclude that

$\frac{3\sqrt{3}}{8}\cdot n\leq\frac{\sqrt{3}}{4}L^{2},$

and thus $n\leq\frac{2}{3}L^{2},$ as claimed.

Remark. The constant $\frac{2}{3}$ cannot be improved. The tessellation in
Figure 4 illustrates how one can achieve
$n\geq\left(\frac{2}{3}-\epsilon\right)L^{2}$, for any $\epsilon>0$.

Figure 4: Tessellation achieving almost $\frac{2}{3}$ density. (In the
illustration on the left, only one of the green hexagons is drawn in. The
version on the right has all the green hexagons.)

Problem 4 $(228;56,28,6,5,9,7,34,83)$; proposed by Brandon Wang. Carina has
three pins, labeled $A$, $B$, and $C$, respectively, located at the origin of
the coordinate plane. In a _move_ , Carina may move a pin to an adjacent
lattice point at distance $1$ away. What is the least number of moves that
Carina can make in order for triangle $ABC$ to have area $2021$?

(A lattice point is a point $(x,y)$ in the coordinate plane where $x$ and $y$
are both integers, not necessarily positive.)

Solution. The answer is $128$. One construction with $128$ moves is given by
moving the pins to the points

$\displaystyle A$ $\displaystyle=(-3,-18),$ $\displaystyle B$
$\displaystyle=(61,0),$ $\displaystyle C$ $\displaystyle=(0,46).$

We can verify that the number of moves is

$3+18+61+0+0+46=128,$

and that the area of the triangle (using, for example, the shoelace formula)
is

$\frac{61\cdot 46+18\cdot 61+3\cdot 46}{2}=2021.$

We now prove that at least $128$ moves are required. Define the bounding box
of triangle $ABC$ to be the smallest axis-parallel rectangle which contains
all three of the vertices $A$, $B$, $C$; an example is illustrated in Figure
5.

Figure 5: The bounding box of a triangle

We first establish the (well known) claim that the area of a triangle is at
most half the area of its bounding box. Note that all four sides of the
bounding box must contain at least one vertex of the triangle, which implies
that at least one vertex, say $A$, is at a corner. We label the other three
corners of the rectangle by $X$, $Y$, and $Z$, so that $Y$ is opposite $A$. If
$B=Y$ or $C=Y$, then $\triangle ABC$ is contained in $\triangle AXY$ or
$\triangle AZY$, so our claim obviously holds. If $B\neq Y$ and $C\neq Y$,
then these two vertices lie on different sides of the line $AY$, and thus one
of them lies on $XY$ and the other on $YZ$; we assume that $B$ is on $XY$ and
$C$ is on $YZ$. Note that $Z$ is at least as far from the line $AB$ as $C$ is,
which means that the area of $\triangle ABC$ is at most as much as the area of
$\triangle ABZ$, implying our claim.

We can now complete our proof, as follows. Suppose that $n$ moves are made,
leading to a bounding box of width $w$ and length $\ell$. Since any move
changes the dimensions of the bounding box by at most $1$, we have $\ell+w\leq
n$, and thus for the area of triangle $ABC$ we have

$2021\leq\frac{1}{2}\ell
w=\frac{1}{8}\left((\ell+w)^{2}-(\ell-w)^{2}\right)\leq\frac{1}{8}(\ell+w)^{2}\leq\frac{1}{8}n^{2},$

which yields $n\geq 128$, as claimed.

Problem 5 $(202;42,11,8,2,2,31,13,93)$; proposed by Carl Schildkraut. A finite
set $S$ of positive integers has the property that, for each $s\in S$, and
each positive integer divisor $d$ of $s$, there exists a unique element $t\in
S$ satisfying $\gcd(s,t)=d$. (The elements $s$ and $t$ could be equal.)

Given this information, find all possible values for the number of elements of
$S$.

Solution. We claim that the possible sizes are 0 and the nonnegative integer
powers of 2. Since $S=\emptyset$ and $S=\\{1\\}$ obviously work, we need to
show that a set $S$ of size $n\geq 2$ satisfying the requirements exists if,
and only if, $n=2^{k}$ for some positive integer $k$.

We start by verifying that these values are indeed possible. For a given
positive integer $k$, we construct a set of size $2^{k}$ as follows. Suppose
that $p_{1},q_{1},p_{2},q_{2},\dots,p_{k},q_{k}$ are $2k$ pairwise distinct
positive primes; for an ordered pair of subsets $(I,J)$ of
$[k]=\\{1,2,\dots,k\\}$, we will use the notation

$s(I,J)=\prod_{i\in I}p_{i}\cdot\prod_{j\in J}q_{j}.$

(Recall that the empty product equals 1.) We then consider the set

$S=\\{s(I,J)\mid I\subseteq[k],J=[k]\setminus I\\}.$

Since the elements of $S$ are then in a bijection with the subsets $I$ of
$[k]$, we see that $|S|=2^{k}$. We need to show that $S$ satisfies the
required property.

Given an element $s(I,J)$ of $S$, we see that its positive divisors are of the
form $s(I_{0},J_{0})$ where $I_{0}\subseteq I$ and $J_{0}\subseteq J$. Note
also that, since $I$ and $J$ form a partition of $[k]$, the sets
$I^{\prime}=I_{0}\cup(J\setminus J_{0})$ and $J^{\prime}=J_{0}\cup(I\setminus
I_{0})$ do as well, and thus $s(I^{\prime},J^{\prime})$ is an element of $S$;
in fact it is the unique element of $S$ whose greatest common divisor with
$s(I,J)$ equals $s(I_{0},J_{0})$. Therefore, the set $S$ we constructed
satisfies the requirement of the problem.

It remains to be shown that if $S$ is a set satisfying the property and it has
size $n\geq 2$, then $n=2^{k}$ for some positive integer $k$. Let $s\geq 2$ be
any element of $S$, and let $p$ be any positive prime divisor of $s$. We can
then write $s=p^{e}\cdot u$ for some positive integers $e$ and $u$ where $u$
is not divisible by $p$. We claim that $e=1$.

Denoting by $d(m)$ the number of positive divisors of a positive integer $m$,
we have $d(s)=(e+1)\cdot d(u)$; in fact, $s$ has exactly $e\cdot d(u)$
positive divisors that are divisible by $p$ and $d(u)$ that are not. By our
assumption, there is an element $t$ of $S$ for which $\gcd(t,s)=p$. Let us
assume that $e\geq 2$; we can then see that $t=p\cdot v$ for some positive
integer $v$ that is not divisible by $p$. Furthermore, $d(t)=2d(v)$, and $t$
has exactly $d(v)$ positive divisors that are divisible by $p$ and also $d(v)$
that are not.

Now according to our requirement for $S$, the positive divisors of any element
$s$ of $S$ are in a one-to-one correspondence with the elements of $S$ via the
map $a\mapsto\gcd(a,s)$, and thus $|S|=d(s)$. (In particular, all elements of
$S$ must have the same number of positive divisors.) Furthermore, for any
prime divisor $p$ of $s$, we have $p|a$ if and only if $p|\gcd(a,s)$.
Therefore, $S$ has exactly $e\cdot d(u)$ elements that are divisible by $p$
and $d(u)$ that are not. With the same reasoning, $S$ has exactly $d(v)$
elements that are divisible by $p$ and $d(v)$ that are not. But then
$d(u)=d(v)$ and $e\cdot d(u)=d(v)$, from which $e=1$.

This establishes the fact that each element of $S$ is a product of the same
number of pairwise distinct prime numbers. If $s\in S$ is the product of $k$
distinct primes, then $|S|=d(s)=2^{k}$. This completes our proof.

Problem 6 $(147;16,2,0,0,4,21,2,102)$; proposed by Mohsen Jamaali. Let $n\geq
4$ be an integer. Find all positive real solutions to the following system of
$2n$ equations:

$\displaystyle a_{1}=$ $\displaystyle\frac{1}{a_{2n}}+\frac{1}{a_{2}},$
$\displaystyle a_{2}$ $\displaystyle=a_{1}+a_{3},$ $\displaystyle a_{3}=$
$\displaystyle\frac{1}{a_{2}}+\frac{1}{a_{4}},$ $\displaystyle a_{4}$
$\displaystyle=a_{3}+a_{5},$ $\displaystyle a_{5}=$
$\displaystyle\frac{1}{a_{4}}+\frac{1}{a_{6}},$ $\displaystyle a_{6}$
$\displaystyle=a_{5}+a_{7},$ $\displaystyle\vdots$ $\displaystyle\vdots$
$\displaystyle a_{2n-1}=$ $\displaystyle\frac{1}{a_{2n-2}}+\frac{1}{a_{2n}},$
$\displaystyle a_{2n}$ $\displaystyle=a_{2n-1}+a_{1}.$

First solution. It is easy to see that a solution is provided by
$a_{1}=a_{3}=\cdots=a_{2n-1}=1$ and $a_{2}=a_{4}=\cdots=a_{2n}=2$; we prove
that there are no others.

Taking indices modulo $2n$ and eliminating terms of odd indices, for each
$i=1,2,\dots,n$ we have

$\displaystyle a_{2i}=\frac{1}{a_{2i-2}}+\frac{2}{a_{2i}}+\frac{1}{a_{2i+2}}.$
(1)

Adding up these equations yields

$\displaystyle\sum_{i=1}^{n}a_{2i}=\sum_{i=1}^{n}\frac{4}{a_{2i}}.$ (2)

According to the harmonic mean–arithmetic mean inequality,

$\displaystyle\frac{n}{\sum_{i=1}^{n}\frac{1}{a_{2i}}}\leq\frac{\sum_{i=1}^{n}a_{2i}}{n},$
(3)

so by (2) we get $\sum_{i=1}^{n}a_{2i}\geq 2n$.

Now dividing both sides of (1) by $a_{2i}$ yields

$\displaystyle
1=\frac{1}{a_{2i-2}a_{2i}}+\frac{2}{a_{2i}^{2}}+\frac{1}{a_{2i}a_{2i+2}},$ (4)

and adding the equations results in

$\displaystyle
n=\sum_{i=1}^{n}\left(\frac{1}{a_{2i}}+\frac{1}{a_{2i+2}}\right)^{2}.$ (5)

Now we use the quadratic mean–arithmetic mean inequality, which gives

$\displaystyle\frac{1}{n}\cdot\sum_{i=1}^{n}\left(\frac{1}{a_{2i}}+\frac{1}{a_{2i+2}}\right)^{2}\geq\left(\frac{\sum_{i=1}^{n}\left(\frac{1}{a_{2i}}+\frac{1}{a_{2i+2}}\right)}{n}\right)^{2},$
(6)

so by (5) and (2) we get

$1\geq\left(\frac{\sum_{i=1}^{n}\frac{2}{a_{2i}}}{n}\right)^{2}=\left(\frac{\frac{1}{2}\cdot\sum_{i=1}^{n}a_{2i}}{n}\right)^{2}$

and thus $\sum_{i=1}^{n}a_{2i}\leq 2n$.

This means that each of our inequalities is an equality, and therefore
$a_{2i}=2$ for all $i$. This in turn implies that $a_{2i-1}=1$ for all $i$, as
claimed.

Second solution. We write $m=\min\\{a_{2i}\mid 1\leq i\leq n\\}$ and
$M=\max\\{a_{2i}\mid 1\leq i\leq n\\}$, and assume that $m=a_{2j}$ and
$M=a_{2k}$. Then equation (1) from the first solution yields

$m=\frac{1}{a_{2j-2}}+\frac{2}{m}+\frac{1}{a_{2j+2}}\geq\frac{1}{M}+\frac{2}{m}+\frac{1}{M}$

and

$M=\frac{1}{a_{2k-2}}+\frac{2}{M}+\frac{1}{a_{2k+2}}\leq\frac{1}{m}+\frac{2}{M}+\frac{1}{m}.$

Therefore,

$m\geq\frac{2}{m}+\frac{2}{M}\geq M,$

which can only occur when $m=M$. Therefore, all $a_{2i}$ are equal, from which
$a_{2}=a_{4}=\cdots=a_{2n}=2$ and $a_{1}=a_{3}=\cdots=a_{2n-1}=1$, as claimed.

Béla Bajnok<EMAIL_ADDRESS>MR author ID 314851, ORCID number
0000-0002-9498-1596) is the director of the American Mathematics Competitions
program of the MAA. He is also a professor of mathematics at Gettysburg
College, and the author of two books: An Invitation to Abstract Mathematics,
now in second edition by Springer, and Additive Combinatorics: A Menu of
Research Problems, published by CRC Press.

Evan Chen<EMAIL_ADDRESS>MR author ID 1158569, ORCID number
0000-0001-9550-5068) is a mathematics Ph.D. student at the Massachusetts
Institute of Technology. He is the Co-Editor-in-Chief of the USA(J)MO
Editorial Board and a coach of the USA team at the International Mathematical
Olympiad. His book Euclidean Geometry in Mathematical Olympiads was published
by the MAA.


# Approximate Robust NMPC using Reinforcement Learning

Hossein Nejatbakhsh Esfahani, Arash Bahari Kordabad, Sébastien Gros Authors
are with Department of Engineering Cybernetics, Norwegian University of
Science and Technology (NTNU), Trondheim, Norway. {hossein.n.esfahani,
Arash.b.kordabad<EMAIL_ADDRESS>
###### Abstract

We present a Reinforcement Learning-based Robust Nonlinear Model Predictive
Control (RL-RNMPC) framework for controlling nonlinear systems in the presence
of disturbances and uncertainties. An approximate Robust Nonlinear Model
Predictive Control (RNMPC) of low computational complexity is used in which
the state trajectory uncertainty is modelled via ellipsoids. Reinforcement
Learning is then used in order to handle the ellipsoidal approximation and
improve the closed-loop performance of the scheme by adjusting the MPC
parameters generating the ellipsoids. The approach is tested on a simulated
Wheeled Mobile Robot (WMR) tracking a desired trajectory while avoiding static
obstacles.

##  I INTRODUCTION

Nonlinear Model Predictive Control (NMPC) is an optimization based control
approach operating in a receding horizon [1], which is often adopted for its
ability to handle linear/nonlinear state and input constraints. MPC offers
rigorous theoretical properties (such as recursive feasibility, constraint
satisfaction and stability), assuming that an accurate model of the plant is
available. There are many autonomous systems (such as ground vehicles, marine
robots and unmanned aerial vehicles) for which NMPC-based algorithms have been
adopted [2], [3], [4].

Reinforcement Learning (RL) is a powerful tool for tackling Markov Decision
Processes (MDP) without depending on a model of the probability distributions
underlying the state transitions of the real system [5]. Indeed, most RL
methods rely purely on observed state transitions, and realizations of the
stage cost in order to increase the performance of the control policy.

