2
$\begingroup$

Let

  • $d\in\left\{1,\ldots,4\right\}$
  • $\Lambda\subseteq\mathbb R^d$ be bounded, nonempty and open
  • $\mathcal V:=\left\{\phi\in C_c^\infty(\Lambda,\mathbb R^d):\nabla\cdot\phi=0\right\}$, $V:=\overline{\mathcal V}^{\left\|\;\cdot\;\right\|_{H^1(\Lambda,\:\mathbb R^d)}}$ and $H:=\overline{\mathcal V}^{\left\|\;\cdot\;\right\|_{L^2(\Lambda,\:\mathbb R^d)}}$
  • $\mathfrak a(u,v):=\sum_{i=1}^d\langle\nabla u_i,\nabla v_i\rangle_{L^2}$ for $u,v\in V$
  • $\mathfrak b(u,v,w):=\langle((u\cdot\nabla)v,w\rangle_{L^2}$ for $u,v,w\in V$
  • $\nu>0$
  • $f\in L^2(\Lambda,\mathbb R^d)$

I've found proofs showing that there is a $u\in V$ with $$\tilde{\mathfrak a}(u,v):=\nu\mathfrak a(u,v)+\mathfrak b(u,u,v)=\langle f,v\rangle_{L^2(\Lambda,\:\mathbb R^d)}=:\ell(v)\;\;\;\text{for all }v\in V\tag1$$ using the (Leray-)Schauder fixed-point theorem.

However, doesn't the existence (and even uniqueness) of such a $u$ simply follow from the Lax-Milgram theorem?

It's well-known that

  1. $\mathfrak a$ is a bounded and coercive bilinear form on $H_0^1(\Lambda,\mathbb R^d)$
  2. $\mathfrak b$ is a bounded trilinear form on $H_0^1(\Lambda,\mathbb R^d)$ with $$\mathfrak b(u,v,w)+\mathfrak b(u,w,v)=-\langle\left(\nabla\cdot u\right)v,w\rangle_{L^2(\Lambda,\:\mathbb R^d)}\;\;\;\text{for all }u,v,w\in H_0^1(\Lambda,\mathbb R^d)\tag2$$

Since $$\tilde{\mathfrak a}(u,u)=\nu\mathfrak a(u,u)+\underbrace{\mathfrak b(u,u,u)}_{=\:0}=\nu\mathfrak a(u,u)\;\;\;\text{for all }u\in V\tag3\;,$$ $\tilde{\mathfrak a}$ is a bounded and coercive bilinear form on $V$. Moreover, $\ell$ is a bounded linear functional on $V$. So, there should be a unique $u\in V$ with $(1)$ by the Lax-Milgram theorem.

What am I missing?

$\endgroup$
2
$\begingroup$

What I was missing is that $$V\times V\ni(u,v)\mapsto\mathfrak b(u,u,v)$$ (and hence $\tilde{\mathfrak a}$) is not bilinear. It's obvious, but I missed that. However, we can use the Lax-Milgram theorem to show that for all $u\in V$ there is a unique $\Phi(u)\in V$ with $$\nu\mathfrak a(\Phi(u),v)+\mathfrak b(u,\Phi(u),v)=\ell(v)\tag4\;.$$ Now, the Schauder fixed-point theorem can be used to show that $\Phi$ has a fixed-point.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.