This document summarizes an academic paper about controlling a hyperbolic Stefan problem with memory using thermostats. The problem models heat transfer in a two-phase system using integrodifferential equations. Three types of thermostats are considered to control the system: an ideal switch, relay switch, and Preisach hysteresis operator. For each thermostat type, the control problem is formulated as a hyperbolic Stefan problem with a nonlinear and nonlocal boundary condition. Existence of solutions is proved for each problem, and uniqueness is also shown for some cases. The document provides mathematical context and definitions to formally define the control problems and state the main results.
MATLAB sessions: Laboratory 6
MAT 275 Laboratory 6
Forced Equations and Resonance
In this laboratory we take a deeper look at second-order nonhomogeneous equations. We will concentrate
on equations with a periodic harmonic forcing term. This will lead to a study of the phenomenon known
as resonance. The equation we consider has the form
d2y
dt2
+ c
dy
dt
+ ω20y = cosωt. (L6.1)
This equation models the movement of a mass-spring system similar to the one described in Laboratory
5. The forcing term on the right-hand side of (L6.1) models a vibration, with amplitude 1 and frequency
ω (in radians per second = 12π rotation per second =
60
2π rotations per minute, or RPM) of the plate
holding the mass-spring system. All physical constants are assumed to be positive.
Let ω1 =
√
ω20 − c2/4. When c < 2ω0 the general solution of (L6.1) is
y(t) = e−
1
2 ct(c1 cos(ω1t) + c2 sin(ω1t)) + C cos (ωt− α) (L6.2)
with
C =
1√
(ω20 − ω2)
2
+ c2ω2
, (L6.3)
α =
⎧
⎨
⎩
arctan
(
cω
ω20−ω2
)
if ω0 > ω
π + arctan
(
cω
ω20−ω2
)
if ω0 < ω
(L6.4)
and c1 and c2 determined by the initial conditions. The first term in (L6.2) represents the complementary
solution, that is, the general solution to the homogeneous equation (independent of ω), while the second
term represents a particular solution of the full ODE.
Note that when c > 0 the first term vanishes for large t due to the decreasing exponential factor.
The solution then settles into a (forced) oscillation with amplitude C given by (L6.3). The objectives of
this laboratory are then to understand
1. the effect of the forcing term on the behavior of the solution for different values of ω, in particular
on the amplitude of the solution.
2. the phenomena of resonance and beats in the absence of friction.
The Amplitude of Forced Oscillations
We assume here that ω0 = 2 and c = 1 are fixed. Initial conditions are set to 0. For each value of ω, the
amplitude C can be obtained numerically by taking half the difference between the highs and the lows
of the solution computed with a MATLAB ODE solver after a sufficiently large time, as follows: (note
that in the M-file below we set ω = 1.4).
1 function LAB06ex1
2 omega0 = 2; c = 1; omega = 1.4;
3 param = [omega0,c,omega];
4 t0 = 0; y0 = 0; v0 = 0; Y0 = [y0;v0]; tf = 50;
5 options = odeset(’AbsTol’,1e-10,’RelTol’,1e-10);
6 [t,Y] = ode45(@f,[t0,tf],Y0,options,param);
7 y = Y(:,1); v = Y(:,2);
8 figure(1)
9 plot(t,y,’b-’); ylabel(’y’); grid on;
c⃝2011 Stefania Tracogna, SoMSS, ASU 1
MATLAB sessions: Laboratory 6
10 t1 = 25; i = find(t>t1);
11 C = (max(Y(i,1))-min(Y(i,1)))/2;
12 disp([’computed amplitude of forced oscillation = ’ num2str(C)]);
13 Ctheory = 1/sqrt((omega0^2-omega^2)^2+(c*omega)^2);
14 disp([’theoretical amplitude = ’ num2str(Ctheory)]);
15 %----------------------------------------------------------------
16 function dYdt = f(t,Y,param)
17 y = Y(1); v = Y(2);
18 omega0 = param(1); c = param(2); omega = param(3);
19 dYdt = [ v ; cos(omega ...
On an Optimal control Problem for Parabolic Equationsijceronline
International Journal of Computational Engineering Research (IJCER) is dedicated to protecting personal information and will make every reasonable effort to handle collected information appropriately. All information collected, as well as related requests, will be handled as carefully and efficiently as possible in accordance with IJCER standards for integrity and objectivity.
MATLAB sessions: Laboratory 6
MAT 275 Laboratory 6
Forced Equations and Resonance
In this laboratory we take a deeper look at second-order nonhomogeneous equations. We will concentrate
on equations with a periodic harmonic forcing term. This will lead to a study of the phenomenon known
as resonance. The equation we consider has the form
d2y
dt2
+ c
dy
dt
+ ω20y = cosωt. (L6.1)
This equation models the movement of a mass-spring system similar to the one described in Laboratory
5. The forcing term on the right-hand side of (L6.1) models a vibration, with amplitude 1 and frequency
ω (in radians per second = 12π rotation per second =
60
2π rotations per minute, or RPM) of the plate
holding the mass-spring system. All physical constants are assumed to be positive.
Let ω1 =
√
ω20 − c2/4. When c < 2ω0 the general solution of (L6.1) is
y(t) = e−
1
2 ct(c1 cos(ω1t) + c2 sin(ω1t)) + C cos (ωt− α) (L6.2)
with
C =
1√
(ω20 − ω2)
2
+ c2ω2
, (L6.3)
α =
⎧
⎨
⎩
arctan
(
cω
ω20−ω2
)
if ω0 > ω
π + arctan
(
cω
ω20−ω2
)
if ω0 < ω
(L6.4)
and c1 and c2 determined by the initial conditions. The first term in (L6.2) represents the complementary
solution, that is, the general solution to the homogeneous equation (independent of ω), while the second
term represents a particular solution of the full ODE.
Note that when c > 0 the first term vanishes for large t due to the decreasing exponential factor.
The solution then settles into a (forced) oscillation with amplitude C given by (L6.3). The objectives of
this laboratory are then to understand
1. the effect of the forcing term on the behavior of the solution for different values of ω, in particular
on the amplitude of the solution.
2. the phenomena of resonance and beats in the absence of friction.
The Amplitude of Forced Oscillations
We assume here that ω0 = 2 and c = 1 are fixed. Initial conditions are set to 0. For each value of ω, the
amplitude C can be obtained numerically by taking half the difference between the highs and the lows
of the solution computed with a MATLAB ODE solver after a sufficiently large time, as follows: (note
that in the M-file below we set ω = 1.4).
1 function LAB06ex1
2 omega0 = 2; c = 1; omega = 1.4;
3 param = [omega0,c,omega];
4 t0 = 0; y0 = 0; v0 = 0; Y0 = [y0;v0]; tf = 50;
5 options = odeset(’AbsTol’,1e-10,’RelTol’,1e-10);
6 [t,Y] = ode45(@f,[t0,tf],Y0,options,param);
7 y = Y(:,1); v = Y(:,2);
8 figure(1)
9 plot(t,y,’b-’); ylabel(’y’); grid on;
c⃝2011 Stefania Tracogna, SoMSS, ASU 1
MATLAB sessions: Laboratory 6
10 t1 = 25; i = find(t>t1);
11 C = (max(Y(i,1))-min(Y(i,1)))/2;
12 disp([’computed amplitude of forced oscillation = ’ num2str(C)]);
13 Ctheory = 1/sqrt((omega0^2-omega^2)^2+(c*omega)^2);
14 disp([’theoretical amplitude = ’ num2str(Ctheory)]);
15 %----------------------------------------------------------------
16 function dYdt = f(t,Y,param)
17 y = Y(1); v = Y(2);
18 omega0 = param(1); c = param(2); omega = param(3);
19 dYdt = [ v ; cos(omega ...
On an Optimal control Problem for Parabolic Equationsijceronline
International Journal of Computational Engineering Research (IJCER) is dedicated to protecting personal information and will make every reasonable effort to handle collected information appropriately. All information collected, as well as related requests, will be handled as carefully and efficiently as possible in accordance with IJCER standards for integrity and objectivity.
Research Inventy : International Journal of Engineering and Scienceresearchinventy
Research Inventy : International Journal of Engineering and Science is published by the group of young academic and industrial researchers with 12 Issues per year. It is an online as well as print version open access journal that provides rapid publication (monthly) of articles in all areas of the subject such as: civil, mechanical, chemical, electronic and computer engineering as well as production and information technology. The Journal welcomes the submission of manuscripts that meet the general criteria of significance and scientific excellence. Papers will be published by rapid process within 20 days after acceptance and peer review process takes only 7 days. All articles published in Research Inventy will be peer-reviewed.
International journal of engineering and mathematical modelling vol2 no3_2015_2IJEMM
Mixed nite element approximation of reaction front propagation model in porous media is presented. The model consists of system of reaction-diffusion equations coupled with the equations of motion under the Darcy law. The existence of solution for the semi-discrete problem is established. The stability of the fully-discrete problem is
analyzed. Optimal error estimates are proved for both semi-discrete and fully-discrete approximate schemes.
The International Journal of Engineering and Science (The IJES)theijes
The International Journal of Engineering & Science is aimed at providing a platform for researchers, engineers, scientists, or educators to publish their original research results, to exchange new ideas, to disseminate information in innovative designs, engineering experiences and technological skills. It is also the Journal's objective to promote engineering and technology education. All papers submitted to the Journal will be blind peer-reviewed. Only original articles will be published.
IJRET : International Journal of Research in Engineering and Technology is an international peer reviewed, online journal published by eSAT Publishing House for the enhancement of research in various disciplines of Engineering and Technology. The aim and scope of the journal is to provide an academic medium and an important reference for the advancement and dissemination of research results that support high-level learning, teaching and research in the fields of Engineering and Technology. We bring together Scientists, Academician, Field Engineers, Scholars and Students of related fields of Engineering and Technology
I am Grey Nolan. Currently associated with matlabassignmentexperts.com as an assignment helper. After completing my master's from the University of British Columbia, I was in search for an opportunity that expands my area of knowledge hence I decided to help students with their Signals and Systems assignments. I have written several assignments till date to help students overcome numerous difficulties they face in Signals and Systems Assignments.
Ch 04 MATLAB Applications in Chemical Engineering_陳奇中教授教學投影片Chyi-Tsong Chen
The slides of Chapter 3 of the book entitled "MATLAB Applications in Chemical Engineering": Numerical Solution of Ordinary Differential Equations. Author: Prof. Chyi-Tsong Chen (陳奇中教授); Center for General Education, National Quemoy University; Kinmen, Taiwan; E-mail: chyitsongchen@gmail.com.
