How to solve harmonic oscillator-like equation with $\theta$-function?

Question

Suppose second order linear differential equation $$ \frac{d^2 y(t)}{dt^2} + \omega^2(t)y(t) = 0, \quad \omega^2(t) = q^2 + \theta (t-t_{0})m^2 $$ ($\theta (t)$ denotes Heaviside step-function) with boundary condition $$ \lim_{t\to \infty}y(t) = \frac{1}{\sqrt{2\sqrt{q^2 + m^2}}}e^{i \sqrt{m^2+q^2}t} $$ How to solve it?

Answer 1

You have to use the definition of Heaviside function to decompose the equation into two ones corresponding to $t < t_0$ and $t > t_0$ respectively:

$$ \left\{\begin{array}{11} y''(t) + q^2 y(t) = 0,\ t < t_0\\ y''(t) + (q^2 + m^2) y(t) = 0,\ t > t_0\end{array}\right.$$

Each of them is an ordinary harmonic oscillator equation, so one has the following solution: $$y(t) = \left\{\begin{array} {11} A _1 e^{i q (t - t_0)} + A_2 e^{-i q (t - t_0)},\ t < t_0 \\ B_1 e^{i \sqrt{q^2 + m^2} (t - t_0)} + B_2 e^{-i \sqrt{q^2 + m^2} (t - t_0)},\ t > t_0\end{array} \right.$$

Now the most tricky part: find the unknown coefficients. Using the boundary condition it is easy to determine that $B_2 = 0$ and $B_1 = \frac{1}{\sqrt{2\sqrt{q^2+m^2}}} e^{i \sqrt{q^2 + m^2} t_0}.$

The equation should also have a solution at $t = t_0$. The consequence of it is that $y(t)$ is smooth at $t = t_0$ (otherwise $y''(t)$ can't exist): $$\left\{\begin{array}{11} \lim_\limits{t \rightarrow t_0-0}y(t) = \lim_\limits{t \rightarrow t_0 + 0} y(t)\\ \lim_\limits{t \rightarrow t_0-0}y'(t) = \lim_\limits{t \rightarrow t_0 + 0} y'(t) \end{array}\right.$$ So there is a system of two linear equations that makes it possible to determine $A_1$ and $A_2$: $$\left\{\begin{array}{11} A_1 + A_2 = B_1\\ q A_1 - q A_2 = \sqrt{q^2 + m^2} B_1 \end{array}\right.$$