An Inverse Finite Element u/p-Formulation to Predict the Unloaded State of In Vivo Biological Soft Tissues

Abstract

Physically realistic patient-specific biomechanical modelling is of paramount importance for many medical applications, where the geometry of tissues or organs is usually constructed from in vivo images. However, it is common for such biological structures to correspond to a deformed state due to being under external loadings. This necessitates the determination of the stress distribution of the known deformed state through an inverse analysis approach. To achieve this, we propose here a generalised finite element displacement/pressure (u/p)-formulation for evaluating the unloaded configuration of in vivo biological soft tissues that exhibit quasi-incompressible behaviour under finite deformations. Validity and applicability of the proposed numerical framework to practical inverse analysis problems in biomechanics is demonstrated through various numerical examples. The corresponding simulations utilise in vivo measurements of patient-specific geometries derived from different medical imaging modalities, and include recovery of the pressure-free configuration of human aortas and the gravity-free shape of the female breast.

Appendix: Mathematical Derivations

The second Piola-Kirchhoff stress tensor for an elastic solid that is described by the modified stored-energy potential $\bar{W} + Q$ 37 is given by

$$\begin{aligned} \varvec{S} = \varvec{\bar{S}} - \frac{2(\bar{p}-p)}{P(\bar{p})}\,\frac{\partial \bar{p}}{\partial \mathbf{C}} + \frac{(\bar{p}-p)^2}{P^2(\bar{p})}\,\frac{\partial P(\bar{p})}{\partial \mathbf{C}}, \end{aligned}$$

(11)

where $\varvec{\bar{S}} = 2\,{\partial \bar{W}}/{\partial \mathbf{C}}$, $\bar{p} = -\,{\partial \bar{W}}/{\partial J}$, and $P(\bar{p}) = {\partial \bar{p}}/{\partial J}$. Applying the push-forward transformation ($J\,\varvec{\sigma } = \mathbf{F}\cdot \mathbf{S}\cdot \mathbf{F}^T$) on the above formula and considering that: $\mathbf{F}\cdot \left[ {\partial \bullet }/{\partial \mathbf{C}}\right] \cdot \mathbf{F}^T = \mathbf{B}\cdot \left[ {\partial \bullet }/{\partial \mathbf{B}}\right]$, we obtain the Cauchy stress tensor for the forward problem

$$\begin{aligned} \varvec{\sigma } = \varvec{\bar{\sigma }} - \frac{2(\bar{p}-p)}{J P(\bar{p})}\,\mathbf{B}\cdot \frac{\partial \bar{p}}{\partial \mathbf{B}} + \frac{(\bar{p}-p)^2}{J P^2(\bar{p})}\,\mathbf{B}\cdot \frac{\partial P(\bar{p})}{\partial \mathbf{B}}. \end{aligned}$$

(12)

Re-parametrisation of the Cauchy stresses expression to an inverse setting,13 i.e., replacing $J=\jmath ^{-1}$ and $\mathbf{B}=\mathbf{c}^{-1}$, gives

$$\begin{aligned} \varvec{\sigma }&= \varvec{\bar{\sigma }} - \frac{2\jmath (\bar{p}-p)}{P(\bar{p})}\,\mathbf{c}^{-1}\cdot \frac{\partial \bar{p}}{\partial \mathbf{c}^{-1}} + \frac{\jmath (\bar{p}-p)^2}{P^2(\bar{p})}\,\mathbf{c}^{-1}\cdot \frac{\partial P(\bar{p})}{\partial \mathbf{c}^{-1}} \nonumber \\&= \varvec{\bar{\sigma }} - \frac{2\jmath (\bar{p}-p)}{P(\bar{p})}\,\mathbf{c}^{-1}\cdot \frac{\partial \bar{p}}{\partial J}\,\frac{\partial \jmath ^{-1}}{\partial \mathbf{c}^{-1}} + \frac{\jmath (\bar{p}-p)^2}{P^2(\bar{p})}\,\mathbf{c}^{-1}\cdot \frac{\partial P(\bar{p})}{\partial J}\,\frac{\partial \jmath ^{-1}}{\partial \mathbf{c}^{-1}}, \end{aligned}$$

