Uniform convergence of solutions to stochastic hybrid models of gene regulatory networks

Abstract

In a recent paper by Kurasov et al. (Math Biosci 305:170–177, 2018), a hybrid gene regulatory network was proposed to model gene expression dynamics by using a stochastic system of coupled partial differential equations. In more recent work, the existence and strong convergence of the solutions to equilibrium were proven. In this article, we improve upon their result by showing that the convergence rate is independent of the initial state, therefore proving that the solutions converge not only strongly but even uniformly to equilibrium. To this end, we make use of a recent convergence theorem for stochastic, irreducible semigroups that contain partial integral operators.

1 Introduction

Let $a, b, c, d \in \mathbb {R}$ such that $b, d > 0$ and $\tfrac{a}{b} < \tfrac{c}{d}$. Set $I :=\left[ \frac{a}{b}, \frac{c}{d} \right]$ and let $\nu , \mu : I \rightarrow (0, \infty )$ be continuous. Given an initial state $u_0 \in L^1(I; \mathbb {C}^2)$, we are interested in the long-term behaviour of the solutions to the system

$$\begin{aligned} \begin{aligned} \left\{ \begin{aligned} \frac{\text {d}}{\text {d}t} u(t,x)&= - \frac{\text {d}}{\text {d}x} \begin{pmatrix} (a - b x) u_1(t,x) \\ (c - d x) u_2(t,x) \end{pmatrix} + \begin{pmatrix} -\nu (x) &{} \mu (x) \\ \nu (x) &{} -\mu (x) \end{pmatrix} \begin{pmatrix} u_1(t, x) \\ u_2(t, x) \end{pmatrix}, \\ u(0, x)&= u_0(x), \end{aligned} \right. \end{aligned}\end{aligned}$$

(1.1)

subject to the boundary conditions

$$\begin{aligned} u_1 \bigl (t, \tfrac{c}{d} \bigr ) = 0 \quad \text {and} \quad u_2 \bigl (t, \tfrac{a}{b} \bigr ) = 0 \qquad \text {for all } x \in I \text { and } t \ge 0. \end{aligned}$$

In [17], the authors Kurasov, Lück, Mugnolo and Wolf described a hybrid gene regulatory network model aiming to model the evolution of protein counts in cells on a continuous state space by a piecewise deterministic Markov processes, i.e. a system of coupled partial differential equations. Classically, the state space of discrete-stochastic models representing gene regulatory networks is typically high-dimensional due to the combinatorial nature of molecule counts (see [31] for a survey). However, under additional assumptions, exact solutions of the discrete state-space models can be found (see, e.g., [16, 19]). Due to their high-dimensional nature, these models are difficult to solve numerically. Hybrid models of gene regulatory networks emerged as a compromise between the accuracy of deterministic models and the computational efficiency of stochastic models (see [27]). The model in (1.1) follows the same approach and models gene expressions in single cells on a continuous state space. In [18], exact solutions of (1.1) for special choices of parameters are given and proven to converge individually to equilibrium.

Following [17, 18], we discuss the long-term behaviour of solutions of (1.1) by using classical semigroup methods. In [18, Proposition 4.3], it was proven that the problem is well posed and that its solution is given by a stochastic, irreducible semigroup when modelled on a suitable $L^1$-space. Moreover, in [18, Theorem 4.6] the authors prove the existence of equilibrium states and that the solutions to (1.1) converge strongly to equilibrium.

In this paper, we strengthen [18, Theorem 4.6] to operator norm rather than strong convergence. To be precise, we prove the following result.

Theorem 1.1

The solution semigroup $(S(t))_{t \ge 0}$ of (1.1) converges to a rank-1-projection onto its fixed space with respect to the operator norm as $t \rightarrow \infty$.

In other words, the convergence of the solutions to the equilibrium cannot happen arbitrarily slow, while varying over all the initial conditions of the system. A few comments on the above result are in order:

• (a) The first-order differential operator occurring in (1.1) essentially acts as a shift; in particular, it exhibits no smoothing on the solutions unlike diffusion terms. Hereby, the entire mass in the different coordinates is shifted in two opposing directions and concentrated at the opposing ends of the interval I. Hence, one cannot expect the solutions to converge without the action of the potential term.

• (b) The potential term is a multiplication operator and, therefore, has also no particular smoothing properties. However, the potential term causes a mixing of the coordinates resulting in these shifts to counteract each other, ensuring the convergence of solutions. This idea was the gist in [18].

1.1 Methodology and related literature

There exists a substantial number of results in the literature on the long-term behaviour of semigroups on $L^p$-spaces that show strong convergence to an equilibrium as $t \rightarrow \infty$ under the assumption that the semigroup contains an integral operator (see Section 3 for a definition). These results have a long history, going back to [15, Korollar 3.11]. Subsequent works provide similar results (see [3, 7, 9,10,11] for examples). For a brief survey of these results, we refer to [4, Section 3]. Furthermore, in [12] results of this type are obtained under the condition that the semigroup contains an operator that dominates a nonzero integral operator. In [13], a generalization of many of the aforementioned results is presented in a more abstract setting.

Furthermore, in [22,23,24,25, 28] and [26, Section 5] an asymptotic theory of substochastic $C_0$-semigroups that dominate an integral operator on an $L^1$-space is developed. These results all yield the strong convergence of semigroups to equilibrium and have a history of applications to models from mathematical biology, kinetic equations and queueing theory. In the proof of Theorem 1.1, we employ a more recent theorem from [14] that yields the stronger operator norm convergence of stochastic semigroups (see Theorem 3.1).

1.2 Organization of the article

In Section 2, we study the first-order differential operators that appear in (1.1). Moreover, the proof of Theorem 1.1 is prepared by simplifying (1.1) via a state-space transform. This shortens the calculations in Section 3.

In Section 3, the main result is proved. This is achieved by showing the existence of a lower bound for the dual resolvent of the generator of the solution semigroup $(T(t))_{t \ge 0}$ and by proving that the solution semigroup contains a partial integral operator. This allows us to apply the aforementioned result from [14].

1.3 Notation and terminology

Let E be a Banach space. Then, we denote the space of bounded linear operators on E by $\mathcal {L}(E)$. Furthermore, if A is a closed operator on E, then we denote by $\sigma (A)$ the spectrum and by $\rho (A)$ the resolvent set of A. Let $(\Omega , \Sigma , \mu )$ be a measure space. For an element $f \in L^\infty (\Omega )$, we use the notation $f \gg 0$ to say that f is bounded away from 0, i.e. that there exists some $c > 0$ such that $f \ge c \cdot {{\,\mathrm{\mathbbm {1}}\,}}$, where ${{\,\mathrm{\mathbbm {1}}\,}}:\Omega \rightarrow \mathbb {C}, \, x \mapsto 1$. Here, the inequality $\ge$ between elements in $L^\infty (\Omega )$ is to be understood pointwise for almost all points in $\Omega$.

