Initial-value problem with a forcing function that has step discontinuity

Question

Context

I am solving an equation of motion derived from a Lagrangian that has a discontinuous step in its potential at the origin. In other words $$V(x) = \left(V_2-V_1\right)H(x) + V_1.$$ Based on this context, my question regards how to solve an initial-value problem to obtain $\dot{x}$. Bare in mind that I can solve this problem using conservation of energy, but that won't teach me anything new with respect to how to utilize the Lagrangian formalism.

Question

I would like to solve the following initial value problem $$ m\, \ddot{x} = - \left[V_2 - V_1\right]\delta(x)$$ with the initial conditions $$x(t_o) = x_o,~\text{where}~x_0<0,~\text{and}~\dot{x}(t_o) = v_o,~\text{where} ~v_0>0.$$ [In truth, I am interested only in $\dot x(t)$].

My Answer

For all values $x$ excepting at $x=0$, the mass is moving with constant velocity. As a consequence, I may write that $x=v\left(t-t_o\right)+x_o$. Therefore, $$ m\, \ddot{x} = - \left[V_2 - V_1\right]\delta\left(v\left(t-t_o\right)+x_o \right).$$

Now, I integrate both sides with respect to time. I find $$ m\, \int \ddot{x} \,dt = - \left[V_2 - V_1\right] \int \delta\left(v\left(t-t_o\right)+x_0 \right)\,dt.$$ Upon integrating the left-hand side, and re-adjusting the right-hand side, I obtain $$ m\, \dot{x} + k_1 = - \left[V_2 - V_1\right] \int \delta\left(v\,t-\left(v\,t_o-x_o\right) \right)\,dt.$$ Now, I do a change of variables. Namely, that $s = v_o\,t$. I find that $$ m\, \dot{x} + k_1 = - \left[V_2 - V_1\right] \int \delta\left(s-\left(v_o\,t_o-x_o\right) \right)\,\frac{ds}{v_o}.$$ Erroneously taking the integral on the right-hand side, I incorrectly find that $$ m\, \dot{x} + k_1 = - \frac{\left[V_2 - V_1\right]}{v_o} H\left(s-\left(v_o\,t_o-x_o\right) \right) + k_2 .$$ Now, since $s = v_o\,t$, I have that $$ m\, \dot{x} + k_1 = - \frac{\left[V_2 - V_1\right]}{v_o} H\left(v_o\left(t-t_o\right) +x_o \right) + k_2 .$$

Ultimately, for an as yet unknown constant $k$, I arrive at $$ \dot{x}(t) = - \frac{\left[V_2 - V_1\right]}{m\,v_o} H\left(v_o\left(t-t_o\right) +x_o \right) + k .$$

Upon satisfying the satisfy the boundary condition, $$ \dot{x}(t_o) = - \frac{\left[V_2 - V_1\right]}{m\,v_o} H\left(v_o\left(t -t_o\right) +x_o \right) + k = k = v_o .$$

So, I erroneously find that $$ \boxed{ \dot{x}(t) = v_o - \frac{\left[V_2 - V_1\right]}{m\,v_o} H\left(v_o\left(t-t_o\right) +x_o \right) . } $$

Remarks

I am not satisfied with my approach. It seems unwieldy, and is erroneous. If you have a more direct approach to solve the problem, then please consider posting it.

Answer 1

I thought that it might be instructive to present an alternative approach that recasts the problem using a nascent Dirac Delta [1]. To that end we proceed.

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Suppose $x_n(t)$ satisifies the differential equation

$$x_n''(t)=K\delta_n(x)\tag1$$

subject to the conditions $x_n(t_0)=x_0$ and $x'_n(t_0)=v_0$. In $(1)$. $K=-(V_2-V_1)/m$ and $\delta_n(x)$ is the nascent Dirac Delta

$$\delta_{n}(x)=\begin{cases}n&,0<x<1/n\\\\0&,\text{elsewhere}\end{cases}$$

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The solution to $(1)$ is given by

$$x_n(t)=\begin{cases}x_0+v_0(t-t_0)&,x_n<0\\\\\frac{nK}2 (t-t_1)^2+A(t-t_1)&,0<x_n<\frac1n\\\\B(t-t_2)+\frac1n\end{cases}\tag2$$

where $x_n(t_1)=0$, $x_n(t_2)=1/n$, and $A$ and $B$ are integration constants.