Ebook purchase: https://play.google.com/store/books/details/MATLAB_Applications_in_Chemical_Engineering?id=kpxwEAAAQBAJ&hl=en_US&gl=US
Research Inventy : International Journal of Engineering and Scienceresearchinventy
Research Inventy : International Journal of Engineering and Science is published by the group of young academic and industrial researchers with 12 Issues per year. It is an online as well as print version open access journal that provides rapid publication (monthly) of articles in all areas of the subject such as: civil, mechanical, chemical, electronic and computer engineering as well as production and information technology. The Journal welcomes the submission of manuscripts that meet the general criteria of significance and scientific excellence. Papers will be published by rapid process within 20 days after acceptance and peer review process takes only 7 days. All articles published in Research Inventy will be peer-reviewed.
International journal of engineering and mathematical modelling vol2 no3_2015_2IJEMM
Mixed nite element approximation of reaction front propagation model in porous media is presented. The model consists of system of reaction-diffusion equations coupled with the equations of motion under the Darcy law. The existence of solution for the semi-discrete problem is established. The stability of the fully-discrete problem is
analyzed. Optimal error estimates are proved for both semi-discrete and fully-discrete approximate schemes.
The International Journal of Engineering and Science (The IJES)theijes
The International Journal of Engineering & Science is aimed at providing a platform for researchers, engineers, scientists, or educators to publish their original research results, to exchange new ideas, to disseminate information in innovative designs, engineering experiences and technological skills. It is also the Journal's objective to promote engineering and technology education. All papers submitted to the Journal will be blind peer-reviewed. Only original articles will be published.
IJRET : International Journal of Research in Engineering and Technology is an international peer reviewed, online journal published by eSAT Publishing House for the enhancement of research in various disciplines of Engineering and Technology. The aim and scope of the journal is to provide an academic medium and an important reference for the advancement and dissemination of research results that support high-level learning, teaching and research in the fields of Engineering and Technology. We bring together Scientists, Academician, Field Engineers, Scholars and Students of related fields of Engineering and Technology
I am Grey Nolan. Currently associated with matlabassignmentexperts.com as an assignment helper. After completing my master's from the University of British Columbia, I was in search for an opportunity that expands my area of knowledge hence I decided to help students with their Signals and Systems assignments. I have written several assignments till date to help students overcome numerous difficulties they face in Signals and Systems Assignments.
Ch 04 MATLAB Applications in Chemical Engineering_陳奇中教授教學投影片Chyi-Tsong Chen
The slides of Chapter 3 of the book entitled "MATLAB Applications in Chemical Engineering": Numerical Solution of Ordinary Differential Equations. Author: Prof. Chyi-Tsong Chen (陳奇中教授); Center for General Education, National Quemoy University; Kinmen, Taiwan; E-mail: chyitsongchen@gmail.com.
Ebook purchase: https://play.google.com/store/books/details/MATLAB_Applications_in_Chemical_Engineering?id=kpxwEAAAQBAJ&hl=en_US&gl=US
A Strategic Approach: GenAI in EducationPeter Windle
Artificial Intelligence (AI) technologies such as Generative AI, Image Generators and Large Language Models have had a dramatic impact on teaching, learning and assessment over the past 18 months. The most immediate threat AI posed was to Academic Integrity with Higher Education Institutes (HEIs) focusing their efforts on combating the use of GenAI in assessment. Guidelines were developed for staff and students, policies put in place too. Innovative educators have forged paths in the use of Generative AI for teaching, learning and assessments leading to pockets of transformation springing up across HEIs, often with little or no top-down guidance, support or direction.
This Gasta posits a strategic approach to integrating AI into HEIs to prepare staff, students and the curriculum for an evolving world and workplace. We will highlight the advantages of working with these technologies beyond the realm of teaching, learning and assessment by considering prompt engineering skills, industry impact, curriculum changes, and the need for staff upskilling. In contrast, not engaging strategically with Generative AI poses risks, including falling behind peers, missed opportunities and failing to ensure our graduates remain employable. The rapid evolution of AI technologies necessitates a proactive and strategic approach if we are to remain relevant.
Synthetic Fiber Construction in lab .pptxPavel ( NSTU)
Synthetic fiber production is a fascinating and complex field that blends chemistry, engineering, and environmental science. By understanding these aspects, students can gain a comprehensive view of synthetic fiber production, its impact on society and the environment, and the potential for future innovations. Synthetic fibers play a crucial role in modern society, impacting various aspects of daily life, industry, and the environment. ynthetic fibers are integral to modern life, offering a range of benefits from cost-effectiveness and versatility to innovative applications and performance characteristics. While they pose environmental challenges, ongoing research and development aim to create more sustainable and eco-friendly alternatives. Understanding the importance of synthetic fibers helps in appreciating their role in the economy, industry, and daily life, while also emphasizing the need for sustainable practices and innovation.
How to Make a Field invisible in Odoo 17Celine George
It is possible to hide or invisible some fields in odoo. Commonly using “invisible” attribute in the field definition to invisible the fields. This slide will show how to make a field invisible in odoo 17.
2024.06.01 Introducing a competency framework for languag learning materials ...Sandy Millin
http://sandymillin.wordpress.com/iateflwebinar2024
Published classroom materials form the basis of syllabuses, drive teacher professional development, and have a potentially huge influence on learners, teachers and education systems. All teachers also create their own materials, whether a few sentences on a blackboard, a highly-structured fully-realised online course, or anything in between. Despite this, the knowledge and skills needed to create effective language learning materials are rarely part of teacher training, and are mostly learnt by trial and error.
Knowledge and skills frameworks, generally called competency frameworks, for ELT teachers, trainers and managers have existed for a few years now. However, until I created one for my MA dissertation, there wasn’t one drawing together what we need to know and do to be able to effectively produce language learning materials.
This webinar will introduce you to my framework, highlighting the key competencies I identified from my research. It will also show how anybody involved in language teaching (any language, not just English!), teacher training, managing schools or developing language learning materials can benefit from using the framework.
Instructions for Submissions thorugh G- Classroom.pptxJheel Barad
This presentation provides a briefing on how to upload submissions and documents in Google Classroom. It was prepared as part of an orientation for new Sainik School in-service teacher trainees. As a training officer, my goal is to ensure that you are comfortable and proficient with this essential tool for managing assignments and fostering student engagement.
June 3, 2024 Anti-Semitism Letter Sent to MIT President Kornbluth and MIT Cor...Levi Shapiro
Letter from the Congress of the United States regarding Anti-Semitism sent June 3rd to MIT President Sally Kornbluth, MIT Corp Chair, Mark Gorenberg
Dear Dr. Kornbluth and Mr. Gorenberg,
The US House of Representatives is deeply concerned by ongoing and pervasive acts of antisemitic
harassment and intimidation at the Massachusetts Institute of Technology (MIT). Failing to act decisively to ensure a safe learning environment for all students would be a grave dereliction of your responsibilities as President of MIT and Chair of the MIT Corporation.
This Congress will not stand idly by and allow an environment hostile to Jewish students to persist. The House believes that your institution is in violation of Title VI of the Civil Rights Act, and the inability or
unwillingness to rectify this violation through action requires accountability.
Postsecondary education is a unique opportunity for students to learn and have their ideas and beliefs challenged. However, universities receiving hundreds of millions of federal funds annually have denied
students that opportunity and have been hijacked to become venues for the promotion of terrorism, antisemitic harassment and intimidation, unlawful encampments, and in some cases, assaults and riots.
The House of Representatives will not countenance the use of federal funds to indoctrinate students into hateful, antisemitic, anti-American supporters of terrorism. Investigations into campus antisemitism by the Committee on Education and the Workforce and the Committee on Ways and Means have been expanded into a Congress-wide probe across all relevant jurisdictions to address this national crisis. The undersigned Committees will conduct oversight into the use of federal funds at MIT and its learning environment under authorities granted to each Committee.
• The Committee on Education and the Workforce has been investigating your institution since December 7, 2023. The Committee has broad jurisdiction over postsecondary education, including its compliance with Title VI of the Civil Rights Act, campus safety concerns over disruptions to the learning environment, and the awarding of federal student aid under the Higher Education Act.
• The Committee on Oversight and Accountability is investigating the sources of funding and other support flowing to groups espousing pro-Hamas propaganda and engaged in antisemitic harassment and intimidation of students. The Committee on Oversight and Accountability is the principal oversight committee of the US House of Representatives and has broad authority to investigate “any matter” at “any time” under House Rule X.
• The Committee on Ways and Means has been investigating several universities since November 15, 2023, when the Committee held a hearing entitled From Ivory Towers to Dark Corners: Investigating the Nexus Between Antisemitism, Tax-Exempt Universities, and Terror Financing. The Committee followed the hearing with letters to those institutions on January 10, 202
Acetabularia Information For Class 9 .docxvaibhavrinwa19
Acetabularia acetabulum is a single-celled green alga that in its vegetative state is morphologically differentiated into a basal rhizoid and an axially elongated stalk, which bears whorls of branching hairs. The single diploid nucleus resides in the rhizoid.
Operation “Blue Star” is the only event in the history of Independent India where the state went into war with its own people. Even after about 40 years it is not clear if it was culmination of states anger over people of the region, a political game of power or start of dictatorial chapter in the democratic setup.
The people of Punjab felt alienated from main stream due to denial of their just demands during a long democratic struggle since independence. As it happen all over the word, it led to militant struggle with great loss of lives of military, police and civilian personnel. Killing of Indira Gandhi and massacre of innocent Sikhs in Delhi and other India cities was also associated with this movement.
Palestine last event orientationfvgnh .pptxRaedMohamed3
An EFL lesson about the current events in Palestine. It is intended to be for intermediate students who wish to increase their listening skills through a short lesson in power point.
Macroeconomics- Movie Location
This will be used as part of your Personal Professional Portfolio once graded.
Objective:
Prepare a presentation or a paper using research, basic comparative analysis, data organization and application of economic information. You will make an informed assessment of an economic climate outside of the United States to accomplish an entertainment industry objective.
Francesca Gottschalk - How can education support child empowerment.pptxEduSkills OECD
Francesca Gottschalk from the OECD’s Centre for Educational Research and Innovation presents at the Ask an Expert Webinar: How can education support child empowerment?
