I have the biharmonic equation on a 2D rectangular domain $\Omega$ with the following boundary conditions:

$\Delta^2 u = f$ on $\Omega$

$\nabla u \bullet \mathbf{n}=0$ on $\partial \Omega$ (1)

$u_{xy} = 0$ on $\partial \Omega$ (2)

I need the weak form of the equations.

I am familiar with Nitsche's method and I can impose the first condition, if the second condition is a Dirichlet condition (e.g. $u=0$):

$\int_\Omega \Delta u \Delta v \mathrm{dx} -\int_{\partial \Omega } \Delta u \nabla v \ \bullet \mathbf{n} ds - \int_{\partial \Omega } \nabla u \bullet \mathbf{n} \Delta v ds +\eta \int_{\partial \Omega } \nabla u \bullet \mathbf{n} \nabla v \bullet \mathbf{n} = \int_{\Omega} fvdx$

where $\mathbf{n}$ is the element normal and $\eta > 0$ is a penalty parameter.

Question 1: Can I impose the $u_{xy} = 0$ somehow in a similar manner?

Question 2: What if I want to impose $u_x = 0$ instead of $\nabla u \bullet \mathbf{n}=0$?

In specific: I have a plate bending problem and I would like to compute only the quarter of the domain. Therefore I would like to impose symmetry boundary conditions on the blue dashed boundaries. enter image description here

  • 1
    $\begingroup$ Is your domain a box? In other words, is $\partial_x\partial_y$ related to normal and tangential derivatives? $\endgroup$ – Wolfgang Bangerth Jun 10 '17 at 4:01
  • $\begingroup$ Your second condition seems redundant to me. The gradient times the normal vector is just what you would call $u_x$ resp. $u_y$. And if these shall be already zero on the boundary, the condition $u_{xy}=0$ does not add anything to the problem. $\endgroup$ – davidhigh Jun 12 '17 at 9:04
  • $\begingroup$ Are $\partial\Omega(1)$ and $\partial\Omega(2)$ two different parts of the boundary? Then you are missing a second boundary condition on each part. Or did you imply $u=0$ already?If you edit, please change 'Nietsche' to 'Nitsche' $\endgroup$ – Guido Kanschat Jun 12 '17 at 13:32

That's basically two questions in one. The first is how to incorporate boundary conditions, the other how to treat the symmetry. I'll consider only the symmetry here, because the boundary value stuff has been already discussed on this page, see e.g. here.

Parity symmetry

In your case it's about parity symmetry, i.e. the symmetry that possibly occurs under a change of sign of the solution (--the Laplace operator is also rotationally invariant, but the rectangular boundary breaks this symmetry).

So, let's assume your problem supports parity symmetry. One sufficient condition for this is that your inhomogeneity $f$ is an even function, $f(x,y)=f(-x,-y)$, and your $\Omega$ is also symmetric with respect to the origin. If this holds, it follows that the solutions to your problem have definite parity,

$$ u(-x,y) = (-1)^{m_x} \;u(x,y)\\ u(x,-y) = (-1)^{m_y} \;u(x,y) $$

Even $m_x$ (resp. $m_y$) correspond to even parity and thus even solutions with respect to the origin, and odd ones to odd parity. (I used to call these "quantum numbers" in my application to quantum mechanics, but the concept comes from group theory).

The boundary conditions for your problem follow directly. For odd parity, one has $v(x) = -v(-x)$ and thus $v(0) = 0$. For even parity, it's $v(x) = v(-x)$ and therefore $v^\prime(0)=0$. This directly extends to two dimensions, and it allows you to specify the solution only in the upper left quadrant and extend it accordingly to the other ones.

As usual, there are two basic ways to incorporate this symmetry. (i) pick basis functions which already satisfy the symmetry and expand your solution in these, or (ii) explicitly enforce them in the solution of your problem (e.g. via Lagrange multipliers). Again, see the linked thread.

| cite | improve this answer | |

I understand your specific question. I don't understand the other two questions or how they relate to the specific question.

Because your weak form contains second derivatives of $u$, shape functions satisfying $C_1$ continuity are required. This means that that both $u$ and first derivatives of $u$ will be nodal unknowns. So you can satisfy the symmetry conditions on your two symmetry edges simply by specifying Dirichlet conditions on the derivative unknowns. That is how this problem is routinely solved.

Developing a conforming plate bending finite element based on Kirchoff thin plate theory (the biharmonic equation you show) is quite challenging. If you are determined to proceed in this direction, I suggest taking a look at these notes (Felippa plate element notes) to get an idea of the difficulties. Felippa has other sets of notes on this site (his advanced FEM course) that provide more details about how to formulate a workable element.

| cite | improve this answer | |

Only essential boundary conditions are implemented by Nitsche's method in weak form. Since your problem is formulated in $H^2(\Omega)$, these can only involve function values and first derivatives.

For higher order derivatives, look at integration by parts:

\begin{align} \int_\Omega \Delta^2 u v \,dx &= -\int_\Omega \nabla\Delta u\cdot\nabla v\,dx + \int_{\partial\Omega} \partial_n \Delta u v\,ds\\ &= \int_\Omega \Delta u \Delta v \,dx - \int_{\partial\Omega} \Delta u \partial_n v\,ds + \int_{\partial\Omega} \partial_n \Delta u v\,ds. \end{align}

The boundary terms allow you to set boundary values by replacing $\partial_n \Delta u$ and $\Delta u$ by their respective boundary values. Note though, that only normal derivatives are involved in this integration by parts. Therefore, most problems involving tangential derivatives are ill-posed.

Examples for ill-posed boundary conditions involving tangential derivatives:

  1. $u=f$ and $\partial_\tau 0= g$, since this can only work if $g=f'$ where the derivative is taken along the boundary.
  2. $\partial_n u=f$ and $\partial_{n}\partial_\tau u=g$ for the same reason.

You could have oblique derivatives by subtracting the tangential derivative in the weak formulation. Be aware though that this requires sufficient regularity to write the boundary integrals. Otherwise, you will generate boundary layers.

| cite | improve this answer | |

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.