*Article Phragmén-Lindelöf* **Alternative Results for a Class of Thermoelastic Plate**

**Shiguang Luo 1, Jincheng Shi 2,\* and Baiping Ouyang <sup>2</sup>**


**Abstract:** The spatial properties of solutions for a class of thermoelastic plate with biharmonic operator were studied. The energy method was used. We constructed an energy expression. A differential inequality which the energy expression was controlled by a second-order differential inequality is deduced. The *Phragmen*´ -*Lindelo f* ¨ alternative results of the solutions were obtained by solving the inequality. These results show that the Saint-Venant principle is also valid for the hyperbolic–hyperbolic coupling equations. Our results can been seen as a version of symmetry in inequality for studying the *Phragmen*´ -*Lindelo f* ¨ alternative results.

**Keywords:** thermoelastic plate; *Phragmen*´ -*Lindelo f* ¨ alternative; Saint-Venant principle; biharmonic equation

## **1. Introduction**

Saint Venant principle points out that for any equilibrium force system on an elastic body, if its action point is limited to a given ball, the displacement and stress generated by the equilibrium force system at any point where the distance from the load is far greater than the radius of the ball can be ignored. This principle is widely used in engineering mechanics in practice. Many papers in the literature dealt with the study of the Saint-Venant principle. For example, Horgan and Knowles [1] and Horgan [2,3] studied the Saint-Venant principle in different equations and different situations. The traditional characteristic of the Saint-Venant theorem is to derive the energy decay estimates of the solutions. Usually, these decays are exponential with the spatial distance from the finite end to the infinity. In order to have some understandings about the study of the Saint-Venant Principle, one could refer to the papers [4–9]. In recent years, the studies of Saint-Venant principle for hyperbolic or quasihyperbolic equations are abundant. Especially for the studies of the spatial behavior of viscoelasticity equations, we could see papers [10–13]. When the spatial variable tends to infinity, the solution is decreasing. In the research of solution spacial decay estimates, people often need to add the solutions must satisfy some constraints at infinity. Many scholars have begun to study the *Phragmen*´ -*Lindelo f* ¨ alternative results of solutions. The advantage of this situation is that there is no need to add constraints on the solutions at infinity. The classical *Phragmen*´ -*Lindelo f* ¨ theorem states that the solutions of the harmonic equation must grow exponentially or decay exponentially with distance from the finite end of the cylinder to infinity. Payne and Schaefer [14] extended the study from harmonic equation to biharmonic equation. They obtained the *Phragmen*´ -*Lindelo f* ¨ alternative results for biharmonic equation in three different regions. Literatures [15–18] studied the spatial behaviors of biharmonic equations by various methods. In particular, we can see that Liu and Lin [19] studied the spatial properties for time-dependent stokes equation. They transformed the equation to a biharmonic equation and obtained the *Phragmen*´ -*Lindelo f* ¨ results by using a second-order differential inequality. The abovementioned studies from the

*Phragmén-Lindelöf* Alternative Results for a Class of Thermoelastic Plate. *Symmetry* **2021**, *13*, 2256. https:// doi.org/10.3390/sym13122256

**Citation:** Luo, S.; Shi, J.; Ouyang, B.

Academic Editors: Sergei D. Odintsov and Nicusor Minculete

Received: 1 November 2021 Accepted: 23 November 2021 Published: 26 November 2021

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**Copyright:** © 2021 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (https:// creativecommons.org/licenses/by/ 4.0/).

literature all consider a single equation. Recently, there some new results about the studies of the hyperbolic equations or biharmonic Equations have been published (see [20–25]). For studies of other equations using energy method, see [26–29].

The domain we consider in this paper is defined as follows:

$$
\Omega\_0 = \{(\mathbf{x}\_1, \mathbf{x}\_2) | \mathbf{x}\_1 > 0, 0 < \mathbf{x}\_2 < h\},
\tag{1}
$$

with *h* is a given positive number. We now give the following notation:

$$L\_z = \{ (\mathbf{x}\_1, \mathbf{x}\_2) | \mathbf{x}\_1 = z \ge 0, 0 \le \mathbf{x}\_2 \le h \}. \tag{2}$$

In reference [30], the authors studied the coupled system of wave-plate type with thermal effect. They obtained the results of the analytic property and the exponential stability of the *C*0-semigroup. The equations are as follows:

$$\begin{cases} \rho\_1 u\_{,tt} - \Delta u - \mu \Delta u\_{,t} + \lambda \Delta v \text{s.} = 0, \\ \rho\_2 v\_{,tt} + \gamma \Delta^2 v \text{s.} + \lambda \Delta u + m \Delta \theta = 0, \\ \tau \theta\_{,t} - k \Delta \theta - m \Delta v\_{,t} = 0. \end{cases} \tag{3}$$

The model is used to represent the evolution process of a system which contains an elastic membrane and plate. The plate has an elastic force and a thermal effect (see [31]). Here *u* is the vertical deflection of the membrane and *v* is the vertical deflection of the plate. *θ* is the difference of temperature. The coefficients *ρ*1, *ρ*2, *μ*, *λ*, *m*,*τ*, *γ*, and *k* are nonnegative constants. Δ denotes the Laplace operator, and Δ<sup>2</sup> denotes the biharmonic operator.

In the present paper, we consider the case when *τ* = 0. In this case, Equation (3) can be rewritten as:

$$
\rho\_1 \mu\_{,tt} - \Delta u - \mu \Delta u\_{,t} + \lambda \Delta v \text{s.} = 0,\tag{4}
$$

$$
\rho\_2 v\_{,tt} + \gamma \Delta^2 v + \lambda \Delta \mu - \frac{m^2}{k} \Delta v\_{,t} = 0. \tag{5}
$$

We give the following initial and boundary value conditions:

$$\begin{cases} v(\mathbf{x}\_1, 0, t) = u(\mathbf{x}\_1, 0, t) = u\_2(\mathbf{x}\_1, 0, t) = 0, \mathbf{x}\_1 > 0, t > 0, \\ v(\mathbf{x}\_1, h, t) = u(\mathbf{x}\_1, h, t) = u\_2(\mathbf{x}\_1, h, t) = 0, \mathbf{x}\_1 > 0, t > 0, \\ v(0, \mathbf{x}\_2, t) = g\_1(\mathbf{x}\_2, t), 0 \le \mathbf{x}\_2 \le h, t > 0, \\ u(0, \mathbf{x}\_2, t) = g\_2(\mathbf{x}\_2, t), 0 \le \mathbf{x}\_2 \le h, t > 0, \\ u\_1(0, \mathbf{x}\_2, t) = g\_3(\mathbf{x}\_2, t), 0 \le \mathbf{x}\_2 \le h, t > 0, \\ v(\mathbf{x}\_1, \mathbf{x}\_2, 0) = u(\mathbf{x}\_1, \mathbf{x}\_2, 0) = u\_t(\mathbf{x}\_1, \mathbf{x}\_2, 0), 0 \le \mathbf{x}\_2 \le h, \mathbf{x}\_1 > 0, \end{cases} \tag{6}$$

where *gi*(*x*2, *t*), *i* = 1, 2, 3 are the given functions and meet the following compatibility conditions:

$$\begin{cases} \mathcal{g}\_1(0,t) = \mathcal{g}\_1(h,t) = \mathcal{g}\_{1,2}(0,t) = \mathcal{g}\_{1,2}(h,t) = 0, \\ \mathcal{g}\_2(0,t) = \mathcal{g}\_2(h,t) = \mathcal{g}\_{2,2}(0,t) = \mathcal{g}\_{2,2}(h,t) = 0, \\ \mathcal{g}\_3(0,t) = \mathcal{g}\_3(h,t) = \mathcal{g}\_{3,2}(0,t) = \mathcal{g}\_{3,2}(h,t) = 0, \\ \mathcal{g}\_4(\mathbf{x}\_2,0) = \mathcal{g}\_2(\mathbf{x}\_2,0) = \mathcal{g}\_3(\mathbf{x}\_2,0) = 0. \end{cases} \tag{7}$$

We try to establish the *Phragmen*´ -*Lindelo f* ¨ alternative results for the solutions of the biharmonic Equations (4) and (5) under conditions (6) and (7). We firstly define an energy expression of the solutions, then we derive that the energy expression satisfies a second-order differential inequality, and finally we obtain the *Phragmen*´ -*Lindelo f* ¨ alternative results of the solutions by solving the second-order inequality. For the inequality is symmetry, we show the application of symmetry in mathematical inequalities in practice. Since the system is a hyperbolic–hyperbolic coupling system, how to define the appropriate energy function will be the greatest innovation in this paper. How to control the energy

function will be the difficulty of this paper. No similar studies have been found on the spatial properties for the solutions of the biharmonic equations with hyperbolic–hyperbolic coupling equations using the second-order differential inequality. In this paper, we use the comma to represent partial differentiation. , *k* denotes the differentiation with respect to the direction *xk*, thus *u*,*<sup>α</sup>* denotes *<sup>∂</sup><sup>u</sup> <sup>∂</sup>x<sup>α</sup>* , and *<sup>u</sup>*,*<sup>t</sup>* denotes *<sup>∂</sup><sup>u</sup> <sup>∂</sup><sup>t</sup>* . The usual summation convection

is employed with repeated Greek subscripts *<sup>α</sup>* summed from 1 to 2. Hence, *<sup>u</sup>α*,*<sup>α</sup>* <sup>=</sup> <sup>2</sup> ∑ *α*=1 *∂uα <sup>∂</sup>x<sup>α</sup>* . The symbol *dA* = *dx*1*dx*2.

## **2. Energy Expression Φ**(*z*,*t*)

In order to get the *Phragmen*´ -*Lindelo f* ¨ alternative results, we must define an energy expression for the solutions. This expression plays an important role in obtaining our results. The energy expression will be constructed by the following Lemmas.

**Lemma 1.** *Let u and v be classical solutions of problems* (4)*–*(7)*, we define the a function ϕ*1(*z*, *t*) *as:*

$$\varrho\_1(z,t) = \frac{\mu}{2} \int\_0^t \int\_{L\_{\varepsilon}} \exp(-\omega \eta) u\_{\mathcal{\mathcal{J}}}^2 \mathrm{d}x\_2 \mathrm{d}\eta + \lambda \int\_0^t \int\_{L\_{\varepsilon}} \exp(-\omega \eta) u \upsilon\_{\mathcal{J}} \mathrm{d}x\_2 \mathrm{d}\eta. \tag{8}$$

*ϕ*1(*z*, *t*) *can also be expressed as:*

*<sup>ϕ</sup>*1(*z*, *<sup>t</sup>*) = *ωρ*<sup>1</sup> 2 *t* 0 *z* 0 *Lξ exp*(−*ωη*)(*<sup>z</sup>* <sup>−</sup> *<sup>ξ</sup>*)*u*<sup>2</sup> ,*η*d*A*d*η* <sup>+</sup> *<sup>ρ</sup>*<sup>1</sup> 2 *z* 0 *Lξ exp*(−*ωt*)(*<sup>z</sup>* <sup>−</sup> *<sup>ξ</sup>*)*u*<sup>2</sup> ,*t*d*A* <sup>+</sup> *<sup>ω</sup>* 2 *t* 0 *z* 0 *Lξ exp*(−*ωη*)(*z* − *ξ*)*u*,*αu*,*α*d*A*d*η* + 1 2 *z* 0 *Lξ exp*(−*ωt*)(*z* − *ξ*)*u*,*αu*,*α*d*A* + *μ t* 0 *z* 0 *Lξ exp*(−*ωη*)(*z* − *ξ*)*u*,*αηu*,*αη*d*A*d*η* + *λ t* 0 *z* 0 *Lξ exp*(−*ωη*)(*z* − *ξ*)*u*,*αv*,*αη*d*A*d*η* − *t* 0 *z* 0 *Lξ exp*(−*ωη*)*u*,*ηu*,1d*A*d*η* <sup>+</sup> *λω <sup>t</sup>* 0 *z* 0 *Lξ exp*(−*ωη*)(*z* − *ξ*)*u*,*αv*,*α*d*A*d*η* + *λ z* 0 *Lξ exp*(−*ωt*)(*z* − *ξ*)*u*,*αv*,*α*d*A* + *λ t* 0 *z* 0 *Lξ exp*(−*ωη*)*u*,1*v*,*η*d*A*d*η* <sup>+</sup> *λω <sup>t</sup>* 0 *z* 0 *Lξ exp*(−*ωη*)*uv*,1d*A*d*η* + *λ z* 0 *Lξ exp*(−*ωt*)*uv*,1d*A* + *k*1(*z*, *t*), (9)

*where*

$$\begin{split} k\_{1}(z,t) &= z \int\_{0}^{t} \int\_{L\_{0}} \exp(-\omega \eta) u\_{\eta} u\_{1} \mathrm{d}x\_{2} \mathrm{d}\eta + \frac{\mu}{2} \int\_{0}^{t} \int\_{L\_{0}} \exp(-\omega \eta) u\_{\eta}^{2} \mathrm{d}x\_{2} \mathrm{d}\eta \\ &+ z \mu \int\_{0}^{t} \int\_{L\_{0}} \exp(-\omega \eta) u\_{\eta} u\_{1\eta} \mathrm{d}x\_{2} \mathrm{d}\eta + \lambda \int\_{0}^{t} \int\_{L\_{0}} \exp(-\omega \eta) u u\_{\eta} \mathrm{d}x\_{2} \mathrm{d}\eta. \end{split} \tag{10}$$

**Proof.** Multiplying both sides of Equation (4) by *exp*(−*ωη*)(*z* − *ξ*)*u*,*<sup>η</sup>* and integrating, we obtain

