# Wave-front tracking for a quasi-linear scalar conservation law with
hysteresis

Fabio Bagagiolo111Department of Mathematics, University of Trento, Italy,
<EMAIL_ADDRESS>and Stefan Moreti222Department of Mathematics,
University of Trento, Italy<EMAIL_ADDRESS>
###### Abstract

In this article we deal with the Cauchy problem for the quasi-linear scalar
conservation law

$u_{t}+\mathcal{F}(u)_{t}+u_{x}=0,$

where $\mathcal{F}$ is a specific hysteresis operator, namely the Play
operator. Hysteresis models a rate-independent memory relationship between the
input $u$ and its output. Its presence in the partial differential equation
gives a particular non-local feature to the latter allowing us to capture
phenomena that may arise in some application fields. Riemann problems and the
interactions between shock lines are studied and via the so-called Wave-Front
Tracking method a solution to the Cauchy problem with bounded variation
initial data is constructed. The solution found satisfies an entropy-like
condition, making it the unique solution in the class of entropy admissible
ones.

## Introduction

In this work we deal with the Cauchy problem for a conservation law with
hysteresis nonlinearities as follows:

$\begin{cases}u_{t}+w_{t}+u_{x}=0&\quad\text{on }\mathbb{R}\times[0,T),\\\
w(x,t)={\cal F}[u(x,\cdot),w^{0}(x)](t)&\quad\forall\ t\in]0,T[,\ \text{a.e.
}x\in\mathbb{R}\\\ u(x,0)=u_{0}(x)&\quad\text{in }\mathbb{R},\\\
w(x,0)=w_{0}(x)&\quad\text{in }\mathbb{R}.\end{cases}$ (1)

where: $\cal F$ is a so-called hysteresis operator, namely representing a
memory-dependent input-output relationship between the pair of scalar
functions $(t\mapsto u(x,t),t\mapsto w(x,t))$, one for almost every $x$;
$u_{0}$ is the initial datum for the solution $u$; $w^{0}$ is a suitable
space-dependent function for the initial values of the output $w$. The memory
dependence represented by $\cal F$ is rate-independent, which is the main
characterization of the hysteresis phenomena. We are going to consider the
case where $\cal F$ is the so-called Play operator.

The presence of the hysteretic term $w$ gives (1) a particular non-local
feature which, up to the knowledge of the authors, for this kind of equation
was not investigated before in the framework of characteristics, wave-front
tracking and the corresponding limit procedure. More precisely, due to the
rate-independent memory, the functional input-output relationship $u\mapsto
w={\cal F}[u,w^{0}]$ is highly non-linear and non-differentiable and this fact
leads to abrupt changes of characteristics after any possible time $t$ when,
in dependence on $x$, the pair $(u,w)$ reaches suitable regions of the phase-
space $u-w$. In the case of the Play operator, such abrupt changes can be
somehow seen as the case of a discontinuity in the derivative of the flux
function $f$ for a generic conservation law $u_{t}+f(u)_{x}=0$. For the
Riemann problem, (1) is then seen as a conservation law with a piece-wise
linear flux function $f$, where the alternation of the values of $f^{\prime}$
depends on the hysteretic relationship $u\mapsto w$ and is different for
different points $x$.

Our main goal is the study of a generic Cauchy problem for (1) with initial
data $u_{0},w_{0}$ just bounded variation functions ($BV$). We then
approximate $u_{0}$ and $w_{0}$ by piece-wise constant functions, solve the
corresponding Riemann problems and, suitably adapting the wave-front tracking
method, we pass to the limit. In doing that we face several non-standard
problems:

i) with respect to the classical wave-front tracking method in our case the
number of discontinuity waves after an interaction may increase, while the
total variation of the solution remains constant.

ii) for the Riemann problem the solution $u$ is discontinuous in time and this
is in general a problem for the hysteresis relation which is usually defined
for continuous inputs $u$; we need to extend the definition of the Play
operator to time-piece-wise constant inputs in the spirit of the so-called
regulated functions (see Brokate-Sprekels [3], Krejci-Laurencot [13] and
Recupero [18]);

iii) for the Cauchy problem, being the solution only $BV$, we need a further
relaxations of the hysteresis relation writing it as a suitable measure-
dependent variational integral inequality (suitably adapting the one in
Visintin [20]). In particular, such variational inequality will be maintained
to the limit, giving the existence of a weak solution for the Cauchy problem
with hysteresis.

Our main results are the existence of a solution of (1) with $BV$ data and the
uniqueness in the class of functions satisfying an entropy condition.

In the article [20], Visintin studies equation (1) with different hysteresis
operator (namely, the delayed relay). However, the PDE problem is there dealt
by a time discretization method, without focusing on characteristic curves. In
our article instead we use a different approach which is more constructive and
gives a concrete idea on how the discontinuities behave and how the solution
evolves in time.

Hysteresis is a phenomenon often observed in various natural and engineered
systems, typically characterized by a lag or delay in the response of a system
to changes in the input. Well-known examples of hysteresis are in the
behaviour of ferromagnetic materials, stress-strain relationship in plasto-
elastic materials and behaviour of thermostats. For a comprehensive account
for mathematical models for hysteresis and their use in connection with PDEs,
we refer to Krasnoselskii-Pokrovskii [12] and Visintin [19].

Hyperbolic and scalar conservation laws with hsyeresis for some specific
applied motivations were studied in [16], [17], [10], [11], [4], [15], [5],
[8]. In particular, the models studied by Peszynska and Showalter in [16] and
[15] come from applications in transport with adsorption in porous media,
where hysteresis is a common feature. Kordulova in [11] summarizes the results
known for such equations and Kopfova in [10] focuses on entropy conditions.
Corli and Fan in their works [4] and [5] investigate a conservation law with
hysteresis relation in the flux given by a parametric family of curves which
are followed subject to the monotonicity of the input and to the trajectory of
an auxiliary ODE. The model is motivated by application in traffic flow, where
hysteresis is due to a delay in change of drivers behavior. Fan in the recent
article [8] studies the same model as in [4],[5] and considers the Wave-Front
tracking method, proving an estimate on the total variations of the solutions.
However, the limit procedure, in order to have existence of a weak solution
for more general initial data, is not completely performed especially for what
concerns the passage to the limit into the hysteresis relationship.

For general basic theory on scalar conservation laws we refer for example to
the book by Evans [7]. For more specific results, such as the solution of
Riemann problems for scalar conservation laws with piecewise linear flux and
the classical Wave Front Tracking Method, we refer to the books by Bressan [2]
and by Holden-Risebro [9] or to the seminal paper by Dafermos [6].

The article is structured as follows. In Section 1 we introduce the
mathematical formalization of hysteresis and we define the Play operator
$\mathcal{F}$ and its suitable extensions. In Section 2 we recall some results
for scalar conservation laws and in particular we focus on the case when the
flux is piece-wise linear. In Section 3 we introduce a suitable formulation of
(1) and in Section 4 we study the corresponding Riemann problem. In Section 5
we prove existence of weak solutions for the case of $BV$ initial data, in
particular we perform the limit in the wave-front-tracking procedure. In
Section 6 we give an entropy condition, showing that the solution constructed
in the previous section is the only entropy solution.

##  1 The Play Operator for hysteresis and its extensions

Figure 1 represents the so-called input-output Play hysteresis relationship
between a time-dependent scalar input $u$ and a time-dependent scalar output
$w$. Here the amplitude $a>0$ is fixed, and we denote by $\cal L$ the strip
$\\{(u,v)||u-w|\leq a\\}$ which is going to represent the feasible states of
the system. If at certain time $t$ the pair $(u(t),w(t))$ satisfies
$|u(t)-w(t)|<a$, that is if it belongs in the interior of $\cal L$, and if the
input changes in time, then the output will not change, until the pair $(u,w)$
will possibly reach one of the two boundary lines of $\cal L$. If
$w(t)=u(t)-a$, that is the pair $(u,w)$ is on the lower boundary of $\cal L$
and if the input increases, then the output will increase together with $u$;
if instead $u$ decreases then $w$ stays constant and the pair $(u,w)$ enters
the interior of $\cal L$. If $(u,w)$ belongs to the upper boundary of $\cal
L$, then the behavior is symmetric, reversing the role of the monotonicity of
$u$.

$u$$w$$-a$$a$ Figure 1: Play operator

Given an initial state $(u(0),w(0))\in\mathcal{L}$ and the evolution of $u$ we
can then trace the evolution $w$. We can notice that the value of $w(t)$ is
not determined pointwisely by $u(t)$, indeed for fixed $u(t)$ we have more
than one possible value $w$, depending on the past evolution of $u$. Hence
$w(t)$, besides its initial value $w(0)$, is determined by the whole history
$u|_{[0,t]}$, i.e. there is a memory effect involved.
In particular this memory effect is rate-independent which means that the
relation between $u$ and $w$ does not depend on the speed of $u(t)$, i.e. it
does not depend on its derivative. Notice that this requirements is essential
if we want to draw hysteresis relations as in Figure 1 and it is a general
feature of hysteresis phenomena.

When the input $u$ is in $W^{1,1}(0,T)$, for some $T>0$, the Play hysteresis
relationship described above can be characterized by the following variational
inequality for almost every $t\in[0,T]$:

$|u-w|\leq a,\quad w^{\prime}(u-w-v)\geq 0\quad\forall v,|v|\leq a.$ (1.1)

Conditions (1.1) can be interpreted in the following way. If $|u-w|<a$ then
the term $u-w-v$ can have either positive or negative sign depending on $v$.
Since the inequality has to hold true for every $v$ then necessarily
$w^{\prime}=0$. Whereas, if for example $u=w+a$, then $u-w-v$ is always non
negative for all $v$, then also $w^{\prime}$ has to non negative. The opposite
happens when $u=w-a$. In [19] (see also [12]) it is shown how (1.1) well
defines an operator on the space of Sobolev functions which, by a density
argument, can be uniquely extended to an operator which maps continuous inputs
to continuous outputs. Hence the Play operator is proved to be as an operator
continuously (with respect to the uniform convergence) acting as

$\begin{split}\mathcal{F}:~{}&\tilde{\cal L}\subset
C^{0}([0,T])\times\mathbb{R}\to C^{0}([0,T])\\\
&(u(\cdot),w_{0})\to[\mathcal{F}(u,w_{0})](\cdot)=:w(\cdot)\end{split}$

where

$\tilde{\cal L}=\left\\{(u,w_{0})\in
C^{0}([0,T])\times\mathbb{R}\Big{|}(u(0),w_{0})\in{\cal L}\right\\}.$

Moreover such an operator satisfies

  1. i)

Causality: $\forall~{}(u_{1},w_{0}),(u_{2},w_{0})$, such that $u_{1}=u_{2}$ in
$[0,t]$ then

$\mathcal{F}(u_{1},w_{0})](t)=[\mathcal{F}(u_{2},w_{0})](t).$

  2. ii)

Rate-independence: $\forall~{}(u,w_{0})$, $\forall~{}t\in[0,T]$ if
$s:~{}[0,T]\to[0,T]$ is an increasing homeomorphism, then

$\mathcal{F}(u\circ s,w_{0})](t)=[\mathcal{F}(u,w_{0})](s(t)).$ (1.2)

  3. iii)

Semigroup property: $\forall~{}(u,w_{0})$, $\forall~{}t_{1}<t_{2}\in[0,T]$
setting $w(t_{1}):=[\mathcal{F}(u,w_{0})](t_{1})$ then we have that

$[\mathcal{F}(u,w_{0})](t_{2})=[\mathcal{F}(u(t_{1}+\cdot),w(t_{1}))](t_{2}-t_{1}).$
(1.3)

In the following, we will need to consider $\mathcal{F}$ as applied to $u$,
with $u$ solution to a Riemann problem for a conservation law. Hence we have
to extend the definition of $\mathcal{F}$ to piece-wise constant inputs with a
finite number of jumps.

