Brachistochrone Problem Including Friction - Reducing a Differential Equation

Question

I have been investigating the brachistochrone problem with friction and in my derivations, I would like help solving the Euler-Lagrange equation below

$\frac{d}{dx}\frac{\partial F}{\partial y'}=\frac{\partial F}{\partial y}$ where $F=\sqrt{\frac{1+y'^2}{2g(y-\mu x)}}.$

I can get up to

$$\frac{d}{dx}\frac{y'}{\sqrt{2g(y-\mu x)(1+y'^2)}}=-\sqrt{\frac{(1+y'^2)}{2g}}\frac{1}{2(y-\mu x)^\frac32}.$$

But I am unsure how the equation above reduces into $$(1+y'^2)(1+\mu y')+2(y-\mu x)y''=0\tag{29},$$ as seen in equation (29) of this Wolfram page.

I am quite new to calculus and would appreciate a step by step solution.

Answer 1

$\frac{d}{dx}.\frac{y'}{\sqrt{2g(y-ux)(1+y'^2)}}=\frac{y''\sqrt{2g(y-ux)(1+y'^2)}-y'(\sqrt{2g}(\frac{(y'-u)(\sqrt{1+y'^2}}{2\sqrt{y-ux}}+\frac{y'y''(\sqrt{y-ux})}{\sqrt{1+y'^2}}))}{2g(y-ux)(1+y'^2)}$

So:

$(\frac{y''\sqrt{2g(y-ux)(1+y'^2)}-y'(\sqrt{2g}\frac{(y'-u)(\sqrt{1+y'^2}}{2\sqrt{y-ux}}+\frac{y'y''(\sqrt{y-ux})}{\sqrt{1+y'^2}})}{2g(y-ux)(1+y'^2)})(2(y-ux)\sqrt{(y-ux)})= -\sqrt{\frac{(1+y'^2)}{2g}} $

$y''(\sqrt{2g}\sqrt{1+y'^2}(y-ux))-y'(\sqrt{2g}((y'-u)(\sqrt{1+y'^2})+\frac{y'y''(y-ux)}{\sqrt{1+y'^2}}))= -\sqrt{\frac{(1+y'^2)}{2g}}(1+y'^2)(2g) $

$\rightarrow 2y''(y-ux)-y'[(y'-u)+\frac{2y'y''(y-ux)}{1+y'^2}]=-(1+y'^2) $

$2y''(y-ux) -y'^2+y'u-\frac{2y'^2y''(y-ux)}{1+y'^2}+(1+y'^2)=0$

$2y''(y-ux)[1-\frac{y'^2}{1+y'^2}]+1+y'u=0$

$2y''(y-ux)[\frac{1}{1+y'^2}]+(1+y'u)=0$

$2y''(y-ux)+(1+y'^2)(1+y'u)=0$

Answer 2

I have freshly derived and indicate steps involved:

Calculation is simplified by using Beltrami's special case solution when $x$ is not explicitly involved in the functional Equn below given (1). However $x$ is involved with friction,but proceeded nevertheless.

I was curious to find out what would happen to the solution:

$$F= \dfrac{\sqrt{1+y^{'2} } }{\sqrt{y- \mu x}} \tag1$$

$$ F- y' \dfrac{\partial F}{\partial y'}=C_1\tag2$$

$$\dfrac{\sqrt{1+y^{'2} } }{\sqrt{y- \mu x}}-\dfrac{y'}{\sqrt{y- \mu x}}\cdot\dfrac{y'}{\sqrt{1+y^{'2} }}= C_1 \tag3$$ Simplifying $${\sqrt{1+y^{'2} }}\cdot{\sqrt{y- \mu x}}=C_3\tag4$$ Square,differentiate using Chain Rule $$\dfrac{y- \mu x}{1+y^{'2} }= -\dfrac{y'-\mu}{2 y' y''}\tag5$$

finally $$\dfrac{y''(y- \mu x)}{1+y^{'2} }= \dfrac{\mu-y'}{2 y'} \tag6$$

whereas Wolfram Mathworld gives

$$\dfrac{y''(y- \mu x)}{1+y^{'2} }= \dfrac{1+\mu y'}{2 } \tag7$$

which is also the final equation as the one derived by Heroschizen.

However on inspection it is found (7) to be same as (6) with substitution $ y'\rightarrow \dfrac{-1}{y'}$ of an orthogonally intersecting cycloid meaning... when integrated with same BC the cycloid holding water gets inverted into one which spills water.

Right now I am unable to comment why the two procedures with and without friction lead essentially to the same result.

Answer 3

The Beltrami identity $$C~=~\frac{(y-\mu x)(1+y^{\prime 2})}{(1+\mu y^{\prime})^2} \tag{30} $$ is derived in my Phys.SE answer here. Differentiation wrt. $x$ yields the sought-for Euler-Lagrange (EL) equation $$\begin{align} 0~\stackrel{(30)}{=}~&\frac{d}{dx}\left[\frac{(y-\mu x)(1+y^{\prime 2})}{(1+\mu y^{\prime})^2}\right]\cr ~=~&\frac{1}{(1+\mu y^{\prime})^3}[(y^{\prime}-\mu)(1+y^{\prime 2})(1+\mu y^{\prime})\cr &\qquad\qquad +(y-\mu x)2y^{\prime}y^{\prime\prime}(1+\mu y^{\prime})\cr &\qquad\qquad -(y-\mu x)(1+y^{\prime 2}) 2\mu y^{\prime\prime} ]\cr ~=~&\frac{(y^{\prime}-\mu)}{(1+\mu y^{\prime})^3}\left[(1+y^{\prime 2})(1+\mu y^{\prime}) +2(y-\mu x)y^{\prime\prime}\right]. \end{align} \tag{29}$$ For the converse derivation, see e.g. this Math.SE post.