A survey on deep learning for financial risk prediction

Kuashuai Peng; Guofeng Yan; Kuashuai Peng; Guofeng Yan

doi:10.3934/QFE.2021032

Quantitative Finance and Economics

2021, Volume 5, Issue 4: 716-737. doi: 10.3934/QFE.2021032

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Review

A survey on deep learning for financial risk prediction

Kuashuai Peng ,
Guofeng Yan ^,

School of Computer Science and Cyber Engineering, Guangzhou University, Guangzhou 510006, China

Received: 09 December 2021 Accepted: 20 December 2021 Published: 23 December 2021
JEL Codes: C45, C5, C1, C38

The rapid development of financial technology not only provides a lot of convenience to people's production and life, but also brings a lot of risks to financial security. To prevent financial risks, a better way is to build an accurate warning model before the financial risk occurs, not to find a solution after the outbreak of the risk. In the past decade, deep learning has made amazing achievements in the fields, such as image recognition, natural language processing. Therefore, some researchers try to apply deep learning methods to financial risk prediction and most of the results are satisfactory. The main work of this paper is to review the predecessors' work of deep learning for financial risk prediction according to three prominent characteristics of financial data: heterogeneity, multi-source, and imbalance. We first briefly introduced some classical deep learning models as the model basis of financial risk prediction. Then we analyzed the reasons for these characteristics of financial data. Meanwhile, we studied the differences of commonly used deep learning models according to different data characteristics. Finally, we pointed out some open issues with research significance in this field and suggested the future implementations that might be feasible.

Keywords:

Citation: Kuashuai Peng, Guofeng Yan. A survey on deep learning for financial risk prediction[J]. Quantitative Finance and Economics, 2021, 5(4): 716-737. doi: 10.3934/QFE.2021032

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Abstract

1. Introduction

In the present paper, we study the following coupled nonlinear wave equations (see ^[1])

$\begin{align} u_{tt}-div\Big(\rho\Big(|\nabla u|^2\Big)\nabla u\Big)+|u_t|^{m-1}u_t+f_1(u, v)& = 0, \end{align}$

(1.1)

$\begin{align} v_{tt}-div\Big(\rho\Big(|\nabla v|^2\Big)\nabla v\Big)+|v_t|^{n-1}v_t+f_2(u, v)& = 0, \end{align}$

(1.2)

where $u_t = \frac{\partial u}{\partial t}, v_t = \frac{\partial v}{\partial t}$ , $m, n > 1$ , $\nabla$ is the gradient operator, div is the divergence operator and $f_i(., .):\mathbb{R}^2\rightarrow \mathbb{R}, i = 1, 2$ are known functions. For arbitrary solutions of (1.1) and (1.2), the function $\rho$ is supposed to satisfy one or the other of the two conditions:

A1). $0 < \rho^2(q^2)q^2\leq m_1\widetilde{\rho}(q^2),$

A2). $0 < \frac{1}{K_1} < \rho^{-1}(q^2)\leq K_2\widetilde{\rho}(q^2)\frac{1}{q^2},$

where $\widetilde{\rho}(q^2) = \int_0^{q^2}\rho(\zeta)d\zeta, \ q^2 = |\nabla u|^2, m_1, K_1, K_2 > 0.$

Quintanilla ^[2] imposed condition A1 and obtained the growth or decay estimates of the solution to the type Ⅲ heat conduction. The condition A1 can be satisfied easily, e.g., $\rho(q^2) = \frac{1}{\sqrt{1+q^2}}$ or $\rho(q^2) = (1+\frac{b}{p}q^2)^{p-1}, b > 0, 0 < p\leq1.$ The similar condition as A2 was also considered by many papers (see ^[3,4]). $\rho(q^2) = \sqrt{1+q^2}$ satisfies the condition A2.

In addition, we introduce a function $F(u, v)$ which is defined as $\frac{\partial F}{\partial u} = f_1(u, v), \ \frac{\partial F}{\partial v} = f_2(u, v),$ where $F(0, 0) = 0.$

The wave equations have attracted many attentions of scholars due to their wide application, and a large number of achievements have been made in the existence of solutions (see ^{[1,5,6,7,8,9,10,11,12]}). The fuzzy inference method is used to solve this problem. The algebraic formulation of fuzzy relation is studied in ^[13,14]. In this paper, we study the Phragmén-Lindelöf type alternative property of solutions of wave equations (1.1)–(1.2). It is proved that the solution of the equations either grows exponentially (polynomially) or decays exponentially (polynomially) when the space variable tends to infinite. In the case of decay, people usually expect a fast decay rate. The Phragmén-Lindelöf type alternative research on partial differential equations has lasted for a long time (see ^{[2,15,16,17,18,19,20,21,22,23]}).

It is worth emphasizing that Quintanilla ^[2] considered an exterior or cone-like region. Under some appropriate conditions, the growth/decay estimates of some parabolic problems are obtained. Inspired by ^[2], we extends the research of to the nonlinear wave model in this paper. However, different from ^[2], in addition to condition A1 and condition A2, we also consider a special condition of $\rho$ . The appropriate energy function is established, and the nonlinear differential inequality about the energy function is derived. By solving this differential inequality, the Phragmén-Lindelöf type alternative results of the solution are obtained. Our model is much more complex than ^[2], so the methods used in ^[2] can not be applied to our model directly. Finally, a nonlinear system of viscoelastic type is also studied when the system is defined in an exterior or cone-like regions and the growth or decay rates are also obtained.

2. The Phragmén-Lindelöf type alternative result under A1

Letting that $\Omega(r)$ denotes a cone-like region, i.e.,

$\Omega(r) = \Big\{\mathit{\boldsymbol{x}}\Big||\mathit{\boldsymbol{x}}|^2\geq r^2, r\geq R_0 > 0\Big\},$

and that $B(r)$ denotes the exterior surface to the sphere, i.e.,

$B(r) = \Big\{\mathit{\boldsymbol{x}}\Big||\mathit{\boldsymbol{x}}|^2 = r^2, r\geq R_0 > 0\Big\},$

Equations (1.1) and (1.2) also have the following initial-boundary conditions

$\begin{align} u(\mathit{\boldsymbol{x}}, 0) = v(\mathit{\boldsymbol{x}}, 0) = 0, \ & in\ \Omega, \end{align}$

(2.1)

$\begin{align} u(\mathit{\boldsymbol{x}}, t) = g_1(\mathit{\boldsymbol{x}}, t), \ v(\mathit{\boldsymbol{x}}, t) = g_2(\mathit{\boldsymbol{x}}, t), \ & in\ B(R_0)\times(0, \tau), \end{align}$

(2.2)

where $g_1$ and $g_2$ are positive known functions, $\textbf{x} = (x_1, x_2, x_3)$ and $\tau > 0$ .

Now, we establish an energy function

$\begin{align} \mathcal{E}(r, t)& = \int_0^t\int_{B(r)}e^{-\omega\eta}\rho\Big(|\nabla u|^2\Big)\nabla u\cdot \frac{\mathit{\boldsymbol{x}}}{r}u_\eta dSd\eta +\int_0^t\int_{B(r)}e^{-\omega\eta}\rho\Big(|\nabla v|^2\Big)\nabla v\cdot\frac{\mathit{\boldsymbol{x}}}{r}v_\eta dSd\eta\\ &\doteq \mathcal{E}_1(r, t)+\mathcal{E}_2(r, t). \end{align}$

(2.3)

Let $r_0$ be a positive constant which satisfies $r > r_0\geq R_0$ . Integrating $\mathcal{E}(z, t)$ from $r_0$ to $r$ , using the divergence theorem and Eqs (1.1) and (1.2), (2.1) and (2.2), we have

