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Given F:

$$ F(x,y,y\prime) = 2\cdot \pi \cdot y \cdot \sqrt{1+(y\prime)^2} $$

We can derive the following Euler-Lagrange equation (I know how to do this part):

$$ \frac{d}{dx}\left(\frac{y\cdot y\prime}{\sqrt{1+(y\prime)^2}}\right) - \left(\sqrt{1+(y\prime)^2}\right) = 0 $$

Can someone please show me how to use:

$$ \frac{1}{y\prime}\left[\frac{d}{dx}\left(F-\frac{\partial{F}}{\partial{y\prime}}\cdot\frac{dy}{dx}\right)-\frac{\partial{F}}{\partial{x}}\right] = 0$$

To derive:

$$ y\cdot y\prime\prime - \left(y\prime\right)^2 - 1 = 0 $$

Here is my working so far but it is wrong. Can someone please show me the error of my ways and provide a step by step method please?

I have $$ \frac{\partial{F}}{\partial{x}} = 0, $$

$$ \frac{\partial{F}}{\partial{y\prime}} \cdot \frac{dy}{dx} = \left(\frac{y\cdot y\prime}{\sqrt{1+(y\prime)^2}}\right) \cdot y\prime = \left(\frac{y\cdot y\prime^2}{\sqrt{1+(y\prime)^2}}\right), $$

$$ \left(F-\frac{\partial{F}}{\partial{y\prime}}\cdot\frac{dy}{dx}\right) = 2\cdot \pi \cdot y \cdot \sqrt{1+(y\prime)^2} - \left(\frac{y\cdot y\prime^2}{\sqrt{1+(y\prime)^2}}\right) $$

Differntiating with respect to x:

$$ \frac{d}{dx}\left(2\cdot \pi \cdot y\prime \cdot \sqrt{1+(y\prime)^2} - \left(\frac{y\cdot y\prime^2}{\sqrt{1+(y\prime)^2}}\right)\right) = 2 \cdot \pi \cdot y \cdot \sqrt{1+(y\prime)^2} + \frac{2 \cdot \pi \cdot y \cdot y\prime \cdot y\prime\prime}{\sqrt{1+y\prime^2}} - \frac{y\prime^3}{\sqrt{1+y\prime^2}} - \frac{2 \cdot y \cdot y\prime \cdot y\prime\prime}{\sqrt{1+y\prime^2}} + \frac{y \cdot y\prime^2 \cdot y\prime\prime}{(1+y\prime^2)^{3/2}} $$

So...

$$ \frac{1}{y\prime}\left[\frac{d}{dx}\left(F-\frac{\partial{F}}{\partial{y\prime}}\cdot\frac{dy}{dx}\right)-\frac{\partial{F}}{\partial{x}}\right] = 2 \cdot \pi \cdot y\prime \cdot \sqrt{1+(y\prime)^2} + \frac{2 \cdot \pi \cdot y \cdot y\prime \cdot y\prime\prime}{\sqrt{1+y\prime^2}} - \frac{y\prime^3}{\sqrt{1+y\prime^2}} - \frac{2 \cdot y \cdot y\prime \cdot y\prime\prime}{\sqrt{1+y\prime^2}} + \frac{y \cdot y\prime^2 \cdot y\prime\prime}{(1+y\prime^2)^{3/2}} $$

$$ = 2 \cdot \pi \cdot \sqrt{1+(y\prime)^2} + \frac{2 \cdot \pi \cdot y \cdot y\prime\prime}{\sqrt{1+y\prime^2}} - \frac{y\prime^2}{\sqrt{1+y\prime^2}} - \frac{2 \cdot y \cdot y\prime\prime}{\sqrt{1+y\prime^2}} + \frac{y \cdot y\prime \cdot y\prime\prime}{(1+y\prime^2)^{3/2}}$$

So...

How on earth do I get from here to $$ y\cdot y\prime\prime - \left(y\prime\right)^2 - 1 = 0 $$

???

It's at this point I'm failing to continue. I'm sure I'm being daft but I would really appreciate some assistance from this point forward if anyone can help.

Thankyou.

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And now you should differentiate the last expression with respect to $x$. –  Siminore Apr 9 '13 at 13:12

1 Answer 1

Beginning from the Euler-Lagrange equations $$ \dfrac{d}{dx}\left(\dfrac{yy'}{\sqrt{1+(y')^2}}\right)-\sqrt{1+(y')^2}=0 $$

we get that $$ (yy')'\left(1+(y')^2\right)^{-1/2}-yy'\left(1+(y')^2\right)^{-3/2}y'y''-\left(1+(y')^2\right)^{1/2}=0 $$

Since $1+(y')^2>0$ we can multiply both sides by $\left(1+(y')^2\right)^{3/2}$ and obtain $$ (yy')'\left(1+(y')^2\right)-y(y')^2y''-\left(1+(y')^2\right)^2=0 $$ Grouping the first and last term, and expanding the term $(yy')'$ we obtain $$ \left(yy''+(y')^2-1-(y')^2\right)\left(1+(y')^2\right)-y(y')^2y''=0 $$ Simplifying the first term we obtain $$ \left(yy''-1\right)\left(1+(y')^2\right)-y(y')^2y''=0 $$ and after distributing the first term we obtain $$ yy''-1+yy''(y')^2-(y')^2-y(y')^2y''=0 $$ which leads to your desired equation: $yy''-(y')^2-1=0$

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Thankyou for that, very useful. One last question, the term $$ \left(yy\prime \right)\prime $$ How do you expand this? –  Mike Shaw Apr 9 '13 at 18:34
    
Using Leibnitz you get: $(yy')'=y'y'+yy''$ –  Heberto del Rio May 8 '13 at 2:35

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