Furthermore, a $C_0$-semigroup $(T(t))_{t \ge 0}$ on $L^1(\Omega )$ is called positive if $T(t) f \ge 0$ for all $0 \le f \in L^1(\Omega )$. A positive $C_0$-semigroup $(T(t))_{t \ge 0}$ on $L^1(\Omega )$ is called stochastic if

$$\begin{aligned} \Vert T(t) f\Vert = \Vert f\Vert \qquad \text {for all } 0 \le f \in L^1(\Omega ). \end{aligned}$$

A $\mathbb {C}^n$-valued function f is said to satisfy the above conditions if it satisfies them componentwise.

Throughout the paper, we assume knowledge of classical semigroup theory and the basics of the theory of positive operators on $L^1$-spaces. For an introduction to these topics, we refer to [5, 8, 21, 29] and [1, 2, 20, 30, 32], respectively.

2 Setting the stage

Let a, b, c, d as well as the interval I and the functions $\nu$ and $\mu$ be defined as in the introduction and let $\mathring{I} :=(\frac{a}{b}, \frac{c}{d})$. Consider the operator $\mathcal {A} : D(\mathcal {A}) \rightarrow L^1(\Omega ; \mathbb {C}^2)$ defined by

$$\begin{aligned} \begin{aligned} \mathcal {A}\begin{pmatrix} f_1 \\ f_2 \end{pmatrix} (x) :=-\frac{\text {d}}{\text {d}x} \begin{pmatrix} a - b x &{} 0 \\ 0 &{} c - d x \end{pmatrix} \begin{pmatrix} f_1 \\ f_2 \end{pmatrix}(x), \end{aligned}\end{aligned}$$

(2.1)

where

$$\begin{aligned} \begin{aligned} D(\mathcal {A}) :=\left\{ f \in L^1(I; \mathbb {C}^2): \frac{\text {d}}{\text {d}x} \begin{pmatrix} (a - b x) f_1\\ (c - d x) f_2 \end{pmatrix} \in L^1(\mathring{I}; \mathbb {C}^2), \, f_1(\tfrac{c}{d}) = 0, \, f_2 (\tfrac{a}{b}) = 0 \right\} . \end{aligned}\end{aligned}$$

Here, $L^1(I; \mathbb {C}^2)$ is equipped with the norm

$$\begin{aligned} \begin{aligned} \Vert f\Vert _1 :=\Vert f_1\Vert _1 + \Vert f_2\Vert _1 \qquad \text{ for } \text{ all } f \in L^1(I; \mathbb {C}^2). \end{aligned} \end{aligned}$$

This norm renders $L^1(I; \mathbb {C}^2)$ isometrically isomorphic to the $L^1$-space of scalar-valued functions over two disjoint copies of I. In particular, one can treat $L^1(I; \mathbb {C}^2)$ as a scalar-valued $L^1$-space.

Consider the bounded operator $\mathcal {B}:L^1(I; \mathbb {C}^2) \rightarrow L^1(I; \mathbb {C}^2)$ given by

$$\begin{aligned} \begin{aligned} \mathcal {B}\begin{pmatrix} f_1 \\ f_2 \end{pmatrix}(x) :=\begin{pmatrix} -\nu (x) &{} \mu (x) \\ \nu (x) &{} -\mu (x) \end{pmatrix} \begin{pmatrix} f_1 (x) \\ f_2 (x) \end{pmatrix}. \end{aligned} \end{aligned}$$

Using the operators $\mathcal {A}$ and $\mathcal {B}$, we can restate (1.1) as the abstract Cauchy problem

$$\begin{aligned} u'(t) = (\mathcal {A}+ \mathcal {B}) u(t), \quad u(0) = u_0 \end{aligned}$$

(ACP)

on the state space $L^1(I; \mathbb {C}^2)$, where $D(\mathcal {A}+ \mathcal {B}) :=D(\mathcal {A})$.

Via an affine linear state-space transform, we map the interval $I = [\frac{a}{b}, \frac{c}{d}]$ to the interval [0, 1]. The transformed system is then of the form

$$\begin{aligned} \begin{aligned} \left\{ \begin{aligned} \frac{\text {d}}{\text {d}t} u(t,x)&= \frac{\text {d}}{\text {d}x} \begin{pmatrix} b x u_1(t,x) \\ d (x - 1) u_2(t,x) \end{pmatrix} + \begin{pmatrix} -{\tilde{\nu }}(x) &{} {\tilde{\mu }}(x) \\ {\tilde{\nu }}(x) &{} -{\tilde{\mu }}(x) \end{pmatrix} \begin{pmatrix} u_1(t, x) \\ u_2(t, x) \end{pmatrix}, \\ u(0, x)&= u_0(x), \end{aligned} \right. \end{aligned} \end{aligned}$$

(2.2)

subject to the boundary conditions $u_1(t,1) = 0$ and $u_2(t, 0) = 0$. Here, ${\tilde{\nu }}$ and $\tilde{\mu }$ are positive and continuous functions on the interval [0, 1]. It follows that we may choose $a = 0$ and $c = d$ without losing the generality offered by the original system (1.1).

In the simplified form (2.2), two differential operators appear. Namely, the operators

$$\begin{aligned} \begin{aligned} f \mapsto \frac{\text {d}}{\text {d}x}(x f) \quad \text{ and } \quad f \mapsto \frac{\text {d}}{\text {d}x} \big ((1 -x) f \big ). \end{aligned} \end{aligned}$$

We commence by studying these operators. It turns out that it is easiest to consider both operators on a domain in $L^1([0, \infty ))$ and $L^1((- \infty , 0])$, respectively. We later restrict the operators to $L^1([0,1])$. To this end, we define

$$\begin{aligned} \begin{aligned} A_1 f&:=\frac{\text {d}}{\text {d}x}(x f),{} & {} D(A_1) :=\lbrace f \in L^1([0, \infty )) : (x f)' \in L^1((0, \infty ))\rbrace , \\ A_2 f&:=\frac{\text {d}}{\text {d}x}((x - 1) f),{} & {} D(A_2) :=\lbrace f \in L^1((-\infty , 1]) : ((x - 1) f)' \in L^1(({-}\infty , 1))\rbrace , \end{aligned}\end{aligned}$$

(2.3)

where the derivatives have to be understood in the distributional sense.

These operators are closely related in the sense that both operators are similar via the reflection operator

$$\begin{aligned} Q :L^1([0, \infty )) \rightarrow L^1((-\infty , 1]), \quad Qf(x) :=f(1 - x). \end{aligned}$$

More precisely, one has the following lemma. We omit its proof as it is straightforward.

Lemma 2.1

• (i) The reflection operator Q is an isometric isomorphism with

  $$\begin{aligned} Q^{-1} :L^1((-\infty , 1]) \rightarrow L^1([0, \infty )), \quad Q^{-1}f(x) :=f(1 - x). \end{aligned}$$

  In particular, both Q and $Q^{-1}$ are positive.

