Brachistochrone

Introduction

To find the shape of the curve which the time is shortest possible. . .

We use WxMaxima to do the calculus part.

Theory

Variational Calculus and Euler--Lagrange Equation

The time that is needed for sliding from point $P_{a}$ to point $P_{b}$ is $t=\int _{P_{a}}^{P_{b}}{\frac {1}{v}}ds$ where $ds={\sqrt {1+y'{^{2}}}}dx$ is the Pythagorean distance measure and $v$ is determined from the the law of conservation of energy ${\frac {1}{2}}mv^{2}=mgy$ giving $v={\sqrt {2gy}}$. Plugging these in, we get

$t=\int _{P_{a}}^{P_{b}}{\sqrt {\frac {1+y'^{2}}{2gy}}}dx=\int _{P_{a}}^{P_{b}}fdx$,

where $f=f(y,y')$ is the function subject to variational consideration.

According to the Euler--Lagrange differential equation the stationary value is to be found, if E-L equation ${\frac {\partial f}{\partial y}}-{\frac {d}{dx}}{\frac {\partial f}{\partial y'}}=0$ is satisfied.

No Friction

Since $f$ does not depend on $x$, we may use the simplified E--L formula $f-y'{\frac {\partial f}{\partial y'}}=$ Constant. The differentials are easy, and we have

$f-y'{\frac {\partial f}{\partial y'}}={\sqrt {\frac {1+y'{^{2}}}{2gy}}}-{\frac {y'{^{2}}}{{\sqrt {2gy}}{\sqrt {1+y'^{2}}}}}=C$ So we have ${\frac {1+y'{^{2}}}{\sqrt {2gy(1+y'{^{2}})}}}-{\frac {y'{^{2}}}{{\sqrt {2gy}}{\sqrt {1+y'^{2}}}}}={\frac {1}{\sqrt {2gy(1+y'{^{2}})}}}=C$ and multiplying this with the denominator and rearring, we have $y\left(1+y'^{2}\right)={\frac {1}{2gC^{2}}}=k^{2}$ by redefining the constant. The standard solution to this differential equation is given by

${\begin{aligned}x&={\frac {1}{2}}k^{2}(\theta -\sin \theta )\\y&={\frac {1}{2}}k^{2}(1-\cos \theta )\end{aligned}}$

and is the equation of a cycloid.

No Friction: Maxima

The details using WxMaxima:

energy : 1/2*m*v^2 = m*g*y;
v_sol : solve( energy, v);
v_sol : v_sol[2];
EL_f : rhs( sqrt(1+'diff(y,t)^2)/v_sol );
doof_dooyp : diff( EL_f, 'diff(y,t));
EL: EL_f - 'diff(y,t)*doof_dooyp = C;
radcan(%);
EL_dy : solve(EL, y);
ode2(EL_dy[1]^2,y,t);

but the ode2 solver cannot handle the nonlinear differential equation.

Rolling Ball: Angular momentum but no radius

The rotational energy is $E_{\text{rot}}={\frac {1}{2}}I\omega ^{2}$ and by applying non-slipping condition $v=\omega r$ we get $E_{\text{rot}}={\frac {v^{2}}{2r^{2}}}I$. Note that actually the ball is rolling on a curve, and thus the given slipping condition is only an approximation.

For the simplified case, the calculation is similar to the previous one, and using Maxima, we get

energy : 1/2*m*v^2 + 1/2*I*v^2/r^2= m*g*y;
. . .

gives

${\begin{aligned}{\frac {\sqrt {m\,{{r}^{2}}+I}}{{\sqrt {2gmy}}r\,{\sqrt {{{\left({\frac {d}{dt}}y\right)}^{2}}+1}}}}=C&&\Rightarrow &&y(1+y'{^{2}})={\frac {mr^{2}+I}{2gmr^{2}C^{2}}}=k_{1}^{2}\end{aligned}}$

and thus only the constant $k_{1}$ differs from the case with no angular momentum.

Friction

The forces on the path. Actually the sliding particle is infinitemal small.

The normal force follows the path, and thus is given by ${\vec {T}}={\frac {dx}{ds}}{\vec {x}}+{\frac {dy}{ds}}{\vec {y}}$, but The friction depends on the normal force of the path. The normal force is perpendicular to the previous, thus we have

${\vec {N}}=-{\frac {dy}{ds}}{\vec {x}}+{\frac {dx}{ds}}{\vec {y}}$

The conservation of energy does not apply here, but we have Newton's Second Law, ${\vec {F}}=m{\frac {d{\vec {v}}}{dt}}$. We need the components along the curve $s$. Thus we have

${\begin{aligned}{\vec {F}}&={\vec {G}}-{\vec {F}}_{\mu }\\&=mg{\frac {dy}{ds}}-\mu mg{\frac {dx}{ds}}\end{aligned}}$