Recently, the integration of machine learning in MPC has been invesstigated,
with the aim of learning the model of the system, the cost function or even
the control law directly [6, 7]. For computational reasons, simple models are
usually preferred in the MPC scheme. Hence, the MPC model often does not have
the structure required to correctly capture the real system dynamics and
stochasticity. As a result, MPC delivers a reasonable but suboptimal
approximation of the optimal policy. Choosing the MPC parameters that
maximises the closed-loop performance for the selected MPC formulation is a
difficult problem. Indeed, e.g. selecting the MPC model parameters that best
fit the model to the real system is not guaranteed to yield the best closed-
loop performance that the MPC scheme can achieve [8]. In [8, 9], it is shown
that adjusting the MPC model, cost and constraints can be beneficial to
achieve the best closed-loop performances, and Reinforcement Learning is
proposed as a possible approach to perform that adjustment in practice.
Further recent research have focused on MPC-based policy approximation for RL
[10, 11, 12, 13, 14], where it is shown that a single Model Predictive Control
(MPC) scheme can capture the optimal value function, action-value function,
and policy of an MDP, even if the MPC model is inaccurate, hence providing a
valid and generic function approximator for RL.

Model-plant mismatch and disturbances can be treated via Robust NMPC (RNMPC)
techniques. For linear MPC models and polytopic disturbance models and
constraints, tube-based MPC techniques provides computationally effective
techniques [15], [16]. Treating nonlinear MPC models or generic disturbances
and constraints is more challenging [17]. Researchers in [18] proposed to use
a tube-based MPC with a Min-Max differential inequality. Multi-stage or
Scenario-tree NMPC scheme was proposed in [19, 20] as a real-time NMPC that
accounts for the uncertain influence and generates decisions to control a
nonlinear plant in a robust sense. These approaches remain challenging for
problems that are not of small scale.

In this paper we model the propagation of perturbations in the state dynamics
via ellipsoids, based on the linearization of the system dynamics and
constraints on the nominal trajectories and using a Gaussian disturbance
model. We propose to adjust this scheme using the RL method in order to tailor
this inaccurate uncertainty model to the real system and achieve a better
closed-loop performance. A fast convergence of the adjustable parameters of
RNMPC is achieved via a second-order Least Square Temporal Difference
Q-learning (LSTDQ).

This paper is organised as follows. In Section II, the proposed approximate
robust NMPC is formulated. The combination of the RL algorithm and RNMPC is
detailed in Section III. A mobile robot under uncertainties is adopted in
Section IV to be controlled by the proposed RL-RNMPC for a trajectory tracking
scenario associated to elliptic static obstacle avoidance in presence of model
uncertainty. Finally, conclusions and future work are given in Section V.

##  II Ellipsoidal-based Robust NMPC

We consider a model based on the nonlinear dynamics

$\displaystyle\boldsymbol{\mathrm{x}}_{+}=\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{u}},\boldsymbol{\mathrm{d}}\right)$
(1)

where $\boldsymbol{\mathrm{d}}$ is a stochastic variables affecting the state
evolution. We assume that the real system is unknown and imperfectly
represented by (1). A nominal NMPC scheme based on (1) reads e.g. as:

$\displaystyle\min_{\bar{\boldsymbol{\mathrm{x}}},\bar{\boldsymbol{\mathrm{u}}}}$
$\displaystyle\quad
T\left(\bar{\boldsymbol{\mathrm{x}}}_{N}\right)+\sum_{k=0}^{N-1}L\left(\bar{\boldsymbol{\mathrm{x}}}_{k},\bar{\boldsymbol{\mathrm{u}}}_{k}\right)$
(2a) $\displaystyle\mathrm{s.t.}$
$\displaystyle\quad\bar{\boldsymbol{\mathrm{x}}}_{k+1}=\boldsymbol{\mathrm{f}}\left(\bar{\boldsymbol{\mathrm{x}}}_{k},\bar{\boldsymbol{\mathrm{u}}}_{k},\boldsymbol{\mathrm{d}}_{k}\right),\quad\boldsymbol{\mathrm{x}}_{0}=\boldsymbol{\mathrm{s}}$
(2b)
$\displaystyle\quad\boldsymbol{\mathrm{h}}\left(\bar{\boldsymbol{\mathrm{x}}}_{k},\bar{\boldsymbol{\mathrm{u}}}_{k}\right)\leq
0,\quad\boldsymbol{\mathrm{h}}_{\mathrm{f}}\left(\bar{\boldsymbol{\mathrm{x}}}_{N}\right)\leq
0$ (2c)

where
$L\left(\bar{\boldsymbol{\mathrm{x}}}_{k},\bar{\boldsymbol{\mathrm{u}}}_{k}\right)$
and $T\left(\bar{\boldsymbol{\mathrm{x}}}_{N}\right)$ are the stage and
terminal costs, respectively. $\boldsymbol{\mathrm{s}}$ is the current system
state while the nominal NMPC predicted trajectories are
$\bar{\boldsymbol{\mathrm{x}}}$ and $\bar{\boldsymbol{\mathrm{u}}}$. We label
$\boldsymbol{\mathrm{h}}$ and $\boldsymbol{\mathrm{h}}_{f}$ as the nominal
stage and terminal inequality constraints, respectively. We will model the
model error and real system stochasticity via the uncertain sequence
${\boldsymbol{\mathrm{d}}}_{k}$ of disturbances, and we will approximate its
effect on the state dynamics in a computationally effective way. More
specifically, we will model the sequence ${\boldsymbol{\mathrm{d}}}_{k}$ via
an i.i.d Gaussian model with mean $\bar{\boldsymbol{\mathrm{d}}}_{k}$ and
covariance $\Lambda$ i.e.

$\displaystyle\boldsymbol{\mathrm{d}}_{k}\sim\mathcal{N}\left(\bar{\boldsymbol{\mathrm{d}}}_{k},\Lambda\right)$
(3)

The first-order deviation of the model state $\boldsymbol{\mathrm{x}}_{k}$
from its nominal trajectory $\bar{\boldsymbol{\mathrm{x}}}_{k}$ is then given
by:

$\displaystyle\Delta\boldsymbol{\mathrm{x}}_{k+1}=$
$\displaystyle\left.\frac{\partial\boldsymbol{\mathrm{f}}}{\partial\boldsymbol{\mathrm{x}}}\right|_{\boldsymbol{\mathrm{x}}_{k},\boldsymbol{\mathrm{u}}_{k}}\Delta\boldsymbol{\mathrm{x}}_{k}+\left.\frac{\partial\boldsymbol{\mathrm{f}}}{\partial\boldsymbol{\mathrm{u}}}\right|_{\boldsymbol{\mathrm{x}}_{k},\boldsymbol{\mathrm{u}}_{k}}\Delta\boldsymbol{\mathrm{u}}_{k}$
(4)
$\displaystyle\qquad+\left.\frac{\partial\boldsymbol{\mathrm{f}}}{\partial\boldsymbol{\mathrm{d}}}\right|_{\boldsymbol{\mathrm{x}}_{k},\boldsymbol{\mathrm{u}}_{k}}\Delta\boldsymbol{\mathrm{d}}_{k}$

where $\Delta\boldsymbol{\mathrm{x}}_{0}$ is a given deviation of the initial
state and
$\Delta\boldsymbol{\mathrm{d}}_{k}=\boldsymbol{\mathrm{d}}_{k}-\bar{\boldsymbol{\mathrm{d}}}_{k}$.
We observe that for
$\mathbb{E}\left[\Delta\boldsymbol{\mathrm{x}}_{0}\right]=0$,
$\bar{\boldsymbol{\mathrm{d}}}_{k}=\mathbb{E}\left[\boldsymbol{\mathrm{d}}_{k}\right]$
and $\boldsymbol{\mathrm{\Delta}}\boldsymbol{\mathrm{u}}_{k}=0$. The expected
value of the state deviation is
$\mathbb{E}\left[\Delta\boldsymbol{\mathrm{x}}_{k}\right]=0$. We consider a
linear feedback over the state deviations for the input deviation
$\Delta\boldsymbol{\mathrm{u}}_{k}$ as:

$\displaystyle\Delta\boldsymbol{\mathrm{u}}_{k}=-K_{k}\Delta\boldsymbol{\mathrm{x}}_{k}$
(5)

where $K_{k}$ is a control gain matrix typically based on Linear Quadratic
Regulator (LQR) techniques. Then the state deviation dynamics read as:

$\displaystyle\Delta\boldsymbol{\mathrm{x}}_{k+1}=$
$\displaystyle\left.\left(\frac{\partial\boldsymbol{\mathrm{f}}}{\partial\boldsymbol{\mathrm{x}}}-\frac{\partial\boldsymbol{\mathrm{f}}}{\partial\boldsymbol{\mathrm{u}}}K_{k}\right)\right|_{\boldsymbol{\mathrm{x}}_{k},\boldsymbol{\mathrm{u}}_{k}}\Delta\boldsymbol{\mathrm{x}}_{k}+\left.\frac{\partial\boldsymbol{\mathrm{f}}}{\partial\boldsymbol{\mathrm{d}}}\right|_{\boldsymbol{\mathrm{x}}_{k},\boldsymbol{\mathrm{u}}_{k}}\Delta\boldsymbol{\mathrm{d}}_{k}$

Let us define the time-varying matrices as follows:

$\displaystyle
A_{k}=\left.\frac{\partial\boldsymbol{\mathrm{f}}}{\partial\boldsymbol{\mathrm{x}}}-K_{k}\frac{\partial\boldsymbol{\mathrm{f}}}{\partial\boldsymbol{\mathrm{u}}}\right|_{\boldsymbol{\mathrm{x}}_{k},\boldsymbol{\mathrm{u}}_{k}},\quad
B_{k}=\left.\frac{\partial\boldsymbol{\mathrm{f}}}{\partial\boldsymbol{\mathrm{d}}}\right|_{\boldsymbol{\mathrm{x}}_{k},\boldsymbol{\mathrm{u}}_{k}}$

Then, the covariance of the state deviation reads as:

$\displaystyle\Sigma_{k+1}$
$\displaystyle=\mathbb{E}\left[\Delta\boldsymbol{\mathrm{x}}_{k+1}\Delta\boldsymbol{\mathrm{x}}_{k+1}^{\top}\right]$
$\displaystyle=\mathbb{E}\left[\left(A_{k}\Delta\boldsymbol{\mathrm{x}}_{k}+B_{k}\Delta\boldsymbol{\mathrm{d}}_{k}\right)\left(A_{k}\Delta\boldsymbol{\mathrm{x}}_{k}+B_{k}\Delta\boldsymbol{\mathrm{d}}_{k}\right)^{\top}\right]$
$\displaystyle=\mathbb{E}\left[A_{k}\Delta\boldsymbol{\mathrm{x}}_{k}\Delta\boldsymbol{\mathrm{x}}_{k}^{\top}A_{k}^{\top}+B_{k}\Delta\boldsymbol{\mathrm{d}}_{k}\Delta\boldsymbol{\mathrm{d}}_{k}^{\top}B_{k}^{\top}\right]$
$\displaystyle=A_{k}\Sigma_{k}A_{k}^{\top}+B_{k}\Lambda B_{k}^{\top}$

where $\Lambda$ is the covariance of the process noise. We observe that the
matrices $\Sigma_{k}$ describe ellipsoids of state uncertainty propagations
defined by the confidence regions:

$\displaystyle
R_{k}:=\left\\{\,\Delta\boldsymbol{\mathrm{x}}_{k}\,\left|\,\frac{1}{2}\Delta\boldsymbol{\mathrm{x}}_{k}^{\top}\Sigma_{k}^{-1}\Delta\boldsymbol{\mathrm{x}}_{k}\leq\sigma_{k}\right.\right\\}$
(6)

where $\sigma_{k}$ is a Mahalanobis distance (the radius of the ellipsoids as
the confidence regions). More specifically, we observe that for a Gaussian
disturbance $\boldsymbol{\mathrm{d}}_{k}$ as in (3) with $n$-dimensional mean
vector $\bar{\boldsymbol{\mathrm{d}}}_{k}$, the state deviation approximation
$\Delta x_{k}$ is also Gaussian, given by the following density:

$\displaystyle\rho\left(\Delta
x_{k}\right)=\left(2\pi\right)^{-\frac{n}{2}}\left(\mathrm{det}\left(\Sigma_{k}\right)\right)^{-\frac{1}{2}}e^{-\frac{1}{2}\Delta\boldsymbol{\mathrm{x}}_{k}^{\top}\Sigma_{k}^{-1}\Delta\boldsymbol{\mathrm{x}}_{k}}$
(7)

such that at time $k$, a state deviations $\Delta x_{k}$ has a probability
density larger or equal to
$\left(2\pi\right)^{-\frac{n}{2}}\left(\mathrm{det}\left(\Sigma_{k}\right)\right)^{-\frac{1}{2}}e^{-\sigma_{k}}$
to belong to $R_{k}$ in (6).

The probability of the state deviation to belong to the ellipsoid described by
(6) is given by:

$\displaystyle\mathbb{P}[\Delta x_{k}\in R_{k}]=\int_{R_{k}}\rho\left(\Delta
x_{k}\right)\mathrm{d}\Delta
x_{k}=\frac{\gamma_{k}\left(\frac{n}{2},\frac{\sigma^{2}_{k}}{2}\right)}{\Gamma_{k}\left(\frac{n}{2}\right)}$
(8)

where, $\Gamma_{k}$ is the Gamma function and $\gamma_{k}$ is the lower
incomplete Gamma function describing the probability content for a
n-dimensional multinormal distribution [21]. Imposing that the ellipsoids
described by (6) are contained within the MPC feasible set for some sequence
$\sigma_{k}$ then becomes a way to approximately satisfy the MPC constraints
with probability (8). In the next section, we detail how to build an
inexpensive Robust MPC scheme that ensures the ellipsoids described by (6) are
contained within the MPC feasible set.

