Differential equation on $\Bbb R$ We have a differential equation on $\Bbb R$ of the form 
$$\frac {d^2}{dx^2}u = \chi_{[0,1]},$$ 
where $\chi_{[0,1]}$ is the characteristic function of the interval $[0, 1] ⊂ \Bbb R$. I want to find a generalized solution for this differential equation.
I also want to know that will the solution be unique?

Definition: If $\Omega$ is a domain in $\Bbb R^n$. We say that $u\in \mathcal D'(\Bbb R^n)$ is a generalized solution of $$\sum_{|\alpha|\le m} a_\alpha D^\alpha u=f(x)$$ in $\Omega.$ If $u$ satisfies
$$\sum_{|\alpha|\le m} a_\alpha \langle D^\alpha u,\phi\rangle =\langle f(x),\phi\rangle $$ for every $\phi\in\mathcal D(\Omega).$  Here $f\in\mathcal D'(\Bbb R^n)$ and constant coefficients  $a_\alpha\in \Bbb R^n $.

Thank you.
 A: Let's solve the problem in $\mathbb{R}$ i.e. find $u\in \mathcal{D}'(\mathbb{R})$ such that $$\tag{1}\langle u'',\varphi\rangle=\langle \chi_{[0,1]},\varphi\rangle,\ \forall\ \varphi\in \mathcal{D}(\mathbb{R})$$
Suppose that $u\in\mathcal{D}'(\mathbb{R})$ satisfies $(1)$, then for all $\varphi\in \mathcal{D}(\mathbb{R})$
\begin{eqnarray}
 \langle u'',\varphi\rangle &=& \langle \chi_{[0,1]},\varphi\rangle     \nonumber \\
   &=& \int_0^1\varphi \nonumber \\
   &=& -\int_0^1x\varphi'+\varphi(1) \\
   &=& \int_0^1\frac{x^2}{2}\varphi''+\varphi(1)-\frac{\varphi'(1)}{2}
\end{eqnarray}
We conclude that $\langle u,\varphi''\rangle=\langle f(x)+g(x)+h(x),\varphi''\rangle$ for all $\varphi\in \mathcal{D}(\mathbb{R})$ where $$f(x)=\chi_{[0,1]}\frac{x^2}{2}$$
$$
g(x) = \left\{ \begin{array}{cc}
 0 &\mbox{ if $x<1$} \\
  x-1 &\mbox{ otherwise}
       \end{array} \right.$$
$$
h(x) = \left\{ \begin{array}{c}
 0 &\mbox{ if $x<1$} \\
  \frac{1}{2} &\mbox{ otherwise}
       \end{array} \right.
$$
Hence, $u=f(x)+g(x)+h(x)+ax+b$, where $a,b\in\mathbb{R}$ is the solution for your problem.
Remark 1: To find $g$, you have to solve the equation $$\langle g,\varphi''\rangle =\langle \delta_1,\varphi''\rangle$$
where $\delta_1$ is Dirac in the point $1$. To find $h$ you do a similar thing. Remember that the Heaviside function is the "integral" of the Dirac distribution.
Remark 2: If you look for a solution in $[0,1]$, then $u=f(x)+ax+b$ is the solution.