• (ii) The operators $A_1$ and $A_2$ are similar via Q, i.e. we have that $D(A_2) = Q D(A_1)$ and

  $$\begin{aligned} A_2 = Q A_1 Q^{-1}. \end{aligned}$$

Next, we prove that $A_1$ and $A_2$ generate stochastic semigroups on the spaces $L^1([0, \infty ))$ and $L^1((-\infty , 1])$, respectively. For that purpose and for later reference, we give explicit identities for the resolvents of both operators.

Lemma 2.2

The following assertions hold:

• (i) We have $(0, \infty ) \subseteq \rho (A_1)$. For all $\lambda > 0$ and $g \in L^1([0, \infty ))$, the identity

  $$\begin{aligned} R(\lambda , A_1) f(x) = x^{\lambda - 1} \int _x^\infty \frac{f(y)}{y^\lambda } \, \textrm{d}y \end{aligned}$$

  holds for almost all $x \in [0, \infty )$.

• (ii) We have $(0, \infty ) \subseteq \rho (A_2)$. For all $\lambda > 0$ and $g \in L^\infty ((-\infty , 1])$, the identity

  $$\begin{aligned} R(\lambda , A_2) f(x) = (1 - x)^{\lambda - 1} \int _{-\infty }^x \frac{f(y)}{(1 - y)^\lambda } \, \textrm{d}y \end{aligned}$$

  holds for almost all $x \in (-\infty ,1]$.

Proof

(i): For $\lambda > 0$, we define the operator

$$\begin{aligned} \begin{aligned} R_\lambda :L^1 ([0, \infty )) \rightarrow L^1 ([0, \infty )), \quad (R_\lambda f)(x) :=x^{\lambda -1} \int _x^\infty \frac{f(y)}{y^{\lambda }} \, \text {d}y. \end{aligned} \end{aligned}$$

We show that $R_\lambda$ is well defined, i.e. that $R_\lambda f \in L^1([0, \infty ))$ for all $f \in L^1([0, \infty ))$. Fix $f \in L^1([0, \infty ))$. Clearly, $R_\lambda f$ is measurable for each $\lambda > 0$ and by Fubini’s theorem one has

$$\begin{aligned} \Vert R_\lambda f\Vert _1&= \int _0^\infty |R_\lambda f(x)| \, \textrm{d}x \le \int _0^\infty \int _x^\infty x^{\lambda - 1} \frac{|f(y)|}{y^\lambda } \, \textrm{d}y \, \textrm{d}x \\&= \int _0^\infty \frac{|f(y)|}{y^\lambda } \int _0^y x^{\lambda - 1} \, \textrm{d}x \, \textrm{d}y = \int _0^\infty \frac{|f(y)|}{\lambda } \, \textrm{d}y = \frac{1}{\lambda } \Vert f\Vert _1 \end{aligned}$$

and thus $R_\lambda f \in L^1([0, \infty ))$. Furthermore, an elementary calculation shows that for all $\varphi \in \textrm{C}_{\textrm{c}}^\infty ((0, \infty ))$, we have

$$\begin{aligned} \begin{aligned} \langle \varphi ', x \cdot R_\lambda f \rangle = - \langle \varphi , \lambda R_\lambda f - f \rangle , \end{aligned} \end{aligned}$$

which implies that

$$\begin{aligned} \begin{aligned} \frac{\text {d}}{\text {d}x}(x \cdot R_\lambda f) = \lambda R_\lambda f - f \in L^1((0, \infty )), \end{aligned} \end{aligned}$$

and thus, $R_\lambda f \in D(A_1)$ for all $\lambda > 0$. Furthermore, another standard calculation shows that

$$\begin{aligned} \begin{aligned} (\lambda - A_1) R_\lambda f = R_\lambda (\lambda - A_1) f = f \qquad \text{ for } \text{ all } \ f \in D(A_1). \end{aligned} \end{aligned}$$

Hence, the claim follows.

(ii): Since $A_1$ and $A_2$ are similar by Lemma 2.1, it follows that $\sigma (A_1) = \sigma (A_2)$, and thus $(0, \infty ) \subseteq \rho (A_2)$. Also by similarity, $R(\lambda , A_2) = Q R(\lambda , A_1) Q^{-1}$ (see, e.g. [8, Section II.2.1]) and thus

$$\begin{aligned} R(\lambda , A_2) f(x) = (1 - x)^{\lambda - 1} \int _{1 - x}^\infty \frac{f(1 - y)}{y^\lambda } \, \textrm{d}y = (1 - x)^{\lambda - 1} \int _{-\infty }^x \frac{f(y)}{(1 - y)^\lambda } \, \textrm{d}y \end{aligned}$$

for all $\lambda > 0$ and $f \in L^1([0, \infty ))$. $\square$

Now we are in a position to show that $A_1$ and $A_2$ generate stochastic semigroups on the spaces $L^1([0, \infty ))$ and $L^1((-\infty , 1])$, respectively.

Proposition 2.3

The following assertions hold:

• (i) The operator $A_1$ is the generator of a stochastic $C_0$-semigroup $({\widetilde{T}}_1(t))_{t \ge 0}$ on $L^1([0, \infty ))$, given by

  $$\begin{aligned} {\widetilde{T}}_1(t) f(x) = \textrm{e}^t f(x \textrm{e}^t) \end{aligned}$$

  for all $f \in L^1([0, \infty ))$.

• (ii) The operator $A_2$ is the generator of a stochastic $C_0$-semigroup $({\widetilde{T}}_2(t))_{t \ge 0}$ on $L^1((-\infty , 1])$, given by

  $$\begin{aligned} {\widetilde{T}}_2(t) f(x) = \textrm{e}^t f(1 - (1 - x) \textrm{e}^t) \end{aligned}$$

  for all $f \in L^1((-\infty , 1])$.

Proof

(i): It is elementary to show that $({\widetilde{T}}_1(t))_{t \ge 0}$ satisfies the semigroup law and that it is strongly continuous. So, it is just left to show that $A_1$ is indeed the generator of $({\widetilde{T}}_1(t))_{t \ge 0}$. To this end, let us denote the generator of $({\widetilde{T}}_1(t))_{t \ge 0}$ by A. Let $f \in D(A_1)$. Then, ${\widetilde{T}}_1(t) f \in D(A_1)$ and the classical Sobolev embedding implies that $T_1(t) f$ is continuous. In particular, we have

$$\begin{aligned} R(\lambda , A) f(x)&= \int _0^\infty \textrm{e}^{-\lambda t} {\widetilde{T}}_1(t) f(x) \, \textrm{d}t = \int _0^{\infty } \textrm{e}^{-\lambda t} \textrm{e}^t f(x \textrm{e}^t) \, \textrm{d}t \\&= x^{\lambda - 1} \int _x^\infty \frac{f(y)}{y^\lambda } \, \textrm{d}y = R(\lambda , A_1) f(x) \end{aligned}$$

by substitution for all $\lambda > 0$ and $x \in [0, \infty )$. Since $R(\lambda , A)$ and $R(\lambda , A_1)$ coincide on D(A) and D(A) is dense in $L^1([0, \infty ))$, it follows that $A = A_1$.

