Some Properties of Solutions of a Fourth-Order Parabolic Equation for Image Processing

Abstract

In this paper, for the IBVP of a fourth-order nonlinear parabolic equation, which is related to image analysis, we studied the existence and uniqueness of weak solutions. Moreover, we also considered the asymptotic behavior and the regularity of solutions of such problem.

1 Introduction

In this paper, we investigate the following fourth-order parabolic equation

$$\begin{aligned} u_t+\Delta \left( \ln \left( \Delta u+\sqrt{1+(\Delta u)^2}\right) \right) +\lambda |u|^{p-2}u=0, \quad \hbox {in}\quad \Omega _T, \end{aligned}$$

(1.1)

where $\lambda >0$, $p>2$, $\Omega _T=\Omega \times (0,T)$ and $\Omega \subset {{\mathbb {R}}}^2$ is a bounded domain with smooth boundary.

On the basis of physical consideration, as usual Eq.  (1.1) is supplemented with the natural boundary value conditions

$$\begin{aligned} u=\Delta u=0,\quad x\in \partial \Omega , \quad t>0, \end{aligned}$$

(1.2)

and the initial value condition

$$\begin{aligned} u(x,0)=u_0(x),\quad x\in \Omega . \end{aligned}$$

(1.3)

Here, inspired by the ideas described in Wei [11], we give a sketch of the formulation of Eq. (1.1) from the image restoration. Wei [11] proposed a real-valued, bounded edge enhancing functional, which leads to a generalized Perona–Malik equation

$$\begin{aligned} \frac{\partial u}{\partial t}=\hbox {div}({d}(u,|\nabla u|)\nabla u)+e(u,|\nabla u|). \end{aligned}$$

In image systems, the distribution of image pixels can be highly inhomogeneous. Hence, the generalized Perona–Malik equation can be made more efficient for image segmentation and noise removing by incorporating an edge sensitive super diffusion operator [11]

$$\begin{aligned} \frac{\partial u}{\partial t}=\hbox {div}(d_1(u,|\nabla u|)\nabla u)+\hbox {div}(d_2(u,|\nabla u|,\Delta u)\nabla \Delta u)+e(u,|\nabla u|). \end{aligned}$$

(1.4)

Here $d_1, d_2$ are edge sensitive diffusion functions. The typical cases of $d_2$ are $d_2(u,|\nabla u|, \Delta u)=-g(|\nabla u|)$ or $=-g(\Delta u)$. The g(s) is a nonincreasing function satisfying the following ([8])

$$\begin{aligned} g(0)=1,\quad g(s)>0,\quad \lim _{s\rightarrow \infty }\frac{\hbox {d}^n}{\hbox {d}s^n}g(s)=0,\;\hbox {for each integer}\; n\ge 0. \end{aligned}$$

An example typically used in applications is [3, 8]

$$\begin{aligned} g(s)=\frac{1}{1+s^2}. \end{aligned}$$

The $g(s)=\frac{1}{\sqrt{1+s^2}}$ is reasonable for Eq.  (1.4). If taking $d_1=0$, $e(u,|\nabla u|)=-\lambda |u|^{p-2}u$ and $d_2=-\frac{1}{\sqrt{1+|\Delta u|^2}}$, we obtain Eq. (1.1). Equation (1.1) is original, which has not been studied by others so far.

Taking $d_1=0$, $e(u,|\nabla u|)=0$ and $d_2=-\frac{1}{1+|\Delta u|^2}$, Eq. (1.4) becomes the fourth-order Perona–Malik analogue [9]

$$\begin{aligned} u_t+\nabla \left( \frac{\nabla \Delta u}{1+|\Delta u|^2}\right) =0. \end{aligned}$$

(1.5)

Wang et al. [10] considered the low-curvature equation

$$\begin{aligned} u_t+\Delta (\arctan \Delta u)=0, \end{aligned}$$

which is exactly the equation of (1.5). They established the existence and uniqueness of weak solutions.

Wei [11] introduced the following equation by taking $d_1=0$, $e(u,|\nabla u|)=0$ and $d_2=-\frac{1}{1+|\nabla u|^2}$ for highly inhomogeneous images,

$$\begin{aligned} u_t+\nabla \left( \frac{\nabla \Delta u}{1+|\nabla u|^2}\right) =0. \end{aligned}$$

Other fourth-order partial differential equations are also proposed in image analysis. You and Kaveh [13] introduced a different form of the fourth-order diffusion,

$$\begin{aligned} u_t+\Delta [g(\Delta u)\Delta u]=0. \end{aligned}$$

This form is derived from a variational formulation. Osher et al. [7] employed a new model

$$\begin{aligned} u_t+\frac{1}{2\lambda }\Delta \left( \hbox {div}\left( \frac{\nabla u}{|\nabla u|}\right) \right) =f-u, \end{aligned}$$

for image decomposition and image restoration into cartoon and texture. The relevant fourth-order parabolic equations have also been studied in [1, 4, 6, 12].

Now we give the definition of the solution in a weak sense of problems (1.1)–(1.3).

Definition 1.1

A function u is a weak solution of problems (1.1)–(1.3), if the following conditions are satisfied

• (1) $u \in C([0,T];L^2(\Omega ))\cap L^\infty (0,T;H_0^1(\Omega ))\cap L^2 (0,T;H^2(\Omega ))$ with $\ln (\Delta u+\sqrt{1+(\Delta u)^2}) \in L^2 (0,T;H_0^1(\Omega ))$;

• (2) For any $\varphi \in C^2(\overline{\Omega }_T)$ with $\varphi (x,T)=0$ and $\varphi (x,t)\mid _{\partial \Omega }=0$, we have

  $$\begin{aligned}&-\int _{\Omega }\varphi (x,0)u_0(x)\hbox {d}x- \int _0^T \int _{\Omega }u\varphi _t\hbox {d}x\hbox {d}t\nonumber \\&\quad +\int _0^T \int _{\Omega }\ln \left( \Delta u+\sqrt{1+(\Delta u)^2}\right) \Delta \varphi \hbox {d}x\hbox {d}t\nonumber \\&\quad +\lambda \int _0^T\int _{\Omega }|u|^{p-2}u\varphi \hbox {d}x\hbox {d}t=0. \end{aligned}$$

  (1.6)

In this paper, we will study a general equation as described in (1.1). Our method for investigating the existence of weak solutions is based on the difference and variation methods to construct an approximate solution. By means of the uniform estimates on solutions of the time difference equations, we prove the existence of weak solutions. Based on a suitable integral equality and the energy techniques, we also obtain the asymptotic behavior and regularity of solutions.

This paper is organized as follows. We investigate the existence and uniqueness of weak solutions of problems (1.1)–(1.3) in Sect. 3. Using energy techniques, we also proved the asymptotic behavior and regularity of solutions subsequently.

2 Existence of solutions

In this section, we are going to prove the existence of weak solutions.

Theorem 2.1

Assume $u_0 \in H_0^1(\Omega )$, problems (1.1)–(1.3) admit a unique weak solution satisfying Definition 1.1.

To prove Theorem 2.1, we first consider the following elliptic problem

$$\begin{aligned} \left\{ \begin{aligned}&\frac{u-u_0}{h}+\Delta \left( \varepsilon \Delta u +\ln \left( \Delta u+\sqrt{1+(\Delta u)^2}\right) \right) +\lambda |u|^{p-2}u=0,\\&u|_{\partial \Omega }=0,\quad \Delta u|_{\partial \Omega }=0, \end{aligned}\right. \end{aligned}$$

(2.1)

where $h=T/n$, $\varepsilon >0$ and $u_0$ is the initial value.

Theorem 2.2

Assume $u_0 \in H_0^1(\Omega )$, there exists a unique weak solution $u_1 \in H_0^1(\Omega )\cap H^2(\Omega )$ with $\Delta u_1\in H_0^1(\Omega )$ for initial-boundary value problem (2.1).

Proof

We will prove the existence of weak solutions by variation methods.