###  II-A Inclusion constraint

Consider the constraint

$\displaystyle\boldsymbol{\mathrm{h}}_{i}\left({\boldsymbol{\mathrm{x}}}_{k},{\boldsymbol{\mathrm{u}}}_{k}\right)\leq
0$ (9)

for some index $i$ and suppose that we want to formulate a computationally
tractable constraint requiring that (9) is satisfied for all state deviations
in the ellipsoid (6). For nonlinear constraints, this is unfortunately very
difficult. As a result, in line with the philosophy above, we will consider
the approximation resulting from linearizing the constraint on the nominal
trajectory
$\bar{\boldsymbol{\mathrm{x}}}_{k},\bar{\boldsymbol{\mathrm{u}}}_{k}$ of the
MPC scheme. More specifically, we will consider imposing the constraint:

$\displaystyle\boldsymbol{\mathrm{h}}_{i}\left(\bar{\boldsymbol{\mathrm{x}}}_{k},\bar{\boldsymbol{\mathrm{u}}}_{k}\right)+$
$\displaystyle\left.\left(\frac{\partial\boldsymbol{\mathrm{h}}_{i}}{\partial\boldsymbol{\mathrm{x}}}-\frac{\partial\boldsymbol{\mathrm{h}}_{i}}{\partial\boldsymbol{\mathrm{u}}}K_{k}\right)\right|_{\bar{\boldsymbol{\mathrm{x}}}_{k},\bar{\boldsymbol{\mathrm{u}}}_{k}}\Delta\boldsymbol{\mathrm{x}}_{k}\leq
0,$ (10) $\displaystyle\forall\,\Delta\boldsymbol{\mathrm{x}}_{k}\in R_{k}$

Fortunately, this requirement takes a closed form. Indeed, we can simply seek
the $\Delta\boldsymbol{\mathrm{x}}_{k}\in\eqref{eq:Ellipsoid}$ that gives the
worst (highest) value to $\boldsymbol{\mathrm{h}}_{i}$. Let us label

$\displaystyle\boldsymbol{\mathrm{b}}_{i}=\left(\frac{\partial\boldsymbol{\mathrm{h}}_{i}}{\partial\boldsymbol{\mathrm{x}}}-\frac{\partial\boldsymbol{\mathrm{h}}_{i}}{\partial\boldsymbol{\mathrm{u}}}K_{k}\right)^{\top}$

Then (10) can be (tightly) enforced by imposing that:

$\displaystyle\boldsymbol{\mathrm{h}}_{i}+\sqrt{2\sigma_{k}}\left(\boldsymbol{\mathrm{b}}_{i}^{\top}\Sigma_{k}\boldsymbol{\mathrm{b}}_{i}\right)^{\frac{1}{2}}\leq
0$ (11)

which is a constraint mixing $\boldsymbol{\mathrm{h}}$ and its derivative, the
dynamics of $\Sigma_{k}$ and the given $\sigma_{k}$.

###  II-B Robust MPC formulation

We can now formulate an approximate robust NMPC scheme:

$\displaystyle V_{\theta}(\boldsymbol{\mathrm{s}})=$
$\displaystyle\min_{\bar{\boldsymbol{\mathrm{x}}},\bar{\boldsymbol{\mathrm{u}}},\boldsymbol{\mathrm{\zeta}},\Sigma}\quad\gamma^{N}\left(T_{\theta}\left(\bar{\boldsymbol{\mathrm{x}}}_{N}\right)+\boldsymbol{\mathrm{w}}_{f}^{\top}\boldsymbol{\mathrm{\zeta}}_{N}\right)$
$\displaystyle+\sum_{k=0}^{N-1}\gamma^{k}\left(l_{\theta}\left(\bar{\boldsymbol{\mathrm{x}}}_{k},\bar{\boldsymbol{\mathrm{u}}}_{k}\right)+\boldsymbol{\mathrm{w}}^{\top}\boldsymbol{\mathrm{\zeta}}_{k}\right)$
(12a) $\displaystyle\mathrm{s.t.}$
$\displaystyle\quad\bar{\boldsymbol{\mathrm{x}}}_{k+1}=\boldsymbol{\mathrm{f}}_{\theta}\left(\bar{\boldsymbol{\mathrm{x}}}_{k},\bar{\boldsymbol{\mathrm{u}}}_{k},\boldsymbol{\mathrm{d}}_{k}\right),\quad\bar{\boldsymbol{\mathrm{x}}}_{0}=\boldsymbol{\mathrm{s}}$
(12b) $\displaystyle\quad\Sigma_{k+1}=A_{k}\Sigma_{k}A_{k}^{\top}+B_{k}\Lambda
B_{k}^{\top},\quad\Sigma_{0}=S_{0}$ (12c)
$\displaystyle\quad\boldsymbol{\mathrm{h}}_{i}\left(\bar{\boldsymbol{\mathrm{x}}}_{k},\bar{\boldsymbol{\mathrm{u}}}_{k}\right)+\sqrt{2\sigma_{k}}\left.\left(\boldsymbol{\mathrm{b}}_{i}^{\top}\Sigma_{k}\boldsymbol{\mathrm{b}}_{i}\right)^{\frac{1}{2}}\right|_{\bar{\boldsymbol{\mathrm{x}}}_{k},\bar{\boldsymbol{\mathrm{u}}}_{k}}\leq\boldsymbol{\mathrm{\zeta}}_{k}$
(12d)
$\displaystyle\quad\boldsymbol{\mathrm{h}}_{i}^{\mathrm{f}}\left(\bar{\boldsymbol{\mathrm{x}}}_{N}\right)+\sqrt{2\sigma_{N}}\left.\left(\boldsymbol{\mathrm{b}}_{i}^{\top}\Sigma_{N}\boldsymbol{\mathrm{b}}_{i}\right)^{\frac{1}{2}}\right|_{\bar{\boldsymbol{\mathrm{x}}}_{N}}\leq\boldsymbol{\mathrm{\zeta}}_{N}$
(12e) $\displaystyle\quad\boldsymbol{\mathrm{\zeta}}_{0,\ldots,N}\geq 0$ (12f)

where, $0<\gamma\leq 1$ is a discount factor and $T_{\theta}$ is an adjustable
terminal cost function. The slack variables and initial guess for the
covariance matrix of the state deviation are labeled by
$\boldsymbol{\mathrm{\zeta}}$ and $S_{0}$, respectively. Although we
initialize $S_{0}$ at zero in this paper, One can use an observer scheme to
estimate this initial matrix. The adjustable stage cost function in the above
RNMPC scheme is defined as follows:

$\displaystyle
l_{\theta}\left(\bar{\boldsymbol{\mathrm{x}}}_{k},\bar{\boldsymbol{\mathrm{u}}}_{k}\right)=L\left(\bar{\boldsymbol{\mathrm{x}}}_{k},\bar{\boldsymbol{\mathrm{u}}}_{k}\right)+\varphi_{\theta}\left(\bar{\boldsymbol{\mathrm{x}}}_{k},\bar{\boldsymbol{\mathrm{u}}}_{k},\Sigma_{k}\right),$
(13)

where,
$\varphi_{\theta}\left(\bar{\boldsymbol{\mathrm{x}}}_{k},\bar{\boldsymbol{\mathrm{u}}}_{k},\Sigma_{k}\right)=\mathrm{Tr}\left(\frac{\partial^{2}L\left(\bar{\boldsymbol{\mathrm{x}}}_{k},\bar{\boldsymbol{\mathrm{u}}}_{k}\right)}{\partial\bar{\boldsymbol{\mathrm{x}}}^{2}_{k}}M\Sigma_{k}\right)$
is proposed to adopt as a cost modification term in the stage cost
$L\left(\bar{\boldsymbol{\mathrm{x}}}_{k},\bar{\boldsymbol{\mathrm{u}}}_{k}\right)$
of the RNMPC scheme. This term is considered to deliver the impact of the
uncertainty on the adopted stage cost. We propose to use a quadratic form for
this term, which includes the adjustable matrix $M$ as an RL parameter.
$\boldsymbol{\mathrm{f}}_{\theta}$ is the nonlinear dynamics, which could be
adjusted by RL through the model bias parameters. Because the robust NMPC
scheme (12) uses approximations and a possibly inaccurate model of the system
dynamics and disturbances, it may become infeasible. To address that issue, an
$\ell_{1}$ relaxation of the inequality constraints is introduced in (12f). If
the weights $\boldsymbol{\mathrm{w}},\boldsymbol{\mathrm{w}}_{f}$ are chosen
large enough, then the solution of the RNMPC scheme (12) respects the
constraints when it is feasible to do so.

We ought to specify here that the proposed approximate RNMPC may be
interpreted either as a genuine approximate robust NMPC scheme or an
approximate stochastic NMPC scheme depending on the nature of the real system
disturbances. Indeed, even if the real system does not yield any approximation
in (12), i.e. the disturbances are Gaussian and the system dynamics and
constraints are linear, then this scheme ensures the constraint satisfaction
in the sense of the probability (8), and shall therefore be seen as a
stochastic NMPC scheme. If the disturbances are bounded and properly covered
by the ellipsoidal model used in (12), then the proposed scheme can be seen as
a robust NMPC scheme. In both cases, the choice of parameters in (12) to model
the ellipsoids so as to achieve the given control objectives is difficult.

The NMPC parameters
$\theta=\left\\{M,\bar{\boldsymbol{\mathrm{d}}}_{k},\Lambda,\sigma_{k}\right\\}$
will be adjusted using LSTDQ-learning. We propose to use (12) as an
approximator for the true value function $V_{\star}$, i.e. we will seek the
robust MPC parameters $\boldsymbol{\mathrm{\theta}}$ that best achieve
$V_{\boldsymbol{\mathrm{\theta}}}\approx V_{\star}$. Let us define the control
policy as:

$\displaystyle\pi_{\theta}(\boldsymbol{\mathrm{s}})=\bar{\boldsymbol{\mathrm{u}}}_{0}^{\star}$
(14)

where, $\bar{\boldsymbol{\mathrm{u}}}_{0}^{\star}$ is the first element of the
input sequence
$\bar{\boldsymbol{\mathrm{u}}}_{0}^{\star},\cdots,\bar{\boldsymbol{\mathrm{u}}}_{N-1}^{\star}$
solution of (12). We next consider this optimal policy delivered by the RNMPC
scheme as an action $\boldsymbol{\mathrm{a}}$ in the context of reinforcement
learning where it is selected according to the above policy with the possible
addition of exploratory moves [5]. As a result in [8], a parameterized NMPC
scheme can be used as an (action)-value function approximator in the context
of the reinforcement learning. Then, we propose to use the parameterized RNMPC
scheme (12) to deliver the value function needed in the proposed second-order
LSTDQ-learning algorithm. Then, the action-value function in (15) results from
solving the same RNMPC scheme with its first input constrained to the
delivered action $\boldsymbol{\mathrm{a}}$.

$\displaystyle
Q_{\theta}(\boldsymbol{\mathrm{s}},\boldsymbol{\mathrm{a}})=\min_{\bar{\boldsymbol{\mathrm{x}}},\bar{\boldsymbol{\mathrm{u}}},\boldsymbol{\mathrm{\zeta}},\Sigma}$
$\displaystyle\quad\eqref{eq:V}$ (15a) $\displaystyle\mathrm{s.t.}$
$\displaystyle\quad\eqref{eq:dyn}-\eqref{eq:ineq}$ (15b)
$\displaystyle\quad\bar{\boldsymbol{\mathrm{u}}}_{0}=\boldsymbol{\mathrm{a}}$
(15c)

##  III RL-based Robust NMPC

In this section, we propose to use RL to tune the approximate robust NMPC
scheme. Indeed, the proposed robust NMPC scheme is based on coarse
approximations of the state uncertainty propagation, which are likely to yield
suboptimality and even constraints violations in practice. Finding the
parameters that yield the best closed-loop performance for the real system in
terms of cost and constraints violations is a difficult problem. We propose to
use RL to find these parameters. More specifically, RL is used to tune the
sequences of $\sigma_{k}$ and the expected value of the noise
$\bar{\boldsymbol{\mathrm{d}}}_{k}$, the process noise $\Lambda$ and the
matrix $M$ in the cost modification $\varphi_{\theta}$ in order to improve the
Robust NMPC scheme. RL allows one to perform this tuning based on observed
state transition, and without any knowledge of the real system stochasticity.

In this section, we first present the algorithmic details needed to implement
a second order LSTDQ-learning algorithm on the ellipsoidal RNMPC scheme. Then
the sensitivity analysis needed in the reinforcement learning scheme is
described.

###  III-A Second-Order LSTDQ Learning

LSTDQ-based reinforcement learning methods make an efficient use of data and
tend to converge faster than more basic temporal-difference learning methods.
Let us form the least squares of Bellman residual error w.r.t $\theta$ as:

$\displaystyle\min_{\theta}\mathbb{E}\left[\|Q_{\star}(\boldsymbol{\mathrm{s}},\boldsymbol{\mathrm{a}})-{Q}_{\theta}(\boldsymbol{\mathrm{s}},\boldsymbol{\mathrm{a}})\|^{2}\right]$
(16)

where, ${Q}_{\theta}$ obtained from the RNMPC scheme in (15) is an
approximation of the true action-value function $Q_{\star}$. Let us consider
the following approximation of the Bellman optimality equation, which is used
in the Temporal Difference (TD)-based learning approaches.

$\displaystyle
Q_{\star}(\boldsymbol{\mathrm{s}},\boldsymbol{\mathrm{a}})\approx
L(\boldsymbol{\mathrm{s}},\boldsymbol{\mathrm{a}})+\gamma
Q_{\theta}(\boldsymbol{\mathrm{s}}_{+},\pi(\boldsymbol{\mathrm{s}}_{+}))$ (17)

where, $L(\boldsymbol{\mathrm{s}},\boldsymbol{\mathrm{a}})$ is the baseline
cost in the RL scheme. Substituting (17) into (16), the least square problem
(16) for the first-order LSTDQ-learning can be solved for some data acquired
from the transitions
$\boldsymbol{\mathrm{s}}_{i},\boldsymbol{\mathrm{a}}_{i}\rightarrow\boldsymbol{\mathrm{s}}_{i+1}$
as:

$\displaystyle\mathbb{E}\left[\delta\nabla_{\theta}Q_{\theta}(\boldsymbol{\mathrm{s}}_{i},\boldsymbol{\mathrm{a}}_{i})\right]=0,$
(18a)
$\displaystyle\delta=L(\boldsymbol{\mathrm{s}}_{i},\boldsymbol{\mathrm{a}}_{i})+\gamma
V_{\theta}(\boldsymbol{\mathrm{s}}_{i+1})-Q_{\theta}(\boldsymbol{\mathrm{s}}_{i},\boldsymbol{\mathrm{a}}_{i})$
(18b)

where, $\delta$ is the temporal difference error. Both value and action-value
functions, respectively,
$V_{\theta}(\boldsymbol{\mathrm{s}}_{i+1})=Q_{\theta}(\boldsymbol{\mathrm{s}}_{i+1},\pi(\boldsymbol{\mathrm{s}}_{i+1}))$
and $Q_{\theta}(\boldsymbol{\mathrm{s}}_{i},\boldsymbol{\mathrm{a}}_{i})$ are
obtained from the RNMPC schemes detailed in the last section. In this paper,
we adopt a Newton method to solve (18) and extract the newton step for the
second-order LSTDQ scheme as follows:

$\displaystyle\theta\leftarrow\theta-\alpha A^{-1}b,$ (19a) $\displaystyle
A=\mathbb{E}\left[\delta\nabla^{2}_{\theta}Q_{\theta}+\nabla_{\theta}Q_{\theta}\nabla_{\theta}^{\top}\delta\right],\quad
b=\mathbb{E}\left[\delta\frac{\partial Q_{\theta}}{\partial\theta}\right]$
(19b)

where, the scalar $\alpha>0$ is the step size.

