What-is-the-general-expression-of-the-divergence-divV-

Question Number 157199 by Fresnel last updated on 20/Oct/21

W h a t i s t h e g e n e r a l e x p r e s s i o n o f t h e d i v e r g e n c e (d i v \overset{⇢}{V})

Answered by TheHoneyCat last updated on 21/Oct/21

It depends on your level :

1) If (t o y o u) \vec{V} is a tridimentionnal vector

\vec{V} = V_{x} {\vec{e}}_{x} + V_{y} {\vec{e}}_{y} + V_{z} {\vec{e}}_{z}

then : div \vec{V} = \frac{\partial V_{x}}{\partial x} + \frac{\partial V_{y}}{\partial y} + \frac{\partial V_{z}}{\partial z}

2) \vec{V} is a function from R^{n} to R^{n}

with : denoting V_{i} its coordinate - functions

(i e t h e f u n c t i o n V_{i} f r o m R^{n} t o R s u c h t h a t

\forall i \in [∣ 1, n ∣] \vec{V} . {\vec{e}}_{i} = V_{i})

Then : div \vec{V} = \underset{i = 1}{\sum^{n}} \frac{\partial V_{i}}{\partial x_{i}}

3) If \vec{V} is a differentiable map of R^{n} \to R^{n}

(or any set isomorphic to it)

its divergence is defined as above

i t m i g h t n o t l o o k l i k e t h e r e i s a d i f f e r e n c e

f r o m 2) b u t t h e f a c t t h a t i t i s d i f f e r e n t a b l e

a s s u r e s t h a t t h e d i v e r g e n c e d o e s n o t d e p e n d

o n t h e c h o i c e o f c o o r d i n a t e s, h e n c e i t i s a

g e n e r a l o p e r a t o r o n t h e s e t o f d i f f e r e n t i b l e

f u n c t i o n s, a n d n o l o n g e r a f o r m u l a t h a t r e f e r s

t o “ {a c t u a l}^{″} c o o r d i n a t e s

Commented by TheHoneyCat last updated on 21/Oct/21

Some people define

\vec{▽} := \underset{i = 1}{\sum^{n}} \frac{\partial}{\partial x_{i}} {\vec{e}}_{i}

and say that div \vec{F} := \vec{▽} . \vec{F}

this would be equivalent to 2)

but a tiny bit more “ {p o w e r f u l}^{″} from a

computationnal point of vue, for you can

learn its “ e x p r e s s i o n i n o t h e r {c o o r d i n a t e s}^{″}

even some improper ones .