How does soft information about small business lending affect bank efficiency under capital regulation?

Jyh-Horng Lin; Shi Chen; Jeng-Yan Tsai; Jyh-Horng Lin; Shi Chen; Jeng-Yan Tsai

doi:10.3934/QFE.2019.1.53

Quantitative Finance and Economics

2019, Volume 3, Issue 1: 53-74. doi: 10.3934/QFE.2019.1.53

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Research article

How does soft information about small business lending affect bank efficiency under capital regulation?

1.
Department of International Business, Tamkang University, New Taipei City, 25137, Taiwan
2.
School of Economics, Southwestern University of Finance and Economics, Chengdu, 611130, China

Received: 24 November 2018 Accepted: 27 February 2019 Published: 07 March 2019
JEL Codes: G21, G28

In this paper, we develop a capped call option model to evaluate the equity of a bank under capital regulation. A capped type of credit risk from the performance of relationship borrowing firms is explicitly considered, captured by a mechanism through borrower soft information that the bank produces and accumulates. We study the impact of soft information for small business lending on the optimal bank interest margin, i.e., the spread between the loan rate and the deposit rate of a bank. Our findings show that favorable soft information increases small business lending at a reduced loan rate (and thus at a reduced margin), and further lowers bank equity risk and enhances efficiency gains from soft information acquisition when borrower dependent on bank financing is heavy. Moreover, we account for the capital regulatory environment stringent to small business lending that increases bank equity risk and decreases soft information efficiency gain. Our results have important bank interest margin implications in terms of achieving efficiency gain and lower risk exposure, which might conflict with capital regulation aiming to promote financial stability.

Keywords:

Citation: Jyh-Horng Lin, Shi Chen, Jeng-Yan Tsai. How does soft information about small business lending affect bank efficiency under capital regulation?[J]. Quantitative Finance and Economics, 2019, 3(1): 53-74. doi: 10.3934/QFE.2019.1.53

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Abstract

1. Introduction

Rotating blades (thin-walled beam) are important structures widely used in mechanical and aerospace engineering as aviation engine blades, various cooling fans, windmill blades, helicopter rotor blades, airplane propellers etc. The study of the dynamics of rotating blades is important to design purposes, optimization, and control.

If the shear effect is not considered, the Euler-Bernoulli beam equation is used to model vibration of thin-walled beam. Chen et al.[6] studied the boundary feedback stabilization of a linear Euler-Bernoulli beam equation, they proved that the total energy of the equation decays uniformly and exponentially with $t$ , when the beam is clamped at the left end and subjected to a feedback boundary conditions at the right end. Then Guo et al.[12] proved the well-posedness and stability of the system proposed by Chen et al.[14] considered a nonlinear Euler-Bernoulli beam equation with a feedback force applied at the free end, the existence of the weak and classical solutions were proved. Other relevant studies are referred to Refs. [4,18,19,17,13]. Note that the above literatures studied the longitudinal vibration in one direction.

A very extensive work devoted to the longitudinal vibration in two directions were done by Librescu and Song[21,32,33] and their co-workers[25,28]. Under assumption of the cross-section to be rigid in its own plane, they modelled the rotating blades by 1-D linear governing equations. The influence of many factors on rotating blades, such as the anisotropy and heterogeneity of constituent materials, functionally graded materials (FGM), temperature, shear effects, primary and secondary warping phenomena (Vlasov effect), centrifugal and Coriolis forces etc have been taken into account in the 1-D linear governing equations.

Following Librescu's approach, various blades models were derived. Georgiades et al.[11] modelled a rotating blades by means of linear strain-displacement relationships, considering arbitrary pitch (presetting) angle and non-constant rotating speed. Choi et al.[8] studied bending vibration control of the pre-twisted rotating composite blades, who emphasized the important of piezoelectric effect in single cell composite blades. Fazelzadeh et al.[9] considered a thin-walled blades made of FGM which is used in turbomachinery under aerothermodynamics loading. In the paper, quasi-steady aerodynamic pressure loadings was determined by the first-order piston theory, and steady beam surface temperature was obtained from gas dynamics theory. Fazelzadeh et al.[10] studied the governing equations which included the effects of the presetting angle and the rotary inertia. The effects of steady wall temperature and quasi-steady aerodynamic pressure loadings due to flow motion were also taken into account.

The models in Refs.[21,32,33,25,28,11,8,9,10] are linear. When the engine blades rotate at a low speed, the linear approximation can completely meet the needs of practical application. However, when the blade rotate at a high speed, the simple linear approximation can not accurately describe the dynamic behavior of the system. So the non-linear analysis of rotating blades has attracted considerable attention.

The nonlinear governing equations of a rotating blades at constant angular velocity was presented by Anderson[1], and the author linearized the equation under the assumption that a small perturbed motion occurred at an initially stressed equilibrium configuration. Chen et al.[7] considered the effects of geometric non-linearity, shear deformation and rotary inertia. Arvin et al.[2] builded a nonlinear governing equation for rotating blades considering centrifugal forces by means of von-Karmans strain-displacement relationships under assumption of the constant rotation speed and zero pitch (presetting) angle. Yao et al.[39] employed the Hamilton's principle to derive the nonlinear governing equations with periodic rotating speed, arbitrary pitch (presetting) angle and linear pre-twist angle. Under the assumption that the location of shear centre is different from the centre of gravity, Avramov et al.[3] obtained results of the investigations on flexural-flexural-torsional nonlinear vibrations of twisted rotating blades described by the model of three nonlinear integro-differential equations. Other nonlinear models can be found in Refs.[34,29,27,37,15,31,36].

To the best of the author's knowledge, all the above literatures about the longitudinal vibration in two directions skipped the existence proof of solutions to the governing equations, and directly used the finite element method to study the influence of various parameters on blades vibration. To address this situation, we first try to model a governing equations of the blades with arbitrary rotating speed, arbitrary pre-setting angle and pre-twist angle. In the process of building the model, we take into account the free vibration at the right end of the blade and the nonlinear relationship between stress and strain. In the paper we aim to investigate the well-posedness and regularity of the governing equations. The well-posedness of other nonlinear blade vibration models can be found in Refs. [5,38,26,35,16].

2. Model building

Let us consider a slender, straight blades mounted on a rigid hub of radius $R_0$ rotating with the angular velocity $\omega(t)$ . The length of the blades is denoted by $l$ , its wall thickness by $h$ , the length and the width of the cross section of the blades are $a$ and $b$ , respectively. The sum of pre-twist angle $\alpha(x)$ and pre-setting angle $\beta$ is defined by $\theta(x)$ , as shown in Figure 1 and 2.

Figure 1. Rotating blades.

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Figure 2. The cross-section of rotating blades.

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2.1. Assumptions

To derive the model of the rotating blades, the following kinematic and static assumptions are postulated:

(ⅰ) The blades is perfectly elastic bodies, the blades material is isotropic and is not affected by temperature,

(ⅱ) The cross section of the blades is rectangular and all its geometrical dimensions remain invariant in its plane,

(ⅲ) The ratio of wall thickness $h$ to the radius of curvature $r$ at any point of the blade wall is negligibly small while compared to unity,

(ⅳ) The transverse shear effect of the cross section is neglected,

(ⅴ) The axial displacement $w$ is much smaller than $u$ or $v$ and the derivatives of $w$ can be neglected in the strain-displacement relations. where $u, v, w$ represent the displacement of the middle line of cross-sections along the $y, z, x$ axis, respectively.

2.2. Four coordinate systems are defined to describe the motion of the blades

Inertial Cartesian coordinate systems $(X, Y, Z)$ attached to the center of the hub $O$ , the unit vectors of the inertial coordinate systems $(X, Y, Z)$ is defined as $(I, J, K)$ .

Rotating coordinate systems $(x, y, z)$ located at the blade root, the origin $o$ of the rotating coordinate systems is set at the center of the beam cross section, the unit vectors of the inertial coordinate systems $(x, y, z)$ is defined as $(i, j, k)$ .

Transformation between $XYZ$ and $xyz$ frames can be written in the form

$\begin{eqnarray} (X, Y, Z) = (x, y, z)+(R_0, 0, 0). \end{eqnarray}$

(1)

Local coordinates systems $(x^p, y^p, z^p)$ also located at the blade root and oriented with respect to plane of rotation $(y, z)$ at angle $\omega(t)$ . So $(x, y, z)' = \mathrm B (x^p, y^p, z^p)'$ , where the rotation matrix $\mathrm B$ is

$\begin{equation*} \left( \begin{array}{ccc} 1&0&0\\ 0&\mathrm {cos} \omega & -\mathrm {sin} \omega \\ 0&\mathrm {sin} \omega & \mathrm {cos} \omega \\ \end{array} \right). \end{equation*}$

Local, curvilinear coordinate systems $(x, n, s)$ related to blade cross section (Figure 2). Its origin is set conveniently at the point on a mid-line contour. $s$ and $n$ are the circumferential and thickness coordinate variables, the unit vectors of axis $n$ and $s$ are defined as $\mathbf e_n, \mathbf e_t$ , respectively.

In order to determine the relationship between the two coordinate systems $(x, y, z)$ and $(x, n, s)$ , one defines the position vector $\mathbf r(\equiv \mathbf r(s, x))$ from the reference $x$ -axis of the blades to an arbitrary point $A$ located on the middle surface as

$\begin{eqnarray} \mathbf r = x \mathbf i+ y(s)\mathbf j+z(s)\mathbf k. \end{eqnarray}$

(2)

The position vector $\mathbf{r^*}$ of an arbitrary point $A^*$ off the mid-surface of the blades can be expressed as

$\begin{eqnarray} \mathbf{r^*} = \mathbf r +n\mathbf e_n, \end{eqnarray}$

(3)

As a result

$\begin{eqnarray} \mathbf e_t = \frac{\mathrm d \mathbf r}{\mathrm d s} = \frac{\mathrm d y(s)}{\mathrm d s}\mathbf j+\frac{\mathrm d z(s)}{\mathrm d s}\mathbf k, \end{eqnarray}$

(4)

$\begin{eqnarray} \mathbf e_n = \mathbf e_t\times \mathbf i = \frac{\mathrm d z(s)}{\mathrm d s}\mathbf j-\frac{\mathrm d y(s)}{\mathrm d s}\mathbf k. \end{eqnarray}$

(5)

Moreover, in order to avoid confusion, the notation $(x, y, z)$ represents the points associated with the middle surface, the notation $A^* = (x^*, y^*, z^*)$ represents the points off the middle surface, the two notations $(x, y, z)$ and $(x^*, y^*, z^*)$ are presented as follows:

$\begin{eqnarray} x^* = x, \ \ y^* = \mathbf{r^*} \cdot \mathbf j = y+n\frac{\mathrm d z}{\mathrm d s}, \ \ z^* = \mathbf{r^*} \cdot \mathbf k = z-n\frac{\mathrm d y}{\mathrm d s}. \end{eqnarray}$

(6)

2.3. Displacement field

Based on the assumptions (ⅳ) and (ⅴ), the axial displacement $D_x, D_y, D_z$ (see Figure 3) are expressed as (see Ref. [21])

$\begin{eqnarray} D_x = \phi_y(x, t)(z(s)-n\frac{\mathrm d y}{\mathrm d s})+\phi_z(x, t)(y(s)+n\frac{\mathrm d z}{\mathrm d s}), \; D_y = u, \; D_z = v. \end{eqnarray}$

(7)

Figure 3. Position vectors of ponit

$A^*$ in reference frames.

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where $\phi_y, \phi_z$ denote rotation of the cross-section about $y$ and $z$ axis. For non-shearable blades, The expression of $\phi_y$ and $\phi_z$ can be written in the form

$\begin{eqnarray} \phi_y = -v_x, \phi_z = -u_x. \end{eqnarray}$

(8)

where $u_x, v_x$ denote derivatives with respect to $x$ .

2.4. Strains and stress

Based on the assumptions (ⅴ), the displacement-strain relationships is expressed as follows[20]:

$\begin{eqnarray} \varepsilon_{xx} = \frac{\partial D_x}{\partial x}+\frac{1}{2}\Big[\Big(\frac{\partial D_y}{\partial x}\Big)^2+\Big(\frac{\partial D_z}{\partial x}\Big)^2\Big]. \end{eqnarray}$

(9)

Thanks to (8), we can get

$\begin{eqnarray*} \varepsilon_{xx} = \bar{\varepsilon}_{xx}+\bar{\bar{\varepsilon}}_{xx}n, \end{eqnarray*}$

where

$\begin{eqnarray} \bar{\varepsilon}_{xx} = \frac{1}{2}(u^2_x+v^2_x)-(u_{xx}y(s)+v_{xx}z(s)), \ \ \bar{\bar{\varepsilon}}_{xx} = -u_{xx}\frac{\mathrm d z}{\mathrm d s}+v_{xx}\frac{\mathrm d y}{\mathrm d s}. \end{eqnarray}$

(10)

The shear strain $\varepsilon_{nx}$ can be expressed as

$\begin{eqnarray} \varepsilon_{nx} = \frac{1}{2}\Big(\frac{\partial U_n}{\partial x}+\frac{\partial w}{\partial n}\Big), \end{eqnarray}$

(11)

where $U_n$ represents the components of $\big(D_x, D_y, D_z\big)$ along the $n$ axis,

$\begin{eqnarray} U_n = \big(D_x, D_y, D_z\big)\cdot \mathbf e_n = u\frac{\mathrm dz}{\mathrm ds}-v\frac{\mathrm dy}{\mathrm ds}. \end{eqnarray}$

(12)

Substituting (7) and (12) into (11), we deduce $\varepsilon_{nx} = 0.$ Using the coordinate transformation of strain component, taking into account assumptions (iv), the shear strain $\varepsilon_{sx}$ can be expressed as

$\begin{eqnarray} \varepsilon_{sx} = \frac{1}{2}\big(\frac{\mathrm dy}{\mathrm ds}\gamma_{xy}+\frac{\mathrm dz}{\mathrm ds}\gamma_{xz}\big) = 0. \end{eqnarray}$

(13)

where $\gamma_{xy}, \gamma_{xz}$ are the transverse shear effect of the cross section.

Since these materials are isotropic, the corresponding thermoelastic constitutive law adapted to the case of structures is expressed as

$\left( \begin{array}{l} \sigma_{ss} \\ \sigma_{xx} \\ \sigma_{xn} \\ \sigma_{ns} \\ \sigma_{sx} \\ \end{array} \right) = \left( \begin{array}{ccccc} Q_{11} & Q_{12} & 0 & 0 & 0 \\ Q_{21} & Q_{22} & 0 & 0 & 0 \\ 0 & 0 & Q_{44} & 0 & 0 \\ 0 & 0 & 0 & Q_{55} & 0 \\ 0 & 0 & 0 & 0 & Q_{66} \\ \end{array}\right)\left( \begin{array}{l} \varepsilon_{ss} \\ \varepsilon_{xx} \\ \varepsilon_{xn} \\ \varepsilon_{ns} \\ \varepsilon_{sx} \\ \end{array}\right),$

Herein, the reduced thermoelastic coefficients are defined as:

$Q_{11} = Q_{22} = \frac{E}{1-\nu^2}, Q_{12} = Q_{21} = \frac{E\nu}{1-\nu^2}, Q_{44} = Q_{55} = \frac{k^2 E}{2(1+\nu)}, Q_{66} = \frac{E}{2(1+\nu)},$

where $E$ is Young's modulus, $\nu$ is Poisson's ratio, $k^2$ is the transverse shear correction factor.

