Normalized Differentiation

Take the derivative with respect to a,

$\displaystyle \frac{d}{da} \left( a^2f(ar)=2\pi f(r)\int^{aR}_0f(u)udu \right)\Rightarrow 2af(ar)+a^2 rf' (ar)=2\pi R(1-a)f(r)F(aR)+2\pi aR^2 f(r)f(aR)$ .

If we set $a=1$ the following differentiable equation can be obtained,

$\displaystyle 2f(r)+rf'(r)=2\pi R^2f(r)f(R)$ .

To solve first simplify to

$\displaystyle f'(r)+\frac{2-2f(R)\pi R^2}{r}f(r)=0$

whose solution is

$\displaystyle f(r)=Ce^{(2f(R)\pi R^2-2)\ln{r}}=Cr^{2f(R)\pi R^2-2}$ .
$f(r)=Cr^{n-2}, n=2f(R)\pi R^2$

To solve for C, use the regularizing function to obtain

$\displaystyle 1=C\int^{2\pi}_0\int^R_0 r^{n-1}dr d\theta\Rightarrow C=\frac{n}{2\pi R^n}$ .

Thus,

$\displaystyle f(r)=\frac{nr^{n-2}}{2\pi R^n}, 0<n<\infty$