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Heshy
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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)$$$\mathcal{L}=-2X'(t)\ln{x'(t)}-2Y'(t)\ln{y'(t)}+\Lambda y(t)x'(t)$$

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$$$\dfrac{\partial\mathcal{L}}{\partial x}-\dfrac{d}{dt}\dfrac{\partial\mathcal{L}}{\partial x^\prime}=0$$

$-\dfrac{d}{dt}\left(-2\dfrac{X^\prime}{x^\prime}+\Lambda y\right)=0$$$-\dfrac{d}{dt}\left(-2\dfrac{X^\prime}{x^\prime}+\Lambda y\right)=0$$

$2\dfrac{X^\prime}{x^\prime}-\Lambda y=c_1$$$2\dfrac{X^\prime}{x^\prime}-\Lambda y=c_1$$

$\Lambda y x^\prime + c_1 x^\prime - 2X^\prime=0$$$\Lambda y x^\prime + c_1 x^\prime - 2X^\prime=0$$

And for $y$:

$\dfrac{\partial\mathcal{L}}{\partial y}-\dfrac{d}{dt}\dfrac{\partial\mathcal{L}}{\partial y^\prime}=0$$$\dfrac{\partial\mathcal{L}}{\partial y}-\dfrac{d}{dt}\dfrac{\partial\mathcal{L}}{\partial y^\prime}=0$$

$\Lambda x^\prime-\dfrac{d}{dt}\left(-2\dfrac{Y^\prime}{y^\prime}\right)=0$$$\Lambda x^\prime-\dfrac{d}{dt}\left(-2\dfrac{Y^\prime}{y^\prime}\right)=0$$

$2\dfrac{Y^\prime}{y^\prime}+\Lambda x=c_2$$$2\dfrac{Y^\prime}{y^\prime}+\Lambda x=c_2$$

$-\Lambda x y^\prime + c_2 y^\prime - 2Y^\prime=0$$$-\Lambda x y^\prime + c_2 y^\prime - 2Y^\prime=0$$

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?

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?

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?

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Heshy
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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}(-2\dfrac{X^\prime}{x^\prime}+\Lambda y)=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}(-2\dfrac{Y^\prime}{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?

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 $x$:

$\dfrac{\partial\mathcal{L}}{\partial x}-\dfrac{d}{dt}\dfrac{\partial\mathcal{L}}{\partial x^\prime}=0$

$-\dfrac{d}{dt}(-2\dfrac{X^\prime}{x^\prime}+\Lambda y)=0$

$2\dfrac{X^\prime}{x^\prime}-\Lambda y=c_1$

$\Lambda y x^\prime + c_1 x^\prime - 2X^\prime=0$

And for $y$:

$\dfrac{\partial\mathcal{L}}{\partial y}-\dfrac{d}{dt}\dfrac{\partial\mathcal{L}}{\partial y^\prime}=0$

$\Lambda x^\prime-\dfrac{d}{dt}(-2\dfrac{Y^\prime}{y^\prime})=0$

$2\dfrac{Y^\prime}{y^\prime}+\Lambda x=c_2$

$-\Lambda x y^\prime + c_2 y^\prime - 2Y^\prime=0$

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?

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?

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