Finally, by using the same substitution as above, one obtains

$$\begin{aligned} \Vert {\widetilde{T}}_1(t) f\Vert _1 = \Vert f\Vert _1 \qquad \text {for all } 0 \le f \in D(A_1), \end{aligned}$$

which implies by density of $D(A_1)$ that $({\widetilde{T}}_1(t))_{t \ge 0}$ is a stochastic semigroup.

(ii): The claim that $A_2$ generates a stochastic semigroup follows immediately from Lemma 2.1. Furthermore, Lemma 2.1 implies

$$\begin{aligned} {\widetilde{T}}_2(t) f(x) = Q {\widetilde{T}}_1(t) Q^{-1} f(x) = \textrm{e}^t f(1 - (1 - x) \textrm{e}^t) \end{aligned}$$

for all $f \in L^1((-\infty , 1])$. $\square$

From the explicit formulas of the resolvents of $A_1$ and $A_2$ in Lemma 2.2, we now derive formulas for the resolvents of the dual operators.

Corollary 2.4

The following assertions hold:

• (i) We have $(0, \infty ) \subseteq \rho (A_1')$ and for all $\lambda > 0$ and all $g \in L^\infty ([0, \infty ))$ that

  $$\begin{aligned} R(\lambda , A_1') g(y) = \frac{1}{y^\lambda } \int _0^y g(x) x^{\lambda - 1} \, \textrm{d}x \end{aligned}$$

  for almost all $y \in [0,\infty )$.

• (ii) We have $(0, \infty ) \subseteq \rho (A_2')$ and for all $\lambda > 0$ and all $g \in L^\infty ((-\infty , 1])$ that

  $$\begin{aligned} R(\lambda , A_2') g(y) = \frac{1}{(1 - y)^\lambda } \int _y^1 g(x) (1 - x)^{\lambda - 1} \, \textrm{d}x \end{aligned}$$

  for almost all $y \in (-\infty ,1]$.

Proof

(i): For $\lambda > 0$, we define the operator

$$\begin{aligned} \begin{aligned} R_\lambda ' :L^\infty ([0, \infty )) \rightarrow L^\infty ([0, \infty )), \quad (R_\lambda ' f)(y) :=\frac{1}{y^\lambda } \int _0^y f(x) x^{\lambda - 1} \, \text {d}x. \end{aligned} \end{aligned}$$

Then, $R_\lambda '$ is well defined with $\Vert R_\lambda '\Vert = \frac{1}{\lambda }$. Now let $f \in L^1([0, \infty ))$ and $g \in L^\infty ([0, \infty ))$. Then, Fubini’s theorem yields

$$\begin{aligned} \langle R(\lambda , A_1) f, g \rangle&= \int _0^\infty \int _x^\infty x^{\lambda - 1} \frac{f(y)}{y^\lambda } g(x) \, \textrm{d}y \, \textrm{d}x \\&= \int _0^\infty f(y) \frac{1}{y^\lambda } \int _0^y g(x) x^{\lambda - 1} \, \textrm{d}x \, \textrm{d}y = \langle f, R_\lambda ' g \rangle \end{aligned}$$

and, thus, $R(\lambda , A_1') = R(\lambda , A_1)' = R_\lambda '$ for all $\lambda > 0$.

(ii): Since $A_1$ and $A_2$ are similar via Q, we have $R(\lambda , A_2') = Q'^{-1} R(\lambda , A_1') Q'$ for all $\lambda > 0$, where

$$\begin{aligned} \begin{aligned} Q' :L^\infty ((-\infty , 1]) \rightarrow L^\infty ([0, \infty )), \quad Q'g(x) :=g(1 - x) \end{aligned} \end{aligned}$$

denotes the dual of the reflection operator Q. Thus, a simple calculation yields the claim. $\square$

Since Theorem 1.1 is concerned with semigroups on the space $L^1([0, 1])$, one needs to restrict the semigroups ${\tilde{T}}_1$ and ${\tilde{T}}_2$ generated by $A_1$ and $A_2$, respectively, to this smaller space in a meaningful way. This is what is done next.

In order to study these restrictions of the semigroups to $L^1([0, 1])$, we consider the following obvious candidates for their generators. We define the operators $C_1$ and $C_2$ by

$$\begin{aligned} \begin{aligned} C_1&:=\frac{\text {d}}{\text {d}x}(x f), \ \ {}{}{} & {} D(C_1) :=\lbrace f \in L^1([0, 1]) : (x f)' \in L^1((0, 1)), \, f(1) = 0\rbrace , \\ C_2&:=\frac{\text {d}}{\text {d}x}((x-1) f),{} & {} D(C_2) :=\lbrace f \in L^1([0, 1]) : ((x - 1) f)' \in L^1((0, 1)), \, f(0) = 0\rbrace . \end{aligned} \end{aligned}$$

(2.4)

The following proposition collects some identities related to the operators $C_1$ and $C_2$. These are used later on to obtain lower bounds for the resolvent of the generator of the semigroup $(T(t))_{t \ge 0}$.

Proposition 2.5

The following assertions hold:

• (i) The operator $C_1$ is the generator of a stochastic $C_0$-semigroup $(T_1(t))_{t \ge 0}$ on $L^1([0, 1])$, given by

  $$\begin{aligned} \begin{aligned} T_1(t) f(x) = \left\{ \begin{array}{ll} \text {e}^t f(x \text {e}^t), \quad &{} \text {if} \ t \le - \ln x, \\ 0, \quad &{} \text {else,} \end{array}\right. \end{aligned}\end{aligned}$$

  for all $f \in L^1([0, 1])$. Moreover, for all $\lambda > 0$ and $f \in L^1([0, 1])$ the identity

  $$\begin{aligned} R(\lambda , C_1) f(x) = x^{\lambda - 1} \int _x^1 \frac{f(y)}{y^\lambda } \, \textrm{d}y \end{aligned}$$

  holds for almost all $x \in [0,1]$.

• (ii) The operator $C_2$ is the generator of a stochastic $C_0$-semigroup $(T_2(t))_{t \ge 0}$ on $L^1([0, 1])$, given by

  $$\begin{aligned} \begin{aligned} T_2(t) f(x) = \left\{ \begin{array}{ll} \text {e}^t f(1 - (1 - x) \text {e}^t), \quad &{} \text {if } t \le - \ln (1 - x), \\ 0, \quad &{}\text {else,} \end{array}\right. \end{aligned}\end{aligned}$$

  for all $f \in L^1([0, 1])$. Moreover, for all $\lambda > 0$ and $f \in L^1([0, 1])$ the identity

  $$\begin{aligned} R(\lambda , C_2) f(x) = (1 - x)^{\lambda - 1} \int _0^x \frac{f(y)}{(1 - y)^\lambda } \, \textrm{d}y \end{aligned}$$

  holds for almost all $x \in [0,1]$.

