Nonlinear Analysis 97 (2014) 210–227

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Nonlinear Analysis

journal homepage: www.elsevier.com/locate/na

Existence and uniqueness results for a nonlinear evolutionequation arising in growing cell populationsJ. Garcia-Falset a, K. Latrach b,∗, A. Zeghal ca Departament d’Anàlisi Matemàtica, Universitat de València, Dr. Moliner 50, 46100, Burjassot, València, Spainb Université Blaise Pascal, Laboratoire de Mathématiques, UMR 6620 - CNRS, Complexe des Cézeaux, BP 80026, 63171 Aubière Cedex,Francec Université Sultan Moulay Slimane, Laboratoire de Mathématiques et Applications, B.P. 523, 23000 Beni-Mellal, Morocco

a r t i c l e i n f o

Article history:Received 31 May 2013Accepted 29 November 2013Communicated by Enzo Mitidieri

MSC:47H0634A1235F20

Keywords:Evolution equationLocal and nonlocal boundary conditionsQuasi-accretive operatorsLocal and global solutions

a b s t r a c t

The present paper is concerned with a nonlinear initial–boundary value problem derivedfrom a model introduced by Rotenberg (1983) describing the growth of a cell population.Each cell of this population is distinguished by two parameters: its degree of maturity µand its maturation velocity v. At mitosis, the daughter cells and mother cells are related bya general reproduction rule. We prove existence and uniqueness results in the case wherethe total cross-section and the boundary conditions are depending on the total density ofpopulation. Local and nonlocal reproduction rules are discussed.

© 2013 Elsevier Ltd. All rights reserved.

1. Introduction

In 1983, Rotenberg [1] presented a model for growing cell population in which each cell is distinguished by twoparameters. The first parameter is the degree of maturity µ and a cell in the process evolution has a degree of maturityµ ∈ [0, 1]. Thus, during each cell mitotic, the degree of maturity of a mother cell is µ = 1 and its daughter cells is µ = 0.The second parameter is the velocity of maturation v. The positivity of velocities comes from the fact that a cell may notbecome less mature with the time and then v ∈ (a, b) (0 < a < b < ∞). If the functionψ(t, µ, v) represents the density ofthe population with respect to the maturationµ and the velocity of maturation v at time t , Rotenberg derived the followingpartial differential equation

∂

∂tψ(t, µ, v)+ v

∂

∂µψ(t, µ, v) = −σ(µ, v)ψ(t, µ, v)+

b

aκ(µ, v, v′)ψ(t, µ, v′)dv′

ψ(0, µ, v) = ψ0(µ, v).

(1)

The kernel κ(µ, v, v′) is the transition rate. It specifies the transition of cells from the maturation velocity v′ to v whileσ(µ, v) =

ba κ(µ, v

′, v)dv′ denotes the total transition cross section. It is the rate of cell mortality or cell loss due to othercauses than division. During each cell division, it can be assumed that there is a correlation between thematuration velocity

∗ Corresponding author. Tel.: +33 473 407 896.E-mail addresses: garciaf@uv.es (J. Garcia-Falset), Khalid.Latrach@math.univ-bpclermont.fr (K. Latrach), a.zeghal@fstbm.ac.ma (A. Zeghal).

0362-546X/$ – see front matter© 2013 Elsevier Ltd. All rights reserved.http://dx.doi.org/10.1016/j.na.2013.11.027

J. Garcia-Falset et al. / Nonlinear Analysis 97 (2014) 210–227 211

of mother cells v′ and that of daughter cells v. This correlation is governed by a transition biological rule mathematicallydescribed by the boundary condition

vψ(t, 0, v) = β

b

ak(v, v′)ψ(t, 1, v′)v′dv′, (2)

where β is the average number of daughter cells viable per mitotic (see [1]).It is also possible that there are cells which degenerate and in this case there is a total inheritance of the maturation

velocity between mother and its daughters. This biological rule is mathematically described by

ψ(t, 0, v) = αψ(t, 1, v), (3)

where α ≥ 0 is the average number of daughter cells viable per mitotic. However, in reality there are always cells whichdegenerate and other cellswhich respect the previous correlationwith its daughter cells. One of themost observed biologicalconsideration is described by the boundary conditions (see, for instance, [2] or [3, p. 476]).

ψ(t, 0, v) = αψ(t, 1, v)+β

v

b

ak(v, v′)ψ(t, 1, v′)v′dv′. (4)

Rotenberg discussed essentially the Fokker–Planck approximation of the Problem (1)–(2) for which he obtainednumerical solutions. Using the eigenfunction technique, Van der Mee and Zweifel [4] obtained analytical solutions of (1) fora variety of boundary conditions. Using Lebowitz and Rubinow’s boundary conditions, it was established in [3, Chapter XIII,Section 5] that the associated Cauchy problem to (1) is governed by a positive C0-semigroup and an estimate of the type ofthe semigroup was derived on the space L1 which allowed to describe the time asymptotic behavior of the solution of theCauchy problem. Similar results were obtained for various boundary conditions (see, for example, [5,2,6]). In [7] a detailedspectral analysis of the Problem (1) supplemented with general (linear) transition rule relating mother and daughter cellsat mitosis covering, in particular, all classical ones considered in [3,8,1,4], was given.

In the introduction of his paper, Rotenberg has pointed out that the adequate formulation of the model (1) seems tobe a nonlinear one. Actually, the cells under consideration are in contact with a nutrient environment which is not part ofthe mathematical formulation. Fluctuations in nutrient concentration and other density-dependent effects such as contactinhibition of growth make the transition rates σ(·, ·) and κ(·, ·, ·) functions of the population density, thus creating anonlinear problem. On the other hand, the biological boundary atµ = 0 andµ = 1 is fixed and tightly coupled throughoutmitosis. The conditions present at the boundaries are felt throughout the system and cannot be removed. This phenomenasuggest that at mitosis daughter cells and parent cells are related by a nonlinear nonlocal reproduction rule. Motivated bythese observations, in [9], the authors have considered a modified nonlinear stationary version of the Rotenberg model. Thetransition rate and the total cross-sectionwere assumed to depend on the density of population and, atmitosis, the daughterand mother cells are related by a nonlinear rule which describes the boundary conditions. More precisely, they consideredthe problem

v∂ψ

∂µ(µ, v)+ λψ(µ, v)+ σ(µ, v, ψ(µ, v)) =

c

0κ(µ, v, v′, ψ(µ, v′)) dv′, (5)

whereσ(·, ·, ·) and κ(·, ·, ·, ·) are nonlinear functions ofψ andλ is a complex number. The boundary conditions aremodeledby

ψ|Γ0= K(ψ|Γ1

), (6)

where Γ0 = 0 × [a, b] and Γ1 = 1 × [a, b], ψ|Γ0(resp. ψ|Γ1

) denotes the restriction of ψ to Γ0 (resp. Γ1). The functionK is a nonlinear operator from a suitable function space on Γ1 to a similar one on Γ0.

In [9] several existence theorems for Eq. (5) supplemented with the boundary conditions (6) were obtained in Lp spaceswith 1 < p < ∞. The analysis was carried out via topological arguments and uses compactness results, established inLp-spaces with 1 < p < ∞, of the operators involved in the problem. Due to the lack of compactness of these operatorson the space L1, in [10], the second and third authors extended the analysis to the case p = 1 using essentially the specificproperties of weakly compact sets on L1-spaces. Despite these works, to our knowledge, the time-dependent Problem (5)even for simple reproduction laws has not yet been investigated. The main purpose of this work is to fill this gap. The paperis organized as follows:

– Introduction,– Main results,– Proofs and consequences,– Comments,– Annex.

212 J. Garcia-Falset et al. / Nonlinear Analysis 97 (2014) 210–227

Section 2 is devoted to introduce the main assumptions and theorems. In Section 3 we study the initial–boundary valueproblem

∂

∂tψ(t, µ, v)+ v

∂

∂µψ(t, µ, v) = −σ(µ, v, ψ(t, µ, v))+

b

aκ(µ, v, v′, ψ(t, µ, v′))dv′

ψ(0, µ, v) = ψ0(µ, v)ψ(t, 0, v) = [Kψ(t, 1, ·)](v),

(7)

where σ(·, ·, ·) and κ(·, ·, ·, ·) are nonlinear functions while K denotes a nonlinear operator on suitable trace spacesmodeling the transition biological rule.

