# Laplacian of a Function depending on r in Polar Coordinates

From a bank of exams:

Let $u(x,y) = f(r)$ be a smooth function in the plane that depends only on $r = \sqrt{x^2 + y^2}$. Compute $\Delta u = u_{xx} + u_{yy}$ in terms of $f$ and its derivatives.

Wikipedia states that the Laplace operator in polar coordinates is $$\Delta f = \frac{1}{r}\frac{\partial f}{\partial r} \left( r \frac{\partial f}{\partial r} \right) + \frac{1}{r^2}\frac{\partial^2 f}{\partial \theta^2},$$ which I suppose I could memorize directly, but I thought there might be an easier way.

I tried to prove this directly, by thinking that $$u_{xx} = \frac{d^2f}{dr^2} \frac{\partial r}{\partial x} + \frac{df}{dr} \frac{\partial ^2r}{\partial x^2}$$ and $$u_{yy} = \frac{d^2f}{dr^2} \frac{\partial r}{\partial y} + \frac{df}{dr} \frac{\partial ^2r}{\partial y^2}.$$ But then I get stuck at $$u_{xx} + u_{yy} = \frac{d^2f}{dr^2} \frac{x+y}{\sqrt{x^2+y^2}} + \frac{df}{dr}\frac{1}{\sqrt{x^2+y^2}} = \frac{d^2f}{dr^2} \frac{r(\cos \theta + \sin \theta)}{r} + \frac{df}{dr}\frac{1}{r}.$$ Any idea on where I'm going wrong? It looks like I need $\displaystyle{\frac{r(\cos \theta + \sin \theta)}{r} = 1}$.

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Sorry I can't follopwup for a while, but you can see this as an issue of finding the representation of/for a function in a given coordinate system, i.e., functions have expressions depending on the manifold charts where the function takes its values. See, e.g., John Lee's 1st Chapter on Smooth Manifolds book.HTH –  gary Apr 4 '11 at 8:04

we have $$u_x=f_rr_x, u_{xx}=f_rr_{xx}+r_x(f_{rr}r_x)\text{ similarly for } y,$$ $$u_{xx}+u_{yy}=f_rr_{xx}+r_x(f_{rr}r_x)+f_rr_{yy}+r_y(f_{rr}r_y) =f_{rr}(r_x^2+r_y^2)+f_r(r_{xx}+r_{yy}).$$ with $$r_x=\frac{x}{\sqrt{x^2+y^2}}, r_{xx}=\frac{y^2}{(x^2+y^2)^{3/2}} \text{ similarly for } y$$ we have $$u_{xx}+u_{yy}=f_{rr}+\frac{1}{r}f_r.$$ of course if $f$ depends on $\theta$ it gets more complicated
You missed the "inner derivative" in one factor of the product when taking the second derivative. It should be $$u_{xx} = \frac{d^2f}{dr^2} \left(\frac{\partial r}{\partial x}\right)^2 + \frac{df}{dr} \frac{\partial ^2r}{\partial x^2}.$$
Aha! so I forgot to use the chain rule and introduce a second $\partial r/\partial x$ when I was differentiating $df/dx$. Thank you, much appreciated. I wish I could mark both answers as correct... –  Michael Chen Apr 4 '11 at 15:24