0 = *t* 0 *z* 0 *Lξ exp*(−*ωη*)(*z* − *ξ*)*u*,*η*(*ρ*1*u*,*ηη* − *u*,*αα* − *μu*,*ααη* + *λv*,*αα*)d*A*d*η* <sup>=</sup> *ωρ*<sup>1</sup> 2 *t* 0 *z* 0 *Lξ exp*(−*ωη*)(*<sup>z</sup>* <sup>−</sup> *<sup>ξ</sup>*)*u*<sup>2</sup> ,*η*d*A*d*η* <sup>+</sup> *<sup>ρ</sup>*<sup>1</sup> 2 *z* 0 *Lξ exp*(−*ωt*)(*<sup>z</sup>* <sup>−</sup> *<sup>ξ</sup>*)*u*<sup>2</sup> ,*t*d*A* <sup>+</sup> *<sup>ω</sup>* 2 *t* 0 *z* 0 *Lξ exp*(−*ωη*)(*z* − *ξ*)*u*,*αu*,*α*d*A*d*η* + 1 2 *z* 0 *Lξ exp*(−*ωt*)(*z* − *ξ*)*u*,*αu*,*α*d*A* − *t* 0 *z* 0 *Lξ exp*(−*ωη*)*u*,*ηu*,1d*A*d*η* + *z t* 0 *L*0 *exp*(−*ωη*)*u*,*ηu*,1d*x*2d*η* + *μ t* 0 *z* 0 *Lξ exp*(−*ωη*)(*z* − *ξ*)*u*,*αηu*,*αη*d*A*d*η* − *μ* 2 *t* 0 *Lz exp*(−*ωη*)*u*<sup>2</sup> ,*η*d*x*2d*η* <sup>+</sup> *<sup>μ</sup>* 2 *t* 0 *L*0 *exp*(−*ωη*)*u*<sup>2</sup> ,*η*d*x*2d*η* + *zμ t* 0 *L*0 *exp*(−*ωη*)*u*,*ηu*,1*η*d*x*2d*η* + *λ t* 0 *z* 0 *Lξ exp*(−*ωη*)(*z* − *ξ*)*u*,*αv*,*αη*d*A*d*η* <sup>+</sup> *λω <sup>t</sup>* 0 *z* 0 *Lξ exp*(−*ωη*)(*z* − *ξ*)*u*,*αv*,*α*d*A*d*η* + *λ z* 0 *Lξ exp*(−*ωt*)(*z* − *ξ*)*u*,*αv*,*α*d*A* + *λ t* 0 *z* 0 *Lξ exp*(−*ωη*)*u*,1*v*,*η*d*A*d*η* − *λ t* 0 *Lz exp*(−*ωη*)*uv*,*η*d*x*2d*η* + *λ t* 0 *L*0 *exp*(−*ωη*)*uv*,*η*d*x*2d*η* <sup>+</sup> *λω <sup>t</sup>* 0 *z* 0 *Lξ exp*(−*ωη*)*uv*,1d*A*d*η* + *λ z* 0 *Lξ exp*(−*ωt*)*uv*,1d*A*. (11)

By combining Equations (8) and (11), we can get (9). The proof of Lemma 1 is finished.

**Lemma 2.** *We suggest u and v are the classical solutions of problems* (4)*–*(7)*, and we define a function ϕ*2(*z*, *t*) *as:*

$$\begin{split} \rho\_{2}(z,t) &= \int\_{0}^{t} \int\_{0}^{z} \int\_{L\_{\xi}} \exp(-\omega\eta)(z-\xi)(u\_{\text{ad}})^{2} \mathrm{d}A d\eta + \mu\omega \int\_{0}^{t} \int\_{0}^{z} \int\_{L\_{\xi}} \exp(-\omega\eta)(z-\xi)(u\_{\text{ad}})^{2} \mathrm{d}A d\eta \\ &+ \mu \int\_{0}^{z} \int\_{L\_{\xi}} \exp(-\omega\eta)(z-\xi)(u\_{\text{ad}})^{2} \mathrm{d}A - \rho\_{1} \int\_{0}^{t} \int\_{0}^{z} \int\_{L\_{\xi}} \exp(-\omega\eta)(z-\xi)u\_{\text{ad}}u\_{\text{d}} \eta \mathrm{d}A d\eta \\ &+ \frac{\rho\_{1}}{2} \int\_{0}^{t} \int\_{0}^{z} \int\_{L\_{\xi}^{z}} \exp(-\omega\eta)u\_{\text{ad}}^{2} \mathrm{d}A d\eta - \omega\rho\_{1} \int\_{0}^{t} \int\_{0}^{z} \int\_{L\_{\xi}} \exp(-\omega\eta)(z-\xi)u\_{\text{d}}\eta \mathrm{d}A d\eta \\ &- \rho\_{1} \int\_{0}^{z} \int\_{L\_{\xi}} \exp(-\omega t)(z-\xi)u\_{\text{d}}\wp \mathrm{d}A - \lambda \int\_{0}^{t} \int\_{0}^{z} \int\_{L\_{\xi}} \exp(-\omega\eta)(z-\xi)u\_{\text{d}}\wp \mathrm{d}A d\eta \\ &+ k\_{2}(z,t), \end{split} \tag{12}$$

*ϕ*2(*z*, *t*) *can also be expressed as:*

$$\mathfrak{gl}\_2(z,t) = -\frac{\rho\_1}{2} \int\_0^t \int\_{L\_x} \exp(-\omega \eta) u\_{,\eta}^2 \mathrm{d}x\_2 \mathrm{d}\eta,\tag{13}$$

*where*

$$k\_2(z,t) = \frac{\rho\_1 z}{2} \int\_0^t \int\_{L\_0} \exp(-\omega \eta) u\_{,\eta}^2 \mathrm{d}x\_2 \mathrm{d}\eta - \rho\_1 z \int\_0^t \int\_{L\_0} \exp(-\omega \eta) u\_{,\eta} u\_{,1\eta} \mathrm{d}x\_2 \mathrm{d}\eta.$$

**Proof.** Multiplying both sides of Equation (4) by *exp*(−*ωη*)(*z* − *ξ*)*u*,*ββ* and integrating, we can obtain

0 = *t* 0 *z* 0 *Lξ exp*(−*ωη*)(*z* − *ξ*)*u*,*ββ*(*ρ*1*u*,*ηη* − *u*,*αα* − *μu*,*ααη* + *λv*,*αα*)d*A*d*η* = *ρ*<sup>1</sup> *t* 0 *z* 0 *Lξ exp*(−*ωη*)(*z* − *ξ*)*u*,*βηu*,*βη*d*A*d*η* − *ρ*<sup>1</sup> *t* 0 *z* 0 *Lξ exp*(−*ωη*)(*z* − *ξ*)*u*,*ηu*,1*η*d*A*d*η* + *ρ*1*z t* 0 *L*0 *exp*(−*ωη*)*u*,*ηu*,1*η*d*x*2d*η* + *ωρ*<sup>1</sup> *t* 0 *z* 0 *Lξ exp*(−*ωη*)(*z* − *ξ*)*u*,*ββu*,*η*d*A*d*η* + *ρ*<sup>1</sup> *z* 0 *Lξ exp*(−*ωt*)(*z* − *ξ*)*u*,*ββu*,*t*d*A* − *t* 0 *z* 0 *Lξ exp*(−*ωη*)(*<sup>z</sup>* <sup>−</sup> *<sup>ξ</sup>*)(*u*,*αα*)2d*A*d*<sup>η</sup>* <sup>−</sup> *μω <sup>t</sup>* 0 *z* 0 *Lξ exp*(−*ωη*)(*<sup>z</sup>* <sup>−</sup> *<sup>ξ</sup>*)(*u*,*αα*)2d*A*d*<sup>η</sup>* − *μ z* 0 *Lξ exp*(−*ωt*)(*<sup>z</sup>* <sup>−</sup> *<sup>ξ</sup>*)(*u*,*αα*)2d*<sup>A</sup>* + *λ t* 0 *z* 0 *Lξ exp*(−*ωη*)(*z* − *ξ*)*u*,*ββv*,*αα*d*A*d*η*. (14)

By combining Equations (12) and (14), we can get (13). The proof of Lemma 2 is finished.

**Lemma 3.** *We suggest u and v are classical solutions of problems* (4)*–*(7)*, and we define a function ϕ*3(*z*, *t*) *as:*

*<sup>ϕ</sup>*3(*z*, *<sup>t</sup>*) = *ωρ*<sup>2</sup> 2 *t* 0 *z* 0 *Lξ exp*(−*ωη*)(*<sup>z</sup>* <sup>−</sup> *<sup>ξ</sup>*)*v*<sup>2</sup> ,*η*d*A*d*η* <sup>+</sup> *<sup>ρ</sup>*<sup>2</sup> 2 *z* 0 *Lξ exp*(−*ωt*)(*<sup>z</sup>* <sup>−</sup> *<sup>ξ</sup>*)*v*<sup>2</sup> ,*t*d*x*2d*η* <sup>+</sup> *<sup>m</sup>*<sup>2</sup> *k t* 0 *z* 0 *Lξ exp*(−*ωη*)(*z* − *ξ*)*v*,*αηv*,*αη*d*A*d*η* <sup>+</sup> *γω* 2 *t* 0 *z* 0 *Lξ exp*(−*ωη*)(*z* − *ξ*)*v*,*αβv*,*αβ*d*A*d*η* <sup>+</sup> *<sup>γ</sup>* 2 *z* 0 *Lξ exp*(−*ωt*)*v*,*αβv*,*αβ*d*A* − *λ t* 0 *z* 0 *Lξ exp*(−*ωη*)(*z* − *ξ*)*v*,*αηu*,*α*d*A*d*η* + *λ t* 0 *z* 0 *Lξ exp*(−*ωη*)*v*,*ηu*,1d*A*d*η* − 2*γ t* 0 *z* 0 *Lξ exp*(−*ωη*)*v*,*αηv*,*<sup>α</sup>*1d*A*d*η* + 2*γκρ*2*ω m*<sup>2</sup> *t* 0 *z* 0 *Lξ exp*(−*ωη*)*v*,*ηv*,1d*A*d*η* + 2*γκρ*<sup>2</sup> *m*<sup>2</sup> *z* 0 *Lξ exp*(−*ωη*)*v*,*tv*,1d*A* <sup>−</sup> <sup>2</sup>*κλγ m*<sup>2</sup> *t* 0 *z* 0 *Lξ exp*(−*ωη*)*u*,*αv*,1*α*d*A*d*η* + *k*3(*z*, *t*). (15)

*ϕ*3(*z*, *t*) *can also be expressed as:*

$$\begin{split} q\_{3}(z,t) &= \frac{m^{2}}{2k} \int\_{0}^{t} \int\_{L\_{z}} \exp(-\omega \eta) v\_{\gamma \text{d}}^{2} \text{d}x \text{d}\eta \\ &\quad + 2\gamma \int\_{0}^{t} \int\_{L\_{z}} \exp(-\omega \eta) v\_{1\eta} v\_{1\eta} \text{d}x\_{2} \text{d}\eta \\ &\quad - \frac{2k\lambda\gamma}{m^{2}} \int\_{0}^{t} \int\_{L\_{z}} \exp(-\omega \eta) u\_{1} v\_{1\eta} \text{d}x\_{2} \text{d}\eta \\ &\quad - \frac{k\gamma^{2}}{m^{2}} \int\_{0}^{t} \int\_{L\_{z}} \exp(-\omega \eta) v\_{\omega \beta} v\_{\omega \beta} \text{d}x\_{2} \text{d}\eta \\ &\quad + \frac{2k\gamma^{2}}{m^{2}} \int\_{0}^{t} \int\_{L\_{z}} \exp(-\omega \eta) v\_{1\alpha} v\_{\gamma \text{d}\beta} \text{d}x\_{2} \text{d}\eta \\ &\quad - \frac{2k\gamma^{2}}{m^{2}} \int\_{0}^{t} \int\_{L\_{z}} \exp(-\omega \eta) v\_{1\beta} v\_{1\beta\beta} \text{d}x\_{2} \text{d}\eta, \end{split} \tag{16}$$

*with*

$$\begin{split} k\_{3}(z,t) &= \left(\frac{m^{2}}{2k} + \frac{k\rho\_{2}\gamma}{m^{2}}\right) \int\_{0}^{t} \int\_{L\_{0}} \exp(-\omega\eta) v\_{\gamma\delta}^{2} \mathrm{d}x\_{2} \mathrm{d}\eta \\ &\quad + \frac{m^{2}}{k} z \int\_{0}^{t} \int\_{L\_{0}} \exp(-\omega\eta) v\_{\gamma\delta} v\_{\gamma 1\eta} \mathrm{d}x\_{2} \mathrm{d}\eta \\ &\quad + \gamma z \int\_{0}^{t} \int\_{L\_{0}} \exp(-\omega\eta) v\_{\gamma 3\eta} v\_{\gamma 11} \mathrm{d}x\_{2} \mathrm{d}\eta \\ &\quad - \gamma z \int\_{0}^{t} \int\_{L\_{0}} \exp(-\omega\eta) v\_{\gamma 3} v\_{\gamma 1\beta\delta} \mathrm{d}x\_{2} \mathrm{d}\eta \\ &\quad + 2\gamma \int\_{0}^{t} \int\_{L\_{0}} \exp(-\omega\eta) v\_{\gamma 1\eta} v\_{\gamma 1} \mathrm{d}x\_{2} \mathrm{d}\eta \\ &\quad - \frac{2k\gamma}{m^{2}} \int\_{0}^{t} \int\_{L\_{0}} \exp(-\omega\eta) v\_{\gamma 3\delta} v\_{\gamma 11} \mathrm{d}x\_{2} \mathrm{d}\eta \\ &\quad - \frac{k\gamma^{2}}{m^{2}} \int\_{0}^{t} \int\_{L\_{0}} \exp(-\omega\eta) v\_{\gamma 4\delta} v\_{\gamma 3\delta} \mathrm{d}x\_{2} \mathrm{d}\eta \\ &\quad + \frac{2k\gamma^{2}}{m^{2}} \int\_{0}^{t} \int\_{L\_{0}} \exp(-\omega\eta) v\_{\gamma 1\delta} v\_{\gamma 1\delta} \mathrm{d}x\_{2} \mathrm{d}\eta. \end{split}$$

