I have a lagrangian \begin{equation} L=\frac{1}{2}\dot{q}^{2}-\frac{1}{2}q\dot{q}-aq^{2} \end{equation} $a>0$, and I can't realise what physical system it describes, since I guess potential energy $U$ must be equal to $\frac{1}{2}q\dot{q}+aq^{2}$, and I don't know any potential having that form. Writing Lagrange equation I end up with: \begin{equation} \ddot{q}=-2aq \end{equation} which is clearly a SHO, but does this lagrangian actually describes a SHO?
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Related: https://physics.stackexchange.com/q/174137/2451 and links therein. – Qmechanic Jun 22 '20 at 11:34
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Lagrangians with $A(q)\dot q$ terms describe particles interecting with a magnetic field. In this case, being one dimensional motion, the field has no effect.

mike stone
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Yes, a property of the Lagrangian is that it is not unique.
$L_*=L+\frac{df(\textbf{q},t)}{dt}$, for a function f, will also describe the system and satisfy Lagrangian equation.

PseudoYousef
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If the equation of motion is one of a simple harmonic oscillator then yes, the lagrangian describes a simple harmonic oscillator.
Remember that there won't be a one to one correspondence between an equation of motion and a lagrangian since lagrangians which differ by a constant $$L^\prime = \lambda L\qquad L^\prime = L+\lambda$$ or by a total time derivative $$L = L+\frac{d\Lambda}{dt}$$ will give the same eom.

Davide Morgante
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