Editing Tautochrone curve (section)

== {{Anchor|Abel problem}}Abel's solution ==
[[Niels Henrik Abel]] attacked a generalized version of the tautochrone problem (''Abel's mechanical problem''), namely, given a function <math>T(y)</math> that specifies the total time of descent for a given starting height, find an equation of the curve that yields this result. The tautochrone problem is a special case of Abel's mechanical problem when <math>T(y)</math> is a constant.

Abel's solution begins with the principle of [[conservation of energy]] – since the particle is frictionless, and thus loses no energy to [[heat]], its [[kinetic energy]] at any point is exactly equal to the difference in [[gravitational energy|gravitational potential energy]] from its starting point.  The kinetic energy is <math display="inline">\frac{1}{2} mv^2</math>, and since the particle is constrained to move along a curve, its velocity is simply <math>{d\ell}/{dt}</math>, where <math>\ell</math> is the distance measured along the curve. Likewise, the gravitational potential energy gained in falling from an initial height <math>y_0</math> to a height <math>y</math> is <math>mg(y_0 - y)</math>, thus:

{{block indent|1=<math>
\begin{align}
\frac{1}{2} m \left ( \frac{d\ell}{dt} \right ) ^2 & = mg(y_0-y) \\
\frac{d\ell}{dt} & = \pm \sqrt{2g(y_0-y)} \\
dt & = \pm \frac{d\ell}{\sqrt{2g(y_0-y)}} \\
dt & = - \frac{1}{\sqrt{2g(y_0-y)}} \frac{d\ell}{dy} \,dy
\end{align}
</math>}}

In the last equation, we have anticipated writing the distance remaining along the curve as a function of height (<math>\ell(y))</math>, recognized that the distance remaining must decrease as time increases (thus the minus sign), and used the [[chain rule]] in the form <math display="inline">d\ell = \frac{d\ell}{dy} dy</math>.

Now we integrate from <math>y = y_0</math> to <math>y = 0</math> to get the total time required for the particle to fall:

{{block indent|1=<math>
T(y_0) = \int_{y=y_0}^{y=0} \, dt = \frac{1}{\sqrt{2g}} \int_0^{y_0} \frac{1}{\sqrt{y_0-y}} \frac{d\ell}{dy} \, dy
</math>}}

This is called '''Abel's integral equation''' and allows us to compute the total time required for a particle to fall along a given curve (for which <math>{d\ell}/{dy}</math> would be easy to calculate). But Abel's mechanical problem requires the converse – given <math>T(y_0)\,</math>, we wish to find <math>f(y) = {d\ell}/{dy}</math>, from which an equation for the curve would follow in a straightforward manner.  To proceed, we note that the integral on the right is the [[convolution]] of <math>{d\ell}/{dy}</math> with <math>{1}/{\sqrt{y}}</math> and thus take the [[Laplace transform]] of both sides with respect to variable <math>y</math>:

{{block indent|1=<math>
\mathcal{L}[T(y_0)] = \frac{1}{\sqrt{2g}} \mathcal{L} \left [ \frac{1}{\sqrt{y}} \right ]F(s)
</math>}}

where <math>F(s) = \mathcal{L} {\left[ {d\ell}/{dy} \right ]}</math>. Since <math display="inline">\mathcal{L} {\left[ {1}/{\sqrt{y}} \right]} = \sqrt{{\pi}/{s}}</math>, we now have an expression for the Laplace transform of <math>{d\ell}/{dy}</math> in terms of the Laplace transform of <math>T(y_0)</math>:

{{block indent|1=<math>
\mathcal{L}\left [ \frac{d\ell}{dy} \right ] = \sqrt{\frac{2g}{\pi}} s^{\frac{1}{2}} \mathcal{L}[T(y_0)]
</math>}}

This is as far as we can go without specifying <math>T(y_0)</math>. Once <math>T(y_0)</math> is known, we can compute its Laplace transform, calculate the Laplace transform of <math>{d\ell}/{dy}</math> and then take the inverse transform (or try to) to find <math>{d\ell}/{dy}</math>.

For the tautochrone problem, <math>T(y_0) = T_0\,</math> is constant.  Since the Laplace transform of 1 is <math>{1}/{s}</math>, i.e., <math display="inline">\mathcal{L}[T(y_0)] = {T_0}/{s}</math>, we find the shape function <math display="inline">f(y) = {d\ell}/{dy}</math>:

{{block indent|1=<math>
\begin{align}
F(s) = \mathcal{L} {\left [ \frac{d\ell}{dy} \right ]}
& = \sqrt{\frac{2g}{\pi}} s^{\frac{1}{2}} \mathcal{L}[T_0] \\
& = \sqrt{\frac{2g}{\pi}} T_0 s^{-\frac{1}{2}}
\end{align}
</math>}}

Making use again of the Laplace transform above, we invert the transform and conclude:

{{block indent|1=<math>\frac{d\ell}{dy} = T_0 \frac{\sqrt{2g}}{\pi}\frac{1}{\sqrt{y}}</math>}}

It can be shown that the cycloid obeys this equation. It needs one step further to do the integral with respect to <math>y</math> to obtain the expression of the path shape.

(''Simmons'', Section 54).