######  Definition 1.1.

A function $g:[0,T]\to\mathbb{R}$ is piece-wise constant if we can find
$0=t_{0}<t_{1}<\dots<t_{N}=T$ and $g_{i}\in\mathbb{R}$ such that

$g(x)=\sum_{i=1}^{N}g_{i}\mathds{1}_{(t_{i-1},t_{i})}.$ (1.4)

We denote by $PC([0,T])$ the space the piece-wise constant functions on
$[0,T].$

Let us consider the simple case of a function $u:[0,T]\to\mathbb{R}$
consisting of two constant states, $u_{-},u_{+}$, separated by discontinuity
at $t=t^{*}.$ and let us suppose that $u_{-}<u_{+}$. The idea is to suitably
approximate $u$ with continuous functions $u_{\varepsilon}$ and then apply the
operator $\mathcal{F}$ to these functions and let $\varepsilon$ go to $0$. So,
for $\varepsilon>0$, let us define

$u_{\varepsilon}(t)=\begin{cases}u_{-}\quad&0\leq t<t^{*}-\varepsilon\\\
f_{\varepsilon}(t)\quad&t^{*}-\varepsilon\leq t\leq t^{*}+\varepsilon\\\
u_{+}\quad&t^{*}+\varepsilon<t\leq T,\end{cases}$ (1.5)

with

$\begin{split}&f_{\varepsilon}:[t^{*}-\varepsilon,t^{*}+\varepsilon]\to[u_{-},u_{+}]\
\text{continuous and increasing and}\\\
&f_{\varepsilon}(t^{*}-\varepsilon)=u_{-}\quad\quad
f_{\varepsilon}(t^{*}+\varepsilon)=u_{+}.\end{split}$ (1.6)

We can notice that, by the rate-independence (1.2), for every $\varepsilon>0$
the value of output $w_{\varepsilon}(t^{*}+\varepsilon)$ does not depend on
the choice of the function $f_{\varepsilon}$ satisfying (LABEL:eq:_f_epsilon)
(see also Remark 1.2). So the limit $\lim_{\varepsilon\to
0}w_{\varepsilon}(t^{*}+\varepsilon):=w^{*}$ exists and we can define the
output $w$ for the discontinuous input $u$ as

$w(t)=\begin{cases}w^{0}\quad&0\leq t<t^{*}\\\ w^{*}\quad&t^{*}<t\leq
T\end{cases}$ (1.7)

which of course coincides for $t\neq t^{*}$ with

$w(t):=\lim_{\varepsilon\to 0}w_{\varepsilon}(t)=\lim_{\varepsilon\to
0}[\mathcal{F}(u_{\varepsilon},w_{0})](t).$

Note that the output $w$ is also a piece-wise constant function with at most
one jump in $t^{*}$, where both $u$ and $w$ do not need to be defined.

Then with the above construction it is straightforward to extend $\mathcal{F}$
to the piece-wise constant function so

$\mathcal{F}:~{}PC([0,T])\times\mathbb{R}\to PC([0,T]).$

$t$$u$$t^{*}$$T$$t$$w$$t^{*}$$T$$u$$w$$(u_{-},w_{0})$$(u_{+},w^{*})$$(u_{\varepsilon}(\cdot),w_{\varepsilon}(\cdot))$
Figure 2: An explicit example of the operator $\mathcal{F}$ applied to
piecewise constant $u$. Here $a=1$, $u_{-}=w_{0}=0$ and $u_{+}=2$ consequently
$w^{*}=1$. In red we highlight the path followed by the couple
$(u_{\varepsilon},w_{\varepsilon})$ which, after the limit procedure,
collapses to $(u_{-},w_{0})$ for $t<t^{*}$ and $(u_{+},w^{*})$ for $t>t^{*}.$

######  Remark 1.2.

The monotonicity property required to $f_{\varepsilon}$ as in
(LABEL:eq:_f_epsilon) is not indeed necessary, due to some suitable properties
of the Play hysteresis operator (see [19]).

Also for piece-wise constant inputs and outputs the semigroup property holds
as: $\forall~{}(u,w_{0})$, $\forall~{}t_{1}<t_{2}\in[0,T]$ such that
$w(t_{1}):=[\mathcal{F}(u,w_{0})](t_{1})$ is defined, then (1.3) holds as
equality for piece-wise constant functions.

The way to extend the Play operator to piecewise continuous inputs, as we
sketched above, can be seen as a special case of a more general issue
concerning the extension to the so-called regulated functions (see[3] and
[18]).

In the sequel of the paper we will also perform a limit of solutions of
approximating Riemann problems, and hence we will need an extension of the
Play operator to even less regular inputs, as $BV$. In view of this fact, here
we give a further characterization of $\mathcal{F}$ for piecewise constant
inputs which will inspire a weak formulation for $BV$ inputs (see Remark 3.2).

######  Proposition 1.3.

Let $w_{0}\in\mathbb{R}$ and $u,w:[0,T]\to\mathbb{R}$ be piecewise constant
functions with a finite number of discontinuities. Moreover, denote by
$\tilde{u}$ and $\tilde{w}$ the right-continuous representatives of $u$ and
$w$ respectively and suppose $u(0)=\tilde{u}(0)$ and $w(0)=\tilde{w}(0)$. Then
the following are equivalent:

  1. 1.

$w=\mathcal{F}[u,w_{0}]$ for almost every $t$;

  2. 2.

$|\tilde{w}-\tilde{u}|\leq a$ for every $t$ and

$\int\limits_{(0,t)}(\tilde{u}-\tilde{w})d(Dw)\geq a|Dw|((0,t))\quad\text{ for
every }t\in(0,T],$ (1.8)

where $Dw$ denotes the measure associated to distributional derivative of $w$,
and $|Dw|$ its total variation.

###### Proof.

First notice that, since $\tilde{w}$ is of the following form
$\tilde{w}=\sum_{i=1}^{N}w_{i}\mathds{1}_{[t_{i-1},t_{i})}$ then its
distributional derivative (and the distributional derivative of $w$) can be
represented by a finite sum of Dirac’s delta. In particular

$Dw=\sum_{i=1}^{N-1}(w_{i+1}-w_{i})\delta_{t_{i}},$ (1.9)

and

$|Dw|((0,t))=\sum_{\\{i~{}|~{}t_{i}<t\\}}|w_{i+1}-w_{i}|.$ (1.10)

$(1\Rightarrow 2)$ By definition of $\mathcal{F}$ of course
$|\tilde{w}-\tilde{u}|\leq a$ everywhere. Let us now consider $t\in(0,t_{1}],$
then $|Dw|((0,t))=Dw((0,t))=0$ hence (1.8) holds trivially. Now if
$t\in(t_{1},t_{2}]$ we have that

$\int\limits_{(0,t)}(\tilde{u}-\tilde{w})~{}d(Dw)=(\tilde{u}(t_{1})-\tilde{w}(t_{1}))(w_{2}-w_{1})$

and

$a|Dw|((0,t))=a|w_{2}-w_{1}|.$

Now since $w_{2}-w_{1}\not=0$, i.e. $w$ did not remain constant, then the
arriving point after the jump, $(\tilde{u}(t_{1}),w_{2})$, must be on the
boundary of $\cal L$ (see Figure 2), hence
$\tilde{u}(t_{1})-\tilde{w}(t_{1})=\pm a$ (recall that by definition
$\lim_{t\to t_{1}^{+}}u(t)=\tilde{u}(t_{1})$ and $\lim_{t\to
t_{1}^{+}}w(t)=\tilde{w}(t_{1})$). In particular we can also write
$\tilde{u}(t_{1})-\tilde{w}(t_{1})=sign(w_{2}-w_{1})a$ and so (1.8) holds as
an equality. Now we can can extend recursively the proof on the whole interval
$(0,T]$.
$(2\Rightarrow 1)$ Consider first $t\in(0,t_{1}]$ then we have $w\equiv w_{1}$
in $(0,t)$ and so the constraint $|\tilde{w}-\tilde{u}|\leq a$ tells us that
$|u-w_{1}|\leq a$ for almost every $t\in(0,t_{1})$ so even if $u$ may change
values it can not exceed the threshold $w_{1}\pm a$ which would imply a change
in $w$. Hence the hysteresis relation $\mathcal{F}[u,w_{0}]$ holds. If instead
we consider $t\in(t_{1},t_{2}]$ then (1.8) tells us that

$(\tilde{u}(t_{1})-\tilde{w}(t_{1}))(w_{2}-w_{1})\geq a|w_{2}-w_{1}|$

which, because of $|\tilde{w}-\tilde{u}|\leq a$, is equivalent to

$(\tilde{u}(t_{1})-\tilde{w}(t_{1}))=sign(w_{2}-w_{1})a.$

So we conclude that if a jump in $w$ occurs then we reach the upper or lower
boundary of $\cal L$, depending whether the jump of $w$ is increasing or
decreasing, and of course, due also to the constraint $(u,w)\in{\cal L}$, $u$
must jump with the same monotonicity sign. Hence an approximation
$u_{\varepsilon}$ as in (1.5) is easily constructed and the corresponding
output $w_{\varepsilon}$ almost everywhere converge to $w$, showing that
$w={\cal F}(u)$ (see again Figure 2). ∎

######  Remark 1.4.

In the proof we saw how (1.8) actually holds as an equality. Sometimes, in the
sequel, it will be convenient to write it as an inequality, nevertheless,
because of the necessary condition $|\tilde{w}-\tilde{u}|\leq a$ the two
formulation are equivalent.

