In the classic elasticity theory, a solid body is considered as a three-dimensional Euclidean space, points of which coincide with body particles [1] [2] . Interaction between parts of this solid body model occurs through central forces only; in particular, a surface element $\text{d}A$ load is transmitted by the main $P\text{d}A$ vector only. This assumption leads us to a symmetric, stressed, and deformed condition. Classic elasticity theory very well explains events taking place in bodies under pressure in all cases, when, let’s say, a “granular” structure of a real body under consideration doesn’t count as an essential characteristic of this event. Though the classic elasticity theory doesn’t satisfactorily explain some events in those cases, the material microstructure plays an essential role.

Even Voigt W. has attempted to correct this deficiency of the classic elasticity theory using the following additional assumption that a surface element load is transmitted not only by $P\text{d}A$ vector, but by principal $\text{d}A$ moment—this assumption leads us to the fact that not only ${\sigma }_{ij}$ stress, but couple-stress ${\mu }_{ij}$ acts on a volume element. In general, this new theory was developed by the Cosserat brothers (Cosserat E. and Cosserat F.). While classic elasticity theory considers a material particle as a point and describes the deformed state by means of point positions (coordinates), in contradistinction, the Cosserat brothers consider each body particle as a small solid body. Deformation of such a body is described not only by the displacement vector, but also by the rotation vector. The mentioned theory, which considers a new model of an elastic body, is called the moment theory of elasticity. There is no general consensus in the moment theory of elasticity when describing the displacement of a solid body particle. The concept of classic elasticity theory is kept unchanged in the works of [3] Mindlin R., Kuvshinsky Е., and Aero E., who assume that displacement and small rigid rotation are completely described, if we take that $\Omega =\frac{1}{2}rotU$. This version of the moment theory of elasticity is called a first version (variant). [1] Nowacki W., Palmov N. et al., alongside a similar displacement field of particles of the body under consideration, introduce a vector field, kinematically dependent on it, which characterises small rotations of body particles—this version is called the second version (variant) in the moment theory of elasticity.

We consider the first version of the moment theory of elasticity. The basic equations of equilibrium in the moment theory of elasticity in terms of displacement components are as follows:

$\Delta U+\frac{1}{1-2\nu }graddivU+l^2\Delta rot\Omega =0$

In case of stationary oscillation, we will have

$\Delta U+\frac{1}{1-2\nu }graddivU+l^2\Delta rot\Omega +k^{2}U=0,$ (1)

where $\nu$ is Poisson ratio, $l$ is new elasticity constant (when $l=0$, we obtain equations of the classic elasticity theory), $U$ is a displacement vector, and $\Omega =\frac{1}{2}rotU$ is rotation vector. Let us write the vector equality (1) in scalar form:

$\Delta u+\frac{1}{1-2\nu }\frac{\partial \theta }{\partial x}+l^2\frac{\partial \Delta \omega }{\partial y}+k^{2}u=0$

$\Delta v+\frac{1}{1-2\nu }\frac{\partial \theta }{\partial y}-l^2\frac{\partial \Delta \omega }{\partial x}+k^{2}v=0$ (2)

where

$\theta =divU\left(u,v\right)=\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y},\omega =\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}$ (3)

Let’s consider the problem: assume that $y=0$ at the boundary of endless $y>0$ semiplane. There is given $u\left(x,0\right)=u_0$, $v\left(x,0\right)=v_0$, $\omega \left(x,0\right)={\omega }_0$, where $u_0,$ $v_0$, ${\omega }_0\in L\left(-\infty ;+\infty \right)$. Let us determine the displacement damping at infinity.

