Global existence and blow up of solutions for the Cauchy problem of some nonlinear wave equations

Abstract

In this paper, we study the global existence and blow up for the Cauchy problem for some hyperbolic system

$$\begin{aligned} u_{ktt}+\delta u_{kt}-\phi \Delta u_{k}+f_k(u_1,u_2)=\lambda |u_k|^{\beta -1}u_k. \quad k=1,2. \end{aligned}$$

Under certain conditions we prove the global existence of solutions by adapting the method of modified potential well in a functional setting of generalized Sobolev spaces, and we prove that the solution decays exponentially by introducing an appropriate Lyapunov function. By the concave method, we discuss the blow-up behavior of weak solution with certain conditions and give some estimates for the lifespan of solutions.

1 Introduction

In the paper, we study the following Cauchy problem

$$\begin{aligned} {\left\{ \begin{array}{ll} u_{ktt}+\delta u_{kt}-\phi \Delta u_{k}+f_k(u_1,u_2)=\lambda |u_k|^{\beta -1}u_k&{} in\ (0,T)\times R^N\\ u(0,x)=\varphi (x)\quad u_t(0,x)=\psi (x) &{}in\ R^N \end{array}\right. }\quad k=1,2.\nonumber \\ \end{aligned}$$

(1.1)

$N\ge 1, T>0$, $u=(u_1,u_2)$ is unknown, $\delta$, $\lambda$ is constant and $\delta >0$, $\lambda >0$, $\beta \ge 2$, $f_k$ is a known continuously differentiable function, $\phi (x)>0$ is a known function. We study the behavior of solutions to (1.1) in the space ${\mathcal {X}}_0\times {\mathcal {X}}_0$, where ${\mathcal {X}}_0=D^{1,2}(R^N)\times L^2_g(R^N)$. Models of this type are of interest in applications in various areas of mathematical physics (see [9, 10, 23, 31]), as well as in geophysics and ocean acoustics, where, for example, the coefficient $\phi (x)$ represents the speed of sound at the point $x\in R^N$ (see [10]), and makes impossible the treatment of the system in the classical Sobolev space setting.

In the paper, we assume that the functions $\phi (x)$ and $g: R^N\rightarrow R$ satisfy the following condition:

• (G) $\quad \phi (x)>0$, $(\phi (x))^{-1}=: g(x)\in {\mathcal {C}}^{0,\gamma }(R^N)$, $\gamma \in (0,1)$ and $g\in L^{N/2}(R^N)\cap L^\infty (R^N)$.

Karahalios and Stavrakakis [9] consider the following initial value problem:

$$\begin{aligned} {\left\{ \begin{array}{ll}u_{tt}+\delta u_t-\phi (x)\triangle u=\lambda u|u|^{\beta -1}&{} x\in R^N,\quad t\ge 0\\ u(x,0)=u_0(x), \quad u_t(x,0)=u_1(x)&{} x\in R^N \end{array}\right. } \end{aligned}$$

(1.2)

The questions of the Cauchy problem for nonlinear wave equations have been treated by many authors, see [5, 6, 9, 11, 20, 22, 25]. In general, global existence happens, when the damping terms dominate over the source terms, while blow-up appears in the opposite situation and under the assumption that the initial data is sufficiently large. In the papers [19] the problem is considered in $R^N$ and the method of modified potential well is used to construct the global solutions. In the works [18, 33] decay properties of solutions of wave equations are discussed. Recently, Levine, Park, Pucci, Serrin and G.Todorova in [12,13,14, 21, 28, 29] studied global existence and nonexistence of solutions for unbounded domain cases and nonlinear damping. In [12, 29] nonexistence occurs for all negative initial energies. In [21] nonexistence results for abstract evolution equations have been obtained.

Aliev and Yusifova [2] studied the Cauchy problem for a system of semilinear hyperbolic equations.

$$\begin{aligned} u_{ktt} -\Delta u_{k} +u_{k} +\gamma u_{kt} =f_{k} (u_1 ,\dots ,u_{m} )\quad k=1,2,\dots ,m\, , \end{aligned}$$

(1.3)

The system describes the model of interaction of various fields with single masses [4]. The absence of global solutions with positive arbitrary initial energy for systems of semilinear hyperbolic equations was investigated in [1].

Reed [22] proposed an interesting problem for the following system of equations:

$$\begin{aligned} {\left\{ \begin{array}{ll} (u_1)_{tt}-\Delta u_1+m_1^2u_1=-4\lambda (u_1 + \alpha u_2)^3-2\beta u_1u_2^2\\ (u_2)_{tt}-\Delta u_2+m_2^2u_2=-4\alpha \lambda (u_1 + \alpha u_2)^3-2\beta u_1^2u_2 \end{array}\right. } \end{aligned}$$

(1.4)

This system defines the motion of charged masses in an electro-magnetic field which was introduced by Segal [26]. Later, Jörgens [7], Makhankov [17] studied such systems to find the existence of weak solutions in a bounded domain. Further generalizations are also given in [15] by using Galerkin methods.

The presentation of this paper is as follows: In Sect. 2 we discuss some useful properties of the homogeneous Sobolev space and imbedding relations with some weighted spaces. In Sect. 3 we discuss the global solutions for (1.1). In Sect. 4 we obtain blow-up results for the solutions of the problem (1.1).

Notation: We will denote by $B_R$ the open ball of $R^N$ with center 0 and radius R. Sometimes for simplicity we use the symbols $L^p$, $1\le p\le \infty$ and $D^{1,2}$, for the spaces $L^p(R^N)$ and $D^{1,2}(R^N)$, respectively; $\Vert \cdot \Vert _p$ for the norm $\Vert \cdot \Vert _{L^p(R^N)}$.

2 Preliminary

In the section, u is a scalar function. The space for the initial conditions and the solutions of the problem is the product space ${\mathcal {X}}_0=D^{1,2}(R^N)\times L^2_g(R^N)$. The space $D^{1,2}(R^N)$ is defined as the closure of $C^\infty _0 (R^N)$ functions with respect to the energy norm $\Vert u\Vert _{D^{1,2}}^2=: \int _{R^N}|\nabla u|^2\,dx$. It is well known that

$$\begin{aligned} D^{1,2}(R^N)=\left\{ u\in L^{\frac{2N}{N-2}}(R^N):\nabla u \in (L^2(R^N))^N\right\} \end{aligned}$$

and that $D^{1,2}$ is embedded continuously in $L^{\frac{2N}{N-2}}$, there exists $k>0$ such that

$$\begin{aligned} \Vert u\Vert _{\frac{2N}{N-2}}\le k\Vert u\Vert _{D^{1,2}} \end{aligned}$$

(2.1)

We shall use the following generalized Poincaré inequality

$$\begin{aligned} \int _{R^N}|\nabla u|^2dx\ge \alpha \int _{R^N}gu^2dx \end{aligned}$$

(2.2)

for all $u\in C^\infty _0$ and $g\in L^{N/2}$, where $\alpha =: k^{-2}\Vert g\Vert ^{-1}_{N/2}$ (see [3]. It has been shown that $D^{1,2}(R^N)$ is a separable Hilbert space. The space $L^2_g(R^N)$ is defined to be the closure of $u\in C^\infty _0(R^N)$ functions with respect to the inner product

$$\begin{aligned} (u,v)_{L^2_g(R^N)}=:\int _{R^N}guv\,dx \end{aligned}$$

(2.3)

Clearly, $L^2_g(R^N)$ is a separable Hilbert space.

Lemma 1

(see [8]) Let $g\in L^{N/2}(R^N\cap L^\infty (R^N))$, then the embedding $D^{1,2}\subset L_g^2$ is compact.

Hence we are able to construct the evolution triple, which is necessary for our problem, namely

$$\begin{aligned} D^{1,2}(R^N)\subset L^2_g (R^N)\subset D^{-1,2}(R^N) \end{aligned}$$

(2.4)

where all the embeddings are compact and dense.

Lemma 2

Let $g\in L^{\frac{2N}{2N-pN+2p} }(R^N)$. Then we get the following continuous embedding $D^{1,2}(R^N)\subset L^p_g (R^N)$, for all $1\le p \le \frac{2N}{N-2}$ .

Proof

$$\begin{aligned} \begin{aligned} \int _{R^N}gu^pdx \!\le \! \left( \int _{R^N}g^adx\right) ^{\frac{1}{a}}\left( \int _{R^N}|u|^{pb}dx\right) ^{\frac{1}{b}} \le \left( \int _{R^N}g^adx\right) ^{\frac{1}{a}}\left( \int _{R^N}|\nabla u|^2dx\right) ^{\frac{p}{2}} \end{aligned} \end{aligned}$$

where $a=\frac{2N}{2N -pN + 2p}$, $b = \frac{2N}{(N-2)p}$. $\square$

Lemma 3

Let g satisfy condition (G) and $1\le q< p < p^*= \displaystyle \frac{2N}{N-2}$, then there exists $C_0 > 0$ such that the inequality

$$\begin{aligned} \Vert u\Vert _{L^p_g}\le C_0\Vert u\Vert _{L^q_g}^{1-\theta }\Vert u\Vert _{D^{1,2}}^\theta \end{aligned}$$

(2.5)

holds for all $\theta \in (0,1)$ which satisfy $\frac{1}{p}=\frac{1-\theta }{q}+ \frac{\theta }{p^*}$.