$\begin{align} \mathcal{E}(r, t)&-\mathcal{E}(r_0, t) = \int_0^t\int_{r_0}^r\int_{B(\xi)}e^{-\omega \eta}\nabla\cdot\Big[\rho\Big(|\nabla u|^2\Big)\nabla uu_\eta\Big] dsd\xi d\eta \\ &+\int_0^t\int_{r_0}^r\int_{B(\xi)}e^{-\omega \eta}\nabla\cdot\Big[\rho\Big(|\nabla v|^2\Big)\nabla vv_\eta\Big] dsd\xi d\eta \\ &+\int_0^t\int_{r_0}^r\int_{B(\xi)}e^{-\omega \eta}\Big[u_{\eta\eta}+|u_\eta|^{m+1}+f_1(u, v)\Big]u_\eta dsd\xi d\eta \\ &+\int_0^t\int_{r_0}^r\int_{B(\xi)}e^{-\omega \eta}\Big[v_{\eta\eta}+|v_\eta|^{n+1}+f_2(u, v)\Big]v_\eta dsd\xi d\eta \\ &+\frac{1}{2}\int_0^t\int_{r_0}^r\int_{B(\xi)}e^{-\omega \eta}\frac{\partial}{\partial \eta}\widetilde{\rho}\Big(|\nabla u|^2\Big) dsd\xi d\eta +\frac{1}{2}\int_0^t\int_{r_0}^r\int_{B(\xi)}e^{-\omega \eta}\frac{\partial}{\partial \eta}\widetilde{\rho}\Big(|\nabla v|^2\Big) dsd\xi d\eta \\ & = \frac{1}{2}e^{-\omega t}\int_{r_0}^r\int_{B(\xi)}\Big[|u_t|^2+|v_t|^2+\widetilde{\rho}\Big(|\nabla u|^2\Big)+\widetilde{\rho}\Big(|\nabla v|^2\Big)+2F(u, v)\Big]dsd\xi \\ &+\frac{1}{2}\omega\int_0^t\int_{r_0}^r\int_{B(\xi)}e^{-\omega\eta}\Big[|u_\eta|^2+|v_\eta|^2+\widetilde{\rho}\Big(|\nabla u|^2\Big)+\widetilde{\rho}\Big(|\nabla v|^2\Big)+2F(u, v)\Big]dSd\xi d\eta \\&+\int_0^t\int_{r_0}^r\int_{B(\xi)}e^{-\omega\eta}\Big[|u_\eta|^{m+1}+|v_\eta|^{n+1}\Big]dsd\xi d\eta, \end{align}$

(2.4)

from which it follows that

$\begin{align} \frac{\partial}{\partial r}\mathcal{E}(r, t)& = \frac{1}{2}e^{-\omega t}\int_{B(r)}\Big[|u_t|^2+|v_t|^2+\widetilde{\rho}\Big(|\nabla u|^2\Big)+\widetilde{\rho}\Big(|\nabla v|^2\Big)+2F(u, v)\Big]ds \\ &+\frac{1}{2}\omega\int_0^t\int_{B(r)}e^{-\omega\eta}\Big[|u_\eta|^2+|v_\eta|^2+\widetilde{\rho}\Big(|\nabla u|^2\Big)+\widetilde{\rho}\Big(|\nabla v|^2\Big)+2F(u, v)\Big]dS d\eta \\&+\int_0^t\int_{B(r)}e^{-\omega\eta}\Big[|u_\eta|^{m+1}+|v_\eta|^{n+1}\Big]dsd\eta, \end{align}$

(2.5)

where $\omega$ is positive constant.

Now, we show how to bound $\mathcal{E}(r, t)$ by $\frac{\partial}{\partial r}\mathcal{E}(r, t)$ . We use the Hölder inequality, the Young inequality and the condition A1 to have

$\begin{align} \Big|\mathcal{E}_1(r, t)\Big|&\leq\Big[\int_0^t\int_{B(r)}e^{-\omega\eta}\rho^2\Big(|\nabla u|^2\Big)|\nabla u|^2ds d\eta\cdot\int_0^t\int_{B(r)}e^{-\omega\eta}u_\eta^2ds d\eta\Big]^\frac{1}{2}\\ &\leq\sqrt{m_1}\Big[\int_0^t\int_{B(r)}e^{-\omega\eta}\widetilde{\rho}\Big(|\nabla u|^2\Big)ds d\eta\cdot\int_0^t\int_{B(r)}e^{-\omega\eta}u_\eta^2ds d\eta\Big]^\frac{1}{2}\\ &\leq\frac{\sqrt{m_1}}{2}\Big[\int_0^t\int_{B(r)}e^{-\omega\eta}\widetilde{\rho}\Big(|\nabla u|^2\Big)ds d\eta+\int_0^t\int_{B(r)}e^{-\omega\eta}u_\eta^2ds d\eta\Big], \end{align}$

(2.6)

and

$\begin{align} \Big|\mathcal{E}_2(r, t)\Big|&\leq\frac{\sqrt{m_1}}{2}\Big[\int_0^t\int_{B(r)}e^{-\omega\eta}\widetilde{\rho}\Big(|\nabla v|^2\Big)ds d\eta+\int_0^t\int_{B(r)}e^{-\omega\eta}v_\eta^2dsd\eta\Big]. \end{align}$

(2.7)

Inserting (2.6) and (2.7) into (2.3) and combining (2.5), we have

$\begin{align} \Big|\mathcal{E}(r, t)\Big|\leq \frac{\sqrt{m_1}}{\omega}\Big[\frac{\partial}{\partial r}\mathcal{E}(r, t)\Big]. \end{align}$

(2.8)

We consider inequality (2.8) for two cases.

I. If $\exists r_0 > R_0$ such that $\mathcal{E}(r_0, t)\geq0$ . From (2.5), we obtain $\mathcal{E}(r, t)\geq\mathcal{E}(r_0, t)\geq0, r\geq r_0$ . Therefore, from (2.8) we have

$\begin{align} \mathcal{E}(r, t)\leq \frac{\sqrt{m_1}}{\omega}\Big[\frac{\partial}{\partial r}\mathcal{E}(r, t)\Big], r\geq r_0. \end{align}$

(2.9)

Integrating (2.9) from $r_0$ to $r$ , we have

$\begin{align} \mathcal{E}(r, t)\geq\Big[\mathcal{E}(r_0, t)\Big]e^{\frac{\omega}{\sqrt{m_1}}(r-r_0)}, r\geq r_0. \end{align}$

(2.10)

Combing (2.4) and (2.10), we have

$\begin{align} \lim\limits_{r\rightarrow \infty}&\Big\{e^{-\frac{\omega}{\sqrt{m_1}}r}\Big[\frac{1}{2}e^{-\omega t}\int_{r_0}^r\int_{B(\xi)}\Big[|u_t|^2+|v_t|^2+\widetilde{\rho}\Big(|\nabla u|^2\Big)+\widetilde{\rho}\Big(|\nabla v|^2\Big)+2F(u, v)\Big]dsd\xi \\ &+\frac{1}{2}\omega\int_0^t\int_{r_0}^r\int_{B(\xi)}e^{-\omega\eta}\Big[|u_\eta|^2+|v_\eta|^2+\widetilde{\rho}\Big(|\nabla u|^2\Big)+\widetilde{\rho}\Big(|\nabla v|^2\Big)+2F(u, v)\Big]dsd\xi d\eta \\&+\int_0^t\int_{r_0}^r\int_{B(\xi)}e^{-\omega\eta}\Big[|u_\eta|^{m+1}+|v_\eta|^{n+1}\Big]dsd\xi d\eta\Big]\Big\} \\ &\geq \mathcal{E}(R_0, t)e^{-\frac{\omega}{\sqrt{m_1}}R_0}. \end{align}$

(2.11)

II. If $\forall r > R_0$ such that $\mathcal{E}(r, t) < 0$ . The inequality (2.8) can be rewritten as

$\begin{align} -\mathcal{E}(r, t)\leq \frac{\sqrt{m_1}}{\omega}\Big[\frac{\partial}{\partial r}\mathcal{E}(r, t)\Big], r\geq R_0. \end{align}$

(2.12)

Integrating (2.12) from $r_0$ to $r$ , we have

$\begin{align} -\mathcal{E}(r, t)\geq\Big[-\mathcal{E}(R_0, t)\Big]e^{-\frac{\omega}{\sqrt{m_1}}(r-R_0)}, r\geq R_0. \end{align}$