Let us consider the following functional on the space $V=H_0^1(\Omega )\cap H^2(\Omega )$,

$$\begin{aligned} J(v)&=\int _\Omega \frac{|v-u_0|^2}{2h}\hbox {d}x+\int _\Omega \frac{\varepsilon |\Delta v|^2}{2}\hbox {d}x\nonumber \\&\quad +\int _\Omega \int _0^{\Delta v}\ln \left( s+\sqrt{1+s^2}\right) \hbox {d}s\hbox {d}x+\lambda \int _\Omega \frac{|v|^p}{p}\hbox {d}x. \end{aligned}$$

(2.2)

In addition, letting $f(t)=\int _0^t\ln (s+\sqrt{1+s^2})\hbox {d}s$, we know that $f'(t)=\ln (t+\sqrt{1+t^2})$, $f''(t)=\frac{1}{\sqrt{1+t^2}}>0$ and $f(0)=f'(0)=0$. Hence, $f(t)\ge f(0)=0$. It is obvious that

$$\begin{aligned} \int _\Omega \int _0^{\Delta v}\ln \left( s+\sqrt{1+s^2}\right) \hbox {d}s\hbox {d}x\ge 0. \end{aligned}$$

Therefore, we see that

$$\begin{aligned} J(v)\ge&\int _\Omega \frac{\varepsilon |\Delta v|^2}{2}\hbox {d}x+\lambda \int _\Omega \frac{|v|^p}{p}\hbox {d}x. \end{aligned}$$

By $\lambda >0$, $p>2$ and the Poincaré inequality, we know that $J(v)\rightarrow +\infty$, as $\Vert v\Vert _{H^2}\rightarrow +\infty$. Hence, J(v) satisfies the coercive condition. On the other hand, since $\int _0^t \ln (s+\sqrt{1+s^2})\hbox {d}s$ is a convex function, J(v) is weakly lower semi-continuous on V. So, it follows from the theory in [2] that there exists $u_1\in V$ such that

$$\begin{aligned} J(u_1)=\inf _{v\in V}J(v), \end{aligned}$$

which implies that $u_1 \in H_0^1(\Omega )\cap H^2(\Omega )$ is a minimizer of the functional J(v) in V.

Now for every $\varphi \in C_0^\infty$ and every $\varepsilon \in \mathbb {R}$, since $u_1+\varepsilon \varphi \in V$, $F(0)\le F(\varepsilon )$, where

$$\begin{aligned} F(\varepsilon )=J(u_1+\varepsilon \varphi ). \end{aligned}$$

Thus, we get $F'(0)=0$, which is

$$\begin{aligned}&\int _\Omega \frac{u_1-u_0}{h}\varphi \hbox {d}x+\int _\Omega \Delta \left( \varepsilon \Delta u_1 +\ln \left( \Delta u_1+\sqrt{1+(\Delta u_1)^2}\right) \right) \varphi \hbox {d}x \nonumber \\&\quad +\,\lambda \int _\Omega |u_1|^{p-2}u_1\varphi \hbox {d}x=0. \end{aligned}$$

(2.3)

Therefore, the function $u_1$ is a weak solution of the corresponding Euler–Lagrange equation of J(v), which is problem (2.1). For every $\eta \in C_0^\infty (\Omega )$, there exists a unique $\varphi \in H_0^1(\Omega )\cap H^2(\Omega )$ such that $-\Delta \varphi =\eta$. Let $w\in H_0^1(\Omega )\cap H^2(\Omega )$ be the unique solution for equation

$$\begin{aligned} -\Delta w=\frac{u_1-u_0}{h}+\lambda |u_1|^{p-2}u_1. \end{aligned}$$

By $u_1-u_0, |u_1|^{p-2}u_1\in H^1_0(\Omega )$, we know that $w\in H_0^1(\Omega )\cap H^3(\Omega )$. Hence by (2.3), we have

$$\begin{aligned} \int _\Omega (-\Delta w)\varphi \hbox {d}x=\int _\Omega \left( \varepsilon \Delta u_1 +\ln \left( \Delta u_1+\sqrt{1+(\Delta u_1)^2}\right) \right) \eta \hbox {d}x. \end{aligned}$$

On the other hand, we know that

$$\begin{aligned} \int _\Omega (-\Delta w)\varphi \hbox {d}x=\int _\Omega w(-\Delta \varphi ) \hbox {d}x=\int _\Omega w\eta \hbox {d}x. \end{aligned}$$

Therefore, we derive

$$\begin{aligned} f_\varepsilon (\Delta u_1)=\varepsilon \Delta u_1 +\ln (\Delta u_1+\sqrt{1+(\Delta u_1)^2})=w. \end{aligned}$$

For function $f_\varepsilon (t)=\varepsilon t +\ln (t+\sqrt{1+t^2})$, we know that

$$\begin{aligned} \varepsilon <f'_\varepsilon (t)=\varepsilon +\frac{1}{\sqrt{1+t^2}}\le \varepsilon +1. \end{aligned}$$

So its inverse function $g_\varepsilon (t)=f^{-1}_\varepsilon (t)$ exists and satisfies

$$\begin{aligned} \frac{1}{1+\varepsilon }\le g'_\varepsilon (t)<\frac{1}{\varepsilon }. \end{aligned}$$

Hence, we obtain

$$\begin{aligned} \Delta u_1=g_\varepsilon (w)\in H_0^1(\Omega ). \end{aligned}$$

So we complete the proof of the existence.

Now we prove the uniqueness. Suppose that there exists another weak solution $\widetilde{u}_1$ of problem (2.1). Then, it follows from (2.3) that, for every $\varphi \in H_0^1(\Omega )\cap H^2(\Omega )$,

$$\begin{aligned}&\int _\Omega \frac{\widetilde{u}_1-u_0}{h}\varphi \hbox {d}x+\int _\Omega \Delta \left( \varepsilon \Delta \widetilde{u}_1 +\ln \left( \Delta \widetilde{u}_1+\sqrt{1+\left( \Delta \widetilde{u}_1\right) ^2}\right) \right) \varphi \hbox {d}x\\&\quad +\,\lambda \int _\Omega |\widetilde{u}_1|^{p-2}\widetilde{u}_1\varphi \hbox {d}x=0. \end{aligned}$$

So that,

$$\begin{aligned}&\int _\Omega \frac{\widetilde{u}_1-u_1}{h}\varphi \hbox {d}x+\int _\Omega \varepsilon \Delta (\widetilde{u}_1-u_1)\Delta \varphi \hbox {d}x+\lambda \int _\Omega (|\widetilde{u}_1|^{p-2}\widetilde{u}_1-|u_1|^{p-2}u_1)\varphi \hbox {d}x \\&\quad +\,\int _\Omega \left( \ln \left( \Delta \widetilde{u}_1+\sqrt{1+(\Delta \widetilde{u}_1)^2}\right) -\ln \left( \Delta u_1+\sqrt{1+(\Delta u_1)^2}\right) \right) \Delta \varphi \hbox {d}x =0. \end{aligned}$$

Choosing $\varphi =\widetilde{u}_1-u_1$, we have

$$\begin{aligned}&\int _\Omega \frac{(\widetilde{u}_1-u_1)^2}{h} \hbox {d}x+\int _\Omega \varepsilon \Delta ^2 (\widetilde{u}_1-u_1)(\Delta \widetilde{u}_1-\Delta u_1) \hbox {d}x\\&\quad +\int _\Omega \left( \ln \left( \Delta \widetilde{u}_1+\sqrt{1+(\Delta \widetilde{u}_1)^2}\right) -\ln \left( \Delta u_1+\sqrt{1+(\Delta u_1)^2}\right) \right) (\Delta \widetilde{u}_1-\Delta u_1) \hbox {d}x\\&\quad +\,\lambda \int _\Omega (|\widetilde{u}_1|^{p-2}\widetilde{u}_1-|u_1|^{p-2}u_1)(\widetilde{u}_1-u_1)\hbox {d}x=0. \end{aligned}$$

Since function $\ln (t+\sqrt{1+t^2})$ is increasing, we know that every term on the left-hand side is nonnegative. Therefore, we conclude that $u_1=\widetilde{u}_1$ a.e. in $\Omega$ and complete the proof of the uniqueness. $\square$

Next, we discuss the parabolic problem

$$\begin{aligned} \left\{ \begin{array}{ll} u_t+\Delta (\varepsilon \Delta u+\ln (\Delta u+\sqrt{1+(\Delta u)^2}))+\lambda |u|^{p-2}u=0,\quad &{} \quad \hbox {in} \quad \Omega _T,\\ u=0,\quad \Delta u=0, &{} \quad \hbox {on} \quad \partial \Omega ,\\ u(x,0)=u_0(x), \end{array} \right. \end{aligned}$$

(2.4)

where $\varepsilon >0$.

