It might be very basic, but I am curious about the calculation of $(V\nabla V)\cdot n_{\rm out}$ where $V$ is defined on $\Omega$ and $n_{\rm out}$ is unit normal tangent vector on $\partial \Omega$.

  1. Like $\nabla \cdot (V\nabla V)=|\nabla V|^2+V\Delta V$, distribute the inner product and get $(V\nabla V)\cdot n_{\rm out}=(V\cdot n_{\rm out})\nabla V+V(\nabla V\cdot n_{\rm out})$.
  2. Simply write $V\nabla V\cdot n_{\rm out}$.

I know it is a very simple and easy question, but I am a little bit confused now.
Could you help me?

  • $\begingroup$ You can't have an inner product between $V$ and $n_{out},$ $V$ is a scalar. It would just be $V\nabla V \cdot n_{out},$ or $V(\nabla V \cdot n_{out})$ $\endgroup$ May 16 at 23:39
  • $\begingroup$ Sorry $V$ is a scalar function on $\Omega$? $\endgroup$ May 16 at 23:40
  • $\begingroup$ If it's a scalar then $(V\cdot n_{out})\nabla V$ doesn't make sense $\endgroup$ May 16 at 23:42
  • $\begingroup$ Oh, sorry! I forgot that $V$ is a scalar function. Thanks a lot! $\endgroup$
    – okw1124
    May 16 at 23:46


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