## 1 Introduction

In their recent paper [1], O. Arino and R. Rudnicki considered a model of phytoplankton at the level of aggregates of cells. The aggregates are structured by their size and the phytoplankton system consists of aggregates of all possible sizes. The aggregate size can change due to the usual birth and death of individual cells, but also there are two other mechanisms acting at the level of aggregate: splitting of an aggregate into several parts and combining of two or more aggregates into a bigger one. The latter two are known in physics and chemical engineering as fragmentation–coagulation processes and describe a variety of phenomena ranging from polymerization/polymer degradation, droplets break-up and build-up, through rock crushing and grinding, solid drugs break-up in organisms, to blood cell aggregation and fragmentation. In phytoplankton, the major role in fragmentation and coagulation processes is played by the substance called TEP (Transparent Exopolymer Particles) that is a by-product of the growth of phytoplankton, and its stickiness causes the cells to remain together [2–5]. On the contrary, a low level of concentration of TEP results in fragmentation of the aggregate due to external causes, like currents or turbulence on one hand, and internal unspecified forces of biotic nature on the other.

In [1], the authors considered a relatively simple model of binary fragmentation and coagulation with bounded fragmentation and coagulation rates, as their aim was to investigate the long-time behaviour of the solution, and they succeeded in proving the existence of a time-invariant distribution to which the population of aggregates converges as time tends to infinity, whatever the initial population might be.

Our aim in this paper is to analyze more closely the inter-relation between the growth and fragmentation of aggregates so that we shall disregard the coagulation part. By the very nature of the model, the fragmentation process itself should be conservative, that is, the total amount (mass, the number of particles or cells) of the described quantity, say Q, contained in all the aggregates before and after a fragmentation event should be the same. Thus, if in some system the fragmentation occurs alongside another process of growth or decay determined by a certain law, then the evolution of the total amount of Q should follow this law due to the conservativity of the fragmentation process. If this is the case, then such a process is said to be honest. However, for pure fragmentation models and models combining fragmentation of clusters with their dissolution in the surrounding solute, it has been known for some time [6–9], that if the fragmentation rate of small clusters is large enough, then there appears an unexpected leakage of Q from the system, that is, the amount of Q in the system is strictly smaller than predicted by the laws of nature used to build the model.

In the existing physical literature, op. cit., this unaccounted for loss of Q (in this case, mass-loss), termed shattering fragmentation, is attributed to a phase transition and formation of a ‘dust’ of particles with zero size and non-zero mass (a similar but in some sense opposite process of forming an ‘infinitely large’ particle is known in coagulation as a gelation). For some relatively simple models, shattering fragmentation was analyzed in [4,7] by probabilistic methods. In a series of recent papers [10–14], the shattering and non-shattering fragmentation was fully characterized by the properties of the generator of the semigroup describing the evolution and the theory was applied to a wide range of processes providing a comprehensive classification of fragmentation models.

In particular, in [10], a model where fragmentation occurs together with a continuous mass loss due to dissolving of the substance has been analyzed and conditions ensuring conservativity and shattering have been provided. A crucial rôle in the analysis is played by the theory of substochastic, that is, positivity preserving and contractive semigroups. In this paper, we shall show that the model introduced by Arino and Rudnicki, though obviously not substochastic due to the appearance of the growth term, can be nevertheless transformed into one, and treated by a generalization of the theory developed in [10] yielding similar results, that is, the process is honest for rates of fragmentation bounded at 0, otherwise shattering fragmentation occurs irrespective of the growth rate (within the limits of the model).

It is, however, fair to admit that shattering fragmentation, as related to the creation of infinitesimally small aggregates, is not really a biological (or physical) phenomenon as in the real world there is always a lowest size of objects beyond which we cannot reach without encountering quantum effects. If one adopts such a point of view, then our results can be restated as saying that the models with fragmentation rates that are unbounded at 0 are non-biological.

## 2 The model

Following [1], we consider the following fragmentation model with mass loss:

$${\partial}_{t}u(x\text{,}t)=-{\partial}_{x}[b(x)u(x\text{,}t)]-d(x)u(x\text{,}t)-p(x)u(x\text{,}t)+\underset{x}{\overset{\infty}{\int}}p(y)k(x\text{,}y)u(y\text{,}t)\phantom{\rule{0.2em}{0ex}}\mathrm{d}y$$ | (2.1) |

$$X={L}_{1}({\mathbb{R}}_{+}\text{,}x\phantom{\rule{0.2em}{0ex}}\mathrm{d}x)=\{u\text{;}\phantom{\rule{0.25em}{0ex}}\Vert u\Vert \text{:}=\underset{0}{\overset{\infty}{\int}}|u(x)|x\phantom{\rule{0.2em}{0ex}}\mathrm{d}x<\infty \}$$ |

$$0<b(x)\u2a7d\tilde{b}x\text{,}\phantom{\rule{1em}{0ex}}x>0$$ | (2.2) |

$${b}^{\prime}(0)>0$$ | (2.3) |

The fragmentation is characterized by two functions: p and k. The function p is the fragmentation rate, that is, the number of fragmentation events of aggregates of size x per unit time. We assume that $p\in {L}_{\infty \text{,}\mathrm{loc}}({\mathbb{R}}_{+})$ and $p\u2a7e0$ a.e. Further, k is a non-negative measurable function that describes the distribution of particle masses x spawned by the fragmentation of a particle of mass y. Formal balance of mass in fragmentation requires:

$$\underset{0}{\overset{y}{\int}}xk(x\text{,}y)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x=y$$ | (2.4) |

$$\underset{0}{\overset{y}{\int}}k(x\text{,}y)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x={M}_{y}$$ | (2.5) |

The typical choices for k used in the literature are: the power law $(\nu +2){x}^{\nu}/{y}^{\nu +1}$ with $-2<\nu \u2a7d0$, and its generalization

$$k(x\text{,}y)=\frac{1}{y}h\left(\frac{x}{y}\right)$$ | (2.6) |

Integrating (2.1) multiplied by x, we obtain the formal equation governing the evolution of the total size of the system:

$$\frac{\mathrm{d}}{\mathrm{d}t}\underset{0}{\overset{\infty}{\int}}u(x\text{,}t)x\phantom{\rule{0.2em}{0ex}}\mathrm{d}x=-\underset{0}{\overset{\infty}{\int}}d(x)u(x\text{,}t)x\phantom{\rule{0.2em}{0ex}}\mathrm{d}x+\underset{0}{\overset{\infty}{\int}}b(x)u(x\text{,}t)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x$$ | (2.7) |

