I'd like to find the weak formulation of the problem
$-u''+au=f$ on $(0,1)$
$u(0)=0$
$u'(1)=b$
$a>0$
and show that there exists a unique solution using Lax-Milgram.
What I did: By multiplying the first term with a test function $v$ and doing partial integration I get that \begin{align} -\int_{0}^{1}u''(t)v(t)dt=-u'(1)v(1)+u'(0)v(0)+\int_{0}^{1}u'(t)v'(t)dt. \end{align} Since I want the first part of the equation to be zero, I set $v \in H_{0}^{1}$. So the formulation of the problem is
\begin{align} \int_{0}^{1}u'(t)v'(t)dt+a\int_{0}^{1}u(t)v(t)dt=\int_{0}^{1}f(t)v(t)dt \end{align} for all $v \in H^{1}_{0}$
Is that part correct? I'm confused where the boundary condition $u'(1)=b$ is taken into account. Is $u$ in $H^{1}_{0}$ as well or do I need to define a new space
$ V=\{v \in L^2(0,1), v' \in L^2(0,1), v'(1)=b, v(0)=0\}$?
For the second part of the question (uniqueness + existence of solution): Using Cauchy Schwarz I get for the bilinear form $a(u,v)$ \begin{align} |a(u,v)|=\int_{0}^{1}u'(t)v'(t)+au(t)v(t)dt \leq ||u'||_{L^2}||v'||_{L^2}+a||u||_{L^2}||v||_{L^2}\leq||u'||_{H^{1}_{0}}||v||_{H^{1}_{0}}+a||u||_{H^{1}_{0}}||v||_{H^{1}_{0}} \end{align} So this shows continuity of $a$. I'm not sure though which norm to use for $u$ since I'm not sure which space to choose. Using Poincare I get \begin{align} |a(u,u)|=||u'||_{L^2}^2+a||u||_{L^2}^2 \geq ||u'||_{L^2}^2 \geq C ||u||_{L^2}^2 \end{align} showing that the bilinear form is coercive. The last thing I need to show is that $F$, my linear functional, is continuous (bounded). This follows from \begin{align} |F(v)| \leq ||f||_{L^2}||v||_{L^2} \leq ||f||_{L^2}||v||_{H^1_0} \end{align} Hence, using Lax Milgram I've shown existence and uniqueness of a solution.
Is it correct what I am doing? I am very new to this topic so I am not so sure about some steps, expecially how to choose the space for my $u$.
You almost got it!
An essential thing to remember when attempting to use Lax-Milgram is that the solution space and the test function space have to be the same! Which seems you did not achieve here. Here is a better approach.
You use $v \in V$ so you end up with a weak form: find $u \in V$ such that $$B(u,v) = L(v), \ \ \ \ \forall v \in V.$$ Such that \begin{align*} &B(u,v) = \int u' v' + a \int u v,\\ &L(v) = \int fv + bv(1) \\ &V=\{v \in H^1, \ \ v(0)=0\}. \end{align*} You have already proved that $B(\cdot,\cdot)$ is bounded and in this case you use $H^1$ norm. So you get $$|B(u,v)| \leq (1+a)||u||_{H^1} ||v||_{H^1}.$$ For the coercivity, you do not need Poincare here. You can use $$|B(u,u)| = \int (u')^2 + a \int (u)^2 \geq \min(1,a) ||u||^2_{H^1}.$$ For the boundedness of $L(\cdot)$ $$|L(v)| \leq \int |fv| + |bv(1)| \leq ||f||_{L^2} ||v||_{L^2} + |b| \||v||_{\infty} \leq ||f||_{L^2} ||v||_{L^2} + |b| \ ||v'||_{L^2} \leq M ||v||_{H^1}$$ Where M = 2 $ \max(||f||_{L^2}, |b|)$. From here, Lax-Milgram can be used.