Objective conformation positivity and stored-energy dissipation

Since $u$ and $x$ are $C^1$, $L(t)=\mathrm{\nabla }u(t,x(t))$ is continuous on the compact interval $[0,T]$. Applying [F1] gives a unique $C^1$ solution $Z^{\mathrm{\prime }}=LZ$, $Z(0)=I_3$ on all of $[0,T]$; by [F2], each $Z(t)$ is invertible.

Set $x^{\ast }(t)=Q(t)x(t)+c(t)$, and on the open set $\{(t,y):t\in J,Q(t{)}^{\mathsf{T}}(y-c(t))\in \mathrm{\Omega }\}$ define $u^{\ast }(t,y)=Q^{\mathrm{\prime }}(t)z+Q(t)u(t,z)+c^{\mathrm{\prime }}(t)$, where $z=Q(t{)}^{\mathsf{T}}(y-c(t))$. These are $C^1$ maps, and product differentiation gives $(x^{\ast }{)}^{\mathrm{\prime }}=u^{\ast }(t,x^{\ast })$. At fixed time the chain rule gives ${\mathrm{\nabla }}_y{u}^{\ast }=(Q^{\mathrm{\prime }}+Q\mathrm{\nabla }u)Q^{\mathsf{T}}$, which along $x^{\ast }$ is precisely the stated $L^{\ast }$. Differentiating $Q{Q}^{\mathsf{T}}=I_3$ shows that $Q^{\mathrm{\prime }}{Q}^{\mathsf{T}}$ is skew-symmetric, so $D^{\ast }:=(L^{\ast }+(L^{\ast }{)}^{\mathsf{T}})\mathrm{/}2=QD{Q}^{\mathsf{T}}$. The path $C^{\ast }=QC{Q}^{\mathsf{T}}$ satisfies $(C^{\ast }{)}^{\mathrm{\prime }}=Q{C}^{\mathrm{\prime }}{Q}^{\mathsf{T}}+Q^{\mathrm{\prime }}C{Q}^{\mathsf{T}}+QC(Q^{\mathrm{\prime }}{)}^{\mathsf{T}}$. Substitution cancels the observer-derivative terms and gives $(C^{\ast }{)}^{\mathrm{\prime }}-L^{\ast }{C}^{\ast }-C^{\ast }(L^{\ast }{)}^{\mathsf{T}}=Q(C^{\mathrm{\prime }}-LC-C{L}^{\mathsf{T}})Q^{\mathsf{T}}=-(C^{\ast }-I_3)\mathrm{/}\lambda$. Thus the trajectory equation is covariant at the stated regularity.

By [F3] and the chain rule along $Z$, $(Z^{-1}{)}^{\mathrm{\prime }}=-Z^{-1}{Z}^{\mathrm{\prime }}{Z}^{-1}=-Z^{-1}L$; transposing gives $(Z^{-\mathsf{T}}{)}^{\mathrm{\prime }}=-L^{\mathsf{T}}{Z}^{-\mathsf{T}}$.

Define $Y(t)=e^{t\mathrm{/}\lambda }Z(t{)}^{-1}C(t)Z(t{)}^{-\mathsf{T}}$. Product differentiation using [F4], then substitution of the equation for $C^{\mathrm{\prime }}$ and the identities in step 2.1, cancels the $LC$ and $C{L}^{\mathsf{T}}$ terms and yields $Y^{\mathrm{\prime }}(t)=\frac{1}{\lambda }{e}^{t\mathrm{/}\lambda }Z(t{)}^{-1}Z(t{)}^{-\mathsf{T}}$.

Fix a nonzero $v\in {\mathbb{R}}^3$ and set $f_v(t)=v^{\mathsf{T}}Y(t)v$. This function is $C^1$ on $[0,T]$ and step 3.1 gives $f_v^{\mathrm{\prime }}(t)=e^{t\mathrm{/}\lambda }\mathrm{\mid }Z(t{)}^{-\mathsf{T}}v{\mathrm{\mid }}^2\mathrm{/}\lambda >0$. For $t>0$, [F11] on $[0,t]$ gives $f_v(t)>f_v(0)=v^{\mathsf{T}}{C}_{0}v>0$; the same positivity at $t=0$ is the initial hypothesis. As $Y$ is symmetric and $v$ was arbitrary, $Y(t)$ is positive definite for every $t\in [0,T]$. The identity $C=e^{-t\mathrm{/}\lambda }ZY{Z}^{\mathsf{T}}$ and invertibility of $Z$ then give $v^{\mathsf{T}}Cv=e^{-t\mathrm{/}\lambda }(Z^{\mathsf{T}}v{)}^{\mathsf{T}}Y(Z^{\mathsf{T}}v)>0$ for every nonzero $v$.

