# Euler-Lagrange Query

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

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 Apr 9 '13 at 18:34
Using Leibnitz you get: $(yy')'=y'y'+yy''$ – Heberto del Rio May 8 '13 at 2:35