Decay of solution for a 2D irrotational Mindlin–Timoshenko plate system

Abstract

This paper discusses the Cauchy problem for a semilinear homogeneous, isotropic Mindlin–Timoshenko plate in ${\mathbb {R}}^2$ with irrotational filament angle field. Applying Kawashima’s energy method in the frequency space, the decay rate for the semigroup associated with the linearized system is established. Moreover, the existence and uniqueness of local solution are proved with Banach’s fixed point theorem. Finally, the global existence and uniqueness of classical solution along with polynomial decay are proved via the barrier method.

1 Introduction and main result

The two-dimensional linear system for frictional damping isotropic plate can be written as follows

$$\begin{aligned} \left\{ \begin{array}{llll} \psi _{tt}-\frac{D}{A}\left( \psi _{xx}+\frac{1-\mu }{2}\psi _{yy}+\frac{1+\mu }{2}\varphi _{xy}\right) +\frac{K}{A}(\psi +w_x)+\alpha \psi _t= 0, &{}&{} &{} \\ \varphi _{tt}-\frac{D}{A}\left( \varphi _{yy}+\frac{1-\mu }{2}\varphi _{xx}+\frac{1+\mu }{2}\psi _{xy}\right) + \frac{K}{A}(\varphi +w_y)+\alpha \varphi _t= 0, &{}&{}&{} \\ w_{tt}-K[(\psi +w_x)_x+(\varphi +w_y)_y]+\beta w_t= 0, &{}&{}&{} \end{array}\right. \end{aligned}$$

(1.1)

which describes the vibration of an elastic plate. Here, $\varvec{x}=(x,y)\in {\mathbb {R}}^2$ and $t\in {\mathbb {R}}_+$ denote the space and time variables, respectively. $D>0$ is the modulus of flexural rigidity, $A=\frac{\rho h^3}{12}$ and $\rho , h, \alpha>0, \beta >0, 0<\mu <1$ are constants. K is called the shear modulus, more details can be found in [1, 2, 14].

In [14], Lagnese considered a bounded domain with Lipschitz boundary, under some geometric conditions, the exponentially stable was achieved without any restrictions on the coefficients. To our best knowledge, Mu$\tilde{\textrm{n}}$oz Rivera and Racke [18] first considered the dissipative nonlinear Timoshenko system which has the heat conduction effect,

$$\begin{aligned} \left\{ \begin{array}{llll} &{} \rho _1\varphi _{tt}-\sigma (\varphi _x-\psi )_x=0, &{}&{} (t,x)\in {\mathbb {R}}_+\times (0,L),\\ &{} \rho _2\psi _{tt}-b\psi _{xx}-k(\varphi _x-\psi )+l\theta _x=0, &{}&{} (t,x)\in {\mathbb {R}}_+\times (0,L),\\ &{} \rho _3\theta _t-k\theta _{xx}+l\psi _{tx}=0, &{}&{} (t,x)\in {\mathbb {R}}_+\times (0,L), \end{array}\right. \end{aligned}$$

(1.2)

where $\rho _1,\rho _2,\rho _3,b,k, l$ are positive constants. They obtained the global existence of small, smooth solutions as well as the exponential stability. Then, Fernández Sare [4] studied the stability of Mindlin–Timoshnko plates and proved that the system was not exponentially stable independent of any relations between the constants of the system, the polynomial decay rates also was obtained. Grobbelaar–van Dalsen did extensive research on the decay estimates for the Mindlin–Timoshenko system with different kinds of conditions, which can be found in [6,7,8,9,10]. For more related boundary value results, we refer to [3, 5, 21,22,23,24,25, 29] and references were cited.

For the Cauchy problem, Rack et al. [28] studied a semilinear Timoshenko system with two damping effects, the optimal decay rates and the global well-posedness for small data were obtained. Recently, Li and Xue [16] considered semilinear Mindlin–Timoshenko system with an irrotationality condition for the angle variables, and the decay estimate was shown. More results for Cauchy problem can be found in [26, 30] and the references therein. Motivated by the papers mentioned above, in this paper, we focus on the decay estimate and the global existence of solution for the Cauchy problem to a semilinear Mindlin–Timoshenko system with full-friction terms. By adding one irrotationality condition for the angle terms, i.e., $\psi _y=\varphi _x$, the system is simplified and the polynomial decay rate is obtained.

For system (1.1), based on the irrotationality condition, we can derive the following system

$$\begin{aligned} \left\{ \begin{array}{llll} \partial _x\left( f_{tt}-\frac{D}{A}\Delta f + \frac{K}{A}(f+w) +\alpha f_t \right) = 0, &{}&{}&{} {\varvec{x}}\in {\mathbb {R}}^2, t\in {\mathbb {R}}_{+},\\ \partial _y\left( f_{tt}-\frac{D}{A}\Delta f + \frac{K}{A}(f+w) +\alpha f_t \right) = 0, &{}&{}&{} {\varvec{x}}\in {\mathbb {R}}^2, t\in {\mathbb {R}}_{+},\\ w_{tt}-K\Delta (f+w) +\beta w_t= 0, &{}&{}&{} {\varvec{x}}\in {\mathbb {R}}^2, t\in {\mathbb {R}}_{+},\\ (f,f_x,f_{xt},f_y,f_{yt},w,w_t)(0,\varvec{x})=(f_0,\varphi _0,\varphi _1,\psi _0,\psi _1,w_0,w_1)(\varvec{x}),&{}&{}&{} {\varvec{x}}\in {\mathbb {R}}^2, \end{array}\right. \qquad \end{aligned}$$

(1.3)

where $A=\frac{\rho h^3}{12},$ f be a function derived from $\varphi ,\psi$ which will be introduced in Sect. 2, ${\varvec{x}}=(x,y)$ and $\alpha ,\beta >0.$

In this paper, for the sake of clearing the influence of the frictional damping terms and semilinearity on the properties of solutions, we construct the following more complex semilinear problem of the system (1.3). More models can be found in [13].

$$\begin{aligned} \left\{ \begin{array}{llll} \partial _x\left( f_{tt}-\frac{D}{A}\Delta f + \frac{K}{A}(f+w) +\alpha f_t \right) = |f_x|^r, &{}&{}&{} {\varvec{x}}\in {\mathbb {R}}^2, t\in {\mathbb {R}}_{+},\\ \partial _y\left( f_{tt}-\frac{D}{A}\Delta f + \frac{K}{A}(f+w) +\alpha f_t \right) = |f_y|^r, &{}&{}&{} {\varvec{x}}\in {\mathbb {R}}^2, t\in {\mathbb {R}}_{+},\\ w_{tt}-K\Delta (f+w) +\beta w_t= 0, &{}&{}&{} {\varvec{x}}\in {\mathbb {R}}^2, t\in {\mathbb {R}}_{+},\\ (f,f_x,f_{xt},f_y,f_{yt},w,w_t)(0,\varvec{x})=(f_0,\varphi _0,\varphi _1,\psi _0,\psi _1,w_0,w_1)(\varvec{x}),&{}&{}&{} {\varvec{x}}\in {\mathbb {R}}^2. \end{array}\right. \qquad \end{aligned}$$

(1.4)

Define

$$\begin{aligned} U:= \left( f_{xt},-\sqrt{\frac{D}{A}}{{\tilde{\Lambda }}} f_x,f_{yt},-\sqrt{\frac{D}{A}}{{\tilde{\Lambda }}} f_y,w_t,\sqrt{K}{{\tilde{\Lambda }}}(f+w)\right) (t,\varvec{x}), \end{aligned}$$

(1.5)

and

$$\begin{aligned} U_0(\varvec{x}):=\left( \varphi _1(\varvec{x}),-\sqrt{\frac{D}{A}}{{\tilde{\Lambda }}} \varphi _0(\varvec{x}),\psi _1(\varvec{x}),-\sqrt{\frac{D}{A}}{{\tilde{\Lambda }}}\psi _0(\varvec{x}),w_1(\varvec{x}), \sqrt{K}{{\tilde{\Lambda }}}(f_0(\varvec{x})+w_0(\varvec{x}))\right) ,\nonumber \\ \end{aligned}$$

(1.6)

where

$$\begin{aligned} {{\tilde{\Lambda }}}=i\Lambda ,\ \ \ \ \ \ \ i=\sqrt{-1}, \end{aligned}$$

and the symbol of $\Lambda$ is $|\varvec{\xi }|$, i.e., ${\mathcal {F}}[\Lambda f]=|\varvec{\xi }|{{\hat{f}}}$.

Then, our main result can be described by the following theorem.

Theorem 1.1

(Global existence) Assume $U_0\in (L^1({\mathbb {R}})\cap L^2({\mathbb {R}}^2)))^6$, $\displaystyle \int _{{\mathbb {R}}^2}U_0d(x,y)=0$, $r>8$, then there exists a constant $\varepsilon >0$, such that if

$$\begin{aligned} I_0+\Vert U_0\Vert _1+\Vert U_0\Vert _2<\varepsilon , \end{aligned}$$

(1.7)

then there exists a unique global classical solution U to (1.4) satisfying

$$\begin{aligned} \Vert U\Vert _2\le C(1+t)^{-1/4}(I_0+\Vert U_0\Vert _1+\Vert U_0\Vert _2), \end{aligned}$$

(1.8)

where $I_0^2=\displaystyle \int _{{\mathbb {R}}^2}e^{\gamma (x^2+y^2)}U_0^2d(x,y)$, the constant $\gamma$ satisfies that

$$\begin{aligned} 0<\gamma <\min \left\{ \frac{\alpha A}{4D},\frac{\beta }{4K}\right\} . \end{aligned}$$

Remark 1.1

Note that the result in [16] focuses on the simplifies model of equation (1.3) and obtained the polynomial decay rate $(1+t)^{-\frac{1}{4}}$ and global existence of solutions. Compared with the results in [16], the result in this paper is concerning a more complex models under full frictional damping, we also get the same decay rate.

This paper is organized as follows. In Sect. 2, the definition of classical solution for the semilinear problem (1.4) is given. In Sect. 3, we give the estimates of linearized system in Fourier space by using the Kawashima energy method. In Sect. 4, the local existence theorem for the semilinear system (1.4) is shown based on the Banach’s fixed point theorem. In Sect. 5, under the smallness assumption, we prove that there exists a unique global in time classical solution to the semilinear system (1.4), where a weighted a priori estimate is given here.

In this paper, we use the following notations. Let ${\hat{g}}(\varvec{\xi })$ be the Fourier transform of $g(\varvec{x})$,

$$\begin{aligned} {\hat{g}}(\varvec{\xi }):={\mathcal {F}}(g)( \varvec{\xi })=\int _{{\mathbb {R}}^2}g(\varvec{x})e^{-2\pi i \varvec{x}\cdot \varvec{\xi }}d\varvec{x}, \end{aligned}$$

(1.9)

where $\varvec{\xi }=(\xi _1,\xi _2)$, $\varvec{x}=(x,y)$, and ${\mathcal {F}}^{-1}({\hat{g}})$ be the inverse Fourier transform of ${{\hat{g}}}$. Let $\Vert \cdot \Vert _{p}$ denote the norm of the Lebesgue space $L^p({\mathbb {R}}^2)$, $1\le p\le \infty$, and $\Vert \cdot \Vert _{H^s}$ the norm of the Sobolev space $H^s({\mathbb {R}}^2)$ for $s\ge 0$. Denote the symbol of the operator $\Lambda$ as $|\varvec{\xi }|$, $\varvec{\xi }\in {\mathbb {R}}^2$, which means ${\mathcal {F}}(\Lambda f)=|\varvec{\xi }|{\hat{f}}$, and we define operator ${{\tilde{\Lambda }}} = i\Lambda$. The operator $\Lambda ^s,s \in {\mathbb {R}}$ be defined as

$$\begin{aligned} \Lambda ^s f = \int _{{\mathbb {R}}^2}|\varvec{\xi }|^s{\hat{f}}e^{2\pi i\varvec{x}\cdot \varvec{\xi }}d\varvec{\xi }. \end{aligned}$$

(1.10)

We give some standard lemmas which are quite useful in this paper.

Lemma 1.1

([17, 27]) Let $a>0$ and $b>0$ be given. If $\max (a,b)>1$, there exists a constant $C>0$ such that for all $t\ge 0,$

$$\begin{aligned} \int _0^t(1+t-\tau )^{- a}(1+\tau )^{-b}d\tau \le C(1+t)^{-\min (a,b)}. \end{aligned}$$

Lemma 1.2

([20]) Let u and f be nonnegative continuous functions defined for $t\ge 0$. If

$$\begin{aligned} u^2(t) \le c^2 + 2\int _0^t f(s)u(s)ds \end{aligned}$$

for $t\ge 0$, where $c\ge 0$ is a constant, then for $t\ge 0$,

$$\begin{aligned} u(t)\le c + \int _0^tf(s)ds. \end{aligned}$$

Lemma 1.3

(Gagliardo–Nirenberg’s inequality [19]) Let $1\le p,q,r\le \infty$ and let j, k be two integers such that $0\le j<m$. If

$$\begin{aligned} \frac{1}{p}=\frac{j}{n}+l\left( \frac{1}{r}-\frac{m}{n}\right) +\frac{1-l}{q} \end{aligned}$$

for some $l\in [\frac{j}{m},1] (l<1$ if $r>1$ and $m-j-\frac{n}{r}=0)$, then there exists constant $C=C(n,m,j,l,q,r)$ such that

$$\begin{aligned} \sum \limits _{|\beta |=j}\Vert \nabla ^\beta u\Vert _{p}\le C\left( \sum \limits _{|\beta |=m}\Vert \nabla ^\beta u\Vert _{r}\right) ^{l}\Vert u\Vert _{q}^{1-l} \end{aligned}$$

for every $u\in C_c^\infty ({\mathbb {R}}^n)$.

Lemma 1.4

(Plancherel Theorem [15]) Let $g\in L^1({\mathbb {R}}^n)\cap L^2({\mathbb {R}}^n)$, then ${{\hat{g}}}\in L^2({\mathbb {R}}^n)$ and satisfies

$$\begin{aligned} \Vert {{\hat{g}}}\Vert _2=\Vert g\Vert _2. \end{aligned}$$

2 Preliminary remarks

In this section, we first give the derivation of the system (1.3) and then, denote the classical solution for the semilinear system (1.4).

2.1 Derivation of system (1.3)

By Green’s theorem, let $\varphi , \psi \in C^1({\mathbb {R}}^2)$ and if $\psi _y=\varphi _x$, there exists a function $f\in C^1({\mathbb {R}}^2)$, satisfying

$$\begin{aligned} df=\psi dx+\varphi dy,\quad \quad \psi =f_x,\quad {\text {and}}\quad \varphi =f_y. \end{aligned}$$

Then, the original problem (1.1) can be rewritten as

$$\begin{aligned} \left\{ \begin{array}{llll} \partial _x\left( f_{tt}-\frac{D}{A}\Delta f + \frac{K}{A}\big (f+w\big ) + \alpha f_t\right) =0, &{}&{}&{}\varvec{x}\in {\mathbb {R}}^2, t\in {\mathbb {R}}_{+},\\ \partial _y\left( f_{tt}-\frac{D}{A}\Delta f + \frac{K}{A}(f+w) + \alpha f_t\right) = 0, &{}&{}&{}\varvec{x}\in {\mathbb {R}}^2, t\in {\mathbb {R}}_{+},\\ w_{tt}-K\Delta (f+w) + \beta w_t = 0, &{}&{}&{}\varvec{x}\in {\mathbb {R}}^2, t\in {\mathbb {R}}_{+} \\ (f,f_x,f_{xt},f_y,f_{yt},w,w_t)(0,\varvec{x})=(f_0,\varphi _0,\varphi _1,\psi _0,\psi _1,w_0,w_1)(\varvec{x}),&{}&{}&{} {\varvec{x}}\in {\mathbb {R}}^2. \end{array} \right. \qquad \end{aligned}$$

(2.1)

2.2 Classical solution of the semilinear system (1.4)

In order to obtain estimates of the solution for the semilinear system (1.4), the change of variables method is applied so that the system (1.4) is reduced to a first-order one.