######  Remark 1.5.

When we introduce the operator $\mathcal{F}$ in a PDE, we have to define it on
functions $u$ both depending on a space variable $x\in\mathbb{R}^{n}$, for
some $n$, and on time $t$. In this case, for every fixed $x\in\mathbb{R}^{n}$,
we see $u(x,\cdot)$ as a function of time only, and then we define the output
as

$w(x,t):=[\mathcal{F}(u(x,\cdot),w_{0}(x))](t),\quad\text{a.e}\ x,\forall\ t,$
(1.11)

where the initial state of the output is now a given function depending on
$x$.

##  2 Solution to scalar Riemann problems with piecewise linear flux

In this section we will briefly recall some results for the Riemann problem
for a conservation law with piecewise linear flux function which will be used
in Section 4. Let us deal with the Cauchy problem

$\begin{cases}u_{t}+f(u)_{x}=0&\quad(x,t)\in\mathbb{R}\times[0,+\infty),\\\
u(x,0)=u_{0}(x)&\quad x\in\mathbb{R},\end{cases}$ (2.1)

with $f:\mathbb{R}\to\mathbb{R}$ continuous and piecewise linear and

$u_{0}(x)=\begin{cases}u_{l}\quad x<0,\\\ u_{r}\quad x>0.\end{cases}$

With $u_{0}$ as above, (2.1) is usually called Riemann problem.

A weak solution for (2.1), even for more general $f$ and $u_{0}$ (at least
integrable), is here defined.

######  Definition 2.1.

We say that $u:\mathbb{R}\times[0,+\infty)\to\mathbb{R}$, $u\in L^{1}_{loc}$,
is a weak solution of (2.1) if

$\int\limits_{0}^{+\infty}\int\limits_{-\infty}^{+\infty}[u\phi_{t}+f(u)\phi_{x}]~{}dx~{}dt+\int\limits_{-\infty}^{+\infty}u_{0}(x)\phi(x,0)~{}dx=0$
(2.2)

for every $C^{1}$ function $\phi:\mathbb{R}\times[0,+\infty)\to\mathbb{R}$
with compact support.

It is known that a weak solution with a jump discontinuity along a curve $C$
parametrized by $t\mapsto s(t)$ with $s$ derivable, should satisfy the so
called Rankine-Hugoniot condition (see e.g. [7]):

$f(u_{-}(t))-f(u_{+}(t))=s^{\prime}(t)(u_{-}(t)-u_{+}(t))$ (2.3)

on $C,$ where $u_{-}(t),u_{+}(t)$ are respectively the limits from the left
and right parts of the domain separated by discontinuity curve $C$.

In general for conservation laws there is not uniqueness of the weak solution.
The introduction of the so-called entropy condition allows us to select a
unique solution. In particular one can require that $u$ satisfies

$\int_{0}^{+\infty}\int_{-\infty}^{+\infty}\\{|u-k|\phi_{t}+sign(u-k)(f(u)-f(k))\phi_{x}\\}~{}dxdt\geq
0,$ (2.4)

for every $k\in\mathbb{R}$ and for every non negative smooth $\phi$ with
compact support. It is proven, [14] (see also [2], [9]), that for a scalar
conservation law there exists one and only one weak solution satisfying (2.4).
Such unique solution is usually called the entropy solution.

For piece-wise constant weak solutions, using also the Rankine-Hugoniot
condition (2.3), the entropy condition (2.4) simply reduces to (see [2])

$\frac{f(u)-f(u_{+})}{u-u_{+}}\leq\frac{f(u_{+})-f(u_{-})}{u_{+}-u_{-}}\leq\frac{f(u)-f(u_{-})}{u-u_{-}}$
(2.5)

for every value $u$ between $u_{-}$ and $u_{+}$.

For the case of piecewise linear flux, the entropy solution of the Riemann
problem can be explicitly constructed, see [6], [2] and [9]. Such solution is
piecewise constant with a finite number of values. For the purposes of next
sections we sketch here such construction.

We recall that $f:\mathbb{R}\to\mathbb{R}$ is piecewise linear if $f$ is
continuous and there is a finite family of disjoint intervals, covering
$\mathbb{R}$, where $f$ is affine. Moreover
$u:\mathbb{R}\times[0,+\infty[\to\mathbb{R}$ is piecewise constant if the
domain can be partioned in a finite family of subsets where $u$ is constant.

In (2.1), let us suppose $u_{l}<u_{r}$, and consider $f_{c}$ the greatest
convex minorant of $f$ on the interval $[u_{l},u_{r}]$, which we recall to be

$f_{c}:=\sup\\{h\text{ convex in }[u_{l},u_{r}]~{}|~{}h\leq f\\}.$ (2.6)

It is easy to see that $f_{c}$ is a piecewise linear function hence the graph
is a polygonal curve. Let us denote the vertices in the following way:
$(u_{l},f(u_{l}))$,$(u_{1},f(u_{1}))$, …, $(u_{k},f(u_{k}))$,
$(u_{r},f(u_{r}))$ where $u_{l}<u_{1}<\dots<u_{r}$ and $(u_{i},f(u_{i}))$, by
construction, are some of the vertices of $f$. Because of the convexity we
have

$\frac{f(u_{1})-f(u_{l})}{u_{1}-u_{l}}<\frac{f(u_{2})-f(u_{1})}{u_{2}-u_{1}}<\dots<\frac{f(u_{r})-f(u_{k})}{u_{r}-u_{k}},$
(2.7)

hence we define the following function, see Figure 3,

$u(x,t)=\begin{cases}u_{l}\quad&-\infty<\frac{x}{t}<\frac{f(u_{1})-f(u_{l})}{u_{1}-u_{l}}\\\
u_{1}\quad&\frac{f(u_{1})-f(u_{l})}{u_{1}-u_{l}}<\frac{x}{t}<\frac{f(u_{2})-f(u_{1})}{u_{2}-u_{1}}\\\
\quad\dots\\\ \quad\dots\\\
u_{r}\quad&\frac{f(u_{r})-f(u_{k})}{u_{r}-u_{k}}<\frac{x}{t}<+\infty.\end{cases}$
(2.8)

Now by construction the Rankine-Hugoniot condition holds, hence we have a weak
solution (as it is known, it is easy to see that a piecewise constant function
satisfying the Rankine-Hugoniot condition is a weak solution). Moreover for
every $u\in[u_{i},u_{i+1}]$ the entropy condition (2.5) holds as an equality,
due to the piecewise linearity. When $u_{l}>u_{r}$ we consider instead the
least concave majorant, which again is a piecewise linear function and by
considering the vertices of its graph we construct the solution similarly as
above.

$u$$f_{c}(u)$$u_{l}$$u_{1}$$u_{2}$$u_{3}$$u_{r}$$x$$t$$u_{l}$$u_{1}$$u_{2}$$u_{3}$$u_{r}$
Figure 3: An example of solution with piecewise linear flux

##  3 Weak formulation for the Cauchy problem with Play hysteresis

In this section we consider the following problem

$\begin{cases}u_{t}+w_{t}+u_{x}=0\quad\text{in }\mathbb{R}_{T},\\\
w=[\mathcal{F}(u,w_{0})],\\\ u(x,0)=u_{0}(x)\quad\text{in }\mathbb{R},\\\
w(x,0)=w_{0}(x)\quad\text{in }\mathbb{R},\end{cases}$ (3.1)

where $\mathbb{R}_{T}:=\mathbb{R}\times[0,T)$, and the state-dependent
hysteresis operator is defined as in Remark 1.5, starting from the Play
operator (with amplitude $a>0$) applied to $t\mapsto u(x,t)$ for almost every
$x$ fixed.

######  Definition 3.1.

A couple of $L^{1}_{loc}$ functions $(u,w)$ with
$u,w:\mathbb{R}_{T}\to\mathbb{R}$ is a weak solution to (3.1) if:

  1. i)

it satisfies the following weak formulation of the PDE

$\int\limits_{0}^{+\infty}\int\limits_{-\infty}^{+\infty}[(u+w)\phi_{t}+u\phi_{x}]~{}dx~{}dt+\int\limits_{-\infty}^{+\infty}(u_{0}(x)+w_{0}(x))\phi(x,0)~{}dx=0,$
(3.2)

for every $C^{1}$ function $\phi$ with compact support in $\mathbb{R}_{T}$;

  2. ii)

for almost every $(x,t)\in\mathbb{R}_{t}$ it holds

$|u(x,t)-w(x,t)|\leq a;$ (3.3)

  3. iii)

for almost every $x$, the distributional derivative $\frac{\partial
w}{\partial t}$ is a measure on $\mathbb{R}_{T}$ (denoted in the same way)
that satisfies

$\frac{1}{2}\int_{\mathbb{R}}(u(x,t)^{2}-u_{0}(x)^{2})~{}dx+\frac{1}{2}\int_{\mathbb{R}}(w(x,t)^{2}-w_{0}(x)^{2})~{}dx\leq-a\Big{|}\frac{\partial
w}{\partial t}\Big{|}(\mathbb{R}\times(0,t)),$ (3.4)

for almost every $t\in(0,T)$.

######  Remark 3.2.

Similarly to [20], equation (3.4) can be interpreted as an equivalent
formulation of (1.8), in the case of $H^{1}$ functions which also depend on
the state $x$. Indeed suppose that both $u,w$ are in $H^{1}(\mathbb{R}_{T})$
solution of the PDE (3.1) then (3.4) reads as follows

$\int_{-\infty}^{+\infty}\int_{0}^{t}u_{t}u+w_{t}w~{}dtdx+a\int_{-\infty}^{+\infty}\int_{0}^{t}|w_{t}|~{}dtdx\leq
0$

so, by the PDE,

$\int_{-\infty}^{+\infty}\int_{0}^{t}-(w_{t}u+u_{x}u)+w_{t}w~{}dtdx+a\int_{-\infty}^{+\infty}\int_{0}^{t}|w_{t}|~{}dtdx\leq
0.$

Since for almost every $t$, $u$ is in $H^{1}(\mathbb{R})$ so

$\int_{-\infty}^{+\infty}u_{x}u~{}dx=0\text{ for almost every }t,$

we can conclude that

$\int_{-\infty}^{+\infty}\int_{0}^{t}(w-u)w_{t}~{}dtdx+a\int_{-\infty}^{+\infty}\int_{0}^{t}|w_{t}|~{}dtdx\leq
0,$

which is indeed (1.8) extended to space-dependent $H^{1}$ functions. Finally,
as in Section 1 for time-dependent piecewise constant functions, and arguing
as in [21] and [20], (3.4) can be seen to be a suitable weak extension of the
Play operator to time-space dependent integrable functions.