Proof

Using the interpolation inequality

$$\begin{aligned} \Vert u\Vert _{L^p_g}\le \Vert u\Vert _{L^q_g}^{1-\theta }\Vert u\Vert _{L_g^{p^*}}^\theta \end{aligned}$$

(see [24]) and (2.1). Here $C_0 = k^\theta$. $\square$

Lemma 4

Assume that $g\in L^1(R^N)\cap L^\infty (R^N)$, then the following continuous embedding $L^p_g(R^N)\subset L^q_g(R^N)$ is valid for any $1\le q\le p<\infty$.

Proof

Using Hölder inequality, we get

$$\begin{aligned} \int _{R^N}gu^q dx \le \left( \int _{R^N}(g^\sigma )^a dx\right) ^{\frac{1}{a}}\left( \int _{R^N}(g^\tau |u|^q)^b dx\right) ^{\frac{1}{b}} \end{aligned}$$

where $a = \frac{p}{p-q}$, $b = \frac{p}{q}$. Hence for $\sigma =\frac{p-q}{p}$, $\tau =\frac{q}{p}$, we obtain the embedding inequality $\Vert u\Vert _{L^q_g}\le C_*\Vert u\Vert _{L^p_g}$ , where $C_* = \Vert g\Vert _1^{\frac{p-q}{pq}}$. $\square$

Lemma 5

Assume that $1<a, b, c < \infty$, $s\in [0, c^{-1})$ and $\frac{1}{a}+ \frac{1}{b}+ \frac{1}{c}= 1$. Then for every $u\in L^a_g$ , $v\in L^b_g$, $w\in L^c_g$ and every $K > 0$ we have the inequality

$$\begin{aligned} \left| \int _{R^N}guvw dx\right| \le K^{s-\frac{1}{c}}\Vert w\Vert _{L^c_g} (\Vert u\Vert ^a_{L^a_g} + \Vert v\Vert ^b_{L^b_g} + K)^{1-s} \end{aligned}$$

Proof

The proof is a direct application of Levine et al. [12].

In order to deal with (1.1), We consider the equation

$$\begin{aligned} -\phi (x)\Delta u(x) = \eta (x), \quad x\in R^N \end{aligned}$$

(2.6)

Since for every u, v in $C_0^\infty (R^N)$

$$\begin{aligned} (-\phi u, v)_{L^2_g}= \int _{R^N} \nabla u \nabla v dx \end{aligned}$$

(2.7)

and $L^2_g(R^N)$ is defined as the closure of $C_0^\infty (R^N)$ with respect to the inner product (2.3), we may consider equation (2.6) as an operator equation

$$\begin{aligned} A_0u =\eta ,\quad A_0: D(A_0)\subseteq L^2_g (R^N)\longrightarrow L^2_g (R^N),\ for\ any \ \eta \in L^2_g (R^N) \end{aligned}$$

(2.8)

Relation (2.7) implies that the operator $A_0 = -\phi \Delta$ with domain of definition $D(A_0) = C_0^\infty (R^N)$ is symmetric. Let us note that the operator $A_0$ is not symmetric in the standard Lebesgue space $L^2(R^N)$, because of the appearance of $\phi (x)$. From (2.2) and (2.7), we have

$$\begin{aligned} (A_0u, u)_{L^2_g}\ge \alpha \Vert u\Vert ^2_{L^2_g} , \ for\ all \ u\in D(A_0) \end{aligned}$$

(2.9)

From (2.7) and (2.9), we conclude that $A_0$ is a symmetric, strongly monotone operator on $L^2_g (R^N)$. Hence, the Friedrichs extension theorem (see [32]) is applicable. The energy scalar product given by (2.7) is

$$\begin{aligned} (u, v)_E = \int _{R^N} \nabla u \nabla vdx \end{aligned}$$

and the energy space is the completion of $D(A_0)$ with respect to $(u, v)_E$. It is obvious that the energy space $X_E$ is the homogeneous Sobolev space $D^{1,2}(R^N)$. The energy extension $A_E = -\phi \Delta$ of $A_0$, namely

$$\begin{aligned} -\phi \Delta : D^{1,2}(R^N)\longrightarrow D^{-1,2}(R^N) \end{aligned}$$

(2.10)

is defined to be the duality mapping of $D^{1,2}(R^N)$. For every $\eta \in D^{-1,2}(R^N)$ the Eq. (2.6) has a unique solution. Define D(A) to be the set of all solutions of the Eq. (2.6), for arbitrary $\eta \in L^2_g (R^N)$. The Friedrichs extension A of $A_0$ is the restriction of the energy extension $A_E$ to the set D(A). The operator A is self-adjoint and therefore graph-closed. Its domain, D(A), is a Hilbert space with respect to the graph scalar product

$$\begin{aligned} (u, v)_{D(A)} = (u, v)_{L^2_g} + (Au, Av)_{L^2_g} , \quad \mathrm{for\ all}\quad u, v\in D(A). \end{aligned}$$

The norm induced by the scalar product $(u, v)_{D(A)}$ is

$$\begin{aligned} \Vert u\Vert _{D(A)} = \left\{ \int _{R^N} g|u|^2 dx + \int _{R^N} \phi |\nabla u|^2\,dx \right\} ^\frac{1}{2} \end{aligned}$$

which is equivalent to the norm

$$\begin{aligned} \Vert Au\Vert _{L^2_g} =\left\{ \int _{R^N} \phi |\nabla u|^2 dx \right\} ^\frac{1}{2} \end{aligned}$$

A consequence of the compactness of the embeddings in (2.4) is that for the eigenvalue problem .

$$\begin{aligned} -\phi (x)\Delta u = \mu u,\ x \in R^N \end{aligned}$$

(2.11)

there exists a complete system of eigensolutions $\{w_n,\mu _n\}$ with the following properties

$$\begin{aligned} {\left\{ \begin{array}{ll} -\phi \Delta w_j=\mu _jw_j,\ j=1,2,\cdots ,\ w_j\in D^{1,2}(R^N)\\ 0<\mu _1\le \mu _2\le \cdots ,\ \mu _j\longrightarrow \infty , \ as\ j\longrightarrow \infty . \end{array}\right. } \end{aligned}$$

(2.12)

It can be shown, as in [3], that every solution of (2.11) is such that

$$\begin{aligned} u(x)\longrightarrow 0, \ as |x|\longrightarrow \infty \end{aligned}$$

(2.13)

uniformly with respect to x. Finally, we give the definition of weak solutions for the problem (1.1). $\square$

Definition 1

A weak solution of (1.1) is a function $u(x,t)=(u_1(x,t),u_2(x,t))$ such that

• (i) $u_k\!\in \! L^2[0,T;D^{1,2}(R^N)]$, $u_{kt}\!\in \! L^2[0,T; L^2_g(R^N)]$, $u_{ktt}\!\in \! L^2[0, T;D^{-1,2}(R^N)]$,

• (ii) for all $v\in C^\infty _0([0,T]\times R^N)$, u satisfies the generalized formula

  $$\begin{aligned}&\int ^T_0(u_{ktt}(\tau ), v(\tau ))_{L^2_g}d\tau +\delta \int ^T_0(u_{kt}(\tau ), v(\tau ))_{L^2_g}d\tau \nonumber \\&\quad +\,\int ^T_0\int _{R^N}\nabla u_k(\tau )\nabla v(\tau )dxd\tau -\lambda \int ^T_0(f(u(\tau )), v(\tau ))_{L^2_g}d\tau =0,\qquad \quad \end{aligned}$$

  (2.14)

  where $f(s)=|s|^{\beta -1}s$, and

• (iii) u satisfies the initial conditions

  $$\begin{aligned} u(x,0)=\varphi (x)\in D^{1,2}(R^N)\times D^{1,2}(R^N),\quad u_t(x,0)=\psi (x)\in L^2_g(R^N)\times L^2_g(R^N). \end{aligned}$$

3 Global existence

Now we make the following assumptions.

• (A1)
  $$\begin{aligned} \varphi (x)= & {} (\varphi _1(x),\varphi _2(x))\in D^{1,2}(R^N)\times D^{1,2}(R^N),\\ \psi (x)= & {} (\psi _1(x),\psi _2(x))\in L^2_g(R^N)\times L^2_g(R^N) \end{aligned}$$

• (A2) $f_k:R^2\rightarrow R$ is continuously differentiable such that for each $u=(u_1,u_2)\in {\mathcal {X}}_0\times {\mathcal {X}}_0$, $u_kf_k(u)\in L^1(\Omega ),k= 1,2$, $F(u)\in L^1(R^N)$, where $F(u) =\displaystyle \int _0^{u_1}f_1(s,u_2)ds + \int _0^{u_2}f_2(0,s)ds$.