Francesca Gottschalk - How can education support child empowerment.pptx
Automatic Control Via Thermostats Of A Hyperbolic Stefan Problem With Memory
1. Automatic control via thermostats of a
hyperbolic Stefan problem with memory
Pierluigi Colli1 Maurizio Grasselli2 J
urgen Sprekels3
November 11, 1996
1991 Mathematics Subject Classi cation. 35R35, 35R70, 45K05, 93C20.
Keywords. Feedback control, Stefan problems, memory kernels, hyperbolic heat conduc-
tion.
1 Dipartimento di Matematica, Universit
a di Torino, Via Carlo Alberto 10, I{10123 Torino, Italy
2 Dipartimento di Matematica, Politecnico di Milano, Via Bonardi 9, I{20133 Milano, Italy
3 Weierstrass Institute for Applied Analysis and Stochastics, Mohrenstrasse 39, D{10117 Berlin, Germany
0
2. Abstract. A hyperbolic Stefan problem based on the linearized Gurtin{Pipkin heat conduction law
is considered. Temperature and free boundary are controlled by a thermostat acting on the boundary.
This feedback control is based on temperature measurements performed by real thermal sensors located
into the domain containing the two{phase system and/or at its boundary. Three di erent types of
thermostats are analyzed: ideal switch, relay switch, and Preisach hysteresis operator. The resulting
models lead to formulate integrodi erential hyperbolic Stefan problems with nonlinear and nonlocal
boundary conditions. In all the cases, existence results are proved. Uniqueness is also shown, unless in
the situation corresponding to the ideal switch.
1. Introduction
Consider a two{phase system which occupies a bounded domain R
N (N 1) at
any time t 2 [0;T] (T 0): Letting QT := (0;T); we denote by # : QT ! R the
relative temperature (rescaled in order # = 0 be the critical temperature at which the
two phases can coexist) and by : QT ! [0;1] the concentration of the more energetic
phase (e.g., water in a water{ice system). Within the framework of the study of memory
e ects in heat conduction phenomena, the following integrodi erential model has been
proposed (cf. [4])
@t('0# + ' # + ) ;k # = f in QT; (1:1)
2 H(#) in QT; (1:2)
where '0 is a positive constant, '; ; k : [0;+1) ! R are smooth relaxation (or
memory) kernels, and indicates the time convolution product on (0;t): Besides,
f : QT ! R is a known function which depends both on the heat supply and on the
past histories of # and up to t = 0 (supposed to be given), while H stands for the
Heaviside graph, that is,
H(s) =
8
:
f0g if s 0;
[0;1] if s = 0;
f1g if s 0:
System (1.1{2) endowed with suitable initial and boundary conditions produces a
hyperbolic Stefan problem provided that k(0) 0 (cf. [4, Section 2]). In the case of
homogeneous Dirichlet boundary conditions, this problem has been already investigated
in some detail, proving both strong and weak well{posedness among other things (see
[3{6]). Here we are interested in analyzing the heat exchange at the boundary ; := @
under the in uence of a thermostat, and to this aim we have to deal with a boundary
condition of the third type. Then, by taking the initial condition
('0#)(;0) = e0 in (1:3)
(the datum e0 expressing an enthalpy density), we are going to play with the following
relation
;k #n = (#; ;#e) on T := ; (0;T); (1:4)
1
3. where #n is the outward normal derivative of # on ;; denotes a positive constant,
#; is the trace of # on the boundary, and #e represents the external temperature
coupled with another term depending on the prescribed past history of # up to t = 0:
As we shall see, #e plays a crucial role in the control problems described below.
Suppose that we are able to measure the temperature # by a real system of thermal
sensors placed in some xed positions inside the body or on its surface. This fact can
be made schematic by assuming that the quantity
M(#)(t) :=
Z
0
#(x;t)!I(x)dx +
Z
;0
#;(y;t)!S(y)d; (1:5)
is known at any time t 2 [0;T]: Here 0 and ;0 ; are the involved sets, with
positive Lebesgue domain and surface measures, respectively. Moreover, the notation
R is used to indicate the mean value, and !I : 0 ! [0;+1); !S : ;0 ! [0;+1) are
weight functions determined by the characteristics of the sensors.
To control the evolution of the free boundary, a thermostat device acts modifying #e
on account of M(#): According to [11] (see also [8, 12]), the behavior of the thermostat
is described by
#e(y;t) = u(t)#A(y;t)+ #B(y;t); (y;t) 2 T; (1:6)
u : [0;T] ! R being a heating (or cooling) device whose dynamics obeys
u0
+ u = W(M(#)) + #C in [0;T]; (1:7)
u(0) = u0: (1:8)
The functions #A; #B : T ! R and #C : [0;T] ! R are given (with a sign property
for #A); is a positive parameter, W models the action of the thermostat, and
u0 2 R: We have to specify the operator W yet. Here, referring to [7{8, 11{14], we
are going to consider three di erent cases W1; W2; W3:
(A) Simple switch
A critical time{dependent value %(t) provides a jump discontinuity so that, for r 2
C0([0;T]) and t 2 [0;T];
W1(r)(t) := H(r(t) ;%(t)) =
8
:
+1 if r(t) %(t);
w(t) if r(t) = %(t);
;1 if r(t) %(t);
(1:9)
where w 2 L1
(0;T) ful lls ;1 w(t) 1 for a.a. t 2 (0;T): Obviously, % 2
C0([0;T]) is xed inside the thermostat. As we will postulate later, the selection of w
is purely random and ;H is nothing but the opposite of the maximal graph resulting
from the sign function.
2
4. (B) Relay switch
In this case there are two thresholds %L; %U 2 C0
([0;T]) with %L %U and such that,
for any r 2 C0
([0;T]); W2(r) changes its value at time tc from ;1 to +1 or vice
versa according to the rule
W2(r)(tc) :=
+1 if r(tc) = %L(tc) and W2(r)(t) = ;1 just before,
;1 if r(tc) = %U (tc) and W2(r)(t) = +1 just before.
(1:10)
The meaning of just before is made precise later (see Section 4), as well as the de nition
of W2(r) in the case when r coincides with one of the two quantities %L or %U on
an open subset of [0;T]:
(C) Hysteresis operator of Preisach type
To introduce W3; we partly follow [13, Chapter IV]. For r 2 C([0;T]) and for any pair
(%1;%2) 2 R2
satisfying %1 %2; we set
H(%1;%2)(r;)(0) :=
8
:
+1 if r(0) %1;
(%1;%2) if %1 r(0) %2;
;1 if r(0) %2
with : (%1;%2) 7! (%1;%2) 2 f;1;+1g being a given Borel measurable function. In
addition, if t 2 (0;T] we let
Tt := f 2 (0;t] : r() = %1 or r() = %2g
and
H(%1;%2)(r;)(t) :=
8
:
H(%1;%2)(r;)(0) if Tt = ;;
+1 if Tt 6= ; and r(max Tt) = %1;
;1 if Tt 6= ; and r(max Tt) = %2:
One can easily check that the functions z = H(%1;%2)(r;) are continuous on the right
in [0;T) and have nite total variation on [0;T]; i.e., z 2 C0
r ([0;T)) BV (0;T): We
are thus led to consider the mapping H(%1;%2)(;) : C0
([0;T]) ! C0
r ([0;T))BV (0;T)
which is called delayed relay operator (compare with (B)). Now, if is a nonnegative
Borel measure on the plane P := f(%1;%2) 2 R2
: %1 %2g; the associated Preisach
operator is speci ed by
W3(r)(t) :=
Z
P
H(%1;%2)(r;)(t)d(%1;%2): (1:11)
The main properties of such transformation will be recalled in Section 5.
Next, let us come to our control problems. They can be roughly formulated saying
that we are looking for a triplet (#;;u) ful lling (1.1{4), (1.6{8) with M prescribed
3
5. by (1.5) and W Wj; j = 1; 2; 3: Problems of this kind have been studied by several
authors (see, e.g., [7, 8, 11, 12] and the references therein). Nevertheless, the present
paper reports the rst attempt to investigate the thermostat control of a hyperbolic
Stefan problem with memory e ects.
It is convenient to put the feedback control problems we have just described in a
more general form (see [8]). Indeed, looking at (1.7{8), it is easy to observe that u is
given by a Volterra operator, namely
u(t) =
Z t
0
e;(t;)= (Wj(M(#))()+ #C())d + u0e;t= (1:12)
for any t 2 [0;T] and for j = 1; 2; 3: On account of (1.12), equation (1.6) can be
written as
#e = F[Wj(M(#))] on T; (1:13)
where, for any r 2 L2
(0;T);
F[r](y;t) :=
Z t
0
E(y;t;)r()d + E0(y;t); (y;t) 2 T; (1:14)
with (in the special case of (1.12))
E(;t;) = e;(t;)= #A(;t);
E0(;t) =
Z t
0
e;(t;)= #C()d + u0e;t=
#A(;t) + #B(;t)
almost everywhere on ;; t and both varying in [0;T]: Consequently, the feedback
control problems reduce to hyperbolic integrodi erential Stefan problems with a non-
linear and nonlocal boundary condition. More precisely, for j = 1; 2; 3 we shall deal
with
Problem (Pj). Find a pair (#;) satisfying (1.1{3) and
;k #n = (#; ;F[Wj(M(#))]) on T: (1:15)
Taking advantage of the xed{point techniques used in [11] (see also [8, 12]), we
prove the existence of a solution to (P1). Uniqueness is not expected in this framework,
due to the random behavior of the model. Regarding (P2), the inductive argument
developed in [11] allows to show existence and uniqueness. In the case of (P3), we
can apply the Schauder xed{point theorem to derive existence of solutions. Besides,
under further restrictions on the measure and for !S 0 in (1.5) (no boundary
measurements) we deduce uniqueness via suitable contracting estimates.
4
6. In order to prove these results, we rst need a careful analysis of the well{posedness
of the Stefan problem (1.1{4). Also, we have to state Problems (Pj ), j = 1; 2; 3; in a
more rigorous way. Let us present the plan of the paper. In Section 2, we give a precise
formulation of (1.1{4) and of the related results, which are shown and fully detailed in
Sections 6{8. In addition, we establish the crucial properties of the operators M and F :
The subsequent Sections 3, 4, 5 are devoted to Problems (P1), (P2), (P3), respectively.
In each of these sections, the feedback control problem is settled and discussed up to
the proof of the related main result.