(2.13)

Inequality (2.13) shows that $\lim_{r\rightarrow \infty}\Big[-\mathcal{E}(r, t)\Big] = 0$ . Integrating (2.5) from $r$ to $\infty$ and combining (2.13), we obtain

$\begin{align} \frac{1}{2}&e^{-\omega t}\int_{r}^\infty\int_{B(\xi)}\Big[|u_t|^2+|v_t|^2+\widetilde{\rho}\Big(|\nabla u|^2\Big)+\widetilde{\rho}\Big(|\nabla v|^2\Big)+2F(u, v)\Big]dsd\xi \\ &+\frac{1}{2}\omega\int_0^t\int_{r}^\infty\int_{B(\xi)}e^{-\omega\eta}\Big[|u_\eta|^2+|v_\eta|^2+\widetilde{\rho}\Big(|\nabla u|^2\Big)+\widetilde{\rho}\Big(|\nabla v|^2\Big)+2F(u, v)\Big]dsd\xi d\eta \\&+\int_0^t\int_{r}^\infty\int_{B(\xi)}e^{-\omega\eta}\Big[|u_\eta|^{m+1}+|v_\eta|^{n+1}\Big]dsd\xi d\eta \\ &\leq \Big[-\mathcal{E}(r_0, t)\Big]e^{-\frac{\omega}{\sqrt{m_1}}(r-R_0)}. \end{align}$

(2.14)

We summarize the above result as the following theorem.

Theorem 2.1. Let $(u, v)$ be solution of the (1.1), (1.2), (2.1), (2.2) in $\Omega(R_0)$ , and $\rho$ satisfies condition A1. Then for fixed $t$ , $(u, v)$ either grows exponentially or decays exponentially. Specifically, either (2.11) holds or (2.14) holds.

3. The Phragmén-Lindelöf type alternative result under A2

If the function $\rho$ satisfies the condition A2, we recompute the inequality (2.6) and (2.7). Therefore

$\begin{align} \Big|\mathcal{E}_1(r, t)\Big|&\leq\Big[\int_0^t\int_{B(r)}e^{-\omega\eta}\rho^2\Big(|\nabla u|^2\Big)|\nabla u|^2ds d\eta\cdot\int_0^t\int_{B(r)}e^{-\omega\eta}u_\eta^2ds d\eta\Big]^\frac{1}{2}\\ &\leq K_1\sqrt{K_1}\Big[\int_0^t\int_{B(r)}e^{-\omega\eta}\rho^{-1}\Big(|\nabla u|^2\Big)ds d\eta\cdot\int_0^t\int_{B(r)}e^{-\omega\eta}u_\eta^2ds d\eta\Big]^\frac{1}{2}\\ &\leq\frac{K_1\sqrt{K_1K_2}}{2}\Big[\int_0^t\int_{B(r)}e^{-\omega\eta}\widetilde{\rho}\Big(|\nabla u|^2\Big)ds d\eta+\int_0^t\int_{B(r)}e^{-\omega\eta}u_\eta^2ds d\eta\Big], \end{align}$

(3.1)

and

$\begin{align} \Big|\mathcal{E}_2(r, t)\Big|&\leq\frac{K_1\sqrt{K_1K_2}}{2}\Big[\int_0^t\int_{B(r)}e^{-\omega\eta}\widetilde{\rho}\Big(|\nabla v|^2\Big)ds d\eta+\int_0^t\int_{B(r)}e^{-\omega\eta}v_\eta^2dsd\eta\Big]. \end{align}$

(3.2)

Inserting (3.1) and (3.2) into (2.3) and combining (2.5), we have

$\begin{align} \Big|\mathcal{E}(r, t)\Big|\leq \frac{K_1\sqrt{K_1K_2}}{\omega}\Big[\frac{\partial}{\partial r}\mathcal{E}(r, t)\Big]. \end{align}$

(3.3)

By following a similar method to that used in Section 2, we can obtain the Phragmén-Lindelöf type alternative result.

Theorem 3.1. Let $(u, v)$ be solution of the (1.1), (1.2), (2.1), (2.2) in $\Omega(R_0)$ , and $\rho$ satisfies condition A1. Then for fixed $t$ , $(u, v)$ either grows exponentially or decays exponentially.

Remark 3.1. Clearly, the rates of growth or decay obtained in Theorems 2.1 and 3.1 depend on $\omega$ . Because $\omega$ can be chosen large enough, the rates of growth or decay of the solutions can become large as we want.

Remark 3.2. The analysis in Sections 2 and 3 can be adapted to the single-wave equation

$\begin{align} u_{tt}-div\Big(\rho\Big(|\nabla u|^2\Big)\nabla u\Big)+|u_t|^{m-1}u_t+f(u) = 0 \end{align}$

(3.4)

and the heat conduction at low temperature

$\begin{align} au_{tt}+bu_t-c\Delta u+\Delta u_t = 0, \end{align}$

(3.5)

where $a, b, c > 0$ .

4. The Phragmén-Lindelöf type alternative result when $\rho(q^2)=b_1+b_2q^{2\beta}$

In this section, we suppose that $\rho$ satisfies $\rho(q^2) = b_1+b_2q^{2\beta}$ , where $b_1, b_2$ and $\beta$ are positive constants. Clearly, $\rho(q^2) = b_1+b_2q^{2\beta}$ can not satisfy A1 or A2. In this case, we define an "energy" function

$\begin{align} \mathcal{F}(r, t)& = \int_0^t\int_{B(r)}e^{-\omega\eta}\Big(b_1+b_2|\nabla u|^{2\beta}\Big)\nabla u\cdot\frac{\mathit{\boldsymbol{x}}}{r}u_\eta dsd\eta \\ &+\int_0^t\int_{B(r)}e^{-\omega\eta}\Big(b_1+b_2|\nabla v|^{2\beta}\Big)\nabla v\cdot\frac{\mathit{\boldsymbol{x}}}{r}v_\eta dsd\eta \\ &\doteq \mathcal{F}_1(r, t)+\mathcal{F}_2(r, t). \end{align}$

(4.1)

Computing as that in (2.4) and (2.5), we can get

$\begin{align} \mathcal{F}(r, t)& = \mathcal{F}(r_0, t)+\frac{1}{2}e^{-\omega t}\int_{r_0}^r\int_{B(\xi)}\Big[|u_t|^2+b_1|\nabla u|^2+\frac{1}{\beta+1}b_2|\nabla u|^{2(\beta+1)}\Big]dsd\xi\\ &+\frac{1}{2}e^{-\omega t}\int_{r_0}^r\int_{B(\xi)}\Big[|v_t|^2+b_1|\nabla v|^2+\frac{1}{\beta+1}b_2|\nabla v|^{2(\beta+1)}+F(u, v)\Big]dsd\xi\\ &+\frac{1}{2}\omega\int_0^t\int_{r_0}^r\int_{B(\xi)}e^{-\omega\eta}\Big[|u_\eta|^2+b_1|\nabla u|^2+\frac{1}{\beta+1}b_2|\nabla u|^{2(\beta+1)}\Big]ds d\xi d\eta\\ &+\frac{1}{2}\omega\int_0^t\int_{r_0}^r\int_{B(\xi)}e^{-\omega\eta}\Big[|v_\eta|^2+b_1|\nabla v|^2+\frac{1}{\beta+1}b_2|\nabla v|^{2(\beta+1)}+F(u, v)\Big]dsd\xi d\eta\\ &+\int_0^t\int_{r_0}^r\int_{B(\xi)}e^{-\omega\eta}|u_\eta|^{m+1}dsd\xi d\eta+\int_0^t\int_{B(r)}e^{-\omega\eta}|v_\eta|^{n+1}dsd\xi d\eta, \end{align}$

