I am a bit confused by the following question and I hope someone could help me out.

Let $u$ be the solution of the following initial value problem
$$
u''(t) = g(t) \; \text{ in } (0,\infty), \quad\quad u(0)=a, \quad u'(0)=0. \quad\quad\quad\quad\quad (1)
$$
Let $U$ and $G$ be the extensions of $u$ and $g$ as zero to $(-\infty, \infty)$, that is,
$$
U(t) = \left\{
\begin{array}{ll}
u(t) & t \geq 0 \\
0 & t < 0,
\end{array}
\right.
\quad\quad\quad
G(t) = \left\{
\begin{array}{ll}
g(t) & t \geq 0 \\
0 & t < 0,
\end{array}
\right.
$$
For any compactly supported smooth function $\varphi$, $(U'',\varphi) = (U,\varphi'') = \int^\infty_0 u(t)\varphi''(t) \,dt = - a \varphi'(0) + (g,\varphi)$. This means
$$
U'' = a \delta' + G \quad \text{ as distributions in } \mathbb{R}. \quad\quad\quad\quad\quad\quad\quad\quad \quad\quad  (2)
$$
It is clear that (2) restricted to $(0,\infty)$ yields the equation in (1), but what conditions are needed to restore the initial conditions (other than the trivial condition that $U(0)=a$ and $U'(0)=0$)? I feel that the initial condition should have been included in (2) as the constant $a$ appears, but I don't know how to restore them.

Thank you.