We discuss the existence and uniqueness of a solution to Problem (7) in the same way as it was done in [11] forLebowitz–Rubinow’s model [12] (also see [13]). It is shown under reasonable assumptions that the problem has a uniquemild solution on appropriate weight Lp spaces where 1 ≤ p < ∞, and, if p > 1, this solution is also a weak solution of theproblem. Further, if the initial data belongs to the domain of a certain operator involved in the problem, then the solution isa strong solution (see Theorem 2.1). In this same section we are also concerned with the problem

∂ψ

∂t(t, µ, v)+ v

∂ψ

∂µ(t, µ, v)+ σ1(µ, v, ⟨ψ⟩(t))ψ(t, µ, v) =

b

aκ1(µ, v, v

′, ψ(t, µ, v′)) dv′, (8)

where σ1(·, ·, ·) and κ1(·, ·, ·, ·) are nonlinear functions and

⟨ψ⟩(t) =

1

0

b

aψ(t, µ, v)dµdv.

We allow the transition rate and the total transition cross section to depend on the density of population. In this case, theboundary conditions are modeled by

ψ(t, 0, v) =

b

aκ2(v, v

′, ⟨ψ⟩(t), ψ(t, 1, v′))v′dv′+ σ2(v, ⟨ψ⟩(t), ψ(t, 1, v)), (9)

where σ2(·, ·, ·) and κ2(·, ·, ·, ·) are nonlinear functions of ψ(1, ·)(v) and ⟨ψ⟩. After giving some preparatory results,we discuss, under appropriate conditions, the existence of local and global solutions of the problem on Lp-spaces with1 < p < ∞ (see Theorem 2.2). The analysis of the problem in the space L1 is performed in Theorem 2.3. Our analysisuses the approach developed for the semilinear evolution equation in [14].

In Section 5 we recall the main results concerning accretive operators as well as gather some facts from the functionalanalysis which are required in the remainder of the paper.

Mathematical models for nonlinear age structured dynamics with nonlocal boundarywere already considered in [14,15].Let us point out that the book [16] provides a detailed account for transport equations in Biology where the mathematicalanalysis of various kinds of such equations was derived.

2. Main results

2.1. Local boundary conditions

In this section we are concerned with the existence and uniqueness result for Problem:∂

∂tψ(t, µ, v)+ v

∂

∂µψ(t, µ, v) = −σ(µ, v, ψ(t, µ, v))+

b

aκ(µ, v, v′, ψ(t, µ, v′))dv′

ψ(0, µ, v) = ψ0(µ, v)ψ(t, 0, v) = [Kψ(t, 1, ·)](v).

(10)

For our subsequent analysis, we need the following hypotheses:

(A1) The operator K : Lp([a, b], vdv) → Lp([a, b], vdv) satisfies a Lipschitz condition, i.e.; there exists LK > 0 such that

∥K(f1)− K(f2)∥(∗,p) ≤ LK∥f1 − f2∥(∗,p),

for all f1, f2 ∈ Lp([a, b], vdv). (In general LK will be considered bigger or equal to 1.)(A2) The function σ(·, ·, ·), is measurable and satisfies the following condition: there existsΛ > 0 such that

|σ(µ, v, η1)− σ(µ, v, η2)| ≤ Λ|η1 − η2|.

(A3) The function κ(·, ·, ·, ·) is measurable and satisfies

|κ(µ, v, v′, ψ1(µ, v′))− κ(µ, v, v′, ψ2(µ, v

′))| ≤ |ρ(µ, v, v′)|ψ1(µ, v

′)− ψ2(µ, v′) ,

where ψ1, ψ2 ∈ Lp(Ω) and ρ ∈ L∞([0, 1] × [a, b] × [a, b]).

J. Garcia-Falset et al. / Nonlinear Analysis 97 (2014) 210–227 213

Next, using κ we define a nonlinear operator B byB : Lp(Ω) → Lp(Ω),

ψ →

b

aκ(µ, v, v′, ψ(µ, v′))dv′

and, for all ψ ∈ Lp(Ω), we set

Fpψ := Bψ − σ(·, ·, ψ).

Using the trace theorem (see, for example, [3, Proposition 3.3, p. 379]) it is possible to give a sense to the following operatorTK : D(TK ) ⊆ Lp(Ω) → Lp(Ω)

ϕ → (TKϕ)(µ, v) = v∂ϕ

∂µ(µ, v),

where D(TK ) := ϕ ∈ Wp(Ω) : γ0(ϕ) = K(γ1(ϕ)).Taking u(t) := ψ(t, ·) ∈ Lp(Ω), we may interpret and rewrite Problem (10) as follows:

u′(t)+ TK (u(t)) = Fp(u(t))u(0) = ψ0 ∈ Lp(Ω).

(11)

Theorem 2.1. If the assumptions (A1)–(A3) hold, then

(a) Problem (11) has a unique mild solution,(b) if p > 1, this mild solution is, in fact, a weak solution,(c) if p > 1 and the initial data ψ0 ∈ D(TK ), then the solution is a strong solution.

2.2. Nonlocal boundary conditions

In this section we are concerned with the existence and uniqueness result for Problem:∂

∂tψ + v

∂

∂µψ = −σ1(µ, v, ⟨ψ⟩)ψ +

b

aκ1(µ, v, v

′, ψ(t, µ, v′))dv′

ψ(0, µ, v) = ψ0(µ, v)

ψ(t, 0, v) =

b

aκ2(v, v

′, ⟨ψ⟩, ψ(t, 1, v′))dv′+ σ2(v, ⟨ψ⟩, ψ(t, 1, v)),

(12)

where ⟨ψ⟩(t) = 10

ba ψ(t, µ, v)dµdv is the mass of the population. For our subsequent analysis, we need the following

hypotheses:

(A4) The function σ1(·, ·, ·) is measurable and it satisfies the condition: for any α > 0, there existsΛα > 0 such that

|σ1(µ, v, η1)− σ1(µ, v, η2)| ≤ Λα|η1 − η2|,

for a.e. (µ, v) ∈ [0, 1] × [a, b], η1, η2 ∈ [−α, α].(A5) The functions σ2(·, ·, ·) and κ2(·, ·, ·, ·) are measurable and satisfy the conditions: there exists a constantΛ > 0

|σ2(v, η1, ξ1)− σ2(v, η2, ξ2)| ≤ Λ(|η1 − η2| + |ξ1 − ξ2|),

|κ2(v, v′, η1, ξ1)− κ2(v, v

′, η2, ξ2)| ≤ Λ(|η1 − η2| + |ξ1 − ξ2|),

for a.e. (v, v′) ∈ [a, b] × [a, b], η1, η2 ∈ R and ξ1, ξ2 ∈ R.(A6) There exist two constants σ 1 ∈ R and σ 1 > 0 such that

σ 1 ≤ σ1(µ, v, η) ≤ σ 1,

for a.e. (µ, v) ∈ [0, 1] × [a, b], η ≥ 0.

The function κ1(·, ·, ·, ·) is measurable and defines a nonlinear operator B1 by

B1: Lp(Ω) → Lp(Ω), ψ →

b

aκ1(µ, v, v

′, ψ(µ, v′))dv′.

For our subsequent analysis we need the following hypothesis

(A7) |κ1(µ, v, v′, ψ1(µ, v

′))− κ1(µ, v, v′, ψ2(µ, v

′))| ≤ |ρ(µ, v, v′)||ψ1 − ψ2|(µ, v′),

where ψ1, ψ2 ∈ Lp and ρ ∈ L∞([0, 1] × [a, b] × [a, b]).

214 J. Garcia-Falset et al. / Nonlinear Analysis 97 (2014) 210–227

We set

F 1p ψ := B1(ψ)− σ1(·, ·, ⟨ψ⟩)ψ, ψ ∈ Lp([0, 1] × [a, b]). (13)

We define the free streaming operator TK with domain including the nonlocal boundary conditionsTK : D(TK ) ⊂ Lp(Ω) −→ Lp(Ω)

ψ −→ TKψ(µ, v) = v∂ψ

∂µ(µ, v)

D(TK ) =ψ ∈ Wp(Ω) such that γ0(ψ) = K(γ1ψ,ψ)

,

where K denotes the following boundary operator

K : Lp([a, b], vdv)× Lp(Ω) → Lp([a, b], vdv)(u, ψ) → K(u, ψ)

and

K(u, ψ)(v) =

b

aκ2(v, v

′, ⟨ψ⟩, u(v′))v′dv′+ σ2(v, ⟨ψ⟩, u(v)). (14)

It is well known that −T0 (κ2(·, ·, ·) ≡ 0 and σ2(·, ·, ·) ≡ 0) generates a C0-semigroup of contractions on Lp(Ω) and that, forall λ > 0, the operator (I + λT0)−1 is positive (in the lattice sense).

Taking u(t) := ψ(t, ·) ∈ Lp(Ω), we may interpret and rewrite Problem (12) as follows:u′(t)+ TK (u(t)) = F 1

p (u(t))u(0) = ψ0 ∈ Lp(Ω).

