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Divergence

{title2=$\nabla\cdot\mathbf u=\frac1r\partial_r(ru_r)+\frac1r\partial_\theta u_\theta$}

For $\mathbf u=u_r\mathbf e_r+u_\theta\mathbf e_\theta$,
$$
\nabla\cdot\mathbf u
=\frac1r\frac{\partial(ru_r)}{\partial r}
+\frac1r\frac{\partial u_\theta}{\partial\theta}.
$$


Divergence in polar coordinates

Past exam of the mathematics course of the University of Cambridge / 2018 / ib / Paper 1 / 5D / Solution

Divergence in polar coordinates ( $\nabla \cdot u = \frac{1}{r} \partial_{r} (r u_{r}) + \frac{1}{r} \partial_{θ} u_{θ}$ )

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Divergence in polar coordinates (∇⋅u=r1​∂r​(rur​)+r1​∂θ​uθ​)

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Divergence in polar coordinates ( $\nabla \cdot u = \frac{1}{r} \partial_{r} (r u_{r}) + \frac{1}{r} \partial_{θ} u_{θ}$ )

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