(4.2)

and

$\begin{align} \frac{\partial}{\partial r}\mathcal{F}(r, t)& = \frac{1}{2}e^{-\omega t}\int_{B(r)}\Big[|u_t|^2+b_1|\nabla u|^2+\frac{1}{\beta+1}b_2|\nabla u|^{2(\beta+1)}\Big]ds\\ &+\frac{1}{2}e^{-\omega t}\int_{B(r)}\Big[|v_t|^2+b_1|\nabla v|^2+\frac{1}{\beta+1}b_2|\nabla v|^{2(\beta+1)}+F(u, v)\Big]ds\\ &+\frac{1}{2}\omega\int_0^t\int_{B(r)}e^{-\omega\eta}\Big[|u_\eta|^2+b_1|\nabla u|^2+\frac{1}{\beta+1}b_2|\nabla u|^{2(\beta+1)}\Big]ds d\eta\\ &+\frac{1}{2}\omega\int_0^t\int_{B(r)}e^{-\omega\eta}\Big[|v_\eta|^2+b_1|\nabla v|^2+\frac{1}{\beta+1}b_2|\nabla v|^{2(\beta+1)}+F(u, v)\Big]ds d\eta\\ &+\int_0^t\int_{B(r)}e^{-\omega\eta}|u_\eta|^{m+1}dsd\eta+\int_0^t\int_{B(r)}e^{-\omega\eta}|v_\eta|^{n+1}ds d\eta. \end{align}$

(4.3)

Using the Hölder inequality and Young's inequality, we have

$\begin{align} \Big|\mathcal{F}_1(r, t)\Big|&\leq b_1\Big[\int_0^t\int_{B(r)}e^{-\omega\eta}|\nabla u|^2dsd\eta \int_0^t\int_{B(r)}e^{-\omega\eta}u_\eta^2 dsd\eta\Big]^\frac{1}{2} \\ &+b_2\Big[\int_0^t\int_{B(r)}e^{-\omega\eta}|\nabla u|^{2(\beta+1)}dsd\eta\Big]^\frac{2\beta+1}{2(\beta+1)} \int_0^t\int_{B(r)}e^{-\omega\eta}|u_\eta|^{m+1} dsd\eta\Big]^\frac{1}{m+1} |t2\pi r|^{\frac{1}{2(\beta+1)}-\frac{1}{m+1}} \\ &\leq \frac{\sqrt{b_1}}{R_0\omega}\Big[\frac{b_1}{2}r\omega\int_0^t\int_{B(r)}e^{-\omega\eta}|\nabla u|^2dsd\eta+ \frac{1}{2}r\omega\int_0^t\int_{B(r)}e^{-\omega\eta}u_\eta^2 dsd\eta\Big] \\ &+b_2|2t\pi|^{\frac{1}{2(\beta+1)}-\frac{1}{m+1}}R_0^{-\frac{\beta}{\beta+1}-\frac{2}{m+1}}\Big[\frac{\frac{2\beta+1}{2(\beta+1)}r}{\frac{2\beta+1}{2(\beta+1)} +\frac{1}{m+1}}\int_0^t\int_{B(r)}e^{-\omega\eta}|\nabla u|^{2(\beta+1)}dsd\eta \\ &+\frac{\frac{1}{m+1}r}{\frac{2\beta+1}{2(\beta+1)} +\frac{1}{m+1}} \int_0^t\int_{B(r)}e^{-\omega\eta}|u_\eta|^{m+1} dsd\eta\Big]^{\frac{2\beta+1}{2(\beta+1)} +\frac{1}{m+1}}, \end{align}$

(4.4)

and

$\begin{align} \Big|\mathcal{F}_2(r, t)\Big|&\leq\frac{\sqrt{b_1}}{R_0\omega}\Big[\frac{b_1}{2}r\omega\int_0^t\int_{B(r)}e^{-\omega\eta}|\nabla v|^2dsd\eta+ \frac{1}{2}r\omega\int_0^t\int_{B(r)}e^{-\omega\eta}v_\eta^2 dsd\eta\Big] \\ &+b_2|2t\pi|^{\frac{1}{2(\beta+1)}-\frac{1}{n+1}}R_0^{-\frac{\beta}{\beta+1}-\frac{2}{n+1}} \Big[\frac{\frac{2\beta+1}{2(\beta+1)}r}{\frac{2\beta+1}{2(\beta+1)} +\frac{1}{n+1}}\int_0^t\int_{B(r)}e^{-\omega\eta}|\nabla v|^{2(\beta+1)}dsd\eta \\ &+\frac{\frac{1}{n+1}r}{\frac{2\beta+1}{2(\beta+1)} +\frac{1}{n+1}} \int_0^t\int_{B(r)}e^{-\omega\eta}|v_\eta|^{n+1} dsd\eta\Big]^{\frac{2\beta+1}{2(\beta+1)} +\frac{1}{n+1}}, \end{align}$

(4.5)

where we have chosen $\frac{1}{2(\beta+1)} > \frac{1}{n+1}$ . Inserting (4.4) and (4.5) into (4.1), we obtain

$\begin{align} \Big|\mathcal{F}(r, t)\Big|&\leq c_1\Big[r\frac{\partial}{\partial r}\mathcal{F}(r, t)\Big]+c_2\Big[r\frac{\partial}{\partial r}\mathcal{F}(r, t)\Big]^{\frac{2\beta+1}{2(\beta+1)} +\frac{1}{m+1}} +c_3\Big[r\frac{\partial}{\partial r}\mathcal{F}(r, t)\Big]^{\frac{2\beta+1}{2(\beta+1)} +\frac{1}{n+1}}, \end{align}$

(4.6)

where $c_1 = \frac{\sqrt{b_1}}{R_0\omega}, c_2 = b_2|2t\pi|^{\frac{1}{2(\beta+1)}-\frac{1}{n+1}}R_0^{-\frac{\beta}{\beta+1}-\frac{2}{m+1}}, c_3 = b_2|2t\pi|^{\frac{1}{2(\beta+1)}-\frac{1}{n+1}}R_0^{-\frac{\beta}{\beta+1}-\frac{2}{n+1}}$ .

Next, we will analyze Eq (4.6) in two cases

I. If $\exists r_0\geq R_0$ such that $\mathcal{F}(r_0, t)\geq0$ , then $\mathcal{F}(r, t)\geq\mathcal{F}(r_0, t)\geq0, r\geq r_0.$ Therefore, (4.6) can be rewritten as

$\begin{align} \mathcal{F}(r, t)&\leq c_1\Big[r\frac{\partial}{\partial r}\mathcal{F}(r, t)\Big]+c_2\Big[r\frac{\partial}{\partial r}\mathcal{F}(r, t)\Big]^{\frac{2\beta+1}{2(\beta+1)} +\frac{1}{m+1}} \\ &+c_3\Big[r\frac{\partial}{\partial r}\mathcal{F}(r, t)\Big]^{\frac{2\beta+1}{2(\beta+1)} +\frac{1}{n+1}}, \quad\quad\quad\quad\quad r\geq r_0. \end{align}$

(4.7)

Using Young's inequality, we have

$\begin{align} \Big[r\frac{\partial}{\partial r}\mathcal{F}(r, t)\Big]^{\frac{2\beta+1}{2(\beta+1)} +\frac{1}{m+1}}&\leq\Big(\frac{4\beta+3}{4(\beta+1)}+\frac{1}{2(m+1)}\Big) \Big[r\frac{\partial}{\partial r}\mathcal{F}(r, t)\Big]^\frac{1}{2} \\ &+\Big(\frac{1}{4(\beta+1)}-\frac{1}{2(m+1)}\Big) \Big[r\frac{\partial}{\partial r}\mathcal{F}(r, t)\Big], \end{align}$

(4.8)

$\begin{align} \Big[r\frac{\partial}{\partial r}\mathcal{F}(r, t)\Big]^{\frac{2\beta+1}{2(\beta+1)} +\frac{1}{n+1}}&\leq\Big(\frac{4\beta+3}{4(\beta+1)}+\frac{1}{2(n+1)}\Big) \Big[r\frac{\partial}{\partial r}\mathcal{F}(r, t)\Big]^\frac{1}{2} \\ &+\Big(\frac{1}{4(\beta+1)}-\frac{1}{2(n+1)}\Big) \Big[r\frac{\partial}{\partial r}\mathcal{F}(r, t)\Big]. \end{align}$

