On a singular limit as $\theta \rightarrow 0$ for a model for the evolution of morphogens in a growing tissue

Abstract

This paper is devoted to the singular limit of a model for the regulation of growth and patterning in developing tissues by diffusing morphogens. The model is governed by a system of nonlinear PDEs. The arguments are based on energy estimates and a change of variable that reduces the system into a nonlinear PDE with singular diffusion.

1 Introduction

The differentiation and growth of embryonic cells are mainly regulated by morphogens (see [1, 8, 9, 12]). Experimental evidences show that morphogens develop from a localized source spreading in concentration gradients that control the behavior of surrounding cells as a function of their distance from the source, see Wartlick, Mumcu, Kicheva, Bitting, Seum, Jülicher, and González-Gaitán in [10, 11].

The experimental observations mentioned in [10, 11] have been implemented in the mathematical model proposed by Averbukh, Ben-Zvi, Mishra and Barkai in [2], in which a growth law based on a parameter $\theta$ is formulated. It takes into account the fact that a cell divides when it detects that the relative morphogens concentration increases by a factor of $1+ \theta$.

The model developed in [2] is the following one

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _tM+\partial _x(uM)+\alpha M=D\partial ^2_{xx}M,&{}\quad t>0,\>0<x<L(t),\\ \displaystyle \partial _tM+u\partial _xM-\frac{\theta }{\log 2} M\partial _xu=0,&{}\quad t>0,\>0< x< L(t),\\ \displaystyle \partial _xM(t,0)=-\frac{\eta }{D},\quad \partial _xM(t,L(t))=0,&{}\quad t>0,\\ M(0,x)=M_0(x),&{}\quad 0<x<L_0,\\ u(t,0)=0,&{}\quad t>0,\\ L'(t)=u(t,L(t)),&{}\quad t>0,\\ L(0)=L_0,&{}{} \end{array}\right. } \end{aligned}$$

(1.1)

where the unknowns are

$$\begin{aligned} M=M_\theta (t,x),\qquad u=u_\theta (t,x),\qquad L=L_\theta (t), \end{aligned}$$

and

$$\begin{aligned} L_0>0;\quad 0< c_*\le M_0(x)\le c^*,\quad 0\le x\le L_0. \end{aligned}$$

Here $M_\theta (t,x)$ is the morphogen concentration in the one-dimensional growing tissue $[0,L_\theta (t)]$, $L_\theta (t)$ is the length of the tissue, $u_\theta (t,x)$ is the (local) flow rate of the growing tissue with $\partial _xu_\theta (t,x)$ being the cell proliferation rate, and $\alpha ,\,D, \,\eta$ are positive parameters that correspond to the morphogen degradation rate, diffusion rate and incoming morphogen flux rate, respectively. The evolution of the morphogens concentration in (1.1) is described by the first equation, which is a nonlinear advection–reaction–diffusion PDE. The second equation gives the expression of the cell division rule due to morphogens proliferation and flow rates. Finally, the tissue length L(t) obeys an ODE flow type. The two PDEs are augmented with suitable initial data and non-homogeneous Neumann-type boundary conditions.

Here we are interested in the analysis of $(M_\theta (t,x), u_\theta (t,x), L_\theta (t))$ as

$$\begin{aligned} \theta \rightarrow 0^+, \end{aligned}$$

as a consequence in the following we will always assume

$$\begin{aligned} 0<\theta <\log 2. \end{aligned}$$

Under this condition we have established in [4] the well-posedness (existence, uniqueness, and stability) and in [6] the asymptotic behavior as $t\rightarrow \infty$ of $(M_\theta (t,x),$ $u_\theta (t,x),$ $L_\theta (t))$. We will often recall some results of these papers, and therefore, we will assume that the hypotheses assumed therein are satisfied, also here. Before stating them explicitly, we point out that the hypothesis in [6]:

$$\begin{aligned} DM_0''(x)-\alpha M_0(x),\,\,0\le x\le L_0, \text {has constant sign}, \end{aligned}$$

in this paper it is assumed by formulating two alternative conditions:

$$\begin{aligned} DM_0''(x)-\alpha M_0(x)\ge 0\quad \text {or}\quad DM_0''(x)-\alpha M_0(x)\le 0. \end{aligned}$$

The results in the two cases are different while retaining a certain “symmetry”. Having said that, in this paper we assume that the following hypotheses are satisfied

$$\begin{aligned} 0< c_*\le M_0(x)\le c^*,\quad 0\le x\le L_0;\quad M_0\in H^2(0,L_0), \end{aligned}$$

(1.2)

and one within the following

$$\begin{aligned}&M'_0(x)\le 0;\quad D M''_0(x)-\alpha M_0(x)\ge 0,\quad 0\le x\le L_0, \end{aligned}$$

(1.3)

$$\begin{aligned}&M'_0(x)\le 0;\quad D M''_0(x)-\alpha M_0(x)\le 0,\quad 0\le x\le L_0. \end{aligned}$$

(1.4)

The difference between the two cases is further highlighted by the different initial mean morphogens concentrations, indeed

$$\begin{aligned}&DM_0''(x)-\alpha M_0(x)\ge 0,\,\,0\le x\le L_0\,\,\Rightarrow \,\, \left\| M_0 \right\| _{L^1(0,L_0)}\le \frac{\eta }{\alpha }, \\&DM_0''(x)-\alpha M_0(x)\le 0,\,\,0\le x\le L_0\,\,\Rightarrow \,\, \left\| M_0 \right\| _{L^1(0,L_0)}\ge \frac{\eta }{\alpha }. \end{aligned}$$

Key tool for the analysis of the well-posedness (see [3, 4, 7]) and of the asymptotic behavior as $t\rightarrow \infty$ (see [5, 6]) is the definition of a suitable family of “characteristic” curves which start at the points of $[0, L_0]$ and “cover” $\{(t, x) \,| \,t \ge 0,0 \le x \le L_\theta (t) \}$. Let us briefly recall them because they are also useful in this paper.

Let $(M_\theta (t,x), u_\theta (t,x), L_\theta (t))$ be the solution of (1.1), for every $y\in [0,L_0],$ let $X_\theta (t,y)$ be the solution of

$$\begin{aligned} {\left\{ \begin{array}{ll} \displaystyle \frac{d}{\textrm{d}t} X_\theta (t,y)=u_\theta (t,X_\theta (t,y)),&{}\\ X_\theta (0,y)=y.&{} \end{array}\right. } \end{aligned}$$

(1.5)

Thanks to (1.1), it is clear that 0 solves (1.5) in correspondence of $y=0$ and $L_\theta (t)$ solves (1.5) in correspondence of $y=L_0.$ The image of $X_\theta (t,\cdot )$ is $[0, L_\theta (t)]$. $X_\theta (t,\cdot )$ is invertible; its inverse $Y_\theta (t,\cdot )$ is defined on $[0, L_\theta (t)]$ and its image is $[0,L_0]$ (see [3, 4, 7]).

The main results of this paper are the following.

Theorem 1.1

If $\displaystyle \left\| M_0 \right\| _{L^1(0,L_0)}\le \frac{\eta }{\alpha }$ and the assumptions (1.2), (1.3) hold, we have that

i):

$\displaystyle \lim \limits _{\theta \rightarrow 0}\int _0^{L_0} |M_\theta (t,X_\theta (t,y))-M_0(y)|^r\textrm{d}y=0,\quad 1\le r<\infty ,$ uniformly with respect to t on every compact set [0, T];

ii):

$0\le u_\theta (t,x);\quad \displaystyle \limsup \limits _{\theta \rightarrow 0}u_\theta (t,x)\le e^{-\alpha t}\,\,\frac{\eta -\alpha \left\| M_0 \right\| _{L^1(0,L_0)}}{M_0(L_0)};$

iii):

$\displaystyle L_0e^{-\alpha t}+\sqrt{\alpha D}\tanh \Big ({L_0\sqrt{\frac{\alpha }{D}}}\Big ) \frac{1-e^{-\alpha t}}{\alpha }\le \liminf \limits _{\theta \rightarrow 0} L_{\theta }(t)$ $\qquad \quad \displaystyle \le \limsup \limits _{\theta \rightarrow 0} L_{\theta }(t)\le L_0e^{-\alpha t}+\frac{\eta }{M_0(L_0)}\frac{1-e^{-\alpha t}}{\alpha }.$

Theorem 1.2

If $\displaystyle \left\| M_0 \right\| _{L^1(0,L_0)}\ge \frac{\eta }{\alpha }$ and the assumptions (1.2), (1.4) hold, we have that

i):

$\displaystyle \lim \limits _{\theta \rightarrow 0}\int _0^{L(t)} |M_\theta (t,x)-M_0(Y_\theta (t,x)|^r\textrm{d}y=0,\quad 1\le r<\infty ,$ uniformly with respect to t on every compact set [0, T];

ii):

$\displaystyle -e^{-\alpha t}\,\, \frac{\alpha \left\| M_0 \right\| _{L^1(0,L_0)}-\eta }{M_0(L_0)}0$ $\le \displaystyle \liminf \limits _{\theta \rightarrow 0}u_\theta (t,x);\quad u_\theta (t,x)\le 0;$

iii):

$\displaystyle L_0e^{-\alpha t}+\frac{\eta }{M_0(0)} \frac{1-e^{-\alpha t}}{\alpha }$ $\le \liminf \limits _{\theta \rightarrow 0} L_{\theta }(t)\le \limsup \limits _{\theta \rightarrow 0} L_{\theta }(t)\le$ $\qquad \quad \le \displaystyle L_0e^{-\alpha t}+\frac{\eta }{M_0(0)}\frac{1-e^{-\alpha t}}{\alpha }+\sqrt{\alpha D}\,\Big (\log \frac{M_0(0)}{M_0(L_0)}\Big ) \frac{1-e^{-\alpha t}}{\alpha }.$

The paper is organized as follows. In Sect. 2 we recall some preliminary results. Section 3 is devoted to some a priori estimates on the sign of the derivatives of the unknowns. Theorems 1.1 and 1.2 are proved in Sects. 4 and 5, respectively.