## 3 Transport semigroup

In this section we consider the differential part of Eq. (2.1), that is, the Cauchy problem:

$${\partial}_{t}u(x\text{,}t)=-{\partial}_{x}[b(x)u(x\text{,}t)]-d(x)u(x\text{,}t)-p(x)u(x\text{,}t)\text{,}\phantom{\rule{1em}{0ex}}x>0\text{,}\phantom{\rule{0.25em}{0ex}}t>0\phantom{\rule{1em}{0ex}}u(x\text{,}0)=g(x)$$ | (3.1) |

$$\begin{array}{c}{\partial}_{t}u(x\text{,}t)=-{\partial}_{x}[b(x)u(x\text{,}t)]\text{,}\phantom{\rule{1em}{0ex}}x>0\text{,}\phantom{\rule{0.25em}{0ex}}t>0\hfill \\ u(x\text{,}0)=g(x)\hfill \end{array}$$ | (3.2) |

$$[{T}_{B}u](x)=-{(b(x)u(x))}_{x}$$ |

$$D({T}_{B})=\{u\in X\text{;}\phantom{\rule{0.25em}{0ex}}bu\phantom{\rule{0.25em}{0ex}}\text{is a.a.c. and}\phantom{\rule{0.25em}{0ex}}{\left(bu\right)}_{x}\in X\}$$ |

Denoting by B a fixed antiderivative of $1/b$, say, $B(x)={\int}_{1}^{x}\frac{\mathrm{d}s}{b(s)}\text{,}$ we see, due to $0<b(x)<\tilde{b}x$ for $x>0$, that:

$$\underset{x\to \infty}{\mathrm{lim}}B(x)=+\infty \text{,}\phantom{\rule{2em}{0ex}}\underset{x\to 0}{\mathrm{lim}}B(x)=-\infty $$ | (3.3) |

$$[{S}_{{T}_{B}}(t)g(\cdot )](x)=\frac{b(Y(t\text{,}x))g(Y(t\text{,}x))}{b(x)}$$ |

$$\Vert {S}_{{T}_{B}}(t)u\Vert \u2a7d{\mathrm{e}}^{\tilde{b}t}\Vert u\Vert $$ | (3.4) |

$$\Vert R(\lambda \text{,}{T}_{B})g\Vert \u2a7d\frac{1}{\lambda -\tilde{b}}\Vert g\Vert $$ | (3.5) |

Using the above we can prove the following result for the the semigroup solving (3.1).

#### Proposition 3.1

The operator T defined by the formal expression:

$$[Tu](x)=-{(b(x)u(x))}_{x}-a(x)u(x)$$ |

$$D(T)=\{u\in X\text{,}au\in X\text{,}bu\phantom{\rule{0.25em}{0ex}}\text{is a.a.c. and}\phantom{\rule{0.25em}{0ex}}{\left(bu\right)}_{x}\in X\}$$ |

$$\Vert {S}_{T}(t)u\Vert \u2a7d{\mathrm{e}}^{\tilde{b}t}\Vert u\Vert $$ | (3.6) |

#### Proof

Let us consider the resolvent equation of (3.1):

$${(b(x)u(x))}_{x}+a(x)u(x)+\lambda u(x)=f(x)$$ |

$$[R(\lambda )g](x)=\frac{{\mathrm{e}}^{-\lambda B(x)-A(x)}}{b(x)}\underset{0}{\overset{x}{\int}}{\mathrm{e}}^{\lambda B(y)+A(y)}g(y)\phantom{\rule{0.2em}{0ex}}\mathrm{d}y$$ |

$$\Vert R(\lambda )g\Vert \u2a7d\underset{0}{\overset{\infty}{\int}}\left(\frac{{\mathrm{e}}^{-\lambda B(x)-A(x)}}{b(x)}\underset{0}{\overset{x}{\int}}{\mathrm{e}}^{\lambda B(y)+A(y)}|g(y)|\phantom{\rule{0.2em}{0ex}}\mathrm{d}y\right)x\phantom{\rule{0.2em}{0ex}}\mathrm{d}x\u2a7d\frac{1}{\lambda -\tilde{b}}\Vert g\Vert $$ |

$$\frac{a(x)}{b(x)}{\mathrm{e}}^{-\lambda B(x)-A(x)}=-\frac{\lambda}{b(x)}{\mathrm{e}}^{-\lambda B(x)-A(x)}-\frac{\mathrm{d}}{\mathrm{d}x}{\mathrm{e}}^{-\lambda B(x)-A(x)}$$ | (3.7) |

$$\Vert aR(\lambda )g\Vert \u2a7d\underset{0}{\overset{\infty}{\int}}\left(\frac{{\mathrm{e}}^{\lambda B(y)+A(y)}}{y}\underset{y}{\overset{\infty}{\int}}\frac{xa(x){\mathrm{e}}^{-\lambda B(x)-A(x)}}{b(x)}\phantom{\rule{0.2em}{0ex}}\mathrm{d}x\right)|g(y)|y\phantom{\rule{0.2em}{0ex}}\mathrm{d}y\u2a7d\underset{0}{\overset{\infty}{\int}}(1+\frac{{\mathrm{e}}^{\lambda B(y)+A(y)}}{y}\underset{y}{\overset{\infty}{\int}}{\mathrm{e}}^{-\lambda B(x)-A(x)}\phantom{\rule{0.2em}{0ex}}\mathrm{d}x)\times y|g(y)|\phantom{\rule{0.2em}{0ex}}\mathrm{d}y\u2a7d(1+{(\lambda -\tilde{b})}^{\mathrm{-1}})\Vert g\Vert $$ |

