How many point we can fix value to have a compatible problem

2011-12-10T19:44:02Z

Consider a 2D Poisson equation in a rectangular domain $\Omega $with the given Neumann boundary conditions on its boundaries $\partial \Omega$.

$\nabla.(\lambda \nabla U)=-f(x,y) \quad in \quad \Omega$

$\lambda \nabla U = g(x,y) \quad on \partial \Omega$

In order to solve the problem above with standard Finite volume method, it is necessary to set the U to arbitrary value (usually zero) at least a node inside the domain.

In a special case, we need to impose the known value of U at corners of the rectangular domain. This means that the boundary conditions of the domain are the Mixed boundary condition (Dirichlet at corners ($\partial\partial \Omega$)+ Neumann boundary conditions on the edges of the domain $\quad \partial \Omega$).

$\nabla.(\lambda \nabla U)=-f(x,y) \quad in \Omega$

$\lambda \nabla U = g(x,y) \quad on \quad \partial \Omega$

$\lambda \nabla U = \delta_{i,j} \quad at \quad \partial\partial \Omega$ Where $\delta_{i,j}$ is Dirac delta function. In fact,at corners of the domain, the values of U is set to zero except one corner. Is there a unique solution for the problem above? In other words, the values that are imposed at corner of the domain are compatible with the Poisson equations and the Neumann boundary condition. Another problem is how we can include the effect of corner value in standard cell centered finite volume method discretization.

How many point we can fix value to have a compatible problem - MathOverflow

How many point we can fix value to have a compatible problem