2 Preliminary results

We transform (1.1) into a problem equivalent to it, in the sense that the well-posedness of one of them implies the well-posedness of the other one and from the solution of one of them we obtain at the solution of the other one. Defining

$$\begin{aligned} \beta :=\frac{\log 2}{\theta }, \qquad N_\beta (t,y):=M_\theta (t,X_\theta (t,y)), \end{aligned}$$

(1.1) is equivalent to the following problem

$$\begin{aligned} {\left\{ \begin{array}{ll} \displaystyle \partial _tN_\beta +a N_\beta =d\Big (\frac{M_0(y)}{|N_\beta |}\Big )^\beta \partial _y\Big (\Big (\frac{M_0(y)}{|N_\beta |}\Big )^\beta \partial _yN_\beta \Big ) ,&{}\, t>0,\,\,0<y<L_0,\\ \displaystyle \Big (\frac{M_0(0)}{|N_\beta (t,0)|}\Big )^\beta \partial _yN_\beta (t,0)=-\frac{\eta }{D},\> \Big (\frac{M_0(L_0)}{|N_\beta (t,L_0)|}\Big )^\beta \partial _yN_\beta (t,L_0)=0,&{} t>0,\\ N_\beta (0,y)=M_0(y),&{} 0<y<L_0, \end{array}\right. } \end{aligned}$$

(2.1)

where

$$\begin{aligned} a:=\frac{\alpha }{\beta +1};\quad d:=\frac{D}{\beta +1}. \end{aligned}$$

(2.2)

For every $0\le T<\infty$ we will use the following notation

$$\begin{aligned} E_T:=]0,T[\times ]0,L_0[;\quad \overline{E}_T:=\,\,\text {closure of}\,\, E_T. \end{aligned}$$

We define in an analogous way $E_\infty$ and $\overline{E}_\infty .$

Let us recall some properties of $N_\beta (t, y)$ useful in the next sections.

Theorem 2.1

(Existence, uniqueness, and regularity of $N_\beta (t,y)$ [4, Theorem 2.1], [6, Theorem 2.3]) If (1.2) holds, then, for every $\beta$ and $T>0$, (2.1) admits a unique solution $N_\beta (t,y)$ such that

i):

$c_* e^{-\frac{\alpha }{\beta +1}t}\le N_\beta (t,y), \,\,\, t\ge 0,\,\, 0\le y\le L_0\,\,\,$ and $N_\beta \in L^\infty (E_\infty );$

ii):

$\displaystyle \partial _tN_\beta ,\,\,\,\partial _yN_\beta ,\,\,\,\partial _y\left( \frac{M_0^\beta }{N_\beta ^\beta }\partial _yN_\beta \right) \in L^2(E_T),$ $\partial ^2_{yy}N\in L^1(E_T);$

iii):

here exists $c(T)>0$ such that for every $(t_1,y_1), (t_2,y_2)\in E_T$ $|N_\beta (t_1,y_1)- N_\beta (t_2,y_2)|\le c(T)\big (\sqrt{|t_1-t_2|}+|y_1-y_2|\big )^\frac{1}{4}$;

iv):

$\partial _yN_\beta \in C(]0,\infty [\times [0,L_0]);\,\,\,\partial _tN_\beta ,\,\partial ^2_{yy}N_\beta \in C(E_\infty ).$

Let us show how to pass from $N_\beta (t,y)$ to $(M_\theta (t,x),\,u_\theta (t,x),\,L_\theta (t))$ and vice versa. Thanks to the properties of $N_\beta (t,y)$, the function $(t,y)\mapsto \Big (\frac{M_0(y)}{N_\beta (t,y)}\Big )^\beta$ is positive and Hölder continuous in every $\overline{E}_T$ (see [7, Theorem 2.1], [3, Theorem 2.1], and [6, Theorem 2.3]). As a consequence

$$\begin{aligned} \frac{\textrm{d}Y_\theta }{\textrm{d}x}=\Big (\frac{M_0(Y)}{N_\beta (t,Y)}\Big )^\beta ,\,\, Y_\theta (t, 0)=0\, \end{aligned}$$

(¹) admits a unique (maximal) solution $Y_\theta (t,\cdot ).$

Let $[0,L_\theta (t)]$ be the (maximal) existence interval of $Y_\theta (t,\cdot )$. We have $Y_\theta (t,L_\theta (t))=L_0,$ and defining

$$\begin{aligned} M_\theta (t,x)&=N_\beta (t,Y(t,x));\\ u_\theta (t,x)&=\beta \int _0^{Y_\theta (t,x)} \frac{N_\beta (t,y)^{\beta -1}\partial _tN_\beta (t,y)}{M_0(y)^\beta } \textrm{d}y;\\ L_\theta (t)&=\int _0^{L_0}\big (\frac{N_\beta (t,y)}{M_0(y)}\big )^\beta \textrm{d}y, \end{aligned}$$

$(M_\theta (t,x),\,u_\theta (t,x)\,L_\theta (t))$ is a solution of (1.1). (²) As a first step in our analysis, we begin by studying the behavior of $N_\beta (t,y)$ as

$$\begin{aligned} \beta \rightarrow \infty , \end{aligned}$$

from now on we assume that

$\beta >1$ (³).

Let us also briefly recall the results on the asymptotic behavior for $t\rightarrow \infty$ of $N_\beta (t,y)$ (see [6, Theorem 2.1]) and of $(M_\theta (t, x), \, u_\theta (t, x), \, L_\theta (t))$ (see [6, Theorem1.1]). If the assumptions (1.2) and (1.3) or (1.2) and (1.4) hold, then the function $N_\beta (\cdot ,y)$ is monotone and its limit

$$\begin{aligned} \overline{N}_\beta (y):=\lim \limits _{t\rightarrow \infty }N_\beta (t,y) \end{aligned}$$

belongs to $C^2([0,L_0])$, is positive, decreasing and solves the stationary problem

$$\begin{aligned} {\left\{ \begin{array}{ll} \displaystyle \alpha \overline{N}_\beta =D\Big (\frac{M_0(y)}{\overline{N}_\beta }\Big )^\beta \Big [\Big (\frac{M_0(y)}{\overline{N}_\beta }\Big )^\beta \overline{N}_\beta '\Big ]',\quad 0\le y\le L_0,&{}\\ \displaystyle \Big (\frac{M_0(0)}{\overline{N}_\beta (0)}\Big )^\beta \overline{N}_\beta '(0)=-\frac{\eta }{D};\quad \Big (\frac{M_0(L_0)}{\overline{N}_\beta (L_0)}\Big )^\beta \overline{N}_\beta '(L_0)=0. &{} \end{array}\right. } \end{aligned}$$

(2.3)

Moreover, $y\mapsto N_\beta (t,y)$ converges to $\overline{N}_\beta (y)$ uniformly with respect to y as $t\rightarrow \infty$.

The triplet $(M_\theta (t,x),\,u_\theta (t,x),\,L_\theta (t))$ satisfies the following statements.

i):

$L_\theta (t)$ converges to $\overline{L}_\theta$ as $t\rightarrow \infty ,$ and

$$\begin{aligned} |L_\theta (t)-\overline{L}_\theta |\le ce^{-\alpha t}, \end{aligned}$$

for some constant c independent on t.

ii):

$\lim \limits _{t\rightarrow \infty }u_\theta (t,x)= 0$ uniformly with respect to x.

iii):

$M_\theta (t,x)$ converges to $\overline{M}_\theta \in C^2([0,\overline{L}_\theta ])$ as $t\rightarrow \infty ,$ $M_\theta (t,\xi L_\theta (t))$ converges to $\overline{M}_\theta (\xi \overline{L}_\theta )$ uniformly with respect to $0\le \xi \le 1.$ Moreover, $\overline{M}_\theta (x)$ satisfies

$$\begin{aligned} {\left\{ \begin{array}{ll} \alpha \overline{M}_\theta =D\overline{M}_\theta '',\,\, \text { in } [0,\overline{L}_\theta ],\\ \overline{M}_\theta '(0)=-\frac{\eta }{\overline{D}};\,\,\, \overline{M}_\theta '(\overline{L}_\theta )=0, \end{array}\right. } \end{aligned}$$

and its explicit expression is

$$\begin{aligned} \overline{M}_\theta (x)=\frac{\eta }{\sqrt{\alpha D}}\frac{\cosh {\big [(x-\overline{L}_\theta )}\sqrt{\frac{\alpha }{D}}\big ]}{\sinh \big [{\overline{L}_\theta }\sqrt{\frac{\alpha }{D}}\big ]},\quad 0\le x\le \overline{L}_\theta . \end{aligned}$$

We conclude this section recalling that

$$\begin{aligned} \overline{N}_\beta (0)=\overline{M}_\theta (0)=\frac{\eta }{\sqrt{\alpha D}} \coth {\Big (\overline{L}_\theta \sqrt{\frac{\alpha }{D}}\Big )}. \end{aligned}$$

3 On the signs of $\partial _tN_\beta (t,y)$ and $\partial _yN_\beta (t,y)$

On the sign of $\partial _tN_\beta (t,y)$, we proved the following result.

Theorem 3.1

(Sign of $\partial _tN_\beta (t,y)$ [6, Theorem 2.5]) For every $t> 0,\,\,0\le y\le L_0$, we have that

i):

$DM_0^{''}(y)-\alpha M_0(y)\ge 0\,\,\Rightarrow \,\, \partial _tN_\beta (t,y)\ge 0,$

ii):

$DM_0^{''}(y)-\alpha M_0(y)\le 0\,\,\Rightarrow \,\, \partial _tN_\beta (t,y)\le 0.$

To clarify the link between the hypotheses (1.3), (1.4) and the initial mean morphogens concentration, i.e., $\left\| M_0 \right\| _{L^1(0, L_0)}$, the following lemma is needed.