• (A3) $f_k:{\mathcal {X}}_0\times {\mathcal {X}}_0\rightarrow L^2(R^N), k=1,2$, satisfies a local Lipschitz condition, i.e., for any $\delta >0$, there exists a positive constant $C(\delta )$ such that

  $\Vert f_k(u)-f_k(v)\Vert \le C(\delta )\Vert u-v\Vert _{{\mathcal {X}}_0\times {\mathcal {X}}_0}$ for $u,v\in {\mathcal {X}}_0\times {\mathcal {X}}_0$ with

  $\Vert u\Vert _{{\mathcal {X}}_0\times {\mathcal {X}}_0}\le \delta$, $\Vert v\Vert _{H_0^1\times H_0^1}\le \delta$.

• (A4) $\displaystyle \frac{\partial f_1}{\partial u_2}=\frac{\partial f_2}{\partial u_1}$.

• (A5) $u_1f_1+u_2f_2\ge F(u)\ge 0,\forall u_1,u_2\in R$.

• (A6) There exists a positive constant $n_0\ge {1/p}$ such that

  $\displaystyle \max \{u_1f_1+u_2f_2,0\}\le \frac{1}{n_0}F(u),\forall u_1,u_2 \in R$.

We may consider the $C^1$ functionals $I_k,J_k:H_0^1(R^N)\rightarrow R$ defined by

$$\begin{aligned}&I_k(u_k)=:\Vert u_k\Vert ^2_{D^{1,2}}-\lambda \Vert u_k\Vert _{L_g^{\beta +1}}^{\beta +1} \end{aligned}$$

(3.1)

$$\begin{aligned}&J_k(u_k)=:\frac{1}{2}\Vert u_k\Vert ^2_{D^{1,2}}-\frac{\lambda }{\beta +1} \int _{R^N}g(x)|u(t)|^{\beta +1}dx\\&N_k=\{u_k\in H_0^1(R^N)\backslash \{0\}\big |\big . I_k(u_k)=0\}\nonumber \end{aligned}$$

(3.2)

The mountain pass value of $J_k$ (also known as potential well depth) is defined as

$$\begin{aligned} d_k=\inf \limits _{u\in H_0^1(\Omega )\backslash \{0\}}\max \limits _{\lambda \ge 0}J_k(\lambda u) \end{aligned}$$

It is readily seen (see [30]) that the mountain pass level $d_k$ may also be characterized as

$$\begin{aligned} d_k=\inf _{u\in N_k}J_k(u) \end{aligned}$$

Finally, we consider the energy functional E(t) The energy of the problem (1.1) is defined as

$$\begin{aligned} E(t)= & {} \sum _{k=1}^2\left( \frac{1}{2}\Vert u_{kt}(t)\Vert _{L_g^2}^2+\frac{1}{2}\Vert u_k\Vert _{D^{1,2}}^2 -\frac{\lambda }{\beta +1}\int _{R^N}g|u_k|^{\beta +1}dx\right) +\int _{R^N}gF(u)dx\nonumber \\&E(t)+\sum _{k=1}^2\int _s^t\delta \Vert u_{kt}(\tau )\Vert _{L_g^2}^2d\tau =E(s),\quad 0\le s<t<T_{max}\nonumber \\ \end{aligned}$$

(3.3)

F(u) will be defined in the following (A3).

From equation (1.1), we can have

$$\begin{aligned} E(t)=E(0)-\sum _{k=1}^2\delta \int ^t_0\Vert u_{kt}(\tau )\Vert ^2_{L^2_g}d\tau \end{aligned}$$

(3.4)

Under certain assumptions on the initial data, solutions exist globally in the energy space ${\mathcal {X}}_0$. In addition to the principal condition (G) in the introduction, we shall use the following additional hypotheses for the function g and the nonlinearity exponent $\beta$.

• (G1) $g\in L^1(R^N)$ and $1<\beta \le \frac{N}{N-2}$ , for all $N\ge 3$.

• (G2) $N\ge 3$ and $\frac{N+2}{N}\le \beta \le \frac{N}{N-2}$.

• (G3) $N=3,4$ and $\frac{N+4}{N}\le \beta \le \frac{N}{N-2}$.

Let us note that since $g\in L^{N/2}(R^N)\cap L^\infty (R^N)$ by hypothesis (G), then any g satisfying hypothesis (G1) belongs to all spaces $L^p(R^N)$, for $p\in [1,+\infty )$.

Proposition 1

Let the assumptions (A1)-(A5) be fulfilled. Then problem (1.1) admits a unique weak solution $(u_1,u_2)$ defined on $[0, T_{max})$, and at least one of the following statements is valid:

• (1) $T_{max}=\infty$;

• (2) $T_{max}<\infty$, and

  $$\begin{aligned} \lim \limits _{t\rightarrow T_{max}}\sum _{k=1}^2\Vert u_{kt}(t)\Vert ^2_2+\Vert \nabla u_k\Vert _2^2=\infty \\ E(t)+ \sum _{k=1}^2\int _0^t\Vert u_{kt}(\tau )\Vert _*^2d\tau =E(0) \end{aligned}$$

Lemma 6

Let g, $\beta$, N satisfy conditions (G1) or (G2). Suppose that the constants $\delta >0$, $\lambda <\infty$ and the initial conditions

$$\begin{aligned} u_0\in D^{1,2}(R^N) \quad \mathrm{{and}} \quad u_1\in L^2_g(R^N). \end{aligned}$$

(3.5)

are given. Then for sufficiently small $T>0$ the problem (1.1) admits a unique (weak) solution such that

$$\begin{aligned} u\in C[0,T;D^{1,2}(R^N)] \quad \mathrm{{and}} \quad u_t\in C[0,T;L^2_g(R^N)]. \end{aligned}$$

(3.6)

Proof

(a) Local Existence of the Restricted Problem on $B_R$

First we prove an existence result for the problem

$$\begin{aligned} {\left\{ \begin{array}{ll}u_{ktt}+\delta u_{kt}-\phi \Delta u_{k}+f_k(u)=\lambda |u_k|^{\beta -1}u_k&{} in\ (0,T)\times B_R\\ u(0,x)=\varphi (x)\qquad u_t(0,x)=\psi (x) &{}in\ B_R\\ u(t,x)=0&{} on\ (0,T)\times \partial B_R \end{array}\right. }k=1,2.\qquad \end{aligned}$$

(3.7)

where $\varphi _k\in D^{1,2}(B_R)$ and $\psi _k\in L^2_g(B_R)$. Let $Z_k=\{z_k, z_{kt}\}\in C[0,T; \chi _0(B_R)]$ be given. In order to obtain solutions for (3.7) we first consider the following non-homogeneous problem

$$\begin{aligned} {\left\{ \begin{array}{ll}u_{ktt}+\delta u_{kt}-\phi \Delta u_{k}=\lambda |Z_k|^{\beta -1}Z_k-f_k(Z)&{} in\ (0,T)\times B_R\\ u(0,x)=\varphi (x)\qquad u_t(0,x)=\psi (x) &{}in\ B_R\\ u(t,x)=0&{} on\ (0,T)\times \partial B_R \end{array}\right. }\quad k=1,2.\qquad \end{aligned}$$

(3.8)

where $Z=(Z_1,Z_2)$ , $\varphi _k\in D^{1,2}(B_R)$ and $\psi _k\in L^2_g (B_R)$. Existence of a unique (weak) solution for the problem (3.8) can be obtained by using Faedo–Galerkin approximations (see [16]).

For $Z_k\in C[0,T; \chi _0(B_R)]$ we define the mapping

$$\begin{aligned} {\mathbb {T}} : \bigg (C[0,T; \chi _0(B_R)]\bigg )^2\longrightarrow \bigg (C[0,T; \chi _0(B_R)]\bigg )^2 \end{aligned}$$

by $U = {\mathbb {T}} (Z)$, where $U =(u_1, u_2)$ is the unique solution of equation (3.8). It is clear that the map ${\mathbb {T}}$ is well defined. Next, we show that ${\mathbb {T}}$ maps the ball $B_M$ to itself, where

$$\begin{aligned} B_M =: \left\{ \Psi \in \chi _{0,T} : \sup _{0\le t\le T} \Vert \Psi (. t)\Vert \le M\right\} \end{aligned}$$

and the space $\chi _{0,T}$ is defined by

$$\begin{aligned} \chi _{0,T} =: \{ \Psi \in C[0, T; \chi _0(B_R)] : \Psi (0,\cdot ) = {\varphi _k,\psi _k, \}\in \chi _0(B_R)} \end{aligned}$$

For $Z_k\in B_M$, we multiply Eq. (3.8) by $gu_t$ and integrate with respect to time and space on the set $(0, t)\times B_R$, for some $t\in (0,T]$, to obtain

$$\begin{aligned}&\frac{1}{2}\Vert u_{kt}(.,t)\Vert _{L_g^2}^2+\frac{1}{2}\Vert u_k(.,t)\Vert _{D^{1,2}}^2 -\frac{1}{2}\Vert u_{kt}(.,0)\Vert _{L_g^2}^2-\frac{1}{2}\Vert u_k(.,0)\Vert _{D^{1,2}}^2\nonumber \\&\quad +\,\delta \int _0^t\Vert u_{kt}(., s)\Vert ^2_{L^2_g}ds +\int _0^t\int _{B_R}gf_k(Z)u_{kt}\le \lambda \int _0^t\int _{B_R}g|z_k|^\beta u_{kt} \end{aligned}$$

