jedishrfu said:
Yes, we'd be curious how you came up with the equation.
Ok. I was reading about
Hausdorff dimension, and I wanted to find a continuous, compact surface, without holes, with dimension 1 \leq r \leq 2.
So, the area function A(x), to have a Hausdorff dimension r, must satisfy this equation: \epsilon A_{(\frac{x}{\epsilon})} = \epsilon^r A_{(x)}.
Equivalently:
A_{(\frac{x}{\epsilon})} = \epsilon^{r-1} A_{(x)} [
I]
The derivative of A(x) is
A'_{(x)}=\lim_{\epsilon \to 1}\frac{A_{(x.\epsilon)}-A_{(x)}}{x.\epsilon-x} [
II]
Because [
I]
A_{(x.\epsilon)}=\epsilon^{1-r} A_{(x)}
So [
II] is
A'_{(x)}=A_{(x)} \lim_{\epsilon \to 1}\frac{\epsilon^{1-r}-1}{x.\epsilon-x} = A_{(x)} \frac{1-r}{x}
Solving it: A_{(x)}=c_1 x^{(1-r)}
I think that is the only solution for [
I], but do not have proof.
I now want this area to continuously transform from the real line (dimension r=1) to the complex plane (r=2), so I want a 2D surface curved in 3D space with area A(x).
I want to find the function radius R(x) whose
surface of revolution is A(x).
My equation is A_x =2\pi \int R_{(x)} \sqrt{1+\left( R'_{(x)}\right)^2} \, dx [
IV]
I will deal with integration limits later.
Deriving [
IV] and squaring both sides, I get:
\left (\frac {c_1 (1-r)}{2 \pi} \right)^2 x^{-2r} = R_{(x)} ^2 + \left ( R_{(x)} R'_{(x)} \right )^2
Replacing
R_{(x)}^2 =2y
R_{(x)}R'_{(x)} =y'
a= - \left (\frac {c_1 (1-r)}{2 \pi} \right)^2
b= - 2r
I get the differential equation I want to solve:
2y + y'^2 + a x^b=0
I had been vaguely told that should look on
Weierstrass[/PLAIN] p function, \wp and that it may have a closed solution for b \in \mathbb{Z}.
I found
this paper, about using the \wp function for solving differential equations, but I can't figure how to use it.