Lemma 3.1

([6, Theorem 2.1. ii]) For every $\beta \ge 1$ and $t\ge 0$, we have that

$$\begin{aligned} \int _0^{L_0}\frac{N_\beta (t,y)^{\beta +1}}{M_0(y)^\beta }\textrm{d}y=\frac{\eta }{\alpha }+ e^{-\alpha t}\Big (\left\| M_0 \right\| _{L^1(0,L_0)}-\frac{\eta }{\alpha }\Big ). \end{aligned}$$

Proof

Let us quickly sketch the proof of [6, Theorem2.1.ii)]. It is not difficult to rewrite the equation of (2.1) as follows

$$\begin{aligned} \partial _t\Big (e^{\alpha t}\frac{N_\beta (t,y)^{\beta +1}}{M_0(y)^\beta } \Big )= D\partial _y\Big [e^{\alpha t}\Big (\frac{M_0(y)}{N_\beta (t,y)}\Big )^\beta \partial _yN_\beta (t,y)\Big ]. \end{aligned}$$

We integrate both sides in y on $[0,L_0]$, thanks to the boundary and initial data in (2.1),

$$\begin{aligned} \partial _t\Big (e^{\alpha t}\int _0^{L_0}\frac{N_\beta (t,y)^{\beta +1}}{M_0(y)^\beta }\textrm{d}y \Big )=De^{\alpha t}\frac{\eta }{D} \end{aligned}$$

and then

$$\begin{aligned} e^{\alpha t}\int _0^{L_0}\frac{N_\beta (t,y)^{\beta +1}}{M_0(y)^\beta }\textrm{d}y= \int _0^{L_0}M_0(y)\textrm{d}y+\frac{\eta }{\alpha }(e^{\alpha t}-1), \end{aligned}$$

that gives the claim. $\square$

The relation between the assumptions (1.3), (1.4) and $\left\| M_0 \right\| _{L^1(0,L_0)}$ is clarified in the next statements

$$\begin{aligned}&1.3 \,\,\Rightarrow \,\,\displaystyle DM_0^{''}(\cdot )-\alpha M_0(\cdot )\ge 0\,\,\Rightarrow \,\, \left\| M_0 \right\| _{L^1(0,L_0)}\le \frac{\eta }{\alpha }, \end{aligned}$$

(3.1)

$$\begin{aligned}&1.4 \,\,\Rightarrow \,\,\displaystyle DM_0^{''}(\cdot )-\alpha M_0(\cdot )\le 0\,\,\Rightarrow \,\, \left\| M_0 \right\| _{L^1(0,L_0)}\ge \frac{\eta }{\alpha }. \end{aligned}$$

(3.2)

We prove only (3.1), because the same argument works also for (3.2)

$$\begin{aligned} \displaystyle DM_0^{''}(y)-\alpha M_0(y)\ge 0\,\,&\Rightarrow \,\,\partial _tN_\beta (t,y)\ge 0\,\,\Rightarrow \,\,\partial _t\int _0^{L_0}\frac{N_\beta (t,y)^{\beta +1}}{M_0(y)^\beta }\textrm{d}y\ge 0\,\,\Leftrightarrow \\&\Leftrightarrow \,\,\partial _t\left\{ \frac{\eta }{\alpha }+e^{-\alpha t}\left( \left\| M_0 \right\| _{L^1(0,L_0)} -\frac{\eta }{\alpha }\right) \right\} \ge 0\,\,\Leftrightarrow \\&\Leftrightarrow \,\,\alpha e^{-\alpha t}\left( \frac{\eta }{\alpha }-\left\| M_0 \right\| _{L^1(0,L_0)}\right) \ge 0\,\,\Leftrightarrow \,\,\left\| M_0 \right\| _{L^1(0,L_0)}\le \frac{\eta }{\alpha }. \end{aligned}$$

To determine the sign of $\partial _yN_\beta (t, y)$, it is convenient to consider a reformulation of (2.1) useful for partially camouflaging the cumbersome initial datum $M_0 (y)$. We will use the following notations, given $0 \le T <\infty ,$

$$\begin{aligned} Q_T: =]0,T[\times ]0,1[; \quad \overline{Q}_T: = \, \, \text {closure} \, \, Q_T. \end{aligned}$$

Similarly, we define $Q_\infty$ and $\overline{Q}_\infty .$

Due to the assumptions on $M_0$, we can consider the function

$$\begin{aligned} z=Z_0(y):=\frac{1}{\mu _0}\int _0^y M_0(\xi )^{-\beta }\textrm{d}\xi ,\,\,\, 0\le y\le L_0, \end{aligned}$$

where

$$\begin{aligned} \mu _0=\left\| M_0^{-\beta } \right\| _{L^1(0,L_0)}. \end{aligned}$$

If $Y_0(z)$ is the inverse of $Z_0(y)$, we define

$$\begin{aligned} n_\beta (t,z):=N_\beta (t,Y_0(z)). \end{aligned}$$

Passing from the unknown $N_\beta (t,y)$ to $n_\beta (t,z)$, we simplify (2.1) in the following way

$$\begin{aligned} {\left\{ \begin{array}{ll} \displaystyle \partial _tn+an=\frac{A}{|n|^\beta } \partial _z\Big (\frac{\partial _zn}{|n|^\beta }\Big ),&{}\quad (t,z)\in Q_\infty ,\\ \displaystyle \frac{\partial _zn(t,0)}{|n(t,0)|^\beta }=-B,\quad \frac{\partial _zn(t,1)}{|n(t,1)|^\beta }=0, &{}\quad t>0,\\ n(0,z)=M_0(Y_0(z)),&{}\quad 0< z<1, \end{array}\right. } \end{aligned}$$

(3.3)

where

$$\begin{aligned} A:=\frac{d}{\mu _0^2}=\frac{D}{(\beta +1) \mu _0^2};\qquad B:=\frac{\eta \mu _0}{D}. \end{aligned}$$

(3.4)

Theorem 3.2

(Sign of $\partial _yN_\beta (t,y)$) Let $\beta >0$ be given. If

$$\begin{aligned} M'_0(y)\le 0,\,\,\,0\le y\le L_0, \end{aligned}$$

then

$$\begin{aligned} \partial _yN_\beta (t,y)\le 0,\,\,\, t>0,\,0\le y\le L_0. \end{aligned}$$

Proof

In order to keep the presentation simple and clear, we start considering (3.3) and proving

$$\begin{aligned} \partial _zn_\beta (t,z)\le 0,\,\,\,t>0,\,\,0< z< 1.\,\, \end{aligned}$$

(3.5)

(⁴) We begin by assuming $\beta \not =1.$ Consider the functions

$$\begin{aligned} v(t,z)=n_\beta (t,z)^{1-\beta },\qquad \omega =\partial _zv_\beta , \end{aligned}$$

$\omega$ satisfies in the weak sense the following identity

$$\begin{aligned} A\partial _z(v^m\partial _z\omega )+a(\beta -1)\omega =\partial _t\omega \quad \text {in }\,\,Q_\infty , \end{aligned}$$

(3.6)

with

$$\begin{aligned} m=\frac{2\beta }{\beta -1} \,\,\\ |v^m\partial _z\omega |=&|(n^{1-\beta })^\frac{2\beta }{\beta -1}\partial ^2_{zz}v|=\big |\frac{1}{n^{2\beta }}\partial _z\big ((1-\beta )\frac{\partial _zn}{n^\beta }\big )\big |\le \\ \le&\frac{|\beta -1|}{c_*^\frac{\alpha \beta t}{\beta +1}} \big |\frac{1}{n^{\beta }}\partial _z\big (\frac{\partial _zn}{n^\beta }\big )\big |\in L^2(Q_t),\quad \forall t>0;\\ |\omega |=&|\partial _zv|=|\partial _zn^{1-\beta }|=|(1-\beta )\frac{\partial _zn}{n^\beta }|\le \frac{|\beta -1|}{c_*^\frac{\alpha \beta t}{\beta +1}}|\partial _zn|\in L^2(Q_t);\\ |v^m\partial _z\omega \partial _z\omega |=&(v^\frac{m}{2}\partial _z\omega )^2=\big (\frac{1}{n^\beta }(1-\beta )\partial _z\big (\frac{\partial _zn}{n^\beta }\big )\big )^2\in L^1(Q_t). \end{aligned}$$

.

(⁵) Moreover, the following are satisfied in the sense of traces

$$\begin{aligned} \omega (t,0)=B(\beta -1);\quad \omega (t,1)=0,&\quad t>0,\\ \omega (0,z)=\mu _0 (1-\beta ) M'_0(Y_0(z)),&\quad 0<z<1. \end{aligned}$$

Let us distinguish two cases $0<\beta <1$, $1<\beta$.

$\underline{0<\beta <1}.$ We multiply (3.6) by

$$\begin{aligned} \omega (t,z)^+e^{\lambda t},\quad \text {whit}\>\>\lambda =-2a\beta ,\>\omega ^+:= \frac{1}{2}(|\omega |+\omega ), \end{aligned}$$

and integrate over $Q_t,\,\,t>0.$ Being

$$\begin{aligned} \omega (t,0)^+=\omega (t,1)^+=0, \end{aligned}$$

we have

$$\begin{aligned} \begin{aligned} -A\int _{Q_t}v^m (\partial _z\omega )^2\frac{{\textrm{sign}}\left( \omega \right) +1}{2}e^{\lambda \tau }\textrm{d}\tau \textrm{d}z+a(\beta -1)\int _{Q_t}\omega \omega ^+e^{\lambda \tau }\textrm{d}\tau \textrm{d}z =\int _{Q_t}\partial _\tau \omega \omega ^+e^{\lambda \tau }\textrm{d}\tau \textrm{d}z&. \end{aligned} \end{aligned}$$

(3.7)

Since

$$\begin{aligned} \partial _\tau \omega \cdot \omega ^+e^{\lambda \tau }&=\partial _\tau (\omega \omega ^+ e^{\lambda \tau })-\omega e^{\lambda \tau }\Big (\partial _\tau \omega \frac{{\textrm{sign}}\left( \omega \right) +1}{2}+\lambda \omega ^+\Big )\\&=\partial _\tau (\omega \omega ^+ e^{\lambda \tau })-\partial _\tau \omega \cdot \omega ^+e^{\lambda \tau }-\lambda \omega \omega ^+ e^{\lambda \tau }, \end{aligned}$$

we have

$$\begin{aligned} \partial _\tau \omega \cdot \omega ^+e^{\lambda \tau }= \frac{1}{2}\partial _\tau (\omega \omega ^+e^{\lambda \tau })-\frac{\lambda }{2}\omega \omega ^+ e^{\lambda \tau }, \end{aligned}$$

and using (3.7)

$$\begin{aligned} -A\int _{Q_t}&v^m(\partial _z\omega )^2\frac{{\textrm{sign}}\left( \omega \right) +1}{2}e^{\lambda \tau }\textrm{d}\tau \textrm{d}z+\,\,a(\beta -1)\int _{Q_t} \omega \omega ^+e^{\lambda \tau }\textrm{d}\tau \textrm{d}z\\ =&\frac{1}{2}\int _0^1 \omega (t,z)\omega (t,z)^+e^{\lambda t}\textrm{d}z-\frac{1}{2}\int _0^1 \omega (0,z)\omega (0,z)^+ \textrm{d}z-\frac{\lambda }{2}\int _{Q_t}\omega \omega ^+e^{\lambda \tau }\textrm{d}\tau \textrm{d}z. \end{aligned}$$

