Fluid Dynamics Proof

Question

By considering the Product Rule for $$\frac{\partial}{\partial x_j}\left(F_iF_j\right)$$ Where $F_i = [\underline{F}]_i$, show that if $\underline{\nabla}\cdot\underline{F} = 0$ then $$\iiint_{\Sigma}\left(\underline{F}\cdot\underline{\nabla}\right)\underline{F}\,dV=\iint_S \left(\underline{F}\cdot\underline{\hat{n}}\right)\underline{F}\,dA$$ For the first part, using the product rule, I have an answer of $$F_i \frac{\partial F_j}{\partial x_j} + \frac{\partial F_i}{\partial x_j}F_j = \left(\underline{\nabla}\cdot\underline{F}\right)\underline{F} + \left(\underline{F}\cdot\underline{\nabla}\right)\underline{F}$$ I don't know what else I need to do for this question so can I please have some guidence?

Answer 1

Hint. One of the terms from the product rule is 0 by assumption. Hence the integral you're looking for is actually an integral of a divergence, so use the Divergence Theorem.

In full: For each $i$, $$ ∭_{\Sigma} (F\cdot \nabla F)_i \, dV = ∭_{\Sigma} \nabla \cdot (FF_i)\, dV = ∬_SF_i F\cdot n \, dA $$ which is the result.

Answer 2

Okay I have used your answer to help me achieve this: $$\iiint_\Sigma (\underline{F}\cdot\underline{\nabla})\underline{F}\,dV = \underbrace{\iiint_\Sigma \frac{\partial F_i}{\partial x_j}F_j\,dV = \iint_S F_iF_jn_j\,dA}_{\Large\text{By Divergence Theorem}} = \iint_S (\underline{F}\cdot\underline{\hat{n}})\underline{F}\,dA$$