Which functions have the same arc length to the origin at every point as the *signed* curvature at the point?

Question

Which functions $y\left( x \right)$ have the same arc length $s$ to the origin at every point as the signed curvature $\kappa$ at the point?

We know the formulas for the arc length $s$ $$ s\left( a, b, \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right) = \int_{a}^{b} \sqrt{1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2}} \operatorname{d}x $$ and signed curvature $\kappa$ $$ \kappa\left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x}, \frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}x^{2}} \right) = \frac{\frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}x^{2}}}{\left( 1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2} \right)^{\frac{3}{2}}} \text{.}\\ $$

Since we are talking about the length of the function $y\left( x \right)$ from the origin to every point, here $a = 0$ and $b = x$, giving us a nonlinear, higher-order ordinal differential equation: $$ \begin{align*} s\left( 0, x, \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right) &= \kappa\left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x}, \frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}x^{2}} \right)\\ \int_{0}^{x} \sqrt{1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2}} \operatorname{d}x &= \frac{\frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}x^{2}}}{\left( 1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2} \right)^{\frac{3}{2}}}\\ \end{align*} $$

Now comes the hard part: Solving the ODE for y(x).

(From here only my failed attempts follow...)

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My first attempt at solving the ODE

Since I can't think of a specific solution, I would try a bit of reshaping first. $$ \begin{align*} \int_{0}^{x} \sqrt{1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2}} \operatorname{d}x &= \frac{\frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}x^{2}}}{\left( 1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2} \right)^{\frac{3}{2}}}\\ \int_{0}^{x} \sqrt{1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2}} \operatorname{d}x &= \frac{\frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}x^{2}}}{\left( 1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2} \right)^{3 \cdot \frac{1}{2}}}\\ \int_{0}^{x} \sqrt{1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2}} \operatorname{d}x &= \frac{\frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}x^{2}}}{\left( \sqrt{1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2}} \right)^{3}} \quad\mid\quad u\left( x \right) := \sqrt{1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2}}\\ \int_{0}^{x} u\left( x \right) \operatorname{d}x &= \frac{\frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}x^{2}}}{u\left( x \right)^{3}} \quad\mid\quad \cdot u\left( x \right)^{3} \tag{1.}\\ \int_{0}^{x} u\left( x \right) \operatorname{d}x \cdot u\left( x \right)^{3} &= \frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}x^{2}}\\ \\ u\left( x \right) &= \sqrt{1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2}} \quad\mid\quad \left( \right)^{2}\\ u\left( x \right)^{2} &= 1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2} \quad\mid\quad -1\\ u\left( x \right)^{2} - 1 &= \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2} \quad\mid\quad \sqrt{}\\ \sqrt{u\left( x \right)^{2} - 1} &= \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \quad\mid\quad \frac{\operatorname{d}}{\operatorname{d}x}\\ \frac{\frac{\operatorname{d}u\left( x \right)}{\operatorname{d}x}}{2 \cdot \sqrt{u\left( x \right) - 1}} &= \frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}x^{2}} \tag{2.}\\ \\ \int_{0}^{x} u\left( x \right) \operatorname{d}x \cdot u\left( x \right)^{3} &= \frac{\frac{\operatorname{d}u\left( x \right)}{\operatorname{d}x}}{2 \cdot \sqrt{u\left( x \right) - 1}}\\ \end{align*} $$

And I'm stuck aigan.

My second attempt at solving the ODE

Next, I would try to transform the higher-order nonlinear ODE into a third-order nonlinear ODE, because maybe that will help. Since the substitution simplified the equation in my opinion, I use it again. $$ \begin{align*} \int_{0}^{x} \sqrt{1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2}} \operatorname{d}x &= \frac{\frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}x^{2}}}{\left( 1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2} \right)^{\frac{3}{2}}} \quad\mid\quad \text{useing } \left( 1. \right) \text{ and } \left( 2. \right)\\ \int_{0}^{x} u\left( x \right) \operatorname{d}x &= \frac{\frac{\operatorname{d}u\left( x \right)}{\operatorname{d}x}}{2 \cdot u\left( x \right)^{3} \cdot \sqrt{u\left( x \right) - 1}} \quad\mid\quad \frac{\operatorname{d}}{\operatorname{d}x}\\ \\ u\left( x \right) &= \dots \end{align*} $$

Time to give up... Since it looks harder now.

