Laplace Equation on the Corners and Boundary of a Rectangle?

Question

Consider for some rectangle $[a,b] \times [c,d] \in \mathbb{R}^2$, we have a generic boundary value problem: \begin{equation*} \begin{cases} \frac{\partial ^2 u}{\partial x ^2}+\frac{\partial ^2 u}{\partial y ^2}=0, & a<x<b \textrm{ and } c<y<d\\ u(x,c)=g_c(x), & a<x<b \\ u(x,d)=g_d(x), & a<x<b \\ u(a,y)=h_a(y), & c<y<d \\ u(b,y)=h_b(y), & c<y<d \end{cases} \end{equation*} which describes, say, the steady state heat distribution of a 2-dimensional rectangle.

I noticed that the partial differential equation is specified for $a<x<b$ and $c<y<d$, not $a\le x \le b$ and $c \le y \le d$. And similarly for the boundary conditions, which don't include their endpoinds. I have two questions:

1. Does this boundary value problem say anything about the corners of the rectangle? For example, does the BVP necessarily imply that $g_c(a)=h_a(c)$, $g_c(b)=h_b(c)$, $g_d(a)=h_a(d)$ and $g_d(b)=h_b(d)$?
2. Does the Laplace Equation $\frac{\partial ^2 u}{\partial x ^2}+\frac{\partial ^2 u}{\partial y ^2}=0$ only hold on the interior of the rectangle, i.e where $a<x<b$ and $c<y<d$ as stated in the problem? Or, does the Laplace equation also hold for $x$ and $y$ on the boundary, so that $\frac{\partial ^2 u}{\partial x ^2}+\frac{\partial ^2 u}{\partial y ^2}=0$ for all $a \le x \le b$ and $c \le y \le d$? I would like to say it holds for all $(x,y) \in [a,b] \times [c,d]$, but again this depends if it actually holds on the boundary.

Thank you!

Answer 1

The problem as stated doesn't imply anything about the corners; for instance, it doesn't imply that $g_c(a)$ is even defined, and even if it does happen to be defined, nothing requires it to equal $h_a(c)$.

Normally the solution $u$ is just considered to be a function on $[a,b] \times [c,d]$. As such it does not make sense to speak of, say, the $y$ partial derivatives of $u$ at a point with $y=c$, because the derivative depends on the values of $u$ at points outside the rectangle, where it is undefined. So we can't say that the Laplace equation holds on the boundary, because the statement doesn't even make sense.

Answer 2

I think there are three main things to say here. The first is simple: the differential equation of a BVP on a domain is assumed to hold on the interior of the domain. On the boundary there is often no way to sensibly define all of the derivatives in the problem. There is usually a way to define some of the derivatives on the boundary; for example if the function is twice continuously differentiable on the interior, then its gradient has a well-defined extension to the boundary but its second derivatives generally do not. (Note that this extension is not really a derivative in the usual sense, instead it is just the limit of the derivative as you approach the boundary.)

Second, you should expect bad behavior where the boundary is not smooth. For instance in this problem you can sensibly formulate a boundary condition which, if extended to the corners, would require the function to take on two different values at the corners themselves. The situation is even worse for a Neumann condition, in that the normal vector to the boundary will not even be defined where the boundary is not smooth, which makes it impossible to impose a pointwise flux condition there.

Third, this is all natural in terms of the weak formulation. Roughly speaking the corners contribute nothing at all to the weak formulation because they have no length, so they do not contribute to the relevant integrals. Since I am not entirely sure of your level, I will expand on this if and only if I am asked about it in the comments.