(3.9)

The positivity of the quantity $\delta \int _0^t\Vert u_{kt}(.,s)\Vert ^2_{L^2_g}ds$ implies that

$$\begin{aligned}&\frac{1}{2}\Vert u_{kt}(\cdot ,t)\Vert _{L_g^2}^2+\frac{1}{2}\Vert u_k(\cdot ,t)\Vert _{D^{1,2}}^2 -\frac{1}{2}\Vert u_{kt}(\cdot ,0)\Vert _{L_g^2}^2-\frac{1}{2}\Vert u_k(\cdot ,0)\Vert _{D^{1,2}}^2\nonumber \\&\quad +\,\int _0^t\int _{B_R}gf_k(Z)u_{kt}\le \lambda \int _0^t\int _{B_R}g|z_k|^\beta u_{kt} \end{aligned}$$

(3.10)

and

$$\begin{aligned}&\int _0^t\int _{B_R}gf_k(Z)u_{kt}\le \int _0^t\Vert f_k\Vert _{\chi _0{B_R}}\Vert u_{kt}\Vert _{\chi _0{B_R}} \le CMT+\int _0^t\Vert u_{kt}\Vert ^2_{\chi _0{B_R}}\\&\int _0^t\int _{B_R}g|z_k|^\beta u_{kt}\le \int _0^t\Vert z_k\Vert ^\beta _{\chi _0{B_R}}\Vert u_{kt}\Vert _{\chi _0{B_R}} \le CM^\beta T+\int _0^t\Vert u_{kt}\Vert ^2_{\chi _0{B_R}} \end{aligned}$$

We use the assumption on Z and relation (3.10) to obtain

$$\begin{aligned} \Vert U(\cdot , t)\Vert _{C[0,T ; \chi _0(B_R)]}\le C\Vert U(\cdot , 0)\Vert _{C[0,T ; \chi _0(B_R)]} + CM^\beta T \end{aligned}$$

Choosing T sufficiently small and M sufficiently large, depending on the norm of the initial data, we have

$$\begin{aligned} \sup _{0\le t\le T}\Vert U(., t)\Vert _{C[0,T ; \chi _0(B_R)]}\le M,\ i.e.\ U\in B_M \end{aligned}$$

The next step is to show that ${\mathbb {T}}$ is a contraction. Let $Z, Z^*\in \chi _{0,T}$ such that $U={\mathbb {T}}(Z)$, $U^*={\mathbb {T}} (Z^*)$ and consider the difference $W =: U-U^*$, which satisfies the equation

$$\begin{aligned} w_{ktt}\!+\!\delta w_{kt}\!-\!\phi \Delta w_{k}\!=\!\lambda (|Z_k|^{\beta -1}Z_k\!-\!|Z_k^*|^{\beta -1}Z_k^*)\!-\!(f_k(Z)\!-\!f_k(Z^*)), \quad (x, t)\!\in \! B_R\times (0, T)\nonumber \\ \end{aligned}$$

(3.11)

Following the procedure above, for the right hand side of equation (3.11), we get the estimates

$$\begin{aligned}&\left| \int _{B_R}g(x)(|Z_k|^{\beta -1}Z_k-|Z_k^*|^{\beta -1}Z_k^*)(u_{kt}-u_{kt}^*)\right| \nonumber \\&\quad \le C\int _{B_R}|g(x)||Z_k-Z_k^*|(|Z_k|^{\beta -1}+|Z_k^*|^{\beta -1})|u_{kt}-u_{kt}^*|dx\nonumber \\&\quad \le C\Vert g\Vert _\infty ^{\frac{N-2}{2N}}\Vert Z_k-Z_k^*\Vert _{\frac{2N}{N-2}}\Vert u_{kt}-u_{kt}^*\Vert _{L^2_g (B_R)}\nonumber \\&\times \left\{ \Vert Z_k\Vert _{L_g^{N(\beta -1)}(B_R)}^{\beta -1}+\Vert Z_k^*\Vert _{L_g^{N(\beta -1)}(B_R)}^{\beta -1}\right\} \nonumber \\&\quad \le C\Vert Z_k-Z_k^*\Vert _{D^{1,2}}\Vert u_{kt}-u_{kt}^*\Vert _{L^2_g (B_R)} \left\{ \Vert Z_k\Vert _{D^{1,2}(B_R)}^{\beta -1}+\Vert Z_k^*\Vert _{D^{1,2}(B_R)}^{\beta -1}\right\} \end{aligned}$$

(3.12)

$$\begin{aligned}&\int _{B_R}(f_k(Z)-f_k(Z^*))g(x)(u_{kt}-u_{kt}^*)\le C\Vert Z_k-Z_k^*\Vert _{D^{1,2}}\Vert u_{kt}-u_{kt}^*\Vert _{L^2_g (B_R)}\nonumber \\ \end{aligned}$$

(3.13)

From relations (3.11)–(3.13), we have

$$\begin{aligned} \begin{aligned}&\Vert W(\cdot , t)\Vert ^2_{\chi _0(B_R)}\le C(\lambda )\int _0^t \left( \Vert Z_k\Vert _{\chi _0(B_R)}^{\beta -1}+\Vert Z_k^*\Vert _{\chi _0(B_R)}^{\beta -1}\right) \Vert W(., s)\Vert _{\chi _0}\\&\times \Vert Z(\cdot , s)-Z^*(\cdot , s)\Vert _{\chi _0(B_R)}ds \end{aligned} \end{aligned}$$

which is equivalent to the inequality

$$\begin{aligned} \Vert {\mathbb {T}}(Z)-{\mathbb {T}}(Z^*)\Vert _{C[0,T ;\chi _0(B_R)]}\le 2C(\lambda ) M^{\beta -1}T\Vert Z-Z^*\Vert _{C[0,T ;\chi _0(B_R)]} \end{aligned}$$

For $T < C^{-1}(\lambda ) M^{1-\beta }$ the map ${\mathbb {T}}$ is a contraction. Then the result of existence for (3.7) is a direct consequence of the contraction mapping theorem.

(b) Extension of Solutions to $R^N$. For $R>R_0$, $R\in N$, with ${\varphi _k, \psi _k}\in C_0^\infty (B_R)\times C_0^\infty (B_R)$ such that $supp(\varphi _k)\subset B_{R_0}$ and $supp(\psi _k)\subset B_{R_0}$ , we consider the approximating problem

$$\begin{aligned} {\left\{ \begin{array}{ll}u_{ktt}^R+\delta u_{kt}^R-\phi \Delta u_{k}^R+f_k(u^R)=\lambda |u_k^R|^{\beta -1}u_k^R&{} in\ (0,T)\times B_R\\ u^R(0,x)=\varphi (x)\qquad u_t^R(0,x)=\psi (x) &{}in\ B_R\\ u^R(t,x)=0&{} on\ (0,T)\times \partial B_R \end{array}\right. }k=1,2.\nonumber \\ \end{aligned}$$

(3.14)

The existence result in (a) holds for (3.14). We get that $u_k^R$ is bounded in $C[0, T; D^{1,2}(B_R)]$ and $u^R_{kt}$ is bounded in $C[0, T; L^2_g (B_R)]$, independently of R. Since, for any Banach space X, the following continuous embedding $C[0, T; X] \subset L^p[0, T; X]$ is valid, for all $1\le p <\infty$, we have $u_k^R$, $u^R_{kt}$ remain bounded in $L^2[0, T; D^{1,2}(B_R)]$ and in $L^2[0, T; L^2_g (B_R)]$, respectively. We extend $u_k^R$, as

$$\begin{aligned} {\widetilde{u}}_k^R(x, t) =: {\left\{ \begin{array}{ll} u_k^R(x, t) &{}if\ |x|\le R\\ 0 &{}otherwise \end{array}\right. } \end{aligned}$$

So that ${\widetilde{u}}_k^R$, ${\widetilde{u}}_{kt}^R$ remain bounded in the above spaces with $B_R$ replaced by $R^N$. Using the assumptions on $\beta$, we may easily check that $f(u^R)$ is bounded in $L^2[0, T; L^2_g (R^N)]$. From the relations (2.10) and (3.14) we obtain (as in [16], Remark 8.2, p. 265), that $u^R{ktt}$ is bounded in $L^2[0, T; D^{-1,2}(B_R)]$. Lemma 2.1 applied to [27] implies that ${\widetilde{u}}_k^R$ is relatively compact in $C[0, T; L^2_g (R^N)]$. Therefore we get

$$\begin{aligned} {\widetilde{u}}_k^R\longrightarrow {\widetilde{u}}_k \ in\ L^2[0, T; L^2_g(R^N)] \end{aligned}$$

Hence we may extract a subsequence of ${\widetilde{u}}_k^R$, denoted by ${\widetilde{u}}_k^{R_m}$, such that

$$\begin{aligned} \begin{aligned}&{\widetilde{u}}_k^{R_m}\longrightarrow {\widetilde{u}}_k&in\&L^2[0,T;D^{1,2}(R^N)]\\&{\widetilde{u}}_{kt}^{R_m}\longrightarrow {\widetilde{u}}_{kt}&in\&L^2[0,T;L^2_g(R^N)]\\&{\widetilde{u}}_{ktt}^{R_m}\longrightarrow {\widetilde{u}}_{ktt}&in\&L^2[0,T;D^{-1,2}(R^N)]\\&f_k({\widetilde{u}}^{R_m})\longrightarrow f_k({\widetilde{u}})&in\&L^2[0,T;L^2_g(R^N)] \end{aligned} \end{aligned}$$

(3.15)

Following the arguments in [8] we may see that ${\widetilde{u}}$ defines a unique weak solution of (1.1) with initial data satisfying (3.5). $\square$

Definition 2

Let $T_{max}=\sup \{T>0:u=u(x,t)$ exists on $[0,T)\}$. If $T_{max}<\infty$, we say that the solution to (1.1) blows up and that $T_{max}$ is the blow up time. If $T_{max}=\infty$, we say that the solution is global.