(13)

where after some algebra it can be deduced that

$$\begin{aligned} \varvec{\sigma } = \varvec{\bar{\sigma }} - \frac{\bar{p}-p}{P(\bar{p})}\,\frac{\partial \bar{p}}{\partial J}\,\mathbf{I} + \frac{(\bar{p}-p)^2}{2 P^2(\bar{p})}\,\frac{\partial P(\bar{p})}{\partial J}\,\mathbf{I}. \end{aligned}$$

(14)

In the forward setting, the Cauchy stress tensor is defined by Eq. (1), where the potential stored-energy is a function of the modified strain invariants, i.e., $\bar{W}(J_1,J_2,J)$. However, in the particular case of a transversely isotropic hyperelastic solid having one family of fibres—described by a single direction unit vector ($\hat{\mathbf{N}}_{0}$) in the reference state—the potential stored-energy can be extended to include the modified pseudo-invariants:16 $J_4=J^{-2/3}\,\hat{\mathbf{N}}_{0}\cdot \mathbf{C}\cdot \hat{\mathbf{N}}_{0}$ and $J_5=J^{-4/3}\,\hat{\mathbf{N}}_{0}\cdot \mathbf{C}^2\cdot \hat{\mathbf{N}}_{0}$. Apparently more pseudo-invariants can be incorporated in $\bar{W}$ if more preferred directions of isotropy are considered in the constitutive model.

Using chain rule differentiation in Eq. (1), substituting the tensor derivatives of the modified invariants into Eq. (1) and making use of the Cayley–Hamilton theorem23 for $\mathbf{B}$, i.e., $\mathbf{B}^2 = I_1\,\mathbf{B} - I_2\,\mathbf{I} + J^2\,\mathbf{B}^{-1}$, produces

$$\begin{aligned} \varvec{\bar{\sigma }}&= \frac{\partial \bar{W}}{\partial J_1} \left( 2J^{-5/3}\,\mathbf{B}-{2J_1}/{3J} \mathbf{I}\right) + \frac{\partial \bar{W}}{\partial J_2} \left( {2J_2}/{3J} \mathbf{I}-2J^{-1/3}\,\mathbf{B}^{-1}\right) + \frac{\partial \bar{W}}{\partial J}\, \mathbf{I} \nonumber \\&\quad+ \frac{\partial \bar{W}}{\partial J_4} \left( 2 J^{-5/3}\,{\mathbf{A}_0}\right) + \frac{\partial \bar{W}}{\partial J_5} \left( 2 J^{-7/3}\,{\mathbf{A}_0}\cdot \mathbf{B} +2 J^{-7/3}\,\mathbf{B}\cdot {\mathbf{A}_0}\right). \end{aligned}$$

(15)

where ${\mathbf{A}_0} = \mathbf{F}\cdot (\hat{\mathbf{N}}_{0}\otimes \hat{\mathbf{N}}_{0})\cdot \mathbf{F}^T\!$. However, it is straightforward to extend this constitutive equation for multi-fibrous anisotropic hyperelastic materials by incorporating more pseudo-invariants in the isochoric part of the potential function.

Re-parametrisation of the above Cauchy stresses expression to an inverse setting can give the analytic expression

$$\begin{aligned} \varvec{\bar{\sigma }}&= \frac{\partial \bar{W}}{\partial J_1} \left( 2\jmath ^{5/3}\,\mathbf{c}^{-1}-{2}/{3} \jmath_2\,\jmath \,\,\mathbf{I}\right) + \frac{\partial \bar{W}}{\partial J_2} \left( {2}/{3} \jmath_1\,\jmath \mathbf{I}-2\jmath ^{1/3}\,\mathbf{c}\right) + \frac{\partial \bar{W}}{\partial J}\, \mathbf{I} \nonumber \\&\quad+ \frac{\partial \bar{W}}{\partial J_4} \left( 2\,\jmath_4^{-1}\,\jmath {\mathbf{a}_0}\right) + \frac{\partial \bar{W}}{\partial J_5} \left( 2\,\jmath_4^{-1}\,\jmath ^{5/3}\,{\mathbf{a}_0}\cdot \mathbf{c}^{-1} +2\,\jmath_4^{-1}\,\jmath ^{5/3}\,\mathbf{c}^{-1}\!\cdot {\mathbf{a}_0}\right), \end{aligned}$$