2. Preliminaries
Here and in the sequel RN
(N 1) is an open, bounded, and connected set
with a smooth boundary ; (for instance, ; of class C
2 ). We put H := L2( ) and
V := H1( ); recalling that V ,! H ,! V0 with dense and compact injections provided
H is identi ed with its dual space H0: The duality pairing between V0 and V and
the scalar product in H are both denoted by h ; i; while ( ; ) stands for the scalar
product in HN
: The norm either in H or in HN
is simply indicated by k k : Also, let
k k
; be the norm in L2(;):
We remind that v; stands for the trace of a function v 2 V on ;; the normal
derivative on ; being represented by vn whenever v is regular enough. Moreover,
prime speci es the derivatives of functions depending only on time, whereas the position
(1 z)( ;t) :=
R t
0 z( ;s)ds holds for any t 2 [0;T] and any z 2 L1(0;T;V0):
Now, we introduce the assumptions on the data (cf. [4, Section 2]). Assume that
' 2 W1;1(0;T); (2:1)
2 W2;1(0;T); (0) 0; (2:2)
k 2 W2;1(0;T); k(0) 0; (2:3)
f 2 W1;1(0;T;H); (2:4)
e0 2 V; (2:5)
#e 2 H1(0;T;L2(;)); (2:6)
#e( ;0) = ';1
0 e0; a.e. on ;: (2:7)
Consequently, accounting for the identity k # = k(0)(1 #) + k0 (1 #); we can
rigorously formulate the hyperbolic Stefan problem (1.1{4).
Problem (SP). Find the pair (#;) such that # 2 L1(0;T;V) W1;1(0;T;H) and
2 L1(0;T;H) satisfy
1 # 2 L1(0;T;H); (2:8)
5
7. #n 2 L2(T ); (2:9)
#; 2 H1(0;T;L2(;)); (2:10)
@t('0# + ' # + ) ;k # = f a.e. in QT ; (2:11)
2 H(#) a.e. in QT ; (2:12)
('0#)(;0) = e0 a.e. in ; (2:13)
;k #n = (#; ;#e) a.e. on T ; (2:14)
where Qt := (0;t) and t := ; (0;t) for t 2 [0;T]:
Remark 2.1. Note that 2 L1(QT ) thanks to (2.12). Owing to (2.8), the normal
trace #n has a a meaning in H;1(0;T;H;1=2(;)) and (2.9) asserts that is even a
square integrable function. By (2.10) it turns out that (2.14) holds almost everywhere
on T :
Existence and uniqueness for (SP) are given by
Theorem 2.1. Let (2.1{7) hold. Then Problem (SP) has a unique solution (#;):
Moreover, there exists a positive constant 1 such that
k#kL1(0;T ;V)W 1;1(0;T ;H) + k1 #kL1(0;T ;H) + k#;kH1(0;T ;L2(;))
+ k#nkL2(T ) 1
n
1 + kfkW 1;1(0;T ;H) + ke0kV + k#ekH1(0;T ;L2(;))
o
; (2:15)
this constant depending only on ; ;; T; '0; k(0); ; and on the norms k'kW 1;1(0;T ) ;
k kW 2;1(0;T ) ; kkkW 2;1(0;T ) :
A further useful result concerns the Lipschitz continuity of the map associating #e
with the solution component #; namely
Theorem 2.2. Let (2.1{5) hold. Consider #i
e; i = 1; 2; ful lling (2.6{7) and denote
by (#i;i) the corresponding solution to Problem (SP). Then there exists a positive
constant 2 such that for any t 2 [0;T] one has
k#1 ;#2kC0([0;t];H) + k1 (#1 ;#2)kC0([0;t];V)
+ k(#1 ;#2);kL2(t) 2 #1
e ;#2
e L2(t) : (2:16)
Moreover, 2 depends only on T; '0; (0); k(0); ; k'kW 1;1(0;T ) ; k kW 2;1(0;T ) ; and
kkkW 2;1(0;T ) :
The last part of this section is devoted to establish some basic properties of the
already introduced operators M; de ned by (1.5), and F; taken as in (1.14) for suitable
functions E(y;t;); E0(y;t) (cf. (2.18) below) not necessarily coinciding with those of
the Introduction. We require that
!I 2 L2( 0;[0;+1)); !S 2 L2(;0;[0;+1)) (2:17)
6
8. and that
E; Et; E 2 L2(; (0; T )2); E0 2 H1(0; T ;L2(;)); (2:18)
noticing that Et; E represent the partial derivatives of E with respect to the time
variables. We also set the compatibility condition (cf. (1.13) and (2.7))
'0E0(; 0) = e0; a.e. on ;: (2:19)
Here are two statements concerning M and F; respectively.
Proposition 2.1. Under the assumption (2.17), M : V ! R is linear and continuous
and can be naturally extended to spaces of functions from (0; T ) to V: In particular,
for all v1; v2 2 V there holds
jM(v1) ;M(v2)j k!IkL2( 0) kv1 ;v2k+ k!SkL2(;0) k(v1 ;v2);k; : (2:20)
Moreover, whenever v 2 L1(0; T ;V)W 1;1(0; T ;H) and v; 2 H1(0; T ;L2(;)); then
M(v) belongs to C0;1=2([0; T ]) and satis es
jM(v)(t);M(v)()j 3
n
kvtkL2(QT ) + k@tv;kL2(T )
o
jt ;j1=2 (2:21)
for any pair (t; ) 2 [0; T ]2; where, for instance, 3 = k!IkL2( 0) + k!SkL2(;0) :
Proof. As (2.20) is a trivial consequence of the de nition of M; it suces to check
(2.21). Letting t ; from (2.20) it turns out that
jM(v)(t) ;M(v)()j = jM(v(; t));M(v(; ))j
3
Z
t
(kvt(; s)k+ k@tv;(; s)k;)ds
and the H
older inequality allows us to infer (2.21).
Proposition 2.2. Under the assumptions (2.18{19), F is a continuous operator from
L1(0; T ) to H1(0; T ;L2(;)) such that
F[r](; 0) = ';1
0 e0; a.e. on ;; 8 r 2 L1(0; T ):
Moreover, one can determine a constant 4; depending only on T; kEkL2(;(0;T)2) ;
kEtkL2(;(0;T)2) ; kE0kH1(0;T;L2(;)) ; on the L2(T ) norm of the (diagonal) trace func-
tion (y; t) 7! E(y; t; t); and on the quantity max0tT kE(; t; )kL2(T ) ; such that
kF[r]kH1(0;T;L2(;)) 4
n
1 + krkL1(0;T)
o
; (2:22)
k(F[r1] ;F[r2])(t)k; 4 kr1 ;r2kL2(0;t) (2:23)
7
9. for any t 2 [0; T ] and for all r; r1; r2 2 L1(0; T ):
Proof. In view of (1.14) and (2.18), it is clear that
@t(F[r])(y; t) = E(y; t; t)r(t)+
Z t
0
Et(y; ; t)r()d + @tE0(y; t)
for a.a. (y; t) 2 T : Hence, observing that
Z T
0
Z t
0
Et(; t; )r()d
2
;
dt
Z T
0
krk2
L2(0;t)
Z t
0
kEt(; t; )k2
; d dt
T krk2
L1(0;T ) kEtk2
L2(;(0;T )2) ;
it is a standard matter to recover (2.22). Regarding (2.23), by the H
older inequality we
have that
k(F[r1] ;F[r2])(t)k;
Z t
0
kE(; t; )k; j(r1 ;r2)()jd
kE(; t; )kL2(t) kr1 ;r2kL2(0;t)
and easily achieve the proof.
3. Problem (P1): simple switch
As we have already mentioned in the Introduction (see (A)), we prescribe that at the
times when r reaches the threshold %; then W1(r) takes some value between ;1 and
+1 in a purely random way. We assume that
% 2 C0([0; T ]) (3:1)
and consider the set{valued convexi cation of the ranges in (1.9) putting
H(s) :=
8
:
f+1g if s 0;
[;1; 1] if s = 0;
f;1g if s 0;
(3:2)
so that ;H : R ! 2R is a maximal monotone graph. Now, let W1 be the multivalued
operator from C0([0; T ]) to L1(0; T ) de ned by
w 2 W1(r) if w(t) 2 H(r(t) ;%(t)) for a.a. t 2 (0; T ): (3:3)
It is straightforward to verify that W1(r) is nonempty for any r 2 C0([0; T ]); since the
existence of a measurable selection w (obviously bounded) is ensured by the continuity
of r and %:
8
10. Hence, Problem (P1) can be precisely formulated as
Problem (P1). Find # 2 L1(0;T;V) W1;1(0;T;H); 2 L1(0;T;H); and z 2
L1(0;T) satisfying (2.8{13) and
;k #n = (#; ;F[z]) a.e. on T ; (3:4)
z 2 W1(M(#)): (3:5)
The result we are going to prove is
Theorem 3.1. Let (2.1{5), (2.17{19), and (3.1{2) hold. Then there exists a solution
to Problem (P1).
Proof. We introduce the set
YT :=
2 H1(0;T;L2(;)) : (;0) = ';1
0 e0; a.e. on ; (3:6)
and the mapping
S1 : YT ! 2YT ; S1() := fF[w]; w 2 W1(M(#()))g; (3:7)
where
(#();()) represents the solution of (SP) with #e = : (3:8)
These positions are plainly justi ed by Theorem 2.1, Propositions 2.1{2, and (3.3).
Moreover, from (2.22) and (3.2) it follows that
kkH1(0;T ;L2(;)) 24 8 2 S1(); 8 2 YT : (3:9)
Thus, setting
UT :=
n
2 YT : kkH1(0;T ;L2(;)) 24
o
; (3:10)
we have that S1 is a multivalued operator from UT to UT : Thanks to (3.4{5), (3.8),
and (2.14), we realize that any xed point of S1 makes the pair in (3.8) solve (P1).
Therefore, to show Theorem 3.1 we just need to nd 2 UT such that 2 S1():
Reasoning as in [8], we use the Glicksberg xed{point theorem (see [9]). This
tool works for set{valued mappings under suitable convexity, compactness, and closure
hypotheses. Let us omit the statement here and check carefully the assumptions in our
frame.