(15)

Theorem 2.2. Let p ∈ (1,+∞) and assume that the hypotheses (A4)–(A7) are satisfied. Then Problem (15) has a local weaksolution for each ψ0 ∈ Lp(Ω). Further, if ψ0 ∈ D(TK ), then Problem (15) has a local strong solution. Finally, if we suppose thatκ1(·, ·, ·, 0) = 0, then Problem (15) has a unique global strong solution for each ψ0 ∈ D(TK ) and it has a unique global weaksolution if ψ0 ∈ D(TK ).

Theorem 2.3. Let p = 1 and assume that the hypotheses (A4)–(A7) are satisfied. Then Problem (15) has a local mild solutionfor each ψ0 ∈ L1(Ω). If further, we add the conditions

(a) κ1(·, ·, ·, 0) = 0,(b) κ2(·, ·, 0, 0) = 0 and σ2(·, 0, 0) = 0,(c) b∥ρ∥∞e

δa ≤ σ 1,

then Problem (15) has a unique global mild solution for each ψ0 ∈ L1(Ω).

3. Proofs of the main results

We refer to Section 5.2, where the reader will find the notation and the functional spaces used in this paper.

3.1. Proof for the local boundary condition

In this section we are concernedwith the existence and uniqueness result for Problem (10). Wemay consider the Banachspace Lp,δ where δ = 0 if K is a nonexpansive mapping and δ is a fixed number bigger than b ln(LK ) if K is LK -Lipschitzianmapping with LK > 1. Thus, in what follows we will take a fixed δ ≥ max0, b ln(LK ).

Proposition 3.1. If the condition (A1) holds true, then TK is a δ-m-accretive operator on Lp,δ .

Proof. Given g ∈ Lp,δ and λ ∈ (0, 1δ) (here if δ = 0, 1

δmeans +∞). We wish to see that there exists a function ϕ ∈ D(TK )

such that ϕ + λTK (ϕ) = g . Thus, we seek the solution of the following differential equation

ϕ(µ, v)+ λv∂ϕ

∂µ(µ, v) = g(µ, v). (16)

Since Eq. (16) is linear with respect to the variable µ the general solution is given by:

ϕ(µ, v) = e−µvλ χ(v)+

1vλ

µ

0g(s, v)e−

µ−svλ ds. (17)

J. Garcia-Falset et al. / Nonlinear Analysis 97 (2014) 210–227 215

Using the operators Qλ andΞλ we get

ϕ(µ, v) = Qλχ(v)+ (Ξλg)(µ, v).

Let us show that if χ ∈ Lp([a, b], vdv), then ϕ ∈ Wp,δ(Ω). Indeed, using that 0 < λδ < 1, we have

∥vϕ∥p,δ =

1

0

b

ae−δ

1−µv pvp

e−µvλ χ(v)+

1vλ

µ

0g(s, v)e

s−µvλ ds

p dvdµ1p

≤

1

0

b

avp−1v|χ(v)|pdv

dµ 1

p

+

1λp

1

0

b

ae−δ

1−µv p

µ

0|g(s, v)|pe

s−µvλ

pdsdvdµ 1

p

≤ b1q ∥χ∥(∗,p) +

1λ

b

a

1

0

µ

0|g(s, v)|pe−pδ 1−s

v dsdµdv 1

p

≤ b1q ∥χ∥(∗,p) +

1λ

∥g∥p,δ. (18)

On the other hand, we find that,v ∂ϕ∂µp,δ

=1λ

∥g − ϕ∥p,δ ≤1λ

∥g∥p,δ +

1a∥vϕ∥p,δ

. (19)

It is clear that if χ ∈ Lp([a, b], vdv) from inequalities (18) and (19) we infer that ϕ ∈ Wp,δ(Ω).Now, our next step will be to prove that there exists a function ϕ ∈ D(TK )which satisfies Eq. (17). In this case γ0(ϕ) = χ .

The following condition should be satisfied

γ1(ϕ) = γ1(Qλ(χ))+ γ1(Ξλ(g)) = e−1vλ γ0(ϕ)+ γ1(Ξλ(g)).

Since ϕ ∈ D(TK )we know that γ0(ϕ) = K(γ1(ϕ)). Consequently, we have

γ1(ϕ)(v) = Pλ(K(γ1(ϕ)))(v)+ γ1(Ξλ(g))(v).

At this moment we may introduce the operator Sλ : Lp([a, b], vdv) → Lp([a, b], vdv) defined by

Sλ(ρ)(v) = Pλ(K(ρ))(v).

We claim that Sλ is a strict contraction. Indeed, using (33) and (A1), we obtain

∥Sλ(ρ1)− Sλ(ρ2)∥(∗,p) = ∥Pλ(K(ρ1)− K(ρ2))∥(∗,p) ≤ e−1bλ LK∥ρ1 − ρ2∥(∗,p).

Since we have taken δ ≥ b ln(LK ) and λ ∈ (0, 1δ), it is clear that e−

1bλ LK < 1 which proves our claim. In this case, it is

not difficult to prove that the operator I − Sλ : Lp([a, b], vdv) → Lp([a, b], vdv) is bijective and moreover (I − Sλ)−1 iscontinuous (for instance see [17]), then given γ1(Ξλ(g)) ∈ Lp([a, b], vdv) there exists a unique functionψ ∈ Lp([a, b], vdv)such that (I − Sλ)(ψ) = γ1(Ξλ(g)). So we can define the function

ϕ(µ, v) := (I + λTK )−1(g)= Qλ(K(I − Sλ)−1γ1(Ξλ(g)))(v)+ Ξλ(g)(µ, v) ∈ Wp,δ(Ω).

This yields that R(I + λTK ) = Lp,δ whenever λ ∈ (0, 1δ).

Next we shall prove that TK is a δ-accretive operator on Lp,δ . In order to see this, we first prove the result in the space L1,δ .Indeed, let g1, g2 ∈ L1,δ , by the above argument we get ϕ1, ϕ2 ∈ D(TK ) such that

ϕ1 = (I + λTK )−1(g1), ϕ2 = (I + λTK )−1(g2).

Hence, using (16), (A1) and δ ≥ b ln(LK ), we obtain

∥ϕ1 − ϕ2∥1,δ =

Ω

sgn0(ϕ1 − ϕ2)

(g1 − g2)− λv

∂

∂µ(ϕ1 − ϕ2)

hδ(µ, v)dµdv

≤ −λ

Ω

sgn0(ϕ1 − ϕ2)hδv∂(ϕ1 − ϕ2)

∂µdµdv +

Ω

sgn0(ϕ1 − ϕ2)hδ(g1 − g2)dµdv

≤ −λ

b

a

1

0v∂

∂µ|hδ(ϕ1 − ϕ2)|dµdv + δλ

Ω

|hδ(ϕ1 − ϕ2)|dµdv

+

Ω

hδ(µ, v)|g1(µ, v)− g2(µ, v)|dµdv

216 J. Garcia-Falset et al. / Nonlinear Analysis 97 (2014) 210–227

≤ λ

b

a

e−δ/vv|K(γ1(ϕ1))− K(γ1(ϕ2))| − v|γ1(ϕ1)− γ1(ϕ2)|

dv + δλ∥ϕ1 − ϕ2∥1,δ + ∥g1 − g2∥1,δ

≤ λ(e−δ/bα − 1)∥γ1(ϕ1)− γ1(ϕ2)∥Y1 + δλ∥ϕ1 − ϕ2∥1,δ + ∥g1 − g2∥1,δ

≤ δλ∥ϕ1 − ϕ2∥1,δ + ∥g1 − g2∥1,δ

which leads to

∥(I + λTK )−1(g1)− (I + λTK )−1(g2)∥1,δ = ∥ϕ1 − ϕ2∥1,δ ≤1

1 − δλ∥g1 − g2∥1,δ.

Now, we consider the space Lp,δ with p > 1. To this end, let ϕ1, ϕ2 be two elements of D(TK ). We claim that TK is δ-accretive.To see this, set ψ := ϕ1 − ϕ2. Hence we have

[TK (ϕ1)− TK (ϕ2), ψ]s ≥ ∥ψ∥1−pp,δ

b

a

1

0hpδ |ψ |

p−1v∂

∂µ(ψ)sgn0(ψ)dµ

dv

=∥ψ∥

1−pp,δ

p

b

a

1

0hpδv

∂

∂µ(|ψ |

p)

dµdv

=∥ψ∥

1−pp,δ

p

b

a

1

0v

∂

∂µ(hδ|ψ |)p

dµdv

−∥ψ∥

1−pp,δ

p

b

a

1

0v|ψ |

p∂

∂µhpδ

dµdv. (20)

Using the fact that the trace mappings are linear and continuous, the last term of inequality (20) is equal to

∥ψ∥1−pp,δ

p×

b

av|γ1(ϕ1)(v)− γ1(ϕ2)(v)|

p− e−

δpv |γ0(ϕ1)(v)− γ0(ϕ2)(v)|

pdv − δ∥ψ∥p,δ. (21)

Since ϕ1, ϕ2 ∈ D(TK ), we obtain that expression (21) is greater than or equal to

∥ψ∥1−pp,δ

p

b

av|γ1(ϕ1)(v)− γ1(ϕ2)(v)|

p− e−

δpv |K(γ1(ϕ1))(v)− K(γ1(ϕ2))(v)|

pdv − δ∥ψ∥p,δ.