Next we observe that for $f\in X$,

$$b(x)u(x)={\mathrm{e}}^{-\lambda B(x)-A(x)}\underset{0}{\overset{x}{\int}}{\mathrm{e}}^{\lambda B(y)+A(y)}f(y)\phantom{\rule{0.2em}{0ex}}\mathrm{d}y$$ |

$$-{(b(x)u(x))}_{x}=(\lambda +a(x))\frac{{\mathrm{e}}^{-\lambda B(x)-A(x)}}{b(x)}\underset{0}{\overset{x}{\int}}{\mathrm{e}}^{\lambda B(y)+A(y)}f(y)\phantom{\rule{0.2em}{0ex}}\mathrm{d}y-f(x)=(\lambda +a(x))u(x)-f(x)\in X$$ |

$${(b(x)u(x))}_{x}+a(x)u(x)+\lambda u(x)=0$$ |

$${\mathrm{e}}^{-\lambda B(x)}={\mathrm{e}}^{-\lambda {\int}_{1}^{x}\frac{\mathrm{d}s}{b(s)}}={\mathrm{e}}^{\lambda {\int}_{x}^{1}\frac{\mathrm{d}s}{b(s)}}\u2a7e{\mathrm{e}}^{-\frac{\lambda}{\tilde{b}}\mathrm{ln}x}={x}^{-\frac{\lambda}{\tilde{b}}}$$ |

$$\Vert {u}_{\lambda}\Vert =\underset{0}{\overset{\infty}{\int}}\frac{{\mathrm{e}}^{-\lambda B(x)-A(x)}}{b(x)}x\phantom{\rule{0.2em}{0ex}}\mathrm{d}x\u2a7ec\underset{0}{\overset{\alpha}{\int}}\frac{{\mathrm{e}}^{-\lambda B(x)}}{b(x)}x\phantom{\rule{0.2em}{0ex}}\mathrm{d}x\u2a7e\frac{c}{\tilde{b}}\underset{0}{\overset{\alpha}{\int}}{x}^{-\frac{\lambda}{\tilde{b}}}\phantom{\rule{0.2em}{0ex}}\mathrm{d}x=\infty $$ | (3.8) |

From this proposition it follows that the operator

$$(\tilde{T}\text{,}D(T))=(T-\tilde{b}I\text{,}D(T))$$ | (3.9) |

$${S}_{\tilde{T}}(t)u={\mathrm{e}}^{-\tilde{b}t}{S}_{T}(t)u$$ | (3.10) |

## 4 Substochastic semigroups

In this section we shall summarize relevant facts from substochastic semigroup theory as developed in [10]. To avoid confusion, we shall use the same notation for the abstract operators as for the particular application discussed in this paper, however the theory is fairly general and requires only that the assumptions (A1)–(A3) be satisfied.

Let $(\Omega \text{,}\mu )$ be a measure space and let $X={L}_{1}(\Omega \text{,}\mu )$. If $Z\subset X$ is a subspace, then ${Z}_{+}$ denotes the cone of nonnegative elements of Z and for $f\in X$ the symbols ${f}_{\pm}$ denote the positive and negative part of f, that is, ${f}_{+}=\mathrm{max}\{f\text{,}0\}$ and ${f}_{-}=-\mathrm{min}\{f\text{,}0\}$. Let $(S{(t))}_{t\u2a7e0}$ be a strongly continuous semigroup on X. We say that $(S{(t))}_{t\u2a7e0}$ is a substochastic semigroup if for any $t\u2a7e0$, $S(t)\u2a7e0$ and $\Vert S(t)\Vert \u2a7d1$, and a stochastic semigroup if additionally $\Vert S(t)f\Vert =\Vert f\Vert $ for $f\in {X}_{+}$.

Accordingly, we consider linear operators in X: $\tilde{T}\subset {T}_{B}+\tilde{A}$ with $D(\tilde{T})\subset D({T}_{B})\cap D(\tilde{A})$, and K, that have the following properties:

- (A1) $(\tilde{T}\text{,}D(\tilde{T}))$ generates a substochastic semigroup $({S}_{\tilde{T}}{(t))}_{t\u2a7e0}$;
- (A2) $D(K)\supset D(\tilde{T})$ and $Ku\u2a7e0$ for $u\in D{\left(\tilde{T}\right)}_{+}$;
- (A3)
for all $u\in D{\left(\tilde{T}\right)}_{+}$
$$\underset{\Omega}{\overset{}{\int}}(\tilde{T}u+Kf)\phantom{\rule{0.2em}{0ex}}\mathrm{d}\mu \u2a7d0$$ (4.1)

#### Theorem 4.1 [11,17]

Under the above assumptions, there exists a smallest substochastic semigroup$({S}_{\tilde{G}}{(t))}_{t\u2a7e0}$generated by an extension$\tilde{G}$of the operator$\tilde{T}+K$. This semigroup, for arbitrary$u\in D(\tilde{G})$and$t>0$, satisfies:

$$\frac{\mathrm{d}}{\mathrm{d}t}{S}_{\tilde{G}}(t)u=\tilde{G}{S}_{\tilde{G}}(t)u$$ | (4.2) |

The generator $\tilde{G}$ of $({S}_{\tilde{G}}{(t))}_{t\u2a7e0}$ is characterized by:

$${(\lambda I-\tilde{G})}^{\mathrm{-1}}f=\sum _{n=0}^{\infty}{(\lambda I-\tilde{T})}^{\mathrm{-1}}{\left[K{(\lambda I-\tilde{T})}^{\mathrm{-1}}\right]}^{n}f\phantom{\rule{1em}{0ex}}f\in X$$ | (4.3) |

Formula (4.3) does not provide any explicit information as to how large an extension of $\tilde{T}+K$ the generator $\tilde{G}$ is and this problem is closely related to the behaviour of $({S}_{\tilde{G}}{(t))}_{t\u2a7e0}$. To make this remark precise, we adapt the concept of honesty and dishonesty from the theory of Markov processes [18].

