I'm looking for a curve that has a steadily increasing curvature, that is for $\gamma: I \subset \mathbb{R}^+ \rightarrow \mathbb{R}^2$ and $C>0$ constant.
\begin{equation}\label{eq:1} \frac{d}{dt} || \gamma^{\prime \prime}(t) || = C, \end{equation} with $|| \gamma^{\prime}(t) || = 1 $ and some initial conditions (say $\gamma(0) = (0,0), \gamma^{\prime}(0) = (1,0), \gamma^{\prime \prime} = (0,0)$).
I know that such a problem most likely has not a closed form, but numerically I could try to search for something. This is what I got to:
Lets name $\gamma(t) = (\gamma_x(t), \gamma_y(t)) = (\gamma_x, \gamma_y)$. By integrating the curvature equation up to time $t$ and taking the square of the norm we get $$ (\gamma_x^{\prime\prime})^2 + (\gamma_y^{\prime\prime})^2 = C^2t^2.$$
Consider the equation for the derivative of $\gamma$, by taking the square and differentiating we obtain: \begin{align} (\gamma_x^\prime)^2 + (\gamma_y^\prime)^2 = 1 \\ \Rightarrow \gamma_x^\prime \gamma_x^{\prime \prime} + \gamma_y^{\prime} \gamma_y^{\prime \prime} = 0 \\ \Rightarrow \gamma_x^{\prime \prime} = - \frac{\gamma_y^{\prime} \gamma_y^{\prime \prime}}{\gamma_x^\prime} = - \frac{\gamma_y^{\prime} \gamma_y^{\prime \prime}}{\sqrt{1 - (\gamma_y^{\prime})^2}} \end{align}
So returning to the equation derived from the curvature, we get a nonlinear differential equation for $\gamma_y$:
\begin{align} \frac{(\gamma_y^{\prime}\gamma_y^{\prime \prime})^2}{1-(\gamma_y^{\prime})^2} + (\gamma_y^{\prime \prime})^2 = C^2 t^2 \\ \Rightarrow (\gamma_y^{\prime \prime})^2 = C^2t^2(1 - (\gamma_y^{\prime})^2) \\ \Rightarrow \gamma_y^{\prime \prime} = - Ct \sqrt{1 - (\gamma_y^{\prime})^2 } \end{align}
Where in the last step I just took one of the solutions. Now I don't know how to proceed as I'm not familiar with solving nonlinear differential equations. How would you suggest to proceed? or is there any alternative way to obtain such a curve?