If the weak solution has a jump discontinuity on a curve $(s(t),t)$, e.g. say
$u_{-}(t)\not=u_{+}(t)$ or $w_{-}(t)\not=w_{+}(t)$ then we get the following
extended Rankine-Hugoniot condition

$\frac{u_{-}(t)-u_{+}(t)}{u_{-}(t)-u_{+}(t)+w_{-}(t)-w_{+}(t)}=s^{\prime}(t).$
(3.5)

The entropy condition (2.4) is also extended to the following one

$\int_{\mathbb{R}_{T}}(|u-k|+|w-\hat{k}|)\phi_{t}+|u-k|\phi_{x}~{}dxdt\geq 0,$
(3.6)

for every $\phi$ non negative, smooth and with compact support in
$\mathbb{R}\times(0,T)$ and for every couple $(k,\hat{k})\in{\cal L}$.

##  4 The Riemann problem with Play hysteresis

We study the Riemann problem for (3.1) where $u_{0}(x)$, $w_{0}(x)$ consist of
two constant states separated by a discontinuity in the origin. We recall
that, for regular input and output, if $(u,w)$ belongs to the interior of
$\cal L$ then $w_{t}=0$, whereas when $(u,w)$ is on the boundary of $\cal L$
then $w_{t}=u_{t}$ whenever $w_{t}\neq 0$. Note that on the boundary, we may
have $w_{t}=0$ and $u_{t}\neq 0$ only for negligible times $t$ (because in
such a case, the pair would immediately enter the interior of $\cal L$ or move
along the boundary).

By the cases discussed here above, we can rewrite (3.1) as follows

$\begin{cases}u_{t}+u_{x}=0,\quad&|u-w|<a,\\\
u_{t}+\frac{1}{2}u_{x}=0,\quad&|u-w|=a,\\\ w=[\mathcal{F}(u,w_{0})],\\\
u(x,0)=u_{0}(x),\\\ w(x,0)=w_{0}(x).\end{cases}$ (4.1)

Of course the initial data must satisfy $|u_{0}(x)-w_{0}(x)|\leq a$ for almost
all $x$. Our goal is to rewrite (4.1) as a unique conservation law with piece-
wise linear flux. We consider two different cases, a) $|u_{0}(x)-w_{0}(x)|<a$
and b) $|u_{0}(x)-w_{0}(x)|=a$.

a) $|u_{0}(x)-w_{0}(x)|<a$ a.e. $x$
Let us consider initial data

$u_{0}(x):=\begin{cases}u_{l}\quad x<0,\\\ u_{r}\quad x>0\end{cases}$ (4.2)

with $u_{l}<u_{r}$ and

$w_{0}(x):=\begin{cases}w_{l}\quad x<0,\\\ w_{r}\quad x>0.\end{cases}$ (4.3)

hence in this case it is $|u_{l}-w_{l}|<a$ and $|u_{r}-w_{r}|<a$. At the
intial time $t=0$ the PDE is then

$u_{t}+u_{x}=0,\quad\forall x\in\mathbb{R}.$ (4.4)

If there was no hysteresis the PDE above is solved by the travelling wave
solution $u(x,t)=u_{0}(x-t).$ Hence, if we fix $x\in\mathbb{R}$ we are
expecting our solution $u$ either to remain equal to $u_{l}$ for $x<0$ or to
decrease from $u_{r}$ to $u_{l}$ after a time $t=x$, when $x>0$.

We have two subcases.

a1) If $a$ is such that $w_{r}-a\leq u_{l}$ then the jump from $u_{r}$ to
$u_{l}$ is such that the couple $(u,w)$ will always remain in the internal
region of hysteresis, and so $w_{t}=0$ for all $x$ and $t$. Hence we do not
experience the effect of hysteresis and so we get the trivial solution,
$u(x,t)=u_{0}(x-t)$ and $w(x,t)\equiv w_{0}(x).$

a2) Suppose instead $w_{r}-a>u_{l}$. For $x<0$, $u$ remains constantly equal
to $u_{l}$ and so does $w$. For $x>0$, we expect our solution to decrease, so
if we look at the hysteresis diagram we notice that for $w_{r}-a<u<u_{r}$ we
have $w_{t}=0$, whereas for $u_{l}<u<w_{r}-a$ we are on the boundary and so
$w_{t}=u_{t}$. See Figure 4.

$u$$w$$u_{r}$$u_{l}$$w_{r}-a$$w_{r}$$u$$w$$u_{r}$$u_{l}$$w_{r}-a$$w_{r}$
Figure 4: In the first picture we see that if $w_{r}-a<u_{l}$ then we remain
inside the hysteresis region when jumping from $u_{r}$ to $u_{l}$; in the
second picture, when $w_{r}-a>u_{l}$, instead we hit the boundary, so we have
$w_{t}=0$ initially and then $w_{t}=u_{t}.$

Let us define the following function

$h(x,u):=\begin{cases}1\quad&x<0\text{ or }(x>0\text{ and
}u_{r}>u>w_{r}-a),\\\ \frac{1}{2}\quad&x>0\text{ and
}u_{l}<u<w_{r}-a.\end{cases}$ (4.5)

As already pointed out, for $x<0$, the solution is constantly equal to $u_{l}$
(and the output to $w_{l}$), hence we can focus on $h$ defined only when
$x>0$, which is then only dependent on $u\in[u_{l},u_{r}]$. Moreover, the
switching rule (4.5) for $h$ actually encodes the hysteresis behavior between
$u$ and $w$. We can then rewrite our initial problem as follows

$\begin{cases}u_{t}+h(u)u_{x}=0,\\\ u(x,0)=u_{0}(x).\end{cases}$ (4.6)

We can see $h$ as the weak derivative of some piecewise linear flux
$g:~{}[u_{l},u_{r}]\to\mathbb{R}$ of the following form

$g(u):=\begin{cases}\frac{1}{2}u&\quad u_{l}\leq u\leq w_{r}-a,\\\ u+c&\quad
w_{r}-a\leq u\leq u_{r},\end{cases}$ (4.7)

where the constant $c$ makes the flux continuous. Also note that $g$ is
convex. Our PDE then reads as the following conservation law with piecewise
linear flux (compare with (2.1))

$u_{t}+g(u)_{x}=0,$ (4.8)

for which we consider the Riemann problem with datum (4.2). As in Section 2,
we find the solution to (4.8)

$u(x,t)=\begin{cases}u_{l}&\quad\frac{x}{t}<\frac{1}{2},\\\
w_{r}-a&\quad\frac{1}{2}<\frac{x}{t}<1,\\\ u_{r}&\quad
1<\frac{x}{t}.\end{cases}$ (4.9)

We then compute $w(x,t)=[\mathcal{F}(u(x,\cdot),w_{0}(x))](t)$ that, as
explained in Section 1 is, see also Figure 5.

$w(x,t)=\begin{cases}w_{l}(x)&\quad x<0,\\\ w_{r}(x)&\quad
0<\frac{x}{t}<\frac{1}{2},\\\
u_{l}+a&\quad\frac{1}{2}<\frac{x}{t}.\end{cases}$ (4.10)

$x$$t$$u:$$u_{l}$$w_{r}-a$$u_{r}$$x$$t$$w_{l}$$u_{l}+a$$w:$$w_{r}$ Figure 5:
The solution $u$ and corresponding $w$

Note that the pair $(u,w)$ as above constructed, satisfies the Rankine-
Hugoniot condition (3.5). Indeed we have two discontinuities on $u$: the one
from $u_{r}$ to $w_{r}-a$ is such that $w$ remains constant so

$s^{\prime}(t)=\frac{u_{r}-(w_{r}-a)}{u_{r}-(w_{r}-a)}=1;$

the one from $w_{r}-a$ to $u_{l}$ is such that also $w$ jumps from $w_{r}$ to
$u_{l}-a$ hence

$s^{\prime}(t)=\frac{w_{r}-a-u_{l}}{w_{r}-a-u_{l}+w_{r}-(u_{l}+a)}=\frac{1}{2}.$

We also have a discontinuity on $w$ with slope $0$ between $w_{l}$ and
$u_{l}+a$ but it still satisfies the Rankine-Hugoniot condition indeed $u$ is
continuous implying that $s^{\prime}(t)=0.$ Being the Rankine-Hugoniot
condition satisfied by the piece-wise constant pair $(u,w)$, then the latter
satisfies the weak formulation of (3.1), that is (3.2).

######  Remark 4.1.

The argumentation above on the Rankine-Hugoniot conditions, works well not
only because we have posed $w={\cal F}(u)$ but mainly because $u$ solves (4.8)
where, as already pointed out, hysteresis is somehow encoded in the flux $g$.

As said, $(u,w)$ can be seen as a weak solution of the Riemann problem (3.1),
(4.2), (4.3). Also note that, as we already know, the relation
$w(x,\cdot)=\mathcal{F}(u(x,\cdot),w_{0}(x))$, for piece-wise constant
functions $t\mapsto(u(x,t),w(x,t))$, is equivalent to (1.8). In the proof of
Theorem 5.4, we will show that (1.8) for piece-wise constant functions,
together with the hypothesis of integrability on $\mathbb{R}\times[0,T)$
(which is not satisfied by the solution of the Riemann problem), implies
condition (3.4), giving a weak solution in the sense of Definition 3.1.

######  Remark 4.2.

If the initial datum satisfies $u_{l}>u_{r}$, then we end up with a similar
situation as above but this time the piecewise linear flux $g$ will be
concave, coherently with what said in Section 2. The construction of the
solution then goes similarly as above.

b) $|u_{0}(x)-w_{0}(x)|=a$ a.e. $x$
Consider the initial states (4.2),(4.3) with $u_{l}<u_{r}$. We have two
subcases.

b1) Suppose $w_{r}=u_{r}+a$, that is $(u_{r},w_{r})$ belongs to the upper
boundary of $\cal L$. Here we expect the solution $u$ to decrease from $u_{r}$
to $u_{l}$ for $x>0$. Then we can notice that the pair $(u,w)$ will always
remain on the upper boundary $w=u+a$, see Figure 7. The solved PDE is then

$u_{t}+\frac{1}{2}u_{x}=0.$ (4.11)

The solution $u(x,t)=u_{0}(x-1/2t)$, and its output $w$, see Figure 6 are then

$u(x,t)=\begin{cases}u_{l}\quad&0<\frac{x}{t}<\frac{1}{2}\\\
u_{r}\quad&\frac{1}{2}<\frac{x}{t}\end{cases},\quad
w(x,t)=\begin{cases}w_{l}\quad&0<x\\\
u_{r}+a\quad&0<\frac{x}{t}<\frac{1}{2}\\\
u_{l}+a\quad&\frac{1}{2}<\frac{x}{t}.\end{cases}$ (4.12)

$x$$t$$u:$$u_{l}$$u_{r}$$x$$t$$w_{l}$$u_{l}+a$$w:$$u_{r}+a$ Figure 6: The
solution $u$ and corresponding $w$ for the subcase b1)
$u$$w$$u_{l}$$u_{r}$$w_{r}$ Figure 7: If we start on the upper boundary and we
jump from $u_{r}$ to $u_{l}$ then we will always remain on that boundary.

