Let us consider a smooth bounded domain $\Omega \subset \mathbb R^n$ and the problem $$ \begin{cases} -\Delta u = 0 & x \in \Omega \\ u = 1 & x \in \partial \Omega \end{cases} $$ Does it make sense to use the change of variables $v = u-\mathbf{1}_{\partial \Omega} $ to reduce the problem to the following one (with a source term but homogeneous boundary data)? $$ \begin{cases} -\Delta v = \Delta \mathbf{1}_{\partial \Omega} & x \in \Omega \\ v = 0 & x \in \mathbb R^n \setminus \Omega \end{cases} $$ How can this change of variable be made rigorous in the context of viscosity solutions?

Because of the answer below, let us consider the same question for $$ \begin{cases} -\Delta u +\lambda u= 0 & x \in \Omega \\ u = 1 & x \in \partial \Omega \end{cases} $$ with $\lambda >0$.

This question is motivated by Viscosity solutions of $(-\Delta)^s u = 0$ in $\Omega $ with non-homogeneous data $u = 1$ in $\mathbb R^n \setminus \Omega$


There seems to be a weakness to the approach with your (generalized) differential equation for $v$, because the Laplacian of the indicator function $\Delta\mathbf{1}_{\partial\Omega}$ is not actually singular enough to produce a meaningful boundary condition on the surface $\partial\Omega$. We have that $\Delta\mathbf{1}_{\partial\Omega}=0$ wherever it is defined, and where it is undefined (that is, on $\partial\Omega$) it is merely undefined, not singular like a $\delta$-distribution (or derivative thereof). You can see this by integrating $\Delta\mathbf{1}_{\partial\Omega}$ times a test function and applying Cauchy's second formula (which is integration by parts twice), $$\int_{\mathcal{V}}dV\, \phi\left(\Delta\mathbf{1}_{\partial\Omega}\right)= \int_{\mathcal{V}}dV\, (\Delta\phi)\mathbf{1}_{\partial\Omega} +\int_{\partial\mathcal{V}}dA\left(\phi\frac{\partial\mathbf{1}_{\partial\Omega}}{dn}-\mathbf{1}_{\partial\Omega}\frac{\partial\phi}{\partial n} \right)=0,$$ where $\mathcal{V}$ is a larger volume containing $\Omega$. The integral is exactly the same as if $\mathbf{1}_{\partial\Omega}$ were replaced by $0$.

If the source term on the right-hand-side of the Poisson equation contained a singular distribution, that could yield a condition on a discontinuity of $v$ or its derivatives. However, with only $\Delta\mathbf{1}_{\partial\Omega}$, we are left with merely, $$\Delta v=0\quad \textrm{almost everywhere},$$ to which the generalized solution satisfying the boundary condition that $v|_{\partial\Omega}=0$ is just $v(x)=0$ everywhere. This is not actually giving a wrong answer; it just corresponds to the fact that the original boundary value problem for $u$ has the unique solution $u(x)=1$ on $\Omega$. However, your revised source is not actually contributing anything to the solution, so it is questionable whether this approach can be usefully generalized.

  • $\begingroup$ Thanks. What if we consider $-\Delta u +\lambda u= 0$? Then $u \equiv 1$ is no longer a solution. $\endgroup$
    – Zac
    Nov 3 '21 at 1:02

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