1 Introduction

In this paper, we consider the nonlinear evolution equations with damping and diffusion:

$$ \textstyle\begin{cases} \psi_{t}^{\beta}=-( \sigma-\alpha)\psi^{\beta}-\sigma\theta _{x}^{\beta}+ \alpha\psi^{\beta}_{xx}, \\ \theta_{t}^{\beta}=-(1-\beta)\theta^{\beta}+\mu\beta \psi^{\beta}_{x}+2\psi^{\beta}\theta_{x}^{\beta}+ \beta\theta^{\beta}_{xx},\quad 0< x< 1, t>0, \end{cases} $$
(1.1)

with the initial-boundary conditions

$$ \begin{aligned} &\bigl(\psi^{\beta}, \theta^{\beta}\bigr) (x,0)=(\psi_{0},\theta_{0}) (x),\quad 0 \leq x\leq1, \\ &\bigl(\psi^{\beta},\theta^{\beta}\bigr) (1,t)=\bigl( \psi^{\beta},\theta^{\beta}\bigr) (0,t)=(0,0),\quad t \geq0, \end{aligned} $$
(1.2)

where \(\sigma, \alpha, \beta\), and μ are positive constants with \(\alpha<\sigma\) and \(0<\beta<1\). The corresponding problem of zero diffusion limit as \(\beta\rightarrow0\) is the following:

$$ \textstyle\begin{cases} \psi_{t}^{0}=-( \sigma-\alpha)\psi^{0}-\sigma\theta_{x}^{0}+\alpha \psi^{0}_{xx}, \\ \theta_{t}^{0}=-\theta^{0}+2\psi^{0} \theta_{x}^{0},\quad 0< x< 1, t>0, \end{cases} $$
(1.3)

with the initial-boundary conditions

$$ \begin{aligned} &\bigl(\psi^{0}, \theta^{0}\bigr) (x,0)=(\psi_{0},\theta_{0}) (x),\quad 0 \leq x\leq1, \\ & \psi^{0}(1,t)=\psi^{0}(0,t)=0,\quad t \geq0. \end{aligned} $$
(1.4)

The system (1.1) was originally proposed by Hsieh in [2] to observe the nonlinear interaction between ellipticity and dissipation. In [3], Hsieh et al. established a link between this interaction and chaos. We also refer to [4, 5] for the physical background of (1.1). Some similar problems were studied in [6, 7] and the references therein.

Our main purpose is to estimate the thickness of boundary layer for problem (1.1)-(1.2) as \(\beta\rightarrow0\). Before stating the main result, we first recall the concept of BL-thickness in the sprit of [8].

Definition 1.1

A function \(\delta(\beta)\) is called a BL-thickness for problem (1.1)-(1.2) with vanishing diffusion if \(\delta(\beta)\downarrow0\) as \(\beta\downarrow0\), and

$$ \begin{aligned} &\lim_{\beta\rightarrow0}\bigl\Vert \psi^{\beta}-\psi^{0}\bigr\Vert _{L^{\infty}(0,T;L^{\infty}[0,1])}=0, \\ &\lim_{\beta\rightarrow0}\bigl\Vert \theta^{\beta}- \theta^{0}\bigr\Vert _{L^{\infty}(0,T;L^{\infty}[\delta,1-\delta])}=0, \\ &\mathop{\inf\lim} _{\beta\rightarrow0}\bigl\Vert \theta^{\beta}- \theta^{0}\bigr\Vert _{L^{\infty}(0,T;L^{\infty}[0,1])}>0, \end{aligned} $$

for any \(T>0\), where \((\psi^{\beta}, \theta^{\beta})\) (resp. \((\psi^{0}, \theta^{0})\)) is the solution for problem (1.1)-(1.2) (resp. problem (1.3)-(1.4)).

The theory of boundary layers is one of the most fundamental and important issues in fluid dynamics (cf. [9, 10]) since the seminal work by Prandtl in 1904. There are a number of papers dedicated to the questions of boundary layers for the Navier-Stokes equations; see for instance [8, 11–17] and the references therein. Moreover, the boundary layer problem also arises in the theory of hyperbolic systems when parabolic equations with small viscosity are applied as perturbations; see for instance [18–23] and the references therein.