b2) Suppose $(u_{r},w_{r})$ to be on the lower boundary of $\cal L$, that is
$w_{r}=u_{r}-a$. In this case for $x>0$ we have $u$ decreasing from $u_{r}$ to
$u_{l}$ so the couple $(u(x,\cdot),w(x,\cdot))$ will move internally possibly
reaching the upper boundary and following it until $u=u_{l}$, see Figure 9. We
then have two further subcases.

b2i) Suppose $u_{l}\geq w_{r}-a$, that is the left-case in Figure 9. By (4.1),
when $u=u_{r}$ or $u_{l}<u<u_{r}$ the PDE is, respectively,

$u_{t}+\frac{1}{2}u_{x}=0\quad\text{or}\quad u_{t}+u_{x}=0.$ (4.13)

So if we define

$h(u):=\begin{cases}\frac{1}{2}&\quad u=u_{r}\\\ 1&\quad u_{l}\leq
u<u_{r},\end{cases}$ (4.14)

and as in the case a2) consider a conservation law with $g$ piecewise linear
flux, then the solution is $u_{0}(x-t)$, that is

$u(x,t)=\begin{cases}u_{l}\quad&\frac{x}{t}<1,\\\
u_{r}\quad&\frac{x}{t}>1,\end{cases}$ (4.15)

which generates the constant output $w(x,t)\equiv w_{0}(x)$.

b2ii) Suppose now $u_{l}<w_{r}-a$ so for $x>0$ when $u$ decreases from $u_{r}$
to $u_{l}$ we hit also the upper boundary at $u=w_{r}-a$, see the right-case
in Figure 9. Again by (4.1) we have

$h(u)=\begin{cases}\frac{1}{2}\quad&u=u_{r},\\\ 1\quad&w_{r}-a<u<u_{r}\\\
\frac{1}{2}\quad&u_{l}<u<w_{r}-a,\end{cases}$ (4.16)

which is the derivative of the flux. Then, by solving as in case a2) and
subcase b2i) a conservation law with piece-wise linear flux and after
computing $w=\mathcal{F}(w,w_{0})$, see Figure 8, we find the pair

$u(x,t)=\begin{cases}u_{l}\quad&\frac{x}{t}<\frac{1}{2},\\\
w_{r}-a\quad&\frac{1}{2}<\frac{x}{t}<1,\\\
u_{l}\quad&\frac{x}{t}>1,\end{cases}\quad
w(x,t)=\begin{cases}w_{l}\quad&x<0,\\\
u_{l}+a\quad&0<\frac{x}{t}<\frac{1}{2},\\\
u_{r}-a\quad&\frac{x}{t}>\frac{1}{2}.\end{cases}$ (4.17)

$x$$t$$u:$$u_{l}$$w_{r}-a$$u_{r}$$x$$t$$w_{l}$$u_{l}-a$$w:$$u_{r}+a$ Figure 8:
The solution $u$ and corresponding $w$ in subcase b2ii)
$u$$w$$u_{r}$$u_{l}$$w_{r}-a$$w_{r}$$u$$w$$u_{r}$$u_{l}$$w_{r}-a$$w_{r}$
Figure 9: The two possible cases whether $w_{r}-a\leq u_{l}$ or
$w_{r}-a>u_{l}.$

As for the case a), Rankine-Hugoniot conditions can be also derived for case
b) and moreover similar remarks as Remark 4.1 and Remark 4.2 hold.

######  Remark 4.3.

Since for $x<0$, $u(x,t)\equiv u_{l}$ and so $w(x,t)\equiv w_{l}$, then the
conditions for the cases a) and b) reduce respectively to $|u_{r}-w_{r}|<a$
and $|u_{r}-w_{r}|=a$. Consequently, the study above covers all possible
cases.

######  Remark 4.4.

For the subcases a1) and b2i), as said, the solution is just a rigid movement
of the initial data. Instead for the subcases a2) and b2ii) we easily see that
$u_{l}<w_{r}-a\leq u_{r}$ and also $u_{l}+a$ is between $w_{r}$ and $w_{l}$,
since $w_{l}\in[u_{l}-a,u_{l}+a]$. For the subcase b1) again holds that
$u_{l}+a$ is between $w_{r}$ and $w_{l}$. These facts are important as they
imply that for every case $Var(u(\cdot,t))=u_{r}-u_{l}=Var(u_{0}(\cdot))$ and
$Var(w(\cdot,t))=w_{r}-w_{l}=Var(w_{0}(\cdot))$ for every $t$, where $Var$
denotes the total variation on $\mathbb{R}$ with respect to the spacial
variable. The fact that the total variation does not increase in time is
important and will give a compactness tool to prove the existence to the
general Cauchy problem as in Section 5.

##  5 The general initial data Cauchy problem

The goal of this section is to construct weak solution in the sense of
Definition 3.1. by using the Wave Front Tracking Method.

As already said, such method consist of approximating initial data with piece-
wise constant function, so we consider first

$u_{0}(x)=\sum_{i=1}^{N}u_{i}\mathds{1}_{(x_{i-1},x_{i})}$ (5.1)

and

$w_{0}(x)=\sum_{i=1}^{N}w_{i}\mathds{1}_{(x_{i-1},x_{i})}$ (5.2)

such that $|u_{0}(x)-w_{0}(x)|\leq a$ where
$-\infty=x_{0}<x_{1}<\dots<x_{N}=+\infty$.

In order to solve the Cauchy problem (3.1), (5.1), (5.2), we give the
following heuristics idea, which differs from a classical one by the presence
of the hysteretic term $w$, see Figure 10:

  1. 1.

We solve $N-1$ Riemann problems at time $t=0$ centered in the points
$x_{1},\dots,x_{N-1}$, which generate discontinuity lines for $u$ and $w$; by
the analysis in Section 4, we know that the discontinuity lines for $u$ are
lines with slopes $1$ or $1/2$ (in the plane $t-x$), whereas the ones for $w$
with slope $0$ or $1/2$; moreover the discontinuity lines of slope $1/2$ are
always discontinuity lines for both $u$ and $w$.

  2. 2.

Let us denote by $\tau_{1}$ the first time that either a discontinuity line of
$u$ impinges a discontinuity line of $w$ with slope $0$ or two discontinuity
lines of $u$ impinge themselves. The pair $(u,w)$ found in the previous point
is a weak solution of (3.1) for $t\leq\tau_{1}$ (i. e. (3.2) and $w={\cal
F}(u)$); note that $\tau_{1}>0$ exists since the number of constancy intervals
is finite;

  3. 3.

we consider $u(\cdot,\tau_{1})$ and $w(\cdot,\tau_{1})$ as new initial
conditions, since they are still piecewise constant, we can solve again a
finite number of Riemann problems at $\\{t=\tau_{1}\\}$. So we extended $u$
and $w$ for small times after $\tau_{1}$;

  4. 4.

we proceed in way extending at each step the solution for larger times.

$x$$t$$\tau_{1}$$\tau_{2}$$u:$$w:$$x_{1}$$x_{2}$$x_{3}$$x_{4}$ Figure 10: A
possible solution generated by the above procedure, where we have interactions
at time $\tau_{1}$ and $\tau_{2}$. In black the discontinuity lines of slope
$1$ of $u$, in red ones of $w$ with slope $0$, and both colors for the common
lines with slope $1/2$.

Next lemma formalizes the heuristics above.

######  Lemma 5.1.

For the Cauchy problem (3.1), (5.1), (5.2), with $T=+\infty$, arguing as
above, we can construct two piece-wise constant functions
$u,w:\mathbb{R}\times[0,+\infty)$ such that:

  1. i)

The couple $(u,w)$ satisfies the weak formulation of the PDE (3.2) and the
relation $w=\mathcal{F}(u,w_{0})$ for almost every $x$.

  2. ii)

For fixed $t\in[0,+\infty)$ we have that

$Var(u(\cdot,t))\leq Var(u_{0}(\cdot)),\quad Var(w(\cdot,t))\leq
Var(w_{0}(\cdot)),$ (5.3)

where $Var$ denotes the total variation on $\mathbb{R}$ of the functions with
respect to the space variable.

  3. iii)

For $t,t^{\prime}\in[0,+\infty)$ the following estimates hold

$\int_{-\infty}^{+\infty}|u(x,t)-u(x,t^{\prime})|~{}dx\leq
Var(u_{0}(\cdot))|t-t^{\prime}|$ (5.4)

and

$\int_{-\infty}^{+\infty}|w(x,t)-w(x,t^{\prime})|~{}dx\leq\frac{1}{2}Var(w_{0}(\cdot))|t-t^{\prime}|.$
(5.5)

###### Proof.

At $t=0$, we solve $N-1$ Riemann problems as described previously and we
compute the couple $(u,w)$ for $t$ smaller then $\tau_{1}.$ By the results of
Section 4, we know that (3.2) holds and $w=\mathcal{F}(u,w_{0})$ for every $x$
possibly except $x_{1},\dots,x_{N-1}$. By Remark 4.4, (5.3) holds for
$t\in[0,\tau_{1})$ as an equality. Inequality (5.4) also holds for
$t,t^{\prime}\in[0,\tau_{1})$. Again, a possible proof of it differs from a
standard case ([6], [2], [9]) by the presence of the hysteresis in the
equation. We give here an idea of that for the single Riemann problem centered
at $0$ with values $u_{l}$ and $u_{r}$ as in Section 4. Considering the set
$R:=\\{(x,s)~{}|~{}1/2\leq x/s\leq 1,~{}0\leq s\leq\tau_{1}\\}$, we can notice
that $u(\cdot,t)$ and $u(\cdot,t^{\prime})$ are equal to $u_{l}$ at the left
of $R$ and to $u_{r}$ at the right of $R.$ Internally they may be equal either
to $u_{l}$, $u_{r}$ or $w_{1}\pm a$ (see for example (4.9) or (4.12)), here we
will denote this internal value with $u^{*}$. Note that, in any case, $u^{*}$
is in between $u_{l}$ and $u_{r}$. It is easy to see that if we only have one
discontinuity $u_{-}\not=u_{+}$, with slope $1$ then

$\int_{-\infty}^{+\infty}|u(x,t)-u(x,t^{\prime})|~{}dx=|t-t^{\prime}|~{}|u_{-}-u_{+}|.$
(5.6)

If instead we have only one discontinuity with slope $1/2$ then

$\int_{-\infty}^{+\infty}|u(x,t)-u(x,t^{\prime})|~{}dx=\bigg{|}\frac{t}{2}-\frac{t^{\prime}}{2}\bigg{|}~{}|u_{-}-u_{+}|\leq|t-t^{\prime}|~{}|u_{-}-u_{+}|.$
(5.7)

Since we may have at most $2$ discontinuities then (5.4) follows from (5.6)
and (5.7), indeed

$\int_{-\infty}^{+\infty}|u(x,t)-u(x,t^{\prime})|~{}dx\leq|t-t^{\prime}|(|u_{r}-u^{*}|+|u^{*}-u_{l}|)=|t-t^{\prime}|~{}|u_{r}-u_{l}|,$

where $|u_{l}-u_{r}|=Var(u_{0})$.

