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A dissipation potential is a function used to generate dissipative rates or their conjugate driving forces. In a force-based convention, a potential $\Phi (X)$ specifies an internal variable rate through $\dot{\alpha }={\partial }_X\Phi$. For example, $\Phi =X^2/(2\eta )$ with $\eta >0$ gives $\dot{\alpha }=X/\eta$ and dissipation $X\dot{\alpha }=X^2/\eta \ge 0$.

If $\Phi$ is differentiable and convex with a minimum at $X=0$, the convexity inequality gives $X{\partial }_X\Phi \ge \Phi (X)-\Phi (0)\ge 0$. This supplies a direct route to nonnegative dissipation. A rate-based convention instead generates forces from rates, so its variables must be stated explicitly. The potential is useful for organizing constitutive equations, but the complete model must still satisfy the Clausius-Duhem inequality, including thermal terms and the chosen energetic sign convention.

The dissipation potential $\Omega =\alpha \Vert Q{\Vert }^{n+1}/(n+1)-\mathrm{tr}(AQ)/\tau$, with the Frobenius norm, gives $\dot{A}+A/\tau =\alpha \Vert Q{\Vert }^{n-1}{Q}^T$. An integrating factor proves the displayed memory formula. For $n>0$, its nonlinear driving term is defined continuously as zero at $Q=0$. Mechanical dissipation is $\mathrm{tr}(Q\dot{A})=\alpha \Vert Q{\Vert }^{n+1}-\mathrm{tr}(AQ)/\tau$, which must be nonnegative on thermodynamically admissible histories.

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.