Calculate divergence of vector fields

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Newcastle University Mathematics and Statistics 9 years, 1 month ago

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Name	Status	Author	Last Modified
Calculate divergence of vector fields	draft	Newcastle University Mathematics and Statistics	20/11/2019 14:51
Harry's copy of Calculate divergence of vector fields	draft	Harry Flynn	14/03/2018 14:19
Harry's copy of Calculate divergence of vector fields	draft	Harry Flynn	09/03/2018 14:27
Ugur's copy of Calculate divergence of vector fields	Ready to use	Ugur Efem	07/06/2023 16:47

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Question statement

For each of the following vector fields $\boldsymbol{u}$ , find the divergence $\boldsymbol{\nabla\cdot u}$ .

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			Name	Type	Generated Value

p1	integer	5
a1	number	6
a3	number	4
a2	number	8
b1	number	4
b3	number	-4
c3	number	-8
c1	number	-5

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"Part b)" - prompt

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Part c) Gap-fill
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$\boldsymbol{u}=\pmatrix{(\simplify[std]{{a1}*x+{b1}*y+{c1}*z})(\simplify[std]{{b1}*y-{c1}*z}),(\simplify[std]{{a1}*y+{b1}*z+{c1}*x})(\simplify[std]{{b1}*z-{c1}*x}),(\simplify[std]{{a1}*z+{b1}*x+{c1}*y})(\simplify[std]{{b1}*x-{c1}*y})}$ .

$\boldsymbol{\nabla\cdot u}=$ Unnamed gap.

Advice

The divergence of a vector field $\boldsymbol{u}=\pmatrix{u_1,u_2,u_3}$ is given by

$\boldsymbol{\nabla\cdot u}=\frac{\partial u_1}{\partial x}+\frac{\partial u_2}{\partial y}+\frac{\partial u_3}{\partial z}.$

The variables $x$ , $y$ , and $z$ appear in a cyclical manner in each of the three components of $\boldsymbol{u}$ . Once you have calculated $\frac{\partial u_1}{\partial x}$ , you can use cyclic permutations to determine the other two derivatives. Hence

$\begin{align}\frac{\partial u_1}{\partial x}&=\frac{\partial}{\partial x}\left(\simplify{({a1}*x+{b1}*y+{c1}*z)*({b1}*y-{c1}*z)}\right)\\&=\simplify{{a1}*({b1}*y-{c1}*z)},\end{align}$

and so, cyclically permuting the variables,

$\frac{\partial u_2}{\partial y}=\simplify{{a1}*({b1}*z-{c1}*x)}$

and

$\frac{\partial u_3}{\partial z}=\simplify{{a1}*({b1}*x-{c1}*y)}.$

Finally, adding the components together gives the divergence

$\boldsymbol{\nabla\cdot u}=\simplify[std]{{a1}*({b1}*y-{c1}*z)+{a1}*({b1}*z-{c1}*x)+{a1}*({b1}*x-{c1}*y)}=\simplify{{a1*b1-a1*c1}*x+{a1*b1-a1*c1}*y+{a1*b1-a1*c1}*z}.$

As in part a) the variables $x$ , $y$ , and $z$ appear cyclically in each component of $\boldsymbol{u}$ , so we only need calculate one derivative explicitly, then use cyclic permutations to calculate the other two. Hence

$\begin{align}\frac{\partial u_1}{\partial x}&=\frac{\partial}{\partial x}\left\{\left(x^\var{p1}+y^\var{p1}\right)\left(\simplify{{a2}*y^{p1-1}-{a2}*z^{p1-1}}\right)\right\}\\&=\simplify{{p1}*x^{p1-1}}\left(\simplify{{a2}*y^{p1-1}-{a2}*z^{p1-1}}\right),\end{align}$

and by symmetry

$\frac{\partial u_2}{\partial y}=\simplify{{p1}*y^{p1-1}}\left(\simplify{{a2}*z^{p1-1}-{a2}*x^{p1-1}}\right),$

and

$\frac{\partial u_3}{\partial z}=\simplify{{p1}*z^{p1-1}}\left(\simplify{{a2}*x^{p1-1}-{a2}*y^{p1-1}}\right).$

Finally, adding the derivatives together gives the divergence

$\boldsymbol{\nabla\cdot u}=\simplify{{p1}*x^{p1-1}}\left(\simplify{{a2}*y^{p1-1}-{a2}*z^{p1-1}}\right)+\simplify{{p1}*y^{p1-1}}\left(\simplify{{a2}*z^{p1-1}-{a2}*x^{p1-1}}\right)+\simplify{{p1}*z^{p1-1}}\left(\simplify{{a2}*x^{p1-1}-{a2}*y^{p1-1}}\right)=0.$

First note that $f_1$ does not depend on $z$ , $f_2$ does not depend on $y$ , and $f_3$ does not depend on $z$ . This makes the differentiation very straight forward, and hence

$\begin{align}\boldsymbol{\nabla\cdot u}&=\frac{\partial}{\partial x}\left(\simplify{{a3}*x}+f_1(y,z)\right)+\frac{\partial}{\partial y}\left(\simplify{{b3}*y}+f_2(x,z)\right)+\frac{\partial}{\partial z}\left(\simplify{{c3}*z}+f_3(x,y)\right)\\&=\simplify[all,!collectNumbers]{{a3}+{b3}+{c3}}\\&=\var{a3+b3+c3}.\end{align}$

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Calculate divergence of vector fields

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Newcastle University Mathematics and Statistics 9 years, 1 month ago

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Generated value: integer

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