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.


1 Answer 1


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.

  • $\begingroup$ 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! $\endgroup$
    – demlevi33
    Nov 14, 2021 at 19:57
  • $\begingroup$ Wait so $f$ should also be periodic on the boundary? (the way I defined periodic) $\endgroup$
    – demlevi33
    Nov 14, 2021 at 20:18
  • $\begingroup$ Also, I don't see how $u\in H^2_{\text{loc}}(\mathbb{R}^2)$ directly implies that $u_x,u_y\in X$. $\endgroup$
    – demlevi33
    Nov 14, 2021 at 20:20
  • $\begingroup$ @demlev You are asking why if $f(\cdot +n)=f(\cdot)$ for every $n$ then the same hold for $Df$? $\endgroup$
    – username
    Nov 14, 2021 at 20:24
  • $\begingroup$ Yes. How did you deduce this? $\endgroup$
    – demlevi33
    Nov 14, 2021 at 20:27

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