calculuspowerup.com

The Brachistochrone

April 9th, 2009 |

Author: Merrin

Probably the original challenge problem of calculus of variations is the brachistochrone problem or curve of steepest decent. It was posed by Bernoulli to all the greatest mathematicians and physicists of his time. Newton solved the problem and sent it back anonymously, to which Bernoulli is to have said, “I know the lion by his paw.”

Think of a rigid wire that is bent into a function between two points. On the wire a bead with a hole drilled through the middle slides in a uniform gravitational field. The problem is how to bend the wire to get the bead to fall from point A to point B the quickest. The integral to be minimized which is the total travel time can be found from some simple physics. Assume the particle starts at a height y=0 and then falls a total height y while traveling across. It starts with zero initial velocity and then falls at which point it has some kinetic energy. By balancing the energy we find

${\cfrac{1}{2} mv^{2} =mgy} \\ {v=\sqrt{2gy} }$

We can find the total time of the trip by the following equation

$T=\displaystyle \int \cfrac{ds}{v} =\displaystyle \int \cfrac{ds}{\sqrt{2gy} } =\displaystyle \int _{x_{1} }^{x_{2} }\cfrac{\sqrt{1+\dot{y}^{2} } }{\sqrt{2gy} } dx =\displaystyle \int _{y_{0} }^{0}\cfrac{\sqrt{1+\dot{x}^{2} } }{\sqrt{2gy} } dy$

Again we can do our trick of reversing the variables so we have simpler Euler-Lagrange equation to work with. Also we won’t need to worry about the factor of the square root of (2g) which just multiplies the functional integral by a constant

$\cfrac{d}{dy} \cfrac{\partial f}{\partial \dot{x}} =\cfrac{\partial f}{\partial x}$

The equation

$\cfrac{\partial f}{\partial x} =0$

gives an integral of the motion.

$\cfrac{\partial f}{\partial \dot{x}} =c=\cfrac{1}{\sqrt{y} } \cfrac{\dot{x}}{\sqrt{1+\dot{x}^{2} } }$

Anticipating the solution we can write in a value of the constant in the final form

${\cfrac{y}{2R} =\cfrac{\dot{x}^{2} }{1+\dot{x}^{2} } \to \dot{x}=\cfrac{\sqrt{y} }{\sqrt{2R-y} } =\cfrac{y}{\sqrt{2Ry-y^{2} } } } \\ {x=\displaystyle \int \cfrac{ydy}{\sqrt{2Ry-y^{2} } } =\displaystyle \int \cfrac{ydy}{\sqrt{R^{2} -(y-R)^{2} } } }$

This integral can be solved by adding and subtracting R to the numerator. One integral gives a square root and the other gives an inverse sine.

$x=\sqrt{R^{2} -(y-R)^{2} } -R\arcsin (y/R-1)+C$

This form is not much help in elucidating the shape of the final functional form which turns out to be a cycloid. Anticipating the solution it is useful to make a special substitution which is similar to the form of the argument of the arcsin.

$y=R(1-\cos \theta )$

Now we can solve for x with this substitution.

$\int \cfrac{ydy}{\sqrt{R^{2} -(y-R)^{2} } } =\displaystyle \int \cfrac{R(1-\cos \theta )R\sin \theta d\theta }{\sqrt{R^{2} -R^{2} (\cos ^{2} \theta )} } =\displaystyle \int R(1-\cos \theta )d\theta } \\ {x(\theta )_{}^{_{_{}^{} }^{} } =R(\theta -\sin \theta )+C\quad x(0)=0\quad C=0$