Theorem 2.3

Assume $u_0 \in H_0^1(\Omega )$, problem (2.4) admits a unique weak solution $u_\varepsilon \in C([0,T];L^2(\Omega ))\cap L^\infty (0,T;H_0^1(\Omega ))\cap L^2 (0,T;H^3(\Omega ))$ with $\Delta u_\varepsilon \in L^2 (0,T;H_0^1(\Omega ))$, which satisfies the following estimates

$$\begin{aligned}&\frac{1}{2}\int _\Omega u_\varepsilon ^2\mathrm{d}x+\int _0^t\int _\Omega \left( \varepsilon \Delta u_\varepsilon +\ln \left( \Delta u_\varepsilon +\sqrt{1+\left( \Delta u_\varepsilon \right) ^2}\right) \right) \Delta u_\varepsilon \mathrm{d}x\mathrm{d}\tau \nonumber \\&\quad +\lambda \int _0^t\int _\Omega |u_\varepsilon |^p\mathrm{d}x\mathrm{d}\tau \le \frac{1}{2}\int _\Omega u_0^2\mathrm{d}x, \end{aligned}$$

(2.5)

and

$$\begin{aligned}&\frac{1}{2}\int _\Omega |\nabla u_\varepsilon |^2\mathrm{d}x+\int _0^t\int _\Omega \left( \varepsilon +\frac{1}{\sqrt{1+|\Delta u_\varepsilon |^2}}\right) |\nabla \Delta u_\varepsilon |^2 \mathrm{d}x\mathrm{d}\tau \nonumber \\&\quad +\lambda \int _\Omega (p-1)|u_\varepsilon |^{p-2}|\nabla u_\varepsilon |^2\mathrm{d}x\le \frac{1}{2}\int _\Omega |\nabla u_0|^2\mathrm{d}x. \end{aligned}$$

(2.6)

Proof

By Theorem 2.2, we define $u_k\in H_0^1(\Omega )\cap H^2(\Omega ),k=1,2,\ldots ,n$ to be the weak solution of the following elliptic problems

$$\begin{aligned} \left\{ \begin{aligned}&\frac{u_k-u_{k-1}}{h}+\Delta \left( \varepsilon \Delta u_k+\ln \left( \Delta u_k+\sqrt{1+(\Delta u_k)^2}\right) \right) +\lambda |u_k|^{p-2}u_k=0,\\&u_k|_{\partial \Omega }=0,\quad \Delta u_k|_{\partial \Omega }=0. \end{aligned}\right. \end{aligned}$$

(2.7)

Therefore, for every $\varphi \in C_0^\infty (\Omega )$,

$$\begin{aligned}&\int _\Omega \frac{u_k-u_{k-1}}{h}\varphi \hbox {d}x+\int _\Omega \Delta \left( \varepsilon \Delta u_k+\ln (\Delta u_k+\sqrt{1+\left( \Delta u_k\right) ^2})\right) \varphi \hbox {d}x\\&\quad +\lambda \int _\Omega |u_k|^{p-2}u_k\varphi \hbox {d}x=0. \end{aligned}$$

Choosing $\varphi =\Delta u_k$, we have

$$\begin{aligned}&\frac{1}{h}\int _\Omega (\nabla u_k-\nabla u_{k-1})\nabla u_k \hbox {d}x+\int _\Omega \left( \varepsilon +\frac{1}{\sqrt{1+|\Delta u_k|^2}}\right) |\nabla \Delta u_k|^2 \hbox {d}x\\&\quad +\lambda \int _\Omega (p-1)|u_k|^{p-2}|\nabla u_k|^2 \hbox {d}x=0, \end{aligned}$$

that is

$$\begin{aligned}&\frac{1}{2h}\int _\Omega |\nabla u_k|^2\hbox {d}x+\int _\Omega \left( \varepsilon +\frac{1}{\sqrt{1+|\Delta u_k|^2}}\right) |\nabla \Delta u_k|^2 \hbox {d}x \nonumber \\&\quad +\lambda \int _\Omega (p-1)|u_k|^{p-2}|\nabla u_k|^2 \hbox {d}x\le \frac{1}{2h}\int _\Omega |\nabla u_{k-1}|^2\hbox {d}x. \end{aligned}$$

(2.8)

Next, we construct an approximate solution $u_h$ of problem (2.4) by defining

$$\begin{aligned} u_h(x,t)=\left\{ \begin{aligned}&u_0(x),\quad t=0,\\&u_1(x),\quad 0<t\le h,\\&\cdots ,\cdots ,\\&u_j(x),\quad (j-1)h<t\le jh,\\&\cdots ,\cdots ,\\&u_n(x),\quad (n-1)h<t\le nh=T. \end{aligned}\right. \end{aligned}$$

(2.9)

For every $t\in [0,T]$, (2.8) implies

$$\begin{aligned} \Vert \nabla u_h(x,t)\Vert _{L^2(\Omega )}^2\le \Vert \nabla u_0\Vert _{L^2(\Omega )}^2. \end{aligned}$$

From the above inequality, we see that

$$\begin{aligned} \Vert \nabla u_h(x,t)\Vert _{L^\infty (0,T;L^2(\Omega ))}\le \Vert \nabla u_0\Vert _{L^2(\Omega )}^2. \end{aligned}$$

(2.10)

Summing up the inequalities in (2.8), we derive that

$$\begin{aligned}&\int _0^T\int _\Omega \left( \varepsilon +\frac{1}{\sqrt{1+|\Delta u_h|^2}}\right) |\nabla \Delta u_h|^2\hbox {d}x\hbox {d}\tau \le \Vert \nabla u_0\Vert _{L^2(\Omega )}^2, \end{aligned}$$

(2.11)

$$\begin{aligned}&\lambda \int _0^T\int _\Omega (p-1)|u_h|^{p-2}|\nabla u_h|^2\hbox {d}x\hbox {d}\tau \le \Vert \nabla u_0\Vert _2^2. \end{aligned}$$

(2.12)

Thus,

$$\begin{aligned} \Vert \Delta u_h\Vert _{L^2(0,T;H_0^1(\Omega ))}+\Vert \ln \left( |\Delta u_h|+\sqrt{1+|\Delta u_h|^2}\right) \Vert _{L^2(0,T;H_0^1(\Omega ))}\le C. \end{aligned}$$

(2.11) implies that

$$\begin{aligned} \int _0^T\int _\Omega |\nabla \left( \ln \left( \Delta u_h+\sqrt{1+(\Delta u_h)^2}\right) \right) |^2\hbox {d}x\hbox {d}\tau&= \int _0^T\int _\Omega \frac{|\nabla \Delta u_h|^2}{1+|\Delta u_h|^2}\hbox {d}x\hbox {d}\tau \\&\le \int _0^T\int _\Omega \frac{|\nabla \Delta u_h|^2}{\sqrt{1+|\Delta u_h|^2}}\hbox {d}x\hbox {d}\tau \le C. \end{aligned}$$

By $\Delta u_h|_{\partial \Omega }=0$, we know that

$$\begin{aligned}&\Vert u_h\Vert _{L^\infty (0,T;H_0^1(\Omega ))}+\Vert u_h\Vert _{L^2(0,T;H^3(\Omega ))}\\&\quad +\,\Vert \ln \left( \Delta u_h+\sqrt{1+\left( \Delta u_h\right) ^2}\right) \Vert _{L^2(0,T;H_0^1(\Omega ))}\le C. \end{aligned}$$

Therefore, we may choose a subsequence (we also denote it by the original sequence for simplicity) such that

$$\begin{aligned} \begin{aligned} u_h \rightharpoonup u_\varepsilon ,\qquad&\hbox {weakly-}^* \hbox {in} \quad L^\infty (0,T;H_0^1(\Omega )),\\ u_h \rightharpoonup u_\varepsilon ,\qquad&\hbox {weakly in} \quad L^2(0,T;H^3(\Omega )),\\ \Delta u_h \rightharpoonup \Delta u_\varepsilon ,\qquad&\hbox {weakly in} \quad L^2(0,T;H_0^1(\Omega )),\\ \ln \left( \Delta u_h+\sqrt{1+(\Delta u_h)^2}\right) \rightharpoonup \xi _\varepsilon ,\qquad&\hbox {weakly in} \quad L^2(0,T;H_0^1(\Omega )), \end{aligned} \end{aligned}$$

(2.13)

which follows that ([5], Chapter 2)

$$\begin{aligned} \Vert u_\varepsilon \Vert _{L^\infty (0,T;H_0^1(\Omega ))}+\Vert u_\varepsilon \Vert _{L^2(0,T;H^3(\Omega ))}+\Vert \xi _\varepsilon \Vert _{L^2(0,T;H_0^1(\Omega ))}\le C. \end{aligned}$$

(2.14)

For each $\varphi \in C^1(\overline{\Omega }_T)$ with $\varphi (\cdot ,T)=0$ and for every $k\in \{1,2,\ldots ,n\}$, taking $\varphi (x,kh)$ as a test function in (2.7), we know that

$$\begin{aligned}&\frac{1}{h}\int _\Omega u_k\varphi (x,kh)\hbox {d}x-\frac{1}{h}\int _\Omega u_{k-1}\varphi (x,kh)\hbox {d}x+\lambda \int _\Omega |u_k|^{p-2}u_k\varphi (x,kh)\hbox {d}x \nonumber \\&\quad -\int _\Omega \nabla \left( \varepsilon \Delta u_k+\ln \left( \Delta u_k+\sqrt{1+(\Delta u_k)^2}\right) \right) \nabla \varphi (x,kh)\hbox {d}x=0. \end{aligned}$$

