Consider some differential equation on a rectangular domain with $0 \leq x \leq L_1$ and $0 \leq y \leq L_2$, and consider the boundary conditions $u(x, 0) = 0$, $u(0, y) = 0$, $u(L_1, y) = k_1$ and $u(x, L_2) = k_2$
Does this not mean that $u(L_1, L_2) = k_1 = k_2$ as it is in the corner and must satisfy both of the end boundary conditions? Furthermore, would this not mean that $u(L_1, 0) = 0$ as it lies in the boundary condition of $u(x, 0) = 0$ but also then $0 = k_1$ and therefore $k_1 = k_2 = 0$?
Is a discontinuity in the corners somehow acceptable?
When solving PDE with boundary conditions, it is reasonable to use boundary conditions that does not contradict each other. This is called 'compatibility condition'.
If this condition fails, as in your case when $k_1,k_2\neq 0$, one can usually still find a solution that will fail to be continuous, at least on the boundary.
Some equations, like Laplace and heat equations, have the property that discontinuity on the boundary does not affect the solution inside the domain, and the solution could even be $C^{\infty}$ there.