Firstly, note that (4.1) can be written as:

$$\underset{\Omega}{\overset{}{\int}}(\tilde{T}+K)u\phantom{\rule{0.2em}{0ex}}\mathrm{d}\mu =-c(u)\text{,}\phantom{\rule{1em}{0ex}}u\in D{\left(\tilde{T}\right)}_{+}$$ | (4.4) |

$$c(u)=\underset{\Omega}{\overset{}{\int}}\varsigma (x)u(x)\phantom{\rule{0.2em}{0ex}}\mathrm{d}{\mu}_{x}$$ | (4.5) |

#### Definition 4.1

We say that a substochastic semigroup $({S}_{\tilde{G}}{(t))}_{t\u2a7e0}$ (generated by an extension $\tilde{G}$ of the operator $\tilde{T}+K$) is honest if c is finite on $D(\tilde{G})$, and, for any $0\u2a7d\stackrel{\u25cb}{u}\in D(\tilde{G})$, the solution $u(t)={S}_{\tilde{G}}(t)\stackrel{\u25cb}{u}$ of (4.2) satisfies:

$$\frac{\mathrm{d}}{\mathrm{d}t}\underset{\Omega}{\overset{}{\int}}u(t)\phantom{\rule{0.2em}{0ex}}\mathrm{d}\mu =\frac{\mathrm{d}}{\mathrm{d}t}\Vert u(t)\Vert =-c(u(t))$$ | (4.6) |

#### Remark 4.1

The definition of honesty is not restricted to contractive semigroups and is valid even if c in (4.4) is of undetermined sign. In fact, for the original model (2.1) we shall be using this definition with a positive right-hand side in (4.6). However, for a general c, the existence part of the theory is usually not a trivial matter and this is why we prefer to present a complete theory for substochastic semigroups, and then apply it to a wider class of models that can be transformed to a substochastic case.

It can be proved that the honesty of $({S}_{\tilde{G}}{(t))}_{t\u2a7e0}$, (4.6) is equivalent to its integral version: $({S}_{\tilde{G}}{(t))}_{t\u2a7e0}$ is honest if and only if for any $f\in {X}_{+}$ and $t\u2a7e0$:

$$\Vert {S}_{\tilde{G}}(t)f\Vert =\Vert f\Vert -c(\phantom{\rule{0.2em}{0ex}}\underset{0}{\overset{t}{\int}}{S}_{\tilde{G}}(s)f\phantom{\rule{0.2em}{0ex}}\mathrm{d}s)$$ | (4.7) |

$${\eta}_{f}(t)=\Vert {S}_{\tilde{G}}(t)f\Vert -\Vert f\Vert +\underset{0}{\overset{t}{\int}}c({S}_{\tilde{G}}(s)f)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s$$ | (4.8) |

#### Theorem 4.2

For any fixed$\lambda >0$, there is$0\u2a7d{\beta}_{\lambda}\in {X}^{*}$with$\Vert {\beta}_{\lambda}\Vert \u2a7d1$such that:

$$\lambda \Vert R(\lambda \text{,}\tilde{G})f\Vert =\Vert f\Vert -\langle {\beta}_{\lambda}\text{,}f\rangle -c(R(\lambda \text{,}\tilde{G})f)$$ | (4.9) |

The properties of ${\eta}_{f}$ and its relation to ${\beta}_{\lambda}$ are summarized in the proposition below.

#### Proposition 4.1

The following holds:

- (i) for any $f\in {X}_{+}$ , ${\eta}_{f}$ is a non-positive and a non-increasing function for $t\u2a7e0$ ;
- (ii)

hence$({S}_{\tilde{G}}{(t))}_{t\u2a7e0}$is honest if and only if${\beta}_{\lambda}\equiv 0$for any (some) $\lambda >0$;$$\underset{0}{\overset{\infty}{\int}}{\mathrm{e}}^{-\lambda t}{\eta}_{f}(t)\phantom{\rule{0.2em}{0ex}}\mathrm{d}t=-\frac{1}{\lambda}\langle {\beta}_{\lambda}\text{,}f\rangle $$ - (iii)
if$({S}_{\tilde{G}}{(t))}_{t\u2a7e0}$is dishonest, then for some$f\in {X}_{+}$and any$t>0$:
$$\Vert {S}_{\tilde{G}}(t)f\Vert <\Vert f\Vert +\underset{0}{\overset{t}{\int}}c({S}_{\tilde{G}}(s)f)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s$$

An important characterization of honesty is given in the following theorem.

#### Theorem 4.3

The following are equivalent:

- (a) The semigroup $({S}_{\tilde{G}}{(t))}_{t\u2a7e0}$ is honest;
- (b) $\tilde{G}=\overline{\tilde{T}+K}$ ;
- (c)
For any
$u\in R(\lambda \text{,}\tilde{G}){X}_{+}$
, where
$\lambda >0$
is arbitrary, we have:
$$\underset{\Omega}{\overset{}{\int}}\tilde{G}u\phantom{\rule{0.2em}{0ex}}\mathrm{d}\mu \u2a7e-c(u)$$ (4.10)

The problem with the characterization results given above is that they require the knowledge of the generator itself and therefore they are not immediately useful. To circumvent this problem, we shall be using certain extensions of the involved operators, that are defined below.

Define by $\mathsf{E}$ the set of measurable functions that are defined on Ω and take values in the extended set of real numbers and by ${\mathsf{E}}_{f}$ the subspace of $\mathsf{E}$ consisting of functions that are finite almost everywhere. $\mathsf{E}$ is a vector lattice with respect to the usual relation: ⩽ almost everywhere, $X\subset {\mathsf{E}}_{f}\subset \mathsf{E}$ with X and ${\mathsf{E}}_{f}$ being sublattices of $\mathsf{E}$.

In what follows, we shall denote by $\tilde{\mathcal{T}}\text{,}\mathcal{K}\text{,}\tilde{\mathcal{G}}$ and ${\tilde{\mathcal{L}}}_{\lambda}$ extensions of the operators $\tilde{T}$, K, $\tilde{G}$ and $R(\lambda \text{,}\tilde{T})$, respectively. By $\mathcal{L}$ we abbreviate ${\mathcal{L}}_{1}$. At this moment, we shall require only that all the extensions have domains and ranges in ${\mathsf{E}}_{f}$, that $\mathcal{K}\text{,}\tilde{\mathcal{L}}$ and ${\tilde{\mathcal{L}}}_{\lambda}$ are positive operators on their domains and that $\tilde{\mathcal{G}}\subset \tilde{\mathcal{T}}+\mathcal{K}$.

We shall present here a theorem giving a sufficient condition for dishonesty in terms of these extensions.

