I'm studying variational problems and finite element method, and I'm trying to solve the following equation with boundary conditions:
for $u: ]0,1[ \to \mathbb{R}$,
$-((1 + x)u')'= x,$ for $x \in ]0,1[,$
$u(0) = 0, $ and $u'(1) + u(1) = 1 $
I'm trying to write this in its variational form with the help of a $v(x) \in V = \{ v \in H^1(]0,1[) | v(0) = 0 \}$
But I think I am getting stuck somewhere. For now, I have tried the traditional method of rewriting it with the help of an auxiliary v(x) and solving by integration by parts.
$-((1 + x)u')'v(x) = x v(x) $
$\int_0^1 -((1 + x)u')'v(x) dx = \int_0^1 x v(x) dx $
by IPP, finding for the left-hand side:
$-((1 + x)u')v(x)|^1_0 + \int_0^1 -((1 + x)u')v'(x) dx $
I tried solving the second integral also by IPP, but this time for 3 functions, but I'm not sure if my method is correct. I have an idea that since $1$ is $x'$, there is some sort of trick I could use.
I split up the integral in two parts and solving for both by triple IPP I found for $\int u'(x)v'(x) dx$, I found, with $a' = (1 + x), b = u'(x), c = v'(x)$:
$\int (1+x)u'(x) v'(x) dx = (1+ x) u(x) v'(x) | _0^1 - \int_0^1 u' (x) v'(x) dx - \int_0^1 (1+x) u(x) v''(x) dx $
But I'm not sure how to proceed from here now. Doing another IPP for the third integral didn't help me. Does anyone have any tips in how to from here? Thank you so much already in advance!