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.

  • 1
    $\begingroup$ What do you mean by "restore the initial conditions"? (I am not clear on what you are trying to accomplish with this construction of $U$. For example, why do you choose to extend $u$ by $0$ instead of extending it by $a$, which would make your life simpler?) $\endgroup$ Mar 23 at 2:49
  • $\begingroup$ @Willie Wong: Thank you for the prompt response. by "restore the initial condition" I mean to specify conditions on U so that its restriction to $[0,\infty)$ satisfies (1). Extending $u$ by $a$ is a good idea. It seems the distributional equation becomes $U''=G$, but this equation has infinite solutions. I was wondering if we could specify some conditions on $U$ so that its restriction on $[0,\infty)$ satisfies (1). $\endgroup$
    – user460242
    Mar 23 at 4:08
  • $\begingroup$ @Willie Wong: To put it another way, let us extend $u$ by $a$ as you suggested. My question is, if we start with $U''=G$ (which has infinite solutions), is it possible to specify conditions on $U$ to exclude other solutions and make it precisely the extension of the solution of (1). Thank you. $\endgroup$
    – user460242
    Mar 23 at 4:19
  • $\begingroup$ It is perhaps worth recalling the general principle at work. Suppose that you have a piecewise smooth function with jumps at the origin (i.e. the function and its derivatives have right and left limits there which may be distinct). Then its derivative coincides with the pointwise one plus a delta function at the origin times the jump of the function there. Similar formulae hold for higher derivatives. The proofs are elementary and involves adding suitable multiples of the Heaviside function for which the result is clear. $\endgroup$ Mar 23 at 15:37
  • $\begingroup$ @ bathalf15320: Thank you for the suggestion. $\endgroup$
    – user460242
    Mar 23 at 19:21

1 Answer 1


$\newcommand{\R}{\mathbb R}$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 $(-\infty,0)$.

  • $\begingroup$ Thank you for the very detailed explanation. I am a bit confused by the definition of $V$: if it is another distribution defined by (3), it would be identical to $U$. Did you mean $V$ is another distribution defined by (4) (instead of (3))? $\endgroup$
    – user460242
    Mar 23 at 18:41
  • $\begingroup$ @user460242 : Yes, of course, it is (4), instead of (3). This is now fixed. $\endgroup$ Mar 23 at 18:52
  • $\begingroup$ @losif Pinelis: This answers my question. Thank you so much! $\endgroup$
    – user460242
    Mar 23 at 19:19

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