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I'm looking for a general solution for $f(t)$ given an unrelated function $g(t)$ in

$$f(t)^2 - 2g(t)f(t)\sin(t) - 2f'(t) + g(t)^2 - 2g(t)\cos(t) + 1 = 0$$

Is it possible to solve without knowing $g(t)$?

Here is the motivation:

Square wheels Now the the explanation. I wanted to verify that the humps really were catenaries because that wasn't intuitive to me. I then proceeded to accidentally prove that if the rate of rotation is constant with respect to the rate of forward motion and the forward motion was in a straight line, the only hump that fits the tangent lines created by this motion requires that the wheel do some significant slipping.

Since slipping is obviously not an option, I'm starting by defining the arc length of the curve to be the equal to the arc length of the equivalent position on a side of the wheel with respect to time, which should be the definition of zero slippage.

The general form of the curve is given by:
    $x(t) = f(t)\cos(t) - \sin(t) + p(t)$
    $y(t) = -f(t)\sin(t) - \cos(t)$

By creating a line for a side $x(t) = f(t), y(t) = -1$, rotating it by $t$ (rate of rotation is arbitrary constant with time) and moving forward by $p(t)$. ($g(t)$ in my original setup was $p'(t)$.)
Thus $f(t)$ is the position function of the point on the side and $p(t)$ is the position function of the center of the wheel in the x direction. Which I'd actually like to find later, if I get through all this. As you can see I assumed no movement in the y direction for the center of the wheel (bumpy ride).

So this question is all about restricting $f(t)$ to equal the arc length of the curve it's on for zero slippage. So $(f'(t))^2 = (x'(t))^2 + (y'(t))^2$. Substituting the definitions of $x(t)$ and $y(t)$, I simplified. A lot. Rotational translation usually involves lots of canceling and the trig identity $\sin^2(t) + \cos^2(t) = 1$. The result is the subject of the above question.

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Periodic coefficients, huh? Looks rather difficult. Is there really nothing special about $g$? – J. M. Oct 21 '11 at 15:19
There will be a further constraint that the moving line be tangent to the curve which I am expecting that will define $g(t)$, but I don't think I can find any help there at this point. – jnm2 Oct 21 '11 at 15:23
I think it might be a better idea to post the actual problem that led you to this DE... – J. M. Oct 21 '11 at 15:24
@J.M. You'll like this. – jnm2 Oct 21 '11 at 16:09
I believe I do. :) Did you already check Hall and Wagon's paper? How about these two articles? Or this webpage (en français)? – J. M. Oct 21 '11 at 16:14
up vote 1 down vote accepted

$$f(t)^2 - 2g(t)f(t)\sin(t) - 2f'(t) + g(t)^2 - 2g(t)\cos(t) + 1 = 0$$

This is a Riccati ODE. So, the usal change of function is : $f(t)=-2\frac{y'(t)}{y(t)}$


$4\left(\frac{y'}{y}\right)^2 + 4g\frac{y'}{y}\sin(t) - 2\left(-2\frac{y''}{y}+2\frac{(y')^2}{y^2}\right)+ g^2 - 2g\cos(t) + 1 = 0$

$4g\frac{y'}{y}\sin(t) + 4\frac{y''}{y}+ g^2 - 2g\cos(t) + 1 = 0$

Knowing $g(t)$, one can compute $A(t)$ and $B(t)$ \begin{cases} A(t)=g\:\sin(t) \\ B(t)=\frac{1}{4}\left( g^2 - 2g\cos(t) + 1 \right) \\ \end{cases}

$$y''+A(t)y'+ B(t)y = 0$$

Of course, without knowing $g(t)$, it is not possible to solve the second order linear ODE. In case of known $g(t)$ leading to a solvable form for the ODE, the general solution $y(t)$ can be explicitely expressed. Then the wanted solution $f(t)$ can be derived : $$f(t)=-2\frac{y'(t)}{y(t)}$$

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We need to derive the natural/intrinsic equation and indeed if we translate (s,R) to ( x,y) it is a straight line.

Catenary has natural equation,

answer to be continued.....

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