How to find maximum minimum radii and inflection point of a curve with the polar symmetry given by the following ode ? Primed on arc,the angle $\psi$ is made between curve and radial lines. $\theta \text{=const.}$
$$ 2a \psi'=\left(1-\dfrac{T^2}{r^2}\right);\;r'= \cos \psi \; \tag1 $$
Patterns above are drawn for constants $ a=1,T= (4,2) ,\theta_i=\pi/4, r_i=1,s_{max=}=320 a, $ after numerical integration.
The outer maximum radii are $ (\approx \sqrt2,2)$ which could not be found analytically.
In polar coordinates ... is the following true for local straightness ?
$$\psi^{'}_{inflection}= {0}. \tag 2 $$
Pictorially it does not agree for the first pattern (no inflection), but seems a better tally for the second case where:
$$ r_{inflection} =T \;$$
To find radial extrema
$$\cos \psi=0,\; \psi= \frac{\pi}{2}, \frac{3\pi}{2}..\; \tag 3$$
Differentiating (1)
$$ 2a \psi{''}=\dfrac{2T^2\cos \psi}{r^3} \tag4$$
from (3)
$$ \psi^{''}=0 \tag5 $$
Is it then a correct criterion for finding max/min radii in general ? And further on how can we find $ (r, \theta ) ? $
Towards a solution for $\psi$
Let $ u=\dfrac{\sin \psi}{r} \to u'= ( r \cos \psi\psi^{'} -\sin \psi \cos \psi)/r^2 \tag6$
Eliminating $ \psi^{'}$ between (1) and (6) and simplifying we obtain
$$\dfrac{u'}{\sqrt{1-u^2r^2}}+ \dfrac{u}{r}= \dfrac{1-T^2/r^2}{2ar} \tag 7 $$
whose integral analytical solution $ u=f(r) $ is required.
Thanking you for the attention.