Discretization in cylindrical coordinates, unit thickness for azimuth?

geetar_king
Messages
26
Reaction score
0
I am setting up a numerical simulation from a 2D discretization of the heat equation in cylindrical coordinates.

my spatial variables are radius (r), height (z), and azimuth (ø).

The assumption is that there is no gradient along the azimuth direction (if temperature is T then dT/dø = 0)

My problem is that I do not know how to handle the thickness Δø of my element. If I were to instead have a problem with dT/dz = 0 I know that I would use Δz = 1, but for this problem do I do the same and use Δø=1 or should it be a thickness based on r?
 
Physics news on Phys.org
I'm sorry you are not generating any responses at the moment. Is there any additional information you can share with us? Any new findings?
 
I've realized the azimuth drops out of the differential equation so my question no longer applies.
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
Back
Top