Shifting polar functions vertically

[member 428835]

Hi PF!

I have a function that looks like this $$f(r,\theta) = \sinh (\omega \log (r))\cos(\omega(\theta - \beta))$$

You'll notice ##f## is harmonic and satisfies the BC's ##f_\theta(\theta = \pm \beta) = 0##. Essentially ##f## has no flux into the wall defined at ##\theta = \pm \beta##. So we can imagine the domain is a sort of groove with it's vertex centered at the origin. However, this groove actually needs to be centered at a different vertical location, ##(h,0)##.

I calculate the new function to be $$f(r,\theta) = \sinh \left(\frac{\omega}{2} \log (r^2-2hr\cos\theta + h^2)\right)\cos\left[\omega\left(\arctan \left(\frac{r\sin\theta}{r \cos \theta-h}\right)- \beta\right)\right]$$
but it's so messy. Any clean-up ideas or different approaches?

 

[haruspex]

I don't get the bit about f_θ(.,-β) being zero. Have I misunderstood the notation?

 

[member 428835]

I don't get the bit about f_θ(.,-β) being zero. Have I misunderstood the notation?

Sorry, what I mean is $$\left.\frac{\partial f}{\partial \theta} \right|_{\theta = \pm \beta} = 0$$
The vertex has angle ##2\beta##.

 

[haruspex]

Sorry, what I mean is $$\left.\frac{\partial f}{\partial \theta} \right|_{\theta = \pm \beta} = 0$$
The vertex has angle ##2\beta##.

Yes, that's what I thought you meant, but you seem to be saying this follows from

##f(r,\theta) = \sinh (\omega \log (r))\cos(\omega(\theta - \beta))##

I get ##\left.\frac{\partial f}{\partial \theta} \right|_{\theta = -\beta} = -\omega\sinh (\omega \log (r))\sin(\omega( - 2\beta))##

 

[member 428835]

Yes, that's what I thought you meant, but you seem to be saying this follows from

I get ##\left.\frac{\partial f}{\partial \theta} \right|_{\theta = -\beta} = -\omega\sinh (\omega \log (r))\sin(\omega( - 2\beta))##

Oh shoot, sorry, I forgot to say ##\omega = k \pi / \beta : k\in \mathbb N##.

 

[FactChecker]

Is there a reason that you can't just introduce new variables for the shift and leave the original equations essentially as-is? I think it would be much more clear. If I were to document or program that problem, I think that is how I would do it.

 

[member 428835]

Is there a reason that you can't just introduce new variables for the shift and leave the original equations essentially as-is? I think it would be much more clear. If I were to document or program that problem, I think that is how I would do it.

Can you elaborate?

See, this is one component to a very large problem I'm working on. The issue is, I am in two different coordinate systems that do not have the same center. Obviously in order to have them both work I need to choose one of the two coordinate systems. I know this one is by far easier (namely, I need to shift the vertex to ##(h,0)##).

I also think that when I shift the coordinate system, the BC's are no longer valid at ##\theta = \beta##. Instead, the boundaries are now valid along a line not passing through the origin. Specifically, the upper boundary line is parameterized as $$r = h \csc(\beta - \theta) \sin\beta$$
What do you think? Am I making an error by thinking I can simply shift ##f## vertically, evaluate ##f_\theta## along the shifted line, and still get a good answer? I tried doing that and I'm not getting 0.

 

[FactChecker]

My suggestion is just a cosmetic difference which I think would make it easier to understand what is going on. I would define the shifted coordinate ##(r',\theta') = (\sqrt {r^2-2hr\cos \theta + h^2}, \frac {r \sin \theta}{r \cos \theta - h})## and then use ##f(r', \theta')##.
I think that combining it all into one equation makes it messy and more difficult to understand. And I don't see any benefit from combining it.