On the diffraction of nonstationary SH-wave on semi-infinite crack in porous elastic medium (Q2729322)

The authors study the diffraction problem for a plane SH shock wave on a semi-infinite crack in a porous medium for the case of energy loss due to intercomponent friction.NEWLINENEWLINE An unbounded porous isotropic medium in the space \(\mathbb R^3\) is considered which is characterized by the elastic constants \(\lambda\), \(\mu\), \(\alpha\), friction coefficient \(\kappa\), partial density \(\rho_s\) of the elastic body, and partial density \(\rho_l\) of the liquid. It is assumed that the medium under consideration has a crack (defect) of the form NEWLINE\[NEWLINE z = 0,\qquad 0\leq x <\infty, \qquad -\infty < y < \infty. NEWLINE\]NEWLINE An SH-wave is incident on this crack where the wave is defined by the formula NEWLINE\[NEWLINE u^0(t,x,z) = \tilde{\varphi}\left(t - \frac{x\sin\alpha - z\cos\alpha}{c_t} \right),\qquad c_t^2 = \mu\rho_s^{-1}. NEWLINE\]NEWLINE Here \(\tilde{\varphi}(\xi)_{\xi < 0} \equiv 0\) (the incident wave \(\tilde{\varphi}\) has a leading edge), \(\alpha\) is the angle of incidence of the wave. In the absence of a crack, this wave causes the following tangential stresses in the porous isotropic medium: NEWLINE\[NEWLINE \tau^{0}_{yz}(t,x,z) = \mu\frac{\partial u^0}{\partial z} = \mu\cos\alpha \left|\frac{d\tilde{\varphi}(\xi)}{d\xi}\right|_{\xi = t - (x\sin\alpha -z\cos \alpha)/c_t}. NEWLINE\]NEWLINE The following problem can be posed: Find the distributions of stresses and displacements of the elastic body under consideration, and the quantities \(\tau_{yz}(t,x,z)\), \(u(t,x,z)\), \(v(t,x,z)\) after the incidence of the SH-wave on the crack where the functions \(u\), \(v\) satisfy the system NEWLINE\[NEWLINE \begin{alignedat}{2} \frac{1}{c_t^2}\frac{\partial^2u}{\partial t^2} - \frac{\partial^2u}{\partial x^2} - \frac{\partial^2u}{\partial z^2} + \frac{\chi}{c_t^2}\frac{\rho_l^2}{\rho_s}\left (\frac{\partial u}{\partial t} - \frac{\partial v }{\partial t}\right) &= 0, &&\qquad | x| < \infty,\;z\neq 0, \;t > 0, \\ \frac{\partial^2v}{\partial t^2} - \chi\rho_l\left (\frac{\partial u}{\partial t} - \frac{\partial v }{\partial t}\right) &= 0, &&\qquad | x| < \infty,\;z\neq 0, \;t > 0 \end{alignedat} NEWLINE\]NEWLINE under zero initial conditions and with the following conditions on the crack: NEWLINE\[NEWLINE \begin{aligned} & \langle u(t,x,0)\rangle = u(t,x,-0) - u(t,x,+0) = \varphi(t,x) \neq 0,\qquad x > 0, \\ & \langle v(t,x,0)\rangle = v(t,x,-0) - v(t,x,+0) = \psi(t,x) \neq 0,\qquad x > 0,\\ & \tau_{yz}(t,x,\pm 0) = 0,\qquad x \geq 0. \end{aligned} NEWLINE\]NEWLINE To solve this problem, the authors use the integral Laplace transform with respect to the time variable. Then, a discontinuous solution is constructed to the equations obtained for the crack. As a result, the authors expose formulas for the wave scattering in terms of the Laplace transform.