I have the following calculus of variations problem:
$$\mathcal{L}=-2X'(t)\ln{x'(t)}-2Y'(t)\ln{y'(t)}+\Lambda y(t)x'(t)$$
where $x(t)$ and $y(t)$ are the functions I'm interested in, and $X(t)$ and $Y(t)$ are given as input (I'd like them to remain general).
I know that this has a closed-form solution, because I tried plugging it into EulerEquations in Mathematica and then feeding the result directly (which included second derivatives) to DSolve, and to my surprise it gave me a closed-form mess.
Needs["VariationalMethods`"] EulerEquations [-2X'[t] Log[x'[t]]-2Y'[t] Log[y'[t]]+\[Lambda] y[t] x'[t],{x[t],y[t]},t] DSolve[%,{x[t],y[t]},{t}] I then tried to do the first steps by hand, in the hopes of getting a less ugly closed form solution.
The Euler-Lagrange equations are, for $x$:
$$\dfrac{\partial\mathcal{L}}{\partial x}-\dfrac{d}{dt}\dfrac{\partial\mathcal{L}}{\partial x^\prime}=0$$
$$-\dfrac{d}{dt}\left(-2\dfrac{X^\prime}{x^\prime}+\Lambda y\right)=0$$
$$2\dfrac{X^\prime}{x^\prime}-\Lambda y=c_1$$
$$\Lambda y x^\prime + c_1 x^\prime - 2X^\prime=0$$
And for $y$:
$$\dfrac{\partial\mathcal{L}}{\partial y}-\dfrac{d}{dt}\dfrac{\partial\mathcal{L}}{\partial y^\prime}=0$$
$$\Lambda x^\prime-\dfrac{d}{dt}\left(-2\dfrac{Y^\prime}{y^\prime}\right)=0$$
$$2\dfrac{Y^\prime}{y^\prime}+\Lambda x=c_2$$
$$-\Lambda x y^\prime + c_2 y^\prime - 2Y^\prime=0$$
These two equations are nice-looking and symmetric, but I tried plugging them into DSolve and it didn't work (just returned the initial DSolve).
DSolve[{\[Lambda] y[t]D[x[t],t]+c1 D[x[t],t]-2Xdot[t]==0,-\[Lambda] x[t]D[y[t],t]+c2 D[y[t],t]-2Ydot[t]==0},{x[t],y[t]},t] Can anyone help me proceed from here?