Inequality (5.5) is treated in the same way with the difference that
discontinuities of $w$ have either slope $0$ or $1/2$ so the maximum is $1/2.$

The lemma is proven for $t\in[0,\tau_{1})$ now we need to define $u$ and $w$
for $t=\tau_{1}$ and use them as new initial data. As a standard non-
hysteresis case, the piece-wise constant functions $u(\cdot,\tau_{1})$,
$w(\cdot,\tau_{1})$ can be constructed as limits as $t\to\tau_{1}^{-}$ without
increasing the total variations (see again [6]). Moreover, for
$x\not\in\\{x_{1},...,x_{N-1}\\}\cup\\{y|(y,\tau_{1})\ \text{belongs to
discontinuity lines started at $t=0$}\\}$, it holds $w(x,\tau_{1})=[{\cal
F}(u(x,\cdot),w_{0}(x)](\tau_{1})$ as $u$ and $w$ are constant in $(x,t)$ for
$t<\tau_{1}$ sufficiently close to $\tau_{1}$, see Figure 10. Also note that
the excluded values of $x$ are a finite quantity.

Taking the piece-wise constant functions $u(\cdot,\tau_{1})$,
$w(\cdot,\tau_{1})$ as initial conditions at $\tau_{1}$ and repeating the
reasoning above, we are able to find a new instant $\tau_{2}>\tau_{1}$ and to
extend $u,w$ for $t\in[\tau_{1},\tau_{2})$, satisfying (5.3), (5.4) and (5.5)
in $[0,\tau_{2})$. Also note that, by the semigroup property (see Remark 1.2),
$w=\mathcal{F}(u,w_{0})$ in $[0,\tau_{2})$ for every $x$ different from the
one excluded previously.

Then we apply such extension procedure recursively, getting a sequence of
instants $\tau_{1}<\tau_{2}<\tau_{3}<\dots$. Next two lemmas imply that the
number of such instants (and hence of recursive cases) is finite and hence we
actually may perform the extension up to whole time-line $[0,+\infty)$.
Finally note that at each step we exclude a finite number of $x$ such that
$w\not=\mathcal{F}(u,w_{0})$, hence $w=\mathcal{F}(u,w_{0})$ in $[0,+\infty)$
for almost every $x$ ∎

######  Lemma 5.2.

The number of discontinuity lines of $u$ intersecting each other is finite.

###### Proof.

Let us denote by $u_{1},\dots,u_{N}$ and $w_{1},\dots,w_{K}$ the initial
values of $u$ and $w$ respectively. At every step-instant $\tau_{k}$, by
solving the Riemann problems, we end up with $u$ taking values in the sets
$\\{\text{initial values of $u$}\\}$ and $\\{w\pm a\\}$ remaining always
between $u_{m}:=\min u_{i}$ and $u_{M}:=\max u_{i}$, see e.g. (4.9) or (4.12).
Similarly, $w$ takes values in the sets $\\{\text{initial values of $w$}\\}$
and $\\{u\pm a\\}$. We conclude that, at every time-step $\tau_{k}$, for
almost every $(x,t)\in\mathbb{R}\times[0,\tau_{k})$ $u(x,t)$ belongs to

$\bigcup_{\begin{subarray}{c}i=1,\dots,N\\\
j=1,\dots,K\end{subarray}}\bigcup_{k\in\mathbb{N}}\left(\\{u_{i}\pm
ka\\}\cup\\{w_{j}\pm ka\\}\right)~{}\bigcap~{}[u_{m},u_{M}]=:IM(u)$

which does not depend on the time-step $\tau_{k}$ and necessarily consists of
a finite number of points. In particular

$\delta:=\min\left\\{|x-y|~{}\bigg{|}~{}x,y\in IM(u),x\not=y\right\\}>0.$
(5.8)

If two discontinuity lines of $u$ cross each other, then the (lower) one must
have slope $1/2$ and the (upper) one slope $1$ (see Figure 11-right). We
denote by $u_{l}$, $u_{r}$ and $u^{*}$ the values of $u$ on the left, on the
right and between the two curves, respectively. Such setting tell us that the
first jump between $u_{r}$ and $u^{*}$ implies a jump for the corresponding
pair $(u,w)$ between two points belonging to the same boundary of $\cal L$,
and in particular that $w$ changes accordingly to $u$; whereas the second one,
from $u^{*}$ to $u_{l}$, implies a jump for $(u,w)$ between two points with
the same ordinate $w$ (see Figure 11-left). From this we conclude that $u^{*}$
can not be between $u_{l}$ and $u_{r}$. Also note that the Riemann problem
solved at the intersecting point results in a single discontinuity wave
between $u_{l}$ and $u_{r}$ with either slope $1/2$ or $1$, depending on the
mutual position of $u_{r}$ and $u_{l}$ and on their distance. In particular,
by what said above, we have that the total variation of the solution $u$, in
$x$ for $t$ fixed, after the intersection point decreases at least by
$2\delta$, with $\delta$ as in (5.8). Indeed, before it was
$|u_{r}-u^{*}|+|u^{*}-u_{l}|$ and after it is $|u_{r}-u_{l}|$, being
$|u_{r}-u^{*}|+|u^{*}-u_{l}|-|u_{r}-u_{l}|\geq 2\delta>0$.

Since the total variation of $u$ at time $t=0$ is finite and it is non
increasing in time (see (5.3)) then we cannot have an infinite number of such
interactions.

We finally note that the corresponding jump of $w$ before the intersection was
a jump between two values $w_{r}$ and $w_{l}$ and after becomes a jump between
$w_{r}$ and a new intermediate value $w^{*}$ (see Figure 11). In particular
new $0$-slope discontinuity in $w$ may arise. ∎

######  Lemma 5.3.

The number of discontinuity lines of $u$ intersecting the discontinuity lines
of $w$ is finite.

###### Proof.

Since the discontinuity lines for $w$ with slope $1/2$ are also discontinuity
lines for $u$, we restrict ourselves to the discontinuity lines for $w$ with
slope $0$.

The discontinuities of $w$ with slope $0$ starting at time $t=0$ are finite
and a new discontinuity with slope $0$ for $w$ may arise only by the
interactions of two discontinuity lines for $u$, as in Lemma 5.2. Hence the
total number of discontinuities of $w$ with slope $0$ that may be generated
during the whole process is a-priori bounded. Now when a discontinuity line of
$u$ crosses a $0$-slope discontinuity of $w$, we may generate a new
discontinuity of $u$ but the number of $0$-ones for $w$ does not increase, see
the example in Figure 12.

Since the discontinuity lines for $u$ have positive slope, it is not possible
that one of them, say $r_{1}$, intersects a vertical discontinuity line for
$w$ that has already generated discontinuity lines from which, after a finite
number of new intersections, $r_{1}$ itself is generated.

Concluding, since at time $0$ the number of discontinuities for $u$ is finite,
then there is a finite number of intersection with $w$-discontinuity of $0$
slope. ∎

$u$$w$$u^{*}$$u_{r}$$u_{l}$$w_{r}$$w_{l}$$w^{*}$$\tau_{k}$$u_{l}$$u_{r}$$u^{*}$$w_{l}$$w_{r}$$w^{*}$
Figure 11: Explaining figure for Lemma 5.2. We see how $u^{*}$ cannot be
between $u_{l}$ and $u_{r}$ so after the interaction the total variation of
$u$ decreases. Moreover we generate a $0$-slope discontinuity in $w$ as
$w^{*}\not=w_{l}$.
$u$$w$$u_{r}$$u_{l}$$w^{*}$$w_{r}$$\tau_{k}$$u_{l}$$u_{r}$$u^{*}$$w_{l}$$w_{r}$$w^{*}$
Figure 12: Explaining figure for Lemma 5.3. We see how the number of
discontinuities of $u$ can increase while the total variation
$Var(u(\cdot,t))$ remains constant in time.

Now, we prove the existence of a solution for the Cauchy Problem (3.1), with
general initial conditions $u_{0}$, $w_{0}$.

######  Theorem 5.4.

Consider (3.1) with $T>0$ and $u_{0},w_{0}\in BV(\mathbb{R})\cap
L^{1}(\mathbb{R})$, satisfying $|u_{0}(x)-w_{0}(x)|\leq a$ for almost every
$x$. Then there exists a couple $(u,w)\in L^{1}(\mathbb{R}_{T})$ weak solution
in the sense of Definition 3.1.

###### Proof.

Since $u_{0}$,$w_{0}\in BV(\mathbb{R})$ then they are bounded
($|u_{0}(x)|,|w_{0}(x)|\leq M$ a.e. $x$), their limits at infinity exist and
such limits are null as $u_{0}$, $w_{0}$ are also in $L^{1}(\mathbb{R})$.
Moreover, there is a sequences $\\{u_{0}^{(n)}\\}_{n}$ and
$\\{w_{0}^{(n)}\\}_{n}$ of piece-wise constant functions approximating $u_{0}$
and $w_{0}$ respectively in $L^{1}(\mathbb{R})$ with

$Var(u_{0}^{(n)})\leq Var(u_{0})\quad\text{and}\quad Var(w_{0}^{(n)})\leq
Var(w_{0})\quad\forall n\in\mathbb{N}.$ (5.9)

Let $(u_{n},w_{n})$ be the weak solution to the problem with initial data
$u_{0}^{(n)}$ and $w_{0}^{(n)}$ as in Lemma 5.1. Since $u_{0}^{(n)}$ and
$w_{0}^{(n)}$, in order to approximate $u_{0}$ and $w_{0}$ in
$L^{1}(\mathbb{R})$, must vanish at infinity (actually, being the number of
pieces finite, their last interval of constancy is a semi-line where they are
null, they vanish outside a compact set), hence it is easy to see that also
$u_{n}(\cdot,t)$ and $w_{n}(\cdot,t)$ has the same property, for every fixed
$t$. Applying standard procedures for the wave-front tracking method, (see
[6], [2]) we get the existence of a function $u\in L^{1}(\mathbb{R}_{T})$ such
that (up to a subsequence)

$u_{n}\longrightarrow u,\quad\text{in }L^{1}(\mathbb{R}_{T}),\text{ as
}n\to\infty.$ (5.10)

By the same argument, there exists $w\in L^{1}(\mathbb{R}_{T})$ such that (up
to a subsequence) $w_{n}\to w$ in $L^{1}(\mathbb{R}_{T})$.