###  III-B Sensitivity Analysis

The gradient and Hessian of the function $Q_{\theta}$ needed in (19) require
one to compute the sensitivities of the optimal value of NLP (15). Let us
define the Lagrange function $\mathcal{L}_{\theta}$ associated to the RNMPC
problem (15) as follows:

$\displaystyle\mathcal{L}_{\theta}=\Phi_{\theta}+\boldsymbol{\mathrm{\lambda}}^{\top}G_{\theta}+\boldsymbol{\mathrm{\mu}}^{\top}H_{\theta}$
(20)

where $H_{\theta}$ gathers the inequality constraints of (15) and
$\Phi_{\theta}$ is the cost of the RNMPC optimization problem. Variable
$\boldsymbol{\mathrm{\lambda}}$ is the Lagrange multiplier vector associated
to the equality constraints $G_{\theta}$ of the RNMPC. Variable
$\boldsymbol{\mathrm{\mu}}$ is the Lagrange multiplier vector associated to
the inequality constraints of the RNMPC scheme. Let us label the primal
variables for the RNMPC scheme as
$\boldsymbol{\mathrm{p}}=\left\\{\boldsymbol{\mathrm{X}},\boldsymbol{\mathrm{U}}\right\\}$,
where $\boldsymbol{\mathrm{X}},\boldsymbol{\mathrm{U}}$ are the state and
control input trajectories predicted by (15). Then, the primal-dual variables
read as
$\boldsymbol{\mathrm{z}}=\left\\{\boldsymbol{\mathrm{p}},\boldsymbol{\mathrm{\lambda}},\boldsymbol{\mathrm{\mu}}\right\\}$.

The sensitivity analysis of optimization problems detailed in [22] delivers
the Gradient $\nabla_{\theta}Q_{\theta}$ and Hessian
$\nabla^{2}_{\theta}Q_{\theta}$ ($H(Q_{\theta})$) terms needed in (19) as
follows:

$\displaystyle\frac{\partial
Q_{\theta}}{\partial\theta}=\frac{\partial\mathcal{L}_{\theta}(\boldsymbol{\mathrm{s}},\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{z}}^{\star})}{\partial\theta},$
(21a) $\displaystyle
H(Q_{\theta})=\frac{D}{D\theta}\left(\frac{\partial\mathcal{L}_{\theta}(\boldsymbol{\mathrm{s}},\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{z}}^{\star})}{\partial\theta}\right)\approx\frac{\partial^{2}\mathcal{L}_{\theta}(\boldsymbol{\mathrm{s}},\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{z}}^{\star})}{\partial\theta^{2}}$
(21b)

where $\boldsymbol{\mathrm{z}}^{\star}$ is the primal-dual solution of (15)
and $D$ is the total derivative. The Hessian term $H(Q_{\theta})$ can improve
the search direction using the curvature information and thus make more
progress per step in comparison with the first-order LSTDQ learning relying
only on the gradient calculation.

###  III-C Constrained RL steps

In the proposed RL-RNMPC scheme, there is a matrix $\Lambda$ in the dynamics
(12c), which is adjusted using the reinforcement learning method. As a
requirement, this disturbance covariance matrix must be positive semidefinite.
However, the RL steps delivered by second-order LSTDQ learning do not
necessarily respect this requirement, and we need to enforce it via
constraints on the RL steps throughout the learning process. To address this
requirement, we formulate a Semi-Definite Program (SDP) as a least squares
optimization problem:

$\displaystyle\min_{\Delta\theta}$
$\displaystyle\quad\frac{1}{2}\left\|\Delta\theta\right\|^{2}-F^{\top}\Delta\theta$
(22a) $\displaystyle\mathrm{s.t.}$
$\displaystyle\quad\Lambda({\theta+\Delta\theta})\geq 0$ (22b)

where we assume that the above matrix is linear function of $\theta$ and the
gradient term $F=-\alpha A^{-1}b$ is delivered by the second-order LSTDQ-
learning algorithm (19).

##  IV Numerical Example

To illustrate proposed RL-RNMPC, we consider a simulated Wheeled Mobile Robot
(WMR) tracking trajectories while avoiding static obstacles avoidance, and
affected by model uncertainties and disturbances. Let us define the WMR model
as:

$\displaystyle\boldsymbol{\mathrm{f}}(\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{u}})=\begin{bmatrix}\cos(\psi)&0\\\
\sin(\psi)&0\\\ 0&1\end{bmatrix}\boldsymbol{\mathrm{u}}$ (23)

where $\boldsymbol{\mathrm{x}}=\left[x,y,\psi\right]^{\top}$ and
$\boldsymbol{\mathrm{u}}=\left[v,\omega\right]^{\top}$ are the state and
control input vectors, respectively. The position coordinates of the WMR are
labeled $x,y$ and $\psi$ is the robot orientation angle. The initial position
of the mobile robot is
$\boldsymbol{\mathrm{x}}_{0}=\left[-1,2,0\right]^{\top}$. The control inputs
$v$ and $\omega$ are the linear and angular velocities, respectively. To
discretize the above continuous model, we adopt a fourth-order Runge-Kutta
(RK4) integrator providing discretized function $\boldsymbol{\mathrm{f}}_{d}$
of the WMR model. We will consider that the real system evolves according to
the dynamics:

$\displaystyle\boldsymbol{\mathrm{x}}(k+1)=\boldsymbol{\mathrm{f}}_{d}\left(\boldsymbol{\mathrm{x}}(k),\boldsymbol{\mathrm{u}}(k)+\boldsymbol{\mathrm{\Gamma}}_{1}(k)\right)+\boldsymbol{\mathrm{\Gamma}}_{2}(k)$
(24)

where the two variables
$\boldsymbol{\mathrm{\Gamma}}_{1},\boldsymbol{\mathrm{\Gamma}}_{2}$ in the
above equations model the uncertainties as:

$\displaystyle\boldsymbol{\mathrm{\Gamma}}_{1}(k)=v(k)\begin{bmatrix}d_{1}(k)\\\
d_{2}(k)\\\
\end{bmatrix},\quad\boldsymbol{\mathrm{\Gamma}}_{2}(k)=T_{s}v(k)\begin{bmatrix}0\\\
0\\\ d_{2}(k)\end{bmatrix}$ (25)

where $d_{1}(k)\sim\mathcal{N}\left(0,\Sigma_{1}^{2}\right)$ and
$d_{2}(k)\sim\mathcal{N}\left(0,\Sigma_{2}^{2}\right)$. The sampling time is
$T_{s}$= 0.2 $s$ and disturbance variances $\Sigma_{1}$ and $\Sigma_{2}$ are
set as 0.2 and 0.4, respectively. We initialize the adjustable process noise
covariance matrix ($3\times 3$) as
$\Lambda=diag(\Sigma_{1}^{2},\Sigma_{2}^{2},\Sigma_{2}^{2})$. The expected
value of the process noise $\bar{d}$ and the matrix $M$ as an RL parameter in
the cost modification term $\varphi_{\theta}$ are initialized at zero.
According to (8), the probability of the uncertainties in the ellipsoids can
be obtained from the Gamma functions. Since the simulated mobile robot has
three states, there is a 3-dimensional multinormal distribution $n=3$. To
ensure a large enough probability of the uncertainty propagation in the
ellipsoids (confidence regions), we set $\sigma_{k}=2.65$. Therefore, with
pair $n=3,\quad\sigma_{k}=2.65$ we will have the probability of the state
deviations to belong to the ellipsoids $\mathbb{P}[\Delta x_{k}\in
R_{k}]\approx 93\%$.

The RMPC model will be based on the real system dynamics, assuming that we
know them perfectly. More specifically, the nominal model used in the RNMPC
scheme is based on the expected value of the exact distribution of the
disturbances $d_{1},d_{2}$, and the structure according to which the
disturbances enter the system is assumed known. We further observe that by
assuming the disturbances are Gaussian, we are considering an ideal setup for
the approximate robust MPC to perform well. Hence any gain of performance
achieved by RL in this simulation setup is done through handling the
approximations introduced in the robust MPC scheme. As it is proposed to adopt
the RL approach in order to adjust the RNMPC parameters, we formulate a
second-order LSTDQ scheme with a batch size equal to 100 data samples and let
RL to update the parameters based on the collected data in the batch. In this
simulation, the number of transitions (iterations) is 8000 and thus the number
of RL steps is 80. The step size is set as $\alpha=10^{-6}$. We adopt a
baseline stage cost used in the LSTDQ scheme as:

$\displaystyle
L(\boldsymbol{\mathrm{x}}_{k},\boldsymbol{\mathrm{u}}_{k})=l(\boldsymbol{\mathrm{x}}_{k},\boldsymbol{\mathrm{u}}_{k})+\boldsymbol{\mathrm{w}}^{\top}\max(0,\boldsymbol{\mathrm{h}}(\boldsymbol{\mathrm{x}}_{k}))$
(26)

where $l(\boldsymbol{\mathrm{x}}_{k},\boldsymbol{\mathrm{u}}_{k})$ can be
expressed as a quadratic function of the state and action deviations from
their desired values. The second term in the above baseline is considered to
cope with the violations, where $\boldsymbol{\mathrm{h}}\geq 0$ is pure
inequality vector of constraints induced by the obstacles. The penalty weights
are $\boldsymbol{\mathrm{w}}=[30,30,30]$.

The obstacles are represented as ellipsoids (see Fig. 1). The reference
trajectory shown in Fig. 1 is an eight-shaped path, which intersects the
obstacles or comes very close to them.

In Fig. 2, we compare the closed-loop performance of the nominal NMPC scheme
(2), the approximate robust NMPC without learning, and with learning. The
second-order LSTDQ-learning algorithm improves the performance of the robust
NMPC scheme in comparison with the classic NMPC and RNMPC without learning.

Figure 1: A comparative study: Trajectory tracking and obstacle avoidance for
a WMR under uncertainties Figure 2: Average Closed-Loop performance index for
25 elapsed trajectories (laps).

Since the approximate RNMPC scheme is built based on an exact knowledge of the
disturbance statistics and structure, and system model, the improvement of
performance observed in Fig. 2 results purely on improving the approximation
performed in the RNMPC scheme via modifying the robust NMPC parameters
$\theta=\left\\{M,\bar{\boldsymbol{\mathrm{d}}}_{k},\Lambda,\sigma_{k}\right\\}$.
For majority of the ellipsoids, the dimension of the ellipsoid-shaped
confidence regions $R_{k}$ is changed by RL to be modelled as larger than the
originally selected $\sigma_{k}$ in order to reduce the risk of constraints
violation, see Fig. 5. For some situations that the risk of hitting is not
high, this dimension is decreasing using the RL. Indeed, there is a trade-off
made by the RL between the avoidance and tracking accuracy in this proposed
RL-RNMPC.

Figure 3: Matrix $M$ is adjusted as a RL parameter $\theta$ in the cost
modification term
$\varphi_{\theta}\left(\bar{\boldsymbol{\mathrm{x}}}_{k},\bar{\boldsymbol{\mathrm{u}}}_{k},\Sigma_{k}\right)$.
Figure 4: Tuning the expected values of the process noises applied to the
three states for a prediction horizon $N=15$. Figure 5: Tuning the ellipsoidal
confidence region (adjustment of the radius)

The adjustment of the RNMPC can also contribute the mobile robot to avoid
hitting the obstacles 1,2 due to possible drifting (as an uncertainty) in the
road bend shown in Fig. 1. The RL scheme additionally modifies the uncertainty
model via $\bar{d}$ and $\Lambda$, see Fig. 4, in order to gain performance.

##  V Conclusion

In the context of the Robust Nonlinear Model Predictive Control (RNMPC), the
formal RNMPC techniques are difficult to implement on nonlinear systems, and
thus it is common to use approximate RNMPC methods instead. In this paper we
proposed to describe the propagation of the uncertainties using the
ellipsoidal tubes in which the disturbances and state deviations are modeled
as a Gaussian noise. However, the approximated models and constraints used in
this kind of RNMPC can affect the closed-loop performance and thus we adopted
an LSTDQ learning as a fast RL algorithm to adjust some crucial parameters of
the approximate RNMPC. In this paper, we used the proposed RNMPC as a value
function approximator for the LSTDQ algorithm. AS a future work, we will
propose to embed a Linear Quadratic Regulator (LQR) in the proposed RNMPC in
order to deliver the first guess of the control gain matrix for the adopted
linear feedback over the state deviation. Furthermore, we will investigate the
proposed RL-RNMPC for a large-scale application.

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# Uniqueness of conservative solutions to the the modified Camassa-Holm
equation via Characteristics

Zhen $\mbox{He}^{1}$ 111Email<EMAIL_ADDRESS>Zhaoyang
$\mbox{Yin}^{1,2}$222E-mail<EMAIL_ADDRESS>
${}^{1}\mbox{Department}$ of Mathematics, Sun Yat-sen University, Guangzhou
510275, China
${}^{2}\mbox{School}$ of Science,
Shenzhen Campus of Sun Yat-sen University, Shenzhen 518107, China

###### Abstract

In this paper,for a given conservative solution, we introduce a set of
auxiliary variables tailored to this particular solution, and prove that these
variables satisfy a particular semilinear system having unique solutions. In
turn, we get the uniqueness of the conservative solution in the original
variables.

2020 Mathematics Subject Classification: 35Q30, 35Q84, 76B03, 76D05.

Keywords: A modified Camassa-Holm equation; Globally weak solutions ;
Uniqueness.

