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 $$V\nabla V\cdot n_{\rm out}$$.
• 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 $V(\nabla V \cdot n_{out})$ May 16 at 23:39
• Sorry $V$ is a scalar function on $\Omega$? May 16 at 23:40
• If it's a scalar then $(V\cdot n_{out})\nabla V$ doesn't make sense May 16 at 23:42
• Oh, sorry! I forgot that $V$ is a scalar function. Thanks a lot! May 16 at 23:46