i'm just working through Calculus of variations problems and I'd like to cement my understanding and see if theres anywhere i'm going wrong. so if possibe could someone please provide a check on the workings. Thank you in advance.
Find the general solution of the Euler-Lagrange equation associated with the function $$F(y) = \int_{a}^{b} \sqrt{x}\sqrt{1+|y'|^{2}} dx $$
- Take the lagrangian to be $$\Lambda(x,y,y') = \sqrt{x}\sqrt{1+|y'|^{2}} $$
- then the Euler-Lagrange equation is $$\frac{\partial \Lambda}{\partial y} - \frac{d}{dx}\frac{\partial \Lambda}{\partial y'} = 0$$
- Substituting $$\frac{\partial \Lambda}{\partial y} = 0$$ with $$\frac{\partial \Lambda}{\partial y'} = \frac{y'\sqrt{x}}{\sqrt{1+y'^{2}}} $$ then $$\frac{d}{dx}\frac{\partial \Lambda}{\partial y'} = \frac{2xy'' + y'[1+y'^2]}{2x(1+y'^2)^{3/2}} \implies$$ $$\frac{d}{dx}\frac{\partial \Lambda}{\partial y'} - \frac{\partial \Lambda}{\partial y} = \frac{2xy'' + y'[1+y'^2]}{2x(1+y'^2)^{3/2}} = 0$$
- substitute $y' = v \implies y'' = v'$ and we have $$\frac{2xv' + v[1+v^2]}{2x(1+v^2)^{3/2}} = 0 \implies$$ $$\frac{v'}{(1+v^2)^{3/2}} = \frac{-v[1+v^{2}]}{2x(1+v^2)^{3/2}} \implies$$ $$v' = \frac{-v[1+v^{2}]}{2x} \implies$$ $$2 \frac{dv}{dx} = \frac{-v[1+v^{2}]}{2x} \implies$$ $$2 \frac{dv}{-v(1+v^2)} = \frac{dx}{x} \implies$$ $$ln(v^{2}+1)-2ln(v)=ln x+c \implies$$ $$ ln \frac{v^{2}+1}{v^2} = ln x + c \implies$$ $$1 + \frac{1}{v^2} = e^{c}x \implies$$ $$ v = \pm \frac{1}{\sqrt{xe^{c}-1}} \implies$$ $$\frac{dy}{dx} = \pm\frac{1}{\sqrt{xe^{c}-1}} \implies$$ $$y = \pm \int \frac{1}{\sqrt{xe^{c}-1}} = \pm \frac{2 \sqrt{xe^{c}-1}}{e^{c}} + C$$ As a general solution:
If there's any mistakes please, point them out. I appreciate it greatly and thanks for taking the time to read this.
If your function doesn't depend on $y$ then Euler-Lagrange equation is realy simple. You can rewrite this as $\frac{\partial \Lambda}{\partial y'}=C$