In view of the definition of U and $U_0$ in (1.5), the system (1.4) becomes

$$\begin{aligned} \left\{ \begin{array}{llll} \partial _t U(t,\varvec{x})+L({\mathcal {D}})U(t,\varvec{x})=F(t,\varvec{x}),&{} \\ U(0,\varvec{x})=U_0(\varvec{x}),&{} \\ F(t,\varvec{x})= (g_1,0,g_2,0,0,0),&{} \end{array} \right. \end{aligned}$$

(2.2)

where $g_1=|f_x|^r$, $g_2=|f_y|^r$, and $L({\mathcal {D}})$ be defined as the operator whose symbol is ${{\hat{L}}}(\varvec{\xi })$

$$\begin{aligned} {\hat{L}}(\varvec{\xi }) = \left( \begin{array}{cccccc} \alpha &{}i|\varvec{\xi }|\sqrt{\frac{D}{A}}&{}0&{}0&{}0&{}\frac{\sqrt{K}}{A}\frac{\xi _1}{|\varvec{\xi }|}\\ i|\varvec{\xi }|\sqrt{\frac{D}{A}}&{}0&{}0&{}0&{}0&{}0\\ 0&{}0&{}\alpha &{}i|\varvec{\xi }|\sqrt{\frac{D}{A}}&{}0&{}\frac{\sqrt{K}}{A}\frac{\xi _2}{|\varvec{\xi }|}\\ 0&{}0&{}i|\varvec{\xi }|\sqrt{\frac{D}{A}}&{}0&{}0&{}0\\ 0&{}0&{}0&{}0&{}\beta &{}-i|\varvec{\xi }|\sqrt{K}\\ -\frac{\xi _1}{|\varvec{\xi }|}\sqrt{K}&{}0&{}-\frac{\xi _2}{|\varvec{\xi }|}\sqrt{K}&{}0&{}-i|\varvec{\xi }|\sqrt{K}&{}0\\ \end{array} \right) . \end{aligned}$$

Then, the linearized system (2.2) is

$$\begin{aligned} \left\{ \begin{array}{llll} \partial _t U_1(t,\varvec{x})+L({\mathcal {D}})U_1(t,\varvec{x})=0,&{}\\ U_1(0,\varvec{x})=U_0(\varvec{x}).&{} \end{array} \right. \end{aligned}$$

(2.3)

It is easy to conclude that the fundamental solution $G(t,\varvec{x})$ of the Cauchy problem for the linear system (2.3) satisfies

$$\begin{aligned} \left\{ \begin{array}{llll} \partial _t G(t,\varvec{x})+L({\mathcal {D}})G(t,\varvec{x})=0,&{} \\ G(0,\varvec{x})=\delta I_6,&{} \end{array} \right. \end{aligned}$$

(2.4)

where $\delta =\delta _x\otimes \delta _y$ is the Dirac data function and $I_6$ is the sixth-order identity matrix.

The Fourier transform equation (2.4) can be written as

$$\begin{aligned} \left\{ \begin{array}{llll} \partial _t {{\hat{G}}}(t,\varvec{\xi })+{{\hat{L}}}(\varvec{\xi }){{\hat{G}}}(t,\varvec{\xi })=0,&{} \\ {{\hat{G}}}(0,\varvec{\xi })=I_6, \end{array} \right. \end{aligned}$$

(2.5)

hence, ${{\hat{G}}}(t,\varvec{\xi })=e^{-t{{\hat{L}}}(\varvec{\xi })}$.

Thus, the solution of the system (2.3) can be written as

$$\begin{aligned} U_1=G(t,\varvec{x})*U_0(\varvec{x}), \end{aligned}$$

and, by Duhamel’s formula, the solution of the system (2.2) can be expressed as

$$\begin{aligned} U=G(t,\varvec{x})*U_0(\varvec{x})+\int _0^t G(t-s,\cdot )*F(s,\cdot )ds. \end{aligned}$$

(2.6)

3 Decay estimates for the linear system

In this section, we derive the decay estimate for the linear system of (1.4) by using energy method in the Fourier space as in [11].

By Fourier transform of the linearized equation (1.3), we have

$$\begin{aligned} \left\{ \begin{array}{llll} &{} i\xi _1\left( {\hat{f}}_{tt}+\frac{D}{A} |\varvec{\xi }|^2{\hat{f}} + \frac{K}{A}({\hat{f}}+{\hat{w}}) + \alpha {\hat{f}}_t\right) = 0,\\ &{} i\xi _2\left( {\hat{f}}_{tt}+\frac{D}{A} |\varvec{\xi }|^2{\hat{f}} + \frac{K}{A}({\hat{f}}+{\hat{w}}) + \alpha {\hat{f}}_t\right) = 0,\\ &{} {\hat{w}}_{tt} + K |\varvec{\xi }|^2({\hat{f}}+{\hat{w}})+\beta {\hat{w}}_t= 0, \end{array} \right. \end{aligned}$$

(3.1)

where $|\varvec{\xi }|^2 = \xi _1^2+\xi _2^2.$

Let

$$\begin{aligned} {\hat{U}} =&\ \left( i\xi _1{\hat{f}}_t,\sqrt{\frac{D}{A}}\xi _1|\varvec{\xi }|{\hat{f}},i\xi _2{\hat{f}}_t,\sqrt{\frac{D}{A}}\xi _2|\varvec{\xi }|{\hat{f}},{\hat{w}}_t,i\sqrt{K}|\varvec{\xi }|({\hat{f}}+{\hat{w}})\right) \\ :=&\ \left( {\hat{u}},{\hat{v}},{\hat{W}},{\hat{X}},{\hat{Y}},{\hat{Z}}\right) . \end{aligned}$$

Then, equation (3.1) can be represented as follows

$$\begin{aligned} \left\{ \begin{array}{llll} &{} {\hat{u}}_t+i|\varvec{\xi }|\sqrt{\frac{D}{A}}{\hat{v}}+\frac{\sqrt{K}}{A}\frac{\xi _1}{|\varvec{\xi }|}{\hat{Z}} + \alpha {\hat{u}}= 0,\\ &{} {\hat{v}}_t+i|\varvec{\xi }|\sqrt{\frac{D}{A}}{\hat{u}}=0,\\ &{} {\hat{W}}_t+i|\varvec{\xi }|\sqrt{\frac{D}{A}}{\hat{X}}+\frac{\sqrt{K}}{A}\frac{\xi _2}{|\varvec{\xi }|}{\hat{Z}} + \alpha {\hat{W}}= 0,\\ &{} {\hat{X}}_t+i|\varvec{\xi }|\sqrt{\frac{D}{A}}{\hat{W}}=0,\\ &{} {\hat{Y}}_{t}-i|\varvec{\xi }|\sqrt{K}{\hat{Z}}+\beta {\hat{Y}}= 0,\\ &{} {\hat{Z}}_t -i|\varvec{\xi }|\sqrt{K}{\hat{Y}}-\frac{\xi _1}{|\varvec{\xi }|}\sqrt{K}{\hat{u}}-\frac{\xi _2}{|\varvec{\xi }|}\sqrt{K}{\hat{W}}=0. \end{array} \right. \end{aligned}$$

(3.2)

Using the definition of ${{\hat{L}}}(\varvec{\xi })$ in Sect. 2, system (3.2) is written as

$$\begin{aligned} \partial _t{\hat{U}}(t,\varvec{\xi })+{\hat{L}}(\varvec{\xi }){\hat{U}}(t,\varvec{\xi })=0. \end{aligned}$$

(3.3)

We prove the following lemma using the energy estimate in the Fourier space method as in Ide, Haramoto and Kawashima [11].

Lemma 3.1

For all $t>0$ and $\varvec{\xi }\in {\mathbb {R}}^2$, the matrix $e^{-t{\hat{L}}(\varvec{\xi })}$ satisfies

$$\begin{aligned} |e^{-t{\hat{L}}(\varvec{\xi })}|\le Ce^{-c\rho (\varvec{\xi })t}, \end{aligned}$$

where $\rho ({\varvec{\xi }})=\frac{|\varvec{\xi }|^2}{(1+|\varvec{\xi }|^2)^2}$, and C, c are positive constants.

Proof

By Kawashima’s multiplier method in the Fourier space, we first multiply the equations in (3.2) by $\bar{{\hat{u}}},\bar{{\hat{v}}},\bar{{\hat{W}}},\bar{{\hat{X}}},\frac{\bar{{\hat{Y}}}}{A},\frac{\bar{{\hat{Z}}}}{A}$, respectively, sum up the results and take the real part of both sides to get

$$\begin{aligned} \frac{1}{2}\partial _t\left( |{\hat{u}}|^2+|{\hat{v}}|^2+|{\hat{W}}|^2+|{\hat{X}}|^2+\frac{1}{A}|{\hat{Y}}|^2 +\frac{1}{A}|{\hat{Z}}|^2\right) +\alpha |{\hat{u}}|^2+\alpha |{\hat{W}}|^2+\frac{\beta }{A}|{\hat{Y}}|^2=0. \end{aligned}$$

Second, we multiply the first equation and second equation of (3.2) by $-i|\varvec{\xi }|\bar{{\hat{v}}},i|\varvec{\xi }|\bar{{\hat{u}}}$, multiply the second and fifth equation in (3.2) by $\frac{\xi _1}{A|\varvec{\xi }|}\bar{{\hat{Y}}},\frac{\xi _1}{A|\varvec{\xi }|}\bar{{\hat{v}}}$, sum up the results, the real parts satisfy

$$\begin{aligned}&\text {Re}\left\{ -i|\varvec{\xi }|{\hat{u}}_t\bar{{\hat{v}}}\right\} +|\varvec{\xi }|^2\sqrt{\frac{D}{A}}|{\hat{v}}|^2 +\text {Re}\left\{ -i\alpha |\varvec{\xi }|{\hat{u}}\bar{{\hat{v}}}\right\} +\text {Re}\left\{ i|\varvec{\xi }|{\hat{v}}_t\bar{{\hat{u}}}\right\} -|\varvec{\xi }|^2\sqrt{\frac{D}{A}}|{\hat{u}}|^2\\&+\text {Re}\left\{ \frac{\xi _1}{A|\varvec{\xi }|}{\hat{Y}}_t\bar{{\hat{v}}}\right\} +\text {Re}\left\{ \frac{\beta \xi _1}{A|\varvec{\xi }|}{\hat{Y}}\bar{{\hat{v}}}\right\} +\text {Re}\left\{ \frac{\xi _1}{A|\varvec{\xi }|}{\hat{v}}_t\bar{{\hat{Y}}}\right\} +\text {Re}\left\{ i\frac{\sqrt{D}\xi _1}{A\sqrt{A}}{\hat{u}}\bar{{\hat{Y}}}\right\} = 0. \end{aligned}$$

This indicates that

$$\begin{aligned} \partial _t\bigg \{\text {Re}\{-i|\varvec{\xi }|{\hat{u}}\bar{{\hat{v}}}\}&+\text {Re}\Big \{\frac{\xi _1}{A|\varvec{\xi }|}{\hat{Y}}\bar{{\hat{v}}}\Big \}\bigg \} +|\varvec{\xi }|^2\sqrt{\frac{D}{A}}\Big (|{\hat{v}}|^2-|{\hat{u}}|^2\Big ) +\text {Re}\{-i\alpha |\varvec{\xi }|{\hat{u}}\bar{{\hat{v}}}\}\\&+\text {Re}\Big \{\frac{\beta \xi _1}{A|\varvec{\xi }|}{\hat{Y}}\bar{{\hat{v}}}\Big \} +\text {Re}\Big \{i\frac{\sqrt{D}\xi _1}{A\sqrt{A}}{\hat{u}}\bar{{\hat{Y}}}\Big \} = 0. \end{aligned}$$

By Young’s inequality,

$$\begin{aligned}&\partial _t\bigg \{\text {Re}\{-i|\varvec{\xi }|{\hat{u}}\bar{{\hat{v}}}\} +\text {Re}\Big \{\frac{\xi _1}{A|\varvec{\xi }|}{\hat{Y}}\bar{{\hat{v}}}\Big \}\bigg \} +|\varvec{\xi }|^2\sqrt{\frac{D}{A}}\Big (|{\hat{v}}|^2-|{\hat{u}}|^2\Big )\\&\quad \le 2\varepsilon |\varvec{\xi }|^2|{\hat{v}}|^2+C_{\varepsilon }\alpha ^2|{\hat{u}}|^2 +C_{\varepsilon }\frac{\beta ^2}{|\varvec{\xi }|^2}|{\hat{Y}}|^2 +\varepsilon |\varvec{\xi }|^2|{\hat{u}}|^2+C_{\varepsilon }|{\hat{Y}}|^2, \end{aligned}$$

where $0<\varepsilon<1<C_{\varepsilon }$ are constants. This implies that

$$\begin{aligned}&\partial _t\bigg \{\text {Re}\{-i|\varvec{\xi }|{\hat{u}}\bar{{\hat{v}}}\} +\text {Re}\Big \{\frac{\xi _1}{A|\varvec{\xi }|}{\hat{Y}}\bar{{\hat{v}}}\Big \}\bigg \} +\Big (\sqrt{\frac{D}{A}}-2\varepsilon \Big )|\varvec{\xi }|^2|{\hat{v}}|^2\nonumber \\&\quad \le C_{\varepsilon }\Big (1+\alpha ^2+\beta ^2+|\varvec{\xi }|^2+\frac{1}{|\varvec{\xi }|^2}\Big )\Big (|{\hat{u}}|^2+|{\hat{Y}}|^2\Big ). \end{aligned}$$

(3.4)

Third, similarly, we multiply the third equation and fourth equation of (3.2) by $-i|\varvec{\xi }|\bar{{\hat{X}}},i|\varvec{\xi }|\bar{{\hat{W}}}$, multiply the fourth and fifth equation of (3.2) by $\frac{\xi _2}{A|\varvec{\xi }|}\bar{{\hat{Y}}},\frac{\xi _2}{A|\varvec{\xi }|}\bar{{\hat{X}}}$, sum up the results. Then, the real part satisfies

$$\begin{aligned}&\text {Re}\{-i|\varvec{\xi }|{\hat{W}}_t\bar{{\hat{X}}}\}+|\varvec{\xi }|^2\sqrt{\frac{D}{A}}|{\hat{X}}|^2 +\text {Re}\{-i\alpha |\varvec{\xi }|{\hat{W}}\bar{{\hat{X}}}\}+\text {Re}\{i|\varvec{\xi }|{\hat{X}}_t\bar{{\hat{W}}}\} -|\varvec{\xi }|^2\sqrt{\frac{D}{A}}|{\hat{W}}|^2\\&\quad +\text {Re}\Big \{\frac{\xi _2}{A|\varvec{\xi }|}{\hat{Y}}_t\bar{{\hat{X}}}\Big \} +\text {Re}\Big \{\frac{\beta \xi _2}{A|\varvec{\xi }|}{\hat{Y}}\bar{{\hat{X}}}\Big \} +\text {Re}\Big \{\frac{\xi _2}{A|\varvec{\xi }|}{\hat{X}}_t\bar{{\hat{Y}}}\Big \} +\text {Re}\Big \{i\frac{\sqrt{D}\xi _2}{A\sqrt{A}}{\hat{W}}\bar{{\hat{Y}}}\Big \} = 0, \end{aligned}$$

which implies

$$\begin{aligned}&\partial _t\bigg \{\text {Re}\{-i|\varvec{\xi }|{\hat{W}}\bar{{\hat{X}}}\} +\text {Re}\Big \{\frac{\xi _2}{A|\varvec{\xi }|}{\hat{Y}}\bar{{\hat{X}}}\Big \}\bigg \} +|\varvec{\xi }|^2\sqrt{\frac{D}{A}}\Big (|{\hat{X}}|^2-|{\hat{W}}|^2\Big ) +\text {Re}\{-i\alpha |\varvec{\xi }|{\hat{W}}\bar{{\hat{X}}}\}\\&\quad +\text {Re}\Big \{\frac{\beta \xi _2}{A|\varvec{\xi }|}{\hat{Y}}\bar{{\hat{X}}}\Big \} +\text {Re}\Big \{i\frac{\sqrt{D}\xi _2}{A\sqrt{A}}{\hat{W}}\bar{{\hat{Y}}}\Big \} = 0. \end{aligned}$$

