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?


\begin{eqnarray} \int_{S(t)} {\rm d}\textbf{S} \cdot (\sigma \cdot\textbf{u}) &=& \int_{S(t)} {\rm d}S\;\textbf{n} \cdot (\sigma \cdot\textbf{u}) = \int_{V(t)}{\rm d}V\;\nabla\cdot(\sigma\cdot\textbf{u}) \\ &=&\int_{V(t)}{\rm d}V\; (\nabla\cdot\sigma)\cdot\mathbf{u} + \sigma:\nabla\mathbf{u} \end{eqnarray}

  • $\begingroup$ What is that extra nabla doing there? $\endgroup$ – user197848 Nov 15 '16 at 12:26
  • $\begingroup$ @HMPARTICLE Sorry, editing error $\endgroup$ – caverac Nov 15 '16 at 12:32
  • $\begingroup$ 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]? $\endgroup$ – user197848 Nov 15 '16 at 14:24
  • $\begingroup$ Absolutely, it is a matter of notation $\tau \mathbf{u} \equiv \tau \cdot \mathbf{u}$ $\endgroup$ – caverac Nov 15 '16 at 14:34
  • $\begingroup$ thanks! I was reluctant to use the identity for this very reason. $\endgroup$ – user197848 Nov 15 '16 at 19:40

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