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An exercise made me consider the following Lagrangian $$L = \dot{x}_1^2+\dot{x}_2^2+2 \dot{x}_1 \dot{x}_2 + x_1^2+x_2^2.\tag{1}$$ If I didn't make a mistake the equations of motion should be given by: $$2x_1 = 2 \ddot{x}_1 + 2 \ddot{x}_2\tag{2}$$ $$2x_2 = 2 \ddot{x}_2 +2 \ddot{x}_1.\tag{3}$$ But this already implies that

$$x_1 = x_2. \tag{4}$$

So this equations of motion seem to impose an additional constraint. Moreover, there should not be a solution for initial conditions, where $x_1 \neq x_2.$ How can it be that the equations of motion already pose a constraint on the initial conditions? Is there some deeper theory behind that as to when this happens? And what does it mean that the lagrangian only has a solutions for very specific constraints?

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    $\begingroup$ We should remember this question as an example of a “good” homework-and-exercises question. $\endgroup$ Apr 25, 2022 at 3:43
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    $\begingroup$ @ZeroTheHeroDone! $\endgroup$ Apr 25, 2022 at 23:42

3 Answers 3

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with

$$L = \dot{x}_1^2+\dot{x_2}^2+2 \dot{x_1} \dot{x_2} + x_1^2+x_2^2$$

you can obtain the "mass matrix"

$$M= \left[ \begin {array}{cc} {\frac {\partial ^{2}}{\partial {{\dot{x}}_{ {1}}}^{2}}}L \left( {\dot{x}}_{{1}},{\dot{x}}_{{2}} \right) &{\frac { \partial ^{2}}{\partial {\dot{x}}_{{2}}\partial {\dot{x}}_{{1}}}}L \left( {\dot{x}}_{{1}},{\dot{x}}_{{2}} \right) \\ { \frac {\partial ^{2}}{\partial {\dot{x}}_{{2}}\partial {\dot{x}}_{{1}}}} L \left( {\dot{x}}_{{1}},{\dot{x}}_{{2}} \right) &{\frac {\partial ^{2}} {\partial {{\dot{x}}_{{2}}}^{2}}}L \left( {\dot{x}}_{{1}},{\dot{x}}_{{2}} \right) \end {array} \right] =\left[ \begin {array}{cc} 2&2\\ 2&2\end {array} \right] $$

thus the determinate of the masse matrix is zero this means that you have "constraint" in the equations of motion (the accelerations are linearly dependent ).


the equations of motion are:

$$\mathbf M\, \begin{bmatrix} \ddot{x}_1 \\ \ddot{x}_2 \\ \end{bmatrix}=\begin{bmatrix} 2\,{x}_1 \\ 2\,{x}_2 \\ \end{bmatrix}\quad\Rightarrow\\ \begin{bmatrix} \ddot{x}_1 \\ \ddot{x}_2 \\ \end{bmatrix}=\mathbf M^{-1}\,\begin{bmatrix} 2\,{x}_1 \\ 2\,{x}_2 \\ \end{bmatrix}$$

and because the determinant of M is zero you don't get any solution.

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    $\begingroup$ Wow these are really good answers. Thank you very much, hose are really great answers $\endgroup$
    – jabru
    Apr 24, 2022 at 16:51
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    $\begingroup$ @jabru If that's the case, you should choose one and make it the accepted answer. $\endgroup$
    – garyp
    Apr 24, 2022 at 19:10
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    $\begingroup$ This is elegant! $\endgroup$ Apr 25, 2022 at 3:56
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First, for dimensional reasons, let's introduce a parameter $\omega$ with dimensions of frequency and rewrite your original Lagrangian as \begin{equation} L = \dot{x}_1^2 + \dot{x}_2^2 + 2 \dot{x}_1 \dot{x}_2 + 2 \omega^2 x_1^2 + 2\omega^2 x_2^2 \end{equation}

Now let's define the new variable \begin{eqnarray} y = x_1 + x_2 \end{eqnarray} Then your Lagrangian becomes \begin{equation} L = \dot{y}^2 + 2 \omega^2 y^2 - 4 \omega^2 y x_2 + 4 \omega^2 x_2^2 \end{equation} If we vary with respect to $x_2$ we get the equation \begin{equation} x_2 = \frac{1}{2} y \end{equation} Since this is an algebraic equation (no time derivatives), we can plug the solution back into the Lagrangian, yielding \begin{equation} L = \dot{y}^2 + \omega^2 y^2 \end{equation} In this form, it is clear that the system really only describes one degree of freedom, and was simply written in a confusing way before with a redundant set of variables.

Just as a sanity check, if we vary this with respect to $y$, we get \begin{equation} \ddot{y} = \omega^2 y \end{equation} Or, in terms of the original variables, \begin{equation} \ddot{x_1} + \ddot{x}_2 = \omega^2 \left(x_1 + x_2\right) \implies \ddot{x_1} + \ddot{x}_2 = 2 \omega^2 x_1 \implies 2 \ddot{x_1} + 2 \ddot{x}_2 = 4 \omega^2 x_1 \end{equation} where I used $x_1=x_2$ (which follows from $y=x_1+x_2$ and $x_2=\frac{1}{2}y$). Setting $\omega^2=\frac{1}{2}$ yields the form of the equation of motion that you originally derived.

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If we change coordinates

$$x_{\pm}~=~x_1 \pm x_2, \tag{A}$$

then OP's 2D Lagrangian (1)

$$ L~=~L_+ + L_- \tag{B} $$

decouples into two independent 1D subsystems:

  1. An inverted harmonic oscillator $$L_+~=~T_+-V_+~=~\dot{x}_+^2 +\frac{1}{2}x_+^2, \tag{C} $$ which is unstable.

  2. An unstable static system $$L_-~=~-V_-~=~ \frac{1}{2}x_-^2 ,\tag{D} $$ consisting of a negative potential and no kinetic term. Since the Lagrangian (D) is of zeroth order, no boundary conditions are needed, cf. OP's title. The unique stationary solution$^1$ $$x_-~\approx~0\tag{E}$$ [that OP found in their eq. (4)] is unstable.

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$^1$ The $\approx$ symbol means equality modulo EOMs.

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