Denote by H1
w
(0;T;L2(;)) the space H1(0;T;L2(;)) endowed with the weak
topology. It is a locally convex topological vector space. Due to the boundedness
property in (3.10), UT is a nonempty, convex, and compact subset of H1
w
(0;T;L2(;)):
In addition, UT is also sequentially compact. For an arbitrary 2 UT we claim that
9
11. S1() is nonempty (because of (3.7) and (3.3)) and convex. Indeed, if 1; 2 2 S1();
then there are w1; w2 2 L1(0; T) such that i = F[wi] and wi 2 W1(M(#()))
for i = 1; 2: On the other hand, by construction W1(M(#())) is convex so that
w1+(1;)w2 2 W1(M(#())) whenever 0 1: As F is an ane transformation,
we conclude that F[w1 + (1 ;)w2] = 1 + (1 ;)2 2 S1() for any 2 [0; 1]:
It remains to deduce that the graph of S1;
G(S1) := f(; ) : 2 UT ; 2 S1()g; (3:11)
is closed in the product space (H1
w
(0; T;L2
(;)))2
: To this aim, take a Moore{Smith
sequence f(a; a)ga2A; A being a directed set, ful lling (a; a) 2 G(S1) for any
a 2 A and
lim
a2Aa = 1; lim
a2Aa = 1
for some pair (1; 1) 2 (H1
(0; T;L2
(;)))2
: Owing to (3.9{10), f(a; a)ga2A is
bounded, hence weakly sequentially compact, in (H1
(0; T;L2
(;)))2
: Therefore, we can
extract a subsequence f(j ; j )gj2N such that
j ! 1; j ! 1 weakly in H1
(0; T;L2
(;)) (3:12)
as j % 1: Remark that 1 and 1 are both in UT : Since j 2 S1(j ); we can
x zj 2 L1(0; T) satisfying j = F[zj ] and zj 2 W1(M(#(j ))) for any j 2 N: By
(3.2{3) we have that kzj kL1(0;T )
1; and consequently a subsequence of fzj g admits
a weak star limit z1; i.e.,
zj ! z1 weakly star in L1(0; T): (3:13)
On account of (3.12{13) and Proposition 2.2, the (strong and weak) continuity of F
implies that 1 = F[z1]: At this point, to obtain (1; 1) 2 G(S1) it suces to
prove that (cf. (3.7{8))
z1 2 W1(M(#(1))): (3:14)
By virtue of (3.10), (2.15) and (2.12) there exist #1 and 1 such that, possibly taking
subsequences,
#(j ) ! #1 weakly star in L1(0; T;V)W1;1(0; T;H); (3:15)
1 #(j ) ! 1 #1 weakly star in L1(0; T;H); (3:16)
#;(j ) ! #1; weakly in H1
(0; T;L2
(;)); (3:17)
#n(j ) ! #1n weakly in L2
(T ); (3:18)
(j ) ! 1 weakly star in L1(QT ) (3:19)
10
12. as j % 1: The convergence (3.15) and the generalized Ascoli theorem entail
#(j) ! #1 strongly in C0([0; T ];H): (3:20)
With the help of (3.12), (3.15{20), and (2.1{3) it is not dicult to infer that (#1; 1)
solves Problem (SP) for #e = 1; so that (by Theorem 2.1 and (3.8)) #1 = #(1):
In fact, from (3.19{20) it results that
ZZ
QT
(j)#(j) !
ZZ
QT
1#1; (3:21)
and then (3.21) and the maximal monotonicity of H (as induced operator in L2(QT ))
ensure that (see, e.g., [2, Lemma 1.3, p. 42]) 1 and #1 ful ll the nonlinear con-
dition (2.12). Now, (3.15), (3.17), and Proposition 2.1 enable us to conclude that
fM(#(j))gj2N is a bounded and equicontinuous (cf. (2.21)) subset of C0([0; T ]); and
thus M(#(j)) converges to M(#(1)) not only weakly star in L1(0; T ) but strongly
in C0([0; T ]): Finally, referring to (3.13), (3.2{3) and arguing as above, one can exploit
the maximal monotonicity of ;H to get (3.14).
Hence, S1 has a closed graph in (H1
w(0; T ;L2(;)))2: The Glicksberg theorem now
applies and yields the existence of at least one solution to (P1) .
4. Problem (P2): relay switch
The response of the thermostat in this case ((B) in the Introduction) needs to be better
speci ed following [11, Section 5]. First of all, consider the two critical functions
%L; %U 2 C0([0; T ]); (4:1)
which are supposed to satisfy
%U (t) ;%L(t) 0 8 t 2 [0; T ]; (4:2)
for some xed bound : Then, in view of (1.10), (1.15) and (1.3{5), we can assume, for
instance, that
M(e0) =
Z
0
e0!I +
Z
;0
e0;!S '0%L(0); (4:3)
and the relay is initially switched on, namely
W2(M(#))(0) = +1: (4:4)
Here the notion of solution for (P2) is made precise .
11
13. Problem (P2). Find # 2 L1(0;T;V) W1;1(0;T;H); 2 L1(0;T;H); z 2
L1(0;T) and a nite sequence fthgm
h=0 of switching times with
0 =: t0 t1 ::: tm = T
satisfying (2.8{13), (4.4),
;k #n = (#; ;F[z]) a.e. on T; (4:5)
z(t) = (;1)h if t 2 [th;th+1[; (4:6)
th+1 is exactly the in mum of the set fTg[Kh+1; (4:7)
where
Kh+1 :=
t 2 (th;T] : M(#)(t) =
%U(t) if h is even,
%L(t) if h is odd
for h = 0;:::;m ;1:
Remark 4.1. Note that Kh+1 may be empty for some h and in this case th+1 = T
and m = h+1: On the other hand, as the di erence M(#);%L is uniformly continuous
in [0;T] (see Proposition 2.1 and (4.1)), the switching times are at most nitely many
owing to (4.2).
In analogy with [11, Theorem 5.1], we have
Theorem 4.1. Let (2.1{5), (2.17{19), (4.1{3) hold. Then there exists a unique solu-
tion to Problem (P2).
Before proving Theorem 4.1, we establish a preliminary result about the number of
switching times. The next assertion improves that of Remark 4.1, taking advantage of
the fact that there is a modulus of continuity for M(#) uniform with respect to any
admissible #:
Lemma 4.1. Under the same hypotheses as in Theorem 4.1, for
w 2 B :=
n
r 2 L1(0;T) : krkL1(0;T) 1
o
let (#(w);(w)) be the unique solution to (SP) with #e = F[w]: Then there exists some
number 0 such that, for any w 2 B and for all times tL; tU 2 [0;T] ful lling
M(#(w))(tL) = %L(tL); M(#(w))(tU) = %U(tU); (4:8)
there holds
jtL ;tUj ;
and consequently the function M(#(w)) commutes at most [T= ] times between the
threshold functions %L and %U ([T= ] denoting the integer part of T= ):
12
14. Proof. Since w lies in B; thanks to (2.22), (2.15), and (2.21) there is a constant 5
such that
jM(#(w))(tL) ;M(#(w))(tU )j 5jtL ;tU j1=2
; (4:9)
where 5 depends only on 1; 3; 4; kfkW 1;1(0;T ;H) ; and ke0kV : Recalling (4.1),
let 1 0 be such that
j%L(t) ;%L()j
2 whenever t; 2 [0; T]; jt ;j 1:
Therefore, from (4.8{9) it may happen that either jtL ;tU j 1 or
5jtL ;tU j1=2
j%L(tL) ;%U (tU )j %U (tU ) ;%L(tU ) ;j%L(tL) ;%L(tU )j
2
because of (4.2). Hence we can choose := min
1; 2
=(25)2
: Concerning the
last part of the statement we also remind (2.13) and (4.3), which imply that, provided
t1 T; there is some 0 2 [t0; t1) with M(#(w))(0) = %L(0):
Proof of Theorem 4.1. We basically reproduce the inductive argument devised in [11,
Proof of Theorem 5.1]. On account of (4.4), we begin picking
w0(t) = +1 8 t 2 [0; T]
and taking the triplet (#(w0); (w0); w0); where (#(w0); (w0)) solves Problem (SP)
when #e = F[w0] (cf. Theorem 2.1). Consider the set
D1 := ft 2 (0; T] : M(#(w0))(t) = %U (t)g:
If D1 ;; then (#(w0); (w0); w0) provides the unique solution to (P2). Otherwise,
t1 := inf D1 is a minimum due to the continuity of M(#(w0)) and Lemma 4.1 yields
t1 : Further, we de ne
w1(t) :=
w0(t) if t 2 [0; t1);
;1 if t 2 [t1; T];
the solution (#(w1); (w1)) of (SP) with #e = F[w1]; and the set
D2 := ft 2 (tL; T] : M(#(w1))(t) = %L(t)g:
Reasoning as above, we conclude that (#(w1); (w1); w1) gives the unique solution to
(P2) unless D2 6= ;: In this alternative situation we introduce t2 := inf D2 and
w2(t) :=
w1(t) if t 2 [0; t2);
+1 if t 2 [t2; T]:
13
15. Similarly, we realize that t2 2 D2 and t2 2 : Then we can start again by considering
(#(w2);(w2);w2): Proceeding by induction, it is clear that there exist m 2 N; a triplet
(#(wm);(wm);wm); and a sequence of switching times fthgm
h=0 such that m T= ;
tm = T; and the triplet represented by # = #(wm); = (wm); z = wm uniquely
satis es (2.8{13) and (4.4{6).
Remark 4.2. As one can easily observe, our analysis covers the degenerate cases when
M(#) just touches one of the thresholds %L; %U or coincides with one of them on a
time interval.
5. Problem (P3): hysteresis operator of Preisach type
The case (C) of the Introduction is characterized by position (1.11) for suitable choices
of the measure and of the two{valued function : The latter has to ful ll
: P ! f;1;1g is Borel measurable. (5:1)
Regarding the former, in order to prove an existence result for Problem (P3) (which is
stated precisely below) we require that
is a nonnegative Borel measure with bounded density; (5:2)
(f%1gR) = (Rf%2g) = 0 8 (%1;%2) 2 R2
: (5:3)
Within this framework, the operator W3 enjoys two important properties.
Proposition 5.1. Under the assumptions (5.1{3), there hold
kW3(r)kL1(0;T) (P) +1 8 r 2 C0
([0;T]); (5:4)
W3 is strongly continuous from C0
([0;T]) to C0
([0;T]): (5:5)
Moreover, denoting by ` the bidimensional Lebesgue measure, if
(A) `(A) for all Lebesgue measurable sets A P; (5:6)
for some constant ; then there exists a constant 6; depending only on (P) and
; such that, for all r1; r2 2 C0
([0;T]);
j(W3(r1) ;W3(r2))(t)j 6 kr1 ;r2kC0([0;t]) 8 t 2 [0;T]: (5:7)
Proof. As H(%1;%2)(r;) 1 a.e. in (0;T) (see the Introduction, above (1.11)),
(5.4) is a straightforward consequence of (5.2). Concerning (5.5), we just notice that it
follows from (5.3), referring to [14, Section IV.3, Theorem 3.2] for details. The stronger
14
16. Lipschitz continuity (5.7) is ensured by assumption (5.6) (which implies (5.3), of course)
thanks to, e.g., [13, formula (1.14) and Proposition 8]. More general conditions yielding
(5.7) are examined in [14, Section IV.3].