Clearly, for the left hand side of NII we have $m{\frac {dv}{dt}}=m{\frac {dv}{ds}}{\frac {ds}{dt}}=m{\frac {dv}{ds}}v=mv{\frac {dv}{ds}}=m{\frac {1}{2}}{\frac {dv^{2}}{ds}}$, and by including the differential part only, we have

${\begin{aligned}{\frac {1}{2}}v^{2}&=g(y-\mu x)\\v&={\sqrt {2g(y-\mu x)}}\end{aligned}}$

and $f$ for the Euler--Lagrange equation is $f={\sqrt {\frac {1+y'^{2}}{2g(y-\mu x)}}}$

Euler--Lagrange

NII : 1/2*v^2 = g*(y(x)-mu*x);

v_sol : solve( NII, v);
v_sol : v_sol[2];

EL_f : rhs( sqrt(1+'diff(y(x),x)^2)/v_sol );
df_dy : diff(EL_f, y(x));
df_dyp : diff(EL_f, 'diff(y(x),x));
d_dx : diff( df_dyp, x);


EL : df_dy - d_dx = 0; 
Elrad : radcan( EL );

num( lhs(ELrad) )/sqrt(2)/sqrt(y(x)-mu*x)=0;
ratsimp(%);

${\begin{aligned}\left(2\mu x-2\operatorname {y} (x)\right)\,\left({\frac {{d}^{2}}{d{{x}^{2}}}}\operatorname {y} (x)\right)-\mu {{\left({\frac {d}{dx}}\operatorname {y} (x)\right)}^{3}}-{{\left({\frac {d}{dx}}\operatorname {y} (x)\right)}^{2}}-\mu \left({\frac {d}{dx}}\operatorname {y} (x)\right)-1=0\\2\left(y-\mu x\right)y''+\left(1+\left(y'\right)^{2}\right)\left(1+\mu y'\right)=0\end{aligned}}$

$\int _{a}^{b}\!f(x)\,dx\,$

Reduction

Remember that $y''dy=y'd(y')$. Then, note that ${\frac {d}{dx}}\left(y-\mu x\right)=y'-\mu$. Thus, we multiply EL equation by $y'-\mu$ to obtain

$-{\frac {y'-\mu }{y-\mu x}}={\frac {2(y'-\mu )y''}{(1+y'^{2})(1+\mu y')}}$

The left hand side can be integrated:

$-\int {\frac {y'-\mu }{y-\mu x}}dx=-\ln |y-\mu x|+C_{1}$

The right hand side can be integrated using partial fraction decompisition

$\int {\frac {2(y'-\mu )y''}{(1+y'^{2})(1+\mu y')}}dx=\int {\frac {2y'}{1+y'^{2}}}y''-{\frac {2\mu }{1+\mu y'}}y''dx=\ln |1+y'^{2}|-2\ln |1+\mu y'|+C_{2}$

Together we have

$-\ln |y-\mu x|+C_{1}=\ln |1+y'^{2}|-2\ln |1+\mu y'|+C_{2}$

that can be written as

$\ln {\frac {1}{|y-\mu x|}}=\ln {\frac {1+y'^{2}}{(1+\mu y')^{2}}}+C_{3}$

and it gives finally

${\begin{aligned}{\frac {C}{y-\mu x}}={\frac {1+y'^{2}}{(1+\mu y')^{2}}}\iff (1+\mu y')^{2}=C(y-\mu x)(1+y'^{2})\end{aligned}}$

depends(y,x );
EL: 2*( y - mu*x )*diff( y,x,2) + (1 + diff(y,x)^2)*(1+mu*diff(y,x)) = 0;

factor( ratsimp(solve(EL, diff(y,x,2))*(diff(y,x)-mu)*2/(1+diff(y,x)^2)/(1+mu*diff(y,x))) );

eq1 : integrate( rhs( EL_2[1]),x) + log(C);
eq2 : integrate( partfrac( lhs( EL_2[1]), diff(y,x) ), x);

exp(eq1)=exp(eq2);

Solution

The solution can be obtained by setting $y'={\tfrac {dy}{dx}}=\cot({\tfrac {1}{2}}\theta )$ which implies $dx=\tan {\tfrac {1}{2}}\theta dy$ and we have $1+y'^{2}=\sin ^{-2}{\tfrac {\theta }{2}}$.

We solve for $x$, and get

$y-\mu x=C{\frac {(1+\mu y')^{2}}{1+y'^{2}}}=C\left(\sin {\frac {\theta }{2}}+\mu \cos {\frac {\theta }{2}}\right)^{2}$

If $\mu =0$ which means no friction, we get $y=C\sin ^{2}(\theta /2)=k(1-\cos(\theta ))$, which is the result obtained earlier.