(2.15)

Summing up all the equalities and using $\varphi (\cdot ,T)=\varphi (\cdot ,nh)=0$, we deduce

$$\begin{aligned}&h\sum _{k=1}^{n-1}\int _\Omega \frac{u_h(x,kh)[\varphi (x,kh)-\varphi (x,(k+1)h)]}{h}\hbox {d}x-\int _\Omega u_0\varphi (x,h)\hbox {d}x \nonumber \\&\quad -h\sum _{k=1}^n \int _\Omega \nabla \left( \varepsilon \Delta u_h+\ln \left( \Delta u_h+\sqrt{1+(\Delta u_h)^2}\right) \right) \nabla \varphi (x,kh)\hbox {d}x \nonumber \\&\quad +h\lambda \sum _{k=1}^n \int _\Omega |u_h|^{p-2}u_h\varphi (x,kh)\hbox {d}x=0. \end{aligned}$$

(2.16)

Passing to the limits as $h\rightarrow 0$, we obtain from (2.13), (2.14), (2.16) that

$$\begin{aligned}&-\int _0^T\int _\Omega u_\varepsilon \frac{\partial \varphi }{\partial t}\hbox {d}x\hbox {d}\tau -\int _\Omega u_0\varphi (x,0)\hbox {d}x \nonumber \\&\quad -\int _0^T \int _\Omega \nabla (\varepsilon \Delta u_\varepsilon +\xi _\varepsilon )\nabla \varphi \hbox {d}x\hbox {d}\tau +\lambda \int _0^T \int _\Omega |u_\varepsilon |^{p-2}u_\varepsilon \varphi \hbox {d}x\hbox {d}\tau =0. \end{aligned}$$

(2.17)

In addition, if $\varphi \in C_0^\infty (\Omega _T)$, we obtain

$$\begin{aligned}&-\int _0^T\int _\Omega u_\varepsilon \frac{\partial \varphi }{\partial t}\hbox {d}x\hbox {d}\tau \nonumber \\&\quad =\int _0^T \int _\Omega \nabla (\varepsilon \Delta u_\varepsilon +\xi _\varepsilon )\nabla \varphi \hbox {d}x\hbox {d}\tau -\lambda \int _0^T \int _\Omega |u_\varepsilon |^{p-2}u_\varepsilon \varphi \hbox {d}x\hbox {d}\tau . \end{aligned}$$

(2.18)

Noticing that $u_\varepsilon \in L^2(0,T;H^3(\Omega ))$ and $\xi _\varepsilon \in L^2(0,T;H_0^1(\Omega ))$, we see that

$$\begin{aligned} \frac{\partial u_\varepsilon }{\partial t}\in L^2(0,T;H^{-1}(\Omega )). \end{aligned}$$

As $u_\varepsilon \in L^2(0,T;H^1_0(\Omega ))$, it follows from the compact imbedding relation

$$\begin{aligned} H_0^1(\Omega )\hookrightarrow L^2(\Omega )\hookrightarrow H^{-1}(\Omega ), \end{aligned}$$

that

$$\begin{aligned} u_\varepsilon \in C([0,T];L^2(\Omega )). \end{aligned}$$

As the function $u_\varepsilon$ satisfies (2.17), we only need to show that $\xi _\varepsilon =\ln (\Delta u_\varepsilon +\sqrt{1+(\Delta u_\varepsilon )^2})$ a.e. in $\Omega _T$ to prove the existence of weak solutions. Taking $u_\varepsilon$ as a test function in (2.4), we have

$$\begin{aligned}&\int _\Omega \frac{1}{2}|u_\varepsilon (T)|^2\hbox {d}x-\int _\Omega \frac{1}{2}|u_0|^2\hbox {d}x +\int _0^T \int _\Omega (\varepsilon \Delta u_\varepsilon +\xi _\varepsilon )\Delta u_\varepsilon \hbox {d}x\hbox {d}\tau \nonumber \\&\quad +\lambda \int _0^T \int _\Omega |u_\varepsilon |^p\hbox {d}x\hbox {d}\tau =0. \end{aligned}$$

(2.19)

Choosing $u_k$ as a test function in (2.7), we have

$$\begin{aligned}&\frac{1}{2}\int _\Omega u_k^2\hbox {d}x+\int _\Omega h\Delta \left( \varepsilon \Delta u_k+\ln \left( \Delta u_k+\sqrt{1+(\Delta u_k)^2}\right) \right) u_k\hbox {d}x\\&\quad +h\lambda \int _\Omega |u_k|^p\hbox {d}x\le \frac{1}{2}\int _\Omega u^2_{k-1}\hbox {d}x. \end{aligned}$$

Summing up the above equalities for $k=1,2,\ldots ,n$, we derive that

$$\begin{aligned}&\frac{1}{2}\int _\Omega u^2_h(T)\hbox {d}x+\int _0^T\int _\Omega \left( \varepsilon \Delta u_h+\ln \left( \Delta u_h+\sqrt{1+(\Delta u_h)^2}\right) \right) \Delta u_h\hbox {d}x\hbox {d}\tau \nonumber \\&\quad +\lambda \int _0^T\int _\Omega |u_h|^p\hbox {d}x\hbox {d}\tau \le \frac{1}{2}\int _\Omega u^2_0\hbox {d}x. \end{aligned}$$

(2.20)

Using the fact

$$\begin{aligned} \left( \ln \left( \xi +\sqrt{1+\xi ^2}\right) -\ln \left( \eta +\sqrt{1+\eta ^2}\right) \right) (\xi -\eta )\ge 0, \end{aligned}$$

for all $\xi ,\eta \in \mathbb {R}$, we easily know that

$$\begin{aligned} \int _0^T\int _\Omega \left( \ln \left( \Delta u_h+\sqrt{1+(\Delta u_h)^2}\right) -\ln \left( \Delta v+\sqrt{1+(\Delta v)^2}\right) \right) (\Delta u_h-\Delta v)\hbox {d}x\hbox {d}\tau \ge 0, \end{aligned}$$

for every $v\in L^2(0,T;H^2(\Omega ))$. Thus, from (2.20) we have that

$$\begin{aligned}&\frac{1}{2}\int _\Omega u^2_h(T)\hbox {d}x-\frac{1}{2}\int _\Omega u^2_0\hbox {d}x +\int _0^T\int _\Omega \varepsilon |\Delta u_h|^2\hbox {d}x\hbox {d}\tau +\lambda \int _0^T\int _\Omega |u_h|^p\hbox {d}x\hbox {d}\tau \\&\quad +\int _0^T\int _\Omega \left( \left( \ln \left( \Delta u_h+\sqrt{1+(\Delta u_h)^2}\right) \right) \Delta v +\left( \ln \left( \Delta v+\sqrt{1+(\Delta v)^2}\right) \right) \Delta u_h\right) \hbox {d}x\hbox {d}\tau \\&\quad -\int _0^T\int _\Omega \left( \ln \left( \Delta v+\sqrt{1+(\Delta v)^2}\right) \right) \Delta v\hbox {d}x\hbox {d}\tau \le 0. \end{aligned}$$

Passing to limits as $h\rightarrow 0$ and noticing

$$\begin{aligned} \Vert u_\varepsilon (T)\Vert _{L^2(\Omega )}\le \liminf _{h\rightarrow 0}\Vert u_h(T)\Vert _{L^2(\Omega )},\\ \Vert \Delta u_\varepsilon \Vert _{L^2(\Omega _T)}\le \liminf _{h\rightarrow 0}\Vert \Delta u_h\Vert _{L^2(\Omega _T)}, \end{aligned}$$

we obtain

$$\begin{aligned}&\frac{1}{2}\int _\Omega u_\varepsilon ^2(T)\hbox {d}x-\frac{1}{2}\int _\Omega u^2_0\hbox {d}x+\int _0^T\int _\Omega \varepsilon |\Delta u_\varepsilon |^2\hbox {d}x\hbox {d}\tau +\lambda \int _0^T\int _\Omega |u_\varepsilon |^p\hbox {d}x\hbox {d}\tau \nonumber \\&\quad +\int _0^T\int _\Omega \xi _\varepsilon \Delta v\hbox {d}x\hbox {d}\tau +\int _0^T\int _\Omega \left( \ln \left( \Delta v+\sqrt{1+(\Delta v)^2}\right) \right) \Delta u_\varepsilon \hbox {d}x\hbox {d}\tau \nonumber \\&\quad -\int _0^T\int _\Omega \left( \ln \left( \Delta v+\sqrt{1+(\Delta v)^2}\right) \right) \Delta v\hbox {d}x\hbox {d}\tau \le 0. \end{aligned}$$