After these preliminaries, let us come to the formulation of (P3).
Problem (P3). Find # 2 L1(0;T;V)W1;1(0;T;H) and 2 L1(0;T;H) satisfying
(2.9{13) and
;k #n = (#; ;F[W3(M(#))]) a.e. on T : (5:8)
By exploiting just (5.5) (and not (5.7)), we are able to show existence for (P3).
More precisely, we have
Theorem 5.1. Let (2.1{5), (2.17{19), (5.1{3) hold. Then there exists a solution to
Problem (P3).
Proof. Let us rst put Problem (P3) in a convenient xed{point setting by introducing
the operator
S3 : C0
([0;T]) ! C0
([0;T]); S3(r) := M(#(r)); (5:9)
where
(#(r);(r)) is the unique solution to (SP) with #e = F[W3(r)]: (5:10)
On account of Propositions 2.1{2 and Theorem 2.1, one can readily check that S3 is
well de ned. Besides, observe that solving (P3) is equivalent to nding a xed point for
S3: By means of (5.4), (2.22), (5.10), (2.15), and (2.21) we deduce that S3(C0
([0;T]))
C0;1=2
([0;T]) and we are able to nd a constant 7; depending on (P) and on the
same quantities as 5 does (cf. (4.9)), such that
kS3(r)kC0;1=2([0;T ]) 7 8 r 2 C0
([0;T]): (5:11)
Hence, due to the Ascoli theorem, S3 maps C0
([0;T]) (which is nonempty, closed,
and convex) into a relatively compact subset of C0
([0;T]): Therefore, if we prove the
continuity of S3; then we achieve the proof by the Schauder theorem. To this aim,
let frj gj2N be a sequence converging to some r in C0
([0;T]): Then (5.5), (2.23),
(5.10), (2.16), (2.20), and (5.9) allow us to infer, step by step, that W3(rj ) ! W3(r) in
C0
([0;T]); F[W3(rj )] ! F[W3(r)] in C0
([0;T];L2
(;)); #(rj ) ! #(r) in C0
([0;T];H)
and #;(rj ) ! #;(r) in L2
(0;T;L2
(;)); and nally
S3(rj ) ! S3(r) in L2
(0;T) (5:12)
as j % 1; all convergences being strong. On the other hand, in view of (5.11), by
compactness there are a subsequence fjhgh2N and an element w such that
S3 (rjh
) ! w in C0
([0;T]) (5:13)
15
17. as h % 1: Combining (5.12) and (5.13), the uniqueness of the limit entails w = S3(r)
and
S3(rj ) ! S3(r) in C0([0;T])
for the whole sequence, that is the desired conclusion.
Remark 5.1. In the study of Problem (P1) we could not argue this way since the
operator W1 was set{valued and, further, without strong continuity properties.
Under additional assumptions on and on the location of the thermostat sensors,
we can also prove uniqueness.
Theorem 5.2. Let (2.1{5), (2.17{19), (5.1{3), (5.6), and
!S 0 (5:14)
hold. Then Problem (P3) admits a unique solution.
Proof. By contradiction assume that there are two pairs (#i;i); i = 1;2; solving (P3).
By comparing (5.8) and (2.14), we put #i
e = F[W3(M(#i))]); i = 1;2; and use (2.23),
(5.7), and (2.20) to obtain
(#1
e ;#2
e)(t) 2
;
Z t
0
j4(W3(M(#1)) ;W3(M(#2)))(s)j2
ds
(46)2
Z t
0
kM(#1) ;M(#2)k2
C0([0;s]) ds
46 k!I kL2( 0)
2Z t
0
k#1 ;#2k2
C0([0;s];H) ds
for any t 2 [0;T]: Now, we can invoke (2.16) and get
(#1
e ;#2
e)(t) 2
; 8
Z t
0
(#1
e ;#2
e)(s) 2
; ds 8 t 2 [0;T];
with, for instance, 8 =
;
246 k!I kL2( 0)
2
T: Then the Gronwall lemma enables
us to establish that #1
e = #2
e a.e. in T ; whence the thesis is an outcome of Theo-
rem 2.1.
Remark 5.2. The uniqueness theorem works in a restricted framework essentially
because of (5.14). We do not know whether or not this restriction can be removed,
having at our disposal only an inequality like (5.7) for the mapping W3: In fact, the
reader may check that if (5.7) was replaced by
j(W3(r1) ;W3(r2))(t)j 6 j(r1 ;r2)(t)j; (5:15)
16
18. or more generally by
k(W3(r1) ;W3(r2))kL2(0;t) 6 j(r1 ;r2)jL2(0;t) ; (5:16)
then the above argument could be suitably modi ed in order to apply to the stan-
dard situation where !S 2 L2
(;0;[0;+1)): On the other hand, a Preisach hysteresis
operator is not expected to enjoy properties like (5.15) or (5.16).
6. Auxiliary parabolic problems
The last part of the paper is concerned with the proof of Theorems 2.1 and 2.2. To
sum up, we have to show the well{posedness of Problem (SP) and, in particular, to
derive estimates (2.15) and (2.16). Existence of a solution is proved passing through
a parabolic regularization of (SP). The aim of this section is that of preparing some
technical results for two related problems, one linear and the other nonlinear.
Therefore, we x a parameter 0 (subject to tend to 0 elsewhere) and, letting
the data '; F; #0; #e ful ll
' 2 H1
(0;T); (6:1)
F 2 H1
(0;T;H); F(;0) 2 V; (6:2)
#0 2 V; #0 2 V; (6:3)
#e 2 W2;1
(0;T;L2
(;)); (6:4)
;#0n = (#0; ;#e(;0)) a.e. on ;; (6:5)
we consider
Problem (LP). Find # 2 C1
([0;T];V)H2
(0;T;H) satisfying
# 2 H1
(0;T;H); (6:6)
#n 2 C1
([0;T];L2
(;)); (6:7)
@t('0# + ' #) ;# ;k # = F a.e. in QT ; (6:8)
#(;0) = #0 a.e. in ; (6:9)
;#n ;k #n = (#; ;#e) a.e. on T : (6:10)
Regarding this linear problem, we can state
Theorem 6.1. Under the assumptions (2.3) and (6.1{5), Problem (LP) has a unique
solution.
17
19. Proof. Taking (6.10) into account, let us rewrite equation (6.8) as
'0h#t;vi+ (r#;rv) +
Z
;
#;v; + hR(#);vi
= hF;vi+
Z
;
#ev; 8 v 2 V; a.e. in(0;T); (6:11)
where the mapping R is speci ed by
hR();vi := h@t(' );vi+ (k r;rv)
for any 2 L2(0;T;V): Recalling the well{known formulas
a b = a(0)(1 b) + at 1 b; @t(a b) = a(0)b + at b; (6:12)
which hold whenever they make sense, by (6.1) and (2.3) one easily veri es that R is
linear and continuous from C0([0;T];V)H1(0;T;H) to L2(0;T;H)+W1;1(0;T;V0):
Then, an application of [1, Teorema 6.1] allows us to deduce that there exists one and
only one function # 2 C0([0;T];V) H1(0;T;H) solving the Cauchy problem (6.11),
(6.9). From (6.11) we plainly recover (6.8) in the sense of distributions, and so the
condition
# + k # 2 L2(0;T;H) (6:13)
follows. Thanks to the classical theory of Volterra integral equations, it turns out that
# 2 L2(0;T;H): Thus, (6.13) ensures the validity of (6.8) and of the equality for the
traces, that is,
;#n ;k #n = (#; ;#e) in L2(0;T;H;1=2(;)): (6:14)
Moreover, owing to (6.4) and to the regularity of #; (6.14) implies
#n 2 C0([0;T];L2(;)) (6:15)
and consequently (6.10) is ful lled.
It only remains to infer the further smoothness of #: Observe that u := @t#
formally satis es (see (6.12))
'0hut;vi+ (ru;rv) +
Z
;
u;v; + hR(u);vi = hFt ;'0#0;vi
;k(r#0;rv) +
Z
;
(@t#e)v; 8 v 2 V; a.e. in (0;T) (6:16)
in addition to
u(;0) = ';1
0 (F(;0) ;'(0)#0 + #0) a.e. in : (6:17)
18
20. We note that (6.17) has been obtained by reading (6.8) at the initial time and using
(6.9). Due to (6.1{4), the abstract result in [1, Teorema 6.1] applies to (6.16{17) as well.
Therefore, this initial value problem has one and only one solution, say e
u; belonging to
C0([0;T];V)H1(0;T;H): Besides, arguing as before (cf. (6.13{15)) we achieve that
e
u 2 L2(0;T;H) and e
un 2 C0([0;T];L2(;)) (6:18)
(here (6.5) plays a role). At this point, it is not dicult to check that # #0 + 1 e
u
whence # 2 C1([0;T];V)H2(0;T;H) and e
u u @t#: Since (6.18) and (6.3) entail
(6.6{7), the proof is completed.
Next, we show a similar result for a nonlinear version of (LP). More precisely, let
G be a (possibly nonlinear and nonlocal) operator such that
G : H1(0;t;H) ! H1(0;t;H) 8 t 2 (0;T]; (6:19)
G[](;0) 2 V whenever 2 H1(0;T;H) and (;0) 2 V: (6:20)
A constant 9 is supposed to exist in order that
kG[]kL2(0;T;H) 9; (6:21)
kG[1] ;G[2]kL2(0;t;H) 9 k1 ;2kL2(0;t;H) (6:22)
for any t 2 (0;T] and for all ; 1; 2 2 H1(0;T;H): In this setting we introduce
Problem (NP). Find # 2 C1([0;T];V) H2(0;T;H) satisfying (6.6{7), (6.9{10),
and
@t('0# + ' #) ;# ;k # = G[#] a.e. in QT : (6:23)
We are still able to prove
Theorem 6.2. Let (2.3), (6.1), and (6.3{5) hold. Moreover, let G be as in (6.19{22).
Then there exists a unique solution to Problem (NP).
