## 3 Nonlinear Material Equations and the Energy-Based Method for the Inverse Analysis
Moreover, in linear rheology, the strain rate is approximated as equal to the velocity gradient, \(\epsilon_{ij}^{*}=d_{ij}\), such that a linear response:  
\[\epsilon_{12}=\epsilon_{0}\exp(\mathrm{i}(\omega t+\delta))\, \tag{5}\]  
allows us to rewrite:  
\[\epsilon_{12}=\epsilon_{0}\exp(\mathrm{i}(\omega t+\delta))=\frac{\sigma_{0}} {\sigma_{0}}\epsilon_{0}\exp(\mathrm{i}\omega t)\exp(\mathrm{i}\delta)=\frac {\epsilon_{0}}{\sigma_{0}}\exp(\mathrm{i}\delta)\sigma_{12}\, \tag{6}\]  
and propose a relation between stress and strain:  
\[\sigma_{12}=G^{*}\epsilon_{12}\,,\ G^{*}=\frac{\sigma_{0}}{\epsilon_{0}}\exp (-\mathrm{i}\delta)=-\frac{\sigma_{0}}{\epsilon_{0}}\big{(}\cos(\delta)+ \mathrm{i}\sin(\delta)\big{)}=G^{\prime}+\mathrm{i}G^{\prime\prime}. \tag{7}\]  
For a linear material, \(G^{\prime}\) relates to the elastic modulus; \(G^{\prime\prime}/\omega\) relates to the viscosity since:  
\[\epsilon_{12}^{*}=\mathrm{i}\omega\epsilon_{0}\exp(\mathrm{i}(\omega t+ \delta))=\mathrm{i}\omega\epsilon_{12}\, \tag{8}\] \[\sigma_{12}=G^{*}\epsilon_{12}=(G^{\prime}+\mathrm{i}G^{\prime \prime})\epsilon_{12}=G^{\prime}\epsilon_{12}+\frac{G^{\prime\prime}}{\omega} \epsilon_{12}\.\]  
In order to determine \(G^{\prime}\) and \(G^{\prime\prime}\) from the energy, we use:  
\[E=\int_{\tau}\int_{\mathcal{B}}\sigma_{ij}d_{ij}\,\mathrm{d}V\,\mathrm{d}t= \int_{\tau}\int_{\mathcal{B}}2\sigma_{12}\epsilon_{12}^{*}\,\mathrm{d}V\, \mathrm{d}t\, \tag{9}\]since the velocity gradient in Equation (3) consists of only shear components. For a detailed analysis by using spherical coordinates, we refer to [27] (Section 8). By using a cone-plate or a small plate-plate geometry, we can justify the assumption of constant stress and strains within the sample such that we obtain the following energy density (energy per volume):  
\[e=\frac{E}{V}=\int_{\tau}2\sigma_{12}\epsilon_{12}^{*}\,\mathrm{d}t\, \tag{10}\]  
where \(V\) is the volume of the sample. Since the measurement is a shear experiment, we assume that the volume remains the same throughout the experiment. We use a sinusoidal excitation, \(\sigma_{0}\sin(\omega t)=\,\mathcal{I}\mathrm{m}(\sigma_{12})\), meaning that we need to employ the imaginary part,  
\[\tilde{e}=\int_{\tau}2\mathcal{I}\mathrm{m}(\sigma_{12})\,\mathcal{I}\mathrm{ m}(\epsilon_{12})\,\mathrm{d}t. \tag{11}\]  
As we have the relation, \(\sigma_{12}=G^{*}\epsilon_{12}\), we acquire:  
\[ \begin{split}&\epsilon_{12}=\frac{\sigma_{12}^{*}}{G^{*}}= \frac{\sigma_{0}\mathrm{i}\omega\exp(\mathrm{i}\omega t)}{G^{\prime}+\mathrm{ i}G^{\prime\prime}}=\frac{\sigma_{0}\mathrm{i}\omega\big{(}\cos(\omega t)-\mathrm{i} \sin(\omega t)\big{)}(G^{\prime}-\mathrm{i}G^{\prime\prime})}{(G^{\prime})^{ 2}+(G^{\prime\prime})^{2}}\\ &=\frac{\sigma_{0}\omega\big{(}G^{\prime}\sin(\omega t)+G^{\prime \prime}\cos(\omega t)+\mathrm{i}(G^{\prime}\cos(\omega t)+G^{\prime\prime} \sin(\omega t))\big{)}}{(G^{\prime})^{2}+(G^{\prime\prime})^{2}}\end{split}  \tag{12}\]  
such that:  
\[\mathcal{I}\mathrm{m}(\epsilon_{12}^{*})=\frac{\sigma_{0}\omega\big{(}G^{ \prime}\cos(\omega t)+G^{\prime\prime}\sin(\omega t)\big{)}}{(G^{\prime})^{2} +(G^{\prime\prime})^{2}}. \tag{13}\]  
For notational simplicity, we introduce:  
\[J^{\prime}=\frac{G^{\prime}}{(G^{\prime})^{2}+(G^{\prime\prime})^{2}}\,\ J^{ \prime\prime}=\frac{G^{\prime\prime}}{(G^{\prime})^{2}+(G^{\prime\prime})^{2}}. \tag{14}\]  
Now, we can calculate the energy density under the aforementioned sinusoidal stress input,  
\[ \begin{split}\tilde{e}&=\int_{\tau}2\sigma_{0}^{2} J^{\prime}\omega\sin(\omega t)\cos(\omega t)\,\mathrm{d}t+\int_{\tau}2\sigma_{0}^{2} J^{\prime\prime}\sin^{2}(\omega t)\,\mathrm{d}t\\ &=-\sigma_{0}^{2}J^{\prime}\cos^{2}(\omega t)\Big{|}_{\tau}+ \sigma_{0}^{2}J^{\prime\prime}\bigg{(}\omega t-\frac{1}{2}\sin(2\omega t) \bigg{)}\bigg{|}_{\tau}\.\end{split}