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The mathematical description of elastic bodies whose stress-free material state changes through growth or remodelling. A multiplicative decomposition of the deformation gradient separates the growth tensor $\mathbf{G}$ from the elastic accommodation $\mathbf{A}$. The relaxed grown configuration need not be globally compatible, so even an unloaded body can develop internal stress.

Write $\lambda =L_1/L_0>1$. The one-dimensional multiplicative decomposition of the deformation gradient reads $\ell /L_0=\alpha \gamma$. Before contact, $\alpha =1$; afterwards, the imposed fixed length givesSubstituting in the linear constitutive equation yieldsHere negative axial stress means compression. The two branches agree at $t_c$ because $1+c{t}_c=\lambda$. At late times,Thus the law permits arbitrarily severe elastic shortening at a finite compressive stress. It is a small-strain approximation, appropriate near $\alpha =1$, and is not a credible material law as $\alpha \to 0$. The one-dimensional constraint rules out buckling in this particular model; the finite limiting stress is a defect of extrapolating the linear constitutive equation, even with that constraint retained.

The specific Helmholtz free energy $\psi =U-\theta \eta$ satisfies $d\psi ={\rho }_0^{-1}{P}_{Ii}d{F}_{iI}-\eta d\theta -{\rho }_0^{-1}{f}_{r}d{\xi }_r$. For the multiplicative decomposition of the deformation gradient $F=F_{\ast }A$, adopt $P_{Ji}^{\ast }={\rho }_0\partial \psi /\partial (F_{\ast }{)}_{iJ}$, consistently with the material-first nominal stress tensor. Differentiating $F_{\ast }=F{A}^{-1}$ givesConsequentlyorThese relations follow from the chain rule; the transpose convention matters.

Differentiating the dissipation potential, with the Frobenius norm of $Q$, givesAn integrating factor yields the exponential memory from tensor relaxation:For $n>0$, the driving term is continuously zero at $Q=0$.

Taking the product $T=PF$ explicitly defined in the question, $F=F_{\ast }A$ gives $T=QA$, and hence its requested history functional isThere is a naming error in the PDF: with its $P_{Ii}$ convention, the actual second Piola-Kirchhoff stress tensor is $S=P{F}^{-T}$, while $PF=SC$ is the transpose of the Mandel stress tensor, where $C=F^{T}F$. Thus $S=QA{C}^{-1}$ if that stress measure is required. This agrees with the stress push-forward used in Question 5; the printed product is fully determined by the history above, whereas recovering $S$ also requires the deformation or an invertible elastic constitutive relation.

The multiplicative decomposition of the deformation gradient is used on histories where $A$ stays invertible. Thermodynamic admissibility additionally requires $\mathrm{tr}(Q\dot{A})=\alpha \Vert Q{\Vert }^{n+1}-\mathrm{tr}(AQ)/\tau \ge 0$: positivity of the constants alone does not ensure this for arbitrary prescribed histories.