$H^{1}$ or $H^{-1}$?

It is hard to know exactly what you are referring to without an example but let me give you a general explanation.

Suppose that we are trying to come up with a weak formulation of the elliptic boundary value problem $$\tag{$\ast$} Lu:=-a^{ij}D_{ij}u = f \qquad \text{in } \Omega$$ with boundary condition $$ u=0 \qquad \text{on } \partial \Omega.$$ (Here we are using the summation convention, $L$ is uniformly elliptic, $a^{ij},f,\partial \Omega$ are smooth, and $\Omega$ is bounded). Supposing (for now) that $u$ is smooth, multiplying ($\ast$) by a smooth test function $v\in C^\infty_0(\Omega)$, integrating over $\Omega$ then integrating by parts we get $$ \int_\Omega a^{ij} D_iuD_jv \, dx = \int_\Omega f v \, dx. \tag{$\ast\ast$}$$ We note straight away that for ($\ast\ast$) to make sense we only need $u$ to have one derivative not two as in ($\ast$). In fact, since ($\ast\ast$) is all contained inside an integral, we only need $u$ to once weakly differentiable. Hence, it seems the appropriate space for $u$ to be in is $H^1(\Omega)$. However we also want to incorporate the zero Dirichlet boundary data which means the actual appropriate space for $u$ to be in is $H^1_0(\Omega)$. (Recall $H^1_0(\Omega)$ can be understood to be those functions in $H^1(\Omega)$ that vanish on the boundary in the trace sense). Thus, we say $u \in H^1_0(\Omega)$ is a weak solution of the original BVP if $u$ satisfies ($\ast\ast$) for all $v \in C^\infty_0(\Omega)$. What I'm trying to emphasise here is that we take the solution $u$ to be $H^1_0(\Omega)$ because $H^1_0(\Omega)$ is pretty much the most general space for which ($\ast\ast$) makes sense and $u$ has the correct boundary conditions. To illustrate this point, if we replaced '$u=0$ on $\partial \Omega$' with '$u=g$ on $\partial \Omega$' then we would assume $u \in H^1(\Omega)$ and $u-g \in H^1_0(\Omega)$.

Now let's try understand why $H^{-1}(\Omega)$ comes in to the picture. So far we've be trying to understand what functional space $u$ can be in for ($\ast\ast$) to make sense but we can ask the same question about $f$. Of course ($\ast\ast$) makes sense if $f$ is smooth but we can see that if $f\in L^2(\Omega)$ then ($\ast\ast$) still makes sense since the RHS of ($\ast\ast$) is just $(f,v)_{L^2(\Omega)} $ (the $L^2$ inner product of $f$ and $v$). Now when we go to prove the existence and uniqueness via Lax-Milgram say the fact that the RHS of ($\ast\ast$) is $(f,v)_{L^2(\Omega)} $ is not important - the important part is that $v\mapsto (f,v)_{L^2(\Omega)} $ is a bounded linear functional. Hence, we can generalise ($\ast\ast$) to $$\int_\Omega a^{ij} D_i u D_j v \, dx = \langle f,v \rangle \tag{$\ast\ast\ast$} $$ where $f$ is any bounded, linear functional on $H^1_0(\Omega)$ and $\langle f , v \rangle = f(v)$. Note that in the original case with $f\in L^2(\Omega)$ we have made an identification between $f$ and the linear functional $v \mapsto (f,v)_{L^2(\Omega)}$. Thus, ($\ast\ast\ast$) is an even more general form of ($\ast\ast$) and all we require is that $f$ is a bounded linear functional on $H^1_0(\Omega)$. But $H^{-1}(\Omega)$ is defined to be the dual space of $H^1_0(\Omega)$. Thus, what we really require is that $f\in H^{-1}(\Omega)$ for ($\ast\ast\ast$) to make sense.