Proof

(i): Consider the closed, $({\widetilde{T}}_1(t))_{t \ge 0}$-invariant subspace

$$\begin{aligned} F :=\lbrace f \in L^1([0, \infty )) : f(x) = 0 \text { for almost all } x \ge 1\rbrace . \end{aligned}$$

Then, the part $A_{1 |}$ of $A_1$ in F is given by

$$\begin{aligned} \begin{aligned} A_{1 |} f = \frac{\text {d}}{\text {d}x}(xf), \quad D(A_{1 |})&= \lbrace f \in D(A_1) \cap F : A_1 f \in F\rbrace \\ {}&= \{ f \in L^1([0,\infty )) : (x f)' \in L^1([0,\infty )), \\ {}&\qquad \qquad f(x) = 0 \text{ for } \text{ all } x \ge 1\}. \end{aligned} \end{aligned}$$

and generates the subspace semigroup $({\widetilde{T}}_1(t)_|)_{t \ge 0}$ on F (see [8, Section II.2.3]). Clearly, F is isomorphic $L^1([0, 1])$ and the generator of the subspace semigroup $({\widetilde{T}}_1(t)_|)_{t \ge 0}$ is isomorphic to $C_1$. The above resolvent identity is another direct consequence of this isomorphy.

(ii): This follows either from the fact that $(T_2(t))_{t \ge 0}$ is similar to $(T_1(t))_{t \ge 0}$ via the restricted reflection operator

$$\begin{aligned} Q :L^1([0, 1]) \rightarrow L^1([0, 1]), \quad Qf(x) = f(1 - x) \end{aligned}$$

which is an isometric isomorphism with $Q^{-1} = Q$ or, alternatively, by analogous arguments as used in (i). $\square$

Analogously to Corollary 2.4, the following result is a consequence of Proposition 2.5.

Corollary 2.6

The following assertions hold:

• (i) We have $(0, \infty ) \subseteq \rho (C_1')$ and for all $\lambda > 0$ and all $g \in L^\infty ([0,1])$ that

  $$\begin{aligned} R(\lambda , C_1') g(y) = \frac{1}{y^\lambda } \int _0^y g(x) x^{\lambda - 1} \, \textrm{d}x \end{aligned}$$

  for almost all $y \in [0,\infty )$.

• (ii) We have $(0, \infty ) \subseteq \rho (C_2')$ and for all $\lambda > 0$ and all $g \in L^\infty ([0,1])$ that

  $$\begin{aligned} R(\lambda , C_2') g(y) = \frac{1}{(1 - y)^\lambda } \int _y^1 g(x) (1 - x)^{\lambda - 1} \, \textrm{d}x \end{aligned}$$

  for almost all $y \in [0,\infty )$.

3 Long-term behaviour

In this section, we prove our main result (Theorem 1.1). Namely, we show that the solutions to (1.1) converge uniformly to the equilibrium of the equation with respect to the operator norm. In Sect. 2, we observed that this is equivalent to studying the long-term behaviour of the solutions to the abstract Cauchy problem

$$\begin{aligned} u'(t) = (\mathcal {A}+ \mathcal {B}) u(t), \quad u(0) = u_0, \end{aligned}$$

(ACP)

in the space $L^1([0, 1]; \mathbb {C}^2)$. Here, $\mathcal {A}$ and $\mathcal {B}$ are given as at the beginning of Sect. 2. Recall that by the discussion in the introduction of Sect. 2 we may, and shall, assume that $a = 0$ and $c = d$. It was already shown in [18, Proposition 4.3] that the operator $\mathcal {A}+ \mathcal {B}$ generates a stochastic, irreducible semigroup $(T(t))_{t \ge 0}$ on $L^1([0, 1])$.

We first recall the notion of integral operators. Let $(\Omega , \Sigma , \mu )$ be a $\sigma$-finite measure space. Then, a bounded operator T on $L^1(\Omega )$ is called an integral operator if there exists a measurable function $k :\Omega \times \Omega \rightarrow \mathbb {R}$ such that for each $f \in L^1(\Omega )$ the function $y \mapsto k(x, y) f(y)$ is in $L^1(\Omega )$ for almost every $x \in \Omega$ and

$$\begin{aligned} T f = \int _\Omega k(\mathord {\,\cdot \,}, y) f(y) \, \textrm{d}y \quad \text {for all } f \in L^1(\Omega ). \end{aligned}$$

Furthermore, an operator T on $L^1(\Omega )$ is called a partial integral operator if there exists a nonzero integral operator $K \in \mathcal {L}(L^1(\Omega ))$ such that $0 \le K \le T$. The following result (see [14, Corollary 1.2]) is our main tool for the proof of Theorem 1.1.

Theorem 3.1

Let $(\Omega , \Sigma , \mu )$ be a $\sigma$-finite measure space and let ${(T(t))}_{t \ge 0}$ be a stochastic and irreducible $C_0$-semigroup on $L^1(\Omega )$. Suppose that there exists $t_0 > 0$ such that $T(t_0)$ is a partial integral operator. Then, the following statements are equivalent:

• (i) ${(T(t))}_{t \ge 0}$ converges with respect to the operator norm as $t \rightarrow \infty$.

• (ii) For each nonzero $0 \le g \in L^\infty (\Omega )$, there exists $\lambda > 0$ such that $R(\lambda , A)' g \gg 0$.

So to apply Theorem 3.1 to the semigroup $(S(t))_{t \ge 0}$ generated by $\mathcal {A}+ \mathcal {B}$, we need to show that the following two assertions hold:

• (a) For each nonzero $0 \le g \in L^\infty ([0, 1]; \mathbb {C}^2)$ there exists some $\lambda > 0$ and some $c > 0$ such that

  $$\begin{aligned} R(\lambda , \mathcal {A}+ \mathcal {B})' g \ge c \cdot {{\,\mathrm{\mathbbm {1}}\,}}. \end{aligned}$$

  This is shown in Sect. 3.1.

• (b) There exists a time $t_0 > 0$ such that $S(t_0)$ is a partial integral operator. This is shown in Sect. 3.2.

Note that, as discussed at the beginning of Sect. 2, the space $L^1([0, 1]; \mathbb {C}^2)$ is isometrically isomorphic to a scalar-valued $L^1$-space due to the choice of the norm. In particular, the above theorem is applicable.

3.1 The dual resolvent improves positivity

The next proposition shows that condition (a) is true.

Proposition 3.2

For each nonzero $0 \le g \in L^\infty ([0, 1]; \mathbb {C}^2)$, there exists some $\lambda > 0$ and some $c > 0$ such that $R(\lambda , \mathcal {A}+ \mathcal {B})' g \ge c {{\,\mathrm{\mathbbm {1}}\,}}$.

The following lemma simplifies estimates later in the proof of Proposition 3.2.

Lemma 3.3

There exist $\varepsilon > 0$, $\gamma > 0$ and some $\lambda _0 > 0$ such that the following assertions hold:

• (i) $R(\lambda , \mathcal {A}+ \mathcal {B}) \ge R(\lambda + \gamma , \mathcal {A}+ \mathcal {E})$ for all $\lambda > \lambda _0$.