Recently, Ruan and Zhu [1], Theorem 1.3, discussed the existence and zero diffusion limit for problem (1.1)-(1.2), and proved that the thickness of boundary layer is of the order \(O(\beta^{\gamma})\) with \(0<\gamma<1/2\) if \(\frac{(\sigma+\mu\beta)^{2}}{4(1-\beta)}< \alpha<\sigma\) and if the initial data satisfy \(\psi_{0}\in H^{2}([0,1]), \theta_{0}\in H^{3}([0,1]), (\psi_{0},\theta_{0})(1)=(\psi_{0},\theta_{0})(0)=(0, 0)\), and \(\Vert (\psi_{0},\theta _{0})\Vert _{2}\) is sufficiently small. Here \(H^{l}([0, 1])\) denotes the usual lth order Sobolev space with the norm \(\Vert f\Vert _{l}= (\sum_{i=0}^{l}\int_{0}^{1} \vert \partial_{x}^{i} f\vert ^{2}\,dx )^{1/2}\). In the present paper, we improve the result by extending the range of γ to \((0, 1)\). Our main result can be stated as follows.

Theorem 1.2

Let \(0<\beta<1\) and \(\frac{(\sigma+\mu\beta)^{2}}{4(1-\beta)}< \alpha<\sigma\). Assume that the initial data satisfy \(\psi_{0}\in H^{2}([0,1]), \theta_{0}\in H^{3}([0,1]), (\psi_{0},\theta_{0})(1)=(\psi_{0},\theta_{0})(0)=(0, 0)\), and \(\Vert (\psi_{0},\theta_{0})\Vert _{2}\) is sufficiently small. Then any function \(\delta(\beta)\) satisfying \(\delta(\beta) \downarrow 0\) and \(\frac{\beta}{\delta(\beta)} \rightarrow0\) as \(\beta\downarrow 0\) is a BL-thickness such that

$$ \begin{aligned} \bigl\Vert \theta^{\beta}- \theta^{0}\bigr\Vert _{L^{\infty}(0,T;L^{\infty}[\delta,1-\delta])} \leq C \sqrt{\frac{\beta}{\delta}},\quad \forall\delta\in(0, 1/2), \end{aligned} $$
(1.5)

where \(T>0\), and C is a positive constant independent of β and δ.

The proof of Theorem 1.2 will be given in the next section.

2 Proof of Theorem 1.2

To prove Theorem 1.2, we need the following result, which can be found in [1], Lemmas 2.2, 2.4, 2.5 and 3.1.

Lemma 2.1

Let the assumptions of Theorem 1.2 hold. Then there exists a positive constant independent of β such that

$$\begin{aligned} & \int_{0}^{1} \bigl[\bigl(\psi^{\beta}_{x} \bigr)^{2}+\bigl(\theta^{\beta}_{x} \bigr)^{2}+\bigl(\psi^{\beta}_{xx} \bigr)^{2}+\beta\bigl(\theta^{\beta}_{xx} \bigr)^{2} \bigr]\,dx \leq C, \end{aligned}$$
(2.1)
$$\begin{aligned} & \int_{0}^{1} \bigl[\bigl(\psi^{0}_{x} \bigr)^{2}+\bigl(\theta^{0}_{x}\bigr)^{2} +\bigl(\theta^{0}_{xx}\bigr)^{2}+\bigl(\psi ^{0}_{xx}\bigr)^{2} \bigr]\,dx \leq C \end{aligned}$$
(2.2)

and

$$ \int_{0}^{1} \bigl[ \bigl(\psi^{\beta}- \psi^{0}\bigr)^{2}+\bigl(\theta^{\beta}- \theta^{0}\bigr)^{2} \bigr]\,dx + \int_{0}^{T} \int _{0}^{1} \bigl(\psi^{\beta}- \psi^{0}\bigr)_{x}^{2} \,dx\,dt \leq C\beta. $$
(2.3)

Proof of Theorem 1.2

It suffices to prove (1.5). Set

$$u^{\beta}=\psi^{\beta}-\psi^{0}, \qquad v^{\beta}= \theta^{\beta}-\theta^{0}. $$

Then it follows from the equation of \(\theta^{\beta}\) that

$$ v^{\beta}_{t}=-(1- \beta)v^{\beta}+2\psi^{\beta}v^{\beta}_{x}+2u^{\beta}\theta^{0}_{x} +\beta v^{\beta}_{xx}+\beta \bigl(\mu\psi^{\beta}_{x}+\theta^{0}+ \theta^{0}_{xx}\bigr). $$

Differentiating the equation, we see that \(z:=v^{\beta}_{x}\) satisfies

$$ z_{t}=-(1-\beta)z+2\bigl( \psi^{\beta}z\bigr)_{x}+2\bigl(u^{\beta}\theta^{0}_{x}\bigr)_{x}+\beta z_{xx}+ \beta\bigl(\mu\psi^{\beta}_{xx}+\theta^{0}_{x}+ \theta^{0}_{xxx}\bigr). $$
(2.4)

