Journal of Function Spaces

Volume 2014 (2014), Article ID 748792, 10 pages

http://dx.doi.org/10.1155/2014/748792

## Spectral Distribution of Transport Operator Arising in Growing Cell Populations

Department of Mathematics, Shangrao Normal University, Shangrao, Jiangxi 334001, China

Received 27 May 2014; Accepted 12 August 2014; Published 25 August 2014

Academic Editor: Leszek Olszowy

Copyright © 2014 Hongxing Wu et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

Transport equation with partly smooth boundary conditions arising in growing cell populations is studied in space. It is to prove that the transport operator generates a semigroup and the ninth-order remainder term of the Dyson-Phillips expansion of the semigroup is compact, and the spectrum of transport operator consists of only finite isolated eigenvalues with finite algebraic multiplicities in a trip . The main methods rely on theory of linear operators, comparison operators, and resolvent operators approach.

#### 1. Related Knowledge

In this paper, we are concerned with the following transport equation, which was proposed by Rotenberg in [1]: with the initial condition and the general biological rule where is maturity degree of cells, , ; the degree of maturation is then defined in the manner that at the birth and at the death, and their maturation velocity , , describes the number density of cell population as a function of the degree of maturation ; the maturation velocity and the time , denote the total transition cross-section while the function represents the transition rate at which cells change their velocities from to , is initial condition, is linear operator in boundary space and is known as transition rule in biology.

It is well known that the streaming operator generates a strongly continuous semigroup () (see, e.g., [2–5]). If the collision operator is bounded, then the classical perturbation theory (see, [6–8]) shows that the transport operator generates also a strongly continuous semigroup () (see, e.g., [9]) given by the Dyson-Phillips expansion: where and where represents disturbance operators, and the remainder term is given by As is well known, the solution of (1)-(2) exists and is unique for all ; here represents transport operator with boundary condition .

For the past few years, there are many research works on the spectrum analysis of (1), and some of them had been discussed in [6, 10–16]. Jeribi et al. [10] discussed Rotenberg’s model of cell population with general compact boundary conditions and proved that the transport semigroup was irreducible, and a spectral decomposition of the solution into an asymptotic term was derived. Latrach and Megdiche [11] discussed the large time behavior of the solution to the Cauchy problem governed by a transport equation with Maxwell compact boundary conditions arising in growing cell population in spaces and proved that the remainder term was compact and got Wang and Cheng [13] had proved that the transport operator with compact integral boundary conditions generated a semigroup and the existence of the eigenvalues in space in L-R’s model. So, it is natural to set the following question: what happens about the transport operator spectral distribution in Rotenberg’s model when boundary conditions are partly compact? In this paper, we discuss the transport equation with partly smooth boundary conditions arising in growing cell populations in space. It is to prove that the transport operator generates a semigroup and the ninth-order remainder term of the Dyson-Phillips expansion of the semigroup is compact, and the spectrum of transport operator consists of only finite isolated eigenvalues with finite algebraic multiplicities in a trip .

Now, we introduce different notions and notations. Let us first make precisely the functional setting of the problem. Let whose norm is We denote by the following boundary spaces: where In the sequel, and will often be identified with . We define the partial Sobolev space as follows: We define the disturbance operators and streaming operator by So we may define the transport operator by Let be the real defined by Consider now the resolvent equation for operator : where is a given function of . For the solution is formally given by Accordingly, for , we get Now, let us define the transport operators , , , and by Clearly, for satisfying , the operators , , , and are bounded and positive, and it is not difficult to check that Because of (18), we can write Let be a real number defined by Clearly, for , we get Because of (17) and (23) we get Accordingly, for , the resolvent of the operator is given by

In the following, we assume that and satisfy the following:(O_{1}), where is bounded and positive operator and is compact and positive operator;(O_{2}) is regular operator in . So it can be approximated in the uniform operator topology by operators; then
where , , , , and is finite set.

Indeed, let us first observe that if we replace in the definition of the function by the real , we obtain a new streaming operator which we denote by . Arguing as above we can define the operators , , , and , which satisfy, for any ,

Lemma 1. *For any , , then
**
where and denote the nth remainder term of the Dyson-Phillips expansion of the semigroup generated by and .*

*Proof. *For all , a simple calculation shows that
For (25), (27), and (29) we get
Let be a fixed real; by (30), it is obvious that, for all integer and all such that , we have
Consequently,
Next, in view of the exponential formula for strongly continuous semigroups, we have
where and denote the strongly continuous semigroup generated by and . The positivity of and (30) imply that
Now, we define transport operator by
Because
therefore
The exponential formula for strongly continuous semigroups leads to
where and denote the strongly continuous semigroup generated by and . Because of (12),
As a consequence of (12)–(33)–(38)-(39), we have (28) immediately.