Finally, remembering that K is LK -Lipschitzian on Lp([a, b]vdv) and using that δ ≥ b ln(LK ), we infer that

[TK (ϕ1)− TK (ϕ2), ψ]s ≥∥ψ∥

1−pp,δ

p

1 − LpK e

−δpb

∥γ1(ϕ1)− γ1(ϕ2)∥

pYp − δ∥ψ∥p,δ

≥ −δ∥ψ∥p,δ.

This together with Proposition 5.1 proves that TK is δ-accretive as claimed.

Remark 3.1. It should be noticed that, if K : Lp([a, b]vdv) → Lp([a, b]vdv) is nonexpansive, i.e., LK ≤ 1, then we can takeδ = 0 and therefore we can work with the spaces Lp(Ω)without using any weight.

Proposition 3.2. D(TK )Lp(Ω)

= Lp(Ω).

Proof. To obtain this result, it is enough to prove that C∞

0 (Ω) ⊆ D(TK )Lp(Ω)

. Let u be an element of C∞

0 (Ω). For each n ∈ N,define the function:

un(µ, v) =

u(µ, v), (µ, v) ∈ Ω, µ ∈ [2/n, 1]

u(µ, v)µ−

1n

n + K(u(1, v))

2n

− µ

n (µ, v) ∈ Ω, µ ∈

1n,2n

K(u(1, ·))(v), (µ, v) ∈ Ω µ ∈ [0, 1/n[.

It is clear that |un(µ, v)| ≤ |u(µ, v)| + |K(u(1, v))| and both functions belong to Lp(Ω). It is also clear that upn → up a.e. in

Ω . Thus, by the dominated convergence theorem we have that ∥un − u∥p → 0. On the other hand,

∂un

∂µ=

∂u∂µ(µ, v), (µ, v) ∈ Ω µ ∈ [2/n, 1]

∂u∂µ(µ, v)

µ−

1n

n + u(µ, v)− nK(u(1, v)), (µ, v) ∈ Ω, µ ∈

1n,2n

0, (µ, v) ∈ Ω µ ∈ [0, 1/n[.

J. Garcia-Falset et al. / Nonlinear Analysis 97 (2014) 210–227 217

Since u ∈ C∞

0 (Ω), it is clear that un ∈ Wp(Ω). Moreover, γ1(un)(v) = u(1, v), and γ0(un)(v) = K(u(1, v)) = K(γ1(un)(v))which means that un ∈ D(TK ).

Lemma 3.1. Let p ∈ [1,∞) and assume that hypotheses (A2) and (A3) hold true. Then there exists a constant C > 0 such that

∥Fpψ1 − Fpψ2∥p ≤ C∥ψ1 − ψ2∥p.

That is, Fp is globally Lipschitz.

Proof. For all (ψ1, ψ2) ∈ Lp(Ω)× Lp(Ω)we have

Fpψ1 − Fpψ2 = B(ψ1)− B(ψ2)−

σ(·, ·, ψ1)− σ(·, ·, ψ2)

.

It follows from (A3) thatFpψ1 − Fpψ2 ≤ ∥ρ∥L∞

b

a

ψ1(µ, v′)− ψ2(µ, v

′) dv′

+Λ |(ψ1 − ψ2)| .

Using Hölder’s inequality we get

Fpψ1 − Fpψ2 ≤ ∥ρ∥L∞ b

1q

b

a

ψ1(µ, v′)− ψ2(µ, v

′)p dv′

1p

+Λ |(ψ1 − ψ2)| .

A simple calculation using (A2) and (A3) givesFpψ1 − Fpψ2Lp

≤ b ∥ρ∥L∞ ∥ψ1 − ψ2∥p +Λ ∥ψ1 − ψ2∥p

≤ C ∥ψ1 − ψ2∥p

where C := b ∥ρ∥L∞ +Λ.

Remark 3.2. At this point, applying Lemmas 3.1 and 5.1, one sees that

∥Fp(ϕ1)− Fp(ϕ2)∥p,δ ≤ ∥Fp(ϕ1)− Fp(ϕ2)∥p

≤ C∥ϕ1 − ϕ2∥p

≤ eδa C∥ϕ1 − ϕ2∥p,δ,

which means that Fp : Lp,δ → Lp,δ is a globally Lipschitzian mapping.

Proof of Theorem 2.1. Propositions 3.1 and 3.2 show that the operator TK : D(TK ) → Lp,δ is quasi-m-accretive, sinceLemma 3.1 and Remark 3.2 yield that the operator Fp : Lp,δ → Lp,δ is globally Lipschitzian, then it is not difficult to see thatTK − Fp is also quasi-m-accretive on Lp,δ . Thus, we derive Theorem 2.1(a) by using [18, Corollary 4.1].

On the other hand, if p > 1, since Lp,δ has the Radon–Nikodym property, by Theorem 5.3 this mild solution is, in fact,a weak solution. Moreover, if the initial data belongs to D(TK ), by Theorem 5.1, we infer that the weak solution is a strongsolution.

Next result shows that the solution depends continuously on the initial data. To this end, let us introduce the Banach spaceC([0, T ], Lp,δ) endowed with its usual norm

∥u∥∞ := sup∥u(t)∥p,δ : 0 ≤ t ≤ T .

Proposition 3.3. Let u1, u2 ∈ C([0, T ], Lp,δ), with p ≥ 1, be two mild solutions of Problem (11). Given ε > 0, there exists δ > 0such that if ∥u1(0)− u2(0)∥p,δ ≤ δ, then ∥u1 − u2∥∞ ≤ ε.

Proof. Since TK is a δ-m accretive operator on Lp,δ and Fp : Lp,δ → Lp,δ is r-Lipschitzian, where r := eδa C (see Remark 3.2),

it is easy to see that TK − Fp is a δ + r-m-accretive operator on Lp,δ . So, given i ∈ 1, 2, ui is the unique solution of theproblem:

u′(t)+ TK (u(t))− Fp(u(t)) = 0u(0) = ui(0) ∈ Lp,δ.

(22)

Hence, by (32) we infer that

∥u1(t)− u2(t)∥p,δ ≤ e(δ+r)t∥u1(0)− u2(0)∥p,δ.

The above inequality implies that, for every t ∈ [0, T ],

∥u1(t)− u2(t)∥p,δ ≤ e(δ+r)T∥u1(0)− u2(0)∥p,δ,

218 J. Garcia-Falset et al. / Nonlinear Analysis 97 (2014) 210–227

and therefore

∥u1 − u2∥∞ ≤ e(δ+r)T∥u1(0)− u2(0)∥p,δ.

To end the proof, it suffices to take δ =ε

e(δ+r)T .

3.2. Proof for the nonlocal boundary condition

In this section we shall construct a solution to the initial–boundary value problem (12) in the sense of Theorem 5.2.

Lemma 3.2. Let p ∈ [1,+∞) and assume that the hypotheses (A4)–(A7) are satisfied. Then,(i) for each α > 0, there exists a constant Cα > 0 such that

∥F 1p ψ1 − F 1

p ψ2∥p ≤ Cα∥ψ1 − ψ2∥p,

for all ψi ∈ Bp,δα :=

u ∈ Lp,δ : ∥u∥p,δ ≤ α

, i ∈ 1, 2.

(ii) There exists a constant C := C(p) such that

∥K(u1, ψ1)− K(u2, ψ2)∥(∗,p) ≤ C(∥u1 − u2∥(∗,p) + ∥ψ1 − ψ2∥p),

for all ψi ∈ Lp(Ω) and ui ∈ Lp([a, b], vdv), i ∈ 1, 2.

Proof. (i) For all (ψ1, ψ2) ∈ (Bp,δα )

2 we have

F 1p ψ1 − F 1

p ψ2 = B1(ψ1 − ψ2)−

σ1(·, ·, ⟨ψ1⟩)ψ1 − σ1(·, ·, ⟨ψ2⟩)ψ2

.

It follows from (A7) thatF 1p ψ1 − F 1

p ψ2 ≤ ∥ρ∥L∞

b

a

ψ1(µ, v′)− ψ2(µ, v

′) dv′

+ |σ1(µ, v, ⟨ψ1⟩)(ψ1 − ψ2)|

+ |(σ1(µ, v, ⟨ψ1⟩)− σ1(µ, v, ⟨ψ2⟩))ψ2| .