**Proof.** Multiplying both sides of Equation (5) by *exp*(−*ωη*)(*z* − *ξ*)*v*,*<sup>η</sup>* and integrating, we can obtain

0 = *t* 0 *z* 0 *Lξ exp*(−*ωη*)(*<sup>z</sup>* <sup>−</sup> *<sup>ξ</sup>*)*v*,*η*(*ρ*2*v*,*ηη* <sup>+</sup> *<sup>γ</sup>v*,*ααββ* <sup>+</sup> *<sup>λ</sup>u*,*αα* <sup>−</sup> *<sup>m</sup>*<sup>2</sup> *<sup>k</sup> <sup>v</sup>*,*ααη*)d*A*d*<sup>η</sup>* <sup>=</sup> *ωρ*<sup>2</sup> 2 *t* 0 *z* 0 *Lξ exp*(−*ωη*)(*<sup>z</sup>* <sup>−</sup> *<sup>ξ</sup>*)*v*<sup>2</sup> ,*η*d*A*d*η* <sup>+</sup> *<sup>ρ</sup>*<sup>2</sup> 2 *z* 0 *Lξ exp*(−*ωt*)(*<sup>z</sup>* <sup>−</sup> *<sup>ξ</sup>*)*v*<sup>2</sup> ,*t*d*A* − *λ t* 0 *z* 0 *Lξ exp*(−*ωη*)(*z* − *ξ*)*v*,*αηu*,*α*d*A*d*η* + *λ t* 0 *z* 0 *Lξ exp*(−*ωη*)*v*,*ηu*,1d*A*d*η* + *γ t* 0 *z* 0 *Lξ exp*(−*ωη*)(*z* − *ξ*)*v*,*ηv*,*ααββ*d*A*d*η* <sup>+</sup> *<sup>m</sup>*<sup>2</sup> *k t* 0 *z* 0 *Lξ exp*(−*ωη*)(*z* − *ξ*)*v*,*αηv*,*αη*d*A*d*η* <sup>−</sup> *<sup>m</sup>*<sup>2</sup> 2*k t* 0 *Lz exp*(−*ωη*)*v*<sup>2</sup> ,*η*d*x*2d*η* <sup>+</sup> *<sup>m</sup>*<sup>2</sup> 2*k t* 0 *L*0 *exp*(−*ωη*)*v*<sup>2</sup> ,*η*d*x*2d*η* <sup>+</sup> *<sup>m</sup>*<sup>2</sup> *k z t* 0 *L*0 *exp*(−*ωη*)*v*,*ηv*,1*η*d*x*2d*η*. (17)

Next, we begin to deal with the term *γ t* 0 *z* 0 *<sup>L</sup><sup>ξ</sup> exp*(−*ωη*)(*<sup>z</sup>* − *<sup>ξ</sup>*)*v*,*ηv*,*ααββ*d*A*d*η*.

$$\begin{split} &\gamma \int\_{0}^{t} \int\_{0}^{z} \int\_{L\_{\xi}} \exp(-\omega \eta) (z - \xi) v\_{\eta} v\_{\omega \alpha \beta \beta} \mathrm{d}A \mathrm{d}\eta \\ &= \frac{\gamma \omega}{2} \int\_{0}^{t} \int\_{0}^{z} \int\_{L\_{\xi}} \exp(-\omega \eta) (z - \xi) v\_{\omega \beta} v\_{\omega \beta} \mathrm{d}A \mathrm{d}\eta \\ &+ \frac{\gamma}{2} \int\_{0}^{z} \int\_{L\_{\xi}} \exp(-\omega \eta) (z - \xi) v\_{\omega \beta} v\_{\omega \beta} \mathrm{d}A \\ &- 2\gamma \int\_{0}^{t} \int\_{0}^{z} \int\_{L\_{\xi}} \exp(-\omega \eta) v\_{\omega \beta} v\_{\omega 1} \mathrm{d}A \mathrm{d}\eta \\ &+ \gamma z \int\_{0}^{t} \int\_{L\_{0}} \exp(-\omega \eta) v\_{\omega \eta} v\_{\omega 1} \mathrm{d}x\_{2} \mathrm{d}\eta \\ &- \gamma z \int\_{0}^{t} \int\_{L\_{0}} \exp(-\omega \eta) v\_{\eta} v\_{\omega \beta} \mathrm{d}x\_{2} \mathrm{d}\eta. \end{split} \tag{18}$$

Now let us deal with −2*γ t* 0 *z* 0 *<sup>L</sup><sup>ξ</sup> exp*(−*ωη*)*v*,*αηv*,*<sup>α</sup>*1d*A*d*η*.

$$\begin{split} & -2\gamma \int\_{0}^{t} \int\_{0}^{z} \int\_{L\_{\xi}} \exp(-\omega \eta \eta) v\_{,\alpha \eta} v\_{,\alpha 1} \mathbf{d} A \mathbf{d} \eta \\ & = 2\gamma \int\_{0}^{t} \int\_{0}^{z} \int\_{L\_{\xi}} \exp(-\omega \eta) v\_{,\alpha \alpha \eta} v\_{,1} \mathbf{d} A \mathbf{d} \eta \\ & - 2\gamma \int\_{0}^{t} \int\_{L\_{\xi}} \exp(-\omega \eta) v\_{,1 \eta} v\_{,1} \mathbf{d} x\_{2} \mathbf{d} \eta \\ & + 2\gamma \int\_{0}^{t} \int\_{L\_{0}} \exp(-\omega \eta) v\_{,1 \eta} v\_{,1} \mathbf{d} x\_{2} \mathbf{d} \eta. \end{split} \tag{19}$$

Using the Equation (5), we can get

2*γ t* 0 *z* 0 *Lξ exp*(−*ωη*)*v*,*ααηv*,1d*A*d*η* = 2*γ t* 0 *z* 0 *Lξ exp*(−*ωη*)( *<sup>k</sup> <sup>m</sup>*<sup>2</sup> *<sup>ρ</sup>*2*v*,*ηη* <sup>+</sup> *k <sup>m</sup>*<sup>2</sup> *<sup>γ</sup>v*,*ααββ* <sup>+</sup> *k <sup>m</sup>*<sup>2</sup> *<sup>λ</sup>u*,*αα*)*v*,1d*A*d*<sup>η</sup>* <sup>=</sup> <sup>−</sup>*kρ*2*<sup>γ</sup> m*<sup>2</sup> *t* 0 *Lz exp*(−*ωη*)*v*<sup>2</sup> ,*η*d*x*2d*η* + *kρ*2*γ m*<sup>2</sup> *t* 0 *L*0 *exp*(−*ωη*)*v*<sup>2</sup> ,*η*d*x*2d*η* + 2*kγρ*2*ω m*<sup>2</sup> *t* 0 *z* 0 *Lξ exp*(−*ωη*)*v*,*ηv*,1d*A*d*η* + 2*kγρ*<sup>2</sup> *m*<sup>2</sup> *z* 0 *Lξ exp*(−*ωη*)*v*,*tv*,1d*A* + 2*kγ*<sup>2</sup> *m*<sup>2</sup> *t* 0 *z* 0 *Lξ exp*(−*ωη*)*v*,1*v*,*ααββ*d*A*d*η* <sup>−</sup> <sup>2</sup>*kλγ m*<sup>2</sup> *t* 0 *z* 0 *Lξ exp*(−*ωη*)*u*,*αv*,1*α*d*A*d*η* + 2*kλγ m*<sup>2</sup> *t* 0 *Lz exp*(−*ωη*)*u*,1*v*,11d*x*2d*η* <sup>−</sup> <sup>2</sup>*kλγ m*<sup>2</sup> *t* 0 *L*0 *exp*(−*ωη*)*u*,1*v*,11d*x*2d*η*. (20)

Now, let us deal with the term <sup>2</sup>*kγ*<sup>2</sup> *m*<sup>2</sup> *t* 0 *z* 0 *<sup>L</sup><sup>ξ</sup> exp*(−*ωη*)*v*,1*v*,*ααββ*d*A*d*η*.

2*kγ*<sup>2</sup> *m*<sup>2</sup> *t* 0 *z* 0 *Lξ exp*(−*ωη*)*v*,1*v*,*ααββ*d*A*d*η* <sup>=</sup> <sup>−</sup>2*kγ*<sup>2</sup> *m*<sup>2</sup> *t* 0 *z* 0 *Lξ exp*(−*ωη*)*v*,1*αv*,*αββ*d*A*d*η* + 2*kγ*<sup>2</sup> *m*<sup>2</sup> *t* 0 *Lz exp*(−*ωη*)*v*,1*v*,1*ββ*d*x*2d*η* <sup>−</sup> <sup>2</sup>*kγ*<sup>2</sup> *m*<sup>2</sup> *t* 0 *L*0 *exp*(−*ωη*)*v*,1*v*,1*ββ*d*x*2d*η* <sup>=</sup> *<sup>k</sup>γ*<sup>2</sup> *m*<sup>2</sup> *t* 0 *Lz exp*(−*ωη*)*v*,*αβv*,*αβ*d*x*2d*<sup>η</sup>* <sup>−</sup> *<sup>k</sup>γ*<sup>2</sup> *m*<sup>2</sup> *t* 0 *L*0 *exp*(−*ωη*)*v*,*αβv*,*αβ*d*x*2d*η* <sup>−</sup> <sup>2</sup>*kγ*<sup>2</sup> *m*<sup>2</sup> *t* 0 *Lz exp*(−*ωη*)*v*,1*αv*,1*α*d*x*2d*η* + 2*kγ*<sup>2</sup> *m*<sup>2</sup> *t* 0 *L*0 *exp*(−*ωη*)*v*,1*αv*,1*α*d*x*2d*η* + 2*kγ*<sup>2</sup> *m*<sup>2</sup> *t* 0 *Lz exp*(−*ωη*)*v*,1*v*,1*ββ*d*x*2d*<sup>η</sup>* <sup>−</sup> <sup>2</sup>*kγ*<sup>2</sup> *m*<sup>2</sup> *t* 0 *L*0 *exp*(−*ωη*)*v*,1*v*,1*ββ*d*x*2d*η*. (21)

A combination of (15) and (17)–(21), we obtain

$$\begin{split} \rho\_{3}(z,t) &= \frac{m^{2}}{2k} \int\_{0}^{t} \int\_{\mathcal{L}\_{z}} \exp(-\omega\eta) v\_{\gamma\eta}^{2} \mathrm{d}x\_{2} \mathrm{d}\eta + 2\gamma \int\_{0}^{t} \int\_{\mathcal{L}\_{z}} \exp(-\omega\eta) v\_{,1\eta} v\_{,1} \mathrm{d}x\_{2} \mathrm{d}\eta \\ &- \frac{2k\lambda\gamma}{m^{2}} \int\_{0}^{t} \int\_{\mathcal{L}\_{z}} \exp(-\omega\eta) u\_{,1} v\_{,11} \mathrm{d}x\_{2} \mathrm{d}\eta - \frac{k\gamma^{2}}{m^{2}} \int\_{0}^{t} \int\_{\mathcal{L}\_{z}} \exp(-\omega\eta) v\_{,\mu\beta} v\_{,\mu\beta} \mathrm{d}x\_{2} \mathrm{d}\eta \\ &+ \frac{2k\gamma^{2}}{m^{2}} \int\_{0}^{t} \int\_{\mathcal{L}\_{z}} \exp(-\omega\eta) v\_{,1a} v\_{,1a} \mathrm{d}x\_{2} \mathrm{d}\eta - \frac{2k\gamma^{2}}{m^{2}} \int\_{0}^{t} \int\_{\mathcal{L}\_{z}} \exp(-\omega\eta) v\_{,1} v\_{,1\beta\beta} \mathrm{d}x\_{2} \mathrm{d}\eta. \end{split}$$

The proof of Lemma 3 is finished.