Lemma 2 (see [9]). *Let B be the generator of a strongly continuous semigroup on a Banach space , and denote the bounded linear operators in . Assume that there exist and satisfying the following:*(1)* is compact for all such that ;*(2)*for every satisfying , then
* *Then is compact on for each .*

Lemma 3 (see [12]). *Let be a positive measure space and let be bounded linear operators on . If is compact and , then is compact.*

#### 2. Main Result

In this section we are ready to prove the main result of this paper. Let

Theorem 4. *If assumptions O _{1} and O_{2} are satisfied, for , then
*

*uniformly on .*

*Proof. *Now we are going to divide the proof into several steps.*Step **O**ne*. For any , prove that
uniformly on . For all ,
Using the change of variable , for all , we have
Now, we define a bounded linear operator by
For all , clearly, converges in the operator topology to uniformly on as . So, it suffices to show that, for ,
uniformly on . Putting
we get
Let , where
Since and are bounded operators, it suffices to prove that
uniformly on . For all ,
Applying the Hölder inequality we get
Then we have
Putting , , and , we get
so we have
For any ,
and we get
So, it remains to show that
uniformly on . We define now the functions
such that , , , and . So, it suffices to establish that
uniformly on . Because the functions and are simple functions, we fix , and for all , we have
where , , is nonnegative constant and . We get
where
This implies that
Consequently, we have
where
Finally, we get
So we can get (43) immediately. *Step **Two*. For any , for , prove that
uniformly on . Since is positive and compact operator, so it suffices to establish the result for a positive operator of rank one; that is,
where , , and . So for any , , we get
Let
where
Since and are bounded operators,
so it suffices to prove that
uniformly on . For any , , let
we defined a bounded linear operator by
Let and ; then
where and
So, we can assume that
where . Clearly, using the Lebesgue dominated convergence theorem, we get the nets of operators . And for all , it suffices to prove that
uniformly on . Because
we get
And we get
Applying the Hölder inequality we get
And we have
So we can get (69) immediately. *Step Three*. For any , for , prove that
uniformly on . For all , easy calculations show that
Let
where
Because and are bounded operators, it suffices to prove that
uniformly on . In the same principle as (75), we can get (87) immediately.*Step Four*. For any , because of (30) and assumption O_{2}, we have
According to (25)–(27)–(29)-(30)–(43), it suffices to establish that, for , prove that
uniformly on . Note that and do not commute, so we have
where each is the product of n factors involving both and except the term
So, for , the operator appears at least one time in the expression of . So, there exists , such that
where is uniformly bounded on . Now using the inequality
and (53), for , , we get
uniformly on . On the other hand, for , we have
Using (69) again, it remains to prove that
uniformly on . This follows from (87). This ends the proof.

Theorem 5. *If assumptions O _{1} and O_{2} are satisfied and , then the spectrum of transport operator consists of only finite isolated eigenvalues which have finite algebraic multiplicities in trip ; furthermore a real exists such that , where denotes point spectrum set of transport operator .*

*Proof. *
Consider the following.*Step One*. Prove that the spectrum of the transport operator consists of, at most, finite isolated eigenvalues which have finite algebraic multiplicities in trip . By Lemmas 1–3, it suffices to prove that is compact operator on .

On one hand, because of hypothesis , we have and is compact operator on . Let
Since is bounded operator and
we get that is compact operator on .

On the other hand, because of
thanks to (29), and for , we have
So for , together with (102), (103), and (104) we have
By Theorem 4, we get
uniformly on . Now, applying Lemma 3 we conclude that, for each , is compact operator on , together with [3]; we end Step one. Now we consider again the resolvent equation which is equivalent to solving in the following one:
If , then the operator is bounded invertible and (104) becomes
where
*Step Two*. Prove that exists for in the half plane
with sufficiently large.

Let be an element of
and set
where and are the dual operators of and . Easy calculations show that the dual of the operator is . Let
be a sequence in such that
We consider ; then (112) show that
By the Riemann-Lebesgue theorem, we obtain
a.e. on . Furthermore, for every integer , we have
By the dominated convergence theorem, we get
for all .

Prove that the family , , is collectively compact. Let be the unit ball of the space and let be a sequence in ; then there exists a sequence in such that
Since is compact and is bounded in the space , there exists a converging subsequence of in . This is the wanted.

The use of and together with Anselone’s Proposition [17, Proposition 3.1] gives
uniformly on . Then
uniformly on . Since
therefore, we obtain
uniformly on . This ends Step two by . *Step **Three*. We will prove that .

First, because of
there exists for and we get
Hence
Because is compact on and
we get that