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We determine $A$ and $B$, and $t_1$ and $t_2$ in $(2)$ by enforcing continuity of position and velocity at $x_n=0$ and $x_n=1/n$. Proceeding, we find that for $v_0^2+2K\ge0$

$$\begin{align} t_1&=t_0-x_0/v_0\\\\ A&=v_0\\\\ t_2&=t_1-\frac{v_0}{nK}+\frac{\sqrt{v_0^2+2K}}{nK}\\\\ B&=\sqrt{v_0^2+2K} \end{align}$$

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Finally, letting $n\to \infty$, we find that for $v_0^2+2K\ge0$, $x(t)=\lim_{n\to\infty}x_n(t)$ is given by

$$x(t)=\begin{cases} x_0+v_0(t-t_0)&, x<0\\\\ \sqrt{v_0^2-2(V_2-V_1)/m}\,(t-t_1)&,x>0 \end{cases}$$

And we are done!

References

[1] https://en.wikipedia.org/wiki/Dirac_delta_function#nascent_delta_function

Answer 2

One way to solve the equation is to first multiply with $\dot x(t)$: $$ m\,\ddot{x}\dot{x} = - \left[V_2 - V_1\right]\delta(x)\dot{x}. $$ Both sides can then be written as derivatives: $$ \frac{d}{dt}\left(\frac12 m\dot{x}^2\right) = -\frac{d}{dt}\left(\left[V_2 - V_1\right]H(x)\right) $$ so $$ \frac12 m\dot{x}^2 = C - \left[V_2 - V_1\right]H(x) $$ but this practically leads to conservation of energy which you didn't want to use.

Answer 3

Conservation of energy is the way to go. Since you mention Lagrangian mechanics, the discontinuous potential $H(x)$ means you'd either need to formulate the equations of motion in a weak sense, or let $V(x)$ be the limit of some regularized potential. However; since you insist, let us study directly

$$\tag{1} mx''(t)=-\alpha\delta(x(t)) $$

The solution will be linear away from $x=0$. With the given initial condition, $x(t)$ will be zero exactly once, say at $t=t^*$. The piecewise solution is

$$\tag{2} x(t)=\cases{v_0(t-t^*) &$t<t^*$\\ A(t-t^*) &$t>t^*$} $$

We want to find the constant $A$. If we try to integrate (1) wrt $t$ we meet a nontrivial issue: $\delta(x(t))=\frac{\delta(t-t^*)}{|x'(t^*)|}$ (using the composition of delta with a function property, see here), but $x$ is not differentiable at $t^*$. I propose integrating (1) wrt $x$ by making the change of variables $u(x)=x'(t)$. Then (1) reads

$$\tag{3} (u^2)'=-\frac{2\alpha}{m}\delta(x) $$

Which we can integrate wrt $x$ across the singularity at $x=0$. Some care is required based on the sign of $A$. If $A>0$ the particle is 'transmitted', and by $\lim_{\epsilon\to 0}\int\limits_{-\epsilon}^\epsilon dx$ of (3) we have

$$\tag{4} A^2-v_0^2=-\frac{2\alpha}{m} $$

Exactly as you would find using conservation of energy. Thus

$$\tag{5} A=+ \sqrt{v_0^2-\frac{2\alpha}{m}} $$

Substituting into (2) you'd find that

$$ \tag{6} x(t)=\cases{v_0(t-t^*) &$t<t^*$\\ \sqrt{v_0^2-\frac{2\alpha}{m}}(t-t^*) &$t>t^*$}. $$

Upon taking the derivative, you'd have $$ \tag{7} v(t)=\cases{v_0 &$t<t^*$\\ \sqrt{v_0^2-\frac{2\alpha}{m}} &$t>t^* \qquad,\quad v_0^2>2\alpha/m $} $$

If $A=0$ then $v_0^2=2\alpha/m$ and (1) will not be satisfied for $t>t^*$, this corresponds to the particle becoming 'trapped' at $x=0$.

The case $A<0$ remains, which corresponds to 'reflection', and where we cannot integrate (3) in the manner $\int\limits_{-\epsilon}^\epsilon dx$. Idea: we integrate $\int\limits_{-\epsilon}^{-\epsilon} dx$ where from $-\epsilon \to 0$ we integrate $u_{t<t^*}$ and from $0 \to -\epsilon$ we integrate $u_{t>t^*}$. This leads to

$$\tag{8} A^2-v_0^2=-\frac{2\alpha}{m}H(0) $$

I assert that the physical value of $H(0)$ in this problem is $0$, in which case we have

$$ v(t)=\cases{v_0 & $t<t^*$\\-v_0 & $t>t^*$} $$

as expected. The justification is entirely post-hoc, but should not trouble us: as we should not expect to get unique solutions to the differential equation (1).