To obtain global existence, we adapt the method of modified potential well, as developed by Payne and Sattinger [20] and generalized to all of $R^N$ by Nakao and Ono [19]. To this end we consider the potential well

$$\begin{aligned} W_k=: Int \left\{ u_k\in D^{1,2}(R^N) : I_k(u_k)=:\Vert u_k\Vert ^2_{D^{1,2}}-\lambda \Vert u_k\Vert _{L_g^{\beta +1}}^{\beta +1}\ge 0\right\} \end{aligned}$$

where IntB denotes the interior of set B. It is easily seen that 0 is in W. Indeed, from Lemma 3, the Poincaré inequality (2.2) and hypothesis (G1) we have

$$\begin{aligned} \begin{aligned} \Vert u_k\Vert _{L_g^{\beta +1}}^{\beta +1}&\le C\Vert u_k\Vert _{L_g^2}^{(1-\theta )(\beta +1)}\Vert u_k\Vert _{D^{1,2}}^{\theta (\beta +1)}\\&\le C\Vert u_k\Vert _{L_g^2}^{(1-\theta )(\beta +1)}\Vert u_k\Vert _{D^{1,2}}^{\theta (\beta +1)-2}\Vert u_k\Vert ^2_{D^{1,2}}\\&\le \frac{C_0}{\alpha }\Vert u_k\Vert _{D^{1,2}}^{\beta -1}\Vert u_k\Vert ^2_{D^{1,2}} \end{aligned} \end{aligned}$$

Therefore, for any $\lambda \in R^+$ we obtain

$$\begin{aligned} I_k(u_k)\ge \left( 1-\frac{\lambda C_0}{\alpha }\Vert u_k\Vert _{D^{1,2}}^{\beta -1}\right) \Vert u_k\Vert ^2_{D^{1,2}} \end{aligned}$$

It is obvious that, if $\Vert u_k\Vert _{D^{1,2}}$ is sufficiently small, then $I_k(u_k)\ge 0$ and 0 is in W. By the definition of W we have that

$$\begin{aligned} J_k(u_k)\ge \frac{\beta -1}{2(\beta +1)}\Vert u_k\Vert ^2_{D^{1,2}}\quad for \ every \ u_k\in W \end{aligned}$$

(3.16)

Multiply equation (1.1) by $gu_{kt}$ and integrate over $R^N$ to obtain

$$\begin{aligned} \begin{aligned}&\sum _{k=1}^2\frac{1}{2}\frac{d}{dt}\Vert u_{kt}(t)\Vert _{L_g^2}^2+ \sum _{k=1}^2\delta \Vert u_{kt}(\tau )\Vert _{L_g^2}^2+ \sum _{k=1}^2\frac{1}{2}\frac{d}{dt}\Vert u_k\Vert _{D^{1,2}}^2\\&\qquad -\,\sum _{k=1}^2\frac{\lambda }{\beta +1}\frac{d}{dt}\Vert u_k\Vert _{L_g^{\beta +1}}^{\beta +1} +\frac{d}{dt}\int _{R^N}gF(u)dx=0 \end{aligned} \end{aligned}$$

The energy of the problem is defined as

$$\begin{aligned} E(t)\!=\!\sum _{k=1}^2\left( \frac{1}{2}\Vert u_{kt}(t)\Vert _{L_g^2}^2+\frac{1}{2}\Vert u_k\Vert _{D^{1,2}}^2 -\frac{\lambda }{\beta +1}\int _{R^N}g|u_k|^{\beta +1}dx\right) \!+\!\int _{R^N}gF(u)dx \end{aligned}$$

$$\begin{aligned} E(t)+\sum _{k=1}^2\int _s^t\delta \Vert u_{kt}(\tau )\Vert _{L_g^2}^2d\tau =E(s),\qquad 0\le s<t<T_{max} \end{aligned}$$

(3.17)

Let us note that $E(t)\ge 0$ if $u_k\in {\bar{W}}$ and $u_k\notin {\bar{W}}$ if $E(t)<0$. Lemma 2 and Proposition 1 imply that the functional E(t) is well defined. From (3.17), it is easy to obtain that $E'(t)= -\sum _{k=1}^2\delta \Vert u_{kt}(t)\Vert _{L_g^2}^2\le 0$. Therefore E(t) is a nonincreasing function of t, i.e.,

$$\begin{aligned} E(t)\le E(0)\quad for \ every\ t\in [0, T). \end{aligned}$$

(3.18)

The global existence result is given in the following theorem.

Theorem 1

Let condition (G3) be satisfied and $u_0\in W$. Assume that the initial data satisfy (3.5) and they are sufficiently small in the sense

$$\begin{aligned} E(0)< \left( \frac{1}{C_0\lambda \mu _0^{p_1}}\right) ^{\frac{1}{p_2}} \end{aligned}$$

(3.19)

where $p_1=\frac{2(\beta +1)-N(\beta -1)}{2}$, $p_2=\frac{N\beta -N-4}{4}$ , then the (weak) solution of (1) is such that

$$\begin{aligned} u\in C[0,\infty ;D^{1,2}(R^N)] \quad \mathrm{{and}} \quad u_t\in C[0,\infty ;L^2_g(R^N)]. \end{aligned}$$

(3.20)

And there exist two positive constants ${\widehat{C}}$ and $\xi$, independent of t such that:

$$\begin{aligned} 0<E(t)\le {\widehat{C}}e^{-\xi t}\qquad \mathrm{{for\ all}}\quad t\ge 0 \end{aligned}$$

(3.21)

Proof

We shall show that the local solution given by Proposition 1, is in the modified potential well W, as long as it exists. We argue by contradiction. Assume that there exists some time $T^* > 0$ , such that $u_k(t)\in W$, where $0\le t<T$ and $u_k(T^*)\in \partial W$. Then $I_k(u_k(T^*)) = 0$ and $u(T^*)\ne 0$. We multiply equation (1.1) by $gu_k$ and integrate over $R^N$, to get the equation

$$\begin{aligned}&\frac{d}{dt}(u_k,u_{kt})_{L^2_g}-\Vert u_{kt}\Vert ^2_{L^2_g}+\frac{\delta }{2}\frac{d}{dt} \Vert u_k\Vert ^2_{L^2_g} + \Vert u_k\Vert ^2_{D^{1,2}}-\lambda \int _{R^N} g|u_k|^{\beta +1}dx\nonumber \\&\quad +\int _{R^N}gu_kf_kdx= 0 \end{aligned}$$

(3.22)

We integrate over [0, t], for some $t\in [0, T)$, to get the inequality

$$\begin{aligned} \delta \Vert u_k(t)\Vert ^2_{L^2_g}\le & {} \delta \Vert u_k(0)\Vert ^2_{L^2_g}\nonumber \\&+\,2|(u_k(t),u_{kt}(t))_{L^2_g}|+ 2(\varphi _k, \psi _k)_{L^2_g}+ 2\int _0^t\Vert u_{kt}(s)\Vert ^2_{L^2_g}ds\nonumber \\\le & {} \delta \Vert u_k(0)\Vert ^2_{L^2_g} +2\left( \frac{\delta }{4}\Vert u_k(t)\Vert _{L^2_g}^2+\frac{1}{\delta }\Vert u_{kt}(t)\Vert _{L^2_g}^2\right) \nonumber \\&+\, 2(\varphi _k, \psi _k)_{L^2_g}+ 2\int _0^t\Vert u_{kt}(s)\Vert ^2_{L^2_g}ds \end{aligned}$$

(3.23)

by applying Young inequality for $\epsilon = \delta /2$. Since $u_k(t)$ is in W, we have from (3.16) and (3.17)

$$\begin{aligned} \frac{1}{2}\Vert u_{kt}(s)\Vert ^2_{L^2_g}+ \delta \int _0^t\Vert u_{kt}(s)\Vert ^2_{L^2_g}ds\le E(0) \end{aligned}$$

(3.24)

