Differential equation with Dirac Delta function

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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 $v_1$.