Solutions to $f''(x)=-f(x)^2$

Question

Suppose that $f''(x)=-f(x)^2$ and $f(0) = f(1) = 0$. How does one show that $f(1-x)=f(x)$? Is it sufficient to show that $f'(0.5)=0$?

I'm also curious about the number of solutions to this differential equation. Thanks for helping.

Answer 1

Obviously, if you can find a solution for the boundary conditions $f(0)=0$ and $f'(0.5)=0$, $f(0.5)\ne 0$, then you can extend this to the full interval via $f(1-x)=f(x)$ and get a solution for the original BVP.

So with $f'(x)^2+\frac23f(x)^3=C$ you get $C=v_0^2$, $v_0=f'(0)$, and have to solve $$ \frac12=\int_0^{\sqrt[3]{\frac32v_0^2}}\frac{df}{\sqrt{v_0^2-\frac23f^3}} =\frac{\sqrt[3]{\frac32v_0^2}}{|v_0|}\int_0^1\frac{ds}{\sqrt{1-s^3}} $$ which can be solved for $|v_0|≈33.0822$.