Then from (3.23) and (3.24) we get the estimate

$$\begin{aligned} \Vert u_{kt}(s)\Vert ^2_{L^2_g}\le \frac{2}{\delta }\left\{ \delta \Vert u_k(0)\Vert ^2_{L^2_g}+ 2(\varphi _k, \psi _k)_{L^2_g} + \frac{4}{\delta }E(0)\right\} =:\mu _0^2 \end{aligned}$$

(3.25)

Using Lemma 3 and relation (3.25) we obtain the inequality

$$\begin{aligned} \Vert u_k(t)\Vert _{L_g^{\beta +1}}^{\beta +1}\le & {} C_0\mu _0^{(\beta +1)(1-\theta )}\Vert u_k(t)\Vert _{D^{1,2}}^{(\beta +1)\theta }\nonumber \\\le & {} C_0\mu _0^{(\beta +1)(1-\theta )}J_k(u_k(t))^{\frac{(\beta +1)\theta }{2}-1}\Vert u_k(t)\Vert _{D^{1,2}}^2\nonumber \\\le & {} C_0\mu _0^{(\beta +1)(1-\theta )}E(0)^{\frac{(\beta +1)\theta }{2}-1}\Vert u_k(t)\Vert _{D^{1,2}}^2 \end{aligned}$$

(3.26)

where $\theta =\frac{N(\beta -1)}{2(\beta +1)}$ according to Lemma 3, $p_1=(\beta +1)(1-\theta )$, $p_2=\frac{(\beta +1)\theta }{2}-1$ and $p_1$, $p_2$ are positive by hypothesis (G3). Setting $\delta _1 = C_0\mu _0^{p_2}E(0)^{p_1}$ then inequality (3.26) implies, for $t = T^*$, that

$$\begin{aligned} I_k(u_k(T^*))\ge (1-\lambda \delta _1)\Vert u_k(T^*)\Vert _{D^{1,2}}^2> 0 \end{aligned}$$

under the assumption that $\lambda < \displaystyle \frac{1}{\delta _1}$ [which is equivalent to the relation (3.19)] and the contradiction is achieved.

From (3.2) and (3.3), we have

$$\begin{aligned} E(t)=\sum \left( \frac{1}{2}\Vert u_{kt}(t)\Vert ^2_{L^2_g}+J_k(u_k)\right) +\int gF(u)dx \end{aligned}$$

$\forall t\ge 0$, $u(t)\in W$, so we have

$$\begin{aligned} 0<E(t) \quad \mathrm{{for\, all}}\quad t\ge 0. \end{aligned}$$

(3.27)

The proof of the other inequality relies on the construction of a Lyapunov functional by performing a suitable modification of the energy. To this end, for $\varepsilon >0$, to be chosen later, we define

$$\begin{aligned} L(t)=E(t)+\varepsilon \sum _{k=1}^2\int gu_ku_{kt}dx \end{aligned}$$

(3.28)

It is straightforward to see that L(t) and E(t) are equivalent in the sense that there exist two positive constants $\beta _1$ and $\beta _2$ depending on $\varepsilon$ such that for $t\ge 0$,

$$\begin{aligned} \beta _1E(t)\le L(t)\le \beta _2E(t) \end{aligned}$$

(3.29)

By taking the time derivative of the function L defined above , using problem (1.1), and performing several integration by parts, we get:

$$\begin{aligned} \frac{dL(t)}{dt}= & {} -\delta \sum _{k=1}^2\Vert u_{kt}\Vert _{L_g^2}^2+\varepsilon \sum _{k=1}^2\int gu^2_{kt}dx+\varepsilon \sum _{k=1}^2\int gu_ku_{ktt}dx\nonumber \\= & {} -\delta \sum _{k=1}^2\Vert u_{kt}\Vert _{L_g^2}^2+\varepsilon \sum _{k=1}^2\Vert u_{kt}\Vert _{L_g^2}^2+\varepsilon \sum _{k=1}^2\int gu_k\lambda u_k|u_k|^{\beta -1}dx\nonumber \\&-\,\varepsilon \sum _{k=1}^2\int gu_k\delta u_{kt}dx+\varepsilon \sum _{k=1}^2\int gu_k\phi \Delta u_kdx\nonumber \\= & {} -\delta \sum _{k=1}^2\Vert u_{kt}\Vert _{L_g^2}^2+\varepsilon \sum _{k=1}^2\Vert u_{kt}\Vert _{L_g^2}^2 +\varepsilon \lambda \sum _{k=1}^2\Vert u_k\Vert _{L_g^{\beta +1}}^{\beta +1}\nonumber \\&-\,\varepsilon \sum _{k=1}^2\Vert \nabla u_k\Vert _2^2-\varepsilon \delta \sum _{k=1}^2\int gu_ku_{kt}dx \end{aligned}$$

(3.30)

By using Young inequality and Sobolev inequality, we obtain, for any $\gamma >0$,

$$\begin{aligned} \int gu_ku_{kt}dx\le \frac{1}{4\gamma }\int gu_{kt}^2dx+\gamma \int gu_k^2dx \le \frac{1}{4\gamma }\int gu_{kt}^2dx+\gamma \alpha \Vert \nabla u_k\Vert _2^2\nonumber \\ \end{aligned}$$

(3.31)

here $\alpha$ is the Sobolev constant.

Consequently, inserting (3.31) into (3.30), we have

$$\begin{aligned} \frac{dL(t)}{dt}\le \left( \varepsilon +\frac{\varepsilon \delta }{4\gamma }-\delta \right) \sum _{k=1}^2\Vert u_{kt}\Vert _{L_g^2}^2 +\bigg (\varepsilon \lambda c_0\mu _0^{p_1}E(0)^{p_2}-\varepsilon +\varepsilon \delta \gamma \alpha \bigg )\sum _{k=1}^2\Vert \nabla u_k\Vert _2^2\nonumber \\ \end{aligned}$$

(3.32)

By the condition $E(0)^{p_2}c_0\lambda \mu _0^{p_1}<1$, let us choose $\gamma$ small enough such that

$$\begin{aligned} \varepsilon (\lambda c_0\mu _0^{p_1}E(0)^{p_2}-1+\delta \gamma \alpha )<0 \end{aligned}$$

(3.33)

From (3.32) we may find $\eta >0$, which depends only on $\gamma$ such that

$$\begin{aligned} \frac{dL(t)}{dt}\le \left( \varepsilon \left( \frac{\delta }{4\gamma }+1\right) -\delta \right) \sum _{k=1}^2\Vert u_{kt}\Vert _{L_g^2}^2 -\varepsilon \eta \sum _{k=1}^2\Vert \nabla u_k\Vert _2^2 \end{aligned}$$

(3.34)

Consequently, using the definition of the energy (3.3), for any positive constant M, we obtain

$$\begin{aligned} \frac{dL(t)}{dt}\le & {} -M\varepsilon E(t)+\left( \varepsilon \left( \frac{\delta }{4\gamma }+1+\frac{M}{2}\right) -\delta \right) \sum _{k=1}^2\Vert u_{kt}\Vert _{L_g^2}^2\nonumber \\&+\varepsilon (M\alpha -\eta )\sum _{k=1}^2\Vert \nabla u_k\Vert _2^2 \end{aligned}$$

(3.35)

Choose $M\alpha <\eta$, and $\varepsilon$ small enough such that

$$\begin{aligned} \varepsilon \left( \frac{\delta }{4\gamma }+1+\frac{M}{2}\right) -\delta <0 \end{aligned}$$

(3.36)

inequality (3.35) becomes

$$\begin{aligned} \frac{dL(t)}{dt}\le -M\varepsilon E(t)\qquad \mathrm{{for\ all}}\quad t\ge 0 \end{aligned}$$

(3.37)

On the other hand, by virtue of (3.29), setting $\xi =\displaystyle \frac{M\varepsilon }{\beta _2}$, the last inequality becomes

$$\begin{aligned} \frac{dL(t)}{dt}\le -\xi L(t)\quad \mathrm{{for\ all}}\quad t\ge 0 \end{aligned}$$

(3.38)

Integrating the previous differential inequality (3.28) between 0 and t gives the following estimate for the function L

$$\begin{aligned} L(t)\le Ce^{-\xi t}\qquad \mathrm{{for\ all}}\quad t\ge 0 \end{aligned}$$

(3.39)

Consequently, by using (3.29) once again, we conclude

$$\begin{aligned} E(t)\le {\widehat{C}}e^{-\xi t}\quad \mathrm{{for\ all}}\quad t\ge 0 \end{aligned}$$

(3.40)

