Our task is to solve the equation a=\tau \dot{a}+\frac{U_{0}}{m}[-\delta(x)+\delta(x-L)] , subject to the boundary conditions
(1) x continuous at x = 0 and x = L;
(2) v continuous at x = 0 and x = L;
(3) \left.\Delta a=\pm U_{0} / m \tau v \text { (plus at } x=0, \text { minus at } x=L\right).
The third of these follows from integrating the equation of motion:
\int \frac{d v}{d t} d t=\tau \int \frac{d a}{d t} d t+\frac{U_{0}}{m} \int[-\delta(x)+\delta(x-L)] d t,
\Delta v=\tau \Delta a+\frac{U_{0}}{m} \int[-\delta(x)+\delta(x-L)] \frac{d t}{d x} d x=0,
\Delta a==\frac{U_{0}}{m \tau} \int \frac{1}{v}[-\delta(x)+\delta(x-L)] d x=\pm \frac{U_{0}}{m \tau v}.
In each of the three regions the force is zero (it acts only at x = 0 and x = L), and the general solution is
a(t)=A e^{t / \tau} ; \quad v(t)=A \tau e^{t / \tau}+B ; \quad x(t)=A \tau^{2} e^{t / \tau}+B t+C.
(I’ll put subscripts on the constants A, B, and C, to distinguish the three regions.)
Region iii (x > L): To avoid the runaway we pick A_{3}=0 ; \text { then } a(t)=0, v(t)=B_{3}, x(t)=B_{3} t+C_{3} . Let the final velocity be v_{f}\left(=B_{3}\right) , set the clock so that t = 0 when the particle is at x = 0, and let T be the time it takes to traverse the barrier, so x(T)=L=v_{f} T+C_{3}, \text { and hence } C_{3}=L-v_{f} T . Then
a(t)=0 ; \quad v(t)=v_{f}, \quad x(t)=L+v_{f}(t-T), \quad(t<T) .
Region ii (0<x<L): a=A_{2} e^{t / \tau}, v=A_{2} \tau e^{t / \tau}+B_{2}, x=A_{2} \tau^{2} e^{t / \tau}+B_{2} t+C_{2}.
(3) \Rightarrow 0-A_{2} e^{T / \tau}=-\frac{U_{0}}{m \tau v_{f}} \Rightarrow A_{2}=\frac{U_{0}}{m \tau v_{f}} e^{-T / \tau}.
(2) \Rightarrow v_{f}=A_{2} \tau e^{T / \tau}+B_{2}=\frac{U_{0}}{m v_{f}}+B_{2} \Rightarrow B_{2}=v_{f}-\frac{U_{0}}{m v_{f}}.
(1) \Rightarrow L=A_{2} \tau^{2} e^{T / \tau}+B_{2} T+C_{2}=\frac{U_{0} \tau}{m v_{f}}+v_{f} T-\frac{U_{0} T}{m v_{f}}+C_{2}=v_{f} T+\frac{U_{0}}{m v_{f}}(\tau-T)+C_{2} \Rightarrow
C_{2}=L-v_{f} T+\frac{U_{0}}{m v_{f}}(T-\tau).
\begin{aligned}&a(t)=\frac{U_{0}}{m \tau v_{f}} e^{(t-T) / \tau} \\&v(t)=v_{f}+\frac{U_{0}}{m v_{f}}\left[e^{(t-T) / \tau}-1\right] ; \\&x(t)=L+v_{f}(t-T)+\frac{U_{0}}{m v_{f}}\left[\tau e^{(t-T) / \tau}-t+T-\tau\right] ;\end{aligned} (0 < t < T).
[Note: if the barrier is su!ciently wide (or high) the particle may turn around before reaching L, but we’re interested here in the r´egime where it does tunnel through.]
In particular, for t = 0 (when x = 0):
0=L-v_{f} T+\frac{U_{0}}{m v_{f}}\left[\tau e^{-T / \tau}+T-\tau\right] \Rightarrow L=v_{f} T-\frac{U_{0}}{m v_{f}}\left[\tau e^{-T / \tau}+T-\tau\right] . qed
Region i (x<0): a=A_{1} e^{t / \tau}, v=A_{1} \tau e^{t / \tau}+B_{1}, x=A_{1} \tau^{2} e^{t / \tau}+B_{1} t+C_{1} . \text { Let } v_{i} be the incident velocity (\text { at } t \rightarrow-\infty) \text {; then } B_{1}=v_{i} . Condition (3) says
\frac{U_{0}}{m \tau v_{f}} e^{-T / \tau}-A_{1}=\frac{U_{0}}{m \tau v_{0}},
where v0 is the speed of the particle as it passes x = 0. From the solution in region (ii) it follows that v_{0}=v_{f}+\frac{U_{0}}{m v_{f}}\left(e^{-T / \tau}-1\right) . But we can also express it in terms of the solution in region (i): v_{0}=A_{1} \tau+v_{i}.
Therefore
v_{i}=v_{f}+\frac{U_{0}}{m v_{f}}\left(e^{-T / \tau}-1\right)-A_{1} \tau=v_{f}+\frac{U_{0}}{m v_{f}}\left(e^{-T / \tau}-1\right)+\frac{U_{0}}{m v_{0}}-\frac{U_{0}}{m v_{f}} e^{-T / \tau}
=v_{f}-\frac{U_{0}}{m v_{f}}+\frac{U_{0}}{m v_{0}}=v_{f}-\frac{U_{0}}{m v_{f}}\left(1-\frac{v_{f}}{v_{0}}\right)=v_{f}-\frac{U_{0}}{m v_{f}}\left\{1-\frac{v_{f}}{v_{f}+\left(U_{0} / m v_{f}\right)\left[e^{-T / \tau}-1\right]}\right\}
=v_{f}-\frac{U_{0}}{m v_{f}}\left\{1-\frac{1}{1+\left(U_{0} / m v_{f}^{2}\right)\left[e^{-T / \tau}-1\right]}\right\}. qed
\text { If } \frac{1}{2} m v_{f}^{2}=\frac{1}{2} U_{0} , then
L=v_{f} T-v_{f}\left[\tau e^{-T / \tau}+T-\tau\right]=v_{f}\left[T-\tau e^{-T / \tau}-T+\tau\right]=\tau v_{f}\left(1-e^{-T / \tau}\right);
v_{i}=v_{f}-v_{f}\left[1-\frac{1}{1+e^{-T / \tau}-1}\right]=v_{f}\left(1-1+e^{T / \tau}\right)=v_{f} e^{T / \tau}.
Putting these together,
\frac{L}{\tau v_{f}}=1-e^{-T / \tau} \Rightarrow e^{-T / \tau}=1-\frac{L}{\tau v_{f}} \Rightarrow e^{T / \tau}=\frac{1}{1-\left(L / \tau v_{f}\right)} \Rightarrow v_{i}=\frac{v_{f}}{1-\left(L / v_{f} \tau\right)} . qed
In particular, for
L=v_{f} \tau / 4, v_{i}=\frac{v_{f}}{1-1 / 4}=\frac{4}{3} v_{f}, \text { so } \frac{K E_{i}}{K E_{f}}=\frac{\frac{1}{2} m v_{i}^{2}}{\frac{1}{2} m v_{f}^{2}}=\left(\frac{v_{i}}{v_{f}}\right)^{2}=\frac{16}{9} \Rightarrow
K E_{i}=\frac{16}{9} K E_{f}=\frac{16}{9} \frac{1}{2} U_{0}=\frac{8}{9} U_{0}.