(4.9)

Inserting (4.8) and (4.9) into (4.7), we have

$\begin{align} \mathcal{F}(r, t)&\leq c_4\Big[r\frac{\partial}{\partial r}\mathcal{F}(r, t)\Big]^\frac{1}{2}+c_5\Big[r\frac{\partial}{\partial r}\mathcal{F}(r, t)\Big], \quad r\geq r_0. \end{align}$

(4.10)

where $c_4 = c_2\Big(\frac{4\beta+3}{4(\beta+1)}+\frac{1}{2(m+1)}\Big)+c_3\Big(\frac{4\beta+3}{4(\beta+1)}+\frac{1}{2(n+1)}\Big)$ and $c_5 = c_1+c_2\Big(\frac{1}{4(\beta+1)}-\frac{1}{2(m+1)}\Big)+c_3\Big(\frac{1}{4(\beta+1)}-\frac{1}{2(n+1)}\Big)$ . From (4.10) we have

$\begin{align} \frac{\mathcal{F}(r, t)}{c_5}&\leq \Big[\sqrt{r\frac{\partial}{\partial r}\mathcal{F}(r, t)}+\frac{c_4}{2c_5}\Big]^2 -\frac{c^2_4}{4c^2_5}, \quad r\geq r_0 \end{align}$

$\begin{align} \frac{\frac{\partial}{\partial r}\mathcal{F}(r, t)}{\Big[\sqrt{\frac{\mathcal{F}(r, t)}{c_5}+\frac{c^2_4}{4c^2_5}}-\frac{c_4}{2c_5}\Big]^2} \geq \frac{1}{r}, \quad r\geq r_0 \end{align}$

(4.11)

Integrating (4.11) from $r_0$ to $r$ , we get

$\begin{align} 2c_5\Big[ln\Big(\sqrt{\frac{\mathcal{F}(r, t)}{c_5}+\frac{c^2_4}{4c^2_5}}-\frac{c_4}{2c_5}\Big) &-ln\Big(\sqrt{\frac{\mathcal{F}(r_0, t)}{c_5}+\frac{c^2_4}{4c^2_5}}-\frac{c_4}{2c_5}\Big)\Big] \\ -c_4\Big\{\Big[\sqrt{\frac{\mathcal{F}(r, t)}{c_5}+\frac{c^2_4}{4c^2_5}}-\frac{c_4}{2c_5}\Big]^{-1} &-\Big[\sqrt{\frac{\mathcal{F}(r_0, t)}{c_5}+\frac{c^2_4}{4c^2_5}}-\frac{c_4}{2c_5}\Big]^{-1}\Big\} \\ &\geq ln\Big(\frac{r}{r_0}\Big), \quad r\geq r_0 \end{align}$

(4.12)

Dropping the third term on the left of (4.12), we have

$\begin{align} \sqrt{\frac{\mathcal{F}(r, t)}{c_5}+\frac{c^2_4}{4c^2_5}}-\frac{c_4}{2c_5}\geq Q_1(r_0, t)\Big(\frac{r}{r_0}\Big)^\frac{1}{2c_5}, \end{align}$

(4.13)

where $Q_1(r_0, t) = \Big(\sqrt{\frac{\mathcal{F}(r_0, t)}{c_5}+\frac{c^2_4}{4c^2_5}}-\frac{c_4}{2c_5}\Big)\exp\Big\{-\frac{c_4}{2c_5} \Big[\sqrt{\frac{\mathcal{F}(r_0, t)}{c_5}+\frac{c^2_4}{4c^2_5}}-\frac{c_4}{2c_5}\Big]^{-1}\Big\}$ .

In view of

$\sqrt{\frac{\mathcal{F}(r, t)}{c_5}+\frac{c^2_4}{4c^2_5}}\leq\sqrt{\frac{\mathcal{F}(r, t)}{c_5}}+\frac{c_4}{2c_5},$

we have from (4.13)

$\begin{align} \mathcal{F}(r, t)\geq c_5Q_1^2(r_0, t)\Big(\frac{r}{r_0}\Big)^\frac{1}{c_5}. \end{align}$

(4.14)

Combining (4.2) and (4.14), we have

$\begin{align} \lim\limits_{r\rightarrow \infty}&\Big\{r^{-\frac{1}{c_5}}\Big[\frac{1}{2}e^{-\omega t}\int_{r_0}^r\int_{B(\xi)}\Big[|u_t|^2+b_1|\nabla u|^2+\frac{1}{\beta+1}b_2|\nabla u|^{2(\beta+1)}\Big]dsd\xi\\ &+\frac{1}{2}e^{-\omega t}\int_{r_0}^r\int_{B(\xi)}\Big[|v_t|^2+b_1|\nabla v|^2+\frac{1}{\beta+1}b_2|\nabla v|^{2(\beta+1)}+F(u, v)\Big]dsd\xi \\ &+\frac{1}{2}\omega\int_0^t\int_{r_0}^r\int_{B(\xi)}e^{-\omega\eta}\Big[|u_\eta|^2+b_1|\nabla u|^2+\frac{1}{\beta+1}b_2|\nabla u|^{2(\beta+1)}\Big]ds d\xi d\eta\\ &+\frac{1}{2}\omega\int_0^t\int_{r_0}^r\int_{B(\xi)}e^{-\omega\eta}\Big[|v_\eta|^2+b_1|\nabla v|^2+\frac{1}{\beta+1}b_2|\nabla v|^{2(\beta+1)}+F(u, v)\Big]dsd\xi d\eta \\ &+\int_0^t\int_{r_0}^r\int_{B(\xi)}e^{-\omega\eta}|u_\eta|^{m+1}dsd\xi d\eta+\int_0^t\int_{B(r)}e^{-\omega\eta}|v_\eta|^{n+1}dsd\xi d\eta\Big]\Big\} \\ &\geq c_5Q_1^2(r_0, t)r_0^{-\frac{1}{c_5}}. \end{align}$

(4.15)

II. If $\forall r\geq R_0$ such that $\mathcal{F}(r, t) < 0$ , then (4.6) can be rewritten as

$\begin{align} -\mathcal{F}(r, t)&\leq c_1\Big[r\frac{\partial}{\partial r}\mathcal{F}(r, t)\Big]+c_2\Big[r\frac{\partial}{\partial r}\mathcal{F}(r, t)\Big]^{\frac{2\beta+1}{2(\beta+1)} +\frac{1}{m+1}} \\ &+c_3\Big[r\frac{\partial}{\partial r}\mathcal{F}(r, t)\Big]^{\frac{2\beta+1}{2(\beta+1)} +\frac{1}{n+1}}, \quad\quad\quad\quad\quad r\geq R_0. \end{align}$

(4.16)

Without losing generality, we suppose that $m > n > 1$ .

