I have the following problem in Finite Element Method

$$ -(\alpha u')' + \beta u' + \gamma u = f$$

with $ \Omega = (0, 1)$, $ u(0) = 0 $ and $ u'(1) = 3 $

to be able to write the weak formulation of the problem do I need a lifting function?

i.e., do I need to define something like

$$ \tilde u = u - R$$

I'm stuck after integrating by parts, I got something like:

$$ - \Big [(\alpha u' v) _0^1 - \int_0^1 \alpha u' v' \Big ] + \beta \int_0^1 u' v + \gamma \int_0^1 u v = \int_0^1 f v $$

$ \forall v \in V = H_{\Gamma_D}^1 (0,1)$

To get rid of the first term $$ (\alpha u' v) _0^1 $$ and have the bilinear form $a(u,v)$, when $$v(0) = 0$$ the lower limit dissapears, but for the upper limit I get $$ - \alpha u' (1) v(1)$$ and it does not disappear, how to proceed?

  • $\begingroup$ Hi @BRabbit27, and welcome to scicomp! I noticed the strong form of your problem seems incomplete... It seems to be missing a right hand side to make it a true equation. $\endgroup$ – Paul Nov 13 '12 at 18:46
  • $\begingroup$ Sorry, editing and correcting it. $\endgroup$ – BRabbit27 Nov 13 '12 at 18:51
  • 1
    $\begingroup$ It doesn't need to disappear, because you can just substitute $u'(1)=3$ in it. You can then move the term to the right hand side. $\endgroup$ – Christian Clason Nov 13 '12 at 18:57
  • $\begingroup$ So first of all, no need to have a lifting function $ \tilde u = u - R $. Second of all, the weak formulation will be $ \alpha \int_0^1 u' v' + \beta \int_0^1 u' v + \gamma \int_0^1 u v = \int_0^1 f v + 3 \alpha v $, am I right? $\endgroup$ – BRabbit27 Nov 13 '12 at 19:02
  • $\begingroup$ That's right (except it should be $3\alpha v(1)$). You only need lifting functions for nonhomogeneous Dirichlet conditions. Also, this shows that the proper Neumann conditions should be $\alpha u'(1) = g$ (this makes a real difference in $2$ and more dimensions). $\endgroup$ – Christian Clason Nov 13 '12 at 19:52

For the second term $\alpha u'(1)v(1)$, you can insert the Neumann boundary condition $u'(1) = 3$ to obtain $3\alpha v(1)$. Since this term can be evaluated (for a given test function $v$), it becomes part of the right hand side: The weak formulation is to find $u\in H^1_{\Gamma_D}(0,1)$ such that $$ \int_0^1 u'(x)v'(x) dx + \beta \int_0^1 u'(x)v(x) dx + \int_0^1 u(x)v(x) dx = \int_0^1 f(x) v(x) dx + 3\alpha v(1) $$ for all $v\in H^1_{\Gamma_D}(0,1)$.

Note that in two and more dimensions, the term arising from integration by parts is $$ \alpha(x)\nabla u(x)\cdot \nu(x) v(x), \qquad x\in\Gamma_N, $$ where $\alpha$ can be a matrix and $\nu(x)$ is the unit outward normal at $x$. To handle that term, you need to specify $$ \alpha\nabla u\cdot \nu = g $$ on $\Gamma_N$, and not just the normal derivative $\nabla u\cdot \nu$.


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.