Given a curve $\gamma:\mathbb R\to\mathbb R^2$ with $\Vert\gamma'\Vert=1$ and curvature $\kappa(s)=\frac{c}{\cosh s}$, $c\in\mathbb R$, how can I show that $\gamma$ is an elastic curve for some suitable choice of $c$
This is what I have done so far: We know what an arc-length parametrized curve is elastic iff $$\kappa''(s)+\frac{1}{2}\kappa^3(s)+\alpha\kappa(s)=0 $$ for some $\alpha\in\Bbb R$ and for all $s$. The derivatives are $\kappa'(s)=\frac{c\sinh(s)}{\cosh^2(s)}$ and $\kappa''(s)=\frac{c\cosh^3(s)+2c\cosh(s)\sinh^2(s)}{\cosh^4(s)}.$ Hence $$\kappa''(s)+\frac{1}{2}\kappa^3(s)+\alpha\kappa(s) =\frac{c\cosh^3(s)+2c\cosh(s)\sinh^2(s)}{\cosh^4(s)} +\frac{c^3}{2\cosh^3(s)}+\frac{\alpha c}{\cosh s}=\frac{c(2c\cosh(2s)+2\alpha\cosh^2(s)+c^2)}{2\cosh^3(s)}$$
Obviously, for $c=0$ we obtain $\kappa''(s)+\frac{1}{2}\kappa^3(s)+\alpha\kappa(s)=0 $ for all $s$. What did I do wrong? The solution doesn't seem to be very promising..
Ideas and hints would be much appreciated!
I don't know what an elastic curve is, but after introducing the $\alpha$ things look much better. I started with $\kappa(s):=c/\cosh s$ and let Mathematica compute $$\kappa''(s)+{1\over2}\kappa^3(s)+\alpha\kappa(s)={c\bigl(-3+\alpha+c^2+(1+\alpha)\cosh(2s)\bigr)\over2\cosh^3 s}\ .$$ It is now obvious that $\alpha:=-1$ and $c:=\pm2$ make the right hand side of this equation identically zero.