$\begin{align} \Big[r\frac{\partial}{\partial r}\mathcal{F}(r, t)\Big]^{\frac{2\beta+1}{2(\beta+1)} +\frac{1}{n+1}}&\leq\frac{\frac{1}{n+1}-\frac{1}{m+1}}{\frac{2\beta+1}{2(\beta+1)}-\frac{1}{m+1}} \Big[r\frac{\partial}{\partial r}\mathcal{F}(r, t)\Big] \\ &+\frac{\frac{2\beta+1}{2(\beta+1)}-\frac{1}{m+1}}{\frac{2\beta+1}{2(\beta+1)}-\frac{1}{m+1}} \Big[r\frac{\partial}{\partial r}\mathcal{F}(r, t)\Big]^{\frac{2\beta+1}{2(\beta+1)} +\frac{1}{m+1}}, \end{align}$

(4.17)

$\begin{align} \Big[r\frac{\partial}{\partial r}\mathcal{F}(r, t)\Big]&\leq\frac{2\beta(m+3)+4}{(2\beta+1)(m+2)+1} \Big[r\frac{\partial}{\partial r}\mathcal{F}(r, t)\Big]^{\frac{2\beta+1}{2(\beta+1)} +\frac{1}{m+1}} \\ &+\frac{m+1-2(\beta+1)}{(2\beta+1)(m+2)+1} \Big[r\frac{\partial}{\partial r}\mathcal{F}(r, t)\Big]^{2\Big[\frac{2\beta+1}{2(\beta+1)} +\frac{1}{m+1}\Big]}. \end{align}$

(4.18)

Inserting (4.17) and (4.18) into (4.16), we get

$\begin{align} -\mathcal{F}(r, t)&\leq c_6\Big[r\frac{\partial}{\partial r}\mathcal{F}(r, t)\Big]^{\frac{2\beta+1}{2(\beta+1)} +\frac{1}{m+1}} +c_7\Big[r\frac{\partial}{\partial r}\mathcal{F}(r, t)\Big]^{2\Big[\frac{2\beta+1}{2(\beta+1)} +\frac{1}{m+1}\Big]}, \ r\geq R_0, \end{align}$

(4.19)

where $c_6 = \Big[c_1+c_3\frac{\frac{1}{n+1}-\frac{1}{m+1}}{\frac{2\beta+1}{2(\beta+1)}-\frac{1}{m+1}}\Big] \frac{2\beta(m+3)+4}{(2\beta+1)(m+2)+1}+c_3\frac{\frac{2\beta+1}{2(\beta+1)}-\frac{1}{m+1}}{\frac{2\beta+1}{2(\beta+1)}-\frac{1}{m+1}}, c_7 = c_3\frac{m+1-2(\beta+1)}{(2\beta+1)(m+2)+1}$ . From (4.19) we obtain

$\begin{align} r\frac{\partial}{\partial r}\mathcal{F}(r, t)&\geq \Big[\sqrt{\frac{-\mathcal{F}(r, t)}{c_7}+\frac{c_6^2}{4c_7^2}}-\frac{c_6}{2c_7}\Big]^{\frac{1}{\frac{2\beta+1}{2(\beta+1)} +\frac{1}{m+1}}}, \ r\geq R_0 \end{align}$

$\begin{align} 2c_7\frac{\sqrt{\frac{-\mathcal{F}(r, t)}{c_7}+\frac{c_6^2}{4c_7^2}}}{\Big[\sqrt{\frac{-\mathcal{F}(r, t)}{c_7}+\frac{c_6^2}{4c_7^2}}-\frac{c_6}{2c_7}\Big]^{\frac{1}{\frac{2\beta+1}{2(\beta+1)} +\frac{1}{m+1}}}}d\Big\{\sqrt{\frac{-\mathcal{F}(r, t)}{c_7}+\frac{c_6^2}{4c_7^2}}-\frac{c_6}{2c_7}\Big\}&\leq-\frac{1}{r}, \ r\geq R_0 \end{align}$

(4.20)

Integrating (4.20) from $R_0$ to $r$ , we obtain

$\begin{align} &\frac{2c_7\Big(\frac{2\beta+1}{2(\beta+1)}+\frac{1}{m+1}\Big)}{\frac{\beta}{2(\beta+1)}+\frac{2}{m+1}} \Big[\sqrt{\frac{-\mathcal{F}(r, t)}{c_7}+\frac{c_6^2}{4c_7^2}}-\frac{c_6}{2c_7}\Big]^\frac{\frac{\beta}{2(\beta+1)}+\frac{2}{m+1}}{\frac{2\beta+1}{2(\beta+1)}+\frac{1}{m+1}} \\ &-\frac{2c_7\Big(\frac{2\beta+1}{2(\beta+1)}+\frac{1}{m+1}\Big)}{\frac{\beta}{2(\beta+1)}+\frac{2}{m+1}} \Big[\sqrt{\frac{-\mathcal{F}(R_0, t)}{c_7}+\frac{c_6^2}{4c_7^2}}-\frac{c_6}{2c_7}\Big]^\frac{\frac{\beta}{2(\beta+1)}+\frac{2}{m+1}}{\frac{2\beta+1}{2(\beta+1)}+\frac{1}{m+1}} \\ &-\frac{c_6\Big(\frac{2\beta+1}{2(\beta+1)}+\frac{1}{m+1}\Big)}{\frac{1}{2(\beta+1)}-\frac{1}{m+1}} \Big[\sqrt{\frac{-\mathcal{F}(r, t)}{c_7}+\frac{c_6^2}{4c_7^2}}-\frac{c_6}{2c_7}\Big]^{-\frac{\frac{1}{2(\beta+1)}-\frac{1}{m+1}}{\frac{2\beta+1}{2(\beta+1)}+\frac{1}{m+1}}} \\ &+\frac{c_6\Big(\frac{2\beta+1}{2(\beta+1)}+\frac{1}{m+1}\Big)}{\frac{1}{2(\beta+1)}-\frac{1}{m+1}} \Big[\sqrt{\frac{-\mathcal{F}(R_0, t)}{c_7}+\frac{c_6^2}{4c_7^2}}-\frac{c_6}{2c_7}\Big]^{-\frac{\frac{1}{2(\beta+1)}-\frac{1}{m+1}}{\frac{2\beta+1}{2(\beta+1)}+\frac{1}{m+1}}} \\ &\leq ln\Big(\frac{R_0}{r}\Big). \end{align}$

(4.21)

Dropping the first and fourth terms on the left of (4.21), we obtain

$\begin{align} \sqrt{\frac{-\mathcal{F}(r, t)}{c_7}+\frac{c_6^2}{4c_7^2}} \leq\Big[c_8ln\Big(\frac{r}{R_0}\Big) -Q_2(R_0, t)\Big]^{-\frac{\frac{2\beta+1}{2(\beta+1)}+\frac{1}{m+1}}{\frac{1}{2(\beta+1)}-\frac{1}{m+1}}} +\frac{c_6}{2c_7}, \end{align}$

(4.22)

where $Q_2(R_0, t) = \frac{2c_7\Big(\frac{1}{2(\beta+1)}-\frac{1}{m+1}\Big)}{c_6\Big(\frac{\beta}{2(\beta+1)}+\frac{2}{m+1}\Big)} \Big[\sqrt{\frac{-\mathcal{F}(R_0, t)}{c_7}+\frac{c_6^2}{4c_7^2}}-\frac{c_6}{2c_7}\Big]^\frac{\frac{\beta}{2(\beta+1)}+\frac{2}{m+1}}{\frac{2\beta+1}{2(\beta+1)}+\frac{1}{m+1}}$ and $c_8 = \frac{\frac{1}{2(\beta+1)}-\frac{1}{m+1}}{c_6\Big(\frac{2\beta+1}{2(\beta+1)} +\frac{1}{m+1}\Big)}$ .

Squaring (4.22) we have

$\begin{align} -\mathcal{F}(r, t)& \leq c_7\Big[c_8ln\Big(\frac{r}{R_0}\Big) -Q_2(R_0, t)\Big]^{-\frac{\frac{2\beta+1}{\beta+1}+\frac{2}{m+1}}{\frac{1}{2(\beta+1)}-\frac{1}{m+1}}} \\ &+c_6\Big[c_8ln\Big(\frac{r}{R_0}\Big) -Q_2(R_0, t)\Big]^{-\frac{\frac{2\beta+1}{2(\beta+1)}+\frac{1}{m+1}}{\frac{1}{2(\beta+1)}-\frac{1}{m+1}}}. \end{align}$

(4.23)

From (4.15) and (4.23) we can obtain the following theorem.

Theorem 4.1. Let $(u, v)$ be the solution of (1.1), (1.2), (2.1) and (2.2) with $\rho(q^2) = b_1+b_2q^{2\beta}$ , where $\frac{1}{2(\beta+1)} > \max\Big\{\frac{1}{m+1}, \frac{1}{n+1}\Big\}$ . Then for fixed $t$ , when $r\rightarrow \infty$ , $(u, v)$ either grows algebraically or decays logarithmically. The growth rate is at least as fast as $z^\frac{1}{c_5}$ and the decay rate is at least as fast as $\Big(ln r\Big)^{-\frac{\frac{2\beta+1}{2(\beta+1)}+\frac{1}{m+1}}{\frac{1}{2(\beta+1)}-\frac{1}{m+1}}}$ .