According to assumption (ⅱ), the cross section is rigid, then we can derive $\sigma_{ns} = 0$ . Considering the assumption of the hoop stress $\sigma_{ss}$ to be negligible, we can get

$(\sigma_{ss}, \sigma_{xx}, \sigma_{xn}, \sigma_{ns}, \sigma_{sx}) = (0, E\varepsilon_{xx}, 0, 0, 0).$

2.5. The centrifugal force and the perturbed gas pressure

The centrifugal force can be represented as

$F_c = \int_x^l \rho\bar A \omega^2(R_0+x)\mathrm dx = \rho \bar A \omega^2 R(x),$

where $\rho$ is the density of the blades, $\bar A$ is the cross section area of the blades, $R(x) = R_0(l-x)+\frac{1}{2}(l^2-x^2)$ .

In the paper, we use the first-order piston theory (see Ref.[24]) to evaluate the perturbed gas pressure. The pressure on the principal plane of the blade can be obtained as

$\begin{eqnarray} P_{y^p} = C_{\infty}\rho_{\infty}\Big(\frac{\partial u^p}{\partial t}+U_{y^p}^t\frac{\partial u^p}{\partial x}\Big), P_{z^p} = C_{\infty}\rho_{\infty}\Big(\frac{\partial v^p}{\partial t}+U_{z^p}^t\frac{\partial v^p}{\partial x}\Big), \end{eqnarray}$

(14)

where

$\begin{eqnarray} U_{y^p}^t = U_{\infty}\cos \theta, \ \ U_{z^p}^t = U_{\infty}\sin \theta, \end{eqnarray}$

(15)

$C_{\infty}$ represents the speed of sound, $\rho_{\infty}$ and $U_{\infty}$ , respectively, denote the density and the velocity of the free stream air, $U_{y^p}^t$ and $U_{z^p}^t$ are, respectively, the tangential components of the fluid velocity on the positive $y^p$ and $z^p$ plane, and $u^p$ and $v^p$ denote the displacement components along the principal axes $y^p$ and $z^p$ , respectively.

The transformation relationship between $(u^p, v^p)$ and $(u, v)$ is given by

$\begin{eqnarray} u^p = u\cos \theta +v \sin \theta, \ v^p = -u\sin \theta +v \cos \theta. \end{eqnarray}$

(16)

Therefore, the external forces per unit axial length in the $y$ direction and the $z$ direction can be obtained as

$\begin{eqnarray} p_y = aP_{z^p}\sin \theta -bP_{y^p} \cos \theta, \ p_z = -aP_{z^p}\cos \theta -bP_{y^p} \sin \theta. \end{eqnarray}$

(17)

Combining (14), (15), (16), $p_y, p_z$ can be expressed as a linear function of $u, v, u_t, v_t, u_x, v_x$ , respectively.

$\begin{eqnarray} &&p_y = b_1 u_x+b_2 v_x+b_3 u+b_4 v +b_5 u_t+b_6 v_t \end{eqnarray}$

(18)

$\begin{eqnarray} &&p_z = e_1 u_x+e_2 v_x+e_3 u+e_4 v +e_5u_t+e_6 v_t \end{eqnarray}$

(19)

where

$\begin{eqnarray*} &&b_1 = -{C_\infty }{\rho _\infty}U_{\infty}(a{\sin ^3}\theta + b{\cos ^3}\theta)\\ &&b_2 = -{C_\infty }{\rho _\infty}U_{\infty}\sin \theta\cos \theta(-a\sin \theta+ b\cos \theta)\\ &&b_3 = -{C_\infty }{\rho _\infty }U_{\infty}\theta_x\sin \theta \cos \theta(a\sin \theta -b\cos \theta)\\ &&b_4 = -{C_\infty }{\rho _\infty }U_{\infty}\theta_x(a{\sin ^3}\theta + b{\cos^3}\theta)\\ &&b_5 = -{C_\infty }{\rho _\infty}(a{\sin ^2}\theta +b{\cos ^2}\theta)\\ &&b_6 = {C_\infty }{\rho _\infty }\sin\theta\cos \theta(a - b)\\ &&e_1 = -{C_\infty }{\rho _\infty }U_{\infty}\sin \theta\cos \theta(- a\sin \theta + b\cos \theta)\\ &&e_2 = -{C_\infty }{\rho _\infty }U_{\infty}\sin \theta\cos \theta(a\cos \theta +b\sin \theta)\\ &&e_3 = {C_\infty }{\rho _\infty }U_{\infty}\theta_x\sin \theta\cos \theta(a\sin \theta +b\cos \theta)\\ &&e_4 = -{C_\infty }{\rho _\infty }U_{\infty}\theta_x\sin \theta\cos \theta(-a\sin \theta +b\cos \theta) \end{eqnarray*}$

$\begin{eqnarray*} &&e_5 = {C_\infty }{\rho _\infty}\sin\theta\cos \theta(a - b)\\ &&e_6 = -{C_\infty }{\rho _\infty }(a{\cos ^2}\theta +b{\sin ^2}\theta) \end{eqnarray*}$

2.6. The velocity vector and the acceleration

In order to calculate the kinetic energy, the velocity vector and the acceleration should be given first. Based on the assumption (ⅴ), the position vector $\mathbf R$ of a point $A^{**}$ of the deformed blades is expressed in the form

$\mathbf R(X, Y, Z, t) = x^* \mathbf i + (y^*+u)\mathbf j +(z^*+v)\mathbf k + R_0 \mathbf i .$

Keep in mind that the rotation takes place solely in the $XY$ plan, it results

$\mathbf i_t = \omega \mathbf j;\ \ \mathbf j_t = -\omega \mathbf i;\ \ \mathbf k_t = 0.$

Then the velocity vector and the acceleration of an arbitrary point $A^{**}$ are obtained:

$\begin{eqnarray} &&\mathbf R_{t} = -\omega(y^*+u)\mathbf i +(\omega(R_0+x)+u_t)\mathbf j+v_t\mathbf k, \end{eqnarray}$

(20)

$\begin{eqnarray} &&\mathbf R_{tt} = -[2\omega u_t+\omega_t(y^*+u)+\omega^2(R_0+x)]\mathbf i \\ &&\ \ \ \ \ \ \ \ \ +[u_{tt}-\omega_t(R_0+x)-\omega^2(y^*+u)] \mathbf j + v_{tt} \mathbf k, \end{eqnarray}$

(21)

where subscript $t$ denotes derivatives with respect to the time.

2.7. Rotating blades model

In order to derive the blades model and the associated boundary conditions, the extended Hamilton's principle is used. This can be formulated as

$\begin{eqnarray} \int_{t_1}^{t_2} (\delta K-\delta U +\delta W) \mathrm dt = 0, \; \; \delta u = 0, \delta v = 0 \; \mathrm{at}\; t = t_1, t_2. \end{eqnarray}$

(22)

where $K$ and $U$ , respectively, denote the kinetic energy and the strain energy, $W$ is the virtual work of external forces, $\delta$ is the variation operator.

Thanks to the cross section of the blades is rectangular, we get

$\begin{eqnarray} \oint y(s)\mathrm ds = \oint z(s)\mathrm ds = 0. \end{eqnarray}$

(23)

Utilizing (23), the kinetic energy is obtained

$\begin{eqnarray*} \begin{aligned} K = &\frac{1}{2}\int_{\tau}\rho \mathbf R_t^2 \mathrm d\tau\\ = &\frac{1}{2}\rho\int_{\tau} u_t^2+v_t^2+2\omega(R_0+x)u_t\\ &\ \ \ \ \ \ \ \ +\omega^2\big(u^2+2(y+n\frac{ dz}{ds})u+y+n\frac{dz}{ds}+(R_0+x)^2\big)\mathrm d\tau\\ = &\frac{1}{2}\rho\bar A \int_0^l u_t^2+v_t^2+2\omega(R_0+x)u_t\\ &\ \ \ \ \ \ \ \ +\omega^2\big(u^2+2\frac{h \oint y \mathrm d s }{\bar A}u+\frac{h \oint y \mathrm d s }{\bar A}+(R_0+x)^2\big)\mathrm dx\\ = &\frac{1}{2}\rho\bar A \int_0^l u_t^2+v_t^2+2\omega(R_0+x)u_t+\omega^2\big(u^2+(R_0+x)^2\big)\mathrm dx, \end{aligned} \end{eqnarray*}$

where $\mathrm d \tau$ denotes the differential volume element. Then

$\begin{eqnarray} \int_{t_1}^{t_2}\delta K \mathrm dt = -\rho \bar A \int_{t_1}^{t_2}\int _0^l\big\{[u_{tt}+\omega_t(R_0+x)-u\omega^2]\delta u +v_{tt}\delta v\big\}\mathrm dx \mathrm dt, \end{eqnarray}$

(24)

Due to the rotating motion of the blades, the total strain energy consists of two parts. The strain energy caused by the centrifugal force can be obtained as

$\begin{eqnarray} U_1 = \frac{1}{2}\int_0^l\oint\int_{-\frac{h}{2}}^{\frac{h}{2}}\int_{x}^l \rho\omega^2(R_0+\varsigma) \varepsilon_{xx}\mathrm d\varsigma\mathrm dn\mathrm ds\mathrm dx \end{eqnarray}$

(25)

where $\rho$ is the bulk density of the blades, $\int_{x}^l \rho\omega^2(R_0+\varsigma) \mathrm d\varsigma$ is the centrifugal force per unit cross section. Then,

$\begin{equation} \begin{split} \delta U_1 = &\frac{1}{2}\int_0^l\oint\int_{-\frac{h}{2}}^{\frac{h}{2}}\int_{x}^l \rho\omega^2(R_0+\varsigma)\delta \varepsilon_{xx}\mathrm d\varsigma\mathrm dn\mathrm ds\mathrm dx \\ = & \frac{1}{2}\int_0^l\oint\int_{-\frac{h}{2}}^{\frac{h}{2}}\int_{x}^l \rho\omega^2(R_0+\varsigma)\delta \bar{\varepsilon}_{xx}\mathrm d\varsigma\mathrm dn\mathrm ds\mathrm dx \\ = &\rho \bar A \omega^2 \int_0^l R(x)(u_x \delta u_x + v_x \delta v_x)\mathrm dx\\ &-\frac{\rho\omega^2h}{2}\int_0^l\int_{x}^l (R_0+\varsigma)(\oint y(s)\mathrm ds\delta u_{xx}+\oint z(s)\mathrm ds\delta v_{xx})\mathrm d\varsigma\mathrm dx, \end{split} \end{equation}$

(26)

where $\bar A$ is the area of the cross section of the blades, $R(x) = R_0(l-x)+\frac{1}{2}(l^2-x^2).$

Thanks to (23), the variation of the strain energy caused by the centrifugal force can be rewritten as

$\begin{eqnarray} \delta U_1 = \rho \bar A \omega^2 \big[R(x)(u_x \delta u + v_x \delta v)\big]\Big|_0^l-\int_0^l (R(x)u_x)_x \delta u + (R(x)v_x)_x \delta v\mathrm dx \end{eqnarray}$

(27)

The strain energy induced by the deformation of the rotating blades can be expressed as

$\begin{eqnarray} U_2 = \frac{1}{2}\int_0^l\oint\int_{-\frac{h}{2}}^{\frac{h}{2}}\sigma_{xx} \varepsilon_{xx}\mathrm dn\mathrm ds\mathrm dx. \end{eqnarray}$

(28)

Substituting the expressions of the stress and strain resultants into (28) yields

$\begin{equation} \begin{split} \delta U_2 = &E\int_0^l\oint\int_{-\frac{h}{2}}^{\frac{h}{2}}\varepsilon_{xx}\delta \varepsilon_{xx}dndsdx\\ = &\frac{Eh}{4}\int_0^l \oint (u^2_x+v^2_x)\delta (u^2_x+v^2_x) dsdx\\ &-\frac{Eh}{2}\int_0^l\oint (u^2_x+v^2_x) \delta (u_{xx}y(s)+v_{xx}z(s)) dsdx\\ &-\frac{Eh}{2}\int_0^l\oint (u_{xx}y(s)+v_{xx}z(s)) \delta (u^2_x+v^2_x) dsdx\\ &+Eh\int_0^l\oint (u_{xx}y(s)+v_{xx}z(s)) \delta (u_{xx}y(s)+v_{xx}z(s)) dsdx\\ &+\frac{Eh^3}{12}\int_0^l \oint (-u_{xx}\frac{\mathrm d z}{\mathrm d s}+v_{xx}\frac{\mathrm d y}{\mathrm d s})\delta (-u_{xx}\frac{\mathrm d z}{\mathrm d s}+v_{xx}\frac{\mathrm d y}{\mathrm d s}) ds dx \end{split} \end{equation}$

(29)

$\begin{equation} \begin{split} \delta U_2& = E\int_0^l\oint\int_{-\frac{h}{2}}^{\frac{h}{2}}\varepsilon_{xx}\delta \varepsilon_{xx}\mathrm dn\mathrm ds\mathrm dx\\ & = \rho\bar A\Bigg\{ \int_0^l[-\frac{a_5}{2}((u^2_x+v^2_x)u_x)_x+(a_6u_{xx}-a_3v_{xx})_{xx}\\ &\ \ \ \ \ \ \ \ \ \ -\frac{a_1}{2}(u^2_x+v^2_x)_{xx}+(a_1u_{xx}+a_2v_{xx})u_x)_x] \mathrm dx \Bigg\}\delta u\\ &\; +\rho\bar A\Bigg\{ \int_0^l[-\frac{a_5}{2}((u^2_x+v^2_x)v_x)_x+(a_4v_{xx}-a_3u_{xx})_{xx}\\ &\ \ \ \ \ \ \ \ \ \ -\frac{a_2}{2}(u^2_x+v^2_x)_{xx} +(a_1u_{xx}+a_2v_{xx})v_x)_x] \mathrm dx \Bigg\}\delta v\\ &\; +\rho\bar A\Bigg\{ [(a_6u_{xx}-a_3v_{xx})-\frac{a_1}{2}(u^2_x+v^2_x)] \delta u_x \Bigg\}\Bigg|_0^l\\ &\; +\rho\bar A\Bigg\{ [(a_4v_{xx}-a_3u_{xx})-\frac{a_2}{2}(u^2_x+v^2_x)] \delta v_x \Bigg\}\Bigg|_0^l\\ &\; +\rho\bar A\Bigg\{ [\frac{a_5}{2}(u^2_x+v^2_x)u_x-(a_6u_{xx}-a_3v_{xx})_x\\ &\ \ \ \ \ \ \ \ \ \ +\frac{a_1}{2}(u^2_x+v^2_x)_x+ a_1u_{xx}+a_2v_{xx}u_x] \delta u\Bigg\}\Bigg|_0^l\\ &\; +\rho\bar A\Bigg\{ [\frac{a_5}{2}(u^2_x+v^2_x)v_x-(a_4v_{xx}-a_3u_{xx})_x\\ &\ \ \ \ \ \ \ \ \ \ +\frac{a_2}{2}(u^2_x+v^2_x)_x+ a_1u_{xx}+a_2v_{xx}v_x] \delta v\Bigg\}\Bigg|_0^l\\ \end{split} \end{equation}$

(30)

where

$\begin{eqnarray*} a_1 = \frac{Eh}{\bar A\rho} \oint y \mathrm ds = 0;\\ a_2 = \frac{Eh}{\bar A\rho} \oint z \mathrm ds = 0;\\ a_3(x) = \frac{Eh}{\bar A\rho}\oint \frac{h^2}{12}\frac{\mathrm d y}{\mathrm d s}\frac{\mathrm d z}{\mathrm d s}-yz {\mathrm d s};\\ a_4(x) = \frac{Eh}{\bar A\rho}\oint \frac{h^2}{12}(\frac{\mathrm d z}{\mathrm d s})^2 +y^2 {\mathrm d s};\\ a_5 = \frac{E}{\rho}\\ a_6(x) = \frac{Eh}{\bar A\rho}\oint \frac{h^2}{12}(\frac{\mathrm d y}{\mathrm d s})^2+z^2{\mathrm d s}. \end{eqnarray*}$

