Euler-Lagrange equation alternative form

Question

I have the following exercise:

Let $a,b,A,B\in \mathbb{R},a<b,f\in C^2\left(\left[a,b\right]\times \mathbb{R}\right),$ and $J:\mathcal{M}\rightarrow\mathbb{R}$ given by $$J\left(y\right)=\int_a^bf\left(x,y\left(x\right)\right)\cdot\sqrt{1+y'^2\left(x\right)}\text{ dx}$$ where $$\mathcal{M}=\left\lbrace w\in C^1\left(\left[a,b\right]\right)\mid w\left(a\right)=A \wedge w\left(b\right)=B\right\rbrace$$

By notation, $f=f\left(x,u\right)$. Show that Euler-Lagrange's equation can be rewritten as: $$f_u-f_x\cdot y'-f\cdot \frac{y''}{1+y'^2}=0.$$ And this is my attempt so far:

Let $F\left(x,u,\xi\right)=f\cdot\sqrt{1+\xi^2}$, we know that the Euler-Lagrange's equation is given by the following expression: $$F_u\left(x,y\left(x\right),y'\left(x\right)\right)-\frac{d}{dx}F_{\xi}\left(x,y\left(x\right),y'\left(x\right)\right)=0.$$ Thus, I try to get both terms separately and then its substraction. Therefore, $$F_u\left(x,u,\xi\right)=\sqrt{1+\xi^2}\cdot f_u\implies F_u\left(x,y\left(x\right),y'\left(x\right)\right)=\sqrt{1+y'^2\left(x\right)}\cdot f_u$$ $$F_\xi\left(x,u,\xi\right)=\frac{\xi}{\sqrt{1+\xi^2}}\cdot f\implies F_{\xi}\left(x,y\left(x\right),y'\left(x\right)\right)=\frac{y'\left(x\right)}{\sqrt{1+y'^2\left(x\right)}}\cdot f$$ $$\frac{d}{dx}F_{\xi}\left(x,y\left(x\right),y'\left(x\right)\right)=\frac{y'\left(x\right)}{\sqrt{1+y'^2\left(x\right)}}\cdot f_x+\frac{y''\left(x\right)}{\sqrt{1+y'^2\left(x\right)}}\cdot f \left(1-\frac{y'\left(x\right)}{1+y'^2\left(x\right)}\right)$$ So the Euler-Lagrange's equation becomes $$\sqrt{1+y'^2\left(x\right)}\cdot f_u-\frac{y'\left(x\right)}{\sqrt{1+y'^2\left(x\right)}}\cdot f_x-\frac{y''\left(x\right)}{\sqrt{1+y'^2\left(x\right)}}\cdot f \left(1-\frac{y'\left(x\right)}{1+y'^2\left(x\right)}\right)=0$$

I am stuck here, I do not know how to proceed to get the required expression. Any help or hint is welcomed. Thanks.

Answer 1

Your mistake is in computing $\dfrac{d}{dx}\left( \dfrac{\partial F}{\partial \xi}(x, y(x), y'(x))\right)$. It should be \begin{align} & \qquad \dfrac{d}{dx}\left( \dfrac{\partial F}{\partial \xi}(x, y(x), y'(x))\right) \\ &= \dfrac{d}{dx} \left( f(x,y(x)) \cdot \dfrac{y'(x)}{\sqrt{1 + y'(x)^2}} \right) \\ &= \left( \dfrac{\partial f}{\partial x}(x, y(x)) + \dfrac{\partial f}{\partial u}(x, y(x)) \cdot y'(x) \right) \cdot \dfrac{y'(x)}{\sqrt{1 + y'(x)^2}} \\ &+ f(x,y(x)) \cdot \dfrac{d}{dx} \left( \dfrac{y'(x)}{\sqrt{1 + y'(x)^2}} \right) \end{align}

For the second term, I'm sure you can apply the quotient rule appropriately. The term you missed out was the one involving $f_u$. Simplifying, I got

\begin{align} \text{LHS} &= \left( \dfrac{\partial f}{\partial x}(x, y(x)) + \dfrac{\partial f}{\partial u}(x, y(x)) \cdot y'(x) \right) \cdot \dfrac{y'(x)}{\sqrt{1 + y'(x)^2}} \\ &+ f(x,y(x)) \cdot \dfrac{y''(x)}{\left( 1 + y'(x)^2 \right)^{3/2}} \end{align}

Substituting this into the Euler Lagrange equations yields the desired answer.