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In solving Laplace equation $\Delta u=0$, every textbook will tell you to transform $\Delta=\partial_{xx}+\partial_{yy}$ into polar coordinate form $\Delta_p$(What it looks like doesn't matter here). Then it sets $\Delta_p v(r,\theta)=0$. How can you be sure that $u(x,y)=v(r(x,y),\theta(x,y))$?

I think the problem does not only exist in Laplacian operator case. So my question is "is Del operator coordinate free, in what sense and why?"

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    That's the first derivative. I still have doubts about higher derivative. Laplacian is second derivative.2012-05-01

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