to one of those variables while holding the others constant:
@f[
\frac{\partial f\left(x, y\right)}{\partial x}
- = \frac{d f\left(x, y\right)}{d x} \vert_{y} .
+ = \frac{d f\left(x, y\right)}{d x} \Big\vert_{y} .
@f]
More specific to what's encoded in the dissipation inequality (with the very general
free energy density function $\psi_{0}$ with its parameterization yet to be formalized),
understood by comparing the following two statements:
@f{align*}
\frac{\partial f\left(x, y\left(x\right)\right)}{\partial x}
- &= \frac{d f\left(x, y\left(x\right)\right)}{d x} \vert_{y} \\
+ &= \frac{d f\left(x, y\left(x\right)\right)}{d x} \Big\vert_{y} \\
\frac{d f\left(x, y\left(x\right)\right)}{d x}
- &= \frac{d f\left(x, y\left(x\right)\right)}{d x} \vert_{y}
- + \frac{d f\left(x, y\left(x\right)\right)}{d y} \vert_{x} \frac{d y\left(x\right)}{x} .
+ &= \frac{d f\left(x, y\left(x\right)\right)}{d x} \Big\vert_{y}
+ + \frac{d f\left(x, y\left(x\right)\right)}{d y} \Big\vert_{x} \frac{d y\left(x\right)}{x} .
@f}
Returning to the thermodynamics of the problem, we next exploit the arbitrariness
For the case of rate-dependent materials, this expands to
@f{align*}{
\mathcal{H}^{\text{tot}} \left( \mathbf{C}, \mathbf{C}_{v}^{i}, \boldsymbol{\mathbb{H}} \right)
- &= 2 \frac{d^{2} \psi_{0} \left( \mathbf{C}, \mathbf{C}_{v}^{i}, \boldsymbol{\mathbb{H}} \right)}{d \mathbf{C} \otimes \partial \mathbf{C}} , \\
+ &= 4 \frac{d^{2} \psi_{0} \left( \mathbf{C}, \mathbf{C}_{v}^{i}, \boldsymbol{\mathbb{H}} \right)}{\partial \mathbf{C} \otimes d \mathbf{C}} , \\
\mathbb{D} \left( \mathbf{C}, \mathbf{C}_{v}^{i}, \boldsymbol{\mathbb{H}} \right)
- &= -\frac{d^{2} \psi_{0} \left( \mathbf{C}, \mathbf{C}_{v}^{i}, \boldsymbol{\mathbb{H}} \right)}{d \boldsymbol{\mathbb{H}} \otimes \partial \boldsymbol{\mathbb{H}}} , \\
+ &= -\frac{d^{2} \psi_{0} \left( \mathbf{C}, \mathbf{C}_{v}^{i}, \boldsymbol{\mathbb{H}} \right)}{\partial \boldsymbol{\mathbb{H}} \otimes d \boldsymbol{\mathbb{H}}} , \\
\mathfrak{P}^{\text{tot}} \left( \mathbf{C}, \mathbf{C}_{v}^{i}, \boldsymbol{\mathbb{H}} \right)
- &= - 2 \frac{d^{2} \psi_{0} \left( \mathbf{C}, \mathbf{C}_{v}^{i}, \boldsymbol{\mathbb{H}} \right)}{d \boldsymbol{\mathbb{H}} \otimes \partial \mathbf{C}} , \\
+ &= - 2 \frac{d^{2} \psi_{0} \left( \mathbf{C}, \mathbf{C}_{v}^{i}, \boldsymbol{\mathbb{H}} \right)}{\partial \boldsymbol{\mathbb{H}} \otimes d \mathbf{C}} , \\
\left[ \mathfrak{P}^{\text{tot}} \left( \mathbf{C}, \mathbf{C}_{v}^{i}, \boldsymbol{\mathbb{H}} \right) \right]^{T}
- &= - 2 \frac{d^{2} \psi_{0} \left( \mathbf{C}, \mathbf{C}_{v}^{i}, \boldsymbol{\mathbb{H}} \right)}{d \mathbf{C} \otimes \partial \boldsymbol{\mathbb{H}}} ,
+ &= - 2 \frac{d^{2} \psi_{0} \left( \mathbf{C}, \mathbf{C}_{v}^{i}, \boldsymbol{\mathbb{H}} \right)}{\partial \mathbf{C} \otimes d \boldsymbol{\mathbb{H}}} ,
@f}
while for rate-independent materials the linearizations are
@f{align*}{
\mathcal{H}^{\text{tot}} \left( \mathbf{C}, \boldsymbol{\mathbb{H}} \right)
- &= 2 \frac{d^{2} \psi_{0} \left( \mathbf{C}, \boldsymbol{\mathbb{H}} \right)}{d \mathbf{C} \otimes d \mathbf{C}} , \\
+ &= 4 \frac{d^{2} \psi_{0} \left( \mathbf{C}, \boldsymbol{\mathbb{H}} \right)}{d \mathbf{C} \otimes d \mathbf{C}} , \\
\mathbb{D} \left( \mathbf{C}, \boldsymbol{\mathbb{H}} \right)
&= -\frac{d^{2} \psi_{0} \left( \mathbf{C}, \boldsymbol{\mathbb{H}} \right)}{d \boldsymbol{\mathbb{H}} \otimes d \boldsymbol{\mathbb{H}}} , \\
\mathfrak{P}^{\text{tot}} \left( \mathbf{C}, \boldsymbol{\mathbb{H}} \right)
// @cite Pelteret2018a, equation 29). Thereafter we can compute the
// dissipative component of the energy density function; its expression
// is stated in @cite Pelteret2018a (equation 28), which is a
- // straight-forward extension of an energy density function formulated by
+ // straight-forward extension of an energy density function formulated in
// @cite Linder2011a (equation 46).
const Differentiation::SD::Expression f_mu_v_sd =
1.0 +
// This means that deriving any function $f = f(\mathbf{C}, \mathbf{Q})$
// with respect to $\mathbf{C}$ will return partial derivatives
// $\frac{\partial f(\mathbf{C}, \mathbf{Q})}{\partial \mathbf{C}}
- // \Big\vert_{\mathbf{C}_{v}}$ as opposed to the total derivative
+ // \Big\vert_{\mathbf{Q}}$ as opposed to the total derivative
// $\frac{d f(\mathbf{C}, \mathbf{Q}(\mathbf{C}))}{d \mathbf{C}} =
// \frac{\partial f(\mathbf{C}, \mathbf{Q}(\mathbf{C}))}{\partial
- // \mathbf{C}} \Big\vert_{\mathbf{C}_{v}} + \frac{\partial f(\mathbf{C},
- // \mathbf{Q}(\mathbf{C}))}{\partial \mathbf{C}_{v}}
+ // \mathbf{C}} \Big\vert_{\mathbf{Q}} + \frac{\partial f(\mathbf{C},
+ // \mathbf{Q}(\mathbf{C}))}{\partial \mathbf{Q}}
// \Big\vert_{\mathbf{C}} : \frac{d \mathbf{Q}(\mathbf{C}))}{d
// \mathbf{C}}$.
//