By Young’s inequality, we get

$$\begin{aligned}&\partial _t\bigg \{\text {Re}\{-i|\varvec{\xi }|{\hat{W}}\bar{{\hat{X}}}\} +\text {Re}\Big \{\frac{\xi _2}{A|\varvec{\xi }|}{\hat{Y}}\bar{{\hat{X}}}\Big \}\bigg \} +|\varvec{\xi }|^2\sqrt{\frac{D}{A}}\Big (|{\hat{X}}|^2-|{\hat{W}}|^2\Big )\\&\quad \le 2\varepsilon |\varvec{\xi }|^2|{\hat{X}}|^2+C_{\varepsilon }\alpha ^2|{\hat{W}}|^2 +C_{\varepsilon }\frac{\beta ^2}{|\varvec{\xi }|^2}|{\hat{Y}}|^2+\varepsilon |\varvec{\xi }|^2|{\hat{W}}|^2 +C_{\varepsilon }|{\hat{Y}}|^2, \end{aligned}$$

where $0<\varepsilon<1<C_{\varepsilon }$ are constants. This leads to

$$\begin{aligned}&\partial _t\bigg \{\text {Re}\{-i|\varvec{\xi }|{\hat{W}}\bar{{\hat{X}}}\} +\text {Re}\Big \{\frac{\xi _2}{A|\varvec{\xi }|}{\hat{Y}}\bar{{\hat{X}}}\Big \}\bigg \} +\Big (\sqrt{\frac{D}{A}}-2\varepsilon \Big )|\varvec{\xi }|^2|{\hat{X}}|^2\nonumber \\&\quad \le C_{\varepsilon }\Big (1+\alpha ^2+\beta ^2+|\varvec{\xi }|^2+\frac{1}{|\varvec{\xi }|^2}\Big )\Big (|{\hat{W}}|^2+|{\hat{Y}}|^2\Big ). \end{aligned}$$

(3.5)

Last but not least, we multiply the fifth equation and sixth equation of (3.2) by $i|\varvec{\xi }|\bar{{\hat{Z}}},-i{|\varvec{\xi }|}\bar{{\hat{Y}}},$ sum up the results. Then, the real part satisfies

$$\begin{aligned}&\text {Re}\{i|\varvec{\xi }|{\hat{Y}}_t\bar{{\hat{Z}}}\}+|\varvec{\xi }|^2\sqrt{K}|{\hat{Z}}|^2 +\text {Re}\{i|\varvec{\xi }|\beta {\hat{Y}}\bar{{\hat{Z}}}\}\\&+\text {Re}\{-i{|\varvec{\xi }|}{\hat{Z}}_t\bar{{\hat{Y}}}\}-|\varvec{\xi }|^2\sqrt{K}|{\hat{Y}}|^2 +\text {Re}\{i\xi _1\sqrt{K}{\hat{u}}\bar{{\hat{Y}}}\}\\&+\text {Re}\{i\xi _2\sqrt{K}{\hat{W}}\bar{{\hat{Y}}}\}=0, \end{aligned}$$

which leads to

$$\begin{aligned}&\partial _t\text {Re}\{i|\varvec{\xi }|{\hat{Y}}\bar{{\hat{Z}}}\} +\sqrt{K}|\varvec{\xi }|^2\Big (|{\hat{Z}}|^2-|{\hat{Y}}|^2\Big )+\text {Re}\{i|\varvec{\xi }|\beta {\hat{Y}}\bar{{\hat{Z}}}\}\\&\quad +\text {Re}\{i\xi _1\sqrt{K}{\hat{u}}\bar{{\hat{Y}}}\}+\text {Re}\{i\xi _2\sqrt{K}{\hat{W}}\bar{{\hat{Y}}}\}=0. \end{aligned}$$

By Young’s inequality,

$$\begin{aligned} \partial _t\text {Re}\{i|\varvec{\xi }|{\hat{Y}}\bar{{\hat{Z}}}\}&+\sqrt{K}|\varvec{\xi }|^2(|{\hat{Z}}|^2-|{\hat{Y}}|^2)\\&\le \varepsilon |\varvec{\xi }|^2|{\hat{Z}}|^2+C_{\varepsilon }\beta ^2|{\hat{Y}}|^2 +\varepsilon |\xi _1|^2|{\hat{u}}|^2+\varepsilon |\xi _2|^2|{\hat{W}}|^2+2C_{\varepsilon }|{\hat{Y}}|^2. \end{aligned}$$

Hence,

$$\begin{aligned}&\partial _t\text {Re}\{i|\varvec{\xi }|{\hat{Y}}\bar{{\hat{Z}}}\}+(\sqrt{K}-\varepsilon )|\varvec{\xi }|^2|{\hat{Z}}|^2\nonumber \\&\quad \le C_{\varepsilon }(1+\alpha ^2+\beta ^2+|\varvec{\xi }|^2)(|{\hat{u}}|^2+|{\hat{W}}|^2+|{\hat{Y}}|^2). \end{aligned}$$

(3.6)

Let

$$\begin{aligned} E_1(t)&:=\frac{1}{2}\Big (|{\hat{u}}|^2+|{\hat{v}}|^2+|{\hat{W}}|^2+|{\hat{X}}|^2+\frac{1}{A}|{\hat{Y}}|^2 +\frac{1}{A}|{\hat{Z}}|^2\Big ), \end{aligned}$$

(3.7)

$$\begin{aligned} E_2(t)&:=\text {Re}\{-i|\varvec{\xi }|{\hat{u}}\bar{{\hat{v}}}\} +\text {Re}\Big \{\frac{\xi _1}{A|\varvec{\xi }|}{\hat{Y}}\bar{{\hat{v}}}\Big \}, \end{aligned}$$

(3.8)

$$\begin{aligned} E_3(t)&:=\text {Re}\{-i|\varvec{\xi }|{\hat{W}}\bar{{\hat{X}}}\} +\text {Re}\Big \{\frac{\xi _2}{A|\varvec{\xi }|}{\hat{Y}}\bar{{\hat{X}}}\Big \}, \end{aligned}$$

(3.9)

$$\begin{aligned} E_4(t)&:=\text {Re}\{i|\varvec{\xi }|{\hat{Y}}\bar{{\hat{Z}}}\}. \end{aligned}$$

(3.10)

Then, the Lyapunov function

$$\begin{aligned} E_T(t): = CE_1(t)+\tau _1\frac{|\varvec{\xi }|^2}{(1+|\varvec{\xi }|^2)^2}E_2(t)+\tau _2\frac{|\varvec{\xi }|^2}{(1+|\varvec{\xi }|^2)^2}E_3(t) +\tau _3\frac{E_4(t)}{1+|\varvec{\xi }|^2}, \end{aligned}$$

(3.11)

satisfies

$$\begin{aligned}&\partial _tE_T(t)+\Big (C\alpha -(\tau _1+\tau _3)C_{\varepsilon }\Big )|{\hat{u}}|^2 +\tau _1\Big (\sqrt{\frac{D}{A}}-2\varepsilon \Big )\frac{|\varvec{\xi }|^4}{(1+|\varvec{\xi }|^2)^2}|{\hat{v}}|^2\nonumber \\&\quad +\Big (C\alpha -(\tau _2+\tau _3)C_{\varepsilon }\Big )|{\hat{W}}|^2 +\tau _2\Big (\sqrt{\frac{D}{A}}-2\varepsilon \Big )\frac{|\varvec{\xi }|^4}{(1+|\varvec{\xi }|^2)^2}|{\hat{X}}|^2\nonumber \\&\quad +\Big (\frac{C\beta }{A}-(\tau _1+\tau _2+\tau _3)C_{\varepsilon }\Big )|{\hat{Y}}|^2 +\tau _3\left( \sqrt{K}-\varepsilon \right) \frac{|\varvec{\xi }|^2}{1+|\varvec{\xi }|^2}|{\hat{Z}}|^2\le 0. \end{aligned}$$

(3.12)

First, we choose $\varepsilon$ small enough such that $\varepsilon <\min \{\frac{\sqrt{D}}{2\sqrt{A}},\sqrt{K}\},$ and fix small positive $\tau _1,\tau _2$ and $\tau _3$, then we take large positive C such that

$$\begin{aligned} C>\max \left\{ \frac{(\tau _1+\tau _3)C_\varepsilon }{\alpha },\frac{(\tau _2+\tau _3)C_\varepsilon }{\alpha }, \frac{(\tau _1+\tau _2+\tau _3)C_\varepsilon A}{\beta }\right\} . \end{aligned}$$

(3.13)

Therefore, the inequality (3.12) can be rewritten as

$$\begin{aligned} \partial _tE_T{t}+cF(t)\le 0, \end{aligned}$$

(3.14)

where c is positive constant, and

$$\begin{aligned} F(t):=&|{\hat{u}}|^2+\frac{|\varvec{\xi }|^4}{(1+|\varvec{\xi }|^2)^2}|{\hat{v}}|^2+|{\hat{W}}|^2+\frac{|\varvec{\xi }|^4}{(1+|\varvec{\xi }|^2)^2}|{\hat{X}}|^2 +|{\hat{Y}}|^2+\frac{|\varvec{\xi }|^2}{1+|\varvec{\xi }|^2}|{\hat{Z}}|^2. \end{aligned}$$

(3.15)

Since there exist positive constants $C_i,i=1,2,3,4,$ such that

$$\begin{aligned} C_1|{\hat{U}}(\varvec{\xi },t)|^2\le E_T(t) \le C_2|{\hat{U}}(\varvec{\xi },t)|^2, \quad F(t)\ge C_3\rho (\varvec{\xi })|{\hat{U}}(\varvec{\xi },t)|^2\ge C_4\rho (\varvec{\xi })E_T(t),\nonumber \\ \end{aligned}$$

(3.16)

where $\rho (\varvec{\xi }) = \frac{|\varvec{\xi }|^4}{(1+|\varvec{\xi }|^2)^2}$. Therefore,

$$\begin{aligned} \partial _tE_T(t)+c\rho (\varvec{\xi })E_T(t) \le 0. \end{aligned}$$

(3.17)

Hence,

$$\begin{aligned} E_T(t) \le e^{-c\rho (\varvec{\xi })t}E_T(0). \end{aligned}$$

(3.18)

By inequality (3.16), we have

$$\begin{aligned} |{\hat{U}}(t,\varvec{\xi })| \le Ce^{-c\rho (\varvec{\xi })t}|{\hat{U}}(0,\varvec{\xi })|. \end{aligned}$$

(3.19)

The choice of ${{\hat{U}}}(0,\varvec{\xi })$ can be arbitrary, hence,

$$\begin{aligned} e^{-t{{\hat{L}}}(\varvec{\xi })}\le Ce^{-c\rho (\varvec{\xi })t}. \end{aligned}$$

So, we complete the proof of Lemma 3.1. $\square$

Remark 3.1

Notice that $\frac{\xi _1}{A|\varvec{\xi }|}\bar{\hat{Y}}$ is singularity when $\varvec{\xi }=0$, the energy method in the Fourier space does not immediately apply in this case, we can apply the fundamental solution method to get the behavior of $e^{-t{{\hat{L}}}(\varvec{\xi })}$ as the result in Lemma 3.1, for more details can be referred as [16, 28, 30]. In addition, the Fourier splitting method has been employed to investigate the time decay rate of the incompressible third grade fluid equations [31], the Navier–Stokes–Voigt equations [32], the MHD equations and generalized MHD equations [12, 33].

Lemma 3.2

Let $e^{-tL}$ be the semigroup associated with the system (2.2) which is defined as

$$\begin{aligned} (e^{-tL}\varphi )(\varvec{x})={\mathcal {F}}^{-1}[e^{-t{\hat{L}}}{\hat{\varphi }}(\cdot )](\varvec{x}). \end{aligned}$$

Then, we have the following decay estimate:

$$\begin{aligned} \Vert e^{-tL}\varphi \Vert _2\le C(1+t)^{-\frac{1}{4}}\Vert \varphi \Vert _1+Ce^{-ct}\Vert \varphi \Vert _2, \end{aligned}$$

(3.20)

where C and c are positive constants.

Proof

Based on Lemma 3.1 and Plancherel Theorem (Lemma 1.4), we have

$$\begin{aligned} \Vert e^{-tL}\varphi \Vert _2^2 =&\ \ (2\pi )^{-2}\int _{{\mathbb {R}}^2}\left| e^{-t{\hat{L}}}{{\hat{\varphi }}}(\varvec{\xi })\right| ^2d\varvec{\xi }\nonumber \\ \le&\ \ C\int _{{\mathbb {R}}^2}\left| e^{-c\rho (\varvec{\xi })t}{{\hat{\varphi }}}(\varvec{\xi })\right| ^2d\varvec{\xi }\nonumber \\ \le&\ \ C\int _{|\varvec{\xi }|\le 1}\left| e^{-c\rho (\varvec{\xi })t}{{\hat{\varphi }}}(\varvec{\xi })\right| ^2d\varvec{\xi }+C\int _{|\varvec{\xi }|\ge 1}\left| e^{-c\rho (\varvec{\xi })t}{{\hat{\varphi }}}(\varvec{\xi })\right| ^2d\varvec{\xi }\nonumber \\ :=&\ \ I_L+I_H, \end{aligned}$$

(3.21)

where $I_L$ and $I_H$ denote the low and high frequency parts, respectively.

For the low-frequency part, we have $|\varvec{\xi }|\le 1$, then $\rho (\varvec{\xi })\ge c|\varvec{\xi }|^4$. According to the Hausdorff–Young inequality, we obtain

$$\begin{aligned} I_L\le C\Vert {\hat{\varphi }}\Vert _\infty ^2\int _{|\varvec{\xi }|\le 1}e^{-c|\varvec{\xi }|^4t}d\varvec{\xi }\le C(1+t)^{-\frac{1}{2}}\Vert \varphi \Vert _1^2. \end{aligned}$$

(3.22)

For the high-frequency part, we have $|\varvec{\xi }|\ge 1$, then $\rho (\varvec{\xi })\ge c$. Moreover, we get

$$\begin{aligned} I_H\le Ce^{-ct}\int _{|\varvec{\xi }|\ge 1}|{{\hat{\varphi }}}(\varvec{\xi })|^2d\varvec{\xi }\le Ce^{-ct}\Vert \varphi \Vert _2^2. \end{aligned}$$

(3.23)

Combining the inequalities (3.22) and (3.23), we complete the proof of Lemma 3.2. $\square$

4 Local existence for the semilinear system

In this section, Banach’s fixed point theorem [28] is applied to prove the local existence theorem.