• (ii) $\Vert \mathcal {E}R(\lambda , \mathcal {A})\Vert < 1$ for all $\lambda > \lambda _0$.

Here, $\mathcal {E}\in \mathcal {L}(L^1([0, 1]; \mathbb {C}^2))$ is given by

$$\begin{aligned} \mathcal {E}f :=\varepsilon \begin{pmatrix} 1 &{} 1 \\ 1 &{} 1 \end{pmatrix} f \qquad \text {for all } f \in L^1([0, 1]; \mathbb {C}^2). \end{aligned}$$

Proof

Fix $\gamma > \max \lbrace \Vert \nu \Vert _\infty , \Vert \mu \Vert _\infty \rbrace$ and pick $\varepsilon > 0$ such that

$$\begin{aligned} \begin{aligned} \varepsilon< \gamma - \max \lbrace \Vert \nu \Vert _\infty , \Vert \mu \Vert _\infty \rbrace \quad \text{ and } \quad \varepsilon < \min \bigg \lbrace \min _{x \in [0, 1]} \nu (x), \min _{x \in [0, 1]} \mu (x) \bigg \rbrace . \end{aligned} \end{aligned}$$

Such an $\varepsilon$ exists, since $\nu$ and $\mu$ are strictly positive and continuous. Now pick $\lambda _0 > \max \lbrace \gamma , 2 \varepsilon \rbrace$ (we may later enlarge $\lambda _0$ further if necessary). We show that the assertions (i) and (ii) hold for our choices of $\varepsilon$ and $\lambda _0$.

(i): By the choice of $\varepsilon$ and $\gamma$, one has

$$\begin{aligned} \mathcal {A}+ \mathcal {B}+ \gamma I = \mathcal {A}+ \mathcal {E}+ (\mathcal {B}+ \gamma I - \mathcal {E}), \end{aligned}$$

where $\mathcal {B}+ \gamma I - \mathcal {E}$ is a positive operator. Moreover, since $\mathcal {A}$ is the generator of a stochastic semigroup and $\mathcal {E}$ is a positive operator, it follows from the Trotter product formula $\textrm{e}^{t(\mathcal {A}+ \mathcal {E})}f = \lim _{n \rightarrow \infty } \textrm{e}^{\tfrac{t}{n} \mathcal {A}} \textrm{e}^{\tfrac{t}{n} \mathcal {E}}f$ for all $f \in L^1((0,1);\mathbb {C}^2)$ (see, e.g., [8, Corollary III.5.8]) that the semigroup generated by $\mathcal {A}+ \mathcal {E}$ is positive. Hence, the integral representation of the resolvent implies that $R(\lambda , \mathcal {A}+ \mathcal {E})$ is positive for all $\lambda > \lambda _0$ with $\lambda _0$ large enough. Moreover, it follows from the Hille–Yosida theorem and by choosing $\lambda _0$ even larger that

$$\begin{aligned} \Vert (\mathcal {B}+ \gamma {{\,\textrm{id}\,}}- \mathcal {E}) R(\lambda , \mathcal {A}+ \mathcal {E})\Vert&\le \Vert (\mathcal {B}+ \gamma {{\,\textrm{id}\,}}- \mathcal {E})\Vert \Vert R(\lambda , \mathcal {A}+ \mathcal {E})\Vert \\&\le \Vert (\mathcal {B}+ \gamma {{\,\textrm{id}\,}}- \mathcal {E})\Vert \cdot \frac{1}{\lambda - \omega } < 1 \end{aligned}$$

for all $\lambda > \lambda _0$, where $\omega \in \mathbb {R}$ is the growth bound of the semigroup generated by $\mathcal {A}+ \mathcal {E}$. Thus, by a consequence of the Neumann series expansion (see, e.g. [29, Theorem 4.5]), we obtain that

$$\begin{aligned} R(\lambda , \mathcal {A}+ \mathcal {B})&= R(\lambda + \gamma , \mathcal {A}+ \mathcal {B}+ \gamma I) = R(\lambda + \gamma , \mathcal {A}+ \mathcal {E}+ (\mathcal {B}+ \gamma I - \mathcal {E})) \\&= R(\lambda + \gamma , \mathcal {A}+ \mathcal {E}) \sum _{k = 0}^\infty \big ( (\mathcal {B}+ \gamma I - \mathcal {E}) R(\lambda + \gamma , \mathcal {A}+ \mathcal {E}) \big )^k \\&\ge R(\lambda + \gamma , \mathcal {A}+ \mathcal {E}) \end{aligned}$$

for all $\lambda > \lambda _0$.

(ii): Since $\mathcal {A}$ is the generator of a stochastic semigroup, we have

$$\begin{aligned} \Vert \mathcal {E}R(\lambda , \mathcal {A})\Vert \le \Vert \mathcal {E}\Vert \cdot \Vert R(\lambda , \mathcal {A})\Vert \le \frac{2 \varepsilon }{\lambda }< \frac{2 \varepsilon }{\lambda _0} < 1 \end{aligned}$$

for all $\lambda > \lambda _0$ by the Hille–Yosida theorem. $\square$

We are now in the position to prove Proposition 3.2.

Proof of Proposition 3.2

By Lemma 3.3, there exist $\varepsilon > 0$, $\gamma > 0$ and $\lambda _0 > \gamma$ such that

$$\begin{aligned} R(\lambda - \gamma , \mathcal {A}+ \mathcal {B}) \ge R(\lambda , \mathcal {A}+ \mathcal {E}), \end{aligned}$$

and thus,

$$\begin{aligned} R(\lambda - \gamma , \mathcal {A}+ \mathcal {B})' \ge R(\lambda , \mathcal {A}+ \mathcal {E})', \end{aligned}$$

and $\Vert R(\lambda , \mathcal {A})' \mathcal {E}'\Vert = \Vert \mathcal {E}R(\lambda , \mathcal {A})\Vert < 1$ for all $\lambda > \lambda _0$. We show that there exists $\lambda > \lambda _0$ such that $R(\lambda , \mathcal {A}+ \mathcal {E})' \gg 0$. So, by [29, Theorem 4.5], one has for $\lambda > \lambda _0$ that

$$\begin{aligned} R(\lambda , \mathcal {A}{+} \mathcal {B})' \ge R(\lambda , \mathcal {A}+ \mathcal {E})' {=} \sum _{k {=} 0}^\infty (R(\lambda , \mathcal {A})' \mathcal {E}')^k R(\lambda , \mathcal {A})' \ge (R(\lambda , \mathcal {A})' \mathcal {E}')^2 R(\lambda , \mathcal {A})'. \end{aligned}$$

