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Suppose I have the boundary condition $u_y(x,0) = 0$. How can I transform it into polar coordinates?

$$ x= r\cos(\theta), y = r\sin(\theta), $$ so $$ \frac{\partial u}{\partial y} = ? $$

I'm a little confused since I don't think I can use the chain rule here.

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  • $\begingroup$ Real my answer to your previous question. $\endgroup$
    – Julián Aguirre
    Feb 5, 2015 at 16:34

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By the chain rule you have $$\frac{\partial u}{\partial y} = \frac{\partial u}{\partial r} \cdot \frac{dr}{dy} + \frac{\partial u}{\partial \theta} \cdot \frac{d\theta}{dy}$$

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  • $\begingroup$ so $\frac{\partial r}{\partial y} = \left(\frac{\partial y}{\partial r}\right)^{-1} = \frac{1}{\sin\theta}$? $\endgroup$
    – user90593
    Feb 5, 2015 at 16:33
  • $\begingroup$ Yes. In the end, you'll have an equation of the form $A \frac{\partial u}{\partial r} + B \frac{\partial u}{\partial \theta} = 0$. This equation holding on the boundary of the problem (which you can no longer simply describe as the line $y = 0$) is equivalent to the original boundary condition. $\endgroup$
    – nukeguy
    Feb 5, 2015 at 22:02

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