#### Theorem 4.4

Assume that there exists $u\in D{\left(\tilde{\mathcal{G}}\right)}_{+}$ such that

- (i) $[{\tilde{\mathcal{L}}}_{\lambda}(\lambda I-\tilde{\mathcal{T}})u](x)=u(x)$ , a.e., for some $\lambda >0$ ,
- (ii) for some $\lambda >0$ , $\lambda u(x)-[\tilde{\mathcal{G}}u](x)=g(x)\in {X}_{+}$ ,
- (iii)
$c(u)$
is finite and
$$\underset{\Omega}{\overset{}{\int}}\tilde{\mathcal{G}}u\phantom{\rule{0.2em}{0ex}}\mathrm{d}\mu <-c(u)$$ (4.11)

## 5 Back to the growth–fragmentation equation

Let us look at the problem (2.1) from the point of view of the developed theory. Let us recall that we consider the operator K defined by the expression:

$$[Ku](x)=\underset{x}{\overset{\infty}{\int}}p(y)k(x\text{,}y)u(y\text{,}t)\phantom{\rule{0.2em}{0ex}}\mathrm{d}y$$ |

$$\underset{0}{\overset{\infty}{\int}}(\tilde{T}u+Ku)x\phantom{\rule{0.2em}{0ex}}\mathrm{d}x=-\underset{0}{\overset{\infty}{\int}}(\tilde{b}x-b(x))u(x)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x-\underset{0}{\overset{\infty}{\int}}d(x)u(x)x\phantom{\rule{0.2em}{0ex}}\mathrm{d}x$$ | (5.1) |

#### Proposition 5.1

There is an extension G of $T+K$ given by $(G\text{,}D(G))=(\tilde{G}+\tilde{b}I\text{,}D(\tilde{G}))$ that generates a positive semigroup $({S}_{G}{(t))}_{t\u2a7e0}=({\mathrm{e}}^{\tilde{b}t}{S}_{\tilde{G}}{(t))}_{t\u2a7e0}$ . Moreover, the generator G is characterized by:

$${(\lambda I-G)}^{\mathrm{-1}}f=\sum _{n=0}^{\infty}{(\lambda I-T)}^{\mathrm{-1}}{\left[K{(\lambda I-T)}^{\mathrm{-1}}\right]}^{n}f$$ | (5.2) |

#### Proof

The operator $\tilde{T}$ was constructed from T by subtracting the bounded operator $\tilde{b}I$. Let us consider the approximating semigroups $({S}_{r}{(t))}_{t\u2a7e0}$, mentioned in Theorem 4.1. They are generated by $(T-\tilde{b}I+rK\text{,}D(T))$, $0<r<1$ and

$$\underset{r\to {1}^{-}}{\mathrm{lim}}{S}_{r}(t)f={S}_{\tilde{G}}(t)f$$ | (5.3) |

Formula (5.2) follows immediately from (4.3) by noting that since $\lambda I-G=(\lambda -\tilde{b})I-\tilde{G}$, we have ${(\lambda I-G)}^{\mathrm{-1}}={({\lambda}^{\prime}I-\tilde{G})}^{\mathrm{-1}}$ for $\lambda >\tilde{b}$ and the same holds for the resolvent of T. □

Formula (5.1) for $T+K$ takes the form:

$$\underset{0}{\overset{\infty}{\int}}(Tu+Ku)x\phantom{\rule{0.2em}{0ex}}\mathrm{d}x=\underset{0}{\overset{\infty}{\int}}b(x)u(x)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x-\underset{0}{\overset{\infty}{\int}}d(x)u(x)x\phantom{\rule{0.2em}{0ex}}\mathrm{d}x=\text{:}C(u)$$ | (5.4) |

$$\frac{\mathrm{d}}{\mathrm{d}t}\underset{0}{\overset{\infty}{\int}}u(t)x\phantom{\rule{0.2em}{0ex}}\mathrm{d}x=\frac{\mathrm{d}}{\mathrm{d}t}\Vert u(t)\Vert =C(u(t))$$ | (5.5) |

$$\underset{0}{\overset{\infty}{\int}}Gux\phantom{\rule{0.2em}{0ex}}\mathrm{d}x\u2a7eC(u)$$ | (5.6) |

To proceed, we have to specify the extensions of the operators which we will be working with. Possibly the most general choice is as follows. For $u\in D(\mathcal{T})\text{:}=\{u\in {L}_{1}([0\text{,}\infty )\text{,}x\phantom{\rule{0.2em}{0ex}}\mathrm{d}x)\text{;}\phantom{\rule{0.25em}{0ex}}bu\in \text{a.a.c.}\}$ we denote:

$$[\mathcal{T}u](x)=-{(b(x)u(x))}_{x}-a(x)u(x)$$ | (5.7) |

$$[\mathcal{K}u](x)=\underset{x}{\overset{\infty}{\int}}p(y)k(x\text{,}y)u(y)\phantom{\rule{0.2em}{0ex}}\mathrm{d}y$$ | (5.8) |

$$[\mathcal{G}u](x)\text{:}=[\mathcal{T}u](x)+[\mathcal{K}u](x)$$ | (5.9) |

$$[{\mathcal{L}}_{\lambda}f](x)\text{:}=\frac{{\mathrm{e}}^{-\lambda B(x)-A(x)}}{b(x)}\underset{0}{\overset{x}{\int}}{\mathrm{e}}^{\lambda B(y)+A(y)}f(y)\phantom{\rule{0.2em}{0ex}}\mathrm{d}y$$ | (5.10) |

We illustrate the usefulness of the concept of extensions in the following observation.

#### Proposition 5.2

Any function$u\in D(G)$is continuous on$(0\text{,}\infty )$.

#### Proof

Let first $f\in {X}_{+}$ and $\lambda >\tilde{b}$. Since $\lambda I-T$ extends to a positive integral operator ${\mathcal{L}}_{\lambda}$ on $\mathsf{E}$, by (5.2) the element ${\overline{u}}_{+}={(\lambda I-G)}^{\mathrm{-1}}f={\mathcal{L}}_{\lambda}g\text{,}$ where $g={\sum}_{n=0}^{\infty}{\left[K{(\lambda I-T)}^{\mathrm{-1}}\right]}^{n}f$, is a well-defined element of $\mathsf{E}$ as the series is increasing. However, as ${\overline{u}}_{+}\in D(G)\subset X$, it must be finite almost everywhere. From (5.10) we have:

$$u(x)=\frac{{\mathrm{e}}^{-\lambda B(x)-A(x)}}{b(x)}\underset{0}{\overset{x}{\int}}{\mathrm{e}}^{\lambda B(y)+A(y)}g(y)\phantom{\rule{0.2em}{0ex}}\mathrm{d}y$$ |

The following technical result can be proved as in [10 (Lemma 4.2)]

