The problem is formulated as follows: Let $\Omega\subset\mathbb{R}^2$ a bounded Lipschitz domain. For $u\in L_2(\Omega)$ one can define the one dimensional weak dirivative $\partial_1 u$. Now define the space $M:=\{u\in L_2(\Omega):\partial_1 u\in L_2(\Omega)\}$. Then $M$ with the norm $\|u\|_{L_2}+\|\partial_1 u\|_{L_2}$ is a Hilbert space. It is also clear that $C_0^\infty(\Omega)\subset M$. Now debote by $M_0$ the closure of $C_0^\infty(\Omega)$ w.r.t. the special norm.

Let's consider the following equation: Given some $f\in L_2(\Omega)$. Find $u\in M_0$ such that \begin{equation} \int_\Omega f\partial_1 vdx=\int_\Omega\partial_1u\partial_1vdx \end{equation} is valid for all $v\in M_0$. From Lax-Milgram and one dimensional Poincare's inequality one can deduce that such $u$ exists and is unique. Now the question is, if $f\in H_{loc}^1(\Omega)$, is $\partial_1 u$ in general also in $H_{loc}^1(\Omega)$?

Update: So far I have proved that if $f\in H_0^1(\Omega)$, then $\partial_1\alpha$ has to be in $H^1_{loc}(\Omega)$. The trick is to use the discrete Fourier transform.

Let me explain the trick I used (I am also not quite sure if I was mistaken, since I am not very familiar with the Fourier transform method). Since $f$ and $\alpha$ can be approximated by smooth functions with compact support, we can extend them to $\mathbb{R}^2$ by setting their value equal to zero outside of the region and they belong to $H^1(\mathbb{R}^2)$ and $M(\mathbb{R}^2)$. In particular, they and their derivatives can be written as the discrete Fourier sum (I will write also scalar components by $c$, since they don't essentially influence the result), for example, \begin{align} f(x)&=c\sum_{n\in\mathbb{Z}^2}Ff(n)e^{in\cdot x},\\ \partial_2f(x)&=c\sum_{n\in\mathbb{Z}^2}n_2Ff(n)e^{in\cdot x},\\ \partial_1\alpha(x)&=c\sum_{n\in\mathbb{Z}^2}n_1F\alpha(n)e^{in\cdot x} \end{align}

and so on. In fact, using the standard differnce quotient method one derive \begin{equation} f(x)-\partial_1\alpha(x)=G(x_2) \end{equation} for some $G\in L_2(\Omega)$. Now for $g(x):=G(x_2)$ we also have its Fourier sum \begin{equation} G(x)=c\sum_{n_2\in\mathbb{Z}}Fg(n_2)e^{in_2\cdot x_2}. \end{equation}

Sum up all, we have \begin{equation} \partial_2f(x)=c\sum_{n\in\mathbb{Z}^2}n_1n_2F\alpha(n)e^{in\cdot x}+c\sum_{n_2\in\mathbb{Z}}n_2Fg(n_2)e^{in_2x_2}. \end{equation} Since $\partial_2 f\in L_2$, we have \begin{equation} \sum_{n\in\mathbb{Z}^2}|n_1n_2F\alpha(n)|^2<\infty. \end{equation} Then define $v(x)=c\sum_{n\in\mathbb{Z}^2}n_1n_2F\alpha(n)e^{in\cdot x}$, and $v$ is in $L_2$ and using the definition of weak derivative, it is actually equal to $\partial_2\partial_1\alpha$.

  • $\begingroup$ As far as I understand it, the problem is one-dimensional. For each $x_2$ you have a separate ODE $\partial_1^2 u = \partial_1 f$ on the domain $\{ x_1 : (x_1,x_2) \in \Omega \}$ with zero boundary conditions, hence you can solve it explicitly. $\endgroup$ – Michał Miśkiewicz Dec 11 '16 at 16:00

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