(16)

where $J_1=\jmath_2$, $J_2=\jmath_1$, $J_4=\jmath_4^{-1}$, $J_5=\jmath_5\,\jmath_4^{-1}\,\jmath ^2$, $J=\jmath {-1}$ ($\jmath =\det {\mathbf{f}}$), and $\jmath_1=\imath_1\,\jmath ^{-2/3}\!$, $\jmath_2=\imath_2\,\jmath ^{-4/3}\!$, $\jmath_4=\imath_4\,\jmath ^{-2/3}\!$, $\jmath_5=\imath_5\,\jmath ^{-4/3}\!$, while ${\mathbf{a}_0}=\hat{\mathbf{n}}_{0}\otimes \hat{\mathbf{n}}_{0}$. The first and second invariants of the inverse deformation tensor $\mathbf{c}$ are denoted with $\imath_1$ and $\imath_2$ respectively, while $\imath_4$, $\imath_5$ are the corresponding Eulerian pseudo-invariants of the fibre.20 Also, the fibre tangent unit vector in the deformed state, $\hat{\mathbf{n}}_{0}$, is related to the corresponding one in the reference frame, $\hat{\mathbf{N}}_{0}$, through the relationship:

$$\begin{aligned} \mathbf{F}\cdot \hat{\mathbf{N}}_{0} = I_4^{1/2} \hat{\mathbf{n}}_{0} \,\Leftrightarrow \, \hat{\mathbf{N}}_{0} = I_4^{1/2} \mathbf{F}^{-1}\!\cdot \hat{\mathbf{n}}_{0} \,\Leftrightarrow \, \hat{\mathbf{N}}_{0} = \imath_4^{-1/2} \mathbf{f}\cdot \hat{\mathbf{n}}_{0} \,\Leftrightarrow \, \hat{\mathbf{N}}_{0} = \jmath_4^{-1/2}\,\jmath ^{-1/3} \mathbf{f}\cdot \hat{\mathbf{n}}_{0}, \end{aligned}$$

In order to arrive at Eq. (2), one has to substitute the partial derivatives of the stored-energy function with respect to the inverse invariants, i.e., ${\partial \bar{W}}/{\partial J_1} = {\partial \bar{w}}/{\partial \jmath_2}$, ${\partial \bar{W}}/{\partial J_2} = {\partial \bar{w}}/{\partial \jmath_1}$, ${\partial \bar{W}}/{\partial J} = -\jmath ^2\,\,{\partial \bar{w}}/{\partial \jmath }$, ${\partial \bar{W}}/{\partial J_4} = -\jmath_4^2\,\,{\partial \bar{w}}/{\partial \jmath_4}-\jmath_4\,\jmath_5\,\,{\partial \bar{w}}/{\partial \jmath_5}$ and ${\partial \bar{W}}/{\partial J_5} = \jmath_4\,\,\jmath ^2\,\,{\partial \bar{w}}/{\partial \jmath_5}$, in Eq. (16).