#### Lemma 5.1

Let $\mathcal{K}$ and ${\mathcal{L}}_{\lambda}$ be the extensions introduced above. If for some $g\in D{\left(\mathcal{L}\right)}_{+}$ , both g and ${\mathcal{KL}}_{\lambda}g$ belong to ${L}_{1}([\alpha \text{,}N]\text{,}x\phantom{\rule{0.2em}{0ex}}\mathrm{d}x)$ , where $0\u2a7d\alpha <N\u2a7d\infty $ , then:

$$\underset{\alpha}{\overset{N}{\int}}(-g(x)+[{\mathcal{KL}}_{\lambda}g](x)+\lambda [{\mathcal{L}}_{\lambda}g](x))x\phantom{\rule{0.2em}{0ex}}\mathrm{d}x=\alpha b(\alpha )[{\mathcal{L}}_{\lambda}g](\alpha )-\underset{\alpha}{\overset{N}{\int}}p(y)[{\mathcal{L}}_{\lambda}g](y)(\phantom{\rule{0.2em}{0ex}}\underset{0}{\overset{\alpha}{\int}}k(x\text{,}y)x\phantom{\rule{0.2em}{0ex}}\mathrm{d}x)\phantom{\rule{0.2em}{0ex}}\mathrm{d}y-Nb(N)[{\mathcal{L}}_{\lambda}g](N)+\underset{N}{\overset{\infty}{\int}}p(y)[{\mathcal{L}}_{\lambda}g](y)(\phantom{\rule{0.2em}{0ex}}\underset{\alpha}{\overset{N}{\int}}k(x\text{,}y)x\phantom{\rule{0.2em}{0ex}}\mathrm{d}x)\phantom{\rule{0.2em}{0ex}}\mathrm{d}y+\underset{\alpha}{\overset{N}{\int}}b(x)[{\mathcal{L}}_{\lambda}g](x)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x-\underset{\alpha}{\overset{N}{\int}}d(x)[{\mathcal{L}}_{\lambda}g](x)x\phantom{\rule{0.2em}{0ex}}\mathrm{d}x$$ | (5.11) |

A crucial rôle in the following considerations is played by the next theorem.

#### Theorem 5.1

If $u\in D(G)$ , then there are sequences ${\alpha}_{k}\to {0}^{+}$ and ${N}_{k}\to \infty $ as $k\to \infty $ such that:

$$\underset{0}{\overset{\infty}{\int}}[Gu](x)x\phantom{\rule{0.2em}{0ex}}\mathrm{d}x\phantom{\rule{1em}{0ex}}=\underset{k\to \infty}{\mathrm{lim}}(-\underset{{\alpha}_{k}}{\overset{{N}_{k}}{\int}}p(y)u(y)(\phantom{\rule{0.2em}{0ex}}\underset{0}{\overset{{\alpha}_{k}}{\int}}k(x\text{,}y)x\phantom{\rule{0.2em}{0ex}}\mathrm{d}x)\phantom{\rule{0.2em}{0ex}}\mathrm{d}y+\underset{{N}_{k}}{\overset{\infty}{\int}}p(y)u(y)(\phantom{\rule{0.25em}{0ex}}\underset{{\alpha}_{k}}{\overset{{N}_{k}}{\int}}k(x\text{,}y)x\phantom{\rule{0.2em}{0ex}}\mathrm{d}x)\phantom{\rule{0.2em}{0ex}}\mathrm{d}y)+\underset{0}{\overset{\infty}{\int}}b(x)u(x)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x-\underset{0}{\overset{\infty}{\int}}d(x)u(x)x\phantom{\rule{0.2em}{0ex}}\mathrm{d}x$$ | (5.12) |

#### Proof

Using a similar argument to Proposition 5.2 we see that if $g\in {\mathsf{E}}_{+}$ is such that ${\mathcal{L}}_{\lambda}g\in X$, then $g\in {L}_{1}([\alpha \text{,}N]\text{,}x\phantom{\rule{0.2em}{0ex}}\mathrm{d}x)$ for any $0<\alpha <N<\infty $. Following [10 (Corollary 4.1)], we observe that if $u=R(\lambda \text{,}G)f$, $f\in {X}_{+}$ there is $g\in {\mathsf{E}}_{f\text{,}+}$, constructed as in the proof of Proposition 5.2, such that $u={\mathcal{L}}_{\lambda}g$ and:

$$Gu=\lambda {\mathcal{L}}_{\lambda}g-g+{\mathcal{KL}}_{\lambda}g$$ |

$$\underset{0}{\overset{\infty}{\int}}[Gu](x)x\phantom{\rule{0.2em}{0ex}}\mathrm{d}x=\underset{k\to \infty}{\mathrm{lim}}({\alpha}_{k}b({\alpha}_{k})u({\alpha}_{k})-\underset{{\alpha}_{k}}{\overset{{N}_{k}}{\int}}p(y)u(y)\times (\phantom{\rule{0.2em}{0ex}}\underset{0}{\overset{{\alpha}_{k}}{\int}}k(x\text{,}y)x\phantom{\rule{0.2em}{0ex}}\mathrm{d}x)\phantom{\rule{0.2em}{0ex}}\mathrm{d}y-{N}_{k}b({N}_{k})u({N}_{k})+\underset{{N}_{k}}{\overset{\infty}{\int}}p(y)u(y)(\phantom{\rule{0.25em}{0ex}}\underset{{\alpha}_{k}}{\overset{{N}_{k}}{\int}}k(x\text{,}y)x\phantom{\rule{0.2em}{0ex}}\mathrm{d}x)\phantom{\rule{0.2em}{0ex}}\mathrm{d}y)+\underset{0}{\overset{\infty}{\int}}b(x)u(x)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x-\underset{0}{\overset{\infty}{\int}}x\phantom{\rule{0.2em}{0ex}}\mathrm{d}(x)u(x)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x$$ |

Since we know that $u\in {L}_{1}([0\text{,}\infty )\text{,}x\phantom{\rule{0.2em}{0ex}}\mathrm{d}x)\cap C(0\text{,}\infty )$, we have ${\mathrm{lim\hspace{0.17em}inf}}_{x\to \infty}{x}^{2}|u(x)|=0$. Thus, there is a sequence ${\left({N}_{k}\right)}_{k\in \mathbb{N}}$ converging to ∞ such that ${\mathrm{lim}}_{k\to \infty}{N}_{k}^{2}|u({N}_{k})|=0$. Similarly, we obtain a sequence ${\left({\alpha}_{k}\right)}_{k\in \mathbb{N}}$ that converges to 0 as $k\to \infty $, such that ${\mathrm{lim}}_{k\to \infty}{\alpha}_{k}^{2}|u({\alpha}_{k})|=0$. Since $b(x)\u2a7d\tilde{b}x$ for $x>0$, we obtain the thesis. □

