How do I transform a cylindrical problem into a rectangle problem?

Question

Hi guys,
I have a rotation symmetrical PDE problem.
That means

$a \cdot \triangle u = b \cdot \frac{du}{dt}$

becomes to:

$\frac{a}{r} \cdot \dfrac{\text{d}(r \cdot \frac{\text{d}u}{\text{d}r})}{\text{d}r} + a \frac{\text{d}²u}{\text{d}z²} = b \cdot \frac{\text{d}u}{\text{d}t}$ because $\frac{\text{d} u}{\text{d}\phi} = 0$

My question: How does that change the weak formulation? In the tutorial carthesian coordinates are used which means

$ -\iiint\limits_\Omega v\triangle u\, \text{d}x \, \text{d}y \, \text{d}z = \iiint\limits_\Omega \nabla v \cdot \nabla u \, \text{d}x \, \text{d}y \, \text{d}z - \iint\limits_{\partial \Omega} \frac{\partial u}{\partial z} \cdot v\, \text{d}x \, \text{d}y$ (if the derivative of u to x and y on the boundaries is zero)

Unfortunately I dont really know how to use a testfunction on something that is not $\triangle u$.

Your help would be highly appreciated.

Comments

Hi Max,

please use Latex to format your question. E.g. $\triangle u$ will give you $\triangle u$.

Then, it looks like you first have to come to terms with the coordinates. It is not the problem that changes the Laplacian. It is (YOUR) choice of the coordinate system you describe your problem in.

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Hi Jan, i changed it to latex format. Thanks for your advice. I know it's my choice, but it is way faster to compute a 2D-solution than a 3D-solution. That's why I want to solve the problem in zylindrical coordinates, I just dont know how to transform the equation above.

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Finally I understood what I was doing there, could fix it on my own:

$\triangle_{\text{cylindrical}(r,z)} u = \frac{\partial^2 u}{\partial r^2} + \frac{\partial^2 u}{\partial z^2} + \frac{1}{r} \cdot \frac{\partial u}{\partial r} = \triangle_{\text{carthesian}(r,z)} u + \frac{1}{r} \cdot \frac{\partial u}{\partial r}$ and therefore
$-\iint_{\Omega} v \cdot (\frac{\partial^2 u}{\partial r^2} + \frac{\partial^2 u}{\partial z^2} + \frac{1}{r} \cdot \frac{\partial u}{\partial r}) \, \text{d}r \, \text{d}z = \iint_{\Omega} \frac{\partial v}{\partial r} \cdot \frac{\partial u}{\partial r} + \frac{\partial v}{\partial z} \cdot \frac{\partial u}{\partial z} - \frac{v}{r} \cdot \frac{\partial u}{\partial r}\, \text{d}r \, \text{d}z + \int_{\partial \Omega} v \cdot \frac{\partial u}{\partial n} \text{d}n$.