Boundary condition inserted in the local PDE

Question

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.

Answer 1

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 $$ \partial_x^2 \bar{u} = \partial_x u \, \chi_{(0,L)} + \partial_x u(0+) \, \delta_0 - \partial_x u(L-) \, \delta_L $$

Answer 2

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.

Answer 3

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 $[0,L]$.