I am studying PDEs and Sobolev spaces. In some examples, the solution and/or initial data lies in the Sobolev spaces $H^{1}$ or $H^{-1}$. Sincerely, I do not understand the reason why some solutions need to lie in the Sobolev space with a negative exponent.
Let me ask:
- Why sometimes the functional space is $H^{1}$ and others it is $H^{-1}$?
- What are the motivations to define solutions in the Sobolev spaces with negative exponents?
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.