#### Theorem 5.2

If

$$\underset{x\to {0}^{+}}{\mathrm{lim}}p(x)+d(x)<+\infty $$ | (5.13) |

#### Proof

As in the previous proof, it is enough to consider $u=R(\lambda \text{,}G)f$, $f\in {X}_{+}\text{,}\lambda >\tilde{b}$; for such f we have also $u={\mathcal{L}}_{\lambda}g$ for some $g\in {\mathsf{E}}_{+}$. Since $u\in X$, by (5.10) and Tonelli's theorem, we obtain:

$$\underset{0}{\overset{\infty}{\int}}({\mathcal{L}}_{\lambda}g)(x)x\phantom{\rule{0.2em}{0ex}}\mathrm{d}x=\underset{0}{\overset{\infty}{\int}}yg(y)\left(\frac{{\mathrm{e}}^{\lambda B(y)+A(y)}}{y}\underset{y}{\overset{\infty}{\int}}\frac{x{\mathrm{e}}^{-\lambda B(x)-A(x)}}{b(x)}\phantom{\rule{0.2em}{0ex}}\mathrm{d}x\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}y=\underset{0}{\overset{\infty}{\int}}yg(y)\psi (y)\phantom{\rule{0.2em}{0ex}}\mathrm{d}y$$ |

$$\underset{y\to {0}^{+}}{\mathrm{lim}}\psi (y)=\underset{y\to {0}^{+}}{\mathrm{lim}}\frac{1}{-\frac{b(y)}{y}+\lambda +p(y)+d(y)}>0$$ |

$$\underset{0}{\overset{\infty}{\int}}[Gu](x)x\phantom{\rule{0.2em}{0ex}}\mathrm{d}x=\underset{N\to +\infty}{\mathrm{lim}}(\phantom{\rule{0.2em}{0ex}}\underset{N}{\overset{\infty}{\int}}p(y)u(y)(\phantom{\rule{0.2em}{0ex}}\underset{0}{\overset{N}{\int}}k(x\text{,}y)x\phantom{\rule{0.2em}{0ex}}\mathrm{d}x)\phantom{\rule{0.2em}{0ex}}\mathrm{d}y)+\underset{0}{\overset{\infty}{\int}}b(x)u(x)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x-\underset{0}{\overset{\infty}{\int}}xd(x)u(x)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x\u2a7eC(u)$$ |

The theorem on dishonesty below is intended primarily as an example so that the regularity assumptions on the coefficients are not optimal. We shall also put $d\equiv 0$ as adding or subtracting a bounded operator does not change the domain of the generator; hence $a\equiv p$. Moreover, we restrict our attention to k given by (2.6): $k(x\text{,}y)={y}^{\mathrm{-1}}h(x/y)$ and satisfying:

$$-\underset{0}{\overset{1}{\int}}zh(z)\mathrm{ln}z\phantom{\rule{0.2em}{0ex}}\mathrm{d}z<+\infty $$ | (5.14) |

#### Theorem 5.3

Assume that $b\in {C}^{1}([0\text{,}\infty ))$ with ${\mathrm{inf}}_{0\u2a7dx<\infty}{b}^{\prime}(x)>-\infty $ ,

$$\frac{1}{xp(x)}\in {L}_{1}([0\text{,}\eta ])\text{,}\phantom{\rule{2em}{0ex}}\frac{1}{{x}^{k}p(x)}\in {L}_{1}([N\text{,}\infty ))$$ | (5.15) |

$$\underset{x\in [0\text{,}\infty )}{\mathrm{sup}}\left|\frac{x{p}^{\prime}(x)}{p(x)}\right|=L<+\infty $$ | (5.16) |

#### Proof

To simplify notation, we put $\eta =1$. We use Theorem 4.4 so that we work with the operator extensions introduced at the beginning of this section and construct $u\in \mathcal{D}{\left(\mathcal{G}\right)}_{+}$ satisfying the assumptions of this theorem. Let us define:

$$u(x)=\{\begin{array}{cc}\frac{1}{{x}^{2}p(x)}\hfill & \text{for}\phantom{\rule{0.25em}{0ex}}0<x<1\hfill \\ \frac{1}{{x}^{2+m}p(x)}\hfill & \text{for}\phantom{\rule{0.25em}{0ex}}x\u2a7e1\hfill \end{array}$$ | (5.17) |

$$\underset{0}{\overset{\infty}{\int}}[\mathcal{G}u](x)x\phantom{\rule{0.2em}{0ex}}\mathrm{d}x\phantom{\rule{1em}{0ex}}=-\underset{k\to \infty}{\mathrm{lim}}\underset{{\alpha}_{k}}{\overset{\infty}{\int}}p(y)u(y)(\phantom{\rule{0.2em}{0ex}}\underset{0}{\overset{{\alpha}_{k}}{\int}}k(x\text{,}y)x\phantom{\rule{0.2em}{0ex}}\mathrm{d}x)\phantom{\rule{0.2em}{0ex}}\mathrm{d}y+\underset{0}{\overset{\infty}{\int}}b(x)u(x)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x$$ | (5.18) |

