Equations of motion only have a solution for very specific initial conditions

Question

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?

Answer 1

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 ).

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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.

Answer 2

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.

Answer 3

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.