Differential equation with Dirac Delta function While studying thermoionic emission from metals I wanted to get a feeling for the problem with classical mechanics before delving into quantum mechanics. The potential used to model the situation is this one:
$$V(x)=V_0 \Theta(x)$$
Where $\Theta (x)$ is the Heaviside step function. If we want the classical force for this potential we differentiate:
$$F_x = - \frac{dV}{dx}= - V_0 \delta(x)$$
Where $\delta(x)$ is the Dirac delta function. This gives an equation of motion of the type:
$$m \ddot{x} = -V_0 \delta(x)$$
With $m$ and $V_0$ positive parameters and the dots denote differentiation with respect to time.
My question is: how to treat this equation? It turns out that the problem is much simpler in quantum mechanics if we try to solve the time independent Schroedinger equation.
Thanks in advance.
 A: As in the QM case, the usual way to solve differential equations involving delta-functions is to solve them piecewise on each domain.  We first re-cast the equation to solve for the speed $v = dx/dt$ as a function of $x$:
$$
\frac{d^2 x}{dt^2} = \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = v \frac{dv}{dx} = \frac{1}{2} \frac{d}{dx} \left(v^2 \right).
$$
Our equation is now 
\begin{equation}
\frac{m}{2} \frac{d}{dx} \left(v^2 \right) = - V_0 \delta(x). \qquad \qquad (1)
\end{equation}
We can now note that for the regions $x < 0$ and $x > 0$, we have
$$
\frac{m}{2} \frac{d}{dx} \left(v^2 \right) = 0,
$$
which implies that the solution is
$$
v(x) = \begin{cases} v_- & x < 0 \\ v_+ & x>0 \end{cases}
$$
To find the relationship between $v_-$ and $v_+$, we integrate equation (1) in a small interval $[-\epsilon, \epsilon]$ around 0:
\begin{align*}
\frac{m}{2} \int_{-\epsilon}^{\epsilon} \frac{d}{dx} \left(v^2 \right) \, dx &= - V_0 \int_{-\epsilon}^{\epsilon} \delta(x) \, dx \\
\frac{m}{2} \left[ v^2 \right]_{-\epsilon}^{\epsilon} &= - V_0 \\
\frac{m}{2} \left(v_+^2 - v_-^2 \right) &= - V_0.
\end{align*}
This latter equation can be recognized as energy conservation across the boundary $x = 0$:  $\Delta KE = - \Delta PE$.
The solution for $v(x)$ is then
$$
v(x) = \begin{cases} v_0 & x < 0 \\ \sqrt{v_0 - 2V_0/m} & x > 0 \end{cases}
$$
If you want the solution for $x(t)$, you can then integrate this with respect to time.
Alternately, if you want to skip the step of finding $v(x)$, you can instead use the identity 
$$
\delta(x(t)) = \sum_i \frac{1}{|\dot{x}(t_i)|} \delta(t - t_i)
$$
where the sum runs over the zeroes of the function $x(t)$.  This then allows us to recast this equation solely in terms of $x$ as a function of $t$.  One can oncesagain solve this piecewise between successive zeroes of the function $x(t)$, and integrate over small intervals of $t$ surrounding these zeros to "patch" the piecewise solutions together.  In this case, the solutions for $x(t)$ "between" the zeroes will be simply linear functions of $t$, which means that you will only have one zero for $x(t)$, and applying the above techniques will yield the same sort of solution.
A: There are two approaches.  One is just to think about energy conservation.  To the left of the step the energy is $\frac 12mv_0^2$, so as long as that is greater than the step, the velocity will reduce to keep the energy constant and the new velocity will be $\sqrt{v_0^2-\frac 2mV}$.  If you want to solve the equation, you integrate across the time the particle crosses the step.  The delta integrates to a step function, the acceleration integrates to the velocity, and you get a step function in velocity just as before.  The second approach is like what you would do in quantum mechanics.
A: How to treat this equation?
Well, in the usual manner: as the equation of motion is time independent we write down the first integral - energy conservation
$$E = \frac{m v^2}{2}+V_0 \theta (x)=\frac{m v_0^2}{2}$$
Here $v_0$ is the velocity of the particle for $x<0$
For definiteness let us assume $V_0 > 0$.
We have to distinguish two cases of initial conditions $(1)\; v_0 < 0$, 
$(2)\; v_0 > 0$.
In the first case the particle moves indefinitely along the negative x-axis with velocity $v_0$.
In the second case the particle is either reflected at the potential barrier, if
$$(2a)\;0 < v_0 < v_c $$
where the critical velocity is defined as
$$v_c = \sqrt{\frac{2 V_0}{m}}$$
and moves further on as in case (1), or, for suffiently high velocity
$$(2b) \; v_0 > v_c $$
the particle is decelerated to
$$v_1 = \sqrt{v_0^2 - v_c^2}$$
and continues the motion into the region $x>0$ with verlocity $v_1$.
