I have been working on solving Euler-Lagrange Equation problems in differential equations, specifically in Calculus of Variations, but this one example has me stuck. I am probably making mistakes in my integration.
I am supposed to solve the Euler-Lagrange equation given that $f(t,x(t),x^{\prime}(t))=f(t,u,v)=\frac{\sqrt{1+v^2}}{u}$.
I know that $f_u=-\frac{\sqrt{1+v^2}}{u^2}$ and $f_v=\frac{v\sqrt{1+v^2}}{u(1+v^2)}$. Plugging this into the Euler-Lagrange Equation, $f_u-\frac{d}{dt}f_v$ gives
$$\frac{\sqrt{1+(x^{\prime})^2}}{x^2}-\left(\frac{x^{\prime}\sqrt{1+(x^{\prime})^2}}{x(1+(x^{\prime})^2)}\right)^{\prime}=0.$$
I cannot figure out from here how to solve it. Have I made a mistake up to now? Could someone help me out in actually solving this for x?
Best Answer
You are on the right track. Write $$ f_v = \frac{v}{u\sqrt{1+v^2}} \,. $$ Then $$ \frac{df_v}{dt} = \frac{v}{\sqrt{1+v^2}}\frac{d}{dt}\left(\frac{1}{u}\right) + \frac{1}{u}\frac{d}{dt}\left(\frac{v}{\sqrt{1+v^2}}\right) $$ Expand: $$ \frac{df_v}{dt} = -\frac{v}{u^2\sqrt{1+v^2}}\frac{du}{dt} +\left[\frac{1}{u\sqrt{1+v^2}}-\frac{v^2}{u(1+v^2)^{3/2}}\right]\frac{dv}{dt} $$ Equate to $f_u$ to get $$ -\frac{1+v^2}{u} = -\frac{v}{u}\frac{du}{dt} +\frac{1}{1+v^2}\frac{dv}{dt} $$ Use $ v = du/dt$: $$ -\frac{1}{u} = \frac{1}{1+v^2}\frac{dv}{dt} $$ The ODE becomes $$ u\frac{d^2u}{dt^2} + \left[\frac{du}{dt}\right]^2 +1 = 0 $$ Use ODE solution techniques to solve.