(2.21)

Combining (2.19) with (2.21), we have, for every $v\in L^2(0,T;H^2(\Omega ))$,

$$\begin{aligned}&\int _0^T\int _\Omega \xi _\varepsilon \Delta u_\varepsilon \hbox {d}x\hbox {d}\tau -\int _0^T\int _\Omega \xi _\varepsilon \Delta v\hbox {d}x\hbox {d}\tau +\int _0^T\int _\Omega \left( \ln \left( \Delta v+\sqrt{1+(\Delta v)^2}\right) \right) \Delta u_\varepsilon \hbox {d}x\hbox {d}\tau \\&\quad -\int _0^T\int _\Omega \left( \ln \left( \Delta v+\sqrt{1+(\Delta v)^2}\right) \right) \Delta v\hbox {d}x\hbox {d}\tau \ge 0, \end{aligned}$$

which is

$$\begin{aligned} \int _0^T\int _\Omega \left( \xi _\varepsilon -\ln \left( \Delta v+\sqrt{1+(\Delta v)^2}\right) \right) (\Delta u_\varepsilon -\Delta v) \hbox {d}x\hbox {d}\tau \ge 0. \end{aligned}$$

For each $\gamma >0,\quad \omega \in C_0^\infty (\Omega _T)$, we choose $v\in L^2(0,T;H^2(\Omega )\cap H_0^1(\Omega ))$ to be the solution of $\Delta v=\Delta u_\varepsilon -\lambda \omega$ in the above inequality to have

$$\begin{aligned} \int _0^T\int _\Omega \omega (\xi _\varepsilon -\ln ((\Delta u_\varepsilon -\gamma \omega )+\sqrt{1+(\Delta u_\varepsilon -\gamma \omega )^2})) \hbox {d}x\hbox {d}\tau \ge 0. \end{aligned}$$

Passing to limits as $\gamma \rightarrow 0$ and using Lebesgue’s dominated convergence theorem, we get

$$\begin{aligned} \int _0^T\int _\Omega \omega (\xi _\varepsilon -\ln (\Delta u_\varepsilon +\sqrt{1+(\Delta u_\varepsilon )^2})) \hbox {d}x\hbox {d}\tau \ge 0, \end{aligned}$$

for every $\omega \in C_0^\infty (\Omega _T)$ and conclude that $\xi _\varepsilon =\ln (\Delta u_\varepsilon +\sqrt{1+(\Delta u_\varepsilon )^2})$ a. e. in $\Omega _T$. By approximation, we use (2.20) to obtain (2.5) and use (2.8) to obtain (2.6). Therefore, we finish the proof of the existence of weak solutions.

The proof of the uniqueness of weak solutions is similar to the proof of uniqueness of problem (2.1), so we omit the details. Thus, we complete the proof of Theorem 2.3. $\square$

Proof of Theorem 2.1

First, by Theorem 2.3, we know that

$$\begin{aligned}&\Vert u_\varepsilon \Vert _{L^\infty (0,T;H_0^1(\Omega ))}+\varepsilon \Vert u_\varepsilon \Vert _{L^2(0,T;H^3(\Omega ))} +\Vert \ln (\Delta u_\varepsilon \\&\quad +\sqrt{1+(\Delta u_\varepsilon )^2})\Vert _{L^2(0,T;H_0^1(\Omega ))}\le C. \end{aligned}$$

Therefore, we can extract a subsequence from $\{u_\varepsilon \}$, denoted also by $\{u_\varepsilon \}$, such that

$$\begin{aligned} u_\varepsilon \rightharpoonup u,\qquad&\hbox {weakly-}^* \hbox {in} \quad L^\infty (0,T;H_0^1(\Omega )),\\ \sqrt{\varepsilon }\nabla \Delta u_\varepsilon \rightharpoonup \zeta ,\qquad&\hbox {weakly in} \quad L^2(\Omega _T),\\ \xi _\varepsilon =\ln (\Delta u_\varepsilon +\sqrt{1+(\Delta u_\varepsilon )^2}) \rightharpoonup \xi ,\qquad&\hbox {weakly in} \quad L^2(0,T;H_0^1(\Omega )), \end{aligned}$$

which follows that ([5], Chapter 2)

$$\begin{aligned} \Vert u\Vert _{L^\infty (0,T;H_0^1(\Omega ))}+\Vert \zeta \Vert _{L^2(\Omega _T)}+\Vert \xi \Vert _{L^2(0,T;H_0^1(\Omega ))}\le C. \end{aligned}$$

(2.22)

Using (2.17), we deduce that

$$\begin{aligned}&-\int _0^T\int _\Omega u_\varepsilon \frac{\partial \varphi }{\partial t}\hbox {d}x\hbox {d}\tau -\int _\Omega u_0\varphi (x,0)\hbox {d}x\\&\quad -\int _0^T \int _\Omega \nabla (\varepsilon \Delta u_\varepsilon +\xi _\varepsilon )\nabla \varphi \hbox {d}x\hbox {d}\tau + \lambda \int _0^T \int _\Omega |u_\varepsilon |^{p-2}u_\varepsilon \varphi (x,0)\hbox {d}x\hbox {d}\tau =0. \end{aligned}$$

Letting $\varepsilon \rightarrow 0$, we see that

$$\begin{aligned}&-\int _0^T\int _\Omega u\frac{\partial \varphi }{\partial t}\hbox {d}x\hbox {d}\tau -\int _\Omega u_0\varphi (x,0)\hbox {d}x \nonumber \\&\quad -\int _0^T \int _\Omega \nabla \xi \nabla \varphi \hbox {d}x\hbox {d}\tau + \lambda \int _0^T \int _\Omega |u|^{p-2}u\varphi (x,0)\hbox {d}x\hbox {d}\tau =0. \end{aligned}$$

(2.23)

Choosing $\varphi \in C_0^\infty (\Omega _T)$, we get

$$\begin{aligned} -\int _0^T\int _\Omega u\frac{\partial \varphi }{\partial t}\hbox {d}x\hbox {d}\tau -\int _0^T \int _\Omega \nabla \xi \nabla \varphi \hbox {d}x\hbox {d}\tau =0, \end{aligned}$$

which implies that

$$\begin{aligned} \frac{\partial u}{\partial t}\in L^2(0,T;H^{-1}(\Omega )). \end{aligned}$$

As $u\in L^2(0,T;H^1_0(\Omega ))$, it follows from the compact imbedding relation

$$\begin{aligned} H_0^1(\Omega )\hookrightarrow L^2(\Omega )\hookrightarrow H^{-1}(\Omega ), \end{aligned}$$

that

$$\begin{aligned} u\in C([0,T];L^2(\Omega )). \end{aligned}$$

On the other hand, (2.6) implies that

$$\begin{aligned}&\int _\Omega |\nabla u_\varepsilon |^2\hbox {d}x\le \int _\Omega |\nabla u_0|^2\hbox {d}x, \end{aligned}$$

(2.24)

$$\begin{aligned}&\int _0^t\int _\Omega \frac{1}{\sqrt{1+|\Delta u_\varepsilon |^2}}|\nabla \Delta u_\varepsilon |^2 \hbox {d}x\hbox {d}\tau \le \frac{1}{2}\int _\Omega |\nabla u_0|^2\hbox {d}x. \end{aligned}$$

(2.25)

Denote

$$\begin{aligned} \omega _\varepsilon =\Delta u_\varepsilon . \end{aligned}$$

Noticing that $|\nabla |\omega _\varepsilon ||\le |\nabla \omega _\varepsilon |$, we conclude that

$$\begin{aligned} \int _0^t\int _{\Omega }|\nabla \ln \left( |\omega _\varepsilon |+\sqrt{1+|\omega _\varepsilon |^2}\right) |^2\hbox {d}x\hbox {d}\tau&=\int _0^t\int _{\Omega }\left| \frac{\nabla |\omega _\varepsilon |}{\root 4 \of {1+|\omega _\varepsilon |^2}}\right| ^2\hbox {d}x\hbox {d}\tau \\&\le \int _0^t\int _{\Omega }\frac{|\nabla \omega _\varepsilon |^2}{\sqrt{1+|\omega _\varepsilon |^2}}\hbox {d}x\hbox {d}\tau \\&\le \frac{1}{2}\int _{\Omega }|\nabla u_0|^2\hbox {d}x\le C. \end{aligned}$$