${\begin{aligned}du&={\frac {c}{2}}{\frac {\sin \theta +2\mu \cos \theta -\mu ^{2}\sin \theta }{\cot {\tfrac {1}{2}}\theta -\mu }}d\theta \\dy&={\frac {c}{2}}{\frac {\cos {\tfrac {1}{2}}\theta (\sin \theta +2\mu \cos \theta -\mu ^{2}\sin \theta )}{\cos {\tfrac {1}{2}}\theta -\mu \sin {\tfrac {1}{2}}\theta }}d\theta \end{aligned}}$.

oo: (1+mu*cot(t/2))^2/(1+cot(t/2)^2);
oo1 : oo, t/2 = s;
trigrat(oo1);
trigexpand(%);
trigsimp(%);
oo2 : expand(%);
part(oo2,1) + part(oo2,2) + trigsimp( part(oo2, 3) + part(oo2,4) );
oo3 : factor(%);
oo3, s = t/2;

Maxima:

• trigreduce : product of sinuses and cosines as a Fourier sum (with terms containing only a single sin or cos).
• trigexpand : no multiple angles. Uses sum-of-angles formulas
• trigsimp : Pythagorean identity
• substitution, eg. eq1, 2*x = y;
• trigrat : does many things?

Rolling Ball with radius

$\omega ={\frac {r_{\text{curve}}+r}{r}}{\frac {d\alpha }{dt}}-{\frac {d\alpha }{dt}}={\frac {\rho }{r}}{\frac {d\alpha }{dt}}={\frac {v}{r}}$

The conservation of energy:

$mgy={\frac {1}{2}}mv^{2}+{\frac {1}{2}}{\frac {2}{5}}mr^{2}v^{2}/r^{2}$

Beltrami Indentity

E-L states: ${\frac {\partial L}{\partial y}}={\frac {d}{dx}}{\frac {\partial L}{\partial y'}}$, but ${\frac {dL}{dx}}=y'{\frac {\partial L}{\partial y}}+y''{\frac {\partial L}{\partial y'}}+{\frac {\partial L}{\partial x}}$, and now $\partial L/\partial x=0$ and by substituting the first result, we have

${\begin{aligned}{\frac {dL}{dx}}-(y'{\frac {d}{dx}}{\frac {\partial L}{\partial y'}}+y''{\frac {\partial L}{\partial y'}})=0\\\Leftrightarrow \\{\frac {dL}{dx}}-{\frac {d}{dx}}\left(y'{\frac {\partial L}{\partial y'}}\right)={\frac {d}{dx}}\left(L-y'{\frac {\partial L}{\partial y'}}\right)=0\end{aligned}}$

and thus Beltrami follows.

References

https://mathworld.wolfram.com/BrachistochroneProblem.html

https://physicscourses.colorado.edu/phys3210/phys3210_sp20/lecture/lec04-lagrangian-mechanics/

http://hades.mech.northwestern.edu/images/e/e6/Legeza-MechofSolids2010.pdf

https://www.tau.ac.il/~flaxer/edu/course/computerappl/exercise/Brachistochrone%20Curve.pdf

https://mate.uprh.edu/~urmaa/reports/brach.pdf The Nonlinear Brachistochrone Problem with Friction Pablo V. Negr´on–Marrero∗ and B´arbara L. Santiago–Figueroa

https://medium.com/cantors-paradise/the-famous-problem-of-the-brachistochrone-8b955d24bdf7

https://wiki.math.ntnu.no/_media/tma4180/2015v/calcvar.pdf BASICS OF CALCULUS OF VARIATIONS MARKUS GRASMAIR

http://www.doiserbia.nb.rs/img/doi/0354-5180/2012/0354-51801204697M.pdf

http://info.ifpan.edu.pl/firststep/aw-works/fsV/parnovsky/parnovsky.pdf Some Generalisations of Brachistochrone Problem. A.S. Parnovsky

[https://arxiv.org/pdf/1604.03021.pdf Tautochrone and Brachistochrone Shape Solutions for Rocking Rigid Bodies. Patrick Glaschke April 12, 2016]

https://issuu.com/nameou/docs/math_seminar_paper A complete detailed solution to the brachistochrone problem. N. H. Nguyen.

https://arxiv.org/pdf/1908.02224.pdf Brachistochrone on a velodrome. GP Benham, C Cohen, E Brunet and C Clanet

https://arxiv.org/pdf/1712.04647.pdf On the brachistochrone of a fluid-filled cylinder. Srikanth Sarma Gurram, Sharan Raja, Pallab Sinha Mahapatra and Mahesh V. Panchagnula.

https://arxiv.org/pdf/1001.2181.pdf A Detailed Analysis of the Brachistochrone Problem R.Coleman

https://math.stackexchange.com/questions/3068293/euler-lagrange-equation-for-the-brachistochrone-problem-with-friction

https://math.stackexchange.com/questions/3685969/brachistochrone-problem-including-friction-reducing-a-differential-equation

https://www.jstor.org/stable/2974953?seq=1#metadata_info_tab_contents Exploring the Brachistochrone Problem. LaDawn Haws and Terry Kiser

https://math.stackexchange.com/questions/3077935/solving-the-euler-lagrange-equation-for-the-brachistochrone-problem-with-frictio.