Timeline for An ordinary differential equation
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 5, 2017 at 13:09 | vote | accept | Martin Chow | ||
| May 5, 2017 at 0:52 | comment | added | Robert Bryant | @RobertIsrael: One thing you can say is that, when $a\not=b$, the curve $\bigl(x(t),y(t)\bigr)$ is an algebraic curve of genus $1$: This is because its parametrized as $(x,y)=\bigl(at/u+bt/v, \ c-(u+v)/(uv)\bigr)$ where $t$, $u$, and $v$ satisfy the relations $$u^2-at^2-1=v^2-bt^2-1=0,$$ so that this $tuv$-curve is the intersection of two quadrics in $3$-space that are in (projectively) general position whenever $ab(a{-}b)\not=0$, so that they define a space curve of genus $1$ and degree $4$. Consequently, I think that, whenever $ab(a{-}b)\not=0$, the $xy$-curve of degree $8$ is irreducible. | |
| May 4, 2017 at 19:05 | comment | added | Robert Israel | And here is an animation of the curve for $a=1, c=0$ with varying $b$. | |
| May 4, 2017 at 17:38 | comment | added | Robert Israel | @RobertBryant Yes, that's what the Groebner basis says. If, for example, $b=a$, then you get $ \left( a{c}^{2}-2\,acy+a{y}^{2}+{x}^{2} \right) \left( a{c}^{ 2}-2\,acy+a{y}^{2}+{x}^{2}-4\,a \right)$. | |
| May 4, 2017 at 16:34 | comment | added | Robert Bryant | @RobertIsrael: It's quite possible that, for certain values of $a$ and $b$, the polynomial of degree 8 will be reducible, so that there are solutions of lower degree, but I think that, for the generic values of $a$ and $b$, the curve will be irreducible of degree $8$. | |
| May 4, 2017 at 16:03 | comment | added | Robert Israel | You might find this graph of the curve in the case $a=1$, $b=2$, $c=3$ interesting. | |
| May 4, 2017 at 15:52 | comment | added | Robert Israel | We have the following polynomial relations among $x, y, t, u = \sqrt{1+at^2}$ and $v = \sqrt{1+bt^2}$: $u^2 - (1+at^2)=0$, $v^2 - (1+bt^2)=0$, $uvx - but - avt = 0$, $uvy - cuv+u+v=0$. Eliminate $u,v,t$ by computing a "plex" Groebner basis of the ideal generated by the left sides of these equations. The first element of the basis is a rather complicated polynomial of degree $8$ in $x$ and $y$. | |
| May 4, 2017 at 14:51 | comment | added | Willie Wong | Can you explain that final comment? I think I can work out why the curve is algebraic, but how did you constrain the degree to be 8? | |
| May 4, 2017 at 14:43 | comment | added | Robert Bryant | @WillieWong: Thanks! That's a neat demo. FYI: The solution curve is algebraic, of course. It might be worth mentioning that it is an algebraic curve in the $xy$-plane of degree 8 in general. | |
| May 4, 2017 at 14:33 | comment | added | Willie Wong | To see what the solution looks like: desmos.com/calculator/xsuj2he8ts | |
| May 4, 2017 at 14:13 | history | answered | Robert Bryant | CC BY-SA 3.0 |