###### Contents

  1. 1 Introduction
  2. 2 Basic definitons and results
  3. 3 Preliminary lemmas
  4. 4 Proof of Theorem2.3

##  1 Introduction

In this paper, we consider the Cauchy problem of the following modified
Camassa-Holm (MOCH) equation [26]

(1.1)
$\left\\{\begin{array}[]{l}\gamma_{t}=\lambda(v_{x}-\gamma-\frac{1}{\lambda}v\gamma)_{x},\quad
t>0,\ x\in\mathbb{R},\\\ v_{xx}=\gamma_{x}+\frac{\gamma^{2}}{2\lambda},\quad
t\geq 0,\ x\in\mathbb{R},\\\ \gamma(0,x)=\gamma_{0}(x),\quad
x\in\mathbb{R},\end{array}\right.$

which was called by Gorka and Reyes[16]. Let $G={\partial_{x}}^{2}-1,n=Gv$

The equation (1.1) can be rewritten as

(1.2)
$\left\\{\begin{array}[]{l}\gamma_{t}+G^{-1}n\gamma_{x}=\frac{\gamma^{2}}{2}+\lambda
G^{-1}n-\gamma G^{-1}n_{x},\quad t>0,\ x\in\mathbb{R},\\\
n=\gamma_{x}+\frac{\gamma^{2}}{2\lambda},\quad t\geq 0,\ x\in\mathbb{R},\\\
\gamma(0,x)=\gamma_{0}(x),\quad x\in\mathbb{R}.\end{array}\right.$

The equation $\eqref{eq0}$ was first studied through the geometric approach
in[6] [18]. Conservation laws and the existence and uniqueness of weak
solutions to the modified Camassa–Holm equation were presented in [16]. We
observe that if we solve (1.2), then $m$ will formally satisfy the following
physical form of the Camassa–Holm

(1.3) $\displaystyle n_{t}=-2vn_{x}-nv_{x}+\lambda v_{x}$

If $\lambda=0$, it is known as the well-known Camassa–Holm (CH) equation. As
far as we know, the CH equation has many properties, such as: integrability
[7, 10, 11], Hamiltonian structure, infinitely many conservation laws [7, 15].
The local well-posedness of CH equation has been proved in Sobolev spaces
$H^{s},s>\frac{3}{2}$ and in Besov space $B^{s}_{p,r}$ with
$s>\max\\{\frac{3}{2},1+\frac{1}{p}\\}$ or $s=1+\frac{1}{p},p\in[1,2],r=1$ in
[8, 9, 13, 14, 17, 19, 20]. In [12, 25], the authors showed the existence and
uniqueness of global weak solutions for the CH equation. In addition, Bressan
and Constantin studied the existence of the global conservative solutions [3]
and global disspative solutions [4] in $H^{1}(\mathbb{R}).$ Later, Bressan and
his collaborators used the similar technique to a variational wave equation[5]
and obtain the uniqueness [2]. Utilizing the good structure of the semilinear
ODE system proposed in [5], Bressan and Chen proved that, For an open dense
set of $C^{3}$ initial data, we prove that the solution is piecewise smooth in
the t-x plane, while the gradient $u_{x}$ can blow up along finitely many
characteristic curves[1]. Later, Li and Zhang studied similar property in [21]

Luo, Qiao and Yin studied the locally well-posedness in
$B^{s}_{p,r},s\textgreater\max\\{\frac{1}{2},\frac{1}{p}\\}$ or
$s=\frac{1}{p},1\leq p\leq 2,r=1$, blow up phenomena, global existence for
periodic MOCH and global conservative solution[22, 23, 24].

The remainder of the paper is organized as follows. In Section 2 we review
basic definitions and state our main uniqueness result Theorem 2.3 . Section 3
establishes the key technical tool (Lemma 3.2), determining a unique
characteristic curve through each initial point. In Section 4 we conclude the
proof of the main theorem. We can easily deduce that
$\gamma=(\partial_{x}-1)m$ is a unique solution to (1.1).

##  2 Basic definitons and results

Firstly we think about the initial problem

(2.1) $\left\\{\begin{array}[]{l}x_{t}=u(t,x)\\\
x(0)=\bar{x}.\end{array}\right.$

Now, let us consider a transformation
$m=(\partial_{x}+1)G^{-1}\gamma=(\partial_{x}-1)^{-1}\gamma$. Then we have
$\gamma=(\partial_{x}-1)m$, and therefore, equation (1.2) is changed to

(2.2) $m_{t}+um_{x}=F,$

where

$\displaystyle F=mu-P_{4}-P_{4x}+\frac{1}{2}(-m^{2}+P_{3}+P_{3x})+\lambda
P_{1x}+\frac{1}{2}(P_{5}+P_{5x}-P_{2})$ $\displaystyle P_{1}=G^{-1}m=p\ast
m,~{}~{}~{}~{}~{}P_{2}=p\ast m^{2},~{}~{}~{}~{}~{}P_{3}=p\ast m_{x}^{2},$
$\displaystyle
u=m-P_{1x}+P_{1}+\frac{1}{2\lambda}(P_{3}+P_{2}-P_{2x}),~{}~{}~{}~{}~{}P_{4}=p\ast
H,$ $\displaystyle P_{5}=p\ast
P_{3},~{}~{}~{}~{}~{}~{}H=um_{x}-um,~{}~{}~{}~{}~{}~{}p=\frac{1}{2}e^{-|x|}.$

For the initial data, we have

(2.3) $m(0,x)=\bar{m}(x)=(\partial_{x}-1)^{-1}\bar{\gamma}(x).$

For smooth solutions, differentiating (2.2) w.r.t. x one obtains

(2.4) $m_{xt}+u_{x}m_{x}+um_{xx}=F_{x},$

multiplying (2.4) with $m_{x}$, we obtain

(2.5) $\frac{d}{2dt}m_{x}^{2}+\frac{1}{2}(um_{x}^{2})_{x}=E,$

where

$\displaystyle
E=(mu)_{x}m_{x}-P_{4x}m_{x}-Hm_{x}-P_{4}m_{x}+\frac{1}{2}(-2mm_{x}^{2}+P_{3x}m_{x}+m_{x}^{3}+P_{3}m_{x})$
$\displaystyle~{}~{}~{}~{}~{}~{}+\lambda(m+P_{1})m_{x}+\frac{1}{2\lambda}(P_{5x}+P_{5}+P_{3}-P_{2}+m^{2})m_{x}-\frac{1}{2}u_{x}m_{x}^{2}$
$\displaystyle~{}~{}~{}=(mu)_{x}m_{x}-P_{4x}m_{x}-Hm_{x}-P_{4}m_{x}+\frac{1}{2}(P_{3x}m_{x}+P_{3}m_{x})+\lambda(m+P_{1})m_{x}$
$\displaystyle~{}~{}~{}~{}~{}~{}+\frac{1}{2\lambda}(P_{5x}+P_{5}+P_{3}-P_{2}+m^{2})m_{x}$
(2.6)
$\displaystyle~{}~{}~{}~{}~{}~{}-\frac{1}{2}(mm_{x}^{2}-P_{1}m_{x}^{2}+m_{x}^{2}P_{1x}-\frac{1}{2\lambda}(P_{3}+P_{2}-P_{2x})m_{x}^{2}).$

We denote $m_{x}^{2}$ by $w$, then

(2.7) $\frac{d}{dt}w+(uw)_{x}=2E.$

Because of (2.7), the characteristic curve $t\rightarrow x(t)$ satisies the
additional eqaution

(2.8)
$\frac{d}{dt}\int_{-\infty}^{x(t)}m_{x}^{2}(t,x)dx=\int_{-\infty}^{x(t)}2E(t,x)dx.$

And we introduce new coordinates (t,$\beta$), where $\beta$ is implictly
defined as

(2.9) $\displaystyle
x(t,\beta)+\int_{-\infty}^{x(t,\beta)}m_{x}^{2}(t,\xi)d\xi=\beta.$

At a time t where the measure $\mu_{(t)}$ is not absolutely continuous w.r.t
Lebesgue measure, we can define x(t,$\beta$) to be the unique point x such
that

(2.10) $\displaystyle x(t,\beta)+\mu_{(t)}(]-\infty,x[)\leq\beta\leq
x(t,\beta)+\mu_{(t)}(]-\infty,x])$

Then we shall prove Lipshcitz continuous of t and u as functions of the
variables t,$\beta$. Before providing our main results in this paper, let us
first introduce some definitions of the globally conservative solutions for
(1.1) and (2.2)

######  Definition 2.1.

Let $\bar{\gamma}\in L^{2}(\mathbb{R})$. We say that $\gamma(t,x)$ is locally
conservative solution to the Cauchy problem (1.1) if $\gamma$ satisies the
following equality:

$\displaystyle\int_{\mathbb{R}^{+}}\int_{\mathbb{R}}\big{(}\gamma\psi_{t}+(\gamma\partial_{x}G^{-1}\gamma+\frac{1}{2\lambda}\gamma
G^{-1}\gamma^{2})\psi_{x}\big{)}(t,x)dxdt$
$\displaystyle=-\int_{\mathbb{R}^{+}}\int_{\mathbb{R}}\big{(}(\frac{\gamma^{2}}{2}+\lambda\partial_{x}G^{-1}\gamma+\frac{1}{2}G^{-1}\gamma^{2})\psi\big{)}(t,x)dxdt-\int_{\mathbb{R}}\bar{\gamma}(x)\psi(0,x)dx,$

for every test function $\phi\in C_{c}^{\infty}(\mathbb{R}^{2})$

######  Definition 2.2.

A solution m=m(t,,x) is conservative if $w=m_{x}^{2}$ provides a
distributional solution to the balance law (2.5), namely

(2.11)
$\displaystyle\int_{0}^{\infty}\int[m_{x}^{2}\phi_{t}+um_{x}^{2}\phi_{x}+2E\phi]dxdt+\int
m_{0,x}^{2}\phi(0,x)dx=0$

for every test function $\phi\in C_{c}^{1}(\mathbb{R}^{2})$

Our main theorem is stated as follows.

######  Theorem 2.3.

For any initial data $\bar{m}\in H^{1}(\mathbb{R})$, the Cauchy problem (2.2)
has a unique conservative solution

######  Corollary 2.4.

For any initial data $\gamma_{0}\in L^{2}(\mathbb{R})$, the Cauchy problem
(1.1) has a unique conservative solution

##  3 Preliminary lemmas

######  Lemma 3.1.

Let m=m(t,x) be a conservative solution of equation (2.2). Then, for every
$t\geq 0$, the maps $\beta\rightarrow x(t,\beta)$ and $\beta\rightarrow
m(t,\beta)\triangleq m(t,x(t,\beta))$ implicitly defined by (2.10) are
Lipschitz continuous with constant 1. The map t $\rightarrow x(t,\beta)$ is
also Lipschitz continuous with constant 1. The map $t\rightarrow x(t,\beta)$
is also Lipschitz continuous with a constant deppending only on
$\|m_{0}\|_{H^{1}}$

###### Proof.

Fix any time t$\geq 0$. Then the map

(3.1) $\displaystyle x\rightarrow\beta(t,x)\triangleq
x+\int_{-\infty}^{x}m_{x}^{2}(t,y)dy$

is right continuous and strictly increasing. Hence it has a well defined,
continuous, nondecreasing inverse $\beta\rightarrow x(t,\beta)$. If
$\beta_{1}<\beta_{2}$, by (2.10) we obtain

(3.2) $\displaystyle
x(t,\beta_{2})-x(t,\beta_{1})+\mu_{(t)}(]-\infty,x(\beta_{2})[)-\mu_{(t)}(]-\infty,x(\beta_{1})])\leq\beta_{2}-\beta_{1},$

which implies

(3.3) $\displaystyle x(t,\beta_{2})-x(t,\beta_{1})\leq\beta_{2}-\beta_{1}$

showing that the map $\beta\rightarrow x(t,\beta)$ is a contraction.

According to (3.2)

$\displaystyle|m(t,x(t,\beta_{2}))-m(t,x(t,\beta_{1}))|$
$\displaystyle\leq\int_{x(t,\beta_{1})}^{x(t,\beta_{2})}|m_{x}|dx\leq\int_{x(t,\beta_{1})}^{x(t,\beta_{2})}\frac{1}{2}|1+m_{x}^{2}|dx$
$\displaystyle=\frac{1}{2}(x(t,\beta_{2})-x(t,\beta_{1}))+\int_{x(t,\beta_{1})}^{x(t,\beta_{2})}\frac{1}{2}|m_{x}^{2}|dx$
$\displaystyle\leq\frac{1}{2}(x(t,\beta_{2})-x(t,\beta_{1}))+\frac{1}{2}\mu_{(t)}(]x(t,\beta_{1}),x(t,\beta_{2})[)$
(3.4) $\displaystyle\leq\frac{1}{2}(\beta_{2}-\beta_{1}).$

Next we prove the Lipschitz continuity of the map $t\rightarrow x(t,\beta)$.
By Sobolev’s inequality, we have

(3.5) $\displaystyle\|m\|_{L^{\infty}}$ $\displaystyle\leq\|m\|_{H^{1}}.$

Using Young’s inequality, we get

(3.6) $\displaystyle\|P_{1}\|_{L^{\infty}}$
$\displaystyle\leq\|e^{-|\cdot|}\|_{L^{1}}\|m\|_{L^{\infty}}\leq
C\|m\|_{H^{1}},~{}~{}~{}\|P_{1x}\|_{L^{\infty}}\leq\|sgn(\cdot)e^{-|\cdot|}\|_{L^{1}}\|m\|_{L^{\infty}}\leq
C\|m\|_{H^{1}},$ (3.7) $\displaystyle\|P_{2}\|_{L^{\infty}}$
$\displaystyle\leq\|e^{-|\cdot|}\|_{L^{\infty}}\|m^{2}\|_{L^{1}}\leq
C\|m\|_{H^{1}},~{}~{}~{}\|P_{2x}\|_{L^{\infty}}\leq\|sgn(\cdot)e^{-|\cdot|}\|_{L^{\infty}}\|m^{2}\|_{L^{1}}\leq
C\|m\|_{H^{1}},$ (3.8) $\displaystyle\|P_{3}\|_{L^{\infty}}$
$\displaystyle\leq\|e^{-|\cdot|}\|_{L^{\infty}}\|m_{x}^{2}\|_{L^{1}}\leq
C\|m\|_{H^{1}},~{}~{}~{}\|P_{3x}\|_{L^{\infty}}\leq\|sgn(\cdot)e^{-|\cdot|}\|_{L^{\infty}}\|m_{x}^{2}\|_{L^{1}}\leq
C\|m\|_{H^{1}},$

From (3.6)-(3.8), we can deduce that

(3.9) $\displaystyle\|u\|_{L^{\infty}}\leq C_{\lambda}\|m\|_{H^{1}}\triangleq
C_{\infty}.$

Similarly

(3.10) $\displaystyle\|P_{5}\|_{L^{\infty}}$
$\displaystyle\leq\|e^{-|\cdot|}\|_{L^{1}}\|P_{3}\|_{L^{\infty}}\leq
C\|m\|_{H^{1}},~{}~{}~{}\|P_{5x}\|_{L^{\infty}}\leq\|sgn(\cdot)e^{-|\cdot|}\|_{L^{1}}\|P_{3}\|_{L^{\infty}}\leq
C\|m\|_{H^{1}}.$

To estimate $\|P_{4}\|_{L^{\infty}}$, we can see

(3.11) $\displaystyle\|P_{4}\|_{L^{\infty}}=\|p\ast
H\|_{L^{\infty}}\leq\|e^{|\cdot|}\|_{L^{\infty}}\|um_{x}\|_{L^{1}}+\|e^{|\cdot|}\|_{L^{1}}\|um\|_{L^{\infty}}.$

Then the main diffculty that confronts us is the estimation of the term
$\|um_{x}\|_{L^{1}}$.