Based on the definition of weak solution, the operator $L({\mathcal {D}})$ in system (2.2) with domain $D(L):=(H^1({\mathbb {R}}^2))^6\subset (L^2({\mathbb {R}}^2))^6\rightarrow (L^2({\mathbb {R}}^2))^6$ is, as mentioned in the linear part, the generator of a contraction semigroup $e^{-tL}$, and for $U_0\in D(L)$ and $g_1\in C^1([0,\infty ),L^2({\mathbb {R}}^2))$, $g_2\in C^1([0,\infty ),L^2({\mathbb {R}}^2))$, we have a classical solution

$$\begin{aligned} U\in C^1([0,\infty ),L^2({\mathbb {R}}^2))\cap C^0([0,\infty ),H^1({\mathbb {R}}^2)) \end{aligned}$$

satisfying

$$\begin{aligned} U=e^{-tL}U_0+\int _0^t e^{-(t-s)}F(s)ds. \end{aligned}$$

(4.1)

A weak solution is given by an approximation process. Letting $(\delta _n^1)$ and $(\delta _n^2)$ be fixed two Dirac sequences of mollifiers with respect to x and t, respectively, we define $U_0\in L^2({\mathbb {R}}^2)$ and $F\in C^0([0,\infty ),L^2({\mathbb {R}}^2))$, approximations $U_{0,n}:=\delta _n^1*U_0$ and $F_n:=\delta _n^2*F$ satisfying

$$\begin{aligned}&U_{0,n} \rightarrow U_0\ \ \textrm{in}\ \ (L^2({\mathbb {R}}^2))^6, \end{aligned}$$

(4.2)

$$\begin{aligned}&F_n \rightarrow F\ \ \textrm{in}\ \ (C^0([0,\infty ),L^2({\mathbb {R}}^2)))^6. \end{aligned}$$

(4.3)

Then, we get a sequence of classical solutions $U_n$ from (4.1) with $U_{0,n}$ and $F_n$. Due to (4.2) and (4.3), $U_n$ converges to some U in $C^0([0,\infty ),L^2({\mathbb {R}}^2))$ and U satisfies (4.1). This U is called a weak solution.

Define

$$\begin{aligned} \phi (t,x,y) = \frac{\gamma (x^2+y^2)}{1+t}, \end{aligned}$$

(4.4)

where $\gamma$ is a small constant to be defined later, for simplicity denote $\phi = \phi (t,x,y)$.

Apparently

$$\begin{aligned} \left\{ \begin{array}{llll} &{} \phi _t(t,x,y) = \frac{-\gamma (x^2+y^2)}{(1+t)^2}<0,\\ &{} \phi _x(t,x,y) = \frac{2\gamma x}{1+t},\\ &{} \phi _y(t,x,y) = \frac{2\gamma y}{1+t}, \end{array} \right. \end{aligned}$$

(4.5)

hence,

$$\begin{aligned} \frac{\phi _x^2(t,x,y)+\phi _y^2(t,x,y)}{\phi _t(t,x,y)} = -4\gamma , \end{aligned}$$

(4.6)

and

$$\begin{aligned} -\phi _t(t,x,y) = \frac{\phi (t,x,y)}{1+t}. \end{aligned}$$

(4.7)

Notation. For any $t>0$, $f\in H^{1}_{\kappa \phi (t,\cdot ,\cdot )}({\mathbb {R}}^2)$ means that

$$\begin{aligned} \Vert e^{(\kappa +\varepsilon )\phi (t,\cdot ,\cdot )}f\Vert _2^2+\Vert e^{(\kappa +\varepsilon )\phi (t,\cdot ,\cdot )}f_x\Vert _2^2 +\Vert e^{(\kappa +\varepsilon )\phi (t,\cdot ,\cdot )}f_y\Vert _2^2<\infty , \end{aligned}$$

where $\varepsilon >0$ is small constant.

By Gagliardo–Nirenberg’s inequality (Lemma 1.3), we can obtain the following lemma.

Lemma 4.1

Let $\theta (m) = 1-\frac{2}{m}$, where $m>0$, $0\le \theta (m)\le 1, 0\le \nu \le 1,$ and if $w\in H_{\nu \phi }^{1}({\mathbb {R}}^2)$, then

$$\begin{aligned} \Vert e^{\nu \phi }w\Vert _m\le C_{\nu ,\gamma }(1+t)^{(1-\theta (m))/2}\left( \Vert w_x\Vert _2^{1-\nu }\Vert e^{\nu \phi }w_x\Vert _2^{\nu }+\Vert w_y\Vert _2^{1-\nu }\Vert e^{\nu \phi }w_y\Vert _2^{\nu }\right) . \end{aligned}$$

Before giving the proof of the above Lemma, we first discuss the following estimates.

Lemma 4.2

Let $\nu >0,w\in H_{\nu \phi }^{1}({\mathbb {R}}^2)$. Then,

$$\begin{aligned} 2\nu \gamma (1+t)^{-1}\Vert e^{\nu \phi }w\Vert _2^2+\Vert \partial _x(e^{\nu \phi }w)\Vert _2^2\le \Vert e^{\nu \phi }w_x\Vert _2^2,\\ 2\nu \gamma (1+t)^{-1}\Vert e^{\nu \phi }w\Vert _2^2+\Vert \partial _y(e^{\nu \phi }w)\Vert _2^2\le \Vert e^{\nu \phi }w_y\Vert _2^2. \end{aligned}$$

Proof

From the definition of $\phi$. Denoting $g=e^{\nu \phi }w$, we have

$$\begin{aligned} e^{\nu \phi }w_x=g_x-\nu \phi _x g. \end{aligned}$$

Thus,

$$\begin{aligned} \Vert e^{\nu \phi }w_x\Vert _2^2&= \int _{{\mathbb {R}}^2}|g_x-\nu \phi _x g|^2d(x,y)\\&= \int _{{\mathbb {R}}^2}g_x^2d(x,y)+\nu ^2\int _{{\mathbb {R}}^2}g^2\phi _x^2d(x,y)-2\nu \int _{{\mathbb {R}}^2}gg_x\phi _xd(x,y)\\&= \int _{{\mathbb {R}}^2}g_x^2d(x,y)+\nu ^2\int _{{\mathbb {R}}^2}g^2\phi _x^2d(x,y)+\nu \int _{{\mathbb {R}}^2}g^2\phi _{xx}d(x,y)\\&\ge \int _{{\mathbb {R}}^2}g_x^2d(x,y)+\nu \int _{{\mathbb {R}}^2}g^2\frac{2\gamma }{1+t}d(x,y)\\&= \Vert \partial _xg\Vert _2^2+2\nu \gamma (1+t)^{-1}\Vert g\Vert _2^2. \end{aligned}$$

Similarly, since

$$\begin{aligned} e^{\nu \phi }w_y=g_y-\nu \phi _yg, \end{aligned}$$

we can derive that

$$\begin{aligned} \Vert e^{\nu \phi }w_x\Vert _2^2\ge \Vert \partial _yg\Vert _2^2+2\nu \gamma (1+t)^{-1}\Vert g\Vert _2^2. \end{aligned}$$

This completes the proof of Lemma 4.2. $\square$

Proof of Lemma 4.1

For $w\in H^1_{\nu \phi (t,\cdot ,\cdot )}({\mathbb {R}}^2)$, $\nu \in (0,1]$, we have

$$\begin{aligned} \Vert e^{\nu \phi }w_x\Vert _2^2&= \int _{{\mathbb {R}}^2}|w_x|^{2(1-\nu )}e^{2\nu \phi }|w_x|^{2\nu }dxdy\\&\le \Vert w_x^{2(1-\nu )}\Vert _{\frac{1}{1-\nu }}\Vert e^{2\nu \phi }w_x^{2\nu }\Vert _{\frac{1}{\nu }}\\&= \Vert w_x\Vert _2^{2(1-\nu )}\Vert e^{\phi }w_x\Vert _2^{2\nu }. \end{aligned}$$

Similarly,

$$\begin{aligned} \Vert e^{\nu \phi }w_y\Vert _2^2\le \Vert w_y\Vert _2^{2(1-\nu )}\Vert e^{\phi }w_y\Vert _2^{2\nu }. \end{aligned}$$

Let $g=e^{\nu \phi }w$, according to the Gagliardo–Nirenberg’s inequality, we have

$$\begin{aligned} \Vert g\Vert _m\le C\Vert g\Vert _2^{1-\theta (m)}\Vert \nabla g\Vert _2^{\theta (m)},\ \ \ \theta (m)=1-\frac{2}{m}. \end{aligned}$$

Thus, combining with Lemma 4.2 can yield that

$$\begin{aligned} \Vert \nabla g\Vert _2\le \Vert e^{\nu \phi }w_x\Vert _2+\Vert e^{\nu \phi }w_y\Vert _2, \end{aligned}$$

and

$$\begin{aligned} 2\Vert g\Vert _2\le \frac{1}{\sqrt{2\nu \gamma }}(1+t)^{\frac{1}{2}}(\Vert e^{\nu \phi }w_x\Vert _2+\Vert e^{\nu \phi }w_y\Vert _2). \end{aligned}$$

Then, we have

$$\begin{aligned} \Vert g\Vert _m \le&C_{\nu ,\gamma }(1+t)^{(1-\theta (m))/2}(\Vert e^{\nu \phi }w_x\Vert _2+\Vert e^{\nu \phi }w_y\Vert _2)\\ \le&C_{\nu ,\gamma }(1+t)^{(1-\theta (m))/2}(\Vert w_x\Vert _2^{1-\nu }\Vert e^{\nu \phi }w_x\Vert _2^\nu +\Vert w_y\Vert _2^{1-\nu }\Vert e^{\nu \phi }w_y\Vert _2^\nu ). \end{aligned}$$

The proof of Lemma 4.1 is finished. $\square$

Now, let us give the local existence theorem, which can be described as follows.

Theorem 4.1

(Local existence) Given (f, w) satisfies $U_0\in {\mathcal {H}}=(H^1({\mathbb {R}}^2))^6,f_x(0,\cdot ,\cdot )\in L^2({\mathbb {R}}^2), f_y(0,\cdot ,\cdot )\in L^2({\mathbb {R}}^2)$, where $U_0$ is the initial data as (1.6), and

$$\begin{aligned} E_0:=\Vert e^{\phi (0,\cdot ,\cdot )}U_0\Vert _2+\Vert e^{\phi (0,\cdot ,\cdot )}f_x(0,\cdot ,\cdot )\Vert _2+\Vert e^{\phi (0,\cdot ,\cdot )}f_y(0,\cdot ,\cdot )\Vert _2<\infty . \end{aligned}$$

(4.8)

Then, there exists a time $T_m: = T_m(E_0)>0$, such that the system (2.2) exists a unique solution (f, w) satisfying

$$\begin{aligned} \sup _{t\in [0,T]}\{\Vert e^{\phi (t,\cdot ,\cdot )}U\Vert _2+\Vert e^{\phi (t,\cdot ,\cdot )}f_x\Vert _2 +\Vert e^{\phi (t,\cdot ,\cdot )}f_y\Vert _2\}<\infty , \end{aligned}$$

(4.9)

where $0\le T<T_m.$ If $T_m<\infty ,$ then

$$\begin{aligned} \limsup _{t\rightarrow T_m}\{\Vert e^{\phi (t,\cdot ,\cdot )}U\Vert _2+\Vert e^{\phi (t,\cdot ,\cdot )}f_x\Vert _2+\Vert e^{\phi (t,\cdot ,\cdot )}f_y\Vert _2\}=\infty . \end{aligned}$$

(4.10)

Proof

Define

$$\begin{aligned} B_{\phi ,T}^K: =&\Big \{V=(f,w):(f_{xt},-\sqrt{\frac{D}{A}}{{\tilde{\Lambda }}} f_x, f_{yt}, -\sqrt{\frac{D}{A}}{{\tilde{\Lambda }}} f_y, w_t,\sqrt{K}{{\tilde{\Lambda }}}(f+w))\\&\in (C([0,T],L^2({\mathbb {R}}^2)))^6, \text {and}\ \Vert V\Vert _{\phi ,T}\le K\Big \}, \end{aligned}$$

where

$$\begin{aligned} \Vert V\Vert _{\phi ,T} :=&\Vert (f,w)\Vert _{\phi ,T}\\ :=&\sup _{t\in [0,T]}\Big \{\Vert e^\phi f_{xt}\Vert _2+\Vert e^\phi f_{yt}\Vert _2+\Vert e^\phi w_t\Vert _2+\Vert e^\phi f_{xx}\Vert _2+\Vert e^\phi f_{yy}\Vert _2+\Vert e^\phi f_{xy}\Vert _2\\&+\Vert e^\phi f_x\Vert _2+\Vert e^\phi f_y\Vert _2+\Vert e^\phi (f_x+w_x)\Vert _2+\Vert e^\phi (f_y+w_y)\Vert _2\Big \}. \end{aligned}$$

Letting

$$\begin{aligned} \Omega :=\Big \{(f,w):\big (f_{xt},-\sqrt{\frac{D}{A}}{{\tilde{\Lambda }}} f_x,f_{yt}, -\sqrt{\frac{D}{A}}{{\tilde{\Lambda }}} f_y, w_t,\sqrt{K}{{\tilde{\Lambda }}}(f+w)\big )\in (C([0,T],L^2({\mathbb {R}}^2)))^6\Big \}, \end{aligned}$$

then $\Omega$ with norm $\Vert \cdot \Vert _{\phi ,T}$ is a Banach space.

We fix the initial data $U_0\in {\mathcal {H}}=(H^1({\mathbb {R}}^2))^6,f_x(0,\cdot ,\cdot )\in L^2({\mathbb {R}}^2),f_y(0,\cdot ,\cdot )\in L^2({\mathbb {R}}^2)$. For a fixed ${{\bar{V}}}=({{\bar{f}}},0)^\tau \in B_{\phi ,T}^K$, we define $\Psi :B_{\phi ,T}^K\rightarrow \Omega ,\Psi ({{\bar{V}}}):=(f,w)^\tau$, where $(f,w)^\tau$ is the weak solution to the following system

$$\begin{aligned} \left\{ \begin{array}{llll} &{} \partial _x(f_{tt}-\frac{D}{A}\Delta f+\frac{K}{A}(f+w)+\alpha f_t)=|{{\bar{f}}}_x|^r, &{}&{} \varvec{x}\in {\mathbb {R}}^2,t\in {\mathbb {R}}_+,\\ &{} \partial _y(f_{tt}-\frac{D}{A}\Delta f+\frac{K}{A}(f+w)+\alpha f_t)=|{{\bar{f}}}_y|^r,&{}&{} \varvec{x}\in {\mathbb {R}}^2,t\in {\mathbb {R}}_+,\\ &{} w_{tt}-K\Delta (f+w)+\beta w_t=0,&{}&{} \varvec{x}\in {\mathbb {R}}^2,t\in {\mathbb {R}}_+,\\ &{} (f_x,f_{xt},f_y,f_{yt},w,w_t)(0,\varvec{x}) =(\varphi _0,\varphi _1,\psi _0,\psi _1,w_0,w_1)(\varvec{x}), &{}&{} \varvec{x}\in {\mathbb {R}}^2. \end{array} \right. \qquad \end{aligned}$$

(4.11)

Multiplying the three equations of (4.11) by $f_{xt},f_{yt},\frac{1}{A}w_{t}$, respectively, we have

$$\begin{aligned}&\frac{1}{2A}\partial _t\Big [(Af_{xt}^2+Af_{yt}^2+w_t^2) +D(f_{xx}^2+f_{yy}^2+2f_{xy}^2)+K(f_x+w_x)^2 +K(f_y+w_y)^2)\Big ]\nonumber \\&\quad \quad \quad -\frac{1}{A}\partial _x\Big [(Df_{xt}f_{xx}+Df_{yt}f_{xy} +Kw_t(f_x+w_x)\Big ]\nonumber \\&\quad \quad \quad -\frac{1}{A}\partial _y\Big [Df_{xt}f_{xy}+Df_{yt}f_{yy} +Kw_t(f_y+w_y)\Big ]\nonumber \\&\quad \quad \quad +\alpha (f_{xt}^2+f_{yt}^2)+\frac{\beta }{A}w_{t}^2 = |\bar{f}_x|^rf_{xt}+|{{\bar{f}}}_y|^rf_{yt}. \end{aligned}$$