Remark 4.1. Obviously, in this case of $\rho(q^2) = b_1+b_2q^{2\beta}$ , the decay rate obtained by Theorem 4.1 is slower than that obtained by Theorem 2.1 and Theorem 3.1.

5. A nonlinear system of viscoelastic type

In this section, we concern with a system of two coupled viscoelastic equations

$\begin{align} u_{tt}&-\Delta u+\int_0^th_1(t-\eta)\Delta u(\eta)d\eta+f_1(u, v) = 0, \end{align}$

(5.1)

$\begin{align} v_{tt}&-\Delta v+\int_0^th_2(t-\eta)\Delta v(\eta)d\eta+f_2(u, v) = 0, \end{align}$

(5.2)

which describes the interaction between two different fields arising in viscoelasticity. In (5.1) and (5.2), $0 < t < T$ and $h_1, h_2$ are differentiable functions satisfying $h_1(0), h_2(0) > 0$ and

$\begin{align} 2\Big(&\int_0^T h_1^2(T-\eta)d\eta\Big), \ \frac{4T}{h_1(0)}\Big[\int_0^T \Big(h_1'(T-\eta)-\omega h_1(T-\eta)\Big)^2d\eta\Big] \leq h_1(0), \end{align}$

(5.3)

$\begin{align} 2\Big(&\int_0^T h_2^2(T-\eta)d\eta\Big), \ \frac{4T}{h_2(0)}\Big[\int_0^T \Big(h_2'(T-\eta)-\omega h_2(T-\eta)\Big)^2d\eta\Big] \leq h_2(0). \end{align}$

(5.4)

Messaoudi and Tatar ^[24] considered the system (5.1) and (5.2) in a bounded domain and proved the uniform decay for the solution when $t\rightarrow \infty$ . For more special cases, one can refer to ^[25,26,27]. They mainly concerned the well-posedness of the solutions and proved that the solutions decayed uniformly under some suitable conditions. However, the present paper extends the previous results to Eqs (5.1) and (5.2) in an exterior region. We consider Eqs (5.1) and (5.2) with the initial-boundary conditions (2.1) and (2.2) in $\Omega$ .

We define two functions

$\begin{align} G_1(r, t)& = \int_0^t\int_{B(r)}e^{-\omega\eta}\nabla u\cdot\frac{\mathit{\boldsymbol{x}}}{r}u_\eta dsd\eta -\int_0^t\int_{B(r)}e^{-\omega\eta}\Big(\int_0^\eta h_1(\eta-s)\nabla uds\Big)\cdot\frac{\mathit{\boldsymbol{x}}}{r}u_\eta dsd\eta \\ &\doteq I_1+I_2, \end{align}$

(5.5)

$\begin{align} G_2(r, t)& = \int_0^t\int_{B(r)}e^{-\omega\eta}\nabla v\cdot\frac{\mathit{\boldsymbol{x}}}{r}v_\eta dsd\eta -\int_0^t\int_{B(r)}e^{-\omega\eta}\Big(\int_0^\eta h_2(\eta-s)\nabla vds\Big)\cdot\frac{\mathit{\boldsymbol{x}}}{r}v_\eta dsd\eta \\ &\doteq J_1+J_2. \end{align}$

(5.6)

Integrating (5.5) from $r_0$ to $r$ and using (5.1), (5.2), (2.1), (2.2) and the divergence theorem, we have

$\begin{align} G_1(r, t)& = G_1(r_0, t)+\frac{1}{2}\int_{r_0}^r\int_{B(\xi)}e^{-\omega t}\Big[|u_t|^2+|\nabla u|^2\Big]dsd\xi+\frac{1}{2}\omega\int_0^t\int_{r_0}^r\int_{B(\xi)}e^{-\omega \eta}|u_\eta|^2dsd\xi d\eta\\ &+\frac{1}{2}\omega\int_0^t\int_{r_0}^r\int_{B(\xi)}e^{-\omega \eta}|\nabla u|^2dsd\xi d\eta+\int_0^t\int_{r_0}^r\int_{B(\xi)}e^{-\omega \eta}h_1(0)|\nabla u|^2dsd\xi d\eta\\ &-\int_{r_0}^r\int_{B(\xi)}e^{-\omega t}\Big(\int_0^t h_1(t-\tau)\nabla ud\tau\Big)\cdot\nabla udsd\xi\\ &+\int_0^t\int_{r_0}^r\int_{B(\xi)}e^{-\omega \eta}\Big[\int_0^\eta \Big(h_1'(\eta-\tau)-\omega h_1(\eta-\tau)\Big)\nabla ud\tau\Big]\cdot\nabla udsd\xi d\eta\\ &+\int_0^t\int_{r_0}^r\int_{B(\xi)}e^{-\omega \eta}f_1(u, v)u_\eta dsd\xi d\eta. \end{align}$

(5.7)

From (5.7) it follows that

$\begin{align} \frac{\partial}{\partial z}G_1(r, t)& = \frac{1}{2}\int_{B(r)}e^{-\omega t}\Big[|u_t|^2+|\nabla u|^2\Big]ds+\frac{1}{2}\omega\int_0^t\int_{B(r)}e^{-\omega \eta}|u_\eta|^2dsd\eta \\ &+\frac{1}{2}\omega\int_0^t\int_{B(r)}e^{-\omega \eta}|\nabla u|^2dsd\eta+\int_0^t\int_{B(r)}e^{-\omega \eta}h_1(0)|\nabla u|^2dsd\eta \\ &-\int_{B(r)}e^{-\omega t}\Big(\int_0^t h_1(t-\tau)\nabla ud\tau\Big)\cdot\nabla uds\\ &+\int_0^t\int_{B(r)}e^{-\omega \eta}\Big[\int_0^\eta \Big(h_1'(\eta-\tau)-\omega h_1(\eta-\tau)\Big)\nabla ud\tau\Big]\cdot\nabla uds d\eta\\ &+\int_0^t\int_{B(r)}e^{-\omega \eta}f_1(u, v)u_\eta dsd\eta. \end{align}$

(5.8)

By the Young inequality and the Hölder inequality, we have

$\begin{align} \Big|&-\int_{B(r)}e^{-\omega t}\Big(\int_0^t h_1(\eta-\tau)\nabla ud\tau\Big)\cdot\nabla uds\Big|\\ &\leq\int_{B(r)}e^{-\omega t}\Big[\Big(\int_0^t h_1(t-s)\nabla u(s)ds\Big)^2+\frac{1}{4}|\nabla u|^2\Big]ds\\ &\leq\Big(\int_0^t h_1^2(t-\tau)d\tau\Big)\int_0^t\int_{B(r)}e^{-\omega \eta}|\nabla u|^2dsd\eta+\frac{1}{4}\int_{B(r)}e^{-\omega t}|\nabla u|^2ds, \end{align}$

(5.9)

and

$\begin{align} \Big|&\int_0^t\int_{B(r)}e^{-\omega \eta}\Big[\int_0^\eta \Big(h_1'(\eta-\tau)-\omega h_1(\eta-\tau)\Big)\nabla ud\tau\Big]\cdot\nabla uds d\eta\Big|\\ &\leq \frac{t}{h_1(0)}\Big[\int_0^t \Big(h_1'(\eta-\tau)-\omega h_1(\eta-\tau)\Big)^2d\tau\Big]\int_0^t\int_{B(r)}e^{-\omega \eta}|\nabla u|^2dsd\eta\\ &+\frac{1}{4}\int_0^t\int_{B(r)}e^{-\omega \eta}h_1(0)|\nabla u|^2dsd\eta. \end{align}$

(5.10)