Firstly, by Young’s inequality

$\displaystyle\|m_{x}P_{3}\|_{L^{1}}$ $\displaystyle\leq
C\|(1+\frac{m_{x}^{2}}{2})G^{-1}m_{x}^{2}\|_{L^{1}}\leq
C(\|G^{-1}m_{x}^{2}\|_{L^{1}}+\|\frac{m_{x}^{2}}{2}G^{-1}m_{x}^{2}\|_{L^{1}})$
$\displaystyle\leq
C\|e^{-|\cdot|}\|_{L^{\infty}}\|m_{x}^{2}\|_{L^{1}}+\|\frac{m_{x}^{2}}{2}\|_{L^{1}}\|G^{-1}m_{x}^{2}\|_{L^{\infty}}$
$\displaystyle\leq
C(\|m_{x}^{2}\|_{L^{1}}+\|m_{x}^{2}\|_{L^{1}}\|e^{|\cdot|}\|_{L^{\infty}}\|m_{x}^{2}\|_{L^{1}})$
(3.12) $\displaystyle\leq C(\|m\|_{H^{1}}+\|m\|_{H^{1}}^{2}).$

Similarly we have

(3.13)
$\displaystyle\|m_{x}P_{2}\|_{L^{1}},\|m_{x}P_{2x}\|_{L^{1}},\|m_{x}P_{1}\|_{L^{1}},\|m_{x}P_{1x}\|_{L^{1}}\leq
C\|m\|_{H^{1}}^{2}.$

Then one obtains

(3.14) $\displaystyle\|um_{x}\|_{L^{1}}\leq
C(\|m\|_{H^{1}}+\|m\|_{H^{1}}^{2}).$

Using the same mothed, we can deduce that

(3.15) $\displaystyle\|P_{4}\|_{L^{\infty}}\leq
C(\|m\|_{H^{1}}+\|m\|_{H^{1}}^{2}),~{}~{}~{}\|P_{4x}\|_{L^{\infty}}\leq
C(\|m\|_{H^{1}}+\|m\|_{H^{1}}^{2}).$

Next, by Young’s inequality, we have

(3.16)
$\displaystyle\|P_{1}\|_{L^{1}},~{}\|P_{2}\|_{L^{1}},~{}\|P_{3}\|_{L^{1}},~{}\|P_{3x}\|_{L^{1}},~{}\|P_{5}\|_{L^{1}},~{}\|P_{5x}\|_{L^{1}}\leq
C\|m\|_{H^{1}},$

and

(3.17) $\displaystyle\|P_{2}\|_{L^{2}}\leq
C\|e^{-|\cdot|}\|_{L^{2}}\|m^{2}\|_{L^{1}}\leq C\|m\|_{H^{1}}$ , (3.18)
$\displaystyle\|P_{2x}\|_{L^{2}}\leq
C\|sgn(\cdot)e^{-|\cdot|}\|_{L^{2}}\|m^{2}\|_{L^{1}}\leq C\|m\|_{H^{1}}$
(3.19) $\displaystyle\|P_{3}\|_{L^{2}}\leq C\|m\|_{H^{1}}$ .

From (3.18) and (3.19), one obtains

(3.20) $\displaystyle\|u\|_{L^{2}}\leq C\|m\|_{H^{1}}.$

Similarly,

(3.21) $\displaystyle\|P_{4}\|_{L^{1}},\|P_{4x}\|_{L^{1}}\leq\|H\|_{L^{1}}\leq
C(\|m\|_{H^{1}}+\|m\|_{H^{1}}^{2}).$

By (3.16)-(3.21), we can deduce that there exists a constant C depending on
$\lambda$ such that

(3.22) $\displaystyle\|F\|_{L^{1}}\leq
C(\|m\|_{H^{1}}+\|m\|_{H^{1}}^{2})\triangleq C_{S}.$

Assuming that $t>\tau$ and y=x($\tau$,$\beta$), we obtain

$\displaystyle\mu_{(t)}(]-\infty,y-C_{\infty}(t-\tau)[)$
$\displaystyle=\mu_{(\tau)}((-\infty,y))-\mu_{(\tau)}((-\infty,y))+\mu_{(\tau)}((-\infty,x(t,\beta)))$
$\displaystyle~{}~{}~{}+\mu_{(t)}((-\infty,y))-\mu_{(t)}((-\infty,y))-\mu_{(t)}((y-C_{\infty}(t-\tau),y))$
$\displaystyle\leq\mu_{(\tau)}((-\infty,y))+\mu_{(\tau)}((-\infty,x(t,\beta)))-\mu_{(\tau)}(]-\infty,y[)$
$\displaystyle~{}~{}~{}+\mu_{(t)}((x(t,\beta),y))-\mu_{(t)}\mu_{(t)}((y-C_{\infty}(t-\tau),y))$
(3.23)
$\displaystyle\leq\mu_{(\tau)}((-\infty,y))+\int_{\tau}^{t}\|F\|_{L^{1}}dt^{\prime}\leq\mu_{(\tau)}((-\infty,y))+C_{S}(t-\tau).$

Let $y^{-}(t)\triangleq y-(C_{\infty}+C_{S})(t-\tau)$, we get

$\displaystyle y^{-}+\mu_{(t)}(]-\infty,y^{-}(t)[)$ $\displaystyle\leq
y-(C_{\infty}-C_{S})(t-\tau)+\mu_{(\tau)}+C_{S}(t-\tau)$ (3.24)
$\displaystyle\leq y-\mu_{(t)}(]-\infty,y[)\leq\beta$

Which implies

$x(t,\beta)\geq y^{-}(t)=y-(C_{\infty}+C_{S})(t-\tau).$

Likewise, if we define $y^{+}(t)\triangleq y+(C_{\infty}+C_{S})(t-\tau)$, we
can deduce that

$x(t,\beta)\leq y^{+}(t)=y+(C_{\infty}+C_{S})(t-\tau).$

∎

######  Lemma 3.2.

Let $m(t,x)\in H^{1}(\mathbb{R})$ be a conservative solution of the Cauchy
problem (2.2). Then, for any $\bar{y}\in\mathbb{R}$, there exists a unique
Lipschitz continuous map $t\mapsto x(t)$ which satisfies both (2.8) and (2.1).
Moreover, for any $0\leq t\leq\tau$ one has

(3.25) $\displaystyle m(t,x(t))-m(\tau,x(\tau))=-\int_{\tau}^{t}F(s,x(s))ds.$

###### Proof.

  * 1.

Using the adapted coordinates $(t,\beta)$ as in (2.8), we asumme that x(t) is
the charateristic starting at $\bar{y}$ in the form of $t\mapsto
x(t)=x(t,\beta(t))$, where $\beta(t)$ is a map to be determined. We sum (2.8)
and (2.1), and integrating w.r.t. time we obtain

(3.26) $\displaystyle
x(t)+\int_{-\infty}^{x(t)}m_{x}^{2}(t,x)dx=\bar{y}+\int_{-\infty}^{\bar{y}}m_{0,x}^{2}dx+\int_{0}^{t}(u(t,x(t))+\int_{-\infty}^{x(t)}Edx)dt.$

Introducing the function

(3.27) $\displaystyle
G(t,\beta)\triangleq\int_{-\infty}^{x(t,\beta)}u_{x}+Edx$

and the constant

(3.28)
$\displaystyle\bar{\beta}=\bar{y}+\int_{-\infty}^{\bar{y}}m_{0,x}^{2}dx,$

we can rewrite the equation (3.26) as the form of

(3.29) $\displaystyle\beta(t)=\bar{\beta}+\int_{0}^{t}G(s,\beta(s))ds.$

  * 2.

Next, we cliam that For each fixed $t\geq 0$, the function $\beta\mapsto
G(t,\beta)$ defined at (3.27) is uniformly bounded and absolutely continuous.
Moreover,

(3.30) $\displaystyle
G_{\beta}=[u_{x}+E]x_{\beta}=\frac{u_{x}+E}{1+m_{x}^{2}}\in[-C,C]$

for some constant C depending only on the norm of $m$ and $\lambda$.

To prove the claim, we notice that

(3.31)
$\displaystyle\|G_{\beta}\|_{L^{\infty}}=\|\frac{u_{x}+E}{1+m_{x}^{2}}\|_{L^{\infty}}\leq\|\frac{u_{x}}{1+m_{x}^{2}}\|_{L^{\infty}}+\|\frac{E}{1+m_{x}^{2}}\|_{L^{\infty}}.$

By (3.6)-(3.8), we can deduce that

(3.32) $\displaystyle\|\frac{u_{x}}{1+m_{x}^{2}}\|_{L^{\infty}}\leq C.$

Using the fact that

(3.33)
$\displaystyle\frac{m_{x}}{1+m_{x}^{2}},\frac{m_{x}^{2}}{1+m_{x}^{2}}\leq 1,$

it is not hard to check that for any $i=1,2,3,4,5$, the following inequalities
holds

(3.34)
$\displaystyle\|\frac{m_{x}P_{i}}{1+m_{x}^{2}}\|_{L^{\infty}},\|\frac{m_{x}P_{ix}}{1+m_{x}^{2}}\|_{L^{\infty}}\leq
C,$

and by (3.6)-(3.8) and (3.13),we can deduce that

$\displaystyle\|\frac{mu_{x}m_{x}}{1+m_{x}^{2}}\|_{L^{\infty}}$
$\displaystyle\leq
C\|m\|_{L^{\infty}}\|\frac{m_{x}^{2}-mm_{x}+\frac{1}{2\lambda}(P_{2x}+P_{3}-m^{2}-P_{2})m_{x}}{1+m_{x}^{2}}\|_{L^{\infty}}$
$\displaystyle\leq
C\|m\|_{L^{\infty}}(\|\frac{m_{x}^{2}}{1+m_{x}^{2}}\|_{L^{\infty}}+\|m\|_{L^{\infty}}\|\frac{m_{x}}{1+m_{x}^{2}}\|_{L^{\infty}}$
$\displaystyle~{}~{}~{}+\frac{1}{2\lambda}\|P_{2x}+P_{3}-m^{2}-P_{2}\|_{L^{\infty}}\|\frac{m_{x}}{1+m_{x}^{2}}\|_{L^{\infty}}$
(3.35) $\displaystyle\leq C.$

Moreover, one can derive that

$\displaystyle\|\frac{m_{x}(P_{5x}+P_{5xx})}{1+m_{x}^{2}}\|_{L^{\infty}}$
$\displaystyle=\|\frac{m_{x}(\partial_{x}^{2}+\partial_{x})G^{-1}G^{-1}m_{x}^{2}}{1+m_{x}^{2}}\|_{L^{\infty}}$
$\displaystyle\leq
C\|G^{-1}m_{x}^{2}+G^{-1}G^{-1}m_{x}^{2}+\partial_{x}G^{-1}G^{-1}m_{x}^{2}\|_{L^{\infty}}$
(3.36) $\displaystyle\leq C\|m_{x}^{2}\|_{L^{1}}\leq C,$

and

$\displaystyle\|\frac{Hm_{x}}{1+m_{x}^{2}}\|_{L^{\infty}}$
$\displaystyle\leq\|\frac{um_{x}^{2}}{1+m_{x}^{2}}\|_{L^{\infty}}+\|\frac{umm_{x}}{1+m_{x}^{2}}\|_{L^{\infty}}$
(3.37) $\displaystyle\leq
C(\|u\|_{L^{\infty}}+\|u\|_{L^{\infty}}\|m\|_{L^{\infty}}).$

Combining (3.34)-(3.37), we conclude that

(3.38) $\displaystyle\|\frac{E}{1+m_{x}^{2}}\|_{L^{\infty}}\leq C,$

which completed the proof of our claim.

Hence the function $G$ in (3.27) is uniformly Lipschitz continuous w.r.t.
$\beta$.

  * 3.

Thanks to the Lipschitz continuity of the function G, the existence of a
unique solution to the integral equation (3.29) can be proved by a standard
fixed point argument. Namely, consider the Banach space of all continuous
functions $\beta:\mathbb{R}_{+}\mapsto\mathbb{R}$ with weighted norm

$\|\beta\|_{\star}\triangleq\sup_{t\geq 0}e^{-2Ct}|\beta(t)|.$

On this space, we claim that the Picard map

(3.39)
$\displaystyle\mathscr{P}\beta(t)\triangleq\bar{\beta}+\int_{0}^{t}G(\tau,\beta(\tau))d\tau$

is a strict contraction. Indeed, assume $\|\beta-\tilde{\beta}\|=\delta>0$.
This implies

(3.40) $\displaystyle|\beta(\tau)-\beta(\tau)|\leq\delta e^{2C\tau}$

Hence

$\displaystyle|\mathscr{P}\beta(t)-\mathscr{P}\tilde{\beta}(t)|$
$\displaystyle=|\int_{0}^{t}(G(\tau,\beta(\tau))-G(\tau,\tilde{\beta(\tau)}))d\tau|$
(3.41) $\displaystyle\leq
C\int_{0}^{t}|\beta(\tau)-\tilde{\beta(\tau)}|d\tau\leq\int_{0}^{t}C\delta
e^{2C\tau}d\tau\leq\frac{\delta}{2}e^{2Ct}.$

Then we conclude
$\|\mathscr{P}\beta-\mathscr{P}\tilde{\beta}\|_{\star}\leq\frac{\delta}{2}$.The
contraction mapping principle guarantees that (3.29) has a unique solution.
Thanks to the arbitrary of the $T$, we infer that the integral equation (3.29)
has a unique solution on $\mathbb{R}_{+}$.