(4.12)

Multiplying the above equation by $e^{2\phi (t,x,y)}$, we obtain

$$\begin{aligned}&\partial _t\Big [\frac{e^{2\phi }}{2}(f_{xt}^2+f_{yt}^2+\frac{1}{A}w_t^2 +\frac{D}{A}(f_{xx}^2+f_{yy}^2+2f_{xy}^2)+\frac{K}{A}(f_x+w_x)^2 +\frac{K}{A}(f_y+w_y)^2)\Big ]\nonumber \\&\quad \quad \quad -e^{2\phi }{\phi _t}\Big [f_{xt}^2+f_{yt}^2+\frac{1}{A}w_t^2 +\frac{D}{A}(f_{xx}^2+f_{yy}^2+2f_{xy}^2)+\frac{K}{A}(f_x+w_x)^2 +\frac{K}{A}(f_y+w_y)^2\Big ]\nonumber \\&\quad \quad \quad -\partial _x\Big [e^{2\phi }(\frac{D}{A}f_{xt}f_{xx}+\frac{D}{A}f_{yt}f_{xy} +\frac{K}{A}w_t(f_x+w_x))\Big ]\nonumber \\&\quad \quad \quad -\partial _y\Big [e^{2\phi }(\frac{D}{A}f_{xt}f_{xy}+\frac{D}{A}f_{yt}f_{yy} +\frac{K}{A}w_t(f_y+w_y))\Big ]\nonumber \\&\quad \quad \quad +2e^{2\phi }\phi _x\Big [\frac{D}{A}f_{xt}f_{xx} +\frac{D}{A}f_{yt}f_{xy}+\frac{K}{A}w_t(f_x+w_x)\Big ]\nonumber \\&\quad \quad \quad +2e^{2\phi }\phi _y\Big [\frac{D}{A}f_{xt}f_{xy}+\frac{D}{A}f_{yt}f_{yy} +\frac{K}{A}w_t(f_y+w_y)\Big ] +\alpha e^{2\phi }(f_{xt}^2+f_{yt}^2)+\frac{\beta e^{2\phi }}{A}w_{t}^2 \nonumber \\&\quad \quad \quad = e^{2\phi }|{{\bar{f}}}_x|^rf_{xt}+e^{2\phi }|{{\bar{f}}}_y|^rf_{yt}. \end{aligned}$$

(4.13)

Since

$$\begin{aligned}&e^{2\phi }\phi _x\left( \frac{D}{A}f_{xt}f_{xx}+\frac{D}{A}f_{yt}f_{xy} +\frac{K}{A}w_t(f_x+w_x)\right) \nonumber \\&\quad = \frac{e^{2\phi }}{2\phi _t}\left[ -\frac{D}{A}(\phi _xf_{xt}-\phi _tf_{xx})^2 -\frac{D}{A}(\phi _xf_{yt}-\phi _tf_{xy})^2-\frac{K}{A}(\phi _xw_{t} -\phi _t(f_{x}+w_x))^2\right] \nonumber \\&\quad \quad +e^{2\phi }\frac{\phi _t}{2}\left[ \frac{D}{A}f_{xx}^2+\frac{D}{A}f_{xy}^2 +\frac{K}{A}(f_{x}+w_x)^2\right] +e^{2\phi }\frac{\phi _x^2}{2\phi _t}\left[ \frac{D}{A}f_{xt}^2 +\frac{D}{A}f_{yt}^2+\frac{K}{A}w_t^2\right] \end{aligned}$$

(4.14)

and

$$\begin{aligned}&e^{2\phi }\phi _y\left( \frac{D}{A}f_{xt}f_{xy}+\frac{D}{A}f_{yt}f_{yy} +\frac{K}{A}w_t(f_y+w_y)\right) \nonumber \\&\quad = \frac{e^{2\phi }}{2\phi _t}\left[ -\frac{D}{A}(\phi _yf_{xt}-\phi _tf_{xy})^2 -\frac{D}{A}(\phi _yf_{yt}-\phi _tf_{yy})^2 -\frac{K}{A}(\phi _yw_{t}-\phi _t(f_{y}+w_y))^2\right] \nonumber \\&\quad \quad +e^{2\phi }\frac{\phi _t}{2}\left[ \frac{D}{A}f_{xy}^2+\frac{D}{A}f_{yy}^2 +\frac{K}{A}(f_{y}+w_y)^2\right] +e^{2\phi }\frac{\phi _y^2}{2\phi _t}\left[ \frac{D}{A}f_{xt}^2 +\frac{D}{A}f_{yt}^2+\frac{K}{A}w_t^2\right] . \end{aligned}$$

(4.15)

Equation (4.13) can be written as

$$\begin{aligned}&\partial _t\left[ \frac{e^{2\phi }}{2}(f_{xt}^2+f_{yt}^2+\frac{1}{A}w_t^2+\frac{D}{A}(f_{xx}^2+f_{yy}^2+2f_{xy}^2)+\frac{K}{A}(f_x+w_x)^2 +\frac{K}{A}(f_y+w_y)^2)\right] \nonumber \\&\quad -e^{2\phi }{\phi _t}\left( f_{xt}^2+f_{yt}^2+\frac{1}{A}w_t^2\right) -\partial _x\left[ e^{2\phi }(\frac{D}{A}f_{xt}f_{xx}+\frac{D}{A}f_{yt}f_{xy}+\frac{K}{A}w_t(f_x+w_x))\right] \nonumber \\&\quad -\partial _y\left[ e^{2\phi }(\frac{D}{A}f_{xt}f_{xy}+\frac{D}{A}f_{yt}f_{yy}+\frac{K}{A}w_t(f_y+w_y))\right] \nonumber \\&\quad + \frac{e^{2\phi }}{\phi _t}\left[ -\frac{D}{A}(\phi _xf_{xt}-\phi _tf_{xx})^2 -\frac{D}{A}(\phi _xf_{yt}-\phi _tf_{xy})^2-\frac{K}{A}(\phi _xw_{t}-\phi _t(f_{x}+w_x))^2\right] \nonumber \\&\quad + \frac{e^{2\phi }}{\phi _t}\left[ -\frac{D}{A}(\phi _yf_{xt}-\phi _tf_{xy})^2 -\frac{D}{A}(\phi _yf_{yt}-\phi _tf_{yy})^2-\frac{K}{A}(\phi _yw_{t}-\phi _t(f_{y}+w_y))^2\right] \nonumber \\&\quad +e^{2\phi }\frac{\phi _x^2+\phi _y^2}{\phi _t}\left[ \frac{D}{A}f_{xt}^2+\frac{D}{A}f_{yt}^2+\frac{K}{A}w_t^2\right] +\alpha e^{2\phi }(f_{xt}^2+f_{yt}^2)+\frac{\beta e^{2\phi }}{A}w_{t}^2 \nonumber \\&\quad \quad = e^{2\phi }|{{\bar{f}}}_x|^rf_{xt}+e^{2\phi }|{{\bar{f}}}_y|^rf_{yt}. \end{aligned}$$

(4.16)

Together with $e^{2\phi }>0,\phi _t<0,$ this equation leads to

$$\begin{aligned}&\partial _t \left[ \frac{e^{2\phi }}{2}(f_{xt}^2+f_{yt}^2+\frac{1}{A}w_t^2+\frac{D}{A}(f_{xx}^2+f_{yy}^2+2f_{xy}^2)+\frac{K}{A}(f_x+w_x)^2 +\frac{K}{A}(f_y+w_y)^2)\right] \nonumber \\&\quad -\partial _x\left[ e^{2\phi }(\frac{D}{A}f_{xt}f_{xx}+\frac{D}{A}f_{yt}f_{xy}+\frac{K}{A}w_t(f_x+w_x))\right] \nonumber \\&\quad -\partial _y\left[ e^{2\phi }(\frac{D}{A}f_{xt}f_{xy}+\frac{D}{A}f_{yt}f_{yy}+\frac{K}{A}w_t(f_y+w_y))\right] \nonumber \\&\quad +e^{2\phi }\frac{\phi _x^2+\phi _y^2}{\phi _t}\left[ \frac{D}{A}f_{xt}^2+\frac{D}{A}f_{yt}^2+\frac{K}{A}w_t^2\right] +\alpha e^{2\phi }(f_{xt}^2+f_{yt}^2)+\frac{\beta e^{2\phi }}{A}w_{t}^2\nonumber \\&\quad \quad \le e^{2\phi }|{{\bar{f}}}_x|^rf_{xt}+e^{2\phi }|{{\bar{f}}}_y|^rf_{yt}. \end{aligned}$$

(4.17)

Suppose $\gamma$ in the weight function $\phi$ satisfies $0<\gamma <\min \{\frac{\alpha A}{4D},\frac{\beta }{4K}\},$ such that

$$\begin{aligned} \alpha +\frac{D}{A}\frac{\phi _x^2+\phi _y^2}{\phi _t}\ge 0,\ \ \ \ \frac{\beta }{A}+\frac{K}{A}\frac{\phi _x^2+\phi _y^2}{\phi _t}\ge 0. \end{aligned}$$

(4.18)

Then, we have

$$\begin{aligned}&\partial _t \left[ {e^{2\phi }}(f_{xt}^2+f_{yt}^2+\frac{1}{A}w_t^2+\frac{D}{A}(f_{xx}^2+f_{yy}^2+2f_{xy}^2)+\frac{K}{A}(f_x+w_x)^2 +\frac{K}{A}(f_y+w_y)^2)\right] \nonumber \\&\quad -\partial _x\left[ e^{2\phi }(\frac{D}{A}f_{xt}f_{xx}+\frac{D}{A}f_{yt}f_{xy}+\frac{K}{A}w_t(f_x+w_x))\right] \nonumber \\&\quad -\partial _y\left[ e^{2\phi }(\frac{D}{A}f_{xt}f_{xy}+\frac{D}{A}f_{yt}f_{yy}+\frac{K}{A}w_t(f_y+w_y))\right] \nonumber \\&\quad \quad \le e^{2\phi }|{{\bar{f}}}_x|^rf_{xt}+e^{2\phi }|{{\bar{f}}}_y|^rf_{yt}. \end{aligned}$$

(4.19)

Integrating both sides over $[0,t)\times {\mathbb {R}}^2$, the $-\partial _x$ and $-\partial _y$ terms become 0. Therefore,

$$\begin{aligned} E_{f,w}^{\phi }(t) \le E_{f,w}^{\phi }(0) + \int _0^t\int _{{\mathbb {R}}^2}(e^{2\phi (s,x,y)}|\bar{f}_x|^rf_{xs}+e^{2\phi (s,x,y)}|{{\bar{f}}}_y|^rf_{ys})d(x,y)ds, \end{aligned}$$

(4.20)

where

$$\begin{aligned} E_{f,w}^{\phi }(t)&= \frac{1}{2}\Big (\Vert e^{\phi (t,\cdot ,\cdot )}f_{xt}\Vert _2^2+\Vert e^{\phi (t,\cdot ,\cdot )}f_{yt}\Vert _2^2 +\frac{1}{A}\Vert e^{\phi (t,\cdot ,\cdot )}w_t\Vert _2^2\\&\quad +\frac{D}{A}(\Vert e^{\phi (t,\cdot ,\cdot )}f_{xx}\Vert _2^2+\Vert e^{\phi (t,\cdot ,\cdot )}f_{yy}\Vert _2^2\\&\quad +2\Vert e^{\phi (t,\cdot ,\cdot )}f_{xy}\Vert _2^2) +\frac{K}{A}\Vert e^{\phi (t,\cdot ,\cdot )}(f_x+w_x)\Vert _2^2 +\frac{K}{A}\Vert e^{\phi (t,\cdot ,\cdot )}(f_y+w_y)\Vert _2^2\Big ). \end{aligned}$$

By Hölder’s inequality,

$$\begin{aligned} E_{f,w}^{\phi }(t) \le E_{f,w}^{\phi }(0) + \int _0^t\Vert e^{\phi (s,\cdot ,\cdot )}({\bar{f}}_x)^r\Vert _2\Vert e^{\phi (s,\cdot ,\cdot )}{f}_{xs}\Vert _2ds\nonumber \\ + \int _0^t\Vert e^{\phi (s,\cdot ,\cdot )}({\bar{f}}_y)^r\Vert _2\Vert e^{\phi (s,\cdot ,\cdot )}{f}_{ys}\Vert _2ds. \end{aligned}$$

(4.21)

By definition of $E_{f,w}^{\phi }(t)$,

$$\begin{aligned} E_{f,w}^{\phi }(t) \le&E_{f,w}^{\phi }(0) + \sqrt{2}\int _0^t(\Vert e^{\phi (s,\cdot ,\cdot )}({\bar{f}}_x)^r\Vert _2+\Vert e^{\phi (s,\cdot ,\cdot )}({\bar{f}}_y)^r\Vert _2)(E_{f,w}^{\phi }(s))^{1/2}ds. \end{aligned}$$

(4.22)

Lemma 1.2 gives that

$$\begin{aligned} (E_{f,w}^{\phi }(t))^{1/2} \le&(E_{f,w}^{\phi }(0))^{1/2} +\frac{1}{\sqrt{2}} \int _0^t(\Vert e^{\phi (s,\cdot ,\cdot )}({\bar{f}}_x)^r\Vert _2+\Vert e^{\phi (s,\cdot ,\cdot )}({\bar{f}}_y)^r\Vert _2)ds. \end{aligned}$$

(4.23)

Take $\nu = 1/r, m = 2r$. Since $r>1$, Lemma 4.1 gives that

$$\begin{aligned} \Vert e^{\phi (s,\cdot ,\cdot )}({{\bar{f}}}_x)^r\Vert _2^2 =&\Vert e^{\phi (s,\cdot ,\cdot )/r}({\bar{f}}_x)\Vert _{2r}^{2r}\\ \le&C(1+s)\left( \Vert \bar{f}_{xx}\Vert _2^{(1-1/r)}\Vert e^\phi {\bar{f}}_{xx}\Vert _2^{1/r} +\Vert \bar{f}_{xy}\Vert _2^{(1-1/r)}\Vert e^\phi {{\bar{f}}}_{xy}\Vert _2^{1/r}\right) ^{2r}. \end{aligned}$$

Similarly,

$$\begin{aligned} \Vert e^{\phi (s,\cdot ,\cdot )}({{\bar{f}}}_y)^r\Vert _2^2 =&\Vert e^{\phi (s,\cdot ,\cdot )/r}({\bar{f}}_y)\Vert _{2r}^{2r}\\ \le&C(1+s)\left( \Vert \bar{f}_{yx}\Vert _2^{(1-1/r)}\Vert e^\phi {\bar{f}}_{yx}\Vert _2^{1/r} +\Vert \bar{f}_{yy}\Vert _2^{(1-1/r)}\Vert e^\phi {{\bar{f}}}_{yy}\Vert _2^{1/r}\right) ^{2r}. \end{aligned}$$

Since $e^{\phi }>1,$ we have $\Vert {\bar{f}}_{xx}\Vert _2\le \Vert e^{\phi }{\bar{f}}_{xx}\Vert _2.$ Hence,

$$\begin{aligned} (E_{f,w}^{\phi }(t))^{1/2}&\le (E_{f,w}^{\phi }(0))^{1/2} +C \int _0^t(1+T)^{1/2}K^rds\nonumber \\&\le (E_{f,w}^{\phi }(0))^{1/2} +CT(1+T)^{1/2}K^r . \end{aligned}$$

