Steady compliance (Linear viscoelasty)

Boltzmann superposition: assume that there is no stress when $t\le 0$, so the integral starts from $0$ and $\sigma(0^{-})=0$:

$$\begin{equation} \gamma(t)=\int_0^t J(t-t^\prime)\dot{\sigma}(t^\prime)\mathrm{d}t^\prime \end{equation}$$

 Performing Laplace Transform yields: 

$$\begin{align} \hat{\gamma}(s)&=\hat{J}(s)\hat{\dot{\sigma}}(s)\\ &=\hat{J}(s)\left(s\hat{\sigma}(s)-\sigma(0^{-})\right) \\ &=\hat{J}(s)\left(s\hat{G}(s)\hat{\dot{\gamma}}(s)-\sigma(0^{-})\right)\\ &=\hat{J}(s)\left(s\hat{G}(s)(s\hat{\gamma}(s)-\gamma(0^-))-\sigma(0^{-})\right)\\ &=s^2\hat{J}(s)\hat{\gamma}(s)\hat{G}(s)  \end{align}$$

Here we let $\hat{f}(s):=\mathcal{L}\lbrace f(t)\rbrace(s)$ be the Laplace transformation and since the stress starts at time $0$, $\gamma(0^-)$ and $\sigma(0^-)$ are simply $0$. The 2nd to 3rd step is derived from the convolution relation $\sigma(t)=\int_0^t G(t-t^\prime)\dot{\gamma}(t^\prime)\mathrm{d}t^\prime$. Cancelling out the $\hat{\gamma}(s)$ and rearranging the last equation give

$$\begin{equation} \frac{1}{s^2}=\hat{J}(s)\hat{G}(s) \end{equation}$$

The inverse Laplace of above equation gives the convolution relation

$$\begin{equation} t=\int_0^{t}J(t-t^\prime)G(t^\prime)\mathrm{d}t^\prime \end{equation}$$

The last step is substituting $J(t)=J_e + t/\eta$ into above convolution relation and making $t\to\infty$: 

$$\begin{equation} \begin{aligned} \color{blue}{t} &=\int_0^{t}\left(J_e+\frac{t-t^\prime}{\eta}\right)G(t^\prime)\mathrm{d}t^\prime\\ &= J_e\int_0^\infty G(t^\prime)\mathrm{d}t^\prime + {\color{blue}{\frac{t}{\eta}\int_0^\infty G(t^\prime)\mathrm{d}t^\prime}} - \int_0^\infty G(t^\prime)\frac{t^\prime}{\eta}\mathrm{d}t^\prime \end{aligned} \end{equation}$$

Since $\eta:=\int_0^\infty G(t^\prime)\mathrm{d}t^\prime$, the blue terms cancelled out, which gives \begin{equation} J_e\int_0^\infty G(t^\prime)\mathrm{d}t^\prime = J_e\eta= \int_0^\infty G(t^\prime)\frac{t^\prime}{\eta}\mathrm{d}t^\prime = \frac{1}{\eta} \int_0^\infty G(t^\prime)t^\prime\mathrm{d}t^\prime \end{equation} Therefore we have 

$$J_e = \frac{1}{\eta^2}\int_0^\infty tG(t)\mathrm{d}t$$

$Q.E.D.$

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