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what does $\sum _{j=1}^{\frac{1}{\text{lambda}}-1} (i+1)^j$ mean? 1/lambda is not in general an integer. Also, for $e$ close to $0$ or $1$, $1/\lambda$ diverges so most expressions become meaningless.
@IosifPinelis Right, one should disregard that part of my answer. That curve simply gives the condition for the ODE to have a unique solution instead of a pair (the curve is the discriminant of the quadratic in $a'$). But you are absolutely correct in that the curve does not necessarily solve the ODE generically. For example there is no choice of initial condition for which $c+u\pm u^2/2$ collapses to a single function.
@yarchik Your initial ODE has a unique solution but your manipulations introduced a second, fictitious solution (as I mentioned above, your reduced system has two solutions). I only wrote one of the two solutions and I suspect that I chose the wrong one. What I wrote above is most likely the fictitious solution, not the one you want. I need to think whether it is possible to find the other solution as well.
@yarchik Thank you for the nice words! I'm not sure I understand the relation between this post and the one on overflow. If you take the solution here and expand around $u=\infty$ you find $a\sim -1/u+\cdots$ which is different from the asymptotics $a\sim-1.024/u+\cdots$. The two systems do not seem to be equivalent, no?
@cvgmt Not really. The equations in the OP are compatible only along the line $$\frac{1}{2}(x^2+y^2)+\frac{\pi (x+y) (\pi -\arccos(-\frac{1}{2} (x+y)))}{\sqrt{4-(x+y)^2} \arccos(-\frac{1}{2} (x+y))^3}=0$$You can check this using the code {a'[L], b'[L], c'[L]} == {-(b[L]^2 + c[L]^2), (2 - 4 a[L]) b[L], (2 - a[L]) c[L]} /. {a -> (\[Pi]/(2 ArcCos[-(x[#] + y[#])/2]) &), b -> ((x[#] + y[#])/2 &), c -> ((x[#] - y[#])/2 &)} // Thread %[[1]] /. Solve[%[[2 ;; 3]], {x'[L], y'[L]}][[1]] // Simplify
Note that $a=\pi/(2 \arccos(-b))$ so the system is overdetermined. Equivalently the variables $x,y$ are not independent but satisfy a constraint $f(x,y)=0$. So what exactly do you mean by a 2d plot?
@ppp Works fine on my laptop, see i.imgur.com/eXn7NeO.png Did you try on a fresh kernel? What version are you using? Can you run the code line by line and let me know which specific part makes it hang?
