## 4 Verification and Discussion of the Example
### Viscoelastic Displacement of the Rod Subject to Uniaxial Tension
#### 4.1.2 Analytical Solution  
The analytical solution can be obtained by solving the general expression of the differential constitutive equation of linear viscoelastic materials as follows:  
\[P(D)\sigma=Q(D)\varepsilon \tag{16}\]  
where \(P(D)=\sum\limits_{k=0}^{m}p_{k}\frac{\partial^{k}}{\partial t^{k}}\), \(Q(D)=\sum\limits_{k=0}^{m}q_{k}\frac{\partial^{k}}{\partial t^{k}}\), and \(D=\frac{\partial}{\partial t}\) are differential operators over time. The generalized Maxwell model, composed of a spring element and a Maxwell element in parallel, degenerates into the traditional Poynting-Thomson model, commonly known as the \(H\,|\,M\) model (\(H\) represents Hookean spring, \(M\) represents a Maxwell element, and \(H\,|\,M\) represents a spring and a Maxwell element in parallel). Its differential constitutive equation is as follows:  
\[\sigma+\frac{\eta_{1}}{E_{1}}\dot{\sigma}=E_{\infty}\varepsilon+\left(\eta_{ 1}+\frac{E_{\infty}\eta_{1}}{E_{1}}\right)\dot{\varepsilon} \tag{17}\]  
where \(E_{1}\) and \(\eta_{1}\) are the elastic and viscous coefficients of the Maxwell elements, respectively; \(E_{\infty}\) is the elastic coefficient of the spring, representing the elastic modulus of the material as time approaches infinity. Referring to the general expression (16), for the \(H\,|\,M\) model, we have the following:  
\[P(D)=1+\frac{\eta_{1}}{E_{1}}D,\ \ \ Q(D)=E_{\infty}+\frac{E_{1}+E_{\infty}}{E_{1 }}\eta_{1}D \tag{18}\]  
Rewriting Equation (16) yields the following:  
\[\sigma=\frac{Q(D)}{P(D)}\varepsilon \tag{19}\]  
It can be seen that \(\frac{Q(D)}{P(D)}\) is equivalent to the elastic modulus \(E\) in elastic materials. According to the principle of elasticity-viscoelasticity correspondence, the solutions to linear viscoelastic problems are exactly the same as those of elastic problems under identical boundary conditions in Laplace space. Simply substitute \(E\) in the elastic so  
Figure 1: Axial rod of the linear viscoelastic material. lution with the Laplace space expression \(\frac{Q(s)}{P(s)}\) corresponding to \(\frac{Q(D)}{P(D)}\) in the viscoelastic constitutive equation. Here, \(Q(s)\) is the Laplace transform of \(Q(D)\), and \(P(s)\) is the Laplace transform of \(P(D)\). For the \(H\,|\,M\) model used in the given example  
\[P(s)=1+\frac{\eta_{1}}{E_{1}}s,\ \ Q(s)=E_{\infty}+\frac{E_{1}+E_{\infty}}{E_{1}} \eta_{1}s \tag{20}\]  
Therefore,  
\[\frac{Q(s)}{P(s)}=\frac{E_{\infty}+\frac{E_{1}+E_{\infty}}{E_{1}}\eta_{1}s}{1 +\frac{E_{1}}{E_{1}}s}=E_{\infty}+\frac{E_{1}\eta_{1}s}{E_{1}+\eta_{1}s} \tag{21}\]  
For the axial tension rod, its elastic displacement solution is \(u=\frac{PL}{EA}\). According to the principle of elasticity-viscoelasticity correspondence, the linear viscoelastic displacement solution retains the same form in Laplace space. To obtain the viscoelastic displacement solution in Laplace space, simply substitute \(E\) with \(\frac{Q(s)}{P(s)}\) in Equation (21), thus resulting in the expression shown below:  
\[\widetilde{u}(s)=\frac{PL}{sA\left(E_{\infty}+\frac{E_{1}\eta_{1}s}{E_{1}+ \eta_{1}s}\right)} \tag{22}\]  
Organize Equation (22) to obtain a form that is easy to perform inverse Laplace transform with as follows:  
\[\widetilde{u}(s)=\frac{PL}{A}\times\frac{E_{1}}{\eta_{1}(E_{\infty}+E_{1}) \cdot s\left[s+\frac{E_{\infty}E_{1}}{\eta_{1}(E_{\infty}+E_{1})}\right]}+ \frac{PL}{A}\times\frac{1}{(E_{\infty}+E_{1})\cdot\left[s+\frac{E_{\infty}E_{ 1}}{\eta_{1}(E_{\infty}+E_{1})}\right]} \tag{23}\]  
After this arrangement, each term in Equation (23) can be identified using the Laplace transform table to find its corresponding inverse transformation form.