**Lemma 4.** *We suggest u and v are classical solutions of problems* (4)*–*(7)*, and we define a function ϕ*4(*z*, *t*) *as :*

$$\begin{split} q\_{4}(z,t) &= -\frac{\rho\_{2}}{2} \int\_{0}^{t} \int\_{L\_{\varepsilon}} \exp(-\omega \eta) \boldsymbol{\upsilon}\_{,\eta}^{2} \mathrm{d}\mathbf{x}\_{2} \mathrm{d}\eta + \gamma \int\_{0}^{t} \int\_{L\_{\varepsilon}} \exp(-\omega \eta) \boldsymbol{\upsilon}\_{,1\alpha\beta} \boldsymbol{\upsilon}\_{,\alpha\beta} \mathrm{d}\mathbf{x}\_{2} \mathrm{d}\eta \\ & \quad - \frac{\gamma}{2} \int\_{0}^{t} \int\_{L\_{z}} \exp(-\omega \eta) \boldsymbol{\upsilon}\_{,1\alpha} \boldsymbol{\upsilon}\_{,1\alpha} \mathrm{d}\mathbf{x}\_{2} \mathrm{d}\eta + \frac{\gamma}{2} \int\_{0}^{t} \int\_{L\_{z}} \exp(-\omega \eta) \boldsymbol{\upsilon}\_{,11}^{2} \mathrm{d}\mathbf{x}\_{2} \mathrm{d}\eta \\ & \quad - \frac{\gamma}{2} \int\_{0}^{t} \int\_{L\_{z}} \exp(-\omega \eta) \boldsymbol{\upsilon}\_{,12}^{2} \mathrm{d}\mathbf{x}\_{2} \mathrm{d}\eta. \end{split} \tag{22}$$

*Then, ϕ*4(*z*, *t*) *can also be expressed as:*

*ϕ*4(*z*, *t*) = −*ρ*<sup>2</sup> *t* 0 *z* 0 *Lξ exp*(−*ωη*)(*<sup>z</sup>* <sup>−</sup> *<sup>ξ</sup>*)*v*<sup>2</sup> ,1*η*d*A*d*<sup>η</sup>* <sup>+</sup> *<sup>m</sup>*<sup>2</sup> 2*k z* 0 *Lξ exp*(−*ωt*)(*z* − *ξ*)*v*,1*αv*,1*α*d*A* + *γ t* 0 *z* 0 *Lξ exp*(−*ωη*)*v*,1*αβv*,1*αβ*d*A*d*η* − *ρ*2*ω t* 0 *z* 0 *Lξ exp*(−*ωη*)(*z* − *ξ*)*v*,11*v*,*η*d*A*d*η* − *ρ*<sup>2</sup> *z* 0 *Lξ exp*(−*ωt*)(*z* − *ξ*)*v*,11*v*,*t*d*A* + *λ t* 0 *z* 0 *Lξ exp*(−*ωη*)(*z* − *ξ*)*v*,11*αu*,*α*d*A*d*η* − *λ t* 0 *z* 0 *Lξ exp*(−*ωη*)*v*,11*u*,1d*A*d*η* + *t* 0 *z* 0 *Lξ exp*(−*ωη*)*v*,12*v*,2*η*d*A*d*η* + *k*4(*z*, *t*), (23)

*with*

$$\begin{split} k\_{4}(z,t) &= -\frac{\rho\_{2}}{2} \int\_{0}^{t} \int\_{L\_{0}} \exp(-\omega \eta) v\_{\gamma\eta}^{2} \mathrm{d}x\_{2} \mathrm{d}\eta \\ &\quad - \rho\_{2} z \int\_{0}^{z} \int\_{L\_{0}} \exp(-\omega \eta) v\_{\gamma 1\eta} v\_{\gamma 3\eta} \mathrm{d}x\_{2} \mathrm{d}\eta \\ &\quad + \frac{m^{2}}{k} z \int\_{0}^{t} \int\_{L\_{0}} \exp(-\omega \eta) v\_{\gamma 12} v\_{\gamma 2\eta} \mathrm{d}x\_{2} \mathrm{d}\eta \\ &\quad + \gamma \int\_{0}^{t} \int\_{L\_{0}} \exp(-\omega \eta) v\_{\gamma 1 \alpha\beta} v\_{\gamma 3\delta} \mathrm{d}x\_{2} \mathrm{d}\eta \\ &\quad - \frac{\gamma}{2} \int\_{0}^{t} \int\_{L\_{0}} \exp(-\omega \eta) v\_{\gamma 1 \alpha} v\_{\gamma 1 \alpha} \mathrm{d}x\_{2} \mathrm{d}\eta \\ &\quad - \gamma z \int\_{0}^{t} \int\_{L\_{0}} \exp(-\omega \eta) v\_{\gamma 1 \alpha}^{2} v\_{\gamma 1 \delta} \mathrm{d}x\_{2} \mathrm{d}\eta \\ &\quad + \frac{\gamma}{2} \int\_{0}^{t} \int\_{L\_{0}} \exp(-\omega \eta) v\_{\gamma 1 \delta}^{2} \mathrm{d}x\_{2} \mathrm{d}\eta \\ &\quad + \gamma z \int\_{0}^{t} \int\_{L\_{0}} \exp(-\omega \eta) v\_{\gamma 1 \delta} v\_{\gamma 1 \delta} \mathrm{d}x\_{2} \mathrm{d}\eta \\ &\quad + \lambda z \int\_{0}^{t} \int\_{L\_{0}} \exp(-\omega \eta) v$$

**Proof.** Multiplying both sides of Equation (5) by *exp*(−*ωη*)(*z* − *ξ*)*v*,11 and integrating, we can obtain

$$\begin{split} 0 &= \rho\_2 \int\_0^t \int\_0^z \int\_{L\_\zeta^z} \exp(-\omega \eta) (z - \xi) v\_{,11} v\_{,\eta \eta} \mathbf{d} A \mathbf{d} \eta \\ &+ \gamma \int\_0^t \int\_0^z \int\_{L\_\zeta^z} \exp(-\omega \eta) (z - \xi) v\_{,11} v\_{,\mu \alpha \beta \beta} \mathbf{d} A \mathbf{d} \eta \\ &+ \lambda \int\_0^t \int\_0^z \int\_{L\_\zeta^z} \exp(-\omega \eta) (z - \xi) v\_{,11} u\_{,\mu \alpha} \mathbf{d} A \mathbf{d} \eta \\ &- \frac{m^2}{k} \int\_0^t \int\_0^z \int\_{L\_\zeta^z} \exp(-\omega \eta) (z - \xi) v\_{,11} v\_{,\mu \alpha \beta} \mathbf{d} A \mathbf{d} \eta. \end{split} \tag{24}$$

Using the divergence theorem, the first term on the right of Equation (22) can be rewritten as

$$\begin{split} &\rho\_{2}\int\_{0}^{t}\int\_{0}^{z}\int\_{L\_{\zeta}}\exp(-\omega\eta)(z-\xi)v\_{,11}v\_{,\eta\eta}\mathrm{d}A\mathrm{d}\eta \\ &=\rho\_{2}\int\_{0}^{t}\int\_{0}^{z}\int\_{L\_{\zeta}}\exp(-\omega\eta)(z-\xi)v\_{,1\eta}v\_{,1\eta}\mathrm{d}A\mathrm{d}\eta - \frac{\rho\_{2}}{2}\int\_{0}^{t}\int\_{L\_{\zeta}}\exp(-\omega\eta)v\_{,\eta}^{2}\mathrm{d}x\_{2}\mathrm{d}\eta \\ &+\frac{\rho\_{2}}{2}\int\_{0}^{t}\int\_{L\_{0}}\exp(-\omega\eta)v\_{,\eta}^{2}\mathrm{d}x\_{2}\mathrm{d}\eta + \rho\_{2}z\int\_{0}^{t}\int\_{L\_{\zeta}}\exp(-\omega\eta)v\_{,1\eta}v\_{,\eta}\mathrm{d}x\_{2}\mathrm{d}\eta \\ &+\rho\_{2}\omega\int\_{0}^{t}\int\_{0}^{z}\int\_{L\_{\zeta}}\exp(-\omega\eta)(z-\xi)v\_{,11}v\_{,\eta}\mathrm{d}A\mathrm{d}\eta \\ &+\rho\_{2}\int\_{0}^{z}\int\_{L\_{\zeta}}\exp(-\omega t)(z-\xi)v\_{,11}v\_{,\zeta}\mathrm{d}A. \end{split} \tag{25}$$

Similarly, the second term on the right of Equation (22) can be rewritten as

*γ t* 0 *z* 0 *Lξ exp*(−*ωη*)(*z* − *ξ*)*v*,11*v*,*ααββ*d*A*d*η* = −*γ t* 0 *z* 0 *Lξ exp*(−*ωη*)*v*,1*αβv*,1*αβ*d*A*d*η* + *γ t* 0 *Lz exp*(−*ωη*)*v*,1*αβv*,*αβ*d*x*2d*η* − *γ t* 0 *L*0 *exp*(−*ωη*)*v*,1*αβv*,*αβ*d*x*2d*<sup>η</sup>* <sup>−</sup> *<sup>γ</sup>* 2 *t* 0 *Lz exp*(−*ωη*)*v*,1*αv*,1*α*d*x*2d*η* <sup>+</sup> *<sup>γ</sup>* 2 *t* 0 *L*0 *exp*(−*ωη*)*v*,1*αv*,1*α*d*x*2d*η* + *γz t* 0 *L*0 *exp*(−*ωη*)*v*,11*αv*,*<sup>α</sup>*1d*x*2d*η* <sup>+</sup> *<sup>γ</sup>* 2 *t* 0 *Lz exp*(−*ωη*)*v*<sup>2</sup> ,11d*x*2d*<sup>η</sup>* <sup>−</sup> *<sup>γ</sup>* 2 *t* 0 *L*0 *exp*(−*ωη*)*v*<sup>2</sup> ,11d*x*2d*η* − *γ* 2 *t* 0 *Lz exp*(−*ωη*)*v*<sup>2</sup> ,12d*x*2d*<sup>η</sup>* <sup>+</sup> *<sup>γ</sup>* 2 *t* 0 *L*0 *exp*(−*ωη*)*v*<sup>2</sup> ,12d*x*2d*η* − *γz t* 0 *L*0 *exp*(−*ωη*)*v*,11*v*,1*ββ*d*x*2d*η*. (26)

The third term on the right of Equation (22) can be rewritten as

$$\begin{split} &\lambda \int\_{0}^{t} \int\_{0}^{z} \int\_{L\_{\tilde{\xi}}} \exp(-\omega \eta) (z - \tilde{\xi}) v\_{,11} u\_{,\text{ad}} \mathrm{d}A \mathrm{d}\eta \\ &= -\lambda \int\_{0}^{t} \int\_{0}^{z} \int\_{L\_{\tilde{\xi}}} \exp(-\omega \eta) (z - \tilde{\xi}) v\_{,11a} u\_{,a} \mathrm{d}A \mathrm{d}\eta \\ &+ \lambda \int\_{0}^{t} \int\_{0}^{z} \int\_{L\_{\tilde{\xi}}} \exp(-\omega \eta) v\_{,11} u\_{,1} \mathrm{d}A \mathrm{d}\eta \\ &- \lambda z \int\_{0}^{t} \int\_{L\_{0}} \exp(-\omega \eta) v\_{,11} u\_{,1} \mathrm{d}x\_{2} \mathrm{d}\eta. \end{split} \tag{27}$$

The fourth term on the right of Equation (22) can be rewritten as

$$\begin{split} & -\frac{m^2}{k} \int\_0^t \int\_0^z \int\_{L\_\xi} \exp(-\omega \eta) (z - \xi) v\_{,11} v\_{,aa\eta} \mathrm{d}A \mathrm{d}\eta \\ & = -\frac{m^2}{2k} \int\_0^z \int\_{L\_\xi} \exp(-\omega t) (z - \xi) v\_{,1a} v\_{,1a} \mathrm{d}A \\ & -\frac{m^2}{k} z \int\_0^t \int\_{L\_0} \exp(-\omega \eta) (z - \xi) v\_{,12} v\_{,2\eta} \mathrm{d}x\_2 \mathrm{d}\eta \\ & - \int\_0^t \int\_0^z \int\_{L\_\xi} \exp(-\omega \eta) v\_{,12} v\_{,2\eta} \mathrm{d}A \mathrm{d}\eta. \end{split} \tag{28}$$

A combination of (24)–(28) gives (23). The proof of Lemma 4 is finished.

**Lemma 5.** *We define new energy expressions ϕ*(*z*, *t*) *and* Φ(*z*, *t*) *as follows:*

$$
\varrho(z,t) = \varrho\_1(z,t) + k\_1\varrho\_2(z,t) + \varrho\_3(z,t) + k\_2\varrho\_4(z,t),\tag{29}
$$

*and*

$$
\Phi(z,t) = \int\_0^t \varphi(z,s)ds. \tag{30}
$$

*The following second-order partial differential inequality holds*

$$|\Phi(z,t)| \le k\_3 \frac{\partial^2 \Phi(z,t)}{\partial z^2},\tag{31}$$

*where k*<sup>1</sup> *k*<sup>2</sup> *and k*<sup>3</sup> *are positive constance to be defined later.*

**Proof.** From (9), we can get

*∂*2*ϕ*1(*z*, *t*) *<sup>∂</sup>z*<sup>2</sup> <sup>=</sup> *ωρ*<sup>1</sup> 2 *t* 0 *Lz exp*(−*ωη*)*u*<sup>2</sup> ,*η*d*x*2d*η* <sup>+</sup> *<sup>ρ</sup>*<sup>1</sup> 2 *Lz exp*(−*ωt*)*u*<sup>2</sup> ,*t*d*x*<sup>2</sup> <sup>+</sup> *<sup>ω</sup>* 2 *t* 0 *Lz exp*(−*ωη*)*u*,*αu*,*α*d*x*2d*η* + 1 2 *Lz exp*(−*ωt*)*u*,*αu*,*α*d*x*<sup>2</sup> − *λ t* 0 *Lz exp*(−*ωη*)*u*,1*ηu*,1d*x*2d*η* − *λ t* 0 *Lz exp*(−*ωη*)*u*,*ηu*,11d*x*2d*η* + *λ t* 0 *Lz exp*(−*ωη*)*u*,11*v*,*η*d*x*2d*η* + *λ t* 0 *Lz exp*(−*ωη*)*u*,1*v*,1*η*d*x*2d*η* <sup>+</sup> *λω <sup>t</sup>* 0 *Lz exp*(−*ωη*)*u*,1*v*,1d*x*2d*η* <sup>+</sup> *λω <sup>t</sup>* 0 *Lz exp*(−*ωη*)*uv*,11d*x*2d*η* + *μ t* 0 *Lz exp*(−*ωη*)*u*,*αηu*,*αη*d*x*2d*η* + *λ t* 0 *Lz exp*(−*ωη*)*u*,*αv*,*αη*d*x*2d*η* + *λ Lz exp*(−*ωt*)*u*,1*v*,1d*x*<sup>2</sup> + *λ Lz exp*(−*ωt*)*uv*,11d*x*<sup>2</sup> <sup>+</sup> *λω <sup>t</sup>* 0 *Lz exp*(−*ωη*)*u*,*αv*,*α*d*x*2d*η* + *λ Lz exp*(−*ωt*)*u*,*αv*,*α*d*x*2.

