I have the following problem in Finite Element Method

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

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?

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?