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I am curious if there are any examples of functions that are solutions to second order differential equations, that also parametrize an algebraic curve.

I am aware that the Weierstrass $\wp$ - Elliptic function satisfies a differential equation. We can then interpret this Differential Equation as an algebraic equation, with solutions found on elliptic curves. However this differential equations is of the first order.

So, are there periodic function(s) $F(x)$ that satisfy a second order differential equation, such that we can say these parametrize an algebraic curve?

Could a Bessel function be one such solution?

T. Poindexter
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1 Answers1

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The functions $\sin(x)$ and $\cos(x)$ solve a second-order differential equation, namely $$u'' = -u,$$ and $(\cos(x), \sin(x))$ parametrizes the algebraic curve $x^2 + y^2 = 1$.

Daniel McLaury
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  • Yeah i was aware of that one thanks for the response. I was thinking of a complicated 2nd order linear equation like the following. ay'' + by' + cy = 0. I cant really find an instance where such a ODE can also be interpreted as an algebraic expression. – T. Poindexter May 07 '15 at 01:47
  • In the case of the $\wp$ function, you have an ODE of the form $$(\wp')^2 = 4\wp^3 - g_2 \wp - g_3,$$ so you get a solution of the form $\wp(t)$, and then you can consider the function sending $$t \mapsto (\wp(t), \wp'(t))$$ as tracing out a curve, and then this curve has to satisfy $$y^2 = 4 x^3 - g_2 x - g_3$$ just by the form of the ODE. – Daniel McLaury May 07 '15 at 01:56
  • In the case of your equation $$a y'' + by'+ c y = 0,$$ you get a two-dimensional family of solutions $$y = C_1 e^{r_1 t} + C_2 e^{r_2 t},$$ where the $r_i$ are constants depending on $a,b,c$. Supposing you take such a solution and consider the function $$t \mapsto (y''(t), y'(t), y(t)),$$ then you get a curve which has to satisfy the equation $$a x + b y + c z = 0$$ just by the form of the equation. However, in this case that doesn't describe a curve but rather just a plane that your curve lies on. – Daniel McLaury May 07 '15 at 02:01
  • However, it's actually not hard to see that the curve parametrized by $$t \mapsto (y''(t), y'(t), y(t))$$ here is in fact algebraic. – Daniel McLaury May 07 '15 at 02:07
  • YUP! That was kind of what i was thinking. Fantastic answer Thanks! – T. Poindexter May 07 '15 at 03:03