Using Lemma 5.1 one sees that |⟨ψi⟩| ≤ ∥ψi∥1 ≤ αeδa (b)

1q , for i = 1, 2, and ∥ψ2∥p ≤ e

δa α. So, a simple calculation using

(A4) and (A6) givesF 1p ψ1 − F 1

p ψ2Lp

≤ b ∥ρ∥L∞ ∥ψ1 − ψ2∥p + σ 1 ∥ψ1 − ψ2∥p + αΛβe2δa ∥ψ1 − ψ2∥p

≤ Cα ∥ψ1 − ψ2∥p

where Cα := b ∥ρ∥L∞ + σ 1 + αΛβ eδa and β = αe

δa (b)

1q .

(ii) According to (14), we find that

∥K(u1, ψ1)− K(u2, ψ2)∥(∗,p) ≤

b

a

b

a|κ2(v, v

′, ⟨ψ1⟩, u1(v′))− κ2(v, v

′, ⟨ψ2⟩, u2(v′))|v′dv′

pvdv

1p

+

b

a|σ2(v, ⟨ψ1⟩, u1(v))− σ2(v, ⟨ψ2⟩, u2(v))|

pvdv 1

p

= I1 + I2.

Using the hypothesis (A5), we get

I1 =

b

a

b

a|κ2(v, v

′, ⟨ψ1⟩, u1(v′))− κ2(v, v

′, ⟨ψ2⟩, u2(v′))|v′dv′

pvdv

1p

≤ Λ

b2

2

1p b

a|⟨ψ1⟩ − ⟨ψ2⟩| vdv +

b

a|u1 − u2| vdv

≤ Λ

b2

2

1p

b2

2

∥ψ1 − ψ2∥1 +

b

a|u1 − u2| vdv

.

It follows from Hölder’s inequality that b

a|u1(v)− u2(v)| vdv = (b2/2)

1q ∥u1 − u2∥(∗,p) .

J. Garcia-Falset et al. / Nonlinear Analysis 97 (2014) 210–227 219

Now using Lemma 5.1

∥ψ1 − ψ2∥1 ≤ (b)1/q ∥ψ1 − ψ2∥p , (23)

so we infer

I1 ≤ Λ

b2

2

1p

b2

2

(b)1/q ∥ψ1 − ψ2∥p +

b2

2

1q

∥u1 − u2∥(∗,p)

≤ C1

∥ψ1 − ψ2∥p + ∥u1 − u2∥(∗,p)

where C1 = max(Λ(b2/2)1+

1p (b)1/q, b2

2 Λ).On the other hand, Assumption (A5), Eq. (23) and the inequality (|a| + |b|)p ≤ 2p

|a|p + |b|p

lead to

I2 ≤

b

a|σ2(v, ⟨ψ1⟩, u1(v))− σ2(v, ⟨ψ2⟩, u2(v))|

pvdv 1

p

≤ 2Λ

b

a|⟨ψ1⟩ − ⟨ψ2⟩|

pvdv +

b

a|u1 − u2|

p vdv

1/p

≤ 2(b2/2)1/pΛ ∥ψ1 − ψ2∥1 + 2Λ ∥u1 − u2∥(∗,p)

≤ 2(b2/2)1/pΛ (b)1/q ∥ψ1 − ψ2∥p + 2Λ ∥u1 − u2∥(∗,p)

≤ C2

∥ψ1 − ψ2∥p + ∥u1 − u2∥(∗,p)

where C2 = max(2Λ, 2(b2/2)1/pΛ (b)1/q). Putting C = max(C1, C2)we get

∥K(u1, ψ1)− K(u2, ψ2)∥(∗,p) ≤ C ∥(u1, ψ1)− (u2, ψ2)∥(∗,p)×Lp

which concludes the proof.

Remark 3.3. The result (i) in Lemma 3.2 remains true if ∥ψi∥p ≤ α.

The following lemma asserts that TK is quasi-accretive on Lp,δ whenever δ ≥ max0, b ln(2C) where C is the constant ofLemma 3.2(ii).

Lemma 3.3. Let p ∈ [1,+∞) and assume that the hypotheses (A4)–(A7) are satisfied. Then, there exist two constantsω(p) > 0and δ ≥ 0 such that

∥(ψ1 + λTKψ1)− (ψ2 + λTKψ2)∥p,δ ≥ (1 − λω(p))∥ψ1 − ψ2∥p,δ

for all λ ∈ (0, ω(p)−1) and ψi ∈ D(TK ), i ∈ 1, 2.

Proof. Let ψi ∈ D(TK ), i = 1, 2. It suffices to prove that

⟨TKψ1 − TKψ2, ϕδ⟩δ ≥ −ω(p)∥ψ1 − ψ2∥p

for some ϕδ ∈ Fδ(ψ1 − ψ2). To do so, we assume that ψ1 = ψ2 and we put ω(p) = δ +1p where δ is any nonnegative real

number such that δ ≥ b(ln(2C)) and let C be the constant given in Lemma 3.2(ii). Let us write ψ = ψ1 − ψ2, and

ϕδ(µ, v) = ∥ψ∥1−pp,δ exp

−

pδ(1 − µ)

v

|ψ(µ, v)|p−1sign0(ψ(µ, v)).

An integration by parts yields

⟨TKψ1 − TKψ2, ϕδ⟩δ =∥ψ∥

1−pp,δ

p

1

0

b

aexp

−

pδ(1 − µ)

v

v∂

∂µ|ψ(µ, v)|pdµdv

= −∥ψ∥

1−pp,δ

p

1

0

b

a

pδvv

exp

−pδ(1 − µ)

v

|ψ(µ, v)|pdµdv

+

b

aexp

−

pδv

|γ0(ψ)(v)|

pvdv −

b

a|γ1(ψ)(v)|

pvdv

.

220 J. Garcia-Falset et al. / Nonlinear Analysis 97 (2014) 210–227

Using Lemma 3.2, we can write

⟨TKψ1 − TKψ2, ϕδ⟩δ ≥ −δ∥ψ∥p,δ −∥ψ∥

1−pp,δ

p

×

exp

−

pδb

∥K(γ1ψ1, ψ1)− K(γ1ψ2, ψ2)∥

p(∗,p) − ∥γ1ψ1 − γ1ψ2∥

p(∗,p)

≥ −δ∥ψ∥p,δ −1p

2pCp exp

−

pδb

∥ψ∥

1−pp,δ ∥ψ∥

pp

−1p

2pCp exp

−

pδb

− 1

∥ψ∥

1−pp,δ ∥γ1ψ1 − γ1ψ2∥

p(∗,p).

Since 2pCpe−pδb ≤ 1, applying Lemma 5.1 we infer that

⟨TKψ1 − TKψ2, ϕδ⟩δ ≥ −ω(p)∥ψ1 − ψ2∥p,δ,

which concludes the proof.

In the following we seek for a solution to the resolvent equation for the operator TK , (I + λTK )ψ = ϕ, where ϕ is a givenfunction of Lp(Ω) and the unknown ψ must be sought in D(TK ).

So, for λ > 0, by the argument developed in the proof of Proposition 3.1, the solution is formally given by

ψ(µ, v) = e−µλv K(γ1(ψ), ψ)(v)+

1λv

µ

0e−

µ−µ′

λv ϕ(µ′, v)dµ′. (24)

In particular, for µ = 1, we get

ψ(1, v) := γ1(ψ)(v) = e−1λv K(γ1(ψ), ψ)(v)+

1λv

1

0e−

1−µ′

λv ϕ(µ′, v)dµ′. (25)

Using the fact that ψ must satisfy the boundary conditions, Eqs. (24) and (25) may be written asψ = QλK(γ1ψ,ψ)+ Ξλϕ,γ1ψ = PλK(γ1ψ,ψ)+Πλϕ.

Therefore, (γ1ψ,ψ) is a solution of the fixed point equation (u, g) = Jλ(u, g) on the product space Lp([a, b], vdv)× Lp(Ω),where

Jλ(u, g) = (PλK(u, g)+Πλϕ, QλK(u, g)+ Ξλϕ), (u, g) ∈ Lp([a, b], vdv)× Lp(Ω).

Lemma 3.4. Let p ∈ [1,+∞) and assume that the hypotheses (A5)–(A7) are satisfied. Then there exists a constant λp > 0 suchthat, for each λ ∈ (0, λp) and ϕ ∈ Lp(Ω), there exists ψ ∈ D(TK ) such that

ψ + λTK (ψ) = ϕ. (26)

Proof. As we have seen above, in order to solve Eq. (26) it suffices to prove that the fixed point problem Jλ(u, ψ) = (u, ψ)has a unique solution in Lp([a, b], vdv)× Lp(Ω)where

Jλ(u, ψ) = (J1λ,J

2λ)(u, ψ) = (PλK(u, ψ)+Πλ(ϕ),QλK(u, ψ)+ Ξλ(ϕ)) .