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My numerical attempt

We know from calculus $\frac{\operatorname{d}f\left( x \right)}{\operatorname{d}x} = \frac{f\left( x + h \right) - f\left( x \right)}{h} + \mathcal{O}\left( h \right)$ or $\frac{\operatorname{d}f\left( x \right)}{\operatorname{d}x} \approx \frac{f\left( x + h \right) - f\left( x \right)}{h}, \text{for small } \left| h \right|$ (Euler method) aka:

$$ \begin{align*} \int_{0}^{x} \sqrt{1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2}} \operatorname{d}x &= \frac{\frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}x^{2}}}{\left( 1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2} \right)^{\frac{3}{2}}} \quad\mid\quad \text{useing } \left( 1. \right) \text{ and } \left( 2. \right)\\ \int_{0}^{x} u\left( x \right) \operatorname{d}x &= \frac{\frac{\operatorname{d}u\left( x \right)}{\operatorname{d}x}}{2 \cdot u\left( x \right)^{3} \cdot \sqrt{u\left( x \right) - 1}}\\ \int_{0}^{x} u\left( x \right) \operatorname{d}x &\approx \frac{\frac{u\left( x + h \right) - u\left( x \right)}{h}}{2 \cdot u\left( x \right)^{3} \cdot \sqrt{u\left( x \right) - 1}}\\ \int_{0}^{x} u\left( x \right) \operatorname{d}x &\approx \frac{u\left( x + h \right) - u\left( x \right)}{2 \cdot u\left( x \right)^{3} \cdot \sqrt{u\left( x \right) - 1} \cdot h} \quad\mid\quad \frac{\operatorname{d}}{\operatorname{d}x}\\ u\left( x \right) &\approx \frac{\operatorname{d}}{\operatorname{d}x} \frac{u\left( x + h \right) - u\left( x \right)}{2 \cdot u\left( x \right)^{3} \cdot \sqrt{u\left( x \right) - 1} \cdot h}\\ u\left( x \right) &\approx \frac{\frac{u\left( x + 2 \cdot h \right) - u\left( x + h \right)}{2 \cdot u\left( x + h \right)^{3} \cdot \sqrt{u\left( x + h \right) - 1} \cdot h} - \frac{u\left( x + h \right) - u\left( x \right)}{2 \cdot u\left( x \right)^{3} \cdot \sqrt{u\left( x \right) - 1} \cdot h}}{h}\\ u\left( x \right) &\approx \frac{u\left( x + 2 \cdot h \right) - u\left( x + h \right)}{2 \cdot u\left( x + h \right)^{3} \cdot \sqrt{u\left( x + h \right) - 1} \cdot h^{2}} - \frac{u\left( x + h \right) - u\left( x \right)}{2 \cdot u\left( x \right)^{3} \cdot \sqrt{u\left( x \right) - 1} \cdot h^{2}}\\ u\left( x \right) &\approx \frac{u\left( x + 2 \cdot h \right) \cdot u\left( x \right)^{3} \cdot u\left( x \right)^{3} - u\left( x + h \right) \cdot u\left( x \right)^{3}}{2 \cdot u\left( x + h \right)^{3} \cdot \sqrt{u\left( x + h \right) - 1} \cdot h^{2}} - \frac{u\left( x + h \right) \cdot u\left( x + h \right)^{3} - u\left( x \right) \cdot u\left( x + h \right)^{3}}{2 \cdot u\left( x + h \right)^{3} \cdot u\left( x \right)^{3} \cdot \sqrt{u\left( x \right) - 1} \cdot h^{2}}\\ u\left( x \right) &\approx \frac{u\left( x + 2 \cdot h \right) \cdot u\left( x \right)^{3} \cdot u\left( x \right)^{3} - u\left( x + h \right) \cdot u\left( x \right)^{3} - \left( u\left( x + h \right) \cdot u\left( x + h \right)^{3} - u\left( x \right) \cdot u\left( x + h \right)^{3} \right)}{2 \cdot u\left( x + h \right)^{3} \cdot u\left( x \right)^{3} \cdot \sqrt{u\left( x \right) - 1} \cdot h^{2}} \quad\mid\quad \cdot \left( 2 \cdot u\left( x + h \right)^{3} \cdot u\left( x \right)^{3} \cdot \sqrt{u\left( x \right) - 1} \cdot h^{2} \right)\\ u\left( x \right) \cdot 2 \cdot u\left( x + h \right)^{3} \cdot u\left( x \right)^{3} \cdot \sqrt{u\left( x \right) - 1} \cdot h^{2} &\approx u\left( x + 2 \cdot h \right) \cdot u\left( x \right)^{3} \cdot u\left( x \right)^{3} - u\left( x + h \right) \cdot u\left( x \right)^{3} - \left( u\left( x + h \right) \cdot u\left( x + h \right)^{3} - u\left( x \right) \cdot u\left( x + h \right)^{3} \right)\\ u\left( x \right)^{4} \cdot 2 \cdot u\left( x + h \right)^{3} \cdot \sqrt{u\left( x \right) - 1} \cdot h^{2} &\approx u\left( x + 2 \cdot h \right) \cdot u\left( x \right)^{6} - u\left( x + h \right) \cdot u\left( x \right)^{3} - u\left( x + h \right)^{4} + u\left( x \right) \cdot u\left( x + h \right)^{3}\\ \end{align*} $$ Time to give up, since this dosn't look solvable for me...

