Change of variable and boundary data for Laplace equation

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 $$\lambda >0$$.

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.
• Thanks. What if we consider $-\Delta u +\lambda u= 0$? Then $u \equiv 1$ is no longer a solution.