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I'm a little confused about this: if $f, g: \mathbb{R}^{3} \rightarrow \mathbb{R}$ are differentiable functions, calculate $\mathrm{div}(f \cdot \mathrm{grad} \hspace{2pt} g)$. I try expanding definitons but I get stuck at $\mathrm{div} \hspace{2pt} (\sum_{i } f \frac{\partial g}{\partial x_{i}} e_{i})$. How can I proceed?

EDIT:

I'm sorry I could not add a comment below.

To "Chris K": Notice that both functions are scalar valued.

EDIT 2:

To "Chris K" and "Spencer": You're right, I edited the title so the dot disappeared. Thanks.

Lucas Willy
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  • So, $f$ is a vector-valued function and $g$ is a scalar function... but then you're taking the divergence of a scalar, which is nonsense. – Christopher K Nov 21 '13 at 01:36
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    f and g are both scalar valued functions (Their image is in R) so what he wrote isn't nonsense. A scalar times a gradient is a vector. – Spencer Nov 21 '13 at 01:41
  • @Spencer, I thought he meant the "dot" product. In fact, the "dot" there is kind of confusing... – Christopher K Nov 21 '13 at 01:42
  • @Chris K, yea I thought so too at first maybe the question should be edited to avoid the confusion with notation. – Spencer Nov 21 '13 at 01:43

1 Answers1

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$\mathrm{div}(f \, \nabla g) = \sum_i \frac{\partial}{\partial x_i} (f \, \frac{\partial g}{\partial x_i})= \sum_i \frac{\partial f}{\partial x_i} \frac{\partial g}{\partial x_i}+ f \frac{\partial^2 g}{\partial x_i^2} = \nabla f \cdot \nabla g + f \, \Delta g$

sjvega
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