Denote \(\varphi_{\varepsilon}\) for \(\varepsilon\in(0, 1)\) and \(\xi_{\delta}\) for \(\delta\in(0, 1/2)\) by

$$ \varphi_{\varepsilon}(s)=\sqrt{s^{2}+ \varepsilon^{2}},\qquad \xi_{\delta}(x)= \textstyle\begin{cases}x,&0\leq x\leq\delta, \\ \delta,& \delta\leq x\leq1-\delta, \\ 1-x,&1-\delta\leq x \leq1. \end{cases} $$

It is easy to check that \(\varphi_{\varepsilon}\) satisfies

$$ \textstyle\begin{cases} \vert s\vert \leq \vert \varphi_{\varepsilon}(s)\vert \leq \vert s\vert +1, \\ \vert \varphi_{\varepsilon}'(s)\vert \leq1,\quad 0\leq s\varphi_{\varepsilon}'(s)\leq \varphi_{\varepsilon}(s), \\ \varphi_{\varepsilon}''(s)\geq0,\quad s^{2} \varphi_{\varepsilon}''(s) \geq 0, \end{cases} $$

and \(\xi_{\delta}\) satisfies

$$0\leq \xi_{\delta}\leq\delta,\qquad \xi_{\delta}(1)=\xi_{\delta}(0)=0. $$

Multiplying (2.4) by \(\varphi_{\varepsilon}'(z)\xi_{\delta}\) and integrating it over \((0, 1)\times(0, t)\), we have

$$\begin{aligned} & \int_{0}^{1}\varphi_{\varepsilon}(z) \xi_{\delta}\,dx- \varepsilon \int_{0}^{1}\xi _{\delta}\,dx \\ & \quad =-(1-\beta) \int_{0}^{t} \int_{0}^{1} z \varphi_{\varepsilon}'(z) \xi_{\delta}\,dx\,d\tau+2 \int_{0}^{t} \int_{0}^{1}\bigl(\psi^{\beta}z \bigr)_{x}\varphi_{\varepsilon}'(z)\xi _{\delta}\, dx\,d\tau \\ &\qquad{} +2 \int_{0}^{t} \int_{0}^{1} \bigl(u^{\beta}\theta^{0}_{x}\bigr)_{x}\varphi_{\varepsilon}'(z) \xi _{\delta}\,dx\,d\tau +\beta \int_{0}^{t} \int_{0}^{1}z_{xx}\varphi_{\varepsilon}'(z) \xi_{\delta}\, dx\,d\tau \\ &\qquad{} +\beta \int_{0}^{t} \int_{0}^{1} \varphi_{\varepsilon}'(z) \xi_{\delta}\bigl(\mu\psi^{\beta}_{xx}+ \theta^{0}_{x}+\theta^{0}_{xxx}\bigr) \,dx \,d\tau=: \sum_{i=1}^{5}E_{j}. \end{aligned}$$
(2.5)

Next we estimate \(E_{i} (i=1,2,3,4,5)\). From \(0\leq s\varphi_{\varepsilon}'(s)\leq\varphi_{\varepsilon}(s)\), we have

$$ \begin{aligned} E_{1}\leq \int_{0}^{t} \int_{0}^{1} \varphi_{\varepsilon}(z) \xi_{\delta}\,dx\,d\tau. \end{aligned} $$
(2.6)

To estimate \(E_{2}\), we note, using integration by parts,

$$\begin{aligned} E_{2}={}&2 \int_{0}^{t} \int_{0}^{1}\psi^{\beta}_{x}z \varphi_{\varepsilon}'(z)\xi_{\delta}\,dx\,d\tau+2 \int_{0}^{t} \int_{0}^{1}\psi^{\beta}z_{x} \varphi_{\varepsilon}'(z)\xi _{\delta}\,dx\,d\tau \\ ={}&2 \int_{0}^{t} \int_{0}^{1}\psi^{\beta}_{x}z \varphi_{\varepsilon}'(z)\xi_{\delta}\,dx\,d\tau - 2 \int_{0}^{t} \int_{0}^{1} \varphi_{\varepsilon}(z) \psi^{\beta}_{x}\xi_{\delta}\,dx\,d\tau \\ &{} - 2 \int_{0}^{t} \int_{0}^{1} \varphi_{\varepsilon}(z) \psi^{\beta}\xi_{\delta}' \,dx\,d\tau \\ =:{}&E_{2}^{1}+E_{2}^{2}+E_{2}^{3}. \end{aligned}$$
(2.7)

