Dear people with an affinity for math,
I am just an engineer approaching the field of viscoelasticity. Currently, I would like to understand the derivation of the generalized Kelvin-Voigt material. It is common practice in viscoelasticity to use the so-called Hereditary integral or memory function to account for a stress history, which gives the following expression for the viscous strain
$\varepsilon_\mathsf{v} = \int_0^t J_{\mathsf{v}}(t-\tau)\dot{\sigma}(\tau)\rm{d}\tau$.
The physical details are not really important. All quantities are scalar values/functions. It is important to note that the strain is expressed in this convolutional integral. In the dissertation of Woldekidan (2011, p. 165), without further explanation or citing, the following is stated
$\varepsilon_\mathsf{v} = \int_0^t J_{\mathsf{v}}(t-\tau)\dot{\sigma}(\tau)\rm{d}\tau = \int_0^t \dot{J}_{\mathsf{v}}(t-\tau)\sigma(\tau)\rm{d}\tau$,
which confuses me. Without the symbolic derivative (upper dot) I read this as
$\varepsilon_\mathsf{v} = \int_0^t J_{\mathsf{v}}(t-\tau)\frac{\rm{d}\sigma(\tau)}{\rm{d}\tau}\rm{d}\tau = \int_0^t \frac{\rm{d}{J}_{\mathsf{v}}(t-\tau)}{\rm{d}(t-\tau)}\sigma\rm{d}\tau$.
Can you just swap the derivative in a product without consequences? Also mind the different variables of derivatives, $\tau$ and $t-\tau$, respectively. First I thought it could be something related to the product rule, but there would need to be another term. I would really be happy if you could explain why this is correct or why it is not and when this interchanging of derivatives is applicable.
I managed to stumble upon an answer, which I found in Viscoelasticity (Flügge, 1975, p. 38).
First, to be precise, the Hereditary integral should be
$\varepsilon(t)=\sigma(0)J(t)+\int^t_0J(t-\tau)\frac{\rm{d}\sigma(\tau)}{\rm{d}\tau}\rm{d}\tau$.
The initial stress input $\sigma(0)J(t)$ is usually assumed to be included in the latter integral, but for the sake of completeness, this should be considered here. Then integrating by parts gives
$\varepsilon(t) = \sigma(0)J(t) + [J(t-\tau)\sigma(\tau)]^t_0 - \int_0^t\frac{\rm{d}J(t-\tau)}{\rm{d}\tau}\sigma(\tau)\rm{d}\tau = \sigma(t)J(0)-\int_0^t\frac{\rm{d}J(t-\tau)}{\rm{d}\tau}\sigma(\tau)\rm{d}\tau$.
Using the chain rule, it can be shown that $\frac{\rm{d}J(t-\tau)}{\rm{d}\tau}=-\frac{\rm{d}J(t-\tau)}{\rm{d}(t-\tau)}$, which finally gives
$\varepsilon(t) = \sigma(t)J(0)+\int_0^t\frac{\rm{d}J(t-\tau)}{\rm{d}(t-\tau)}\sigma(\tau)\rm{d}\tau$.