Proof. Once more we exploit a xed{point technique. Taking 2 H1(0;T;H) with
(;0) 2 V; due to (6.19{20) and Theorem 6.1 we can consider the function
# = #() 2 C1([0;T];V)H2(0;T;H) solving (LP) for F = G[]: (6:24)
Multiply the corresponding equation (6.8) by # and integrate by parts in space and
time over Qt; t 2 (0;T]: With the help of (6.9{10) and (6.12) one easily obtains
'0
2 k#(;t)k2 +
ZZ
Qt
jr#j2 + k(0)
2 k(1 r#)(;t)k2
+
ZZ
t
j#;j2 = '0
2 k#0k2 +
4
P
j=1
Ij(t); (6:25)
19
21. where
I1(t) := ;
ZZ
Qt
('(0)# + '0
#)#; I2(t) := ;
Z t
0
(r(k0
1 #)(;s);r#(;s))ds
I3(t) :=
ZZ
t
#e#;; I4(t) :=
ZZ
Qt
G[]#:
To estimate I1(t) we have recourse to the Young inequality for the convolution product,
namely
ka bkLr (0;T ;X) kakLp(0;T ) kbkLq(0;T ;X) 8 a 2 Lp
(0;T); b 2 Lq
(0;T;X); (6:26)
letting X denote a real Banach space and 1 p; q; r 1 ful ll 1=r = (1=p)+(1=q);1:
It is straightforward to get
jI1(t)j
j'(0)j+ k'0
kL1(0;T )
Z t
0
k#(;s)k2ds: (6:27)
By (6.12) it is a standard matter to verify that
I2(t) = ;((k0
1 r#)(;t);(1r#)(;t))
+
Z t
0
;
k0
(0)k(1 r#)(;s)k2 + ((k00
1 r#)(;s);(1r#)(;s))
ds;
from which we derive
jI2(t)j k(0)
4 k(1 r#)(;t)k2 + 10
Z t
0
k(1 r#)(;s)k2ds; (6:28)
where 10 is a positive constant only depending on k(0); T; and kkkW 2;1(0;T ) : Note
that here (6.26) has been used twice, the rst time for r = 1 and p = q = 2:
Concerning I3(t) and I4(t); we have that
jI3(t)j+ jI4(t)j 2
ZZ
t
;
j#;j2 + j#ej2
+ 1
2
Z t
0
;
kG[](;s)k2 + k#(;s)k2
ds: (6:29)
Combining (6.27{29) with (6.25), then applying the Gronwall lemma, one nds a con-
stant 11 (whose dependences are clear) such that
k#k2
C0([0;t];H) + kr#k2
L2(0;t;HN ) + k1 r#k2
C0([0;t];HN ) + k#;k2
L2(t)
11
n
k#0k2 + k#ek2
L2(t) + kG[]k2
L2(0;t;H)
o
; (6:30)
this inequality obviously holding for any t 2 [0;T]:
20
22. Besides estimate (6.30), we need a higher order estimate which is obtained by
multiplying equation (6.23) by ;# (this is admissible owing to (6.6)) and integrating
over (0;t); with t 2 (0;T]: In this case, (6.9{10) and the Green formula help us
to infer that
'0
2 kr#(;t)k2 +
Z t
0
k#(;s)k2ds + k(0)
2 k(1 #)(;t)k2
= '0
2 kr#0k2 +
8
P
j=5
Ij (t); (6:31)
where
I5(t) :=
ZZ
t
('0@t#; + '(0)#; + '0
#;)#n;
I6(t) := ;
Z t
0
(r('(0)# + '0
#)(;s);r#(;s))ds;
I7(t) := ;
Z t
0
h(k0
1 #)(;s)#(;s)ids; I8(t) := ;
ZZ
Qt
G[]#:
As @t#; = ;@t#n ;k(0)#n ;k0
#n + @t#e a.e. on T because of (6.10) and (6.7),
playing on I5(t) with (6.26) and the elementary Young inequality, it is not dicult to
get
I5(t) ;'0
2 k#n(;t)k2
; + '0
2 k#0nk2
; ; k(0)'0
2 k#nk2
L2(t)
+12
Z t
0
k#nk2
L2(s) ds + k#ek2
L2(t) + k#;k2
L2(t)
(6:32)
for some constant 12 which depends exactly on k(0); '0; ; kk0
kL2(0;T ) ; j'(0)j; and
k'0
kL1(0;T ) : On the other hand, arguing as in the deduction of (6.27{28), we are led to
jI6(t)j (j'(0)j+ k'0
kL1(0;T ))
Z t
0
kr#(;s)k2ds; (6:33)
jI7(t)j k(0)
4 k(1 #)(;t)k2 + 10
Z t
0
k(1 #)(;s)k2ds: (6:34)
It remains to point out that
jI8(t)j
2
Z t
0
k#(;s)k2
ds + 1
2
Z t
0
kG[](;s)k2
ds: (6:35)
Now, we estimate the right hand side of (6.31) with the aid of (6.32{35) and also of
(6.30). Moving the negative terms on the left hand side and invoking the Gronwall
21
23. lemma, we realize that
kr#k2
C0([0;t];HN ) + k#k2
L2(0;t;H) + k1 #k2
C0([0;t];H)
+ k#nk2
C0([0;t];L2(;)) + k#nk2
L2(t)
13
n
k#0k2
V + k#0nk2
; + k#ek2
H1(0;t;L2(;)) + (1 + 1=)kG[]k2
L2(0;t;H)
o
(6:36)
for any t 2 [0;T]; the constant 13 being independent of : Moreover, on account of
(6.30), (6.36), (2.3), and (6.24), a comparison in (6.8) yields an analogous bound for
k#tk2
L2(0;t;H):
Therefore, thanks to (6.21) there is a constant 14; depending only on 9; ; ;; T;
'0; k(0); ; k'kW1;1(0;T) ; kkkW2;1(0;T) ; k#0k2
V ; k#0k2
H ; k#ek2
H1(0;t;L2(;)) ; and (we
remind that is xed in this section) such that (cf. (6.24))
k#()kH1(0;T;H) 14 for all 2 H1(0;T;H) satisfying (;0) 2 V: (6:37)
Then, if we endow the set
XT :=
n
2 H1(0;T;H) : (;0) = #0 a.e. in ; kkH1(0;T;H) 14
o
with the distance function
dXT
(1;2) := k1 ;2kC0([0;T];H) ; 1; 2 2 XT;
it turns out that the operator N : z 7! #(z) acts from XT into itself (by virtue of
(6.9) and (6.37)) and, due to the weak lower semicontinuity of the norm kkH1(0;T;H) ;
XT is a complete metric space. Thus, to prove the theorem it is sucient to show that
either N or a suitable power of it is a contracting mapping. Letting 1; 2 2 XT and
reasoning as for (6.30), we can easily conclude that
kN(1) ;N(2)k2
C0([0;t];H) 11 kG[1] ;G[2]k2
L2(0;t;H) 8 t 2 [0;T]:
Hence, from (6.22) it follows that
jdXT
(N(1);N(2))j2
15
Z t
0
jdXs
(1;2)j2
ds 15 jdXt
(1;2)j2
t (6:38)
for any t 2 [0;T] and for 15 = 11(9)2; the de nition of dXt
being obvious. As it
is known, inequalities like (6.38) allow you to determine some m 2 N such that Nm is
a contraction from XT into itself. Hence an application of the generalized Contracting
Mapping Principle ends the matter.
22
24. 7. Proof of Theorem 2.1
Since the main aim here is to recover the existence of one solution to Problem (SP), we
go directly to implement our approximation procedure in terms of the parameter 0:
Then we will derive estimates independent of and pass to the limit as 0:
Henceforth, we let (2.1{7) hold. For 0 1 consider some regularizing sequences
f'g; ffg; f#
0g; f#
eg ful lling
' 2 H1(0;T) 8 2 (0;1]; (7:1)
' ! ' strongly in W1;1(0;T) as 0; (7:2)
f 2 H1(0;T;H); f(;0) 2 V 8 2 (0;1]; (7:3)
f ! f strongly in W1;1(0;T;H) as 0; (7:4)
e
0 ! e0 weakly in V as 0; (7:5)
#
e 2 W2;1(0;T;L2(;)); #
e(;0) = ';1
0 e
0; 8 2 (0;1]; (7:6)
#
e ! #e weakly in H1(0;T;L2(;)) as 0: (7:7)
Moreover, for any 2 (0;1] we introduce the solution #
0 2 V of the elliptic variational
equality
h#
0 ;';1
0 e
0;vi+ (r#
0;rv) +
Z
;
(#
0; ;#
e(;0))v = 0 8 v 2 V: (7:8)
One readily sees that #
0 solves the boundary value problem
#
0 ;#
0 = ';1
0 e
0 a.e. in ; (7:9)
;#
0n = (#
0; ;#
e(;0)) a.e. on ;: (7:10)
In addition, choosing the test function v =
;
#
0 ;';1
0 e
0
= in (7.8), thanks to (7.5{6)
it is straightforward to infer that
2
#
0 ;';1
0 e
0
2
+ kr#
0k2
+ 2
k#
0; ;#
e(;0)k2
;
r
;
';1
0 e
0
2
16 f1 + ke0kVg
for any 2 (0;1] and for some constant 16 depending only on '0: Consequently, the
convergences
#
0 ! ';1
0 e0 weakly in V and strongly in H as 0 (7:11)
and the boundedness
k#
0nk2
; ( =2)16 f1 + ke0kVg 8 2 (0;1] (7:12)
23
25. are entailed by (7.5) and (7.10) (we remind that V is compactly embedded into H):
Next, let us approximate the Heaviside graph H by (cf. [5, Appendix and especially
Remark 9.1])
H(s) :=
8
:
0 if 0;
;
3s2 ;2s3
=3 if 0 s ;
1 if :
(7:13)
Note that, for any 2 (0;1];
H : R ! [0;1] is a maximal monotone graph in R2; (7:14)
H 2 C1(R) W1;1
(R): (7:15)
Then our regularized version of (SP) reads
Problem (SP). Find # 2 C1([0;T];V)H2(0;T;H) and 2 L1
(QT ) satisfying
# 2 H1(0;T;H); (7:16)
#n 2 C1([0;T];L2(;)); (7:17)
2 C0([0;T];V)W1;1
(0;T;H); (7:18)
@t('0# + ' # + ) ;# ;k # = f a.e. in QT ; (7:19)
= H(#) a.e. in QT ; (7:20)
#(;0) = #
0 a.e. in ; (7:21)
;#n ;k #n = (#; ;#
e) a.e. on T : (7:22)
Referring to the previous section, now we let
G[] := f ;@t( H()) = f ; (0)H() ; 0
H() (7:23)
for 2 H1(0;T;H); and observe that this operator obeys (6.19{22) by virtue of (7.3),
(7.15), and (2.2) (obviously, the constant in (6.22) blows up as 1= does). Thus, on
account of (7.20), Problem (SP) is nothing but a particular case of Problem (NP)
with G given by (7.23). In view of (7.1), (7.5), and (7.9{10), one checks that Theorem
6.2 applies and therefore Problem (SP) admits a unique solution (#;):
The second step of the proof consists in proving some a priori estimates on #
which allow us to pass to the limit in (SP) as 0 and to get a solution to the limit
problem (SP). Let us start by noting that (6.30) and (7.23) directly yield
k#k2
C0([0;T];H) + kr#k2
L2(0;T;HN) + k1 r#k2
C0([0;T];HN) + k#;k2
L2(T)
17
n
1 + kfk2
L2(0;T;H) + k#
0k2 + k#
ek2
L2(T)
o
; (7:24)
24
26. where the constant 17 just depends on 11; T; and k kW1;1(0;T) : Then, multiplying
equation (7.19) by ;# and integrating over Qt (with t 2 (0;T]); we obtain the
identity corresponding to (6.31). However, here we deal with I8(t) in a di erent way.