(4.24)

Moreover, due to the fact that $\phi$ is a monotone decreasing function in t,

$$\begin{aligned} \Vert e^{\phi (t,\cdot ,\cdot )}f_x\Vert _2&\le \Vert e^{\phi (t,\cdot ,\cdot )}f_x(0,\cdot ,\cdot )\Vert _2+\int _0^t\Vert e^{\phi (t,\cdot ,\cdot )}f_{xs}\Vert _2ds\nonumber \\&\le \Vert e^{\phi (0,\cdot ,\cdot )}f_x(0,\cdot ,\cdot )\Vert _2+\sqrt{2}\int _0^t[(E_{f,w}^{\phi }(0))^{1/2}+C(1+T)^{1/2}TK^r]ds\nonumber \\&\le \Vert e^{\phi (0,\cdot ,\cdot )}f_x(0,\cdot ,\cdot )\Vert _2+\sqrt{2}(E_{f,w}^{\phi }(0))^{1/2}T+C(1+T)^{1/2}T^2K^r. \end{aligned}$$

(4.25)

Similarly, we have

$$\begin{aligned} \Vert e^{\phi (t,\cdot ,\cdot )}f_y\Vert _2&\le \Vert e^{\phi (t,\cdot ,\cdot )}f_y(0,\cdot ,\cdot )\Vert _2+\int _0^t\Vert e^{\phi (t,\cdot ,\cdot )}f_{ys}\Vert _2ds\nonumber \\&\le \Vert e^{\phi (0,\cdot ,\cdot )}f_y(0,\cdot ,\cdot )\Vert _2+\sqrt{2}\int _0^t[(E_{f,w}^{\phi }(0))^{1/2}+C(1+T)^{1/2}TK^r]ds\nonumber \\&\le \Vert e^{\phi (0,\cdot ,\cdot )}f_y(0,\cdot ,\cdot )\Vert _2+\sqrt{2}(E_{f,w}^{\phi }(0))^{1/2}T+C(1+T)^{1/2}T^2K^r. \end{aligned}$$

(4.26)

Combining equations (4.24), (4.25) and (4.26), we have

$$\begin{aligned} \Vert (f,w)\Vert _{\phi ,T}&\le (E_{f,w}^{\phi }(0))^{1/2}+C(1+T)^{1/2}TK^r+\Vert e^{\phi (0,\cdot ,\cdot )}f_x(0,\cdot ,\cdot )\Vert _2\\&\quad +\Vert e^{\phi (0,\cdot ,\cdot )}f_y(0,\cdot ,\cdot )\Vert _2+2\sqrt{2}(E_{f,w}^{\phi }(0))^{1/2}T+C(1+T)^{1/2}T^2K^r. \end{aligned}$$

Choosing $K>0$ large enough such that

$$\begin{aligned} (E_{f,w}^{\phi }(0))^{1/2}+\Vert e^{\phi (0,\cdot ,\cdot )}f_x(0,\cdot ,\cdot )\Vert _2+\Vert e^{\phi (0,\cdot ,\cdot )}f_y(0,\cdot ,\cdot )\Vert _2\le K/2, \end{aligned}$$

then choosing $T>0$ small enough such that

$$\begin{aligned} C(1+T)^{1/2}TK^r+2\sqrt{2}(E_{f,w}^{\phi }(0))^{1/2}T+C(1+T)^{1/2}T^2K^r\le K/2, \end{aligned}$$

we get

$$\begin{aligned} \Vert (f,w)\Vert _{\phi ,T}\le K. \end{aligned}$$

Then, we have $(f,w)\in B_{\phi ,T}^K.$

Next, we show the mapping is contractive. Letting $(f,w) = \Psi ({\bar{f}},0) = \Psi ({\bar{V}})$, and $({\check{f}},{\check{w}}) = \Psi (\bar{{\check{f}}},0) = \Psi (\bar{{\check{V}}}).$ Assume that ${\tilde{f}} = f-{\check{f}}, {\tilde{w}} = w-{\check{w}},$ then $({\tilde{f}},{\tilde{w}})$ satisfies

$$\begin{aligned} \left\{ \begin{array}{llll} &{} \partial _x({\tilde{f}}_{tt}-\frac{D}{A}\Delta {\tilde{f}} + \frac{K}{A}({\tilde{f}}+{\tilde{w}}) + \alpha {\tilde{f}}_t) = |{\bar{f}}_x|^r-|\bar{{\check{f}}}_x|^r, &{}&{}\varvec{x}\in {\mathbb {R}}^2, t\in {\mathbb {R}}_{+},\\ &{} \partial _y({\tilde{f}}_{tt}-\frac{D}{A}\Delta {\tilde{f}} + \frac{K}{A}({\tilde{f}}+{\tilde{w}}) + \alpha {\tilde{f}}_t) = |{\bar{f}}_y|^r-|\bar{{\check{f}}}_y|^r, &{}&{}\varvec{x}\in {\mathbb {R}}^2, t\in {\mathbb {R}}_{+},\\ &{} w_{tt}-K\Delta (f+w) + \beta w_t = 0, &{}&{} \varvec{x}\in {\mathbb {R}}^2, t\in {\mathbb {R}}_{+},\\ &{} ({{\tilde{f}}}_{xt},{{\tilde{f}}}_x,{{\tilde{f}}}_{yt},{{\tilde{f}}}_y, \tilde{w}_t, {{\tilde{w}}})(0,x,y) = (0,0,0,0,0,0), &{}&{}\varvec{x}\in {\mathbb {R}}^2. \end{array} \right. \qquad \end{aligned}$$

(4.27)

Similar as the above estimates, we have

$$\begin{aligned} E_{{{\tilde{f}}},{{\tilde{w}}}}^{\phi }(t) \le&\int _0^t\int _{{\mathbb {R}}^2}e^{2\phi (s,x,y)}(|{\bar{f}}_x|^r-|\bar{{\check{f}}}_x|^r){{\tilde{f}}}_{xs}+e^{2\phi (s,x,y)}(|{{\bar{f}}}_y|^r-|\bar{{\check{f}}}_y|^r){{\tilde{f}}}_{ys}d(x,y)ds\nonumber \\ :=&J_a+J_b. \end{aligned}$$

(4.28)

On the strength of inequalities

$$\begin{aligned} \left| |{\bar{f}}_x|^r-|\bar{{\check{f}}}_x|^r\right| \le r|{\bar{f}}_x-\bar{{\check{f}}}_x|(|{\bar{f}}_x|+|\bar{{\check{f}}}_x|)^{r-1} \end{aligned}$$

(4.29)

and

$$\begin{aligned} \left| |{\bar{f}}_y|^r-|\bar{{\check{f}}}_y|^r\right| \le r|{\bar{f}}_y-\bar{{\check{f}}}_y|(|{\bar{f}}_y|+|\bar{{\check{f}}}_y|)^{r-1}, \end{aligned}$$

(4.30)

together with Hölder’s inequality

$$\begin{aligned} J_a&\le r \int _0^t \int _{{\mathbb {R}}^2}e^{2\phi }|\bar{f_x}-\bar{\check{f_x}}|(|\bar{f_x}|+|\bar{\check{f_x}}|)^{r-1} |{\tilde{f}}_{xs}|d(x,y)ds\\&\le C\int _0^t (E_{{\tilde{f}},{\tilde{w}}}^{\phi }(s))^{1/2}\left( \int _{{\mathbb {R}}^2}e^{2\phi }|\bar{f_x}-\bar{\check{f_x}}|^2(|\bar{f_x}|+|\bar{\check{f_x}}|)^{2(r-1)}d(x,y)\right) ^{1/2}ds\\&\le C\int _0^t (E_{{\tilde{f}},{\tilde{w}}}^{\phi }(s))^{1/2}\Vert e^{\phi /2}(\bar{f_x}-\bar{\check{f_x}})\Vert _{2r}\Vert e^{\frac{\phi }{2(r-1)}}(|\bar{f_x}|+|\bar{\check{f_x}}|)\Vert _{2r}^{r-1}ds \end{aligned}$$

and

$$\begin{aligned} J_b&\le r \int _0^t \int _{{\mathbb {R}}^2}e^{2\phi }|\bar{f_y}-\bar{\check{f_y}}|(|\bar{f_y}|+|\bar{\check{f_y}}|)^{r-1} |{\tilde{f}}_{ys}|d(x,y)ds\\&\le C\int _0^t (E_{{\tilde{f}},{\tilde{w}}}^{\phi }(s))^{1/2}\left( \int _{{\mathbb {R}}^2}e^{2\phi }|\bar{f_y}-\bar{\check{f_y}}|^2(|\bar{f_y}|+|\bar{\check{f_y}}|)^{2(r-1)}d(x,y)\right) ^{1/2}ds\\&\le C\int _0^t (E_{{\tilde{f}},{\tilde{w}}}^{\phi }(s))^{1/2}\Vert e^{\phi /2}(\bar{f_y}-\bar{\check{f_y}})\Vert _{2r}\Vert e^{\frac{\phi }{2(r-1)}}(|\bar{f_y}|+|\bar{\check{f_y}}|)\Vert _{2r}^{r-1}ds, \end{aligned}$$

we have

$$\begin{aligned} E_{{{\tilde{f}}},{{\tilde{w}}}}^\phi (t)&\le C\int _0^t\left( E_{\tilde{f},{{\tilde{w}}}}^\phi (s)\right) ^{\frac{1}{2}} \left( \Vert e^{\phi /2}({{\bar{f}}}_x-\bar{{\check{f}}}_x)\Vert _{2r}\Vert e^{\frac{\phi }{2(r-1)}}(|{{\bar{f}}}_x|+|\bar{{\check{f}}}_x|)\Vert _{2r}^{r-1}\right. \\&\quad \left. +\Vert e^{\phi /2}({{\bar{f}}}_y-\bar{\check{f}}_y)\Vert _{2r}\Vert e^{\frac{\phi }{2(r-1)}}(|{{\bar{f}}}_y|+|\bar{\check{f}}_y|)\Vert _{2r}^{r-1}\right) ds. \end{aligned}$$

By Lemma 1.2,

$$\begin{aligned} \left( E_{{{\tilde{f}}},{{\tilde{w}}}}^\phi (t)\right) ^{\frac{1}{2}}&\le C\int _0^t \left( \Vert e^{\phi /2}({{\bar{f}}}_x-\bar{{\check{f}}}_x)\Vert _{2r}\Vert e^{\frac{\phi }{2(r-1)}}(|{{\bar{f}}}_x|+|\bar{{\check{f}}}_x|)\Vert _{2r}^{r-1}\right. \\&\quad \left. +\Vert e^{\phi /2}({{\bar{f}}}_y-\bar{{\check{f}}}_y)\Vert _{2r}\Vert e^{\frac{\phi }{2(r-1)}}(|{{\bar{f}}}_y|+|\bar{{\check{f}}}_y|)\Vert _{2r}^{r-1}\right) ds\\&:= I_a+I_b. \end{aligned}$$

Applying the Minkowski’s inequality, we obtain

$$\begin{aligned} I_a\le C\int _0^t\Vert e^{\phi /2}({{\bar{f}}}_x-\bar{\check{f}}_x)\Vert _{2r}\left( \Vert e^{\frac{\phi }{2(r-1)}}{{\bar{f}}}_x\Vert _{2r} +\Vert e^{\frac{\phi }{2(r-1)}}\bar{{\check{f}}}_x\Vert _{2r}\right) ^{r-1}ds. \end{aligned}$$

(4.31)

By Lemma 4.1 and taking $\nu = \frac{1}{2(r-1)}$ and $m = 2r$, we have

$$\begin{aligned} \Vert e^{\frac{\phi }{2(r-1)}}\bar{f_x}\Vert _{2r}&\le C(1+s)^{\frac{1-\theta (2r)}{2}}\left( \Vert {\bar{f}}_{xx}\Vert _2^{1-\frac{1}{2(r-1)}}\Vert e^{\phi }{{\bar{f}}_{xx}}\Vert _2^{\frac{1}{2(r-1)}} +\Vert {{\bar{f}}}_{xy}\Vert _2^{1-\frac{1}{2(r-1)}}\Vert e^\phi {{\bar{f}}}_{xy}\Vert _2^{\frac{1}{2(r-1)}}\right) \nonumber \\&\le C(1+T)^{\frac{1}{2r}}K. \end{aligned}$$

(4.32)

Similarly, for $\nu = 1/2$ and $m = 2r$,

$$\begin{aligned} \Vert e^{\phi /2}(\bar{f_x}-\bar{\check{f_x}})\Vert _{2r}&\le C(1+T)^{\frac{1}{2r}}\left( \Vert e^{\phi }({\bar{f}}_{xx}-\bar{{\check{f}}}_{xx})\Vert _2 +\Vert e^\phi ({{\bar{f}}}_{xy}-\bar{{\check{f}}}_{xy})\Vert _2\right) . \end{aligned}$$

(4.33)

Plugging (4.32), (4.33) into (4.31), we get

$$\begin{aligned} I_a\le C(1+T)^{1/2}TK^{r-1}\Vert {{\bar{V}}}-\bar{{\check{V}}}\Vert _{\phi ,T}. \end{aligned}$$

Similarly, we have

$$\begin{aligned} I_b\le C(1+T)^{1/2}TK^{r-1}\Vert {{\bar{V}}}-\bar{{\check{V}}}\Vert _{\phi ,T}. \end{aligned}$$

Combining the above two inequalities, we obtain

$$\begin{aligned} (E_{{\tilde{f}},{\tilde{w}}}^{\phi }(t))^{1/2}&\le C(1+T)^{1/2}TK^{r-1}\Vert {\bar{V}}-\bar{{\check{V}}}\Vert _{\phi ,T}. \end{aligned}$$

(4.34)

On the other hand,

$$\begin{aligned} \tilde{f_x} = \int _0^t {\tilde{f}}_{xs}ds,\ \ \ \ \ \tilde{f_y}=\int _0^t{\tilde{f}}_{ys}ds. \end{aligned}$$

Since the function $\phi$ is monotonic decreasing in t,

$$\begin{aligned} \Vert e^{\phi }\tilde{f_x}\Vert _2&\le \int _0^t \Vert e^{\phi (t,x,y)}{\tilde{f}}_{xs}\Vert _2ds\nonumber \\&\le \int _0^t\Vert e^{\phi (s,x,y)}{\tilde{f}}_{xs}\Vert _2ds\nonumber \\&\le C(1+T)^{\frac{1}{2}}T^2K^{r-1}\Vert {\bar{V}}-\bar{{\check{V}}}\Vert _{\phi ,T} \end{aligned}$$

(4.35)

and

$$\begin{aligned} \Vert e^{\phi }\tilde{f_y}\Vert _2&\le \int _0^t \Vert e^{\phi (t,x,y)}{\tilde{f}}_{ys}\Vert _2ds\nonumber \\&\le \int _0^t\Vert e^{\phi (s,x,y)}{\tilde{f}}_{ys}\Vert _2ds\nonumber \\&\le C(1+T)^{\frac{1}{2}}T^2K^{r-1}\Vert {\bar{V}}-\bar{{\check{V}}}\Vert _{\phi ,T}. \end{aligned}$$

(4.36)

Inequalities (4.34), (4.35) and (4.36) imply that

$$\begin{aligned} \Vert (f-{\check{f}},w-{\check{w}})\Vert _{\phi ,T}\le C(1+T)^{\frac{3}{2}}TK^{r-1}\Vert {\bar{V}}-\bar{{\check{V}}}\Vert _{\phi ,T}. \end{aligned}$$

(4.37)

Choose $T>0$ small enough such that

$$\begin{aligned} C(1+T)^{\frac{3}{2}}TK^{r-1} < \frac{1}{2}. \end{aligned}$$

(4.38)

Hence, $\Psi$ is contractive mapping and has a unique fixed point (f, w). With expression of solution, for $U_0 \in {\mathcal {H}},$ we have a unique classical solution U.

