Nabla operations, vector calculus problem

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Homework Help Overview

The discussion revolves around vector calculus, specifically the nabla operator and its application in cylindrical coordinates. The original poster is attempting to calculate the divergence and Laplacian of a velocity field described by a given equation, comparing their results with those of their teacher.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster describes their understanding of the nabla operator and attempts to calculate both the divergence and Laplacian of the velocity field. They express confusion over discrepancies between their results and their teacher's. Other participants question the assumptions made regarding the coordinate system and suggest reviewing the definitions of divergence and Laplacian in cylindrical coordinates.

Discussion Status

Participants are actively exploring the problem, with some providing clarifications about the differences in calculations based on coordinate systems. There is an acknowledgment of the need for further examination of the definitions and the original poster's notation.

Contextual Notes

There is a mention of potential confusion arising from the use of cylindrical versus Cartesian coordinates, as well as the implications of non-constant unit vectors in cylindrical coordinates. The original poster also notes their uncertainty regarding the steps taken by their teacher.

Karl Karlsson
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Homework Statement
A liquid flows in a cylindrical tube with the z-axis as the axis of symmetry and radius R. The flow rate of the liquid is given by $$ \vec v = \frac {p_0} {4lμ} (R^2-ρ^2)$$
where ##\frac {p_0} {l}## is the pressure drop in the pipe per unit length, ##μ## is the viscosity of the liquid and ##ρ## is the radial cylinder coordinate. Calculate ##∇\cdot\vec v## and ##∇^2\cdot\vec v##
Relevant Equations
$$ \vec v = \frac {p_0} {4lμ} (R^2-ρ^2)$$
Here is how my teacher solved this:
Skärmavbild 2020-08-31 kl. 00.10.27.png

I understand what the nabla operator does, ##∇\cdot\vec v## means that I am supposed to calculate ##\sum_{n=1}^3\frac {d\vec v} {dx_n}## where ##x_n## are cylindrical coordinates and ##\vec e_3 = \vec e_z##. I understand why ##∇\cdot\vec v = 0##, I would get the same answer as my teacher if I used ##\sum_{n=1}^3\frac {d\vec v} {dx_n}## to calculate it. However when I calculate ##∇^2\cdot\vec v## I don't get the same answer. I would get ##∇^2\cdot\vec v = (∇\cdot (∇v_1),∇\cdot (∇v_2),∇\cdot (∇v_3)) = \frac {p_0} {4lμ}\cdot (0,0, ∇\cdot (∇(R^2-ρ^2))) =\frac {p_0} {4lμ}\cdot (0,0, \frac {\partial^2 (R^2-ρ^2)} {\partial ρ^2})##. Which is not equal to what my teacher got. Did I miss something? Also could someone explain the steps he took to arrive at his answer?

Thanks in advance!
 
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Karl Karlsson said:
Homework Statement:: A liquid flows in a cylindrical tube with the z-axis as the axis of symmetry and radius R. The flow rate of the liquid is given by $$ \vec v = \frac {p_0} {4lμ} (R^2-ρ^2)$$
where ##\frac {p_0} {l}## is the pressure drop in the pipe per unit length, ##μ## is the viscosity of the liquid and ##ρ## is the radial cylinder coordinate. Calculate ##∇\cdot\vec v## and ##∇^2\cdot\vec v##
Relevant Equations:: $$ \vec v = \frac {p_0} {4lμ} (R^2-ρ^2)$$

Here is how my teacher solved this:
View attachment 268604
I understand what the nabla operator does, ##∇\cdot\vec v## means that I am supposed to calculate ##\sum_{n=1}^3\frac {d\vec v} {dx_n}## where ##x_n## are cylindrical coordinates and ##\vec e_3 = \vec e_z##. I understand why ##∇\cdot\vec v = 0##, I would get the same answer as my teacher if I used ##\sum_{n=1}^3\frac {d\vec v} {dx_n}## to calculate it. However when I calculate ##∇^2\cdot\vec v## I don't get the same answer. I would get ##∇^2\cdot\vec v = (∇\cdot (∇v_1),∇\cdot (∇v_2),∇\cdot (∇v_3))
The last expression above doesn't seem right to me, but I am somewhat rusty with this stuff. The part that bothers me is where you're taking the gradient of the individual components ##v_1, v_2, v_3##.
Karl Karlsson said:
= \frac {p_0} {4lμ}\cdot (0,0, ∇\cdot (∇(R^2-ρ^2))) =\frac {p_0} {4lμ}\cdot (0,0, \frac {\partial^2 (R^2-ρ^2)} {\partial ρ^2})##. Which is not equal to what my teacher got. Did I miss something? Also could someone explain the steps he took to arrive at his answer?

Thanks in advance!
 
Karl Karlsson said:
I understand what the nabla operator does, ##∇\cdot\vec v## means that I am supposed to calculate ##\sum_{n=1}^3\frac {d\vec v} {dx_n}## where ##x_n## are cylindrical coordinates and ##\vec e_3 = \vec e_z##.
Not exactly. If you're using cartesian coordinates, then
$$\nabla \cdot \vec v = \frac{\partial v_x}{\partial x} + \frac{\partial v_y}{\partial y} +\frac{\partial v_z}{\partial z}.$$ If you're using cylindrical coordinates, however, the divergence takes on a more complicated form. See, for example, https://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates.

I understand why ##∇\cdot\vec v = 0##, I would get the same answer as my teacher if I used ##\sum_{n=1}^3\frac {d\vec v} {dx_n}## to calculate it. However when I calculate ##∇^2\cdot\vec v## I don't get the same answer. I would get ##∇^2\cdot\vec v = (∇\cdot (∇v_1),∇\cdot (∇v_2),∇\cdot (∇v_3)) = \frac {p_0} {4lμ}\cdot (0,0, ∇\cdot (∇(R^2-ρ^2))) =\frac {p_0} {4lμ}\cdot (0,0, \frac {\partial^2 (R^2-ρ^2)} {\partial ρ^2})##. Which is not equal to what my teacher got. Did I miss something? Also could someone explain the steps he took to arrive at his answer?
I think if you visit that Wikipedia page and see what the Laplacian is in cylindrical coordinates, it'll answer your question. Your notation is a bit off too. Note that your professor wrote ##\nabla^2 \vec v##, not ##\nabla^2 \cdot \vec v##.
 
This is another consquence of \mathbf{e}_\rho and \mathbf{e}_\phi not being constant.

In this case it's not too difficult to calculate \nabla \cdot \mathbf{v} and \nabla^2 \mathbf{v} in cartesian coordinates using <br /> \nabla (\rho^2) = 2x\mathbf{e}_x + 2y\mathbf{e}_y and <br /> \nabla \cdot (f\mathbf{g}) = f\nabla \cdot \mathbf{g} + \mathbf{g} \cdot \nabla f.<br />
 

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