(32)

From (12), we can get

$$\begin{split} \frac{\partial^2 q\_2(z,t)}{\partial z^2} &= \int\_0^t \int\_{L\_z} \exp(-\omega \eta) (u\_{\mu\alpha})^2 \mathrm{d}x\_2 \mathrm{d}\eta \\ &+ \mu \omega \int\_0^t \int\_{L\_z} \exp(-\omega \eta) (u\_{\mu\alpha})^2 \mathrm{d}x\_2 \mathrm{d}\eta \\ &+ \mu \int\_{L\_z} \exp(-\omega \eta) (u\_{\mu\alpha})^2 \mathrm{d}x\_2 \\ &- \rho\_1 \int\_0^t \int\_{L\_z} \exp(-\omega \eta) u\_{\beta\eta} u\_{\beta\eta} \mathrm{d}x\_2 \mathrm{d}\eta \\ &+ \rho\_1 \int\_0^t \int\_{L\_z} \exp(-\omega \eta) u\_{1\eta} u\_{1\eta} \mathrm{d}x\_2 \mathrm{d}\eta. \end{split} \tag{33}$$

From (15), we can get

*∂*2*ϕ*3(*z*, *t*) *<sup>∂</sup>z*<sup>2</sup> <sup>=</sup> *ωρ*<sup>2</sup> 2 *t* 0 *Lz exp*(−*ωη*)*v*<sup>2</sup> ,*η*d*x*2d*<sup>η</sup>* <sup>+</sup> *<sup>ρ</sup>*<sup>2</sup> 2 *Lz exp*(−*ωt*)*v*<sup>2</sup> ,*t*d*x*<sup>2</sup> <sup>+</sup> *<sup>m</sup>*<sup>2</sup> *k t* 0 *Lz exp*(−*ωη*)*v*,*αηv*,*αη*d*x*2d*<sup>η</sup>* <sup>+</sup> *γω* 2 *t* 0 *Lz exp*(−*ωη*)*v*,*αβv*,*αβ*d*x*2d*η* <sup>+</sup> *<sup>γ</sup>* 2 *Lz exp*(−*ωt*)*v*,*αβv*,*αβ*d*x*<sup>2</sup> − *λ t* 0 *Lz exp*(−*ωη*)*v*,*αηu*,*α*d*x*2d*η* + *λ t* 0 *Lz exp*(−*ωη*)*v*,1*ηu*,1d*x*2d*η* + *λ t* 0 *Lz exp*(−*ωη*)*v*,*ηu*,11d*x*2d*η* + 2*γκρ*2*ω m*<sup>2</sup> *t* 0 *Lz exp*(−*ωη*)*v*,1*ηv*,1d*x*2d*η* + 2*γκρ*2*ω m*<sup>2</sup> *t* 0 *Lz exp*(−*ωη*)*v*,*ηv*,11d*x*2d*η* + 2*γκρ*<sup>2</sup> *m*<sup>2</sup> *Lz exp*(−*ωt*)*v*,1*tv*,1d*x*<sup>2</sup> + 2*γκρ*<sup>2</sup> *m*<sup>2</sup> *Lz exp*(−*ωt*)*v*,*tv*,11d*x*<sup>2</sup> <sup>−</sup> <sup>2</sup>*κλγ m*<sup>2</sup> *t* 0 *Lz exp*(−*ωη*)*u*,1*αv*,1*α*d*x*2d*<sup>η</sup>* <sup>−</sup> <sup>2</sup>*κλγ m*<sup>2</sup> *t* 0 *Lz exp*(−*ωη*)*u*,*αv*,11*α*d*x*2d*η*. (34)

From (23), we can get

*∂*2*ϕ*4(*z*, *t*) *<sup>∂</sup>z*<sup>2</sup> <sup>=</sup> <sup>−</sup>*ρ*<sup>2</sup> *t* 0 *Lz exp*(−*ωη*)*v*<sup>2</sup> ,1*η*d*x*2d*<sup>η</sup>* <sup>+</sup> *<sup>m</sup>*<sup>2</sup> 2*k Lz exp*(−*ωt*)*v*,1*αv*,1*α*d*x*<sup>2</sup> + *γ t* 0 *Lz exp*(−*ωη*)*v*,1*αβv*,1*αβ*d*x*2d*η* − *ρ*2*ω t* 0 *Lz exp*(−*ωη*)*v*,11*v*,*η*d*x*2d*η* + *ρ*<sup>2</sup> *Lz exp*(−*ωη*)*v*,11*v*,*t*d*x*<sup>2</sup> + *λ t* 0 *Lz exp*(−*ωη*)*v*,11*αu*,*α*d*x*2d*η* − *λ t* 0 *Lz exp*(−*ωη*)*v*,111*u*,1d*x*2d*η* − *λ t* 0 *Lz exp*(−*ωη*)*v*,11*u*,11d*x*2d*η* + *t* 0 *Lz exp*(−*ωη*)*v*,112*v*,2*η*d*x*2d*η* + *t* 0 *Lz exp*(−*ωη*)*v*,12*v*,12*η*d*x*2d*η*. (35)

A combination of (8), (13), (21), (22) and (29) leads to

*<sup>ϕ</sup>*(*z*, *<sup>t</sup>*) = *<sup>μ</sup>* 2 *t* 0 *Lz exp*(−*ωη*)*u*<sup>2</sup> ,*η*d*x*2d*η* + *λ t* 0 *Lz exp*(−*ωη*)*uv*,*η*d*x*2d*η* − *k*<sup>1</sup> *ρ*1 2 *t* 0 *Lz exp*(−*ωη*)*u*<sup>2</sup> ,*η*d*x*2d*η* − *k*<sup>2</sup> *γ* 2 *t* 0 *Lz exp*(−*ωη*)*v*<sup>2</sup> ,12d*x*2d*η* <sup>+</sup> *<sup>m</sup>*<sup>2</sup> 2*k t* 0 *Lz exp*(−*ωη*)*v*<sup>2</sup> ,*η*d*x*2d*η* + 2*γ t* 0 *Lz exp*(−*ωη*)*v*,1*ηv*,1d*x*2d*η* <sup>−</sup> <sup>2</sup>*kλγ m*<sup>2</sup> *t* 0 *Lz exp*(−*ωη*)*u*,1*v*,11d*x*2d*<sup>η</sup>* <sup>−</sup> *<sup>k</sup>γ*<sup>2</sup> *m*<sup>2</sup> *t* 0 *Lz exp*(−*ωη*)*v*,*αβv*,*αβ*d*x*2d*η* + 2*kγ*<sup>2</sup> *m*<sup>2</sup> *t* 0 *Lz exp*(−*ωη*)*v*,1*αv*,1*α*d*x*2d*<sup>η</sup>* <sup>−</sup> <sup>2</sup>*kγ*<sup>2</sup> *m*<sup>2</sup> *t* 0 *Lz exp*(−*ωη*)*v*,1*v*,1*ββ*d*x*2d*η* − *k*<sup>2</sup> *ρ*2 2 *t* 0 *Lz exp*(−*ωη*)*v*<sup>2</sup> ,*η*d*x*2d*η* + *k*2*γ t* 0 *Lz exp*(−*ωη*)*v*,1*αβv*,*αβ*d*x*2d*η* − *k*<sup>2</sup> *γ* 2 *t* 0 *Lz exp*(−*ωη*)*v*,1*αv*,1*α*d*x*2d*η* + *k*<sup>2</sup> *γ* 2 *t* 0 *Lz exp*(−*ωη*)*v*<sup>2</sup> ,11d*x*2d*η*. (36)

Combining (32)–(36), we have

*∂*2*ϕ*(*z*, *t*) *<sup>∂</sup>z*<sup>2</sup> <sup>=</sup> *ωρ*<sup>1</sup> 2 *t* 0 *Lz exp*(−*ωη*)*u*<sup>2</sup> ,*η*d*x*2d*<sup>η</sup>* <sup>+</sup> *<sup>ρ</sup>*<sup>1</sup> 2 *Lz exp*(−*ωt*)*u*<sup>2</sup> ,*t*d*x*<sup>2</sup> <sup>+</sup> *<sup>ω</sup>* 2 *t* 0 *Lz exp*(−*ωη*)*u*,*αu*,*α*d*x*2d*η* + 1 2 *Lz exp*(−*ωt*)*u*,*αu*,*α*d*x*<sup>2</sup> + *u t* 0 *Lz exp*(−*ωη*)*u*,*αηu*,*αη*d*x*2d*<sup>η</sup>* <sup>+</sup> *λω <sup>t</sup>* 0 *Lz exp*(−*ωη*)*u*,*αv*,*α*d*x*2d*η* + *λ Lz exp*(−*ωt*)*u*,*αv*,*α*d*x*<sup>2</sup> + (*ρ*1*k*<sup>1</sup> − *λ*) *t* 0 *Lz exp*(−*ωη*)*u*,1*ηu*,1d*x*2d*η* − *λ t* 0 *Lz exp*(−*ωη*)*u*,*ηu*,11d*x*2d*η* + 2*λ t* 0 *Lz exp*(−*ωη*)*u*,11*v*,*η*d*x*2d*η* + 2*λ t* 0 *Lz exp*(−*ωη*)*u*,1*v*,1*η*d*x*2d*<sup>η</sup>* <sup>+</sup> *λω <sup>t</sup>* 0 *Lz exp*(−*ωη*)*u*,1*v*,1d*x*2d*η* <sup>+</sup> *λω <sup>t</sup>* 0 *Lz exp*(−*ωη*)*uv*,11d*x*2d*η* + *λ Lz exp*(−*ωt*)*u*,1*v*,1d*x*<sup>2</sup> + *λ Lz exp*(−*ωt*)*uv*,11d*x*<sup>2</sup> + (1 + *μω*)*k*<sup>1</sup> *t* 0 *Lz exp*(−*ωη*)(*u*,*αα*)2d*x*2d*<sup>η</sup>* + *μk*<sup>1</sup> *Lz exp*(−*ωt*)(*u*,*αα*)2d*x*<sup>2</sup> <sup>−</sup> *<sup>ρ</sup>*1*k*<sup>1</sup> *t* 0 *Lz exp*(−*ωη*)*u*,*βηu*,*βη*d*x*2d*η* <sup>+</sup> *<sup>ρ</sup>*2*<sup>ω</sup>* 2 *t* 0 *Lz exp*(−*ωη*)*v*<sup>2</sup> ,*η*d*x*2d*<sup>η</sup>* <sup>+</sup> *<sup>ρ</sup>*<sup>1</sup> 2 *Lz exp*(−*ωt*)*v*<sup>2</sup> ,*t*d*x*<sup>2</sup> <sup>+</sup> *<sup>m</sup>*<sup>2</sup> *k t* 0 *Lz exp*(−*ωη*)*v*,*αηv*,*αη*d*x*2d*<sup>η</sup>* <sup>+</sup> *ωγ* 2 *t* 0 *Lz exp*(−*ωt*)*v*,*αβv*,*αβ*d*x*2d*η* <sup>+</sup> *<sup>γ</sup>* 2 *Lz exp*(−*ωt*)*v*,*αβv*,*αβ*d*x*<sup>2</sup> + 2*γkρ*2*ω m*<sup>2</sup> *t* 0 *Lz exp*(−*ωη*)*v*,1*ηv*,1d*x*2d*η* + ( <sup>2</sup>*γkρ*2*<sup>ω</sup> <sup>m</sup>*<sup>2</sup> <sup>−</sup> *<sup>ρ</sup>*2*ωk*2) *t* 0 *Lz exp*(−*ωη*)*v*,*ηv*,11d*x*2d*η* + 2*γkρ*<sup>2</sup> *m*<sup>2</sup> *Lz exp*(−*ωt*)*v*,1*tv*,1d*x*<sup>2</sup> + ( <sup>2</sup>*γkρ*<sup>2</sup> *<sup>m</sup>*<sup>2</sup> <sup>+</sup> *<sup>ρ</sup>*2*k*2) *Lz exp*(−*ωt*)*v*,*tv*,11d*x*<sup>2</sup> <sup>−</sup> <sup>2</sup>*kλγ m*<sup>2</sup> *t* 0 *Lz exp*(−*ωη*)*u*,1*αv*,1*α*d*x*2d*η* + (*λk*<sup>2</sup> <sup>−</sup> <sup>2</sup>*kλγ <sup>m</sup>*<sup>2</sup> ) *t* 0 *Lz exp*(−*ωη*)*u*,*αv*,11*α*d*x*2d*η* − *ρ*2*k*<sup>2</sup> *t* 0 *Lz exp*(−*ωη*)*v*<sup>2</sup> ,1*η*d*x*2d*η* <sup>+</sup> *<sup>m</sup>*<sup>2</sup> <sup>2</sup>*<sup>k</sup> <sup>k</sup>*<sup>2</sup> *Lz exp*(−*ωt*)*v*,1*αv*,1*α*d*x*<sup>2</sup> + *γk*<sup>2</sup> *t* 0 *Lz exp*(−*ωη*)*v*,1*αβv*,1*αβ*d*x*2d*η* − *λk*<sup>2</sup> *t* 0 *Lz exp*(−*ωη*)*v*,111*u*,1d*x*2d*η* − *λk*<sup>2</sup> *t* 0 *Lz exp*(−*ωη*)*v*,11*u*,11d*x*2d*η* + *k*<sup>2</sup> *t* 0 *Lz exp*(−*ωη*)*v*,112*v*,2*η*d*x*2d*η* + *k*<sup>2</sup> *t* 0 *Lz exp*(−*ωη*)*v*,12*u*,12*η*d*x*2d*η*. (37)

