Mixed boundary conditions Finite Element Method

Question

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?

Answer 1

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 $\nabla u\cdot \nu$.