Moreover,

$$\begin{aligned} \begin{aligned} R(\lambda , \mathcal {A})' \mathcal {E}'&= \begin{pmatrix} b^{-1} R(b^{-1} \lambda , C_1)' &{} 0 \\ 0 &{} d^{-1} R(d^{-1} \lambda , C_2)' \end{pmatrix} \begin{pmatrix} \varepsilon &{} \varepsilon \\ \varepsilon &{} \varepsilon \end{pmatrix} \\ {}&= \varepsilon \begin{pmatrix} b^{-1} R(b^{-1}\lambda , C_1)' &{} b^{-1} R(b^{-1}\lambda , C_1)' \\ d^{-1} R(d^{-1}\lambda , C_2)' &{} d^{-1} R(d^{-1}\lambda , C_2)' \end{pmatrix} \end{aligned} \end{aligned}$$

and hence

$$\begin{aligned} \begin{aligned} (R(\lambda , \mathcal {A})' \mathcal {E}')^2&= \varepsilon ^2 \begin{pmatrix} b^{-1} R(b^{-1}\lambda , C_1)' &{} b^{-1} R(b^{-1}\lambda , C_1)' \\ d^{-1} R(d^{-1}\lambda , C_2)' &{} d^{-1} R(d^{-1}\lambda , C_2)' \end{pmatrix}^2 \\ {}&\ge \frac{\varepsilon ^2}{b d} \begin{pmatrix} R(b^{-1}\lambda , C_1)' R(d^{-1}\lambda , C_2)' &{} R(b^{-1}\lambda , C_1)' R(d^{-1}\lambda , C_2)' \\ R(d^{-1}\lambda , C_2)' R(b^{-1}\lambda , C_1)' &{} R(d^{-1}\lambda , C_2)' R(b^{-1}\lambda , C_1)' \end{pmatrix} . \end{aligned} \end{aligned}$$

So altogether, we have

$$\begin{aligned} \begin{aligned} R(\lambda , \mathcal {A}+ \mathcal {E})' g \ge \frac{\varepsilon ^2}{b d} \begin{pmatrix} R(b^{-1}\lambda , C_1)' R(d^{-1}\lambda , C_2)' &{} R(b^{-1}\lambda , C_1)' R(d^{-1}\lambda , C_2)' \\ R(d^{-1}\lambda , C_2)' R(b^{-1}\lambda , C_1)' &{} R(d^{-1}\lambda , C_2)' R(b^{-1}\lambda , C_1)' \end{pmatrix} {\widetilde{g}}, \end{aligned} \end{aligned}$$

(3.1)

where ${\widetilde{g}} :=R(\lambda , \mathcal {A})' g \ge 0$ is nonzero, whenever $0 \le g \in L^\infty ([0, 1]; \mathbb {C}^2)$ is nonzero.

So to conclude the proof, it suffices to show that

$$\begin{aligned} R(d^{-1}\lambda , C_2)' R(b^{-1}\lambda , C_1)' h \gg 0 \quad \text {and} \quad R(b^{-1}\lambda , C_1)' R(d^{-1}\lambda , C_2)' h \gg 0 \end{aligned}$$

for each nonzero $0 \le h \in L^\infty ([0,1])$. Let $0 \le h \in L^\infty ([0,1])$ be nonzero and choose $\lambda > \max \{ \lambda _0, b, d, \gamma \}$. Then, by Corollary 2.6, we have

$$\begin{aligned} R(b^{-1}\lambda , C_1)' h(y) = \frac{1}{y^{\lambda /b}} \int _0^y h(x) x^{(\lambda - b)/b} \, \textrm{d}x. \end{aligned}$$

By [6, Lemma 8.2], there exists $\alpha \in (0, 1)$ and $\delta > 0$ such that

$$\begin{aligned} R(b^{-1}\lambda , C_1)' h(y) \ge \frac{\delta }{y^{\lambda /b}} \ge \delta , \qquad \text {for all } y \in [\alpha , 1]. \end{aligned}$$

Using Corollary 2.6, again, we obtain for all $z \in [0,1]$ that

$$\begin{aligned} R(d^{-1}\lambda , C_2)' R(b^{-1}\lambda , C_1)' h(z)&= \frac{1}{(1 - z)^{\lambda /d}} \int _z^1 R(b^{- 1}\lambda , C_1)' h(y) (1 - y)^{\lambda /d - 1} \, \textrm{d}y \\&\ge \frac{1}{(1 - z)^{\lambda /d}} \int _{\alpha \vee z}^1 \delta (1 - y)^{\lambda /d - 1} \, \textrm{d}y \\&= \frac{d \delta }{\lambda } \frac{(1 - (\alpha \vee z))^{\lambda /d}}{(1 - z)^{\lambda /d}} \ge \frac{d \delta }{\lambda } (1 - \alpha )^{\lambda /d}. \end{aligned}$$

So, all in all,

$$\begin{aligned} R(d^{-1}\lambda , C_2)' R(b^{-1}\lambda , C_1)' h \ge d \delta (1 - \alpha )^{\lambda /d} {{\,\mathrm{\mathbbm {1}}\,}}\gg 0. \end{aligned}$$

Completely analogously, one shows that $R(b^{-1}\lambda , C_1)' R(d^{-1}\lambda , C_2)' h \gg 0$. Thus, the claim follows as a consequence of the estimate (3.1). $\square$

3.2 Partial integral operator

To apply Theorem 3.1, it remains to show that the semigroup $(S(t))_{t \ge 0}$ generated by $\mathcal {A}+ \mathcal {B}$ contains a partial integral operator, i.e. that condition (b) is satisfied.

Proposition 3.4

Let ${(S(t))}_{t \ge 0}$ be the semigroup generated by $\mathcal {A}+ \mathcal {B}$. Then, there exists $t_0 > 0$ such that $S(t_0)$ is a partial integral operator.

Proof

• Step 1: We show that there exist $\varepsilon > 0$ and $\gamma > 0$ such that the semigroup $(S(t))_{t \ge 0}$ generated by $\mathcal {A}+ \mathcal {B}$ and the semigroup $(S_\varepsilon (t))_{t \ge 0}$ generated by $\mathcal {A}+ \mathcal {E}$, where $\mathcal {E}$ is defined as in Lemma 3.3, satisfy $\textrm{e}^{\gamma t} S(t) \ge S_\varepsilon (t)$.