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I am grateful for every help, correction and suggestion.

Answer 1

First, notice that the defining condition determines an Euler spiral.

Per OP's comment, here's how to construct an explicit parameterization of an (essentially general) curve $\gamma$ satisfying the condition. In the end we'll exploit the construction to describe a function $y(x)$ satisfying the condition.

The Frenet equations in $2$ dimensions, satisfied by the curvature $\kappa(s)$, tangent vector ${\bf T}(s)$, and normal vector ${\bf N}(s)$, are $$\pmatrix{{\bf T}' & {\bf N}'} = \pmatrix{{\bf T} & {\bf N}} \pmatrix{\cdot&-\kappa\\\kappa&\cdot} .$$ As usual, $s$ is the arc length parameter. In our case, $\kappa(s) = s$, so our differential equation becomes $$\pmatrix{{\bf T}' & {\bf N}'} = \pmatrix{{\bf T} & {\bf N}} s \pmatrix{\cdot&-1\\1&\cdot},$$ and rearranging gives $$\pmatrix{{\bf T} & {\bf N}}^{-1}\pmatrix{{\bf T}' & {\bf N}'} = s \pmatrix{\cdot&-1\\1&\cdot} ,$$ so \begin{align*} \pmatrix{{\bf T} & {\bf N}} &= \pmatrix{{\bf T}(0) & {\bf N}(0)} \exp \int_0^s \left[u\pmatrix{\cdot&-1\\1&\cdot}\right] du\\ &= \pmatrix{{\bf T}(0) & {\bf N}(0)} \exp \left[\pmatrix{\cdot&-1\\1&\cdot} \int_0^s u \,du\right] \\ &= \pmatrix{{\bf T}(0) & {\bf N}(0)} \exp \left[\frac{1}{2} s^2 \pmatrix{\cdot&-1\\1&\cdot}\right] \\ &=\pmatrix{{\bf T}(0) & {\bf N}(0)} \pmatrix{\cos \frac{s^2}{2} & \ast \\ \sin \frac{s^2}{2} & \ast}. \end{align*} If we denote ${\bf T}(0) = (\cos \theta, \sin \theta)$, then $${\bf T} = \pmatrix{\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta} \pmatrix{\cos \frac{s^2}{2} \\ \sin \frac{s^2}{2}}.$$ Integrating and setting $\gamma(0) = 0$ gives $$\gamma(s) = \int_0^s {\bf T}'(u) \,du = \pmatrix{\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta} \pmatrix{\int_0^s \cos \frac{u^2}{2} \,du \\ \int_0^s \sin \frac{u^2}{2} \,du} .$$ So, up to rotation, there is a unique curve satisfying $\gamma(0) = 0$. Taking $\theta = 0$ to choose a representative we find that $$\color{#df0000}{\boxed{\gamma(s) = \pmatrix{\int_0^s \cos \frac{u^2}{2} \,du \\ \int_0^s \sin \frac{u^2}{2} \,du} = \pmatrix{\sqrt{2} C\left(\frac{s}{\sqrt 2}\right) \\ \sqrt{2} S\left(\frac{s}{\sqrt 2}\right)}}},$$ where $C$ and $S$ are the Fresnel integral functions.