Setting $v_\varepsilon =\ln (|\omega _\varepsilon |+\sqrt{1+|\omega _\varepsilon |^2})$ and using $v_\varepsilon |_{\partial \Omega }=0$, we know that $v_\varepsilon \in L^2(0,T;H_0^1(\Omega ))$ and

$$\begin{aligned} \Vert v_\varepsilon \Vert ^2_{L^2(0,T;H_0^1(\Omega ))}\le C\int _0^T\int _\Omega |\nabla v_\epsilon |^2\hbox {d}x\hbox {d}\tau \le C. \end{aligned}$$

(2.26)

By $N=2$, we see that $H_0^1(\Omega )\hookrightarrow L^\varphi (\Omega )$ with $\varphi =\exp t^2-1$. Then, we have $L^2(0,T;H_0^1(\Omega ))\hookrightarrow L^1(0,T;L^\varphi (\Omega ))$, and that there exist two positive numbers $C_1, C_2$ such that

$$\begin{aligned} \int _0^T\int _\Omega \exp \left( \frac{v_\varepsilon }{C_1||\nabla v_\varepsilon ||_{L^2(\Omega _T)}}\right) ^2\hbox {d}x\hbox {d}\tau \le C_2|\Omega _T|. \end{aligned}$$

In addition, for every $\delta >0$,

$$\begin{aligned} e^{2t}\le e^{(\frac{t}{\delta })^2}+e^{4\delta ^2}. \end{aligned}$$

Choosing $t=v_\varepsilon$ and $\delta =C_1\Vert \nabla v_\varepsilon \Vert _{L^2(\Omega _T)}$ in the above inequality, we derive

$$\begin{aligned} \int _0^T\int _\Omega e^{2v_\varepsilon }\hbox {d}x\hbox {d}\tau&\le \int _{\Omega _T}\exp \left( \frac{v_\varepsilon }{C_1\Vert \nabla v_\varepsilon \Vert _{L^2(\Omega _T)}}\right) ^2\hbox {d}x\hbox {d}\tau +e^{4C_1^2\Vert \nabla v_\varepsilon \Vert ^2_{L^2(\Omega _T)}}|\Omega _T|\\&\le C|\Omega _T|, \end{aligned}$$

which further implies that

$$\begin{aligned} \int _0^T\int _\Omega |\omega _\varepsilon |^2\hbox {d}x\hbox {d}\tau =\int _{\Omega _T}|\Delta u_\varepsilon |^2\hbox {d}x\hbox {d}\tau \le C|\Omega _T|\le C. \end{aligned}$$

(2.27)

It follows from $u_\varepsilon \in L^\infty (0,T;H_0^1(\Omega ))$ that

$$\begin{aligned} \Vert u_\varepsilon \Vert _{L^2 (0,T;H^2(\Omega ))}\le C. \end{aligned}$$

(2.28)

Therefore, we can extract a subsequence from $\{u_\varepsilon \}$, denoted also by $\{u_\varepsilon \}$, such that

$$\begin{aligned} u_\varepsilon \rightharpoonup u,\qquad \hbox {weakly in} \quad L^2 (0,T;H^2(\Omega )), \end{aligned}$$

(2.29)

which follows that ([5], Chapter 2)

$$\begin{aligned} \Vert u\Vert _{ L^2 (0,T;H^2(\Omega ))}\le C. \end{aligned}$$

(2.30)

We only need to show that $\xi =\ln (\Delta u+\sqrt{1+(\Delta u)^2})$ a. e. in $\Omega _T$ to prove the existence of weak solutions of problems (1.1)–(1.3).

Taking u as a test function in (2.23), we know that

$$\begin{aligned} \int _\Omega \frac{1}{2}u^2(T)\hbox {d}x-\int _\Omega \frac{1}{2}u^2_0\hbox {d}x+\int _0^T\int _\Omega \xi \Delta u\hbox {d}x\hbox {d}\tau +\lambda \int _0^T \int _\Omega |u|^p\hbox {d}x\hbox {d}\tau =0. \end{aligned}$$

(2.31)

Passing to limits as $\varepsilon \rightarrow 0$ and noticing

$$\begin{aligned} \Vert u(T)\Vert _{L^2(\Omega )}\le \liminf _{\varepsilon \rightarrow 0}\Vert u_\varepsilon (T)\Vert _{L^2(\Omega )},\\ \Vert \Delta u\Vert _{L^2(\Omega _T)}\le \liminf _{\varepsilon \rightarrow 0}\Vert \Delta u_\varepsilon \Vert _{L^2(\Omega _T)}, \end{aligned}$$

we obtain

$$\begin{aligned}&\int _\Omega \left[ \frac{1}{2}u^2(T)-\frac{1}{2}u^2_0\right] \hbox {d}x+\lambda \int _0^T\int _\Omega |u|^p\hbox {d}x\hbox {d}\tau \\&\quad +\int _0^T\int _\Omega \left[ \xi \Delta v+\ln \left( \Delta v+\sqrt{1+(\Delta v)^2}\right) \Delta u\right. \\&\quad \left. -\ln \left( \Delta v+\sqrt{1+(\Delta v)^2}\right) \Delta v\right] \hbox {d}x\hbox {d}\tau \le 0. \end{aligned}$$

Using (2.31), we have

$$\begin{aligned}&\int _0^T\int _\Omega \xi \Delta u\hbox {d}x\hbox {d}\tau -\!\int _0^T\int _\Omega \xi \Delta v\hbox {d}x\hbox {d}\tau \!+\!\int _0^T\int _\Omega \left( \ln \left( \Delta v+\sqrt{1+(\Delta v)^2}\right) \right) \Delta u\hbox {d}x\hbox {d}\tau \\&\quad -\int _0^T\int _\Omega \left( \ln \left( \Delta v+\sqrt{1+(\Delta v)^2}\right) \right) \Delta v\hbox {d}x\hbox {d}\tau \ge 0, \end{aligned}$$

which is

$$\begin{aligned} \int _0^T\int _\Omega \left( \xi -\ln \left( \Delta v+\sqrt{1+(\Delta v)^2}\right) \right) (\Delta u-\Delta v)\hbox {d}x\hbox {d}\tau \ge 0. \end{aligned}$$

For each $\gamma >0, \omega \in C_0^\infty (\Omega _T)$, we choose $v\in L^2(0,T;H^2(\Omega )\cap H_0^1(\Omega ))$ to be the solution of $\Delta v=\Delta u-\gamma \omega$ in the above inequality to have

$$\begin{aligned} \int _0^T\int _\Omega \omega \left( \xi -\ln \left( \Delta u-\gamma \omega +\sqrt{1+(\Delta u-\gamma \omega )^2}\right) \right) \hbox {d}x\hbox {d}\tau \ge 0. \end{aligned}$$

Passing to limits as $\gamma \rightarrow 0$ and using Lebesgue’s dominated convergence theorem, we get

$$\begin{aligned} \int _0^T\int _\Omega \omega \left( \xi -\ln \left( \Delta u+\sqrt{1+(\Delta u)^2}\right) \right) \hbox {d}x\hbox {d}\tau \ge 0, \end{aligned}$$

for every $\omega \in C_0^\infty (\Omega _T)$ and conclude that $\xi =\ln (\Delta u+\sqrt{1+(\Delta u)^2})$ a. e. in $\Omega _T$.

It follows from (2.23) that u is a weak solution of problems (1.1)–(1.3). Therefore, we finish the proof of the existence of weak solutions.

The proof of uniqueness of weak solutions is obvious, so we omit the details. The proof of Theorem 2.1 is complete. $\square$

3 Asymptotic behavior

This section is devoted to the asymptotic behavior of solutions. To this purpose, we first show that

Theorem 3.1

The weak solution u satisfies that for any $0\le \rho \in C^2(\overline{\Omega })$,

$$\begin{aligned}&\frac{1}{2}\int _\Omega \rho (x)| u(x,t)|^2\hbox {d}x-\frac{1}{2}\int _\Omega \rho (x)| u_0(x)|^2\hbox {d}x+\lambda \iint _{Q_t}\rho (x)|u(x,\tau )|^p \hbox {d}x\hbox {d}\tau \nonumber \\&\quad =-\iint _{Q_t}\ln \left( \Delta u+\sqrt{1+(\Delta u)^2}\right) \Delta (\rho (x)u(x,\tau ))\hbox {d}x\hbox {d}\tau , \end{aligned}$$

(3.1)

where $Q_t=\Omega \times (0,t)$.