Let us note that $L\equiv L\left(R\right)$ is a designation of a class of function integrated over $R$ with a norm $\Vert f\Vert ={\displaystyle \int \left|f\left(x\right)\right|\text{d}x}$, $\hat{\phi }\left(t\right)=\frac{1}{\sqrt{2\pi }}{\displaystyle {\int }_{-\infty }^{+\infty }\phi \left(x\right){\text{e}}^{-itx}\text{d}x}$ $\left(-\infty <x<+\infty \right)$. $\phi \left(x\right)\in L$ is a Fourier transformation of a function and $t$ is a Fourier transformation parameter [4] .

By applying of $div$ and $rot$ operations on Equation (1), we obtain

$\Delta \theta +k^{2}h\theta =0,\Delta \left[\omega -k^2\omega \right]+k^2\omega =0$ (4)

where

$h=\frac{1-2\nu }{2\left(1-\nu \right)}$ (5)

Let’s rewrite (2) in the following form

$\Delta u+k^{2}u=-\frac{1}{1-2\nu }\frac{\partial \theta }{\partial x}-l^2\frac{\partial \Delta \omega }{\partial y}$

$\Delta v+k^{2}v=-\frac{1}{1-2\nu }\frac{\partial \theta }{\partial y}+l^2\frac{\partial \Delta \omega }{\partial x}$ (6)

From Equations (5) and (6), using Fourier transformation (in terms of $x$ variable), we obtain

$\frac{{\text{d}}^2\hat{\theta }}{\text{d}{y}^2}-\left(t^2-h{k}^2\right)\hat{\theta }=0$

$l^2\frac{{\text{d}}^4\hat{\omega }}{\text{d}{y}^4}-\left(1+2t^2{l}^2\right)\frac{{\text{d}}^2\hat{\omega }}{\text{d}{y}^2}+\left(t^2+t^4{l}^2-k^2\right)\hat{\omega }=0$ (7)

$\frac{{\text{d}}^2\hat{u}}{\text{d}{y}^2}-\left(t^2-k^2\right)\hat{u}=-\frac{it}{1-2\nu }\hat{\theta }-l^2\left[\frac{{\text{d}}^3\hat{\omega }}{\text{d}{y}^3}-t^2\frac{\text{d}\hat{\omega }}{\text{d}y}\right]$

$\frac{{\text{d}}^2\hat{v}}{\text{d}{y}^2}-\left(t^2-k^2\right)\hat{v}=-\frac{1}{1-2\nu }\frac{\text{d}\hat{\theta }}{\text{d}y}+l^2\left[it\frac{{\text{d}}^2\hat{\omega }}{\text{d}{y}^2}-i{t}^3\hat{\omega }\right]$ (8)

Taking into account that we seek the solution damped at infinity, the general solution of Equation (7) will be represented as follows:

$\hat{\theta }=A{\text{e}}^{{\lambda }_{1}y}$ and $\hat{\omega }=B{\text{e}}^{{\lambda }_{2}y}+C{\text{e}}^{{\lambda }_{3}y}$ (9)

where

${\lambda }_1=-\sqrt{t^2-h{k}^2}$

${\lambda }_2=-\sqrt{\frac{1+2l^2{t}^2+\sqrt{1+4l^2{k}^2}}{2l^2}}$

${\lambda }_3=-\sqrt{\frac{1+2l^2{t}^2-\sqrt{1+4l^2{k}^2}}{2l^2}}$

Taking Equation (9) into account in the right-hand side of Equation (8), we obtain for its general solution:

$\begin{array}{c}\hat{u}=-\frac{A_{i}t}{\left({\lambda }_1^2-t^2+k^2\right)\left(1-2\nu \right)}{\text{e}}^{{\lambda }_{1}y}-\frac{B{\lambda }_2{l}^2\left({\lambda }_2^2-t^2\right)}{{\lambda }_2^2-t^2+k^2}{\text{e}}^{{\lambda }_{2}y}\\ -\frac{C{\lambda }_3{l}^2\left({\lambda }_3^2-t^2\right)}{{\lambda }_3^2-t^2+k^2}{\text{e}}^{{\lambda }_{3}y}+D{\text{e}}^{-\sqrt{t^2-k^2}y}\end{array}$