By (2.1) and the embedding \(W^{1,1}[0, 1]\hookrightarrow L^{\infty}[0, 1]\), we have

$$ \bigl\vert \psi^{\beta}_{x}(x,t) \bigr\vert \leq \int_{0}^{1} \bigl\vert \psi^{\beta}_{x} \bigr\vert \,dx+ \int_{0}^{1}\bigl\vert \psi ^{\beta}_{xx} \bigr\vert \,dx\leq C, $$
(2.8)

where C denotes the generic positive constant independent of \(\beta, \delta\), and ε, so

$$ \begin{aligned} &\bigl\vert \psi^{\beta}(x,t) \bigr\vert \leq \int_{0}^{x} \bigl\vert \psi^{\beta}_{y}(y,t) \bigr\vert \,dy\leq Cx\leq C\xi _{\delta}(x), \quad\forall x\in[0, \delta], \\ &\bigl\vert \psi^{\beta}(x,t)\bigr\vert \leq \int_{x}^{1} \bigl\vert \psi^{\beta}_{y}(y,t) \bigr\vert \,dy\leq C(1-x)\leq C\xi_{\delta}(x), \quad \forall x\in[1-\delta,1]. \end{aligned} $$
(2.9)

By \(0\leq s\varphi_{\varepsilon}'(s)\leq\varphi_{\varepsilon}(s)\) and (2.8), we obtain

$$ \begin{aligned} E_{2}^{1}+E_{2}^{2} \leq C \int_{0}^{t} \int_{0}^{1} \varphi_{\varepsilon}(z) \xi_{\delta}\,dx\,d\tau. \end{aligned} $$
(2.10)

By the definition of \(\xi_{\delta}\) and (2.9), we have

$$ \begin{aligned} E_{2}^{3} &= - 2 \int_{0}^{t} \int_{0}^{\delta}\varphi_{\varepsilon}(z) \psi^{\beta}\,dx\,d\tau+ 2 \int_{0}^{t} \int_{1-\delta}^{1} \varphi_{\varepsilon}(z) \psi^{\beta}\,dx\,d\tau \\ &\leq C \int_{0}^{t} \int_{0}^{\delta}\varphi_{\varepsilon}(z) \xi_{\delta}\,dx\,d\tau+ C \int_{0}^{t} \int_{1-\delta}^{1} \varphi_{\varepsilon}(z) \xi_{\delta}\,dx\,d\tau \\ &\leq C \int_{0}^{t} \int_{0}^{1} \varphi_{\varepsilon}(z) \xi_{\delta}\,dx\,d\tau. \end{aligned} $$
(2.11)

Thus

$$ \begin{aligned} E_{2} \leq C \int_{0}^{t} \int_{0}^{1} \varphi_{\varepsilon}(z) \xi_{\delta}\,dx\,d\tau. \end{aligned} $$
(2.12)

Using integration by parts and noticing \(\varphi_{\varepsilon}''\geq 0\) and \(\vert \varphi_{\varepsilon}'\vert \leq1\), we have

$$ \begin{aligned} E_{4}={}&-\beta \int_{0}^{t} \int_{0}^{1} z_{x}^{2} \varphi_{\varepsilon}''(z)\xi_{\delta}\,dx\,d\tau- \beta \int_{0}^{t} \int_{0}^{1} z_{x} \varphi_{\varepsilon}'(z) \xi_{\delta}' \,dx\,d\tau \\ \leq{}&- \beta \int_{0}^{t} \int_{0}^{1} z_{x} \varphi_{\varepsilon}'(z) \xi_{\delta}' \,dx\,d\tau \\ ={}&-\beta \int_{0}^{t} \int_{0}^{\delta}z_{x} \varphi_{\varepsilon}'(z) \,dx\,d\tau+\beta \int_{0}^{t} \int_{1-\delta}^{1} z_{x} \varphi_{\varepsilon}'(z) \,dx\,d\tau \\ \leq{}&\beta \biggl( \int_{0}^{t} \int_{0}^{\delta} \vert z_{x}\vert \,dx\,d \tau+ \int_{0}^{t} \int _{1-\delta}^{1} \vert z_{x}\vert \,dx\,d \tau \biggr), \end{aligned} $$

and, by Hölder’s inequality and (2.1), we obtain

$$\begin{aligned} E_{4} \leq{}& C\beta \delta^{1/2} \biggl[ \biggl( \int_{0}^{t} \int_{0}^{\delta} \vert z_{x}\vert ^{2}\,dx\,d\tau \biggr)^{1/2}+ \biggl( \int_{0}^{t} \int_{1-\delta}^{1} \vert z_{x}\vert ^{2}\,dx\,d\tau \biggr)^{1/2} \biggr] \\ \leq{}& C\beta^{1/2} \delta^{1/2}. \end{aligned}$$
(2.13)