To do so, we equip Lp([a, b], vdv) × Lp(Ω) with the norm ∥(u, ψ)∥ = ∥u∥(∗,p) + ∥ψ∥p and let λp := supλ > 0 :

(e−1bλ + λ

1p )C < 1

where C is the constant given in Lemma 3.2(ii). Next, we check that the operator Jλ is a contraction

mapping. Indeed, let (u1, ψ1), (u2, ψ2) ∈ Lp([a, b], vdv) × Lp(Ω). Using the estimates (33)–(36) and applying Lemma 3.2we have

∥Jλ(u1, ψ1)− Jλ(u2, ψ2)∥ = ∥J1λ(u1, ψ1)− J1

λ(u2, ψ2)∥(∗,p) + ∥J2λ(u1, ψ1)− J2

λ(u2, ψ2)∥p

≤

e−

1λb + λ

1p

∥K(u1, ψ1)− K(u2, ψ2)∥(∗,p)

≤

e−

1λb + λ

1pC(∥u1 − u2∥(∗,p) + ∥ψ1 − ψ2∥p).

Hence, by the definition ofλp the operatorJλp is a contractionmapping and therefore the problem (26) has a unique solutionin D(TK ).

J. Garcia-Falset et al. / Nonlinear Analysis 97 (2014) 210–227 221

Remark 3.4. Lemmas 3.3 and 3.4 guarantee the existence of δ > 0 such that TK becomes quasi-m-accretive on Lp,δ .Moreover, using the same argument as in the proof of Proposition 3.2 we obtain the density of the domain of TK , that is,D(TK ) = Lp(Ω).

Proof of Theorem 2.2. According to Remark 3.4, TK is a quasi-m-accretive operator in Lp,δ . Since Lp,δ is reflexive byLemma 3.2, F 1

p is locally Lipschitz. Applying Theorem 5.2 we obtain the existence of a local strong solution to Problem(15) whenever the initial data ψ0 belongs to D(TK ).

Now we are going to study the existence of a local weak solution when the initial dataψ0 ∈ D(TK ). Again by Remark 3.4the domain of TK is dense in Lp(Ω), so we may choose a sequence (ψn)n∈N ⊆ D(TK ) such that ψn → ψ0 as n → ∞ and setR := sup∥ψn∥p,δ : n = 0, 1, . . .. Arguing as in the proof of Theorem 5.2 it is easy to see that there exists TR > 0 such that,for each n ∈ N, the problem

u′(t)+ TK (u(t)) = F 1p (u(t))

u(0) = ψn

admits a unique strong solution, say un, on the interval (0, TR). In order to discuss the existence of a local weak solution,we only have to see that the sequence (un)n∈N is a Cauchy sequence in the Banach space C(0, TR; Lp,δ) which is a trivialconsequence of inequality (32).

Now, to check the existence of a unique global solution, we argue as follows: given an element ϕδ,1 belonging to the set

J(ψ) =w ∈ X∗

: ⟨ψ,w⟩δ = ∥ψ∥2p,δ, and ∥w∥(Lp,δ)∗ = ∥ψ∥p,δ

.

According to (A6)

⟨σ1(·, ·, ⟨ψ⟩)ψ, ϕδ,1⟩δ ≥ σ 1∥ψ∥2p,δ, (27)

and

⟨B1ψ, ϕδ,1⟩δ ≤ ∥B1ψ∥p,δ∥ψ∥p,δ ≤ ∥ψ∥p,δ∥B1ψ∥p.

The use of Hölder’s inequality and (A8) gives

|B1ψ | ≤

b

a|ρ(µ, v, v′)ψ(µ, v′)|dv′

+

b

aκ1(µ, v, v

′, 0)dv′

≤ b

1q ∥ρ∥∞

b

a|ψ(µ, v′)|pdv′

1p

+

b

aκ1(µ, v, v

′, 0)dv′

.Therefore, Fubini’s theorem and Minkowski’s inequality lead to

∥B1ψ∥p ≤ b∥ρ∥∞∥ψ∥p + ∥B1(0)∥p

which proves that

⟨B1ψ, ϕδ,1⟩δ ≤ b∥ρ∥∞eδa ∥ψ∥

2p,δ + ∥B1(0)∥p∥ψ∥p,δ. (28)

Thus using (13)–(28) we infer that

⟨−F 1p ψ, ϕδ,1⟩δ = ⟨σ1(·, ·,ψ)ψ, ϕδ,1⟩δ − ⟨B1ψ, ϕδ,1⟩δ

≥ σ 1∥ψ∥2p,δ − b∥ρ∥∞e

δa ∥ψ∥

2p,δ − ∥B1(0)∥p∥ψ∥p,δ

= −(b∥ρ∥∞eδa − σ 1)∥ψ∥

2p,δ − ∥B1(0)∥p∥ψ∥p,δ.

Finally, the assumption κ1(·, ·, ·, 0) = 0 yields that B1(0) = 0. Thus, we have proved that

⟨−F 1p ψ, ϕδ,1⟩δ ≥ −α∥ψ∥

2p,δ,

where α := (b∥ρ∥∞eδa − σ 1). Now the use of Theorem 5.2 ensures the existence of a unique global solution.

Proof of Theorem 2.3. Let us see that Eq. (15) has a local mild solution. Indeed, let ψ0 ∈ L1(Ω) and consider α > 1 suchthat ∥ψ0∥1,δ ≤ α − 1. Let ρ(·) be the function defined by

ρ(ψ) =

ψ, ∥ψ∥1,δ ≤ αα

∥ψ∥1,δψ, ∥ψ∥1,δ ≥ α.

222 J. Garcia-Falset et al. / Nonlinear Analysis 97 (2014) 210–227

According to Lemma 3.2(i) the function F 11 (ρ(·)) is 2Cα-Lipschitz. Therefore, the operator TK − F 1

1 (ρ(·)) is w-m-accretivein L1,δ . Consequently, by [18, Remark 4.1] we obtain that the problem

u′(t)+ TK (u(t)) = F 11 (ρ(u(t))),

u(0) = ψ0,

has a unique mild solution u.Now, sinceD(TK ) is dense in L1(Ω), we can choose an element y0 ∈ D(TK )with∥y0∥1,δ ≤ α and such that∥y0−ψ∥1,δ ≤

12 .

Next put z0 = TKy0 − F 11 (ρ(y0)) = TKy0 − F 1

1 (y0) and apply (31), then we get

∥u(t)− y0∥1,δ ≤ ewt∥ψ0 − y0∥1,δ +

t

sew(t−τ)[−z0, u(τ )− y0]sdτ

≤ ewt

∥ψ0 − y0∥1,δ + ∥z0∥1,δ1w(1 − e−wt)

.

This yields

∥u(t)∥1,δ ≤ ∥y0∥1,δ + 1 ≤ α,

for 0 ≤ t ≤ Tα with Tα > 0 suitably chosen, which allows us to get the conclusion since it means that u is a mild solutionto Problem (15) in the interval [0, Tα].

As we saw in the proof of Theorem 2.2, conditions (a) and (c) yield

⟨F 11 (ψ), ϕ1,δ⟩δ ≤ 0.

On the other hand, condition (b) means that K(0, 0) = 0 and then we have that TK (0) = 0. Thus, we may use[19, Lemma 3.1] in order to obtain the existence of a unique global mild solution.

4. Comments

In the structured population dynamics framework, we have in general a proliferation of the population, then it isappropriate to consider boundary conditions of the type ∥K(u)∥(∗,p) ≥ ∥u∥(∗,p) for all u ∈ Lp([a, b], vdv). Therefore, withrespect to condition (A1), in general, LK will be considered bigger or equal to 1.

Remark 4.1. At this point, wewould like to notice that the biological consideration given in (4) satisfies condition (A1) sinceit is enough to consider the operator K : Lp([a, b], vdv) → Lp([a, b], vdv) defined by

K(ψ)(v) := αψ(v)+β

v

b

ak(v, v′)ψ(v′)v′dv′.

From a biological point of view it seems quite natural that if the density of the population of the cells at mitosis is zero,then the transition for the distribution of mothers to daughters should be also zero, this is mathematically described byassuming that the transition operator K satisfies that K(0) = 0. When we assume that K(0) = 0, we can replace condition(A3) by another type of conditions for describing the rate of cell mortality. In this sense, let us consider

1. σ1(·, ·, ·) as in (A4) and (A7),2. σ(µ, v, ψ(µ, v)) = σ1(µ, v, ⟨ψ⟩)ψ ,3. κ(·, ·, ·, ·) satisfies condition (A3), κ(·, ·, ., 0) = 0 and such that b∥ρ∥∞e

δa ≤ σ 1.