Consider first the interval $(0\text{,}1]$ where we have $u(x)=1/{x}^{2}a(x)$. Using $k(x\text{,}y)=h(x/y)/y$, we have:

$$\underset{\alpha}{\overset{1}{\int}}(\phantom{\rule{0.2em}{0ex}}\underset{0}{\overset{\alpha}{\int}}k(x\text{,}y)x\phantom{\rule{0.2em}{0ex}}\mathrm{d}x)\frac{1}{{y}^{2}}\phantom{\rule{0.2em}{0ex}}\mathrm{d}y=\underset{\alpha}{\overset{1}{\int}}(\phantom{\rule{0.2em}{0ex}}\underset{0}{\overset{r}{\int}}zh(z)\phantom{\rule{0.2em}{0ex}}\mathrm{d}z)\frac{1}{r}\phantom{\rule{0.2em}{0ex}}\mathrm{d}r$$ |

$$-\underset{\alpha \to {0}^{+}}{\mathrm{lim}}\underset{\alpha}{\overset{1}{\int}}(\phantom{\rule{0.2em}{0ex}}\underset{0}{\overset{\alpha}{\int}}k(x\text{,}y)x\phantom{\rule{0.2em}{0ex}}\mathrm{d}x)a(y)u(y)\phantom{\rule{0.2em}{0ex}}\mathrm{d}y=\underset{0}{\overset{1}{\int}}zh(z)\mathrm{ln}z\phantom{\rule{0.2em}{0ex}}\mathrm{d}z<0$$ |

$$\underset{1}{\overset{\infty}{\int}}p(y)u(y)(\phantom{\rule{0.2em}{0ex}}\underset{0}{\overset{\alpha}{\int}}k(x\text{,}y)x\phantom{\rule{0.2em}{0ex}}\mathrm{d}x)\phantom{\rule{0.2em}{0ex}}\mathrm{d}y$$ |

$$(\lambda u(x)+{(b(x)u(x))}^{\prime})+(p(x)u(x)-\underset{x}{\overset{\infty}{\int}}p(y)h\left(\frac{x}{y}\right){y}^{\mathrm{-1}}u(y)\phantom{\rule{0.2em}{0ex}}\mathrm{d}y)={I}_{1}+{I}_{2}$$ |

$${I}_{1}=\frac{1}{{x}^{2}p(x)}(\lambda +{b}^{\prime}(x)-\frac{2b(x)}{x}-\frac{b(x)}{x}\frac{x{p}^{\prime}(x)}{p(x)})$$ | (5.19) |

$$\frac{1}{{x}^{2}}=\frac{1}{{x}^{2}}\underset{0}{\overset{1}{\int}}zh(z)\phantom{\rule{0.2em}{0ex}}\mathrm{d}z=\underset{x}{\overset{\infty}{\int}}\frac{1}{{y}^{3}}h\left(\frac{x}{y}\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}y\u2a7e\underset{x}{\overset{1}{\int}}\frac{1}{{y}^{3}}h\left(\frac{x}{y}\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}y+\underset{1}{\overset{\infty}{\int}}\frac{1}{{y}^{3+m}}h\left(\frac{x}{y}\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}y$$ |

$$0\u2a7d\frac{1}{{x}^{2}}-\underset{x}{\overset{1}{\int}}\frac{1}{{y}^{3}}h\left(\frac{x}{y}\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}y-\underset{x}{\overset{\infty}{\int}}\frac{1}{{y}^{3+m}}h\left(\frac{x}{y}\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}y={I}_{2}\u2a7d\frac{1}{{x}^{2}}-\underset{x}{\overset{1}{\int}}\frac{1}{{y}^{3}}h\left(\frac{x}{y}\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}y=\frac{1}{{x}^{2}}\underset{0}{\overset{x}{\int}}zh(z)\phantom{\rule{0.2em}{0ex}}\mathrm{d}z$$ |

For $x\in [1\text{,}\infty )$ we have similarly to (5.19)

$${I}_{1}=\frac{1}{{x}^{2+m}p(x)}(\lambda +{b}^{\prime}(x)-\frac{(2+m)b(x)}{x}-\frac{b(x)}{x}\frac{x{p}^{\prime}(x)}{p(x)})$$ |

$${I}_{2}=\frac{1}{{x}^{2+m}}-\underset{x}{\overset{\infty}{\int}}\frac{1}{{y}^{3+m}}h\left(\frac{x}{y}\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}y=\frac{1}{{x}^{2+m}}-\frac{1}{{x}^{2+m}}\underset{0}{\overset{1}{\int}}{z}^{1+m}h(z)\phantom{\rule{0.2em}{0ex}}\mathrm{d}z\u2a7e\frac{1}{{x}^{2+m}}(1-\underset{0}{\overset{1}{\int}}zh(z)\phantom{\rule{0.2em}{0ex}}\mathrm{d}z)=0$$ |

$$0\u2a7d{I}_{2}\u2a7d\frac{1}{{x}^{2+m}}(1-\underset{0}{\overset{1}{\int}}{z}^{1+m}h(z)\phantom{\rule{0.2em}{0ex}}\mathrm{d}z)\in {L}_{1}([1\text{,}\infty )\text{,}x\phantom{\rule{0.2em}{0ex}}\mathrm{d}x)$$ |

$$[{\mathcal{L}}_{\lambda}({\left(bu\right)}^{\prime})](x)=\frac{{\mathrm{e}}^{-\lambda B(x)-A(x)}}{b(x)}\underset{0}{\overset{x}{\int}}{\mathrm{e}}^{\lambda B(y)+A(y)}{(b(y)u(y))}^{\prime}\phantom{\rule{0.2em}{0ex}}\mathrm{d}y=u(x)-\frac{{\mathrm{e}}^{-\lambda B(x)-A(x)}}{b(x)}\underset{y\to {0}^{+}}{\mathrm{lim}}b(y){\mathrm{e}}^{\lambda B(y)+A(y)}u(y)-\frac{{\mathrm{e}}^{-\lambda B(x)-A(x)}}{b(x)}\underset{0}{\overset{x}{\int}}{\mathrm{e}}^{\lambda B(y)+A(y)}\times (\lambda +p(y))u(y)\phantom{\rule{0.2em}{0ex}}\mathrm{d}y$$ |

$$b({x}_{n}){\mathrm{e}}^{\lambda B({x}_{n})+A({x}_{n})}u({x}_{n})\u2a7d\frac{\tilde{b}{x}_{n}{x}_{n}^{\raisebox{1ex}{$\lambda $}\!\left/ \!\raisebox{-1ex}{$\tilde{b}$}\right.}}{{x}_{n}^{2}p({x}_{n})}=\tilde{b}{x}_{n}^{\raisebox{1ex}{$\lambda $}\!\left/ \!\raisebox{-1ex}{$\tilde{b}$}\right.-1}\frac{1}{p({x}_{n})}\to 0$$ |

## Acknowledgment

The paper was prepared while the author visited the Department of Mathematics of the University of Franche-Comté in Besançon, France, as an invited professor. The warm hospitality of Prof. Mustapha Mokhtar-Kharroubi and many stimulating discussions with him are greatly appreciated. The visit was partially supported by the National Research Foundation of South Africa under GUN 2053716.