  * 4

By the previous construction, the map $t\mapsto x(t)\triangleq x(t,\beta(t))$
defined by (2.10) provides the unique solution to (3.29). Due to the Lipschitz
continuity of $\beta(t)$ and x(t)=x(t,$\beta(t)$), $\beta(t)$ and $x(t)$ are
differentiable almost everywhere, so we only have to consider the time where
$x(t)$ is differentiable. It suffices to show that (2.1) holds at almost every
time. Assume, on the contrary, $\dot{x}(\tau)\neq u(\tau,x(\tau))$. Without
loss of generality, let

(3.42) $\displaystyle\dot{x}(\tau)=u(\tau,x(\tau))+2\epsilon_{0}$

for some $\epsilon_{0}>0$. The case $\epsilon_{0}<0$ is entirely similar. To
derive a contradiction we observe that for all $t\in(\tau,\tau+\delta)$, with
$\delta>0$ small enough one has

(3.43) $\displaystyle x^{+}(t)\triangleq
x(\tau)+(t-\tau)[u(\tau,x(\tau))+\epsilon_{0}]<x(t).$

We also observe that if $\phi$ is Lipschitz continuous with compact support,
then the approximation argument guarantees that (2.11) remain hold for any
test function $\phi\in H^{1}(\mathbb{R})$ with compact support. For any
$\epsilon>0$ enough small, we give the following functions

(3.44)
$\varrho^{\varepsilon}(s,x)=\left\\{\begin{array}[]{rcl}0,&y\leq-{\varepsilon}^{-1},\\\
x+{\varepsilon}^{-1},&~{}~{}-{\varepsilon}^{-1}\leq y\leq
1-{\varepsilon}^{-1},\\\ 1,&~{}~{}1-{\varepsilon}^{-1}\leq y\leq y^{+}(x),\\\
1-{\varepsilon}^{-1}(x-x(s)),&~{}~{}~{}~{}~{}~{}y^{+}(s)\leq y\leq
y^{+}(s)+\varepsilon,\\\ 0,&~{}~{}~{}~{}~{}~{}y\geq
y^{+}(s)+\varepsilon,\end{array}\right.$ (3.45)
$\chi^{\varepsilon}(s)=\left\\{\begin{array}[]{rcl}0,&~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}s\leq\tau-{\varepsilon}^{-1},\\\
{\varepsilon}^{-1}(s-\tau+\varepsilon),&~{}~{}\tau-{\varepsilon}\leq
s\leq\tau,\\\ 1,&~{}~{}\tau\leq s\leq t,\\\
1-{\varepsilon}^{-1}(s-t),&~{}~{}t\leq s\leq t+\varepsilon,\\\
0,&~{}~{}~{}~{}~{}~{}~{}s\geq t+\varepsilon.\end{array}\right.$

Define

$\displaystyle\phi^{\varepsilon}(s,x)=\min\\{{\varrho^{\varepsilon}(s,x),\chi^{\varepsilon}(s)}\\}.$

Using $\phi^{\epsilon}$ as test function in (2.11) we obtain

(3.46)
$\displaystyle\int_{0}^{\infty}\int[m_{x}^{2}\phi^{\epsilon}_{t}+um_{x}^{2}\phi^{\epsilon}_{x}+2E\phi^{\epsilon}]dxdt+\int
m_{0,x}^{2}\phi^{\epsilon}(0,x)dx=0.$

If $t$ is sufficiently close to $\tau,$ we have

$\displaystyle\lim\limits_{\varepsilon\rightarrow
0}\int_{\tau}^{t}\int_{x^{+}(s)-\varepsilon}^{x^{+}(s)+\varepsilon}m_{x}^{2}\phi_{t}^{\epsilon}+(um_{x}^{2})\phi_{x}^{\epsilon}+2E\phi^{\epsilon}dyds\geq
0.$

Using the fact that $m(s,x)<m(\tau)+{\epsilon}_{0}$ and
${\phi}_{x}^{\epsilon}\leq 0.$ For any $s\in[\tau+\varepsilon,t-\varepsilon],$
we infer that

(3.47) $\displaystyle
0={\phi}_{t}^{\varepsilon}+[u(\tau,x(\tau))+\epsilon_{0}]{\phi}_{x}^{\varepsilon}\leq{\phi}_{t}^{\varepsilon}+(u(s,x(s))){\phi}_{x}^{\varepsilon}.$

Noticing that the family of measures ${\mu}_{t}$ depends continuously on $t$
in the topology of weak convergence, taking the limit of (3.46) as
$\epsilon\rightarrow 0$, we obtain

$\displaystyle 0=$
$\displaystyle\int_{-\infty}^{x(\tau)}m^{2}_{x}(\tau,y)dy-\int_{-\infty}^{x^{+}(t)}m^{2}_{x}(t,x)dx+\int_{\tau}^{t}\int_{-\infty}^{x^{+}(s)}2E\phi^{\varepsilon}(s,x)dxds$
$\displaystyle+\lim\limits_{\varepsilon\rightarrow
0}\int_{\tau}^{t}\int_{{x^{+}}(s)-\varepsilon}^{{x^{+}}(s)+\varepsilon}m_{x}^{2}(\phi_{t}^{\epsilon}+u\phi_{x}^{\epsilon})(s,x)dxds$
(3.48) $\displaystyle\geq$
$\displaystyle\int_{-\infty}^{x(\tau)}m^{2}_{x}(\tau,x)dx-\int_{-\infty}^{x^{+}(t)}m^{2}_{x}(t,x)dx+\int_{\tau}^{t}\int_{-\infty}^{x^{+}(s)}2E(s,x)dxds.$

In turn,(4) implies

(3.49) $\displaystyle\int_{-\infty}^{x^{+}(t)}m^{2}_{x}(t,x)dx$
$\displaystyle\geq\int_{-\infty}^{x(\tau)}m^{2}_{x}(\tau,x)dx+\int_{\tau}^{t}\int_{-\infty}^{x^{+}(s)}2E(s,x)dxds.$
(3.50)
$\displaystyle=\int_{-\infty}^{x(\tau)}m^{2}_{x}(\tau,x)dx+\int_{\tau}^{t}\int_{-\infty}^{x(s)}2E(s,x)dxds+o_{1}(t-\tau).$

Notice that the last term is a higher order infinitesimal, satisfying
$\frac{o_{1}(t-\tau)}{t-\tau}\rightarrow 0$ as $t\rightarrow\tau$. Indeed

(3.51)
$\displaystyle|o_{1}(t-\tau)|=\int_{\tau}^{t}\int_{-\infty}^{x(s)}2E(s,x)dxds.$

For simplicity, we only show the estimate of the term $(mu)_{x}m_{x}$ in E.
Firstly, we have

$\displaystyle\int_{\tau}^{t}\int_{x^{+}(s)}^{x(s)}(um_{x}^{2})dxds$
$\displaystyle\leq\int_{\tau}^{t}\|u\|_{L^{\infty}}\int_{x(s)}^{x^{+}(s)}m_{x}^{2}dxds\leq\|u\|_{L^{\infty}}\int_{x^{+}(s)}^{x(s)}m_{x}^{2}dxds$
(3.52)
$\displaystyle\leq\|u\|_{H^{1}}\int_{\tau}^{t}\|m_{x}^{2}\|_{L^{1}}(x(s)-x^{+}(s))ds\leq
C(t-\tau)^{2}.$

Similarly

$\displaystyle\int_{\tau}^{t}\int_{x(s)}^{x^{+}(s)}(-mu_{x}m_{x})dxds$
$\displaystyle=\int_{\tau}^{t}\int_{x(s)}^{x^{+}(s)}-m[m_{x}-m+\frac{1}{2\lambda}(G^{-1}m_{x}^{2}+G^{-1}m^{2}-\partial
G^{-1}m^{2})]m_{x}dxds$
$\displaystyle\leq\|m\|_{H^{1}}\int_{\tau}^{t}\|m_{x}^{2}\|_{L^{1}}(x(s)-x^{+}(s))ds$
$\displaystyle+(\|m\|_{H^{1}}^{2}+\|m\|_{H^{1}}^{3})\int_{\tau}^{t}\|m_{x}\|_{L^{2}}(x(s)-x^{+}(s))^{\frac{1}{2}}ds$
(3.53) $\displaystyle\leq C(t-\tau)^{\frac{3}{2}}+C(t-\tau)^{2}.$

Thus we obtain that

(3.54) $\displaystyle|\circ_{1}(t-\tau)|\leq
C((t-\tau)^{\frac{3}{2}}+(t-\tau)^{2}).$

For $t$ sufficiently close to $\tau$, we have

$\displaystyle\beta(t)=$
$\displaystyle\beta(\tau)+(t-\tau)[u(\tau,x(\tau))+\int_{-\infty}^{x(\tau)}Fdx]+o_{2}(t-\tau)$
$\displaystyle=$ $\displaystyle
x(t)+\int_{-\infty}^{x(t)}m_{x}^{2}dx=x(t)+\mu_{(t)}(-\infty,x(t))$
$\displaystyle>$ $\displaystyle
x(\tau)+(t-\tau)[u(\tau)+\epsilon_{0}]+\mu_{(t)}(-\infty,x^{+}(t))$ (3.55)
$\displaystyle\geq$ $\displaystyle
x(\tau)+(t-\tau)[u(\tau)+\epsilon_{0}]+\mu_{(t)}(-\infty,x(t))+\int_{\tau}^{t}\int_{-\infty}^{x(s)}Fdxds+o_{1}(t-\tau).$

We can deduce that

$\displaystyle\beta(\tau)$
$\displaystyle+(t-\tau)[u(\tau,x(\tau))+\int_{-\infty}^{x(\tau)}Fdx]+o_{2}(t-\tau)$
(3.56) $\displaystyle\geq$
$\displaystyle[x(\tau)+\int_{-\infty}^{x(\tau)}m_{x}^{2}dx]+(t-\tau)[u(\tau,x(\tau))+\epsilon_{0}]+\int_{\tau}^{t}\int_{-\infty}^{x(s)}Fdxds+o_{1}(t-\tau).$

Subtracting common terms, dividing both sides by $t-\tau$ and letting
$t\rightarrow\tau$, we achieve a contradiction. Therefore, (2.1) must hold.

  * 5

We now prove (3.25). For every test function
$\phi\in\mathbb{C}^{\infty}_{c}(\mathbb{R}^{2})$, one has

(3.57) $\displaystyle\int_{0}^{\infty}\int m\phi_{t}-um_{x}\phi+F\phi
dxdt+\int m_{0}\phi(0,x)dx=0.$

Given any $\psi\in\mathbb{C}_{c}^{\infty}$, let $\phi=\psi_{x}$. Since the map
$x\mapsto m(t,x)$ is absolutely continuous, we can intergrate by parts w.r.t.
x and obtain

(3.58) $\displaystyle\int_{0}^{\infty}\int
m_{x}\psi+um_{x}\psi_{x}+F\psi_{x}dxdt+\int\bar{m}\psi dx=0.$

For any $\epsilon\geq 0$ sufficiently small, we give the following function

$\varrho^{\varepsilon}(s,x)=\left\\{\begin{array}[]{rcl}0,&~{}~{}~{}~{}y\leq-{\varepsilon}^{-1},\\\
x+{\varepsilon}^{-1},&~{}~{}~{}~{}~{}~{}-{\varepsilon}^{-1}\leq x\leq
1-{\varepsilon}^{-1},\\\ 1,&1-{\varepsilon}^{-1}\leq x\leq x(s),\\\
1-{\varepsilon}^{-1}(x-x(s)),&~{}~{}~{}~{}~{}~{}~{}~{}~{}x(s)\leq x\leq
x(s)+\varepsilon,\\\ 0&~{}~{}~{}~{}~{}~{}~{}~{}x\geq
x(s)+\varepsilon,\end{array}\right.$

and

$\displaystyle\psi^{\varepsilon}(s,x)=\min\\{{\varrho^{\varepsilon}(s,x),\chi^{\varepsilon}(s)}\\},$

We now use the test function $\phi=\psi^{\epsilon}$ in (3.58) and let
$\epsilon\mapsto 0$.

(3.59)
$\displaystyle\int_{-\infty}^{x(t)}m_{x}(t,x)dx=\int_{-\infty}^{x(\tau)}m_{x}(\tau,x)dx-\int_{\tau}^{t}F(s,x(s))ds+\lim_{\epsilon\rightarrow
0}\int_{\tau-\epsilon}^{t+\epsilon}\int_{x(s)}^{x(s)+\epsilon}m_{x}(\psi_{t}^{\epsilon}+u\psi_{x}^{\epsilon})dxds$

Hence, it is shown that

$\displaystyle\lim\limits_{\varepsilon\rightarrow
0}\int_{\tau-\varepsilon}^{t+\varepsilon}\int_{{x}(s)}^{{x}(s)+\varepsilon}m_{x}(\psi_{t}^{\epsilon}+u\psi_{x}^{\epsilon})dyds$
(3.60) $\displaystyle=\lim\limits_{\varepsilon\rightarrow
0}\Big{(}\int_{\tau-\varepsilon}^{\tau}+\int_{\tau}^{t}+\int_{t}^{t+\varepsilon}\Big{)}\int_{{x}(s)}^{{x}(s)+\varepsilon}m_{x}(\psi_{t}^{\epsilon}+u\psi_{x}^{\epsilon})dyds=0.$

Taking advantage of Cauchy’s inequality and $m_{x}\in L^{2}$, one has

$\displaystyle|\int_{\tau}^{t}\int_{{x}(s)}^{{x}(s)+\varepsilon}m_{x}[{\psi}_{t}^{\varepsilon}+u{\psi}_{x}^{\varepsilon}]dxds|$
$\displaystyle\leq\int_{\tau}^{t}\Big{(}\int_{{x}(s)}^{{x}(s)+\varepsilon}|m_{x}|^{2}dy\Big{)}^{\frac{1}{2}}\Big{(}\int_{{x}(s)}^{{x}(s)+\varepsilon}[{\psi}_{t}^{\varepsilon}+u{\psi}_{x}^{\varepsilon}]^{2}dx\Big{)}^{\frac{1}{2}}ds.$

For each $\epsilon>0$ consider the function

(3.61)
$\displaystyle\eta_{\epsilon}(s)\triangleq(\sup_{x\in\mathbb{R}}\int_{x}^{x+\epsilon}m_{x}^{2}(s,y)dy)^{\frac{1}{2}}.$

Observe that all functions $\eta_{\epsilon}$ are uniformly bounded. By the
dominated convergence theorem,

(3.62) $\displaystyle\lim_{\epsilon\rightarrow
0}\int_{\tau}^{t}(\int_{x}^{x+\epsilon}m_{x}^{2}(s,y)dy)^{\frac{1}{2}}ds\leq\lim_{\epsilon\rightarrow
0}\int_{\tau}^{t}\eta_{\epsilon}(s)ds=0$

For $s\in[\tau,t]$, we construct the function

(3.63) $\displaystyle\psi^{\epsilon}_{x}(s,y)=-\epsilon^{-1},$

and

(3.64)
$\displaystyle\psi_{t}^{\epsilon}(s,y)+u(s,x(s))\psi_{x}^{\epsilon}(s,y)=0~{}~{}~{}~{},~{}~{}~{}~{}x(s)<y<x(s)+\epsilon.$

This implies

$\displaystyle\int_{x(s)+\epsilon}^{x(s)}|\psi_{t}^{\epsilon}(s,y)+u(s,y)\psi^{\epsilon}_{x}(s,y)|^{2}dy=\epsilon^{-2}\int_{x(s)}^{x(s)+\epsilon}|u(s,y)-u(s,x(s))|^{2}dy$
$\displaystyle\leq\epsilon^{-1}(\max_{x(s)\leq y\leq
x(s)+\epsilon}|u(s,y)-u(s,x(s))|)^{2}\leq\epsilon^{-1}(\int_{x(s)}^{x(s)+\epsilon}|u_{x}(s,y)|dy)^{2}\leq\epsilon^{-1}(\epsilon^{\frac{1}{2}}\|u_{x}(s)\|_{L^{2}})^{2}$
(3.65) $\displaystyle\leq\|m\|_{H^{1}}$

Combining (3.62) and (5), we show the estimate the integral

$\displaystyle(\int_{\tau-\epsilon}^{\tau}+\int_{t}^{t+\epsilon})\int_{x(s)}^{x(s)+\epsilon}m_{x}(\psi_{t}+u\psi)dxds$
$\displaystyle\leq(\int_{\tau-\epsilon}^{\tau}+\int_{t}^{t+\epsilon})(\int_{x(s)}^{x(s)+\epsilon}m_{x}^{2}dy)^{\frac{1}{2}}(\int_{x(s)}^{x(s)+\epsilon}(\psi_{t}+u\psi)^{2}dy)^{\frac{1}{2}}$
(3.66) $\displaystyle\leq
2\epsilon\|m\|_{H^{1}}(\int_{x(s)}^{x(s)+\epsilon}\epsilon^{-2}Cdy)^{\frac{1}{2}}\leq
C\epsilon^{\frac{1}{2}}\rightarrow 0$

as $\epsilon\rightarrow 0$.The above analysis has shown that

(3.67) $\displaystyle\lim_{\epsilon\rightarrow
0}\int_{\tau-\epsilon}^{t+\epsilon}m_{x}(\psi_{t}+u\psi)ds=0.$

Therefore we arrive at (3.25).

