What does the notation Div with an arrow on top mean in PDE studies?

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I encountered this notation in my PDE studies, but I was confused by what it meant. What does the notation Div with an arrow on top (of the word) mean? I figured meant the divergence of something...but am curious about its exact meaning.
 
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island-boy said:
I encountered this notation in my PDE studies, but I was confused by what it meant. What does the notation Div with an arrow on top (of the word) mean? I figured meant the divergence of something...but am curious about its exact meaning.
The divergence operator in Cartesian coordinates is
{\vec \nabla} = {\hat i} {\partial \over \partial x} + {\hat j} {\partial \over \partial y}+{\hat k} {\partial \over \partial z}
 
so div \rightharpoonup = {\vec \nabla} right?
hmm, makes sense. Thanks!
 
Yes- since div f is a vector-valued function, many (but not all) texts use
{\vec \nabla}
to represent it. Other texts use just \nabla.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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