of admissible fields $\mathbf{y}(\mathbf{x},t)$, $\beta (\mathbf{x},t)$, and $l(\mathbf{x},t)$ satisfying the constraints and . Together with the above formula for $\delta I$ and the arbitrariness of $\delta \mathbf{y}$, $\delta \beta $, and $\delta l$ in $\mathcal{V}$ as well as at $\partial _s$, equation yields $$\label{eq:2.15c}
\mathbf{P}\cdot \nabla =0,
\quad
-w_\beta +\nabla \cdot w_{\nabla \beta }=\frac{\partial D}{\partial \dot{\beta }}, \quad \nabla \cdot w_{\nabla l}=\frac{\partial D}{\partial \dot{l}},$$ which are subjected to the kinematic boundary conditions and , and the natural boundary conditions . The constitutive equations remain exactly the same as -. For the rate-independent theory the dissipation function can be assumed in a simple form $$D=K_1|\dot{\beta }|+K_2|\dot{l}|,$$ with $K_1$ and $K_2$ being positive constants. We call $K_1$ the critical resolved shear stress and $K_2$ the Peierls threshold. In this case equations $_{2,3}$ become $$-w_\beta +\nabla \cdot w_{\nabla \beta }=K_1\, {\text{sign}}\dot{\beta }, \quad \nabla \cdot w_{\nabla l}=K_2\, {\text{sign}}\dot{l}$$ for non-vanishing $\dot{\beta}$ and $\dot{l}$. These are the yield conditions for $\beta $ and $l$: $\dot{\beta}$ and $\dot{l}$ are non-zero if and only if $$|-w_\beta +\nabla \cdot w_{\nabla \beta }|=K_1, \quad |\nabla \cdot w_{\nabla l}|=K_2.$$ On the contrary, if the expressions on the left-hand sides are less than $K_1$ and $K_2$, the plastic slip cannot 