Modeling the mechanics of fibrous-porous scaffolds for skeletal muscle regeneration

Abstract

The scaffolds for skeletal muscle regeneration are designed to mimic the structure, stiffness, and strains applied to the muscle under physiologic conditions. The external strains are also used to stimulate myogenesis of the (stem) cells seeded on the scaffold. The time- and location-dependent mechanics inside the scaffold determine the microenvironment for the seeded cells. Here, fibrous-porous cylindrical scaffolds under the action of external cyclic strains are considered. The scaffold mechanics are described as two-phase (poroelastic) where the solid phase is associated with the fibers and the fluid phase is associated with the liquid-containing pores. In response to an applied cyclic strain, pressure oscillates and fluid moves radially toward and away from the axis of the scaffold. We compute the directions and magnitudes of the radial gradients of the poroelastic characteristics (solid-phase displacement, strain, and velocity; fluid-phase pressure and velocity; relative fluid-solid-phase velocity) determined by the boundary conditions and geometry of the scaffold. Several kinds of the external cyclic strain are analyzed and the resulting poroelastic functions are found. The poroelastic characteristics are obtained in closed form which is convenient for further consideration of myogenesis of the seeded cells and ultimately for the design of the scaffolds for skeletal muscle regeneration.

Appendix

Here, we obtain the Laplace transforms of the major poroelastic functions. Rewriting the continuity equation in the coordinate form, we have

$$ \left(\frac{\partial {v}_r^f}{\partial r}+\frac{v_r^f}{r}+\frac{\partial {v}_z^f}{\partial z}\right)+\left({\varnothing}^{-1}-1\right)\frac{\partial }{\partial t}\left(\frac{{\partial u}_r}{\partial r}+\frac{u_r}{r}+\varepsilon \right)=0 $$

(41)

We consider Eq. (41) as an ODE in terms of the radial component of the fluid-phase velocity, ${v}_r^f(r)$, and integrate it assuming velocity is finite at r = 0. As a result, the fluid-phase velocity can be expressed in terms of the solid-phase displacement as

$$ {v}_r^f=-\left({\varnothing}^{-1}-1\right)\frac{\partial {u}_r}{\partial t}-{\varnothing}^{-1}\frac{r}{2}\frac{\partial \varepsilon }{\partial t} $$

(42)

We can also express the fluid-phase pressure via the solid-phase displacement. For that, we consider equilibrium Eq. (2) and substitute the fluid-phase stress and fluid-phase velocity in terms of the fluid-phase pressure (2) and solid-phase displacement (42), respectively. Finally, we reduce the problem to the following differential equation in terms of the solid-phase displacement.

We can also express fluid-phase pressure via the solid-phase displacement. For that, we consider the fluid-phase equilibrium Eq. (2) and substitute the fluid-phase stress in terms of fluid-phase pressure (2) and the fluid-phase velocity in terms of the solid-phase displacement (42). Finally, we reduce the problem to a differential equation in terms of the solid-phase displacement. To derive this differential equation, we substitute the fluid-phase velocity and pressure into the solid-phase equilibrium Eq. (6), which results in the following equation:

$$ \frac{\partial^2{u}_r}{\partial {r}^2}+\frac{1}{r}\frac{\partial {u}_r}{\partial r}-\frac{u_r}{r^2}=\frac{1}{C_{11}k}\frac{\partial }{\mathrm{\partial t}}\left({u}_r+\frac{r}{2}\varepsilon \right) $$

(43)

Then, we apply the Laplace transform using dimensionless time t^′ = t/t_g to Eq. (43) and obtain

$$ \frac{\partial^2{\overline{{\mathrm{u}}^{\prime}}}_{\mathrm{r}}}{\partial {{\mathrm{r}}^{\prime}}^2}+\frac{1}{{\mathrm{r}}^{\prime }}\frac{\partial {\overline{{\mathrm{u}}^{\prime}}}_{\mathrm{r}}}{\partial {\mathrm{r}}^{\prime }}-\left(\frac{1}{{{\mathrm{r}}^{\prime}}^2}+\mathrm{s}\right){\overline{\mathrm{u}}}_{\mathrm{r}}=\mathrm{s}\frac{{\mathrm{r}}^{\prime }}{2}\overline{\upvarepsilon} $$

(44)

where $\overline{\mathrm{X}}$ indicates the Laplace transform of any function X(t) with the parameter s, and ′ indicates the division by the scaffold radius, a. On the external surface of the cylinder, the stresses satisfy the following conditions:

$$ {\sigma}_{rr}^t={\sigma}_{rr}^s+{\sigma}_{rr}^f=p=0 $$

(45)

which can be used as the following boundary (at r = a) condition for Eq. (44).

$$ {C}_{11}\frac{\partial {{\overline{u}}^{\prime}}_r}{\partial {r}^{\prime }}+{C}_{12}\frac{{{\overline{u}}^{\prime}}_r}{r^{\prime }}+{C}_{13}\overline{\varepsilon}=0 $$

(46)

The solution of differential Eq. (44) with boundary condition (45) takes the form

$$ {{\overline{u}}^{\prime}}_r=-\overline{\varepsilon}\frac{r^{\prime }}{2}\left[1-\frac{\left({C}_{11}+{C}_{12}-2{C}_{13}\right){I}_1\left(\sqrt{s}{r}^{\prime}\right)\cdotp \frac{1}{\sqrt{s}{r}^{\prime }}}{C_{11}{I}_0\left(\sqrt{s}\right)-\frac{\left({C}_{11}-{C}_{12}\right){I}_1\left(\sqrt{s}\right)}{\sqrt{s}}}\right] $$

(47)

where I₁ and I₀ are the modified Bessel functions of the first kind.

