Boundary condition inserted in the local PDE Let us consider the following problem:
$$
\begin{align}
-&u_{xx}=0&&\forall x\in(0,L)&&\tag{1}\\
&u(0)=0\tag{2}\\
&u_x(L)=\alpha\tag{3}
\end{align}
$$
It is possible to insert (3) in (1) as follows:
$$
\begin{align}
-&u_{xx}=\alpha\delta (x-L)&&\forall x\in(0,L]\tag{4}\\
&u(0)=0 \tag{5}\\
&u_x(L)=0 \tag{6}
\end{align}
$$
where $\delta(x-L)=\delta_L$ is the Dirac distribution. Are there results showing that these two formulations are equivalent? It is not clear to me whether (6) should be kept or not.
 A: Let $u$ be the solution on $[0,L]$ and let $\bar{u}$ be its extension to $\mathbb{R}$ defined as
$$
\bar{u}(x) = \begin{cases}
0 & \text{if } x < 0 \\
u(x) & \text{if } x \in [0,L] \\
u(L) & \text{if } x > L \\
\end{cases}
$$
Then $\bar{u}$ is continuous and its distributional partial derivative w.r.t $x$ is given by
$$
\partial_x \bar{u}(x) = \begin{cases}
0 & \text{if } x<0 \\
\partial_x u(x) & \text{if } x \in (0,L) \\
0 & \text{if } x>L \\
\end{cases}
$$
This is however discontinuous so
$$
\partial_x^2 \bar{u} = \partial_x u \, \chi_{(0,L)} + \partial_x u(0+) \, \delta_0 - \partial_x u(L-) \, \delta_L
$$
A: I am suggesting a solution below (but I am not convinced). From (1), (2) and (3), it is clear that the sought solution is $u(x)=\alpha x$. Let us try to solve (4), (5) and (6) in the sense of distributions. Integrating (4) twice yields:
$$-u(x)=ax+b+\alpha (x-L)H(x-L)$$
Condition (5) implies $b=0$ and condition (6) implies
$$a+\alpha H(0)=0 \tag{7}$$
If we consider that $H(0)=1$ by definition, (7) becomes $a=-\alpha$ and the exact solution is retrieved on the interval $[0,L]$. However, another definition of $H$ would generate erroneous results, which seems annoying. We also realize that $u(x)=\alpha x$ does no longer satisfy (6), which looks strange.
A: I am suggesting a second answer here (not really convinced either). The idea is to have the $\delta$ as a non-homogeneous term in the PDE and then push it to the boundary. Formally it reads: Solve
$$
\begin{align}
-&u_{xx}=\alpha\delta (x-\beta)&&\forall x\in(0,L)\tag{8}\\
&u(0)=0 \tag{9}\\
&u_x(L)=0 \tag{10}
\end{align}
$$
where $0<\beta<L$. The distributional solution to (8), (9) and (10) is:
$$-u(x)=\alpha\bigl((x-\beta)H(x-\beta)-xH(L-\beta)\bigr)$$
If we take the limit $\beta\to L$ of the above expression together with the definition $H(0)=1$, then the solution $u(x)=\alpha x$ is retrieved on the interval $[0,L]$.