By \(\vert \varphi_{\varepsilon}'\vert \leq1, 0\leq\xi_{\delta}\leq\delta\), Hölder’s inequality, (2.2), and (2.3), we have

$$\begin{aligned} E_{3}={}& 2 \int_{0}^{t} \int_{0}^{1}u^{\beta}_{x} \theta^{0}_{x}\varphi_{\varepsilon}'(z)\xi _{\delta}\,dx\,d\tau+ 2 \int_{0}^{t} \int_{0}^{1}u^{\beta}\theta^{0}_{xx} \varphi _{\varepsilon}'(z)\xi_{\delta}\,dx\,d\tau \\ \leq{}& C\delta \biggl( \int_{0}^{t} \int_{0}^{1}\bigl(u^{\beta}_{x} \bigr)^{2} \,dx\,d\tau \biggr)^{1/2} \biggl( \int_{0}^{t} \int_{0}^{1}\bigl(\theta^{0}_{x} \bigr)^{2} \,dx\,d\tau \biggr)^{1/2} \\ &{}+C\delta \biggl( \int_{0}^{t} \int_{0}^{1}\bigl(u^{\beta}\bigr)^{2} \,dx\,d\tau \biggr)^{1/2} \biggl( \int_{0}^{t} \int_{0}^{1}\bigl(\theta^{0}_{xx} \bigr)^{2} \,dx\,d\tau \biggr)^{1/2} \\ \leq{}&C\delta\beta^{1/2}. \end{aligned}$$
(2.14)

Finally, we estimate \(E_{5}\). By \(\vert \varphi_{\varepsilon}'\vert \leq1, 0\leq \xi_{\delta}\leq\delta\), and Lemma 2.1, we have

$$\begin{aligned} E_{5}&\leq C \beta\delta \int_{0}^{t} \int_{0}^{1} \bigl(\bigl\vert \psi^{\beta}_{xx} \bigr\vert +\bigl\vert \theta ^{0}_{x}\bigr\vert +\bigl\vert \theta^{0}_{xxx}\bigr\vert \bigr) \,dx\,d\tau \\ &\leq C \beta\delta. \end{aligned}$$
(2.15)

Combining (2.6), (2.12)-(2.15) with (2.5) and noticing

$$\varepsilon \int_{0}^{1} \xi_{\delta}\,dx\leq \varepsilon \delta, $$

we obtain

$$ \begin{aligned} \int_{0}^{1} \varphi_{\varepsilon}(z) \xi_{\delta}\,dx \leq C \int_{0}^{t} \int_{0}^{1} \varphi_{\varepsilon}(z) \xi_{\delta}\, dx\,d\tau+ \varepsilon\delta+C\beta^{1/2} \delta^{1/2}, \end{aligned} $$

so an application of Gronwall’s inequality leads to

$$ \begin{aligned} \int_{0}^{1} \varphi_{\varepsilon}(z) \xi_{\delta}\,dx \leq C \bigl( \varepsilon \delta+\beta^{1/2} \delta^{1/2} \bigr). \end{aligned} $$

From this and the definition of \(\xi_{\delta}\) and \(\vert z\vert \leq \varphi_{\varepsilon}(z)\), we obtain

$$ \begin{aligned} \int_{\delta}^{1-\delta} \vert z\vert \,dx \leq C \biggl( \varepsilon+\sqrt{\frac {\beta}{\delta}} \biggr). \end{aligned} $$

Letting \(\varepsilon\rightarrow0\) yields

$$ \begin{aligned} \int_{\delta}^{1-\delta} \vert z\vert \,dx \leq C \sqrt{ \frac{\beta}{\delta}}. \end{aligned} $$
(2.16)

From (2.3), (2.16), and the embedding \(W^{1,1}([\delta, 1-\delta])\hookrightarrow L^{\infty}([\delta, 1-\delta])\) it follows that

$$ \begin{aligned} \bigl\Vert \bigl(\theta^{\beta}- \theta^{0}\bigr) (\cdot, t)\bigr\Vert _{L^{\infty}[\delta,1-\delta]} \leq{} & \int_{0}^{1}\bigl\vert \theta^{\beta}- \theta^{0}\bigr\vert \,dx+ \int_{\delta}^{1-\delta} \vert z\vert \,dx \\ \leq{}& C \sqrt{\frac{\beta}{\delta}}. \end{aligned} $$

Thus (1.5) is proved, and the proof is complete. □