Indeed, (7.23) and standard integrations by parts enable us to deduce
I8(t) = ;h(f ; 0
H(#))(;t);(1#)(;t)i
+
Z t
0
h(@tf ; 0
(0)H(#) ; 00
H(#))(;s);(1#)(;s)ids
;
ZZ
Qt
(0)H0
(#)jr#j2
+
ZZ
t
(0)(H(#));#n: (7:25)
As H0
0 in R; recalling (7.14), (2.2), (6.26), and the elementary Young inequality,
from (7.25) it is not dicult to nd a constant 18; independent of ; such that (see
(6.31{34) for our choice of coecients)
I8(t) k(0)
8 k(1 #)(;t)k2
+ k(0)'0
4 k#nk2
L2(t)
+ 18
1 + kf(;t)k2 +
Z t
0
(1 + k@tf(;s)k)k(1#)(;s)kds
: (7:26)
Using (7.26) in place of (6.35), we argue as for the derivation of (6.36) even though here
we need a generalized version of the Gronwall lemma (like, e.g., the one stated in [1]).
This procedure leads to
kr#k2
L1(0;T;HN ) + k#k2
L2(0;T;H) + k1 #k2
L1(0;T;H)
+ k#nk2
L1(0;T;L2(;)) + k#nk2
L2(T )
19
n
1 + kfk2
W1;1(0;T;H) + k#
0k2
V + k#
0nk2
; + k#
ek2
H1(0;T;L2(;))
o
; (7:27)
with the constant 19 having the same dependences as 1 does. At this point, the
estimates (7.24) and (7.27) entail an analogous bound for k@t#k2
L2(0;T;H) via a com-
parison in (7.19). Then, owing to (7.4), (7.11{12), and (7.7) there exists a constant 20
such that
k#kL1(0;T;V)H1(0;T;H) +
p
k#kL1(0;T;H) + k1 #kL1(0;T;H)
+
p
k#nkL1(0;T;L2(;)) + k#nkL2(T )
20
n
1 + kfkW1;1(0;T;H) + ke0kV + k#ekH1(0;T;L2(;))
o
; (7:28)
where 20 relies on the same quantities as 1 does.
25
27. Since (7.28) and (7.20) hold for any 2 (0;1]; we can pass to the limit along a
subsequence and thus infer the existence of # and such that
# ! # weakly star in L1
(0;T;V) and weakly in H1
(0;T;H); (7:29)
# ! 0 strongly in L2
(QT ); (7:30)
1 # ! 1 # weakly star in L1
(0;T;H); (7:31)
#n ! 0 strongly in L1
(0;T;L2
(;)); (7:32)
#n ! #n weakly in L2
(T ); (7:33)
! weakly star in L1
(QT ) (7:34)
as 0: In particular, (7.29) implies that (cf. (3.20))
# ! # strongly in C0
([0;T];H) as 0: (7:35)
Convergences (7.29{34) combined with (7.2), (7.4), (7.7), and (7.11) allow us to take
the limit in (7.19{22) and recover (2.11{14). The details are either trivial or developed
in [4, Appendix]. Therefore, the pair (#;) solves Problem (SP). Moreover, it satis es
(7.28) because of the weak star lower semicontinuity of norms, whence (2.15) follows by
additionally comparing the terms of (2.11) and (2.14) (refer to (6.12) too).
Finally, in order to complete the proof of Theorem 2.1, we just remark that unique-
ness is a consequence of (2.16) coupled with a proper reasoning on equation (2.11) to
conclude also for the other variable :
8. Proof of Theorem 2.2
We remind that (#1;1) and (#2;2) denote the solutions of (SP) corresponding to
the respective data #1
e and #2
e; which both ful ll (2.6{7). Setting := #1 ; #2 and
X := 1 ;2; from (2.11) one can easily derive
(0)X + 0
X = L() a.e. in QT ; (8:1)
where
L() := ;@t('0 + ' ) + k : (8:2)
It is known that (cf., e.g., [10, Chapter 2, Section 3]) (8.1) can be equivalently rewritten
as (0)X = L() ; L(); that is,
@t('0 + ' ) ;k + (0)X = ; L() a.e. in QT ; (8:3)
the function 2 W1;1(0;T) being the the unique solution to the integral equation
(0) + 0
= 0
in [0;T]:
26
28. We point out that
k kW 1;1(0;T ) 21 (8:4)
for some constant 21 which only depends on (0); T; and k 0
kW 1;1(0;T ) :
Next, multiply equation (8.3) by and integrate in space and time over Qt; for
t 2 (0;T]: On account of (2.13{14) (observe in particular that (;0) = 0 a.e. in );
by the Green formula and (6.12) it is straightforward to check that
'0
2 k(;t)k2 + k(0)
2 k(1 r)(;t)k2 + k;k2
L2(t)
+ (0)
ZZ
Qt
X =
11
P
j=9
Ij (t); (8:5)
where
I9(t) :=
ZZ
Qt
@t( ('0 + ' ) ;' );
I10(t) :=
Z t
0
(r(( k ;k)0
1 )(;s);r(;s))ds;
I11(t) :=
ZZ
t
(e + (; ;e));;
and e := #1
e ;#2
e: In order to estimate I9(t); for
@t( ('0 + ' ) ;' ) = ('0 (0) ;'(0)) + ('0 0
+ 0
' ;'0
) ;
one can exploit (8.4), (2.1), and (6.26) to determine a constant 22; whose dependences
are obvious, such that
jI9(t)j 22
Z t
0
k(;s)k2ds: (8:6)
Observing that k 2 W2;1(0;T) and comparing I10(t) with I2(t); we have that
(cf. (6.28))
jI10(t)j k(0)
4 k(1 r)(;t)k2 + 23
Z t
0
k(1 r)(;s)k2ds; (8:7)
the constant 23 being dependent on k(0); T; k kW 1;1(0;T ) ; and kk0
kW 1;1(0;T ) : On
the other hand, (6.26) and standard inequalities lead to
jI11(t)j 2 k;k2
L2(t) + kek2
L2(t)
+ 2 k k2
L2(0;T )
Z t
0
k;k2
L2(s) + kek2
L2(s)
ds: (8:8)
27
29. Now, we collect (8.5{8) and note that
(0)X 0 a.e. in QT
due to (2.2), (2.12), and to the monotonicity of H: Thus, there is a constant 24;
having the same dependences as 2 does, such that
'0
2 k( ;t)k
2 + k(0)
4 k(1 r)( ;t)k
2 + 2 k;k
2
L2(t)
24
kek
2
L2(t) +
Z t
0
k( ;s)k
2 + k(1 r)( ;s)k
2 + k;k
2
L2(s)
ds
for any t 2 [0;T]: Finally, we just notice that it is not dicult to recover (2.6) taking
advantage of the Gronwall lemma.
References
[1] C. Baiocchi, Sulle equazioni di erenziali astratte lineari del primo e del secondo ordine negli
spazi di Hilbert, Ann. Mat. Pura Appl. (4) 76 (1967) 233{304.
[2] V. Barbu, Nonlinear semigroups and di erential equations in Banach spaces (Noordho ,
Leyden, 1976).
[3] P. Colli, G. Gilardi, and M. Grasselli, Weak solution to hyperbolic Stefan problems with
memory, NoDEA Nonlinear Di erential Equations Appl. (to appear).
[4] P. Colli and M. Grasselli, Hyperbolic phase change problems in heat conduction with mem-
ory, Proc. Roy. Soc. Edinburgh Sect. A 123 (1993) 571{592.
[5] P. Colli and M. Grasselli, Justi cation of a hyperbolic approach to phase changes in mate-
rials with memory, Asymptotic Anal. 10 (1995) 303{334.
[6] P. Colli and M. Grasselli, Convergence of parabolic to hyperbolic phase change models with
memory, Adv. Math. Sci. Appl. 6 (1996), 147-176.
[7] A. Friedman and K.{H. Ho mann, Control of free boundary problems with hysteresis,
SIAM J. Control Optim. 26 (1988) 42{55.
[8] K. Glasho and J. Sprekels, An application of Glicksberg's theorem to set{valued integral
equations arising in the theory of thermostats, SIAM J. Math. Anal. 12 (1981) 477{486.
[9] I.L. Glicksberg, A further generalization of the Kakutani xed point theorem, with appli-
cation to Nash equilibrium points, Proc. Amer. Math. Soc. 3 (1952) 170{174.
[10] G. Gripenberg, S.{O. Londen, and O. Sta ans, Volterra integral and functional equations,
Encyclopedia Math. Appl. 34 (Cambridge Univ. Press, Cambridge, 1990).
[11] K.{H. Ho mann, M. Niezg
odka, and J. Sprekels, Feedback control via thermostats of mul-
tidimensional two{phase Stefan problems, Nonlinear Anal. 15 (1990) 955{976.
[12] J. Sprekels, Automatic control of one{dimensional thermomechanical phase transitions, in:
J. F. Rodrigues, ed., Mathematical Models for Phase Change Problems, Internat. Ser.
Numer. Math. 88 (Birkh
auser, Basel, 1989) 89-98.
[13] A. Visintin, On the Preisach model for hysteresis, Nonlinear Anal. 8 (1984) 977{996.
[14] A. Visintin, Di erential models of hysteresis, Appl. Math. Sci. 111 (Springer, Berlin,
1994).
28