Being $\beta <1$

$$\begin{aligned} \omega (0,z)^+=\Big (\mu _0(1-\beta )M_0'(Y_0(z))\Big )^+=0,\qquad \lambda =-2a\beta , \end{aligned}$$

and then

$$\begin{aligned} -A\int _{Q_t}&v^m(\partial _z\omega )^2\frac{{\textrm{sign}}\left( \omega \right) +1}{2}e^{\lambda \tau }\textrm{d}\tau \textrm{d}z- a\int _{Q_t} \omega \omega ^+e^{\lambda \tau }\textrm{d}\tau \textrm{d}z=\frac{1}{2}\int _0^1 \omega (t,z)\omega (t,z)^+e^{\lambda t}\textrm{d}z. \end{aligned}$$

Since $\omega \omega ^+\ge 0$ the two sides of the identity have different sings. As a consequence, they must vanish and

$$\begin{aligned} \ \int _0^1\omega (t,z)\omega (t,z)^+e^{\lambda t}\textrm{d}z=0,\qquad t>0, \end{aligned}$$

that gives $\omega (t,z)^+=0,$ namely $\omega (t,z)\le 0.$ In light of the definition of $\omega (t,z)$ we have

$$\begin{aligned} \partial _zv(t,z)=\omega (t,z)\le 0\,\,\Leftrightarrow \,\, (1-\beta )n(t,z)^{-\beta }\partial _zn(t,z)\le 0 \,\,\Leftrightarrow \,\,\partial _zn(t,z)\le 0. \end{aligned}$$

$\underline{\beta >1}.$ We argue as before and multiply (3.6) by

$$\begin{aligned} \omega (t,z)^-e^{\lambda t},\qquad \text {with}\>\>\lambda =-2a\beta . \end{aligned}$$

Being $\beta >1$ we have

$$\begin{aligned} \omega (t,0)^-=\omega (t,1)^-=0,\qquad \omega (0,z)^-=\Big (\mu _0(1-\beta )M_0'(Y_0(z))\Big )^-=0, \end{aligned}$$

and then

$$\begin{aligned} -A\int _{Q_t} v^m(\partial _z\omega )^2\frac{{\textrm{sign}}\left( \omega \right) -1}{2}e^{\lambda \tau }\textrm{d}\tau \textrm{d}z- a\int _{Q_t} \omega \omega ^-e^{\lambda \tau }\textrm{d}\tau \textrm{d}z=\frac{1}{2}\int _0^1 \omega (t,z)\omega (t,z)^-e^{\lambda t}\textrm{d}z. \end{aligned}$$

Since $\omega \omega ^-\le 0$, the two sides of the identity have different sings. As a consequence, they must vanish and

$$\begin{aligned} \int _0^1(\omega \omega ^-)(t,z)e^{\lambda t}\textrm{d}z=0,\qquad t>0, \end{aligned}$$

that gives $\omega (t,z)^-=0,$ namely $\omega (t,z)\ge 0.$ As in the previous case

$$\begin{aligned} \partial _zv(t,z)=\omega (t,z)\ge 0\,\,\Leftrightarrow \,\, (1-\beta )n_\beta (t,z)^{-\beta }\partial _zn_\beta (t,z)\ge 0\,\,\Leftrightarrow \,\,\partial _zn_\beta (t,z)\le 0. \end{aligned}$$

$\underline{\beta =1}.$ Define

$$\begin{aligned} v=\log n_\beta ,\qquad \omega =\partial _zv, \end{aligned}$$

$\omega$ satisfies in the weak sense the identity

$$\begin{aligned} A\partial _z\left( \frac{1}{n_\beta ^2}\partial _z\omega \right) =\partial _t\omega \quad \text { in }\quad Q_\infty \end{aligned}$$

(3.8)

and in the sense of traces

$$\begin{aligned} \omega (t,0)=-B;\quad \omega (t,1)=0;\quad \omega (0,z)=\mu _0M_0'(Y_0(z)). \end{aligned}$$

We multiply (3.8) by $\omega ^+$ and integrate over $Q_t.$ Arguing as in the previous cases we obtain

$$\begin{aligned} -A\int _{Q_t}\frac{1}{n_\beta ^2}(\partial _z\omega )^2\frac{{\textrm{sign}}\left( \omega \right) +1}{2}\textrm{d}\tau \textrm{d}z=\frac{1}{2}\int _0^1\omega (t,z)\omega (t,z)^+\textrm{d}z, \end{aligned}$$

that implies $\omega (t,z)^+=0,$ namely $\omega (t,z)\le 0.$ Therefore,

$$\begin{aligned} \partial _zv(t,z)=\omega (t,z)\le 0\,\,\Leftrightarrow \,\, \frac{\partial _zn_\beta (t,z)}{n_\beta (t,z)}\le 0 \,\,\Leftrightarrow \,\,\partial _zn_\beta (t,z)\le 0. \end{aligned}$$

In this way we have proved (3.5).

Finally, being $N_\beta (t,y)=n_\beta (t,Z_0(y)),$ we have

$$\begin{aligned} \partial _yN_\beta (t,y)=\partial _zn_\beta (t,Z_0(y))Z_0'(y)=\frac{\partial _zn_\beta (t,Z_0(y))}{\mu _0M_0(y)^\beta }\le 0, \end{aligned}$$

that concludes the proof. $\square$

4 Proof of Theorem 1.1

We begin this section by proving some a priori estimates on $N_\beta (t,y)$ and $\partial _yN_\beta (t,y)$ independent on $\beta$.

Lemma 4.1

We have that

$$\begin{aligned}{} & {} M_0(y)\le N_\beta (t,y)\le \frac{\eta }{\sqrt{\alpha D}}\coth \Big (L_0\sqrt{\frac{\alpha }{D}}\Big ),&\quad (t,y)\in \overline{E}_\infty ; \end{aligned}$$

(4.1)

$$\begin{aligned}{} & {} -\frac{\eta }{D}\le \Big (\frac{M_0(y)}{N_\beta (t,y)}\Big )^\beta \partial _yN_\beta (t,y)\le 0,&\quad (t,y)\in ]0,\infty [\times [0,L_0]. \end{aligned}$$

(4.2)

Proof

We prove (4.1). The lower bound on $N_\beta (\cdot ,y)$ follows from the monotonicity of $N_\beta (\cdot ,y)$ (see (1.3) and Theorem 3.1) and the identity $N_\beta (0,y)=M_0(y)$. We have to prove the upper bound on $N_\beta (\cdot ,y)$. Since

$$\begin{aligned} N_\beta (t,y)\le \lim \limits _{t\rightarrow \infty }N_\beta (t,y)=\overline{N}_\beta (y),\end{aligned}$$

the monotonicity of $\overline{N}_\beta (y)$ (see [6, Theorem 2.1]) and (3.8) guarantee

$$\begin{aligned} N_\beta (t,y)\le \overline{N}_\beta (0)=\overline{M}_\theta (0)=\frac{\eta }{\sqrt{\alpha D}}\coth \Big (\overline{L}_\theta \sqrt{\frac{\alpha }{D}}\Big ). \end{aligned}$$

Moreover, by observing

$$\begin{aligned} L_\theta (t)=\int _0^{L_0}\bigg (\frac{N_\beta (t,y)}{M_0(y)}\bigg )^\beta \textrm{d}y\ge \int _0^{L_0}\bigg (\frac{N_\beta (0,y)}{M_0(y)}\bigg )^\beta \textrm{d}y=L_0, \end{aligned}$$

we must have

$$\begin{aligned} \overline{L}_\theta \ge L_0,\,\,\, t\ge 0. \end{aligned}$$

Since $\coth$ is nonincreasing,

$$\begin{aligned} N_\beta (t,y)\le \frac{\eta }{\sqrt{\alpha D}}\coth \Big (\overline{L}_\theta \sqrt{\frac{\alpha }{D}} \Big )\le \frac{\eta }{\sqrt{\alpha D}}\coth \Bigg (L_0\sqrt{\frac{\alpha }{D}} \Bigg ), \end{aligned}$$

(4.3)

that proves (4.1).

We have to prove (4.2). Thanks to the assumption (1.3) and Theorem 3.1, we know that $\partial _tN_\beta (t,y)\ge 0.$ Using the equation in (2.1) we have that $\Big (\frac{M_0(y)}{N_\beta (t,y)}\Big )^\beta \partial _yN_\beta (t,y)$ is nondecreasing with respect to y,  for every $t>0$ and $\beta > 1.$ Using the boundary conditions, we gain

$$\begin{aligned} -\frac{\eta }{D}=\Big (\frac{M_0(0)}{N_\beta (t,0)}\Big )^\beta \partial _yN_\beta (t,0)\,\le \Big (\frac{M_0(y)}{N_\beta (t,y)}\Big )^\beta \partial _yN_\beta (t,y)\le \Big (\frac{M_0(L_0)}{N_\beta (t,L_0)}\Big )^\beta \partial _yN_\beta (t,L_0)=0. \end{aligned}$$

Employing (1.2), Theorem 2.1 and the fact

$$\begin{aligned} \Big (\frac{M_0(y)}{N_\beta (t,y)}\Big )^\beta>0,\quad t>0,\,\, 0\le y\le L_0, \end{aligned}$$

we conclude $\partial _yN_\beta (t,y)\le 0, \,\, (t,y)\in ]0,\infty [\times [0,L_0],$ that proves (4.2). $\square$

We continue with the following result on the limit of $\partial _tN_\beta (t,y)$ as $\beta \rightarrow \infty$.