Finally, taking the above equation and after some tedious algebra, the following analytic expression for the material tangent fourth-order tensor of Eq. (10b) can be deduced:

$$\begin{aligned} \bar{\mathsf {C}}_\text{UU} &=\left[ \frac{\partial ^2\bar{W}}{\partial J_1^2} \left( 4\,\jmath ^{5/3}\,\mathbf{c}^{-1}\!\!-{4}/{3} \jmath_2\,\jmath \,\,\mathbf{I}\right) + \frac{\partial ^2\bar{W}}{\partial J_2\partial J_1} \left( {4}/{3} \jmath_1\,\jmath \,\,\mathbf{I}\!-4\jmath ^{1/3}\,\mathbf{c}\right) + \frac{\partial ^2\bar{W}}{\partial J\partial J_1}\,2\, \mathbf{I} \right] \! \,\frac{\partial \jmath_2}{\partial \mathbf{c}} \nonumber \\ &\quad+\left[ \frac{\partial ^2\bar{W}}{\partial J_1\partial J_2} \left( 4\jmath ^{5/3}\,\mathbf{c}^{-1}\!\!-{4}/{3} \jmath_2\,\jmath \,\,\mathbf{I}\right) + \frac{\partial ^2\bar{W}}{\partial J_2^2} \left( {4}/{3} \jmath_1\,\jmath \,\,\mathbf{I}\!-4\jmath ^{1/3}\,\mathbf{c}\right) + \frac{\partial ^2\bar{W}}{\partial J\partial J_2}\,2\, \mathbf{I} \right] \! \,\frac{\partial \jmath_1}{\partial \mathbf{c}} \nonumber \\ &\quad-\, \frac{\partial ^2\bar{W}}{\partial J_1\partial J} \left( 2\jmath ^{2/3}\,\mathbf{c}^{-1}\!\!-{2}/{3} \jmath_2\,\mathbf{I}\right) \,\mathbf{c}^{-1} - \frac{\partial ^2\bar{W}}{\partial J_2\partial J} \left( {2}/{3} \jmath_1\,\mathbf{I}\!-2\jmath ^{-2/3}\,\mathbf{c}\right) \,\mathbf{c}^{-1} \nonumber \\ &\quad+\frac{\partial \bar{W}}{\partial J_1}\, \left( {20}/{3} \jmath ^{2/3}\,\mathbf{c}^{-1}\,\frac{\partial \jmath }{\partial \mathbf{c}} + 4\jmath ^{5/3}\,\mathsf {I}_{c^{-1}} - {4}/{3} \jmath \,\,\mathbf{I}\,\frac{\partial \jmath_2}{\partial \mathbf{c}} - {4}/{3} \jmath_2\,\,\mathbf{I}\,\frac{\partial \jmath }{\partial \mathbf{c}} \right) \nonumber \\ &\quad+\frac{\partial \bar{W}}{\partial J_2}\, \left( {4}/{3} \jmath \,\,\mathbf{I}\,\frac{\partial \jmath_1}{\partial \mathbf{c}} + {4}/{3} \jmath_1\,\,\mathbf{I}\,\frac{\partial \jmath }{\partial \mathbf{c}} - {4}/{3} \jmath ^{-2/3}\,\,\mathbf{c}\,\frac{\partial \jmath }{\partial \mathbf{c}} - 4\jmath ^{1/3}\,\,\mathsf {I} \right) \nonumber \\ &\quad-\, \frac{\partial ^2\bar{W}}{\partial J_4\partial J_4} \left( 4\,\jmath /\jmath_4^3 {\mathbf{a}_0}\,\frac{\partial \jmath_4}{\partial \mathbf{c}}\right) - \frac{\partial ^2\bar{W}}{\partial J_4\partial J_5} \left( 4\,\jmath /(\jmath_4\,\jmath_5^2) {\mathbf{a}_0}\,\frac{\partial \jmath_5}{\partial \mathbf{c}}\right) \nonumber \\ &\quad+\frac{\partial \bar{W}}{\partial J_4} \left( 4/\jmath_4 {\mathbf{a}_0}\,\frac{\partial \jmath }{\partial \mathbf{c}} - 4\,\jmath /\jmath_4^2 {\mathbf{a}_0}\,\frac{\partial \jmath_4}{\partial \mathbf{c}} \right) \nonumber \\ &\quad+\frac{\partial \bar{W}}{\partial J_5} \left( {10}/{3} \jmath ^{5/3}/\jmath_4 \mathbf{N}_{0c^{-1}}\,\mathbf{c}^{-1} - 4\,\jmath ^{5/3}/\jmath_4^2 \mathbf{N}_{0c^{-1}}\,\frac{\partial \jmath_4}{\partial \mathbf{c}} + 4\,\jmath ^{5/3}/\jmath_4 \mathsf {N}_{0c^{-1}} \right) \nonumber \\ &\quad-\, \frac{\partial ^2\bar{W}}{\partial J_5\partial J_4} \left( 4\,\jmath ^{5/3}/\jmath_4^3 \mathbf{N}_{0c^{-1}}\,\frac{\partial \jmath_4}{\partial \mathbf{c}}\right) - \frac{\partial ^2\bar{W}}{\partial J_5\partial J_5} \left( 4\,\jmath ^{5/3}/(\jmath_4\,\jmath_5^2) \mathbf{N}_{0c^{-1}}\,\frac{\partial \jmath_5}{\partial \mathbf{c}}\right) \nonumber \\ &\quad-\, \frac{\partial ^2\bar{W}}{\partial J_4\partial J} \left( 2\,\jmath /\jmath_4 {\mathbf{a}_0}\,\mathbf{c}^{-1}\right) -\, \frac{\partial ^2\bar{W}}{\partial J_5\partial J} \left( 2\,\jmath ^{5/3}/\jmath_4 \mathbf{N}_{0c^{-1}}\,\mathbf{c}^{-1}\right) - \frac{\partial ^2\bar{W}}{\partial J^2}\, \jmath \mathbf{I} \mathbf{c}^{-1}, \end{aligned}$$

