How can we transform a Lagrangian to obtain a new set of equations of motion?

[arpon]

Consider a Lagrangian:
\begin{equation}
\mathcal{L} = \mathcal{L}(q_1\, \dots\, q_n, \dot{q}_1\, \dots\, \dot{q}_n,t)
\end{equation}
From this Lagrangian, we get a set of $n$ equations:
\begin{equation}
\frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot{q}_i} - \frac{\partial \mathcal{L}}{\partial q_i} = 0;~~~~[i = 1,\dots ,n]
\end{equation}
By taking linear combinations of the set of $n$ equations, we can obtain a new set of $n$ equations. And this new set of equations describe the same physical system.
The new set of equations CANNOT be obtained by the Gauge transformation of Lagrangian:
\begin{equation}
\mathcal{L} \rightarrow \mathcal{L} + \frac{dF}{dt},~~~~\text{[$F$ is any arbitrary function]}
\end{equation}
But there exists some other Lagrangian which gives the new set of equations of motion. The following example will make this clear.
Consider the Lagrangian :
\begin{equation}
\mathcal{L} = \frac{m}{2}(a\dot{x}^2 + 2b\dot{x}\dot{y}+c\dot{y}^2) - \frac{k}{2}(ax^2+2bxy+cy^2),~~~~~[b^2 - ac \neq 0]
\end{equation}
By applying Euler-Lagrange equation, we obtain the equations of motion:
\begin{align}
m(a\ddot{x} + b \ddot{y}) &= -k(ax+by)\\
m(b\ddot{x} + c \ddot{y}) &= -k(bx+cy)
\end{align}
By taking linear combinations of these two equations, we can obtain a new set of equations of motion:
\begin{align}
m\ddot{x} &= -kx\\
m\ddot{y} &= -ky
\end{align}
Clearly this system can be described by the new Lagrangian :
\begin{equation}
\mathcal{L}'=\frac{1}{2}m(\dot{x}^2+\dot{y}^2) - \frac{1}{2}k(x^2 + y^2)
\end{equation}
But there exists no transformation of this form:
$$\mathcal{L} \rightarrow \mathcal{L} + \frac{dF}{dt}$$
which will give the new Lagrangian $\mathcal{L}'$.
I also tried the point transformation :
\begin{align}
x &= cu - bv\\
y&=-bu+av
\end{align}
Then the Lagrangian becomes:
\begin{equation}
\mathcal{L}(u,v,\dot{u},\dot{v},t) = \frac{1}{2}m(ac-b^2)(c\dot{u}^2 - 2b\dot{u}\dot{v} +a\dot{v}^2) - \frac{1}{2}k(ac-b^2)(cu^2 - 2buv +av^2)
\end{equation}
And the equations of motion obtained are :
\begin{align}
m(c\ddot{u} - b\ddot{v}) &= -k(cu-bv)\\
m(-b\ddot{u} +a\ddot{v}) &= -k(-bu+av)
\end{align}
Now substituting functions of $x$ and $y$ for $u$ and $v$ [using equation (10) and (11)], we can obtain equation (7) and (8).
Now let us execute the above mentioned point transformation on the Lagrangian defined by equation (9). We obtain:
\begin{equation}
\begin{split}
\mathcal{L}' = \frac{1}{2} m\big[(c^2+b^2)\dot{u}^2 &- 2b(c+a)\dot{u}\dot{v} +(a^2+b^2)\dot{v}^2\big] \\
& -\frac{1}{2}k\left[(c^2+b^2)u^2 - 2b(c+a)uv +(a^2+b^2)v^2\right]
\end{split}
\end{equation}
It is easy to check that the equations of motion obtained for this new Lagrangian are not exactly the same as equation (12) and (13).
So there exists no gauge transformation between the Lagrangian of equation (12) and (15).
How can we define a transformation which will take care of this?

 

[sweet springs]

I think something is wrong with cross term.
For an example Lagrangean
$$L=\frac{m}{2}\dot{x}\dot{y}-\frac{k}{2}xy$$
gives (7) and (8) by applying Lagrange equation. I am afraid action integral coming from this L does not have stationary value. (4) contains this term.

 

[vanhees71]

From the EoM I'd guess that a transformation of the kind
$$x'=a x+ b y, \quad y'=bx + cy$$
does the job.

You can also diagonalize the quadratic form in the Lagrangian. The point is that in this case they are the same for the kinetic and the potential energy in the Lagrangian.