- #1

#### 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?

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?

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