Theorem 4.1

We have that

$$\begin{aligned} \displaystyle \lim \limits _{\beta \rightarrow \infty } \int _{E_T} (\partial _tN_\beta )^2\textrm{d}t\textrm{d}y=0. \end{aligned}$$

The following lemma is needed

Lemma 4.2

For every $T>0$

$$\begin{aligned} \limsup \limits _{\beta \rightarrow \infty }\Bigg \{\int _{E_T}\big (\frac{N_\beta }{M_0}\big )^\beta (\partial _tN_\beta )^2 \textrm{d}t\textrm{d}y+ \frac{D\beta }{2(\beta +1)} \int _{E_T}\big (\frac{M_0}{N_\beta }\big )^\beta (\partial _yN_\beta )^2\frac{\partial _tN_\beta }{N_\beta } \textrm{d}t\textrm{d}y\Bigg \}\le 0. \end{aligned}$$

Proof

We multiply the equation in (2.1) by $\big (\frac{N_\beta }{M_0}\big )^\beta \partial _tN_\beta (t,y):$

and integrate over $E_T:$

$$\begin{aligned} \int _{E_T}&\left( \frac{N_\beta }{M_0}\right) ^\beta (\partial _tN_\beta (t,y))^2\textrm{d}t\textrm{d}y+ \frac{a}{\beta +2} \int _0^{L_0} \left( \frac{N_\beta (T,y)^{\beta +2}}{M_0(y)^\beta }-M_0(y)^2\right) \textrm{d}y\\&=\textrm{d}\int _0^T\left[ \left( \frac{M_0}{N_\beta }\right) ^\beta \,\partial _yN_\beta \, \partial _tN_\beta \right] _0^{L_0}\textrm{d}t-\textrm{d}\int _{E_T}\left( \frac{M_0}{N}\right) ^\beta \partial _yN_\beta \,\partial ^2_{yt} N_\beta \textrm{d}t\textrm{d}y\\&=\frac{\eta }{\beta +1}\int _0^T\partial _tN_\beta (t,0)\textrm{d}t-\frac{d}{2}\int _{E_T} \left( \frac{M_0}{N_\beta } \right) ^\beta \partial _t(\partial _yN_\beta )^2 \textrm{d}t\textrm{d}y\\&=\frac{\eta }{\beta +1}(N_\beta (T,0)-M_0(0))\\&\quad -\frac{d}{2}\int _0^{L_0} \left[ \left( \frac{M_0}{N_\beta }\right) ^\beta (\partial _yN_\beta )^2 \right] _0^T\textrm{d}y- \frac{d}{2}\int _{E_T}\beta \frac{M_0^\beta }{N_\beta ^{\beta +1}}\,\partial _tN_\beta \, (\partial _yN_\beta )^2\textrm{d}t\textrm{d}y\\&=\frac{\eta }{\beta +1}(N_\beta (T,0)-M_0(0))-\frac{D}{2(\beta +1)}\int _0^{L_0} \left( \frac{M_0(y)}{N_\beta (T,y)}\right) ^\beta (\partial _yN_\beta (y))^2 \textrm{d}y\\&\quad +\frac{D}{2(\beta +1)}\int _0^{L_0}(M'_0(y))^2\textrm{d}y- \frac{D\beta }{2(\beta +1)}\int _{E_T}\frac{M_0^\beta }{N_\beta ^{\beta +1}}\,\partial _tN_\beta \, (\partial _yN_\beta )^2\textrm{d}t\textrm{d}y. \end{aligned}$$

Rearranging the terms in the following way

$$\begin{aligned} \int _{E_T}&\left( \frac{N_\beta }{M_0}\right) ^\beta (\partial _tN_\beta (t,y))^2\textrm{d}t\textrm{d}y+ \frac{D\beta }{2(\beta +1)}\int _{E_T}\frac{M_0^\beta }{N_\beta ^{\beta +1}}\, \partial _tN_\beta \, (\partial _yN_\beta )^2\textrm{d}t\textrm{d}y\\&\quad +\frac{\alpha }{(\beta +1)(\beta +2)} \int _0^{L_0}\frac{N_\beta (T,y)^{\beta +2}}{M_0(y)^\beta }\textrm{d}y+\frac{D}{2(\beta +1)}\int _0^{L_0} \left( \frac{M_0(y)}{N_\beta (T,y)}\right) ^\beta (\partial _yN_\beta (T,y))^2 \textrm{d}y\\&=\frac{\alpha }{(\beta +1)(\beta +2)}\left\| M_0 \right\| ^2_{L^2(0,L_0)} +\frac{\eta }{\beta +1}(N_\beta (T,0)-M_0(0))+ \frac{D}{2(\beta +1)}\left\| M'_0 \right\| ^2_{L^2(0,L_0)}, \end{aligned}$$

we get the claim. $\square$

Proof

(Proof of Theorem 4.1) Since $\partial _tN_\beta (t,y)\ge 0,$ by Lemma 4.2,

$$\begin{aligned} \lim \limits _{\beta \rightarrow \infty }\int _{E_T}\left( \frac{N_\beta }{M_0}\right) ^\beta (\partial _tN_\beta (t,y))^2\textrm{d}t\textrm{d}y=0. \end{aligned}$$

Being $M_0(y)\le N_\beta (t,y)$ (see (4.1))

$$\begin{aligned} \int _{E_T} (\partial _tN_\beta (t,y))^2\textrm{d}t\textrm{d}y\le \int _{E_T}\left( \frac{N_\beta }{M_0}\right) ^\beta (\partial _tN_\beta )^2 \textrm{d}t\textrm{d}y, \end{aligned}$$

that proves the claim. $\square$

We continue with the behavior of $N_\beta (t,y)$ as $\beta \rightarrow \infty$.

Theorem 4.2

For every $T>0$ and $1\le r<\infty$

$$\begin{aligned} \lim \limits _{\beta \rightarrow \infty }\int _0^{L_0}|N_\beta (t,y)-M_0(y)|^r\textrm{d}y =0, \end{aligned}$$

uniformly with respect to $t\in ]0,T[.$

Proof

Since

$$\begin{aligned} \int _0^{L_0}&|N_\beta (t,y)-M_0(y)|\textrm{d}y=\int _0^{L_0} |N_\beta (t,y)-N_\beta (0,y)|\textrm{d}y\\&=\int _0^{L_0}\left| \int _0^t \partial _\tau N_\beta (\tau ,y)\textrm{d}\tau \right| \textrm{d}y\le \int _{E_t} \left| \partial _\tau N_\beta (\tau ,y)\right| \textrm{d}\tau \textrm{d}y\le \sqrt{TL_0}\cdot \left\| \partial \tau N_\beta \right\| _{L^2(E_T)}, \end{aligned}$$

thanks to Theorem 4.1

$$\begin{aligned} \lim \limits _{\beta \rightarrow \infty }\int _0^{L_0}|N_\beta (t,y)-M_0(y)|\textrm{d}y =0, \end{aligned}$$

uniformly with respect to $t\in ]0,T[.$

The boundedness of $(N_\beta )_{\beta > 1}$ in $L^\infty (E_T)$ (see (4.1)) and the boundedness of $M_0(y)$ (see (1.2)) imply the claim. $\square$

We are finally ready for the proof of Theorem 1.1.

Proof

(Proof of Theorem 1.1) Since $M_\theta (t,X_\theta (t,y))=N_\beta (t,y)$ and $\beta =\frac{\log 2}{\theta }$, for every , we have

In light of Theorem 4.2, we have i).

Since $\partial _tN_\beta (t,y)\ge 0,\,\,(t,y)\in E_T,$ (see Theorem 3.1.i)) and $\beta >0$,

The monotonicity of $N_\beta (t,y)$ with respect to t, $0\le Y_\theta (t,x)\le L_0$, $N_\beta (0,y)=M_0(y)$ and the definition of $u_\theta (t,x)$ guarantee

The monotonicity assumption on $M_0(y)$ gives

$$\begin{aligned} u_\theta (t,x)\le \frac{\beta }{M_0(L_0)}\int _0^{L_0} \left( \frac{N_\beta (t,y)}{M_0(y)}\right) ^\beta \partial _tN_\beta (t,y)\textrm{d}y, \end{aligned}$$

and using the equation in (2.1)

$$\begin{aligned} u_\theta (t,x)&\le \frac{\beta }{M_0(L_0)}\int _0^{L_0} \left\{ \textrm{d}\partial _y\left( \left( \frac{N_\beta (t,y)}{M_0(y)}\right) ^\beta \partial _tN_\beta (t,y)\right) -a\frac{N_\beta (t,y)^{\beta +1}}{M_0(y)^\beta }\right\} \textrm{d}y\\&=\frac{\beta }{\beta +1}\frac{1}{M_0(L_0)}\left\{ \int _0^{L_0} D\partial _y\left( \left( \frac{N_\beta (t,y)}{M_0(y)}\right) ^\beta \partial _yN_\beta (t,y)\right) \textrm{d}y-\alpha \int _0^{L_0} \frac{N_\beta (t,y)^{\beta +1}}{M_0(y)^\beta }\right\} \textrm{d}y. \end{aligned}$$

The boundary conditions in (2.1) and Lemma 3.1:

$$\begin{aligned} u_\theta (t,x)&\le \frac{\beta }{\beta +1}\frac{1}{M_0(L_0)} \left\{ D\frac{\eta }{D}-\alpha \left[ \frac{\eta }{\alpha }+ e^{-\alpha t}\Big (\left\| M_0 \right\| _{L^1(0,L_0)}-\frac{\eta }{\alpha }\Big )\right] \right\} \\&=\frac{\beta }{\beta +1}\frac{e^{-\alpha t}}{M_0(L_0)}\Big (\eta -\alpha \left\| M_0 \right\| _{L^1(0,L_0)}\Big ). \end{aligned}$$

As a consequence,

$$\begin{aligned} \limsup \limits _{\theta \rightarrow 0}u_\theta (t,x)\le e^{-\alpha t} \frac{\eta -\alpha \left\| M_0 \right\| _{L^1(0,L_0)}}{M_0(L_0)}, \end{aligned}$$

that proves ii)

