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.



Restore initial condition for distributions
$\newcommand{\R}{\mathbb R}$In accordance with comments by Willie Wong and the OP, let us extend $u$ by $a$ to the left of $0$:
\begin{equation\*}
 U(t) := 
\begin{cases}
u(t) & \text{ if }t\ge0, \\
a & \text{ if }t<0.
\end{cases}
\tag{3}\label{3}
\end{equation\*}
Then
\begin{equation\*}
 U''=G, \tag{4}\label{4}
\end{equation\*}
where $U$ and $G$ are identified with the corresponding distributions on $\R$.
Let $V$ be another distribution on $\R$ such that \eqref{4} holds with $V$ in place of $U$. Let $X:=V-U$.


Then $X''=0$. Hence, the distribution $X$ can be identified with an affine function on $\R$ (please let me know if you need details on this claim). We want to impose an additional condition on $X$ to ensure that $X=0$. Since $X$ and its derivatives are distributions, they do not have definite values at any point in $\R$. So, we cannot impose any condition on such values. What we can do instead is require that $X$ be $0$ on some nonempty finite open interval $I\subset\R$ -- i.e., that $(X,\psi)=0$ for all smooth functions $\psi$ with support $S\_\psi\subset I$.


Thus, $U$ will be uniquely determined by condition \eqref{4} together with the condition
\begin{equation\*}
 (U,\psi)=a\int\_\R\psi
\end{equation\*}
for all smooth functions $\psi$ with $S\_\psi\subset I$, where $I$ is a nonempty finite open subinterval of the interval $(-\infty,0)$.