  Indeed, by Lemma 3.3, there exist $\varepsilon > 0$, $\gamma > 0$ and $\lambda _0 > 0$ such that $R(\lambda , \mathcal {A}+ \mathcal {B}) \ge R(\lambda + \gamma , \mathcal {A}+ \mathcal {E}) = R(\lambda , \mathcal {A}+ \mathcal {E}- \gamma )$ for all $\lambda > \lambda _0$. Hence, the Post–Widder inversion formula (see [8, Corollary III.5.5]) yields

  $$\begin{aligned} S(t) \ge \textrm{e}^{-\gamma t} S_\varepsilon (t) \qquad \text {for all } t \ge 0. \end{aligned}$$

  (3.2)

• Step 2: We use a perturbation argument to obtain a lower bound for the stochastic semigroup $(T(t))_{t \ge 0}$ generated by $\mathcal {A}$ (cf. Proposition 2.5): As $\mathcal {E}$ is a bounded operator, the operator $\mathcal {A}_\varepsilon :=\mathcal {A}+ \mathcal {E}$ generates a $C_0$-semigroup $(S_\varepsilon (t))_{t \ge 0}$ and this semigroup is given by the Dyson–Phillips series

  $$\begin{aligned} S_\varepsilon (t)&= \sum _{k = 0}^\infty U_k(t), \qquad \text {for all } t \ge 0 \end{aligned}$$

  (see, e.g. [8, Theorem III.1.10] or [5, Section 11.2]), where

  $$\begin{aligned} U_0(t) :=T(t), \quad U_k(t)&:=\int _0^t T(t - s) \mathcal {E}U_{k - 1}(s) \, \textrm{d}s \qquad \text {for all } t \ge 0, \, k \in \mathbb {N}. \end{aligned}$$

  Clearly, $S_\varepsilon (t) \ge U_1(t)$ for all $t \ge 0$.

• Step 3: Next, we decompose $U_1(t)$ into operators, which we later show to be partial integral operator for some $t > 0$. This then implies that $U_1(t)$ is a partial integral operator, which in turn implies that $S_\varepsilon (t)$, and thus, S(t) are partial integral operators.

  Recall from Proposition 2.5 the definition of $(T_1(t))_{t \ge 0}$ and $(T_2(t))_{t \ge 0}$. Clearly,

  $$\begin{aligned} \begin{aligned} U_1(t) f&= \int _0^t \begin{pmatrix} T_1(b(t-s)) &{}{} 0 \\ 0 &{} T_2 (d(t-s)) \end{pmatrix} \begin{pmatrix} \varepsilon &{}{} \varepsilon \\ \varepsilon &{}{} \varepsilon \end{pmatrix} \begin{pmatrix} T_1(bs) &{}{} 0 \\ 0 &{} T_2 (ds) \end{pmatrix} f \, \text {d}s \\ {}&= \varepsilon \int _0^t \begin{pmatrix} T_1(b(t-s)) &{} T_1(b(t-s)) \\ T_2(d(t-s)) &{} T_2 (d(t-s)) \end{pmatrix} \begin{pmatrix} T_1(bs) &{} 0 \\ 0 &{} T_2 (ds) \end{pmatrix} f \, \text {d}s \\ {}&= \varepsilon \int _0^t \begin{pmatrix} T_1(b(t-s))T_1(bs) &{} T_1(b(t-s))T_2(ds) \\ T_2(d(t-s)) T_1(bs) &{} T_2 (d(t-s))T_2(bs) \end{pmatrix} f \, \text {d}s, \end{aligned}\end{aligned}$$

  dominates both functions

  $$\begin{aligned} \int _0^t \varepsilon \begin{pmatrix} 0 &{} T_1(b(t-s))T_2(ds) \\ 0 &{} 0 \end{pmatrix} f \, \textrm{d}s \quad \text {and} \quad \int _0^t \varepsilon \begin{pmatrix} 0 &{} 0 \\ T_2(d(t-s)) T_1(bs) &{} 0 \end{pmatrix} f \, \textrm{d}s \end{aligned}$$

  for all $f \in L^1([0, 1]; \mathbb {C}^2)$. Thus, it is sufficient to show that either

  $$\begin{aligned} \int _0^t T_1(b(t-s))T_2(ds) \, \textrm{d}s \in \mathcal {L}(L^1([0,1])) \end{aligned}$$

  (3.3)

  or

  $$\begin{aligned} \int _0^t T_2(b(t-s))T_1(ds) \, \textrm{d}s \in \mathcal {L}(L^1([0,1])) \end{aligned}$$

  (3.4)

  is a partial integral operator.

• Step 4: We show that (3.3) is a partial integral operator if $d \le b$. Let $f \in L^1([0,1])$ and $t \ge 0$. Then, by Proposition 2.5, we have

  $$\begin{aligned} \int _0^t T_1(b(t-s)) T_2(d s) f(x) \, \textrm{d}s = \int _0^t \textrm{e}^{b(t-s)} \textrm{e}^{ds} f(1- (1 - x \textrm{e}^{b(t-s)}) \textrm{e}^{ds} ) \, \textrm{d}s \end{aligned}$$

  for almost all $x \in [0, 1]$. As $0 < d \le b$, the mapping

  $$\begin{aligned} \varphi :[0,1] \times [0, 1] \rightarrow \mathbb {R}, \quad (x,s) \mapsto 1- (1 - x \textrm{e}^{b(t-s)}) \textrm{e}^{ds}. \end{aligned}$$

  has the partial derivative

  $$\begin{aligned} \frac{\partial }{\partial s} \varphi (x, s) = d \textrm{e}^{d s} + (b-d)x \textrm{e}^{b(t-s)} \textrm{e}^{ds} > 0. \end{aligned}$$

  Thus, $\varphi (x, \mathord {\,\cdot \,})$ is strictly increasing for each $x \in [0, 1]$. A substitution yields

  $$\begin{aligned} \int _0^t T_1(b(t-s)) T_2(d s) f(x) \, \textrm{d}s&= \int _0^t \textrm{e}^{b(t-s)} \textrm{e}^{ds} f(\varphi (x, s)) \, \textrm{d}s \\&\ge \int _0^t f(\varphi (x, s)) \, \textrm{d}s \\&= \int _{\varphi (x, 0)}^{\varphi (x, t)}\frac{1}{\frac{\partial }{\partial s} \varphi (x, s)} f(s) \, \textrm{d}s. \end{aligned}$$

  Hence, $\int _0^{t} T_1(b(t-s))T_2(ds) \, \textrm{d}s$ is a partial integral operator for each $t > 0$.

• Step 5: If $b \le d$, then the proof showing that (3.4) is a partial integral operator for each $t > 0$ is analogous to the proof presented in the previous step.

In total, we have proven that $U_1(t_0)$, and thus $S(t_0)$, is a partial integral operator for some $t_0 > 0$. $\square$

3.3 Proof of the main result

We have now gathered all the tools to conclude with the proof of Theorem 1.1.

Proof of Theorem 1.1

Without loss of generality, we may assume that $a = 0$ and $c = d$. By [18, Proposition 4.3], the operator $\mathcal {A}+ \mathcal {B}$ generates an irreducible stochastic semigroup $(S(t))_{t \ge 0}$ on $L^1([0, 1]; \mathbb {C}^2)$. Proposition 3.4 shows that there exists a time $t_0 > 0$ such that $S(t_0)$ is a partial integral operator. Moreover, by Proposition 3.2, for each nonzero $0 \le g \in L^\infty ([0, 1]; \mathbb {C}^2)$ there exists some $\lambda > 0$ such that $R(\lambda , \mathcal {A}+ \mathcal {B})' g \gg 0$. So, Theorem 3.1 implies that $(S(t))_{t \ge 0}$ converges with respect to the operator norm as $t \rightarrow \infty$. $\square$