$\begin{array}{c}\hat{v}=-\frac{A{\lambda }_1}{\left({\lambda }_1^2-t^2+k^2\right)\left(1-2\nu \right)}{\text{e}}^{{\lambda }_{1}y}-\frac{itB{l}^2\left({\lambda }_2^2-t^2\right)}{{\lambda }_2^2-t^2+k^2}{\text{e}}^{{\lambda }_{2}y}\\ +\frac{itC{l}^2\left({\lambda }_3^2-t^2\right)}{{\lambda }_3^2-t^2+k^2}{\text{e}}^{{\lambda }_{3}y}+E{\text{e}}^{-\sqrt{t^2-k^2}y}\end{array}$ (10)

A, B, C, D, E are constants in Equation (10), which have to be considered. As far as the dependence (3) remains valid, the mentioned constants are interdependent ones. Fourier transformation of equality (3) will give us:

$\hat{\theta }=it\hat{u}+\frac{\text{d}\hat{v}}{\text{d}y},\hat{\omega }=it\hat{v}-\frac{\text{d}\hat{u}}{\text{d}y}$ (11)

Using Equations (9)-(11), we obtain

$itD-E\sqrt{t^2-k^2}=0,itE+D\sqrt{t^2-k^2}=0$ (12)

Since for homogeneous (12) systems $\Delta =\left|\begin{array}{cc}it& -\sqrt{t^2-k^2}\\ \sqrt{t^2-k^2}& it\end{array}\right|\ne 0$, it will have a trivial solution only, i.e. $D=E=0$. By Fourier transformation of boundary conditions, we obtain

$\begin{array}{c}{\hat{u}}_o=-\frac{it}{k^{2}h}A+\frac{l\left(1+\sqrt{1+4l^2{k}^2}\right)\left(\sqrt{1+2l^2{t}^2+\sqrt{1+4l^2{k}^2}}\right)}{\sqrt{2}\left(1+2l^2{t}^2+\sqrt{1+4l^2{k}^2}\right)}B\\ +\frac{l\left(1-\sqrt{1+4l^2{k}^2}\right)\left(\sqrt{1+2l^2{t}^2-\sqrt{1+4l^2{k}^2}}\right)}{\sqrt{2}\left(1+2l^2{t}^2-\sqrt{1+4l^2{k}^2}\right)}C\end{array}$

${\hat{v}}_o=\frac{\sqrt{t^2-h{k}^2}}{k^{2}h}A+\frac{it{l}^2\left(1+\sqrt{1+4l^2{k}^2}\right)}{1+2l^2{k}^2+\sqrt{1+4l^2{k}^2}}B+\frac{it{l}^2(1+\sqrt{1-\sqrt{1+4l^2{k}^2}}}{1+2l^2{k}^2-\sqrt{1+4l^2{k}^2}}C$

${\hat{\omega }}_o=B+C$ or else, there are three conditions for three unknowns that make it possible to determine these unknowns. It should be noted as well that displacements may be expressed not only by the given function, but by their Fourier transformation, as well.

The case of infinite slab (layer) is also studied, in particular, when displacements and rotation are given ${u|}_{y=\pm h}=f_1^{\pm }$, ${v|}_{y=\pm h}=f_2^{\pm }$, ${\omega |}_{y=\pm h}=f_3^{\pm }$ on the boundary of infinite slab $\left|y\right|\le h$, $-\infty <x<+\infty$ in this case we have not a condition of damping at infinity, but boundary conditions, and taking equalities (9)-(11) into consideration, six equations for six unknowns are obtained.

In conclusion, it should be noted that the method based on Fourier integral transformation is considered effective when solving a wide range of problems of classic elasticity theory and the moment theory of elasticity. The numerical visualizations presented demonstrate the physical significance of the couple stress parameter and validate the mathematical approach. The study discussed in the work confirms this suggestion.