# Problem of the Week #10 - June 4th, 2012

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#### Chris L T521

##### Well-known member
Staff member
Thanks to those who participated in last week's POTW!! Here's this week's problem.

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Problem: Let $f,g : \mathbb{R}^3\rightarrow\mathbb{R}$ be differentiable functions. Given that $\mathbf{F}=\langle F_1,F_2,F_3\rangle$ is a differentiable vector field and $\text{div}\,(\mathbf{F})=\nabla\cdot\mathbf{F} = \dfrac{\partial F_1}{\partial x}+\dfrac{\partial F_2}{\partial y}+\dfrac{\partial F_3}{\partial z}$, show that $\text{div}\,(\nabla f\times\nabla g) = 0$.

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#### Chris L T521

##### Well-known member
Staff member
This problem was correctly answered by Sudharaka. You can find his solution below.

For any two vector fields, $$\mathbf{F}\mbox{ and }\mathbf{G}$$ the following vector identity holds.

$\text{div}\,(\mathbf{F}\times\mathbf{G}) = \operatorname{curl}(\mathbf{F})\cdot\mathbf{G} \;-\; \mathbf{F} \cdot \operatorname{curl}(\mathbf{G})$

$$\mbox{Replace, }\mathbf{F}\mbox{ by }\nabla f\mbox{ and }\mathbf{G}\mbox{ by }\nabla g\mbox{ and we get, }$$

$\text{div}\,(\mathbf{\nabla f}\times\mathbf{\nabla g}) = \operatorname{curl}(\mathbf{\nabla f})\cdot\mathbf{\nabla g} \;-\; \mathbf{\nabla f} \cdot \operatorname{curl}(\mathbf{\nabla g})$

The curl of the gradient of any scalar field is the zero vector. Therefore, $$\operatorname{curl}(\mathbf{\nabla f})=\mathbf{0}\mbox{ and }\operatorname{curl}(\mathbf{\nabla g})=\mathbf{0}$$

$\therefore\text{div}\,(\mathbf{\nabla f}\times\mathbf{\nabla g}) = 0$

Q.E.D.

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