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 $$ u''(t) = g(t) \; \text{ in } (0,\infty), \quad\quad u(0)=a, \quad u'(0)=0. \label{1}\tag{1} $$ 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 $\varphi$, $$ (U'',\varphi) = (U,\varphi'') = \int\limits^\infty_0 u(t)\varphi''(t) \,dt = - a \varphi'(0) + (g,\varphi). $$ This means $$ U'' = a \delta' + G \quad \text{ as distributions in } \mathbb{R}. \label{2}\tag{2} $$ It is clear that \eqref{2} restricted to $(0,\infty)$ yields the equation in \eqref{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 \eqref{2} as the constant $a$ appears, but I don't know how to restore them.

Thank you.

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