Using the results (3.1)–(3.3) in [19], we have

$$\int\_{0}^{t} \int\_{L\_{\pm}} \exp(-\omega \eta) \upsilon\_{,\mathfrak{a}} \upsilon\_{,\mathfrak{a}} \mathrm{d}x\_{2} \mathrm{d}\eta \leq c\_{1} \int\_{0}^{t} \int\_{L\_{\pm}} \exp(-\omega \eta) \upsilon\_{,\mathfrak{a}\beta} \upsilon\_{,\mathfrak{a}\beta} \mathrm{d}x\_{2} \mathrm{d}\eta,\tag{38}$$

$$\int\_{L\_z} \exp(-\omega t) v\_{,\mu} v\_{,\mu} \mathrm{d}x\_2 \le c\_1 \int\_{L\_z} \exp(-\omega t) v\_{,\mu\beta} v\_{,\mu\beta} \mathrm{d}x\_2. \tag{39}$$

$$\int\_{0}^{t} \int\_{L\_{\varepsilon}} \exp(-\omega \eta) u^{2} \mathrm{d}x\_{2} \mathrm{d}\eta \le c\_{2} \int\_{0}^{t} \int\_{L\_{\varepsilon}} \exp(-\omega \eta) u\_{,\mathrm{ad}} \mathrm{u}\_{,\mathrm{ad}} \mathrm{d}x\_{2} \mathrm{d}\eta,\tag{40}$$

$$\int\_{L\_{\varepsilon}} \exp(-\omega t) u^{2} \mathrm{d}x\_{2} \le c\_{2} \int\_{L\_{\varepsilon}} \exp(-\omega t) u\_{\mu a} u\_{\mu a} \mathrm{d}x\_{2},\tag{41}$$

$$\int\_{L\_{\varepsilon}} \exp(-\omega t) \upsilon\_{,1} \upsilon\_{,1} \mathrm{d}x\_{2} \leq c\_{3} \int\_{L\_{\varepsilon}} \exp(-\omega t) \upsilon\_{,1a} \upsilon\_{,1a} \mathrm{d}x\_{2} \mathrm{d}\eta\_{,} \tag{42}$$

with *c*1, *c*2, and *c*<sup>3</sup> are positive constants .

Using the Schwarz inequality, and combining (37)–(42), we obtain

*∂*2*ϕ*(*z*, *t*) *<sup>∂</sup>z*<sup>2</sup> <sup>≥</sup> ( *ωρ*<sup>1</sup> <sup>2</sup> <sup>−</sup> *<sup>λ</sup>* <sup>2</sup> *<sup>ε</sup>*4) *t* 0 *Lz exp*(−*ωη*)*u*<sup>2</sup> ,*η*d*x*2d*<sup>η</sup>* <sup>+</sup> *<sup>ρ</sup>*<sup>1</sup> 2 *Lz exp*(−*ωt*)*u*<sup>2</sup> ,*t*d*x*<sup>2</sup> + [*<sup>ω</sup>* <sup>2</sup> <sup>−</sup> *λω* <sup>2</sup> *<sup>ε</sup>*<sup>1</sup> <sup>−</sup> *<sup>ρ</sup>*1*k*<sup>1</sup> <sup>−</sup> *<sup>λ</sup>* 2*ε*<sup>3</sup> <sup>−</sup> *λε*<sup>6</sup> <sup>−</sup> *λω* <sup>2</sup> *<sup>ε</sup>*<sup>7</sup> <sup>−</sup> *<sup>k</sup>λγ <sup>m</sup>*<sup>2</sup> *<sup>ε</sup>*<sup>15</sup> <sup>−</sup> (*λk*<sup>2</sup> <sup>−</sup> <sup>2</sup>*kλγ <sup>m</sup>*<sup>2</sup> ) *ε*<sup>16</sup> 2 <sup>−</sup> *<sup>λ</sup>k*<sup>2</sup> 2*ε*<sup>17</sup> ] *t* 0 *Lz exp*(−*ωη*)*u*,*αu*,*α*d*x*2d*<sup>η</sup>* + ( <sup>1</sup> <sup>2</sup> <sup>−</sup> *<sup>λ</sup>* <sup>2</sup> *<sup>ε</sup>*<sup>2</sup> <sup>−</sup> *<sup>λ</sup>* <sup>2</sup> *<sup>ε</sup>*9) *Lz exp*(−*ωt*)*u*,*αu*,*α*d*x*<sup>2</sup> + [*<sup>μ</sup>* <sup>−</sup> *<sup>ρ</sup>*1*k*<sup>1</sup> <sup>−</sup> *<sup>λ</sup>* <sup>2</sup> *<sup>ε</sup>*<sup>3</sup> <sup>−</sup> *<sup>ρ</sup>*1*k*1] *t* 0 *Lz exp*(−*ωη*)*u*,*αηu*,*αη*d*x*2d*η* + [(<sup>1</sup> <sup>+</sup> *μω*)*k*<sup>1</sup> <sup>−</sup> <sup>1</sup> 2*ε*<sup>4</sup> <sup>−</sup> *λε*<sup>5</sup> <sup>−</sup> *λωc*<sup>2</sup> <sup>2</sup> *<sup>ε</sup>*<sup>8</sup> <sup>−</sup> *<sup>k</sup>λγ <sup>m</sup>*<sup>2</sup> *<sup>ε</sup>*<sup>14</sup> <sup>−</sup> *<sup>λ</sup>k*<sup>2</sup> 2*ε*<sup>18</sup> ] *t* 0 *Lz exp*(−*ωη*)(*u*,*αη*)2d*x*2d*<sup>η</sup>* + (*μk*<sup>1</sup> <sup>−</sup> *<sup>λ</sup>c*<sup>2</sup> <sup>2</sup> *<sup>ε</sup>*10) *Lz exp*(−*ωt*)(*u*,*αα*)2d*x*<sup>2</sup> + [ *<sup>ρ</sup>*2*<sup>ω</sup>* <sup>2</sup> <sup>−</sup> *<sup>λ</sup> ε*5 − 1 2 ( 2*γkρ*2*ω <sup>m</sup>*<sup>2</sup> <sup>−</sup> *<sup>ρ</sup>*2*ωk*2)*ε*12] *t* 0 *Lz exp*(−*ωη*)*v*<sup>2</sup> ,*η*d*x*2d*η* + [ *<sup>ρ</sup>*<sup>2</sup> <sup>2</sup> <sup>−</sup> <sup>1</sup> 2 ( 2*γkρ*<sup>2</sup> *<sup>m</sup>*<sup>2</sup> <sup>+</sup> *<sup>ρ</sup>*2*k*2)*ε*13] *Lz exp*(−*ωt*)*v*<sup>2</sup> ,*t*d*x*<sup>2</sup> + [*m*<sup>2</sup> *<sup>k</sup>* <sup>−</sup> *<sup>λ</sup> ε*6 <sup>−</sup> *<sup>γ</sup>kρ*2*<sup>ω</sup> <sup>m</sup>*<sup>2</sup> *<sup>ε</sup>*<sup>11</sup> <sup>−</sup> *<sup>k</sup>*<sup>2</sup> 2*ε*<sup>19</sup> − *ρ*2*k*2] *t* 0 *Lz exp*(−*ωη*)*v*,*αηv*,*αη*d*x*2d*η* + [*ωγ* <sup>2</sup> <sup>−</sup> *λωc*<sup>1</sup> 2*ε*<sup>1</sup> <sup>−</sup> *λωc*<sup>1</sup> 2*ε*<sup>7</sup> <sup>−</sup> *λω* 2*ε*<sup>8</sup> <sup>−</sup> *<sup>γ</sup>kρ*2*ωc*<sup>1</sup> *m*2*ε*<sup>11</sup> − 1 2 ( 2*γkρ*2*ω <sup>m</sup>*<sup>2</sup> <sup>−</sup> *<sup>ρ</sup>*2*ωk*2) <sup>1</sup> *ε*<sup>12</sup> <sup>−</sup> *<sup>k</sup>λγ m*<sup>2</sup> 1 *ε*<sup>14</sup> <sup>−</sup> *<sup>λ</sup>k*<sup>2</sup> <sup>2</sup> *<sup>ε</sup>*18] *t* 0 *Lz exp*(−*ωη*)*v*,*αβv*,*αβ*d*x*2d*<sup>η</sup>* + [*<sup>γ</sup>* <sup>2</sup> <sup>−</sup> *<sup>λ</sup>c*<sup>1</sup> 2*ε*<sup>2</sup> ] *Lz exp*(−*ωt*)*v*,*αβv*,*αβ*d*x*<sup>2</sup> + [*m*<sup>2</sup> <sup>2</sup>*<sup>k</sup> <sup>k</sup>*<sup>2</sup> <sup>−</sup> *<sup>λ</sup>c*<sup>3</sup> 2*ε*<sup>9</sup> <sup>−</sup> *<sup>λ</sup>* 2*ε*<sup>10</sup> <sup>−</sup> <sup>1</sup> 2*ε*<sup>13</sup> ( 2*γkρ*<sup>2</sup> *<sup>m</sup>*<sup>2</sup> <sup>+</sup> *<sup>ρ</sup>*2*k*2)] *Lz exp*(−*ωt*)*v*,1*αv*,1*α*d*x*<sup>2</sup> + [*γk*<sup>2</sup> <sup>−</sup> *<sup>k</sup>λγ m*2*ε*<sup>15</sup> <sup>−</sup> (*λk*<sup>2</sup> <sup>−</sup> <sup>2</sup>*kλγ <sup>m</sup>*<sup>2</sup> ) <sup>1</sup> 2*ε*<sup>16</sup> <sup>−</sup> *<sup>λ</sup>k*<sup>2</sup> <sup>2</sup> *<sup>ε</sup>*<sup>17</sup> <sup>−</sup> *<sup>k</sup>*<sup>2</sup> <sup>2</sup> *<sup>ε</sup>*19] *t* 0 *Lz exp*(−*ωη*)*v*,1*αβv*,1*αβ*d*x*2d*η* + *k*2 2 *Lz exp*(−*ωt*)*v*,12*v*,12d*x*<sup>2</sup> + 2*γkρ*<sup>2</sup> *m*<sup>2</sup> *Lz exp*(−*ωt*)*v*,1*tv*,1d*x*2, (43)

where *εi*,(*i* = 1, 2, ··· , 20) are arbitrary positive constants .

In (43), if we choose *ε*<sup>1</sup> = *ε*<sup>2</sup> = *ε*<sup>7</sup> = *ε*<sup>9</sup> = <sup>1</sup> <sup>4</sup>*<sup>λ</sup>* ,*ε*<sup>3</sup> <sup>=</sup> 2,*ε*<sup>4</sup> <sup>=</sup> *ωρ*<sup>1</sup> <sup>2</sup>*<sup>λ</sup>* ,*ε*<sup>5</sup> <sup>=</sup> *ωρ*<sup>2</sup> <sup>8</sup>*<sup>λ</sup>* ,*ε*<sup>6</sup> = *λ*,*ε*<sup>8</sup> = 1, *<sup>ε</sup>*<sup>10</sup> <sup>=</sup> *<sup>μ</sup>k*<sup>1</sup> *λc*<sup>2</sup> ,*ε*<sup>11</sup> = *ε*<sup>12</sup> = *γ*,*ε*<sup>13</sup> = <sup>1</sup> <sup>2</sup> ( <sup>2</sup>*γkρ*<sup>2</sup> *<sup>m</sup>*<sup>2</sup> <sup>+</sup> *<sup>ρ</sup>*2*k*2)−1*ρ*1,*ε*<sup>14</sup> <sup>=</sup> *<sup>γ</sup> <sup>ω</sup>* ,*ε*<sup>15</sup> <sup>=</sup> *<sup>k</sup>λγ <sup>m</sup>*<sup>2</sup> ,*ε*<sup>16</sup> <sup>=</sup> <sup>2</sup>*<sup>λ</sup> <sup>γ</sup>* ,*ε*<sup>17</sup> <sup>=</sup> *<sup>γ</sup>* 2*λ* , *<sup>ε</sup>*<sup>18</sup> = *<sup>ω</sup>*,*ε*<sup>19</sup> = *<sup>γ</sup>* <sup>2</sup> , *<sup>k</sup>*<sup>1</sup> <sup>=</sup> <sup>1</sup> 2*ρ*<sup>1</sup> ( *μ* <sup>2</sup> <sup>+</sup> *<sup>λ</sup>*), *<sup>k</sup>*<sup>2</sup> <sup>=</sup> <sup>4</sup> *<sup>γ</sup>* ; *m* ≥ max{ 5 4*k k*2 [2*λ*2*c*<sup>3</sup> <sup>+</sup> *<sup>λ</sup>*2*c*<sup>2</sup> <sup>2</sup>*μk*<sup>1</sup> <sup>+</sup> (2*γkρ*2+*ρ*2*k*2)<sup>2</sup> *<sup>ρ</sup>*<sup>1</sup> ], 1, 8*γ*2*k*, 4 2*k*(1 + *γ*2*kρ*2*ω* + <sup>4</sup> *<sup>γ</sup>*<sup>2</sup> <sup>+</sup> <sup>4</sup>*ρ*<sup>2</sup> *<sup>γ</sup>* )}, *<sup>ω</sup>* <sup>≥</sup> max{2*k*1*ρ*<sup>1</sup> <sup>+</sup> <sup>16</sup>*λ*<sup>2</sup> <sup>+</sup> <sup>8</sup>*k*2*λ*2*γ*2, 1}, *<sup>μ</sup>* <sup>≥</sup> 4 <sup>8</sup>*<sup>λ</sup> <sup>ω</sup>*<sup>2</sup> <sup>+</sup> *<sup>ρ</sup>*1*ρ*<sup>2</sup> <sup>+</sup> <sup>4</sup>*ρ*1*λc*<sup>1</sup> <sup>+</sup> <sup>8</sup>*ρ*1*kλγ*2, *<sup>γ</sup>* <sup>≥</sup> max{16*λ*2*c*<sup>1</sup> <sup>+</sup> <sup>10</sup>*<sup>λ</sup>* <sup>+</sup> <sup>4</sup>*kρ*2*c*<sup>1</sup> <sup>+</sup> <sup>4</sup>*kρ*<sup>2</sup> <sup>+</sup> <sup>4</sup>*kλ*, 1}, we can get

