# Vector fields for volumetric-deviatoric decomposition

The strain tensor $\epsilon(u) = \frac12 (\nabla u + (\nabla u)^T)$ in linear elasticity can be decomposed additively into volumetric and deviatoric strains \begin{gather*} \epsilon_D(u) &= \epsilon(u) - \tfrac13 \nabla \cdot u \;\mathbb 1 = \begin{cases} \frac12\begin{pmatrix} \partial_1u_1-\partial_2u_2 & \partial_2u_1+\partial_1u_2 \\ \partial_2u_1+\partial_1u_2 & \partial_2 u_2 - \partial_1 u_1 \end{pmatrix} & \text{in two dimensions}, \\ \begin{pmatrix} \frac{2\partial_1u_1 - \partial_2 u_2 - \partial_3 u_3}3 & \frac{\partial_1u_2+\partial_2u_1}2 & \frac{\partial_1u_3+\partial_3u_1}2 \\ \frac{\partial_1u_2+\partial_2u_1}2 &\frac{2\partial_2u_2 - \partial_1 u_1 - \partial_3 u_3}3 & \frac{\partial_2u_3+\partial_3u_2}2 \\ \frac{\partial_1u_3+\partial_3u_1}2 & \frac{\partial_2u_3+\partial_3u_2}2 &\frac{2\partial_3u_3 - \partial_1 u_1 - \partial_2 u_2}3 \end{pmatrix}& \text{in three dimensions}, \end{cases} \end{gather*} with $\epsilon_D(u):\mathbb 1 = 0$. Thus, Lamé-Navier equations in weak form decouple nicely into $$2 \mu (\epsilon_D(u),\epsilon_D(v)) + (2\mu+\lambda) (\nabla \cdot u, \nabla \cdot v) = (f,v)$$

First question: is there a decomposition similar to Helmholtz decomposition of an arbitrary vector function $u\in C^1$ into $u=u_D+u_V$, such that $$\epsilon_D(u_V) = 0 \quad \wedge \quad \nabla\cdot u_D = 0?$$ Second question: is there such a decomposition of $H^1_0(\Omega)$ into closed subspaces for suitable domains $\Omega$?

• If I understand your notation correctly, $\epsilon_D(u_V)=0$ would locally imply that $i$-th component of $u_V$ is a function of $i$-th coordinate only. Thus $\nabla u_V=f(x)+g(y)+h(z)$ and definitely is not enough to cancel an arbitrary $\nabla u$. Commented Mar 22, 2016 at 6:47
• @PeterKravchuk: I just added the deviatoric strain in Cartesian coordinates. Your suggestion eliminates the off-diagonal terms, but then the diagonals imply $f(x)=g(y)=h(z)$ in every point, and thus $u_V$ linear Commented Mar 22, 2016 at 23:57