The equation in (2.1) gives

(4.4)

and then

where

$$\begin{aligned} a^*=a\beta =\frac{\alpha \beta }{\beta +1};\quad \textrm{d}^*=\textrm{d}\beta =\frac{D\beta }{\beta +1}. \end{aligned}$$

Integrating with respect to y on $[0, L_0]$

$$\begin{aligned}&\partial _t\left( e^{a^*t}\int _0^{L_0}\left( \frac{N_\beta }{M_0}\right) ^\beta \textrm{d}y\right) =\textrm{d}^*e^{a^*t}\int _0^{L_0}\frac{1}{ N_\beta }\partial _y\left( \left( \frac{M_0}{N_\beta }\right) ^\beta \partial _yN_\beta \right) \textrm{d}y\\&\qquad =\,\textrm{d}^*e^{a^*t}\left\{ \left[ \frac{1}{ N_\beta }\left( \frac{M_0}{N_\beta }\right) ^\beta \partial _yN_\beta \right] _0^{L_0} + \int _0^{L_0}\left( \frac{M_0}{N_\beta }\right) ^\beta \left( \frac{\partial _yN_\beta }{N_\beta }\right) ^2\textrm{d}y\right\} \\&\qquad =\textrm{d}^*e^{a^*t}\left\{ \frac{1}{ N_\beta (t,0)}\frac{\eta }{D} + \int _0^{L_0}\left( \frac{M_0}{N_\beta }\right) ^\beta \left( \frac{\partial _yN_\beta }{N_\beta }\right) ^2\textrm{d}y\right\} . \end{aligned}$$

Integrating with respect to t on [0, T]

$$\begin{aligned} e^{a^*T} \int _0^{L_0}&\left( \frac{N_\beta (T,y)}{M_0(y)}\right) ^\beta \textrm{d}y-L_0\\&=\frac{\eta \beta }{\beta +1}\int _0^T\frac{e^{a^*t}}{N_\beta (t,0)}\textrm{d}t+ \textrm{d}^*\int _0^Te^{a^*t}\textrm{d}t \int _0^{L_0}\left( \frac{M_0(y)}{N_\beta (t,y)}\right) ^\beta \left( \frac{\partial _yN_\beta (t,y)}{N_\beta (t,y)}\right) ^2\textrm{d}y \end{aligned}$$

and then

$$\begin{aligned} L_\theta (T)&=L_0e^{-a^*T}+\frac{\eta \beta }{\beta +1}\int _0^T \frac{e^{-a^*(T-t)}}{N_\beta (t,0)}\textrm{d}t+ \\&\quad +\textrm{d}^*\int _0^Te^{-a^*(T-t)}\textrm{d}t \int _0^{L_0}\left( \frac{M_0(y)}{N_\beta (t,y)}\right) ^\beta \left( \frac{\partial _yN_\beta (t,y)}{N_\beta (t,y)}\right) ^2\textrm{d}y. \end{aligned}$$

Thanks to Theorem 3.2

$$\begin{aligned} \begin{aligned}&L_0e^{-a^*T}+\frac{\eta \beta }{\beta +1}\int _0^T \frac{e^{-a^*(T-t)}}{N_\beta (t,0)}\textrm{d}t\le L_\theta (T)\\&\quad \le L_0e^{-a^*T}+\frac{\eta \beta }{\beta +1}\int _0^T \frac{e^{-a^*(T-t)}}{N_\beta (t,0)}\textrm{d}t +\textrm{d}^*\int _0^Te^{-a^*(T-t)}\textrm{d}t \int _0^{L_0}\frac{\eta }{D} \frac{|\partial _yN_\beta (t,y)|}{N_\beta (t,y)^2}\textrm{d}y. \end{aligned}\end{aligned}$$

(4.5)

Since $\partial _yN_\beta (t,y)\le 0,\,\, (t,y)\in E_\infty ,$ (see Theorem 3.2), we observe

$$\begin{aligned} \textrm{d}^*\int _0^Te^{-a^*(T-t)}\textrm{d}t \int _0^{L_0}\frac{\eta }{D} \frac{|\partial _yN_\beta (t,y)|}{N_\beta (t,y)^2}\textrm{d}y&= -\frac{\eta \beta }{\beta +1}\int _0^Te^{-a^*(T-t)}\textrm{d}t \int _0^{L_0} \frac{\partial _yN_\beta (t,y)}{N_\beta (t,y)^2}\textrm{d}y\\&=\frac{\eta \beta }{\beta +1}\int _0^Te^{-a^*(T-t)}\left[ \frac{1}{N_\beta (t,y)} \right] _0^{L_0}\textrm{d}t \\&= \frac{\eta \beta }{\beta +1}\int _0^Te^{-a^*(T-t)}\left( \frac{1}{N_\beta (t,L_0)}-\frac{1}{N_\beta (t,0)} \right) \textrm{d}t. \end{aligned}$$

Using (4.5)

Thanks to (4.1)

Since $a^*=\frac{\alpha \beta }{\beta +1}$

Sending $\theta \rightarrow 0^+$, namely $\beta \rightarrow \infty$, we obtain iii). $\square$

5 Proof of Theorem 1.2

We begin by proving some a priori estimates on $N_\beta (t,y)$ and $\partial _yN_\beta (t,y)$ independent on $\beta$.

Lemma 5.1

We have that

$$\begin{aligned}&c_*e^{-\frac{\alpha }{\beta +1}t}\le N_\beta (t,y)\le M_0(0),&\quad (t,y)\in \overline{E}_\infty ; \end{aligned}$$

(5.1)

$$\begin{aligned}{} & {} -M_0(y)\sqrt{\frac{\alpha }{D}}\le \Big (\frac{M_0(y)}{N_\beta (t,y)}\Big )^\beta \partial _yN_\beta (t,y)\le 0;&\quad (t,y)\in ]0,\infty [\times [0,L_0]. \end{aligned}$$

(5.2)

Proof

The lower bound in (5.1) follows from Theorem 2.1.i). For the upper bound in (5.1), we observe that $N_\beta (\cdot ,y)$ and $M_0(y)$ are nonincreasing (see (1.4) and Theorem 3.1); therefore,

$$\begin{aligned} N_\beta (t,y)\le M_0(y)\le M_0(0). \end{aligned}$$

We multiply the equation in (2.1) by $\partial _yN_\beta (t,y)$

$$\begin{aligned} \partial _tN_\beta (t,y)\partial _yN_\beta (t,y)=\frac{d}{2}\partial _y\left( \Big ( \frac{M_0(y)}{N_\beta (t,y)}\Big )^{2\beta }(\partial _yN_\beta (t,y))^2\right) -\frac{a}{2}\partial _yN_\beta (t,y)^2. \end{aligned}$$

(5.3)

Since $\partial _yN_\beta (t,y)\le 0$ (see Theorem 3.2) and $\partial _tN_\beta (t,y)\le 0$ (see Theorem 3.1), thanks to (5.3),

$$\begin{aligned} \textrm{d}\partial _y\left( \Big ( \frac{M_0(y)}{N_\beta (t,y)}\Big )^{2\beta }(\partial _yN_\beta (t,y))^2\right) \ge a\partial _yN_\beta (t,y)^2. \end{aligned}$$

Integrating with respect to y over $[\xi ,L_0],\,\,0\le \xi \le L_0,$

$$\begin{aligned} \textrm{d}\int _\xi ^{L_0}\partial _y\left( \Big ( \frac{M_0(y)}{N_\beta (t,y)}\Big )^{2\beta }(\partial _yN_\beta (t,y))^2\right) \textrm{d}y\ge a\int _\xi ^{L_0} \partial _yN_\beta (t,y)^2 \textrm{d}y, \end{aligned}$$

and using the boundary conditions in (2.1) and (2.2)

$$\begin{aligned} -\left( \Big ( \frac{M_0(\xi )}{N_\beta (t,\xi )}\Big )^{2\beta }(\partial _\xi N_\beta (t,\xi ))^2\right) \ge \frac{\alpha }{D}(N_\beta (t,L_0)^2- N_\beta (t,\xi )^2), \end{aligned}$$

namely

$$\begin{aligned} \left( \Big ( \frac{M_0(\xi )}{N_\beta (t,\xi )}\Big )^{2\beta }(\partial _\xi N_\beta (t,\xi ))^2\right) \le \frac{\alpha }{D}(N_\beta (t,\xi )^2-N_\beta (t,L_0)^2)\le \frac{\alpha }{D}N_\beta (t,\xi )^2. \end{aligned}$$

Since $N_\beta (\cdot ,\xi )$ is nonincreasing

$$\begin{aligned} \Big ( \frac{M_0(\xi )}{N_\beta (t,\xi )}\Big )^{\beta }|\partial \xi N_\beta (t,\xi )| \le \sqrt{\frac{\alpha }{D}} N_\beta (t,\xi )\le \sqrt{\frac{\alpha }{D}} N_\beta (0,\xi )= \sqrt{\frac{\alpha }{D}} M_0(\xi ), \end{aligned}$$

using $\partial _yN_\beta (t,\xi )\le 0$, we have (5.2). $\square$

We continue with the analysis of the behavior of $\partial _tN_\beta (t,y)$ as $\beta \rightarrow \infty$.

Theorem 5.1

For every $T>0$

$$\begin{aligned} \lim \limits _{\beta \rightarrow \infty }\int _{E_T} \Big ( \frac{N_\beta (t,y)}{M_0(y)}\Big )^{\beta }|\partial _tN_\beta (t,y)|\textrm{d}t\textrm{d}y=0. \end{aligned}$$

The following lemma is needed.