The solution curve $\gamma$, $s \in [-32, 32]$.

The Fresnel cosine function $C$ is injective in a neighborhood of $x = 0$, so it admits a local inverse, $C^{-1}$, in which case a solution $y(x)$ to the original o.d.e. is $$\color{#df0000}{\boxed{y(x) = \sqrt{2} S\left(C^{-1}\left(\frac{x}{\sqrt{2}}\right)\right)}} ,$$ in agreement with achille hui's comment.

The function $y(x)$ with maximum domain.

We can verify directly that the truncation of the power series of $y$ agrees with that in my other answer: $$y(x) = \frac16 x^3 + \frac1{105} x^7 + \frac{293}{237600} x^{11} + \cdots .$$

Answer 2

Your DE is simply $w'w=y''$ where $w(x)=\int_0^x\sqrt {1+y'^2(t) }dt$ $$\int_{0}^{x} \sqrt{1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2}} \operatorname{d}x = \frac{\frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}x^{2}}}{\left( 1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2} \right)^{\frac{3}{2}}}\\$$ $$2\int_{0}^{x} \sqrt{1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2}} \operatorname{d}x {\left( 1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2} \right)^{\frac{3}{2}}}=2y''$$ $$\left (\left (\int_{0}^{x} \sqrt{1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2}} \operatorname{d}x \right)^{2}\right)'=2y''$$ I think that you can reduce the order by direct integration. $$\int_0^x\sqrt {1+p^2(t)}dt=\pm \sqrt {2p+C_1}$$ $$\sqrt {1+p^2}=\pm \dfrac {p'}{\sqrt {2p+C_1}}$$ Where $p=y'=\dfrac {dy}{dx}$ and $p'=\dfrac {dp}{dx}$. The last DE is separable maybe not easy to integrate.

Answer 3

I can't see a closed-form solution for an explicit function $y(x)$, but we can make some progress.

Differentiating both sides of the equation and rearranging gives the $3$rd order o.d.e. $$(1 + (y')^2)^3 = y'''(x) \left[(y')^2 + 1\right] - 3 (y'')^2 y' ,$$ and writing $u := y'$ reduces it to a $2$nd order equation, $$(1 + u^2)^3 = u'' (1 + u^2) - 3 (u')^2 u .$$ Swapping the independent and dependent variables, i.e., regarding $x$ as a function of $u$, yields the equation $$x''(u) = -(1 + u^2)^2 x'(u)^3 - \frac{3 u x'(u)}{1 + u^2}.$$ We may now reduce to a $1$st order equation in $z := x'$: $$z' = -(1 + u^2)^2 z^3 - \frac{3 u z}{1 + u^2} .$$ Substituting $z(u) = \frac{1}{(1 + u^2)^{3 / 2} \sqrt{w(u)}}$ transforms the equation to $$w'(u) = \frac{2}{u^2 + 1},$$ or $$w(u) = 2 \arctan u + C.$$ Backsubstituting yields the generic solutions $$z(u) = \pm \frac{1}{(u^2 + 1)^{3 / 2} (\sqrt{2 \arctan u + C})}$$ and thus $$x(u) = \pm \int_a^u \frac{dt}{(t^2 + 1)^{3 / 2} (\sqrt{2 \arctan t + C})} .$$ The integral on the right-hand side likely has no closed form, but the inverse of this $x(u)$ is $u(x) = y'(x)$, and integrating again gives $y(x)$.

Incidentally, we can compute the truncation of a series solution: Setting $y(0) = 0$, $y'(0) = a$, $y''(0) = b$ gives $$y(x) = a x + \frac{1}{2} b x^2 + \frac{1}{6} \left[(1 + a^2)^2 + \frac{3 a b^2}{1 + a^2}\right] x^3 + \cdots .$$ Taking $a = b = 0$ yields the solution with the highest order of tangency to the $x$-axis: $$y(x) = \frac16 x^3 + \frac1{105} x^7 + \frac{293}{237600} x^{11} + \cdots .$$