Am solving a system of pdes in Fenics and need to write the variational form. Everything looks good expect a term like the following $$ \nabla \xi\nabla \cdot U $$ $\mathbf{U} = (U,V) $ is a vector function and $\xi$ is a scalar function in 2d. So I tried the following. Multiply by a vector function $p$ and integrate by part I arrived at the following, assuming that $p$ vanishes on the boundary. $$-\int (\xi \nabla \cdot \mathbf{U})\cdot\nabla p\space dx .$$ But this doesn't look right. Can anybody please help with this. Any help is much appreciated. Thanks.

  • $\begingroup$ Sorry but I could not understand how did you apply the divergence theorem for the first term in 2, can you explain this more? Thanks $\endgroup$ – Rock Apr 16 '16 at 22:37

Multiplying term by a vector test function $\mathbf{p}$ and integrating over the domain $\Omega$ gives:

$$\int_\Omega \nabla\cdot \mathbf{U}\nabla\xi\cdot\mathbf{p}d\Omega\qquad (1)$$

To start with you can write:

$$ \nabla\cdot(\mathbf{p}\xi) = \xi\nabla\cdot\mathbf{p} + \mathbf{p}\cdot\nabla\xi$$

Which gives:

$$ \mathbf{p}\cdot\nabla\xi = \nabla\cdot(\mathbf{p}\xi) - \xi\nabla\cdot\mathbf{p}$$

Plugging this into (1):

$$\int_\Omega \nabla\cdot \mathbf{U}\left(\nabla\cdot(\mathbf{p}\xi) - \xi\nabla\cdot\mathbf{p}\right) d\Omega\qquad (2)$$

Recall the Divergence theorem:

$$ \int_\Omega \nabla\cdot \mathbf{F} d\Omega = \int_\Gamma \mathbf{F}\cdot\mathbf{n}d\Gamma$$ where $\Gamma$ is the boundary of $\Omega$.

Applying this to the first term in (2) gives the final form:

$$\int_\Gamma \nabla\cdot \mathbf{U} \xi(\mathbf{p}\cdot\mathbf{n})d\Gamma - \int_\Omega \nabla\cdot \mathbf{U} \xi\nabla\cdot\mathbf{p}d\Omega$$

The first term is 0 if $\mathbf{p}$ vanishes on the boundary.

| cite | improve this answer | |
  • $\begingroup$ No problem. Hope all the steps are clear. I think this way is slightly longer but also a little bit more clear than just using integration by parts. $\endgroup$ – Lukas Bystricky Nov 10 '15 at 9:48

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.