Given an equation of curvature, how does one find the family of parametric equations that fit?

Question

I've seen a few questions & answers here for special cases on finding the parametric equations for a given curvature. E.g; Find the parametric equation for a curve with given curvature. However I'm afraid I don't understand the general process. Could someone guide me through the process?

I care about parametric equations of the form

$$\gamma(s)=(x(s),y(s))$$

Hence having signed curvature

$$\kappa=\frac{x'y''-y'x''}{(x'^2+y'^2)^\frac{3}{2}}$$

My question is

Given the equation for $\kappa(s)$, how do you find the family of solutions for $\gamma(s)$?

I assume there is a unique curve which satisfies $\kappa(s)$, though the final solution will have three constants, $x_0$, $y_0$, and $\theta$, which will encode an arbitrary translation and rotation (or some equivalents) of such curve, as, intuitively, curvature does not care about translation or rotation of the entire curve.

As a final note, I'm simply an overoptimistic undergrad, and as such I've only academically dealt with first-order differential equations and have only self-taught curvature. Regardless, I do conceptually understand each. As such, I'd appreciate an answer roughly on my level of understanding.

Answer 1

Not only is there an arbitrary rotation and translation, but also a reflection and parametrisation of the curve. So, first of all, take the standard arclength parametrisation in which the definition of the curvature becomes $$\mathbf{t}'(s)=\kappa(s)\mathbf{n}(s)$$ where $\mathbf{t}(s)=(x'(s),y'(s))$ is the tangent vector and $\mathbf{n}(s)=(-y'(s),x'(s))$ is 'the' normal vector. The latter is only defined up to a sign, so one has to choose one of them arbitrarily. This fixes the handedness of the curve, i.e. the reflection.

Hence the differential equation to solve is $$\begin{pmatrix}x''(s)\cr y''(s)\end{pmatrix}=\kappa(s)\begin{pmatrix}-y'(s)\cr x'(s))\end{pmatrix}$$ As a second order equation, this ought to give four constants of integration, but there is the arclength constraint $(x')^2+(y')^2=1$, so in fact only three constants remain: two for translations, and one for rotation.

Answer 2

As I've stated "I've only academically dealt with first-order differential equations", so this answer to my own question might be overrun with flaws, but this is (I believe) the general form I was looking for. Many thanks to Chrystomath for the insight.

If $(x')^2+(y')^2=1$, then

$$\kappa=x'y''-y'x''$$

Also, $(x')^2+(y')^2=1\implies y''=-\frac{x'x''}{y'}$

$$\implies -\kappa y'=x''\implies\kappa^2(y')^2=(x'')^2\implies\kappa^2(1-(x')^2)=(x'')^2$$

Let $u=x'$

$$\implies \kappa^2(1-(x')^2)=(x'')^2=\kappa^2(1-u^2)=(u')^2\implies\kappa\sqrt{1-u^2}=\pm u'$$ $$\kappa\sqrt{1-u^2}=\pm \frac{du}{dt}\implies\pm\kappa dt=\frac{du}{\sqrt{1-u^2}}$$ $$\implies\pm\int\kappa=\sin^{-1}(u)+c_1$$ $$\implies c_1\pm\int\kappa=\sin^{-1}(u)\implies u=\sin(c_1\pm\scriptsize\int\normalsize\kappa)$$ $$\implies x'=\sin(c_1\pm\scriptsize\int\normalsize\kappa)$$ $$\therefore x=\int\sin(c_1\pm\scriptsize\int\normalsize\kappa)dt$$

With similar logic, follows

$$y=\int\cos(c_1\pm\scriptsize\int\normalsize\kappa)dt$$

Therefore, the parametric equation can be found (conventionally swapping $\sin$ and $\cos$) to be

$$\gamma(s)=\begin{pmatrix} x_0+\int_0^s\cos(\theta\pm\scriptsize\int\normalsize\kappa)dt \cr y_0+\int_0^s\sin(\theta\pm\scriptsize\int\normalsize\kappa)dt\end{pmatrix}$$

Lo and behold, as prophesied by Chrystomath: three constants (two for translation and one for rotation), and the reflections (indicated by $\pm$)!