# Why is del operator coordinate free?

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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Yu can think on the "Laplacian" operator as the trace of the Hessian and, as long it is a trace of an operator, it is coordinate free. – matgaio May 1 '12 at 21:12
There is a more intrinsic view of partial (and, more generally, directional) derivatives as linear functional on the space os smooth real functions. You can think this way: for each $p\in\mathbb{R}^n$ and $v\in\mathbb{R}^n$ define a linear functional $\frac{\partial}{\partial v}:C^\infty(\mathbb{R}^n)\rightarrow \mathbb{R}$ as $\frac{\partial}{\partial v}(f)=\frac{d}{dt}(f(\gamma(t)))_{t=0}$ where $\gamma:(-\epsilon,+\epsilon)\rightarrow\mathbb{R}^n$ is a smooth curve with $\gamma(0)=p$ and $\gamma'(0)=v$. Once you check this doesn't depends on the curve, this is intrinsic. – matgaio May 1 '12 at 21:24
Do you imply that del operator will not change under any coordinate change? Even from an orthogonal one to an oblique one? – Tim May 1 '12 at 21:33
The value that is returns when applyed to a function will not depend on the system of coordinates, once it can be calculated as the derivative of a composition. The expression in the coordinates you are using could change, but given $p$, $v$ and a function, the value $\frac{\partial}{\partial v}(f)$ in $p$ will not change – matgaio May 1 '12 at 21:41
That's the first derivative. I still have doubts about higher derivative. Laplacian is second derivative. – Tim May 1 '12 at 21:49