The work of the non-conservative external forces can be obtained as

$\begin{eqnarray} W = \int_0^l p_y u+p_z v \mathrm dx. \end{eqnarray}$

(31)

Then

$\begin{eqnarray} \begin{aligned} \int_{t_1}^{t_2}\delta W \mathrm dt = &\int_{t_1}^{t_2}\int_0^l \tilde{p}_y \delta u+\tilde{p}_z \delta v \mathrm dx\mathrm dt\\ &+ \int_{t_1}^{t_2}\Big\{(b_1u+e_1v)\delta u+ (b_2u+e_2v)\delta v\Big\}\Big|_0^l\mathrm dt. \end{aligned} \end{eqnarray}$

(32)

where

$\begin{eqnarray} &&\tilde{p}_y = 2b_3u+(e_3+b_4)v \end{eqnarray}$

(33)

$\begin{eqnarray} &&\tilde{p}_z = (e_3+b_4)u+2e_4v \end{eqnarray}$

(34)

Inserting variation of potential energy (27) and (30), variation of kinetic energy (24) and variation of external work equation (32) into the extended Hamilton's principle (22), collecting the terms associated with the same variations, invoking the stationarity of the functional within the time interval $[t_0, t_1]$ , and the fact that the variations $(\delta u, \delta v)$ are independent and arbitrary, their coefficients in the two integrands must vanish independently [see [21]]. The partial differential equations with respect to variation of problem's independent variables and the associated boundary conditions are obtained as

$\begin{eqnarray} u_{tt}+\Big(a_6 u_{xx}-a_3 v_{xx}\Big)_{xx}-\frac{a_5}{2}((u^2_x+v^2_x)u_x)_x\ \ \ \ \ \ \ \ \\ -\omega^2(R(x)u_x)_x-\omega^2 u-\overline p_1+\omega_t(R_0+x) = 0, \end{eqnarray}$

(35)

$\begin{eqnarray} v_{tt}+\Big(a_4 v_{xx}-a_3 u_{xx}\Big)_{xx}-\frac{a_5}{2}((u^2_x+v^2_x)v_x)_x - \omega^2(R(x)v_x)_x-\overline p_2 = 0, \end{eqnarray}$

(36)

where

$\begin{eqnarray} \overline{p}_1 = \frac{\tilde{p}_y}{\rho \bar A}, \ \ \overline{p}_2 = \frac{\tilde{p}_z}{\rho \bar A}. \end{eqnarray}$

(37)

The following two types of boundary conditions are generated due to the different design of engine blades.

$\begin{eqnarray} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; u, u_x = 0, v, v_x = 0, &&x = 0, \end{eqnarray}$

(38)

$\begin{eqnarray} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; u, u_x = 0, v, v_x = 0, &&x = l. \end{eqnarray}$

(39)

and

$\begin{eqnarray} u, u_x = 0, v, v_x = 0, &&x = 0, \end{eqnarray}$

(40)

$\begin{eqnarray} a_6u_{xx}-a_3v_{xx} = 0, &&x = l, \end{eqnarray}$

(41)

$\begin{eqnarray} a_4v_{xx}-a_3u_{xx} = 0, &&x = l, \end{eqnarray}$

(42)

$\begin{eqnarray} \frac{a_5}{2}(u^2_x+v^2_x)u_x-(a_6u_{xx}-a_3v_{xx})_x-b_1u-e_1v = 0, &&x = l, \end{eqnarray}$

(43)

$\begin{eqnarray} \frac{a_5}{2}(u^2_x+v^2_x)v_x-(a_4v_{xx}-a_3u_{xx})_x-b_2u-e_2v = 0, &&x = l. \end{eqnarray}$

(44)

The conditions (38)-(39) represent C-C boundary condition. The conditions (40)-(44) represent C-F boundary condition. Now we study the well-posedness and regularity of the solution for C-C and C-F blades.

3. Preliminary

We write $Q = \Omega \times (0, T),$ where $\Omega = (0, l)$ and $T>0,$

$\begin{eqnarray*} H_0^2 = \{\psi\in H^2(\Omega)\; |\; \psi(x) = 0, \psi_x(x) = 0, x\in \partial\Omega\}, \\ H_f^2 = \{\psi\in H^2(\Omega)\; |\; \psi(0) = 0, \psi_x(0) = 0, x = 0\}. \end{eqnarray*}$

We list Gagliardo-Nirenberg inequality for bounded domains (see [23]) to be used in the subsequent sections

Lemma 3.1. Let $\Omega \subset R^n$ be a bounded domain with smooth boundary. Let $1\leq p, q, r \leq \infty$ be real numbers and $j\leq m$ be non-negative integers. If a real number $\alpha$ satisfies

$\begin{eqnarray*} \frac{1}{p}-\frac{j}{n} = \alpha\left(\frac{1}{r}-\frac{m}{n}\right)+(1-\alpha) \frac{1}{q}, \quad \frac{j}{m} \leq \alpha \leq 1 \end{eqnarray*}$

Then

$\begin{eqnarray*} \left\|\mathrm{D}^{j} f\right\|_{L^{p}(\Omega)} \leq C_{1}\left\|\mathrm{D}^{m} f\right\|_{L^{r}(\Omega)}^{\alpha}\|f\|_{L^{q}(\Omega)}^{1-\alpha}+C_{2}\|f\|_{L^{s}(\Omega)} \end{eqnarray*}$

where $s >0$ , and the constants $C_1$ and $C_2$ depend upon $\Omega$ and the indices $p, q, r, m, j, s$ only.

Now, we give the Aubin-Lions Lemma (see [22]).

Lemma 3.2. Suppose $B_0, B, B_1$ are Banach Space, if

(i) $\{u_i\}_{i = 1}^{\infty}$ is bounded in $L^{p_0}(0, T; B_0)$ ;

(ii) $\{u_{i, t}\}_{i = 1}^{\infty}$ is bounded in $L^{p_1}(0, T; B_1)$ ;

(iii) $B_0\hookrightarrow \hookrightarrow B \hookrightarrow B_1$ ,

then $\{u_i\}_{i = 1}^{\infty}$ admits a strongly converging subsequence in $L^{p_0}(0, T; B)$ , provided $p_0< \infty, p_1 > 1.$

4. Mathematical results of C-C blades

Without loss of generality, we assume $a_5 = 1.$ For the sake of brevity, the initial boundary-value problems of C-C blades are rewritten as:

$\begin{equation} \left\{\begin{aligned} &u_{tt}+(a_6u_{xx})_{xx}- (a_3v_{xx})_{xx}- \frac{1}{2}((u_x^2 +v_x^2)u_x)_x&&\\ &\ \ \ \ - {\omega ^2}(Ru_x)_x-p_1 = 0 \; \; &&\mathrm{in} \; Q, \\ &v_{tt}-(a_3u_{xx})_{xx}+ (a_4v_{xx})_{xx}- \frac{1}{2}((u_x^2 +v_x^2)v_x)_x&&\\ &\ \ \ \ - {\omega ^2}(Rv_x)_x-p_2 = 0 \; \; &&\mathrm{in} \; Q, \\ &u, u_x = 0, v, v_x = 0 &&\mathrm{on} \; \partial \Omega\times [0, T], \\ &u = u_{0}(x), u_t = u_{1}(x), v = v_{0}(x), v_t = v_{1}(x) &&\mathrm{on} \; \Omega\times \{t = 0\}. \end{aligned}\right. \end{equation}$

(45)

where

$\begin{eqnarray} p_1 = \overline{p}_1+\omega^2u-\omega_t(R_0+x), \ \ p_2 = \overline{p}_2. \end{eqnarray}$

(46)

Definition 4.1. We say function $(u, v)$ , $u, v\in L^{\infty}(0, T; H_0^2(\Omega))$ , with

$u_t, v_t \in L^{\infty}(0, T; L^2(\Omega)), u_{tt}, v_{tt} \in L^{\infty}(0, T; H^{-2}(\Omega)),$

is a weak solution of the initial boundary value problem (45) provided

(ⅰ)

$\begin{equation} \begin{split} ( u_{tt}, \varphi)+ (a_6u_{xx}, \varphi_{xx})-(a_3v_{xx}, \varphi_{xx}) + \frac{1}{2}((u_x^2 +v_x^2)u_x, \varphi_x)\\ +\omega ^2(R u_x, \varphi_x)-(p_{1}, \varphi ) = 0. \end{split} \end{equation}$

(47)

$\begin{equation} \begin{split} ( v_{tt}, \varphi)- (a_3u_{xx}, \varphi_{xx})+(a_4v_{xx}, \varphi_{xx}) + \frac{1}{2}((u_x^2 +v_x^2)v_x, \varphi_x)\\ +\omega ^2(R v_x, \varphi_x)-(p_{2}, \varphi ) = 0. \end{split} \end{equation}$

(48)

for each $\varphi \in H_0^2(\Omega)$ , and a.e. time $0\leq t\leq T$ , and

(ⅱ)

$\begin{eqnarray} u(0) = u_0, u_t(0) = u_1;v(0) = v_0, v_t(0) = v_1. \end{eqnarray}$

(49)

Remark 1. From the definition, we know $u, v\in C ( [0, T]; L^2(\Omega) )$ and $u_t, v_t \in C ( [0, T]; H^{-2}(\Omega) )$ .

Remark 2. By the product Minkowshi inequality, we can obtain:

$\begin{eqnarray} a_4a_6-a^2_3 > 0, \; x \in \Omega. \end{eqnarray}$

(50)

Theorem 4.2. (Existence for weak solution of (45)) Assume

$\begin{eqnarray} \omega\in C^{1}(0, T), a_3, a_4, a_6 \in L^{\infty}(\Omega), u_0, v_0 \in H_0^2 (\Omega), u_1, v_1 \in L^2 (\Omega). \end{eqnarray}$

(51)

there exists a weak solution of (45).

We now briefly outline the proof of Theorem 4.2 in the following:

Step 1. employing Galerkin's method to construct solutions of certain finite-dimensional approximations to (45) (correspond to Lemma 6.1 in chapter 6);

Step 2. using the energy method to find the uniform estimates of the finite-dimensional approximations solutions (correspond to Lemma 6.2 in chapter 6);

Step 3. using compactness method to obtain the weak solutions of (45).

Now we give the smoothness of weak solutions of (45).

Theorem 4.3. (Improved regularity) Assume

$\begin{align} \left \{ a_3, a_4, a_6 \in L^{\infty}(\Omega), \omega\in C^{1}(0, T), \omega_{tt}\in L^{\infty}(0, T), \\ u_{0}\in H_0^2(\Omega)\cap H^4(\Omega), v_{0}\in H_0^2(\Omega)\cap H^4(\Omega), u_{1}\in H_0^2(\Omega), v_{1}\in H_0^2(\Omega), \right.\label{condition2} \end{align}$

(52)

the weak solution of (45) satisfies

$\begin{eqnarray} u_t, v_t \in L^{\infty}(0, T; H_0^2(\Omega)), u_{tt}, v_{tt} \in L^{\infty}(0, T; L^2(\Omega)). \end{eqnarray}$

(53)

Theorem 4.4. (Interior regularity) Under the condition (52), for any $\varphi(x)\in C_0^{\infty}(\Omega), \varphi(x)>0, x \in \Omega,$ then the weak solution of (45) satisfies

$\varphi u_{xxx}, \varphi v_{xxx} \in L^{\infty}(0, T; L^2(\Omega)).$

Remark 3. Multiplying the first and second equation of (45) by $a_4(x)u_{tt}, a_3(x)v_{tt}$ respectively, integrating with over $\Omega'$ , and summing the two equations, we can obtain that the weak solution of (45) satisfies

$u_{xxxx}, v_{xxxx} \in L^{\infty}(0, T; L^2(\Omega')).$

where $\Omega'$ is a bounded open interval and $\overline{\Omega}'\subset \Omega.$

If the pre-twist angle $\alpha(x)$ is neglected, we can get $\theta(x), a_3(x), a_4(x), a_6(x)$ are all constant. Then we give the following theorem.

Theorem 4.5. (Improved regularity when $\alpha(x) = 0$ ) Under the condition (52), assume $\alpha(x) = 0$ , then the weak solution of (45) satisfies

$u, v \in L^{\infty}(0, T; H^4(\Omega)\cup H_0^2(\Omega)).$

Now, we study the uniqueness and stability of (45), denote operator $\pi$ satisfying

$\pi:\{a_3, a_4, a_6, \omega, \theta, u_0, v_0, u_1, v_1\}\rightarrow \{u, v\},$

and

$W(Q) = \{\psi| \psi \in L^{\infty}(0, T;H_0^2(\Omega)), \psi_t \in L^{\infty}(0, T;L^2(\Omega))\}$

with the norm

$||\psi||_{W(Q)} = ||\psi||_{L^{\infty}(0, T;H_0^2(\Omega))}+||\psi_t||_{L^{\infty}(0, T;L^2(\Omega))} .$

Then $W(Q)$ is Banach space.

Theorem 4.6. Undering the condition (52),

$\pi:\{(L^{\infty}(\Omega))^3 \times W^{1, 1}(0, T)\cap C^1(0, T)\times L^1(\Omega) \times (H_0^2(\Omega))^2 \times (L^2(\Omega))^2\} \rightarrow (W(Q))^2$

is continuous.

5. Mathematical results of C-F blades

The initial boundary-value problems of C-F blades are rewritten as :

$\begin{equation} \left\{\begin{aligned} &u_{tt}+(a_6u_{xx})_{xx}- (a_3v_{xx})_{xx}- \frac{1}{2}((u_x^2 +v_x^2)u_x)_x&&\\ &\ \ \ \ - {\omega ^2}(Ru_x)_x-p_1 = 0\; \; &&\mathrm{in} \; Q, \\ &v_{tt}-(a_3u_{xx})_{xx}+ (a_4v_{xx})_{xx}- \frac{1}{2}((u_x^2 +v_x^2)v_x)_x&&\\ &\ \ \ \ -{\omega ^2}(Rv_x)_x-p_2 = 0\; \; &&\mathrm{in} \; Q, \\ &u, u_x = 0, v, v_x = 0, \; \; &&x = 0, \\ &a_6u_{xx}-a_3v_{xx} = 0, \; \; &&x = l, \\ &a_4v_{xx}-a_3u_{xx} = 0, \; \; &&x = l, \\ &\frac{1}{2}(u^2_x+v^2_x)u_x-(a_6u_{xx}-a_3v_{xx})_x-b_1u-e_1v = 0, \; \; &&x = l, \\ &\frac{1}{2}(u^2_x+v^2_x)v_x-(a_4v_{xx}-a_3u_{xx})_x-b_2u-e_2v = 0, \; \; &&x = l, \\ &u = u_{0}(x), u_t = u_{1}(x), v = v_{0}(x), v_t = v_{1}(x)\; \; &&\mathrm{on} \; \Omega\times \{t = 0\}. \end{aligned}\right. \end{equation}$

(54)

Definition 5.1. We say function $(u, v)$ , $u, v\in L^{\infty}(0, T; H_f^2(\Omega))$ , with

$u_t, v_t \in L^{\infty}(0, T; L^2(\Omega)), u_{tt}, v_{tt} \in L^{\infty}(0, T; H^{-2}(\Omega)),$

is a weak solution of the initial boundary value problem (54) provided

(ⅰ)

$\begin{equation} \begin{split} ( u_{tt}, \varphi)+ (a_6u_{xx}, \varphi_{xx})-(a_3v_{xx}, \varphi_{xx}) + \frac{1}{2}((u_x^2 +v_x^2)u_x, \varphi_x)+\omega ^2(R u_x, \varphi_x)\\ -(p_{1}, \varphi )-\big(b_1u(l)+e_1v(l)\big)\varphi(l) = 0. \end{split} \end{equation}$

(55)

$\begin{equation} \begin{split} ( v_{tt}, \varphi)- (a_3u_{xx}, \varphi_{xx})+(a_4v_{xx}, \varphi_{xx}) + \frac{1}{2}((u_x^2 +v_x^2)v_x, \varphi_x)+\omega ^2(R v_x, \varphi_x)\\ -(p_{2}, \varphi )-\big(b_2u(l)+e_2v(l)\big)\varphi(l) = 0. \end{split} \end{equation}$

(56)

for each $\varphi \in H_f^2(\Omega)$ , and a.e. time $0\leq t\leq T$ , and

(ⅱ)

$\begin{eqnarray*} u(0) = u_0, u_t(0) = u_1;v(0) = v_0, v_t(0) = v_1. \end{eqnarray*}$

Similar as above, we derive the following conclusion of the initial boundary value problem (54).

