Calculus of Variations

Question

In the Calculus of Variations there is a passage from Euler's characteristic equation: $$ \frac {\partial F}{\partial y} - \frac {d}{dx} \left(\frac {\partial F}{\partial y'} \right)=0 $$ in which $$ F=F(x,y,y') $$ If the function F is made explicitly independent of x then: $$ F=F(y,y') $$ Then by integration of Euler's characteristic equation one finds: $$ F - y' \frac{\partial F}{\partial y'}=k $$ I'm having great difficulty trying to do that integration, hence I'd appreciate any help!!!

Answer 1

I guess there are a couple of approaches, but this is an interesting one: we know that $y=y(x)$ is a minimizer (or extremizer) of $\displaystyle \int F(x,y,y')\,dx=\int F(y,y')x'\,dy$ where $x'=\frac{dx}{dy}$, where it has been used that $F$ is independent of $x$. Now we can rewrite this as $\displaystyle\int \hat{F}(y,x')x'\,dy$ where $\hat{F}(y,x')=F(y,\frac{1}{x'})$. Now $y=y(x)$ minimizing the action is equivalent to $x=x(y)$ minimizing this last integral, so the Euler-Lagrange equations in these coordinates where we are treating $y$ as the independent variable is just $\displaystyle\frac{d}{dy}\bigg(\frac{\partial \hat{F}x'}{\partial x'}\bigg)=0\implies \frac{\partial \hat{F}x'}{\partial x'}=C\implies \hat{F}+x'\frac{\partial \hat{F}}{\partial x'}=C\implies \hat{F}+x'\frac{\partial F}{\partial y'}\frac{-1}{x'^2} =C\implies \hat{F}-\frac{1}{x'}\frac{\partial F}{\partial y'}=C\implies F-y'\frac{\partial F}{\partial y'}=C.$

Answer 2

$$\frac{d}{d x}\bigg( F -y' \frac{\partial F}{\partial y'}\bigg) = y' \frac{\partial F}{\partial y} + y'' \frac{\partial F}{\partial y'} - y'' \frac{\partial F}{\partial y'} - y'\frac{d }{d x}\bigg(\frac{\partial F}{\partial y'}\bigg)$$

$$=y'\bigg( \frac{\partial F}{\partial y} - \frac{d}{d x} \bigg( \frac{\partial F}{\partial y'}\bigg)\bigg) = 0$$