We now use Eq. (9) to obtain the expression for the Laplace transform of the fluid-phase velocity. From Eq. (9), we have

$$ {\overline{{\mathrm{v}}^{\prime}}}^{\mathrm{f}}=-\mathrm{s}\left[\left({\varnothing}^{-1}-1\right){\overline{{\mathrm{u}}^{\prime}}}_{\mathrm{r}}+0.5{\mathrm{r}}^{\prime }{\varnothing}^{-1}\overline{\upvarepsilon}\left(\mathrm{s}\right)\right] $$

(48)

which results in the following equation

$$ {{\overline{v}}^{\prime}}^f=-s\cdotp \overline{\varepsilon}(s)\cdotp \frac{r^{\prime }}{2}\left[1+\left({\varnothing}^{-1}-1\right)\cdotp \frac{\left({C}_{11}+{C}_{12}-2{C}_{13}\right){I}_1\left(\sqrt{s}{r}^{\prime}\right)\cdotp \frac{1}{\sqrt{s}{r}^{\prime }}}{C_{11}{I}_0\left(\sqrt{s\ }\right)-\frac{\left({C}_{11}-{C}_{12}\right){I}_1\left(\sqrt{s}\right)}{\sqrt{s}}}\right] $$

(49)

Here the fluid-phase velocity is made dimensionless by dividing by a/t_g. Finally, the Laplace transform of the fluid-phase pressure is given by the following equation:

$$ {\overline{p}}^{\prime }=\frac{C_{11}\left({C}_{11}+{C}_{12}-2{C}_{13}\right)}{\left({C}_{11}-{C}_{12}\right)}\cdotp \frac{I_0\left(\sqrt{s}{r}^{\prime}\right)-{I}_0\left(\sqrt{s}\right)}{C_{11}{I}_0\left(\sqrt{s}\right)-\frac{\left({C}_{11}-{C}_{12}\right){I}_1\left(\sqrt{s}\right)}{\sqrt{s}}}\overline{\varepsilon}(s) $$

(50)

where pressure is made dimensionless by dividing by 0.5 (C₁₁ − C₁₂).

Equations 47, 49, and 50 include function$\overline{\ \upvarepsilon}$, the Laplace transform of the externally applied strain, which for all of our cases of non-harmonic cyclic strain is given by the following sets of equations:

$$ \varepsilon (t)={\sum}_i{\varepsilon}_R\left(t-{\tau}_i^{+}\right)-\sum \limits_i{\varepsilon}_R\left(t-{\tau}_i^{-}\right) $$

(51)

$$ \overline{\varepsilon}(s)={\sum}_i\overline{\varepsilon_R}\left(s,{\tau}_i^{+}\right)-\sum \limits_i\overline{\varepsilon_R}\left(s,{\tau}_i^{-}\right) $$

(52)

where

$$ {\varepsilon}_R\left(t-{\tau}_i^{+}\right)=\left(t-{\tau}_i^{+}\right)H\left(t-{\tau}_i^{+}\right)-\left(t-{t}_0-{\tau}_i^{+}\right)H\left(t-{t}_0-{\tau}_i^{+}\right) $$

(53)

$$ {\varepsilon}_R\left(t-{\tau}_i^{-}\right)=\left(t-{\tau}_i^{-}\right)H\left(t-{\tau}_i^{-}\right)-\left(t-{t}_0-{\tau}_i^{-}\right)H\left(t-{t}_0-{\tau}_i^{-}\right) $$

(54)

$$ \overline{\varepsilon_R}\left(s,{\tau}_i^{+}\right)={e}^{-{\tau}_i^{+}s}\left(1-{e}^{-{t}_0s}\right)/{s}^2 $$

(55)

$$ \overline{\varepsilon_R}\left(s,{\tau}_i^{-}\right)={e}^{-{\tau}_i^{-}s}\left(1-{e}^{-{t}_0s}\right)/{s}^2 $$

(56)

Here, H(t) is the Heaviside step function. Thus, each ${\varepsilon}_R\left(t-{\tau}_i^{+}\right)$ is a function that equals 0 before time ${\tau}_i^{+}$, equals 1 after time ${\tau}_i^{+}+{t}_0$, and is a constantly increasing ramp in between those times, and likewise for ${\varepsilon}_R\left(t-{\tau}_i^{-}\right)$.