This completes the proof. $\square$

4 Blow-up

Theorem 2

Let the assumptions (A1–A4) and (A6) hold. If $I_k(\varphi _k)<0, (k=1,2)$, $E(0)<\min \{d_1, d_2\}$. Then the solution $(u_1, u_2)$ blows up in finite time, i.e., there exists T such that

$$\begin{aligned} \lim _{t\rightarrow T^-_1}\sum _{k=1}^2\Vert u_k(t)\Vert _{L_g^2}^2=\infty \end{aligned}$$

and an upper bound for T is estimated

$$\begin{aligned} \begin{array}{llll} T&{}\le \left\{ \displaystyle \left. \frac{\sum \Vert \varphi _k\Vert _{L_g^2}^2+\gamma s_0^2}{2\theta \left[ \sum (\varphi _k,\psi _k)_{L_g^2}+\gamma s_0\right] -\delta \sum \Vert \varphi _k\Vert _{L_g^2}^2}\right| \ s_0>-\frac{1}{\gamma }\sum (\varphi _k,\psi _k)\right\} \\ &{}=\displaystyle \frac{1}{2\theta ^2\gamma }\Bigg \{\left[ \left( \sum 2\theta (\varphi _k,\psi _k)_{L_g^2} -\delta \sum \Vert \varphi _k\Vert _{L_g^2}^2\right) ^2 +4\theta ^2\gamma \sum \Vert \varphi _k\Vert \right] ^{\frac{1}{2}}\\ &{}\quad \, +\left( \displaystyle \delta \sum \Vert \varphi _k\Vert _{L_g^2}^2-2\theta \sum (\varphi _k,\psi _k)_{L_g^2}\right) \Bigg \} \end{array} \end{aligned}$$

where $\gamma = 2\left( \sum _{k=1}^2d_k-E(0)\right)$, $\theta =\frac{\beta -1}{4}$

Lemma 7

Suppose that $F(s)\ge 0$ with $s=(s_1,s_2)\in R^1\times R^1$. Let $(u_1,u_2)$ defined on $[0,T_{max})$ be a weak solution of problem (1.1). For each $t\in [0,T_{max})$.

• (i) If for all $k\in \{1,2\}$, $I_k(u_k)\ge 0$ , then

  $$\begin{aligned} \sum J_k(u_k)\ge \sum \frac{\beta -1}{2(\beta +1)}\Vert u_k\Vert ^2_{D^{1,2}} \end{aligned}$$

  (4.1)

• (ii) If for all $k\in \{1,2\}$, $I_k(u_k)<0$ , then

  $$\begin{aligned} \frac{\beta -1}{2(\beta +1)}\Vert u_k\Vert ^2_{D^{1,2}}> d_k \end{aligned}$$

  (4.2)

Proof

• (i) If $I_k(u_k)\ge 0$ for each $t\in [0,T_{max})$, we obtain

  $$\begin{aligned} \begin{array}{llll} E(t)&{}\displaystyle \ge \sum J_k(u_k)=\frac{1}{2}\sum \Vert u_k\Vert ^2_{D^{1,2}}-\sum \frac{\lambda }{\beta +1}\int g|u_k|^{\beta +1}\\ &{}\displaystyle \ge \frac{1}{2}\sum \Vert u_k\Vert ^2_{D^{1,2}}-\sum \frac{1}{\beta +1}\Vert u_k\Vert ^2_{D^{1,2}} =\frac{\beta -1}{2(\beta +1)}\sum \Vert u_k\Vert ^2_{D^{1,2}} \end{array} \end{aligned}$$

• (ii) If $I_k(u_k)<0$ for each $t\in [0,T_{max})$, there exists $t_0\in [0,T_{max})$ such that

  $$\begin{aligned} \frac{\beta -1}{2(\beta +1)}\Vert u_1(t_0)\Vert ^2_{D^{1,2}}\le d_1\quad or \quad \frac{\beta -1}{2(\beta +1)}\Vert u_2(t_0)\Vert ^2_{D^{1,2}}\le d_2 \end{aligned}$$

  we then obtain

  $$\begin{aligned} \lambda \Vert u_1(t_0)\Vert _{L_g^{\beta +1}}^{\beta +1}\le \Vert u_1(t_0)\Vert ^2_{D^{1,2}}\quad or\quad \lambda \Vert u_2(t_0)\Vert _{L_g^{\beta +1}}^{\beta +1}\le \Vert u_2(t_0)\Vert ^2_{D^{1,2}} \end{aligned}$$

  A contradiction with $I_k(u_k)<0$.$\square$

Lemma 8

Suppose that $F(s)\ge 0$ with $s=(s_1,s_2)\in R^1\times R^1$. Let $(u_1,u_2)$ defined on $[0,T_{max})$ be a weak solution of problem (1.1). For any $0\le t< T_{max}$.

• (i) If for all $k\in \{1,2\}$, $I_k(\varphi _k)\ge 0$ , $E(0)<\min \{d_1,d_2\}$, then

  $$\begin{aligned} \sum J_k(u_k)\ge \sum \frac{\beta -1}{2(\beta +1)}\Vert u_k\Vert ^2_{D^{1,2}} \end{aligned}$$

  (4.3)

• (ii) If for all $k\in \{1,2\}$, $I_k(\varphi _k)<0$, $E(0)<\min \{d_1,d_2\}$, then

  $$\begin{aligned} \frac{\beta -1}{2(\beta +1)}\Vert u_k\Vert ^2_{D^{1,2}}>d_k \end{aligned}$$

  (4.4)

Proof

(i) Since $I_k(\varphi _k)\ge 0$, by Lemma 7, we get that

$$\begin{aligned} E(0)\ge \sum J_k(\varphi _k)\ge \sum \frac{\beta -1}{2(\beta +1)}\Vert u_k\Vert ^2_{D^{1,2}} \end{aligned}$$

Define

$$\begin{aligned} T^*=\sup \{t\in [0,T]:E(0)\ge \sum \frac{\beta -1}{2(\beta +1)}\Vert u_k(s)\Vert ^2_{D^{1,2}},\ 0\le s<t\} \end{aligned}$$

If $T^*<T$ , then we have

$$\begin{aligned}&\sum \frac{\beta -1}{2(\beta +1)}\Vert u_k(T^*)\Vert ^2_{D^{1,2}}=E(0)\nonumber \\&\sum \frac{\beta -1}{2(\beta +1)}\Vert u_k(t)\Vert ^2_{D^{1,2}}>E(0),\ \forall t\in (T^*,T_{max}) \end{aligned}$$

(4.5)

Since $E(0)\!<\!\min \{d_1,d_2\}$, Hence $I_k(u_k(T^*))\!>\!0$. we get that $\sum \!\frac{\beta -1}{2(\beta +1)}\Vert u_k(T^*)\Vert ^2_{D^{1,2}}\!<d_k$. By the continuity of $\Vert u_k(t)\Vert ^2_{D^{1,2}}$ , there exists an interval $(T^*,{\hat{T}})\subset (T^*,T_{max})$ such that $I_k(u_k(t))>0$, $\forall t\in (T^*,{\hat{T}})$. Again using Lemma 7, it yields

$$\begin{aligned} \sum \frac{\beta -1}{2(\beta +1)}\Vert u_k(t)\Vert ^2_{D^{1,2}}\le E(t)< E(0),\ \forall t\in (T^*,{\hat{T}}) \end{aligned}$$

This contradicts with (4.5). The contradiction implies that $T^*$ meets with T and (4.3) holds.

(i) Let $I_k(\varphi _k)<0$, then by Lemma 7, we get that $\displaystyle \frac{\beta -1}{2(\beta +1)}\Vert u_k(t)\Vert ^2_{D^{1,2}}>d_k,\ k=1,2$. Define

$$\begin{aligned} T_*=\sup \left\{ t\in [0,T]:\frac{\beta -1}{2(\beta +1)}\Vert u_k(s)\Vert ^2_{D^{1,2}}>d_k,\ k=1,2.\ 0\le s<t\right\} \end{aligned}$$

If $T_*<T$, then we have

$$\begin{aligned} \frac{\beta -1}{2(\beta +1)}\Vert u_1(T_*)\Vert ^2_{D^{1,2}}=d_1,\quad \frac{\beta -1}{2(\beta +1)}\Vert u_2(T_*)\Vert ^2_{D^{1,2}}\le d_2 \end{aligned}$$

or

$$\begin{aligned} \frac{\beta -1}{2(\beta +1)}\Vert u_1(T_*)\Vert ^2_{D^{1,2}}\le d_1,\quad \frac{\beta -1}{2(\beta +1)}\Vert u_2(T_*)\Vert ^2_{D^{1,2}}=d_2 \end{aligned}$$

Therefore, $I_k(u_k(T_*))>0,\ k=1,2$. Then by use of Lemma 7, we obtain

$$\begin{aligned} \sum \frac{\beta -1}{2(\beta +1)}\Vert u_k(T_*)\Vert ^2_{D^{1,2}}\le E(T_*) \end{aligned}$$

(4.6)

Combining (3.3), (4.6) and noting that $E(0)<\min \{d_1,d_2\}$, we get that $\frac{\beta -1}{2(\beta +1)}\Vert u_k(T_*)\Vert ^2_{D^{1,2}}\le d_k$. A contradiction. Thus (4.4) is true.

