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For $x=(x_1,x_2) \in \mathbb R^2$, we denote by $$L(x,\dot{x})= \dot{x_{1}}^{2}+x_{1}\, \dot{x_{1}}\, \dot{x_{2}}+\dot{x_{2}}$$ the Lagrangian associated to an Hamiltonian for an unbounded operator $L$.

I would like solve the Euler–Lagrange equation given by $$\frac{d}{ds}\frac{\partial L}{\partial \dot{x}} = \frac{\partial L}{\partial x} .$$ Thank you in advance

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    $\begingroup$ A good start would be to compute the partial derivatives and to write the equation explicitly. $\endgroup$ – lisyarus Oct 27 '16 at 16:16
  • $\begingroup$ @ lisyarus, is what we have, $\frac{\partial}{\partial x} = (\frac{\partial}{\partial x_1},\frac{\partial}{\partial x_2})$ ? $\endgroup$ – Z. Alfata Oct 27 '16 at 16:20
  • $\begingroup$ Yes. You can think of it as of two equations, one for each coordinate. $\endgroup$ – lisyarus Oct 27 '16 at 16:22
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Let us go step by step:

1) the term $\dot x_2$ is a total time-derivative, thus unimportant for the equations of motion. So I will consider the Lagrangian $\tilde L = \dot x_1^2 + x_1 \dot x_1 \dot x_2$

2) $x_2$ is cyclic and thus $$p_2 = \frac{\partial \tilde L}{\partial \dot x_2} = x_1 \dot x_1$$ is conserved

3) as the Lagrangian does not depend explicitly on time, we have that the "energy" $$ H = \sum_{j=1}^2\frac{\partial \tilde L}{\partial \dot x_j} \dot x_j - \tilde L = \dot x_1 (\dot x_1 + x_1 \dot x_2) = \dot x_1^2 +p_2 \dot x_2 $$ is conserved

Now we solve: 2) $$p_2 = x_1 \dot x_1$$ has the solution $$x_1 = \pm\sqrt{p_2 t + \alpha}. \tag{1}$$

Inserting into the expression 3) for $H$, we have the equation $$ H= \frac{p_2^2}{2 (p_2 t +\alpha)} + p_2 \dot x_2$$ with the solution $$ x_2 = \frac{H t} {p_2} + \beta -\frac12 \ln\left(p_2 t + \alpha \right) \tag{2} =\frac{H t} {p_2} + \beta - \ln x_1 .$$

The general solution is given by (1) and (2) where $\alpha,\beta,p_2,H$ are the 4 constants of integration.

Edit: in order that $x_2$ remains reals, we have to choose the `+' sign in (1). Here is the orbit for $H=p_2=1$ and $\alpha=\beta=0$.

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