How to solve a simple ODE using Green's function

Question

How do I find the solution to

$\ddot{x} = f(t) \equiv\begin{cases} a, 0\leq t \leq T, \\ 0, T<t<\infty \end{cases}$ where $a=$ const, subject to $x(0)=0$, $\dot{x}(0)=0$ using a Green's function? This should be a very straightforward problem but I am stuck. So far I have found (using the boundary + jump conditions + continuity) that the Green's function is given by: $G(t,T) = \begin{cases} G_1(t,T) = 0, 0\leq t \leq T, \\ G_2(t,T) = t-T, t>T \end{cases}$. The solution is then \begin{align} x(t) &= \int_0^{\infty} G(t,T) f(T)dT \\ &=\int_0^{t}G_2(t,T)f(T)dT + \int_t^{\infty}G_1(t,T)f(T)dT \\ &=\int_0^t(t-T)f(T)dT \end{align} as $G_1(t,T) \equiv 0$. But now $f(T) =0$ for $T<t$, so isn't the integral above just zero? I'm confused.

Answer 1

Using $y$ as the variable of integration instead of $T$ we have \begin{align*} x(t) &= \int_0^\infty G(t,y)f(y)dy\\ &= \int_0^tG_2(t,y)f(y)dy + \int_t^\infty G_1(t,y)f(y)dy \\&=\int_0^t(t-y)f(y)dy, \end{align*} since $G_1 \equiv 0$. If $t\geq T$, \begin{align*} x(t) &= \int_0^Ta(t-y)dy + \int_T^t0\cdot(t-y)dy \\ &=aT(t-T/2)\end{align*} If $t<T$, we get \begin{align*}x(t) &= \int_0^ta(t-y)dy \\&= \frac{at^2}{2}\end{align*} Hence the solution is $x(t)$ = \begin{cases} \frac{at^2}{2} & \mbox{if }t<T \\ aT(t-T/2) & \mbox{if } t\geq T \end{cases}