We give the estimates in the following. For any $T>0$ we may consider $L:[0,T]\rightarrow R^+$ defined by

$$\begin{aligned} L(t)=\sum _{k=1}^2\left( \Vert u_k(t)\Vert _{L_g^2}^2+\int _0^t\delta \Vert u_k(\tau )\Vert _{L_g^2}^2d\tau +\delta (T-t)\Vert \varphi _k\Vert _{L_g^2}^2\right) +\gamma (t+s_0)^2\nonumber \\ \end{aligned}$$

(4.7)

where $\gamma$ and $s_0>0$ are constants to be determined. $\square$

Lemma 9

Let the assumptions of Theorem 2 be satisfied. then

$$\begin{aligned} L''(t)\ge (\beta +3)\sum _{k=1}^2\Vert u_{kt}\Vert _{L_g^2}^2+2\delta (\beta +1)\sum _{k=1}^2\int _0^t\Vert u_{kt}\Vert _{L_g^2}^2+(\beta +3)\gamma \end{aligned}$$

Proof

$$\begin{aligned} L'(t)= & {} \displaystyle \sum _{k=1}^2\left( 2\int _\Omega gu_ku_{kt}+\delta \Vert u_k\Vert _{L_g^2}^2-\delta \Vert \varphi _k\Vert _{L_g^2}^2\right) +2\gamma (t+s_0)\nonumber \\= & {} \displaystyle \sum _{k=1}^2\left( 2\int _\Omega gu_ku_{kt}+2\delta \int _0^t(u_k(\tau ),u_{kt}(\tau ))_{L_g^2}d\tau \right) +2\gamma (t+s_0)\end{aligned}$$

(4.8)

$$\begin{aligned} L''(t)= & {} \sum _{k=1}^2\left( 2( u_{ktt},u_k)_{L_g^2}+2\Vert u_{kt}\Vert _{L_g^2}^2+2\delta (u_{kt},u_k)_{L_g^2}\right) +2\gamma \end{aligned}$$

(4.9)

By (4.9), we get that

$$\begin{aligned} L''(t)=2\left[ \sum _{k=1}^2\left( \Vert u_{kt}\Vert _{L_g^2}^2-\Vert \nabla u_k\Vert _2^2-\int _{R^N} gu_kf_k(u)dx+\lambda \int _{R^N} g|u_k|dx^{\beta +1}\right) \right] +2\gamma \end{aligned}$$

Using the energy equality and A6 and Lemma 8 , we have

$$\begin{aligned} \begin{array}{llll} L''(t)&{}\displaystyle \ge (\beta +3)\sum _{k=1}^2\Vert u_{kt}\Vert _{L_g^2}^2+(\beta -1)\sum _{k=1}^2\Vert u_k\Vert _{D^{1,2}}^2 +2(\beta +1)\sum _{k=1}^2\int _0^t\delta \Vert u_{kt}(\tau )\Vert _{L_g^2}^2d\tau \\ &{}\displaystyle -2(\beta +1)E(0)+2(\beta +1)\int _{R^N} gF(u)dx-\int _{R^N}gu_kf_k(u)dx+2\gamma \\ &{}\displaystyle \ge (\beta +3)\sum _{k=1}^2\Vert u_{kt}\Vert _{L_g^2}^2+2(\beta +1)\sum _{k=1}^2\int _0^t\delta \Vert u_{kt}(\tau )\Vert _{L_g^2}^2d\tau +(\beta -1)\sum _{k=1}^2\frac{2(\beta +1)}{\beta -1}d_k\\ &{}\quad \displaystyle -\,2(\beta +1)E(0)+2\left[ \frac{1}{n_0}\int _{R^N} gF(u)dx-\sum _{k=1}^2\int _{R^N}gu_kf_k(u)dx\right] +2\gamma \\ &{}\displaystyle \ge (\beta +3)\sum _{k=1}^2\Vert u_{kt}\Vert _{L_g^2}^2+2(\beta +1)\sum _{k=1}^2\int _0^t\delta \Vert u_{kt}(\tau )\Vert _{L_g^2}^2d\tau \\ &{}\quad \displaystyle +\,2(\beta +1)\sum _{k=1}^2d_k-2(\beta +1)E(0)+2\gamma \end{array} \end{aligned}$$

By the assumptions $\sum _{k=1}^2d_k-E(0)>0$. Now taking $\gamma = 2\left( \sum _{k=1}^2d_k-E(0)\right)$

we get that

$$\begin{aligned} L''(t)\ge (\beta +3)\sum _{k=1}^2\Vert u_{kt}\Vert _{L_g^2}^2+2\delta (\beta +1)\sum _{k=1}^2\int _0^t\Vert u_{kt}(\tau )\Vert _{L_g^2}^2d\tau +(\beta +3)\gamma \end{aligned}$$

Assume by contradiction that the solution u is global. Define

$$\begin{aligned}&P=\sum _{k=1}^2\left( \Vert u_k\Vert _{L_g^2}^2+\delta \int _0^t\Vert u_k(\tau )\Vert _{L_g^2}^2d\tau \right) +\gamma (t+s_0)^2\\&Q=\sum _{k=1}^2\left( (u_k,u_{kt})_{L_g^2}+\delta \int _0^t(u_k(\tau ),u_{kt}(\tau ))_{L_g^2}d\tau \right) +\gamma (t+s_0)\\&U=\sum _{k=1}^2\left( \Vert u_{kt}\Vert _{L_g^2}^2+\delta \int _0^t\Vert u_{kt}(\tau )\Vert _{L_g^2}^2d\tau \right) +\gamma \end{aligned}$$

It is easy to check $0<P\le L(t)$, $Q=\displaystyle \frac{1}{2}L'(t)$, $\displaystyle 0<U\le \frac{L''(t)}{2+p}$. For any real pair $(\lambda ,\eta )$ and for all $t\in [0,T]$, we have

$$\begin{aligned} P\lambda ^2+2Q\lambda \eta +U\eta ^2= & {} \sum _{k=1}^2\Vert \lambda u_k+\eta u_{kt}\Vert _{L_g^2}^2\\&+\sum _{k=1}^2\delta \int _0^t\Vert \lambda u_k+\eta u_{kt}\Vert _{L_g^2}^2d\tau +\gamma [(t+s_0)\lambda +\eta ]^2 \end{aligned}$$

Therefore, $PU-Q^2\ge 0$. We infer from the above inequality that

$$\begin{aligned} L(t)L''(t)-\frac{\beta +3}{4}(L'(t))^2\ge 0,\ t\in [0,T] \end{aligned}$$

(4.10)

(4.10) implies that $[L^{-\theta }(t)]''\le 0,\ t\in [0,T]$ where $\theta =\frac{\beta -1}{4}$. Now taking $s_0>-\frac{1}{\gamma }\sum (\varphi _k,\psi _k)_{L_g^2}$, we get that $L'(0)>0$ and $(L^{-\theta })'(0)<0$ . Choosing $T\ge -\frac{(L^{-\theta })(0)}{(L^{-\theta })'(0)}$, then by the concavity Lemma, there exists $T_1$ satisfying

$$\begin{aligned} 0<T_1\le -\frac{(L^{-\theta })(0)}{(L^{-\theta })'(0)} ,\quad L^{-\theta }(T_1)=0 \end{aligned}$$

(4.11)

From (4.11), we see that $\lim _{t\rightarrow T^-_1}L(t)\!=\!\infty$, which implies that $\displaystyle \lim _{t\rightarrow T^-_1}\sum _{k=1}^2\Vert u_k(t)\Vert _{L_g^2}^2\!=\infty$. This leads to a contraction with $T_{max}=\infty$. Now we give the estimate of T. If (4.11) holds, it suffices

$$\begin{aligned} \frac{\sum \Vert \varphi _k\Vert _{L_g^2}^2+\gamma s_0^2}{2\theta \left[ \sum (\varphi _k,\psi _k)_{L_g^2}+\gamma s_0\right] -\delta \sum \Vert \varphi _k\Vert _{L_g^2}^2}\le T \end{aligned}$$

where if it is necessary we may take $s_0$ sufficiently large such that

$$\begin{aligned} 2\theta \left[ \sum (\varphi _k,\psi _k)_{L_g^2}+\gamma s_0\right] -\delta \sum \Vert \varphi _k\Vert _{L_g^2}^2>0 \end{aligned}$$

Therefore, we only need to take

$$\begin{aligned} \begin{array}{llll} T&{}=\left\{ \displaystyle \left. \frac{\sum \Vert \varphi _k\Vert _{L_g^2}^2+\gamma s_0^2}{2\theta \left[ \sum (\varphi _k,\psi _k)_{L_g^2}+\gamma s_0\right] -\delta \sum \Vert \varphi _k\Vert _{L_g^2}^2}\right| \ s_0>-\frac{1}{\gamma }\sum (\varphi _k,\psi _k)\right\} \\ &{}=\displaystyle \frac{1}{2\theta ^2\gamma }\Bigg \{\left[ \left( \sum 2\theta (\varphi _k,\psi _k)_{L_g^2} -\delta \sum \Vert \varphi _k\Vert _{L_g^2}^2\right) ^2 +4\theta ^2\gamma \sum \Vert \varphi _k\Vert \right] ^{\frac{1}{2}}\\ &{}\quad +\,\left( \displaystyle \delta \sum \Vert \varphi _k\Vert _{L_g^2}^2-2\theta \sum (\varphi _k,\psi _k)_{L_g^2}\right) \Bigg \} \end{array} \end{aligned}$$

and so we have completed the proof of Theorem 2. $\square$