Theorem 5.2. (Existence for the weak solutions of (54)) Assume

$\begin{eqnarray} \omega\in C^{1}(0, T), a_3, a_4, a_6 \in L^{\infty}(\Omega), u_0, v_0 \in H_f^2 (\Omega), u_1, v_1 \in L^2 (\Omega), \end{eqnarray}$

(57)

there exists a weak solution of (54).

Theorem 5.3. (Regularity weak solution of (54)) Assume

$\begin{eqnarray} \left \{\begin{aligned} a_3, a_4, a_6 \in L^{\infty}(\Omega), \omega\in C^{1}(0, T), \omega_{tt}\in L^{\infty}(0, T), \\ u_{0}, v_{0}\in H_f^2(\Omega)\cap H^4(\Omega), u_{1}, v_{1}\in H_f^2(\Omega). \end{aligned} \right. \end{eqnarray}$

(58)

Then there exists $T^*< T$ such that, the weak solution of (54) satisfies

$\begin{eqnarray} u_t, v_t \in L^{\infty}(0, T^*; H_f^2(\Omega)), u_{tt}, v_{tt} \in L^{\infty}(0, T^*; L^2(\Omega)). \end{eqnarray}$

(59)

Remark 4. If pre-twist angle $\alpha(x)$ is neglected, according to the first and second equations of (54), we can obtain that the weak solution of (54) satisfies

$\begin{eqnarray} u, v \in L^{\infty}(0, T^*; H^4(\Omega)\cup H_f^2(\Omega)). \end{eqnarray}$

(60)

where $u_{tt}, v_{tt} \in L^{\infty}(0, T^*; L^2(\Omega))$ and Gagliardo-Nirenberg inequality are used.

6. Proof of mathematical results

We construct weak solution of the initial boundary-value problem (45) by first solving a finite dimensional approximation. We thus employ Galerkin's method by selecting smooth functions $\varphi_k = \varphi_k(x)(k = 1, \cdots)$ such that

$\begin{equation} \{\varphi_k\}_{k = 1}^{\infty} \text{is an orthogonal basis of} H_0^2(\Omega), \end{equation}$

(61)

and

$\begin{equation} \{\varphi_k\}_{k = 1}^{\infty} \text{is an orthonormal basis of} L^2(\Omega). \end{equation}$

(62)

Fix a positive integer $m$ , and write

$\begin{eqnarray} u_m = \sum\limits_{k = 1}^{m}d_{0m}^{k}(t)\varphi_k, \ v_m = \sum\limits_{k = 1}^{m}d_{1m}^{k}(t)\varphi_k, \end{eqnarray}$

(63)

where we intend to select the coefficients $d_{0m}^k(t), d_{1m}^k(t)(0\leq t\leq T, k = 1, \cdots, m)$ to satisfy

$\begin{equation} d_{0m}^k(0) = (u_0, \varphi_k), d_{1m}^k(0) = (v_0, \varphi_k), k = 1, \cdots, m, \end{equation}$

(64)

$\begin{equation} d_{0m, t}^k(0) = (u_1, \varphi_k), d_{1m, t}^k(0) = (v_1, \varphi_k), k = 1, \cdots, m, \end{equation}$

(65)

and

$\begin{equation} \begin{split} (u_{m, tt}, \varphi_k)+(a_6u_{m, xx}, \varphi_{k, xx})-(a_3v_{m, xx}, \varphi_{k, xx})+ \frac{1}{2}((u_{m, x}^2 +v_{m, x}^2)u_{m, x}, \varphi_{k, x})\\ +\omega ^2(Ru_{m, x}, \varphi_{k, x})-(p_{1m}, \varphi_k) = 0, \end{split} \end{equation}$

(66)

$\begin{equation} \begin{split} (v_{m, tt}, \varphi_k)-(a_3 u_{m, xx}, \varphi_{k, xx})+(a_4 v_{m, xx}, \varphi_{k, xx} ) + \frac{1}{2}((u_{m, x}^2 +v_{m, x}^2)v_{m, x}, \varphi_{k, x})\\ +\omega^2(R v_{m, x}, \varphi_{k, x} )-(p_{2m}, \varphi_k) = 0. \end{split} \end{equation}$

(67)

where

$\begin{eqnarray*} \begin{aligned} &p_{1m} = \frac{1}{\rho \bar A}\big(2b_3u_m+(e_3+b_4)v_m\big)+\omega ^2u_m -\omega_t (R_0 + x), \\ &p_{2m} = \frac{1}{\rho \bar A}\big((e_3+b_4)u_m+ 2e_4v_m \big). \end{aligned} \end{eqnarray*}$

In order to proof Theorem 4.2, we need the following Lemma.

Lemma 6.1. Under the condition (51), for each integer $m = 1, 2, \cdots,$ there exists a unique function $u_m, v_m$ of the form (63) satisfying (64)-(67) for $0\leq t \leq t_m$ .

Proof. Assuming $u_m, v_m$ to be given by (63), by using (62), equation (66) and (67) become the nonlinear system of ODE

$\begin{eqnarray} \begin{aligned} &d_{0m, tt}^k+ \sum\limits_{j = 1}^m d_{0m}^j(a_6\varphi_{j, xx}, \varphi_{k, xx})-\sum\limits_{j = 1}^m d_{1m}^j(a_3\varphi_{j, xx}, \varphi_{k, xx} )\\ &+\frac{1}{2}\sum\limits_{j = 1}^m d_{0m}^j\Big(\big((\sum\limits_{i = 1}^m d_{0m}^i\varphi_{i, x})^2 +(\sum\limits_{i = 1}^m d_{1m}^i\varphi_{i, x})^2\big)\varphi_{j, x}, \varphi_{k, x}\Big) \\ &+\omega ^2\sum\limits_{j = 1}^m d_{0m}^j(R \varphi_{j, x}, \varphi_{k, x} )-\frac{1}{\rho \bar A}\big(2b_3d_{0m}^k+(e_3+e_4)d_{1m}^k\big)\\ = &\omega ^2d_{0m}^k- \omega_t((R_0 + x), \varphi_k), \end{aligned} \end{eqnarray}$

(68)

$\begin{eqnarray} \begin{aligned} &d_{1m, tt}^k+ \sum\limits_{j = 1}^m d_{1m}^j(a_4\varphi_{j, xx}, \varphi_{k, xx})-\sum\limits_{j = 1}^m d_{0m}^j(a_3 \varphi_{j, xx}, \varphi_{k, xx})\\ &+\frac{1}{2}\sum\limits_{j = 1}^m d_{1m}^j\Big(\big((\sum\limits_{i = 1}^m d_{0m}^i\varphi_{i, x})^2 +(\sum\limits_{i = 1}^m d_{1m}^i\varphi_{i, x})^2\big)\varphi_{j, x}, \varphi_{k, x}\Big)\\ &+\omega ^2\sum\limits_{j = 1}^m d_{1m}^j(R \varphi_{j, x}, \varphi_{k, x} )-\frac{1}{\rho \bar A}\big((e_3+e_4)d_{0m}^k+2e_4d_{1m}^k\big) = 0. \end{aligned} \end{eqnarray}$

(69)

subject to the initial conditions (64), (65). According to standard theory for ODE, there exists unique $C^2$ funcion $\mathbf d_{0m}(t) = (d_{0m}^1, d_{0m}^2, \cdots, d_{0m}^m), \mathbf d_{1m}(t) = (d_{1m}^1, d_{1m}^2, \cdots, d_{1m}^m)$ , satisfying (64), (65), and solving (68), (69) for $0\leq t \leq t_m,$ where $t_m: = t(m)$ is a function of $m$ .

We propose now to send $m$ to infinity and to show a subsequence of the solutions $u_m, v_m$ of the approximate problem (64)-(67) converges to a weak solution of (45). For this we will need the following uniform estimates.

Lemma 6.2. Undering the condition (51), there exists positive constant $C(\Omega, T)$ , such that

$\begin{equation} \begin{split} &|| u_{m, t} ||^2_{L^2(\Omega)}+|| v_{m, t} ||^2_{L^2(\Omega)}+|| u_{m} ||^2_{H_0^2(\Omega)}+|| v_{m} ||^2_{H_0^2(\Omega)}\\ \leq & C(|| u_{1} ||^2_{L^2(\Omega)} +|| v_{1} ||^2_{L^2(\Omega)}+||u_{0} ||^2_{H_0^2(\Omega)}+|| v_{0} ||^2_{H_0^2(\Omega)})+C\\ \end{split} \end{equation}$

(70)

for $m = 1, 2, \cdots.$

Proof. Multiplying equality (66) by $d_{0m, t}^k$ , summing $k = 1, 2, \cdots$ , we deduce

$\begin{equation} \begin{split} &( u_{m, tt}, u_{m, t})+ (a_6u_{m, xx}, u_{m, txx})-(a_3v_{m, xx}, u_{m, txx}) \\ &\; \; +\frac{1}{2}((u_{m, x}^2 +v_{m, x}^2)u_{m, x}, u_{m, tx})+\omega ^2(R u_{m, x}, u_{m, tx})-(p_{1m}, u_{m, t}) = 0 \end{split} \end{equation}$

(71)

for a.e. $0\leq t\leq t_m.$

Multiplying equality (67) by $d_{1m, t}^k$ , summing $k = 1, 2, \cdots,$ we deduce

$\begin{equation} \begin{split} &( v_{m, tt}, v_{m, t})- (a_3u_{m, xx}, v_{m, txx})+(a_4v_{m, xx}, v_{m, txx})\\ &\; \; + \frac{1}{2}((u_{m, x}^2 +v_{m, x}^2)v_{m, x}, v_{m, tx})+\omega ^2(R v_{m, x}, v_{m, tx})-(p_{2m}, v_{m, t}) = 0 \end{split} \end{equation}$

(72)

for a.e. $0\leq t\leq t_m.$

To simplify the equation (71) and (72), we can get

$\begin{equation} \begin{split} &\frac{1}{2}\frac{\mathrm d}{\mathrm dt} \lVert u_{m, t} \rVert_{L^2(\Omega)}^2 +\frac{1}{2}\frac{\mathrm d}{\mathrm dt}\lVert \sqrt{a_6}u_{m, xx} \rVert_{L^2(\Omega)}^2 -(a_3 v_{m, xx}, u_{m, txx} )\\ &\; \; + \frac{1}{2}((u_{m, x}^2 +v_{m, x}^2)u_{m, x}, u_{m, tx})+\frac{\omega ^2}{2}\frac{\mathrm d}{\mathrm dt} \lVert \sqrt{R}u_{m, x} \rVert_{L^2(\Omega)}^2 -(p_{1m}, u_{m, t}) = 0. \end{split} \end{equation}$

(73)

$\begin{equation} \begin{split} &\frac{1}{2}\frac{\mathrm d}{\mathrm dt} \lVert v_{m, t} \rVert_{L^2(\Omega)}^2 +\frac{1}{2}\frac{\mathrm d}{\mathrm dt}\lVert \sqrt{a_4}v_{m, xx} \rVert_{L^2(\Omega)}^2 -(a_3 u_{m, xx}, v_{m, txx} )\\ &\; \; + \frac{1}{2}((u_{m, x}^2 +v_{m, x}^2)v_{m, x}, v_{m, tx})+\frac{\omega ^2}{2}\frac{\mathrm d}{\mathrm dt} \lVert \sqrt{R}v_{m, x} \rVert_{L^2(\Omega)}^2 -(p_{2m}, v_{m, t}) = 0. \end{split} \end{equation}$

(74)

Summing the equations (73) and (74), we discover

$\begin{equation} \begin{split} &\frac{1}{2}\frac{\mathrm d}{\mathrm dt}\{|| u_{m, t} ||_{L^2(\Omega)}^2 +|| v_{m, t} ||_{L^2(\Omega)}^2 +|| \sqrt{a_6}u_{m, xx} ||_{L^2(\Omega)}^2 +|| \sqrt{a_4}v_{m, xx} ||_{L^2(\Omega)}^2 \\ &\; \; +\frac{\omega ^2}{2}(||\sqrt{R}u_{m, x}||_{L^2(\Omega)}^2 +||\sqrt{R}v_{m, x}||_{L^2(\Omega)}^2 )\}-\frac{\mathrm d}{\mathrm dt}|| a_3 u_{m, xx} v_{m, xx} ||_{L^1(\Omega)} \\ &\; \; +\frac{1}{8}\frac{\mathrm d}{\mathrm dt}||u_{m, x}^2 +v_{m, x}^2||_{L^2(\Omega)}^2\\ = &\omega\omega_t(||\sqrt{R}u_{m, x}||_{L^2(\Omega)}^2 +||\sqrt{R}v_{m, x}||_{L^2(\Omega)}^2 )+(p_{1m}, u_{m, t})+(p_{2m}, v_{m, t}). \end{split} \end{equation}$

(75)

Since $\omega \in C^{1}(0, T)$ , the equality (75) implies

$\begin{equation} \begin{split} &\frac{1}{2}\frac{\mathrm d}{\mathrm dt}\{|| u_{m, t} ||_{L^2(\Omega)}^2 +|| v_{m, t} ||_{L^2(\Omega)}^2 +|| \sqrt{a_6}u_{m, xx} ||_{L^2(\Omega)}^2 +|| \sqrt{a_4}v_{m, xx} ||_{L^2(\Omega)}^2 \\ &\; \; +\frac{\omega ^2}{2}(||\sqrt{R}u_{m, x}||_{L^2(\Omega)}^2 +||\sqrt{R}v_{m, x}||_{L^2(\Omega)}^2 )\}-\frac{\mathrm d}{\mathrm dt}|| a_3 u_{m, xx} v_{m, xx} ||_{L^1(\Omega)}\\ &\; \; +\frac{1}{8}\frac{\mathrm d}{\mathrm dt}||u_{m, x}^2 +v_{m, x}^2||_{L^2(\Omega)}^2 \\ \leq & C (||u_{m, t} ||_{L^2(\Omega)}^2 +||v_{m, t} ||_{L^2(\Omega)}^2 +||u_m ||^2_{H^{1}_0(\Omega)}+||v_m ||^2_{H^{1}_0(\Omega)})+C. \end{split} \end{equation}$

(76)

where we used Young inequality.

