Ytukyg

Question

I want to solve this differential equation for $P : mathbb{R} to mathbb{A}^2$, a plane affine curve.

$ P'''(t) = frac{P'(t)}{t^2}$

Someone recognize this equation? Is a famous curve? Is algebraic? I think it will be a curve invariant under the action of a 1-parameter subgroup of the affine group.

Thanks in advance.

Summary

I considered again the results found in

Affine arc length

I started studying some specials curves:

1. $omega^1 = 0$. These are points and are invariant under the action of a 4-parameter subgroup of the affine group. For example for $x^2+y^2=0$
   $$
   begin{bmatrix}
   1 & 0 & 0 \
   0 & a & b \
   0 & c & d \
   end{bmatrix}
   $$




2. $omega_1^2 = 0$. These are straight lines and are invariant under the action of a 4-parameter subgroup of the affine group. For example for $y = 0$
   $$
   begin{bmatrix}
   1 & 0 & 0 \
   x & a & b \
   0 & 0 & d \
   end{bmatrix}
   $$




3. $omega_2^1 = 0$. These are parabolas and are invariant under the action of a 2-parameter subgroup of the affine group. For example for $y = x^2$
   $$
   begin{bmatrix}
   1 & 0 & 0 \
   x & a & 0 \
   x^2 & 2ax & a^2 \
   end{bmatrix}
   $$




4. $k = 0$. These are ellipses and hyperbolas and are invariant under the action of a 1-parameter subgroup of the affine group. For example for $x^2+y^2=1$
   $$
   begin{bmatrix}
   1 & 0 & 0 \
   0 & costheta & sintheta \
   0 & -sintheta & costheta \
   end{bmatrix}
   $$
   and for $xy=1$
   $$
   begin{bmatrix}
   1 & 0 & 0 \
   0 & a & 0 \
   0 & 0 & 1/a \
   end{bmatrix}
   $$




5. $k = K$. I got the differential equation $ P'''(t) = frac{P'(t)}{t^2}$, but I don't know if it is a well-known curve and especially if it is still algebraic.

I relaxed a bit the conditions about the parametrization.

I searched for P such that $P''' = alpha P'' + beta P'$ with $alpha, beta$ constant. So $lambda^{-1} = sqrt{alpha + frac{2}{9}beta^2}$. In particular I can choose $lambda = 1, i$. For example, if I choose $lambda =1$, I get $alpha = 1 - frac{2}{9}beta^2$ and $kappa = frac{beta}{3}$ and the equation $P''' - beta P'' - (1 - frac{2}{9}beta^2)P' = 0$ and that's easy to solve. So I have found:

1. 
   

   $dsigma^2 = 1$

   
   
   
   
   
   • 
     $P = (e^{frac{3k-sqrt{k^2+4}}{2}t}, e^{frac{3k+sqrt{k^2+4}}{2}t})$ for $|k| neq frac{1}{sqrt{2}}$
     
   
   
   • 
     $P = (t, e^{pmfrac{3}{sqrt{2}}t})$ for $|k| = frac{1}{sqrt{2}}$
     
   
   
   
   


2. 
   

   $dsigma^2 = -1$

   
   
   
   
   
   • 
     $P = (e^{frac{3}{2}kt}cos{sqrt{4-k^2}t}, e^{frac{3}{2}kt}sin{sqrt{4-k^2}t}$) for $|k| < 2$
     
   
   
   • 
     $P = (e^{pm3t}, te^{pm3t})$ for $|k| = 2$
     
   
   
   • 
     $P = (e^{frac{3k-sqrt{k^2-4}}{2}t}, e^{frac{3k+sqrt{k^2-4}}{2}t})$ for $|k| > 2$
     
   
   
   
   

So every curve with constant affine curvature is congruent to one of

1. 
   $P = (t, t^{mu})$ with $mu ne frac{1}{2}, 2$ (from 1.1 and 2.3)


2. 
   $P = (t, e^{pm t})$ (from 1.2)


3. 
   $P = (e^t, pm te^t)$ (from 2.2)


4. 
   $P = (e^{lambda t}cos t, e^{lambda t}sin t) $ (from 2.1)

The algebraic one are only of type 1 with $mu in mathbb{Q}$ (or type 4 with $lambda = 0$)