Finally, we prove the boundedness. Applying similar method as used in [26], we prove that for any $t\in [0,T],T<T_{max}$,

$$\begin{aligned} \Vert e^{\phi (t,\cdot ,\cdot )}U(t,\cdot ,\cdot )\Vert _2+\Vert e^{\phi (t,\cdot ,\cdot )}f_x(t,\cdot ,\cdot )\Vert _2+\Vert e^{\phi (t,\cdot ,\cdot )}f_y(t,\cdot ,\cdot )\Vert _2<\infty . \end{aligned}$$

(4.39)

Assume that $V^{(0)}(t,x,y) = (f(0,x,y),0)'$ with $V^{(0)}\in B_{\phi ,T}^K$, and define $(f^{(n)},w^{(n)})'$ satisfying

$$\begin{aligned} \left( \begin{array}{c} f^{(n)}\\ w^{(n)}\\ \end{array} \right) = \Psi \left( \begin{array}{c} f^{(n-1)}\\ w^{(n-1)}\\ \end{array} \right) , \quad \quad n = 1,2,3,\cdot \cdot \cdot \end{aligned}$$

(4.40)

where $(f^{(n)},w^{(n)})'$ is the solution to the following equations

$$\begin{aligned} \left\{ \begin{array}{llll} &{} \partial _x\left( f_{tt}^{(n)}-\frac{D}{A}\Delta f^{(n)} + \frac{K}{A}(f^{(n)}+w^{(n)}) + \alpha f_t^{(n)}\right) = |f_x^{(n-1)}|^r, &{}&{} \varvec{x}\in {\mathbb {R}}^2,t\in {\mathbb {R}}_+,\\ &{} \partial _y\left( f_{tt}^{(n)}-\frac{D}{A}\Delta f^{(n)} + \frac{K}{A}(f^{(n)}+w^{(n)}) + \alpha f_t^{(n)}\right) = |f_y^{(n-1)}|^r, &{}&{}\varvec{x}\in {\mathbb {R}}^2,t\in {\mathbb {R}}_+,\\ &{} w_{tt}^{(n)}-K\Delta (f^{(n)}+w^{(n)}) + \beta w^{(n)}_t = 0, &{}&{}\varvec{x}\in {\mathbb {R}}^2,t\in {\mathbb {R}}_+,\\ &{} (f_{xt}^{(n)}, f_{x}^{(n)},f_{yt}^{(n)}, f_{y}^{(n)},w_t^{(n)}, w^{(n)} ) = I_{f,w}(x,y),&{}&{}\varvec{x}\in {\mathbb {R}}^2, \end{array} \right. \end{aligned}$$

where

$$\begin{aligned} I_{f,w}(x,y)=\left( f_{xt}(0,\cdot ,\cdot ),f_x(0,\cdot ,\cdot ),f_{yt}(0,\cdot ,\cdot ),f_y(0,\cdot ,\cdot ),w_t(0,\cdot ,\cdot ),w(0,\cdot ,\cdot )\right) . \end{aligned}$$

Since $U=\left( f_{xt},-\sqrt{\frac{D}{A}}{{\tilde{\Lambda }}} f_x,f_{yt},-\sqrt{\frac{D}{A}}{{\tilde{\Lambda }}} f_y,w_t,\sqrt{K}{{\tilde{\Lambda }}}(f+w)\right) \in (C([0,T_m),L^2({\mathbb {R}}^2)))^6$ and $f_x,f_y\in C([0,T_m),L^2({\mathbb {R}}^2))$, for any $T<T_m$, as $n\rightarrow \infty$, we have

$$\begin{aligned} U^{(n)}\rightarrow U\in (C([0,T],L^2({\mathbb {R}}^2)))^6, \end{aligned}$$

where

$$\begin{aligned} U^{(n)}:=\left( f^{(n)}_{xt},-\sqrt{\frac{D}{A}}{{\tilde{\Lambda }}} f^{(n)}_x,f^{(n)}_{yt},-\sqrt{\frac{D}{A}}{{\tilde{\Lambda }}} f^{(n)}_y,w^{(n)}_t,\sqrt{K}{{\tilde{\Lambda }}}(f^{(n)}+w^{(n)})\right) . \end{aligned}$$

Also,

$$\begin{aligned} f_x^{(n)}\rightarrow f_x \quad \in C([0,T],L^2({\mathbb {R}}^2)),\\ f_y^{(n)}\rightarrow f_y \quad \in C([0,T],L^2({\mathbb {R}}^2)), \end{aligned}$$

thus, U is the weak solution of (2.2).

The above argument suggests that

$$\begin{aligned}&\Vert e^{\phi }U^{(n)}\Vert _2+\Vert e^{\phi }f_x^{(n)}\Vert _2+\Vert e^{\phi }f_y^{(n)}\Vert _2\\&\quad \le \Vert e^{\phi (0,x,y)}U_0\Vert _2+\Vert e^{\phi (0,\cdot ,\cdot )}f_x(0,\cdot ,\cdot )\Vert _2+\Vert e^{\phi (0,\cdot ,\cdot )}f_y(0,\cdot ,\cdot )\Vert _2+C\Vert e^{\phi (0,x,y)}U_0\Vert _2T\\&\qquad +C\int _0^t(1+T)^{\frac{1}{2}}K^rds+C\int _0^t(1+T)^{\frac{1}{2}}TK^rds\\&\le \Vert e^{\phi (0,x,y)}U_0\Vert _2+\Vert e^{\phi (0,x,y)}f_x(0,\cdot ,\cdot )\Vert _2+\Vert e^{\phi (0,\cdot ,\cdot )}f_y(0,\cdot ,\cdot )\Vert _2+C\Vert e^{\phi (0,x,y)}U_0\Vert _2T\\&\qquad +C(1+T)^{\frac{3}{2}}TK^r. \end{aligned}$$

Choosing $T>0$ small enough such that

$$\begin{aligned}&\Vert e^{\phi }U^{(n)}\Vert _2+\Vert e^{\phi }f_x^{(n)}\Vert _2 +\Vert e^{\phi }f_y^{(n)}\Vert _2\nonumber \\&\quad \le \Vert e^{\phi (0,x,y)}U_0\Vert _2+\Vert e^{\phi (0,x,y)}f_x(0,x,y)\Vert _2+\Vert e^{\phi (0,x,y)}f_y(0,x,y)\Vert _2<K. \end{aligned}$$

(4.41)

By the method used in Racke and Said–Houari [26], taking the limit for n, then

$$\begin{aligned} \Vert e^{\phi }U\Vert _2+\Vert e^{\phi }f_x\Vert _2+\Vert e^{\phi }f_y\Vert _2 < 3K. \end{aligned}$$

(4.42)

Thus, the local existence theorem is proved. $\square$

5 Global existence

In this section, the a priori estimate is given. Together with the local existence Theorem 4.1, we extend our local classical solution globally in time.

Lemma 5.1

Let U(t, x, y) be a local classical solution to Eq. (2.2) on $[0,T_{m})$, $\phi (t,x,y)$ being the weight function defined as in (4.4), then for any $t\in [0,T_{m})$,

$$\begin{aligned}&\Vert e^{\phi (t,x,y)}U\Vert _2^2\le CI_0^2 + C\Big (\sup _{[0,t]}(1+s)^{\delta }\Vert e^{\kappa \phi (s,x,y)}f_x\Vert _{r+1}\Big )^{r+1}\nonumber \\&\quad + C\Big (\sup _{[0,t]}(1+s)^{\delta }\Vert e^{\kappa \phi (s,x,y)}f_y\Vert _{r+1}\Big )^{r+1}, \end{aligned}$$

(5.1)

where $\kappa>\frac{2}{r+1},\delta >0,$ $I_0$ is as defined in Theorem 1.1.

Proof

Multiplying the three equations in (1.4) by $e^{2\phi }f_{xt},e^{2\phi }f_{yt}$ and $\frac{1}{A}e^{2\phi }w_t$, respectively, and using the methods in Theorem 4.1, we have

$$\begin{aligned}&\partial _t\Big [\frac{e^{2\phi }}{2A}(Af_{xt}^2+Af_{yt}^2+w_t^2+{D}(f_{xx}^2+f_{yy}^2+2f_{xy}^2)+{K}(f_x+w_x)^2+{K}(f_y+w_y)^2)\Big ]\nonumber \\&\quad -\frac{1}{A}\partial _x\left( e^{2\phi }({D}f_{xt}f_{xx}+{D}f_{yt}f_{xy}+{K}w_t(f_x+w_x))\right) \nonumber \\&\quad -\frac{1}{A}\partial _y\left( e^{2\phi }(Df_{xt}f_{xy}+Df_{yt}f_{yy}+Kw_t(f_y+w_y))\right) \nonumber \\&\quad \quad \le e^{2\phi }|f_x|^rf_{xt}+e^{2\phi }|f_y|^rf_{yt}. \end{aligned}$$

(5.2)

Integrating both sides over $[0,t)\times {\mathbb {R}}^2$,

$$\begin{aligned} E_{f,w}^\phi (t)&\le E_{f,w}^\phi (0)+\int _0^t\int _{{\mathbb {R}}^2}e^{2\phi (s,x,y)}|f_x|^rf_{xs}+e^{2\phi (s,x,y)}|f_y|^rf_{ys}d(x,y)ds\\&:= I_0^2+{{\bar{I}}}_a+{{\bar{I}}}_b. \end{aligned}$$

Observing that

$$\begin{aligned} {{\bar{I}}}_a \le&-\frac{1}{r+1}\int _{{\mathbb {R}}^2}e^{2\phi (0,x,y)}|f_x(0,x,y)|^rf_x(0,x,y)d(x,y)\\&+\frac{1}{r+1}\int _{{\mathbb {R}}^2}e^{2\phi (t,x,y)}|f_x(t,x,y)|^rf_x(t,x,y)d(x,y)\\&+\frac{2}{r+1}\int _0^t\int _{{\mathbb {R}}^2}e^{2\phi (s,x,y)}(-\phi _s(s,x,y))|f_x(s,x,y)|^rf_x(s,x,y)d(x,y)ds\\ :=&\, {{\bar{I}}}_{a1}+{{\bar{I}}}_{a2}+{{\bar{I}}}_{a3}. \end{aligned}$$

If $\kappa >\frac{2}{r+1}$, one has

$$\begin{aligned} \Vert \phi (s,x,y)e^{(2-\kappa (r+1))\phi (s,x,y)}\Vert _\infty <\infty , \end{aligned}$$

thus for $\delta >0$, we have

$$\begin{aligned} (r+1){{\bar{I}}}_{a3}&\le 2\int _0^t\int _{{\mathbb {R}}^2}e^{2\phi (s,x,y)}(-\phi _s(s,x,y))|f_x(s,x,y)|^{r+1}d(x,y)ds\\&= 2\int _0^t\frac{1}{s+1}\int _{{\mathbb {R}}^2}\phi e^{(2-\kappa (r+1))\phi (s,x,y)}e^{\kappa (r+1)\phi (s,x,y)}|f_x(s,x,y)|^{r+1}d(x,y)ds\\&\le 2\int _0^t\frac{1}{s+1}\Vert \phi (s,x,y)e^{(2-\kappa (r+1))\phi (s,x,y)}\Vert _\infty \Vert e^{\kappa \phi (s,x,y)}f_x(s,x,y)\Vert _{r+1}^{r+1}ds\\&\le C\int _0^t\frac{1}{(s+1)^{1+\delta (r+1)}}\left\{ \sup _{[0,t]}(1+s)^\delta \Vert e^{\kappa \phi (s,x,y)}f_x(s,x,y)\Vert _{r+1}\right\} ^{r+1}ds\\&\le C\left\{ \sup _{[0,t]}(1+s)^\delta \Vert e^{\kappa \phi (s,x,y)}f_x(s,x,y)\Vert _{r+1}\right\} ^{r+1}. \end{aligned}$$

On the other hand,

$$\begin{aligned} (r+1){{\bar{I}}}_{a2}&\le \int _{{\mathbb {R}}^2}e^{2\phi (t,x,y)}|f_x|^{r+1}d(x,y)\\&= \Vert e^{\frac{2}{r+1}\phi (t,x,y)}f_x\Vert _{r+1}^{r+1}\\&\le \Vert e^{\kappa \phi (t,x,y)}f_x\Vert _{r+1}^{r+1}\\&\le \left\{ \sup _{[0,t]}(1+s)^\delta \Vert e^{\kappa \phi (s,x,y)}f_x(s,x,y)\Vert _{r+1}\right\} ^{r+1}. \end{aligned}$$

Similarly, we also have

$$\begin{aligned} (r+1){{\bar{I}}}_{a1}&\le \int _{{\mathbb {R}}^2}e^{2\phi (0,x,y)}|f_x(0,x,y)|^{r+1}d(x,y)\\&= \Vert e^{\frac{2}{r+1}\phi (0,x,y)}f_x(0,x,y)\Vert _{r+1}^{r+1}\\&\le \Vert e^{\kappa \phi (0,x,y)}f_x(0,x,y)\Vert _{r+1}^{r+1}\\&\le \left\{ \sup _{[0,t]}(1+s)^\delta \Vert e^{\kappa \phi (0,x,y)}f_x(0,x,y)\Vert _{r+1}\right\} ^{r+1}. \end{aligned}$$

Combining the above inequalities, we obtain

$$\begin{aligned} {{\bar{I}}}_{a}\le C\left\{ \sup _{[0,t]}(1+s)^\delta \Vert e^{\kappa \phi (s,x,y)}f_x(s,x,y)\Vert _{r+1}\right\} ^{r+1}. \end{aligned}$$

(5.3)

By the same method mentioned above, we have

$$\begin{aligned} {{\bar{I}}}_b\le C\left\{ \sup _{[0,t]}(1+s)^\delta \Vert e^{\kappa \phi (s,x,y)}f_y(s,x,y)\Vert _{r+1}\right\} ^{r+1}. \end{aligned}$$

(5.4)

The inequality (5.3) and inequality (5.4) yield that

$$\begin{aligned} \Vert e^\phi U\Vert _2^2&\le CI_0^2+C\left\{ \sup _{[0,t]}(1+s)^\delta \Vert e^{\kappa \phi (s,x,y)}f_x(s,x,y)\Vert _{r+1}\right\} ^{r+1}\\&\quad +C\left\{ \sup _{[0,t]}(1+s)^\delta \Vert e^{\kappa \phi (s,x,y)}f_y(s,x,y)\Vert _{r+1}\right\} ^{r+1}. \end{aligned}$$

So, we complete the proof of Lemma 5.1. $\square$

Lemma 5.2

Let U(t, x, y) be a local classical solution to equations (2.2) on $[0,T_m)$ with the initial data $U_0$, and $\phi (t,x,y)$ be the weight function as in (4.4). Then, the following estimates hold,

$$\begin{aligned} (1+t)^{1/4}\Vert U(t,\cdot ,\cdot )\Vert _2&\le C(\Vert U_0\Vert _{1}+\Vert U_0\Vert _2)+C(\sup _{[0,t]}(1+\tau )^{\zeta }\Vert e^{\nu \phi }f_x\Vert _{2r})^r\nonumber \\&\quad +C(\sup _{[0,t]}(1+\tau )^{\zeta }\Vert e^{\nu \phi }f_y\Vert _{2r})^r, \end{aligned}$$

(5.5)

where $\zeta = \frac{3/2+\varepsilon }{r}$ and $\varepsilon , \nu$ are small positive constants.

