3
$\begingroup$

Let's consider the FEM discretization of the Laplace operator without boundary conditions, i.e., $$ a(u,v) = \int_\Omega \nabla u \cdot \nabla v - \int_\Gamma (n\cdot \nabla u) v. $$ For one-dimensional domains, the corresponding matrix has two empty rows (the first and the last), generally in $n$ dimensions the matrix has a nullspace of dimension $n+1$. This appears to suggest that, for example in 2D, Dirichlet conditions in only three separate points are necessary to make the operator nonsingular. Numerical experiments with various domains support this suspicion, e.g. a triangular domain with Dirichlet conditions at the corners:

enter image description here


enter image description here

(This is with linear elements; quadratic elements require more points.)

Are there any results on this?

$\endgroup$
  • 1
    $\begingroup$ I think that the nullspace of this matrix operator in dimension $d$ is always 1. For example in 1D for the operator $A$ to be nonsingular, at least the variable $u$ at one point must be known (Essential BC: restriction) the other boundary may be left as it is (Natural BC: equilibrium). The same applies to higher dimensions. $\endgroup$ – HBR Apr 28 '18 at 10:54
  • $\begingroup$ @HBR Imagine an all-zero rhs: All linear functions then fulfill $-\Delta u=0$, and indeed the same is true for the weak form. In $d$ dimensions, linear functions are characterized by $d+1$ parameters, so that's the size of the nullspace. In 1D, it becomes really clear: You have to lines which are all zero. (See the text.) $\endgroup$ – Nico Schlömer Apr 28 '18 at 11:08
  • $\begingroup$ This case you say results in a nullspace of dimension $1$ and its solution is a constant. In that case equilibrium is always imposed in the boundaries and the dunction $u=x$ does not fulfill that equilibrium $\endgroup$ – HBR Apr 28 '18 at 11:20
  • $\begingroup$ There are no boundary conditions here. Linear functions, including $u(x)=x)$, do fulfill the equation. Am I missing something here? $\endgroup$ – Nico Schlömer Apr 28 '18 at 11:24
  • 1
    $\begingroup$ Isn't the first row $1/h[1 -1 0 \cdots 0]$ and the last $1/h[0 \cdots -1 1]$? $\endgroup$ – nicoguaro Apr 28 '18 at 14:46
2
$\begingroup$

You are solving the following equation: $$-\textrm{div}(\vec{grad}\, u) = -\Delta u= 0 \quad \forall x \in\Omega\tag{P}$$ in a bounded domain $\Omega$.

Therefore we must have the following BC: $$u=u_D\quad\forall x\in\partial\Omega_D\qquad \frac{\partial u}{\partial n}=u_N \quad\forall x \in\partial\Omega_N$$

I want to note that differential operators applied to bounded domains do not make sense without BC's.

But for now, let's abandon the BC's

Applying Gauss Th. one sees that $$\int_{\partial \Omega}\frac{\partial u}{\partial n}\,d\sigma = 0\tag{*}$$ This equation is the equilibrium equation that ensures the net "heat" flux through the boundary adds to $0$. This equation is always present in every conservation principle that does not depend on time. This flux is already "fixed". This restriction is the price one has to pay for $\Omega$ to be bounded.

Therefore, at least, if the normal derivative $\partial_n u$ is specified along the whole boundary $\partial\Omega$, the condition $(*)$ must be fulfilled for the problem $(P)$ to be meaningful.

We notice that the solution to the problem $(P)$ with the restriction $(*)$ has a unique solution up to a constant, because the problem is invariant w.r.t. the transformation $u\to u+c$, and therefore we have only one DOF that has to be fixed.

Infinite domain

Here we have no restrictions and the equation $(P)$ holds everywhere. For example, every solution $u$ is of the type of the real or imaginary part of $f(z)$ with $z=x+iy$ being $f$ analytic.

Another point of view

Take the 1D problem and play with matrices. For the moment assign the matrix $$A=\frac{d}{dx}$$ and its transpose $$A^T = -\frac{d}{dx}$$

The operator $A$ (rectangular, because it does not have any imposed BC) is applied to the vector $u$ to obtain the fluxes $q=Au$. In physics there is always an equilibrium of sth. e.g. in heat problems the equilibrium means that at every node fluxes must add to 0: $A^Tq=0$. This is the Laplace equation $(P)$ in discrete space: $$A^TAu=Lu=0$$ Since $L$ is symmetric, you can calculate its eigenvalues easily. You will find that only one eigenvalue is $0$ and the kernel of $L$ is equal to the vector $u=1$. here is your DOF.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.