Solving non-linear second order differential equation: radius of curvature $= k \theta$

Question

I'm trying to find any curve where the radius of curvature increases linearly with angular displacement. So in polar coordinates

radius of curvature $= k \theta$

$$ \frac{(r^2 + r'^2)^{3/2}}{r^2 + 2r'^2 - rr''} = k \theta$$

where k is a constant

I just need a single curve that fits this equation, does not matter if it is defined as an integral or power series. But, best to have a closed form solution. Thank you.

Answer 1

I understand you are looking for any curve satisfying the property $$\rho=k\theta$$ Therefore choose a curve which has the property that the radius vector makes a constant angle $\alpha$ with the tangent vector. In intrinsic form, we have $$\frac {dy}{dx}=\tan\psi$$ The radius of curvature is $$\rho=\frac{ds}{d\psi}$$ Your curve therefore has the property that $\rho=k(\psi-\alpha)$. We can now use the well-known relationships $$x=\int\rho\cos\psi d\psi$$ and $$y=\int\rho\sin\psi d\psi$$ to obtain $x$ and $y$ in terms of $\theta$ and then you can get the curve easily into polar form.

Answer 2

If $\psi $ is angle between radius vector and tangent and primes are taken with respect to arc length, then we need to solve for $ r,\theta $ via $ \psi$ for reciprocal radius of curvature:

$$ \dfrac{d \psi}{ds} + \dfrac{\sin \psi}{r} = k\, \int \dfrac{\sin \psi}{r} ds, \, r^{'} = \cos \psi,\, \theta^{'}= \dfrac{\sin \psi}{r}. $$

EDIT1: As a first step numerical solution is given below for assumed boundary conditions near origin.