Explanation of spaces of functions in PDE

Question

Let's consider following equation:

The problem

$$ \begin{cases} -\operatorname{div}\left( p\left(x\right) \nabla{u} \right) + q(x)u = f \quad\text{... on } \Omega \\ u = h(x) \quad\text{... on } \partial\Omega \end{cases} $$

where

• $u \in W^{1,2}(\Omega)$
• $u-u_0 \in W^{1,2}_0 \text{, so that } u_0 \in W^{1,2}(\Omega) : Tu_0 = h \in L^2(\partial\Omega)$ (T is a trace)
• $f \in L^2(\Omega)$
• $p,q \in L^\infty(\Omega)$

has the following solution:

$$ \forall v \in W^{1,2}_0(\Omega): \int_\Omega \left( p(x) \nabla u \nabla v + q(x)uv \right)\,dx = \int_\Omega f v \, dx $$

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I know that $L^p$ spaces are Lebesgue integrable and they have a following norm:

$$ || \cdot ||_p = \left( \int_\Omega |f|^p d\lambda \right)^{\frac{1}{p}} $$

I also know, that Sobolev spaces are the completion of the $C^\infty(\overline{\Omega})$ and they have the norm:

$$ || \cdot ||_{k,p} = \left( \sum_{|\alpha| \leq k} \int_\Omega |D^\alpha f |^p d\lambda \right)^{\frac{1}{p}} $$

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But, I don't understand, why $u$ belongs to $W^{1,2}$ specifically, why $h$ and f belong to $L^2$ and why $p$ and $q$ belong to $L^\infty$. Could you, please, explain it to me a little intuitively? (e.g. show me some examples where $L^1$ space wouldn't be suitable instead of $L^2$ etc.)

Answer 1

Why not $L^1$:

In studying elliptic equation, it is most convenient to consider $L^p$ space for $1<p<\infty$. The reason is that one does not have nice $L^p$-estimates

$$\|u\|_{W^{2,p}(\Omega)}\le C (\|f\|_{L^p(\Omega)} + \|u\|_{L^p(\Omega)}$$

for $p=1$ (Here we assume that $p(x)$ is nice). Note that the above estimates is crucial in establishing the existence and uniqueness of solution (Fredholm theory) to your system. One can roughly see how this fails when $p=1$ in this question.

Condition on $p(x)$:

Actually in the easiest cases we even assume that $p$ is continuous, so that the above $L^p$-estimates can be established using perturbation theory. On the other hand, it is natural to assume that $p$ is in $L^\infty$, so that the PDE is uniformly elliptic. See more discussions here

The above discussion is in Gilbarg and Trudinger, chapter 9.