Lemma 5.2

We have that

$$\begin{aligned}&\lim \limits _{\beta \rightarrow \infty }\int _0^{L_0}\partial _tN_\beta (t,y) \partial _yN_\beta (t,y)\textrm{d}y=0,\>\>\text {uniformly with respect} to t\in [0,\infty [;\ \end{aligned}$$

(5.4)

$$\begin{aligned}&\lim \limits _{\beta \rightarrow \infty }\int _{E_T}\Big ( \frac{N_\beta (t,y)}{M_0(y)}\Big )^{\beta }(\partial _tN_\beta (t,y))^2\textrm{d}t\textrm{d}y=0,\>\>T>0. \end{aligned}$$

(5.5)

Proof

We multiply the equation in (2.1) by $\partial _yN_\beta (t,y)$

$$\begin{aligned} \partial _tN_\beta (t,y)\partial _yN_\beta (t,y)+\frac{a}{2}\partial _yN_\beta (t,y)^2=\frac{d}{2}\partial _y\left( \Big ( \frac{M_0(y)}{N_\beta (t,y)}\Big )^{2\beta }(\partial _yN_\beta (t,y))^2\right) , \end{aligned}$$

Integrating with respect to y on $[0,L_0]$ and using Theorem 2.1 and the boundary conditions in (2.1)

$$\begin{aligned} \int _0^{L_0} \partial _tN_\beta (t,y)\partial _yN_\beta (t,y)\textrm{d}y+\frac{\alpha }{2(\beta +1)} (N_\beta (t,L_0)^2-N_\beta (t,0)^2)=\frac{D}{2(\beta +1)}\big (-\frac{\eta ^2}{D^2}\big ). \end{aligned}$$

Thanks to (5.1)

$$\begin{aligned} \left| \int _0^{L_0} \partial _tN_\beta (t,y)\partial _yN_\beta (t,y)\textrm{d}y\right| \le \frac{1}{2(\beta +1)}\big (\alpha M_0(0)^2+\frac{\eta ^2}{D}\big ), \end{aligned}$$

and the (5.4).

Lemma 4.2 holds independently on the sign of $DM_0''(y)-\alpha M_(y)$, (1.3) and (1.4). Indeed, its proof uses only (1.2) and Theorem 2.1. Therefore

$$\begin{aligned} \limsup \limits _{\beta \rightarrow \infty }&\Big \{\int _{E_T}\big (\frac{N_\beta (t,y)}{M_0(y)}\big )^\beta (\partial _tN_\beta (t,y))^2 \textrm{d}t\textrm{d}y+\nonumber \\&+\frac{D\beta }{2(\beta +1)} \int _{E_T}\big (\frac{M_0(y)}{N_\beta (t,y)}\big )^\beta (\partial _yN_\beta (t,y))^2\frac{\partial _tN_\beta (t,y)}{N_\beta (t,y)} \textrm{d}t\textrm{d}y\Big \}\le 0. \end{aligned}$$

(5.6)

Since $N_\beta (\cdot ,y)$ is nonincreasing, the second term is negative; as a consequence in order to prove (5.5), it is enough to prove that the second term vanishes as $\beta \rightarrow \infty$.

Thanks to Lemma 5.1

$$\begin{aligned} I_\beta&:=\left| \int _{E_T}\big (\frac{M_0(y)}{N_\beta (t,y)}\big )^\beta (\partial _yN_\beta (t,y))^2\frac{\partial _tN_\beta (t,y)}{N_\beta (t,y)} \textrm{d}t\textrm{d}y\right| \\&=\int _{E_T}\big (\frac{M_0(y)}{N_\beta (t,y)}\big )^\beta | \partial _yN_\beta (t,y) | \frac{|\partial _tN_\beta (t,y) \partial _yN_\beta (t,y) |}{N_\beta (t,y)} \textrm{d}t\textrm{d}y\\&\le \sqrt{\frac{\alpha }{D}} \int _{E_T}\frac{M_0(y)}{N_\beta (t,y)} |\partial _tN_\beta (t,y) \partial _yN_\beta (t,y) | \textrm{d}t\textrm{d}y. \end{aligned}$$

Using (1.2) and Lemma 5.1

$$\begin{aligned} I_\beta \le \sqrt{\frac{\alpha }{D}}\frac{c^*}{c_*e^{-\frac{\alpha }{\beta +1}T}}\int _{E_T} |\partial _tN_\beta (t,y) \partial _yN_\beta (t,y) | \textrm{d}t\textrm{d}y, \end{aligned}$$

since $\partial _yN_\beta (t,y)\le 0$ (see Theorem 3.2) and $\partial _tN_\beta (t,y)\le 0$ (see Theorem 3.1)

$$\begin{aligned} I_\beta \le \sqrt{\frac{\alpha }{D}}\frac{c^*}{c_*e^{-\frac{\alpha }{\beta +1}T}}\int _{E_T} \partial _tN_\beta (t,y) \partial _yN_\beta (t,y) \textrm{d}t\textrm{d}y. \end{aligned}$$

By (5.4)

$$\begin{aligned} \lim \limits _{\beta \rightarrow \infty }I_\beta =0, \end{aligned}$$

that gives (5.5). $\square$

Proof of Theorem 5.1

Since $N_\beta (\cdot ,y)$ in nonincreasing

$$\begin{aligned} \frac{N_\beta (t,y)}{M_0(y)}\le 1,\quad (t,y)\in E_\infty , \end{aligned}$$

and then

$$\begin{aligned} \int _{E_T}\big (\frac{N_\beta (t,y)}{M_0(y)}\big )^\beta |\partial _tN_\beta (t,y)| \textrm{d}t\textrm{d}y\le&\int _{E_T}\big (\frac{N_\beta (t,y)}{M_0(y)}\big )^{\frac{\beta }{2}} |\partial _tN_\beta (t,y)| \textrm{d}t\textrm{d}y\le \\ \le&\sqrt{TL_0} \sqrt{\int _{E_T}\big (\frac{N_\beta (t,y)}{M_0(y)}\big )^\beta (\partial _tN_\beta (t,y))^2\textrm{d}t\textrm{d}y}. \end{aligned}$$

The claim follows from (5.5). $\square$

We study the behavior of $N_\beta (t,y)$ as $\beta \rightarrow \infty$.

Theorem 5.2

For every $0\le T<\infty$

$$\begin{aligned} \lim \limits _{\beta \rightarrow \infty }\int _0^{L_0}\Big (\frac{N_\beta (t,y)}{M_0(y)}\Big )^\beta |N_\beta (t,y)-M_0(y)|^r\textrm{d}y=0,\quad 1\le r<\infty , \end{aligned}$$

uniformly with respect to $t\in [0,T].$

Proof

Consider

$$\begin{aligned} I_\beta (t):=&\int _0^{L_0}\Big (\frac{N_\beta (t,y)}{M_0(y)}\Big )^\beta |N_\beta (t,y)-M_0(y)|\textrm{d}y\\ =&\int _0^{L_0}\Big (\frac{N_\beta (t,y)}{M_0(y)}\Big )^\beta \Big |\int _0^t \partial _\tau N_\beta (\tau ,y)\textrm{d}\tau \Big |\textrm{d}y. \end{aligned}$$

Since $\Big (\frac{N_\beta (\cdot ,y)}{M_0(y)}\Big )^\beta$ is nonincreasing, for every $0\le \tau \le t$

$$\begin{aligned} \Big (\frac{N_\beta (t,y)}{M_0(y)}\Big )^\beta \le \Big (\frac{N_\beta (\tau ,y)}{M_0(y)}\Big )^\beta \end{aligned}$$

and then

$$\begin{aligned} I_\beta (t)\le \int _0^{L_0}\textrm{d}y\int _0^t\Big (\frac{N_\beta (\tau ,y)}{M_0(y)}\Big )^\beta \Big |\partial _\tau N_\beta (\tau ,y)\Big |\textrm{d}\tau . \end{aligned}$$

Given $0\le T<\infty$

$$\begin{aligned} I_\beta (t)\le \int _{E_T} \Big (\frac{N_\beta (\tau ,y)}{M_0(y)}\Big )^\beta \Big |\partial _\tau N_\beta (\tau ,y)\Big |\textrm{d}\tau \textrm{d}y, \qquad 0\le t\le T. \end{aligned}$$

Using Theorem 5.1

$$\begin{aligned} \lim \limits _{\beta \rightarrow \infty } I_\beta (t)=0 \end{aligned}$$

uniformly with respect to $t\in [0,T].$ Finally, since $(N_\beta )_{\beta >1}$ is bounded in $L^\infty (E_T)$ (see Lemma 5.1.i)) and $M_0\in L^\infty (0,L_0)$ (see (1.2)),

$$\begin{aligned} \int _0^{L_0}\Big (\frac{N_\beta (t,y)}{M_0(y)}\Big )^\beta |N_\beta (t,y)-M_0(y)|^r \textrm{d}y\le \big (M_0(0)+c^*\big )^{r-1} I_\beta (t),\qquad 1<r<\infty , \end{aligned}$$

that gives the claim. $\square$

We are finally ready for the proof of Theorem 1.2.

Proof of Theorem 1.2

$$\begin{aligned} \Delta _\beta (t)=\int _0^{L_\theta (t)}|M_\theta (t,x)-M_0(Y_\theta (t,x))|^r\textrm{d}x \end{aligned}$$

and consider the change of variable $y=Y_\theta (t,x).$ For every t, $x=X_\theta (t,y)$ is the inverse of $y=Y_\theta (t,x),$ therefore

$$\begin{aligned} \textrm{d}x=\partial _yX_\theta (t,y)\textrm{d}y=\frac{\textrm{d}y}{\partial _xY_\theta (t,X_\theta (t,y))}=\Big (\frac{N_\beta (t,y)}{M_0(y)}\Big )^\beta \textrm{d}y. \end{aligned}$$

Thanks to the definition of $N_\beta (t,y)$,

$$\begin{aligned} \Delta _\beta (t)&=\int _0^{L_0} |M_\theta (t,X_\theta (t,y))-M_0(y)|^r \Big (\frac{N_\beta (t,y)}{M_0(y)}\Big )^\beta \textrm{d}y\\&=\int _0^{L_0}\Big (\frac{N_\beta (t,y)}{M_0(y)}\Big )^\beta |N_\beta (t,y)-M_0(y)|^r\textrm{d}y, \end{aligned}$$

and, using Theorem 5.2, we get i).