Proof

In the proof of Theorem 2.3, we have

$$\begin{aligned} f(t)=\frac{1}{2}\int _\Omega | u(x,t)|^2\hbox {d}x\in C([0,T]). \end{aligned}$$

Similarly, we can also easily prove that for any $0\le \rho (x)\in C^2(\overline{\Omega })$,

$$\begin{aligned} f_\rho (t)=\frac{1}{2}\int _\Omega \rho (x)| u(x,t)|^2\hbox {d}x\in C([0,T]). \end{aligned}$$

Consider the functional

$$\begin{aligned} \Phi _\rho [v]=\frac{1}{2}\int _\Omega \rho (x)|v(x)|^2\hbox {d}x. \end{aligned}$$

It is easy to see that $\Phi _\rho [v]$ is a convex functional on $L^2(\Omega )$.

For any $\tau \in (0,T)$ and $h>0$, we have

$$\begin{aligned} \Phi _\rho [u(\tau +h)]-\Phi _\rho [u(\tau )]\ge \langle u(\tau +h)-u(\tau ), \rho (x)u(x,\tau )\rangle . \end{aligned}$$

By $\frac{\delta \Phi _\rho [v]}{\delta v}=\rho (x)v$, for any fixed $t_1,t_2\in [0,T],t_1<t_2$, integrating the above inequality with respect to $\tau$ over $(t_1,t_2)$ , we have

$$\begin{aligned} \int _{t_2}^{t_2+h}\Phi _\rho [u(\tau )]\hbox {d}\tau -\int _{t_1}^{t_1+h}\Phi _\rho [u(\tau )]\hbox {d}\tau \ge \int _{t_1}^{t_2}\langle u(\tau +h)-u(\tau ), \rho (x)u\rangle \hbox {d}\tau . \end{aligned}$$

Multiplying the both sides of the above inequality by $\frac{1}{h}$, and letting $h\rightarrow 0$, we obtain

$$\begin{aligned} \Phi _\rho [u(t_2)]-\Phi _\rho [u(t_1)]\ge \int _{t_1}^{t_2}\left\langle \frac{\partial u}{\partial t},\rho (x) u\right\rangle \hbox {d}\tau . \end{aligned}$$

Similarly, we have

$$\begin{aligned} \Phi _\rho [u(\tau )]-\Phi _\rho [u(\tau -h)] \le \langle (u(\tau )-u(\tau -h)),\rho (x) u\rangle . \end{aligned}$$

Thus,

$$\begin{aligned} \Phi _\rho [u(t_2)]-\Phi _\rho [u(t_1)]\le \int _{t_1}^{t_2}\left\langle \frac{\partial u}{\partial t},\rho (x) u\right\rangle \hbox {d}\tau , \end{aligned}$$

and hence,

$$\begin{aligned} \Phi _\rho [u(t_2)]-\Phi _\rho [u(t_1)] =\int _{t_1}^{t_2}\left\langle \frac{\partial u}{\partial t},\rho (x) u\right\rangle \hbox {d}\tau . \end{aligned}$$

Taking $t_1=0,t_2=t$, we get from the definition of solutions that

$$\begin{aligned}&\Phi _\rho [u(t)]-\Phi _\rho [u(0)]\\&\quad =\int _0^t\left\langle -\Delta \left( \ln \left( \Delta u+\sqrt{1+(\Delta u)^2}\right) \right) -\lambda |u|^{p-2}u, \rho (x) u(\tau )\right\rangle \hbox {d}\tau \\&\quad =-\int _0^t \left\langle \ln \left( \Delta u+\sqrt{1+(\Delta u)^2}\right) , \Delta [\rho (x)u(\tau )]\right\rangle \hbox {d}\tau -\int _0^t \langle \lambda |u|^{p-2}u, \rho (x)u(\tau )\rangle \hbox {d}\tau . \end{aligned}$$

The proof is complete. $\square$

Theorem 3.2

Let u be the weak solution of problems (1.1)–(1.3), then

$$\begin{aligned} \int _\Omega |u(x,t)|^2\hbox {d}x \le \frac{1}{(\frac{p-2}{2}|\Omega |^{\frac{2-p}{2}}\lambda t+C_1)^{\alpha }},\quad \alpha =\frac{2}{p-2},C_1>0. \end{aligned}$$

Proof

Taking $\rho (x)=1$ in equality (3.1), we have

$$\begin{aligned}&\frac{1}{2}\int _\Omega | u(x,t)|^2\hbox {d}x-\frac{1}{2}\int _\Omega | u_0(x)|^2\hbox {d}x \nonumber \\&\quad =-\int _0^t\int _\Omega \ln \left( \Delta u+\sqrt{1+(\Delta u)^2}\right) \Delta u \hbox {d}x\hbox {d}t-\lambda \iint _{Q_t}|u|^p \hbox {d}x\hbox {d}t. \end{aligned}$$

(3.2)

Let $f(t)=\frac{1}{2}\int _\Omega |u(x,t)|^2\hbox {d}x$. By (3.2), we have

$$\begin{aligned} f'(t)=-\int _\Omega \ln \left( \Delta u+\sqrt{1+(\Delta u)^2}\right) \Delta u\hbox {d}x-\lambda \int _\Omega |u|^p \hbox {d}x\le 0. \end{aligned}$$

Noticing that $\ln (\Delta u+\sqrt{1+(\Delta u)^2})\Delta u\ge 0$ and using the Hölder inequality, we conclude that

$$\begin{aligned} \int _\Omega | u(x,t)|^2\hbox {d}x\le |\Omega |^{\frac{p-2}{p}}\left( \int _\Omega |u|^p\hbox {d}x\right) ^{2/p}, \end{aligned}$$

that is $f(t)\le |\Omega |^{\frac{p-2}{p}}\lambda ^{-\frac{2}{p}}|f'(t)|^{2/p}$.

Again by $f'(t)\le 0$, we know that $f'(t)\le -|\Omega |^{\frac{2-p}{2}}\lambda f(t)^{p/2}$, and hence,

$$\begin{aligned} \int _\Omega | u(x,t)|^2\hbox {d}x\le \frac{1}{(\frac{p-2}{2}|\Omega |^{\frac{2-p}{2}}\lambda t+C_1)^{\alpha }},\quad \alpha =\frac{2}{p-2},C_1>0. \end{aligned}$$

The proof is complete. $\square$

4 Regularity of solutions

In this section, we consider the regularity of solutions for problems (1.1)–(1.3).

Theorem 4.1

If u is weak solution of problems (1.1)–(1.3), for any $(x_1,t_1)$, $(x_2,t_2)\in Q_T$, we have

$$\begin{aligned} |u(x_1,t_1)-u(x_2,t_2)|\le C(|x_1-x_2|+|t_1-t_2|^{1/2}), \end{aligned}$$

where C is a constant depending only on p.

Proof

Let

$$\begin{aligned} u_\varepsilon (x,t)=J_\varepsilon u(x,t)=\int _0^T\int _{|x-y|<\varepsilon }j_\varepsilon (x-y,t-s)u(y,s)\hbox {d}y\hbox {d}s, \end{aligned}$$

where $j_\varepsilon (x-y,t-s)$ is the mollifier.

For any $x_1,x_2\in \Omega$, we have

$$\begin{aligned}&u_\varepsilon (x_1,t)-u_\varepsilon (x_2,t)\\&\quad =\int _0^T\int _{R^2}j_\varepsilon (x_1-y,t-s)u(y,s)\hbox {d}y\hbox {d}s-\int _0^T\int _{R^2}j_\varepsilon (x_2-y,t-s)u(y,s)\hbox {d}y\hbox {d}s\\&\quad =\int _0^T\int _{R^2}\frac{\partial j_\varepsilon (zx_1+(1-z)x_2-y,t-s)}{\partial z} u(y,s)\hbox {d}z\hbox {d}y\hbox {d}s\\&\quad =\int _0^T\int _{R^2}\int _0^1\nabla _xj_\varepsilon (zx_1+(1-z)x_2-y,t-s)(x_1-x_2) u(y,s)\hbox {d}z\hbox {d}y\hbox {d}s \\&\quad =-\int _0^T\int _{R^2}\int _0^1\nabla _yj_\varepsilon (zx_1+(1-z)x_2-y,t-s)(x_1-x_2) u(y,s)\hbox {d}z\hbox {d}y\hbox {d}s \\&\quad =\int _0^T\int _{R^2}\int _0^1j_\varepsilon (zx_1+(1-z)x_2-y,t-s)\nabla _yu(y,s) \hbox {d}z\hbox {d}y\hbox {d}s(x_1-x_2). \end{aligned}$$