*∂*2*ϕ*(*z*, *t*) *<sup>∂</sup>z*<sup>2</sup> <sup>≥</sup> *ωρ*<sup>1</sup> 4 *t* 0 *Lz exp*(−*ωη*)*u*<sup>2</sup> ,*η*d*x*2d*<sup>η</sup>* <sup>+</sup> *<sup>ρ</sup>*<sup>1</sup> 2 *Lz exp*(−*ωt*)*u*<sup>2</sup> ,*t*d*x*<sup>2</sup> <sup>+</sup> *<sup>ω</sup>* 8 *t* 0 *Lz exp*(−*ωη*)*u*,*αu*,*α*d*x*2d*η* + 1 4 *Lz exp*(−*ωt*)*u*,*αu*,*α*d*x*<sup>2</sup> <sup>+</sup> *<sup>μ</sup>* 2 *t* 0 *Lz exp*(−*ωη*)*u*,*αηu*,*αη*d*x*2d*<sup>η</sup>* <sup>+</sup> *μωk*<sup>1</sup> 2 *t* 0 *Lz exp*(−*ωη*)(*u*,*αα*)2d*x*2d*<sup>η</sup>* <sup>+</sup> *<sup>μ</sup>k*<sup>1</sup> 2 *Lz exp*(−*ωt*)(*u*,*αα*)2d*x*<sup>2</sup> <sup>+</sup> *<sup>ρ</sup>*2*<sup>ω</sup>* 4 *t* 0 *Lz exp*(−*ωη*)*v*<sup>2</sup> ,*η*d*x*2d*η* <sup>+</sup> *<sup>ρ</sup>*<sup>2</sup> 4 *Lz exp*(−*ωt*)*v*<sup>2</sup> ,*t*d*x*<sup>2</sup> <sup>+</sup> *<sup>m</sup>*<sup>2</sup> 2*k t* 0 *Lz exp*(−*ωη*)*v*,*αηv*,*αη*d*x*2d*η* <sup>+</sup> *ωγ* 4 *t* 0 *Lz exp*(−*ωη*)*v*,*αβv*,*αβ*d*x*2d*<sup>η</sup>* <sup>+</sup> *<sup>γ</sup>* 4 *Lz exp*(−*ωt*)*v*,*αβv*,*αβ*d*x*<sup>2</sup> <sup>+</sup> *<sup>m</sup>*<sup>2</sup> <sup>4</sup>*<sup>k</sup> <sup>k</sup>*<sup>2</sup> *Lz exp*(−*ωt*)*v*,1*αv*,1*α*d*x*<sup>2</sup> + *t* 0 *Lz exp*(−*ωη*)*v*,1*αβv*,1*αβ*d*x*2d*η* + *k*2 2 *Lz exp*(−*ωt*)*v*,12*v*,12d*x*<sup>2</sup> + 2*γkρ*<sup>2</sup> *m*<sup>2</sup> *Lz exp*(−*ωt*)*v*,1*tv*,1d*x*<sup>2</sup> = *E*(*z*, *t*). (44)

We now define a new function *F*(*z*, *t*) as

$$F(z,t) = \int\_0^t E(z,s)ds.\tag{45}$$

We get

*<sup>F</sup>*(*z*, *<sup>t</sup>*) = *ωρ*<sup>1</sup> 4 *t* 0 *s* 0 *Lz exp*(−*ωη*)*u*<sup>2</sup> ,*η*d*x*2d*ηds* <sup>+</sup> *<sup>ρ</sup>*<sup>1</sup> 2 *t* 0 *Lz exp*(−*ωs*)*u*<sup>2</sup> ,*s*d*x*2*ds* <sup>+</sup> *<sup>ω</sup>* 8 *t* 0 *s* 0 *Lz exp*(−*ωη*)*u*,*αu*,*α*d*x*2d*ηds* + 1 4 *t* 0 *Lz exp*(−*ωs*)*u*,*αu*,*α*d*x*2*ds* <sup>+</sup> *<sup>μ</sup>* 2 *t* 0 *s* 0 *Lz exp*(−*ωη*)*u*,*αηu*,*αη*d*x*2d*ηds* <sup>+</sup> *μωk*<sup>1</sup> 2 *t* 0 *s* 0 *Lz exp*(−*ωη*)(*u*,*αα*)2d*x*2d*ηds* <sup>+</sup> *<sup>μ</sup>k*<sup>1</sup> 2 *t* 0 *Lz exp*(−*ωs*)(*u*,*αα*)2d*x*2*ds* <sup>+</sup> *<sup>ρ</sup>*2*<sup>ω</sup>* 4 *t* 0 *s* 0 *Lz exp*(−*ωη*)*v*<sup>2</sup> ,*η*d*x*2d*ηds* <sup>+</sup> *<sup>ρ</sup>*<sup>2</sup> 4 *t* 0 *Lz exp*(−*ωs*)*v*<sup>2</sup> ,*t*d*x*2*ds* <sup>+</sup> *<sup>m</sup>*<sup>2</sup> 2*k t* 0 *s* 0 *Lz exp*(−*ωη*)*v*,*αηv*,*αη*d*x*2d*ηds* <sup>+</sup> *ωγ* 4 *t* 0 *s* 0 *Lz exp*(−*ωη*)*v*,*αβv*,*αβ*d*x*2d*ηds* <sup>+</sup> *<sup>γ</sup>* 4 *t* 0 *Lz exp*(−*ωs*)*v*,*αβv*,*αβ*d*x*2*ds* <sup>+</sup> *<sup>m</sup>*<sup>2</sup> <sup>4</sup>*<sup>k</sup> <sup>k</sup>*<sup>2</sup> *t* 0 *Lz exp*(−*ωs*)*v*,1*αv*,1*α*d*x*2*ds* + *t* 0 *s* 0 *Lz exp*(−*ωη*)*v*,1*αβv*,1*αβ*d*x*2d*ηds* + *k*2 2 *t* 0 *Lz exp*(−*ωs*)*v*<sup>2</sup> ,12d*x*2*ds* <sup>+</sup> *ωγkρ*<sup>2</sup> *m*<sup>2</sup> *t* 0 *Lz exp*(−*ωs*)*v*<sup>2</sup> ,1d*x*2*ds* <sup>+</sup> *<sup>γ</sup>kρ*<sup>2</sup> *m*<sup>2</sup> *Lz exp*(−*ωt*)*v*<sup>2</sup> ,1d*x*2. (46)

We can easily get

$$F(z,t) \ge 0.\tag{47}$$

Following the same procedures as (37)–(44), we obtain

$$\frac{\partial^2 \varrho(z,t)}{\partial z^2} \le \frac{3}{2} E(z,t). \tag{48}$$

Inserting (30) into (48), we have

$$\frac{\partial^2 \Phi(z, t)}{\partial z^2} \le \frac{3}{2} F(z, t). \tag{49}$$

We can also get

$$\frac{\partial^2 \Phi(z, t)}{\partial z^2} \ge F(z, t). \tag{50}$$

Combining (47) and (50), we obtain

$$\frac{\partial^2 \Phi(z,t)}{\partial z^2} = \frac{\partial^2 \int\_0^t q(z,s)ds}{\partial z^2} \ge 0. \tag{51}$$

Combining (36), (44) and (51), using the Schwarz's inequality, we can obtain

$$|\Phi(z,t)| \le k\_3 \frac{\partial^2 \Phi(z,t)}{\partial z^2}.$$

where *k*<sup>3</sup> is a computable positive constant. The proof of Lemma 5 is finished.

## **3.** *Phragmén-Lindelöf* **Alternative Results**

Based on Lemmas 1–5, we can get the following Lemmas:

**Lemma 6.** *We suggest u and v are classical solutions of problems* (4)*–*(7) *in the semi-infinite strip* <sup>Ω</sup><sup>0</sup> *defined by* (1)*, if there exists a <sup>z</sup>*<sup>0</sup> <sup>≥</sup> <sup>0</sup> *such that <sup>∂</sup>*Φ(*z*0,*t*) *<sup>∂</sup><sup>z</sup>* > 0*, then the following inequality holds:*

$$\lim\_{z \to \infty} e^{-kz} G(z, t) \ge c\_1(t), \tag{52}$$

*where c*1(*t*) = <sup>2</sup> 3 [ *∂ <sup>∂</sup>z*Φ(*z*1, *<sup>t</sup>*) + *<sup>k</sup>*4Φ(*z*1, *<sup>t</sup>*)]*e*−*k*4*z*<sup>1</sup> *, G*(*z*, *<sup>t</sup>*)*will be defined in* (61)*.*

**Proof.** Since *<sup>∂</sup>*2Φ(*z*0,*t*) *<sup>∂</sup>z*<sup>2</sup> <sup>≥</sup> 0 for all *<sup>z</sup>* <sup>≥</sup> 0, we can get *<sup>∂</sup>*Φ(*z*,*t*) *<sup>∂</sup><sup>z</sup>* > 0 for all *z* ≥ *z*0. We know the fact <sup>Φ</sup>(*z*, *<sup>t</sup>*) <sup>≥</sup> <sup>Φ</sup>(*z*0, *<sup>t</sup>*) + *<sup>∂</sup>*Φ(*z*0,*t*) *<sup>∂</sup><sup>z</sup>* (*z* − *z*0) for all *z* ≥ *z*0. If we let *z* → +∞, we can obtain Φ(*z*, *t*) > 0. So, wo have the following results: There exists a *<sup>z</sup>*<sup>1</sup> <sup>&</sup>gt; *<sup>z</sup>*<sup>0</sup> such that *<sup>∂</sup>*Φ(*z*1,*t*) *<sup>∂</sup><sup>z</sup>* > 0 and Φ(*z*1, *t*) > 0. From (31), we can get

$$\frac{\partial^2 \Phi(z,t)}{\partial z^2} - k\_4^2 \Phi(z,t) \ge 0,\tag{53}$$

with *k*<sup>4</sup> = 4 <sup>1</sup> *k*3

.

Equation (53) can be rewritten as

$$\frac{\partial}{\partial z}\left(e^{-k\_4 z} \left[\frac{\partial}{\partial z}\Phi(z,t) + k\_4 \Phi(z,t)\right]\right) \ge 0,\tag{54}$$

or

$$\frac{\partial}{\partial z}\left(\epsilon^{k4z}\left[\frac{\partial}{\partial z}\Phi(z,t) - k\_4\Phi(z,t)\right]\right) \ge 0. \tag{55}$$

Integrating (54) and (55), we obtain

$$\frac{\partial}{\partial z}\Phi(z,t) + k\_4\Phi(z,t) \ge [\frac{\partial}{\partial z}\Phi(z\_1,t) + k\_4\Phi(z\_1,t)]e^{k\_4(z-z\_1)},\tag{56}$$

or

$$\frac{\partial}{\partial z}\Phi(z,t) - k\_4\Phi(z,t) \ge [\frac{\partial}{\partial z}\Phi(z\_1,t) - k\_4\Phi(z\_1,t)]e^{-k\_4(z-z\_1)},\tag{57}$$

for all *z* ≥ *z*1.