We have to prove that $u$ and $w$ satisfy the weak formulation of the PDE
(3.2) and the hysteresis relations (3.3), (3.4).

Since (3.2) is solved by every $(u_{n},w_{n})$ then it is also true for the
limit because of the $L^{1}$ convergence on $\mathbb{R}_{T}$. Moreover
$w_{n}=\mathcal{F}(u_{n},w_{0})$ for a.e. $x$, consequently $|w_{n}-u_{n}|\leq
a$ a.e. and hence (3.3) comes from a.e. point-wise convergence.

It remains (3.4). Because of Proposition 1.3, since
$w_{n}=\mathcal{F}(u_{n})$, we have that for all $n$ and almost every $x$ and
$t$

$\int\limits_{(0,t)}\left(\tilde{u}_{n}(x,t)-\tilde{w}_{n}(x,t)\right)d\left(\frac{\partial
w_{n}}{\partial t}\right)_{x}\geq a\bigg{|}\left(\frac{\partial
w_{n}}{\partial t}\right)_{x}\bigg{|}((0,t)).$ (5.11)

Here with $\left(\frac{\partial w_{n}}{\partial t}\right)_{x}$ we denote the
measure associated to the distribution $w_{n}^{\prime}(x,\cdot)$, with fixed
$x$ seen as function of time, and with tilde the right continuous with respect
to time representative of $u_{n}$ and $w_{n}$.

Now we fix $t\in(0,T)$ such that (5.11) holds for a.e. $x$. Since for every
fixed $x$,
$\tilde{w}_{n}(x,\cdot)_{|_{[0,t]}}=\sum_{i=1}^{N(x,t)}\tilde{w}_{n}^{(i)}(x)\mathds{1}_{[t_{i-1},t_{i})}$
then

$\int\limits_{(0,t)}\left(\tilde{w}_{n}(x,t)\right)d\left(\frac{\partial
w_{n}}{\partial
t}\right)_{x}=\sum_{i=1}^{N(x,t)-1}\tilde{w}_{n}^{(i+1)}(x)(\tilde{w}_{n}^{(i+1)}(x)-\tilde{w}_{n}^{(i)}(x)).$

Now, if we exclude the $0$ measure set of all the the points $x$ given by the
union of the discontinuity points of $w_{0}^{(n)}$ with the points $x$ such
that $(x,t)$ lies on a discontinuity line of $w_{n}$ then it holds that

$\tilde{w}_{n}^{(1)}(x)=\lim_{s\to 0^{+}}w_{n}(x,s)=w_{0}^{(n)}(x)$

and

$\tilde{w}_{n}^{(N)}(x)=\lim_{s\to t^{-}}w_{n}(x,s)=w_{n}(x,t).$

So by the inequality

$\sum_{i=1}^{N-1}\tilde{w}_{n}^{(i+1)}(x)(\tilde{w}_{n}^{(i+1)}(x)-\tilde{w}_{n}^{(i)}(x))\geq\frac{1}{2}((\tilde{w}_{n}^{(N)})^{2}-(\tilde{w}_{n}^{(1)})^{2})$

and by integrating (5.11) over $\mathbb{R}$ we deduce

$\int_{-\infty}^{+\infty}\int\limits_{(0,t)}\left(\tilde{u}_{n}(x,t)\right)d\left(\frac{\partial
w_{n}}{\partial t}\right)\geq\\\
\frac{1}{2}\int_{-\infty}^{+\infty}(w_{n}(x,t)^{2}-w_{0}^{(n)}(x,t)^{2})~{}dx+a\bigg{|}\frac{\partial
w_{n}}{\partial t}\bigg{|}(\mathbb{R}\times(0,t)).$ (5.12)

Here with $\frac{\partial w}{\partial t}$ we denote the measure on
$\mathbb{R}\times(0,t)$ associated to the distributional derivative of $w$
with respect to $t$ seen as function of both $x$ and $t$. It exists since
$w_{n}$ is a function of bounded variation of two variables (finite number of
constant pieces), moreover it is $\frac{\partial w_{n}}{\partial
t}=\mathcal{L}^{1}\otimes\left(\frac{\partial w_{n}}{\partial t}\right)_{x}$,
see [1].

Now for every $\phi$ smooth with compact support in $\mathbb{R}\times(0,t)$
because of (3.2) we have

$\left\langle\frac{\partial u_{n}}{\partial t}+\frac{\partial w_{n}}{\partial
t}+\frac{\partial u_{n}}{\partial x},\phi\right\rangle=0$

in a distributional sense and, since $u_{n}$, $w_{n}$ are of bounded variation
as functions of two variable, we can see this relation as an equality between
the measures associated to the distributions i.e.

$\frac{\partial w_{n}}{\partial t}=-\frac{\partial u_{n}}{\partial
t}-\frac{\partial u_{n}}{\partial x}$ (5.13)

as measures on $\mathbb{R}\times(0,t)$. We then have

$\begin{split}\int_{-\infty}^{+\infty}&\int\limits_{(0,t)}\tilde{u}_{n}(x,t)~{}d\left(\frac{\partial
w_{n}}{\partial t}\right)=\\\
&=-\int_{-\infty}^{+\infty}\int\limits_{(0,t)}\tilde{u}_{n}(x,t)~{}d\left(\frac{\partial
u_{n}}{\partial
t}\right)-\int_{-\infty}^{+\infty}\int\limits_{(0,t)}\tilde{u}_{n}(x,t)~{}d\left(\frac{\partial
u_{n}}{\partial x}\right)\\\
&=-\int_{-\infty}^{+\infty}\int\limits_{(0,t)}\tilde{u}_{n}(x,t)~{}d\left(\frac{\partial
u_{n}}{\partial
t}\right)_{x}~{}dx-\int\limits_{(0,t)}\int_{-\infty}^{+\infty}\tilde{u}_{n}(x,t)~{}d\left(\frac{\partial
u_{n}}{\partial x}\right)_{t}~{}dt\\\
&\leq-\frac{1}{2}\int_{-\infty}^{+\infty}(u_{n}(x,t)^{2}-u_{0}^{(n)}(x)^{2})~{}dx-\frac{1}{2}\int\limits_{(0,t)}(u_{n}(+\infty,t)^{2}-u_{n}(-\infty,t)^{2})~{}dt\\\
&=-\frac{1}{2}\int_{-\infty}^{+\infty}(u_{n}(x,t)^{2}-u_{0}^{(n)}(x)^{2})~{}dx.\end{split}$
(5.14)

Here we have applied similar steps as above (between (5.11)–(5.12)), also
exploiting the fact that $\tilde{u}$ for fixed $t$ is left continuous with
respect to $x$ and that $u_{n}(+\infty,t)=u_{n}(-\infty,t)=0$.

Then from (5.12) and (5.14) we conclude that for all $n\in\mathbb{N}$ and a.e.
$t\in(0,T)$

$\frac{1}{2}\int_{-\infty}^{+\infty}(u_{n}(x,t)^{2}-u_{0}^{(n)}(x)^{2})~{}dx+\frac{1}{2}\int_{-\infty}^{+\infty}(w_{n}(x,t)^{2}-w_{0}^{(n)}(x)^{2})~{}dx+\\\
+a\bigg{|}\frac{\partial w_{n}}{\partial t}\bigg{|}(\mathbb{R}\times(0,t))\leq
0.$ (5.15)

We know by construction and by the wave-front tracking method we have applied
(see [6], [2]) that for every fixed $t$ the sequences $u_{n}(\cdot,t)$,
$w_{n}(\cdot,t)$, $u_{0}^{(n)}$ and $w_{0}^{(n)}$ converge respectively to
$u(\cdot,t)$, $w(\cdot,t)$, $u_{0}$ and $w_{0}$ in $L^{1}(\mathbb{R})$, and
since they are all equibounded in $L^{\infty}(\mathbb{R})$ we get the
convergence also in $L^{2}(\mathbb{R})$. This in particular means that

$0\leq\limsup_{n\to\infty}\bigg{|}\frac{\partial w_{n}}{\partial
t}\bigg{|}(\mathbb{R}\times(0,t))\leq C<+\infty$

and so, up to a subsequence, there exists a measure weak star limit of the
sequence $\partial w_{n}/\partial t$ which must then coincide with $\partial
w/\partial t$ by the convergence of $w_{n}$ to $w$ in $L^{1}(\mathbb{R}_{T}).$
Finally, thanks to the lower-semicontinuity

$\bigg{|}\frac{\partial w_{n}}{\partial
t}\bigg{|}(\mathbb{R}\times(0,t))\leq\liminf_{n\to\infty}\bigg{|}\frac{\partial
w_{n}}{\partial t}\bigg{|}(\mathbb{R}\times(0,t)),$

taking the liminf in (5.15), we infer our desired condition (3.4). ∎

######  Remark 5.5.

We note that in [20] similar arguments, as in the proof above, are used but
for approximating $H^{1}$ functions, generated by a time-discretization. Here
instead the functions are much less regular, the discretization is in space
for the initial data, and the derivatives are just measures.

##  6 Uniqueness

Let us recall that the couple $(u,w)$ weak solution is an entropy solution if
it satisfies the following inequality

$\int_{\mathbb{R}_{T}}(|u-k|+|w-\hat{k}|)\phi_{t}+|u-k|\phi_{x}~{}dxdt\geq 0,$
(6.1)

for every $\phi$ non negative, smooth and with compact support in
$\mathbb{R}\times(0,T)$ and for every couple $(k,\hat{k})\in\mathcal{L}$.

Now we show how in our problem with hysteresis this entropy condition can be
characterized in the case of piecewise constant solutions (see e.g. [2] for
the case without hysteresis).

######  Proposition 6.1.

Suppose that the couple of functions
$u,w:~{}\mathbb{R}\times[0,+\infty)\to\mathbb{R}$, satisfying $|u-w|\leq a$,
consists of two constant states $(u_{l},w_{l})\not=(u_{r},w_{r})$ separated by
a line of discontinuity $(s(t),t)$. Then $(u,w)$ is an entropy solution if and
only if it is constructed as in Section 4 i.e. the following conditions are
verified:

  1. i)

$s^{\prime}(t)=:\lambda$ is constant and satisfies the generalized $Rankine-
Hugoniot$ condition (3.5);

  2. ii)

$\lambda$ is either $0,1$ or $1/2$;

  3. iii)

if $u_{l}=u_{r}$ then $\lambda=0$ i.e. we have a discontinuity only on $w$
with slope $0$;

  4. iv)

if $w_{r}=w_{l}$ then $\lambda=1$ so the jump in not on the boundary of the
hysteresis region and $w$ remains constant;

  5. v)

if $u_{r}>u_{l}$ and $w_{r}\not=w_{l}$ then $\lambda=1/2$ and $w_{r}=u_{r}+a$,
$w_{l}=u_{l}+a$, i.e. the jump occurs in both $u$ and $w$ and the couple is on
the upper boundary of the hysteresis region;

  6. vi)

if $u_{r}<u_{l}$ and $w_{l}\not=w_{r}$ then $\lambda=1/2$ and $w_{r}=u_{r}-a$,
$w_{l}=u_{l}-a$, i.e. the jump occurs in both $u$ and $w$ and the couple is on
the lower boundary of the hysteresis region;

###### Proof.

First notice that integrating by parts (6.1) with $(u,w)$ piece-wise constant
we can rewrite such entropy condition as follows

$s^{\prime}(t)[(|u_{r}-k|+|w_{r}-\hat{k}|)-(|u_{l}-k|+|w_{l}-\hat{k}|)]-[|u_{r}-k|-|u_{l}-k|]\geq
0.$ (6.2)

So the couple $(u,w)$ is an entropy solution if and only if (6.2) holds for
every $(k,\hat{k})\in\mathcal{L}$.

