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I have a question from classical mechanics, where a particle's Lagrangian is:

$L = \frac{1}{2}\dot{q}^2 + q^2$.

Solving the Euler Lagrange Equations, we get that: $q(t) = A \exp(\sqrt{2} t)+ B \exp(-\sqrt{2}t)$,

which are exponential solutions. I was just wondering if someone could provide a physical explanation of why one gets exponential solutions, which also imply that $q \neq 0$ unless $t \to \infty$. The only thing I could think of was that this particle is sitting on top of a potential hill, but I'm try to understand the physical motivation for these types of solutions.

Thanks.

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If you look at the scenario of a bead that can slide on a rotating rod, you end up with the Lagrangian of (work left to you): $$L=\frac12\dot r^2+\frac12\omega^2r^2$$

where $r$ is the distance from the center of the point of rotation, $\omega$ is the constant angular frequency of the rotation.

This is of the same form as your example. The bead will move outwards with an exponentially increasing $r$ (if it doesn't start at $r=0$ with no velocity). This can be understood as the centripetal force pulling the bead outwards, hence you can interpret the $\frac12\omega^2r^2$ term from the Lagrangian as coming from an effective potential that is the negative of this.

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