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Many years ago, I considered the following non-linear differential equation: $y=y''\cdot(1+y'^{2})^{-3/2}$

This equation expresses the equality between the value of a given function $y\in C^{2}(R)$ and the value of its curvature at the same point.

I almost convinced myself that there is no nontrivial solution to this equation, but couldn't find any rigorous argument to prove it.

I'm thus looking for techniques and or references to get a real proof of this expected result.

Many thanks in advance.

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  • $\begingroup$ You're right, "non trivial". I don't require the solution to be smooth though, just that the second derivative is defined and continuous for all $x$. $\endgroup$ May 29 '16 at 9:44
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Well the standard techniques would take advantage of the fact that the equation doesn't explicitly involve the independent variable $x$ to integrate the equation once, thereby leading to the conservation law $$ y^2 + 2(1+y'^2)^{-1/2} = r^2,\tag1 $$ where, $r>0$ is a constant. Note that we must have $|y| < r$

The relation (1) can then be solved to yield, after separation of variables, $$ \frac{(r^2-y^2)\,dy}{\bigl((2-r^2+y^2)(2+r^2-y^2)\bigr)^{1/2}} = dx.\tag2 $$ The left hand integral can be computed in terms of elliptic functions, of course, but you don't need to do this to do a qualitative analysis.

If $r>\sqrt2$, then we must have $|y|\ge\sqrt{r^2-2}$, so, replacing $y$ by $-y$, we can assume that $\sqrt{r^2-2}\le y(x) < r$. Because the integral on the left hand side of (2) over the interval $\sqrt{r^2-2}\le y \le r$ is finite, it follows that $y(x)$ can only be defined over an $x$-interval of finite length. Thus, there is no entire solution in this case.

If $r = \sqrt2$, then, again, we could only have $y=0$ where $y'=0$. However, the left-hand integral of (2) over the interval $0<y<a$ when $0<a<\sqrt{2}$ is infinite, so we cannot have $y$ vanishing for any finite value of $x$, thus, $y$ cannot vanish and we can assume $y>0$. The above integral then leads to a solution $y(x)$ on an interval of the form $(-\infty,C]$ where $C<\infty$, so there is no entire solution in this case either.

Finally, if $r<\sqrt{2}$, then we have $|y| < r$, but the integral of the lefthand side of (2) over the interval $-r < y < r$ is finite, so, again, there is no solution defined over the entire $x$-axis.

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    $\begingroup$ @SylvainJULIEN: You are welcome. Of course, for geometric purposes, you might want to allow reparametrizations of the curve $\bigl(x,y(x)\bigr)$ in the form $\bigl(x(s),y(s)\bigr)$ where $s$ is arclength. Then the curve can be continued further as a $C^2$ (in fact, $C^\infty$) curve, through places where the tangent to the curve turns vertical. You might want to think about that for your application. $\endgroup$ May 29 '16 at 12:58
  • $\begingroup$ I"m a little confused (I'm not familiar with the technique), but I don't immediately see how you integrated this (or what you even integrated with respect to). Did you integrate with respect to $y$? How did you integrate $y''$ with respect to $y$...? $\endgroup$
    – user541686
    May 29 '16 at 22:53
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    $\begingroup$ @Mehrdad: You multiply both sides of the equation by $2y'$ and then integrate with respect to $x$. $\endgroup$ May 30 '16 at 0:22
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Curves with the property that the curvature at a point is equal to the $y$-coordinate, more generally that it is proportional to some power thereof, have been studied in some detail in the arXiv paper arXiv:1102.1579 (the rather inelegant title is too long to reproduce here), where they are called Maclaurin catenaries. They have several other related properties, for example they are trajectories of particles moving under parallel force laws of the form $y^\alpha$ and are catenaries for such laws.

It is more convenient to use parametrised representations and the catenaries are of the form $(F_d(t),f_d(t))$ with $f_d(t)=(cos(d t))^{1/d}$ and $F_d$ its primitive. (There is a simple relationship between the exponents $\alpha$ and $d$). The crucial property of functions $f$ of this form is that the differential expressions $f+f’’$, $f^2+f’^2$ and $(f f’’-f’^2)/(f^2+f’^2)^{3/2}$, which occur in the calculations, are all proportional to a power of $f$.

The justification for the use of the name of Maclaurin is that the distinguished scottish mathematician considered curves of the form $r f_d(\theta)=1$ with the above $f$ and showed that they have analogous properties for central forces proportional to $r^{\alpha}$. These are now known as Maclaurin spirals. The reason for this analogy between the parallel and the central case is that there are striking similarities between the geometrical and kinetic properties of curves with parametric representation $(F(t),f(t))$ where $F$ a primitive of $f$ and those with polar representation $rf(\theta)=1$. These are elucidated in the above article.

These curves can initially be interpreted as the graphs of functions on an interval about the origin but can be extended in natural ways to the whole real line, either by translation or, as is more relevant to your question, by a glide reflection. These will have recurring singularities at the endpoints, the precise nature of which will depend on the index $d$. This is not discussed in the above article but the explicit solutions given there should make a detailed analysis possible.

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