  * 6.

Using the uniqueness of $\beta$, we can prove uniqueness of $x(t)$.

∎

######  Lemma 3.3.

Let m=m(t,x) be a conservative solution of (2.2). Then the
map$(t,\beta)\mapsto m(t,\beta)\triangleq m(t,x(t,\beta))$ is Lipschitz
continuous, with a constant depending only on the norm $\|m_{0}\|_{H^{1}}$

###### Proof.

Combining (3) and (3.25), we obtain

$\displaystyle|m(t,x(t,\bar{\beta}))-m(\tau,\bar{\beta})|\leq|m(t,x(t,\bar{\beta}))-m(t,\beta(t))|+|m(t,x(t,\beta(t)))-m(\tau,\beta(\tau))|$
(3.68)
$\displaystyle\leq\frac{1}{2}|\beta(t)-\bar{\beta}|+(t-\tau)\|F\|_{L^{\infty}}\leq(t-\tau)(\|G\|_{L^{\infty}}+\|F\|_{L^{\infty}}).$

∎

##  4 Proof of Theorem2.3

We need to seek a good characteristic, and employ how the gradient $m_{x}$ of
a conservative solution varies along the good characteristic, and complete the
proof of uniqueness.

###### Proof.

Step 1. Lemmas 3.1-3.3 ensure that the map $(t,\beta)\mapsto(x,u)(t,\beta)$
and $\beta\mapsto F(t,\beta)$ are Lipschitz continuous. Thanks to Rademacher’s
theorem, the partial derivatives $x_{t},x_{\beta},m_{t},m_{\beta}$ and
$F_{\beta}$ exist almost everywhere. Moreover, $x(t,\beta)$ is the unique
solution to (2.1), and the following holds.

(GC) For $a.e.~{}t\geq 0$, the point $(t,\beta(t,\bar{\beta}))$ is a Lebesgue
point for the partial derivatives $x_{t},x_{\beta},m_{t},m_{\beta}$ and
$F_{\xi}$. Moreover, $x_{\beta}(t,\beta)>0$ for $a.e.~{}t\geq 0.$

If (GC) holds, then $t\rightarrow x(t,\bar{\beta})$ is a good characteristic.

Step 2. We now construct an O.D.E. to describe that the quantities $m_{\beta}$
and $x_{\beta}$ vary along a good characteristic. Supposing that
$t,~{}\tau\notin\mathcal{N},$ and $y(t,\xi)$ is a good characteristic, we then
have

$\displaystyle
x(t,\beta(t,\bar{\beta}))=\bar{x}(t,\bar{\beta})+\int_{\tau}^{t}u(s,x(s,\beta(t;\tau,\bar{\beta})))ds.$

Where $\mathcal{N}$ denotes a null set with meas($\mathcal{N}$)=0 such that
for every $t\notin\mathcal{N}$ the measure $\mu_{(t)}$ is absolutely continous
and has density $m_{x}^{2}(t,\dot{)}$.

Differentiating the above equation with respect to $\bar{\xi}$, we deduce that

(4.1) $\displaystyle
x_{\beta}\frac{\partial}{\partial\bar{\beta}}\beta(t;\tau,\bar{\beta})=x_{\beta}(\tau,\bar{\beta})+\int_{\tau}^{t}u_{\beta}(s,\beta(t;\tau,\bar{\beta}))\frac{\partial}{\partial\bar{\beta}}\beta(t;\tau,\bar{\beta})ds.$

Likewise, we have

(4.2) $\displaystyle
m_{\beta}\frac{\partial}{\partial\bar{\beta}}\beta(t;\tau,\bar{\beta})=m_{\beta}(\tau,\bar{\beta})+\int_{\tau}^{t}F_{\beta}(s,\beta(t;\tau,\bar{\beta}))\frac{\partial}{\partial\bar{\beta}}\beta(t;\tau,\bar{\beta})ds.$

From (4.1)-(4.2), we end up with

(4.3) $\left\\{\begin{aligned}
&\frac{d}{dt}x_{\beta}+G_{\beta}x_{\beta}=u_{\beta},\\\
&\frac{d}{dt}m_{\beta}+G_{\beta}m_{\beta}=F_{\beta}.\end{aligned}\right.$

Step 3. We now return to the original coordinates $(t,x)$ and derive an
evolution equation for the partial derivative $k_{x}$ along a “good”
characteristic curve. For a fixed point $(t,x)$ with $t\notin\mathcal{N}$.
Suppose that $\bar{x}$ is a Lebesgue point for the map $x\rightarrow
m_{x}(t,x)$, and $\bar{\beta}$ satisfies $\bar{x}=x(t,\bar{\beta}),$ and
suppose that $t\rightarrow\beta(t;\tau,\bar{\beta})$ is a good characteristic,
which implies (GC) holds. We observe that

(4.4) $\displaystyle
m_{\beta}(t,x)=\frac{1}{x^{2}_{\beta}(t,\bar{\beta})}-1>0,\
x_{\beta}(t,\bar{\beta})>0,$

which implies that $y_{\xi}(t,\xi)>0.$

Hence, the partial derivative $m_{x}$ and $u_{x}$ can be calculated as shown
below

$\displaystyle
m_{x}(t,x(t,\beta(t;\tau,\bar{\beta})))=\frac{m_{\beta}(t,\beta(t;\tau,\bar{\beta}))}{x_{\beta}(t,\beta(t;\tau,\bar{\beta}))}.~{}~{}~{}~{}~{}u_{x}(t,x(t,\beta(t;\tau,\bar{\beta})))=\frac{u_{\beta}(t,\beta(t;\tau,\bar{\beta}))}{x_{\beta}(t,\beta(t;\tau,\bar{\beta}))}.$

Applying (4.3) to describe the evolution of $m_{\beta}$ and $x_{\beta},$ we
infer that the map $t\rightarrow m_{x}(t,x(t,\beta(t,{\bar{\beta}})))$ is
absolutely continuous. It follows that

$\displaystyle\frac{d}{dt}m_{x}(t,x(t,\beta(t,{\bar{\beta}})))$
$\displaystyle=\frac{d(\frac{m_{\beta}}{x_{\beta}})}{dt}=\frac{x_{\beta}\frac{d}{dt}m_{\beta}-m_{\beta}\frac{d}{dt}x_{\beta}}{{x_{\beta}}^{2}}=\frac{x_{\beta}(F_{\beta}-m_{\beta}G_{\beta})-m_{\beta}(u_{\beta}-G_{\beta}x_{\beta})}{{x_{\beta}}^{2}}$
$\displaystyle=\frac{F_{\beta}x_{\beta}-u_{\beta}m_{\beta}}{x_{\beta}^{2}}=\frac{F_{\beta}}{x_{\beta}}-u_{x}m_{x}.$

Hence, we conclude that as long as $y_{\beta}\neq 0$, the map $t\rightarrow
k_{x}$ is absolutely continuous.

Step 4. Consider the function

(4.5) $v\triangleq\left\\{\begin{aligned} &2\arctan
m_{x}~{}~{}~{}~{}~{}~{}if~{}~{}~{}0<x_{\beta}\leq 1\\\
&~{}~{}~{}~{}~{}~{}~{}~{}~{}\pi~{}~{}~{}~{}~{}~{}~{}~{}~{}if~{}~{}~{}x_{\beta}=0.\\\
\end{aligned}\right.$

Observe that this implies

(4.6) $\displaystyle
x_{\beta}=\frac{1}{1+m_{x}^{2}}=\cos^{2}\frac{v}{2},~{}~{}~{}\frac{m_{x}}{1+m_{x}^{2}}=\frac{1}{2}\sin
v,~{}~{}~{}\frac{m_{x}^{2}}{1+m_{x}^{2}}=\sin^{2}\frac{v}{2}$

From which it follows that

(4.7) $\left\\{\begin{aligned} &\frac{d}{dt}\beta(t,\bar{\beta})=G,\\\
&\frac{d}{dt}x(t,\beta(t,\bar{\beta}))=u(t,\beta(t,\bar{\beta})),\\\
&\frac{d}{dt}m(t,\beta(t,\bar{\beta}))=F(t,\beta(t,\bar{\beta})),\\\
&\frac{d}{dt}v(t,\beta(t,\bar{\beta}))=2\cos^{2}\frac{v}{2}P-N\sin
v-\sin^{2}\frac{v}{2}.\\\ \end{aligned}\right.$

with
$N(t,\beta)=\big{(}-m-P_{1}-P_{1x}+\frac{1}{2\lambda}(-P_{2}+P_{3x}+P_{2x}-m^{2})\big{)}(t,\beta)$
and $P=F+mN+\lambda u+\frac{m^{2}}{2}$ The function $P_{i}$ admits a
representation of the variavle $\beta$, namely

(4.18)
$\displaystyle\left\\{\begin{array}[]{lllll}P_{1}=-\frac{1}{2}\int_{-\infty}^{+\infty}e^{-|\int_{\beta^{\prime}}^{\beta}\cos^{2}\frac{v}{2}ds|}m\cos^{2}\frac{v}{2}d\beta^{\prime},\\\
P_{1x}=-\frac{1}{2}\big{(}\int_{\beta}^{+\infty}-\int_{-\infty}^{\beta}\big{)}e^{-|\int_{\beta^{\prime}}^{\beta}\cos^{2}\frac{v}{2}ds|}m\cos^{2}\frac{v}{2}d\beta^{\prime},\\\
P_{2}=-\frac{1}{2}\int_{-\infty}^{\infty}e^{-|\int_{\beta^{\prime}}^{\beta}\cos^{2}\frac{v}{2}ds|}m^{2}\cos^{2}\frac{v}{2}d\beta^{\prime},\\\
P_{2x}=-\frac{1}{2}\big{(}\int_{\beta}^{+\infty}-\int_{-\infty}^{\beta}\big{)}e^{-|\int_{\beta^{\prime}}^{\beta}\cos^{2}\frac{v}{2}ds|}m^{2}\cos^{2}\frac{v}{2}d\beta^{\prime},\\\
P_{3}=-\frac{1}{2}\int_{-\infty}^{+\infty}e^{-|\int_{\beta^{\prime}}^{\beta}\cos^{2}\frac{v}{2}ds|}\sin^{2}\frac{v}{2}d\beta^{\prime},\\\
P_{3x}=-\frac{1}{2}\big{(}\int_{\beta}^{+\infty}-\int_{-\infty}^{\beta}\big{)}e^{-|\int_{\beta^{\prime}}^{\beta}\cos^{2}\frac{v}{2}ds|}\sin^{2}\frac{v}{2}d\beta^{\prime},\\\
P_{4}=-\frac{1}{2}\int_{-\infty}^{+\infty}e^{-|\int_{\beta^{\prime}}^{\beta}q\cos^{2}\frac{v}{2}ds|}\frac{2}{u}\sin
v-mu\cos^{2}\frac{v}{2}d\beta^{\prime},\\\
P_{4x}=-\frac{1}{2}\big{(}\int_{\beta}^{+\infty}-\int_{-\infty}^{\beta}\big{)}e^{-|\int_{\beta^{\prime}}^{\beta}\cos^{2}\frac{v}{2}ds|}\frac{2}{u}\sin
v-mu\cos^{2}\frac{v}{2}d\beta^{\prime},\\\
P_{5}=-\frac{1}{2}\int_{-\infty}^{+\infty}e^{-|\int_{\beta^{\prime}}^{\beta}\cos^{2}\frac{v}{2}ds|}P_{3}\cos^{2}\frac{v}{2}d\beta^{\prime},\\\
P_{5x}=-\frac{1}{2}\big{(}\int_{\beta}^{+\infty}-\int_{-\infty}^{\beta}\big{)}e^{-|\int_{\beta^{\prime}}^{\beta}\cos^{2}\frac{v}{2}ds|}P_{3}\cos^{2}\frac{v}{2}d\beta^{\prime}.\\\
\end{array}\right.$

For any $\bar{\beta}\in\mathbb{R},$ we deduce that the following initial
conditions

(4.19) $\left\\{\begin{aligned} &\beta(0,\bar{\beta})=\bar{\beta},\\\
&\beta(0,\bar{\beta})=x(0,\bar{\beta}),\\\
&m(0,\bar{\beta})=m_{0}(x(0,\bar{\beta})),\\\ &v(0,\bar{\beta})=2\arctan
u_{0,x}(x(0,\bar{\beta})).\\\ \end{aligned}\right.$

Making use of all coefficients is Lipschitz continuous and the previous steps
again, the system (4.7)-(4.19) has a unique globally solution.

Step 5. Let $m$ and $\tilde{m}$ be two conservative weak solution of (2.2)
with the same initial data $\bar{m}\in H^{1}(\mathbb{R})$ . For $a.e.~{}t\geq
0,$ the corresponding Lipschitz continuous maps $\beta\mapsto
x(t,\beta),{\beta}\mapsto\tilde{x}(t,\beta)$ are strictly increasing. Hence
they have continuous inverses, say
$x\mapsto{\beta}^{-1}(t,x),x\mapsto{\tilde{\beta}}^{-1}(t,x)$. Thus, we deduce
that

$x(t,\beta)=\tilde{x}(t,\beta),\ u(t,x(t,\beta))=\tilde{u}(t,{x}(t,\beta)).$

Moreover, for $a.e.~{}t\geq 0$, we have

$m(t,x)=m(t,y(t,\beta))=\tilde{m}(t,\tilde{x}(t,\beta))=\tilde{m}(t,x).$

Then, we finish the proof of Theorem 2.3. ∎

Acknowledgments This work was partially supported by the National Natural
Science Foundation of China (No.12171493).

Data Availability. The data that support the findings of this study are
available on citation. The data that support the findings of this study are
also available from the corresponding author upon reasonable request.

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