Therefore,

$$\begin{aligned}&|u_\varepsilon (x_1,t)-u_\varepsilon (x_2,t)| \\&\quad \le \int _0^T\int _{R^2}\int _0^1|j_\varepsilon (zx_1+(1-z)x_2-y,t-s)| |\nabla _y u(y,s)|\hbox {d}z\hbox {d}y\hbox {d}s|x_1-x_2|, \end{aligned}$$

and by $u\in L^2(0,T;H^2(\Omega ))$, we obtain

$$\begin{aligned} |u_\varepsilon (x_1,t)-u_\varepsilon (x_2,t)|\le C|x_1-x_2|. \end{aligned}$$

(4.1)

Set $0<\varepsilon<t_1<t_2<T$. Let $\Delta t=t_2-t_1$, $B_\rho =B_{(\Delta t)^{1/2}} (x_0)$, $x_0\in \Omega$, choose $\rho$ sufficiently small, such that $B_\rho \subset \Omega ,\varphi \in C^2_0(B_\rho )$. Therefore, we can obtain

$$\begin{aligned}&\int _{B_\rho }\varphi (x)(u_\varepsilon (x,t_2)-u_\varepsilon (x,t_1))\hbox {d}x \nonumber \\&\quad =\int _{B_\rho }\varphi (x)\int _0^1\frac{\partial u_\varepsilon (x,st_2+(1-s)t_1)}{\partial s}\hbox {d}s\hbox {d}x \nonumber \\&\quad =\Delta t\int _{B_\rho }\varphi (x)\int _0^1\int _0^T\int _{|x-y|<\varepsilon } u(y,\tau )\nonumber \\&\qquad \cdot j_{\varepsilon t}(x-y,st_2+(1-s)t_1-\tau )\hbox {d}y\hbox {d}\tau \hbox {d}s\hbox {d}x \nonumber \\&\quad =-\Delta t\int _{B_\rho }\varphi (x)\int _0^1\int _0^T\int _{|x-y|<\varepsilon } u(y,\tau )\nonumber \\&\qquad \cdot j_{\varepsilon \tau }(x-y,st_2+(1-s)t_1-\tau ) \hbox {d}y\hbox {d}\tau \hbox {d}s\hbox {d}x. \end{aligned}$$

(4.2)

Fixed $(x,t)\in Q_T,\;0<\varepsilon<t<T-\varepsilon$, we have $j_\varepsilon (x-y,t-\tau )\in C_0^2(Q_T)$, from definition of weak solution

$$\begin{aligned}&\int _0^T\int _{|x-y|<\varepsilon } j_{\varepsilon \tau }(x-y,st_2+(1-s)t_1-\tau )u(y,\tau )\hbox {d}y\hbox {d}\tau \\&\quad =\int _0^T\int _{|x-y|<\varepsilon }\ln \left( \Delta _yu+\sqrt{1+(\Delta _yu)^2}\right) \Delta _yj_\varepsilon (x-y,st_2+(1-s)t_1-\tau )\hbox {d}y\hbox {d}\tau \\&\qquad +\lambda \int _0^T\int _{|x-y|<\varepsilon }|u|^{p-2}uj_\varepsilon (x-y,st_2+(1-s)t_1-\tau )\hbox {d}y\hbox {d}\tau , \end{aligned}$$

and hence, (4.2) is converted into

$$\begin{aligned}&\int _{B_\rho }\varphi (x)(u_\varepsilon (x,t_2)-u_\varepsilon (x,t_1))\hbox {d}x\\&\quad =-\Delta t\int _{B_\rho }\varphi (x)\int _0^1\int _0^T\int _{|x-y|<\varepsilon } \ln \left( \Delta _yu+\sqrt{1+(\Delta _yu)^2}\right) \\&\qquad \cdot \Delta _yj_\varepsilon (x-y,st_2+(1-s)t_1-\tau )\hbox {d}y\hbox {d}\tau \hbox {d}s\hbox {d}x \\&\qquad -\lambda \Delta t\int _{B_\rho }\varphi (x)\int _0^1\int _0^T\int _{|x-y|<\varepsilon }|u|^{p-2}u\\&\qquad \cdot j_\varepsilon (x-y,st_2+(1-s)t_1-\tau )\hbox {d}y\hbox {d}\tau \hbox {d}s\hbox {d}x\\&\quad =-\Delta t\int _0^1\int _{B_\rho }\Delta _x\varphi (x)\int _0^T\int _{|x-y|<\varepsilon } \ln \left( \Delta _yu+\sqrt{1+(\Delta _yu)^2}\right) \\&\qquad \cdot j_\varepsilon (x-y,st_2+(1-s)t_1-\tau )\hbox {d}y\hbox {d}\tau \hbox {d}x\hbox {d}s\\&\qquad -\lambda \Delta t\int _{B_\rho }\varphi (x)\int _0^1\int _0^T\int _{|x-y|<\varepsilon }|u|^{p-2}u\\&\qquad \cdot j_\varepsilon (x-y,st_2+(1-s)t_1-\tau )\hbox {d}y\hbox {d}\tau \hbox {d}s\hbox {d}x. \end{aligned}$$

Taking

$$\begin{aligned} \varphi (x)=\varphi _h(x)=\int _{x_0-(\Delta t)^{1/2}e}^x\int _{-h}^{(\Delta t)^{1/2}-|y-x_0|-2h}\delta _h(s)\hbox {d}s\hbox {d}y, \end{aligned}$$

where $e=(1,1)$, $\delta (s)\in C_0^2(R);\;\delta (s)\ge 0;\;\delta (s)=0,$ as $|s|\ge 1$;$\int _R \delta (s)\hbox {d}s=1$. For $h>0$, define $\delta _h(s)=\frac{1}{h}\delta (\frac{s}{h})$.

Hence,

$$\begin{aligned}&\int _{B_\rho }\varphi _h(x)(u_\varepsilon (x,t_2)-u_\varepsilon (x,t_1))\hbox {d}x\\&\quad =-\Delta t\int _0^1\int _{I_\rho }\delta _h((\Delta t)^{1/2}-|x-x_0|-2h)\frac{x_0-x}{|x-x_0|}J_\varepsilon \left( \ln \left( \Delta u+\sqrt{1+(\Delta u)^2}\right) \right) \hbox {d}x\hbox {d}s\\&\qquad -\lambda \Delta t\int _{B_\rho }\varphi (x)\int _0^1\int _0^T\int _{|x-y|<\varepsilon }|u|^{p-2}u\\&\qquad \cdot j_\varepsilon (x-y,st_2+(1-s)t_1-\tau )\hbox {d}y\hbox {d}\tau \hbox {d}s\hbox {d}x. \end{aligned}$$

Noticing that for $x\in B_\rho ,\;\displaystyle \lim _{h\rightarrow 0}\varphi _h(x)=1$, and if $|x-x_0|<(\Delta t)^{1/2}-h$, then $\delta _h((\Delta t)^{1/2}-|x-x_0|-2h)=0$, $\delta _h\le \frac{C}{h}$ and

$$\begin{aligned} m(B_\rho \backslash B_{(\Delta t)^{1/2}-h})\le Ch(\Delta )^{1/2}. \end{aligned}$$

By $J_\varepsilon (\ln (\Delta u+\sqrt{1+(\Delta u)^2}))\le C$ and $u\in L^\infty (0,T;H^1_0(\Omega ))$, therefore

$$\begin{aligned} \left| \int _{B_\rho }\varphi _h(x)(u_\varepsilon (x,t_2)-u_\varepsilon (x,t_1))\hbox {d}x\right| \le C(\Delta t)^{3/2}. \end{aligned}$$

Letting $h\rightarrow 0$, we obtain

$$\begin{aligned} \left| \int _{B_\rho }(u_\varepsilon (x,t_2)-u_\varepsilon (x,t_1))\hbox {d}x\right| \le C(\Delta t)^{3/2}. \end{aligned}$$

Applying the mean value theorem, we see that for some $x^*\in B_\rho$ such that

$$\begin{aligned} |u_\varepsilon (x^*,t_2)-u_\varepsilon (x^*,t_1)|\le C(\Delta t)^{1/2}. \end{aligned}$$

Taking this into account and using (4.1), it follows that

$$\begin{aligned}&|u_\varepsilon (x,t_2)-u_\varepsilon (x,t_1)|\\&\quad \le |u_\varepsilon (x,t_2)-u_\varepsilon (x^*,t_2)|+|u_\varepsilon (x^*,t_2)-u_\varepsilon (x^*,t_1)|+ |u_\varepsilon (x^*,t_1)-u_\varepsilon (x,t_1)|\\&\quad \le C(\Delta t)^{1/2}, \end{aligned}$$

and letting $\varepsilon \rightarrow 0$, we known that u is Hölder continuous. The proof is complete. $\square$