Proof

From the representation of solution U and Lemma 3.2, we have

$$\begin{aligned} \Vert U(t,\cdot ,\cdot )\Vert _2 = \Vert {{\hat{U}}}\Vert _2&\le \Vert e^{-t{{\hat{L}}}}{{\hat{U}}}_0\Vert _2+\int _0^te^{-(t-\tau ){{\hat{L}}}}\Vert {{\hat{F}}}(\tau )\Vert _2d\tau \nonumber \\&\le C(1+t)^{-1/4}\Vert U_0\Vert _1+Ce^{-ct}\Vert U_0\Vert _2\nonumber \\&\quad +C\int _0^t(1+t-\tau )^{-1/4}\Vert F(\tau )\Vert _1d\tau \nonumber \\&\quad +C\int _0^te^{-c(t-\tau )}\Vert F(\tau )\Vert _2d\tau . \end{aligned}$$

(5.6)

Denote

$$\begin{aligned} {{\bar{J}}}_1&:= C\int _0^t(1+t-\tau )^{-1/4}\Vert F(\tau )\Vert _1d\tau \nonumber \\&\le C\int _0^t(1+t-\tau )^{-1/4}\left( \Vert f_x(\tau ,x,y)\Vert _r^r+\Vert f_y(\tau ,x,y)\Vert _r^r\right) d\tau . \end{aligned}$$

(5.7)

Combining Lemma 3.2, we obtain

$$\begin{aligned} \Vert f_x(\tau ,x,y)\Vert _r^r&= \int _{{\mathbb {R}}^2}e^{r\nu \phi (\tau ,x,y)}|f_x(\tau ,x,y)|^re^{-r\nu \phi (\tau ,x,y)}dxdy\\&\le \Vert e^{r\nu \phi (\tau ,x,y)}|f_x(\tau ,x,y)|^r\Vert _2\Vert e^{-r\nu \phi (\tau ,x,y)}\Vert _2\\&= \Vert e^{\nu \phi (\tau ,x,y)}f_x(\tau ,x,y)\Vert _{2r}^r\left( \int _{{\mathbb {R}}^2}e^{-2r\nu \phi (\tau ,x,y)}dxdy\right) ^{1/2}. \end{aligned}$$

Since that

$$\begin{aligned} \int _{{\mathbb {R}}^2}e^{-2r\nu \gamma \frac{x^2+y^2}{1+\tau }}dxdy&= 2\int _0^{2\pi }\int _0^\infty le^{-2r\nu \gamma \frac{l^2}{1+\tau }}dld\theta \\&= 4\pi \int _0^\infty le^{-2r\nu \gamma \frac{l^2}{1+\tau }}dl\\&\le C(1+\tau ). \end{aligned}$$

Thus,

$$\begin{aligned} \Vert f_x(\tau ,x,y)\Vert _r^r\le C(1+\tau )^{1/2}\Vert e^{\nu \phi (\tau ,x,y)}f_x(\tau ,x,y)\Vert _{2r}^r. \end{aligned}$$

Similarly, we have

$$\begin{aligned} \Vert f_y(\tau ,x,y)\Vert _r^r\le C(1+\tau )^{1/2}\Vert e^{\nu \phi (\tau ,x,y)}f_y(\tau ,x,y)\Vert _{2r}^r. \end{aligned}$$

Applying Lemma 1.1 and inequality (5.7), we obtain

$$\begin{aligned} {{\bar{J}}}_1&\le C\int _0^t(1+t-\tau )^{-1/4}(1+\tau )^{1/2}\left( \Vert e^{\nu \phi (\tau ,x,y)}f_x(\tau ,x,y)\Vert _{2r}^r+\Vert e^{\nu \phi (\tau ,x,y)}f_y(\tau ,x,y)\Vert _{2r}^r\right) d\tau \\&\le C\int _0^t(1+t-\tau )^{-1/4}(1+\tau )^{-1-\varepsilon }d\tau \Big (\{\sup _{[0,t]}(1+\tau )^\zeta \Vert e^{\nu \phi (\tau ,x,y)}f_x(\tau ,x,y)\Vert _{2r}\}^r\\&\quad +\{\sup _{[0,t]}(1+\tau )^\zeta \Vert e^{\nu \phi (\tau ,x,y)}f_y(\tau ,x,y)\Vert _{2r}\}^r\Big )\\&\le C(1+t)^{-1/4}\Big (\{\sup _{[0,t]}(1+\tau )^\zeta \Vert e^{\nu \phi (\tau ,x,y)}f_x(\tau ,x,y)\Vert _{2r}\}^r\\&\quad +\{\sup _{[0,t]}(1+\tau )^\zeta \Vert e^{\nu \phi (\tau ,x,y)}f_y(\tau ,x,y)\Vert _{2r}\}^r\Big ), \end{aligned}$$

where $\zeta =\frac{{3}/{2}+\varepsilon }{r}$.

Define

$$\begin{aligned} {{\bar{J}}}_2:= \int _0^te^{-c(t-\tau )}\Vert F(\tau )\Vert _2d\tau \le \int _0^te^{-c(t-\tau )}\left( \Vert f_x(\tau ,x,y)\Vert _{2r}^r+\Vert f_y(\tau ,x,y)\Vert _{2r}^r\right) d\tau .\nonumber \\ \end{aligned}$$

(5.8)

For the boundedness of $\Vert e^{-\nu \phi (\tau ,x,y)}\Vert _\infty$, we have

$$\begin{aligned} \Vert f_x(\tau ,x,y)\Vert _{2r}^r&= \Vert e^{\nu \phi (\tau ,x,y)}f_x(\tau ,x,y)e^{-\nu \phi (\tau ,x,y)}\Vert _{2r}^r\\&\le \Vert e^{-\nu \phi (\tau ,x,y)}\Vert _\infty \Vert e^{\nu \phi (\tau ,x,y)}f_x(\tau ,x,y)\Vert _{2r}^r\\&\le C(1+\tau )^\varepsilon \Vert e^{\nu \phi (\tau ,x,y)}f_x(\tau ,x,y)\Vert _{2r}^r. \end{aligned}$$

Thus,

$$\begin{aligned}&\int _0^te^{-c(t-\tau )}\Vert f_x(\tau ,x,y)\Vert _{2r}^rd\tau \\&\quad = \int _0^{t/2}e^{-c(t-\tau )}\Vert f_x(\tau ,x,y)\Vert _{2r}^rd\tau +\int _{t/2}^te^{-c(t-\tau )}\Vert f_x(\tau ,x,y)\Vert _{2r}^rd\tau \\&\quad \le Ce^{-ct/2}\int _0^{t/2}e^{-c(t/2-\tau )}(1+\tau )^\varepsilon \Vert e^{\nu \phi (\tau ,x,y)}f_x(\tau ,x,y)\Vert _{2r}^rd\tau \\&\qquad +C\int _{t/2}^te^{-c(t-\tau )}(1+\tau )^\varepsilon \Vert e^{\nu \phi (\tau ,x,y)}f_x(\tau ,x,y)\Vert _{2r}^rd\tau \\&\quad \le C \Big (e^{-ct/2}\int _0^{t/2}e^{-c(t/2-\tau )}(1+\tau )^{-3/2}d\tau +\int _{t/2}^te^{-c(t-\tau )}(1+\tau )^{-3/2}d\tau \Big )\\&\quad \quad \cdot \Big \{\sup _{[0,t]}(1+\tau )^\zeta \Vert e^{\nu \phi (\tau ,x,y)}f_x(\tau ,x,y)\Vert _{2r}\Big \}^r\\&\quad \le C(1+t)^{-1/4}\Big \{\sup _{[0,t]}(1+\tau )^\zeta \Vert e^{\nu \phi (\tau ,x,y)}f_x(\tau ,x,y)\Vert _{2r}\Big \}^r. \end{aligned}$$

Similarly, we also have

$$\begin{aligned} \int _0^te^{-c(t-\tau )}\Vert f_y(\tau ,x,y)\Vert _{2r}^rd\tau \le C(1+t)^{-1/4}\Big \{\sup _{[0,t]}(1+\tau )^\zeta \Vert e^{\nu \phi (\tau ,x,y)}f_y\Vert _{2r}\Big \}^r. \end{aligned}$$

Together with inequality (5.6), we can obtain

$$\begin{aligned} (1+t)^{1/4}\Vert U(t,\cdot ,\cdot )\Vert _2&\le C(\Vert U_0\Vert _1+\Vert U_0\Vert _2) +C\Big (\sup _{[0,t]}(1+\tau )^\zeta \Vert e^{\nu \phi }f_x\Vert _{2r}\Big )^r\\&\quad +C\Big (\sup _{[0,t]}(1+\tau )^\zeta \Vert e^{\nu \phi }f_y\Vert _{2r}\Big )^r. \end{aligned}$$

Thus, the proof is completed. $\square$

The proof of Theorem 1.1

Denote

$$\begin{aligned} W(t)=\Vert e^\phi U\Vert _2+(1+t)^{1/4}\Vert U\Vert _2. \end{aligned}$$

According to Lemma 5.1 and Lemma 5.2, we have

$$\begin{aligned} W(t)&\le CI_0+C\left( \sup _{[0,t]}(1+\tau )^\delta \Vert e^{\kappa \phi (\tau ,x,y)}f_x(\tau ,x,y)\Vert _{r+1}\right) ^{\frac{r+1}{2}}\\&\quad +C\left( \sup _{[0,t]}(1+\tau )^\delta \Vert e^{\kappa \phi (\tau ,x,y)}f_y(\tau ,x,y)\Vert _{r+1}\right) ^{\frac{r+1}{2}}\\&\quad +C(\Vert U_0\Vert _1+\Vert U_0\Vert _2)+C\left( \sup _{[0,t]}(1+\tau )^\zeta \Vert e^{\nu \phi (\tau ,x,y)}f_x(\tau ,x,y)\Vert _{2r}\right) ^r\\&\quad +C\left( \sup _{[0,t]}(1+\tau )^\zeta \Vert e^{\nu \phi (\tau ,x,y)}f_y(\tau ,x,y)\Vert _{2r}\right) ^r, \end{aligned}$$

where $\delta ,\nu >0$, $\kappa >\frac{2}{r+1}$, $\zeta =\frac{3/2+\varepsilon }{r}$. Taking $m=2r$ and applying Lemma 4.1, one has

$$\begin{aligned} \Vert e^{\nu \phi (\tau ,\cdot ,\cdot )}f_x\Vert _{2r}&\le C(1+\tau )^{\frac{1-\theta (2r)}{2}}\Big (\Vert f_{xx}\Vert _2^{1-\nu }\Vert e^{\phi (\tau ,\cdot ,\cdot )}f_{xx}\Vert _2^\nu +\Vert f_{xy}\Vert _2^{1-\nu }\Vert e^{\phi (\tau ,\cdot ,\cdot )}f_{xy}\Vert _2^{\nu }\Big )\\&\le C(1+\tau )^{\frac{1-\theta (2r)}{2}}(1+\tau )^{-\frac{1-\nu }{4}}W(\tau )^\nu \Big (\{(1+\tau )^{1/4}\Vert f_{xx}\Vert _2\}^{1-\nu }\\&\quad +\{(1+\tau )^{1/4}\Vert f_{xy}\Vert _2\}^{1-\nu }\Big )\\&\le C(1+\tau )^{\frac{1-\theta (2r)}{2}-\frac{1-\nu }{4}}W(\tau ). \end{aligned}$$

Similarly, let $m=r+1$, $\nu =\kappa$, then we have

$$\begin{aligned} \Vert e^{\kappa \phi (\tau ,\cdot ,\cdot )}f_x\Vert _{r+1}\le C(1+\tau )^{\frac{1-\theta (r+1)}{2}-\frac{1-\nu }{4}}W(\tau ). \end{aligned}$$

The similar results are also hold for $f_y$, we can get

$$\begin{aligned} W(t)&\le C(I_0+\Vert U_0\Vert _1+\Vert U_0\Vert _2) +C\left\{ \sup _{[0,t]}(1+\tau )^{\delta +\frac{1-\theta (r+1)}{2}-\frac{1-\kappa }{4}}W(\tau )\right\} ^\frac{r+1}{2}\\&\quad +C\left\{ \sup _{[0,t]}(1+\tau )^{\zeta +\frac{1-\theta (2r)}{2}-\frac{1-\nu }{4}}W(\tau )\right\} ^r. \end{aligned}$$

Taking $\kappa =\frac{2}{r+1}+\varepsilon _1$, where $\varepsilon _1>0$. Since the definition of $\theta (m)$, we choose $\nu ,\delta ,\varepsilon >0$ and $\varepsilon _1>0$ are small enough, such that

$$\begin{aligned} \delta +\frac{1-\theta (r+1)}{2}-\frac{1-\kappa }{4}=-\frac{1}{4}+\frac{3}{2(r+1)}+\frac{1}{4}\varepsilon _1+\delta <0 \end{aligned}$$

and

$$\begin{aligned} \zeta +\frac{1-\theta (2r)}{2}-\frac{1-\nu }{4} =&\left( \frac{3}{2r}-\frac{1}{4}+\frac{1}{2r}\right) +\frac{\varepsilon }{r}+\frac{1}{4}\nu \\ =&\left( \frac{2}{r}-\frac{1}{4}\right) +\frac{\varepsilon }{r}+\frac{1}{4}\nu <0. \end{aligned}$$

Then,

$$\begin{aligned} \sup _{[0,t]}W(\tau )\le C(I_0+\Vert U_0\Vert _1+\Vert U_0\Vert _2)+C\left( \sup _{[0,t]}W(\tau )\right) ^\frac{r+1}{2}+C\left( \sup _{[0,t]}W(\tau )\right) ^r.\nonumber \\ \end{aligned}$$

(5.9)

Let

$$\begin{aligned} M(t)=\sup _{[0,t]}W(\tau ), \end{aligned}$$

then we have

$$\begin{aligned} M(t)\le C(I_0+\Vert U_0\Vert _1+\Vert U_0\Vert _2)+CM(t)^{\frac{r+1}{2}}+CM(t)^r. \end{aligned}$$

If the initial data $I_0+\Vert U_0\Vert _1+\Vert U_0\Vert _2$ are small enough, then we obtain

$$\begin{aligned} M(t)\le C(I_0+\Vert U_0\Vert _1+\Vert U_0\Vert _2), \ \ \ \forall \ \ t\in [0,T_m). \end{aligned}$$

Since

$$\begin{aligned} f_x(t,x,y)=f_x(0,x,y)+\int _0^tf_{xs}(s,x,y)ds, \end{aligned}$$

and the monotone property of function $\phi$, we have

$$\begin{aligned} \Vert e^{\phi (t,\cdot ,\cdot )}f_x\Vert _2&\le \Vert e^{\phi (t,\cdot ,\cdot )}f_x(0,x,y)\Vert _2+\int _0^t\Vert e^{\phi (t,\cdot ,\cdot )}f_{xs}(s,x,y)\Vert _2ds\\&\le \Vert e^{\phi (0,\cdot ,\cdot )}f_x(0,x,y)\Vert _2+\int _0^t\Vert e^{\phi (s,\cdot ,\cdot )}f_{xs}(s,x,y)\Vert _2ds\\&\le \Vert e^{\phi (0,\cdot ,\cdot )}f_x(0,x,y)\Vert _2+C(I_0+\Vert U_0\Vert _1+\Vert U_0\Vert _2)t. \end{aligned}$$

Assume that $T_m<\infty$, then the above inequality yields that

$$\begin{aligned} \limsup _{t\rightarrow T_m}\left\{ \Vert e^\phi U\Vert _2+\Vert e^\phi f_x\Vert _2+\Vert e^\phi f_y\Vert _2\right\} <+\infty . \end{aligned}$$

This contradicts the inequality (4.10), so we can obtain $T_m=\infty$.

Thus, the proof of global existence Theorem 1.1 is finished. $\square$