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I have a problem with understanding of derivation of stress equation $\sigma(t)$ for Maxwell rheological model. Below is the classic equation:

$$\dot{\sigma}(t) + \sigma(t) \, \frac{E_0}{\eta}=E_0 \,\dot{\varepsilon}(t) \tag{1}$$ which is a differential equation of the following type: $$\dot{y} + p(\tau) \, y = r(\tau) \tag{2}$$

Solution for this kind of equation when $p(\tau) = p_0$ and $ y(\tau_0)=y_0 $ is following : $$y(t) = exp\left[ p_0 \, (\tau_0-t) \right] \int_{\tau_0}^{t} r(\tau) \; exp\left[ p_0 \, (t-\tau_0) \right] d\tau + y_0 \; exp\left[ p_0 \, (\tau_0-t) \right] \tag{3}$$

Here comes my issue. Similarly, for Maxwell model, when $\sigma(\tau_0)=0$ thus $y_0 = 0 $, solution will be: $$\sigma(t) = exp\left[ \frac{E_0}{\eta} \, (\tau_0-t) \right] \int_{\tau_0}^{t} E_0 \, \dot{\varepsilon}(\tau) \; exp\left[ \frac{E_0}{\eta} \, (t-\tau_0) \right] d\tau \tag{4}$$

because $E=E_0 \; exp\left[ \frac{E_0}{\eta} \, (t-\tau_0) \right] $ then it will be:

$$\sigma(t) = exp\left[ \frac{E_0}{\eta} \, (\tau_0-t) \right] \int_{\tau_0}^{t} E(t-\tau_0) \, \dot{\varepsilon} \; d\tau \tag{5}$$

What bothers me is $exp\left[ \frac{E_0}{\eta} \, (\tau_0-t) \right]$ in equation (5), just before integral. I think it shouldn't be left there. Whole above derivation is based on the book "Computational Viscoelasticity" by Marques, Severino P. C. and Creus, Guillermo J. In the mentioned book final equation is:

$$\sigma(t) = \int_{\tau_0}^{t} E(t-\tau_0) \, \dot{\varepsilon} \; d\tau \tag{6}$$

I don't quite understand what happened with $exp\left[ \frac{E_0}{\eta} \, (\tau_0-t) \right]$.

Adalbert
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When I solve it, I get $$\sigma(t)=E_0\int_{t_0}^t{e^{-\frac{(t-\tau)E_0}{\eta}}\frac{de}{d\tau}d\tau}=\int_{t_0}^t{E(t-\tau)\frac{de}{d\tau}}d\tau=\int_{t_0}^t{E(t-\tau)\dot{e}(\tau)d\tau}$$

Here's how to solve the original differential equation: If we multiply both sides of the differential equation by the "integrating factor" $\exp{\left(\frac{E_0t}{\eta}\right)}$, we obtain: $$\left(\dot{\sigma+\frac{E_0}{\eta}\sigma}\right)\exp{\left(\frac{E_0t}{\eta}\right)}=\frac{d}{dt}\left(\exp{\left(\frac{E_0t}{\eta}\right)}\sigma\right)=\exp{\left(\frac{E_0t}{\eta}\right)}E_0\dot{e}(t)$$If we integrate this equation between $t_0$ and t, we obtain:$$\exp{\left(\frac{E_0t}{\eta}\right)}\sigma(t)-\exp{\left(\frac{E_0t_0}{\eta}\right)}\sigma(t_0)=E_0\int_{t_0}^t{\exp{\left(\frac{E_0\tau}{\eta}\right)}\dot{e}(\tau)d\tau}$$where $\tau$ is a dummy variable of integration. If we now multiply both sides of this equation by $\exp{\left(-\frac{E_0t}{\eta}\right)}$, we obtain: $$\sigma(t)=\sigma(t_0)\exp{\left(-\frac{E_0(t-t_0)}{\eta}\right)}+E_0\int_{t_0}^t{\exp{\left(-\frac{E_0(t-\tau)}{\eta}\right)}\dot{e}(\tau)d\tau}$$

Chet Miller
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  • I'm afraid it is not enough to explain what I'm missing. General solution of this type of differential equation is showed on link. What happened to exponential function outside integral? – Adalbert May 20 '18 at 08:28
  • I've added the full solution to my answer, obtained using the simple "integrating factor" method. – Chet Miller May 20 '18 at 11:57
  • Thanks for your help. Because I had calculated definite integral for "integrating factor" it left me with extra tau_0 and "t" instead of "tau". – Adalbert May 20 '18 at 14:59