By step 4.1, $C(t)$ is positive definite for every $t\in [0,T]$. Fix such a time and write $C=C(t)$. By [F6], write $C=U\mathrm{diag}(c_1,c_2,c_3)U^{\mathsf{T}}$ with $c_i>0$. Determinant multiplicativity and $U{U}^{\mathsf{T}}=I_3$ give $\det C=c_1{c}_2{c}_3$; finite-sum trace cyclicity gives $\mathrm{tr}C={\sum }_i{c}_i$; hence [F7] gives $\log \det C={\sum }_i\log c_i$. For each $c>0$, [F5] gives $\log c={\int }_1^{c}dt\mathrm{/}t$: if $c>1$, then $\log c<c-1$ because $1\mathrm{/}t<1$ on the interior of $[1,c]$; if $0<c<1$, then $\log c=-{\int }_c^{1}dt\mathrm{/}t<-(1-c)=c-1$ because $1\mathrm{/}t>1$ on the interior of $[c,1]$; at $c=1$ the integral is zero. Thus $c-\log c-1\ge 0$, with equality exactly at $c=1$, so $w(C)=\frac{G}{2}{\sum }_i(c_i-\log c_i-1)\ge 0$, with equality exactly when all $c_i=1$, that is, $C=I_3$.

By step 4.1, $C(t)$ is positive definite for every $t\in [0,T]$. Fix such a time and write $C=C(t)$. By [F6], write $C=U\mathrm{diag}(c_1,c_2,c_3)U^{\mathsf{T}}$ with $c_i>0$. The unique positive square root and inverse square root in [F8] are $U\mathrm{diag}(\sqrt{c_i})U^{\mathsf{T}}$ and $U\mathrm{diag}(c_i^{-1\mathrm{/}2})U^{\mathsf{T}}$, respectively. Thus $C^{-1}=U\mathrm{diag}(c_i^{-1})U^{\mathsf{T}}$ and $\mathrm{tr}C+\mathrm{tr}{C}^{-1}-6={\sum }_{i=1}^3(c_i+c_i^{-1}-2)={\sum }_{i=1}^3(\sqrt{c_i}-c_i^{-1\mathrm{/}2}{)}^2\ge 0$.

Since $C$ is positive definite by step 4.1 and $\det C>0$ by step 5.1, Jacobi's formula and the derivative of logarithm in [F5] give $\frac{d}{dt}w(C)=\frac{G}{2}\mathrm{tr}((I_3-C^{-1})C^{\mathrm{\prime }})$. Substituting [A1] and multiplying out gives $\mathrm{tr}((I_3-C^{-1})(LC+C{L}^{\mathsf{T}}))=\mathrm{tr}(LC)+\mathrm{tr}(C{L}^{\mathsf{T}})-\mathrm{tr}(C^{-1}LC)-\mathrm{tr}(L^{\mathsf{T}})$. The finite-sum identity $\mathrm{tr}(AB)=\mathrm{tr}(BA)$ and $C^{\mathsf{T}}=C$ give $\mathrm{tr}(C{L}^{\mathsf{T}})=\mathrm{tr}(LC)$, $\mathrm{tr}(C^{-1}LC)=\mathrm{tr}L$, and $\mathrm{tr}(L^{\mathsf{T}})=\mathrm{tr}L$. Also $(C-I_3):D=\mathrm{tr}((C-I_3)L)=\mathrm{tr}(LC)-\mathrm{tr}L$, since the contraction of a symmetric matrix with the skew part of $L$ is zero. Therefore the displayed linear term is $2(C-I_3):D$. Direct multiplication gives $(I_3-C^{-1})(C-I_3)=C+C^{-1}-2I_3$, so its trace is $\mathrm{tr}C+\mathrm{tr}{C}^{-1}-6$. Hence $\frac{d}{dt}w=G(C-I_3):D-\frac{G}{2\lambda }(\mathrm{tr}C+\mathrm{tr}{C}^{-1}-6)$.

From [A1], $({\tau }_s+{\tau }_p):D=2{\eta }_{s}D:D+G(C-I_3):D$. Subtracting the derivative in step 6.1 gives the identity in the Statement; it is nonnegative by ${\eta }_s\ge 0$, $D:D\ge 0$, and step 5.2. Since $D^{\ast }=QD{Q}^{\mathsf{T}}$ and $C^{\ast }=QC{Q}^{\mathsf{T}}$, the stress definitions give ${\tau }_s^{\ast }=Q{\tau }_s{Q}^{\mathsf{T}}$ and ${\tau }_p^{\ast }=Q{\tau }_p{Q}^{\mathsf{T}}$; trace and determinant invariance give $w(C^{\ast })=w(C)$.