Integrating (76) with respect to $t$ , we discover

$\begin{equation} \begin{split} &|| u_{m, t} ||_{L^2(\Omega)}^2 +|| v_{m, t} ||_{L^2(\Omega)}^2 +|| \sqrt{a_6}u_{m, xx} ||_{L^2(\Omega)}^2 +|| \sqrt{a_4}v_{m, xx} ||_{L^2(\Omega)}^2 \\ &-2|| a_3 u_{m, xx} v_{m, xx} ||_{L^1(\Omega)}+\frac{\omega ^2}{2}(||\sqrt{R}u_{m, x}||_{L^2(\Omega)}^2 +||\sqrt{R}v_{m, x}||_{L^2(\Omega)}^2 )\\ &+\frac{1}{4}||u_{m, x}^2 +v_{m, x}^2||_{L^2(\Omega)}^2\\ \leq & C(\Omega, T)\{|| u_{1m} ||_{L^2(\Omega)}^2 +|| v_{1m} ||_{L^2(\Omega)}^2 +||u_{0m} ||^2_{H^{2}(\Omega)}+|| v_{0m} ||^2_{H^{2}(\Omega)}\}\\ &+C \int_0^t ||u_{m, t} ||_{L^2(\Omega)}^2 +||v_{m, t} ||_{L^2(\Omega)}^2 \mathrm dt+||u_m ||^2_{H^{1}_0(\Omega)}+||v_m ||^2_{H^{1}_0(\Omega)}+C. \end{split} \end{equation}$

(77)

Thanks to (50), there exists a constant $C$ , such that

$\begin{equation} \begin{aligned} &|| \sqrt{a_6} u_{m, xx} ||_{L^2(\Omega)}^2 + || \sqrt{a_4} v_{m, xx} ||_{L^2(\Omega)}^2- 2||a_3u_{m, xx}v_{m, xx}||_{L(\Omega)}\\ \geq& C(|| u_{m} ||^2_{H_0^2(\Omega)}+|| v_{m} ||^2_{H_0^2(\Omega)}). \end{aligned} \end{equation}$

(78)

Substituting (78) into the inequality (77), By using Poincaré inequality, we find

$\begin{equation} \begin{split} &|| u_{m, t} ||_{L^2(\Omega)}^2 +|| v_{m, t} ||_{L^2(\Omega)}^2 +|| u_{m} ||^2_{H_0^2(\Omega)}+|| v_{m} ||^2_{H_0^2(\Omega)}\ \\ \leq & C(|| u_{1m} ||_{L^2(\Omega)}^2 +|| v_{1m} ||_{L^2(\Omega)}^2 +||u_{0m} ||^2_{H_0^2(\Omega)}+|| v_{0m} ||^2_{H_0^2(\Omega)})\\ &+C\int_0^t ||u_{m, t} ||_{L^2(\Omega)}^2 +||v_{m, t} ||_{L^2(\Omega)}^2 \mathrm dt+||u_m ||^2_{H_0^2(\Omega)}+||v_m ||^2_{H_0^2(\Omega)}+C. \end{split} \end{equation}$

(79)

Then, by using Gronwall inequality, we obtain

$\begin{equation*} \begin{split} || u_{m, t} ||^2_{L^2(\Omega)} +&|| v_{m, t} ||^2_{L^2(\Omega)}+|| u_{m} ||^2_{H_0^2(\Omega)}+|| v_{m} ||^2_{H_0^2(\Omega)}\\ \leq & C(|| u_{m, t}(0) ||^2_{L^2(\Omega)} +|| v_{m, t}(0) ||^2_{L^2(\Omega)}\\ &+||u_{m}(0) ||^2_{H_0^2(\Omega)}+|| v_{m}(0) ||^2_{H_0^2(\Omega)})+C\\ \leq & C(|| u_1 ||^2_{L^2(\Omega)} +|| v_1 ||^2_{L^2(\Omega)}+||u_0 ||^2_{H_0^2(\Omega)}+|| v_0 ||^2_{H_0^2(\Omega)})+C \end{split} \end{equation*}$

for $m = 1, 2, \cdots.$

Remark 5. From the Energy estimates, we can obtain: $t_m \rightarrow T,$ as $m\rightarrow \infty.$

Thanks to Lemma 6.1 and Lemma 6.2, we can obtain the existence for the weak solutions of the initial boundary value problem (45).

6.1. Proof of Theorem 4.2

Proof. (ⅰ) According to the energy estimates (70), we see that

$\begin{eqnarray} \{u_m\}_{m = 1}^{\infty}, \{v_m\}_{m = 1}^{\infty} &&{\rm \; is \; bounded\; in\; \; } L^{\infty}(0, T;H_0^2(\Omega)); \end{eqnarray}$

(80)

$\begin{eqnarray} \{u_{m, t}\}_{m = 1}^{\infty}, \{v_{m, t}\}_{m = 1}^{\infty} &&{\rm \; is\; bounded\; in\; \; } L^{\infty}(0, T;L^2(\Omega)); \end{eqnarray}$

(81)

$\begin{eqnarray} \{u_{m, tt}\}_{m = 1}^{\infty}, \{v_{m, tt}\}_{m = 1}^{\infty} &&{\rm \; is \; bounded \; in\; \; } L^{\infty}(0, T;H^{-2}(\Omega)). \end{eqnarray}$

(82)

As a consequence there exists subsequence $\{u_{\mu}, v_{\mu}\}_{m = 1}^{\infty}$ and $u, v \in L^{\infty}(0, T; H_0^2(\Omega))$ , with $u_t, v_t \in L^{\infty}(0, T;L^2(\Omega))$ , $u_{tt}, v_{tt} \in L^{\infty}(0, T;H^{-2}(\Omega))$ , such that

$\begin{equation} \left \{ \begin{aligned} &u_{\mu}\rightarrow u, v_{\mu}\rightarrow v &&{\rm \; weakly\; * in\; } \ \ L^{\infty}(0, T;H_0^2(\Omega))\\ &u_{\mu, t}\rightarrow u_t, v_{\mu, t}\rightarrow v_t &&{\rm \; weakly\; * in\; } \ \ L^{\infty}(0, T;L^2(\Omega))\\ &u_{\mu, tt}\rightarrow u_{tt}, v_{\mu, tt}\rightarrow v_{tt} &&{\rm \; weakly\; * in\; } \ \ L^{\infty}(0, T;H^{-2}(\Omega)). \end{aligned}\right. \end{equation}$

(83)

(ⅱ) By Gagliardo-Nirenberg inequality, we can see

$\begin{equation} \begin{split} ||u_{\mu, x}||_{L^{\infty}(Q)}\leq C||u_{\mu, xx}||_{L^{\infty}(0, T; L^{2}(\Omega))}\leq C. \end{split} \end{equation}$

(84)

Otherwise,

$\begin{equation} \begin{split} ||u_{\mu, x}^2 + v_{\mu, x}^2||_{L^{\infty}(0, T; L^{2}(\Omega))}&\leq ||u_{\mu, x}||^2_{L^{\infty}(0, T; L^{4}(\Omega))} + ||v_{\mu, x}||^2_{L^{\infty}(0, T; L^{4}(\Omega))} \\ &\leq C||u_{\mu}||^2_{L^{\infty}(0, T; H_0^{2}(\Omega))} + c||v_{\mu}||^2_{L^{\infty}(0, T; H_0^{2}(\Omega))}\\ &\leq C. \end{split} \end{equation}$

(85)

Combining (84) and (85), we discover

$\begin{equation} \begin{split} ||(u_{\mu, x}^2 + v_{\mu, x}^2)u_{\mu, x}||_{L^{\infty}(0, T; L^{2}(\Omega))} \leq C||u_{\mu, x}^2 + v_{\mu, x}^2||_{L^{\infty}(0, T; L^{2}(\Omega))}\leq C. \end{split} \end{equation}$

(86)

Moreover, there exists a function $\chi \in L^{\infty}(0, T; L^{2}(\Omega))$ satisfy

$\begin{equation} \begin{split} (u_{\mu, x}^2 + v_{\mu, x}^2)u_{\mu, x}\rightarrow \chi \ \ \text{weakly * in} \ \ L^{\infty}(0, T; L^{2}(\Omega)). \end{split} \end{equation}$

(87)

By Lemma 3.2, we can find

$\begin{equation} \begin{split} u_{\mu}\rightarrow u, v_{\mu}\rightarrow v \ \ \text{strongly in} \ \ L^2(0, T; H_0^1(\Omega)) . \end{split} \end{equation}$

(88)

And so

$\begin{equation} \begin{split} u_{\mu, x} \rightarrow u_x, v_{\mu, x} \rightarrow v_x \ \ \text{strongly in}\ \ L^2(Q), \end{split} \end{equation}$

(89)

Thus

$\begin{equation} \begin{split} (u^2_{\mu, x}+v^2_{\mu, x})u_{\mu, x} \rightarrow (u^2_x+v^2_x)u_x . \end{split} \end{equation}$

(90)

Combining (87) and (90), we can obtain

$\begin{equation} \begin{split} (u_{\mu, x}^2 + v_{\mu, x}^2)u_{\mu, x}\rightarrow (u_x^2+v_x^2)u_x \ \ \text{weakly in} \ \ L^2(Q), \end{split} \end{equation}$

(91)

where we used the Lemma 1.3 of Chapter 1 in [22]. Furthermore, we have $\chi = (u_x^2+v_x^2)u_x.$

Meanwhile,

$\begin{equation} \begin{split} ((u_{\mu, x}^2 + v_{\mu, x}^2)u_{\mu, x}, \varphi_{k, x})\rightarrow((u_x^2+v_x^2)u_x, \varphi_{k, x}) \ \ \text{weakly * in} \ \ L^{\infty}(0, T). \end{split} \end{equation}$

(92)

In the same way,

$\begin{equation} \begin{split} ((u_{\mu, x}^2 + v_{\mu, x}^2)v_{\mu, x}, \varphi_{k, x})\rightarrow((u_x^2+v_x^2)v_x, \varphi_{k, x}) \ \ \text{weakly * in} \ \ L^{\infty}(0, T). \end{split} \end{equation}$

(93)

Next fix an integer $k$ , we select $\mu \geq k$ , from (66) we can get

$\begin{equation} \begin{split} &(u_{\mu, tt}, \varphi_k)+(a_6u_{\mu, xx}, \varphi_{k, xx})-(a_3v_{\mu, xx}, \varphi_{k, xx})+ \frac{1}{2}((u_{\mu, x}^2 +v_{\mu, x}^2)u_{\mu, x}, \varphi_{k, x})\\ &\; \; -\omega ^2(Ru_{\mu, x}, \varphi_{k, x})-(p_{1\mu}, \varphi_k) = 0. \end{split} \end{equation}$

(94)

Thanks to (83), we can get

$\left\{ \begin{array}{l} ({u_{\mu ,tt}},{\varphi _k}) \to ({u_{tt}},{\varphi _k}){\rm{weakly * in}}\;\;{L^\infty }(0,T),\\ ({a_6}{u_{\mu ,xx}},{\varphi _{k,xx}}) \to ({a_6}{u_{xx}},{\varphi _{k,xx}}){\rm{weakly * in}}\;\;{L^\infty }(0,T),\\ ({a_3}{v_{\mu ,xx}},{\varphi _{k,xx}}) \to ({a_3}{v_{xx}},{\varphi _{k,xx}}){\rm{weakly * in}}\;\;{L^\infty }(0,T),\\ ({p_{1\mu }},{\varphi _k}) \to ({p_1},{\varphi _k}){\rm{weakly * in}}\;\;{L^\infty }(0,T). \end{array} \right.$

(95)

From (92) and (95), we can discover

$\begin{equation} \begin{split} +v_x^2)u_x, \varphi_{k, x})\\ +\omega ^2(Ru_x, \varphi_{k, x})-(p_1, \varphi_k) = 0 \end{split} \end{equation}$

(96)

for all fixed $k$ .

Note

$\{\varphi_k\}_{k = 1}^{\infty}~~ {\rm{is}}\;\;{\rm{an}}\;\;{\rm{orthogonal}}\;\;{\rm{basis}}\;\;{\rm{of}}\;\;H_0^2(\Omega ),$

then,

$\begin{equation} \begin{split} (u_{tt}, \varphi)+(a_6u_{xx}, \varphi_{xx})-(a_3v_{xx}, \varphi_{xx})+ \frac{1}{2}((u_x^2 +v_x^2)u_x, \varphi_x)\\ +\omega ^2(Ru_x, \varphi_x)-(p_1, \varphi) = 0 \end{split} \end{equation}$

(97)

for arbitrary $\varphi \in H_0^2(\Omega)$ .

In the same way,

$\begin{equation} \begin{split} (v_{tt}, \varphi)+(a_4v_{xx}, \varphi_{xx})-(a_3u_{xx}, \varphi_{xx})+ \frac{1}{2}((u_x^2 +v_x^2)v_x, \varphi_x)\\ +\omega ^2(Rv_x, \varphi_x)-(p_2, \varphi) = 0 \end{split} \end{equation}$

(98)

for arbitrary $\varphi \in H_0^2(\Omega)$ .

Synthesizes the above analysis, there exists $u, v$ satisfying (47), (48), and

$\begin{eqnarray*} u, v &\in& L^{\infty}(0, T; H_0^2(\Omega)), \\ u_t, v_t &\in& L^{\infty}(0, T; L^2(\Omega)), \\ u_{tt}, v_{tt} &\in& L^{\infty}(0, T; H^{-2}(\Omega)). \end{eqnarray*}$

(ⅲ) Now let's prove the initial conditions.

Since, $u_{\mu}, v_{\mu}, u_{\mu, t}, v_{\mu, t}$ are bounded in $L^{2}(Q)$ , by Lemma 1.2 of Chapter 1 in [22], we can discover

$\begin{equation} \begin{split} u_{\mu}(x, 0)\rightarrow u(x, 0) \; \text{ weakly in} \; L^2(\Omega). \end{split} \end{equation}$

(99)

Otherwise,

$\begin{equation} \begin{split} u_m(x, 0)\rightarrow u_0(x)\; \text{ in }\; H_0^2(\Omega). \end{split} \end{equation}$

(100)

Combining identities (99) and (100), we can get

$u(x, 0) = u_0(x).$

Next, according to (83), we can obtain

$\begin{equation*} \begin{split} (u_{\mu, t}, \varphi_k) \rightarrow (u_{t}, \varphi_k) \; \text{ weakly * in }\; L^{\infty}(0, T), \end{split} \end{equation*}$

$\begin{equation*} \begin{split} (u_{\mu, tt}, \varphi_k)\rightarrow (u_{tt}, \varphi_k) \; \text{ weakly * in }\; L^{\infty}(0, T). \end{split} \end{equation*}$

Then, we can discover

$\begin{eqnarray} (u_{\mu, t}(x, 0), \varphi_k) \rightarrow (u_t, \varphi_k)|_{t = 0} = (u_t(x, 0), \varphi_k). \end{eqnarray}$

(101)

Otherwise,

$\begin{eqnarray} (u_{m, t}(x, 0), \varphi_k)\rightarrow (u_1(x), \varphi_k), \end{eqnarray}$

(102)

Comparing identities (101) and (102), we can get

$(u_t(x, 0), \varphi_k) = (u_1(x), \varphi_k), \text{ for arbitrary } k.$

$u_t(x, 0) = u_1(x).$

In the same way, we can obtain

$v(x, 0) = v_0(x), v_t(x, 0) = v_1(x).$

6.2. Proof of Theorem 4.3

Proof. Differentiating the first equation of (45) with respect to $t$ , multiplying by $u_{tt}$ and integrating with respect to $x$ , we discover

$\begin{equation} \begin{split} &(u_{ttt}, u_{tt})+((a_6 u_{txx})_{xx}, u_{tt})-((a_3 v_{txx})_{xx}, u_{tt}) -\frac{1}{2}(((u^2_{x} + v^2_{x}) u_{tx})_x, u_{tt})\\ &\; \; - ((u^2_{x} u_{tx})_x, u_{tt})-((u_{x} v_{x}v_{tx})_x, u_{tt}) - 2\omega\omega_t ((Ru_{x})_x, u_{tt})\\ &\; \; - \omega^2((Ru_{tx})_x, u_{tt}) - (p_{1, t}, u_{tt}) = 0. \end{split} \end{equation}$

(103)

Differentiating the second equation of (45) with respect to $t$ , multiplying by $v_{tt}$ and integrating with respect to $x$ , we discover

