Elliptic equation on square with periodic boundary values for the solution and it's partial derivatives

Suppose that $$\Omega=[0,1]^2$$. I will say that a real valued function $$u$$ on $$\Omega$$ satisfies periodic boundary values if $$u(x,0)=u(x,1), \;u(0,y)=u(1,y),\;\;\;\text{ for all }x,y\in[0,1].$$

Now let $$f\in L^2(\Omega)$$, and suppose that I want to look for a solution of the following problem $$\begin{cases}-\Delta u+u=f, \;\;\text{ on }\Omega=(0,1)^2,\\ u,\; u_x,\text{ and }u_y\text{ satisfy periodic boundary values.}\end{cases}$$

The question is how to show existence (if possible) of $$u$$.

My input to the problem: The weak formulation of the problem with test space taken to be $$X=\{v\in H^1(\Omega):\text{ v satisfies periodic boundary values}\},$$ is the same as the $$H^1_0$$ formulation, ie $$a(u,v):=\int_\Omega\nabla u\cdot \nabla v+\int_\Omega uv =\int_\Omega fv=: F(v),\;\;\;\text{ for all }v\in X.$$ This is because the boundary integral vanishes due to $$u_x$$, $$u_y$$ and $$v$$ being in $$X$$. On the other hand, $$X$$ is a closed subspace of $$H^1(\Omega)$$ and therefore it is a Hilbert space with the $$\|\cdot\|_{H^1}$$ norm. I think we can use the Lax-Milgram theorem on $$X$$, because $$a$$ seems coercive on $$X$$, $$a$$ and $$F$$ are continuous on $$X$$.

However, the Lax-Milgram theorem gives uniqueness, which is weird because unlike $$H^1_0(\Omega)$$, the space $$X$$ does not exactly prescribe the boundary values (it just says that they are periodic). In any case, even if I do get the solution $$u\in X$$ and then by regularity $$u\in X\cap H^2(\Omega)$$, I should still show that $$u_x$$ and $$u_y$$ satisfy periodic boundary values, and this I don't know how to.

You are using the wrong space, that's all. The correct space is $$X=\{u \in H^1_{\textrm{loc}}(\mathbb R^2): u(\cdot+n)=u(\cdot) \quad \forall n=(n_1,n_2)\in \mathbb Z^2\}.$$ Then you apply Lax-Milgram and you are good to go. Look at Asymptotic Analysis for Periodic Structures, by Bensoussan, Lions, Papanicolaou (1979) chapter 1, for example.

Answering your comment. $$f$$ is likewise extended periodically, namely $$f$$ is extended by zero into $$f_0$$, and then identified with $$\sum_{n\in \mathbb Z^2} f_0(\cdot + n)$$ which is periodic on $$\mathbb R^2$$ and $$L^2_{\textrm{loc}} (\mathbb R^2)$$.

Now, on say $$(-2,2)^2$$, $$-\Delta u + u=f$$ as a weak solution, and therefore interior regularity shows $$u\in H^2((-\frac32,\frac32)^2)$$. All in all, $$u\in H^2_{\textrm{loc}} (\mathbb R^2)$$. Consequently, $$\partial_x u \in X$$, and $$\partial_y u \in X$$.

Incidentally, because all coefficients are constant, you can write write $$u$$ explicitely as a (double) Fourier series, its coefficients being that of $$f$$, divided by $$n_1^2+n_2^2+1$$. It saves you the trouble of using Lax-Milgram.

• Thank you for the reference. However, even if I do get $u\in X$ as you defined it, I also need that $u_x,u_y\in X$, which doesn't seem that it follows trivially from the weak formulation! Commented Nov 14, 2021 at 19:57
• Wait so $f$ should also be periodic on the boundary? (the way I defined periodic) Commented Nov 14, 2021 at 20:18
• Also, I don't see how $u\in H^2_{\text{loc}}(\mathbb{R}^2)$ directly implies that $u_x,u_y\in X$. Commented Nov 14, 2021 at 20:20
• @demlev You are asking why if $f(\cdot +n)=f(\cdot)$ for every $n$ then the same hold for $Df$? Commented Nov 14, 2021 at 20:24
• Yes. How did you deduce this? Commented Nov 14, 2021 at 20:27