Inserting (5.9) and (5.10) into (5.8) and using (5.3), we have

$\begin{align} \frac{\partial}{\partial z}G_1(r, t)&\geq\frac{1}{2}\int_{B(r)}e^{-\omega t}\Big[|u_t|^2+\frac{1}{2}|\nabla u|^2\Big]ds+\frac{1}{2}\omega\int_0^t\int_{B(r)}e^{-\omega \eta}|u_\eta|^2ds d\eta\\ &+\frac{1}{2}\omega\int_0^t\int_{B(r)}e^{-\omega \eta}|\nabla u|^2dsd\eta+\int_0^t\int_{B(r)}e^{-\omega \eta}f_1(u, v)u_\eta ds d\eta. \end{align}$

(5.11)

Similar to (5.11), we also have for $G_2(r, t)$

$\begin{align} \frac{\partial}{\partial z}G_2(r, t)&\geq\frac{1}{2}\int_{B(r)}e^{-\omega t}\Big[|v_t|^2+\frac{1}{2}|\nabla v|^2\Big]ds+\frac{1}{2}\omega\int_0^t\int_{B(r)}e^{-\omega \eta}|v_\eta|^2dsd\eta\\ &+\frac{1}{2}\omega\int_0^t\int_{B(r)}e^{-\omega \eta}|\nabla v|^2dsd\eta+\int_0^t\int_{B(r)}e^{-\omega \eta}f_2(u, v)v_\eta dsd\eta. \end{align}$

(5.12)

If we define

$G(r, t) = G_1(r, t)+G_2(r, t),$

then by (5.11) and (5.12) we have

$\begin{align} \frac{\partial}{\partial z}G(r, t)&\geq\frac{1}{2}\int_{B(r)}e^{-\omega t}\Big[|u_t|^2+|v_t|^2+\frac{1}{2}|\nabla u|^2+\frac{1}{2}|\nabla v|^2+2F(u, v)\Big]ds \\ &+\frac{1}{2}\omega\int_0^t\int_{B(r)}e^{-\omega \eta}\Big[|u_\eta|^2+|v_\eta|^2+|\nabla u|^2+|\nabla v|^2+2F(u, v)\Big]dsd\eta. \end{align}$

(5.13)

On the other hand, we bound $G(r, t)$ by $\frac{\partial}{\partial r}G(r, t)$ . Using the Hölder inequality, the AG mean inequality, (5.3) and combining (5.13), we have

$\begin{align} \Big|I_1\Big|+\Big|J_1\Big|&\leq\Big(\int_0^t\int_{B(r)}e^{-\omega\eta}|\nabla u|^2 dAd\eta\cdot\int_0^t\int_{B(r)}e^{-\omega\eta}|u_\eta|^2 dsd\eta\Big)^\frac{1}{2} \\ &+\Big(\int_0^t\int_{B(r)}e^{-\omega\eta}|\nabla v|^2 dAd\eta\cdot\int_0^t\int_{B(r)}e^{-\omega\eta}|v_\eta|^2 dsd\eta\Big)^\frac{1}{2} \\ &\leq\frac{1}{\omega}\Big[\frac{\partial}{\partial r}G(r, t)\Big], \end{align}$

(5.14)

and

$\begin{align} \Big|I_2\Big|&\leq\Big(\int_0^t\int_{B(r)}e^{-\omega\eta}\Big|\int_0^\eta h_1(\eta-\tau)\nabla ud\tau\Big|^2 dsd\eta\cdot\int_0^t\int_{B(r)}e^{-\omega\eta}|u_\eta|^2dsd\eta\Big)^\frac{1}{2} \\ &\leq\Big(\int_0^t\int_{B(r)}e^{-\omega\eta}\Big(\int_0^\eta h^2_1(\eta-\tau)d\tau\Big)\Big(\int_0^\eta |\nabla u|^2d\tau\Big) dsd\eta\cdot\int_0^t\int_{B(r)}e^{-\omega\eta}|u_\eta|^2dsd\eta\Big)^\frac{1}{2} \\ &\leq t\Big(\int_0^t h^2_1(\eta-\tau)d\tau\Big)^\frac{1}{2}\Big(\int_0^t\int_{B(r)}e^{-\omega\eta} |\nabla u|^2 dsd\eta\cdot\int_0^t\int_{B(r)}e^{-\omega\eta}|u_\eta|^2dsd\eta\Big)^\frac{1}{2} \\ &\leq\frac{t}{2}\Big(\int_0^t h^2_1(\eta-\tau)d\tau\Big)^\frac{1}{2}\Big[\int_0^t\int_{B(r)}e^{-\omega\eta} |\nabla u|^2 dsd\eta+\int_0^t\int_{B(r)}e^{-\omega\eta}|u_\eta|^2dsd\eta\Big]\\ &\leq\frac{T}{2\omega}\sqrt{h_1(0)}\Big[\frac{\partial}{\partial r}G(r, t)\Big]. \end{align}$

(5.15)

Similar to (5.15), we have

$\begin{align} \Big|J_2\Big|\leq\frac{T}{2\omega}\sqrt{h_2(0)}\Big[\frac{\partial}{\partial r}G(r, t)\Big]. \end{align}$

(5.16)

Inserting (5.14)–(5.16) into (5.5) and (5.6), we have

$\begin{align} \Big|G(r, t)\Big|\leq c_9\frac{1}{\omega}\Big[\frac{\partial}{\partial r}G(r, t)\Big], \end{align}$

(5.17)

where $c_9 = \frac{T}{2}(\sqrt{h_1(0)}+\sqrt{h_2(0)})+2$ .

We can follow the similar arguments given in the previous sections to obtain the following theorem.

Theorem 5.1. Let $(u, v)$ be the solution of (5.1), (5.2), (2.1) and (2.2) in $\Omega$ , and (5.3) and (5.4) hold. For fixed $t$ ,

(1) If $\exists\ R_0\geq r_0$ , $G(R_0, t)\geq0$ , then

$G(r, t)\geq G(R_0, t)e^{\frac{\omega}{b_9}(r-R_0)}.$

(2) If $\forall\ r\geq r_0$ , $G(r, t) < 0$ , then

$-G(r, t)\leq \Big[-G(r_0, t)\Big]e^{-\frac{\omega}{b_9}(r-r_0)}.$

Again, the rate of growth or decay obtained in this case is arbitrarily large

Remark 5.1. It is clear that the above analysis can be adapted without difficulties to the equation (see ^[28,29])

$u_{tt}-k_0\triangle u+\int_0^tdiv[a(x)h(t-s)\nabla u(s)]ds+b(x)g(u_t)+f(u) = 0$

and the equation (see ^[30])

$|u_{t}|^\sigma u_{tt}-k_0\triangle u-\Delta u_{tt}+\int_0^th(t-s)\Delta u(s)]ds-\gamma\Delta u_t = 0$

with some suitable $g$ and $a(x)+b(x)\geq b_{10} > 0$ and $k_0, \sigma, \gamma > 0.$

6. Conclusions

In this paper, we have considered several situations where the solutions of Eqs (1.1) and (1.2) either grow or decay exponentially or polynomially. We emphasize that the Poincaré inequality on the cross sections is not used in this paper. Thus, our results also hold for the two-dimensional case. On the other hand, there are some deeper problems to be studied in this paper. We can continue to study the continuous dependence of coefficients in the equation as that in ^[31]. These are the issues we will continue to study in the future.

Use of AI tools declaration

The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

The authors express their heartfelt thanks to the editors and referees who have provided some important suggestions. This work was supported by the Guangzhou Huashang College Tutorial Scientific Research Project (2022HSDS09), National Natural Science Foundation of China (11371175) and the Research Team Project of Guangzhou Huashang College (2021HSKT01).

Conflict of interest

The authors declare that there is no conflict of interest.

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2. The Phragmén-Lindelöf type alternative result under A1

3. The Phragmén-Lindelöf type alternative result under A2

4. The Phragmén-Lindelöf type alternative result when ρ(q2)=b1+b2q2β \rho(q^2)=b_1+b_2q^{2\beta}

5. A nonlinear system of viscoelastic type

6. Conclusions

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4. The Phragmén-Lindelöf type alternative result when $\rho(q^2)=b_1+b_2q^{2\beta}$