Thanks to (1.4) and Theorem 3.1.ii), we have $\partial _tN_\beta (t,y)\le 0,\,\,(t,y)\in E_\infty$. Moreover, since $0\le Y_\theta (t,x)\le L_0$,

$$\begin{aligned} u_\theta (t,x)=\beta \int _0^{Y_\theta (t,x)}\frac{N_\beta (t,y)^{\beta -1}\partial _tN_\beta (t,y)}{M_0(y)^\beta }\textrm{d}y\le 0 \end{aligned}$$

(5.7)

and

$$\begin{aligned} |u_\theta (t,x)|\le \beta \int _0^{L_0}\frac{1}{N_\beta (t,y)}\Big (\frac{N_\beta (t,y)}{M_0(y)}\Big )^\beta |\partial _tN_\beta (t,y)|\textrm{d}y. \end{aligned}$$

Due to (5.1)

$$\begin{aligned} |u_\theta (t,x)|\le \frac{\beta }{c_*e^{-\frac{\alpha }{\beta +1}t}} \int _0^{L_0}\Big (\frac{N_\beta (t,y)}{M_0(y)}\Big )^\beta |\partial _tN_\beta (t,y)|\textrm{d}y, \end{aligned}$$

where $c_*$ is defined in (1.2). Since $M_0(y)$ is nonincreasing (see (1.4))

$$\begin{aligned} |u_\theta (t,x)|&\le \frac{\beta e^{\frac{\alpha }{\beta +1}t}}{\inf \limits _{0\le y\le L_0}M_0(y)}\int _0^{L_0}\Big (\frac{N_\beta (t,y)}{M_0(y)}\Big )^\beta |\partial _tN_\beta (t,y)|\textrm{d}y\\&=\frac{\beta e^{\frac{\alpha }{\beta +1}t}}{M_0(L_0)}\int _0^{L_0}\Big (\frac{N_\beta (t,y)}{M_0(y)}\Big )^\beta |\partial _tN_\beta (t,y)|\textrm{d}y. \end{aligned}$$

Being $|\partial _tN_\beta (t,y)|=-\partial _tN_\beta (t,y)$, thanks to the equation in (2.1) and (2.2)

$$\begin{aligned} |u_\theta (t,x)|&\le \frac{\beta e^{\frac{\alpha }{\beta +1}t}}{(\beta +1)M_0(L_0)} \left\{ \alpha \int _0^{L_0}\frac{N_\beta (t,y)^{\beta +1}}{M_0(y)^\beta }\textrm{d}y -D\int _0^{L_0}\partial _y\left( \Big (\frac{N_\beta (t,y)}{M_0(y)}\Big )^\beta \partial _yN_\beta (t,y)\right) \textrm{d}y\right\} . \end{aligned}$$

Lemma 3.1 and the boundary conditions in (2.1) imply

$$\begin{aligned} |u_\theta (t,x)|&\le \frac{\beta }{\beta +1} \frac{e^{\frac{\alpha }{\beta +1}t}}{M_0(L_0)} \left\{ \alpha \Big ( \frac{\eta }{\alpha }+e^{-\alpha t}(\left\| M_0 \right\| _{L^1(0,L_0)} -\frac{\eta }{\alpha })\Big )-D\frac{\eta }{D}\right\} \\&=\frac{\beta }{\beta +1} \frac{e^{\frac{\alpha }{\beta +1}t}}{M_0(L_0)} \left\{ \alpha \left\| M_0 \right\| _{L^1(0,L_0)} -\eta \right\} . \end{aligned}$$

In light of (5.7) we get

$$\begin{aligned} -\frac{\beta }{\beta +1}\frac{e^{\frac{\alpha }{\beta +1}t}}{M_0(L_0)} \left\{ \alpha \left\| M_0 \right\| _{L^1(0,L_0)} -\eta \right\} \le u_\theta (t,x)\le 0, \end{aligned}$$

that gives ii).

We observe that

$$\begin{aligned} L_\theta (t)=&\int _0^{L_0} \Big (\frac{N_\beta (t,y)}{M_0(y)}\Big )^\beta \textrm{d}y \\ =&\underbrace{\int _0^{L_0} \Big (\frac{N_\beta (t,y)}{M_0(y)}\Big )^\beta \frac{M_0(y)-N_\beta (t,y)}{M_0(y)}\textrm{d}y}_{I_\beta (t)}+ \underbrace{\int _0^{L_0} \Big (\frac{N_\beta (t,y)}{M_0(y)}\Big )^\beta \frac{ \textrm{d}y}{M_0(y)}}_{II_\beta (t)}. \end{aligned}$$

Being $N_\beta (\cdot ,y)$ and $M_0(y)$ nonincreasing, we have

$$\begin{aligned} N_\beta (t,y)\le&N_\beta (0,y)=M_0(y),\qquad M_0(L_0)\le M_0(y),\\ 0\le I_\beta (t)\le&\frac{1}{M_0(L_0)} \int _0^{L_0} \Big (\frac{N_\beta (t,y)}{M_0(y)}\Big )^\beta (M_0(y)-N_\beta (t,y))\textrm{d}y, \end{aligned}$$

and using Theorem 5.2

$$\begin{aligned} \lim _{\beta \rightarrow \infty } I_\beta (t)=0,\quad \text {uniformly with respect to} t\in [0,T],\,\,0\le T<\infty . \end{aligned}$$

From the equation in (2.1), we get

$$\begin{aligned} \partial _t\left( e^{\alpha t}\,\frac{N_\beta (t,y)^{\beta +1}}{M_0(y)^\beta }\right) = D\partial _y\left( e^{\alpha t}\, \Big (\frac{M_0(y)}{N_\beta (t,y)}\Big )^\beta \partial _yN_\beta (t,y)\right) . \end{aligned}$$

Multiplying by $1/{M_0(y)}$ and integrating over $[0,L_0]$

$$\begin{aligned} \int _0^{L_0}\partial _t\left( e^{\alpha t}\,\frac{N_\beta (t,y)^{\beta +1}}{M_0(y)^\beta }\frac{1}{M_0(y)}\right) \textrm{d}y=D\int _0^{L_0}\partial _y\left( e^{\alpha t}\, \Big (\frac{M_0(y)}{N_\beta (t,y)}\Big )^\beta \partial _yN_\beta (t,y)\right) \frac{\textrm{d}y}{M_0(y)}, \end{aligned}$$

and then

$$\begin{aligned} \partial _t&\left( e^{\alpha t}\int _0^{L_0}\frac{N_\beta (t,y)^{\beta +1}}{M_0(y)^\beta }\frac{\textrm{d}y}{M_0(y)}\right) \\&=De^{\alpha t}\left[ \Big (\frac{M_0(y)}{N_\beta (t,y)}\Big )^\beta \frac{\partial _yN_\beta (t,y)}{M_0(y)}\right] _0^{L_0} -De^{\alpha t}\int _0^{L_0} \Big (\frac{M_0(y)}{N_\beta (t,y)}\Big )^\beta \partial _yN_\beta (t,y)\Big (-\frac{M'_0(y)}{M_0(y)^2}\Big )\textrm{d}y\\&=De^{\alpha t}\frac{\eta }{D}\frac{1}{M_0(0)}+De^{\alpha t}\int _0^{L_0} \Big (\frac{M_0(y)}{N_\beta (t,y)}\Big )^\beta \frac{ \partial _yN_\beta (t,y) M'_0(y)}{M_0(y)^2}\textrm{d}y. \end{aligned}$$

Using (1.4) and Theorem 3.2, we have $\partial _yN_\beta (t,y)M'_0(y)\ge 0,$ that gives

$$\begin{aligned} e^{\alpha t}\frac{\eta }{M_0(0)}\le&\partial _t\left( e^{\alpha t}\int _0^{L_0}\frac{N_\beta (t,y)^{\beta +1}}{M_0(y)^\beta }\frac{\textrm{d}y}{M_0(y)}\right) \le \\ \le&e^{\alpha t}\frac{\eta }{M_0(0)}+De^{\alpha t}\int _0^{L_0} \Big (\frac{M_0(y)}{N_\beta (t,y)}\Big )^\beta \frac{ \partial _yN_\beta (t,y) M'_0(y)}{M_0(y)^2}\textrm{d}y. \end{aligned}$$

Thanks to (5.2) and (1.4)

$$\begin{aligned} e^{\alpha t}\frac{\eta }{M_0(0)}&\le \partial _t\left( e^{\alpha t}\int _0^{L_0}\frac{N_\beta (t,y)^{\beta +1}}{M_0(y)^\beta }\frac{\textrm{d}y}{M_0(y)}\right) \\&\le e^{\alpha t}\frac{\eta }{M_0(0)}+De^{\alpha t}\sqrt{\frac{\alpha }{D}}\int _0^{L_0} \frac{| M'_0(y)|}{M_0(y)}\textrm{d}y\\&=e^{\alpha t}\frac{\eta }{M_0(0)}-e^{\alpha t}\sqrt{\alpha D}\int _0^{L_0} \frac{ M'_0(y)}{M_0(y)}\textrm{d}y\\&=e^{\alpha t}\frac{\eta }{M_0(0)}+e^{\alpha t}\sqrt{\alpha D}\log \frac{M_0(0)}{M_0(L_0)}. \end{aligned}$$

Integrating with respect to t on [0, T]

$$\begin{aligned} \frac{e^{\alpha T}-1}{\alpha }\frac{\eta }{M_0(0)}&\le e^{\alpha T}\int _0^{L_0} \frac{N_\beta (T,y)^{\beta +1}}{M_0(y)^\beta }\frac{\textrm{d}y}{M_0(y)}-L_0\\&\le \frac{e^{\alpha T}-1}{\alpha }\Big (\frac{\eta }{M_0(0)}+\sqrt{\alpha D}\log \frac{M_0(0)}{M_0(L_0)}\Big ). \end{aligned}$$

Since

$$\begin{aligned} L_\theta (T)-I_\beta (T)=II_\beta (T)=\int _0^{L_0} \frac{N_\beta (T,y)^{\beta +1}}{M_0(y)^\beta }\frac{\textrm{d}y}{M_0(y)}, \end{aligned}$$

we have

$$\begin{aligned} L_0 e^{-\alpha T}+\frac{1-e^{-\alpha T}}{\alpha }\frac{\eta }{M_0(0)}\le L_\theta (T)-I_\beta (T)\le \\ \le L_0 e^{-\alpha T}+\frac{1-e^{-\alpha T}}{\alpha }\frac{\eta }{M_0(0)}+ \frac{1-e^{-\alpha T}}{\alpha }\sqrt{\alpha D}\log \frac{M_0(0)}{M_0(L_0)}. \end{aligned}$$

Sending $\beta \rightarrow \infty$ we get the claim. $\square$