$(\implies)$ Suppose that the pair $(u,w)$ is an entropy solution. Then, for
any $u_{r},u_{l},w_{r},w_{l}$, we can always find a pair
$(k,\hat{k})\in\mathcal{L}$ with $k,\hat{k}$ large enough such that (6.2)
reads as follows

$s^{\prime}(t)[u_{l}+w_{l}-u_{r}-w_{r}]-[u_{l}-u_{r}]\geq 0$ (6.3)

and, by inserting in (6.2) their opposite values, it becomes

$-s^{\prime}(t)[u_{l}+w_{l}-u_{r}-w_{r}]+[u_{l}-u_{r}]\geq 0.$ (6.4)

We then infer that $s^{\prime}(t)$ has to satisfy the Rankine-Hugoniot
condition (3.5) that is

$[u_{l}-u_{r}+w_{l}-w_{r}]s^{\prime}(t)=u_{l}-u_{r},$ (6.5)

hence $s^{\prime}(t)=:\lambda$ has to be constant, so we proved
$\ref{list1})$. Notice that $u_{l}-u_{r}+w_{l}-w_{r}$ cannot be equal to $0$
otherwise from (6.5) we would get also $u_{l}-u_{r}=0$ and so $w_{l}-w_{r}=0$.
This means $u_{l}=u_{r}$, $w_{l}=w_{r}$ which we excluded by hypothesis.

We can now rewrite (6.5) as follows

$\lambda[w_{l}-w_{r}]+(\lambda-1)(u_{l}-u_{r})=0.$ (6.6)

So if $u_{l}=u_{r}$ then $\lambda=0$, and the left-hand-side of (6.2) becomes
$0$, independently from $(k,\hat{k})$. So the couple $(u,w)$ is still an
entropy solution and we imply case $\ref{list3})$.

If $w_{l}=w_{r}$ then $\lambda=1$ so we infer case $\ref{list4})$. Notice that
even in this case (6.2) is trivially satisfied.

Suppose now instead $w_{r}\not=w_{l}$ and $u_{r}\not=u_{l}$ and recall that
$u_{l}-u_{r}+w_{l}-w_{r}$ is either positive or negative. Keeping this in mind
and recalling (6.5) we write (6.2) as follows

$\frac{(u_{l}-u_{r})(|w_{r}-\hat{k}|-|w_{l}-\hat{k}|)+(w_{r}-w_{l})(|u_{r}-k|-|u_{l}-k|)}{u_{l}-u_{r}+w_{l}-w_{r}}\geq
0.$ (6.7)

Now, for $(k,\hat{k})\in\mathbb{R}^{2}$ we study the function

$h(k,\hat{k}):=(u_{l}-u_{r})(|w_{r}-\hat{k}|-|w_{l}-\hat{k}|)+(w_{r}-w_{l})(|u_{r}-k|-|u_{l}-k|),$

which is the numerator appearing in (6.7). We can split the plane
$\mathbb{R}^{2}$ in $9$ regions, representing the cases when $k$ is above,
below or between $u_{l},u_{r}$ combined with the cases $\hat{k}$ above, below
or between $w_{l},w_{r}$, see Figure 13. It is easy to prove that $h$ is
constant in $R_{1},R_{3},R_{7}$ and $R_{9}$ as in these parts both $k$ and
$\hat{k}$ cancel out. In particular, in $R_{1}$ it is $h\equiv
2(u_{l}-u_{r})(w_{l}-w_{r})$ which is strictly positive if the jump between
$u_{l}$ and $u_{r}$ has the same monotonicity of the jump between $w_{l}$ and
$w_{l}$ and strictly negative otherwise. In $R_{9}$ instead
$h\equiv-2(u_{l}-u_{r})(w_{l}-w_{r})$, and in $R_{3}$ and $R_{7}$ it holds
$h\equiv 0$. It is easy so see then that in the other regions $h$ is affine,
connecting continuously the constant parts, so we can deduce that it has the
same constant sign in $R_{2}$ and $R_{4}$ and the opposite one in $R_{6}$ and
$R_{8}$. In $R_{5}$ we have a change of sign on the segment $AB$ connecting
$(\min(u_{l},u_{r}),\min(w_{l},w_{r}))$ to
$(\max(u_{l},u_{r}),\max(w_{l},w_{r}))$.

With this in mind we then conclude that if the jump in $w$ has opposite sign
with respect to the jump in $u$ then this violates our condition (6.2). This
is because both the pairs $(u_{r},w_{r})$ and $(u_{l},w_{l})$ are contained in
$\mathcal{L}$ and, see Figure 13, we can find multiple pairs in $\mathcal{L}$
(e.g. $(u_{l},w_{l})$,$(u_{r},w_{r})$ themselves) such that $h$ is either
positive or negative. Since the denominator $u_{l}-u_{r}+w_{l}-w_{r}\not=0$
has fixed sign this means that the inequality (6.2) cannot be true for all
$(k,\hat{k})\in\mathcal{L}$.

So we can only have jumps in $w$ with the same sign of jumps in $u$. If
$u_{l}>u_{r}$ then the denominator $u_{l}-u_{r}+w_{l}-w_{r}$ is positive,
hence it must be $h\geq 0$ for all $(k,\hat{k})\in\mathcal{L}$. But the only
possibility is that the points $A=(u_{r},w_{r}),B=(u_{l},w_{r})$ (see Figure
13) both belong to the lower boundary of $\cal L$. Otherwise at least one of
them has a neighborhood containing points $(k,\hat{k})$ in $\cal L$ where $h$
is negative. So positive jumps occur only on the lower boundary of $\cal L$.
Similarly, if $u_{l}<u_{r}$, then the points must lie on the upper boundary of
$\cal L$. So negative jumps occur only on the upper boundary of $\cal L$. Also
note that, in both cases, $u$ and $w$ must have a jump of the same amplitude.

So we checked that for $u_{l}\not=u_{r}$ and $w_{l}\not=w_{r}$, (6.2) implies
$\lambda=1/2$ (see (6.5)) and either case $\ref{list5})$ or $\ref{list6})$.

As there are no other possible choices of $(u_{l},w_{l}),(u_{r},w_{r})$ we
also infer $\ref{list2})$.

$(\impliedby)$ Now suppose $\ref{list1}),\dots,\ref{list6})$ to hold. If
$u_{l}=u_{r}$ then we are in case $\ref{list3})$ with $\lambda=0$. As already
said, the left-hand-side of (6.2) becomes $0$, hence the inequality is
trivially satisfies for every $(k,\hat{k})\in\mathcal{L}$.

If $w_{l}=w_{r}$ then we are in case $\ref{list4})$ with $\lambda=1$ and again
(6.2) is trivially satisfies.

Suppose instead $\lambda=1/2$, and notice again, because of $\ref{list1})$,
that $u_{l}-u_{r}+w_{l}-w_{r}\not=0$ otherwise we would get $u_{l}=u_{r}$ and
$w_{l}=w_{r}$ which we excluded by hypothesis. This allows us to rewrite (6.2)
as (6.7). Applying the same reasoning as for the previous implication, it can
be verified that for both the cases $\ref{list5})$ and $\ref{list6})$, (6.7)
holds true for every $(k,\hat{k})\in\mathcal{L}$. ∎

$A$$B$$R_{1}$$R_{2}$$R_{3}$$R_{7}$$R_{8}$$R_{9}$$R_{4}$$R_{6}$$0$$+$$+$$-$$-$$0$$-$$-$$+$$+$
Figure 13: The sign of $h$, when the jump between $u_{l}$ and $u_{r}$ has the
same monotonicity as the jump between $w_{l}$ and $w_{r}$. Here
$A=(\min(u_{l},u_{r}),\min(w_{r},w_{l}))$ and
$B=(\max(u_{l},u_{r}),\max(w_{r},w_{l})).$ Note that if the jump between
$u_{l}$ and $u_{r}$ has different monotonicity to the jump between $w_{l}$ and
$w_{r}$, then the picture remains the same but with opposite signs, in
particular, the segment $AB$ is the same. In this case the pairs
$(u_{l},w_{l})$, $(u_{r},w_{r})$ lie on the highlighted points.

Thanks to Proposition 6.1, it is easy to see that the sequence of piece-wise
constant solutions $(u_{n},w_{n})$ of the Cauchy problem with $u_{0}^{(n)}$
and $w_{0}^{(n)}$ as initial data constructed via the Wave-Front Tracking, is
entropy admissible, so it satisfies (6.1). Finally, since we have
$L^{1}(\mathbb{R}_{T})$ convergence, we can conclude that the solution $(u,w)$
as constructed in the proof of Theorem 5.4 also satisfies (6.1). As a
consequence of the next theorem, we then deduce that such a solution is also
the only entropy solution.

######  Theorem 6.2.

Consider the Cauchy problems with initial conditions $u_{0}^{1},w_{0}^{1}$ and
$u_{0}^{2},w_{0}^{2}$ respectively, where $u_{0}^{i},w_{0}^{i}\in
L^{1}(\mathbb{R})\cap BV(\mathbb{R})$. Let us denote by $(u_{1},w_{1})$ and
$(u_{2},w_{2})$ two entropy solutions to the respective problems. Then it
holds that

$\int_{-\infty}^{+\infty}(|u_{1}-u_{2}|(x,t)+|w_{1}-w_{2}|(x,t))~{}dx\leq\int_{-\infty}^{+\infty}(|u_{0}^{1}-u_{0}^{2}|+|w_{0}^{1}-w_{0}^{2}|)~{}dx,$
(6.8)

for almost every $t\in[0,T]$.

###### Proof.

Adapting the standard method by Kruzkov [14], in [20] an analogous result is
proven for the case of delayed-relay hysteresis. The adaptation to our case
with Play hysteresis is not difficult. ∎

## References

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