# Possible misuse of the Divergence Theorem.

I am trying to show that

$$\int_{S(t)} (\textbf{n} \cdot \sigma ) \cdot\textbf{u} \ dS = \int_{V(t)}(\nabla\cdot\sigma) \cdot \textbf{u} \space + \sigma : \nabla \textbf{u} \space dV$$

where $\textbf{u,n}$ are vectors and $\sigma$ is a symmetric second order tensor.

Using the divergence theorem

$$\int_{S(t)} (\textbf{n} \cdot \sigma ) \cdot\textbf{u} \ dS = \int_{V(t)} (\nabla \cdot \sigma ) \cdot\textbf{u} \ dV$$

Now I have this identity:

$$\tau:\nabla \textbf{u} + \textbf{u}\cdot(\nabla \cdot \tau) = \nabla\cdot(\tau\textbf{u})$$

$\tau$ is a symmetric second order tensor.

This leads me to believe that I have done something wrong. Because I want the two terms on the left inside the volume integral, have I misused the divergence theorem?

• I see that you have used the identity, I was reluctant to use it because we have $\nabla \cdot (\tau \textbf{u})$ not $\nabla \cdot (\tau \cdot \textbf{u})$. What is the difference? $\tau \textbf{u}$ is a vector [matrix-vector multiplication], but I have not seen $\tau \cdot \textbf{u}$ before. Is that also [matrix-vector multiplication]? – user197848 Nov 15 '16 at 14:24
• Absolutely, it is a matter of notation $\tau \mathbf{u} \equiv \tau \cdot \mathbf{u}$ – caverac Nov 15 '16 at 14:34