Checking carefully the proof of Lemma 3.2(i) we may show that the operator

Fpψ := Bψ − σ(·, ·, ψ), ψ ∈ Lp(Ω)

where

B : Lp(Ω) → Lp(Ω), ψ →

b

aκ(µ, v, v′, ψ(µ, v′))dv′,

is locally Lipschitzian as well as by using the same argument developed in the last part of the proof of Theorem 2.2 we mayprove that there exists R > 0 such that for everyψ ∈ Lp,δ with ∥ψ∥p,δ ≥ R, then [Fp(ψ), ψ]s ≤ 0. Therefore, it seems quitenatural to study Problem (11) under the following conditions

(A′) For every s > 0 there exists L(s) > 0 such that

∥Fp(ϕ)− Fp(ψ)∥p,δ ≤ L(s)∥ϕ − ψ∥p,δ

whenever ϕ,ψ ∈ Bp,δs (0) := u ∈ Lp,δ : ∥u∥p,δ ≤ s.

(A′′) There exists R > 0 such that for every ψ ∈ Lp,δ with ∥ψ∥p,δ ≥ R, then [Fp(ψ), ψ]s ≤ 0.

J. Garcia-Falset et al. / Nonlinear Analysis 97 (2014) 210–227 223

Theorem 4.1. Let p ∈ [1,+∞). If 0 ∈ D(TK ), i.e., K(0) = 0 and the hypotheses (A1), (A′) and (A′′) hold true, thenProblem (11) has a unique mild solution.

Proof. Weclaim that Eq. (11) has a uniquemild solution. Indeed, given g ∈ Lp,δ and consider R > 0 such that both conditions∥g∥p,δ ≤ R and (A′′) hold for this R. We now introduce the function

ρ(x) =

x, ∥x∥p,δ ≤ RR

∥x∥p,δx, ∥x∥p,δ ≥ R.

It follows from condition (A′) and [20, p. 364] that the function Fp(ρ(·)) is 2L(R)-Lipschitz. Therefore, by Theorem 2.1, theequation

u′(t)+ TK (u(t)) = Fp(ρ(u(t))),u(0) = g, (29)

has a unique mild solution u. Since we are assuming that 0 = TK (0) and Fp satisfies (A′′) we may apply [19, Lemma 3.1]to get that u ∈ Bp,δ

R (0) =u ∈ Lp,δ : ∥u∥p,δ ≤ R

. So we have ρ(u) = u which leads to a unique mild solution as we

claimed.

We end this section with the following result:

Proposition 4.1. Let p ∈ [1,+∞). Suppose that (A1), (A′) and (A′′) hold true and, for each i ∈ 1, 2, ui ∈ C([0, T ]; Lp,δ) isa mild solution of Eq. (11) with initial data ui(0). Then, for each ε > 0, there exists δ > 0 such that if ∥u1(0) − u2(0)∥p,δ ≤ δ,then ∥u1 − u2∥∞ ≤ ε.

Proof. Since TK is a δ-m accretive operator, by (32) we have

∥u1(t)− u2(t)∥p,δ ≤ eδt∥u1(0)− u2(0)∥p,δ +

t

0eδ(t−s)

∥Fp(u1(s))− Fp(u2(s))∥p,δds.

The above inequality along with the fact e−δs≤ 1 for all s ∈ [0, T ] implies that

∥u1(t)− u2(t)∥p,δ ≤ eδT

∥u1(0)− u2(0)∥p,δ +

t

0∥Fp(u1(s))− Fp(u2(s))∥p,δds

.

Since, for each i ∈ 1, 2, ui is a continuous function, there exists k > R (R is given in (A′′)) such that ∥ui∥∞ ≤ k. In this case,by (A′), we know that Fp is L(k)-Lipschitzian on Bp,δ

k (0). Consequently, putting c = eδT , we get

∥u1(t)− u2(t)∥p,δ ≤ c∥u1(0)− u2(0)∥p,δ +

t

0cL(k)∥u1(s)− u2(s)∥p,δds.

Now, for every t ∈ [0, T ], Gronwall’s inequality implies that

∥u1(t)− u2(t)∥p,δ ≤ c∥u1(0)− u2(0)∥p,δecL(k)T

which means that

∥u1 − u2∥∞ ≤ cecL(k)T∥u1(0)− u2(0)∥p,δ.

Consequently, taking δ =ε

cecL(k)Twe achieve the proof.

5. Annex

5.1. Preliminaries

Let (X, ∥ · ∥) be a real Banach space. An operator A : D(A) ⊆ X → 2X is said to be accretive if the inequality∥x − y + λ(u − v)∥ ≥ ∥x − y∥ holds for all λ ≥ 0, x, y ∈ D(A) and u ∈ Ax, v ∈ Ay. If, in addition, R(I + λA) (i.e., the rangeof the operator I + λA), is for one, hence for all, λ > 0, precisely X , then A is called m-accretive. Accretive operators wereintroduced by F. E. Browder [21] and T. Kato [22] independently.

Finally, A is said to be quasi-accretive (quasi-m-accretive), if there existsw ∈ R such that A+wI is accretive (respectivelym-accretive), in this case we say also that A isw-accretive (w-m-accretive respectively). Notice that A is accretive if and onlyif A is quasi-accretive withw = 0.

The operators which are quasi-m-accretive play an important role in the study of nonlinear partial differential equations.

224 J. Garcia-Falset et al. / Nonlinear Analysis 97 (2014) 210–227

Consider the Cauchy problemu′(t)+ A(u(t)) ∋ f (t), t ∈ (0, T ),u(0) = x0 ∈ D(A),

(30)

where A is quasi-m-accretive on X and f ∈ L1(0, T , X).Given ϵ > 0. An ϵ-discretization on [0, T ] of the equation u′(t) + A(u(t)) ∋ f (t) consists of a partition 0 = t0 ≤ t1 ≤

t2 ≤ · · · ≤ tN of the interval [0, tN ] and a finite sequence (fi)Ni=1 ⊆ X such that

ti − ti−1 < ϵ for i = 1, . . . ,N, T − ϵ < tN ≤ T ,Ni=1

ti

ti−1

∥f (s)− fi∥ds < ϵ.

A DϵA(0 = t0 ≤ t1 ≤ t2 ≤ · · · ≤ tN; f1, . . . , fN) solution to (30) is a piecewise constant function z : [0, tN ] → X whosevalues zi on (ti−1, ti] satisfy the finite difference equation

zi − zi−1

ti − ti−1+ A(zi) ∋ fi, i = 1, 2, . . . ,N.

Such a function z = (zi)Ni=1 is called an ϵ-approximate solution to the Cauchy problem (30) if it further satisfies

∥z(0)− x0∥ ≤ ϵ.

It iswell known (see [18, Corollary 4.1]) that (30) has a uniquemild solution in the sense that there exists a unique continuousfunction u : [0, T ] → D(A) such that u(0) = x0, and moreover, for each ϵ > 0 there is an ϵ-approximate solution z ofu′

+ A(u) ∋ f on [0, T ] such that ∥u(t)− z(t)∥ ≤ ϵ for all t ∈ [0, T ] with u(0) = x0.If u is the mild solution of Problem (30), then for each (x, y) ∈ A and 0 ≤ s ≤ t ≤ T , we have

∥u(t)− x∥ ≤ ew(t−s)∥u(s)− x∥ +

t

sew(t−τ)[f (τ )− y, u(τ )− x]sdτ . (31)

Here the function [·, ·]s : X × X → R is defined by [y, x]s = supx∗(y) : x∗∈ J1(x), where J1 : X → 2X∗

is the dualitymapping on X , i.e., J1(x) = x∗

∈ X∗: x∗(x) = ∥x∥, ∥x∗

∥ = 1. This means that u is an integral solution to Eq. (30) in thesense of Ph. Bénilan [23] and moreover both concepts of the solution coincide under our context.

If u, v are integral solutions of u′(t) + A(u(t)) ∋ f (t) and v′(t) + A(v(t)) ∋ g(t), respectively, with f , g ∈ L1(0, T , X),then

∥u(t)− v(t)∥ ≤ ewt∥u(0)− v(0)∥ +

t

0ew(t−s)

∥f (s)− g(s)∥ds. (32)

A strong solution of Problem (30) is a function u ∈ W 1,∞(0, T ; X), i.e., u is locally absolutely continuous and differentiablealmost everywhere, and u′(t)+ A(u(t)) ∋ f (t) for almost all t ∈ [0, T ].

Concerning the existence of strong solutions, the following theorem is known (see [18, Theorem 4.5] or [24, p. 108]).

Theorem 5.1. If X is a Banach space with the Radon–Nikodym property, A : D(A) ⊆ X → 2X is a quasi-m-accretive operator,and f ∈ BV (0, T ; X), i.e., f is a function of bounded variation on [0, T ], then Problem (30) has a unique strong solution wheneverx0 ∈ D(A).

