Classical motion in delta potential The question about classical motion in delta potential may seem artificial, but it makes sense if you try to calculate the propagator for particle in delta-potential, because you usually need to know classical solutions to pick normalization factor out. We deal with following lagrangian:
\begin{equation}
\mathcal{L} = \dfrac{m\dot{q}^2}{2}-\alpha \delta(q - q_0) \implies E =  \dfrac{m\dot{q}^2}{2}+\alpha \delta(q - q_0) = const.
\end{equation}
So we must solve such an equation:
\begin{equation}
\dot{q}^2 = \dfrac{2E}{m}-\dfrac{2 \alpha}{m}\delta(q-q_0).  
\end{equation}
Unfortunately it looks pretty unsolvable, usual separation of variables doesn't work here because of delta function.
I have an idea to represent delta as a limit, like $$\delta(x) = \lim\limits_{a \rightarrow 0} \dfrac{1}{\pi}\dfrac{a}{a^2+x^2},$$ and take this limit after integration, but the integral is elliptic and even Mathematica can't deal with it.
Does this problem have some solution?
 A: TL;DR: The classical point particle in 1D will transmit as nothing has happened if $\alpha<0$, and reflect if $\alpha>0$.
As a regularization of the Dirac delta distribution we can use the triangular function
$$ \delta_{\epsilon}(x)~=~\frac{1}{\epsilon^2}(\epsilon-|x|)_+ \qquad \Rightarrow\qquad \delta^{\prime}_{\epsilon}(x)~=~-\frac{1}{\epsilon^2}{\rm sgn}(x)\theta(|x|-\epsilon).$$
The mechanical energy conservation
$$ E~=~\frac{m}{2}\dot{x}^2 +V(x), \qquad V(x)~=~\alpha\delta(x),\qquad \delta(x)~=~\lim_{\epsilon\searrow 0^+} \delta_{\epsilon}(x), $$
or equivalently, Newton's 2nd law,
$$ m\ddot{x}~=~-V^{\prime}(x), $$
can be solved exactly for the triangular regularization in each of the space regions 
$$]-\infty, -\epsilon] ,\quad [-\epsilon,0],\quad [0,\epsilon],\quad \text{and}\quad [\epsilon,\infty[.$$
Finally, the exact solutions can be glued together by assuming continuity of position & velocity (but not necessarily acceleration!) as a function of time $t$.
A: Since the system is time independent and the potential vanishes outside the vicinity of $q=q_0$, the kinetic energy of the system must be conserved. This means that a particle which impacts the potential at time $t'$ must look like
$$q(t)=\begin{cases}~~~v(t-t')+q_0 & t<t'\\ \pm v(t-t') +q_0& t>t' \\ \end{cases}$$
If $\alpha<0$ then the particle is unaffected and we pick the $+$ sign, alternatively the particle is reflected for $\alpha>0$ and we pick the $-$ sign.
I think the best approach to derive this from first principles is to use the representation
$$V(q)=\lim_{a\to0}\begin{cases}0 & |q-q_0|>a/2\\ \alpha/a& |q-q_0|<a/2 \\ \end{cases}$$