$\begin{equation} \begin{split} &(v_{ttt}, v_{tt}) + ((a_4 v_{txx})_{xx}, v_{tt})- ((a_3 u_{txx})_{xx}, v_{tt}) -\frac{1}{2}(((u^2_{x} + v^2_{x}) v_{tx})_x, v_{tt})\\ &\; \; - ((v^2_{x} v_{tx})_x, v_{tt})- ((u_{x} v_{x}u_{tx})_x, v_{tt}) - 2\omega\omega_t ((Rv_{x})_x, v_{tt}) \\ &\; \; - \omega^2((Rv_{tx})_x, v_{tt}) -(p_{2, t}, v_{tt}) = 0. \end{split} \end{equation}$

(104)

Summing the equation (103) and (104), we discover after integrating by parts:

$\begin{equation} \begin{split} &\frac{1}{2}\frac{\mathrm d}{\mathrm dt}(||u_{tt}||_{L^2(\Omega)}^2+||v_{tt}||_{L^2(\Omega)}^2) +\frac{1}{2}\frac{\mathrm d}{\mathrm dt}(||\sqrt{a_6}u_{txx}||_{L^2(\Omega)}^2 +||\sqrt{a_4}v_{txx}||_{L^2(\Omega)}^2)\\ &-\frac{\mathrm d}{\mathrm dt}||a_3u_{txx}v_{txx}||_{L^1(\Omega)} +\frac{\omega^2}{2}\frac{\mathrm d}{\mathrm dt}||\sqrt{R(x)} u_{tx}||_{L^2(\Omega)}^2 +\frac{\omega^2}{2}\frac{\mathrm d}{\mathrm dt}||\sqrt{R(x)} v_{tx}||_{L^2(\Omega)}^2\\ = &\frac{1}{2}((u^2_{x} + v^2_{x}) u_{txx}, u_{tt}) + (u^2_{x} u_{txx}, u_{tt}) + (u_{x}v_{x} v_{txx}, u_{tt}) +\frac{1}{2}((u^2_{x} + v^2_{x}) v_{txx}, v_{tt}) \\ &+ (v^2_{x} v_{txx}, v_{tt}) + (u_{x}v_{x} u_{txx}, v_{tt}) +3(u_{x} u_{xx} u_{tx}, u_{tt}) +( v_{x} v_{xx} u_{tx}, u_{tt})\\ &+ (u_{xx} v_{x}v_{tx}, u_{tt}) + ( u_{x} v_{xx} v_{tx}, u_{tt})+(u_{x} u_{xx} v_{tx}, v_{tt}) +3(v_{x} v_{xx}v_{tx}, v_{tt})\\ &+ (u_{xx} v_{x}u_{tx}, v_{tt}) + ( u_{x} v_{xx} u_{tx}, v_{tt})+ 2\omega\omega_t\big( (R_xu_{x}, u_{tt})+ (Ru_{xx}, u_{tt}) \\ &+ (R_xv_{x}, v_{tt})+ (Rv_{xx}, v_{tt})\big)+ (p_{1, t}, u_{tt}) +(p_{2, t}, v_{tt}). \end{split} \end{equation}$

(105)

Obviously, by Young inequality and Hölder inequality, we have

(106)

Next integrate (106) with respect to $t$ ,

$\begin{equation} \begin{split} &||u_{tt}||_{L^2(\Omega)}^2+||v_{tt}||_{L^2(\Omega)}^2+||\sqrt{a_6} u_{txx}||_{L^2(\Omega)}^2 + ||\sqrt{a_4} v_{txx}||_{L^2(\Omega)}^2 \\ &\; \; -2||a_3u_{txx}v_{txx}||_{L(\Omega)}+\omega^2(||\sqrt{R} u_{tx}||_{L^2(\Omega)}^2 + ||\sqrt{R} v_{tx}||_{L^2(\Omega)}^2) \\ \leq&C\int_0^t||u_{t}||^2_{H_0^2(\Omega)}+||v_{t}||^2_{H_0^2(\Omega)}+ ||u_{tt}||_{L^2(\Omega)}^2+||v_{tt}||_{L^2(\Omega)}^2+ ||u||^2_{H_0^2(\Omega)}+||v||^2_{H_0^2(\Omega)}\mathrm dt\\ &\; \; +||u_{tt}(x, 0)||_{L^2(\Omega)}^2 +||u_{tx}(x, 0)||_{L^2(\Omega)}^2+||u_{txx}(x, 0)||_{L^2(\Omega)}^2\\ &\; \; +||v_{tt}(x, 0)||_{L^2(\Omega)}^2 +||v_{tx}(x, 0)||_{L^2(\Omega)}^2+||v_{txx}(x, 0)||_{L^2(\Omega)}^2+C. \end{split} \end{equation}$

(107)

Since $u_1, v_1\in H_0^2(\Omega),$ we have

$\begin{eqnarray} ||u_{txx}(x, 0)||_{L^2(\Omega)}, ||v_{txx}(x, 0)||_{L^2(\Omega)}, ||u_{tx}(x, 0)||_{L^2(\Omega)}, ||v_{tx}(x, 0)||_{L^2(\Omega)}, \end{eqnarray}$

(108)

are bounded. On the other hand, multiplying the first equation and second equation of (45) by $u_{tt}, v_{tt}$ , respectively, and integrating with respect to $x$ , we discover

$||u_{tt}(x, 0)||_{L^2(\Omega)}\leq C(||u_0||_{H^4(\Omega)}+||v_0||_{H^4(\Omega)})+C\leq C,$

$||v_{tt}(x, 0)||_{L^2(\Omega)}\leq C(||u_0||_{H^4(\Omega)}+||v_0||_{H^4(\Omega)})+C\leq C.$

where the condition (52), Young inequality and Poincaré inequality are used.

As before, we have

$\begin{equation} \begin{split} ||u_{tt}||_{L^2(\Omega)}^2&+||v_{tt}||_{L^2(\Omega)}^2+||u_{t}||^2_{H_0^2(\Omega)} + || v_{t}||^2_{H_0^2(\Omega)} \\ &\leq C\int_0^t (||u_{tt}||_{L^2(\Omega)}^2+||v_{tt}||_{L^2(\Omega)}^2+||u_{t}||^2_{H_0^2(\Omega)}+ ||v_{t}||^2_{H_0^2(\Omega)} )\mathrm dt+C. \end{split} \end{equation}$

(109)

Applying the Gronwall inequality to (109) gives

$\begin{equation} \begin{split} u_{t}, v_{t} \ \ \ &\text{is bounded in} \ \ \ L^{\infty}(0, T; H_0^2(\Omega)), \\ u_{tt}, v_{tt} \ \ \ &\text{is bounded in} \ \ \ L^{\infty}(0, T; L^2(\Omega)). \end{split} \end{equation}$

(110)

Moreover, the weak solutions of (45) satisfies

$u_t \in L^{\infty}(0, T; H_0^2(\Omega)),$

$v_t \in L^{\infty}(0, T; H_0^2(\Omega)),$

$u_{tt} \in L^{\infty}(0, T; L^2(\Omega)),$

$v_{tt} \in L^{\infty}(0, T; L^2(\Omega)).$

6.3. Proof of Theorem 4.4

Proof. Multiplying the first equation and the second equation of (45) by $-\varphi^2u_{txx},$ $-\varphi^2v_{txx}$ , respectively, summing the two equations, and integrating with respect to $x$ , we discover

$\begin{equation} \begin{aligned} &(u_{tt}, -\varphi^2u_{txx})+ ( v_{tt}, -\varphi^2v_{txx})+ ((a_6u_{xx})_{xx}, -\varphi^2u_{txx})\\ &\; \; \; \; +((a_4v_{xx})_{xx}, -\varphi^2v_{txx})-((a_3v_{xx})_{xx}, -\varphi^2u_{txx})- ((a_3u_{xx})_{xx}, -\varphi^2v_{txx})\\ &\; \; \; \; -\frac{1}{2}(((u_x^2 +v_x^2)u_x)_x, -\varphi^2u_{txx}) -\frac{1}{2}(((u_x^2 +v_x^2)v_x)_x, -\varphi^2v_{txx})\\ &\; \; \; \; -\omega ^2((R u_x)_x, -\varphi^2u_{txx})-\omega ^2((R v_x)_x, -\varphi^2v_{txx})\\ &\; \; \; \; -(p_{1}, -\varphi^2u_{txx})-(p_{2}, -\varphi^2v_{txx}) = 0. \end{aligned} \end{equation}$

(111)

Thanks to integration by parts and the properties of $\varphi$ , we obtain the following equality

$\begin{eqnarray} I_1+I_2+I_3+I_4+I_5 = 0. \end{eqnarray}$

(112)

where $I_i, i = 1, \cdots, 5,$ are given as follows respectively

$\begin{eqnarray*} I_1 = (u_{tt}, -\varphi^2u_{txx})+ ( v_{tt}, -\varphi^2v_{txx})\\ I_2 = ((a_6u_{xx})_{xx}, -\varphi^2u_{txx})+((a_4v_{xx})_{xx}, -\varphi^2v_{txx})\\ I_3 = -((a_3v_{xx})_{xx}, -\varphi^2u_{txx})- ((a_3u_{xx})_{xx}, -\varphi^2v_{txx})\\ I_4 = -\frac{1}{2}(((u_x^2 +v_x^2)u_x)_x, -\varphi^2u_{txx}) -\frac{1}{2}(((u_x^2 +v_x^2)v_x)_x, -\varphi^2v_{txx})\\ I_5 = -(p_{1}, -\varphi^2u_{txx})-(p_{2}, -\varphi^2v_{txx}). \end{eqnarray*}$

To conclude, we need to estimate each of $I_i, i = 1, \cdots, 5.$ By using integration by parts and conclusion (53), we find

$\begin{equation} \begin{aligned} I_1 = &\frac{1}{2}\frac{\mathrm d}{\mathrm dt}(||\varphi u_{tx}||_{L^2(\Omega)}^2+||\varphi v_{tx}||_{L^2(\Omega)}^2)+2(\varphi \varphi_x u_{tt}, u_{tx})+ 2(\varphi \varphi_x v_{tt}, v_{tx}) \\ \geq& \frac{1}{2}\frac{\mathrm d}{\mathrm dt}(||\varphi u_{tx}||_{L^2(\Omega)}^2+||\varphi v_{tx}||_{L^2(\Omega)}^2)-C \end{aligned} \end{equation}$

(113)

For $I_2$ , we get

$\begin{eqnarray} \begin{aligned} I_2 = &\frac{1}{2}\frac{\mathrm d}{\mathrm dt}(||\sqrt{a_6}\varphi u_{xxx}||_{L^2(\Omega)}^2+||\sqrt{a_4}\varphi v_{xxx}||_{L^2(\Omega)}^2) +(a_{6, x}u_{xx}, \varphi^2 u_{txxx}) \\ &+(a_{4, x}v_{xx}, \varphi^2 v_{txxx})+2(a_6u_{xxx}, \varphi \varphi_x u_{txx})+2(a_4v_{xxx}, \varphi \varphi_x v_{txx})\\ &+2(a_{6, x}u_{xx}, \varphi \varphi_x u_{txx})+2(a_{4, x}v_{xx}, \varphi \varphi_x v_{txx}) \\ \geq&\frac{1}{2}\frac{\mathrm d}{\mathrm dt}(||\sqrt{a_6}\varphi u_{xxx}||_{L^2(\Omega)}^2+||\sqrt{a_4}\varphi v_{xxx}||_{L^2(\Omega)}^2)\\ &-(a_{6, x}u_{xxx}, \varphi^2 u_{txx})-(a_{4, x}v_{xxx}, \varphi^2 v_{txx})\\ &+2(a_6u_{xxx}, \varphi \varphi_x u_{txx})+2(a_4v_{xxx}, \varphi \varphi_x v_{txx})-C \\ \geq&\frac{1}{2}\frac{\mathrm d}{\mathrm dt}(||\sqrt{a_6}\varphi u_{xxx}||_{L^2(\Omega)}^2+||\sqrt{a_4}\varphi v_{xxx}||_{L^2(\Omega)}^2)\\ &-C||\varphi u_{xxx}||_{L^2(\Omega)}^2-C||\varphi v_{xxx}||_{L^2(\Omega)}^2-C \end{aligned} \end{eqnarray}$

(114)

where Lemma 3.1 and (52) are used. Similarly from (114), we deduce

$\begin{eqnarray} \begin{aligned} I_3 = &-(a_3v_{xxx}, \varphi^2 u_{txxx})-(a_3u_{xxx}, \varphi^2 v_{txxx})-(a_{3, x}v_{xx}, \varphi^2 u_{txxx})\\ &-(a_{3, x}u_{xx}, \varphi^2 v_{txxx})-2(a_3v_{xxx}, \varphi \varphi_x u_{txx})-2(a_3u_{xxx}, \varphi \varphi_x v_{txx})\\ &-2(a_{3, x}v_{xx}, \varphi \varphi_x u_{txx})-2(a_{3, x}u_{xx}, \varphi \varphi_x v_{txx})\\ = &-\frac{\mathrm d}{\mathrm dt}||a_3\varphi^2u_{xxx}v_{xxx}||_{L^1(\Omega)}-(a_{3, x}v_{xx}, \varphi^2 u_{txxx})-(a_{3, x}u_{xx}, \varphi^2 v_{txxx})\\ &-2(a_3v_{xxx}, \varphi \varphi_x u_{txx})-2(a_3u_{xxx}, \varphi \varphi_x v_{txx})\\ &-2(a_{3, x}v_{xx}, \varphi \varphi_x u_{txx})-2(a_{3, x}u_{xx}, \varphi \varphi_x v_{txx})\\ = &-\frac{\mathrm d}{\mathrm dt}||a_3\varphi^2u_{xxx}v_{xxx}||_{L^1(\Omega)} +(a_{3, xx}v_{xx}, \varphi^2 u_{txx})+(a_{3, xx}u_{xx}, \varphi^2 v_{txx})\\ &+(a_{3, x}v_{xxx}, \varphi\varphi_x u_{txx})+(a_{3, x}u_{xxx}, \varphi\varphi_x v_{txx}) +2(a_{3, x}v_{xx}, \varphi\varphi_x u_{txx})\\ &+2(a_{3, x}u_{xx}, \varphi\varphi_x v_{txx})-2(a_3v_{xxx}, \varphi \varphi_x u_{txx})-2(a_3u_{xxx}, \varphi \varphi_x v_{txx})\\ &-2(a_{3, x}v_{xx}, \varphi \varphi_x u_{txx})-2(a_{3, x}u_{xx}, \varphi \varphi_x v_{txx})\\ \geq&-\frac{\mathrm d}{\mathrm dt}||a_3\varphi^2u_{xxx}v_{xxx}||_{L^1(\Omega)}-C||\varphi u_{xxx}||_{L^2(\Omega)}^2-C||\varphi v_{xxx}||_{L^2(\Omega)}^2-C \end{aligned} \end{eqnarray}$

(115)

By using Lemma 3.1 and (52), we find

$\begin{eqnarray} I_4 = -C||u_{txx}||_{L^2(\Omega)}^2-C||v_{txx}||_{L^2(\Omega)}^2-C||u_{xx}||_{L^2(\Omega)}^2-c||v_{xx}||_{L^2(\Omega)}^2\geq -C \end{eqnarray}$

(116)

$\begin{eqnarray} I_5 = -C\big(||u_t||_{H_0^2(\Omega)}^2+||v_t|_{H_0^2(\Omega)}^2+||u||_{H_0^2(\Omega)}^2+||v||_{H_0^2(\Omega)}^2\big)-C\geq -C \end{eqnarray}$

(117)

