So, I got the following assignment.

Let $D\subset\mathbb{R}^n$ bounded open set woth smooth boundary. Consider the Neumann boundary problem $$-\Delta u =f \text{, and } \left.\frac{\partial u}{\partial \nu}\right|_{\partial D}=0.$$

(a) Define the notion of weak solution.

(b) Formulate and prove an existance theorem for weak solutions.

So, what I have done so far.

We have $-\int_D\Delta u\; v=\int_D Du\; Dv-\int_{\partial D}\frac{\partial u}{\partial \nu}v$ when ever this makes sense and by enforcing the boundary conditions we get $-\int_D\Delta u\; v=\int_D Du\; Dv$. A solution to the problem satisfies $-\int_D\Delta u\; v=\int_D f\; v$, $\forall v\in L^2(D)$, so we get $$-\int_D\Delta u\; v=\int_D f\; v, \forall v\in H^1 (D).$$

Let $u\sim v \Leftrightarrow u-v=c\in\mathbb{R}$ and let $V$ be the space created of choosing from every equivalent class in $H^1(D)$ the function with the smallest norm. It's eaasy to show that $V$ is Hilbert.

Anyway, I would like to use LAx-Milgram on $V$. So let $B(u,v)=\int_D Du\; Dv$. $B(\cdot,\cdot)$ is clearly bilinear. $|B(u,v)|\le ||Du||_{L^2(D)}||Dv||_{L^2(D)}\le ||u||_{H^1(D)}||v||_{H^1(D)}$.

Then assume that $B(u,u)=0\;\Rightarrow\;\int(Du)^2=0\; \Rightarrow u=c$, but by definition of $V$, $u=0$. So Lax-Milgram theorem can be applied here, which means that for every functional $F(v)=\int_Df\;v$ on $V$ there is a unique element $u$ of $V$ for which we have $B(u,v)=F(v)$.

The problem I have here is that $V$ is too big, in fact I forgot about the boundary condition. Now I would like to define $V_0=\{u\in V:\left.\frac{\partial u}{\partial \nu}\right|_{\partial D}=0\}$ but I am afraid that $\left.\frac{\partial u}{\partial \nu}\right|_{\partial D}$ is not in $L^2(\partial D)$ for all elements of $V$ but rather for only for $V\cap H^2(D)$.

Also there is a hint in the assignment that says to use the fact that if $v\in H^1(D)$ and $\int_D f=0$ then $||f||_L^2(D)\le C ||Df||_{L^2(D)}$ and I cannot see how this is useful to me.

  • $\begingroup$ You integrated by parts the LHS of the equation and got something -- that should be part of your weak formulation and your bilinear form so that you will use the boundary condition. I think you need to show that $B$ is coercive, then Lax-Milgram will give your second question. $\endgroup$
    – soup
    Nov 18, 2012 at 22:42
  • $\begingroup$ @ShuhaoCao how $\int f dx = 0$ ensures that $f$ is consistent with the Neumann boundary? I cannot see that. $\endgroup$
    – rom
    Nov 18, 2012 at 23:28
  • $\begingroup$ @soup $B(u,u)=0\Leftrightarrow u=0$ implies that $\exists c>0$ such that $B(u,u)\ge c ||u||_{L^2(D)}$. $\endgroup$
    – rom
    Nov 18, 2012 at 23:40

2 Answers 2

  • Weak solution $u$ to the homogeneous Neumann boundary value problem: $$\text{Solution space: }u \in H^1(D), \quad \int_{D} u =0$$ such that $$ \int_{D} \nabla u\cdot \nabla v = \int_D fv,\quad \text{ for any } v\in H^1(D).\tag{1} $$ Brief derivation: multiply the equation $-\Delta u = f$ by a test function $v$ and integrating by parts on $D$ $$ \int_{D} -\Delta u v = \int_{D} \nabla u \cdot \nabla v - \color{red}{\int_{\partial D} (\nabla u\cdot \nu)v \,dS} = \int_D fv. $$ Red term vanishes because of the boundary condition. The weak formulation of the Neumann boundary problem is (1), and it is NOT $$\displaystyle \int_{D} -\Delta u \,v = \int_D fv,\tag{2}$$ for now in (1) the lowest regularity $u$ can have is $H^1(D)$, if we write in integration way somehow we are assuming $-\Delta u\in L^2$, a.k.a. strong solution, if we don't abuse our notation using integration to represent the duality pair between $\big(H^1(D)\big)'$ and $H^1(D)$. Now solution to (2) is automatically solution to (1). Conversely, the weak solution, which is the solution of (1), is not the strong solution unless certain conditions are met.

  • Lax-Milgram: You already checked the continuity, the coercity can be checked using the hint which is the Poincaré inequality for zero average functions. Let $\bar{u} = \frac{1}{|D|}\int_D u = 0$, we have $$ \|u\|_{L^2(D)} = \|u - \bar{u}\|_{L^2(D)} \leq C\|\nabla u\|_{L^2(D)}, $$ hence $$ \int_{D} \nabla u\cdot \nabla u \geq c \left(\int_{D} \nabla u\cdot \nabla u + \int_{D} u^2\right) = c\|u\|_{H^1(D)}^2, $$ which is the coercivity. Now you can use Lax-Milgram.

  • Compatibility condition: the right hand side data $f$ has to have average $0$ as well, for $$ \int_D f = \int_D -\Delta u = -\int_D \nabla u\cdot \nu \,dS=0. $$ To guarantee the weak formulation (1) has a solution, this compatibility condition has to be met.

  • $\begingroup$ Just a side question: please do you know the practical meaning of the compatibility condition $\int_Df =0$? I know sometimes $f$ represents the source term and $\nabla u\nu=g, ~~ (g=0)$ stands for the heat flux at the boundary. But I am barely looking for a reference that gives the meaning (in practice or physical ) of the compatibility condition $\int_Df+\int_{\partial D} g =0$ or the meaning (in practice or physical) of the quantity $\int_Df+\int_{\partial D} g $. $\endgroup$
    – Guy Fsone
    Mar 8, 2021 at 10:53
  • $\begingroup$ @GuyFsone Conservation of mass, if the primal variable $u$ represents certain density. In electrostatics, if $u$ is the potential, this compatibility represents conservation of charge (Gauss's law). $\endgroup$
    – Shuhao Cao
    Mar 8, 2021 at 15:04

It actually just hitted me. The space $V$ is defined by taking one function from each equivalent class. We might as well define $V:=\{f\in H^1(D):\int_Df=0\}$. Then $B(\cdot,\cdot)$ can define an equivalent norm on $V$, by using the hint, so there is no need for Lax-Milgram theorem, Reisz representation theorem tells us that indeed there is a unique element $u$ in $V$ such that $B(u,v)=F(v)$. So there is a weak solution of the problem if $\int_Df=0$.

Actually i think that $\{f\in H^1(D):\int_Df=0\}$ and $V$ as defined in the question are in fact the same space but I couldn't figure out a quick proof.

  • $\begingroup$ did you take PDE course in Amsterdam? :) $\endgroup$ Apr 4, 2013 at 17:26
  • $\begingroup$ No, I'm in UK. Why? $\endgroup$
    – rom
    Apr 29, 2013 at 1:36
  • $\begingroup$ It's the same exercise we received as a homework for the PDE course in Amsterdam this spring. $\endgroup$ May 3, 2013 at 14:06

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