(17)

where $\mathbf{N}_{0c^{-1}} = {\mathbf{a}_0}\cdot \mathbf{c}^{-1}+\mathbf{c}^{-1}\cdot {\mathbf{a}_0}$, $\mathsf {N}_{0c^{-1}} = {\mathbf{a}_0}:\mathsf {I}_{c^{-1}}+\mathsf {I}_{c^{-1}}:{\mathbf{a}_0}$, while $\mathsf {I}_{c^{-1}} \equiv {\partial \mathbf{c}^{-1}}/{\partial \mathbf{c}}$, and $\mathsf {I}$ the symmetric identity 4th-order tensor. Also, the partial derivatives of the modified Eulerian invariants are given by ${\partial \jmath_1}/{\partial \mathbf{c}} = \jmath ^{-2/3}\,\mathbf{I} - \jmath_1\,{1}/{3}\,\,\mathbf{c}^{-1}$, ${\partial \jmath_2}/{\partial \mathbf{c}} = \jmath_1\jmath ^{-2/3}\,\mathbf{I} - \jmath ^{-4/3}\,\mathbf{c} - \jmath_2\,{2}/{3}\,\,\mathbf{c}^{-1}$, and ${\partial \jmath_4}/{\partial \mathbf{c}} = \jmath ^{-2/3}\,{\mathbf{a}_0} - \jmath_4\,{1}/{3}\,\,\mathbf{c}^{-1}$, ${\partial \jmath_5}/{\partial \mathbf{c}} = \jmath ^{-4/3}\,({\partial \imath_5}/{\partial \mathbf{c}}) - \jmath_5\,{2}/{3}\,\mathbf{c}^{-1}$, since ${\partial \imath_1}/{\partial \mathbf{c}} = \mathbf{I}$, ${\partial \imath_2}/{\partial \mathbf{c}} = \imath_1\,\mathbf{I}-\mathbf{c}$, ${\partial \jmath }/{\partial \mathbf{c}} = \jmath \,{1}/{2}\,\,\mathbf{c}^{-1}$, ${\partial \imath_4}/{\partial \mathbf{c}} = {\mathbf{a}_0}$ and ${\partial \imath_5}/{\partial \mathbf{c}} = \hat{\mathbf{n}}_{0}\otimes (\mathbf{c}\cdot \hat{\mathbf{n}}_{0}) + (\hat{\mathbf{n}}_{0}\cdot \mathbf{c})\otimes \hat{\mathbf{n}}_{0}$.