Putting (113)-(117) into (112), this yields

$\begin{eqnarray} \begin{aligned} &\frac{1}{2}\frac{\mathrm d}{\mathrm dt}\Big(||\varphi u_{tx}||_{L^2(\Omega)}^2+||\varphi v_{tx}||_{L^2(\Omega)}^2+||\sqrt{a_6}\varphi u_{xxx}||_{L^2(\Omega)}^2+||\sqrt{a_4}\varphi v_{xxx}||_{L^2(\Omega)}^2\\ &\ \ \ \ \ \ \ \ -2||a_3\varphi^2u_{xxx}v_{xxx}||_{L^1(\Omega)}\Big)\\ \leq & C\big(||\varphi u_{xxx}||_{L^2(\Omega)}^2+||\varphi v_{xxx}||_{L^2(\Omega)}^2\big)+C \end{aligned} \end{eqnarray}$

(118)

Integrating with respect to $t \in (0, t)$ , taking into account the condition (50), we discover

$\begin{aligned} &||\varphi u_{tx}||_{L^2(\Omega)}^2+||\varphi v_{tx}||_{L^2(\Omega)}^2+||\varphi u_{xxx}||_{L^2(\Omega)}^2+||\varphi v_{xxx}||_{L^2(\Omega)}^2\nonumber%\\ \end{aligned}$

$\begin{aligned}\leq & C\int_0^t||\varphi u_{xxx}||_{L^2(\Omega)}^2+||\varphi v_{xxx}||_{L^2(\Omega)}^2 \mathrm dt\\ &+ C\big(||u_1||^2_{H^1(\Omega)}+||v_1||^2_{H^1(\Omega)}+||u_0||^2_{H^3(\Omega)}+||v_0||^2_{H^3(\Omega)}\big)+C \end{aligned}$

(119)

Thanks to the Gronwall inequality, we deduce

$\begin{eqnarray} ||\varphi u_{xxx}||_{L^2(\Omega)}^2+||\varphi v_{xxx}||_{L^2(\Omega)}^2 \leq C \end{eqnarray}$

(120)

6.4. Proof of Theorem 4.5

Proof. Utilizing to the first and second equations of (45), we can deduce

$\begin{equation} \begin{split} \big((a_6a_4-a_3^2)u_{xxxx}, u_{xxxx}\big) = &\Big(-a_4u_{tt}-a_3v_{tt}+\omega^2\big(R(a_4u_x+a_3v_x)\big)_x, u_{xxxx}\Big)\\ &+\frac{1}{2}\Big(\big((u_x^2+v_x^2)(a_4u_x+a_3v_x)\big)_x, u_{xxxx}\Big)\\ &+\big((a_4p_1+a_3p_2), u_{xxxx}\big) .\end{split} \end{equation}$

(121)

Thanks to (50) and Hölder inequality, we see that

$\begin{eqnarray} ||u_{xxxx}||_{L^2(\Omega)} \leq C\Big( ||u_{tt}+v_{tt}+ u_t+v_t||_{L^2(\Omega)}+||u+v||_{H_0^2(\Omega)} \Big)\leq C. \end{eqnarray}$

(122)

Furthermore,

$\begin{eqnarray} ||u_{xxx}||_{L^2(\Omega)} \leq C||u_{xxxx}||^{\frac{4}{7}}_{L^2(\Omega)}||u||^{\frac{3}{7}}_{L^2(\Omega)} +C||u||_{L^2(\Omega)}\leq C, \end{eqnarray}$

(123)

where Gagliardo-Nirenberg inequality for bounded domains is used. Similarly, we deduce

$\begin{eqnarray} ||v_{xxx}||_{L^2(\Omega)} \leq C, ||v_{xxxx}||_{L^2(\Omega)} \leq C. \end{eqnarray}$

(124)

Combining with the conclusions in Theorem (4.2) and Theorem (4.3), we can find that

$u, v \in L^{\infty}(0, T; H^4(\Omega)\cup H_0^2(\Omega)).$

6.5. Proof of Theorem 4.6

Proof. Denote

$\{\tilde{u}, \tilde{v}\} = \pi(\{\tilde a_3, \tilde a_4, \tilde a_6, \tilde{\omega}, \tilde{\theta}, \tilde u_0, \tilde v_0, \tilde u_1, \tilde v_1 \}),$

$\eta = u-\tilde u, \ \ \zeta = v-\tilde v.$

Then $\eta, \zeta$ satisfy

$\begin{equation*} \begin{split} &\frac{1}{2}\frac{\mathrm d}{\mathrm dt}\Big(||\eta_t||_{L^2(\Omega)}^2+||\zeta_t||_{L^2(\Omega)}^2 +||\sqrt{a_6}\eta_{xx}||_{L^2(\Omega)}^2\\ &\ \ \ \ \ \ +||\sqrt{a_4}\zeta_{xx}||_{L^2(\Omega)}^2-2||a_3\eta_{xx}\zeta_{xx}||_{L(\Omega)}\Big)\\ = &\frac{1}{2}(((u_x^2 +v_x^2)u_x -(\tilde{u}_x^2 +\tilde{v}_x^2)\tilde{u}_x)_x, \eta_t) +\frac{1}{2}(((u_x^2 +v_x^2)v_x -(\tilde{u}_x^2 +\tilde{v}_x^2)\tilde{v}_x)_x, \zeta_t)\\ &\; \; +\omega ^2(((R(x)\eta_x)_x, \eta_t)+((R(x)\zeta_x)_x, \zeta_t)) + \omega ^2 (\eta, \eta_t) +(\overline{p}_1, \eta_t)+(\overline{p}_2, \zeta_t) \end{split} \end{equation*}$

$\begin{equation} \begin{split}&\; \; -((a_6-\tilde a_6)\tilde u_{xx}, \eta_{txx})-((a_4-\tilde a_4)\tilde v_{xx}, \zeta_{txx}) +((a_3-\tilde a_3)\tilde v_{xx}, \eta_{txx})\\ &\; \; +((a_3-\tilde a_3)\tilde u_{xx}, \zeta_{txx})-(\omega^2-\tilde{\omega}^2)(R \tilde u_x, \eta_{tx}) -(\omega^2-\tilde{\omega}^2)(R \tilde v_x, \zeta_{tx})\\ &\; \; +(\omega^2-\tilde{\omega}^2)(\tilde u, \eta_t)+(\omega_t-\tilde{\omega}_t)(R_0+x, \eta_t). \end{split} \end{equation}$

(125)

In equation (125), the nonlinear term satisfies

$\begin{equation} \begin{split} &((u_x^2 +v_x^2)u_x - (\tilde{u}_x^2 +\tilde{v}_x^2)\tilde{u}_x)_x\\ = &((u_x^2 +v_x^2)\eta_x)_x +((u_x^2 +v_x^2-\tilde{u}_x^2 -\tilde{v}_x^2)\tilde{u}_x)_x\\ = &((u_x^2 +v_x^2)\eta_x)_x +(((u_x+\tilde{u}_x)\eta_x+(v_x+\tilde{v}_x)\zeta_x)\tilde{u}_x)_x\\ = &2(u_xu_{xx} +v_xv_{xx})\eta_x +(u_x^2 +v_x^2)\eta_{xx} +((u_x+\tilde{u}_x)\eta_{xx}+(v_x+\tilde{v}_x)\zeta_{xx})\tilde{u}_x\\ &\; \; +((u_{xx}+\tilde{u}_{xx})\eta_{x} +(v_{xx}+\tilde{v}_{xx})\zeta_x)\tilde{u}_x +((u_x+\tilde{u}_x)\eta_x+(v_x+\tilde{v}_x)\zeta_x)\tilde{u}_{xx}. \end{split} \end{equation}$

(126)

By Hölder inequality and Sobolev inequality, we can obtain

$\begin{equation} \begin{split} &(((u_x^2 +v_x^2)u_x - (\tilde{u}_x^2 +\tilde{v}_x^2)\tilde{u}_x)_x, \eta_t)\\ \leq &C||u_{xx} +v_{xx}||_{L^{2}(\Omega)} ||\eta_x||_{L^{\infty}(\Omega)}||\eta_t||_{L^{2}(\Omega)} +C||\eta_{xx}||_{L^{2}(\Omega)}||\eta_t||_{L^{2}(\Omega)}+||\eta_{xx}\\ &\; \; +(||u_{xx}+\tilde{u}_{xx}||_{L^{2}(\Omega)}||\eta_{x}||_{L^{\infty}(\Omega)} +||v_{xx}+\tilde{v}_{xx}||_{L^{2}(\Omega)}||\zeta_x||_{L^{\infty}(\Omega)})|| \eta_t||_{L^{2}(\Omega)}\\ &\; \; +(|\eta_x||_{L^{\infty}(\Omega)}+||\zeta_x||_{L^{\infty}(\Omega)})||\tilde{u}_{xx}||_{L^{2}(\Omega)} ||\eta_t||_{L^{2}(\Omega)}+\zeta_{xx}||_{L^{2}(\Omega)}||\eta_t||_{L^{2}(\Omega)}\\ \leq &C||\eta||_{H_0^2(\Omega)}||\eta_t||_{L^{2}(\Omega)}+C||\eta||_{H_0^2(\Omega)} ||\eta_t||_{L^{2}(\Omega)}\\ &\; \; +C||\eta||_{H_0^2(\Omega)}||\eta_t||_{L^{2}(\Omega)}+C||\zeta||_{H_0^2(\Omega)}||\eta_t||_{L^{2}(\Omega)}\\ \leq &C(||\eta_t||_{L^2(\Omega)}^2+||\eta||^2_{H_0^2(\Omega)}+||\zeta||^2_{H_0^2(\Omega)}). \end{split} \end{equation}$

(127)

In the same way, we have

$\begin{equation} \begin{split} (((u_x^2 +v_x^2)v_x - (\tilde{u}_x^2 +\tilde{v}_x^2)\tilde{v}_x)_x, \zeta_t)\leq C(||\zeta_t||_{L^2(\Omega)}^2+||\eta||^2_{H_0^2(\Omega)}+||\zeta||^2_{H_0^2(\Omega)}). \end{split} \end{equation}$

(128)

On the other hand, we easily have

$\begin{equation} \begin{split} &(\overline{p}_1, \eta_t)+(\overline{p}_2, \zeta_t)\\ \leq &C\big( ||\theta-\tilde\theta||_{L^1(\Omega)}+||\eta_t||_{L^2(\Omega)}^2+||\zeta_t||_{L^2(\Omega)}^2 +||\eta||^2_{H_0^1(\Omega)}+||\zeta||^2_{H_0^1(\Omega)}\big).\end{split} \end{equation}$

(129)

Substituting (127)-(129) into (125), we deduce

$\begin{equation} \begin{split} &||\eta_t||_{L^2(\Omega)}^2+||\zeta_t||_{L^2(\Omega)}^2+||\eta||^2_{H_0^2(\Omega)}+||\zeta||^2_{H_0^2(\Omega)}\\ \leq&C\int_0^t||\eta_t||_{L^2(\Omega)}^2+||\zeta_t||_{L^2(\Omega)}^2+||\eta||^2_{H_0^2(\Omega)}+||\zeta||^2_{H_0^2(\Omega)}\mathrm dt\\ &\; \; +||u_1-\tilde u_1||_{L^2(\Omega)}^2+||v_1-\tilde v_1||_{L^2(\Omega)}^2 +||u_0-\tilde u_0||^2_{H_0^2(\Omega)}+||v_0-\tilde v_0||^2_{H_0^2(\Omega)}\\ &\; \; +C\Big(\sum\limits_{i = 3}^{5}||a_i-\tilde a_i||_{L^{\infty}(\Omega)}+||\theta-\tilde\theta||_{L^1(\Omega)} +||\omega-\tilde{\omega}||_{W^{1, 1}(0, T)}\Big). \end{split} \end{equation}$

(130)

where we used the inequalities

$\begin{eqnarray*} ||\tilde u||_{L^{\infty}(Q)}\leq C, \; ||\tilde u_x||_{L^{\infty}(Q)}\leq C, ||\tilde v||_{L^{\infty}(Q)}\leq C, \; ||\tilde v_x||_{L^{\infty}(Q)}\leq C, \\ ||\tilde u_{xx}||_{L^{\infty}(0, T, L^2(\Omega))}\leq C, ||\tilde v_{xx}||_{L^{\infty}(0, T, L^2(\Omega))}\leq C, \\ ||\eta_{tx}||_{L^{\infty}(Q)}\leq C, ||\zeta_{tx}||_{L^{\infty}(Q)}\leq C, \end{eqnarray*}$

which are deduced from Theorem 4.2 and Theorem 4.3.

6.6. Proof of Theorem 5.2

Proof. Assume

$\begin{equation} \{\tilde\varphi_k\}_{k = 1}^{\infty} \text{is an orthogonal basis of} H_f^2(\Omega), \end{equation}$

(131)

and

$\begin{equation} \{\tilde\varphi_k\}_{k = 1}^{\infty} \text{is an orthonormal basis of} L^2(\Omega) . \end{equation}$

(132)

Fix a positive integer $m$ , and write

$\begin{eqnarray} u_m = \sum\limits_{k = 1}^{m}\tilde d_{0m}^{k}(t)\tilde\varphi_k, \ v_m = \sum\limits_{k = 1}^{m}\tilde d_{1m}^{k}(t)\tilde\varphi_k, \end{eqnarray}$

(133)

where we intend to select the coefficients $\tilde d_{0m}^k(t), \tilde d_{1m}^k(t)(0\leq t\leq T, k = 1, \cdots, m)$ to satisfy

$\begin{equation} \tilde d_{0m}^k(0) = (u_0, \tilde\varphi_k), \tilde d_{1m}^k(0) = (v_0, \tilde\varphi_k), k = 1, \cdots, m, \end{equation}$

(134)

$\begin{equation} \tilde d_{0m, t}^k(0) = (u_1, \tilde\varphi_k), \tilde d_{1m, t}^k(0) = (v_1, \tilde\varphi_k), k = 1, \cdots, m, \end{equation}$

(135)

and

$\begin{equation} \begin{split} &(u_{m, tt}, \tilde\varphi_k)+(a_6u_{m, xx}, \tilde\varphi_{k, xx})-(a_3v_{m, xx}, \tilde\varphi_{k, xx})+ \frac{1}{2}((u_{m, x}^2 +v_{m, x}^2)u_{m, x}, \tilde\varphi_{k, x})\\ &\; \; \; \; +\omega ^2(Ru_{m, x}, \tilde\varphi_{k, x})-(p_{1m}, \tilde\varphi_k)-\big(b_1u_m(l)+e_1v_m(l)\big)\tilde\varphi(l) = 0, \end{split} \end{equation}$

(136)

$\begin{equation} \begin{split} &(v_{m, tt}, \tilde\varphi_k)-(a_3 u_{m, xx}, \tilde\varphi_{k, xx})+(a_4 v_{m, xx}, \tilde\varphi_{k, xx} ) + \frac{1}{2}((u_{m, x}^2 +v_{m, x}^2)v_{m, x}, \tilde\varphi_{k, x})\\ &\; \; \; \; +\omega^2(R v_{m, x}, \tilde\varphi_{k, x} )-(p_{2m}, \tilde\varphi_k)-\big(b_2u_m(l)+e_2v_m(l)\big)\tilde\varphi(l) = 0. \end{split} \end{equation}$

(137)

As in earlier treatments of C-C boundary condition, we can conclude the following two conclude without difficulty.

(ⅰ) For each integer $m = 1, 2, \cdots,$ there exists a unique Galerkin approximations function $u_m, v_m$ of the from (133) satisfying (134)-(137) for $0\leq t \leq T$ .

