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 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.