Combining (56) and (57), we have

$$\begin{split} \frac{\partial}{\partial z} \Phi(z,t) &\geq \frac{\partial}{\partial z} \Phi(z\_1,t) \frac{e^{k\_4(z-z\_1)} + e^{-k\_4(z-z\_1)}}{2} \\ &+ k\_4 \Phi(z\_1,t) \frac{e^{k\_4(z-z\_1)} - e^{-k\_4(z-z\_1)}}{2} .\end{split} \tag{58}$$

Integrating (49) from *z*<sup>1</sup> to *z*, we obtain

$$\int\_{z\_1}^{z} \frac{\partial^2 \Phi(z, t)}{\partial z^2} d\xi \le \frac{3}{2} \int\_{z\_1}^{z} F(\xi, t) d\xi. \tag{59}$$

Inserting (59) into (58), we have

$$\begin{split} \frac{3}{2} \int\_{z\_1}^{z} F(\xi, t) d\xi &\geq \frac{\partial}{\partial z} \Phi(z\_1, t) \left[ \frac{\epsilon^{k\_4(z - z\_1)} + \epsilon^{-k\_4(z - z\_1)}}{2} - 1 \right] \\ &+ k\_4 \Phi(z\_1, t) \frac{\epsilon^{k\_4(z - z\_1)} - \epsilon^{-k\_4(z - z\_1)}}{2} . \end{split} \tag{60}$$

If we define

$$G(z,t) = \int\_{z\_1}^{z} F(\zeta', t) d\zeta'\_{\prime} \tag{61}$$

Combining (46) and (61), we have

*<sup>G</sup>*(*z*, *<sup>t</sup>*) = *ωρ*<sup>1</sup> 4 *t* 0 *s* 0 *z z*1 *Lξ exp*(−*ωη*)*u*<sup>2</sup> ,*η*d*A*d*ηds* <sup>+</sup> *<sup>ρ</sup>*<sup>1</sup> 2 *t* 0 *z z*1 *Lξ exp*(−*ωs*)*u*<sup>2</sup> ,*s*d*Ads* <sup>+</sup> *<sup>ω</sup>* 8 *t* 0 *s* 0 *z z*1 *Lξ exp*(−*ωη*)*u*,*αu*,*α*d*A*d*ηds* + 1 4 *t* 0 *z z*1 *Lξ exp*(−*ωs*)*u*,*αu*,*α*d*Ads* <sup>+</sup> *<sup>μ</sup>* 2 *t* 0 *s* 0 *z z*1 *Lξ exp*(−*ωη*)*u*,*αηu*,*αη*d*A*d*ηds* <sup>+</sup> *μωk*<sup>1</sup> 2 *t* 0 *s* 0 *z z*1 *Lξ exp*(−*ωη*)(*u*,*αα*)2d*A*d*ηds* <sup>+</sup> *<sup>μ</sup>k*<sup>1</sup> 2 *t* 0 *z z*1 *Lξ exp*(−*ωs*)(*u*,*αα*)2d*Ads* <sup>+</sup> *<sup>ρ</sup>*2*<sup>ω</sup>* 4 *t* 0 *s* 0 *z z*1 *Lξ exp*(−*ωη*)*v*<sup>2</sup> ,*η*d*A*d*ηds* <sup>+</sup> *<sup>ρ</sup>*<sup>2</sup> 4 *t* 0 *z z*1 *Lξ exp*(−*ωs*)*v*<sup>2</sup> ,*t*d*Ads* <sup>+</sup> *<sup>m</sup>*<sup>2</sup> 2*k t* 0 *s* 0 *z z*1 *Lξ exp*(−*ωη*)*v*,*αηv*,*αη*d*A*d*ηds* <sup>+</sup> *ωγ* 4 *t* 0 *s* 0 *z z*1 *Lξ exp*(−*ωη*)*v*,*αβv*,*αβ*d*A*d*ηds* <sup>+</sup> *<sup>γ</sup>* 4 *t* 0 *z z*1 *Lξ exp*(−*ωs*)*v*,*αβv*,*αβ*d*Ads* <sup>+</sup> *<sup>m</sup>*<sup>2</sup> <sup>4</sup>*<sup>k</sup> <sup>k</sup>*<sup>2</sup> *t* 0 *z z*1 *Lξ exp*(−*ωs*)*v*,1*αv*,1*α*d*Ads* + *t* 0 *s* 0 *z z*1 *Lξ exp*(−*ωη*)*v*,1*αβv*,1*αβ*d*A*d*ηds* + *k*2 2 *t* 0 *z z*1 *Lξ exp*(−*ωs*)*v*<sup>2</sup> ,12d*Ads* <sup>+</sup> *ωγkρ*<sup>2</sup> *m*<sup>2</sup> *t* 0 *z z*1 *Lξ exp*(−*ωs*)*v*<sup>2</sup> ,1d*Ads* <sup>+</sup> *<sup>γ</sup>kρ*<sup>2</sup> *m*<sup>2</sup> *z z*1 *Lξ exp*(−*ωt*)*v*<sup>2</sup> ,1d*A*. (62)

we obtain

$$\lim\_{z \to \infty} e^{-k\_4 z} G(z, t) \ge c\_1(t),$$

with *c*1(*t*) = <sup>2</sup> 3 [ *∂ <sup>∂</sup>z*Φ(*z*1, *<sup>t</sup>*) + *<sup>k</sup>*4Φ(*z*1, *<sup>t</sup>*)]*e*−*k*4*z*<sup>1</sup> .

**Lemma 7.** *We suggest u and v are classical solutions of problems* (4)*–*(7) *in the semi-infinite strip* <sup>Ω</sup><sup>0</sup> *defined by* (1)*. If <sup>∂</sup>*Φ(*z*,*t*) *<sup>∂</sup><sup>z</sup>* ≤ 0 *for all z* ≥ 0*, then the following inequality holds:*

$$H(z,t) \le c\_2(t)e^{-k\_4 z}.\tag{63}$$

*where c*2(*t*) = <sup>−</sup> *<sup>∂</sup> <sup>∂</sup>z*Φ(0, *t*) + *k*4Φ(0, *t*)]*, H*(*z*, *t*)*will be defined in* (66)*.*

**Proof.** If we suggest there exists a *<sup>z</sup>*<sup>0</sup> <sup>&</sup>gt; 0, such that <sup>Φ</sup>(*z*0, *<sup>t</sup>*) <sup>&</sup>lt; 0. Since *<sup>∂</sup>*Φ(*z*,*t*) *<sup>∂</sup><sup>z</sup>* ≤ 0 for all *z* ≥ 0, we can get Φ(*z*, *t*) ≤ Φ(*z*0, *t*) < 0 for all *z* ≥ *z*0. From (31), we have *∂*Φ(*z*,*t*) *<sup>∂</sup><sup>z</sup>* <sup>−</sup> *<sup>∂</sup>*Φ(*z*0,*t*) *<sup>∂</sup><sup>z</sup>* <sup>=</sup> *<sup>∂</sup>*2Φ(*ξ*,*t*) *<sup>∂</sup><sup>z</sup>* (*<sup>z</sup>* <sup>−</sup> *<sup>z</sup>*0) ≥ − <sup>1</sup> *k*3 Φ(*ξ*, *t*)(*z* − *z*0), with *z*<sup>0</sup> < *ξ* < *z*. let *z* → ∞, we have *<sup>∂</sup>*Φ(*z*,*t*) *<sup>∂</sup><sup>z</sup>* <sup>&</sup>gt; 0. Which gives a contradiction to *<sup>∂</sup>*Φ(*z*,*t*) *<sup>∂</sup><sup>z</sup>* ≤ 0 for all *z* ≥ 0. So we can conclude Φ(*z*, *t*) ≥ 0 for all *z* ≥ 0.

Integrating (53) from 0 to *z*, we obtain

$$-\frac{\partial \Phi(z,t)}{\partial z} + k\_4 \Phi(z,t) \le c\_2(t)e^{-k\_4 z},\tag{64}$$

with *<sup>c</sup>*2(*t*) = <sup>−</sup> *<sup>∂</sup> <sup>∂</sup>z*Φ(0, *t*) + *k*4Φ(0, *t*)]. Since Φ(*z*, *t*) ≥ 0 for for all *z* ≥ 0, we have

$$-\frac{\partial \Phi(z,t)}{\partial z} \le c\_2(t)e^{-k\_4 z}.\tag{65}$$

From (64), we can get the results <sup>Φ</sup>(*z*, *<sup>t</sup>*) and <sup>−</sup>*∂*Φ(*z*,*t*) *<sup>∂</sup><sup>z</sup>* tend to 0 as *z* → ∞. We thus have

$$\begin{split}-\frac{\partial\Phi(z,t)}{\partial z} &= \int\_{\mathbb{z}}^{\infty} \frac{\partial^2 \Phi(\zeta, t)}{\partial z^2} d\zeta \\ &\geq \int\_{\mathbb{z}}^{\infty} F(\zeta, t) d\zeta \\ &= H(z, t). \end{split} \tag{66}$$

Combining (46) and (66), we have

*<sup>H</sup>*(*z*, *<sup>t</sup>*) = *ωρ*<sup>1</sup> 4 *t* 0 *s* 0 ∞ *z Lξ exp*(−*ωη*)*u*<sup>2</sup> ,*η*d*A*d*ηds* <sup>+</sup> *<sup>ρ</sup>*<sup>1</sup> 2 *t* 0 ∞ *z Lξ exp*(−*ωs*)*u*<sup>2</sup> ,*s*d*Ads* <sup>+</sup> *<sup>ω</sup>* 8 *t* 0 *s* 0 ∞ *z Lξ exp*(−*ωη*)*u*,*αu*,*α*d*A*d*ηds* + 1 4 *t* 0 ∞ *z Lξ exp*(−*ωs*)*u*,*αu*,*α*d*x*2*ds* <sup>+</sup> *<sup>μ</sup>* 2 *t* 0 *s* 0 ∞ *z Lξ exp*(−*ωη*)*u*,*αηu*,*αη*d*A*d*ηds* <sup>+</sup> *μωk*<sup>1</sup> 2 *t* 0 *s* 0 ∞ *z Lξ exp*(−*ωη*)(*u*,*αα*)2d*A*d*ηds* <sup>+</sup> *<sup>μ</sup>k*<sup>1</sup> 2 *t* 0 ∞ *z Lξ exp*(−*ωs*)(*u*,*αα*)2d*Ads* <sup>+</sup> *<sup>ρ</sup>*2*<sup>ω</sup>* 4 *t* 0 *s* 0 ∞ *z Lξ exp*(−*ωη*)*v*<sup>2</sup> ,*η*d*A*d*ηds* <sup>+</sup> *<sup>ρ</sup>*<sup>2</sup> 4 *t* 0 ∞ *z Lξ exp*(−*ωs*)*v*<sup>2</sup> ,*t*d*Ads* <sup>+</sup> *<sup>m</sup>*<sup>2</sup> 2*k t* 0 *s* 0 ∞ *z Lξ exp*(−*ωη*)*v*,*αηv*,*αη*d*A*d*ηds* <sup>+</sup> *ωγ* 4 *t* 0 *s* 0 ∞ *z Lξ exp*(−*ωη*)*v*,*αβv*,*αβ*d*A*d*ηds* <sup>+</sup> *<sup>γ</sup>* 4 *t* 0 ∞ *z Lξ exp*(−*ωs*)*v*,*αβv*,*αβ*d*Ads* <sup>+</sup> *<sup>m</sup>*<sup>2</sup> <sup>4</sup>*<sup>k</sup> <sup>k</sup>*<sup>2</sup> *t* 0 ∞ *z Lξ exp*(−*ωs*)*v*,1*αv*,1*α*d*Ads* + *t* 0 *s* 0 ∞ *z Lξ exp*(−*ωη*)*v*,1*αβv*,1*αβ*d*A*d*ηds* + *k*2 2 *t* 0 ∞ *z Lξ exp*(−*ωs*)*v*<sup>2</sup> ,12d*Ads* <sup>+</sup> *ωγkρ*<sup>2</sup> *m*<sup>2</sup> *t* 0 ∞ *z Lξ exp*(−*ωs*)*v*<sup>2</sup> ,1d*Ads* <sup>+</sup> *<sup>γ</sup>kρ*<sup>2</sup> *m*<sup>2</sup> ∞ *z Lξ exp*(−*ωt*)*v*<sup>2</sup> ,1d*A*. (67)

Inserting (66) into (65), we obtain

$$H(z,t) \le c\_2(t)e^{-k\_4z}.$$

Based on Lemmas 6 and 7, we can get the following theorem.

**Theorem 1.** *We suggest u and v are classical solutions of problems* (4)*–*(7) *in the semi-infinite strip* Ω<sup>0</sup> *defined by* (1)*, then either inequality*

$$\lim\_{z \to \infty} e^{-k4z} G(z, t) \ge c\_1(t)$$

*holds or*

$$H(z,t) \le c\_2(t)e^{-k\_4z}$$

*holds.*

Theorem 1 shows that either the energy expression *G*(*z*, *t*) grows exponentially or the energy expression *H*(*z*, *t*) decays exponentially.

#### **4. Conclusions**

In this paper, we studied the spatial properties of solutions for a class of thermoelastic plate with biharmonic operator in a semi-infinite cylinder in *R*2. The *Phragmen*´ -*Lindelo f* ¨ alternative results were obtained based on a second-order inequality. Our method is also valid for the hyperbolic–parabolic coupling equations. We can only deal with the linear equations. For the case of nonlinear equations, it is difficult to study the spatial properties. The results of these future studies will be of great interest to researchers in our field.

**Author Contributions:** Writing–original draft, S.L., original draft preparation J.S. and review and editing, B.O. All authors have read and agreed to the published version of the manuscript.

**Funding:** The work was supported national natural Science Foundation of China(Grant # 61907010), natural Science in Higher Education of Guangdong, China (Grant # 2018KZDXM048), the General Project of Science Research of Guangzhou (Grant # 201707010126). Guangdong Province Educational Science "Thirteenth Five-Year Plan" 2020 research project approval (NO. 2020JKDY040).

**Data Availability Statement:** This paper focuses on theoretical analysis, not involving experiments and data.

**Acknowledgments:** The authors express their heartfelt thanks to the editors and referees who have provided some important suggestions.

**Conflicts of Interest:** The authors declare no conflict of interest.

## **References**