(ⅱ) $u(x, 0) = u_0(x), v(x, 0) = v_0(x), u_t(x, 0) = u_1(x), v_t(x, 0) = v_1(x).$

Then, we proof the following estimate

$\begin{equation} \begin{array}{c} {\left\|u_{m, t}\right\|_{L^{2}(\Omega)}^{2}+\left\|v_{m, t}\right\|_{L^{2}(\Omega)}^{2}+\left\|u_{m}\right\|_{H_{f}^{2}(\Omega)}^{2} +\left\|v_{m}\right\|_{H_{f}^{2}(\Omega)}^{2}} \\ {\leq C\left(\left\|u_{1}\right\|_{L^{2}(\Omega)}^{2}+\left\|v_{1 }\right\|_{L^{2}(\Omega)}^{2}+\left\|u_{0}\right\|_{H_{f}^{2}(\Omega)}^{2}+\left\|v_{0 }\right\|_{H_{f}^{2}(\Omega)}^{2}\right)+C}\end{array} \end{equation}$

(138)

Similarly from (75), we can deduce

$\begin{equation} \begin{split} \frac{1}{2}\frac{\mathrm d}{\mathrm dt}\{&|| u_{m, t} ||_{L^2(\Omega)}^2 +|| v_{m, t} ||_{L^2(\Omega)}^2 +|| \sqrt{a_6}u_{m, xx} ||_{L^2(\Omega)}^2 +|| \sqrt{a_4}v_{m, xx} ||_{L^2(\Omega)}^2 \\ &+\frac{\omega ^2}{2}(||\sqrt{R}u_{m, x}||_{L^2(\Omega)}^2 +||\sqrt{R}v_{m, x}||_{L^2(\Omega)}^2 )\}-\frac{\mathrm d}{\mathrm dt}|| a_3 u_{m, xx} v_{m, xx} ||_{L^1(\Omega)} \\ &+\frac{1}{8}\frac{\mathrm d}{\mathrm dt}||u_{m, x}^2 +v_{m, x}^2||_{L^2(\Omega)}^2-\frac{1}{2}\frac{\mathrm d}{\mathrm dt}\{b_1u_m^2(l)+e_2v_m^2(l)+2e_1u_m(l)v_m(l)\}\\ = \omega\omega_t(&||\sqrt{R}u_{m, x}||_{L^2(\Omega)}^2 +||\sqrt{R}v_{m, x}||_{L^2(\Omega)}^2 )+(p_{1m}, u_{m, t})+(p_{2m}, v_{m, t}), \end{split} \end{equation}$

(139)

where $b_2 = e_1$ is used.

Then we integrate (139) with respect to $t$ , to discover

$\begin{eqnarray} \begin{aligned} &|| u_{m, t} ||_{L^2(\Omega)}^2 +|| v_{m, t} ||_{L^2(\Omega)}^2 +|| u_{m} ||^2_{H_f^2(\Omega)}+|| v_{m} ||^2_{H_f^2(\Omega)}\\ &-b_1u_m^2(l)-e_2v_m^2(l)-2e_1u_m(l)v_m(l)\\ \leq &C(|| u_{m, t}(0) ||_{L^2(\Omega)}^2 +|| v_{m, t}(0) ||_{L^2(\Omega)}^2 +||u_{m}(t) ||^2_{H_f^2(\Omega)}+|| v_{m}(t) ||^2_{H_f^2(\Omega)})\\ &-b_1u_{0m}^2(l)-e_2v_{0m}^2(l)-2e_1u_{0m}(l)v_{0m}(l)\\ &+C\int_0^t ||u_{m, t} ||_{L^2(\Omega)}^2 +||v_{m, t} ||_{L^2(\Omega)}^2 \mathrm dt+||u_m ||^2_{H_f^2(\Omega)}+||v_m ||^2_{H_f^2(\Omega)}+C. \end{aligned} \end{eqnarray}$

(140)

By simple calculation, we deduce

$\begin{eqnarray} -b_1u_m^2(l)-e_2v_m^2(l)-2e_1u_m(l)v_m(l)\geq 0. \end{eqnarray}$

(141)

where $b_1e_2\geq e_1^2$ is used.

On the other hand, according to $u_{0m}(x), v_{0m}(x) \in H_f^2(\Omega)$ , by Sobolev inequality, we obtain

$\begin{eqnarray*} u_{0m}(x), v_{0m}(x) \in C(\overline\Omega). \end{eqnarray*}$

Thus,

$\begin{eqnarray} u_{0m}(l), v_{0m}(l) \leq C. \end{eqnarray}$

(142)

Substituting (141) and (142) into (140), applying Gronwall inequality, we can deduce (138).

Now we pass to limits in our Galerkin approximations, applying estimate (138), we can discover (92), (93), (95). In order to complete the proof of the theorem, we just have to proof

$\begin{eqnarray} u_{\mu}(l)\rightarrow u(l), v_{\mu}(l)\rightarrow v(l), \; \text{ strongly in }\; L^{\infty}(0, T). \end{eqnarray}$

(143)

where $u_{\mu}, v_{\mu}$ are the convergent subsequence of $u_m, v_m$ , respectively.

To verify this, recalling (138), we observe that

$\begin{eqnarray} u_{\mu} \rightarrow u, v_{\mu} \rightarrow v \ \ \text{strongly in}\ \ C(0, T;H_f^1(\Omega)) \end{eqnarray}$

(144)

where the Corollary 4 of Chapter 8 in [30] is used.

Furthermore, thanks to the conditions $u_m(0) = u(0) = 0$ , we obtain

$\begin{eqnarray} \begin{aligned} &||u_{\mu}(l)-u(l)||_{L^{\infty}(0, T)} = ||(u_{\mu}(l)-u(l))-(u_{\mu}(0)-u(0))||_{L^{\infty}(0, T)}\\ = &\big|\big|\int_0^l \big(u_{\mu}(x)-u(x)\big)_x dx \big|\big|_{L^{\infty}(0, T)} \leq \sqrt l||\big(u_{\mu}(x)-u(x)\big)_x||_{L^{\infty}(0, T;L^2(\Omega))}\\ \leq& \sqrt l||u_{\mu}(x)-u(x)\big||_{L^{\infty}(0, T;H_f^1(\Omega))} \end{aligned} \end{eqnarray}$

(145)

Thanks to (144), we can deduce

$u_{\mu}(l)\rightarrow u(l) \; \text{ strongly in }\; L^{\infty}(0, T),$

Similarly, we have

$v_{\mu}(l)\rightarrow v(l) \; \text{ strongly in }\; L^{\infty}(0, T).$

6.7. Proof of Theorem 5.3

Proof. Similarly as (105), we can get

$\begin{equation} \begin{split} \frac{1}{2}\frac{\mathrm d}{\mathrm dt}E+\frac{1}{2}P_1+P_2 = P_3. \end{split} \end{equation}$

(146)

where

$\begin{eqnarray*} \begin{aligned} E = &||u_{tt}||_{L^2(\Omega)}^2+||v_{tt}||_{L^2(\Omega)}^2 +||\sqrt{a_6}u_{txx}||_{L^2(\Omega)}^2+||\sqrt{a_4}v_{txx}||_{L^2(\Omega)}^2\\ &-2||a_3u_{txx}v_{txx}||_{L(\Omega)}+\omega^2||\sqrt{R(x)} u_{tx}||_{L^2(\Omega)}^2 +\omega^2||\sqrt{R(x)} v_{tx}||_{L^2(\Omega)}^2 \end{aligned} \end{eqnarray*}$

$\begin{eqnarray*} \begin{aligned}P_1 = &\Big(\big((u^2_{x} + v^2_{x}) u_x\big)_t, u_{ttx}\Big)+\Big(\big((u^2_{x} + v^2_{x}) v_x\big)_t, v_{ttx}\Big)\\ P_2 = &-\big(b_1u(l)+e_1v(l)\big)_tu_{tt}(l)-\big(b_2u(l)+e_2v(l)\big)_tv_{tt}(l)\\ P_3 = &2\omega\omega_t\Big( \big((Ru_{x})_x, u_{tt}\big) + \big((Rv_{x})_x, v_{tt})\big)+||\sqrt{R} u_{tx}||_{L^2(\Omega)}^2+||\sqrt{R} v_{tx}||_{L^2(\Omega)}^2 \Big)\\ &+ (p_{1, t}, u_{tt}) +(p_{2, t}, v_{tt}) \end{aligned} \end{eqnarray*}$

By calculation, we can deduce

$\begin{eqnarray} &&P_1 = \frac{1}{2}\frac{\mathrm d}{\mathrm dt}\Big\{ ||\sqrt{u_x^2+v_x^2}u_{tx}||^2_{L^2(\Omega)}+||\sqrt{u_x^2+v_x^2}v_{tx}||^2_{L^2(\Omega)} +2||u_x u_{tx}||^2_{L^2(\Omega)} \\ &&\ \ \ \ \ \ \ \ \ \ \ \ \ \ +2||v_x v_{tx}||^2_{L^2(\Omega)}+4||u_x v_x u_{tx}v_{tx}||_{L^1(\Omega)}\Big\} \\ &&\ \ \ \ \ \ \ \ \ \ \ \ \ \ -3\int_0^l u_x u^3_{tx}+v_x v^3_{tx}+u_x u_{tx}v^2_{tx}+v_x v_{tx}u^2_{tx} \mathrm dx \end{eqnarray}$

(147)

$\begin{eqnarray} &&P_2 = -\frac{1}{2}\frac{\mathrm d}{\mathrm dt}\big\{b_1u_t^2(l)+e_2v_t^2(l)+2e_1u_t(l)v_t(l)\big\} \end{eqnarray}$

(148)

$\begin{eqnarray} &&P_3\leq ||u_{tt}||^2_{L^2(\Omega)}+||v_{tt}||^2_{L^2(\Omega)}+||u||^2_{H_f^2(\Omega)}+||v||^2_{H_f^2(\Omega)} \\ &&\ \ \ \ \ \ \ +||u_{tx}||^2_{L^2(\Omega)}+||v_{tx}||^2_{L^2(\Omega)} \end{eqnarray}$

(149)

Substituting (147), (148) and (149) into (146), we get

$\begin{equation} \begin{split} &\frac{1}{2}\frac{\mathrm d}{\mathrm dt}\Big\{E+ ||\sqrt{u_x^2+v_x^2}u_{tx}||^2_{L^2(\Omega)}+||\sqrt{u_x^2+v_x^2}v_{tx}||^2_{L^2(\Omega)} +2||u_x u_{tx}||^2_{L^2(\Omega)}\\ &\ \ \ \ \ \ \ \ +2||v_x v_{tx}||^2_{L^2(\Omega)}+4||u_x v_x u_{tx}v_{tx}||_{L^1(\Omega)}\\ &\ \ \ \ \ \ \ \ -\big(b_1u_t^2(l)+e_2v_t^2(l)+2e_1u_t(l)v_t(l)\big)\Big\}\\ \leq&||u_{tt}||^2_{L^2(\Omega)}+||v_{tt}||^2_{L^2(\Omega)}+||u||^2_{H_f^2(\Omega)}+||v||^2_{H_f^2(\Omega)} +||u_{tx}||^2_{L^2(\Omega)}+||v_{tx}||^2_{L^2(\Omega)}\\ &\ \ \ \ \ \ \ \ +3\int_0^l u_x u^3_{tx}+v_x v^3_{tx}+u_x u_{tx}v^2_{tx}+v_x v_{tx}u^2_{tx} \mathrm dx \end{split} \end{equation}$

(150)

Then integrate (150) with respect to $t$ , to discover

$\begin{eqnarray*} \begin{aligned} E\leq&\int_0^t||u_{tt}||^2_{L^2(\Omega)}+||v_{tt}||^2_{L^2(\Omega)}+||u||^2_{H_f^2(\Omega)}+||v||^2_{H_f^2(\Omega)} \\ &\ \ \ \ +||u_{tx}||^2_{L^2(\Omega)}+||v_{tx}||^2_{L^2(\Omega)}\mathrm dt\\ &\ \ \ \ +3\int_0^t\int_0^l u_x u^3_{tx}+v_x v^3_{tx}+u_x u_{tx}v^2_{tx}+v_x v_{tx}u^2_{tx} \mathrm dx\mathrm dt\\ &\ \ \ \ +||u_{tt}(x, 0)||_{L^2(\Omega)}^2 +||u_1||_{H_f^2(\Omega)}^2 +||v_{tt}(x, 0)||_{L^2(\Omega)}^2 +||v_1||_{H_f^2(\Omega)}^2. \end{aligned} \end{eqnarray*}$

where

$-b_1u_t^2(l)-e_2v_t^2(l)-2e_1u_t(l)v_t(l)\geq 0$

and

$||u_x u_{tx}||^2_{L^2(\Omega)}+||v_x v_{tx}||^2_{L^2(\Omega)} +2||u_x v_x u_{tx}v_{tx}||_{L^1(\Omega)}\geq 0.$

are used.

Similarly as (108), we discover

$\begin{eqnarray} ||u_{tt}(x, 0)||_{L^2(\Omega)}, ||v_{tt}(x, 0)||_{L^2(\Omega)}\leq c. \end{eqnarray}$

(151)

According to (151), we get

$\begin{eqnarray*} \begin{aligned} E\leq&c\int_0^t||u_{tt}||^2_{L^2(\Omega)}+||v_{tt}||^2_{L^2(\Omega)} +||u_{txx}||^2_{L^2(\Omega)}+||v_{txx}||^2_{L^2(\Omega)}\mathrm dt \\ &\; \; \; \; +3\int_0^t\int_0^l u_x u^3_{tx}+v_x v^3_{tx}+u_x u_{tx}v^2_{tx}+v_x v_{tx}u^2_{tx} \mathrm dx\mathrm dt+c. \end{aligned} \end{eqnarray*}$

By Hölder inequality and Sobolev inequality, we have

$\begin{eqnarray} \begin{aligned} E\leq& c\int_0^t\Big\{||u_{tt}||^2_{L^2(\Omega)} +||u_{txx}||^2_{L^2(\Omega)}+||u_{txx}||^3_{L^2(\Omega)}+||u_{txx}||^4_{L^2(\Omega)}\\ &\; \; \; \; \; \; \; \; ||v_{tt}||^2_{L^2(\Omega)} +||v_{txx}||^2_{L^2(\Omega)}+||v_{txx}||^3_{L^2(\Omega)}+||v_{txx}||^4_{L^2(\Omega)}\Big\}\mathrm dt+c. \end{aligned} \end{eqnarray}$

(152)

Then, we obtain Theorem 5.3 by Gronwall inequality.

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Quantitative Finance and Economics

How does soft information about small business lending affect bank efficiency under capital regulation?

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Abstract

1. Introduction

2. Model building

2.1. Assumptions

2.2. Four coordinate systems are defined to describe the motion of the blades

2.3. Displacement field

2.4. Strains and stress

2.5. The centrifugal force and the perturbed gas pressure

2.6. The velocity vector and the acceleration

2.7. Rotating blades model

3. Preliminary

4. Mathematical results of C-C blades

5. Mathematical results of C-F blades

6. Proof of mathematical results

6.1. Proof of Theorem 4.2

6.2. Proof of Theorem 4.3

6.3. Proof of Theorem 4.4

6.4. Proof of Theorem 4.5

6.5. Proof of Theorem 4.6

6.6. Proof of Theorem 5.2

6.7. Proof of Theorem 5.3

References

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Abstract

1. Introduction

2. Model building

2.1. Assumptions

2.2. Four coordinate systems are defined to describe the motion of the blades

2.3. Displacement field

2.4. Strains and stress

2.5. The centrifugal force and the perturbed gas pressure

2.6. The velocity vector and the acceleration

2.7. Rotating blades model

3. Preliminary

4. Mathematical results of C-C blades

5. Mathematical results of C-F blades

6. Proof of mathematical results

6.1. Proof of Theorem 4.2

6.2. Proof of Theorem 4.3

6.3. Proof of Theorem 4.4

6.4. Proof of Theorem 4.5

6.5. Proof of Theorem 4.6

6.6. Proof of Theorem 5.2

6.7. Proof of Theorem 5.3

References