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.

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 $[0,L]$.