Our results rely on the following theorem.

Theorem 5.2 ([18, p. 150]). Let X be a reflexive Banach space and let A be a quasi-m-accretive operator in X. Let F : X → X belocally Lipschitz. Then, for each y0 ∈ D(A), there is a local strong solution to the problem

u′(t)+ Au(t) ∋ F(u(t))u(0) = y0.

Moreover, if ⟨−Fu, w⟩ ≥ −α ∥u∥2+ β, ∀(u, w) ∈ J where

J(u) = w ∈ X∗: ⟨u, w⟩ = ∥u∥2

X , and ∥w∥X∗ = ∥u∥X ,

then, the solution is global.

On the other hand, we say that u ∈ C(0, T ; X) is a weak solution of Problem (30) if there are sequences (un) ⊆ W 1,∞(0,T ; X) and (fn) ⊆ L1(0, T ; X) satisfying the following four conditions:

1. u′n(t)+ Aun(t) ∋ fn(t) for almost all t ∈ [0, T ], n = 1, 2, . . .;

2. limn→∞ ∥un − u∥∞ = 0;

J. Garcia-Falset et al. / Nonlinear Analysis 97 (2014) 210–227 225

3. u(0) = x0;4. limn→∞ ∥fn − f ∥1 = 0.

With respect to the existence of weak solutions, the following result, which is an easy consequence of Theorem 5.1, isimportant.

Theorem 5.3. Let X be a Banach space with the Radon–Nikodym property. Then Problem (30) admits a unique weak solutionwhich is the unique integral solution of this problem.

Wenowrecall some important facts regarding accretive operatorswhichwill be used in our paper (see, for example, [18]).

Proposition 5.1. Let A : D(A) → 2X be an operator on X. The following conditions are equivalent:• A is anw-accretive operator,• the inequality [u − v, x − y]s ≥ −w∥x − y∥, holds for every x, y ∈ D(A) and u ∈ Ax, v ∈ Ay,• For each 0 < λ < 1

wthe resolvent Jλ := (I + λA)−1

: R(I + λA) → D(A) is a single-valued 11−wλ -Lipschitzian mapping.

Let (X, ∥ · ∥X ) be a Banach space. Let ∥ · ∥X(δ) be a family of norms such that for all δ ≥ 0 there exists a constantm(δ) > 0such that

∥ψ∥X(0) = ∥ψ∥X and m(δ)−1∥ψ∥X(δ) ≤ ∥ψ∥X ≤ m(δ)∥ψ∥X(δ), for all ψ ∈ X .

We denote by X∗(δ) the dual space of X(δ) and ⟨·, ·⟩δ the pairing between X(δ) and X∗(δ) and by Fδ the normalized dualitymap defined by

Fδ(0) = f ∈ X∗(δ) : ∥f ∥X∗(δ) = 1

and, for ψ = 0,

Fδ(ψ) = f ∈ X∗(δ) : ⟨ψ, f ⟩δ = ∥ψ∥X(δ), ∥f ∥X∗(δ) = 1.

We close this section by recalling that if X = Lp,δ, 1 < p < +∞, and ψ ∈ X , then

Fδ(ψ) =

∥ψ∥

1−pp,δ |ψ |

p−2 ψ,

and for p = 1 we havesign0ψ

∈ Fδ(ψ),

where

sign0x =

−1 if x < 00 if x = 01 if x > 0.

5.2. Notations and preparatory results

In this section we fix notations and introduce the functional setting of the problem. Let a, b be two real numbers suchthat 0 < a < b < ∞ and for each p ∈ [1,∞) the real number q denotes the conjugate exponent of p, i.e., q =

pp−1 (if p = 1,

then q = +∞). LetΩ be the set

Ω := (µ, v) such that 0 ≤ µ ≤ 1, a ≤ v ≤ b.

We consider for each p ∈ [1,∞) the Banach space Lp(Ω)with its natural norm which is given by

∥ϕ∥p =

Ω

|ϕ(µ, v)|pdµdv 1

p

=

1

0

b

a|ϕ(µ, v)|pdv

dµ 1

p

.

We consider the partial Sobolev space

Wp(Ω) :=

ϕ ∈ Lp(Ω) : v

∂

∂µϕ ∈ Lp(Ω) and vϕ ∈ Lp(Ω)

which is a Banach space with the norm

∥ϕ∥Wp(Ω) = ∥vϕ∥p +

v ∂ϕ∂µp.

Now we define the trace space Lp([a, b], vdv) endowed with the norm

∥ψ∥(∗,p) :=

b

a|ψ(v)|pvdv

1p

.

226 J. Garcia-Falset et al. / Nonlinear Analysis 97 (2014) 210–227

It is well known (see [25, Theorems 1 and 2, pp. 252–253] or [3, Proposition 3.3, p. 379]) that every function ϕ ∈ Wp(Ω)has traces on the space Lp([a, b], vdv). Moreover, the trace mappings γ0 : ϕ −→ ϕ(0, ·) and γ1 : ϕ −→ ϕ(1, ·) are linearcontinuous mappings fromWp(Ω) into Lp([a, b], vdv).

We next introduce the weighted space Lp,δ = Lp(Ω, hδdµdv)with the norm

∥ϕ∥p,δ =

Ω

|(ϕhδ)(µ, v)|pdµdv 1

p

=

1

0

b

a|ϕhδ|p(µ, v)dv

dµ 1

p

where hδ is given by

hδ(µ, v) = e−δ1−µv .

At this point, it has to be noticed that ∥ · ∥p,δ is an equivalent norm to ∥ · ∥p on Lp(Ω).

Lemma 5.1. Let δ ≥ 0 and ψ ∈ Lp. Then,

∥ψ∥1 ≤ eδa (b)

1q ∥ψ∥p,δ, ∥ψ∥p,δ ≤ ∥ψ∥p ≤ e

δa ∥ψ∥p,δ.

Proof. According to Hölder’s inequality we obtain

∥ψ∥1 =

1

0

b

a|ψ(µ, v)|dµdv

=

1

0

b

aeδ(1−µ)v exp

−δ(1 − µ)

v

|ψ(µ, v)|dµdv

≤

1

0

b

aexp

qδ(1 − µ)

v

dµdv

1q

∥ψ∥δ,p

=

b

a

v

δq

exp

qδv

− 1

dv

1q

∥ψ∥δ,p.

Now, the use of the estimate

exp

qδv

− 1 =

∞k=1

1k!

qδv

k

=qδv

∞k=1

1k!

qδv

k−1

≤qδv

exp

qδv

gives the first inequality. The second inequality is trivial.

We also consider the partial Sobolev space

Wp,δ(Ω) :=

ϕ ∈ Lp,δ : v

∂ϕ

∂µ∈ Lp,δ, vϕ ∈ Lp,δ

.

To allow the abstract formulation, we define the following operators depending on the parameter λ

Pλ : Lp([a, b], vdv) −→ Lp([a, b], vdv), u −→ (Pλu)(v) := u(v)e−1λv ;

Qλ : Lp([a, b], vdv) −→ Lp(Ω), u −→ (Qλu)(µ, v) := u(v)e−µλv ;

Πλ : Lp(Ω) −→ Lp([a, b], vdv),

(Πλg)(v) :=1λv

1

0e−

1−µ′

λv g(µ′, v)dµ′;

and Ξλ : Lp(Ω) −→ Lp(Ω),

(Ξλg)(µ, v) :=1λv

µ

0e−

µ−µ′

λv g(µ′, v)dµ′.

One checks readily that, for λ > 0, these operators are bounded, positive and satisfy

∥Pλ∥L(Lp([a,b],vdv)) ≤ e−1λb , (33)

∥Qλ∥L(Lp([a,b],vdv); Lp(Ω)) ≤ λ1p . (34)

J. Garcia-Falset et al. / Nonlinear Analysis 97 (2014) 210–227 227

Moreover, using Hölder’s inequality we obtain

∥Πλg∥p(∗,p) =

b

a

1(λv)p

1

0e−

1−µ′

λv g(µ′, v)dµ′

p vdv≤

b

a

1(λv)p

1

0e−

1−µ′

λv dµ′

p−1 1

0e−

1−µ′

λv |g(µ′, v)|pdµ′

vdv

=

b

a

1λ

1 − e−

1λv

p−1 1

0e−

1−µ′

λv |g(µ′, v)|pdµ′vdv

≤1λ

∥g∥p

which implies

∥Πλ∥L(Lp(Ω); Lp([a,b],dµdv)) ≤ λ−

1p . (35)

Similar calculations lead to

∥Ξλ∥L(Lp(Ω)) ≤

1aλ

1p

. (36)

Acknowledgments

The first author was partially supported by MTM 2009-10696-C02-02. The authors would like to thank the referees formany useful comments and suggestions and for their contribution to a good presentation of this paper.

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