## 3 Biphasic and Linear Fractional Poroelastic Models
### Biphasic Model--Consolidation Problem  
For the purposes of this study, we restrict our formulation to small strain theory and consider the linear biphasic model from [25]. We assume the solid matrix is incompressible, linear elastic, isotropic, homogeneous and non dissipative, whereas the interstitial fluid is incompressible and non dissipative. The only dissipation comes from the frictional drag due to the relative velocities of the two phases. We denote by \(\lambda_{0}=k/\mu\), the averaged axial permeability of the sample and \(H_{A}\) as the aggregate modulus.  
Under these assumptions and the additional assumptions coming from the confined compression tests setting, the biphasic theory leads to the following unidimensional boundary value problem for the vertical displacement \(u_{z}^{s}\) of the solid phase  
\[\frac{\partial^{2}u_{z}^{s}}{\partial z^{2}}=\frac{1}{H_{A}\lambda_{0}}\frac {\partial u_{z}^{s}}{\partial t}\quad\text{in }(0,h)\times(0,T), \tag{3}\]with the boundary condition at \(z=0\) where the solid skeleton is fixed  
\[u_{z}^{s}(z=0,t)=0\quad\text{on }(0,T), \tag{4}\]  
the initial condition  
\[u_{z}^{s}(z,t=0)=0\quad\text{on }(0,h), \tag{5}\]  
and finally the boundary condition at \(z=h\), which will be different for the creep test and for the stress relaxation test [26]. For the creep test, the condition is  
\[\frac{\partial u_{z}^{s}}{\partial z}_{|z=h}=-\frac{P_{A}}{H_{A}}\quad\text{on }(0,T), \tag{6}\]  
where \(P_{A}\) is the applied compressive stress, and, for the stress relaxation test, the condition is  
\[u(z=h,t)=\left\{\begin{array}{ll}-V_{0}t&\quad\text{if }\;0\leqslant t<t_{0}, \\ -V_{0}t_{0}&\quad\text{if }\;t_{0}\leqslant t\leqslant T,\end{array}\right. \tag{7}\]  
where \(V_{0}\) and \(t_{0}\) are the input data from the stress relaxation test.  
If we denote \(c_{n}=(-1)^{\frac{n-1}{2}}\left(\frac{2h}{n\pi}\right)^{2}\), the solution to Equations (3)-(6), in the case of the creep test, reads  
\[u_{z}(z,t)=\frac{P_{A}}{H_{A}}\left[-z+\frac{2}{h}\sum_{n=1,3}^{\infty}c_{n} \exp\left(-\frac{H_{A}\lambda_{0}n^{2}\pi^{2}t}{4h^{2}}\right)\sin\left(\frac {n\pi z}{2h}\right)\right], \tag{8}\]  
If we denote \(\rho_{n}=\left(\frac{n\pi}{h}\right)^{2}\), the solution to Equations (3)-(5) and (7) [26] is given by  
\[u_{z}^{s}(z,t)=\left\{\begin{array}{ll}-\frac{V_{0}tz}{h}-\frac{2V_{0}}{H_{A }\lambda_{0}h}\sum_{n=1}^{\infty}\frac{(-1)^{n}}{\rho_{n}^{3/2}}\left(1-\exp(- H_{A}\rho_{n}\lambda_{0}t)\right)\sin(\sqrt{\rho_{n}}z)&\quad\text{if }0\leqslant t<t_{0},\\ -\frac{V_{0}t_{0}}{h}-\frac{2V_{0}}{H_{A}\lambda_{0}h}\sum_{n=1}^{\infty} \frac{(-1)^{n}}{\rho_{n}^{3/2}}\exp(-H_{A}\rho_{n}\lambda_{0}t)\big{(}\exp(- H_{A}\rho_{n}\lambda_{0}t)-1\big{)}\sin(\sqrt{\rho_{n}}z)&\quad\text{if }t_{0}\leqslant t \leqslant T.\end{array}\right. \tag{9}\]  
The stress relaxation response \(\sigma_{t}\) is given by \(H_{A}\frac{\partial u}{\partial z}_{|z=h}\), and thus  
\[\sigma_{t}=\left\{\begin{array}{ll}-\frac{V_{0}H_{A}t}{h}+\frac{2V_{0}}{ \lambda_{0}h}\sum_{n=1}^{\infty}\rho_{n}^{-1}\big{(}1-\exp(-H_{A}\rho_{n} \lambda_{0}t)\big{)}&\quad\text{if }0\leqslant t<t_{0},\\ \frac{2V_{0}}{\lambda_{0}h}\sum_{n=1}^{\infty}\rho_{n}^{-1}\exp(-H_{A}\rho_{ n}\lambda_{0}t)\big{(}\exp(-H_{A}\rho_{n}\lambda_{0}t)-1\big{)}&\quad\text{if }t_{0}\leqslant t \leqslant T.\end{array}\right. \tag{10}\]